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MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 21_Boundary_Value_Problems_2.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So today we're going to keep talking about boundary value problems. We'll do that again on Wednesday. On Friday we'll start partial differential equations. And next week we'll work with COMSOL, which I hope you guys all have installed on your laptops. Please check that you have it installed and you can turn it on, your licensing things work and stuff like that. We'll have a COMSOL tutorial I think on Monday in class. And you'll use that for one of the homework problems not immediately but coming up. Speaking of the homework, I received some feedback that the homeworks have been taking an inordinate amount of time. And so I've discussed this with the other people involved in teaching it and we decided to drastically cut the amount of points given for write-ups to try to encourage you not to spend so much time on that. And instead I'd really rather you spent the time and did the reading instead of spending an extra hour doing write-up. You won't get any more points for doing it. So if you want-- I mean, I love beautiful write-ups. I'm sure the graders appreciate the beautiful write-ups. It's good for your thinking to write clearly but it's not worth many, many hours of time. And in general, the instructions at the beginning of course, are on average over the course, your 14 weeks, it should be about nine hours per week of homework. So that's maybe 13 hours per assignment, so you have 10 assignments. And so if it's getting to be 15 hours, you've spent too much time on it and just forget it. Just draw a line and say I'm done, time ran out, and that's fine. This is-- getting the last bug out of your Matlab code is not the main objective of this course. And really, the purpose of homework is to help you learn, not that we need to know what the solution to this problem is. This is now-- I'm getting the solution from somebody else in the class too so this is-- I don't need it from you. The TAs probably already did it already and they have the solution too. So we know the solution. It's not so essential. Its purpose is to help you learn and how to figure it out. And beyond a certain point, it's not that instructive in my experience. Though sometimes the fifteenth and a half hour you suddenly have the great insight and you learn a lot, but most the time not. And by the way the reading is in [? Beer's ?] textbook pages 258 to 311. And there's also some nice short readings by Professor [? Brautz ?] that have been posted, only a few pages long but definitely worth a look. Both on [? PPPs ?] and also in ODEs and [? DAEs. ?] So today what we're going to talk about is relaxation methods. And we talked last time about the shooting method. So that's a good method. At least one famous numerical methods book recommends you always shoot first then relax. So try the shooting method. If it works, you're done. If it doesn't work for you, then you may have to use a relaxation method. And we'll talk a bit now about the relaxation methods. And the general idea is you're going to write your y-- which is an approximation, it's not going to be the real solution-- and you're going to try to write that typically as an expansion to some basis functions. So let's say the nth component of y. And then you're going to vary these coefficients, the d's, to try to make your solution as good as possible. And now we're talking about different definitions of what good is for the solution for a problem. And you have to be aware that almost always, this will not exactly solve the differential equation system. So it's always going to be wrong everywhere, typically. And you can get to decide sort of-- if you wanted to be good at some particular spot, you can do something about that. If you wanted to, on average, be good in some way, you can decide that. And it's sort of how much error you can tolerate. And in general, you have to do a finite sum of a finite basis set. For many, many ODE, [? PPP ?] problems, there's math proofs that if-- and the limit is this goes to infinity if you add an infinite number of functions here. You can always make it work to get the true solution. But you can never afford that so you're always finite truncated basis set is what you're using. And so a lot of the accuracy things have to do with exactly what functions you choose and exactly how many of these terms and sum you include. But typically, these problems will start from the beginning, you'll say I'm only going to include so many, how big my computer is. And so you're going to be stuck with some error, so that's just the way this. Let's recall the problem we have. If we write it in the-- as a first order ODE system, it's dy/dt-- well, there's some function of dy/dt and y and t that's equal to zero which often can be written something like this, dy/dt is equal to an f [INAUDIBLE].. Right? And so if you plug in this approximation for y, then you'll get something here that will be some g of t that's generally not going to equal zero, but you would like it to be zero. If you put the true solution in, you would get zero. If you put your approximate solution, it is not going to be zero. So you'll get g of t and you want this equal to zero for all t. And then in addition, you have boundary conditions. And so you have boundary conditions on the solution and you want those things to be satisfied. And again, you can write them in a form that something has to be zero. Often, you'll exactly satisfy them. So you'll choose your solution to make sure it satisfies the boundary conditions and it won't satisfy in the domain. It won't really satisfy the differential equation. That's the most common thing to do. But it could also be that it doesn't satisfy the boundary conditions either, but you want to be close to satisfy the boundary conditions. We have to think of how are we going to judge if the solution is accurate or not, if our approximate solution is good? And from the way we wrote it, we have our parameters, d. These are numbers we can adjust. And we're going to try to adjust them to make the solution as good as possible. And now we just define what a good means. There are several definitions of good that are widely used. And one of them is called collocation. This is like option number one. And that is you choose a set of ts, of particular time points. And for those particular type points you demand that g of the time point is equal to zero. So you're forcing the residuals-- this is called the residual, it's the error. And you're forcing the error to be zero at some particular time points. And generally between the time points it will not be equal to zero. But you can pick your time points and depending on which ones you pick, you'll get a slightly different optimal choice of the d's. Because the d's will be adjusted to force the residual to be zero at your time points you pick. So that's one option. Another one is called Rayleigh-Ritz. And that one is you minimize overall your d's. The integral in t zero to t final of the norm of g of t. So this means you try to make the average of the square of that deviation to be as small as possible. So it's like sort of like a least squares fit kind of thing. So that's another option. And then a third one that people use a lot is called Galerkin's Method. And his method is you choose some functions-- some of your basis functions typically-- and you integrate them with each of the elements of the residual. And you demand that that has to be equal to zero. Yeah? AUDIENCE: [INAUDIBLE] for [? this method, ?] what do you [INAUDIBLE]? PROFESSOR: The d's, your coefficients. And these ones, I didn't write it out but this g depends implicitly on the d's, so you can write it that way. A lot of d's. And so you optimize the d's, you very the d's to force g to be zero at certain ts. And this one also, the g depends on d's. And so you're going to optimize the d's to force this integral equation to be satisfied. Yeah? AUDIENCE: [INAUDIBLE] all changing d's? PROFESSOR: Yeah. They're all changing d's. And it's just trying to-- your criterion, your error measure, how do you measure or what do you think is good? You want to make something small, some kind of error small, but you have to figure out what are you going to define your error to be. And you'll get different solutions depending on which error measure you use. Is this OK? Now this one's pretty straightforward. I'm just going to write it down a bunch of algebraic equations that depend on my d's and then I'm going to solve them. And this looks a lot like [? an f ?] [? solve ?] problem. This is a Newton problem or something like that. So it's just some algebraic equations. Should be OK? So that one you should be pretty well set up to do right now. Let's just think of how many equations there are. So we have-- say we have our basis set yf t dni fit. That's our basis set. And this is a sum over i equals one to say the number of basis functions. What do you want to call that? N, sound good? So we have n basis functions. And so we have how many unknowns here? Maybe n's not good because n's this one. Let's call it something else. k, OK. All right? In fact, I'll even change this to k too just to make life easier. So there are how many dnk's are there? There's n where n is the dimension, the number of od's or the number of components in y times k. That's how many d's we got that we're going to adjust. We want to have an equal number of equations and unknowns. We have to have as many equations as that. So what equations do we got? How many equations? So if we just have a ODE [? BBP ?] we typically have n boundary conditions. So that many boundary conditions because we need one boundary condition for each differential equation, right? One integration constant for each one. So we took an n boundary conditions. And then we have-- in collocation, we have however many capital M time points we chose. So we have M n equations-- no, that's not right. We have an equation like that for every component of g. So it's M times n equations from the collocation. Is that all right? So just looking at this, it looks like we have n times M plus one equations and we have n times k unknowns that we're trying to adjust. And so therefore, this says we should choose k to be equal to M plus one. So if we choose we want to say we want 100 basis functions, then we need 99 time points to do collocations at it in order to exactly determine everything. Is that OK? How many people think this is OK? OK. He agrees. It's OK. The rest of you, no opinion. This is like the American political system. Only 5% people vote. Everybody else just listens to Donald Trump. So this is how many equations we need. Now sometimes people will choose the basis functions so that, say, some of the boundary condition equations might be satisfied automatically. And I'll talk a little bit in a minute about cleverness in choosing basis functions. So one possible thing is you can try to cleverly choose basis functions so that no matter what values of d's you choose, you're always going to satisfy some of the boundary conditions. And so long as you don't get a full value of n because some of these boundary conditions don't help you determine the d's, any values of d's will work. And so in those cases, you need a few-- might need another time point or something. Get some more equations. So that's collocation. And I think this should be perfectly straightforward. You just evaluate your y's. You need to have your dy/dts. You'll need them, because they appear in g as well. And they're just going to be the summation over k [? of ?] [? dnk ?] v prime k of t. And so you choose basis functions that the analytical derivatives of. So now you know these answers. And now you can evaluate this at any time point tm. So you evaluate your time points. You evaluate these guys at your time points. You plug them all into your g expression over here, and your force is equal to zero, by varying the d's, right? No problem. So that's collocation. That's pretty easy. And because it's so easy, it's kind of pretty widely used as one way to go. All it requires is that you have to know the derivatives of your basis functions. In particular, doesn't require any integrals. When you see the other methods, they're going to involve integrals. So we'll have to figure out how we're going evaluate those. So if you have functions you don't know how to integrate then this is definitely the way to go, to do collocation. And if you use collocation with enough points, you're forcing the error to be zero a lot of points, then probably it won't get that big in between the points. At least you can hope that I won't get that big between the points. And you can try it with different numbers of points. And different size numbers of base functions and see what you can do. See if it converges to something, if you're lucky. All right. So that's one way to go. And it's just f solve problem. And all that's happening is it's just forming a Dracovian Matrix inside there. Now, you still have to choose which basis functions you want. And there's a lot of basis functions you know that you know the derivatives of. So there's some issues here about what choice is best. And we'll talk a little about that in a minute. One thing you have to watch out for, is you don't want your basis functions to be linearly dependent because then you'll end up with indeterminate values a d. Different sets of d's will give you exactly the same y. If these phi's-- if one of these phi's is a linear combination of other phi's in your set, then you could have different values of d's that would actually correspond to exactly the same y. Because of that, when you a Jacobian Matrix as part of it your Newton solve, the Jacobian's going to be singular, and then the whole thing's not going to work. So that's one thing to watch out for. And actually, it doesn't really matter if the functions are orthogonal in the sense of being functions orthogonal, it's really whether the vector, vk evaluated tm, if those are independent of each other or not. I feel like we talked about this before, one time, yes? Yes. All right. So anyway, as long as the vk evaluate of tm, if those things are vectors which are not linerally dependent on each other, this should work fine. OK? All right. Now, really [INAUDIBLE]. This one here often gets to be kind of messy. Part of it is that inside this norm of this vector, it's square, so you end up getting squares of your d's. So, the whole thing is definitely non-linear in d to start with. And then you have to be able to evaluate all the integrals that come up. So this is really sum over n, of g and of td squared that you're trying to integrate. [INAUDIBLE] Right? So, you have a lot of these function squared. You have to add them up. You could do it, but the algebra gets a little complicated. So, this is not so commonly done unless the g has some special form that makes the algebra a little easier. But there's no reason you can't do it in principle, but it just is a little bit of a mess because you get a lot of integrals of cross terms between the g's. And in the d's inside there. And so, not so commonly done. Though one case where it is done a lot is in actually quantum chemistry. Methods called variational methods in quantum chemistry use real [INAUDIBLE] to figure out the coefficients, and like your orbitals, and stuff like that. So, it'd be certain methods. But they've actually gone out of favor recently. So, you probably won't use it that often. But in the past, that was a big deal. Galerkin is used the most of anything, so let's talk about that one for a minute. Now, the concept, where does this equation come from? I guess that's one thing. So maybe we can back up and say, well, where did the collocation come from? Collocation is actually like a special version of this where I choose, instead of these basis functions, I use delta functions. So, if I integrated with delta functions, t minus tm. Another way to look at the collocation is demanding that the integral of delta function t minus tm g n a t dt is equal to zero. OK. So, now instead of using delta functions, I'm using these functions. Basis functions. So I don't know if it's particularly obvious why you'd use one or the other, but anyway, you have a choice. And any time you do this kind of integral with some number of functions, you get some number of equations that can help you determine the d's. This particular choice of the basis functions-- one way to look at it is, you can say, well, suppose my solution g g n, suppose I could write this as a sum of some different expansion coefficients. So I can relate this plus OK, so, there's two terms. One is the expansion of the residuals in the basis that I'm using, and one is all the rest going on to infinity of all the other basis functions in the universe. And if I think my basis set is very complete, that it kind of cover all the kinds of functions I'm ever going to deal with, both in y and in g, in the residual, then this would be a reasonable thing, and you might expect that this term-- if you'd pick big K big enough, this might be small. So that's sort of where the idea is. So then, if I think this is small, then I want to make sure I make this part as close to accurate as possible. And this condition is basically doing that. It's saying that I want the [? error ?] to be orthogonal to the basis functions, in the sense that, for two functions, the integral of the two functions like this, is like an inner product, just like the inner product between two vectors. So I'm saying, like, in the vector space, in the function space that I'm working in, I don't want to have any error. That's what this is saying. But I'm going to have is that there's other g terms which are the rest over here. These guys will not be orthogonal to the first part. Does that make sense? So there's still some error left, in my g. But the part in here, I can make a good [? answer. ?] So that's where this equation comes from conceptually. All right. And we'll come back-- when we do least square [? setting, ?] we'll come back to that same idea. So this is what the equation looks like, and the disadvantage of this one is it involves some integrals. And so you have to be able to evaluate the integrals. All right. So that's the downside of this function. So cleverly choosing using your basis functions to make it easy to evaluate the integrals is like the key thing to make this a good method. And just like we needed analytical expressions for the derivatives here, now we need analytical expressions for integrals. That we're going to get from this guy. OK. So, let's think about what basis functions we can choose that might make it easier to evaluate the integrals. Suppose we can write g explicitly. Like, this. So, g is equal to d gx dyn/dt minus f/n. Suppose. OK? Then what integrals am I going to get? Well, dyn/dt is the sum of the derivatives of the basis functions, and y is just this sum up there. Some of the basis functions. And so, I'm going to have integrals that look like this is like I have an integral phi j summation d phi prime. No, k. t. dt. That will be the integrals from the first term. And then I'll have minus some integrals from the second term here. So, phi j fn ft y, where y is the sum-- it's actually like a matrix. All right. Because this y is a vector. It's all the yns. And so, the whole y vector is d times the phi vector. All right. Where d's that big matrix. The elements are [? dnk. ?] And so, if you cut these integrals, this thing here is a summation of [? dnk. ?] The integral of phi j. Phi k prime. And so, if you choose your basis functions cleverly, you might be able to know all these integrals analytically. OK? But if you don't choose them cleverly, who knows what kind of horrible mess you'll end up here. All right. Ideally, you really want to get these all analytically so you don't have to do numerical integration. As a loop inside, all the rest of the work, you're going to do in this problem. OK? And then, these ones also, now knowing something about this is really important or you may have to do the numerical integration here, because this could be really a mess. You have a lot of phis inside some function, which could be a non-linear function of these guys. And so, in principle this could be really horrible. So you may have to do numerical quadrature for those guys. All right. OK, and so, if you do Galerkin's method, a big part of that is thinking ahead of time, oh, what basis function am I going to use? How can I make a basis function so I can evaluate the integrals easily, and then I might have a chance to do it? All right, questions so far? AUDIENCE: Phi j PROFESSOR: Phi j. Sorry, phi j of t. Here, I'll get rid of the t's. These are both functions of t. You're integrating over t. And these integrals from t0 to to t phi. Your domain. All right. Now in Galerkin's method, in addition to these integral equations, I still have the integrals, the equations, that the boundary conditions. So I still have some equations that look like collocation equations that are evaluated at tm. Do I have that anywhere? Nowhere. I have an equation like this, except it's not g, it's q, it's the boundary condition equations have to be true at the boundary conditions. Right? So they're the same as before, q [? or just ?] before qn of dy/dt evaluated at tn y of tn. tn. This is equal to 0. This is sort of the general way to write a boundary condition. And so there'll be some special ends, the boundaries, where I want to have extra conditions. I'll get some equations like this. These have to be satisfied in Galerkin's method as well and they will not be integrals. They are just that, tn. And so in addition to the integral equations, that you have to solve over here, you want these things to be zero and you also want to satisfy the boundary conditions. So then, all these methods, a big part of it gets to be cleverness of a basis function, a choice of basis functions. And there's kind of two families of approaches. So, one family is global basis functions. So basis functions that are defined on the whole domain. And there are special ones, sines and cosines, Bessel functions, all these functions inferred from your classes, all the special functions. And a lot of them, the integrals are known for a lot of cases. There might be special tricks that make the integrals easy to evaluate. They can satisfy the boundary conditions automatically. And some of them, for example, many of the problems we have with like the heat flow equation, so you have-- actually it won't be d. I'm probably going to get this wrong. It's, like, kappa or alpha. Which one is it? Alpha. Alpha d square of td, x squared minus, there's some source of heat that might depend on the temperature, and this has to be, say, equal to zero. Right, does this seem what you've seen in classes before? So this is a common one. So, in this case, you might try t to be a sum of, say, sines. So [? dnk ?] sine of k something something something. In there, and the cleverness of this is that d squared phi k dx squared is going to be equal to some number times phi k. Because the second derivatives of sines are also sines. Right? So all your differentials will solve and in fact, the derivatives will be really simple. Now, the q terms could still be a horrible mess. For example, this could be an Arrhenius thing where's the t's up in the exponent. And then the whole thing might be horrible. Anyway. But at least the differential part is, like, super easy. OK. And you might have a boundary condition say, at one end, that dt/dx at someplace like d final x final is equal to zero. Right? That might be like, you're up against a x insulator. Or something like that. So, you don't have a heat flow. So that would be a boundary condition you might have. And then, by cleverly choosing your definition of the sines, you could force that all the sines satisfy this derivative condition. OK, so, rescale your coordinates so that ends up with pi over 2, an all sides of everything and pi over two is always zero, the derivative, so you're good. So, you can do some clever trickiness to try to make the problem easier to solve. And so, that's one whole branch of these basis function methods, is clever basis functions to match the special problem you have. OK? So that's one option. Then, the other option is the non-clever approach, where you just say, well, I've got a computer. Who cares about being clever? Let's just brute force it. All right. And instead, you just want to write a general method, any problem you can solve. And this is more or less what COMSOL does. And you're going to do it. And so the distinction here is that this kind of thing, this sine function, has a value everywhere, all across the domain. So this is kind of like a global function. OK? And the alternative is to do local basis. Try to have basis functions that are only defined in little tiny areas. And then, at least when I integrate the integrals, I don't have to integrate over the whole range. I don't have to have to integrate right around my little basis function. So, this global basis function-- this is sort of similar to what we're doing, interpolation? Do you remember you could use high order polynomial to interpolate, or you could alternatively do a little piecewise interpolations, say, with straight lines between your points. And, you know, it's not so clear, actually, which would be the best way to do it. Or maybe you want to do some combination. Do little parabolas between little triples of points, or something. That might be a good interpolation procedure. It's the same thing here. Some problems, you can find a global basis set that works great and you should use it. In other problems you might do better to just break up the domain in little tiny pieces and then do simple little polynomials or something in those little domains. All right. So this is the global basis. Let's see a local basis. It's OK to delete this? So, local basis functions. A really common choice for these guys are the b-splines. And, in particular, first order of b-splines and these functions have the shape phi phi k, it looks like this. So, at zero, all the way up to ti minus 1. Then it goes up to one at ti, and then it goes down to zero again, and goes out. OK. So this function is a b-spline. It's also called a tent function because it looks like a tent. Some people call it a hat function. I don't have a pointy head so it doesn't look like a hat to me, but they call it a hat function so I guess they must have hats like that. And this is a very common basis set. And the nice thing about this basis function is, it's zero except in this little tiny domain around it. OK? And it's the only basis function in the whole set that has a non-zero value of ti. And so if you want the function to equal something in ti, it's going to be equal to the coefficient of this basis function, right? Because we're going to write y n of t is equal to summation [? dnk ?] phi k of t. And so, if I really care about yn evaluated at ti, the only one basis function this whole sum is going to have a non-zero value there. So that's going to be equal to dn dnk, k where this is the special k. k prime matches up to ti. OK, so there's one base function that looks like this. There's another one over here that's the one base center on this point, and there's another one over here. So on this point, it's [INAUDIBLE].. And I have as many basis functions as I have points. So this is like a way I discretize the problem, but I've kept my solution as continuous function because my y-- that's the sum of these guys-- has a value everywhere. It's a continuous function. The way this is written it looks like it might not be differentiable at all the points because it has all these kinks, but there's a clever trick you can do to deal with the kinks. So, actually, it's not a problem. AUDIENCE: So, for these basis functions, how would you define if you had the boundaries? Like t0? PROFESSOR: So, you'll have a basis function at the very end. Suppose this is t0 here. You have one like that. OK, so it's just a half of a tent. Can you get a [? sparse ?] d-matrix? Yes. So locality is really good because it makes the Jacobian matrix sparse, the overall problem. So, when I compute the integrals of this thing, for example, the integral of phi i of t, phi i minus 1 of t dt, this turns out to be equal to 1 or 2 times ti minus ti [INAUDIBLE]. No. I take that back. Just one half. Just half. So it is like a brilliant thing. Is that right? It does have a ti [INAUDIBLE],, not a [INAUDIBLE].. I'll have to double check. Possibly including delta t. I can't remember. All right. But the integrals are very analytical, and only certain ones are non-zero. So only one that the two is when the two is differ by one unit, do they have a non-zero integral? All the rest of them are non-zeros. So, when I write down these equations, I mean, many, many, many, many zeros, so it looks horrible when I write Galerkin's method, I get so many integrals, but actually a zillion of them are zero, and then there's a bunch of special tricks I can do that make it even better than that. OK? So then, the Jacobian sparse. So, that'll save you a lot of time in linear algebra, which allows you use a lot of points. So, you can use a very large basis set because you end up with sparse Jacobians, I mean, it's not going to fill your memory storing all the elements in the Jacobian, even if the number of points is very large. And also, there's vast numerical solution methods for those. So that's the idea of this. I guess should we try to carry one of these out? You guys are [INAUDIBLE] trying to do a lot of algebra on the board? Do you think I can do the algebra on the board? That's the real question. All right. Should I do it for collocation, or you guys confident you can do collocation? You're all right with collocation? You're fine with collocation. Great. OK. So we'll just go right into Galerkin. So, Galerkin. That way, we use a local basis. So, most of the equations we have to solve are this type. Phi j t times the residual function of summation [? dnk ?] phi k prime. Summation [? dnk ?] of phi k and t. Is equal to zero. All right, we have a lot of equations like that. We're trying to find the d's that are going to force all these integrals to be zero. OK? So we have a lot of different j's we're going to try. We want this to be true for all the n's. And then we're going to adjust these [? dnk's ?] to try to force this to be zero. All right? That's the main problem. And then, on top of this, there's some equations with boundary conditions. So, now we have to look at what the form is. If the form is, oh, I erased it, if I can make this explicit in the derivatives, which I can do very often, for example, this could be dyn/dt minus [? fm. ?] So then, dyn/dt is just this. And then I'll have another term, [? fm, ?] which will depend on that [INAUDIBLE]. And so the integral phi j, and I'll just remind you that phi j is not equal to zero. If tj minus 1 is less than tj plus 1. All right. That's the only places where my local basis function is non-zero. That's where the tent is. And all the rest of it's zero. So when I have this integral, originally have it t0 to t-final, but actually I could replace this with tj minus 1 to tj. And it's just the same. Because of all the integral outside that domain is zero. Because this function is zero. All right? So at least I have a small little domain to do the integral over. AUDIENCE: [INAUDIBLE] tj? PROFESSOR: tj plus 1. Thank you. Yes. Yes. Plus 1. Yes. Is that all right? OK. So, I have this integral, and then I have this times the derivative term, so it's dnk dk prime minus the same integral, j minus 1. J plus 1. Phi j sorry [? fm ?] [INAUDIBLE] summation of dn. k phi. k [INAUDIBLE] t. OK. Now, this derivative of phi k, well, I just told you what it was. It's just like the [INAUDIBLE] set functions. So, this integral, you'll know analytically. Like, it's either 0 or it's 1/2. Maybe minus 1/2. Whatever, but it's nothing complicated, and you'll just know it right off the bat. So this is all that can be known. And sparse. It's only going to be when k is equal to j minus 1 or j plus 1 that this will be non-zero, and all the rest of it would be zero. OK? So, that's mostly zeros. This one over here, in principle, I have quite a huge sum here, k equals 1 to k, where I have all my basis functions, which is all my points where I put my little tents down. So I have a domain and I've parked a lot of tents all on the domain, that's my basis functions. And I'm trying to figure out, sort of, how high the tent poles are. On all those tents. And my tunnel functions, the sum of all the heights of those tents. OK? But this guy is only non-zero in this little domain, and these guys, most of them are zero in that domain, because they're mostly tents that are far away from where my special tent phi j is. OK, so I'm going to draw a picture. Here's the domain from t0 to t-final. Over here is my tent corresponding to phi j, which is centered around tj. Hence, this function. And then, these guys are all the other tents. There's a tent here, there's a tent like this, there's another tent like this, another tent like this. All these tents. Those are all the basis functions starting from k equals 1 and going up. All of these guys are zero in this domain because they all have zero tail. So, almost all of these, the f's doesn't really pick up anything from any of those guys in the domain I care about. The only domain I care about is this domain right here. And this whole sum is all zero except for a few special k's when k is equal to j minus 1, j, or j plus 1. So I only have to worry about three terms. The inside contributing to the y in the region of t that I care about. All right? So, when I want to compute the Jacobian with respect to d, the only terms here where the Jacobian's going to be non-zero are for d I have a Jacobian, which is the derivative of this whole thing, ddd nk. Right? And that's only non-zero if k is equal to j minus 1 j, or j plus 1. You guys see that? How many people do not see this? How many people are lying? This is a really important concept for the locality, so this is the advantage of a local basis, that the Jacobian will turn out to be really sparse because it's only non-zero in this special case. And you might have 1,000 points, 1,000 of these little tent, 1,000 basis functions in the sum. But only three of them are non-zero. For four each n. So, actually, it's 3 times n. [INAUDIBLE] non-zero. Is that all right? Because it just depends on the k. Three of the ks are non-zero. There's 1,000 of these guys [INAUDIBLE].. Yep. So that's the big trick of locality. And then, also, because these integration ranges are so small, because you choose your points really finely spaced, then you might get away with simple polynomial expansions of f, for example. Around the points, or [INAUDIBLE] expansions, there are all kinds of little tricks. Or you could even do quadrature and determine some points, do a [INAUDIBLE] quadrature to evaluate the integral. But you only need a few points, because you know it's a little tiny range of dt. And you would hope that you've chosen so many fees, so many time points, that your function doesn't change much from one time point to the next. So, more or less, first order is constant. Your function's constant with respect to t in this little tiny domain, and then maybe has a little slope. And then, if you're really being fancy, you might be able to put a parabola on it. But it's not going to change that much. If it's changing a lot, that's telling you you don't have enough time points and you should go back and put some more basis functions in. And then you can get a good [INAUDIBLE].. Because what we're, really doing here is we're using this basis set. If I add up these guys, what I'm really doing is piecewise linear interpolation between all the points. So I'm approximating my y versus time. It's going to look like this. All right. It's piecewise linear, because that's the only thing you can make from adding up a bunch of straight line segments. A bunch of tents is a bunch of straight lines. And so this is what the approximate function is. Well, you can see, this is really bad if these functions are too different from each other. But if they're all like this-- and you think, well, maybe it's not so bad to use piecewise linear. Yeah. AUDIENCE: [INAUDIBLE] confused at where we're using [INAUDIBLE]. PROFESSOR: Inside [? f ?] solve, what is trying to solve for the d's, it's solving each of this giant set of the equations, a whole lot of equations like this, they all come in to a gigantic f. That's the f of d that need to make equal to to zero. And so, I'm [? burying ?] the d's, I need the Jacobian of that gigantic f. Now, the problem, because it's gigantic, I need a lot of points, a lot of closely spaced points to make my function look smooth. That means the number of base functions is really huge. So the Jacobian principle is huge number squared. I have 1,000 points, say, discretizing my domain. And then it's 1,000 by 1,000 Jacobian. It might be really hard to solve it, or cost a lot of CPU time, or use a lot of memory. But, fortunately, almost all the elements are zero. So, it's like, has a sparsity of 0.3% or something is the occupancy. So, it's almost all completely sparse, and therefore you can solve it even though it's gigantic. Yes. AUDIENCE: [INAUDIBLE] basis function [INAUDIBLE] the Jacobian [INAUDIBLE] should be [INAUDIBLE] PROFESSOR: We're trying to find d, so we're really trying to solve a problem that's f of d is equal to zero. That's our fundamental problem we're trying to solve. We're doing Galerkin. We have something equal to zero, and it depends on d. And we're trying to find the d's. So this is really the function we're trying to solve. In order to evaluate the elements in this, we have to compute a whole bunch of integrals with the Galerkin method. So that's the complexity. But the basis functions, we know what they are. We've pre-specified them. So, the whole question is just what the d's are. And the d's get multiplied by a whole lot of integrals. Is this all right? I think I may have misunderstood the question. So you have f of d is [INAUDIBLE],, so therefore we probably want to know j with respect to d. OK. Time ran out before clarity was achieved. We'll try to achieve clarity on Wednesday morning. |
MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 22_Partial_Differential_Equations_1.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So today we're going to talk about partial differential equations. Let me just give a quick review where we are. So we're going to have a PDE lecture today. Then Professor Swan is going to do a review on Friday. There's no lecture on Monday, but you have the quiz, too, Monday night. Wednesday, next week is Veterans Day so it's a holiday. There's no classes at MIT. So the next time I'll be lecturing to you will be a week from Friday. The homework following the quiz will be posted and so you guys can get started on that. That homework involves a COMSOL problem which, for those of you who had trouble with COMSOL, might take some extra time, since it's the first time you're doing it. So I would suggest you at least give it a try early. I'm sure the TAs would be very happy to help you to force COMSOL to cooperate. Bend COMSOL to your will. All right, so we're going to talk about partial differential equations. And, as it sounds, what's different about a partial differential equation is, it has partial derivatives in it. And so we have a lot of equations like this. So for example, I work a lot on this equation. It's a poor choice, isn't it? OK, so this is the conservation equation for a chemical species in a reacting flow. So the change in the mass fraction of the species z, at a point, xyz at time, t. That differential is given by a diffusion term, a convection term, and a reaction term. So this is a pretty famous equation. This is part of the Navier-Stokes equations for the whole reactive flow system, which I hope you guys have seen already, and for sure you will see, if you haven't seen them yet. And there's a lot of other partial differential equations. I work a lot, also, on the Schrodinger equation. That's another partial differential equation. And what's special about the partial differential equations is that, in this case, this partial derivative is respect to time, holding all the spatial coordinates fixed. And these partial derivatives are respect to space, holding time fixed. A very important thing about partial derivatives, which you probably have encountered in 1040 already, is, it really depends what you hold fixed. And you get different results if you hold different things fixed. Now the implication, when you write an equation like this, is that whatever partials you see-- When I wrote this, if I have this and it has a term that's like d squared, dx squared, is one of the terms in this thing. This says t. This says x. The convention is, oh boy, I better hold x fixed. And here must be, I must hold t fixed. Because there's another one in the same equation. Now, you can change coordinates however you want. Just really watch out that you carefully follow the rules about what you hold fixed, when you change things. Otherwise you can cause all kinds of craziness. Now you guys probably took multivariable calculus at some point in your distant past. And they told you very carefully what the rules were. So follow them. And I suppose, when you're doing thermo right now? Have you started to do all this partial derivative stuff? Yeah, so you've seen how completely confusing it can be with negative signs showing up all over the place. And other things like this. I don't know if you've encountered this yet? At least, I thought it was confusing when I took it. So just be aware. It's the same thing. Now, often you do want to change the equations. So you can write down an equation like this. But, for example, if you have a special symmetry of the problem, like it's cylindrical, for example, then you might want to change the cylindrical coordinates. And then you have to just be careful that you write that correct Laplacian and cylindrical coordinates. Make sure it doesn't mess up anything about what you're holding fixed. Sometimes you might want to be crafty and use some weirdo coordinates. For example, if you wanted to solve the Schrodinger equation for h2 plus, it turns out there's a special coordinate system called the elliptic coordinate system. And in that coordinate system, the partial differential equation is separable. And so, if you do it in that coordinate system, you can actually solve analytical solution, which is pretty unusual for the Schroedinger equation. But it's a pretty goofball coordinate system and you really have to watch out, when you convert, to make sure you do it just right. But it's just the rules, chain rule and the rules about how to keep track of what's being held constant. And when you change, will tell them, how, being held constant, how to change it. Now in principle, this problem is like no different than the ODE BBP problems we were just doing. You just have a few more terms. And so, you can use finite element. You can use finite differences. You can use basis set methods. And it's really no different. So all those methods basically look the same. You just get equations that have more unknowns in them. Because, for example, if you're using finite element methods or basis set methods with a local basis set, you need to put points both in the x direction and the y direction, and in this case, the z direction, and then maybe also in the time direction. So now you have a lot of mesh points in all different directions. And so you get a lot of mesh points. And so fundamentally, there's no problem, but in practice, there's a big problem if the number of unknowns gets to be so large. That's the main problem. So think about it. You have some kind of equation like this. It's in three spatial dimensions and in time. How many points do you think you need to discretize your solution in the spatial coordinate? There might be somewhere, might be 100, I don't know, points might be enough. And so now I have x and y and z. So I'll need, well, just in COMSOL, you saw this on Monday. Suppose I just have two dimensions. Suppose I just have x and z. And I start putting points in. Well, I want to have some points along this way. And I want to have some points along this way. And so, I actually have an unknown here and an unknown here, and one there and one there, and one there and one there. There's quite a few of them. I think of how many points I have. Each of those is a point where I have a little basis function. And each one of those has a magnitude of how much weight I have on that basis function, things they called the d's before in the basis functions. And I have a lot of them. Now, it's going to be, if I have 100 points this way and 100 points this way, I have 10,000 of these little points. That means I have 10,000 unknowns I have to solve for. 10,000 unknowns are a lot harder to solve for than 100 unknowns. and if you saw, on some of the problems we did with the ODE BBP problems, even with 100 unknowns, we were having a lot of trouble, sometimes, to get the real solution. All right? So you can just see how it just gets to be, like, a big numerical problem. So that's two dimensions. The third dimension, you'll get another 100 times as many points. You might have a million unknowns. And once you get up to a million unknowns, we're starting to talk serious math, now. OK? Even for your good computers that you have. In addition, it's not just you have the number of mesh points, but it's the number of mesh points times the number of variables at each mesh point. So in the Navier-Stokes equations, what variables do you have? AUDIENCE: [INAUDIBLE] PROFESSOR: Louder. AUDIENCE: [INAUDIBLE] PROFESSOR: Velocity and pressure, OK. So you have pressure. We have the different components of velocity. What else we got? AUDIENCE: [INAUDIBLE] PROFESSOR: Sorry, what? AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah, though, yeah. Density or temperature. Yeah, temperature maybe, so. So maybe temperature would be what you might use. And then you'd have a state function maybe with these guys. What else would you have? So then you'd have all these species, right,? The mesh fractions are all the species. So you have, like, y, species one. Mesh fraction, species two. However many you got. So that's a pretty big set of variables and you need to know all these numbers at each point. So this might be-- how many have we've got? One, two, three, four, five, plus however many species we have? So this is n species plus 5 times the mesh. And I told you the mesh was what, a million? So maybe I have 10 million unknowns. So the issue, to a large extent, has to do with just, can I solve it? Can I solve a system of 10 million unknowns? 10 million equations, 10 million unknowns. All right, so that's going to be one big issue. Now the other issues, we still have the same problems we had in ordinary differential equations, that we have to worry about numerical stability. We have to worry about actually, is the whole problem even stable to begin with? Physically? Because if it's not stable physically, then probably we're not going to get a stable solution. So you have all those kinds of problems that we had before in the ODE case, and now it's just sort of amplified by the fact, now we have boundary conditions in more than one dimension. So we have multiple chances to screw it up. And we're trying to figure out how to set the boundary conditions correctly. And we have to worry about stability in all the different directions. So those are the different kinds of things we have to worry about. But it's all the same problems you had before, just amplified. Now, because of this problem of the very large number of unknowns, currently, if you have a problem that has two dimensions, you can kind of very routinely solve it. And things like COMSOL on your laptop, you can usually do those. When you get up to three dimensions, you might be able to solve it or might not be able to solve it. And it could be, like, a really big challenge sometimes to solve it. And then in reality, we have problems like the Navier-Stokes is four dimensions, right? There's time, as well. So then that gets to be really hard. And then the Schrodinger equation has three times the number of electrons dimensions in it. So that gets impossibly hard. And so, specialized methods are developed to handle those systems with a lot of dimensionality. So, like, for the Schrodinger equation, people almost always do it with basis set methods. And there's been a huge effort, over decades, to figure out really clever basis sets that are really close to the real solutions, so that you don't need to use too many basis functions to get the good solution. So that's the way to keep the number of variables down. In the Navier-Stokes world, I'll show you what people do to try to deal with that. There's many other problems. I don't know all the problems in the world. But in general, when you have three or more dimensions in your PDE system, you're looking up special tricks. You're looking up, how do people in this field deal with this particular PDE? Special ways that are good for that kind of PDE. All right, so I said, one problem was that you could have, that the PDE system, itself, can be intrinsically unstable. So one example I know like that is, detonation problems. So if I have a system that has some explosive in it, or a mixture of hydrogen and oxygen even, and I have it in a tube or something, if I change my initial conditions, I can go from the case where it's stable, that it just sits there, a hydrogen and oxygen. Or it could turn into a regular burning. Or if I change the conditions a little bit differently, it could change into a detonation, where it actually sends a shockwave down the tube faster than the speed of sound, totally different situation. So those kind of systems can be really sensitive to the physical initial conditions. A very small spark can make a really big difference in a system like that. And also, in your numerical solution, if the thing is physically sensitive, it'll typically be sensitive also to numerical noise. If you get numerical noise at some point, that might be enough to kick it over from, say, not igniting at all to having a detonation, which is a completely different solution. But another case, like in biology, if you're trying to study the growth of a cell culture or the growth of a tumor, as you know, if somebody has a tumor and one of the cancer cells mutates and does something different, it can have a really different outcome of what happens to the patient, what happens to the tumor. The whole morphology can change completely. Everything about it can change. So that's the kind of system, also, that can be very sensitive to small details. A small change in one cell, for example, could be a really big difference. So this is not just in combustion problems you have these kinds of instabilities. All kinds of problems have this kind of thing. You can have just regular chemical reactors, like in multiple steady states, little tiny fluctuations can make it jump from one steady state to another one. That's another very famous kind of problem. There's another class of PDEs where it's stable in some sense, but it causes a problem. We call those hyperbolic. And those are like wave equations. The solutions are propagating waves. So the equations of acoustics, the equations of electromagnetism, are wave equations where, basically, a wave propagates more or less without any damping. So what happens, then, is you have some numerical noise that will make a little wavelet that will propagate, more or less, without any damping, too. And it'll keep bouncing back and forth inside your domain numerically. And if you keep introducing numerical noise, eventually the whole thing is just full of noise. And it doesn't have much relation to the real physical solution. So hyperbolic systems of differential equations need special solution methods. That in the numerical method, it's carefully trying to damp out that kind of propagating wave that's coming from the noise. And they set them up a special way. I'm not going to talk about how you solve them in this class, but there's a special, whole group of solution methods that people use who are solving acoustics equations and solving electromagnetic equations. If you get into solving shockwave equations-- I actually work for Professor [? Besant. ?] He has, like, electrochemical shocks in some of his systems. You have the same kind of thing. You have to use special mathematics to correctly handle that, special numerical tricks. But we're not going to get into that in this class. If you get into those things, you should take the PDE class for hyperbolic equations and they'll tell you just all about the solvers for those kinds of things and special tricks. And there's journal papers all about it. Here, what we're going to focus primarily on two kinds of problems: elliptic problems and parabolic problems. And those problems are ones where there's some dissipation that's very significant. And so, in an elliptic problem, you introduce some numerical noise and as you solve it, the numerical noise kind of goes away. Sort of like intrinsically stable all throughout. Now the real definition of these things, parabolic, hyperbolic, elliptic, has to do with flow of information. So in a hyperbolic system, information-- there's regions of the domain that you can make a change here and it doesn't make any effect on some domain part over there. And so that causes-- because that is the real situation, that also propagates into the numerical solution method. So for example, if I'm modeling a shockwave that's moving faster than the speed of sound, I can introduce any pressure fluctuation I want behind the shockwave, and it has no effect on the shockwave. Because the pressure waves from that pressure fluctuation are traveling at the speed of sound. They won't catch up to the shock. So they will have no effect whatsoever. So if I was judging what was happening by looking at what's happening in the shock, I can introduce any kind of random noise I want back over, downstream of the shock, I guess. Yeah? And it won't make any difference at all. And so, that's part of the reason why I need a special kind of solver. If I have elliptic equations, every point affects every other point. A famous case for that is, like, this steady state heat equation. You have some temperature source here, a heat source here, and you have some coolness source over here. And there's some heat flowing from the hot to the cold. And the heat flow is kind of slow enough that everything sort of affects everything else, since you actually have, the heat is trying to flow by, sort of, every path. And there's some insulation that's resisting its flow. And it has been in the connectivity of flow a certain way. And those kinds of equations are called elliptic. And the methods that we use for the relaxation mesh that we showed are really good for those kinds of methods, for those kind of problems. And then another kind of problem that we get into a lot is like this one. This is called parabolic. And typically, in the cases we see that are parabolic, we have time as part of the problem. Always what we think should happen is that what happens in the future depends on what happened in the past, but not vice versa. So you would expect that when you're computing what's happening at the next time point, it's going to depend on what happened at earlier time points. But if you try to do it the other way around, it would be kind of weird, right? You wouldn't expect that what in the future really affected stuff in the past. So it naturally leads us to sort of pose it as an ODE IVP, or IVP kind of problem. And we know initially, some time zero something, and we're trying to calculate what happens at some later time. And so those kinds of problems we call parabolic. And the methods that we're going to talk about mostly are ones for solving parabolic and elliptic problems. And you can read, in the middle of chapter 6, has a nice discussion. Just has the mathematical definitions of what these things are. Now, even in a steady state problem, the problem can have a sort of directionality to it. So if you have a problem that's very convective. So you have a flow in a pipe. And you have a nice, fast velocity flow down the pipe. What's happening downwind is very much affected by what happened upwind, but not vice versa. So maybe you're doing some reaction upwind. Say you're burning some hydrogen. You're going to get water vapor made upwind and it's going to flow downwind. And you're going to have steam downwind. If I squirt a little methane in downwind, it doesn't actually do anything to what happened upwind. I still have a hydrogen and oxygen flame up there and it's been burning and making steam. Do you see what I mean? So it has a directionality, even though I might do it as a steady state problem and not have time in it at all. And you've probably already done this, where you can convert, like, a flow reactor. You can write it, sort of, as a time thing or as a space thing. Time and space are so related by the velocity of the flow. We'll have similar kinds of phenomenon then. It's almost like a parabolic time system that, if the flow is fast enough, the fusion backup, anything moving upstream is negligible. It's all moving from upstream to downstream. And so then, you'll have the same kind of issues. In fact, you may even try to solve it by just starting it at the upstream end and competing stuff and then propagating what that output from that first one does to the next one downstream, and then do them, one at a time. That would be one possible way to try to solve those kinds of problems that are highly convective. As you slow the velocity down, then the diffusion will start to fight the velocity more and more. If you make the velocity very slow, the Peclet number very low, then you really have to worry about things diffusing back upstream. And then it's a different situation. And that would be more like the elliptic problems we have. Because things downstream would actually affect upstream, if the flow is really small. But typically for our problems, the velocity is always really high. Peclet numbers are really large and so not much goes from downstream to upstream. It's almost all upstream to downstream. This leads to a famous situation. If you look in the textbook at figures 6.7, 6.8, it's kind of a very famous problem. That if you just do this problem. So this is just an ODE problem. So I just have convection and diffusion, no reaction, very simple. This problem. You'd think that I could just use my finite differences, where I would put, I could use the center difference for here and here and change this into this obvious looking thing. You think that that would be a reasonable thing to do, right? And you can do the same one here. You can write-- I'm terrible with minuses, sorry. So these, the finite difference formulas. And so, you can write these like this. This stays equal to zero. And now it's just an algebraic problem to solve. A system of equations, you have equations like this for every value of n. And there's the mesh points. You guys have done this before, yes? Yes. OK, so famous problem is, if you actually try to solve it this way, unless you make delta z really tiny, you get crazy stuff. So on figure 6.7, they actually show what happens for different values of delta z, which is different values of what they call the local Peclet number. So there's a thing called local Peclet number, which is-- OK? If you make the local Peclet number too large, actually anywhere bigger than 2, and you try to solve this equation, what you get is oscillations, unphysical oscillations. They'll make the phi go negative. It's crazy. It looks like the simplest equation in the world, right? It's a linear equation, so what's the problem with this? But it doesn't work. And so, then you could think, well, why doesn't it work? And there's, like, multiple ways to explain this about why this doesn't work. I think that the best way to think about it is about the information flow. So if the local Peclet number is large, that means that phi is just flowing downwind and diffusion is not doing much. Because the Peclet number is big. And so, you really shouldn't use this formula to calculate the flow. Because what happens downwind, at point n plus one, I guess. Here we have the flow this way. So here's n minus one. Here's my point n, n plus 1. I really should not include, is this right, n plus 1 in this formula. Hope I have those signs right. I apologize if I had them backwards. Because I don't think that what happens downstream should really affect this at all. So really, I would do better to change from this center difference formula to what's called the upwind difference formula. Where they would say, OK, instead, let's use phi of n minus, the phi of n minus 1 over delta z. Just use that instead of using this formula. Now, you wouldn't think that that would make much difference. But it makes a gigantic difference. And so if you look at figure 6.8 in the textbook, you see you get perfectly stable solutions when you do that. But where you do this one, you get crazy, oscillatory, unphysical solutions, unless you choose delta z really tiny. Now, there's other ways to look at this. So I think this is the correct way, the best way to think about it is, it has to do with information flow. If you try to make your solution depend on stuff that it doesn't really depend on. So upwind does not really depend on downwind. If you force it to, by choosing to include that information here, you end up with crazy stuff. So that's one way to look at it. Another way to look at it is that, if you really want to compute these derivatives, you've got to keep delta z small. Because we're taking a differential. We're turning it into a finite difference. And if you take your delta z too big, then it's a terrible approximation to do this. And you can look at it. You can see that what we're doing is, when we have this problem where the local Peclet number is getting too large, is we're actually choosing delta z bigger than the, sort of, the thickness of the solution. So if you look at the solution, the analytical solution of this problem, it's like this. Where this has a very, very thin layer at the end. It's like the flow has pushed all your stuff to one side. And if you choose your delta z to be from here to here, that's how big your delta z is, and then you're computing the derivative of this point halfway along from here to where this is. You compute your derivative like that. It's not a very good, not a very accurate representation of what the real derivative is here. And in fact, you really should have a bunch of points here and here and here and here and here to really resolve the sharp edge. And so you did something really crazy to use a big value of delta z to start with. So that's another way to look at this problem. Now, the fix, by doing this upwind differencing, I think the best way to look at this is saying, well, I'm just going to make sure I only depend, I only make the equations depend on what's upwind, with the convective term. Because there's nothing convecting from downwind. So just leave that out. So that's a conceptual way to think about it. Another way to think about it, which is in the text, is that by doing this, you're introducing another thing called numerical diffusion, where you're actually, effectively increasing the diffusivity. Because this is a very poor approximation to the differential. This is a asymmetrical finite difference formula. It's not a very good value estimate of the derivative, if delta z is big. But you can write it out carefully and write this out and say, oh, this actually is sort of like this plus an effective diffusivity chosen very carefully, so the terms cancel out just right. And it turns out, if you look at it that way, the effective diffusion you've added, by using this instead of that, is just enough to make the local Peclet number, if you compute it with the effects of this d effective, to stay less than 2. Which is what you need as a condition to make this numerically stable. So I strongly suggest you read that. It's only two pages long in the textbook. It's very clear. Just read, just look at it if you haven't seen this before. There's a similar thing and it might be relevant to the COMSOL problem that Kristyn showed you on Monday, is you can-- In that case, there was a region of the boundary here that had some concentration c naught, and then there was, over here, the concentration was zero. The boundary is zero. Here, it's some number, 17, whatever number it was. And if you think of how to model what's going on, one way to look at it is, I have a diffusive flux coming in from the drug patch, diffusing the drug into the flow. And I have this big, giant flow, flowing along here. And the flow gets slower as I get closer to the wall because of the friction with the wall. And so if I looked somewhere right in here, I could just draw a little control volume, and look what's happening. I have a diffusive flux coming in here. I don't have any drug up here, so there's no drug coming in here. But I have drug leaving. So basically, it's coming up and it's making a right-hand turn. And so then I could think, well, if I was going to try to figure out, what's the steady state concentration of drug in there, I could say, well, there's a certain amount of drug entering from the diffusive flux, which would be just this times dcdy. This is the y direction. And then this part over here would be the velocity, actually, just times the concentration I have in here, right? So how much is flowing out. And it's probably, my units are screwed up so there's-- Well, maybe not. That's OK. Is that all right? Units are screwed up. You guys will get it right. It's probably an area. Something to do with this size right here. This one here, too. Is that right? So you could compute what the steady state concentration would be if it's just coming in and flowing out. And so, this is the finite volume view of this kind of problem. And so you can write the equations that way, instead. And what you really care about are what the velocity flows and the diffusive fluxes are on these boundaries around the point. So you don't try to compute things right at the point. You compute around it. And one way to look at this is, we just did finite differencing. We were looking at what was happening coming in. And then, this is actually a center difference around this interface, which is halfway between the two points. And so actually, this is a good approximation for the finite volume point of view. This is all right? Now, these are all different ways to write down the equations. All of them are correct in the limit that delta z goes to zero. If you have an infinite number of mesh points, all these are exactly the same. But they lead to really different numerical properties when delta z is large. And they're all inaccurate when delta z is large, so it's all about what happens with the inaccuracies. Now, we can go ahead and take any problem, even really complicated 3-D problems or 4-D problems like this, and we can write them with relaxation methods. And what we're basically doing in a problem like this is turning it into some function of y is equal to zero, where y is the value of all the unknowns. And there's a lot of equations. So I might have 100 million equations and 100 million unknowns that are the values, or every state variable at every state point at all times. And I can write it down, by finding differences, by finding elements, by finding volumes. Any way I want, I can write down an expression like this. And I know in the limit, as I make the number of elements of y go to infinity, it will actually be the real solution. And the problem is, my computer can't handle infinite number of unknowns. In fact, it's even going to have trouble with a 100 million unknowns. And so, I have to figure out what to do. So how would I solve this normally, is I would take the Jacobian of this and I would do, I'd take an initial guess and I'd update it and have the Jacobian times my change from the initial guess to the equal negative of f, right? You guys have done this a million times. As Newton-Raphson is how you'd find improved from an initial guess. You have to provide some initial guess. This is actually pretty hard if you have 100 million unknowns. You have to provide the value of the initial guess for every one of the 100 million. But somehow you did it. And now you want to refine that guess and this is the formula for it. And so, you just use backslash, right? But the problem here is that now f is a 100 million long vector and this is a 100 million squared matrix. And a 100 million squared is pretty big, 10 to 16th elements in it. So I'm not going to just write that matrix down directly like this. Now, we cleverly chose local basis functions, so this is going to be sparse, this Jacobian. So most of the numbers, are those 10 to the 16th numbers in this matrix are zero. So we don't have to represent them. But there still might be quite a few. Probably the diagonal events are non-zero, so that's like 10 to the 8th of them right there. And we've got, probably, a couple of other bands. So I don't know, might be 10 to the 9th non-zero elements inside this thing, which is pretty nutty. And depending on how much memory they provided in the laptop they gave you, you might be filling up the memory already just to write down j. And certainly, if you tried to do Gaussian elimination to solve this, you're going to have a problem because you get what's called fill in. Do you guys remember this? Even if you start from a very sparse Jacobian, as you run the Gaussian elimination steps the number of non-zero entries in the intermediate matrices gets larger and larger. Remember this? And so, even if, initially, you can write down j and store it in your computer, you could easily overwhelm it with the third iteration or something. So you're not going to just use Gaussian elimination. So what do you have to do? You want to find a method. They're called direct. Direct methods to solve this kind of problem, in order to get your increment, which is going to be your update, to get a better estimate of what the solution is. And the direct methods are ones that you don't actually store the gigantic matrix. So you're trading off CPU time for memory. So you don't have enough memory to store the whole thing. Which we, normally you would do this with Gaussian elimination and huge steps. It would solve it perfectly, right? Right, Gaussian elimination's great, backslash. You guys have used it a few times. It's a really nice program. It always works. It's really nice. But you're going to give up that because you can't afford it because the memory, it's consuming too much memory and your cheapo department didn't buy enough RAM in your computer for you. So you're going to instead try a direct method which is, so it's trading off CPU time versus RAM. So you're going to reduce how much memory you actually actively have, but you're going to expect that it's going to cost more CPU time. Because otherwise, everybody would do something else besides Gaussian elimination all the time, and they don't. All right, so what's the method we're going to use? It turns out that variance on the conjugate gradient method turned out really well. So we mentioned this earlier, briefly. What you do is, instead of trying to solve this, you try to solve this. So you just try to minimize that, right? You know at the solution this is going to be zero. Right, jy plus [INAUDIBLE] zero when this is solved. So you try and vary and find the delta y that makes this go to zero by minimizing it. And then this method, it turns out that the conjugate gradient, if everything works great, this is guaranteed in n iterations to go directly to the solution of this kind of problem. Now n is large, now, because we have 10 to the 8th unknowns. So that's 10 to the 8th iterations. You do anything in 10 to the 8th iterations, you're not really sure it's really going to work. So that's a warning. But at least it should get, even after a few iterations, you should get a smaller value of this than you had before. And so this is the method. And what's really nice about this method, if you look into the details of how it works, it never has to store any intermediate matrices. So all it needs is the capability to multiply your matrix times some vector. And from that multiplication, it figures out a step direction. That's the trick of this method. And because it only has to do a forward multiplication, you don't have to actually store j. You can just compute elements of j, as needed, in order to evaluate this top product. This j times v. And you don't have to ever store the whole j at once. So it's really great for RAM to do this method. Now, as I said, when you do 10 to the 8th iterations, things don't always work out so well. And so people have found that when you go more than maybe 10 or 20 iterations like this, you tend to pick up numerical problems. And so, they've worked out better methods. And there's one that people use a lot for these problems called bi cg stab. Which is biconjugate gradient stabilized. The applied mathematicians went crazy trying to figure out better methods like this. And this is the one that, I think currently, people think is the best. Though probably, there's a lot of research on this, so maybe there's even better ones now. But inside Matlab, I think this is the best on they have. And this is a method for doing this. Now, it's iterative. Now, let's remember what we're doing. We're trying to solve this problem. We started with a guess. We're going to break it up into an iterative problem like this, where we're going to do some steps, delta y. This is now doing iterative procedure in order to compute delta y. So we have an interim procedure and another iterative procedure inside the first iterative procedure. So this is going to be a lot more CPU time. So just to warn you. And it's also, it's not guaranteed like Gaussian elimination, to beautifully come to machine precision solution. It's going to get you something. But anyway, this is what people do. So that's one good thing to try. Now, this whole approach is sort of based on Newton-Raphson. We know that doesn't have the greatest radius of convergence. So you need to provide a pretty darn good initial guess. So how are we going to get a good initial guess if you have a 100 million unknowns? So what methods do we know? So one idea is you could do things like homotopy, stuff like that. If you could somehow take your PDE system, get rid of the nonlinear terms. You're really good at solving linear systems of PDEs. You start with that and then you gradually turn the non-linear term on. Maybe you could coax it over. So that's one possibility, if you're really good at PDEs. Another possibility that I've run into a lot is the problem you're trying to solve, for example, might be a steady state problem like this. Where this is zero and you try to figure out what the steady state situation is in a flow reactor, for example. And so, it's actually, your steady state problem came from a time dependent problem. And if you think that your time dependent problem really converges to the steady state problem, sort of without much concern about what the initial guess is, then you could put any random initial guess in and march it along in time enough, and eventually, it should end up at the solution you want. So that's the idea. So that's called a time marching idea. So there's one idea, sort of the homotopy continuation kind of idea. That's one idea about what to do for initial guesses. Another idea is the time marching idea. And the key about it is, you have to really believe that, physically, the system does march to the steady state you want. If it does, then this is a good idea. If it doesn't, then you're not going to end up where you want, because it's not going where you want. And so, you have to watch out. But in some situations, this is true, that things work. So I do a lot of combustion problems. A lot of these things, you light a spark on a stove burner. No matter how you light it, it ends up basically the same flame. So that's a good one. But I have some other ones where I light a candle and I'm in a convective field, a turbulent field, and it flickers. And no matter what, it always flickers. And I never get a steady state, so I'm never going to find a steady state solution with that one. And so I could time march until I retire. My computer's still burning CPU time all that time and it'll never get to a steady state solution. So you really have to know what the real situation is. But if you have one where it's going to converge, then this is a reasonable idea. Now let's talk about the time marching. The time marching, you have, say, dy, dt is equal to some f of y where this is the discretized spatial vergence. I took all the space dimensions and replaced them with the finite difference expressions for the derivatives. Or I did colocation or something to convert this into algebraic equations. So the right-hand side is now algebraic equations, no longer differentials. And now, I'm going to time march this and I just-- This is very long. It's a 100 million. So how are we going to time march that? Well there's kind of two schools of thought for this. One school of thought is, if I can use an explicit method, an explicit ODE solver, those are pretty nice. Because all I have to do is evaluate f and then I can march along and I never have to save any matrices. Because remember, look at those formulas, y, y new, by the explicit method, is equal to some function g of y old. And this is the update formula. It can be forward Euler. It could be RK 2. It could be whatever. But there's just some explicit formula. I plug in the old vector. I compute a new vector. I never had to do any inversions of matrices or anything. So that's good. And this is, indeed, the way that the Navier-Stokes equations are solved currently by the best possible solver. So the best solver in the world discretizes in space and then does ODE IVP using explicit method for marching it. They do problems that are pretty big. So they'll have like 100 million mesh points and then 100 species. So they have 10 to the 10th unknowns. And at each time step, you're computing another vector that's 10 to the 10th long. So it's a pretty big problem. And they use, like, 40% of one of the national supercomputer centers will be running at one time for one job like this. But it works. You never have to store anything that's too big. The biggest thing you have to store are the vectors. Which, you're always going to have to store some object the size of your solution, right, if you're going to get your solution. So that's one possibility. Now, this is really good if an explicit solver can solve it. So that means it's just not stiff, your problem. And also, if you want a time accurate solution. So in that case, they actually want the time accurate solution. But suppose we're trying to really solve the steady state problem over here. Then we really don't care about the time accuracy. We're not going to report anything about our time steps. In the end, we're only going to report our steady state solution. That's all we cared about. So in that case, there's no reason to try to have a really high accuracy, explicit formula and try to make sure we all our time steps are exactly right. We're just going to throw away all those time points anyway. We'll just keep the final one as our initial guess for a Newton-Raphson solve to find the steady state. So in those cases, you might instead want to use-- So this is time accurate. There's another method. I don't know if I should call it time inaccurate. If you don't care, if the solution is really, what the solution yft is, all you care about is where you get to, then you can do other methods. And one of them is the backward Euler. So you can say, well, y new is equal to y old plus delta t times f of y new. Now, the advantage of this is, I can make delta t pretty large. This is guaranteed stable as long as the ODE system is stable. Remember? And so, this will go to this, go to the solution pretty well, to the steady state solution. It won't go there in the right amount of time because this is a very low order of formula. It's not really accurate. But it will end up to the real steady state solution. This is what's actually used a lot. However, the problem with this is, this is now an implicit equation. So we may have to solve it with some method like Newton-Raphson. But we got into this because we couldn't solve our Newton-Raphson steps because we didn't have a good initial guess. So then the question is, can we find a good initial guess for this problem? And in this case, you can because you can use the explicit formulas to again provide the initial guess for this implicit formula. So that's what they do, also. And so now, if we're doing an iterative procedure in order to get the initial guess for our other iterative procedure ever there, which we're going to break up into another iterative procedure inside it to solve every step. But this is currently what people do. So you guys are smart. Maybe you can figure a better way to do it. The world is waiting. I think I'll stop there. By the way, actually just names-- This kind of thing is called method of lines. And if so, if anybody ever says I solved something by method of lines, what they meant was, they discretized in some of the coordinates and they kept one of the coordinates as a differential. And then they solved it with an ODE solver at the end. And in certain systems like this, it's a smart thing to do. You need to be kind of lucky that the set up is right, so you can use it. But if you can, it can be pretty good. All right. See you on Friday. |
MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 11_Unconstrained_Optimization_NewtonRaphson_and_Trust_Region_Methods.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JAMES SWAN: OK. Let's go ahead and get started. We saw a lot of good conversation on Piazza this weekend. So that's good. Seems like you guys are making your way through these two problems on the latest assignment. I would try to focus less on the chemical engineering science and problems that involve those. Usually the topic of interest, the thing that's useful to you educationally is going to be the numerics, right. So if you get hung up on the definition of a particular quantity, yield was one that came up. Rather than let that prevent you from solving the problem, pick a definition and see what happens. You can always ask yourself if the results seem physically reasonable to you not based on your definition. And as long as you explain what you did in your write up, you're going to get full points. We want to solve the problems numerically. If there's some hang up in the science, don't sweat it. Don't let that stop you from moving ahead with it. Don't let it make it seem like the problem can't be solved or there isn't a path to a solution. Pick a definition and go with it and see what happens, right. The root of the second problem is trying to nest together two different numerical methods. One of those is optimization, and the other one is solutions of nonlinear equation, putting those two techniques together, using them in combination. The engineering science problem gives us a solvable problem to work with in that context, but it's not the key element of it. OK, good. So we're continuing optimization, right. Move this just a little bit. We're continuing with optimization. Last time we posed lots of optimization problems. We talked about constrained optimization, unconstrained optimization. We heard a little bit about linear programs. We started approaching unconstrained optimization problems from the perspective of steepest descent. OK, so that's where I want to pick up as we get started. So you'll recall the idea behind steepest descent was all the unconstrained optimization problems we're interested in are based around trying to find minima, OK. And so we should think about these problems as though we're standing on top of a mountain. And we're looking for directions that allow us to descend. And as long as we're heading in descending directions, right, there's a good chance we're going to bottom out someplace and stop. And when we've bottomed out, we've found one of those local minima. That bottom is going to be a place where the gradient of the function we're trying to find the minimum of is zero, OK. And the idea behind steepest descent was well, don't just pick any direction that's down hill. Pick the steepest direction, right. Go in the direction of the gradient. That's the steepest descent idea. And then we did something a little sophisticated last time. We said well OK, I know the direction. I'm standing on top the mountain. I point myself in the steepest descent direction. How big a step do I take? I can take any size step. And some steps may be good and some steps may be bad. It turns out there are some good estimates for step size that we can get by taking a Taylor expansion. So we take our function, right, and we write it at the next iterate is a Taylor expansion. About the current iterate, that expansion looks like this. And it will be quadratic with respect to the step size alpha. If we want to minimize the value of the function here, we want the next iterate to be a minimum of this quadratic function. Then there's an obvious choice of alpha, right. We find the vertex of this quadratic functional. That gives us the optimal step size. It's optimal if our function actually is quadratic. It's an approximate, right. It's an estimation of the right sort of step size if it's not quadratic. And so I showed you here was a function where the contours are very closely spaced. So it's a very steep function. And the minima is in the middle. If we try to solve this with the steepest descent and we pick different steps sizes, uniform step sizes, so we try 0.1 and 1 and 10 step sizes, we'll never find an appropriate choice to converge to the solution, OK. We're going to have to pick impossibly small step sizes, which will require tons of steps in order to get there. But with this quadratic estimate, you can get a reasonably smooth convergence to the root. So that's nice. And here's a task for you to test whether you understand steepest descent or not. In your notes, I've drawn some contours. For function, we'd like to minimize using the method of steepest descent. And I want you to try to draw steepest descent paths on top of these contours starting from initial conditions where these stars are located. So if I'm following steepest descent, the rules of steepest descent here, and I start from these stars, what sort of paths do I follow? You're going to need to pick a step size. I would suggest thinking about the small step size limit. What is the steepest descent path in the small step size limit? Can you work that out, you and your neighbors? You don't have to do all of them by yourself. You can do one, your neighbor could do another. And we'll take a look at them together. OK, the roar has turned into a rumble and then a murmur, so I think you guys are making some progress. What do you think? How about let's do an easy one. How about this one here. What sort of path does it take? Yeah, it sort of curls right down into the center here, right. Remember, steepest descent paths run perpendicular to the contours. So jumps perpendicular to the contour, almost a straight line to the center. How about this one over here? Same thing, right? It runs the other way. It's going downhill 1, 0, minus 1, minus 2. So it runs downhill and curls into the center. What about this one up here? What's it do? Yeah, it just runs to the left, right. The contour lines had normals that just keep it running all the way to the left. So this actually doesn't run into this minimum, right. It finds a cliff and steps right off of it, keeps on going. Steepest descent, that's what it does. How about this one here? Same thing, right, just to the left. So these are what these paths look like. You can draw them yourself. If I showed you paths and asked you what sort of method made them, you should be able to identify that actually, right? You should be able to detect what sort of methodology generated those kinds of paths. We're not always so fortunate to have this graphical view of the landscape that our method is navigating. But it's good to have these 2D depictions. Because they really help us understand when a method doesn't converge what might be going wrong, right. So steepest descent, it always heads downhill. But if there is no bottom, it's just going to keep going down, right. It's never going to find it. Oh, OK. Here's a-- this is a story now that you understand optimization. So let's see, so mechanical systems, conservation of momentum, that's also, in a certain sense, an optimization problem, right. So conservation of momentum says that the acceleration on a body is equal to the sum of the forces on it. And some of those forces are what we call conservative forces. They're proportional to gradients of some energy landscape. Some of those forces are non-conservative, like this one here. It's a little damping force, a little bit of friction proportional to the velocity with which the object moves instead. And if we start some system like this, we give it some initial inertia and let it go, right, eventually it's going to want to come to rest at a place where the gradient in the potential is 0 and the velocity is 0 on the acceleration is 0. We call that mechanical equilibrium. We get to mechanical equilibrium and we stop, right. So physical systems many times are seeking out minimum of an objective function. The objective function is the potential energy. I saw last year at my house we had a pipe underground that leaked in the front yard. And they needed to find the pipe, right. It was like under the asphalt. So they got to dig up asphalt, and they need to know where is the pipe. They know it's leaking, but where does the pipe sit? So the city came out and the guy from the city brought this. Do you know what this is? What is it? Do you know? Yeah, yeah, yeah. It's a dowsing rod. OK, this kind of crazy story right, a dowsing rod. OK, a dowsing rod. How does it work? The way it's supposed to work is I hold it out and it should turn and rotate in point in a direction that's parallel to the flow of the water in the pipe. That's the theory that this is supposed to work on. I'm a scientist. So I expect that somehow the water is exerting a force on the tip of the dowsing rod, OK. So the dowsing rod is moving around as this guy walks around. And it's going to stop when it finds a point of mechanical equilibrium. So the dowsing rod is seeking out a minimum of some potential energy, let's say. That's what the physics says has to be true. I don't know that flowing water exerts a force on the tip of the dowsing rod. The guy who had this believed that was true, OK. It turns out, this is not such a good idea, though, OK. Like in terms of a method for seeking out the minimum of a potential, it's not such a great way to do it. Because he's way up here, and the water's way underground. So there's a huge distance between these things. It's not exerting a strong force, OK. The gradient isn't very big here. It's a relatively weak force. So this instrument is incredibly sensitive to all sorts of external fluctuations. The gradient is small. The potential energy landscape is very, very flat. And we know already from applying things like steepest descent methods or Newton-Raphson that those circumstances are disastrous for any method seeking out minima of potential energies, right. Those landscapes are the hardest ones to detect it in. Because every point looks like it's close to being a minima, right. It's really difficult to see the differences between these. Nonetheless, he figured out where the pipe was. I don't think it was because of this though. How did he know where the pipe was? What's that? STUDENT: Where the ground was squishy? JAMES SWAN: Where the ground was squishy. Well yeah, had some good guesses because it was leaking up a little bit. No, I looked carefully afterwards. And I think it turned out the city had come by and actually painted some white lines on either side of the street to indicate where it was. But he was out there with his dowsing rod making sure the city had gotten it right. It turns out, there's something called the ideomotor effect where your hand has very little, you know, very sensitive little tremors in it. And can guide something like this, a little weight at the end of a rod to go wherever you want it to go when you want it to. It's like a Ouija board, right. It works exactly the same way. Anyway, it's not a good way to find the minimum of potential energy surfaces, OK. We have the same problem with numerical methods. It's really difficult when these potential energy landscapes are flat to find where the minimum is, OK. So fun and games are over. Now we got to do some math. So we talked about steepest descent. And steepest descent is an interesting way to approach these kinds of optimization problems. It turns out, it turns out that linear equations like Ax equals b can also be cast as optimization problems, right. So the solution to this equation Ax equals b is also a minima of this quadratic function up here. How do you know? You take the gradient of this function, which is Ax minus b, and the gradient to 0 to minima. So Ax minus b is 0, or Ax equals b. So we can do optimization on these sorts of quadratic functionals, and we would find the solution of systems of linear equations. This is an alternative approach. Sometimes this is called the variational approach to solving these systems of linear equations. There are a couple of things that have to be true. The linear operator, right, the matrix here, it has to be symmetric. OK, it has to be symmetric, because it's multiplied by x from both sides. It doesn't know that it's transpose is different from itself in the form of this functional. If A wasn't symmetric, the functional would symmetrize it automatically, OK. So a functional like this only corresponds to this linear equation when A is symmetric. And this sort of thing only has a minimum, right, when the matrix A is positive and definite. It has to have all positive eigenvalues, right. The Hessian right, of this functional, is just the matrix A. And we already said that Hessian needs all positive eigenvalues to confirm we have a minima. OK? If one of the eigenvalues is zero, then the problem is indeterminate. The linear problem is indeterminate. And there isn't a single local minimum, right. There's going to be a line of minima or a plane of minima instead. OK? OK, so you can solve systems of linear equations as optimization problems. And people have tried to apply things like steepest descent to these problems. And it turns out steepest descent is kind of challenging to apply. So what winds up happening is let's suppose we don't take our quadratic approximation for the descent direction first. Let's just say we take some fixed step size, right. When you take that fixed step size, it'll always be, let's say good for one particular direction. OK, so I'll step in a particular direction. It'll be good. It'll be a nice step into a local minimum. But when I try to step in the next gradient direction, it may be too big or too small. And that will depend on the eigenvalues associated with the direction that I am trying to step in, OK. How steep is this convex function? Right? How strongly curved is that convex function? That's what the eigenvalues are describing. And so fixed value of alpha will lead to cases where we wind up stepping too far or not far enough. And there'll be a lot of oscillating around on this path that converges to a solution. I showed you how to pick an optimal step size. It said look in a particular direction and treat your function as though it were quadratic along that direction. That's going to be true for all directions associated with this functional, right. It's always quadratic no matter which direction I point in. Right? So I pick a direction and I step and I'll be stepping to the minimal point along that direction. It'll be exact, OK. And then I've got to turn and I've got to go in another gradient direction and take a step there. And I'll turn and go in another gradient direction and take a step there. And in each direction I go, I'll be minimizing every time. Because this step size is the ideal step size. But it turns out you can do even better than that. So we can step in some direction, which is a descent direction, but not necessarily the steepest descent. And it's going to give us some extra control over how we're minimizing this function. I'll explain on the next slide, OK. The first thing you got to do though is given some descent direction, what is the optimal step size? Well, we'll work that out the same way, right. We can write f at the next iterate in terms of f at the current iterate plus all the perturbations, right. So our step method is Xi plus 1 is Xi plus alpha pi, right. So we do a Taylor expansion, and we'll get a quadratic function again. And we'll minimize this quadratic function with respect to alpha i when alpha takes on this value. So this is the value of the vertex of this function. So we'll minimize this quadratic function in one direction, the direction p. But is there an optimal choice of direction? Is it really best to step in the descent direction? Or are there better directions that I could go in? We thought going downhill fastest might be best, but maybe that's not true. Because if I point to a direction and I apply my quadratic approximation, I minimize the function in this direction. Now I'm going to turn, and I'm going to go in a different direction. And I'll minimize it here, but I'll lose some of the minimization that I got previously, right? I minimized in this direction. Then I turned, I went some other way, right. And I minimized in this direction. So this will still be a process that will sort of weave back and forth potentially. And so the idea instead is to try to preserve minimization along one particular direction. So how do we choose an optimal direction? So f, right, at the current iterate, it's already minimized along p, right. Moving in p forward and backwards, this is going to make f and e smaller. That's as small as it can be. So why not choose p so that it's normal to the gradient at the next iterate? OK, so choose this direction p so it's normal to the gradient at the next iterate. And then see if that holds for one iterate more after that. So I move in a direction. I step up to a contour. And I want my p to be orthogonal to the gradient at that next contour. So I've minimized this way, right. I've minimized everything that I could in directions that aren't in the gradient direction associated with the next iterate. And then let's see if I can even do that for the next iteration too. So can it make it true that the gradient at the next iterate is also orthogonal to p? By doing this, I get to preserve all the minimization from the previous steps. So I minimize in this direction. And now I'm going to take a step in a different direction. But I'm going to make sure that as I take that step in another direction, right, I don't have to step completely in the gradient. I don't have to go in the steepest descent direction. I can project out everything that I've stepped in already, right. I can project out all the minimization I've already accomplished along this p direction. So it turns out you can solve, right, you can calculate what this gradient is. The gradient in this function is Ax minus b. So you can substitute exactly what that gradient is. A, this is Xi plus 2 minus b dotted with p, right. This has to be equal to 0. And you can show that means that p transpose A times p has to be equal to 0 as well. You don't need to be able to work through these details. You just need to know that this gives a relationship between the directions on two consecutive iterates, OK. So it says if I picked a direction p on the previous iteration, take how it's transposed by A, and make sure that my next iteration is orthogonal to that vector, OK. Yeah? STUDENT: So does that mean that your p's are all independent of each other, or just that adjacent p is? K, k plus 1 p's are? JAMES SWAN: This is a great question. So the goal with this method, the ideal way to do this would be to have these directions actually be the directions of the eigenvectors of A. And those eigenvectors for symmetric matrix are all orthogonal to each other. OK? And so you'll be stepping along these orthogonal directions. And they would be all independent of each other. OK? But that's a hard problem, finding all the eigenvectors associated with a matrix. Instead, OK, we pick an initial direction p to go in. And then we try to ensure that all of the other directions satisfy this conjugacy condition, right. That the transformation of p by A is orthogonal with the next direction that I choose. So they're not independent of each other. But they are what we call conjugate to each other. It turns out that by doing this, these sets of directions p will belong to-- they can be expressed in terms of many products of A with the initial direction p. That'll give you all these different directions. It starts to look something like the power iteration method for finding the largest eigenvector of a matrix. OK? So you create a certain set of vectors that span the entire subspace of A. And you step specifically along those directions. And that lets you preserve some of the minimization as you step each way. So what's said here is that the direction p plus 1 is conjugate to the direction p. And by choosing the directions in this way, you're ensuring that p is orthogonal to the gradient at i plus 1 and the gradient i plus 2. So you're not stepping in the steepest descent directions that you'll pick up later on in the iterative process. OK? So when you know which direction you're stepping in, then you've got to satisfy this conjugacy condition. But actually, this is a vector in n space, right. This is also a vector n space. And we have one equation to describe all and components. So it's an under-determined problem. So then one has to pick which particular one of these conjugate vectors do I want to step along. And one particular choice is this one, which says, step along the gradient direction, OK, do steepest descent, but project out the component of the gradient along pi. We already minimized along pi. We don't not have to go in the pi direction anymore, right. So do steepest descent, but remove the pi component. So here is a quadratic objective function. It corresponds to a linear equation with coefficient matrix 1 00 10, a diagonal coefficient matrix. And b equals 0. So the solution of the system of linear equations is 00. We start with an initial guess up here, OK. And we try steepest descent with some small step size, right. You'll follow this blue path here. And you can see what happened. That step size was reasonable as we moved along the steepest ascent direction where the contours were pretty narrowly spaced. But as we got down to the flatter section, OK, as we got down to the flatter section of our objective function, those steps are really small. Right? We're headed in the right direction, we're just taking very, very small steps. If you apply this conjugate gradient methodology, well, the first step you take, that's prescribed. You've got to step in some direction. The second step you take though minimizes completely along this direction. So the first step was the same for both of these. But the second step was chosen to minimize completely along this direction. So it's totally minimized. And the third step here also steps all the way to the center. So it shows a conjugate direction that stepped from here to there. And it didn't lose any of the minimization in the original direction that it proceeded along. So that's conjugate gradient. It's used to solve linear equations with order n iterations, right. So A has at most n independent eigenvectors, independent directions that I can step along and do this minimization. The conjugate gradient method is doing precisely that. Doesn't know what the eigendirections are, but it's something along these conjugate directions as a proxy for the eigendirections. So it can do minimization with just n steps for a system of n equations for n unknowns. It requires only the ability to compute the product of your matrix A with some vector, right. All the calculations there are only depended on the product of A with a vector. So don't have to store A, we just have to know what A is. We have some procedure for generating A. Maybe A is a linear operator that comes from a solution of some differential equations instead, right. And we don't have an explicit expression for A, but we have some simulator that produces, take some data, and projects A to give some answer, right. So we just need this product. We don't have to store A exactly. It's only good for symmetric positive definite matrices, right. This sort of free energy functional that we wrote or objective function we wrote only admits symmetric matrices which are positive definite. That's the only way it will have a minimum. And so the only way a steepest descent or descent type procedure is going to get to the optimum. But there are more sophisticated methods that exist for arbitrary matrices. So if we don't want symmetry or we don't care about whether it's positive definite, there are equivalent sorts of methods that are based around the same principle. And it turns out, this is really the state of the art. So if you want to solve complicated large systems of equations, you know Gaussian elimination, that will get you an exact solution. But that's often infeasible for the sorts of problems that we're really interested in. So instead, you use these sorts of iterative methods. Things like Jacobi and Gauss-Seidel, they're sort of the classics in the field. And they work, and you can show that they converge on are lots of circumstances. But these sorts of iterative methods, like conjugate gradient and its brethren other Krylov subspace methods they're called, are really the state of the art, and the ones that you reach to. You already did conjugate gradients in one homework, right. You used this PCG iterative method in Matlab to solve a system of linear equations. It was doing this, right. This is how it works. OK? OK. OK, so that's country ingredients. You could apply it also to objective functions that aren't quadratic in nature. And the formulation changes a little bit. Everywhere where the matrix A appeared there needs to be replaced with the Hessian at a certain iterate. But the same idea persists. It says well, we think in our best approximation for the function that we've minimized as much as we can in one direction. So let's choose a conjugate direction to go in, and try not to ruin the minimizations we did in the direction we were headed before. Of course, these are all linearly convergent sorts of methods. And we know that there are better ways to find roots of non-linear equations like this one, grad f equals zero, namely the Newton-Raphson method, which is quadratically convergent. So if we're really close to a critical point, and hopefully that critical point is a minima in f, right, then we should rapidly converge to the solution of this system of nonlinear equations just by applying the Newton-Raphson method. It's locally convergent, right. So we're going to get close. And we get quadratic improvement. What is the Newton-Raphson iteration, though? Can you write that down? What is the Newton-Raphson iteration that's the iterative math for this system of non-linear equations, grad f equals 0? Can you work that out? What's that look like? Have we got this? What's the Newton-Raphson iterative map look like for this system of non-linear equations? Want to volunteer an answer? Nobody knows or nobody is sharing. OK, that's fine. Right, so we're trying to solve an equation g of x equals 0. So the iterative map is Xi plus 1 is Xi minus Jacobi inverse times g. And what's the Jacobian of g? What's the Jacobian of g? The Hessian, right. So the Jacobian of g is the gradient of g, which is two gradients of f, which is the definition of the Hessian. So really, the Newton-Raphson iteration is Xi plus 1 is Xi minus Hessian inverse times g. So the Hessian plays the role of the Jacobian, the sort of solution procedure. And so everything you know about Newton-Raphson is going to apply here. Everything you know about quasi-Newton-Raphson methods is going to apply here. You're going to substitute for your nonlinear. The nonlinear function you're finding the root for, you're going to substitute the gradient. And for the Jacobian, you're going to substitute the Hessian. Places where the Hessian is, the determent of the Hessian is 0, right it's going to be a problem. Places where the Hessian is singular is going to be a problem. Same as with the Jacobian. But Newton-Raphson has the great property that if our function is quadratic, like this one is, it will converge in exactly one step. So here's steepest descent with a fixed value of alpha, Newton-Raphson, one step for a quadratic function. And why is it one step? STUDENT: [INAUDIBLE] JAMES SWAN: Good. So when we take a Taylor expansion of our f, in order to derive the Newton-Raphson step, we're expanding it out to quadratic order, its function is quadratic. The Taylor expansion is exact. And the solution of that equation, right, gradient f equals 0 or g equals 0, that's the solution of a linear equation, right. So it gives exactly the right step size here to move from an initial guess to the exact solution or the minima of this equation. So for quadratic equations, Newton-Raphson is exact. It doesn't go in the steepest ascent direction, right. It goes in a different direction. It would like to go in the steepest descent direction if the Jacobian were identity. But the Jacobian is a measure of how curved f is. The Hessian, let's say, is a measure of how curved f is. Right? And so there's a projection of the gradient through the Hessian that changes the direction we go in. That change in direction is meant to find the minimum of the quadratic function that we approximate at this point. So as long as we have a good quadratic approximation, Newton-Raphson is going to give us good convergence to a minima or whatever nearby critical point there is. If we have a bad approximation for a quadratic, then it's going to be so good, right. So here's this very steep function. Log of f is quadratic, but f is exponential in x here. So you got all these tightly spaced contours converging towards a minima at 00. And here I've got to use the steepest descent step size, the optimal steepest descent step size, which is a quadratic approximation for the function, but in the steepest descent direction only. And here's the path that it follows. And if I applied Newton-Raphson to this function, here is the path that it follows instead. The function isn't quadratic. So these quadratic approximations aren't-- they're not great, right. But the function is convex, right. So Newton-Raphson is going to proceed downhill until it converges towards a solution anyways. Because the Hessian has positive eigenvalues all the time. Questions about this? Make sense? OK? So you get two different types of methods that you can play with. One of which, right, is always going to direct you down hill. Steepest descent will always carry you downhill, right, towards a minima. And the other one, Newton-Raphson, converges very quickly when it's close to the root. OK, so they each have a virtue. And they're different. They're fundamentally different, right. They take steps in completely different directions. When is Newton-Raphson not going to step down hill? STUDENT: [INAUDIBLE] What's that? STUDENT: [INAUDIBLE] JAMES SWAN: OK, that's more generic an answer than I'm looking for. So there may be circumstances where I have two local minima. That means there must be maybe a saddle point that sits between them. Newton-Raphson doesn't care which critical point it's going after. So it may try to approach the saddle point instead. That's true. That's true. Yeah? STUDENT: When Hessian [INAUDIBLE].. JAMES SWAN: Good, yeah. With the Hessian doesn't have all positive eigenvalues, right. So if all the eigenvalues of the Hessian are positive, then the transformation h times g or h inverse times g, it'll never switch the direction I'm going. I'll always be headed in a downhill direction. Right? In a direction that's anti-parallel to the gradient. OK? But if the eigenvalues of the Hessian are negative, if some of them are negative and the gradient has me pointing along that eigenvector in a significant amount, then this product will switch me around and will have me go uphill instead. It'll have me chasing down a maxima or a saddle point instead. That's what the quadratic approximation of our objective function will look like. It looks like there's a maximum or a saddle instead. And the function will run uphill. OK? So there lots of strengths to Newton-Raphson. Convergence is one of them, right. The rate of convergence is good. It's a locally convergent, that's good. It's got lots of weaknesses, though. Right? It's going to be a pain when the Hessian is singular at various places. You've got to solve systems of linear equations to figure out what these steps are. That's expensive computationally. It's not designed to seek out minima, but to seek out critical points of our objective function. Steepest descent has lots of strengths, right. Always heads downhill, that's good. If we put a little quadratic approximation on it, we can even stabilize it and get good control over the descent. Its weaknesses are it's got the property that it's linearly convergent instead of quadratically convergent when it converges. So it's slower, right. It might be harder to find a minima. You've seen several examples where the path sort of peters out with lots of little iterations, tiny steps towards the solution. That's a weakness of steepest descent. We know that if we go over the edge of a cliff on our potential energy landscape, steepest descent it just going to run away, right. As long as there's one of these edges, it'll just keep running downhill for as long as they can. So what's done is to try to combine these methods. Why choose one, right? We're trying to step our way towards a solution. What if we could craft a heuristic procedure that mixed these two? And when steepest descent would be best, use that. When Newton-Raphson would be best, use that. Yes? STUDENT: Just a quick question on Newton-Raphson. JAMES SWAN: Yes? STUDENT: Would it run downhill also if you started it over there? Or since it seeks critical points, could you go back up to the [INAUDIBLE].. JAMES SWAN: That's a good question. So if there's an asymptote in f, it will perceive the asymptote as a critical point and chase it. OK? And so if there's an asymptote in f, if can perceive that and chase it. It can also run away as it gets very far away. This is true. OK? The contour example that I gave you at the start of class had sort of bowl shape functions superimposed with a linear function, sort of planar function instead. For that one, right, the Hessian is ill-defined, right. There is no curvature to the function. But you can imagine adding a small bit of curvature to that, right. And depending on the direction of the curvature, Newton-Raphson may run downhill or it may run back up hill, right? We can't guarantee which direction it's going to go. Depends on the details of the function. Does that answer your question? Yeah? Good. STUDENT: Sir, can you just go back to that one slide? JAMES SWAN: Yeah. I'm just pointing out, if the eigenvalues of h is further negative, then the formula there for alpha could have trouble too. JAMES SWAN: That's true. STUDENT: Similar to how the Newton-Raphson had trouble. JAMES SWAN: This is true. This is true, yeah. So we chose a quadratic approximation here, right, for our function. We sought a critical point of this quadratic approximation. We didn't mandate that it had to be a minima. So that's absolutely right. So if h has negative eigenvalues and the gradient points enough in the direction of the eigenvectors associated with those negative eigenvalues, then we may have a case where alpha isn't positive. We required early on that alpha should be positive for steepest descent. So we can't have a case where alpha is not positive. That's true. OK. So they're both interesting methods, and they can be mixed together. And the way you mix those is with what's called trust-region ideas, OK. Because it could be that we've had an iteration Xi and we do a quadratic approximation to our functional, which is this blue curve. Our quadratic approximation is this red one. And we find the minima of this red curve and use that as our next best guess for the solution to the problem. And this seems to be working us closer and closer towards the actual minimum in the function. So since quadratic approximation seems good, if the quadratic approximation is good, which method should we choose? STUDENT: Newton-Raphson. JAMES SWAN: Newton-Raphson, right. Could also be the case though that we make this quadratic approximation from our current iteration, and we find a minimum that somehow oversteps the minimum here. In fact, if we look at the value of our objective function at this next step, it's higher than the value of the objective function where we started. So it seems like a quadratic approximation is not so good, right. That's a clear indication that this quadratic approximation isn't right. Because it suggested that we should have had an minima here, right. But our function got bigger instead. And so in this case, it doesn't seem like you'd want to choose Newton-Raphson to take your steps. The quadratic approximation is not so good. Maybe just simple steepest descent is a better choice. OK, so it's done. So if you're at a point, you might draw a circle around that point with some prescribed radius. Call that Ri. This is our iterate Xi. This is our trust-region radius Ri. And we might ask, where does our Newton-Raphson step go? And where does our steepest descent step take us? And then based on whether these steps carry us outside of our trust-region, we might decide to take one or the other. So if I set a particular size Ri, particular trust-region size Ri and the Newton-Raphson step goes outside of that, we might say well, I don't actually trust my quadratic approximation this far away from the starred point. So let's not take a step in that direction. Instead, let's move in a steepest descent direction. If my Newton-Raphson step is inside the trust-region, maybe I'll choose to take it, right. I trust the quadratic approximation within a distance Ri of my current iteration. Does that strategy makes sense? So we're trying to pick between two different methods in order to give us more reliable convergence to a local minima. So here's our Newton-Raphson step. It's minus the Hessian inverse times the gradient. Here's our steepest descent step. It's minus alpha times the gradient. And if the Newton-Raphson step is smaller than the trust-region radius, and the value of the objective function at Xi, plus the Newton-Raphson step is smaller than the current objective function, it seems like the quadratic approximation is a good one, right. I'm within the region in which I trust this approximation, and I've reduced the value of the function. So why not go that way, right? So take the Newton-Raphson step. Else, let's try taking a step in the steepest direction instead. So again, if the steepest ascent direction is smaller than Ri and the value of the function in the steepest descent direction, the optimal steepest descent direction or the optimal step in the steepest ascent direction is smaller than the value of the function at the current point, seems like we should take that step. Right? The Newton-Raphson step was no good. We've already discarded it. But our optimized steepest descent step seems like an OK one. It reduces the value of the function. And its within the trust-region where we think quadratic approximations are valid. If that's not true, if the steepest descent step takes us outside of our trust-region or we don't reduce the value of the function when we take that step, then the next best strategy is to just take a steepest ascent step to the edge of the trust-region boundary. Yeah? STUDENT: Is there a reason here that Newton-Raphson is the default? JAMES SWAN: Oh, good question. So eventually we're going to get close enough to the solution, all right, that all these steps are going to live inside the trust-region ring. Its going to require very small steps to converge to the solution. And which of these two methods is going to converge faster? STUDENT: Newton-Raphson. JAMES SWAN: Newton-Raphson. So we prioritize Newton-Raphson over steepest descent. That's a great question. Its the faster converging one, but its a little unwieldy, right. So let's take it when it seems valid. But when it requires steps that are too big or steps that don't minimize f, let's take some different steps instead. Lets use steepest descent as the strategy. So this is heuristic. So you got to have some rules to go with this heuristic, right. We have a set of conditions under which we're going to choose different steps. We've got to set this trust-region size. This Ri has to be set. How big is it going to be? I don't know. You don't know, right, from the start you can't guess how big Ri is going to be. So you got to pick some initial guess. And then we've got to modify the size of the trust-region too, right. The size of the trust-region is not going to be appropriate. One fixed size is not going to be appropriate all the time. Instead, we want a strategy for changing its size. So it should grow or shrink depending on which steps we choose, right. Like if we take the Newton-Raphson step and we find that our quadratic approximation is a little bit bigger than the actual function value that we predicted, we might want to grow the trust-region. We might be more likely to believe that these Newton-Raphson steps are getting us to smaller and smaller function values, right. The step was even better than we expected it to be. Here's the quadratic approximation in the Newton-Raphson direction. And it was actually bigger than the actual value of the function. So we got more, you know, we got more than we expected out of a step in that direction. So why not loosen up, accept more Newton-Raphson steps? OK, that's a strategy we can take. Otherwise, we might think about shrinking instead, right. So there could be the circumstance where our quadratic approximation predicted a smaller value for the function than we actually found. It's not quite as reliable for getting us to the minimum. These two circumstances are actually these. So this one, the quadratic approximation predicted a slightly bigger value than we found. Say grow the trust-region, right. Try some more Newton-Raphson steps. Seems like the Newton-Raphson steps are pretty reliable here. Here the value of the function in the quadratic approximation is smaller than the value of the function after we took the step. Seems like our trust-region is probably too big if we have a circumstance like that. Should shrink it a little bit, right? We took the Newton-Raphson step, but it actually did worse than we expected it to do with the quadratic approximation. So maybe we ought to shrink the trust-regional a little bit. And you need a good initial value for the trust-region radius. What does Matlab use? It uses 1. OK. It doesn't know. It has no clue. It's just a heuristic. It starts with 1 and it changes it as need be. So this is how fsolve solves systems of nonlinear equations. This is how all of the minimizers in Matlab, this is the strategy they use to try to find minima. They use these sorts of trust-region methods. It uses a slight improvement, which is also heuristic, called a dogleg trust-region method. So you can take a Newton-Raphson step or you can take a steepest descent step. And if you found the steepest descent step didn't quite get you to the boundary of your trust-region, you could then step in the Newton-Raphson direction. Why do you do that? I don't know, people have found that it's useful, right. There's actually no good reason to take these sorts of dogleg steps. People found that for general, right, general objective functions that you might want to find minima of, this is a reliable strategy for getting there. There's no guarantee that this is the best strategy. These are general non-convex functions. These are just hard problems that one encounters. So when you make a software package like Matlab, this is what you do. You come up with heuristics that work most of the time. I'll just provide you with an example here, OK. So you've seen this function now several times. Let's see, so in red covered up back here is the Newton-Raphson path. In blue is the optimal steepest descent path. And in purple is the trust-region method that Matlab uses to find the minima. They all start from the same place. And you can see the purple path is a little different from these two. If I zoom in right up here, what you'll see is initially Matlab chose to follow the steepest descent path. And then at a certain point it decided, because of the value of the trust-region that Newton-Raphson steps were to be preferred. And so it changed direction and it started stepping along the Newton-Raphson direction instead. It has some built in logic that tells it when to make that choice for switching based on the size of the trust-region. And the idea is just to choose the best sorts of steps possible. Your best guess at what the right steps are. And this is all based around how trustworthy we think this quadratic approximation for objective function is. Yeah, Dan? STUDENT: So for the trust-region on the graph what Matlab is doing is at each R trust-region length it's reevaluating which way it should go? JAMES SWAN: Yes. Yes. It's computing both sets of steps, and it's deciding which one it should take, right. It doesn't know. It's trying to choose between them. STUDENT: Why don't you do the Newton-Raphson step through the [? negative R? ?] JAMES SWAN: You can do that as well, actually, right. But if you're doing that, now you have to choose between that strategy and taking a steepest descent step up to R as well, right. And I think one has to decide which would you prefer. It's possible the Newton-Raphson step also doesn't actually reduce f. In which case, you should discard it entirely, right. But you could craft a strategy that does that, right. It's still going to converge, likely. OK? OK. I've going to let you guys go, there's another class coming in. Thanks. |
MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 20_Boundary_Value_Problem_1.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: All right, let's get started. So today we're going to talk about boundary value problems. But I want to try to start back with a motivating example. So unfortunately, like most technologies, the biggest driver for differential equation solutions was from military technology. And the military problem-- I'll tell you in a second. Let me first remind you. So we have ODE-IVPs. And they are-- right. So that's the ODE-IVP is we have n differential equations, and we have n initial conditions. And you guys are really good at these now. Had a lot of practice. But you can have problems where some of the conditions are not specified at the t0. They are specified somewhere else. So if all of them are specified at t0, we call it an IVP. And we call it a BVP is if the conditions are like this. This is not so good. And not all the tm's are the same. If all the tm's are the same, we call it t0, and we're back to IVP. So IVP is like a special case of the ODE-BVP problem. And here I'm ranking them as if everything is explicit. So we know an explicit equation for dy dt. We have an explicit equation for the boundary conditions. In fact, often, we actually only have implicit equations. So I'll show you an example in a minute. But that's it. And we still need to have n equations. Well, I guess I could do this. Take this back. Basic goals, and just be just be careful. These ends are-- it's the same number of them. It's the same number of conditions. This is OK to write it this way, but just be careful. Is that OK? Just to make it confusing. All right. I guess I can call this 0 now. See, but now it's not really 0, right? It's really maybe-- call it star at the boundary. There's some special value-- some special place where you know the value of y. All right. So let's do the motivating example, which comes from military technology. And this is quite an old problem. So, because, in the old days, people didn't have motor vehicles, and, basically, the roads were no good, and there were a lot of bandits out on the roads, people liked to ship everything by water transportation. And so all the towns grew up on rivers, or on ports on the ocean. And so, if you lived in a nice little town, you know, up in the hills there, and then, at the base of the hills, you have your cute little town. You have your house. You have some warehouses where you keep all your nice goods. You have a dock that goes out onto the water. Here's the land. Here, these nice ships come and park at the dock. They unload nice goods. You trade your stuff to them. You make a lot of money. Everyone's happy and prosperous and having a great time living. This is a port city. But then you look out on the horizon in the water, and you see a ship is coming. And sometimes that a ship is coming is really good news for you. But sometimes the ship coming is not good because on the back, it has a flag, and on the flag, you see skull and crossbones. And you realize it's a pirate ship coming. And now you're in serious trouble because you really don't want the pirates to come to your nice little town because who knows what they'll do there. But it won't be good. All right, so the people who live in this town, after this has happened to you a few times, you're really alarmed, and you're really happy, and you don't want to rebuild your town again because it just got burned down the last time and the pirate came and pillaged it and took everything. And you guys all had to run away into the hills. So instead, they've invented cannons or catapults, put one of the cannons, say, at the top of the hill above your town. And then you say, the next time the pirates come, I'm ready for them. So drawing the cannon. And so the question is, you want to shoot the cannonball so it hit the pirate ship at the waterline, when the pirate is distance L from the shore-- hopefully, pretty far, so they're not going to shoot at you first. And you're shooting from up at the top of the hill, so you can shoot further than they can. So you're going to shoot distance L. And you're shooting from an elevation H. And you want to make sure your cannonball hits them, ideally, right at the waterline. Boom, makes a nice hole in their ship, and it sinks. So that's the plan. Now the problem is, getting the cannonball to go from here to there is not so easy. It's a big ocean. There's a lot of places your cannonballs can fall into the ocean, and it might not be where the pirate ship is. And reloading canon is pretty slow. And you guys live in this prosperous port. You're not military guys, and so you don't have that much practice doing it anyway. And you're cheap because you're merchants. And you don't want to spend your money buying lots of cannonballs anyway. So you only get a few of cannonballs up here. And there's a little pile of cannonballs next to the cannon. And then, when you see the pirate ship coming, you guys run up the hill quick, and you get to take a couple of shots, and that's it. And if you hit them, then your town is safe, and, otherwise, everything you love and know in your world is going to be destroyed by the pirates. OK? So this was the motivating sample for that element of differential equation solvers. OK? So let's see if we can write down the equations that correspond to this example. So it's going to be Newton's laws. F equals ma on the cannonball as it travels. So we can write that also as equals-- well what's the F? F is going to be negative mg in the z direction. That's gravity pulling the cannonball down. And then you'll have some air friction on the cannonball. And I'm going to call the air friction coefficient gamma m times v. OK? So that's the simplest model for what the forces are on a cannonball just flying from my cannon, hopefully, to hit the pirate ship. And I can rearrange us by dividing through by m. And acceleration is velocity, so dv dt is equal to negative g. So you have negative gamma-- sorry. My notation is not so good. OK. Everyone is all right with this? Yeah. So then we can write the whole differential equation system, d dt of xz vx vz equals dx dt is vx, dz dt is vz. dvx dt is negative gamma vx. dvz dt is negative g minus gamma v ez. So that's a fine ODE system. It's not too hard to solve. It's actually linear. It's a piece of cake, right? You guys just probably just did this already when you were in high school, or college, anyway, for sure. And let's write down the initial conditions. So we know x0 is equal to 0. z of 0 is equal to h, the height of the canon. vx of 0, oh, what's that equal to? We don't know. vz of 0, we don't know that either. But we bought this canon with a certified muzzle velocity v0, so we actually know that vx of 0 squared plus the vz of 0 squared is equal to v0 squared-- whereas, when I bought the canon, the merchants went out to the local canon dealer and bought a canon, and the guy said, you put this much gunpowder in it, you'll get a muzzle velocity of v0. So they know that that number is. So we have a condition that's not really the way we expect it to be. And we don't have enough conditions. We need four conditions. But of course, the most important condition is what happens at L. We really want to know that x of t final is equal to L and y of t final is equal to 0 at sea level. So that way, our cannonball is going to into the right location at t final. So we have two conditions specified way, way away from where we are. And we have two conditions specified initially. And no you have some oddball thing in here, some oddball fifth condition. Now, OK. One problem here is, we don't actually know what t final is because before we shoot it, we really don't know how long it takes the cannonball to get out there. So this is a kind of weird condition where t final is unknown, whereas, the ones you've done before, I think you've always known t final. Is that right? So let's talk about how to deal with that. So one problem, one way to deal with this is to notice that, if we have an equation system dy dt equals f of y, we could change variables instead of using t. Since we really don't know much about time in this problem, t is not the best variable for us. So we want to change it to some new variable u. So suppose that we want to change to u. And, if we change to u, we can write something like this-- dy du du dt equals f of y. Can you see that? So I can try to get the differentials in terms of u, and I just have to know what my du dt is. And you guys might have even changed the variables before, in calculus sometime, you probably did this-- a long time ago, and chain rule, and stuff like that. All right, so we can sort of pick whatever idea we want, and as long as we put in the du dt in the equation, then we'll be OK. So let's just choose that u is equal to x because we know du dt would then be with the vx. So we could we could write this out this way. We could write d-- I'll keep it as u for a minute, so the notation won't be so confusing. So du dt, dy du is equal to f of y over du dt. And as long as du dt is never zero, we're good. So we could any u we want as long du dt is never zero. Then we're still going to have a nice simple question here. And so I'm going to choose u is equal to x because du dt is vx. And I know that vx is never zero. They're kind of also always moving to the right. So I'm going to divide by something. It's never zero. So I can write it this way-- dy du is equal to-- I can go down that list and divide each of these things by vx. So it's 1, vz over vx, negative gamma, negative g over vx minus gamma vz over vx. That's my new [INAUDIBLE]. OK? OK with this? How about energetic? Keep nodding. Yeah, yeah, sure, keep going faster. It's going way too slow. That's what I want you to say. Yeah? Apparently not. OK. All right. So the advantage to changing here is that now I can write the initial conditions, and all these conditions, in terms of u, which is x. So I really-- I want to go from u is equal to 0, i.e. X is equals to 0, to u is equal to L, which is x equal to L. So now, I could also look at this say, this is kind of stupid. dy du, the first component of y is dy du is equal to one. And that first component of y never appears in the rest of these equations. So that's kind of not a very useful one. So I can change, and I could write a simpler equation, dy tilde du is equal to vz over vx negative gamma, negative g over vx minus gamma vz over vx, where my y tilde is just equal to the last three variables-- z, vx, vz. So I did a lot of algebra, but I ended up simplifying my life because now I have three different equations to solve instead of four. And I can write that y tilde of zero of the first element of y tilde, which is z, is equals to h. So that's a condition I know for sure. And I know that I really want this cannonball to hit the pirate ship. So I want y1 tilde of L to be 0, so I want tilde as z. And I want z to be 0 when x is equal to L. Or that's the point where the cannonball is hitting the ship at the water line there. So those are definitely two conditions I have. And then I need a third condition. And my third condition is this crazy equation here. All right, so now I have three conditions for three differential equations, and one of them is an oddball looking thing. These two look kind of normal. However, this one is a t0, and this one is some goofball value, but at least I know the value. I know what L is because I know how far away the pirate ship is. So far so good? All right. But how am I going to solve this? So how do we know how to solve differential equations right now is we know how to do initial value problems. We have these nice programs like ode45 that work great, integrate differential equations. So we should be able to use that somehow to solve this. But we don't have all the inputs because ode45 would expect us to give values of all three of the y tilde values at 0. I only know one of them. So how can I approach it? Sorry, I'm going to put my beautiful drawing down below here. We'll save it for posterity and look at it later. So I could approach it by saying, well, I don't know what these two initial conditions would be, what vx and vz should be at 0. I'll treat them as unknowns. I want to figure out what they should be. And so, I'm going to vary those unknowns and then, for each value I choose, then I can shoot the cannon with those v's, and I can see where the cannonball went, and if it didn't go to the right location, I can readjust those v's and then try again. Just like, if you were running the cannon, you might shoot at a certain angle, and then adjust the angle of the cannon, and then shoot again. And then, when you change the angle, you change the ratio of vx to vy initially, right? So that's called the shooting method-- for good reason. So the shooting method, that's what we do. We guess the missing initial conditions, then you solve using the guess. You solve the ODE-IVP problem. And then you see what it computes. And if the solution you get doesn't satisfy the conditions you demand, you go back and adjust your guess. And you loop around. So that's the idea. Now, this thing of adjusting some parameter values, some initial conditions, to try to satisfy some condition, that's like an f solve problem. That's a system of nonlinear equations kind of problem. So we're going to have f solve as the outside loop, and that's going to adjust some parameters. And then, what's f solve do? So f solve tries to solve problems of the form g of-- I usually write x, but I'll call it v here. No. V's not good either. What letter have we have not used yet? Any letter. g of w. There we go. g of w equals 0. I want to figure out what w's I need to use. And actually, v is right. Because we're trying to adjust the velocities, right? Sorry, I'm messing up here. It's good to use the eraser in this class-- especially with this nice big eraser. All right, so g of e-- we're trying to adjust the velocities to make something 0. So what's our g that we want? We really want that the z value at L-- we want that to be equal to 0. And this thing is, like, implicitly a function of the v's-- the v0 values we give. If you put the right v0 values in, then z at L would be 0. So this is what we're really trying to solve. Now, to get z of L, we have to solve a system of differential equations. So it's a little complicated to go from v0 to zL, but we can do because we have ode45. So we can write a function to do this. Running out of words. All right so let's try to write the program. So we want to write the program g. So we want to write z of L is equal to g of v0. And let's try to think what this function is. So the meat of this is, we have to solve a differential equation, an IVP. So somewhere, we have an ode45 call. And what does ode45 computing? It's computing the x positions and the z positions. And it's going to use some function f, which was-- do we still have it on the board? Maybe I lost it here. Nope. Sorry, it's under here. I'm a terrible board worker. So this is our function f, f f of y. So this is really f tilde. That's the function that I'm going to integrate. And I can't remember what are the arguments, but it's going to be something like, from 0 to L, and then I have to give it a y0, and then maybe there's some other stuff up there. All right, do you guys remember the ode45? Is that in the right order? I don't remember. OK. So I need to specify the y0. So y0 is going to be equal to the initial value of z, the altitude. So it's going to be H. And then it's the initial value of vx. So I don't know what that is, but it's-- maybe you just want to call it v0. vx0. And then I do have one more condition here, that vz is related to vx. So I can use that. So I can have the last one be the square root of v0 squared minus vx0 squared. So that's the line for what y0 is. And then I solve this. And then I write the number of steps in the time stepping program that did, in order to get here. It was x stepping in this case. Num steps is equal to the length of x vec. And z of L is equal to z vec at N step. That's it, right? So that's the function that you would feed to f solve. And your correct value of vx 0 is the one that makes zL 0, which is what f solve tried to do. Yeah? AUDIENCE: [INAUDIBLE] PROFESSOR: You can, but in the physical problem, usually cannons always shoot upwards. So we'll just shoot upwards. But yeah, it could be either way. Yeah, and, in fact, in this problem, you might do better to change-- instead of using vx0 as your unknown, you might want to change to theta, the angle, the elevation angle of the cannon, and that would be more physical of what the actual user of the cannon is using. They're adjusting theta, not adjusting vx0. It's no problem, actually. From the point of view of the f solves, you can make this theta instead. And then you'd have this be v0 times cos of theta, and this b v0 times sine of theta, and it would work just as well to get the solution. All right, so that's the shooting method. So this function g is the g that's going-- you'll have f solve, so is going to say, vx0 is equal to f solve of g. And then you have to have a guess, right? For f solve. So that's how you would invoke this function. All right, and so if we wanted to figure operation count of how many operations this is to find a solution this way, how's it go? How many iterations does it take to solve an ODE IVP. That depends on the step size, right? It's adaptive. Usually adaptive stepping. So you have some experience with that, maybe 100 steps? Something like that? Is usually enough? Maybe 1,000 if it's not stiff? And then how many iterations does it take f solve to solve something? 10? So you have maybe 1,000 time steps, maybe 10 iterations, means 10 different initial values that you try before you convert to a solution. And then you may have to actually compute the Jacobian of g, inside f solve, if it's going to use the Newton-Raphson step, it's going to need the Jacobian of g. The Jacobian of g takes the number of unknowns times the number of function evaluations to compute it by finite differences-- maybe 2 times that number. So it's going to be something like n squared. It'll be n squared to evaluate the Jacobian. It takes n operations to evaluate f. So order of n, where n is the number of variables, n times that, to the Jacobian, so it would be like order of number variables squared times 10 times 1,000 is kind of an order of the cost. And it's only that cost because we can do an explicit solver. If you had to do an implicit ODE solution, it would be a lot more because you'd have to compute the Jacobians more often. All right. Is this fun? OK. So this is the shooting method. You guys should be able to do this no trouble. Conceptually, this is not so different than the homework problem you did where you did an optimization of the parameter values. Do you remember? So you had some differential equations, and you optimize the parameters to try to find the best solutions. Here, instead of optimizing them, you're doing f solve. You're trying to solve them to make something happen. Now how can this run into troubles. Why would this shooting method not always work? Ready why this might not work? Yeah? AUDIENCE: [INAUDIBLE] PROFESSOR: And a common case where that might happen would be if your ODE system was unstable, Or your ODE solution method was numerically unstable. In either of those cases, who knows what you're getting. And actually, you don't know if the solution is even real. You don't know if the computed final values zL is really what corresponds to that input value of vx0. Because, if you have a big kind of numerical sensitivity and instability. So that's one kind of problem. So this is just like all the ODE IVP problems. If the ODE's problem, intrinsically, is unstable in a big way, you're in trouble because your errors are going to amplify. And then also, if you have a problem where the delta t's are too large compared to this numerical stability of the solver, you'll also have a problem. So those are two kinds of problems that we always have in ODE solutions. And so, you might actually want to use this method for some regular ODE IVP problems. So for example, if you had a problem-- I don't want to wreck my beautiful artwork. You guys know this. All right. If you had a problem that said dy dt is equal to 10 to the 6th times y. Yes? AUDIENCE: I have a question about the Jacobian. When you calculate the Jacobian do you have to solve the ODE system? PROFESSOR: Yes. Well, two ways. You can either solve it for repeated values, different initial values, of v0, and by finite difference, get the Jacobian. Or you could use the sensitivity method that we did in the second problem, a problem set ago, and analytically get the derivatives without respect to the parameters either way. But either way, it's a lot of work. So the from I guess this is a very simple problem, and suppose you wanted to go all the way out to t equals 1. Now, the problem is that the solution of this is Ae times 10 to the 6th times t or something. So this is really not looking so great. As a problem to solve, this is about as unstable as you can get. The tiniest little deviation anywhere along the way is going to make the thing blow up. And so this is not so great and so even if initial condition is that y of t0 is equal to 10 to the minus 12, so actually this is not so bad, this is still hard to solve, numerically. But you could flip it around, and treat it as a shooting problem, and solve it in reverse. So you could change variables that u is equal to negative t. And then you'd have dy du is equal to negative 10 to the 6 times y. And now this is really stable. And now you might want to integrate in. So supposed the original question was, what is y of 1 in the differential equations? You could do it in reverse and say, I'm going to guess, y of 1 is equal to y guess. And I'm going to integrate it and demand that this is my final condition. So now this is actually y-- oh geez, now I'm confusing myself. y of negative 1, I guess. So I just flipped the sign, t. And to make it easier, I would do it this way. 1 minus t. Same derivative. So there we go. y0 is y guess. And y of 1 is 10 to the minus 12. So this is my final condition that I'm trying to shoot to, and I'll vary the y guess. And they're great. And now I have a perfectly stable differential equation I can solve. OK. So this same idea of shooting comes up in multiple applications. And this whole thing about flipping the change in the variables is often an extremely useful trick to just recast the equations in a way that's more numerically stable to solve, has fewer number of equations, you can write explicitly what you want. But this takes some cleverness to know how to recast them. All right, so this motivating example is a good one. Yeah? AUDIENCE: [INAUDIBLE] PROFESSOR: You could, but I was just trying to get back in the form of t0 equals zero. I guess I could have t0 equals negative 1 do it. I just couldn't do the arithmetic on the board. AUDIENCE: [INAUDIBLE] PROFESSOR: y of 0 is known, but then I have to start from t0 equals negative 1. So I could do-- so suppose I did u equals negative t dy du is equal to negative 10 to the 6 y. And I know y of 0 is equal to to-- no, what do I know? Yeah, y of 0 is equal to 10 to the minus 12. And y of negative 1 is equal to y yes, and I can integrate from negative 1 over. So this will t0 if I was using the shooting method, and this would be t final. Or you could shift it over by 1 if you want, and do it the way I did. Either one's fine. Is that all right? Sorry to confuse all the arithmetic. I was getting confused too. All right, now I wrote this-- another situation where this will not work is you can have systems where you have both positive and negative eigenvalues of the Jacobian, and that means that either way you integrate-- in the plus t direction or the negative t direction, either way it's always unstable. So no matter what you do, you're doomed. The IVP method is not going to work because you're numerically unstable in both directions. And so that's really a bummer, and it means our ODE IVP method's not going to work for us, and so then we have to find a different method. But before I go into that, let's just write the more general version of this. OK, so this is-- I think you guys all know this. So let's try a more general version of the ODE IVP problem. And now we'll allow for the fact that things don't have to be explicit, and they have to be as simple as we thought. And you can see it even in the problem I just did-- we had a vx squared plus vz squared had to be equal to something. And so that's not a simple boundary condition like we saw before. So you get a more complicated conditions, and what you really can have is a set of equations, gn of dy dt y, t, and is equal to 0. And corresponding, qn of dy dt evaluated at tn, y of tn, tn is equal to 0. So this is the general ODE system that's written in implicit form, and this has to be true for all t in the domain. All t that are a member of t0 to tf. So all the t's that are inside this interval, the domain we are about, this is the differential equation, written as just as general can be, and n is equal to 1 and N in these equations. AUDIENCE: Is there a reason you switched to q of n for the-- PROFESSOR: I just want to make it clear that this is the conditions, and this is the general equation. I don't know. You can write it anyway you want. Do you like some variable letters instead of q? I can change this. AUDIENCE: No, I was just checking. PROFESSOR: This one, I'm just trying to emphasize. This one is only true at tn only. This equation, this is the condition at one particular time I demand something be true, and this is the general difference equation that has to be true at all times. If I have n differentials, I need n differential equations, and n initial conditions. By the way, you can write the DAEs this way as well-- the differential algebraic equations, and then you just won't have as many initial conditions because every time you have an algebraic equation instead of a differential equation, then you don't need the initial condition because the algebraic equation has its own condition. Does that make sense? All right, so you can write this in general, and just by depending on how many differentials you have, that's how many conditions you need. A lot of times, you can write them explicit. This thing is equivalent to dy dt is equal to f of ty. Many times, you can solve for dy dt, and write this way, but not always. OK All right, and the reason why we need to have n conditions is just like when you're doing integrals, you'd have n constants of integration. Every time you do an integral, you get one, now you have three. We have n differential equations. We'll have n conditions. All right? Now, a bad thing about these problems, in general, is that the solutions depend a lot on the boundary conditions. And in fact, you can write down boundary conditions that are not achievable, which means there's no solution. So this is really a bummer if you don't realize you did it because you could try to solve it all day, and you'll never solve it because it doesn't have a solution. So for example, in this problem, if I set L to be too large, the cannonball can't go that far. I can't set L to be 3,000 kilometers and get a solution because the muzzle velocity is not high enough that the cannonball is going to travel 3,000 kilometers. But when I wrote the problem, you didn't see that, right? You didn't come and tell me, whoa, watch out. There's no way L could be bigger than such as such. And so you might not know it. You might specify a condition that's just not achievable. I used to work an industry, this happened a lot. My manager would specify boundary conditions which were not achievable. I want 99% yield and a 10% cost reduction. Go! And I'm the engineer, I'm trying to work it out, and then sometimes it didn't even come out. So in some ways, it's a lot better to just say max. Please maximize this as best you can, subject to your constraints, instead of specifying this. But of course, I don't know why, there's a school of management that says we weren't definite targets. We don't want to be just the best, we want to be perfect. So the no child left behind. Every single student in this class is going to really achieve. Anyway, that's the way targets gets set. It doesn't have to be achievable. And there's actually a big branch of mathematics about trying to detect if the boundary conditions are indeed consistent so that they can be achieved, and it's not so obvious to do it. So I'm not going to get into that at all. You can take a PDE class. There's a math department, and that's all they talk about all day is about how to figure out are these boundary conditions consistent or not consistent. You also have boundary conditions that are repetitive, and don't have enough information. In those cases, you might have an infinite number of solutions. And I think you can also have cases where you have multiple solutions. So like in the cannonball problem, if you could have a negative vy, and if the ship was close enough, you might be able to shoot it two ways. You can shoot it down at it, or you could just blow up high and come down, and they'll both be solutions that satisfy all the conditions. This is like general non-linear equation problems, that you don't know how many solutions you got, you don't know if you have a solution, you might have an infinite number of solutions, you might have some finite number of solutions, you don't know how many, and it's just that's the way life is, and you'll have to live with it. But it's kind of annoying sometimes. For now, let's just assume that we somehow know that we have a unique solution. There's a unique physical solution. We're just going to try to find that, and we're not going to worry about this other problem about the fact there could actually be multiple solutions, or no solution, or whatever. When you actually run into a case like that, then you'll really worry about it a lot. Is there a question? No? OK. All right, so we had an example in actually the zeroth homework, the very beginning, that had a problem like this. The homework 0P1, we had a problem where it gave you-- Do you remember this? It was a cylinder, and you were trying to compute t of r. Do you remember this? And it gave you some conditions about here. It gave you t when r w was equal to r, and it gave you dt dr-- actually, it gave you some equation. It was actually a function of this and t, all evaluated at r. It was equal to something. Do you remember this? So we gave those two, but then we also told you in the most recent homework, oh, by the way, dt dr has got to be equal to 0 at r equals 0. So this is one of these ones where we should only have two conditions, but we told you three conditions. So were any of you able to achieve all three conditions? No. OK, so this is highlighting the problem. So we gave you three boundary conditions that were apparently incompatible, that you couldn't actually solve them all at the same time. So beware, I guess. Now, in that problem, it might have been more natural to give you this equation, which we are sure is exactly true by the symmetry of the situation as one of the conditions, and then only one of these other conditions outside. That whichever one we were more sure was correct, might be the thing to do. And so we kind of gave you some goofball conditions. I guess, in some ways. Yeah? AUDIENCE: So if all three of those conditions are true, why wouldn't they converge to a solution? PROFESSOR: They could all be true, and if they were all true, then if you solve the numerical exactly, you should get the real solution. It should satisfy all of them. AUDIENCE: So does that mean one of them wasn't true in our case? If we weren't able to do-- PROFESSOR: Yeah, I think one of them wasn't true, and I think it's maybe due to round off. But it's highlights a problem. If you have three conditions that should be true, but they involve numbers in them, then often they won't all be exactly consistent with each other. But there's actually a worse kind of problem because it's close to being consistent to all of those conditions, but you're not exactly consistent. But then if you get into it, trying to figure what's going on, a lot of times you'll get near singular matrices and stuff because you have things that are nearly linearly dependent in some way. So really, it's better if you just have the right number of conditions that you're really sure really specify everything correctly, and just go with those, and treat the ones that you think would be nice if they were true as approximate things that are sort of physical checks to see whether your solution is reasonable. And we did it sort of like that in the problem you did. We said well, dt dr should really be equal to 0, but we didn't demand that in your solution. Your solution didn't actually satisfy that, and we use that as a measure of how far off your solution is. That's actually pretty useful in practice. You get a solution. It's good to have some additional condition that you can test for physical reasonableness of the solution, and particularly because there's a problem that you might have multiple solutions, and maybe one of them is a physical solution, and one of them is some other crazy solution. And so you need to have some way to detect the crazy solution by some other condition that you can test. Does that make sense? All right, so I said you could have cases where the Jacobian of your equation-- so you could be trying to solve this, and it could be, for some problems the Jacobian always has both positive and negative eigenvalues together. And in that case, using IVP is not going to be a smart idea because no matter which direction you integrate in, you're going to have a positive eigenvalue, and so then the IVP is not good. So instead, we're going to use a different method, and we'll talk about that next time, and those are called relaxation methods. And the concept of these is different than it was for shooting. So the relaxation method is you have a wide guess that you're going to have a guess for the entire y throughout the whole domain. Actually, a lot of times you might set this up so that y guess actually exactly satisfies the boundary conditions. So you know it satisfies those equations right from the beginning just from the way you set up y guess, but it won't satisfy the differential equation. And then what you'll do is you'll say how badly does it fail to satisfy that g is equal to 0? Because we really want g is equal 0 for the true solution. And we'll call this failure of our y guess to solve this, we'll call that the residuals. So we can actually plug in y guess in here, and y guess in here, and evaluate this. And normally it will not be 0 because our y guess is not perfect, and so it's going to deviate from 0. But how big it deviates from 0 is going to be our measure of how bad that guess was, and then we're going to try to change y guess to try to make it closer and closer to 0. So the idea is relax it. You get an incorrect solution, and you adjust it to relax it to make it become the correct solution. All right, and so that's the general idea of the relaxation method, and we'll talk about several different forms of that. Normally, what we have to do is you have to write y guess in terms of some parameters. So y guess of t depends on some parameters, and we're going adjust the parameters, and that's going to change what y guess is. And so for example, we could do a basis set expansion as we did in one of the earlier classes. We talked about how you could expand y guess as a sum of some basis functions weighted by p's, and then you can adjust the p's, and find the best possible set of p's to make this g as close to 0 as possible, for example. We'll give examples of that next time. There's several choices about that there. All right. |
MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 6_Singular_Value_Decomposition_Iterative_Solutions_of_Linear_Equations.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JAMES W. SWAN: So this is going to be our last lecture on linear algebra. The first three lectures covered basics. The next three lectures, we talked about different sorts of transformations of matrices. This final lecture is the last of those three. We're going to talk about in another sort of transformation called the singular value decomposition. OK, before we jump in, I'd like to do the usual recap business. I think it's always hopeful to recap or look at things from a different perspective. Early on, I told you that the infinite dimensional equivalent of vectors would be something like a function, which is a map, a unique map maybe from a point to x to some value f of x. And there is an equivalent representation of the eigenvalue eigenvector problem in function space. We call these eigenvalues and eigenfunctions. Here's a classic one where the function is y of x, OK? This is the equivalent of the vector, and equivalent of the transformation or the matrix that's this differential operator this time, the second derivative. So I take the second derivative of this particular function, and the function is stretched. It's multiplied by some fixed value at all points. And it becomes lambda times y. And that operator has to be closed with some boundary conditions as well. We have to say what the value of y is at the edges of some boundary. So there's a one-to-one correspondence between these things. What is the eigenfunction here, or what are the eigenfunctions? And what are the eigenvalues associated with this transformation or this operator? Can you work those out really quickly? You learned this at some point, right? Somebody taught you differential equations and you calculated these things. Take about 90 seconds. Work with the people around you. See if you can come to a conclusion about what the eigenfunction and eigenvalues are. That's enough time. You can work on this on your own later if you've run out of time. Don't worry about it. Does somebody want to volunteer a guess for what the eigenfunctions are in this case? What are they? Yeah? AUDIENCE: [INAUDIBLE] JAMES W. SWAN: OK, so you chose exponentials. That's an interesting choice. That's one possible choice you can make. OK, so we could say-- this is sort of a classical one that you think about when you first learn differential equation. They say, an equation of this sort has solutions that look like exponentials, and that's true. There's another representation for this, which is as trigonometric functions instead, right? Either of those is acceptable. [INAUDIBLE] the trigonometric functions, that representation is a little more useful for us here. We know that the boundary conditions tell us that y of 0 is supposed to be 0. That means that the C1 has to be 0, because cosine of 0 is 1. So C1 has 0 in this case. So that fixes one of these coefficients. And now we're left with a problem, right? Our solutions, our eigenfunctions, cannot be unique. So we don't get to specify C2, right? Any function that's a multiple of this sine should also be an eigenfunction. So instead the other boundary condition, this y of l equals 0, needs to be used to pin down with the eigenvalue is. So the second equation, y of l equals 0, which implies that the square root of minus lambda has to be equal to 2 pi over l, it has to be all the nodes of the sine where the sine is equal to 0. That's the equivalent of our secular characteristic polynomial that prescribes with the eigenvalues are associated with each of the eigenfunctions. So now we know what the eigenvalues are. The eigenvalues are the set of numbers minus 2 pi n over l squared. There's an infinite number of eigenvalues. It's an infinite dimensional space that we're in, so it's not a big surprise that it works out that way. And the eigenvectors then are different scalar multiples of sine of the eigenvalues, square root of the eigenvalues, minus x. There's a one-to-one correspondence between all the linear algebra we've done and linear differential equations or linear partial differential equations. You can think about these things in exactly the same way. I'm sure in 1050, you started to talk about orthogonal functions to represent solutions of differential equations. Or if you haven't, you're going to very soon. This is a part of the course you get to look at the analytical side of some of these things as opposed to the numerical side. But there's a one-to-one relationship between those things. So if you understand one, you understand the other, and you can come at them from either perspective. This sort of stuff is useful. Actually, the classical chemical engineering example comes from quantum mechanics where you think about wave functions and different energy levels corresponding to eigenvalues. That's cool. Sometimes, I like to think about a mechanical analog to that, which is the buckling of an elastic column. So you should do this at home. You should go get a piece of spaghetti and push on the ends of the piece of the spaghetti. And the spaghetti will buckle. Eventually it'll break, but it'll buckle first. It'll bend. And how does it bend? Well, a balance of linear momentum on this bar would tell you that the deflection in the bar at different points x along the bar multiplied by the pressure has to balance the bending moment in the bar itself. So this e is some elastic constant. I has a moment of inertia. And D squared y dx squared is something like the curvature of the bar. So it's the bending moments of the bar that balances the pressure that's being exerted on the bar. And sure enough, this bar will buckle when the pressure applied exceeds the first eigenvalue associated with this differential equation. We just worked that eigenvalue out. We said that that eigenvalue had to be the square root of 2 pi over l squared. And so when the pressure exceeds square root of 2 pi over l squared times the elastic modulus, this column will bend and deform continuously until it eventually breaks, right? It will undergo this linear elastic deformation, then plastic deformation later, and it will break. The Eiffel Tower, actually, is one of the first structures in the world to utilize this principle, right? It's got very narrow beams in it. The beams are engineered so that their elastic modulus is strong enough that they won't buckle. Gustave Eiffel is one of the first applied physicists, somebody who took the physics of elastic bars and applied them to building structures that weren't big and blocky, but used a minimal amount of material. Cool, right? OK, so that's recap. Any questions about that? You've seen these things before. You understood them well before too maybe? Give some thought to this, OK? We talked about eigendecomposition last time that, associated with the square matrix, was a particular eigenvalue or particular set of eigenvalues, stretches and corresponding eigenvectors directions. These were special solutions to the system of linear equations based on a matrix. It was a square matrix. And you might ask, well, what happens if the matrix isn't square? What if A is in the space of real matrices that are n by m, where n and m maybe aren't the same? Maybe they are the same, but maybe they're not. And there is an equivalent decomposition. It's called the singular value decomposition. It's like an eigendecomposition for non-square matrices. So rather than writing our matrix as some w lambda w inverse, we're going to write it as some product U times sigma times V with this dagger. The dagger here is conjugate transpose. Transpose the matrix, and take the complex conjugate of all the elements, OK? I mentioned last time that eigenvalues and eigenvectors could be complex, potentially, right? So whenever we have that case where things can be complex, usually the transposition operation is replaced with the conjugate transpose. What are these different matrices. Well, let me tell you. U is a complex matrix. It maps from the space N to R N to R N, so it's an n by n square matrix. Sigma is a real valued matrix, and it lives in the space of n by n matrices. V is a square matrix again, but it has dimensions m by m. Remember, A maps from R M to R N, so that's what the sequence of products says. B maps from m to m. Sigma maps from m to n. U maps from n to n. So this match from m to n as well. Sigma is like lambda from before. It's a diagonal matrix. It only has diagonal elements. It's just not square, but it only has diagonal elements, all of which will be positive. And then U and V are called the left and right singular vectors. And they have special properties associated with them, which I'll show you right now. Any questions about how this decomposition is composed or made up? It looks just like the eigendecomposition, but it can be applied to any matrix. Yes? AUDIENCE: Quick question. JAMES W. SWAN: Sure. AUDIENCE: Do all matrices have this thing, or is it like the eigenvalues where some do and some don't. JAMES W. SWAN: This is a great question. So all matrices are going to have a singular value decomposition. We saw with the eigenvalue decomposition that there could be a case where the eigenvectors are degenerate, and we can't write that full decomposition. All matrices are going to have this decomposition. So for some properties of this decomposition, U and V are what we call unitary matrices. I talked about these before. Unitary matrices are ones for whom, if they're real valued, their transpose is also their inverse. If they're complex valued, and they're conjugate transpose is the equivalent of their inverse. So U times U conjugate transpose will be identity. V times V conjugate transpose will be identity. Unitary matrices also have the property that they impart no stretch to a matrix-- or to vectors. So their maps don't stretch. They're kind of like rotational matrices, right? They change directions, but they don't stretch things out. If I were to take A conjugate transpose and multiply it by A, that would be the same as taking U sigma V conjugate transpose, and multiplying it by U sigma V. If I use the properties of matrix multiplications and complex conjugate transposes, and work out what this expression is, I'll find out that it's equivalent to V sigma conjugate transpose sigma V conjugate transpose. Well this has exactly the same form as an eigendecomposition. An eigendecomposition of A times A instead of an eigendecomposition of A. So V is the set of eigenvectors of A conjugate transpose A, and sigma squared are the eigenvalues of A conjugate transpose times A. And if I reverse the order of this multiplication-- so I do A times A conjugate transpose-- and work it out, that would be U sigma sigma U. And so U are the eigenvectors of A A conjugate transpose, and sigma squared are still the eigenvalues of A A conjugate transpose. So what are these things U and V? They relate to the eigenvectors of the product of A with itself, this particular product of A with itself, or this particular product of A with itself. Sigma are the singular values. And all matrices possess this sort of a decomposition. They all have a set of singular values and singular vectors. These sigmas are called the singular values of the A. They have a particular name. I'm going to show you how you can use this decomposition to do something you already know how to do, but how to do it formally. What are some properties of the singular value decomposition? So if we take a matrix A and we compute it's singular value decomposition, this is how you do it in Matlab. We'll find out, for this matrix, U is identity. Sigma is identity with an extra column pasted on it. And B is also identity. I mean, this is the simplest possible four by three matrix I can write down. You don't have to know how to compute the singular value decomposition, you just need to know that it can be computed in this way. You might be able to guess how to compute it based on what we did with eigenvalues earlier and eigenvectors. It'll turn out some of the columns of sigma will be non-zero right? There are three non-zero columns of sigma. And the columns of V, they correspond to those columns of sigma, spanned the null space of the matrix A. So the first three columns here are non-zero, the first three columns of V. I'm sorry, the first three columns here are non-zero. The last column is 0. The columns of sigma which are 0 correspond to a particular column in V, this last column here, which lives in the null space of A. So you can see, if I take A and I multiply it by any vector that's proportional to 0, 0, 0, 1, I'll get back 0. So the null space of A is spanned by all these vectors corresponding to the 0 columns of sigma. Some of the columns of sigma are non-zero. These first three columns. And the rows of U corresponding to those three columns span the range of A. So if I do the singular value decomposition of a matrix, and I look at U, V, and sigma and what they're composed of-- where sigma is 0 and non-zero, and the corresponding columns or rows of U and V-- then I can figure out what vectors span the range and null space of the matrix A. Here's another example. So here I have A. Now instead of being three rows by four columns, it's four rows by three columns. And here's the singular value decomposition that comes out of Matlab. There are no vectors that live in the null space of A, and there are no 0 columns in sigma. There's no corresponding columns in V. There are no vectors in the null space of A. The range of A is spanned by the rows corresponding to the non-zero-- the rows of U corresponding to the non-zero columns of sigma. So it's these three columns in the first three rows. And these first three rows, clearly they span-- they describe the same range as the three columns in A. So the singular value decomposition gives us direct access to the null space and the range of a matrix. That's handy. And it can be used in various ways. So here's one example where it can be used. Here I have a fingerprint. It's a bitmap. It's a square bit of data, like a matrix, and each of the elements of the matrix takes on a value describing how dark or light that pixel. Let's say it's grayscale, and it's value's between 0 and 255. That's pretty typical. So I have this matrix, and each element to the matrix corresponds to a pixel. And I do a singular value decomposition. Some of the singular values, the values of sigma, are bigger than others. They're all positive, but some are bigger than others. The ones that are biggest in magnitude carry the most information content about the matrix. So we can do data compression by neglecting singular values that are smaller than some threshold, and also neglecting the corresponding singular vectors. And that's what I've done here. So here's the original bitmap of the fingerprint. I did the singular value decomposition, and then I retained only the 50 biggest singular values and I left all the other singular values out. This bitmap was something like, I don't know, 300 pixels by 300 pixels, so there's like 300 singular values, but I got rid of 5/6 of the information content. I dropped 5/6 of the singular vectors, and then I reconstructed the matrix from the singular values and those singular vectors, and you get a faithful representation of the original fingerprint. So the singular value decomposition says something about the information content in the transformation that is the matrix, right? There are some transformations that are of lower power or importance than others. And the magnitude of these singular values tell you what they are. Does that makes sense? How else can it be used? Well, one way it can be used is finding the least square solution to the equation Ax equals b, where A is no longer a square matrix, OK? You've done this in other contexts before where the equations are overspecified. We have more equations than unknowns, like data fitting. You form the normal equations, you multiply both sides of Ax equals b by A transpose, and then invert A transpose A. You might not be too surprised, then, to think that singular value decomposition could be useful here too. Since we already saw the data in a singular value decomposition corresponds to eigenvectors and eigenvalues of this A transpose A, right? But there's a way to use this sort of decomposition formally to solve problems that are both overspecified and underspecified. Least squares means find the vector of solutions x that minimizes this function phi. Phi is the length of the vector given by the difference between Ax and b. It's one measure of how far an error our solution x is. So let's define the value x which is least in error. This is one definition of least squares. And I know the singular value decomposition of A. So A is U sigma times V. So I have U sigma V times x. I can factor out U, and I've got a factor of U transpose, or U conjugate transpose multiplying by b. So Ax minus b is the same as U times the quantity sigma V conjugate transpose x minus U conjugate transpose b. We want to know the x that minimizes this phi. It's an optimization problem. We'll talk in great detail about these sorts of problems later. This one is so easy to do, we can just work it out in a couple lines of text. We'll define a new set of unknowns, y, which is V transpose times x, and a new right-hand side for a system of equations p, which is U transpose times b. And then we can rewrite our function phi that we're trying to minimize. So phi then becomes U sigma y minus p. U is a unitary vector. It imparts no stretch in the two norms, so this sigma y minus p doesn't get elongated by multiplication with U. So it's length, the length of this, is the same as the length of sigma y minus p. You can prove this. It's not very difficult to show at all. You use the definition of the two norm to prove it. So phi is minimized by y's, which makes this norm smallest, make it closest to 0. Let r be the number of non-zero singular values, the number of those sigmas which are not equal to 0. That's also the rank of A. Then I can rewrite phi as the sum from i equals 1 to r of sigma i i time y i minus p i squared. That's parts of this length, this Euclidean length, for which sigma is non-zero. Plus the sum from r plus 1 to n, the sum over the rest of the values of p, for which the corresponding sigmas are 0. I want to minimize phi, and the only thing that I can change to minimize it is what? What am I free to pick in this equation in order to make phi as small as possible? Yeah? AUDIENCE: y. JAMES W. SWAN: y, so I need to choose the y's that make this phi as small as possible. What value should I choose for the y's? What do you think? AUDIENCE: [INAUDIBLE] JAMES W. SWAN: Perfect, right? Choose y equals p i over sigma i i. Right, y i is p i over sigma i i. Then all of these terms is 0. I can't make this sum any smaller than that. That fixes the value of y i up to r. I can't do anything about this left over bit here. There's no choice of y that's going to make this part and the smaller. It's just left over. It's some remainder that we can't make any smaller or minimize an smaller. There isn't an exact solution to this problem, in many cases. But one way this could be 0 is if r is equal to n. Then there are left over unspecified terms, and then this y i equals p i over sigma i is the exact solution to the problem. So this is what you told me. Choose y i is p i over sigma i i for i bigger than 1 and smaller than r. There are going to be values of y i that go between r plus 1 and m, because A was a vector that mapped from m to n, right? So I have extra values of y that could be specified potentially. If that's true, if r plus 1 is smaller than m, then there's some components of y that I don't get to-- I can't specify, right? My system of equations is somehow underdetermined. I need some external information to show me what values to pick for those y i. I don't know. I can't use them. Sometimes people just set y i equal to 0. That's sort of silly, but that's what's done. It's called the minimum norm least square solution. y has minimum length, when you set all these other components to 0. But the truth is, we can't specify those components, right? We need some external information in order to specify them. Once we know y, we can find x going back to our definition of what y is. So I multiply this equation by V on both sides, and I'll get V y equals x. So I can find my least square solution to the problem from the singular value decomposition. So I can find the least square solution to both overdetermined and underdetermined problems using singular value decomposition. It inherits all the properties you know of solving the normal equations, multiplying by A transpose the entire equation, and solving for a least square solution that way. But that's only good for overdetermined systems of equations. This can work for underdetermined equations as well. And maybe we do have extraneous information that lets us specify these other components somehow. Maybe we do a separate optimization that chooses from all possible solutions where these y i's are free, and picks the best one subject to some other constraint. Does it makes sense? OK, that's the last decomposition we're going to talk about. It's as expensive to compute the singular value decomposition as it is to solve a system of equations. You might have guessed that it's got an order N cubed flavor to it. It's kind of inescapable that we run up against those computational difficulties, order N cubed computational complexity. And there are many problems of practical interest, particularly solutions of PDEs, for which that's not going to cut it. Where you couldn't solve the problem with that sort of scaling in time. You couldn't compute the Gaussian elimination, or the singular value decomposition, or an eigenvalue decomposition. It won't work. And in those cases, we appeal to not exact solution methods, but approximate solution methods. So instead of trying to get an exact solution, we'll try to formulate one that's good enough. We already know the computer introduces numerical error anyways. Maybe we don't need machine precision in our solution or something close to machine precision in our solution. Maybe we're solving engineering problem, and we're willing to accept relative errors on the order of 10 to the minus 3 or 10 to the minus 5, some specified tolerance that we apply to the problem. And in those circumstances, we use iterative methods to solve systems of equations instead of exact methods, elimination methods, or metrics decomposition methods. These algorithms are all based on iterative refinement of an initial guess. So if we have some system of equations we're trying to solve, Ax equals b, we'll formulate some linear map, right? xi plus 1 will be some matrix C times x i plus some little vector c where x i is my last best guess for the solution to this problem, and x i plus 1 is my next best guess for the solution to this problem. And I'm hoping, as I apply this map more and more times, I'm creeping closer to the exact solution to the original system of equations. The map will converge when x i plus 1 approaches x i, when the map isn't making any changes to the vector anymore. And the converged value will be a solution when x i-- which is equal to i minus c inverse times c, if I replace x i was 1 with x i appear, so I say that my map has converged-- when this value is equivalent to A inverse times B, when it's a solution to the original problem, right? So my map may converge. It may not converge to a solution of the problem I like, but if it satisfies this condition, then has converged to be a solution of the problem that I like as well. And so it's all about using this C here and this little c here so that this map converges to solution of the problem I'm after. And there are lots of schemes for doing this. Some of them are kind of ad hoc. I'm going to show you one right now. And then when we do optimization, we'll talk about a more formal way of doing this for which you can guarantee very rapid convergence to a solution. So here's a system of equations I'd like to solve. It's not a very big one. It doesn't really make sense to solve this one iteratively, but it's a nice illustration. One way to go about formulating this map is to split this matrix into two parts. So I'll split it into a diagonal part and an off diagonal part. So I haven't changed the problem at all by doing that. And then I'm going to rename this x x i plus 1, and I'm going to rename this x x i. And then move this matrix vector product to the other side of the equation. And here's my map. Of course, this matrix multiplied doesn't make any-- it's not useful to write it out explicitly. This is just identity. So I can drop this entirely. This is just x i plus one. So here's my map. Take an initial guess, multiply it by this matrix, add the vector 1, 0, and repeat over and over and over again. Hopefully-- we don't really know-- but hopefully, it's going to converge to a solution of the original linear equations. I didn't make up that method. That's a method called Jacobi Iteration. And the strategy is to split the matrix A into two parts-- a sum of its diagonal elements, and it's off diagonal elements-- and rewrite the original equations as an iterative map. So D times x i plus 1 is equal to minus r times x i plus b. Or x i plus 1 is D inverse times minus r x i plus b. If the equations converge, then D plus r times x i has to be equal to b, we will have found a solution. If it converges, right? If these iterations approach a steady value. If they don't change from iteration to iteration. Is The nice thing about the Jacobi method is it turns the hard problem, the order N cubed problem of computing A inverse B, into a succession of easy problems, D inverse times some vector C. How many calculations does it take to compute that D inverse? N, that's right, order N. It's just a diagonal matrix. I invert each of its diagonal elements, and I'm done. So I went from order N cubed, which was going to be hard, into a succession of order N problems. So as long as it doesn't take me order N squared iterations to get to the solution that I want, I'm going to be OK. This is going to be a viable way to solve this problem faster than finding the exact solution. How do you know that it converges? That's the question. Is this thing actually going to converge or not, or are these iterations just going to run on and on forever? Well, one way to check whether it will converge or not is to go back up to this equation here, and substitute b equals Ax, where x is the exact solution to the problem. And you can transform, then, this equation into one that looks like x i plus 1 minus x equal to minus D inverse times r x i minus x. And if I take the norm of both sides and I apply our normal equality-- where the norm of a matrix vector product is smaller than the product of the norms of the matrices of the vectors-- then I can get a ratio like this. That the absolute error in iteration I plus 1 divided by the absolute error in iteration i is smaller than the norm of this matrix. So if I'm converging, then what I expect is this ratio should be smaller than 1. The error in my next approximation should be smaller than the error in my current approximation. That makes sense? So that means that I would hope that the norm of this matrix is also smaller than 1. If it is, then I'm going to be guaranteed to converge. So for a particular coefficient matrix, for a system of linear equations I'm trying to solve, I may be able to find-- I may find that this is true. And then I can apply this method, and I'll converge to a solution. We call this sort of convergence linear. Whatever this number is, it tells me the fraction by which the error is reduced from iteration to iteration. So suppose this is 1/10. Then the absolute error is going to be reduced by a factor of 10 in each iteration. It's not going to be 1/10 usually. It's going to be something that's a little bit bigger than that typically, but that's the idea. You can show-- I would encourage you to try to work this out on your own-- but you can show that the infinity norm of this product-- infinity norm of this product is equal to this. And if I ask that the infinity norm of this product be smaller than 1, that's guaranteed when the diagonal values of the matrix and absolute value are bigger than the sum of the off diagonal values in a particular row or a particular column. And that kind of matrix we call diagonally dominant. The diagonal values are bigger than the sum and absolute value of the off diagonal pieces. So diagonally dominant matrices, which come up quite often, can be-- those linear equations based on those matrices can be solved reasonable efficiency using the Jacobi method. There are better methods to choose. I'll show you one in a second. But you can guarantee that this is going to converge to a solution, and that the solution will be the right solution to the linear equations you were trying to solve. So if the goal is just to turn hard problems into easier to solve problems, then there are other natural ways to want to split a matrix. So maybe you want to split into A lower triangular part which contains the diagonal elements of A, and an upper triangular part which has no diagonal elements of A. We just split this thing apart. And then we could rewrite our system of equations is an iterative map like this, L times x i plus 1 is minus U times x i plus b. All I have to do is invert l to find my next iteration. And how expensive computationally is it to solve a system of equations which is triangular? This is a process we call back substitution. Its order-- AUDIENCE: N squared. JAMES W. SWAN: --N squared. So we still beat N cubed. One would hope that it doesn't require too many iterations to do this. But in principle, we can do this order N squared operations many times. And it'll turn out that this sort of a map converges to the solution that we're after. It converges when matrices are either diagonally dominant as before, or they're symmetric and they're positive definite. Positive definite means all the eigenvalues of the matrix are bigger than 0. So try the iterative method solving some equations and see how we convert. Yes? AUDIENCE: How do you justify ignoring the diagonal elements in that method? JAMES W. SWAN: So the question was, how do you justify ignoring the diagonal elements in this method. Maybe I was going too fast or I misspoke. So I'm going to split A into a lower triangular matrix that has all the diagonal elements, and U is the upper parts with none of those diagonal elements on it. Does that make sense? AUDIENCE: Yeah. JAMES W. SWAN: Thank you for asking that question. I hope that's clear. l holds onto the diagonal pieces and U takes those away. So let's try it. On a matrix like this, the exact solution to this system of equations is 3/4, 1/2, and 1/4. All right, we'll try Jacobi, we'll have to give it some initial guess for the solution, right? We're talking about places where you can derive those initial guesses from later on in the course, but we have to start the iterative process with some guess at the solutions. So here's an initial guess. We'll apply this map. Here's Gauss-Seidel with the same initial guess, and we'll apply this map. They're both linearly convergent, so the relative error will go down by a fixed factor after each iteration. Iteration one, the relative error in Jacobi will be 38%. In Gauss-Seidel, it'll be 40%. If we apply this all the way down to 10 iterations, the relative error Jacobi will be 1.7%, and the relative error in Gauss-Seidel 0.08%. And we can go on and on with these iterations if we want until we get sufficiently converged, we get to a point where the relative error is small enough that we're happy to accept this answer as a solution to our system of equations. So we traded the burden of doing all these calculations to do elimination for a faster, less computationally complex methodology. But the trade off was we don't get an exact solution anymore. We're going to have finite precision in the result, and we have to specify the tolerance that we want to converge to. We're going to see now-- this is the hook into the next part of that class-- we're going to talk about solutions of nonlinear equations next for which there are almost no non-linear equations that we can solve exactly. They all have to be solved using these iterative methods. You can use these iterative methods for linear equations. It's very common to do it this way. In my group, we solve lots of systems of linear equations associated with hydrodynamic problems. These come up when you're talking about, say, low Reynolds number flows, which are linear sorts of fluid flow problems. They're big. It's really hard to do Gaussian elimination, so you apply different iterative methods. You can do Gauss-Seidel. You can do Jacobi. We'll learn about more advanced ones like PCG, which you're applying on your homework now, and you should be seeing that it converges relatively quickly in cases where exact elimination doesn't work. We'll learn, actually, how to do that method. That's one that we apply in my own group. It's pretty common to use out there. Yes? AUDIENCE: One question, is that that Gauss, [INAUDIBLE] JAMES W. SWAN: Order N squared. AUDIENCE: Yeah, that's what I meant. So now we've got an [INAUDIBLE]. So we basically have [INAUDIBLE] iterations, right? JAMES W. SWAN: This is a wonderful question. So this is a pathological problem in the sense that it requires a lot of calculations to get an iterative solution here. We haven't gotten to an end that's big enough that the computational complexities crossover. So for small Ns, probably the factor in front of N-- whatever number that is-- and maybe even the smaller factors, order N squared factors on that order N cubed, play a big role in how long it takes to actually complete this thing. But modern problems are so big that we almost always are running out to Ns that are large enough that we see a crossover. You'll see this in your homework this week. You won't see this crossover at N equals 3. You're going to see it out at N equals 500 or 1,200, big problems. Then we're going to encounter this crossover. That's a wonderful question. So first small system sizes, iterative methods maybe don't buy you much. I suppose it depends on the application though, right? If you're doing something that involves solving problems on embedded hardware, in some sort of sensor or control valve, there may be very limited memory or computational capacity available to you. And you may actually apply an iterative method like this to a problem that that controller needs to solve, for example. It just may not have the capability of storing and inverting what we would consider, today, a relatively small matrix because the hardware doesn't have that sort of capability. So there could be cases where you might choose something that's slower but feasible, versus something that's faster and exact, because there are other constraints. They do exist, but modern computers are pretty efficient. Your cell phone is faster than the fastest computers in the world 20 years ago. We're doing OK. So we've got to get out to big system sizes, big problem sizes, before this starts to pay off. But it does for many practical problems. OK I'll close with this, because this is the hook into solving nonlinear equations. So I showed you these two iterative methods, and they kind of had stringent requirements for when they were actually going to converge, right? I had to have a diagonally dominant system of equations for Jacobi to converge. I had to have diagonal dominance or symmetric positive definite matrices. These things exist and they come up in lots of physical problems, but I had to have it in order for Gauss-Seidel to converge. What if I have a system of equations that doesn't work that way? Or what if I have an iterative map that I like for some reason, but it doesn't appear to converge? Maybe it converges under some circumstances, but not others. Well, there's a way to modify these iterative maps, called successive over-relaxation, which can help promote convergence. So suppose we have an iterative map like this, x i plus 1 is some function of the previous iteration value. Doesn't matter what it is. It could be linear, could be non-linear. We don't actually care. The sought after solution is found when x i plus 1 is equal to x i. So this map is one the convergence to the exact solution of the problem that we want. We've somehow guaranteed that that's the case, but it has to converge. One way to modify that map is to say x i plus 1 is 1 minus some scalar value omega times x i plus omega times f. You can confirm that if you substitute x i plus 1 equals x i into this equation, you'll come up with the same fixed points of this iterative map x i is equal to f of x i. So you haven't changed what value will converge here, but you've affected the rate at which it converges. Here you're saying x i plus 1 is some fraction of my previous solution plus some fraction of this f. And I get to control how big those different fractions. So if things aren't converging well for a map like this, then I could try successive over-relaxation, and I could adjust this relaxation parameter to be some fraction, some number between 0 and 1, until I start to observe convergence. And there are some rules one can use to try to promote convergence with this kind of successive over-relaxation. This is a very generic technique that one can apply. If you have any iterative map you're trying to apply, it should go to the solution you want but it doesn't converge for some reason, then you can use this relaxation technique to promote convergence to the solution. You may slow the convergence way down. It may be very slow to converge, but it will converge. And after all, an answer is better than no answer, no matter how long it takes to get it. So sometimes you've got to get these things by hook or by crook. So for example, you can apply this to Jacobi. This was the original Jacobi map. And we just take that. We add 1 minus omega times x i plus omega times this factor over here. And now we can choose omega so that this solution converges. We always make omega small enough so that the diagonal values of our matrix appear big enough that the matrix looks like it's diagonally dominated. You could go back to that same convergence analysis that I showed you before and try to apply it to this over-relaxation form of Jacobi and see that, while there's always going to be some value of omega that's small enough, that this thing will converge. It will look effectively diagonally dominant, because omega inverse times D will be big enough, or omega times D inverse will be small enough. Does that make sense? You can apply the same sort of damping method to Gauss-Seidel as well. It's very common to do this. The relaxation parameter acts like an effective increase in the eigenvalues of the matrix. So you can think about L. That's a lower triangular matrix. It's diagonal values are its eigenvalues. The diagonal values of L inverse-- well, 1 over those diagonal values are the eigenvalues of L inverse. And so if we make omega very small, then we make the eigenvalues of L inverse very small, or the eigenvalues or L very big. And again, the matrix starts to look diagonally dominated. And you can promote convergence in this way. So even though this may be slow, you can use it to guarantee convergence of some iterative procedures, not just for linear equations, but for non-linear equations as well. And we'll see, there are good ways of choosing omega for certain classes of non-linear equations. We'll apply Newton-Raphson method, and then will damp it using exactly the sort of procedure. And I'll show you how you can choose a nearly optimal value for omega to promote convergence to the solution. Any questions? No, let me address one more thing before you go. We've scheduled times for the quizzes. They are going to be in the evenings on the dates that are specified on the syllabus. We wanted to do them during the daytime. It was really difficult to schedule a room that was big enough for this class, so they have to be from 7:00 to 9:00 in the gymnasium. I apologize for that. We spent several days looking around trying to find a place where we could put everybody so you would all get the same experience in the quiz. I know that the November quiz comes back to back with the thermodynamics exam as well. That's frustrating. Thermodynamics is the next day. That week is tricky. That's AICHE, so most of the faculty have to travel. We won't be able to teach, but you won't have classes one of those days so you have extra time to study. And Columbus Day also falls in that week, so there's no way to put three exams in four days without having them come right back to back. Believe me, we thought about this and tried to get things scheduled as efficiently as we could for you, but sometimes there are constraints that are outside of our control. But the quiz times are set. There's going to be done in October and one in November. They'll be in the evening, and they'll be in the gymnasium. I'll give you directions to it before the exam, just say you know exactly where to go, OK? Thank you, guys. |
MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 9_Homotopy_and_Bifurcation.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK, let's go ahead and get started again. So we had a nice review on Monday of some of the different concepts that we've talked about up to this point in the course. We've covered linear algebra, solutions of systems of nonlinear equations. There's been a lot of fundamentals involved in that. Things are going to get more applied as the sorts of problems we're trying to solve get more complicated. Complicated problems require very sophisticated methods to solve them. It's not going to be our expectation that you're necessarily able to implement all of these sophisticated methods. So for example, you implemented Newton's method on your last homework assignment. That'll be the last time we ask you to do that by hand. You'll utilize tools built into Matlab to solve other systems of nonlinear equations going forward. You understand how it works and what the limitations are. You understand how to build different tools that will be inputs to these functions like the function you're trying to find the root for, or an explicit expression for the Jacobian. You even understand how to set certain parameters associated with these methods, and what those particular parameter values mean. Like setting the function norm tolerance, or setting the step norm tolerance associated with these methods. Setting the number of iterations you're going to let the method go. You've implemented it, so you understand what these things mean, what the consequences of choosing them are, how they might increase the time associated with computations that you're going to conduct. So we've done a lot of fundamentals. Maybe that was difficult for some of you who haven't seen this before. Now things are becoming more about the practice of numerical methods. How do we actually solve things reliably, and that's what today's lecture will talk about. We discussed solving systems of nonlinear equations and needing good initial guesses for our methods. Bad initial guesses can lead to unpredictable results. Good initial guesses, we know the Newton-Raphson method is going to double the number of accurate digits each iteration, and we should converge very rapidly to a solution. So it seems like having good initial guesses is really ideal. And this topic, homotopy and bifurcation, will cover precisely those issues, and the practice of getting these good initial guesses. I think is one of the most useful things you'll learn in this class, actually. It's a really nice way of going about solving complicated problems. Before we begin, I want to address a question I got via email last night. That's always good if these questions go on [? Piaza ?] so everyone can read them, but I thought it was a very nice question about how do you find-- the question was how do you find multiple roots using the Newton-Raphson method. So if I have a polynomial, where the function doesn't cross through the origin, but it just touches the origin, and goes back up. That's a root with multiplicity in the polynomial. And can the Newton-Raphson method find that? Because the derivative of the function is 0 at the root. So who says the Newton-Raphson method will be able to find those roots? What do you think? Yes? And who says it won't be able to find those roots? Or who doesn't know? You don't know. That's OK. That's a good question, right. Do you know or don't you know? So we know the Newton-Raphson method relies on knowing the derivative of the function at different points in order to take our next Newton-Raphson step towards that solution. It may be the case, the derivative is 0 at the root, but nearby the root, the derivative may not be 0. And then the Newton-Raphson steps are going to be fine. They're are always going to keep stepping down towards the root. If you go through an analysis of how the algorithm converges, you'll find out that you only get linear convergence in that case, instead of quadratic convergence. So there will be some penalty to pay for the fact that the derivative is 0 at the root itself, but it's not going to be so bad. The method can still converge to those sorts of solutions. Not all is lost there, but the rate of convergence may be not as good as you expected it to be. And the same is true in the multi-dimensional case as well. It's a little hard to imagine what a multiple root is going to look like. It's something like tangent curves in the solution of plane. So if I have f1 of x1 and x2 equal 0, that's a curve in the x1, x2 plane. And I have f2 in that same plane equal to 0. The place where they touch might be one of these multiple roots. That's a place where the determinant of the Jacobian may be equal to 0. You'll have the same sort of circumstance. Nearby, the Jacobian can be non singular, and you can take steps towards the root, but there's a worst case scenario where you're coming in on a direction that's parallel that belongs to the null space of the Jacobian at the root, and then everything's going to slow down. You can still converge, but it may be very slow as a consequence. I thought that was a really nice question. It's a detailed question about how Newton-Raphson works. It's not necessarily going to fail under these weird circumstances. They don't come up too often. They're going to come up in our discussion today, actually. It's not going to fail, but it may converge very slowly to the solution. OK? Good. I'm going to remind you, the last thing we did with Newton-Raphson methods, we talked about one way of fixing them. In lieu of having good initial guesses, we may not want to take the full Newton-Raphson step. We may want to do these quasi Newton-Raphson methods instead. One of those was this damp Newton-Raphson where we introduced some damping parameter. We don't take the full step, we take some shorter step, and we try to make the absolute value over function become as small as possible. That's an optimization problem. It's as hard as finding the root of the system of equations that we're looking at. So we don't solve that optimization problem exactly. That's crazy. Instead we use this backtracking line search method where we change our damping parameter in some systematic way until we get to a point where we're satisfied. Where we say OK, in the process of changing this, we've actually found a smaller value of our function, and we accept that as our best guess for where the next iterate is supposed to go. And this sort of approach can give you global convergence for the algorithm, which is great, but it's not globally convergent to roots. It's globally convergent to minima, asymptotes, and roots as well. But it's globally convergent. So it'll terminate. The algorithm will stop, and when it stops, you can ask is this a root or not? Do I think this is sufficiently close to a root or not? If it's not sufficiently close to a root, then you know you terminated somewhere like a minima, or you rushed out towards an asymptote, and maybe you need to start with a new initial guess for your solution. These quasi Newton-Raphson methods are very useful. This is what Matlab uses in order to-- it uses a more sophisticated version of this called the Dogleg method, but it's the same principle. It's trying to make sure you're not taking steps that are too big. Trying to make sure you're taking steps in the right direction so that you're approaching the roots of your function. Or at least reducing the absolute value of your function as close to zero as you can get it. Are there any questions about this? Anything we need to follow up on in the previous lecture? No. OK, good initial guesses. This is the key, really, for solving systems of nonlinear equations. It's the key for doing optimization problems too. We need to have some idea of what our solution looks like, and we want guesses that are close to those solutions so that maybe we are in this region of local convergence for the Newton-Raphson method, and everything converges very quickly. So where do they come from? This is an important question, and I'll show you how you can get them. There's another issue too, which is related to this, which is the non-linear equations, they can have multiple roots. If we do optimization problems-- that's our next topic-- they can have multiple minima or maxima that we're trying to seek out. Depending on our initial guess, our numerical method may find one of these roots or another root. It may find one of these minima or another minima. Maybe we're looking for the global minimum, so we need to look at them all. Are there methods that we can use to find them all? So we'll talk about continuation in homotopy, and we'll talk about bifurcation. And it turns out there are methods one can use to find all the roots for a system of non-linear equations. In fact, the homework assignment you have this week will ask you to do ternary phase equilibrium problem. Vapor liquid phase equilibrium with three components. There are five parts to this problem. Four of them, I think, are relatively straightforward. They're about finding the pure component roots or randomly guessing some initial conditions for your nonlinear solver in order to find all of the different azeotropes in the system. There's a fifth part that asks you to actually implement this homotopy method, and try to find all the roots in one go with your algorithm. Like the other homeworks you've had, that part's a little more tricky. Usually these homeworks have one thing that's a little bit more tricky than everything else. That part is a little bit more tricky, but it's really instructive to try to do it. We'll learn how to do it today. You can find all of the coexisting compositions of this ternary phase diagram, simultaneously, with one go through this algorithm. So let's talk about continuation first. This is an interesting idea, and it relates to trying to find roots of these equations, having good initial guesses. Here's a simple example. We have a third degree polynomial, and it has three roots. We can look, graphically, because this is one dimension, and see that one of these roots is near minus 1 and 1/2, another root is near a 1/2, and another root is near 1. And so I can go to my non-linear equation solver, and give it initial guesses, minus 1/2, 1/2, and 1, and I'm probably going to find these roots pretty reliably. But in many dimensions, we can't do these sorts of graphical approaches. You can't look at this function. It may be difficult to actually get enough data to even look at it if it's a two-dimensional set of functions. It's hard to see where these roots are, and so continuation is a method of transforming the problem into an easier one to solve. So let's take this problem, this third order polynomial, and let's transform it. Let's replace the 2 with a 2 times the lambda, and let's try to find the roots of this polynomials as lambda grows from 0 to 1. This is sort of a weird idea, but we actually know when lambda is equal to 0, this goes away, and there's only one root to this polynomial-- one real root-- which is minus 1. We actually know the answer when lambda is equal to 0. If I make lambda a little bit bigger than 0, there will be a root which is close to minus 1. In fact, minus 1 is a good initial guess for that root, and my non-linear equation solver ought to converge very quickly to that value. If I choose lambda to be small, then it should be in that region of local convergence of my Newton-Raphson method, and I very quickly should find the next root. So I can step lambda along. I use the lambda equals 0 root as an initial guess for the next root, and I keep increasing lambda from 0 all the way up to 1, and when lambda is equal to 1, I'm going to find the root to the initial problem I was looking at. Does that makes sense? So I'm continuing from one set of solutions to the next. I vary this parameter continuously, and for previous values of that parameter, they serve as initial guesses for the next solution. Here's what a code that does that would look like. I define my lambdas. We don't know good ways to define the lambda. I've got to figure that out for myself, but here lambda is a vector that goes from 0 to 1, in 1/100 increments. Maybe that's too fine, but that's OK. And my initial guess for the solution when lambda equals 0 will be minus 1, and I loop over all the values of lambda, and I use f0, which is the one-dimensional non-linear equation solver in Matlab, to find the roots of this polynomial function-- x cubed minus 2 lambda x plus 1. Here's the solution and that becomes my initial guess for my next solution. So this loop is going to go round and round until I get lambda equals 1, and the solution that results there should be the solution to the initial problem I was interested in. Here's what that looks like graphically. I started with lambda equals 0. The root there serves as an initial guess for lambda equals 1/100, and so on, and these roots converge eventually, to something that's close to minus 1 and 1/2. So the actual value is a little bit bigger or a little bit smaller than minus 1.6. Here's one of the roots. So rather than having to have a precise initial guess, I transformed my problem from an easy to solve one to a more difficult to solve one, and I used the path along that transformation to give me rapid convergence to better and better initial guesses to the problem that I was interested in. Make sense? There's another way of doing this too. You don't have to go from 0 to 1. Let's go from lambda equals something big back down to 1 instead. So if lambda's big, this term dominates the polynomial equation. This is the biggest term in the polynomial equation, and it can either balance x cubed, or it can balance 1. It can't balance against both of them, it turns out. That So if we have this term large, and it balances against 1, then when lambda's large, an approximation for the root will be 1 over 2 lambda. If x equals 1 over 2 lambda, and these are the two largest terms in the equation, then these two will cancel, and f will be very close to 0. So that's one root when lambda is very large. If I balance these other two terms against each other, I'll see there are two other roots, which are plus or minus the square root of 2 lambda. So let's start with a large lambda instead, and these as initial guesses for the roots-- and let's trace them back to lambda equals 1. When lambda equals 1, we recover the problem we were interested in before. So here, I start with three different initial guesses, and I trace them back, and I'm actually able to find all three roots of the initial equation. So there's something close to minus 1 and 1/2, something close to 1/2, and something close to 1. So in one go, I found all three roots using continuation. Is this clear so far? Sam. STUDENT: Is that one go or is it three separate [INAUDIBLE].. PROFESSOR: That's an excellent question. So I had to initiate this process with three separate guesses, but it turned out in this case, all three guesses converged to three distinct roots in the end. it's one go in the sense that there's one trajectory that I had to follow, one passive lambda as I had to go through, but I had to solve three times to follow each of these three paths. There's actually no reason to stop at lambda equals 1. I stopped there because that's the solution to the problem. I'm interested in, but there's actually something richer to see here. If I keep decreasing lambda, eventually these two branches combine with each other, and then they jump down, and they follow the original branch that I had that went between minus 1 at lambda equals 0, and this root about minus 1.6 at lambda equals 1. So the paths that these trajectories follow can be very, very complicated, actually. A little bit unwieldy or unpredictable. But oftentimes, they can result in predictions of all three roots or multiple roots associated with the system of equations. Yeah. STUDENT: Going back to the first [INAUDIBLE] solve from lambda equals [INAUDIBLE] PROFESSOR: Zero to one? STUDENT: Zero to one. So then would you find only one solution there? PROFESSOR: Yes. So going from 0 to 1, I found one solution. Going from infinity-- 10 is big enough. 10 to 1, I was able to find three. I didn't know beforehand that I'm going to be able to do that. It happened to be the case that we could with this by construction. This is a particular problem where it works out that way. Other questions? So this is neat. Like we've done before with iterative methods, we turn hard to solve problems into a sequence of easier to solve problems. Good initial guesses converge fast with Newton-Raphson. We have a sequence of better and better initial guesses, which all converge very quickly to eventually, the solution that we're looking for evaluated at lambda equals 1. So this is one way of getting good initial guesses. You have a sequence of good initial guesses. Make sense? So we're just here. We had this polynomial problem and we injected lambda in a particular place. We injected it there because I knew it was going to work. I knew the problem could be transformed in this way in order to work out and distinguish this. It's not always obvious how to transform a problem to make it look this way, but oftentimes we have physical problems in which there's a parameter that we can do continuation with that is obvious. So think about fluid mechanics problems. Those are nonlinear partial differential equations. So we're trying to solve say, the Navier-Stokes equations. Rather than jump in and try to solve that right away, we might do continuation. We might find a solution at very small Reynolds numbers, where the equations are almost linear, exact solutions are known in that case, and we might progressively step the Reynolds number up, and use our solution set small Reynolds numbers as initial guesses for increasingly large Reynolds numbers until we get to the one that we're interested in. You can imagine doing that in mass transfer problems where you vary the Peclet number. In multicomponent phase equilibria problems, this might happen by varying temperature or pressure until we get to the temperature or pressure that we're interested in. If we have reaction equilibria problems, we might vary the different reaction rates until we get to the right balance of reaction rates to determine the equilibrium compositions. So they're often natural parameters in the problem that we're free to adjust. We can transform from easy to solve problems to hard to solve problems in a continuous fashion. It's not always going to work out the way that I showed. It's not always going to be so nice and neat. It may be quite a bit more complicated, and sometimes there are problems for which there isn't a natural parameter available to vary. In those cases, there's another strategy one can employ. It's called homotopy. This is the transformation from one problem into another problem in a continuous fashion. So oftentimes, we're seeking a set of roots. x star, which are a function of a parameter lambda, and they're the roots of an equation that looks like this. So they're the roots of this function, h, which is a linear superposition of a function f and a function g. When lambda equals 0, h is equal to g, and so the roots of h are the roots of g. And when lambda equals 1, h is equal to f, and so the roots of h are the roots of f. So we might transform-- we make very lambda continuously between 0 and 1 in a construction like this. So we might transform the roots from the roots of g into the roots of f. Maybe the roots of g are easy to find, but the roots of f are hard for us to find. There's a smooth transformation though, from g to f. When we have those sorts of smooth transformations, there's a chance things will be well-behaved, or at least where we encounter bad behavior-- and I'll show you what I mean by bad behavior in a minute-- there's a reliable way to interpret what's going on there. So we vary lambda in small increments from 0 to 1, and the solution x star for some lambda i is used as the initial guess to find the roots at some lambda, which is a little bit bigger than lambda i-- our next iterate in lambda. Does this idea make sense? Where do f and g come from? So this is a big question-- f is the problem you want to solve, typically, and g is some auxiliary problem. It can be arbitrarily chosen. There's nothing to tell you why one g should be better than another one, but oftentimes, we choose them in a physical way. So there will be a physical connection between the f problem and the g problem, and I'll show you that right now. So here's an example. We looked at this example before. We want to find the roots of the van der Waals equation of state. It's written in terms of the reduced pressure, the reduced molar volume, and the reduced temperature, and let's find those roots for a given pressure and temperature. So this is a function f of the molar volume equals 0, and we're looking for these three roots here. There's a root down here, which is the molar volume is low. Not a lot of volume per number of moles, so maybe this is a liquid like root. Out here, the molar volume is high. We have a lot of volume, the number of moles and materials. This is a vapor like root, and then there's an unstable root in between them. So three roots to find. We'd like to be able to find them in one go, or we'd like to have some reliable mechanisms for finding them. Homotopy is one way to do this. So let's construct a new function h of the molar volume, which is lambda times f. We want to find the roots of f plus 1 minus lambda times g. And as g, let's use something physical. So let's let g be a function that represents the ideal gas equation of state. So PV is nRT, but I've written everything in terms of the reduced pressure, molar volume, and temperature of the van der Waals equation of states. So you pick up this factor of 8/3 there. But this is the ideal gas equation of state. Its roots will be the roots of the ideal gas equation of states. I'm going to transform from something that's easy to solve. I know how to find the molar volume here, there's just one, to something that's hard for us to solve. It has three roots-- the roots of them van der Waals equation of state. So when lambda equals 0, I'll get the roots associated with the ideal gas, and then lambda equals 1, I'll get the roots associated with the van der Waals equation. Yes? STUDENT: If you're starting with lambda as 0 and you get the roots of g, you only get one root, but you're trying to find two roots [INAUDIBLE].. So how does that work? PROFESSOR: I'll show you. Here's what the process might look like. I've got to start with a reduced temperature and pressure. Here's my initial guess for the molar volume. It's the root of our ideal gas equation. Here's the definition of f, and the definition of g, and the definition of h. l times f plus 1 minus l times g, and I'm going to loop over all of my lambdas, solve the equation subject to some initial guess, and update my initial guess every time I update lambda. That's fine. That's the code you. Can implement the code if you want or play around with it. So what happens if I try to carry this out? So I'm going to vary lambda between 0 and 1. I start with my initial ideal gas guess for the root, and as lambda grows to 1, the molar volume shrinks until I get to lambda equals 1, and I've found one root of my equation. One root, good. There's not necessarily any reason to stop at lambda equals 1. That's the root I wanted, but there's no reason I have to stop there. So I could continue. If I keep pushing lambda up higher and higher, eventually I'll get to a point where all of a sudden the molar volume jumps down to a different place. This is the result of the algorithm. I just told lambda to go bigger. But there will be a discontinuity or a cusp here that I jump off of, and I find myself on a different solution branch, which is sort of crazy. So now I can do something clever. I found myself on a different solution branch. Why not try decreasing lambda along that solution branch instead? Maybe I'll find a different solution, and it turns out that's what happens. All right, so i jump down to this other solution branch, and now I try decreasing lambda from 2 back down to 1, and sure enough, I found the vapor molar volume, and now I found the liquid molar volume. There's no reason to stop and lambda equals 1. So if I keep decreasing lambda, eventually I'll hit another cusp, and I'll jump back up to the original solution branch. So I do my homotopy. I vary lambda from 0 to 1, and I find one root, but if I keep going, I might identify a cusp, and jump off that cross. I might reverse my direction with the homotopy, and I could find another root. I might hit another cusp as well. Now, when I hit a cusp, maybe it's best not to jump off the cusp, but instead to reverse direction at the cusp, and try to trace it out the other direction. And if I do that, if I get right to this last point, and I try to reverse direction again, I can trace out one more solution path, and I can find the third root to the equation. So in one go, if I'm paying attention to how my roots are changing-- if they change by too much, I might switch the direction with which I change lambda, and I could find all of these roots at once. So three roots in one go. Yes. STUDENT: Is this like trial and error? How do you know when to change? PROFESSOR: This is a wonderful question. So in this case, you can do it by trial and error. You can detect these jumps. That's even better. You look for jumps in the value of the solution that are of sufficient magnitude, and when you detect them, you can reverse the direction. There's an even more systematic way to do this, which is called arclength continuation. Give me one second, and I'll give you slide on arclength continuation, which is a systematic way of doing exactly this process of tracing out this path in the solution plane. Roots versus homotopy parameter. These cusps here, are sometimes called turning points, and they have a special property, which is the determinant of the Jacobian of the homotopy-- the homotopy function-- is going to be equal to 0. The Jacobian is going to be singular right at these cusps. So you can detect them by tracking what the Jacobian is doing right. The Jacobian is going to be non-singular at most of these other points, but when you hit one of these turning points, the Jacobian will be singular. There isn't a well-defined direction for the homotopy to proceed along. It hits this cusp where the slope is infinite in this plane. There isn't a well-defined direction to step along here. The question was-- OK, it seems a little ad-hoc. You've got this 2D plane, and it's easy to guess when you jump from one solution to another, and trace the path back. That's fine. There's a systematic way of doing this, which is called arclength continuation, which says I know that there's some curve in this solution plane. There is some curve here. I've shown it to you graphically. Let's try to parametrize that curve in terms of a measure of the arclength. Call it s, the distance along the curve. So the roots are a function of lambda, which can be a function of the distance along the curve, and lambda is a function of the distance along the curve, and there is a constraint which says s has to satisfy an arclength formula. This is the formula for the arclength in this plane of roots versus lambda. The derivative of this function with respect to s squared plus the derivative of lambda with respect to s squared equals 1. That defines s as a measure of length along this curve. You learned arclength formulas in the context of integration at one point. This is the differential version of that same formula. So I've got a differential equation. Well, it's actually a differential algebraic equation. This is an algebraic constraint that needs to be satisfied for all values of s, and it depends on how the roots change with s. We'll learn how to solve differential algebraic equations in the two sections from now. So we don't quite know how to solve something like this yet, but if we know how to solve those sorts of equations, then it's clear that we can trace out this entire curve in the solution plane, and then go back and try to find the places where the curve intersects lambda equals 1. So if we follow this path by solving this equation from s equals 0 out to some very large s, then we should be able to find all of the roots that are connected to the path. Yes. STUDENT: If there's a turning point that [INAUDIBLE] then does that imply multiplicity? PROFESSOR: Yes. So it's a real problem. That turning point singularity is actually built into this arclength formula as well. This is the derivative of x with respect to s squared, which takes the direction you're moving in out of the problem. This is a length, not a direction. So when we get to these turning points, the solution methods for these equations are going to have to know that I was headed in a downward direction first. I can get to this turning point, and then the question is which direction do I move in from the turning point? Do I go up or do I go down? Which way does the solution path change? The slope isn't defined there, so it's quite challenging to figure out exactly how to proceed. But it is what you say. It's like the solution has multiplicity at that point. Curves are tangent in the solution plane. It's a great question. OK, so that's the notion of arclength continuation. We may talk about this later when we talk about differential algebraic equations, and we'll talk about some equivalent sets of equations that we know how to solve, but they have to satisfy this constraint that s is a measure of arclength along the curve. But there you go. You can find all the solutions, at least all of the solutions connected to this path defined by the homotopy equation. You can find all of them in one go using some clever methodologies. Oftentimes, we're not just worried about these turning points or cusps, we encounter bifurcations in the solution. So we may have a path that has one root along it, and then that path may split, and there may be more than one root. Here's a simple example. Find the real roots of x cubed minus rx. If r is negative, there will always be one real root, and it'll be x equals 0. If r is equal to 0, there will still be one real root, it's 0. That root will have multiplicity 3 in that case. And if r becomes positive, I'll suddenly have three real roots instead of one. So as I vary the parameter r from a negative number to a positive number, I'll go from one real root to three. In the solution plane, that will look like this. Here are the roots as a function of r. r is less than 0. All the roots are equal to 0. When I hit r equals 0, I have multiplicity associated with the root, and the solution bifurcates, and there will be three possible solutions now. If I was trying to do a homotopy problem where I very r continuously from some negative number to some positive number, I'll hit this point, and I'll want to follow all three paths to find solutions out here at positive r. But I can detect those points. Just like I was said before, at these weird turning points or bifurcations where we have multiplicity in the roots, the Jacobian in is going to be singular. The Jacobian of the homotopy is going to be singular. Its determinant will be equal to 0. So I can detect these points by checking the determinant of the Jacobian. And when I find one, I want to do something smart to try to follow these paths. Your homework this week has a problem like that. You're trying to find all the azeotropes of this ternary vapor liquid equilibrium system, and you'll have a homotopy parameter that you change, and it'll bifurcate. OK, so as you change the parameter, you'll go from one path to two, and then those two paths will split again. You'll be splitting from a single component solution into a two component azeotrope, into a three component azeotrope. And you'll want to detect where these bifurcations occur, and try to follow solution branches off of those bifurcations, and there's some good hints in the problem for how to do that. So occasionally, we'll encounter problems that switch from having one solution to having many solutions. You've seen how this happens in the discontinuous sense with these turning points. Here, for a given value of lambda, all of a sudden it's clear that not only is there one solution here, but there's another solution down there. My path has to choose. Do I want this solution or the one below? I jump off this turning point. That's a sort of bifurcation. These are continuous bifurcations. They're often easier to take care of. So the bifurcations in a homotopy let us find multiple solutions. Every time we detect a bifurcation, if we just track those solution branches, we split our algorithm up to track each of the branches separately, they'll terminate at different roots to the homotopy equation. And when we get to the value of lambda equals 1 in our homotopy, we're there. We found multiple roots to the original equation. These are often of great physical interest. Do you know what sort of things these bifurcations represent? One occurs in the van der Waals equation. So if I start at very high temperatures and a given pressure, the van der Waals equation will have how many real roots? One-- a gas. And then as I turn the temperature down, there is a propensity for liquefaction of that gas phase. The formation of a liquid in coexistence with the vapor. So I'll go from having one root to having three roots all of a sudden, and there will be this bifurcation point where that transition happens in a continuous fashion. STUDENT: Can I ask you a question? PROFESSOR: Yes. STUDENT: [INAUDIBLE] PROFESSOR: Yeah. STUDENT: So as you early put it out that the turning point, the determinant of the Jacobian [INAUDIBLE] including lambda is singular. But here you're writing j of x. Is that [INAUDIBLE] PROFESSOR: Excuse me. This is a typo. So there should be a little h here. This is the determinant of the Jacobian of the homotopy equation at that root. Where there's a bifurcation, the Jacobian will be singular. I apologize. That's a little bit-- STUDENT: And that lambda. PROFESSOR: And that lambda. STUDENT: Just to make it clear, you have n state variables. Like x is how many unknowns you have, and then you're adding another unknown to lambda, another parameter. This Jacobian has n plus 1 dimension. This is also [INAUDIBLE] Is this correct? PROFESSOR: Well, no. Actually, the Jacobian with respect to x only is also singular at these turning and bifurcation points. STUDENT: [INAUDIBLE] as well. PROFESSOR: Yes. It's also true what you said, that if I have the Jacobian, and I take the derivatives with respect to lambda, that will also be singular, but it will be singular because the subspace, which is the derivatives with respect to x, gives you a singular component. The condition is the same as here. So it's the Jacobian of h, taking the derivatives with respect to x-- all the different x's of h. The determinant of that matrix at a particular lambda will be equal to 0, and the value of lambda will be the lambda at which you have this turning point, or which you have a bifurcation. Here's what it looks like in a two-dimensional sense. So I'll have my homotopy equation where I'm trying to find the roots of h. Say it's a two-dimensional-- x is dimensions. So it has x1 and x2, and these curves represent h1 equals 0 and h2 equals 0, and they intersect at one point for a given value of lambda. There's the root. Now I increase the value of lambda. The homotopy equations change. So these curves change shape, and all of a sudden I go from crossing curves to tangent curves, and we know tangent curves in the plane will have a Jacobian which is singular. And this is the bifurcation point. These two curves become tangent, and as I go to a bigger value of lambda, they'll change shape again, and I'll now have multiple roots. I'll bifurcate from one solution to two, or one solution to three, but the bifurcation will occur through this transition where I have tangent curves, and we know there, the determinant of the Jacobian is 0. There's a singularity under those conditions. Does that makes sense? OK, so a couple of things about the practice, and I'll give you a simple example you can try out, which I think is an interesting one, if pathological. So in practice, it's hard to hit this bifurcation point exactly. And I'm telling you that you should be detecting these branches. You should follow these branches, but clearly, I've got to know where this point is in order to follow the branches. That's a problem. If the determinant of the Jacobian is 0 at the bifurcation point, then it's going to change from positive to negative as I cross the bifurcation point with lambda. Actually, I don't necessarily have to find the point where the determinant of the Jacobian is equal to 0. What I really should do is try to track the sign of the determinant of the Jacobian, and see when it goes from positive signed to negative signed. And when it does that, it will have crossed through one of these bifurcation points or one of these cusps. That's the point at which I should stop, and say, OK, there's possibly other solution branches to follow. So as I step from my smaller lambda to my bigger lambda, the sign changes. I want to step in some different directions than the direction I was heading before, and those different directions will trace out the other solution branches. It turns out those different directions belong to the null space of the Jacobian. So if you find vectors that are very close to the null space of the Jacobian, those directions are the directions you should go to find these other solution branches. So it's very interesting. Yeah. STUDENT: Can you have a Jacobian that goes from positive down to 0 and [INAUDIBLE]? PROFESSOR: That may be possible. You won't encounter that in the problem that you're doing. That would be a very difficult problem to do branch detection on. Let's Suppose that was the case and you want to find exactly where that bifurcation point is. These bifurcation points are of physical significance. In the van der Waals equation of state, that bifurcation point is the critical point. It's the critical temperature at which that phase separates. Maybe I want to know that exactly. Well, all I need to do is solve an augmented system of equations. So let's solve a system of equations, which is our original hematology equation equal to 0, and determinant of the Jacobian of this homotopy equation equal to 0. This is if h had dimension n. This is now n plus 1 equations, and let's solve it for n plus 1 unknowns. Let's solve it for x and for lambda. x has dimension n, and we add one more unknown-- the value of the homotopy parameter at which this bifurcation occurred. So we can find that point exactly. So there may be cases where this happens where it doesn't continuously transition from positive to negative. It may hit 0 and go back up, but that's a really challenging case to handle, but you can handle it. You can find precisely the point at which this bifurcation occurs by solving these augmented equations. It's just another non-linear equation to solve. Yes. STUDENT: I'm not sure about the previous slide. So in the previous slide, why is the [INAUDIBLE] PROFESSOR: Let me go back to here. So let's let the blue line here, be the place where the first element of h is equal to 0. So this is all the values of x at which the first element of our homotopy top equation is equal to 0. And the red curve here, let's let that be all the values of x at which the second element of h is equal to 0. And so h itself, is only equal in 0 where both of these curves are equal to 0, and that's their intersection. So they intersect here, and this star is the solution. It is the root of this non-linear equation up here. Now, I increase lambda. I make lambda a little bit bigger. That changes h. h is a function of lambda. So h changes, and these curves change shape in the plane. And at that bifurcation point, these two curves will become tangent to each other. There will be one root, but in this solution plane, the two curves will be tangent. And we saw before that when we get these sort of tangency conditions or the equivalent in higher dimensions, the Jacobian of our function we're trying to find the roots of will be singular. It's like a multiple root. And when we increase lambda further, curves change shape again. I'm not changing the red one because that's complicated. I just changed the blue one for you as a graphical illustration. So the blue one changes shape, and now I've gone from having one solution to having three. So in order to go from one solution to three, in a continuous fashion, I have to go through this tangency state, and that defines the bifurcation point. Does that make sense? Here's a simple example you should try out. So I'll present it to you, and then you should try to solve it in Matlab. You actually know the solution, geometrically, so it's not too hard. So here's a function. The first element to that function, it's a function of two unknowns-- x1 and x2. The first element to that function is an equation for a circle of radius r with center at minus 3 minus 1, and the second element is an equation for a circle of radius r with center at 2, 2. So there's two circles. We want to find the radius where the circles just touch, and that point is a bifurcation point. If r is perfect, these two just touch. If r is a little too big, the two circles overlap, and there are two solutions to this problem. And if r is a little too small, there are no solutions to this problem. So the perfect r for touching is a bifurcation point. It has all the properties of a bifurcation point, just like we discussed. It's a multidimensional equation and it bifurcates. You know the solution to this problem, geometrically. You could work it out in a couple of minutes, but you could also solve it using these augmented equations. So we know that point at which the circles just touch, the value of r at which those circles just touch, will be the place where f of x is equal to 0 at which these equations define a circle, and at which the Jacobian of these equations-- the derivatives of f with respect to x, it's determinant is also equal to 0. So we want to solve this system of three equations for three unknowns. The values of x1 and x2, where the circles just touch, and the value of r, the radius of these circles at which they just touch. That would find the bifurcation point. That will find the touching of the circles. Can you see that, geometrically? All these bifurcation problems work exactly that way. There's some homotopy parameter. In this case, that's r. But there's some parameter we're varying as we pass through that bifurcation. We'd like to know the values of our unknowns and the value of this parameter at that bifurcation point or at that critical point. So you find this point by solving the augmented equations. We have a non-linear equation-- three nonlinear equations for three unknowns, and you can do that with a Newton-Raphson iteration. It's a non-linear equation. We know Newton-Raphson is the way to solve these things. To do Newton-Raphson, that means we need to compute the Jacobian-- this is confusing. We've got to compute the Jacobian of these augmented equations, the derivatives of these augmented equations with respect to our augmented unknowns. So that will be the derivatives of f with respect to x, the derivatives of f with respect to r, the derivatives of the determinant of the Jacobian with respect to x. That's this gradient here. The derivatives of the determinant of the Jacobian with respect to r. This is our augmented Jacobian. The augmented Jacobian times the Newton-Raphson step. That's equal to minus the function we're trying to solve evaluated at the previous iteration. Solve the system of equations, you'll get your Newton-Raphson step, and you can find the solution to the augmented equations. It'll converge quadratically. You should commit yourself, you can figure this out. Don't try to compute all of these derivatives. That's complicated. You're going to make a mistake. I'm going to make a mistake if I try to do that. What do we do instead to compute the Jacobian? How do we compute these derivatives? What's that? Finite difference, right. Don't play games with it. Try to use finite difference to compute the elements of this instead. You know the exact solution too, so you can use that as an initial guess, or something close to the exact solution as an initial guess to see if you can find this bifurcation point. So homotopy bifurcation, there are ways to get good initial guesses by continuously changing our problem. We get a succession of problems that will converge faster because we have good initial guesses, and that will give us a reliable convergence, hopefully, to one of the roots of the equation we're after. If we're able to pick up solution branches, we can track these cusps, or find these bifurcation points, we might be able to find multiple solutions. There are problems for which depending on your choice of g in this homotopy function, maybe some of the solution branches are disconnected from each other. Boy, that's crazy and complicated. We're not going to worry about those sorts of cases, but they can pop up and happen. But the point is there are a robust strategies you can employ to find multiple roots of these equations or multiple minima in optimization problems using precisely this technique. You'll have a chance to practice this on this week's homework assignment, and I think you should take a look at this problem. It's simple. It looks complicated , but if you can understand this from, then you'll understand everything you need to know about bifurcation and homotopy. Any questions? Yes. STUDENT: At a bifurcated point, we go from having one root to three roots, per se. And you said that when that occurs, we need to follow each of these solutions, right? PROFESSOR: Yes. STUDENT: And you said that we need to step in a different direction. So what does that exactly mean? In a 2D sense, we have this stepping forward or stepping backwards. PROFESSOR: This is a wonderful question. So let me go backwards here. Here's the bifurcation point, and when we go through that bifurcation point, there are now going to be three solutions. And I'm going to need some initial conditions for my solver to find each of these three solutions. I can't give them all the same initial condition. I'm going to try to track all three of these branches with different initial conditions. If I give them all the same initial condition, and my algorithm meets the minimum threshold for what an algorithm should be, they're all going to converge to the same root. So that's useless. So I need to give them different initial conditions. So what are those supposed to be? In your problem, I'll give you explicit directions for how to find those. It turns out those initial guesses should be perturbed in directions that belong to the null space of the Jacobian. So I want to perturb from that one solution, which was here, in a direction that's parallel to the tangency of those curves. Those two curves were tangent with each other. In the other solutions, they're going to sit off in the direction that's tangent to those curves. So I've got to find the vector that's in the null space, or very, very close to the null space, and that will tell me how to get better initial guesses for picking up these other solution branches. It's not trivial. It's actually quite challenging. You'll have a chance to practice. It's part five of five parts on this problem. The first four, they're relatively straightforward and easy. Part five is a little bit tricky. You get explicit directions, and the TAs know that that's the most difficult part of that problem. So we'll help you through it. It's something interesting to think about trying to do. You may find you're only able to pick up some of these solution branches, but not all of them. That's OK. There are ways to pick them all up if your algorithm is designed well enough. You may just pick up some of them, and that may be the result you turn in. You can look at the solution to figure out how the other ones are picked up as well. But you're looking at the direction along these tangent curves. That's a vector in the null space of the Jacobian. And if you perturb in those directions, then you can trace out these other roots so you can find the bifurcation. It's a wonderful question. You'll get a chance to practice on it. OK, last question. STUDENT: [INAUDIBLE] PROFESSOR: Yes. STUDENT: Is this supposed to be jrx? PROFESSOR: That's the Jacobian of f. That's the derivatives of f with respect to x. STUDENT: Not including r? PROFESSOR: Yes. Not including derivatives with respect to r. Just derivatives with respect to x. So this is the Jacobian of f. The derivatives of f with respect to x. The determinant of that will be 0 at the bifurcation point. It'll turn out that-- this will be the last thing and then I have to get off the stage here. It'll turn out that this matrix is also singular at the bifurcation point. It will be singular because the subspace j is singular. So it'll turn out to be singular as well. Don't worry about it. The condition for the bifurcation point is that the determine of the Jacobian is equal to 0, and that point is a root of the equations you're looking at. OK, thank you. |
MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 16_ODEIVP_and_Numerical_Integration_4.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: All right, I think we're gathered. So let's get going. We were going to DAEs today, but we decided to postpone for one day and do a little catch up on different things that you should know. And I thought I'd first back up a minute and recall what you've learned so far. So we started out talking a lot about linear algebra and, particularly, the problems of this type where you're given a matrix and you're given a vector, and you want to figure out what the x is that makes this equation true. And you've become extremely familiar with the convenient Matlab backslash function. You've probably used this a lot of times. And if you remember, this is one of the only problems where we often have a unique solution. We know a solution exists. And we can do a finite number of steps in order to get the solution. And actually, the method we have is really good. So it almost always gives us the solution and to machine accuracy. So this is a brilliant method. And so everything is based on this in our whole field, actually. So we learned this first. We learned it takes n cubed operations in order to get the solution, where n is the size of the vectors. And when we do the solution, the x we get for backslash, invariably, it either is a solution, solves this equation to machine accuracy, in the sense that you multiply m times x, you get b to as many digits as you can read. However, we talked about how m could be poorly conditioned so that there might be multiple x's that would solve this equation to machine accuracy. And so that's the problem with ill conditioning. So it's not that it doesn't solve the equation. It's that we're not really sure that's the right x. You could get a bunch of different x's that, basically, solve it as well as we can tell. All right, so do you guys remember this? All right, and so that also implies that if you make a very small change in b, you might get a gigantic change in x. Because even for one b, you have a quite a range of x's that could work, all right. Then we try to solve problems like this. Right, systems of equations, not just linear ones, but nonlinear ones as well. And what we decided to do as one of our best methods is Newton-Raphson, where we keep on doing j backslash f [INAUDIBLE] negative f. Right, so all we did in this method, a lot of times, is we just kept on calling this method over and over again. All right, so this took order of n cubes. This thing is going to take order of n, I don't know what you call it, steps or something. Maybe if you guys have a word for it. What do you call the iterations? Iterations. So it takes the number of iterations times n cubed. Now, if you really can make Newton-Raphson work, and you have a really great initial guess, the number of iterations is really tiny. Because Newton-Raphson is such a fantastically great convergence. However, we're usually not such good guessers. And so it might take a lot of iterations. And in fact, we use the dogleg method to solve this as we've discussed about where you mix up different methods. And you have a trust region and all this baloney. And so it might actually be quite a lot of iterations to get there. And then we went and tried to solve this problem. And what we decided to do for both of these was almost do the same thing. So I said, well, we can really solve this one by instead solving this. Right, that's really what we ended up doing. That's what happens inside fsolve is it converts the problem of finding f of x equals 0 into a minimization problem to try to make the norm of f as small as possible. Because the answer will be when the norm is zero. And if f of x equals 0 has a solution, then the global minimum here is a solution, as long as your initial guess is good enough that you're in the valley that leads to the true minimum. Then this will work. And that's actually what fsolve does. So these two guys get kind of coupled together, the minimization problem and the root-finding problem, or the problem of solving systems of non-linear equations. And in both of them, we're going to call the first method a million times. OK, so that's the way we're working so far. And actually, when you do fmincon, which was the next level, then you actually end up looking for saddle points, which, again, we look for as minimizing the norm of a gradient squared, which gets it back into this kind of form. And so they're all the same thing. And you can, again, work out estimates of how much effort it takes. And it's something times n cubed, Typically, if most your work is this. Now, I should make a comment that, in practice, that's not always true. So in practice, sometimes, although the highest order term is n cubed, the linear algebra programs that do this are so great that the prefactor is tiny. So in order to get the limit where the n cubed term was actually dominate the CPU time, you had to have a gigantic n. And in fact, the limit, it's often the effort to construct the matrix j. And so it really goes as n squared. So although this nominally is like n cubed, in practice, a lot of times, it's n squared. And I say it for these other methods. And then a lot of them get into how good the initial guesses are. Because the number of iterations can get crazy. And in fact, maybe, it never converges if your initial guess is bad. So that can be a really critical thing. [INAUDIBLE] is better. Because you can't change n I mean, how many unknowns you've got is how many unknowns you've got. But you can change the number iterations by, say, more clever initial guessing. And that was the whole concept about continuation and homotopy and all this kind of stuff was trying to help reduce this number. Last time, we talked about implicitly solving ODEs. And so we had an update formula. So we were solving an ODE. The initial value problem, y of-- Right, and we said that, first, we went through the explicit methods. And they were perfectly straightforward to do. And you don't have to solve any systems of nonlinear equations to do them. And so that was great. But then we pointed out that, a lot of times, they're numerically unstable. And so then you have to instead use implicit methods. But the implicit methods generally work this way. They have your new value of y. What you're trying to compute is your current value of y plus delta t sums g. It depends on the size of delta t and y new. Maybe y old also. And these g's are things like the Crank-Nicolson, which you're going to work on in the homework, and we show it in class too. And so there's different update formulas of people suggesting how to do this. The trick of this is that the y new is inside the function. So therefore, this is really another problem like we had before. You can really rewrite this that way by just putting this on the one side. Put them all on one side of zero. And now, I'm trying to figure out what y new is. And I'm going to solve that using this method. And I'm going to solve that by using this method. All right, so now, how is the scaling going to go? At each time step, I have to solve one of these guys. How much effort did that take? We did that, right. Here it is, n [? after ?] times n squared. And then how many times do I have to do it? AUDIENCE: It's h times 7 times [INAUDIBLE].. PROFESSOR: Right it depends on how many times steps I have [INAUDIBLE] my delta t. So it's something like t final minus t 0 over delta t. That's the number of times steps. But I have constant delta t. Times the number of iterations that it takes to do the nonlinear solve times n squared, which is a cost of forming the Jacobian matrix. Now, this can be pretty large. So as we talked about, this might be as big 10 to the eighth in a bad case. Certainly be less than 10 to the eighth because otherwise, you'll have numerical problems. Maybe, it's really going to be 10 to the sixth for a real problem. So you have a million time steps. At each times step, you have to solve the number of iter-- the solve costs this much. Number of iterations might be what? How many iterations do you guys take to solve nonlinear equations? You solve a lot of them now. How many iterations does it take? AUDIENCE: 10. PROFESSOR: 10, does that 10 sound right? AUDIENCE: Yeah, [INAUDIBLE]. PROFESSOR: You've got an initial guess 10. If you have a medium initial guess, maybe 100. It it's more than 100, you're not going to converge, right? AUDIENCE: Yeah. PROFESSOR: Yeah, OK. So this is order of 10 maybe. So the million 10 and then n squared just depends on how big your problem is. So in my group, the problems we typically do, n is approximately 200. So this is like 200 squared. And then you can see that the number of iterations, and this is maybe a little bit overestimated. Maybe I can get away with 10 to the fifth, something like that. But you can see it starts to get pretty big. But for writing it, as you guys are writing software, it's not that big a deal. So I can assign it as part 1 of a three-part homework problem. It's the right ODE solver. It solves implicit equations. And you can write it. And the reason you can is because you've already written methods that solve this. Actually, Matlab did it for you. But you write it yourself. I assigned to the [? 10:10 ?] students. So you write the [? backsub ?] solution to solve the equations. And we we've already wrestled this. And I think you guys wrote one of these yourself. And also, we can use fsolve, which is just another little bit better implementation of the same thing. And then now, you write your next level. So you can see you're doing a composite process, where you're doing something at the top level. It's calling something at a lower level. That's calling something at a lower level. And then what you really have to be concerned about if you're worried about CPU time or whether the thing is going to solve before the homework is due, is how big does this get. How many operations are you really asking the computer to do? Now, remember the computer is fast. Right, so it can do a gigahertz or something. So you can do like 10 to the eighth somethings per second. And what should I say about this? I have not included, but I should that it's-- well, I'm assuming here that the function evaluation is sort of order n of how many variables they have. OK, sometimes, function evaluations are much more difficult than that, primarily because, sometimes, they're actually including further nested stuff inside. So your f is actually coming from some really complicated calculation itself. And then you have another big giant factor of how much it costs to do f. But on this right here, I'm just assuming it goes as n. So actually, let's just multiply this out. So this is what, 10 to the fourth, 10 to the 10th. So this is 10 to the 10th. 10 to the 10th kind of flops. But your computer does 4 times 10 to the ninth per second. Is that right? Is that [? what they ?] told us, four gigahertz? Yeah, so is this really not bad with just a few seconds on your computer. So that's why you can do these homework problems, and you still get them done. And if you have even [? porous ?] limitations, it takes 1,000 times longer. Still, 1,000 [INAUDIBLE] take 20 minutes on your computer. So you're OK, but you can see we're starting to get up there. And it's not that hard to really make it a big problem if you do something wrong. And just a notation thing, a lot of these things we're doing are writing software that takes as input not just numbers or vectors. So those are things that we call functions. Right, things that take a number of vectors and input and give you, say, a function, a number of vectors and output. But oftentimes, we actually want to take the names of functions as inputs. So for example, the root-finding problem, it takes as input what is f. Right, you have to tell it the name of a function for fsolve to be able to call it. So fsolve is an object that takes as input the name of a function. And things that do that are called functionals So for example, this one is x is equal to this root-finding functional, fsolve. And I'm going to write with square brackets. It depends on f. That's it, right? [INAUDIBLE] f. So you give it the function f. And if fsolve's good, it should be able to tell you what the x is. Now, in reality, we help it out. So we give it x 0 [? 2 ?] just give it a better chance. I write the square brackets just indicate that one of the things inside is a function, not just numbers. We do this a lot. So you guys have done this since you were kids. So derivatives Yeah, ddx of f. So we have, I don't know, for example, grad of f. That's a functional, the grad. The symbol means the operator, the functional, it operates on functions. You could write a general gradient function that takes any scalar valued function and computes the gradients with respect to the inputs. This week, we give you one that computes a Jacobian. Give it any f of x. It gives you the Jacobian of f with respect to x. So that's a functional. So anyway, we do a lot with functionals. You've been using them all the time. I'm just trying get you to think about abstractly. And then there's a whole math about functionals. And there's a whole thing called calculus of variations. It's all about, if the objects you're worrying about are functionals, not functions, then you have another whole bunch of theorems and stuff you can do with that, which you'll probably end up doing before you get out of this place. But I'm not going to do that right now. I would comment that one kind of functional that's particularly important to a lot of people in your research is the fact that the energy of any molecule is a functional of the electron density. And this is the basis of what's called density functional theory. And so this is why a lot of people, you and many of you included, are going to end up using computer programs that are trying to find what the electron density shape is, things like molecular orbitals and stuff. You've probably heard these before. And then there's the functional that gives the energy of the molecule. And what you care about, mostly, is energy, because that connects to all your thermo. You're doing [? 1040, ?] it's all about the energies. And once you know all the energies, you can actually compute the entropies and all kinds of stuff too. And so this is one application that's really important. It was a little surprising this was true and able to be proven to be true. And so the guy who did that got the Nobel Prize-- his name was Cohen-- a few years ago. Actually, in liquid phase stuff, you work with Professor Blankschtein. They also have a version of this for properties of solution. And then where it's not the electron density, but it's actually the density of the material in the solution phase. Maybe Professor [? Swan ?] too. You guys do that? Yeah, so if you work with Professor [? Swan, ?] you might learn about that too. So that's one where the actual functional is like a big focus of the entire project. And the key is that this is a function. The density is a function of the position. And then the functional converts that into a number. That's all page one here. Now, I'm going to sort of change topics, but it's related. It's about the connections between numerical integration, interpolation, and basis functions. These things are all intimately tied up together. So I'll try to show you. So we try to solve a numerical integral. And then let's look at what we're really doing. When we're doing this, we typically only do a few function evaluations. So we have a few points where we know the numbers f of tn. We've picked a few t's. We evaluated f. And from those few points, that small number of function evaluations, we want to get a good estimate of this whole integral. Now, it's a little goofy, actually, if you think about what we're trying to do. So suppose you're trying to do an integral from negative 1 to 1. We have some function that has a point here, a point here, a point here. And just for concreteness, let's say this is 1 over 26. And this is 1. And this is 1 over 26. And I have a particular function in mind that has those three points. Now, the first thing to keep in mind is there's an infinite number of functions that go through those three points. Now, what we've been doing is we've been fitting some function to these points and then integrating that function that we fitted. So when you're in high school, you did the rectangle rule and what the function was used was this and this. And you knew the integral of this function and the integral of this function. And you added them up. And you decided from the rectangle rule that the integral was equal to 1 and 1/26. But then some other people in high school are a little smarter. And they said it's not so smart [INAUDIBLE].. This thing was symmetrical to begin with. And now, I'm fitting it with a function that's wildly unsymmetrical. So let's instead not evaluate those points. Let's instead evaluate the midpoints. And maybe we find out that the midpoint values are here and here. And so I would assume that the function is just a flat line. That looks not so good either, because it doesn't get through these points. But anyway, for the two points I knew, if I only used those two points, that would be the midpoint method. And I would get the I midpoint method. And I'd get something. And I did it for this particular function I have in mind. And it was 8 over 29. And then some other people say, well, let's use the trapezoid rule. So let's approximate the function is this over the interval, which looks kind of sensible. And if you do that, it turns out you get the same numbers as you do with the rectangle rule. So I trapezoid. Now, let me just as a warning, if you just do the rectangle rule and then the trapezoid rule, and you got the same number both times, you might be tempted to conclude you have converged. And you have the true solution, because you did two methods of really different orders, and you got exactly the same number. And you're super happy, and you're ready to publish your paper. But of course, the function might not look like this. So the particular function that I always had in mind was f of t is equal to 1 over 1 plus 25 t squared. OK, that was just the function I had in mind. And the way that function looks like is like that. And the real integral of that function, which you guys if you remember how do it from high school, you can actually evaluate that integral. It's nothing to do with either 8/29 or 1 in 126. It's something completely different. Now, those of you who went to really advanced high schools, you also learned Simpson's rule too. And Simpson's rule says let's fit it to a parabola. And so that would give you something like this. And if you do Simpson's rule, I ended up getting this if I did the arithmetic right. 1 and 1/3 plus 2/78. Who knew? OK, which also has little relation to the true value of the integral. Now, if you're smart, and you did trapezoid and then I it to Simpson's, then you say, oh my goodness, these are not really that similar. And so I'm not converged, and I'd better do something else. So I just want you to be aware when we do these rules, what we're actually doing is we're fitting our points to some continuous function that we know how to integrate. And then we do the analytical integral of that function, which is a fine thing to do. But just be aware that it's like extrapolation. It's like we're imagining what the function is doing between these points, when we actually haven't tested it. Now, we could. We could just call more points. We could calculate. But that would cost us more function evals. And we're trying to save our computer time. So we want to finish off before the TG so we can go buy some beer. And so we have to take some choices about what to do. And we can't do an infinite number of points for sure. All right, so in all these methods, what we're doing is we're approximating our true f with some approximate f, which is equal to a sum of some basis functions like this. So we pick some basis functions. We know how to integrate. We fit, somehow determine what these c's are. And then we decide that the integral is equal to the integral of f tilde, not the actual original f. And then that's actually going to be the sum of ck times the integral. And we carefully chose basis functions that we don't how to do this integral; analytically. So we just plug that in, and we just have to figure out that the c's are to get this to work. OK, so that's what you really did when you're doing this trapezoid rule and Simpson's rule, and all that kind of stuff. And it's just different choices of what the basis functions were that you used. Now, there's a lot of choices about how to do that kind of approximation to make f tildes that are sort of like f's with some limited amount of information. And it generally has to do with how many k's we have. So if k is less than n, then it's really an approximation. Because there's no way that we would expect that with just a few parameters, we could get a lot f of tn. So we would expect that f tilde of tn is not equal to f of tn, at least not at all the points. Because if n is-- if we have a lot of points and we only have a few parameters, we don't expect to be able to fit it perfectly. So we know this is what we expect. It might happen if, by some luck, we chose a basis function which was actually the true function f, then it would actually fit perfectly. But most of the time, not. By Murphy's Law, in fact, it's going to be completely wrong. But this is not such a bad idea. We may come back to this later. Another possibility is to choose k equals to n. And then usually, we can force f tilde of tn to equal f of tn. So at the points that we have, if we have enough parameters equal to the number of points, then often, we can get an exact fit. We can force the function to go through all the points. And when we do this, this is called interpolation. Now, when should you do the one or the other? When should I make k as big as n? Or when should I use k less than n? A really crucial thing is whether you really believe your points. If you really believe you know f of tn to a very, very high number of significant figures, then it make sense. You know that's true, so you should force your fitting function to actually match that. However, if you get the f of tn, for example, from an experiment or from a stochastic simulation that has some noise in it, then you might not really want to do this. Because you'll be fitting the noise in the experiment as well. And then you might want to use a smaller number of parameters to make the fits. OK, so you can do either one. You're in charge. Right, you get the problem. If somebody tells you, please give me this integral, they don't care how you do it. They just want you to come back and give them the true value of the integral to some significant figures. So it's up to you to decide what to do. So you can decide what to do. What a lot of people do in the math world is this one, interpolation, where you're assuming that the f of tn's are known exactly. And so your function evaluating function is great. And in that case, you can write down this nice linear equation. So we're trying to make f of tn with a true function to be equal to ck phi n of t. Sorry, phi k of t. Which n in [INAUDIBLE] k is equal to n. And I can rewrite this this way. This is like a set of numbers. So I can write as a vector. I'll call it y. And this thing has two indices on it, so it looks like a matrix. So I'll call that m. And c is my vector of unknown parameters. And wow, I know how to solve this one. So c is equal to m backslash y. Now, for this to work and have a unique solution, I need m not to be singular. And what that means is that the columns of m have to be linearly independent. So what are the columns of m? They are phi k of t1, phi k of t2, phi k of tn. And I want that this column is not a sum of the other columns. So this is kind of requirement for j not equal [? to k ?] This is kind of a requirement if I'm going to do this kind of procedure is I can't choose a set of basis functions that's going to have one of them be a linear combination of the other ones. Because then I'm going to get a singular matrix. My whole process is going to have a big problem right from the beginning. So I have to ensure that this is not true. Yeah, that this is true. These are not equal to some of them. Another way to write that is to choose the the columns that phi to be orthogonal to each other. And so I can write it this way. OK, so I want them to be orthogonal to each other. And then you can see how this kind of generalizes. If I had a lot of t's, this would sort of look like the integral of phi k of t, phi j of t is equal to 0. And so this gives the idea of orthogonal functions. So this is orthogonal functions, and my discrete point's tn. And this is the general concept of orthogonal functions. And then when I define it this way, I need to actually decide my own a's and b's that I want to use. And if you use different ones, you actually have different rules about what's orthogonal to each other. And also, sometimes, people are fishy, and they [? spit ?] a little w of t in here just to do, because they have some special problem where this w keeps showing up or something, and they want to do that. But that's up to them. All right, we won't specify that. But this is the general idea of making basis functions orthogonal to each other. And it's very analogous to making vectors orthogonal to each other, which is what we did here. These are actually vectors. Right, it's just as the vector gets longer and longer, it gets more and more like the continuous function, and we have more t points. And eventually, it looks like that integral as you go to infinity. So anyway, so a lot of times, people will choose functions which are designed to be orthogonal. And one way to do it is just to go with a continuous one to start. Now, we can choose any basis functions we want. Over here, we already chose a whole bunch. You guys have done this already. I don't know if you knew you were doing it. But you chose a lot of different basis functions completely different from each other. You can choose the different ones if you want. And one popular choice is polynomials. But it's not the only choice. And in fact, a little bit later, we're going to, in the homework problem six, I think we're going to have a problem where you have to use a different kind of basis functions. You can use any basis functions you want. And there's kind of ones that match naturally to the problem. However, people know a lot about polynomials. And so they've decided to use polynomials a lot. And so that's the most common choice for numerical integration is polynomials. One thing is polynomials are super easy to integrate, because that was the first thing you learned, right, how to do the intervals of polynomials. And also, they're easy to evaluate. It only takes n operations to evaluate a polynomial of nth order. And there's a ton of theorems, and things are known about polynomials. We have all kinds of special properties. And we have all kinds of convenient tools for handling them. So let's go with polynomials for now. Now, I want to warn you, though, that how you order polynomials are really not very good fitting functions or interpolating functions for a lot of problems. And in particular, for this one I showed here where this is the function. Here, that function, you can try to fit that with higher and higher order polynomials. So I can have my function. The function value at 0 is 1. So the function looks kind of reasonable. Like that, symmetrical, better than I drew it. And you can put some number of points in here. And then you can do an interpolating polynomial that goes through all the points. And if you do it for this one with five points, you'll get something that looks like this. It goes negative, comes up like this, goes negative again. So that's one you get if you use five points across this interval. If you use 10 points, then you get one that looks like this where the peak value here is two where as the max of the original function's just one. And the original function's always positive, but they both go negative. So even within the range of the interpolation, the polynomials can have wild swings. And so this first one was for a fifth order polynomial, or maybe fourth order, five points. And this one is for the next 10th order or 9th order polynomial. I'm not sure which one. Yes. AUDIENCE: What points are these polynomials trying to fit? Or does it seem like [INAUDIBLE]?? PROFESSOR: So I didn't draw very well. When you guys go home, do this. Right, on Matlab it will take you a minute to solve this equation where you put in the function values here. And put in the polynomial values that [INAUDIBLE] here for the monomials, t to the first power, to the second power, to the third power, like that. And just do it, and you can get the [? pot ?] yourself. And just to convince yourself about the problem, and this is very important to keep in mind. Because we're so used to using polynomials that think they're the solution to everything. And Excel does them great. But it's not this solution to everything. Nonetheless, we're going to plow on and keep using them for a few minutes. So one possibility when I'm choosing this basis set is I could choose what are called monomials, which is phi k is equal to x to the k minus 1. And this one is super simple. And so a lot of times it satisfies this orthogonality thing, which I showed you before. But it's a really bad choice usually. And what it is is that the matrix you have to solve for this interpolation, if you have evenly-spaced points and you have x to the k, it turns out that this matrix usually has a condition number that grows exponentially with the number of terms in your basis, the number of how many monomials you include. So condition numbers that grow exponentially is really bad idea. So this is not what people use. So you might be tempted to do this, and maybe you did it once in your life. I won't blame you if you did. But it's not a really smart idea. So instead, a lot of the mathematicians of the 1700s and 1800s spent a lot of time worrying about this. And they'd end up deciding to use other polynomials that have people's names on them. So a guy named Lagrange, the same guy who did the multiplier. He suggested some polynomials. They're kind of complicated. They're given in the book. But what they really are is I'm taking the monomial basis set, and I'm making a new basis set. So I want phi l to be equal to the sum dlk of phi k, where this phi k it the monomials. So really, you can choose any d you want. You're just taking linear combinations of the original monomial basis set to make up polynomials. And if you cleverly choose the d, you can choose it to have special properties. So Lagrange gave a certain choice of the d that is brilliant in the sense that, when you solve the problem m times c is equal to y to try to do your interpolation, it turns out that the c's you want, ck, is equal to f of tk. So you don't have to do any work to solve what the c's are. It's just the f values directly. No chance for condition numbers, no problems, that's it. So that's the Lagrange polynomials. Now, Newton had a different idea. So this is for Lagrange polynomials. If you choose a Lagrangian [? polynomial ?] basis, then this is it. And you do a square system. That's it. If you choose the Newton, even he got into this business too. And he made some polynomials. And he made ones that have a special property that if you solve it-- so he chose a different d. You still have to do some solve. You get the solution. And then you decide that I don't like that interpretation. I want to put some more points in. I'm going to do a few more function evaluations. I'm going to add some more. So from the first solve, I have the values of c1, c2, c3, up to c the number of points I had initially. And now, I want to add one more data point, or add one more f of tn, one more function evaluation. I want to figure out with c of n plus 1 is, what the next one is. It turns out that it's his polynomials. The values these c's don't change when I make this matrix a little bit bigger by adding one more y. And I add one more parameter. And I make one more row in the matrix, one more column in the matrix. When I do that, it's a unique property that none of these other c's change. The solution is going to be the same, except for the n plus one [? level. ?] So then I can write a neat, cute little equation just for the n plus 1 one. I don't have to mess with the previous ones. And so I'm just adding a basis function that sort of polishes up my original fit. So that's a nice property too. Whereas with the Lagrange ones, you have to recompute everything. And all the c's will change. Actually, the c's won't change. I take that back. You add points. The c's won't change, but the actual polynomials will change. The d will change. Because the d depends on the choice of the time points in a complicated way in Lagrange. So now, you start to think, oh my goodness, I have all these choices, I can use Lagrange. I could use Newton. I could use monomials. Maybe, I should go back to bed. And you think, well, could someone just tell me what the best one is? And I'll just do that one. I don't need to learn about all this historical stuff about development of polynomial mathematics. I'd just like you to just tell me what the answer is. So Gauss, typically, figured out what the best on is. And he said, you know, if we're going to do integrals, we'd really like to just write our integrals this way. So our approximate value to be equal to some weight functions times the function evaluations at some times. Right, that's easy to evaluate. That would be good. If I pre-computed the weight functions [? and ?] the times, then I put any function f in here, I could just do this like zip. It's one dot product of the f of tn's I'll have the solution. So I want to do that. So what's the best possible choices of the w's and the t's to make this really be close? Well, the problem is that there's an infinite number of functions that somebody might want to integrate. I can't do them all. So he said, let's just find the one that works for as many monomials as possible to the highest order, as far out as I can go. And it turns out I have nw's. I have nt's. So I have two n parameters that I can fiddle around with. And so I should get something like two n's in my polynomials, maybe 2n minus 1. So what I want is that I want to choose this so that for the monomial 1, and the monomial t, and the monomial t squared, that this i is actually going to be exact. So what I want is for the monomial 1, I want the summation of wn. My f for the monomial 1 is always 1. So it's times 1. I want this to equal the integral from a to b of 1dt, which is b minus a. OK, so that gives me one equation for the w's. Then for the next one, I want the summation wn times tn to equal the integral of a to b of t. dt, which is equal to b squared over a number 2 minus a squared over 2 if I did my calculus correctly. And so there's another question. I have another question for these guys. And I keep going until I get enough equations that determine all the w's and c's. All right, so I do it for t squared. So I can wn tn squared is equal to [INAUDIBLE] b t squared dt, and so on, OK. And so if you do this process, you get the weights and the points to use that will give you what's called the Gaussian quadrature rule. Now, the brilliant thing about this is that by this setup, it's going to be exact for integrating any polynomial up to order 2n minus 1. So if I choose three, here I'll be able to determine-- if I choose three points, I'll have six parameters. So I'll be able determine up to the 11-- 11th order polynomial would integrate exactly. No, that's wrong. n's only 3. I have six parameters. I can't do this. 2 times 3 minus 1, what's that? 5, there we go. To the fifth order polynomial integrate exactly, I'll have an [? error ?] for sixth order polynomial. So this is a rule. So before, if I gave you three points like here, most of you probably would use Simpson's rule. Fit this to a parabola. It has three parameters, a, ax squared plus bx plus c, right, for a second order polynomial. I would fit it. Then I'd integrate that. That would be the normal thing that a lot of people would do. I don't know if you would do that. But a lot of people would do that. So that would be that what you would get. And what [? error ?] would you get? You'd have error that would be the scale of the width of the intervals to the fourth power. Right, because Simpson's rule exact up to the cubic. You guys remember this from high school? I don't know if you remember delta t to the fourth. Yeah, OK. So Simpson's rule has an order of delta t to the fourth. But Gauss's method has order of delta t to the sixth. So this is for three points. In Gauss's method, you can extend it to as many points as you want. All right, so that's kind of nice. You only do three points, you get [? error ?] of the sixth order. So that seems pretty efficient. So people kept going. And so the very, very popular one use, actually, seven points for Gauss. So you'd be able to get to the n minus 1. So you'd be exact for 13. So if you have delta t to the 14th power as your error, that's pretty good. And then what people often do is they'll add eight more points. And they make another method that uses all these points. It's called [? Conrad's ?] method. He made another method that gives you the best possible solution if you had the 15 points, given that 7 of them are already fixed by the Gauss procedure. And so you can get the integral using Gauss's method and using [? Conrad's ?] method. So you can compute the i using Gauss's method, the seven points, the i with the [? Conrad ?] method. I don't know how to spell his name, Actually I think it's like this. 15 points. And then you can take the difference between these two. And if these two are really similar, then you might think your integral's pretty good. Because they're pretty different methods, and if they get the same number, you're good. And people have done a lot of numerical studies in this particular case, because it's used a lot in actual practice. And it turns out that the [? error ?] is almost always less than this quantity to the 3 Gauss power. So you have an error estimate. So you do 15 function evaluations. You get a pretty accurate value for your integral, maybe, if you can be fit well with whatever order polynomial, 13th order polynomial. 13th order polynomial, yeah. And so any function that can be fit pretty well with the 13th order polynomial, you should get pretty good value in the integral. And you can check by comparing to the [? Conrad ?] estimate, which uses a lot more points. And if two of these guys give the same number or pretty close, then you're pretty confident that you're good. And there's even a math proof about this particular one. So this is like really popular, and it's a lot better than Simpson's rule, but it's kind of complicated. And part of it you have to recompute these tn's and wn's for the kind of problem you have. Yeah. AUDIENCE: If you're using 15 points already, why not just split your step size smaller? And if you did that, how would the error [INAUDIBLE] the simplest? Use a trapezoid. PROFESSOR: Trapezoid or Simpson's rule of something like that. Yes. AUDIENCE: How would the error with those 15 steps using trapezoid compare to [? Conrad ?] and Gauss? PROFESSOR: So we should probably do the math. So say we did Simpson's rule, and we have delta t, and we divide it by 15 or something. Then we can figure out what this is, but it's still going to have the scaling to the fourth power. So the prefactor will be really tiny, because it will be 1/15 to the fourth power. But the scaling is bad. So that if I decide that my integral's not good enough, and I need to do more points as I double the number of points to get better accuracy, my thing won't improve that much. Because it's still only going to use the fourth power, whereas the Gauss' one say it's going use the sixth power if I only use the three points. Where we use 15 points, is going to get scale to some 16th power or 15th power or some really high power. So for doing sequences, the Gauss' one is a lot better. However, what people do is they usually pre-solve the system of equations. Because this system of equation does not depend. When you do it with the monomials, it doesn't depend what f is. You can figure out the optimal t's and w's ahead of time. And so they [INAUDIBLE] tables of them depending on where you want to go. People recomputed them, and then you can get them that way. Time is running out. I just want to jump up. Now, for single integrals, I didn't have to tell you and of this. Because you could have just solved it with your ODE solver. So you can just go to ode15s or ode45, and you'll get good enough. It won't be as efficient as Gaussian. But who cares. You still get a good answer. You can report to your boss. It'll take you three minutes to call it e45 to integrate the one-dimensional integral. The real challenge comes with the multidimensional integrals. So in a lot of problems, we have more than one variable we want to integrate over. And so if you have an integral, say, of a to b from the lower bound of x to upper bound of x. So this is dx, dy, f of xy. This is the kind of integral that you might need to do sometimes. OK, because you might have two variables that you want to integrate over. And so how we do this, the best way to think about it, I think, is that f of x is equal to the integral of l of x u of x dy f of xy. And then, if I knew what that was, I would evaluate that say with a Gaussian quadrature or something. So I'd say that this i 2d is approximately equal to the sum of some weights-- I'll label them x just to remind you where they came from-- times f of xn. And I would use, say, Gaussian quadrature to figure out what the best value of the x's and the y's are to use. And then I would actually evaluate this guy with a quadrature two. So I'd say f of xk or xn is equal to the integral from l of xn to u of xn dy f of xn y. And so this itself, I would give it another quadrature. This is going to be equal some other weights. nj f of xn yj. Is that all right? And so this is nested inside this. So now, I have a double loop. So if I took me, I don't know, 15 points, 15 function evaluations to get a good number for my integral in one dimension, then it might take me 15 squared function evaluations to get a pretty good integral for two. 15 might be a little optimistic, so maybe 100 is more like if you really want a good number. So I need 101 dimension, 100 in the second dimension squared. Because I subdivided the integral into six equal pieces, say. And then now, so this is OK. So you can do this. It's only 10,000 function evaluations. No problem, you've got a good gigahertz computer. Now, in reality, in a lot of them, we do have a lot more dimensions than that. So in my life, I do a lot of electronic structure integrals. We have integrals that are dx 1d y1 d z1 for the first electron interacting with a second. And then something that I'm trying to integrate. It depends on, say, the distance between electron one and electron two. OK, so this is actually six integrals. And so instead of being a 2 in the exponent, this is going to be like a 6 in the exponent. So now, this is 10 to the 12th. That means that just evaluating one integral this way might take me 100 seconds if I can evaluate the functions like that. But in fact, the functions I want to evaluate here, these guys will take multiple operations to evaluate the function. And so this turns out I can't do. I can't even do one integral this way with its six integrals deep. And so this is the general problem of multidimensional integration. It's called the curse of dimensionality. Each time you add one more integral, you're nested there. All of a sudden, the effort goes up tremendously and more, a couple of orders of magnitude more. And so we'll come back later in the class about how to deal with cases like this when you have high dimensionality cases to try to get better scaling. Instead of having it being exponential, it was really 100 to the n right now, 100 to the n dimensions. [? Can we ?] figure out some better way to do it. All right, thank you. |
MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 12_Constrained_Optimization_Equality_Constraints_and_Lagrange_Multipliers.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JAMES SWAN: OK. Well, everyone's quieted down, so that means we have to get started. So let me say something here. This will be our last conversation about optimization. So we've discussed unconstrained optimization. And now we're going to discuss a slightly more complicated problem-- but you're going to see it's really not that much more complicated-- constrained optimization. These are the things we discussed before. I don't want to spend much time recapping because I want to take a minute and talk about the midterm exam. So we have a quiz. It's next Wednesday. Here's where it's going to be located. Here's 66. Head down Ames. You're looking for Walker Memorial on the third floor. Unfortunately, the time for the quiz is 7:00 to 9:00 PM. We really did try hard to get the scheduling office to give us something better, but the only way to get a room that would fit everybody in was to do it at this time in Walker. I really don't understand, because I actually requested locations for the quizzes back in April. And somehow I was too early, maybe, and got buried under a pile. Maybe not important enough, I don't know. But it's got to be from seven to nine next Wednesday. Third floor. There's not going to be any class next Wednesday because you have a quiz instead. So you get a little extra time to relax or study, prepare, calm yourself before you go into the exam. There's no homework this week. So you can just use this time to focus on the material we've discussed. There's a practice exam from last year posted on the Steller site, which you can utilize and study from. I'll tell you this. That practice exam is skewed a little more towards some chemical engineering problems that motivate the numerics. I've found in the past that when problems like that are given on the exam, sometimes there's a lot of reading that goes into understanding the engineering problem. And that tends to set back the problem-solving. So I'll tell you that the quiz that you'll take on Wednesday will have less of the engineering associated with it, and focus more on the numerical or computational science. The underlying sorts of questions, the way the questions are asked, the kinds of responses you're expected to give I'd say are very similar. But we've tried to tune the exam so that it'll be less of a burden to understand the structure of the problem before describing how you'd solve it. So I think that's good. It's comprehensive up to today. So linear algebra, systems of nonlinear equations and optimization are the quiz topics. We're going to switch on Friday to ordinary differential equations and initial value problems. So you have two lectures on that, but you won't have done any homework. You probably don't know enough or aren't practiced enough to answer any questions intelligently on the quiz. So don't expect that material to be on there. It's not. It's going to be these three topics. Are there any questions about this that I can answer? Kristin has a question. AUDIENCE: [INAUDIBLE]. JAMES SWAN: OK. So yeah, come prepared. It might be cold. It might be hot. It leaks when it rains a little bit. Yeah, it's not the greatest spot. So come prepared. That's true. Other questions? Things you want to know? AUDIENCE: What can we take to the exam? JAMES SWAN: Ooh, good question. So you can bring the book recommended for the course. You can bring your notes. You can bring a calculator. You need to bring some pencils. We'll provide blue books for you to write your solutions to the exam in. So those are the materials. Good. What else? Same question. OK. Other questions? No? OK. So then let's jump into the topic of the day, which is constrained optimization. So these are problems of the sort. Minimize an objective function f of x subject to the constraint that x belongs to some set D, or find the argument x that minimizes this function. These are equivalent sorts of problem. Sometimes, we want to know one or the other or both. That's not a problem. And graphically, it looks like this. Here's f, our objective function. It's a nice convex bowl-shaped function here. And we want to know the values of x1 and x2, let's say, that minimize this function subject to some constraint. That constraint could be that x1 and x2 live inside this little blue circle. It could be D. It could be that x1 and x2 live on the surface of this circle, right, on the circumference of this circle. That could be the constraint. So these are the sorts of problems we want to solve. D is called the feasible set, and can be described in terms of really two types of constraints. One is what we call equality constraints. So D can be the set of values x such that some nonlinear function c of x is equal to zero. So it's the set of points that satisfy this nonlinear equation. And among those points, we want to know which one produces the minimum in the objective function. Or it could be an inequality constraint. So D could be the set of points such that some nonlinear function h of x is, by convention, positive. So h of x could represent, for example, the interior of a circle, and c of x could represent the circumference of a circle. And we would have nonlinear equations that reflect those values of x that satisfy those sorts of geometries. So equality constrained, points that lie on this circle, inequality constrained, points that lie within this circle. The shape of the feasible set is constrained by the problem that you're actually interested in. So it's easy for me to draw circles in the plane because that's a shape you're familiar with. But actually, it'll come from some sort of physical constraint on the engineering problem you're looking at, like mole fractions need to be bigger than zero and smaller than one, and temperatures in absolute value have to be bigger than zero and smaller than some value because that's a safety factor on the process. So these set up the constraints on various sorts of optimization problems that we're interested in. It could also be true that we're interested in, say, optimization in the domain outside of this circle, too. It could be on the inside, could be on the outside. That's also an inequality constrained sort of problem. You know some of these already. They're familiar to you. So here's a classic one from mechanics. Here's the total energy in a system for, say, a pendulum. So x is like the position of the tip of this pendulum and v is the velocity that it moves with. This is the kinetic energy. This is the potential energy. And we know the pendulum will come to rest in a place where the energy is minimized. Well, the energy can only be minimized when the velocity here is zero, because any non-zero velocity will always push the energy content up. So it comes to rest. It doesn't move. And then there's some value of x at which the energy is minimized. If there is no constraint that says that the pendulum is attached to some central axis, then I can always make the energy smaller by making x more and more negative. It just keeps falling. There is no stopping point. But there's a constraint. The distance between the tip of the pendulum and this central point is some fixed distance out. So this is an equality constrained sort of problem, and we have to choose from the set of v and x the values subject to this constraint that minimize the total energy. And that's this configuration of the pendulum here. So you know these sorts of problems already. We talked about this one, linear sorts of programs. These are optimization problems where the objective function is linear in the design variables. So it's just the dot product between x and some vector c that weights the different design options against each other. So we talked about ice cream. Yes, this is all premium ice cream because it comes in the small containers, subject to different constraints. So those constraints can be things like, oh, x has to be positive because we can't make negative amounts of ice cream. And maybe we've done market research that tells us that the market can only tolerate certain ratios of different types of ice cream. And that may be some set of linear equations that describe that market research that sort of bound the upper values of how much ice cream we can put out on the market. And then we try to choose the optimal blend of pina colada and strawberry to sell. So those are linear programs. This is an inequality constrained optimization. In general, we might write these problems like this. We might say minimize f of x subject to the constraint that c of x is 0 and h of x is positive. So minimize it over the values of x that satisfy these two constraints. There's an old approach that's discussed in the literature. And it's not used. I'm going to describe it to you, and then I want you to try to figure out why it's not used. And it's called the penalty method. And the penalty method works this way. It says define a new objective function, which is our old objective function plus some penalty for violating the constraints. How does that penalty work? So we know that we want values of x for which c of x is equal to 0. So if we add to our objective function the norm of c of x-- this is a positive quantity-- this is a positive quantity-- whenever x doesn't satisfy the constraint, this positive quantity will give us a bigger value for this objective function f than if c of x was equal to 0. So we penalize points which don't satisfy the constraint. And in the limit that this penalty factor mu here goes to zero, the penalties get large, so large that our solution will have to prefer satisfying the constraints. There's another penalty factor over here, which is identical to this one but for the inequality constraint. It says take a heaviside step function for which is equal to 1 when the value of its argument is positive, and it's equal to zero when the value of its argument is negative. So whenever I violate each of my inequality constraints, Hi of x, turn on this heaviside step function, make it equal to 1, and then multiply it by the value of the constraint squared, a positive number. So this is the inequality constraint penalty, and this is the equality constraint penalty. People don't use this, though. It makes sense. I take the limit that mu goes to zero. I'm going to have to prefer solutions that satisfy these constraints. Otherwise, if I don't satisfy these constraints, I could always move closer to a solution that satisfies the constraint, and I'll bring down the value of the objective function. I'll make it lower. So I'll always prefer these lower value solutions. But can you guys take a second and sort of talk to each other? See if you can figure out why one doesn't use this method. Why is this method a problem? OK, I heard the volume go up at some point, which means either you switched topics and felt more comfortable talking about that than this, or maybe you guys were coming to some conclusions, or had some ideas about why this might be a bad idea. Do you want to volunteer some of what you were talking about? Yeah, Hersh. AUDIENCE: Could it be that [INAUDIBLE]?? JAMES SWAN: Well, that's an interesting idea. So yeah, if we have a non-convex optimization problem, there could be some issues with f of x, and maybe f of x runs away so fast that I can never make the penalty big enough to enforce the constraint. That's actually a really interesting idea. And I like the idea of comparing the magnitude of these two terms. I think that's on the right track. Were there some other ideas about why you might not do this? Different ideas? Yeah. AUDIENCE: [INAUDIBLE]. JAMES SWAN: Well, you know, that that's an interesting idea, but actually the two terms in the parentheses here are both positive. So they're only going to be minimized when I satisfy the constraints. So the local minima of the terms in parentheses sit on or within the boundaries of the feasible set that we're looking at. So by construction, actually, we're going to be able to satisfy them because the local minima of these points sits on these boundaries. These terms are minimized by satisfying the constraints. Other ideas? Yeah. AUDIENCE: Do your iterates have to be feasible? JAMES SWAN: What's that? AUDIENCE: Your iterates don't have to be feasible? JAMES SWAN: Ooh, this is a good point. The iterates-- this is an unconstrained optimization problem. I'm just going to minimize this objective function. It's like what Hersh said, I can go anywhere I want in the domain. I'm going to minimize this objective function, and then I'm going to try to take the limit as mu goes to zero. The iterates don't have to be feasible. Maybe I can't even evaluate f of x if the iterates aren't feasible. That's an excellent point. That could be an issue. Anything else? Are there some other ideas? Sure. AUDIENCE: [INAUDIBLE]. JAMES SWAN: I think that's a good point. AUDIENCE: --boundary from outside without knowing what's inside. JAMES SWAN: Short. So you'll see, actually, the right way to do this is to use what's called interior point methods, which live inside the domain. This is an excellent point. There's another issue with this that's I think actually less subtle than some of these ideas, which they're all correct, actually. These can be problems with this sort of penalty method. As I take the limit that mu goes to zero, the penalty function becomes large for all points outside the domain. They can become larger than f for those points. And so there are some practical issues about comparing these two terms against each other. I may not have sufficient accuracy, sufficient number of digits to accurately add these two terms together. So I may prefer to find some point that lives on the boundary of the domain as mu goes to zero. But I can't guarantee that it was a minima of f on that domain, or within that feasible set. So a lot of practical issues that suggest this is a bad idea. This is an old idea. People knew this was bad for a long time. It seems natural, though. It seems like a good way to transform from these constrained optimization problems to something we know how to solve, an unconstrained optimization. But actually, it turns out not to be such a great way to do it. So let's talk about separating out these two different methods from each other, or these two different problems. Let's talk first about equality constraints, and then we'll talk about inequality constraints. So equality constrained optimization problems look like this. Minimize f of x subject to c of x equals zero. And let's make it even easier. Rather than having some vector of equality constraints, let's just have a single equation that we have to satisfy for that equality constraint, like the equation for a circle. Solutions have to sit on the circumference of a circle. So one equation that we have to satisfy. You might ask again, what are the necessary conditions for defining a minimum? That's what we used when we had equality-- or when we had unconstrained optimization. First we had to define what a minimum was, and we found that minima were critical points, places where the gradient of the objective function was zero. That doesn't have to be true anymore. Now, the minima has to live on this boundary of some domain. It has to live in this set of points c of x equals zero. And the gradient of f is not necessarily zero at that minimal point. But you might guess that Taylor expansions are the way to figure out what the appropriate conditions for a minima are. So let's take f of x, and let's expand it, do a Taylor expansion in some direction, d. So we'll take a step away from x, which is small, in some direction, d. So f of x plus d is f of x plus g dot d, the dot product between the gradient of f and d. And at a minimum, either the gradient is zero or the gradient is perpendicular to this direction we moved in, d. We know that because this term is going to increase-- well, will change the value of f of x. It will either make it bigger or smaller depending on whether it's positive or negative. In either case, it will say that this point x can't be a minimum unless this term is exactly equal to zero in the limit that d becomes small. So either the gradient is zero or the gradient is orthogonal to this direction d we stepped in. And d was arbitrary. We just said take a step in a direction, d. Lets take our equality constraint and do the same sort of Taylor expansion, because we know if we're searching for a minima along this curve c of x better be equal to zero. It better satisfy the constraint. And also, c of x plus d, that little step in the direction d, should also satisfy the constraint. We want to study only the feasible set of values. So actually, d wasn't arbitrary. d had to satisfy this constraint that, when I took this little step, c of x plus d had to be equal to zero. So again, we'll take now a Taylor expansion of c of x plus d, which is c of x plus grad of c of x dotted with d. And that implies that d must be perpendicular to the gradient of c of x, because c of x plus d has to be zero and c of x has to be zero. So the gradient of c of x dot d-- it's a leading order has also got to be equal to zero. So d and the gradient in c are perpendicular, and d and the gradient in g have to be perpendicular at a minimum. That's going to define the minimum on this equality constrained set. Does that make sense? c satisfies the constraint, c plus d satisfies the constraint. If this is true, d has to be perpendicular to the gradient of c, g has to be perpendicular to the gradient of d. d is, in some sense, arbitrary still. d has to satisfy condition that it's perpendicular to the gradient of c, but who knows, there could be lots of vectors that are perpendicular to the gradient of c. So the only generic relationship between these two we can formulate is g must be parallel to the gradient of c. g is perpendicular to d, gradient of c is perpendicular to d. In the most generic way, g and gradient of c should be parallel to each other, because d I can select arbitrarily from all the vectors of the same dimension as x. If g is parallel to the gradient of c, then I can write that g minus some scalar multiplied by the gradient of c is equal to zero. That's an equivalent statement, that g is parallel to the gradient of c. So that's a condition associated with points x that solve this equality constrained problem. The other condition is that point x still has to satisfy the equality constraint. But I introduced a new unknown, this lambda, which is called the Lagrange multiplier. So now I have one extra unknown, but I have one extra equation. Let me give you a graphical depiction of this, and then I'll write down the formal equations again. So let's suppose we want to minimize this parabolic function subject to the constraint that the solution lives on the line. So here's the contours of the function, and the solution has to live on this line. So I get to stand on this line, and I get to walk and walk and walk until I can't walk downhill anymore. and I've got to turn and walk uphill again. And you can see the point where I can't walk downhill anymore is the place where this constraint is parallel to the contour, or where the gradient of the objective function is parallel to the gradient of the constraint. So you can actually find this point by imagining yourself moving along this landscape. After I get to this point, I start going uphill again. So that's the method of Lagrange multipliers. Minimize f of x subject to this constraint. The solution is given by the point x at which the gradient is parallel to the gradient of c, and at which c is equal to zero. And you solve this system of nonlinear equations for two unknowns. One is x, and the other is this unknown lambda. How far stretched is the gradient in f relative to the gradient in c? So again, we've turned the minimization problem into a system of nonlinear equations. In order to satisfy the equality constraint, we've had to introduce another unknown, the Lagrange multiplier. It turns out this solution set, x and lambda, is a critical point of something called the Lagrangian. It's a function f of x minus lambda times c. It's a critical point in x and lambda of this nonlinear function called the Lagrangian. It's not a minimum of this function, unfortunately. It's a saddle point of the Lagrangian, it turns out. So we're trying to find a saddle point of the Lagrangian. Does this make sense? Yes? OK. We've got to be careful, of course. Just like with unconstrained optimization, we actually have to check that our solution is a minimum. We can't take for granted, we can't suppose that our nonlinear solver found a minimum when it solved this equation. Other critical points can satisfy this equation, too. So we've got to go back and try to check robustly whether it's actually a minimum. But this is the method. Introduce an additional unknown, the Lagrange multiplier, because you can show geometrically that the gradient of the objective function should be parallel to the gradient of the constraint at the minimum. Does that make sense? Does this picture make sense? OK. So you know how to solve systems of nonlinear equations, you know how to solve constrained optimization problems. So here's f. Here's c. We can actually write out what these equations are. So you can show that the gradient of x minus lambda gradient of c, that's a vector, 2x1 minus lambda and 20x2 plus lambda. And c is the equation for this line down here, so x1 minus x2 minus 3. And that's all got to be equal to zero. In this case, this is just a system of linear equations. So you can actually solve directly for x1, x2, and lambda. And it's not too difficult to find the solution for all three of these things by hand. But in general, these constraints can be nonlinear. The objective function doesn't have to be quadratic. Those are the easiest cases to look at. And the same methodology applies. And so you should check that you're able to do this. This is the simplest possible equality constraint problem. You could do it by hand. You should check that you're actually able to do it, that you understand the steps that go into writing out these equations. Let's just take one step forward and look at a less generic case, one in which we have a vector valued function that gives the equality constraints instead. So rather than one equation we have to satisfy, there may be many. It's possible that the feasible set doesn't have any solutions in it. It's possible that there is no x that satisfies all of these constraints simultaneously. That's a bad problem to have. You wouldn't like to have that problem very much. But it's possible that that's the case. But let's assume that there are solutions for the time being. So there are x's that satisfy the equality constraint. Let's see if we can figure out again what the necessary conditions for defining a minima are. So same as before, let's Taylor expand f of x going in some direction, d. And let's make d a nice small step so we can just treat the f of x plus d as a linearized function. So we can see again that g has to be perpendicular to this direction, d, if we're going to have a minima. Otherwise, I could step in some direction, d, and I'll find either a smaller value of f of x plus d or a bigger value of f of x plus d. So g has to be perpendicular to d. And for the equality constraints, again, they all have to satisfy this equality constraint up there. So c of x has to be equal to zero, and c of x plus d also has to be equal to zero. And so if we take a Taylor expansion of c of x plus d, about x, you'll get c of x plus d plus the Jacobian of c, all the partial derivatives of c with respect to x, multiplied by d. We know that c of x plus d is zero, and c of x is zero, so the directions, d, belong to what set of vectors? The null space. So these directions have to live in the null space of the Jacobian of c. So I can't step in any direction, I have to step in directions that are in the null space of c. g is perpendicular to d, as well. And d belongs to the null space of c. In fact, you know that d is perpendicular to each of the rows of the Jacobian. Right? You know that? I just do the matrix vector product, right? And so each element of this matrix vector product is the dot product of d with a different row of the Jacobian. So those rows are a set of vectors. Those rows describe the range of J transpose, or the row space of J. Remember we talked about the four fundamental subspaces, and I said we almost never use those other ones, but this is one time when we will. So those rows belong to the range of J transpose, or they belong to the left null space of J. I need to find a g, a gradient, which is always perpendicular to d. And I know d is always perpendicular to the rows of J. So I can write g as a linear superposition of the rows of J. As long as g is a linear superposition of the rows, it'll always be perpendicular to d. Vectors from the null space of a matrix are orthogonal to vectors from the row space of that matrix, it turns out. And they're orthogonal for this reason. So it tells us, if Jd is zero, then d belongs to the null space. g is perpendicular to d. That means I could write g as a linear superposition of the rows of J. So g belongs to the range of J transpose, or it belongs to the row space of J. Those are equivalent statements. And therefore, I should be able to write g as a linear superposition of the rows of J. And one way to say that is I should be able to write g as J transpose times some other vector lambda. That's an equivalent way of saying that g is a linear superposition of the rows of J. I don't know the values of lambda. So I introduced a new set of unknowns, a set of Lagrange multipliers. My minima is going to be found when I satisfy this equation, just like before, and when I'm able to satisfy all of the equality constraints. How many Lagrange multipliers do I have here? Can you figure that out? You can talk with your neighbors if you want. Take a couple minutes. Tell me how many Lagrange multipliers, how many elements are in this vector lambda. How many elements are in lambda? Can you tell me? Sam. AUDIENCE: Same as the number of equality constraints. JAMES SWAN: Yes. It's the same as the number of equality constraints. J came from the gradient of c. It's the Jacobian of c. So it has a number of columns equal to the number of elements in x, because I'm taking partial derivatives with respect to each element of x, and has a number of rows equal to the number of elements of c. So J transpose, I just transpose those dimensions. And lambda must have the same number of elements as c does in order to make this product make sense. So I introduce a new number of unknowns. It's equal to exactly the number of equality constraints that I had, which is good, because I'm going to make a system of equations that says g of x minus J transpose lambda equals 0 and c of x equals 0. And the number of equations here is the number of elements in x for this gradient, and the number of elements in c for c. And the number of unknowns is the number of elements in x, and the number of elements in c associated with the Lagrange multiplier. So I have enough equations and unknowns to determine all of these things. So whether I have one equality constraint or a million equality constraints, the problem is identical. We use the method of Lagrange multipliers. We have to solve an augmented system of equations for x and this projection on the row space of J, which tells us how the gradient is stretched or made up, composed of elements of the row space of J. These are the conditions associated with a minima in our objective function on this boundary dictated by the equality constraint. And of course, the solution set is a critical point of a Lagrangian, which is f of x minus c dot lambda. And it's not a minimum of it, it's a critical point. It's a saddle point, it turns out, of this Lagrangian. So we've got to check, did we find a saddle point or not when we find a solution to this equation here. But it's just a system of nonlinear equations. If we have some good initial guess, what do we apply? Newton-Raphson, converge rate towards the solution. If we don't have a good initial guess, we've discussed lots of methods we could employ, like homotopy or continuation to try to develop good initial guesses for what the solution should be. Are there any questions about this? Good. OK. So you go to Matlab and you call fmincon, do a minimization problem, and you give it some constraints. Linear constraints, nonlinear constraints, it doesn't matter actually. The problem is the same for both of them. It's just a little bit easier if I have linear constraints. If this constraining function is a linear function, then the Jacobian I know. It's the coefficient matrix of this linear problem. Now I only have to solve linear equations down here. So the problem is a little bit simpler to solve. So Matlab sort of breaks these apart so it can use different techniques depending on which sort of problem is posed. But the solution method is the same. It does the method of Lagrange multipliers to find the solution. OK? Inequality constraints. So interior point methods were mentioned. And it turns out this is really the best way to go about solving generic inequality constrained problems. So the problems of the sort minimize f of x subject to h of x is positive, or at least not negative. This is some nonlinear inequality that describes some domain and its boundary in which the solution has to live. And what's done is to rewrite as an unconstrained optimization problem with a barrier that's incorporated. This looks a lot like the penalty method, but it's very different. And I'll explain how. So instead, we want to minimize this f of x minus mu times the sum of log of h, each of these constraints. If h is negative, we'll take the log of the negative argument. That's a problem computationally. So the best we could do is approach the boundary where h is equal to zero. And as h goes to zero, the log goes to minus infinity. So this term tends to blow up because I've got a minus sign in front of it. So this is sort of like a penalty, but it's a little different because the factor in front I'm actually going to take the limit as mu goes to zero. I'm going to take the limit as this factor gets small, rather than gets big. The log will always get big as I approach the boundary of the domain. It'll blow up. So that's not a problem. But I can take the limit that mu gets smaller and smaller. And this quantity here will have less and less of an impact on the shape of this new objective function and mu gets smaller and smaller. The impact will only be nearest the boundary. Does that make sense? So you take the limit that mu approaches zero. It's got to approach it from the positive side, not the negative side, so everything behaves well. And this is called an interior point method. So we have to determine the minimum of this new objective function for progressively weaker barriers. So we might start with some value of mu, and we might reduce mu progressively until we get mu down small enough that we think we've converged to a solution. So how do you do that reliably? What's the procedure one uses to solve a problem successively for different parameter values? AUDIENCE: [INAUDIBLE]. JAMES SWAN: Yeah, it's a homotopy, right? You're just going to change the value of this barrier parameter. And you're going to find a minima. And if you make a small change in the barrier parameter, that's going to serve as an excellent initial guess for the next value. And so you're just going to take these small steps. And the optimization routine is going to carry you towards the minimum in the limit that mu goes to zero. So you do this with homotopy. Here's an example of this sort of interior point method, a trivial example. Minimize x subject to x being positive. So we know the solution lives where x equals zero. But let's write this as unconstrained optimization using a barrier. So minimize x minus mu times log x. Here's x minus mu times log x. So out here, where x is big, x wins over log x, so everything starts to look linear. But as x become smaller and smaller, log x gets very negative, so minus log x gets very positive. And here's the log creeping back up. And as I decrease mu smaller and smaller, you can see the minima of this function is moving closer and closer and closer to zero. So if I take the limit that mu decreases from some positive number towards zero, eventually this minimum is going to converge to the minimum of the constrained inequality, constrained optimization problem. Make sense? OK. OK. So we want to do this. You can use any barrier function you want. Any thoughts on why a logarithmic barrier is used? No. OK, that's OK. So minus log is going to be convex. Log isn't convex, but minus log is going to be convex. So that's good. If this function's convex, then their combination's going to be convex, and we'll be OK. But the gradient of the log is easy to compute. Grad log h is 1 over h grad h. So if I know h, I know grad h, it's easy for me to compute the gradient of log h. We know we're going to solve this unconstrained optimization problem where we need to take grad of this objective function equal zero. So the calculations are easy. The log makes it easy like that. The log is also like the most weakly singular function available to us. Out of all the tool box of all problems we can reach to, the log has the mildest sort of singularities. Singularities at both ends, which is sort of funny, but the mildest sort of singularities you have to cope with. So we want to find the minimum of these unconstrained optimization problems where the gradient of f minus mu sum 1 over h, grad h, is equal to zero. And we just do that for progressively smaller values of mu, and we'll converge to a solution. That's the interior point method. You use homotopy to study a sequence of barrier parameters, or continuation to study a sequence of barrier parameters. You stop the homotopy or continuation when what? How are you going to stop? I've got to make mu small, right? I want to go towards the limit mu equals zero. I can't actually get to mu equals zero, I've just got to approach it. So how small do I need to make mu before I quit? It's an interesting question. What do you think? I'll take this answer first. AUDIENCE: So it doesn't affect the limitation. JAMES SWAN: Good. So we might look at the solution and see is the solution becoming less and less sensitive to the choice of mu. Did you have another suggestion? AUDIENCE: [INAUDIBLE]. JAMES SWAN: Set the tolerance. Right, OK. AUDIENCE: [INAUDIBLE]. JAMES SWAN: Mhm. Right, right, right, right. So you-- AUDIENCE: [INAUDIBLE]. JAMES SWAN: Good. So there were two suggestions here. One is along the lines of a step-norm criteria, like I check my solution as I change mu, and I ask when does my solution seem relatively insensitive to mu. When the changes in these steps relative to mu get sufficiently small, I might be willing to accept these solutions as reasonable solutions for the constrained optimization. I can also go back and I can check sort of function norm criteria. I can take the value of x I found as the minimum, and I can ask how good a job does it do satisfying the original equations. How far away am I from satisfying the inequality constraint? How close am I to actually minimizing the function within that domain? OK. So we're running out of time here. Let me provide you with an example. So let's minimize again-- I always pick this function because it's easy to visualize, a nice parabolic function that opens upwards. And let's minimize it subject to the constraint that h of x1 and x2 is equal to 1 minus-- well, the equation for a circle of radius 1, essentially. The interior of that circle. So here's the contours of the function, and this red domain is the constraint. And we want to know the smallest value of f that lives in this domain. So here's a Matlab code. You can try it out. And make a function, the objective function, f, it's x squared plus 10x-- x1 squared plus 10x2 squared. Here's the gradient. Here's the Hessian. Here, I calculate h. Here's the gradient in h. Here's the Hessian in h. I've got to define a new objective function, phi, which is f minus mu log h. This is the gradient in phi and this is the Hessian of phi. Oh, man, what a mess. But actually, not such a mess, because the log makes it really easy to take these derivatives. So it's just a lot of differential sort of calculus involved in working this out, but this is the Hessian of phi. And then I need some initial guess. So I pick the center of my circle as an initial guess for the solution. And I'm going to loop over values of mu that get progressively smaller. I'll just go down to 10 to the minus 2 and stop for illustration purposes here. But really, we should be checking the solution as we go and deciding what values we want to stop with. And then this loop here, what's this do? What's it do? Can you tell? AUDIENCE: Is it Newton? JAMES SWAN: What's that? AUDIENCE: Newton? JAMES SWAN: Yeah, it's Newton-Raphson, right? x is x minus Hessian inverse times grad phi, right? So I just do Newton-Raphson. I take my initial guess and I loop around with Newton-Raphson, and when this loop finishes, I reduce mu, and it'll just use my previous guess as the initial guess for the next value of the loop, until mu is sufficiently small. OK? Interior point method. Here's what that solution path looks like. So mu started at 1, and the barrier was here. It was close to the edge of the circle, but not quite on it. But as I reduced mu further and further and further, you can see the path, the solution path, that was followed works its way closer to the boundary of the circle. And the minimum is found right here. So it turns out the minimum of this function doesn't live in the domain, it lives on the boundary of the domain. Recall that this point should be a point where the boundary of the domain is parallel to the contours of the function, since actually we didn't need the inequality constraint here. We could have used the equality constraint. The equality constrained problem has the same solution as the inequality constrained problem. And look, that actually happened. Here's the contours of the function. The contour of the function runs right along here, and you can see it looks like it's going to be tangent to the circle at this point. So the interpoint method actually solved an equality constrained problem in addition to an inequality constrained problem, which is-- that's sort of cool that you can do it that way. How about if I want to do a combination of equality and inequality constraints? Then what do I do? Yeah. AUDIENCE: [INAUDIBLE]. JAMES SWAN: Perfect. Convert the equality constraint into unknowns, Lagrange multipliers, instead. And then do the interior point method on the Lagrange multiplier problem. Now you've got a combination of equality and inequality constrained. This is exactly what Matlab does. So it converts equality constraints into Lagrange multipliers. Inequality constraints it actually solves using interior point methods. Buried in that interior point method is some form of Newton-Raphson and steepest descent combined together, like dog leg we talked about for unconstrained problems. And it's going to do a continuation. As it reduces the values of mu, it'll have some heuristic for how it does that. It's going to use its previous solutions as initial guesses for the next iteration. So these are very complicated problems, but if you understand how to solve systems of nonlinear equations, and you think carefully about how to control numerical error in your algorithm, you come to a conclusion like this, that you can do these sorts of Lagrange multiplier interior point methods to solve a wide variety of problems with reasonable reliability. OK? Any more questions? No? Good. Well, thank you, and we'll see you on Friday. |
MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 27_Probability_Theory_2.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. WILLIAM GREEN: All right, so I know some of you have succeeded to do the homework and some of you, I think, have not. Is this correct? AUDIENCE: Yeah. WILLIAM GREEN: OK. So I was wondering if someone who has succeeded to do their homework might comment on how small a mesh do you need to converge. AUDIENCE: [INAUDIBLE] WILLIAM GREEN: It's about l? L? OK, so you need something on the order of l to converge. Is that correct? So if you're trying to do the problem using mesh much bigger than l, you should probably try a tighter mesh. Yes? AUDIENCE: [INAUDIBLE] WILLIAM GREEN: All right. Yes? AUDIENCE: [INAUDIBLE] WILLIAM GREEN: Yes. Yes. All right. And has anyone managed to get the [INAUDIBLE] solution to actually be consistent with the [INAUDIBLE] solution? AUDIENCE: Something like 3% or 4% or so. WILLIAM GREEN: 3% or 4%, OK. And I assume that the [INAUDIBLE] is also using a mesh of similar size? Hard to tell? AUDIENCE: I used like a triangular system-- WILLIAM GREEN: Yeah, yeah, but I mean, it's really, really tiny ones at the bottom? If you want me to blow it up, I can just take a look and see to be sure. All right, and is backslash able to handle a million by million matrix? AUDIENCE: Like 10 seconds with [INAUDIBLE].. WILLIAM GREEN: [INAUDIBLE] OK. So you need to-- so do the sparse allocation. And MATLAB is so smart that it just can handle it with a million by million, which is pretty amazing, actually. That's a pretty big matrix. All right, sorry, this is too loud. All right, so last time, we were doing some elementary things about probability. Actually, any more questions about the homework problem before we get started? AUDIENCE: What's the answer? WILLIAM GREEN: What's the answer? You could ask your classmates. Any other questions? All right. So I had you confused a little bit with this formula probability of either A or B. So I asked what the probability of-- I flipped two coins-- that one of them would be a head. And I could see a lot of consternation. The general formula for this is it's the probability of A plus the probability of B minus the probability of A and B. It can't just be the two of them added together because if you have 50% chance for head of the penny and the dime is 50% chance, this would add up to 100% chance that you'll get a head, but you know sometimes it's true. So this is the formula. And then the probability of A and B is often written in terms of the conditional probabilities, the probability of A times the probability that B would happen given that A already happened, which is also equal to the other way around. And this has to be read carefully. It means B already happened, and then you want to know the probability of A given that B already happened. So it's sort of like this way-- I don't know-- the way I think about it. Like, this happened first. and now I'm checking the probability that that's going to happen. Now, a nice little example of this is given in [INAUDIBLE] textbook. And I think it's nice enough that it's worthwhile to spend a few minutes talking about it. So he was-- [INAUDIBLE] who wrote the textbook, was not actually a numerical guy. He was a polymer chemist. And so he gave a nice polymer example. So if you have a polymer and the monomers have some big molecule, and at one side, they have a sort of acceptor group, and the other side, some kind of donor group-- we'll call it D, I guess. And these are the monomers. And so they can link together. The donor can react to the acceptor. So you can end up with things like this and so on. So this is the monomer. This is the dimer. Then you could keep on [INAUDIBLE] like this. And many, many, many of the materials you use every day, the fabrics in the seats that you're sitting on, the backs of the seats, your clothing, the binder holding the chalk together, all this stuff is made from polymers like this. So this is a pretty important, actually, practical problem. And so you start with the monomers, and they react where you have A reacting plus D, over and over again. And we want to understand the statistics of what chain lengths are going to make, maybe what weight percent or what would the average microweight be, something like that would be the kind of things we care about. So a way to think about it is if I've reacted this to some extent and I just grab a random polymer chain, any molecule in there, and I look and find, let's say, the unreacted D end-- so any oligomer is going to have one unreacted D end. You can see no matter how long I make it, there will still be one unreacted D end. And I'm neglecting the possibility this might circle around and make a loop. So assuming no loops, then any molecule I grab is going to have one unreacted D end. So I grab a molecule. I start at the unreacted D end, and I look at the A that's next to it. And I say, is that A reacted or not? So if it's a monomer, I grab the D. I look over here. The A is unreacted. So the probability that it's a monomer is going to be equal sort of like 1 minus P where P is the probability that As react. So it didn't react, so just like that. This one, the one next to it has reacted. So this is just going to be the probability of a dimer is the probability that my nearest neighbor reacted and next neighbor is unreacted, right? Is that OK? So I can write this way. I could say, what's the probability that my nearest reacted times a conditional probability, next unreacted if nearest is reacted? So far, so good? You guys are OK with this? So I grabbed a chain. I'm trying to see if it's a dimer. I'm going to calculate the probability that this next acceptor group has been reacted to a donor group. If it has reacted, then I'm going to check the next one after that. So this is the nearest neighbor. This is the next nearest neighbor. And I want this to be unreacted. If that's both true, then I have a dimer. If either one of those is false, [INAUDIBLE].. Is that OK? So now I need to have a probability. So what's the probability that the nearest one is reacted? There's some probability that things have reacted. So this is going to be my P, probability that things reacted. And I wanted this to be unreacted. Now, there's a question. Are these correlated or not? Now, in reality, everything's correlated to everything. So probably, they're correlated. But if we're trying to make a model and think about it, the fact that this thing reacted at this side doesn't really affect this side if this is a big enough [INAUDIBLE] So to a good approximation, this is independent of whether or not it's reacted or not. So this is still going to have the ordinary probability of being unreacted, which would be 1 minus P. So I could write down that the probability of being a monomer is equal to 1 minus P. The probability of being a dimer is equal to P times 1 minus P. What's the probability of being a trimer? P squared times 1 minus P. And in general, the probability of being an n-mer is equal to P n minus 1 times 1 minus P. So now you guys are statistical polymer chemists. So this derivation was derived by a guy named Flory. He got the Nobel Prize. He's a pretty important guy. If you want to learn a lot about him, I think both Professor Cohen and Professor Rutledge teach classes that are basically, learn what Mr. Flory figured out. Well, maybe that's a little bit too strong, but pretty much. There's another guy named [INAUDIBLE] that did a bit too, so [INAUDIBLE] and Flory. Basically everything about polymers worked out by at these guys. And all they did was just probability theory, so it was a piece of cake. And so this is the probability that you have an n-mer. So now we can compute things like, what is the expectation value of the chain length? How many guys link together? And that's defined to be the sum of n times the probability of n. So that, in this case, is going to be sum of n times P to the n minus 1 times 1 minus P. Now, a lot of these kinds of simple series summations, there's formulas for it. And maybe in high school, you guys might have studied series. I don't know if you remember. And so you can look up. And some of these have analytical formulas that are really simple. But you can just leave it this way too, because you get a value numerically with MATLAB, no trouble. You can also figure out what is the concentration of oligomers with n units in them. And so that's going to be equal to the total concentration of polymers times the probability that it has n. So this one, we just worked out. The total concentration, a way to figure that out is to think about there's one monomer or one monomer-- I'll call this a polymer too. This is a polymer with one unit. There's one polymer molecule per unreacted end, unreacted D end. So it's really, I want to know how many are unreacted. So that's going to be 1 minus P times the amount of monomer I had to start with. It could be A or D. It doesn't matter. It's like, how many of them-- I started with a certain amount of free ends. What fraction of them have reacted based on 1 minus P. Yeah, it's 1 minus P. So as P goes-- well, yeah, it goes backwards. Yeah, as P goes to infinity-- I think that's right. Yeah, when P is-- well, I'm totally confused here now. 1 minus P sound right? Maybe I did the reasoning backwards. This is definitely the right formula. I'm just confusing myself with my language. This is a, at least for me, endemic problem with probability is you could say things very glibly. You've got to think of exactly what you mean. So the concentration of unreacted ends, so initially, this was equal to A. It was all unreacted ends. And as the process proceeds, as P increases, then at the end, it's going to be very small. So this is right. And the concentration of unreacted ends is equal to the total concentration of polymers, the number of polymers [INAUDIBLE].. So it's this times P n minus 1 times [INAUDIBLE].. All right, and this is called the Flory redistribution. And that gives the concentrations of all your oligomers after you do a polymerization if they're all uncorrelated and you don't form any loops. It's often very important to know the width of the distribution. If you make a polymer, you want to make things have as monodisperse as possible. It's because you'd really like to make this pure chemical. There's some polymer chain length which is optimal for your purpose. You want to try to make sure that the average value, average value, this is going to be equal to the value you want. So you want to keep running P up until you reach the point where the average chain length is the chain length that's optimal for your application. If you make the polymer too long, then it's going to be hard to dissolve it. It's going to be hard to handle and it's can be solid. If you make it too short, then it may not have the mechanical properties you need for the polymer to have. So there's some optimal choice. So you typically run the conversion until P reaches a number so that this is your optimal value, but then you care about what's the dispersion about that optimal value. And particularly, the unreacted monomers that are left might be a problem because they might leach out over time because they might still be liquids, or even gases that come out. So this famous problem, people made baby bottles and they have some leftover small molecules in the baby bottles. And then they can leach out into the milk, and the mothers don't appreciate that. So there's a lot of real practical problems about how to do this. So anyway, you'd be interested in the width of the distribution. So we define what's called the variance. And the variance of n is written this way. And it's just defined to be the expectation value of n squared minus the expectation value of n squared. These two are always different or almost always different, so it's not 0. So this is equal to the summation of n squared times the probability of n minus-- all right? And a lot of times in the polymer field, what they'll take is they'll take the square root of this and they'll compare sigma n divided by expectation value of n. This is a dimensionless number because sigma n will have the dimensions. Sigma squared is dimensions n squared. This is dimensions of n, so it's a dimensionless number. And that's-- I think they call it dispersity of polymer, something like that. Now, notice that when we use these [INAUDIBLE],, when we wrote it this way, it's implicitly that these things are divided by the summation of the probability of n. But because these probabilities sum to 1, I can just leave it out. But sometimes, it may be difficult for you to figure out exactly what the probabilities are and you'll need a scaling factor to force this thing to be equal to 1. So sometimes, people leave these in the denominator. There's another thing you might care about, which would be like, what's the weight percent of Pn? So what fraction of the weight of the polymer is my particular oligomer, Pn? [INAUDIBLE] sorry, some special one, Pm. And I want to know its weight percent. So that's going to be equal to the weight of Pm in the mix divided by the total weight. So that's equal to the weight of m times the probability of m divided by the total weight, which is going to be the weight of all these guys, times the probability of each of them. And you can see this is different. This is not the same as-- not equal to, right? It's not the same thing. So just watch out when you do this. And in fact, in the polymer world, they always have to say, I did weight average. I did number average, because they're different. Is this OK? Yeah? So I would-- my general confidence, at least for me, if I skip steps, I always get it wrong when I do probability. So don't skip steps. Do it one by one by one, what you really mean. Then you'll be OK. All right, now, this is a cute little example. It's discrete variables. It's easy to count everything. Very often, we care about probability distributions of continuous variables. And we have to do those probability density functions that I talked about last time which have units in them. And so as we mentioned last time, if you want to know the probability that a continuous variable, that x is a member of this interval, the probability this is true is equal to Px of x hat times dx. And so this quantity has units of 1 over x, whatever the units of x are. And then you have to multiply it by x in order to get the units to be dimensionless, which is what the probability is. And this is like obvious things, like P Px of x prime dx prime. [INAUDIBLE] value as possible of x has got to be equal to 1. It's a probability, which is the same as saying that probability of x is some value anywhere is 1. So there's some [INAUDIBLE] you measure. And you can also have a probability that x is less than or equal to x prime. And that's the integral from negative infinity to x prime of Px of x dx. And the mean is just the integral of x Px. And you can compute the x squared. The average of x and x squared, same thing. You average of anything. You can put these together. You can get sigma x squared is equal to x squared minus the average squared. So that's the variance of x. You can also do this with any function. So you can say that the average value of a function is equal to the integral of f of x Px of x dx. This is an average value of a function of a random variable described by probability density function with P of x. And then you can get things like sigma f squared is equal to the integral of f of x, quantity squared, Px of x minus-- all right? Everything's OK? Yeah. All right, so a lot of times, people are going to say, we do sampling from Px. So sampling from Px means that we have some probability distribution function, Px of x, and we want to have one value of x that we draw from that probability distribution. When we say it that way, we mean that we're more likely to find x's where Px has a high value and we're less likely to draw an x value that Px has a low value. So that's what's sampling from. Now, you can do that mathematically using random number generators in MATLAB, for example, and we'll do that sometimes. But you do it all the time when you do experiments. So the experiment has some probability density function that you're going observe something, you're going to measure something. And you don't know what that distribution is, but every time you make a measurement, you're sampling from that distribution. So that's the key conceptual idea is that there is a Px of x out there for our measurement. So you're trying to measure how tall I am. Every time you measure it, you're drawing from a distribution of experimental measurements of Professor Green's height. And there is some Px of x that exists even though you don't know what it is. And each time you make the measurement, you're drawing numbers from that distribution. And if you draw a lot of them, then you can do an average. And it should be an average that's close to this. If you drew an infinite number of values, then you're sampling this. You can make a histogram plot of the heights you measure of me, and it should have some shape that's similar to Px of x. Does that makes sense? All right. So actually, everyday you're drawing from probability distributions. You just didn't know it. It's like [INAUDIBLE] street. The probability the bus is going to hit me or not and the bus driver is going to stop, I think there's a high probability, but I'm always a little worried, actually. Good. I'm drawing from-- it's a particular instance of that probability distribution about whether the bus driver's really going to stop or not. And if I sample enough times, I might be dead. But anyway, all right. Often we have multiple variables. So you can write down-- you can define Px hat. So now I have multiple x's. It's like more than one variable of x. And I wanted the probability density function of them. I'm going to measure this and this and this and this, all right? And this is equal to the probability that x1 is a member of the set, x1, x1 plus dx1, and x2 is a member of x2, x2 plus dx2, and that. That's what probability density function means with multiple variables. So this is very common for us because we often measure and experiment more than one thing, right? So you measure the flow rate and the temperature. You measure the yield and the absorption at some wavelength that corresponds to an impurity. Usually when you experiment, you often measure multiple things. And so you're sampling from multiple observable simultaneously. And implicitly, you're sampling from some complicated PDF like this even though you don't know the shape of the PDF usually to start with. And so then when you have this multiple variable case, you can define a thing called the covariance matrix where the elements of the matrix Cij are equal to xi xj, the mean of that product, minus xi xj. And so you can see that, for example, sigma i squared is equal to Cii, but the diagonal elements are just the variances. But now we have the covariances because we measured, let's say, two things. All right, so suppose we do n measurements and we compute the average of our repeats. So we'd just repeat the same measurements over and over. So suppose you measure my height and my weight. Every time I go to the medical clinic, they always measure my height, my weight, my blood pressure. You've got three numbers. And I could go back in there 47 times, and they'll do it 47 times. And if a different technician measured it using different [INAUDIBLE] and a different scale, I might get a different number. Sometimes, I forget to take my shoes off so I'm a little bit taller than I would have been. So the numbers go up and down. They fluctuate, right? You'd expect that, right? If you looked at my medical chart, it's not the same number every time. But you'd think, if everything's right in the world, that I'm an old guy. I've been going to the medical clinic for a long time. If I look at my chart and average all those numbers, it should be somewhere close to the true value of those numbers. So I should have that the average values experimentally, which I just define to be the averages-- this is the number of experiments. OK, so I can have these averages. And I would expect that as n goes to infinity, I hope that my experimental values go to the same value of x that I would have gotten from the true probability distribution function. If I knew what Px of x is and I evaluated the integral and I got x, I think it should be the same as the experiment as long as I did enough repeats. So this is almost like an article of faith here, yeah? It's what you'd expect. Now, the interesting thing about this-- I mean, probably you've done this a lot. You probably did experiments and you've averaged some things before, right? If everybody in the class tried to measure how tall I was, you guys all wouldn't get the same number. But you'd think that if you took the average of the whole classroom, it might be pretty close to my true height, right? So the key idea here is that the sigma squared of the x measurement experimental, which we define to be this-- maybe we should do this one at a time. [INAUDIBLE] Then I can have a vector of these guys for all the different measurements. So there's some error in my height. There's some error in my weight. There's some different error in my blood pressure measurement, but each should have their own variances. I can have the covariances. OK, so these are all the experimental quantities. You guys maybe even computed all these before in your life. And we expect that this should go like this as n goes to infinity. Now what's going to happen to these guys as n goes to infinity? That's the really important question. So there's an amazing theory called the central limit theorem of statistics. And what this theorem says, that as n gets large and if trials are uncorrelated and the x's aren't correlated, which is the same as saying that Cij is equal to 0 off the diagonal, then the probability of making the measurement x is proportional to the Gaussian, the bell curve. All right? So this is only true as n gets very large. It doesn't specify exactly how large has to be, but it's true for any Px, any distribution function, probability distribution function. So everything becomes a bell curve if you look at the averages. And sigma i squared in that limit goes to 1 over n sigma xi squared experimental. And this is really important. So what this says is that the width of this Gaussian distribution gets narrower and narrower as you increase the number of repeated experiments or increase the number of samples. So this is really saying that the uncertainty in the mean is scaling as 1 over root n where n is the number of samples or number of experiments that's repeated. Now, sigma, the variance, is not like that at all. So this quantity, actually as you increase n, just goes to a constant. It goes to whatever the real variance is, which if you're measuring me, it might how good your ruler or something. It'll tell you roughly what the real variance is. And that number does not go to 0 as the number of repeats happens. I mean, I could get the whole student body to measure how tall I am at MIT, and they're still not going to have 0 variance. It's going to still be some variance, right? So this quantity stays constant as n increases or goes to a constant value once it sort of stabilizes. You have to have enough samples. But this quantity, the uncertainty in the mean value, gets smaller and smaller and smaller as the square of n. Now, this is only true in the limit as n is large. Now, this is a huge problem because experimentalists are lazy, and you don't want to do that many measurements. And it's hard to do a measurement. So for example, the Higgs boson was discovered, what, a year and a half ago, two years ago? And I think altogether, they had like nine observations or something when they reported it, OK? So nine is not infinity. And so they don't have infinitely small error bars on that measurement. And in fact, who knows if it really looks like a Gaussian distribution from such a small sample, but they still reported 90% confidence interval using the Gaussian distribution formula to figure out confidence intervals. So everybody does this. If n is big, it should be right. And you could prove mathematically it's right, but the formula doesn't really tell you how big is big. So this is like a general problem. And it leads to us oftentimes misestimating how accurate our results are because we're going to use formulas that are based on-- assuming that we've averaged enough repeats that we're in this limit where we can use the Gaussian formulas and get this nice limit formula. But in fact, we haven't really reach that because we haven't done enough repeats. So anyway, this is just the way life is. That's the way life is. And I think there's even discussions in statistics journals and stuff about how to make corrections and use slightly better forms that get the fact that your distribution of the mean doesn't narrow down to a beautiful Gaussian so fast. It has some stuff in the tails. People talk about that, like low probability events out in the tails of distributions, stuff like that. So that's a big field of statistics. I don't know too much about it, but it's like-- I mean, it's very practical because-- now unfortunately, oftentimes in chemical engineering, we make so few repeats that we have no chance to figure out what the tails are doing, maybe [INAUDIBLE] our tails. And so this is a big problem for trying to make sure you really have things right. So I would say in general, this is an optimistic estimate of what the uncertainty in the mean is. Uncertainties are usually bigger. So you shouldn't be surprised if your data doesn't match a model brilliantly well as predicted by this formula. Now, if it's off by some orders of magnitude, you might be a little alarmed. And that might be the normal situation, too. But anyway, if it's just off by a little bit, I wouldn't sweat it because you probably haven't done enough repeats to be entitled to such a beautiful result as this. We can write a similar-- actually, so here I assumed that the x's are uncorrelated. That's almost never true. If you actually numerically evaluate the C's, usually they have off-diagonal elements. For example, my weight and my blood pressure are probably correlated. And so you wouldn't expect them to be totally uncorrelated. And so there's another formula like this. It's given in the notes by Joe Scott that includes the covariance. And you just get a different form of what you'd expect, OK? And the covariance should also converge roughly as 1 over n if you have enough samples. So you should eventually get some covariance. You can write very similar formulas like this for functions. So if I have a function f of x and that's really what I care about-- remember, I said that I have the average value of f is equal to f of x Px of x dx. And I could make this vectors if I want. And I could repeat my function, and I'd get some number. And I could repeat the variance. I have a sigma f. And this is something I like to do a lot of times. Then if we do experimental delta f-- so we don't know what the probability distribution function is usually, or often. So we'll try to evaluate this experimentally. This is going to be 1 over n, the values f of x little n, the n-th trial. And we could write a similar thing for sigma f, which I just did right there. You can do the same thing. Just make these experimental values now. The sigma f squared experimental should go to 1 over n times the variance. And this was the sigma in the mean of f, 1 over n times the variance of the sigma. All right, so this is the same beautiful thing, that the uncertainty in the mean value of f narrows with the number of trials. So you have some original variance that you computed here, either experimentally or from the PDF. Experimentally is fine. And then now you want to know the uncertainty in the mean value, and that drops down with the number of trials in the number of things you average. So this all leads in two directions. What we're going to talk about first is about comparing models versus experiments where we're sampling by doing the experiment. So that's one really important direction, maybe the most important one. But it also suggests ways you could do numerical integration. So if I wanted to evaluate an integral that looks like this, f of x P of x dx, and if I had some way to sample from Px, then one way to evaluate this numerical integral would be to-- sorry, I made this vector [INAUDIBLE] a lot of species there, a lot of directions. If I want to evaluate this multiple integral-- it's a lot of integrals for every dimension of x-- that would be very hard to do, right? We talked about in [INAUDIBLE],, if you get more than about three or four of these integral signs, usually you're in big trouble to evaluate the integral. But you can do it by what's called Monte Carlo sampling where you sample from P of x and just evaluate the value of f at some particular x points you pull as a sample and just repeat their average. And the average of those things should converge, according to this formula, as you increase the number of samples. And so that's the whole principle of Monte Carlo methods, and we'll come back to that a little bit later. And you can apply that to a lot of problems. Basically, any problem you have in numerics, you have a choice. You can use deterministic methods or stochastic methods. Deterministic methods, if you can do them, usually are the fastest and more accurate, but stochastic ones are often very easy to program and sometimes are actually the fastest way to do it. In particular, in this kind of case, we have lots of dimensions, many, many x's. It turns out that stochastic ones are pretty good way to do it. But we're going to talk mostly about [INAUDIBLE] data because that's going to be important to all of you in your research. So let's talk about that for a minute. I'll just comment, there's really good notes posted on the [INAUDIBLE] website for all this material, so you should definitely read it. And the textbook has a lot of material. It's maybe not so easy to read as the notes are, but plenty to learn, for sure. So we generally have a situation where we have an experiment. And what do we have in the experiment? We have some knobs. These are things that we can change. So we can change some valve positions. We can change how much electricity goes into our heaters. We can change the setting on our back pressure regulator. We can change the chemicals we pour into the system. So there's a lot of knobs that we control. And I'm going to call the knobs x. And then we have parameters. And these are other things that affect the result of the experiment that we don't have control over. And I'm going to call those theta. So for example, if I do a kinetics experiment, it depends on the rate coefficients. I have no control of the rate coefficients. They're going to [INAUDIBLE] by God, as far as I know. So they're some numbers, but they definitely affect the result. And if the rate coefficient had a different value, I would get a different result in the kinetic experiment. The molecular weight of sulfur, I have no control over that. That's just a parameter. But if I weigh something and it has a certain number of atoms of sulfur, it's going to be a very important parameter in determining the result. So we have these two things. And then we're going to have some measurables, things that we can measure. Let's call them y. And in general, we think that if we set the x value and we know the theta values, we should get some measurable values. And so there's a y that the model says that's a function of the x's and the thetas. Now, I write this as a simple function like this. This might be really complicated. It might have partial differential equations embedded inside it. It might have all kinds of horrible stuff inside it. But you guys already know how to solve all these problems already because you've done it. You've been in this class through seven homeworks already. And so no problem, right? So if I give you something-- I give you some knobs. I give you some parameters-- you can compute it, all right? And so then the question is-- that's what the model says. So we could predict the forward prediction of what the model should say if I knew what the parameter values were, if I knew what the knob values were. And I want to-- oftentimes what I measure, y data, which is a function of the knobs, it's implicitly a function of the parameters. I have no control of them, so I'm not going to even put them in here. So I set the knobs I want. I get some data. I want these two to match each other. I think they should be the same thing if my model is true, yeah? So this is my model, really. But I don't think they should be exactly the same. I mean, just like when you try to measure my height, you don't get exactly the same numbers. So these y data are not going to be exactly the same numbers as my model would say. So now I have to cope with the fact that I have deviations between the data and the model. And how am I going to handle that, all right? And also, we have a set of these guys, typically do some repeats. So we have like several numbers for each setting in the x's, and they don't even agree with each other because they're all different. Every time I repeated the experiment, I got some different result-- that's my y's-- for each x. And then I change the x a few times at different knob settings. Then I make some more measurements. And I have a whole bunch of y values that are all scattered numbers that maybe scatter around this model possibly, if I'm lucky, if the model's right. Often, usually I also don't know if the model's correct. So that's another thing to hold in the back of your mind is like, we're going to this whole comparison assuming the model's correct. And then we might, at the end, decide, hmm, maybe the model's not really right. I may have to go make a new model. So that's just a thing to keep in the back your mind. But we'll be optimistic to start with, and we'll assume that the model is good. And our only challenge is we just don't have the right values of the thetas, maybe, in my model. And this is another thing, too. So the thetas are things like rate coefficients and molecular weights and viscosities and stuff that are like properties of the universe, and they're real numbers, maybe. They're also things like the length of my apparatus and stuff like that. But I don't know those numbers to perfect precision, right? The best number I can find, if I look in the database is, you know, you could find like the speed of light to like 11 significant figures, but I don't know it to the 12th significant figure. So I don't know any of the numbers perfectly. And a lot of numbers I don't even know at all. So like there's some rate coefficients that no one has ever measured or calculated in the history of the world. And my students have to deal with that a lot in the Green group. So a lot of these are quite uncertain. But there are some that are pretty certain. You have quite a big variance, actually, of how certainly you know the parameter values. So one idea, a very popular idea, is to say, you know, I have this deviation between the model and the experiment. So I want to sort of do a minimization by varying, say, parameter values of some measure of the error between the model and the data. Somehow, I want to minimize that. And I have to think about, well, what should I really minimize? And the popular thing to minimize is these guys squared and actually to weight them by some kind of sigma for each one of these guys. So this is-- we should change the notation, make this clearer. These guys-- one model, and it's the i-th measurement that corresponds to that n-th experiment. So I think that the difference between what I measured and what the model calculated should be sort of scaled by the variance, right? So I would expect that this sum has a bunch of numbers that are sort of order of one because I expect the deviation to be approximately scaled of the variance of my measurements. And if these deviations are much larger than the variance, then I think my model's not right and what I'm going to try to do right here is I'm going to try to adjust the thetas, the parameters, to try to force the model to agree better to my experiment. And this form looks a lot like this. Do you see this? You see I have a sum of the deviations between the experiment and a theoretical sort of thing divided by some variance? And so this is the motivation of where this comes from, is that I want to make the probability that I would make this observation experimentally would be maximum if this quantity in the exponent is as small as possible. So I'm going to try to minimize that quantity, and that's exactly what I'm doing over here. Is that all right? OK, so next time when we come back, I'll talk more about how we actually do it. |
MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 26_Partial_Differential_Equations_2.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. WILLIAM GREEN, JR: All right, let's get going. So today, I'll say a few more words about PDEs, and then we'll leave that topic for a while. I think I'll actually come back and have a little lecture about that right before Thanksgiving, for those of you still around. And then we'll start talking about probability, and then that will lead into several lectures about models versus data, which is a very important topic for all of you who either plan to generate data or plan to generate models during your stay here, which probably is everybody. So we'll talk about that for a while. So PDEs. I guess the first comment is, homework 7, have any of you looked at this yet? AUDIENCE: [INTERPOSING VOICES] [LAUGHTER] WILLIAM GREEN, JR: Not a single one. AUDIENCE: [INAUDIBLE] [LAUGHTER] WILLIAM GREEN, JR: So homework 7 is the same problem that Kristin showed in class in the demo for COMSOL, and I want you to solve it both ways. So solve it with COMSOL and then solve it writing your own finite volume code. And I want to warn you. This problem has a characteristic length that's way smaller than the dimensions of the problem. And so in principle, you might need to use an incredibly fine mesh to resolve the gradients in the problem. So just remember, in case you haven't looked at it, the problem is we have a drug patch. It has some concentration of the drug. The drug is diffusing slowly out of the patch into the flow, and we have some flow here like that. And the characteristic length is at the-- this is a velocity boundary layer. So the velocity in the x-direction is equal to y times something, dv dy, I guess. This is some number, and so this has units of-- what is that-- per second, so like a strain rate, OK? And the diffusion here is controlled by diffusivity D, and that has units of, say, centimeter squared per second. And so the D over dvx dy, this thing gives a characteristic length squared, which is sort of the natural length scale of this problem. And the problem is that for the drug molecule, it's a big molecule. It has a very small diffusivity. And so therefore, this is a really tiny ratio, and so the L is very small. And similarly, if you look at it from the point of view in the x-direction, the Peclet number is really gigantic. And so both of those will tell you that you have to watch out, there might be very sharp gradients in the problem. And if you just think of it physically, right over here we think the concentration's 0, somewhere over here. And all of a sudden right here, the concentration is going to be close to the concentration in the patch. So there's almost a discontinuity in the concentration. So there's a really sharp gradient on the upstream edge. And then something funky is going to happen down here at the end of the patch, too. It won't be quite as abrupt, but could be pretty strange. All right, so you got it? OK. And in the problem, we want you to figure out the drug diffusing all the way over to here somewhere, way far over there. And so you may need quite a few mesh points in the y-direction as well. All right. And this kind of problem, this is a very simple case, right? There's no reactions. The velocity's just in one direction, and this is not a very hard case. But you'll see it's actually still pretty tricky to get the right solution. So don't just believe what the code tells you. Just run COMSOL and just-- don't believe it's showing the truth. And don't believe you just write down some finite volumes that you'll get the truth. So mess around with it and try to convince yourself it's really converged, and you really have the real physical solution. Because we expect a sharp gradient, say, in the upstream edge, you might want to play with using a finer mesh in the x-direction here than you would down here, because down here presumably the gradients in x-direction are much smaller. So you don't have to use square finite volumes. You could use rectangles. Yes? AUDIENCE: I don't understand what you have written [INAUDIBLE] as vx equal y dvx dy. WILLIAM GREEN, JR: Yeah. AUDIENCE: So what-- WILLIAM GREEN, JR: So that's because the velocity-- the vx is 0 at the wall, and the velocity increases with y. So vx increases the further you get from y. This is y equals 0, always going up that way. As you can increase the height, the flow gets faster. And that's typical, because you have a no-slip boundary condition at the wall, right? Is that OK, or did I misunderstand your question? Is that good? OK? AUDIENCE: So y is a constant. WILLIAM GREEN, JR: Yeah, in this particular problem this is just a number that we tell you in the problem. AUDIENCE: [INAUDIBLE] WILLIAM GREEN, JR: Right. If you had a flow in a cylindrical pipe, it wouldn't be-- wouldn't be a number. Be more complicated, yeah, all right? So this is-- I mean, again, this is like the simplest you got. It's only 2D. It's really simple. But you're going to see here that even here you could have a lot of trouble. And if you just don't think about it-- the computer will give you an answer. Whatever-- you put in some finite difference equations or finite element equations, finite volume equations, fsolve, whatever, it's going to solve and give you some numbers. It doesn't mean it has any relation to the physical reality. So this is a problem to really pay attention to whether you're really converged. All right. And so what I suggest you do in this problem is go through a sequence where you vary-- the real problem wants you to do this all the way out to-- this is 1 centimeter. But I suggest you instead solve a simpler problem, where you put the-- here's a simpler problem. Put your boundary right here, really close, and then solve that problem. You won't need as many mesh points across to get from here to here. And this should be-- this is-- suppose you choose this is c equal 0 boundary condition for this wall, then this should be an upper bound. Because if you put an absorbing layer here, it should drive the diffusion faster, right? So as you increase this distance-- let's call it little h-- as you increase little h, you should converge to the true solution that you want sort of from above. Does that makes sense? Yeah? Is this OK? All right. And similarly, you can vary the mesh, how big your boxes are, say, in delta x and delta y. And again, for each of those, you could think about how should things converge. And so look and see if it's actually converging the way you think it should be converging, right? Any more questions about this? OK, so this problem might-- this homework problem might look like a MATLAB coding problem. It has MATLAB coding, but it's not really-- that's not really what it is. It's really like a-- it's a conceptual problem about what you're doing. All right. All right, in this problem, I want you to use finite volumes. So let's talk about finite volumes for a minute. So finite volumes is to imagine that we have little control volumes, and each one of them has a little propeller in it, stirs it up, like little CSTRs. And so we do it just like you did back in your intro ChemE class and mass/energy balances a million years ago. You know that you have some flow coming in here, and maybe some flow going out there. And maybe you have some flux come in here and maybe something out there. And you can add up all the flows in and all the flows out, and then that-- the net of all these fluxes has got to be equal to the accumulation. And if we're doing a steady-state problem, then there's no time derivative, so the accumulation term should be 0. And you did a lot of these problems a long time ago, right? OK, so you just do it like that. The only problem is now we have a million of these little boxes all coupled to each other. So you get one equation for each box. And in this problem, we're going to-- when you do this method, what people assume is that you have a uniform concentration sort of across here or the number you use as the average of the concentration in the cell as [INAUDIBLE]. And so it's not really exactly realistic. So that's where the approximation is. What you're doing at the boundaries is very realistic, because you're actually computing the fluxes across control volumes is exactly the way you should do it. So there's different methods, right, finite element, finite difference, finite volume. The nice thing about finite volume is you're really treating exactly what's happening across the boundaries of the mesh area, the mesh volume. And we'll find that a lot of ChemE problems, this is how people prefer to do it. And because you're treating the fluxes exactly, if you have your mesh volume sitting on the wall or on top of the drug patch, suppose if we're sitting right on top of the drug patch here. Then if you're sitting-- first, let's do a wall. Suppose you're sitting on an impermeable wall, then we just know that the flux here will be 0 if it's a wall. So that's easy. No flux here. So we only have to worry about this one, this one, and this one. That one's not doing anything. This case, if you say in the drug patch there is a flux, stuff's coming in and you'll need to figure out how to write that boundary condition. So the boundary condition as written is that C, the drug, is equal to some number, C drug, whatever is in the patch. But that's not so easy to impose here. And so there's two ways to look at it. One way is people compute this flux by considering-- suppose I know this is C drug here. I'm trying to really figure out what's the average concentration here. And so one way to compute this is to say it's the diffusivity times C drug minus C the middle over delta y over 2. That would be the flux. So that's one way to look at it. Another way people look at it is they draw what they call a ghost volume. And so here, here's my C that I care about, C middle. Here's C ghost. And here is the line between them. Now, I can use the same equation here that they would have used before. But I don't know what C ghost is, because there's no real cell here. This is below the patch. Here's the patch. But imagine the patch is not there for a second, and I write down the same equation I would have here. And the flux diffusively would have been D C ghost minus C middle over delta y. That's what you would have written as the diffusive flux. And I don't know what C ghost is. So now, I have to think about how do I estimate what C ghost is. I can say, well, let's do a linear interpolation from this concentration to that concentration. So let's say that the C boundary is equal to C ghost, the average of these two, plus C mid over 2. That's what we got if we did a linear interpolation between these two guys to figure out what it is here. But here we know what it is. We know what a C boundary is. That's the concentration of the drug patch so that we can solve for what c ghost is. AUDIENCE: How do you know what C middle is? WILLIAM GREEN, JR: C middle is the unknown. That's what we're going to compute. We just want an equation that involves C middle, because we have c middle as an unknown. We need an equation the involves C middle, just like we would for-- if we have a-- If we have finite volume that is in the interior somewhere, we have c of this grid point, Cij. We just want equations that involve Cij. But we need an equation that somehow connects it to the boundary conditions. Yeah. AUDIENCE: When would you use that second [INAUDIBLE] if you know C boundary [INAUDIBLE]?? WILLIAM GREEN, JR: Right. And actually, I think if you do it this way, in this case, you get the same formula. You're running out of C. AUDIENCE: So when would you use the C ghost? WILLIAM GREEN, JR: The C ghost thing is handy when these conditions-- actually, in the finite volumes, this is it. I don't think you need to do it. Maybe if you had a flow, velocity flow here. I don't know. But you won't in a wall. So I don't know if, in the finite difference method, you can use those ghost points as well to do the flux gutter conditions. And that's often useful to do the ghost thing for if you have a symmetry-imposed flux boundary condition. Yeah. AUDIENCE: [INAUDIBLE] WILLIAM GREEN, JR: That's the flux. The flux is coming from across this wall. And we'll have to add that with the flux coming this way and the flux coming this way and the flux coming this way to get our total flux, which is going to net out to 0 instead of C. Is that all right? But the other ones, you all know how to write them? They're just velocity flows. The nice thing about the finite volume method is that when you're trying to consider the velocity, the velocity that matters is actually the velocity here. It's the flux that couples the two finite values together. So your velocity is being evaluated halfway between the two mesh point centers. This turns out to make the whole procedure more numerically stable. You compute the flux by, say, the difference between these two. And you're implicitly evaluating right at halfway between. Other methods, like finite difference, you're trying to get the velocity or the flux at the same point here. And it doesn't really make much sense, from the point of view of trying to compute how much material's close to here, to use the velocity right there. And so when you work out equations, and particularly the equations that involve pressure, so suppose you try to discretize the Navier-Stokes equations and you have a pressure gradient that's driving the flow, then you have to figure out where are you going to evaluate the pressures and where are you going to evaluate the velocities. And when you try to evaluate them at the same point, it turns out that you get numerically unstable problems. But if you do it by this funny volume method, then it just works out naturally. It's fine. And then you need special methods, if you're going to try to do it where you have the velocity of the mesh at the same point as the pressure mesh. You'll notice in most of the problems we give you, we just leave pressure out. We try to rewrite the equations so no pressure ever appears. And that's because pressure in general is a problem, because you can have acoustic waves physically. And if you do the equations, which allow for acoustic waves, you'll get them in the numerical solution as well. But usually, we don't care about that. So you have sound waves racing around your solution, and that causes a lot of trouble numerically. So a lot of times, people rewrite the equations to try to remove the pressure from the equation. So you write down the Navier-Stokes equations. You guys did this transport. Maybe you'll study that today in the class, in the test. I don't know. Anyway, there's a pressure term if you just write naturally. But oftentimes, you can remove that by an equation of state, for example. And then you can get rid of it. And that turns out to be better from the numerical solution. All right. Yes. AUDIENCE: I have a question. So you said that for this problem, [INAUDIBLE] was very small. So that means that these finite volume cells [INAUDIBLE] you have to do a lot of them. [INAUDIBLE] gets around this by having a variable volume. Do we have to address this on that lab that have some variable volume? WILLIAM GREEN, JR: OK, so from the point of view of writing a code, it's a lot easier to write it with fixed mesh because all equations look exactly the same. So I suggest you start that way. And what you do, if you use a very small value of each, then I think you'll be able solve it, no problem. You'll have enough mesh. Then, once you figure out how to do that, now you might be able to see, OK, can my solver solve this? And then that gets into another question, actually. What solver are you going to use? So any ideas about this problem? AUDIENCE: [INAUDIBLE] WILLIAM GREEN, JR: So you could use fsolve But I'm telling you, you're going to have to use a lot of mesh points. That means that fsolve's going to have to solve for a lot of variables. AUDIENCE: WILLIAM GREEN, JR: Backslash it might be. So the key thing, the nice thing about this problem is it's a linear differential equation. There's no nonlinear term today. So when you rewrite the equations of the finite volumes, it's all going to be linear in the unknowns. And what else is nice about this problem, when you use local finite volumes as your-- It's going to be super-sparse. So the matrix that comes in is going to be really sparse. And so you'll want to use some method that can handle gigantic sparse matrices. So you wrote a code like that earlier, so that would be one possibility. If you know how to use the right flags in MATLAB to help their built-in solvers handle sparsity, then that should be good. If you just ask it to solve it by dense method, by, like, LEU or something, you're going to have lot trouble, once you've put your mesh points in there. But you can just experiment. Try bigger and bigger matrices. And then at some point, if you have backslash, it will give you a warning or something, unless you tell it that you're sparse. All right? Any more questions that about this? How would people solve it do you think professionally? Suppose I was doing a problem like this in 3D instead of 2D. How would people solve it? What solver would they use? AUDIENCE: [INAUDIBLE] WILLIAM GREEN, JR: Not fsolve. No. AUDIENCE: [INAUDIBLE] gradient. WILLIAM GREEN, JR: Yeah, so they would use conjugate gradient. So probably-- I think it's called this, BiCGSTAB. That's the program that we would use if you have a really, really gigantic sparse matrix. So that's the conjugate gradient. And Professor Swan talked about, the advantage of that is you never have to actually store the whole matrix. You only need to evaluate the matrix elements, and then you can throw them away. And so if you have a very sparse matrix, that's pretty cheap to do. So it's a really good code. I think in this 2D problem, you can probably get away with other solvers. You don't have to use this. But this is a definite possibility. This is a built-in MATLAB program as well. Be warned, though, this is an iterative solver. It's not just going to be one solve, boom. And It might have troubles. So you might want to go with the other ones, but anyway. May I ask you a question? How about, do you need initial guess? What do you think? AUDIENCE: Depends. WILLIAM GREEN, JR: Depends on the solver. So if you solve it with backslash, do you need an initial guess? If you can solve it with fsolve, do you have to initial guess? If you can solve it via BiCGSTAB, do you have to use initial guess? AUDIENCE: Yes. WILLIAM GREEN, JR: Yes. OK. So then you have to think about how you really get your initial guess, too. So this is things to think about. All right. What else to tell you about? One last thing about PDEs, and we'll come back to this later, so far we haven't done really anything that's very time-dependent. But a lot of real world PDEs have a time dependent in them. And there's is a very important concept, a thing called the CFL number. And this is named after a 1928 paper, and I'll write the guys' names down. And what they showed was that, if you're trying to solve the PDE system, where you're discretizing in both x and time, that you have a number that they defined as delta t times the velocity and the extraction divided by delta x. So that's a dimensionless number. So that's a CFL number. You see a lot of papers that will say what CFL number they used. What that means is the ratio of their time mesh compared to their space mesh. And conceptually, let's think about what's happening here. So suppose we have a flow flowing in an upwards direction, and we have a bunch of little finite volumes. So we've discretized the delta x already. And this is x. And there's a flow here. And I've already decided, somehow, what length scale I want to use. So I've decided my delta x. And now, I'm trying to figure out what delta t I should use. Now, from the point of view of saving CPU time I want the delta to be as giant as possible, because I want to be able to zoom along or predict for long periods of time what's going to happen in my system. But if I make delta t really large, let's think about what happens. Suppose I choose delta t to be 10 times delta x divided by u. So it means that in my one time step, I have some guess or some current value of the concentrations in all these finite values. And then I wait through a time step that's 10 times delta x over ux. So I had some stuff that was here. Where is that going to be 10 times later? 10 time steps later. 10 blocks up, right? So it's going to be, like, way up here somewhere. And so what's going to happen there is that my numerical methods are all computing stuff locally from the spatial derivatives. But it's crazy if, between my time steps, this stuff completely left the picture. It's already convected all the way off the screen. And some new stuff, which was way down here before, is now the stuff that's here. Should be there if I was physical. Numerically, who knows what will happen if you try to do this. But it won't be good. So the condition is that you need this number to be less than 1, or same order of magnitude as 1, and you try to make this much bigger than 1, then you're doing something crazy because you're convecting stuff over multiple mesh points. And so that turns out to be a very serious limitation if you try to do simulations for, let's say, a reacting flow for a long period of time, because you might have to use a really tiny delta t. And then people have developed all different fancy methods to try to get around that. But if you just do the obvious things to do, you'll always run into this limitation. Then you need to choose the time steps small. And also, it's bad, because as you make delta x smaller, which improves your accuracy, You'll have to make your delta t's smaller, too. But of course, making delta x smaller increases your CPU time because you have more finite volumes to compute. And then you'll also have to make double t smaller, which means you'll have to do more time steps, too. So it's even like a double whammy. So getting more accuracy is going to really cost you badly. And so this another reason why people used to always refer to color fluid dynamics. You can make a pretty picture, but it might not be physical at all. Because you can make it solve equations that maybe, for example, didn't impose this, then who knows what kind of crazy stuff you'll get. You'll got something. It'll compute something, but it may have no relation to the real problem. I think that's all I was say about PDEs. Are there any questions about PDEs before I start talking about probability? You got it totally down. I'm looking forward to some really awesome solutions. How about that? Just one last comment. If you decide you wanted to do adaptive meshing and you want to change your mesh size, you can choose measures like this if you want. And you can even do things like this, where you have a bigger mesh, and then maybe have two smaller meshes underneath it in the next trial. So you can just have stuff flowing here, stuff flowing here stuff flowing there. So you can do all kinds of crazy stuff like. This can really help improve the accuracy of the solution a lot. But it's, I would say, very prone to bugs. So if you do this, be really careful and don't do it too often, I would say. You might have a few boundaries where you do something funky like that to change that mesh size. But don't go crazy with it. [INAUDIBLE] is smart. It has a really nice way of doing the meshing for you. So that's its advantage, that somebody very carefully coded how to handle this kind of stuff in a general case. But if you guys are doing it for the first time, it might not be so good. So that's enough of PDEs. Let's talk about probability. So probability is everywhere, except in undergraduate ChemE homework problems. So when you do problems as an undergraduate, they always had some nice solution. It was 2 pi, it was 3.0. Everything was, like, deterministic, is definite. The grader could go through. Oh, no, you're off by 0.1. It couldn't possibly be right. That's, like, not reality. So any time you actually make a measurement, you always had measurement noise. And if you try to repeat a measurement, you don't get the same result. So that's, like, completely different than an undergraduate problem. But this is the reality. So the reality is the world is, like, not so nice as you think. But actually, it's even more fundamental. It's nothing about-- I mean, there's one problem about how good an experimental those people are, and how fancy an apparatus you bought that can make things exactly reproducible. But even if you do that perfectly, the equations we use really don't correspond to the real physical reality. So we always use the continuum equations. You guys probably are studying them a lot in 1040 and 1050, especially 1050, I guess. But those equations are really all derived from averages over ensembles or little finite volumes or something if you look at the derivation of the equations. And reality is that the world is full of molecules. And so they're all wiggling around. And if you look in a little finite volume, you look right now, you'll see that there's 27 molecules in there. If you look a second later, there might be 28. Then a little later, maybe 26. It's always fluctuating around. But according to our average equations that we use, like in the Navier-Stokes equations, it always 27.3. But of course, there's not 0.3 of a molecule. So, I mean, explicitly, it's the average is what we're computing, and the reality is fluctuating around the edges. Same thing in thermo. We say that such-and-such has a certain amount of energy. But you guys have some [INAUDIBLE] already, yes? So you saw that's not true? So really, all that's saying is that's, like, the probability, the average. If you've had many, many ensembles that were exactly the same and you average them all, you get some number and that's your average energy. But for any actual realization, it has some different value of the energy. And it's even worse than that. That's because we have a lot of particles. You can even go down to, like, the microscopic level, where you have one molecule. And you calculate things about that, it turns out you have to use the Schrodinger equation for that. And the Schrodinger equation explicitly only gives you probability densities. So it just tells you the probability the molecule might be somewhere, the electron might be somewhere, the energy might be something. But it's not actually whether it really is. It's just saying a probability distribution. And every time you do the experiment on the molecule, you get a different result. Now, this is super-annoying, but it's the way life is. Einstein got so annoyed, he has a famous saying, and it's "God does not play dice." He was, like, just completely annoyed at these equations. But it's the way it is. So the reality is that things, all we know about, really, are probability distributions. In most of our work, in our lives, we always talk about, like, the mean or the median. And we're talking as if it's a real number. But really, it's always some distribution. So it's time, I guess, you're in graduate school, it's time to, like, face up to this. And that's what we're going to talk about for the next week or two. AUDIENCE: [INAUDIBLE] WILLIAM GREEN, JR: This also makes it makes you wonder what you're doing when you make a measurement. So if you make a measurement, first of all, the fact that when you measure something repeatedly, you're not going to get the same number, that's alarming. Because I want to say I'm 5 foot 9", this should be pretty common, I'm always 5 foot 9". But actually, if you measure me multiple times, sometimes you'll get 5' 9 and 1/4" Some people will get 5' 8 and 3/4" So it might make you worry, did I change? Did I grow between the measurements? So that's one issue. Because our experiments are not repeatable and we get different numbers every time we make a measurement, then we have a big problem. Somebody says, well, I really want to know how tall Professor Green is. I've always wondered, how tall is he? And everybody's told me a different number. And when you go measure it again, you get a different number again. What's going on? And so you'll have to then-- then we have, like, a concept that there is a true height of Professor Green and we just don't know what it is. And then we'll try maybe to make repeated measurements of my height, and then maybe take the average. That would be the obvious thing to do. And we take the average and report that. We'll tell the boss, we'll lie to him, say, oh, Professor Green's 5 foot 9". When really, we never actually measured 5' foot 9". Every time we measured, it was 5' 9 and 1/8", 5' 8 and 3/4". Every time, it was something different. But we just say, OK, it's 5' 9". And the boss, he doesn't want to know about all this complexity, anyway, so he just believes you. So it takes your average number. But you know that you're not really sure I'm exactly 5' 9" And so you have a probability distribution, and you're doing your best guess. And in fact, if you're honest to your boss, you'll give him an error bar. You'll say he's 5' 9", plus or minus 1/2 an inch. And that way, what you're saying is you're pretty sure the true height of Professor Green is somewhere in that range. Is this OK? Yeah. Now it might be that I'm actually changing height. I get a good night's rest, I lie down for a long time. Maybe when I stretch out a little bit. When I stand up in front of lecture here for a long time, I'm shrinking. My vertebrae are being compressed by standing here so long. So it's, like, a combination of things. One is your measurement system is not perfect, and one is that I actually might be fluctuating. I had a big breakfast, I'm growing. So is it true with everything? Every experiment you do is like this, that there is a real fluctuation of the real, physical apparatus of the thing that you're trying to measure because mostly, things we're trying to measure have fluctuations intrinsically. And then on top of that, your measurement device is fluctuating, which means-- and then the combination is what you're measuring. It gives you fluctuation. If you have a very nice instrument, the fluctuation of your instrument is smaller than the fluctuation in the real system and you'll go down to the limit. Anybody with a really good instrument should measure approximately the same probability distribution, which is actually the real fluctuation. My height, for example. So if you bought a laser interferometer and mounted a mirror on my head, and measured my height to the wavelength of light, you're pretty sure that it's pretty good. It's within a wavelength of light. So the error bar there is just due to the fact that I slouch sometimes. Is that OK? So let's talk about some basic things about probability. So we're always saying there's a probability of an event. And so we want to give a number. So for example, I'm flipping coins. I flip a penny, it could be heads or tails. I'll say the probability of heads is approximately equal to 1/2. So I flipped the coin, and if you flip the coin 100 times, you might expect to see 50 heads. Now, it could be 49 heads, it could be 51 heads. So you have to worry about, like, exactly how precisely you know it. But you think it's something like a half. Now, just to warn you, I actually didn't specify any more significant figures here, and it might be really hard to figure out what those significant figures are. And this is related to the measurement problem. So we think that a coin has about a 50/50 chance of being heads or tails. But if you really wanted to prove it, that might be really hard to do. You can have joint probabilities. So suppose I have a penny and I have a dime. And I try to think, if I flipped them both, what could happen? I could have that they both come up heads. I could have this guy come up heads, this guy tails. This guy tails, this heads, tails. So there's, like, four possible outcomes, and we think they all have about approximately equal probabilities. So the probability of any of these things happening should be about 1/4. So you can write the probability of event 1 and event 2. So this would be, for example, the probability that I got heads for penny and heads for dime, and I think that this is about 1/4. Now, I could also say the probability of heads, and heads is equal to the probability of heads for the penny times the probability of heads for the dime if I got heads for the penny. Now, if these two coin flips are completely uncorrelated, then the probability of heads on the dime and heads on the penny is the same. It's just the probability of the heads on the dime. So they don't matter. But many things that we'll study are correlated. That the probability of something happens depends on whether something else happened. So this kind of expression is very important. Now, this is just an equality. It's like a definition of what these are, right? And I'll just notice you can write that the other way around. So it's the probability of heads on the dime times the probability of heads on the penny. This is OK? So these two guys are equal to each other, and you can rearrange that equation any way you want. And we'll come back to a very famous way to rewrite that equation. It's called Bayes' theorem, and that turns out to be really important in model versus data comparisons. Instead of doing AND, do OR. So maybe you can say, then, what's the probability that I see at least one head? So I flip my two coins, and I have the probability of, I see at least one head. So we know we intuitively the answer is 3/4. But let's try to think of where does that really come from. So-- [HIGH-PITCHED SOUND] What is that? Sorry. Really threw me there. So probability of at least one head, it's not equal to the probability of head for the penny plus the probability of head for the dime, because we know this is really 3/4, and this is 1/2, this is 1/2. You add them up, 1/2 plus 1/2 does not equal to 3/4. So be careful. There's a lot of things you can say quickly that are not true. So anyway, you really have to consider the whole thing. And in the best case, if you can enumerate what's going to happen, it's very simple to add up things. Otherwise, you have to be very careful with the algebra to make sure you add all things correctly. Let's see at least what this should be equal to. So this should be equal to the probability that a head on a penny times the probability that I have head for the dime if I had a head for the penny plus probability of the tail for the dime or for the penny plus the probability that I have for the dime and then times something like this, too. And in a case like this, where I'm summing over all of the possibilities. So in this case, I'm either going to get a head for the dime or a tail for the dime. It's not going to balance on its end. I'm assuming that that chance is 0. Then these two things just add up to 1. Is that all right? It's really saying the probability of the dime will do something if I have a head for the penny. Is this all right? So you want to practice doing little algebra things like this to make sure you know how to do it. Yes? No? Maybe? Let's see what else I've got here. AUDIENCE: Professor. WILLIAM GREEN, JR: Yes AUDIENCE: What you just wrote, [INAUDIBLE] WILLIAM GREEN, JR: Yeah, what's the correct thing to write in here? What is the right thing? AUDIENCE: Tails. WILLIAM GREEN, JR: Yeah. So it's tricky. Yeah. AUDIENCE: The probability [INAUDIBLE] WILLIAM GREEN, JR: Yeah, it's the easy way to do it, for this case. And so there's, like, a lot of-- for some cases, it might be easier to write it this way. In some cases, write it that way. It depends on how many different options there are. So anyway, I'm just trying to warn you by writing this out. It's like, you might write down stuff quickly without thinking about it. And it's easy to double count. Like for example, if you put the other term in here, you double count. Because you already have, they had head, you already had the head case here. Yeah, maybe-- Did I write the-- I'll just write the base theorem down the way you usually see it. So this is the general expression, which is always true. The base theorem way to write it is the probability of A given B is equal to the probability of B given A times the probability of A over the probability of B. And that's just rearranging this equality. Rearrange again. And we'll come back to this with the situation that's like, what's the probability that Professor Green is 5' 9", given our measurements, is related to the probability that if Professor Green was 5' 9", we would have made the measurements we got. So this is like that way to invert that statement. And if the thing that's here is exclusive, it means there's, like, many possible things that could happen. This is the probability that one of them happened. There's a lot of other things that could have happened. For instance, like heads and tails, it's either heads or it's tails. It's exclusive. If it's like that, you could rewrite that probability of B is equal to probability of Aj, probability of B given Aj summed over j. This is something, whatever A measurement it is, if it ends up with B, that's probability B. Is that all right? So you can put that into the denominator here. You can substitute that in so you can rewrite this as probability of Ai given B is equal to the probability of B given Ai times the probability of Ai divided by the sum of the probability of Aj [INAUDIBLE] Aj. And this is the form that you'll normally see based here. A lot of times, we have a continuous variable instead of discrete events like this. And so then we talk about probability distributions. And so instead of having a sum there, we might have an integral. And this is a topic that also is quite confusing to many people. So suppose I had a Maxwell-Boltzmann distribution. And I have, like, the probability density of having a certain velocity in the extraction instead of the particle. And so that's going to be something like e to the negative 1/2 mvx squared over keT divided by something. And maybe it actually is [INAUDIBLE].. I'm not sure. Maybe there's a vx here. Something like that. So you have an expression that you get for [INAUDIBLE] for probability density. Now, what does this mean? We want this thing to be the integral of P of vx dvx over from negative infinity to infinity. We want this to equal something. What do we want this to equal? 1. So that means that the units of this, this is units of centimeters per second, what's the units of this? AUDIENCE: [INAUDIBLE] WILLIAM GREEN, JR: Centimeters per second now minus 1. It's, like, it's per the unit to this to get a dimensionless number there. So this has units of seconds per centimeter, which probably none of you thought until I just said that to you. So probability densities are tricky, and they always have to be multiplied by a delta. When I talk about this, I really need to talk about P of vx, delta vx. I need to have something here to make this look like a probability again. And so the issue is that the probability that the velocity is exactly something is, like, 0. It's really the probability that's in a certain range, plus or minus something. Then you get a nonzero probability. There's another quantity you'll see a lot called the cumulative probability distribution. Let's see, what letter do I use? Call it F. And this would be like the integral from negative infinity to vx prime, or vx of P over vx prime, dvx prime. And this is the probability that the particle has vx less than something. So this is F is equal to the probability that vx-- let's call this vx star-- vx is less than or equal to vx star. And that can be quite an important property. So for example, you're designing a supersonic nozzle. You want to know what gas molecules are going to come out at a certain speed. You really need to know that probability. How many of them are going to be bigger than that speed, how many less than that speed? So these are two different ways to express a similar thing. This is, like, the probably that it does have that speed within a certain range. This is the probability that it has anything less than or equal to that speed. And in a completely different way, this is an integral. This has units of dimensionless. This has units of per velocity. All right? All right, we'll pick up more of this on Monday. |
MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 28_Models_vs_Data_1.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. WILLIAM GREEN: So today we're going to talk about Bayesian prior estimation and prior estimation in general. So the last time we were writing down the expressions for the probability of observing a mean measurement if you know what the model is. So let's try to do that again. So suppose I have a model that predicts some observable and it depends on some knobs, and it depends on some parameters. And suppose because I have great powers of faith that I believe this model is 100% correct with every core of my being. And also because I have tremendous confidence in all the people who built my apparatus, and the knobs that I turn actually correspond to the real values, and I have tremendous confidence in all the literature that reports the parameter values. And so I'm absolutely certain that this is the truth. So we'll start from a position of absolute certainty, and then we'll degrade into doubt as the collector goes on. So let's start from the position of someone who has absolute faith that this is the truth. Is true. So I have a model, I really believe this model. So for example, I believe that the kilogram weight in the SI Institute in Paris weighs exactly one kilogram. I believe that with every core of my being. I'm completely confident that model is correct. So there are some things I'm really confident on. That's one. And maybe guys have some things you really believe, too. So let's go with things we really believe. So I plan to conduct some experiments that measure this observable and are related to this model. And so I'm going to do 10 repeats of measuring y. So I'm going to get to the kilogram blob that's in Paris, and I'm going to stick it on my really expensive scale that I really believe is great, and I'm going to measure its weight. And then I'm going to put it back, and then put it back, and put it back, and put it back. And I'm going to get another really great scale that I really believe is great, and I'm going to measure it there, too. So I've got a lot of repeats of measuring the weight of this kilogram, and I believe it's really a kilogram. But the stupid measurements don't say a kilogram. They say, you know, 1.0003, 0.99995, all kinds of numbers not equal to one kilogram. So now I'm going to try to figure out what the probability is that it would have measured some particular value y. So what is the probability that my experimental is between some value, say, y and y plus dy. So that's a question for you. So what's the probability? Sorry, what? [INAUDIBLE] OK, so we think that the probability that y is in this interval given that the model is true. And I know the theta values perfectly, and I know the x values perfectly, is equal to some integral of what? The bounds integral, probably y to y plus dy, believe that? What's the integrand? Sorry, what? AUDIENCE: The probability [INAUDIBLE] AC function of y. WILLIAM GREEN: Right, so what is it? AUDIENCE: [INAUDIBLE] You wrote it down last time, I think. So this [INAUDIBLE] is large? Standard normal, right? So it should be one over sigma root of 2 pi. Does that sound OK? I mean here. It's probably the same, it's fine. Yep. AUDIENCE: What does that notation mean, if your model is true? WILLIAM GREEN: So this means, given that the model is true, and I know these data values are exactly certain numbers, and the x values are actually certain numbers, what's the probability that I would make a measurement whose average would fall in this interval? So this line means given that this is true, what's the probability of that? OK, is this right? Is this surprising? This is OK? So this is what I mean. So we say that our probability distribution converges to a Gaussian distribution, this is what we expect. So we expect it to have been large enough for this to be true. Yeah? This is very important. This is like the whole course, actually. This is the whole section, is this one equation. So I just wanted to make sure you really get what this says. And if you don't like the integral, you can make dy really small, and then it's just this times dy. OK? Actually, this notation's like [INAUDIBLE] I think I should do this way. I should do this. This is a number. Let's get rid of the integral. Let's make dy really small. I'll make it [INAUDIBLE]. That all right? So this is the probability density that we would observe, this is the experimental value y that we observe from the mean, and this is the little with of our tiny little interval. Is that all right? Yes? AUDIENCE: So is sigma the [INAUDIBLE] on there? WILLIAM GREEN: Ah, what is sigma? That's a great question. We didn't write down what sigma was. What is sigma? AUDIENCE: Standard deviation? WILLIAM GREEN: It's not the standard deviation exactly. Standard deviation of the mean, right? So there's two sigmas. We have the sigma of y, of the measurements, and that's equal to average value of y squared minus. So we just figure that for how many experiments we do, we just compute the average of y squared, the average y, subtract them. That's the variance. And then sigma that I used in that equation there is 1 over n times sigma y. And we call this the variation of the mean, it's the uncertainty in the mean value of y. And the central limits theorem said that as long as n gets really large, we expect that this should converge to this. And we talked last time about how when n get bigger, these averages don't really change when it gets big. They're just the average. But this number declined as n gets big because of this one over n formula. And to understand that, suppose I measure the weight, and I measure, it should be around one kilogram. But in fact my measurements are all over here. Lots of measurements. So they have a variance something like this. But if I make a plot of-- as I run, I compute the running average. So when I run the first two points, I get some average value here. After I run 27 points more, the average value is here. After I run 1,000 repeats, the average value is here. It's getting pretty close to this, and the uncertainty in this number's getting smaller and smaller as I'm doing better and better averages. The average more and more repeats. Does that make sense? OK. So from this key equation, I can derive a lot of things. And it depends what you want to do. So one thing people do a lot is it was called model validation. And what does this mean? It means I have a model, I believe it's true. I have some parameters, I believe they're true. But there are some foolish skeptics out there who don't have the faith that I do. And they think that my model's baloney, or my parameter values are wrong, or something. And so to prove I'm right, I'm going to make some experiments. And I'm going to show that I make a plot that looks like the experiment and model agree. Some of you might have done this in your life, yes? Everybody might make a parity plot or something. You've seen these things before. Now, this is like a confidence builder. You're trying to get the skeptics out there to believe that there's some evidence to back up your faith that this model is perfect. And what you really want to know is like, if the measurement that I measure, the average for my 10,000 repeated measurements, I expect that this quantity should be pretty big somehow in some way. By then quantitatively saying what that means exactly what's a good fit, what's a bad fit, this is actually kind of a difficult question, and we'll come back to this one. But that's a very common use of this equation is to try to do validation. Now because it's kind of complicated, most people don't actually do it. So instead what they do is they just plot some data points, and they plot your model curve. And as long as they look good, then you're done. So that's the normal way that it's done in the literature currently. But of course, that's completely unquantitative. It doesn't really say whether the model and the data really agree, it just means they look sort of like each other. So that's like a human qualitative thing. Now, if the purpose of validation is just to convince humans, then you've done the purpose. Now, if your purpose is to try to quantitatively say something, then you really have to get into this equation, which usually is not done but would be the right thing to do for validation. Now the alternative view is disproving a model. But I just say that there's several ways this can happen. You can try to disprove a model, but you might also show that the theta values are incorrect. Or you might show that the experiment is wrong. These are all possibilities, reasons why the model and the data might not agree with each other. So this equation, it only holds if the model is really true, if the parameter values are all perfectly correct, if we know exactly what all the knob values are perfectly. If any of those things are not true, then you should have some discrepancy, and there should be a way to show it. And really what you're showing is that you'd observed some y that is very unlikely to be observed. So probably observing that y is very extremely unlikely if all these other things were true. So if all these things are true, and you compute this value, and this value's very tiny, then it makes you think that it's unlikely that you would have observed that. And therefore, you might try to use that as an argument to say that something must be wrong. The model's wrong, parameters are wrong, the knobs are wrong, something's wrong. My y values are wrong. It could be any of those things. So this is often the most exciting papers to publish. You publish a paper, you take some model that a lot of people believe. You tell them they're full of baloney, it's completely wrong. My great experiment shows you are completely wrong. And so you'll see a lot of these in Nature. I should warn you, a lot of those get retracted later, a very high retraction rate in Nature. Because they want to publish papers like that that show that the common view is incorrect, and sometimes it's true. But oftentimes the common view is actually correct, and there's something wrong with the experiment, or the interpretation, or how they computed this equation, or whatever. And so actually it turns out the common view is perfectly fine, and it's just that the foolish authors went off on a tangent. And then they have to six months later publish a retraction, by the way, sorry, paper was completely wrong. And so you see a lot of that. So that's a second kind of thing. And we'll talk more about that a little bit later, too. And then another thing is I'll relax my assumptions. So I'll say, well, I'm sure that the model is true, and I'm sure that my knob settings are perfect, and I know what they are. But I'm not really sure about all the parameters. And therefore I want to use the experiment to try to refine parameter values. So I'm trying to take my y's that I measure and somehow infer something about the thetas. And this is a very common thing to do. So in my group we've tried to measure the rate coefficient for a reaction. We believe there is value of that theta, and in fact, we probably have an estimate of what it is. But we're not sure of the exact number, and we'd like to do an experiment to refine the number and get it more accurately determined. So that's another useful thing to do. And this leads into two somewhat different points of view about this. One you've probably done already called least squares fitting. That's one view. And the other is this Bayesian view that I'll tell you about next. So there's sort of A and B. There's one that I'll call Bayesian, and one I'll call least squares. They're sort of related to each other, but not exactly the same conceptually. So I'll try to explain that. So the Bayesian view is probabilistic, so it's actually pretty straightforward to write down. Remember that we wrote that the probability of A and B is equal to the probability of A times the probability of B given A, and it's also equal to the probability of B times the probability of A given B. And what we have here is one of these conditional probabilities, if the thetas have a certain value, this is a certain probability. So I should be able to use that formula somehow. So I can write down that the probability of measuring y given theta is equal to the probability of y times the probability of theta given y divided by the probability of theta. So I just took this formula, and I plugged in y's and thetas instead of A's and B's. So I said, these two are equal to each other. Rearranged it so then I can rewrite this. This is the way we have it in here, probably of measuring y given theta. Let's flip it around. So probability of theta given that we measured y is equal to the probability of theta times the probability of observing y if theta was true divided by the probability of y. Terrible handwriting there. That's just algebra. So this is what we want to know. We want to know, what's the probability distribution of the parameter values y? Because some of them are uncertain. Now, before we started the experiment, we had some idea of what the ranges were for all the primary values. Like I'm trying to measure a rate coefficient. I know from experience with other similar reactions, from a quantum chemistry calculation, from some indirect evidence, from some other more complicated experiment, I have some idea that this rate coefficient has to be in a certain range. Now, it could be pretty uncertain. It might be five orders of magnitude uncertain. But I know it's not less than 0. I know it can't be faster than the diffusion limit, how fast things can come together. So for sure I know some range, and oftentimes I know a much narrower range than that. So I have some information about these parameter values before I even start. Some of the parameters of the model I know perfectly, or pretty well. So you know maybe there's a Planck's constant, or the heat of formation of one of my chemicals or something like that shows up in the numbers, and I might know that parameter really pretty accurately. Whereas the particular rate coefficient I care about is the thing I really don't know very well. So some of these have tight probability distributions ahead of time, and some of them have loose ones. And this thing has a name. It's called the prior. And it's our prior information before we did the experiment. And this one, after we've done the experiment, we're going to change it. So we're going to say previously people thought that the parameters all lied in these certain ranges. And now I'm going to get a tighter range, because I have some additional experimental information. So this is called the posterior. This means before, it means after. So this is what I know about parameter values before and after the experiment. This is the formula that I have over there. It's a probability that if the thetas had a certain value, probably would have observed what I saw. Yeah? AUDIENCE: Which one refers to which? WILLIAM GREEN: Sorry, this is the prior, this is the posterior. And those of you who are paying attention to notation realize I'm not doing this very nicely. Because these are continuous variables, and I'm writing capital PRs, and they should be not be capital PRs, it should be probably density functions instead. So let's rewrite it nicely. So the probability of theta given y, probability density is equal to the probability density of theta initially times the probability of y given theta, [INAUDIBLE] density divided by the probability density of-- all right? And what I just basically did was this is the correct equation the previous other one was all multiplied by d theta dy, it shouldn't be done that way. So this is OK? Now this is the prior information I have about the parameter values. I know that they have to fall into some ranges. And really all I'm doing is I'm correcting that information. I'm improving the information to tighten the distribution. So initially I know that my rate constant, here's my rate constant. I know that it's got to be greater than zero, and I don't think it's really down there at zero anyway. I think it's somewhere in here. I really don't know much. And I really don't think it's all up at the diffusion limit, and no way it's higher than the diffusion limit. So that's my initial information that I have about the probability distribution of K. So it's the rate coefficient I want to know, and I know it's bigger than zero, and I know it's less than infinity. And actually I know there's some physical limit, it can't be higher than something or other. And you can do this for any problem, right? I give you any parameter, you should be able to tell me something about it. You might be uncertain by 20 orders of magnitude, but at least you have an error bar some width. It can't be anything, right? A lot of parameters have to be positive, for example. You know that. And you usually know something. you might not think you know anything, but you do, you actually do know before you start. So you actually know, this is the P of theta to start with. And after I've done the experiment, hopefully you're going to know more information about it. I might know that this quantity here is going to be like a Gaussian distribution. It might have a kind of goofball dependence on theta. I should comment that. Notice how theta appears inside F. So theta's up in the exponent. It's sort of inside a Gaussian, but it's like processed by F and so the observable might have a pretty goofball dependence on this rate coefficient. So this thing could be some weird thing. But for sure, when I change theta so this changes a lot, it's going to make a pretty big difference. Because up inside the exponent of a Gaussian, so it's going to drop off a lot somewhere. So I should get something that looks something like this maybe for my experiment. So this one is P of K initially, the prior. This one is P of yk. And what this equation says is I want to multiply those two together. And so I'm going to multiply this times this, and I'm going to get some new thing that's something like that when I multiply this time that. Is that OK? And so that's my new numerator of this equation. Now this denominator doesn't make too much sense. This says, what's the probability that I measured the mean I measured, given nothing? So this is sort of like the prior probability that I would have measured it or something. I don't know what this is. So instead what people do, is they say, forget this. But instead, let's multiply this by a constant that's going to normalize it to make it probability density so that it integrates to one. So that's the way Bayes' theorem is used. This is called Bayesian analysis. And so what it's telling you is how to take your experimental information as expressed in this formula and use all your previous information about the parameters, put them all together, now we have a cumulative information about everything. So we have some parameters that came into our problem into my experiment, but from previous work, I also knew something about those parameters. Now I put it all together and I get a new value of probability distribution of those parameters. And if my expert was really good, it would make this really tight [WHOOSHING SOUND].. And then when I multiply these two together, it's going to make this really sharp, and we have a really good value of k. So that's like the ideal case if I have a really great, well-designed experiment executed perfectly with great precision, then I can do this. More generally, when I don't think about it, I get some distribution like this. I still learn something compared to what I had before, but it might not be much. So now I can end up with some distribution that's a little tighter than before. So is this OK so far? All right, now this is super simple. I didn't have to solve anything, all I had to do was multiply two distributions together. So in some respects, this is what you should always do. All you do is you take your experiment, you multiply the probability distribution corresponds to your experiment times the prior, and you get some posterior, and that's why new information about the distribution. And if I have a distribution like this, suppose this is my new distribution here, I can still get it central value, that's my mean value, k. I can get an estimate of the range of k. So I end up with a k plus or minus dk maybe, from just looking at the plot. In fact, I never even have to evaluate what this constant is in order to do this. I can just go look at the plot, see where the peak is, figure out the width, and I can report now because in my experiment, k plus or minus dk is more precisely determined than it was before. Now, a practical challenge with this is that theta is usually a lot of parameters. And I only drew the plot here in one dimension, but really it's a multi-dimensional plot. So really what looked like, suppose I had two parameters. I had my k I care about, and I have some other parameter, theta 2, that also it shows up in my model. And say, before I started, I knew theta 2 fell in this kind of range, and I knew k fell in this kind of range. So really before I started, if I think what it looks like, I really had sort of a blobby rectangular contour plot, where I think it's more likely that the k value and the theta 2 value are somewhere in this range. And the most likely one is maybe somewhere in the middle there. But I really didn't know much. So it could be anywhere in this whole blob. Now, when I do the experiment, the experimental value depends of both k and theta 2. And commonly what'll happen is that the distribution from the experiment-- need color chalk here. Let's get rid of these guys. So this is my probably distribution, there's my prior. If I do the experiment, maybe I'll have something like this. That the experiment says that the guys have to be somewhere in the contour plot like this. Because I can get pretty good fits of the data with different values of k as long as I compensate with the value theta 2. Now I multiply these two dimensional functions. The original is a blob function, and this is a stretched out blob. And I multiply a stretched out blob times a fat blob, I get some stretched out blob that looks something like the intersection of these guys. And so I end up with some kind of blob like that. I'll draw it really thick. So this is my posterior, some kind of blob like this. So now I know a little bit more about these two parameters than I did before I started because of my experiment. Is this OK? I really can't say I know what the real value of k is, or the value of theta 2. But I know that combinations of k and theta 2 that are sort of in this range, all of them will give me pretty good fits to my data, and also be consistent with all the previous information I have about those parameter values. Is that all right? Now, I drew it with two parameters. In a lot of models we have, we have five parameters, six parameters, seven parameters, nine parameters, 14 parameters. We have a lot of parameters. And so then we try to make this plot, even how to display the plot is going to be a little problematic. But it's there, right? And somehow, we still narrowed down the hypervolume in the parameter space from whatever it was to begin with to now we know something a little bit better. We have a narrower range of the parameters that would be consistent with all the information available, including my new experiment. And then the next guy does his experiment, and he does an experiment that shows that these guys have to be somewhere in this range in order to be consistent with his experiment. And so now I can narrow down the range to be something like that. And the next person does their experiment, and they get something else, and something else, and something else. And eventually by 2050, we have a pretty nice determination of the parameter values. So that's the advance of science, as drawn in chalk by Professor Green at the board. So this is a very important way to think about it, is what you're doing when you do experiments, is you're generally restricting the range of parameter space that's still consistent with everything. And when we mean consistent, we mean that the probability that you would have observed what you did observe is reasonably high. We'll still have to come back to quantitatively figure out what reasonably high means. Now, when you did this before when you were kids, nobody mentioned the word Bayes, or Bayesian, or conditional probabilities, right? So they just said, oh, just do a least squares fit. How many of you did that before? So somebody told me before, forget this stuff, we're going to never even mention this stuff. We're just going to do a least squares fit. Now, where did the least squares fit idea came from? It came from looking at this formula and saying, you know, this is the deviations between the experiment and the model prediction, and I weigh them somehow, and I have the square. And that's the thing I want to make small. If I have a high probability that what I observed really happened, or the probably I'm going to observe this, it's got to be that these guys have to be reasonably close to each other. They're really different. And it's going to be very small, because it's inside an exponential. And if those guys are really different, and the squared thing is really large, then the probability is incredibly small that I would have observed that. So we think that this thing should be small. And in fact, if I want to get the very best fit I can get, which means like the probability was the highest of what I observed in the real observation or something, then if I'm free to adjust one of these thetas, I can adjust the theta, try to make this thing like equal to zero, or small as I can. So that's where the concept of least squares comes from. Now, when you're doing least squares, you almost always have multiple parameters, and therefore you're going to have to have multiple data. And they can't just be a repeat of one number. Can't be your data, it's not sufficient to determine the parameters. So normally when you do an experiment, you have to change the knobs. We have to make measurements in a couple different conditions. Like for example, kinetics. You often want the Arrhenius A factor and the EA. And so I got to run the experiment in more than one temperature or I'm never going to be able to figure that out. So I have to change the temperature in my reactor. Make some measurements at one temperature, and make some measurements at a different temperature. And for almost everything in life that you want to measure, you're going to have to do this. You vary the concentration of your enzyme if you want to see how the enzyme kinetics depends on something. you can't just keep running exactly the same condition over and over. You'll get that number really precise, but it's not enough information to really fill out the parameters in your model. So you're going to have to run several different experiments with different knob settings. Also, normally we don't just measure one quantity, one observable in each experiment. We usually try to measure as many things as we can. So we actually have several observables at each knob setting, and we have several knob settings, so we have quite a lot of data. And each one of those is repeated a whole bunch of times so that we're confident that we can use this Gaussian formula. And so what we really have is the pi-th observable measured at the l-th knob position. Well, I'm sorry, l's not good either, it's used in your notes for something else. M, there you go. The m-th knob position. Now, normally you have several knobs, so that's a factor. And we have a lot of observables we can make at each position. So this thing is a measurement. And we repeated this multiple times so I can get the average. And we're also going to have a corresponding sigma I M, which is the variance of the mean. So it's variance, that's going to be divided by the square root of the number of repeats for that particular experiment and that particular observable. So this is your incoming data set, and you also have your model which predicts y model, it predicts the observable i for the sequel to fi of xk theta. So if you have certain knob settings, like certain temperature, and you have your parameter values, then you can calculate what the model thinks should be the observable value, and then you can actually measure it and measure its variance. so that's the normal situation. And now you want to figure out, are there some values of the theta that make the model and the data agree? And that's the least squares fitting thing. So what we can define as a new quantity, weight of the residual vector EJ, which is defined to be jk, to be consistent with Joe Scott's notes. AUDIENCE: Is k the same as m? WILLIAM GREEN: M is in oppositions, I'll tell you what k is in a second. m, sorry. Too many indices. OK, so this is the residual between the model position. And now-- oh man, I'm sorry, [INAUDIBLE].. K is an index over i and m. So k is just going to list all the data you got. Some of the data came from the same knob settings, some came from different knob settings. Yeah? AUDIENCE: So is x the m the y model i [INAUDIBLE]?? WILLIAM GREEN: Thank you, yes. y model i, I guess this is now k. And so k is one of these indices that carry-- you can bind two indices together and put them together just like you did in your PD problems. All right. Now, I wrote down this sigma. But actually if you're measuring multiple things at the same experiment, you should expect them to be correlated. So really what we should worry about is the c, the covariance matrix, that we defined last time. So you should also compute that thing. And so what you should expect is the probability density that we would measure any particular residuals if the model is true. And if we have these certain parameters, theta, this should be equal to 2 pi negative k over 2 the determinant of c negative 1.2 exponential of negative 1/2, epsilon transpose c inverse epsilon. So this is the multi measurement version of the same equation here. So this is the quantity that we think should be small if we have good parameter values and we did a good experiment. Actually, even when we did bad experiments, still should be small if we have good parameter values. And that's because the c's, if we did a bad experiment, we'll have a high variance or something, then we should see the c's will give us weightings that will reflect that. Yeah? AUDIENCE: [INAUDIBLE] WILLIAM GREEN: Is that-- AUDIENCE: So you have the next [INAUDIBLE] K [INAUDIBLE].. WILLIAM GREEN: Oh I'm sorry, this is the capital K, this is the number of data points. So little k is equal to 1 to capital K. AUDIENCE: So does capital K count for both experiments? WILLIAM GREEN: It's the number of distinct data values after you've already averaged over repeats. So you do m experiments, at each experiment you measure capital I observables. So it's like m times I, so K. If you measured everything in every experiment, it's equal to I times m. Now there's two ways that people approach this in literature. The fancy way is you say, you know, this covariance matrix comes in in a pretty important way into this probability distribution function. And so maybe I need to worry a lot about whether I really know the covariance matrix. And my uncertainty in the mean drops pretty fast as I do averaging, but I'm not so confident that my answer in the covariance matrix was small. So what people do sometimes is they'll try to vary both c and theta, and try to get a best fit where they're varying c. But then they have additional constraints on c that c has to satisfy the equations you gave last time about how you calculate the covariance matrix from data. And so I was saying, well, I want this c to personify these equations pretty well, but true covariance of the world of the system is not the same as what I actually measure by just measuring, say, five repeats of an experiment. And so I might want to vary the c. You try to vary the c, turns out to be kind of complicated math, so not many people do it. Even though conceptually it makes some sense, you should worry about the fact you're not really sure about the covariance. So what a lot of people do is they say, let's just use the c that's computed from the formulas I gave you last time experimentally. So just say, let's just take c experimental, put them in here. And now this is a constant. And now the only thing that varies in this problem is thetas which come into the epsilons. Because the epsilons depend on theta. And so in that case, I can just try to maximize this probability. And what that happens to do is to minimize this quantity in the exponent. And so all I need to do is say, for example, theta best is equal to arg [INAUDIBLE] theta epsilon of theta c epsilon. And so this is the least squares fitting problem that you guys have probably done before. And probably what you did was you assumed I had perfectly uncorrelated data, and all my errors were the same. And so c, which is the identity matrix, and I took it out. Probably did that before? Yeah, OK. That's pretty dangerous to do, I'd say. What people do a lot, which is a little bit less dangerous, is at least say, well, you know, when I measure the concentration of species x by GC, I have an error bar of plus or minus 5%. And when I measure the temperature with my thermocouple, I have an error bar of plus or minus 2 degrees. And so the variances of these guys should be a lot different, temperature and GC signal. And therefore I definitely need to weight my deviation somehow. And it's really what you do is you keep the diagonal entries of this. That's often done. And we just forget the fact that they might be covariant. But if you've done the experiments, you actually do have enough information to compute this thing anyway, so you might as well just use the experimental value. So this is the least squares thing. And let's think, what the heck is this doing? We're saying, all of a sudden we grabbed all the parameters in the model, which might include things like the molecular weight of hydrogen or something. And we can find the very best values that would make our data match the experiment as best as possible. And in some sense, that's great, we know the best values and parameters for our experiment. But of course, if we vary the molecular weight of hydrogen, it's going to screw up somebody else's experiment. Because somebody else did some other experiment that depended on the molecular weight of hydrogen, and they had to get some other value to match their experiment. So in these parameter set, anything I do to vary those parameters, I got to watch out that maybe some of those parameters are involved with somebody else's model and [INAUDIBLE] some other experiments. And I'm not really free to vary them all freely. So this is the idea from the Bayesian of having the prior intimation is so you know some of the ranges on these thetas already, and some of them you might know really sharp distributions, like the molecular weight of hydrogen. You might know that to a lot of decimal places. And so when people do this, normally you don't vary all of the thetas. Usually what you do is you select a set of thetas that you feel free to vary because they're so uncertain, and other thetas that you think, oh, I better not touch them. Because if I adjust them, I may go to crazy values that are inconsistent with somebody else's experiment. So a lot of times like the molecular weights, you would not touch them. You would just say, I got to just stick to the recommended values and the tables. I'm not free to vary the molecular weight of hydrogen, even though if I did it would make my data match my experiment better. Makes my model and the experiment match more precisely. So deciding which parameters to vary in this is a really crucial thing. And that's a lot of the art of doing this has to do with that issue. Also, you don't have to keep the thetas in the form you have them. You could do a transformation. So you could change to, say, W's that's equal to, say, some matrix times the thetas, and I could express the equation in terms of the W's. So I could transform my original representational parameters as some other parameters. And often times, your experiment might be really good at determining some of these W's, even if it might be incapable of determining any of the thetas. So you often might know some linear combination of parameters, or maybe not linear combinations, some non-linear combination of parameters might actually be determinable very well from your experiment, even though you can't determine things separately. And this gets into the idea of dimensionless numbers. So your experiment might depend on some dimensionless number very sensitively. And you can be quite confident from your external data what the value of that dimensionless number must be. But if you look inside the dimensionless number, it depends on a lot of different things. And you might not have any information about them separately. All you know is about your experiment just tells you the value of that one parameter very accurately. So this is another big part of the art of doing the model versus data is setting up your model in terms of parameters that you really can determine, and getting out all the ones you can't determine and fixing them. So we're really going to generally change do this kind of thing. But we're going to say that some thetas are fixed, and also we might change to a different representation, change to W's instead. Yeah? AUDIENCE: Can you explain where this transform-- I don't really know what's up with-- WILLIAM GREEN: Yeah, let's get an example. Suppose I was doing a reactor that had A equilibrium of B. And I was really interested in kf, the forward rate for A going to B. I'm a kineticist, I love to know A goes to B. However, if I setup the experiment wrong, it might be that this readction ran all the way to equilibrium. And what I see in the products is actually just the equilibrium ratio of A to B. So what I'm measuring might be something that's dependent really on kf over kr, and that might be the quantity I can really determine. Because that's equilibrium constant. If I didn't think about it, I could just try to have the model fitting procedure, just optimize to find the very best value of kf. And in that situation, it might have a lot of trouble, because it might be quite indeterminant what the kf is, because really all that matters is the ratio. Also I think about this some more, suppose I run at short times, and I measure the time dependence. What I'm really measuring is kf plus kr. Do you remember we did the analysis of A goes to B, one of the early homework problems? The time cost was actually kf plus kr, not kf separately. And so if I measure the exponential decay time constant, I'm really determining kf plus kr, I might be able to determine that very well. Actually, in my lab, I can do a great job for this. I have an instrument that can measure the time constant of the exponential of k really precisely, but it's determining the sum. It's not determining either one of them separately. And I might have to do a separate experiment, say a thermo-experiment to get the ratio. And then from the two I can put them together and get the two values distinctly. So this will be an example of this would be a W. My W is kf plus kr, the matrix would be 1001, something like that. 1-1, something like that. Where I add these two guys, kf, kr. These are my two parameters, 1 plus 1. And I can determine W1 now very accurately. sorry, this is m, this is W. So now in terms of W, this has two parameters now, W1 and W2. I can't determine W2 from my experiment, but I can determine W1 really well. So then when I do the least squares fitting, I should vary W1. I can fix it it for my experimental data, and just leave W2 fixed at some value. I can't do anything about it. That all right? Now, do you get the difference in these two points of view? This is like, two completely different ways to look at the problem. You can think about it as, these parameters are free for me to vary, and I just have to be careful to select the ones I'm really free to vary. And that's the least squares fitting point of view. Or I could say, I'm not really determining anything in particular, all I'm doing is taking the whole range of uncertainty that we have about parameters, and by my experiment, I narrowed it down in the Bayesian view. So it's the two different points of view. To do this one, I need to make sure I have enough data to determine something. So I have to have enough determine some parameter, at least one, otherwise there's no point in doing this. This one I can do even if I can't determine anything, because I could still narrow down the range of parameters. But this might be harder to report in a table. Because all I have at the end is a new probably density function of multiple parameters. All right? OK, we're done. See you guys on Friday. |
MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 25_Review_Session.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JAMES SWAN: OK, shall we begin? So on the syllabus, today's class is supposed to be a review for the next quiz. And I'm happy to answer any questions you have about the material or about the quiz to the best of my ability. And when we've exhausted those questions, I have some notes on another method for solving partial differential equations. We didn't really cover it in class, but I think it's quite interesting and applicable for chemical engineers in particular because of the classes of problems that we study. And then notes on how to construct a simulation of a PDE that follow from what Professor Green discussed during your last PDE lecture. And we'll do an example where we actually develop a code to solve the PDE and maybe refine it so that it runs more quickly by noticing certain things about the structure of the problem. OK? I have your gift cards. I remembered them. So I'll hand those out at the end the class, just in case more people trickle in. But it doesn't look like there's 53 people here, so maybe the TAs are going to be eating burritos for a while. There's going to be a lot left over. We'll see. I'll hold onto them. Try to make sure everyone gets their reward for responding to the class survey. So let's start with questions about the quiz, about the material that we've covered up to this point. I'm really going to leave it to you to point out things that you feel like you need help with, you think need to be reviewed. I know you've had a review session with the TAs already. Maybe that's been sufficient. OK? Questions. It's up to you. Yeah. AUDIENCE: Will the test contain chemical engineering material like last year's test, or will it be like Professor Green's test. JAMES SWAN: Oh, good question. So I'll tell you that I believe the problems on the quiz this year are motivated by chemical engineering problems, OK? The reason for that is, unlike linear algebra problems or optimization problems, those can be crafted out of whole cloth. But workable, right, boundary value problems, workable initial value problems really need to stem from something physical. You are not asked to derive anything from your extensive chemical engineering knowledge. This is not a qualifier for reactors or thermo or transport. We provide you with those details. OK? But they are-- I think they are of a chemical engineering origin. So you have equations that are reasonable to work with. OK? More questions. Yes. AUDIENCE: I would love to briefly go over some boundary value problems, specifically like the collocation and-- JAMES SWAN: OK. AUDIENCE: --Galerkin methods, maybe. JAMES SWAN: Sure. So OK. So let's just discuss boundary value problems first. Turns out PDEs are a class of boundary value problem, too, largely, at least the ones that chemical engineers are interested in. So if we understand boundary value problems in general, we should understand how to solve partial differential equations too. Many classes of partial differential equations that we're interested in can also be solved in these ways. So we talked about relaxation methods in general, which were these methods that rely on somehow discretizing the continuous function that we're trying to describe in this boundary value problem. So usually, you've got some-- linear function, some linear operator that operates on some variable that we're trying to understand in a boundary value problem. Maybe it's temperature, right, equals 0. So it's going to be something like thermal diffusion. This equation could represent that. Could represent some advection diffusion problem in some bounded domain. Could be a PDE. It could just be an ordinary differential equation, right, but a boundary value problem. And with this class of relaxation methods, we try not to exactly satisfy this equation. We relax that assumption. We try to somehow discretize this function. It could be temperature, could be energy, could be velocity or momentum. It could be lots of different things. Try to relax the constraint that this is satisfied everywhere. And we try to satisfy in some approximate sense instead. So we discussed collocation methods, where we take the domain we're interested in and we divide it up into a series of points. And we try to satisfy this equation at those points. And we have some approximation for the function, its representation at those points. That approximation could be a finite difference approximation. That approximation could be a basis set expansion. It could be a finite volume approximation. We could expand in terms of local basis functions. That's the finite element method. We covered all these different methods. That involves chopping the domain up into a series of points and trying to satisfy the function at each of those points. There was an alternative method, this sort of Galerkin method, where instead of trying to satisfy the function at each of those points, we have some basis set expansion. So we have some approximate representation of our function in terms of these basis sets-- in terms of this basis set. And instead, we try to make it be the case that the integral of our operator applied to our approximation-- I'm going to break chalk-- multiplied by each of the basis functions and then integrated over this entire domain is equal to 0. So we try to satisfy this equation with respect to each of these basis functions, essentially, right? There's some-- this isn't going to exactly satisfy the equation. This is an approximation. It's not going to exactly satisfy the boundary value problem we were after. But the projection of the residual, or how far off we are with this operator, on each of the basis functions, we're going to try to force to be 0. So rather than trying to get a pointwise solution of the problem, we're going to try to sort of smooth out the error with respect to the basis functions and make this equation is satisfied as it possibly can with respect to these basis functions. And that's the Galerkin view. So it's different. It's like you're trying to minimize sort of a global error with respect to the basis functions, as opposed to a pointwise error, OK? And again, these basis functions can be chosen in a lot of different ways. They can be chosen as sort of global functions. They can be things like sines and cosines. We may know that the solution to this equation looks a lot like sines and cosines, so that makes sense. Or we may choose local basis functions instead, little tent functions in 1D. They're actually tent functions in many dimensions as well. You can imagine sort of extending that to look like a tent in many dimensions. That's the finite element method. So we can choose those basis functions however we want. And the goal is to just minimize this residual with respect to each of the basis functions that we include in our approximate solution. And we would call that a reasonable approximation. Presumably, as the number of basis functions goes to infinity, this is going to converge. That's the applied mathematician's job, to show us that it's going to converge to the true solution of the equations. OK? So those are the methods that we-- those are the methods that we apply. Good choices of basis function are important. So there was this discussion about quantum chemistry, where you're trying to solve the Schrodinger equation for n electrons, there's 3n degrees of freedom. So a very high-dimensional space. How are you going to do that? You do it with basis set expansions. And the choice of those basis functions is really important. So if you read a quantum chemistry paper, you'll see they give a long series of digits and numbers that are meant to indicate which basis set expansion they used. And for certain classes of problem, some are better than others. And they make the problem actually solvable when it's not, if you were to just choose your basis functions without care for the physics of the problem. Sam. AUDIENCE: The last homework, in the Galerkin method, where you multiplied by the complex, how did you get one basis set, basis functions? Where does that fit into this? JAMES SWAN: Well, it should have been right here. So technically, the basis functions don't have to be real-valued functions. They can be complex-valued functions. And then this is, in some sense, a dot product, an inner product in the space of these basic functions. And the proper form for an inner product when we have complex-valued things is to take the real part of one and multiply it with the complex part of another. So that should have been right here. This should be the complex conjugate. OK? Yeah. AUDIENCE: So when we do the Galerkin method, we necessarily have to approximate these derivatives and things using basis functions. JAMES SWAN: Yes. AUDIENCE: But in the collocation method, there was lots of other ways that we could do it, like find differences. I know you mentioned a few of them earlier. JAMES SWAN: Yes. Yes. AUDIENCE: OK. JAMES SWAN: That's right. AUDIENCE: That [? LOT term ?] kind of contains also the basis function in the Galerkin method, or in the collocation method. JAMES SWAN: Well, in the-- if we have an expansion of the sort of objective that we're trying to calculate in terms of those basis functions, L is the real L, right? It is what it is. This is the model equation to be solved. We operate on our approximate solution with the real L. In the collocation method, it may be the real L, or it may be some approximation for L. If we do finite differences, we're approximating all these derivatives. If we do a basis set expansion and we try to satisfy that at a certain number of points, we're not approximating L. We're approximating T instead. So they're a little bit different. More questions? So that's BVPs. This picture applies for PDEs, too. So this is the integral over the domain of interest, the domain that's bounded by our boundary conditions. These basis functions don't have to be in one dimension. They can be in any number of dimensions we want. We just want them to be good basis functions in principle. So for finite elements, you use little tents. They're multi-dimensional tents, but they're still tents. The same as these hat functions that Professor Green talked about in 1D. Just do them in many dimensions. They work exactly the same way. They're easy to take the derivatives of. These intervals are, for the most part, simple to calculate. And that's why COMSOL, for example, uses that methodology. It's based on this sort of Galerkin picture. So PDEs, BVPs look very similar. Usually, when we're talking about spatial dimensions, they look very similar. When we go to the time domain, things get a little more confusing. That mixes together what we know from ODE IVPs with what we know from boundary value problems now. So you talked about the method of lines last time as one way of studying, for example, parabolic partial differential equations. You have the spatial discretization like this. And then in time, you have some ODE IVP technique that you integrate forward. You rely on these robust ODE methods that know how to control the numerical error, can detect, maybe, when there's some sort of an instability and try to damp it out by altering the time step so that things stay stable. Yes? AUDIENCE: I can't visualize the local basis. Can you give me an [INAUDIBLE] expression for the local basis? JAMES SWAN: Sure. So first of all, there are a lot of local basis functions we could write down, OK? AUDIENCE: This is just one of them. JAMES SWAN: There are many of them. So yes, let's do the hat. OK? Let's suppose I have this line in x, and I target in my collocation method a series of points, column x i minus 1, xi, x i plus 1, and they extend well beyond here. It doesn't matter where they are. I choose where they are based on the physics of the problem. I'll put them where I want them to go. And in fact, you might even try to optimize where they go. So you might start with some distribution of them, and then move them around in some way to improve the quality of your solution, right? Maybe you distribute them uniformly at first, and then you notice, oh, my solution is changing very quickly over here. I would like more points in this region. In MATLAB, there's a function called bvp4c, which is a boundary value problem solver. It solves 1D boundary value problems. And it does precisely this. It uses finite differences and collocation. And the good thing about it is it's trying to adaptively change where these points are positioned. So it solves it, it looks where things are changing quickly, and tries to shift its points around and resolve it in order to improve the quality of the solution, hit certain targets for the error. So what are these-- what would these local basis functions look like? So we talked about these hats. They're 1 here. They're going to be 0 on either side of this point. We could call this our hat i, which is centered at point xi. And it's going to be a piecewise sort of function. It's going to be an increasing linear function between x i minus 1 and xi. So for x between xi and x i minus 1, it'll have one form. Then it's going to switch. It's going to be a decreasing function on the other side. So between xi and x i plus 1, it'll have a different form. So here, it's going to increase between 0 and 1. So it's going to have to look something like x minus x i minus 1 divided by xi minus x i minus 1. So it goes from 0 to 1. And then on the other side, it's going to have to have the opposite sign. So this is going to have to look like-- let's see, x minus x i plus 1 divided by xi minus x i plus 1. Will that do it? I think so. OK? You want me to make it look like this one. So we could have plus 1 xi, and then put a minus sign in front of it if you want. You decide how you want to write it. But it describes this line here and this line here. But there are other choices. So it doesn't have to be piecewise linear necessarily. A lot of times, people choose different types of splines, different types of interpolating functions. They might not be localized to one point. They might be localized over a few points. Maybe something about the physics of a particular problem that tell you, you can't get away with these piecewise linear functions. They're not good enough to describe, say, higher-order derivatives in your equation. So you might instead interpolate with, I don't know, like a quadratic basis function that spans a few more points. That'd be fine. It would be easy to do the integration in your Galerkin method. You just do these integrals out. They're complicated, but you do them once for a generic point xi, and you know their value for every point in the domain. So you choose them. You choose the basis set. But that'll give me an equation for the tent function centered at xi. That's the tent function centered at any point in the domain, in principle. In many dimensions, right-- so if I have two dimensions, you see in COMSOL, you have a sort of a triangular mesh. And what does my tent function do? It sort of stands up, stands up above the triangle. It's a real tent now. It's coming out of the board. Can you see it? Put on your 3D glasses. You'll see it. But it's coming out of the board. It has height 1 above this triangular domain. It sits at the center of the triangle, for example. It's one way to draw these elements. There are actually many ways to draw these elements. This is just one way of drawing these elements. OK? Good. Other questions? AUDIENCE: Could you please explain why central differencing is second-order accurate and not-- JAMES SWAN: Oh, good question. Good question. So why is central differencing second-order accurate? So the central difference formula says that the derivative of a function x could be written as-- let's see, f at x i plus 1 minus f at x i minus 1, divided by 2 delta x. Or let's say these aren't equally spaced points. Well, let's make these equally spaced points for now. Let's make it easy. So let's say x i plus delta x and x i minus-- it'll make our lives easier if we do it this way. It can be generalized, but let's do it the easiest possible way. And this is the derivative at the point xi. Graphically, here's x, here's xi. Here's my function. I want to know what the slope here is. And I'm going to approximate it by looking one point upstream and one point downstream. And I'm going to connect those points with a line. You would agree that as the spacing gets smaller and smaller, it will be a good approximation for what the derivative is. So why is it second-order accurate? So you can show that. It's not too hard. You can show that using Taylor series expansion. Like everything else in this course, the Taylor series is really a key. So let's look at it. So f at x i plus delta x. I want to write this as a Taylor series expansion about the point xi. So tell me the first term. f of xi. OK, tell me the second term. AUDIENCE: f prime at xi. JAMES SWAN: f prime at xi. Times? AUDIENCE: Delta x. JAMES SWAN: Delta x. Tell me the third term. AUDIENCE: 1/2 f double prime. JAMES SWAN: 1/2 f double prime at? AUDIENCE: Xi. JAMES SWAN: Times? AUDIENCE: Delta x squared. JAMES SWAN: How big is the third term? AUDIENCE: [INAUDIBLE]. JAMES SWAN: OK. Is this always true, by the way? This? Not always, right? There are lots of functions, non-analytic functions that don't have these sorts of expansions. So we're sort of-- we've got to have a function that has these derivatives. They need to exist at the point xi. Let's assume that's true, OK? Good. OK, now I'm going to do something really simple. Let's write f at x i minus delta x. And let's just point out the differences between these two formulas. So this is the same. This switches sign because delta x switched signs. It was plus delta x. Now it's minus delta x. This is the same, delta x squared, same as minus delta x squared. Doesn't matter. This one down here-- well, OK. It's order delta x cubed. It'll actually look the same but have the opposite sign. OK? We can't resolve it, though. That's sort of the tricky part, right? We don't know. We don't know for sure what value that takes on. This just gives us a bound for how big the error is in this expansion. So let's now compute what's in that formula. So if you take the difference between this and this, these two will cancel. These two will combine. I'll get 2 f prime xi times delta x. These two will cancel. So I'll get 0. These two, I can't rely on them necessarily canceling each other out. In fact, they want to combine in principle, right? We've just bounded the error. Here, we don't really know what it is, but it gives me something that's order delta x cubed in size. So this is the-- this is 1 minus 2. It's 2 f prime xi times delta x plus the error, which is order delta x cubed. If I divide by 2 delta x-- I know you guys hate it when the instructor does this, but you know, you can't help it. Divide by delta x, the error becomes order delta x squared. So if the derivative is this thing plus some order delta x squared piece, it's second-order accurate, we would say, right? The error is going to be reduced like the square of the spacing. Now you can imagine all the finite difference formulas out there in the world are constructed in this way. You use Taylor series expansion. You've got as many terms as you need to go for the level of accuracy you're trying to achieve for different points based in different ways. And you try to dream up some linear combination of these things so that you get a certain level of accuracy. They're also not unique. I can construct these in a lot of different ways and have the same sort of error level out here, but different coefficients associated with the error, but the same overall scaling. And then you might choose different approximations for other reasons. Sam. AUDIENCE: What would stop me from going on further and then just subtracting f prime from both sides? I mean, it wouldn't really tell me anything, except maybe the order of [INAUDIBLE] approximately 0. JAMES SWAN: Well, I mean, yeah. So let's say you go out further. What's the next term here? It's going to be 1 over 6 f triple prime xi delta x cubed. And here, you'll have minus that because I cubed the delta x. So when I combine them, I'll actually get 1 over 3 f triple prime. So I could do a little better. If I knew what my function was, I can tell you exactly how big the error was. I can tell you that it was as big as the third derivative of f. My function doesn't have a third derivative at that point. The error might be huge. And it's possible to construct functions that lack that particular property, OK? So you can go further, and you can use it to sort of quantify exactly how many you think the error might be in the approximation. But it's all constructed from linear combinations of different Taylor expansions about nearby points. And the related problem, actually, the equivalent problem is polynomial interpolation. So the sort of-- if you can somehow, in your mind, transpose this problem, it looks a lot like polynomial interpolation. So these two are intimately related to each other. OK? Yes. AUDIENCE: How much do we have to know about polynomial interpolation? JAMES SWAN: How much do you have to know about polynomial interpolation? AUDIENCE: Yeah, like, do we have to be able to compute the primes and polynomials? JAMES SWAN: Say that again. AUDIENCE: Do we have to be able to compute the primes of polynomials? [INAUDIBLE] JAMES SWAN: What do I want to say about that? I wouldn't expect that level of detail. I think it's quite difficult to construct Lagrange polynomial expansions by hand. I don't know anyone who does that professionally or for pleasure. I think that's a bad idea. You should understand polynomial interpolation. You should understand the context in which it's important. What is polynomial interpolation important for, numerically? Do you recall? Hersch. AUDIENCE: Numerical integration. JAMES SWAN: Numerical integration, right. So this was really the key to numerical integration, was trying to construct appropriate approximations of our function from a finite set of data and figure out the area under the curve. So you should understand how one uses polynomial interpolation to do numerical integration. But I wouldn't expect us to ask you to actually compute a polynomial interpolant and numerically integrate it by hand. I think that's too much. OK? AUDIENCE: OK. JAMES SWAN: OK. More questions. Hersch. AUDIENCE: Now do you mind showing how the trapezoid rule is over delta t squared accurate? JAMES SWAN: Sure. AUDIENCE: For ODE IVPs. JAMES SWAN: For ODE-- AUDIENCE: IVPs. JAMES SWAN: ODE IVP, OK. OK, so we have to have an ODE we want to solve. Can I just do-- can I do a one-dimensional unknown? Do I need to do a vector here? Can I just-- would you be satisfied if I wrote this as f of y and t and that it's sufficiently general? That way, you don't have to blame me when I forget to underscore things along the way. And we don't have to think about the Jacobian or anything like that. OK. So oh, time derivative, right? So there's the ODE we're going to solve. The trapezoid rule. OK, so let's do this the following way. It's actually going to wind up looking very similar to what we just did, which is useful. I might even be able to pull that board down and reuse it at a certain point. So that'll be good. Let's do this. So what is-- ultimately, what is ODE IVP solving like? What are you trying to do when you solve one of these things? AUDIENCE: Extrapolate. JAMES SWAN: It's extrapolation, right? I know the value of y initially. I'm trying to figure out where the hell this thing is going. I've got to extrapolate for it. And you should be very uncomfortable with the notion that I can do that, right? How good a job can I do predicting the future knowing how things are now? It's really quite complicated. In fact, it's so complicated, there are certain circumstances in which it's not really possible. Discovered here at MIT, actually. So there was weather research going on at a certain point at MIT. It's a very famous case. A researcher, a very famous researcher was doing numerical simulations of a weather model in an ODE IVP. He ran it for a while, and he stopped. He looked at his end state, the last value of y in his simulation. He put in three digits, used that as the initial condition, and started his simulation again, and it went wacky. Didn't look anything like the initial solution. OK? How did that happen? It seemed fine up to that point. Solution's cranking along. He stopped. He took three digits accuracy out of the computer, and he started again. And it, I don't know, did something else. So there are problems that are inherently chaotic and very sensitive to their initial conditions. They're hard to resolve. So the entire point of doing this sort of analysis of ODE IVPs is, at least in the best-case scenarios, we want to try to manage the error associated with these kinds of extrapolation. In some case, it's really not very possible. You don't know the initial conditions well enough to predict what the final state is. So there are problems that are inherently unstable or sensitive to their initial conditions. They don't have to blow up either. They can find themselves moving through the phase space in some way that's very, very sensitive. And two points that start very close together can wind up very far apart from each other. So that's deterministic chaos. But outside of those problems, we try to do this extrapolation and figure out sort of a best approximation for where we think our target function is going. So what are we trying to do? In principle, we know y at a certain point. And we want to know where it's going to go. And so you might say, well, maybe I know dy dt initially. And so I should just move a small increment along that slope. And that's a good approximation. That's the forward Euler approximation. Trapezoid rule-- or equivalently, the midpoint rule-- says, well, maybe that initial slope isn't really the right one to use. Maybe y does something like this, and we should actually look sort of halfway between the point we're trying to extrapolate to and the current point, and use that value as the best approximation of the derivative and extrapolate for it along that. And it turns out that's more accurate. So what does that look like? So one typically says, you know, y i plus 1 is equal to yi plus delta t times f y i plus 1/2 comma. And we won't worry about the t here, but it's going to have to be t i plus 1/2. So I look halfway forward in time. And I say, that's the slope, that's the derivative of y. So let me step forward along that line. It turns out this is going to be reasonably accurate if I can get a good approximation for y i plus 1/2. OK? OK. So let's-- [COUGH]. Excuse me. Yeah. AUDIENCE: Is this the midpoint method or-- JAMES SWAN: They're functionally equivalent to each other. They work in precisely the same fashion. So whether it's the trapezoidal rule or the midpoint method, they have fundamentally the same pictures there. OK? Here, if you do forward Euler, you're doing numerical integration, assuming the slope is the value at this point. If you're doing the midpoint method or the trapezoid rule, you're trying to get a better approximation for the slope, right? So you're trying to build a trapezoid and integrate the area under that trapezoid. So they're fundamentally the same thing. Now the question becomes, I don't know f at y i plus 2. So I need to use a Taylor expansion of this quantity in terms of the value of f that I know. So I need to expand this in terms of f at yi. And I'll know the derivative of y with respect to t at yi. I know it from this formula. So if I take this Taylor expansion and I carry it out to the first term I don't know, which is the order delta t squared term, and I carry through all of the errors, I'll find out that there's a numerical error proportional to order delta t squared. So I need to do exactly this and substitute in terms of quantities I know and quantities I don't know. And the leading-order contribution of the quantities I don't know will give me this order delta t squared piece. And that'll give me the local truncation error in the solution. Does that make sense? So it's still Taylor expansions. Good. Other questions. Yes. AUDIENCE: Why is the central difference 0, not [INAUDIBLE]?? JAMES SWAN: Why is the central difference-- AUDIENCE: No, why's the outlook [INAUDIBLE] the one we did [INAUDIBLE]? That one is not implicit, right? JAMES SWAN: I don't think I said anything like that. So implicit versus explicit really depends on what you're trying to do. So if I'm trying to use a central difference formula of some sort to solve an ODE which progresses forward in time-- it has a direction associated with it that things always progress in-- then this might be an explicit formula. Because I could solve for f at x i plus 1 in terms of what's going on in the past. If we tried to do finite differences in a boundary value problem instead, and that boundary value problem doesn't have some direction associated with it anymore-- so it's not like space has some preferred direction. Time goes in a preferred direction. Space does not. You may find that the pointwise equations you write down cannot be solved strictly in terms of one of these. You can't just move this to the other side of the equation and solve in terms of everything else. So you'll have a coupled set of equations that can't be diagonalized, can't be simply reduced. They won't have the form of, say, of triangular matrix, for example. This will happen. And in that case, you would say this formula is implicit. So it really depends on which problem you're trying to solve, whether you label them explicit or implicit. Explicit is going to mean that the data I want can be found with a simple equation solve in terms of the data I know. Implicit is going to mean I've got to do some complicated equation-solving methodology. I've got to go do Gaussian elimination, or I need to go to fsolve and solve this thing. The data that I want is not available through some simple scheme where I can just write explicitly a thing I want equals some function of things I know. OK? Other questions. OK. Yeah. Yeah, yeah. AUDIENCE: [INAUDIBLE],, how many questions will there be? And for each question, how many subquestions? Could you just roughly-- JAMES SWAN: Oh, man. The question was, how many questions and how many subquestions on the quiz? Yeah. The quiz is-- in my opinion, it's now shorter than the last one that you took. We've been pretty sensitive to time concerns. We know that was something that was an issue on the previous quiz. I think there's only going to be two problems on it. They have many parts, but we've really tried to refine the questions in those parts. That's what you should expect. It's comprehensive. It Covers all the things from this section. OK? Yes. AUDIENCE: Also, DAEs problem-- JAMES SWAN: Yeah. AUDIENCE: It says that as index increases, we have less degrees of freedom. JAMES SWAN: Yes. AUDIENCE: Could you explain? JAMES SWAN: Sure. Sure. So higher-index DAEs come up all the time. And the degrees of freedom refers to your ability to choose initial conditions for the problem. And we saw several examples in which, if we tried to convert that DAE to an ODE, we found along the way that there were certain hidden conditions that all the initial conditions would need to satisfy in order for this ODE, the system of ODEs that we built from the DAEs, to follow the same solution path or solution manifold as the original DAE system would follow. The higher the index is, the more of these hidden constraints you find. And those hidden constraints tell you that the initial conditions have to be very particular. So there are-- I would encourage you to look at the specific examples in the notes. And you'll see there was a simple CSTR example that was index 1 and almost unconstrained, basically. There was an index 2 DAE in which the consistent initial conditions, you actually couldn't specify any of the initial conditions. They were fixed by the equations themselves. But remember, that sort of happened because certain DAEs are pathological in nature. They often ask us to do something we can't do physically, like try to control the input by-- or try to predict-- let's see, how do I say this? It's easy for us to predict the output knowing the input. It's very hard for us to predict the input knowing the output. And these higher-index DAEs were essentially asking us to do that. There's a lag in how we measure things. And it's easier to go forward through a process than backwards through a process and make these predictions. There's inherent sensitivity in doing the former. It's the same sort of sensitivity you get from these systems of ODEs, in fact. A small change in the initial conditions can lead to big changes in the outlet. So if I measure the outlet, it's hard to predict the inlet. It's just very inherently difficult to do. And so formulating a model which has this high index, you will naturally find that your choice of initial conditions for the solution of that model are constrained. And they're constrained because of the way you formulated the model. If you're actually trying to do control of a chemical process, it's going to be very difficult to try to somehow control the-- try to make the input be a certain thing by controlling the output. That doesn't make sense almost, right? It seems very, very strange. You would really want to change the input in order to hit a target in the output. How do you change the output and hit a target in the input? The process doesn't actually care about things in that way. And you saw that. Because in that particular CSTR example, you couldn't fix any of the initial conditions, actually, right? There was no free choice in the problem. Everything was automatically fixed, and you had to have perfect knowledge of everything in the system. I wasn't actually able to do control in a sense there, right? I could just measure the output, and I can tell you what the input was supposed to be. And it was super sensitive to my measurement. So these high-index problems tend to come up when we want to constrain things in a way that's, let's say, physically quite challenging. High-index problems are incredibly common in mechanical engineering, things like robotics where you want to make sure the robot's arm doesn't whack someone in the head. This is important. And so you're trying to constrain all these degrees of freedom and the paths that it's going to move on. And all those constraints-- I mean, there's actually a very narrow envelope in which this thing gets to operate. So accurately simulating the motion and then making the robot move that way is very empowering. In chemical processes, that's true, too. But oftentimes, we can make choices in model formulation, right? Like how you try to control the process that can ease those constraints. If I had some control or it doesn't control in an instantaneous fashion, even if it's a very high-bandwidth control, there's still a finite time for signals to propagate. There's dynamics associated with the controller that can convert that DAE system into a system of ODEs, for example. You don't have-- they may be stiff ODEs. DAEs are the limit of sort of infinite stiffness in ODEs. But they may not be as troublesome to solve as the DAE system would be. OK? It's about model formulation. OK, can I talk briefly about PDEs? Is that OK? I'll tell you a little bit about it. I like this finite volume method. I think it's very interesting. You've got some notes on it. I think I like it because I think it comes very naturally, actually, to chemical engineers. Because the problems we like to solve are of the type that finite volume eats for breakfast. These are problems of conservation. So we have a conserved quantity b. Is that what Dean uses in his book? The new version still uses b as the conserved quantity, right? So the time rate of change of b is equal to minus the gradient in the flux plus some, I don't know, some volumetric change. Maybe a reaction, maybe there's some injections somewhere of mass or momentum, or you know, some source. And our numerical solution, oftentimes, we would like it to also conserve b, right? Conserve mass, conserve momentum, conserve energy, conserve species. These collocation methods in general, or relaxation methods in general, don't do that. They're not designed to do that. You might be able to construct, for example, basis functions that have that property that they will satisfy conservation. You might be able to construct certain meshing methods, certain collocation methods-- and this is one of them, the finite volume method-- that respects conservation. And here's how you do it. So you worked really hard in transport to derive these differential forms of the transport equations. But finite volume actually likes the integral form. So you've got to go back. So you've got to take the integral of these equations over some volume V. And then the time derivative of the amount of b in this volume is related to the flux of b through the surfaces of that volume plus the amount of generation or consumption of the conserved quantity within that volume V, right? And this volume is my finite volume. I take the domain of interest, and I chop it up into all these little volumes. And then I just need to conserve b within that volume. So I have the first equation you learned in chemical engineering, right? The amount of accumulation of the conserved quantity is related to the fluxes in and out of the control volume plus the amount of generation or consumption within the control volume. This comes very naturally to chemical engineers. It fits perfectly in the problems we like to solve. Yeah. AUDIENCE: Is the nominal vector pointing outwards or inwards? JAMES SWAN: Well, it all depends on how we choose to write j, actually, right? So is j the flux into or out of this volume? So I won't prescribe that here. It depends on what you chose as the flux. So maybe, I don't know, what does-- Dean writes the stress tensor with the opposite sign of Bird, Stewart, and Lightfoot. So these guys, that's the flux of momentum, right? They choose a sign. Then the normals have to respect that sign. Doesn't really matter. The physics doesn't care. You just have to choose the flux in the right direction. If we think of j as the flux into the volume, n should point in. OK, so you have this conservation equation. You've got to solve for each of your finite volumes. And the finite volumes are coupled together, because a flux out of one volume is a flux into another volume, right? So you can think of these volumes as being little CSTRs. They're all sitting next to each other. And they're all linked together by these volumetric flows from one to another. And that's, in fact, what they are. The key is, how do I model these fluxes, basically? The accuracy of the method is going to be set by how big a volume am I going to model. Can't have infinitely small volumes. There are going to be too many of them in my domain to solve the problem accurately. And then, do I have good models for these fluxes? And you can think about these terms in this way. That sometimes is useful. So this big B, the amount of b in the volume, is like the volume times the density of b, the volumetric density of b in the volume. The average density in the volume. R is like the volume multiplied by the average rate of generation or consumption. F is the flux in or out. It's the sum over all the faces of the flux through each of the faces, which is the area of the face multiplied by the average flux through the face. So this is the normal to the face, and this is the flux through the face. And so I just-- I'm going to solve this equation. I try to figure out, in each volume, what's the average b? That's what this tells me. On average, what is b? On average, what is the concentration? On average, what is the amount of momentum? On average, what is the total energy? That's fine. So finite volume, very simple. What's nice about it? Well, you know, it's really suitable to any sort of geometry you want to look at. We want to solve this equation, say, in this square domain. We divide the domain up into a bunch of little square volumes. And in each volume, we just make sure that the amount of this b balances the fluxes coming through the volume and the amount of consumption or generation of that quantity within the volume. But it didn't have to be a square domain, right? A square domain, that's OK. But maybe I want to solve it on some circular domain. Well, I can divide that up into volumes, too. And actually, these equations are the same. They look identical. I've just got to calculate the flux through each of the faces. OK? But maybe I'm interested in, I don't know, pollution in one of the Great Lakes. And I want to know how concentration of some pollutant is transformed. I'd divide this up into a bunch of little volumes, too. And now I've got to calculate the flux through each of these faces. And that links all these domains together. Maybe I'm interested in transport within some porous medium-- say, a bone. I think it's someone's hip joint here. OK, well, I divide that up into a bunch of volumes. And I still solve the exact same conservation equation. So I've sort of decoupled the problem of meshing and equations, the governing equations associated with the meshes. As long as I have a way to divide the domain up. It may be very complicated. I'm going to go ask a buddy in applied math to do that. But I'm going to come to him with the right engineering equations for each of those domains, a little equation for a CSTR. OK? Turns out this-- you know, one of the most common fluid mechanical codes out there that people use now is OpenFOAM. So it's very common. Lots of people use it. It's a very general solver. Does finite volume. That's the method it chooses to use. Why does it choose to use it? Because it wants to conserved momentum? A guaranteed momentum conservation. Can't lose momentum, can't lose energy. Can't lose mass. I just have to pass the right amount between each of these faces. As long as my calculation of that is accurate, I've conserved the quantity. Never evaporates. Finite difference, finite element, different basis set expansions-- they will have this problem that at low levels of accuracy, you can have sort of artificial consumption of, say, the quantity, or artificial production of the conserved quantity of interest. So this can be a very nice way to solve these problems. It's another relaxation method. It's a type of collocation. I divide the domain up. And then I try to satisfy conservation equations on each of these little volumes. Hence finite volume. OK? OK. So how do you solve PDEs by hand or write your own code? There's a nice little tutorial here. We'll see how far we're able to get through it before I've got to give you guys your cards. But I'll illustrate some key points that I think are really important. So step one, no matter what method you choose, whether you choose collocation, whether you choose Galerkin, whether you're using basis sets or you're using finite elements or finite volumes, you've got some domain you're trying to solve your PDE in. And you've got to discretize it. You've got to figure out how to decompose the domain. So for finite differences, you might decompose it into a series of nodes at which you want to know the value of your function, whatever that function is. For finite volume, you would decompose it into cells in which you would like to know the average value of your conserved quantity. For finite elements, you divide the domain up into some points, across which local basis functions span the region. We would like to know the weighting of these local basis functions near each of these points. But you've got to divide the domain up. And the most important thing is you always choose the spacing of the nodes, or the dimensions of the cells, to match the physics. A very natural thing to want to do is to pick-- say I've got this square grid-- to pick the number of grid points to discretize with. I say, I'm going to use 100 points in this dimension and 100 points in that dimension. Actually, that tells me nothing. That's an irrelevant number. The physics has no idea how many grid nodes you put in there. Doesn't care. There are length scales that are intrinsic to the problem you're trying to study. And the discretization should resolve those physically relevant length scales. So it might be the case that I only need a resolution of one part in 10 along this dimension and one part in 10 along this dimension because the physics are such that that's good to get a description of the solution to this problem. It might be that I need one part in a million in this dimension and one part in 10 in that dimension because the physics are that way. So you really want to think about length scales, not numbers of nodes. It's a very common mistake. You say, well, 100 and 100, that makes sense, or 10 and 10. Or you know, you pick these randomly. But you want to think about physics instead. Here's a good example. This is my fault. You'll solve a problem like this. You already saw it in some sense in the finite element or COMSOL tutorial. But there's a little drug patch here near this red zone. It's leaking drug into a sheer flow, and it's all being advected down. Here's my solution to the problem. I think it's a pretty good one. Here's a solution from when I taught of course two years ago that the TAs produced. And I didn't talk with the TAs about this issue with length scales. And it sort of got the right structure. You can see, here's the patch, and there's some advection and diffusion. But it's not a very stable solution, it turns out. It's also not very accurate. It doesn't reflect the essential physics. And the point is actually the scales. So here, the patch was 1 centimeter in diameter. And so here, you got many centimeters long this direction, where you get advection. Here, you've got 2 centimeters in this direction. But actually, the solution is really confined down to this small space. The physics all happens at fractions of a centimeter. You took Transport, so you know that this is a problem in which there's a boundary layer. And all the change in concentration is confined to that little boundary layer. So really, when you do your numerical solution, you'll do like was done up here. Now the lengths are made dimensionless on the width of the patch, but they're-- you know, you can think of these as measured in centimeters. But look at the dimension here. We need to resolve the solution down at the 0.1 centimeter and smaller scale. I want to put all my little discretization or my nodes down here so that I can see how the concentration field changes as a function of distance. I have to know something about the physics to do that, though, OK? Can't just solve this blindly. It may or may not be the case that COMSOL recognizes the physical situation and remeshes your problem to put mesh points down where they matter. It may do that. It may not do that. It just doesn't know. People build heuristics to try to solve these remeshing problems, but they may not always apply. So it's good to understand the physics of the problem. This drug patch problem, this is a pretty common transport problem. I mean, did Professor Bazant give you guys a talk during your visit weekend? Was that this last year? Was that two years ago? It was two years ago. He does problems with bromine flow batteries. And they worked exactly this way. You have flow of bromine in. An electrochemical reaction takes place at the surface. This entire thing is premised on the fact that you have a very thin boundary layer in which the electrochemical-- in which the reagents are confined. And if that boundary layer gets too thick and reaches the other electrode, it screws up the reaction and doesn't work as a battery anymore. So being able to resolve concentration gradients in this small domain, it's not just important for this drug problem. It's important in general. It's a common sort of transport-y type problem. And the numerical solution is really going to hinge on your ability to recognize the physics and resolve the solution over the length scales that are important in the problem. OK? So that's the number one most important thing for doing these PDEs. The rest of the notes will take you through the construction of a numerical solution to the diffusion equation in 2D, and some different scenarios that you might try to employ in MATLAB to solve them, and how you can accelerate the solution using different MATLAB functions. It's fun to try out and play with. You'll see that I can make the solution to this problem go very fast for a domain that's quite highly discretized by recognizing certain forms of the equations, choosing certain solvers to solve with. OK? Good. All right, I'll meet you guys in back, and we'll hand out these gift cards. And good luck on your quiz. |
MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 36_Final_Lecture.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So this is our last 10.34 lecture of the year. And we're just going to use it for review. So I'm going to give a brief recap of the course. We've done a lot, actually. I hope you've learned a lot over the course of the term. We've had lots of opportunities to practice with different numerical methods. I've checked in from time to time on the homework submissions. I've been really impressed with the quality of the work that the class has produced as a whole. It tells me that you guys do understand what you're doing, that you can apply these methods to problems of engineering interests. And I hope that the things you've learned over the course of the term can be applied to your research down the road. Right, that's sort of the goal here. Whether you do experiments or modeling or theory, at some point, computation is going to play a role. You should feel comfortable after that things that we've done in this course on reaching back to these tools and utilizing them to either solve simple transport models of a drug delivery problem that you're working on, or fitting data reliably, doing hypothesis testing of models versus data. You have the tools now to do that. The homework assignments that you've completed over the course of the term sort of lay out the framework in which that can be done. And you seem very competent. We'll test that competence on the final exam, which is Monday from 9:00 to 12:00 AM. It's not in Walker, though, OK. It's in 56-154. I'll remind you of that again at the end. But it's not in Walker, so don't show up there at 9:00 AM. You'll be taking the wrong final exam. It's in 56. There's going to be a review session with the TAs on Friday. It's going to be in this room. It's going to be from 10:00 to 12:00 AM. And the TAs asked me to ask you to review your grade on Stellar. Make sure that everything is entered accurately. So check that your assignment grades match the grades that are entered there. Check that your quiz grades match the grades that are entered there. You're going to take your final on Monday. And we're going to grade it on Tuesday and submit your final grades, because some of us have to travel right after the final exam period. So we want to make sure everything is as accurate as possible when we enter the final data. OK? Good. OK, so a course recap. We've done a lot this term, actually, right? You should be really pleased with yourself for having made it this far. Any one of these subjects could comprise an entire course in and of itself. And you've gotten a broad overview of all of them. So I'll remind you. We started with linear algebra. I told you the first day that linear algebra would be the underpinning of all the numerical methods that we utilized along the way. And that was true. I didn't lie. You really wanted to understand linear algebra well if you were going to solve any problems numerically. We moved on to things that were more interesting from an engineering context, systems of nonlinear equations and optimization. So being able to solve engineering problems, being able to do engineering design by optimizing over various uncontrolled degrees of freedom is important. And we did both constrained and unconstrained optimization. Then we moved on to dynamics. We did ODE initial value problems and DAEs, which are the combination of nonlinear equations and initial value problems. We discussed boundary value problems and partial differential equations, which largely are a class of boundary value problems as well, at least the ones that chemical engineers are typically interested in. And we switched a little bit. We went into probability and models and data. And then we came back to more modeling in the form of stochastic simulation. And all along, this was threaded through with lots and lots of MATLAB. So hopefully, you learned to do programming if you didn't know that before. MATLAB's actually a very useful programming tool to know. It's not so specific. There aren't specific quirks and details you need to understand like you would with a lower level programming language like C or Fortran. It's very high level, and the concepts that you utilize there can be translated to other programming languages as well. So you may find at some point in your research you want to do a lot of data handling. MATLAB's not great for handling data actually. Parsing large data sets is quite difficult in MATLAB. But there are tool kits that are built into programming environments like Python that are very efficient at doing that. And Python is also underpinned by similar numerical methods that are readily accessible. Everything you learned in MATLAB can be applied there as well. So you should have a broad understanding of how to draft an algorithm, how to modularize code, how to integrate several different numerical methods together to form a solution to a more complex problem. So we did linear algebra. This was the basis of all the problem solving we did. Principally, it was concerned with the solution of systems of linear equations, though oftentimes, that's sort of disguised. It isn't a direct system of linear equations we're trying to solve. But somehow, we need to decompose a matrix. And in the process of decomposing it, we're essentially doing solution of systems of linear equations. One thing we're really concerned with was how linear algebra operations scale with the size of the problem that we're interested in. The bigger the system of equations gets, the slower the calculation is going to be. And how slow can that get? This will introduce fundamental limits on what we can do computationally. When we try to solve partial differential equations, and we decompose them into linear equations, and we have huge numbers of states that we'd like to be able to resolve with a partial differential equation, the problem can get so complex so fast that a computer can't approach it. So being able to estimate the computational cost of these operations is critical for deciding whether a problem is feasible or unfeasible. You'd like to know that before you start writing down a numerical solution to the problems. We focused a lot on that. There was fundamentals like arithmetic, solving linear equations numerically using Gaussian elimination, for example. We discussed decomposition in the form of eigenvalues and eigenvectors and, more generally, the singular value decomposition. And we discussed briefly iterative solutions of linear equations. That was sort of our segue into iterative solutions of nonlinear equations as well. So that was things like the Jacobi method or the Gauss-Seidel method, which were ways of the really operator-splitting methods of the same sort of class as we talked about in PDEs, where we took that matrix and we split it into different parts. We split it in a way that would make it convenient to solve a simpler system of linear equations. And depending on the properties of the matrix, that solution might be achievable or not achievable. The iterative method may not converge. But for certain classes of problems, we could guarantee that it would converge, and it might do it very quickly. Things we didn't cover, but that are important, I think. There's a notion called linear operator theory, which is an extension of linear algebra to infinite dimensions. You've seen some of this already in your transport class, for example, where you solve differential equations with sets of functions. You have some infinite set of functions that you solve a differential equation with, an ordinary differential equation or a partial differential equation. And all of the underpinnings of that fall under linear operator theory, which is an extension of what you've learned already for linear algebra. We talked about Gaussian elimination and the LU decomposition. You'll find in the literature there are lots of matrix decomposition methods. And the ones that are chosen are chosen for often problem-specific reasons. So there's things like the QR decomposition or the Cholesky decomposition or the Schur decomposition. These are technical terms. You can look them up. And you should feel confident that, if you read about them, you should be able to understand them given the basis in linear algebra you have now. But we didn't look at those in particular. And the state of the art in linear algebra and solving systems of linear equations lies in what I refer to as Krylov subspace methods. We talked about conjugate gradient as an iterative method for linear equations. There's a broader view of that that says Krylov subspace method. You may see that in the literature from time to time. It's a more nuanced discussion of the linear dependence of columns in a matrix or linear independence and how you can take advantage of that to rapidly generate iterative solutions to those equations. Any questions about linear algebra? No? Things are clear. You've done a lot of it this term. Systems of nonlinear equations. So these are, of course, essential for any engineering problems of practical importance. Linear equations tend to be easy to solve overall. Most engineering problems that are interesting are nonlinear by nature. That's what makes them interesting. And we also saw later on that solving systems of nonlinear equations was intimately associated with optimization. And so we discussed fundamentals like convergence of sequences and the rate of convergence of those sequences, stopping criteria for iterative methods. The best sort of nonlinear equation solvers with multiple equations and multiple unknowns are Newton-Raphson-type methods. They're the ones that are going to, if we're close enough to a solution, give us quadratic convergence, which is often more than we deserve. It's exceptionally fast, but it's not a globally convergent sort of methodology. So we also discuss quasi-Newton-Raphson methods that can help enhance the convergence of Newton-Raphson. And then we knew that we needed good initial guesses for solving these problems. We don't know how many solutions there are. We have no idea where they're located. And so we talked about homotopy and bifurcation as ways of generating those initial guesses and ways of tracking multiple routes or multiple solutions to the system of equations. I mentioned briefly during that lecture the concept of arclength continuation, which is a more advanced sort of homotopy method. And we didn't get to discuss that in detail because, actually, the underlying equation for arclength continuation is a DAE. OK, it's a fully implicit DAE that one has to solve in order to track the solution path as a function of the homotopy parameter. So we didn't cover this. We didn't try to solve any arclength continuation problems. So it's a very common method that's applied to solutions of nonlinear equations. Then we moved on to optimization. So we need this for design. We use this for model fitting. And we discussed both unconstrained and constrained optimization. In the unconstrained context, we derived optimality conditions. How do we know that we found the solution to this problem, at least a locally optimal solution? And then we discuss steepest descent methods that could get us to that solution with linear conversions. We discussed Newton-Raphson methods that can get us to that solution with quadratic convergence. We discussed dogleg methods that are heuristic by nature, but meant to mix the properties of steepest descent and Newton-Raphson to give us reliable convergence no matter how far we are from the solution. We moved on to constrained optimization. Those problems came in two types, right, equality constraints and inequality constraints. And we discussed the method of Lagrange multipliers to handle equality constraints. We discussed interior point methods to handle inequality constraints and their mixture to handle a combination of equality and inequality constraints. We left out a lot. Optimization is a huge field. So there are special classes of problems called linear programs, which are solved in a particular fashion. These are problems that have a linear dependence on the design variables. There's problems called integer programs where the design variables can only take on integer values. These might be very important. You can imagine trying to design a plant and asking what integer number of heat exchangers or separators is optimal. Associated with constrained optimization and inequality constraints, there is an equivalent set of optimality conditions that we didn't try to derive. Their called the KKT conditions. So this comes up a lot in literature. So it's sort of the fundamentals of knowing that you found the optimum to the inequality constrained problem. They're not difficult to derive. But we have a limited amount of time in the class. It's something fundamental that's sort of missing. And then these sorts of methods of all find local optima. But oftentimes, we're interested in global optimization instead. And that's not something that we discussed at all. That's quite complicated and beyond the scope of the course. But things like genetic algorithms, that's a term that will come up a lot. That's one way of trying to seek out global optima in a landscape. ODE-IVP, right, so now we want to move from static engineering situations to dynamics. And we were really concerned, when modeling dynamics, with both the accuracy and the stability of these methods. We were concerned about stability with other algorithms, too, for solutions of nonlinear equations. We were concerned with whether the Newton-Raphson method would converge or not. That's a stability issue in a sense. OK, so we're concerned with stability there. But here we really focused on it in detail, in part because some of these differential equations are inherently unstable. In engineering circumstances, that happens. And we would be unwilling to accept unstable solutions in situations where the equations themselves are stable. In the case of Newton-Raphson, we can do these sorts of quasi-Newton-Raphson methods to change the path the solution follows. And we didn't really care what path it followed. But here, the path is important. We're not free to change the path. So the method itself has to be inherently stable. So we discussed numerical integration. We discussed two classes of method, explicit methods and implicit methods. So explicit methods, we mentioned these explicit Runge-Kutta formulas that have been derived for all sorts of accuracy levels, both local truncation error and global truncation error. Explicit methods, unfortunately you can't guarantee that they're always stable. They lack a property called A stability. Implicit methods, on the other hand, can be made to be stable under all circumstances. So one of those is the backward Euler method. That's a nice one to use. The penalty for using these implicit methods is you likely have to solve a nonlinear equation at each time step. So there is a cost to pay associated with enhancing stability. We discussed local and global truncation error. And we talked about stiffness and linear stability criteria. So linearizing the ordinary differential equation, assessing what the eigenvalues associated with the Jacobian of that linearization is. And using those eigenvalues to tell us, for example, what kind of time steps we need with our method in order to achieve stable integration in time. Things we didn't cover that are sort of interesting and maybe important in engineering context, one of those is event location. You may want to know the time at which something happens. So I may want to know when does the solution hit a certain point in time. Maybe, I even want to change my dynamics when the solution hits a particular point in time. There's a broader class of methods that can do that. They're built into MATLAB too, actually. You can set up these events and have the dynamics change or have a report that an event occurred. But we didn't really discuss that at all. And in numerical computing, parallelization is important for studying classes of big problems. It's sort of funny, but you can even do parallelization in time. So this is something that's been discussed for a long time. And for certain classes of problems, it offers computational advantages over serial integration of the dynamics an ODE-IVP. So you can imagine how do you parallelize in time. So you can take the time window that you're interested in and cut it up into different slices. And you could try to solve in parallel the dynamics over each of those slices. But for each slice, you need an initial condition that has to be precisely prescribed. And you don't know those initial conditions. That's what the serial integration would tell you. One way of doing parallel in time is to say, well, those initial conditions are unknown. And they match up to the terminal conditions of the previous slice. So why not wrap this serial integration process in a Newton-Raphson solver to determine each of these points? With certain classes of problems, certain ODE-IVPs that are inherently stable, you can do that. And you can get a computational speedup over serial integration just by doing the integration for each of the slices on different computers and bringing all of the results together. The bigger the farm of computers you have, the more slices you can utilize. That's sort of interesting, right? Parallel in time, it's bizarre, because we usually think of time integration as being sort of causal. I have to know where I was to predict where I'm going. But you can actually divide these things up in more sophisticated ways. We talked about BVPs and PDEs. We treated them as the same class of problem broadly. Usually, when chemical engineers are looking at these problems, we're interested in spatial variation of some sort of a conserved quantity, so momentum or energy or concentration of a species or mass. And here we are also concerned with both the accuracy and stability of method stability now for when we had also time integration of the solution, in addition to spatial variation. And so for BVPs, we talked about shooting methods. So can I recast a boundary value problem in one dimension as an initial value problem with an unknown? That's a very common way of solving these problems. And they are sometimes stable and sometimes not. When they're unstable, you can do what's called multiple shooting, where you divide the domain up into subdomains, and you shoot from the initial condition of each of those subdomains. It looks exactly like the parallel in time integration that I described for you before. And then you try to match up the initial conditions with the terminal conditions to get a continuous solution. Then we discussed relaxation methods for solution of BVPs and PDEs. So collocation, Galerkin. Collocation, you'll recall, was we want to try to satisfy the governing equation at various points in the solution domain. So we might write our solution as the superposition of a set of functions. And then try to satisfy at various points that we've laid out in the domains. Maybe, there's even an optimal place to put all these points in order to minimize the error in the solution. Galerkin, instead, we tried to make the projection of the error in our equations on different orthogonal functions that represented the solution be zero. So that was sort of a global error that we tried to minimize over different orthogonal functions. Whereas collocation is sort of a local error that we're trying to minimize, a local truncation error that we're trying to minimize. We discussed finite difference and finite volume methods as well. So ways of discretizing in the equations. Finite difference, I would say, is the easiest one to reach for. If you had to pick a quick and dirty method to try to get a solution, finite difference can be really easy to go to. It's easy to figure out how big the errors are in your finite difference method. But it may not be the optimal approach to the problem. For certain geometries, finite volume is really far more preferable, because it's easy to estimate the fluxes through surfaces of funny shapes where your grid shape now conforms to your geometry. Finite volume also had the advantage of being strictly conservative. So I can never gain or lose a conserved quantity in the finite volume method, at least to within the numerical error. And that isn't true of finite difference or finite element. So if I'm concerned about those things, finite volume is really the approach to use. And you'll find one fluid mechanical solver that's freely available. You can go download and use it today, and it is widely used in research, is OpenFOAM. And that's all based on the finite volume method. So they're trying to conserve momentum, conserve mass, and conserve energy associated with the fluid. OpenFOAM is good at drawing bizarre grids of the domain and giving you very accurate integration of fluid mechanical equations. And it's good at doing that because it uses finite volume. You discussed the method of lines, which was a way of discretizing space, but leaving the time dimension undiscretized and then applying ODE-IVP methods for solving in time. It's an incredibly useful way to approach those problems. Because it leverages expertise in two different techniques. There are specialized methods for doing time integration with some spatial discretization that are vetted for their stability properties and their accuracy. And those are fine, but I would say they're a little antiquated. They're reliable for particular classes of problems. But for general problems, you may not have any idea whether they're going to work or not. Whereas method of lines may be a really good approach to a parabolic partial differential equation you're working with where you can just rely on the adaptive time stepping and the air control and an ODE-IVP solver to keep everything stable in time as you integrate forward. We did application of commercial software. So you used COMSOL to solve a problem. I don't know. Would you say was COMSOL easier to use than writing your own MATLAB code? Harder to use than writing your own MATLAB code, do you think? No. AUDIENCE: Easier. PROFESSOR: Easier to use. But the result, I would say, wasn't as good without some careful refinement of the COMSOL solution. So for that particular problem, if you wanted a solution that looked like your MATLAB solution that was carefully resolved in space, you really had to go in and refine the grid in particular places. So it gets harder and harder the more detail you want. But it's an easy way to get an answer. You should check whether that answer is right by comparing it with other methods. You can't guarantee that it's giving you the right result either. To give you an example, I had mentioned an unsteady advection diffusion problem to one of my grad students. And he's very good with COMSOL. I mentioned that it was a hard problem because there were boundary layers. And the boundary layers changed thickness and time, and resolving those things can be quite challenging without any physical insight. And he says, well, I think I can just do it in COMSOL. And he tried it, and COMSOL was happy to report a solution in time and space, but the solution was nonsensical. We could visualize it and see that it didn't look like what other numerical solutions look like, and in certain limits, didn't correspond to what those solutions are known to be in those limits. So the black box is great in that it reports an answer. But it can also be really problematic as well. So maybe, you have expertise with COMSOL. You can get a solution to converge for a difficult problem. But you should really have other methodologies that let you assess the quality of that solution. For simple problems, simple elliptic problems and simple parabolic problems, it's going to do great, because it's built to eat those for breakfast. But for complicated engineering problems, you've got to vet your solutions. Oh, what didn't we cover? We didn't talk about hyperbolic equations. Not really. We talked a little bit about advection equations, which have a hyperbolic character associated with them. So you have like a pulse of [? solute, ?] and it's advected along in time. And you want that advection to be stable, so you want to use things like upwind differencing to ensure that stability. But there's a broader class of hyperbolic problems that relate to things like the motion of waves in elastic solids. We don't typically encounter those in chemical engineering, but chemical engineering is a broad discipline. And maybe you're engineering soft materials for some sort of device. And you want to look at the elastic response of that material. The PDEs that govern that are a completely different class than the ones that we typically look at. They're not parabolic. They're not elliptic. They're hyperbolic, and they are these wave-like solutions associated with them. They require their own particular solution methods in order to be stable and accurate. And then there are all sorts of specialized methods that are dreamed up to apply to, maybe, a broad class of problems, but it's kind of hard to ensure that they're going to work. But nonetheless, you'll see a lot of this out in the literature. So there are things called multi-grid methods. They try to discretize a partial differential equation with different levels of fineness. And you use course solutions as initial guesses for finer and finer solutions in space. For some problems, this works great. And in fact, you can show that it can work as well as is possible. So depending on the size of the problem, say you have n points at which you want to find the solution. You can find it in order n time with some of these methods depending on the particular problem you're looking at. That's good. But oftentimes, engineering problems aren't of that sort. There's something more complicated going on. And it's hard to ensure that you get those sorts of rates of convergence. But nonetheless, they exist and something that we didn't cover you should be aware of. We did DAEs. So now, we coupled ODE-IVPs with nonlinear algebraic equations. Most dynamic engineering problems are of this type. We showed that DAEs are really an issue with model formulation ultimately. We write down conservation equations, and we write down certain constraints on those conservation equations. The constraints can be specifications like we want particular flow rates in certain places. Or they can be fundamental. They can be thermodynamic constraints on the coexistence of different phases of a particular material in a particular unit operation. If we could solve those constraints for certain unknowns and substitute them in the equations, we might be able to produce ODE-IVPs. Probably we can't come up with those solutions. So instead, we've always got this coupled system of initial value problems with nonlinear equations. So you have that sort of a circumstance. How should you think about that? One way to think about it was to say that the algebraic equations are essentially infinitely stiff. They have to be imposed exactly at every time step. That they never relax in time with the finite time constant. And so we knew that those sorts of methods are hard to resolve with typical ODE solvers. And we tried to assess that by computing something called the differential index. So we asked how many times do I have to take derivatives of the algebraic equations or any new algebraic constraints in order to get a system of ordinary differential equations to represent the same model. I may want to solve that system of ordinary differential equations. Or I may just want to know the differential index so that I can choose a particular method to solve this problem. We saw the higher the index was, the more sensitive the solution was to perturbations in different input variables. So the differential index played an important role in telling us something physical about the problem and the responses that could be elicited by DAE systems. We also talked about consistent initialization. I need to prescribe the initial conditions for all the differential and algebraic variables. Can I prescribe any of those independently? Well, the answer is actually no. Depending on the differential index, there are underlying algebraic constraints that have to be satisfied at all points in time. And if they're not, then that can break the method that's trying to integrate the DAE. So you have to prescribe the initial conditions consistently. If they're not prescribed consistently, and you go to a piece of commercial software, it may give an error. It may tell you nothing. You may get a reasonable result or an unreasonable result. It really depends on the methodology. So it's up to you to know in advance that this is an important aspect of the problem. We solved index-1 DAEs typically. Index-2 DAEs can be converted into index-1 DAEs and solved pretty easily. Index-3 and bigger DAEs are difficult because of their high sensitivities. So there's special software that's out there. I mentioned some of it during our DAE lecture. And that's really what you want to reach for if you have a high index DAE. Examples, we were able to craft all sorts of examples where we tried to do funny things with causality, where you tried to change the input by affecting the output, which doesn't make a lot of sense. But that led to very high index DAEs. Mechanical systems often have very high index DAEs associated with the mechanical constraints, typically lead to index-3 DAEs. So these pop up in places that are important. And being able to look at the model and assess the index of it is kind of essential. We did probability. So the physical world has inherent uncertainty. It's not just uncertainty in our ability to measure things, but fundamental, physical uncertainty is built into the world around us. So measurements have uncontrolled or even uncontrollable sensitivities to this uncertainty. And we wanted to know how to handle it. So we talked about things like sources of randomness. We did fundamentals. Maybe this overlapped with 10.40. But I think overlap is good, because humans tend to have bad intuition about probability in general. So we discuss probability densities, means, covariances, expected values, joint probabilities, conditional probabilities. We talked about common probability densities. You covered the central limit theorem. You talked about the difference between sample averages and the population mean. One thing that's important that I don't think that gets covered much in detail, but if you're going to be working with randomness computationally, oftentimes you have to generate random numbers. And you'd like to generate good random numbers. Turns out a computer doesn't generate proper random numbers, only pseudo random numbers. And these generators have various properties associated with them. And some of them are good. And some of them are bad. And you want to use good ones. So if you have to generate random numbers for your research, you'd like to know that they are high quality, that the sequence of numbers that's being generated doesn't repeat over the time that you're using it, that it won't overlap over many different uses of this methodology. So high quality random numbers are important. We discussed models versus data. So we wanted to regress, estimate parameters from the data. We wanted to do things like hypothesis testing. I have a model. Is the model consistent with the data, or is it not consistent with the data? So you covered things like linear and nonlinear least squares, maximum likelihood estimates, confidence intervals found using the chi-square distribution. We did Bayes' theorem and Bayesian parameter estimation as well, so a way of using prior information about parameter values to better estimate the probability that those parameters will take on values given the data. We didn't cover design of experiments. That's a 10.551 topic. But there are specific ways that one can design the experiment. Which data values do I take in order to make the fitting of parameters optimal? And there are other problems besides regression that are important. But we don't typically utilize those in chemical engineering. So one of those is classification. So if I have a series of points, and I do regression, I'm in some sense asking for like what's the best curve due to this model that can be drawn through those points, right? What parameter values give me the curve that's closest to all these points, that's least squares. Classification might ask instead, what's the curve that divides these points most evenly, or sits furthest from all these points in an equal amount instead. Which side of this line do the points sit on? Are they of type A or type B? It's kind of a related problem to regression. But it's a different sort of problem. The same sorts of methods can be applied to classification that you apply to doing regression and parameter estimation. It's an important problem. And then we finished up with stochastic simulation. So you learned that sampling of random processes can be used for engineering calculations. And they're even inherently random physical processes that you might want to model reliably. So we did Metropolis Monte Carlo. This was a way of basically computing high dimensional integrals. That's what it boiled down to, where we wanted to sample from different places in the domain, weighted by some probability density function. You did kinetic Monte Carlo, which is really a sort of event-driven algorithm. You are expecting that certain events are going to occur with certain rates. They're actually nonstochastic versions of that, the same sort of event-driven algorithm. I know certain things should happen at certain points in time. And rather than integrating the dynamics of the system all along these points in time, I just advance the dynamics until the next event occurs. That event occurs. I keep a list of what the next event is going to be. And I follow those events along. So it could be something like billiard balls on a table. I know the trajectories of the billiard balls, given their position and their momenta. And all I need to do is keep track of when they collide and where they're going to go next and what the next events are going to be. Well, kinetic Monte Carlo was the same thing, except the events are now stochastic. They occur with prescribed rates. And so I just need the sample from this distribution of rates to get the right sequence of events. There's funny things about kinetic Monte Carlo. So applied to chemical kinetics, you can estimate these rates reasonably well. You've got to know the rates for this process to look right. If you don't have the rates correct, or you don't have the relative magnitudes of the rates, then the simulation is kind of meaningless. The result is going to be nonsense or garbage. And so there's going to be limitations on your ability to do this based upon how accurately you can estimate these rates. It also means that if you know that there's disparities in the rates, say one rate is very high and one rate is very low, you can make certain assumptions about the process. So things that happen very quickly you may treat as though there are equilibrated and not try to resolve them at all and look more at infrequent processes that occur, slower rates. So the precision with which you have to know these things is important. The relative rates are what's really important in these problems. And you want to get them right in order to resolve these physical processes correctly. Then we discussed molecular dynamics after that, which is another type of stochastic simulation methodology. Here you just integrate the equations of motion associated with whatever particles you're interested in. Kinetic Monte Carlo and molecular dynamics are sort of fundamentally far from equilibrium, but may relax towards equilibrium, depending on what constraints you put on the particular rate processes that you're modeling. Metropolis Monte Carlo is usually sampling an underlying-- if we're applying it in the statistical mechanical sense, it's usually sampling an underlying equilibrium Boltzmann distribution. So they often get used in fundamentally different ways as well. We didn't cover stochastic differential equations. So depending on which degrees of freedom you want to represent in a model like molecular dynamics, there may be some degrees of freedom you simply don't care about. But they have an influence on the problem. So in my research, for example, we model nanoparticles and particulates. We're not very interested in the details of the solvents surrounding these particles. There's a huge disparity in length scales associated with nanoparticles and the molecular dimensions of the solvent that surrounds them. The solvent has an influence nonetheless. It's fluctuating all the time. Solvent molecules are colliding with these particles, and they're imparting momentum. One way of resolving a system like that without having to look explicitly at the solvent is to try to integrate out the solvent degrees of freedom and treat those collisions of the solvent with this particle in a stochastic sense. So at various points in time, there's momentum that's randomly transferred to particles dispersed in a solvent. The physical manifestation of that is Brownian motion. When you look at small particles in the fluid, you see them diffuse around, and that's where that's coming from. So you can simulate processes like that without having to resolve molecular scales. So as you go up in scale, you integrate out degrees of freedom, but oftentimes, that introduces inherent stochasticity in the underlying dynamics of whatever you're trying to model. That's very interesting. And then for Monte Carlo methods, there's advanced sampling methods. And we didn't discuss any of these in detail. But you saw some of this with the model you saw with the problem you approached in your homework where you had two populations or two peaks in the probability distribution that were widely separated. And it may be hard for a Monte Carlo method to traverse between those peaks. And umbrella sampling as a way of biasing the probability distribution to make those paths from one peak to another more probable. So it's just a way of seeking out rare events like the transitions between these two different populations. And it's pretty important, ultimately, for exploring complex high dimensional spaces. But you have to understand something about the underlying physics in order to figure out the biases needed to enhance the sampling. So it's so quite complicated. So I'll remind you. Your final exam, right, it's Monday. It's in 56-154 at 9:00 AM. Don't show up to Walker. It's not going to be there. It's three hours. It's going to be comprehensive. You can expect roughly one problem each thematic topic. We've written it now. I think it's nine problems long. So we're in the process of revising it for difficulty and time. It's going to be short answer format. We sort of aim at 10 to 15 minutes per problem. I think that's the sort of frame you should be looking at. The problems are structured in such a way that that's about how long they should take. You should use that to guide your temporal budget on the exam. If it's taking too long, switch to something else. Don't let yourself get hung up focusing on one thing. Move around and come back to it later. There will be some fundamentals. There's going to be some translation of engineering into numerics. And there's going to be some MATLAB. So it'll look like a blend of the first exam and the second exam. Yeah. AUDIENCE: When you say it's on MATLAB, do you mean [INAUDIBLE] MATLAB? PROFESSOR: Yeah. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah. AUDIENCE: No computers. PROFESSOR: No computers. It's just write a little script. We've asked problems like that before. It'll be of the same type. Yes. AUDIENCE: You said nine problems. [INAUDIBLE] PROFESSOR: I think there's nine, yeah. Well, that's how many we've written now. I'm not planning on writing anymore. AUDIENCE: Plus or minus one. PROFESSOR: Plus or minus one, yeah. AUDIENCE: [INAUDIBLE]. PROFESSOR: Oh sure. Listen, if we gave you nine problems of the same length as on the first two quizzes, you wouldn't get through the problem. So they're not written to be so long. But there are multiple parts. But multiple parts are usually intended to guide you to the solution we're looking for. Usually, we're testing for a particular understanding. It also helps you out when there are multiple parts. So there's lots of partial credit available. You don't have to get the whole thing. You've just got to get two out of the three parts. You did pretty good. OK, more questions? AUDIENCE: So when you say short answer, does that just mean that similar types of problems as what we had on the first two quizzes. They're just shorter. PROFESSOR: Yes, that's right. Yeah. AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, they're not multiple choice, true or false. They're short answer. You want to hit a level of detail that shows us you understand what you're talking about, right. We don't need pages of information on these things. But we need to have enough information that we can see you know what you're doing. We're trying to assess your understanding of the material. That's all. More questions? No more questions, OK. Well, good. It's been a real pleasure teaching you guys. I'm really pleased with the outcome of this course. This is my third time teaching it. I think this is the strongest group that I've seen come through 10.34. So you guys should be really proud of yourselves. Good, all right, good luck. |
MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 34_Stochastic_Chemical_Kinetics_1.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. WILLIAM GREEN JR: All right today, I would like to do a Kinetic Monte Carlo example with you in class and, also, talk a little bit about the connection between the master equations, master equation solution, the normal kinetic equations we use, and the-- how you handle these trajectories, you'll get how to Kinetic Monte Carlo. All right, so let's just pick a simple example. As we mentioned, in the real world, it's pretty easy for these examples to get completely out of hand, because the number of possible states can be so gigantic. Let's all do a really super, super simple one here. To try to keep that underhand, not worry about that. When you do the real problems, that will be a major issue-- is if he number of states get out of hand. So the simple example I'm gonna do is a real one from inside your body. So inside your body, you have a thing called low density lipoprotein. And there's little vesicles of fat, lipids, inside your body, actually, in your blood vessels. And these are bad. If you go to the doctor, they measure your LDL and your HDL. And if your LDL number is too high then they yell at you for eating too much fat food. And they make you exercise and stuff like that. This has happened to me, so I know. And there's a reason for this, is because the LDL is susceptible to oxidation. And if too many of those lipid molecules peroxidize, it causes inflammation in your blood vessels. And that can cause your blood vessels to constrict and stuff that eventually can lead to heart attacks. And so that's why your doctor is alarmed if you have too much LDL, because you have a lot more material available that could be oxidized, that could kill you. There's actually a really complicated, interesting story about what vitamin E does in the LDL, and how it interacts with vitamin C. And if you take my kinetics class, I'll tell you about that. But we'll do the simplest case now. There's no vitamin E, there's nobody Vitamin C. You just have some lipid, and it's oxidizing. And let's model what's happening. So what you have is a little bubble of fat. So out here is water. And here's my fat which I'll politely call lipid. That sounds like not quite as fat, if you call it lipid, slimy. And the lipid is susceptible to attack by this reaction. So if I have some peroxy radical, and I have the lipid molecules, which are some hydrocarbon, they can react and make peroxide plus a radical. And in your blood vessels, if you're not under the water or being killed, you have a lot of oxygen. And so the carbon set of radicals you form immediately react with oxygen to make the peroxy radical back. And so the net process-- we can kind of combine these steps, because this is super fast. So it's like a microsecond times scale. For that to happen in your body, because the oxygen concentration is so high in the blood. And so this reaction is really roo, plus RAH, plus O2, to rooh, plus roo. So the radical is just catalyzing the oxidation of your lipid into the peroxide. When this level gets too high, then your body will detect it. There's an inflation, and you'll have all kinds of trouble. OK, so we like to understand this. Now, these vesicles are very small, so there's a probability you might have only a very, very, small number of radicals in there. Also, your body has a lot of antioxidants in it, like vitamin C and vitamin E and some other ones too. And the antioxidants are trying to get rid of radicals. So the number of radicals you'd expect the concentration-- background concentration is very tiny. And because of that, this is tiny and the concentration is tiny. So the concentration times the volume might also be tiny. So the total number of radicals might be very, very small. And so therefore, you might need to worry about the stochastic kinetics being different than the continuum kinetics. OK, so that's the idea. Now, on top of that, you also have a lot of these vesicles in your body. So you might be able to somehow work up some continuum model, treat them all. And in fact, I think that's what people did originally, is they just said well, let's take the total amount of LDL you got in your body and do a continuum model. But as you will see, it's different, because the fact that the vesicles are small, each one might only have, say, one radical in it. And that will actually behave quite differently than if you had a lot of radicals. All right so, this is the kinetics, and the oxygen is almost always high concentration, so I'm not going to worry about that. And I'm going to assume that we're not going let your lipid oxidize so much that the amount of lipid changes significantly. So let's just, for now, we can just ignore this. OK, maybe later we might go back and figure out how you'd account for the fact that these cells are being consumed, which would change the rates a little bit. So we'll just assume that concentration is constant. So you have a tiny amount of radicals. And if you had the radicals in your vesicle, they're creating roh. And you see, the ro is catalytic. And so what we're gonna see is basically D. If you did it in classical kinetics, you'd write DR dt this is basically equals some constant times roo. And then, I lumped these concentrations into this constant. [INAUDIBLE] That all right? All right, so that's the import process. We also have another important process, is that if you get two radicals in there, and they bump into each other, they can kill each other off. So two peroxy radicals can combine and make stable molecules. And we don't care what stable molecules they really are. What's important is that they get rid of the radicals, because the radicals are catalyzing the unfavorable oxidation. In this program I started writing, I called this k3. This one is k4, reaction four. There's two other things happening. Out here, I had some peroxy radicals. They're floating around the water. Some of them might go into my lipid. I'll call that process K1. There's some time constant, which new radicals arrive. And then, I also have the possibility these guys could diffuse out. So I call that k2. And k1, I'll just treat as a constant. So as soon as a background concentration radicals that's constant, and so this is a sum-- every 10 seconds or something, a radical arrives in a vesicle. And this will depend on the concentration, the number of ro's. So the more you have inside here, the faster the rate which they go out. That all right? And this just mass transfer. But from the point of view of the equations, it doesn't care if it's mass transfer or chemical reaction. They arrive in the same way. All right, so that's the system. And so let's look at the-- I should try to write the MATLAB code for this. And maybe, we can try to finish this together in class. So I have four processes-- arrival of new radical, diffusion of the radical and the vesicle, reaction to make the peroxide, and self-destruction of the radicals. And I wrote down what each of these does. So the first one increases the number of radicals by one. The second one gets rid of a radical because it's diffused away. The third one increases the amount of peroxide. The last one gets rid of two radicals. So we're going to try to compute a trajectory by the Kinetic Monte Carlo method. And trajectory is going to be some time points, the number of radicals at that time point, the number of peroxides I have at that time point. And as I run, I'm going to get many, many, different time points. And each one will have a different number of radicals or a different number of peroxides. And that's what I want to compute. And then, at the end, after I have trajectories like that, big tables of these things, I want to somehow put it together to figure out on average what happened or something, try to understand it. And so let's see, so my inputs are the initial numbers of radicals and peroxides in the vesicle, the vector of rate coefficients, the max amount of time I want to run the trajectory for-- not clock time, but time in the simulation. So it's like how many microseconds or milliseconds or seconds I look at my lipid protein and see what it's doing. And the maximum number of steps, because I don't want to get a trajectory list that's 10 to the ninth long, because over a while, my memory, my computer would cause me trouble. So here they are, all the setups. Everything is fine so far. So here's the loop. So the outermost loop is just to keep track of how many steps we have, which is going to be the number of entries we have in the trajectory table. And the second one is really-- what we care about-- is well, the time is less than the maximum time of the simulation. First, we want to figure out the time [INAUDIBLE] until something happens. So following Joe Scott's notes, we compute this quantity called A, which is the sum of all the rates. And that's equal to the rate of the first process. And then, they're in the second process, which depends on the number of radicals. And the rate of the third process all depends on the radicals. And the rate of the fourth process, which is a number radicals times the number of radicals minus one. Now k4 has to have units of per second. But bi-molecular rates would normally have units of volume per second, per mole even-- volume is per molecule per second. And so that k4 has to really be a normal kind of k divided by the volume. And we're going to have to figure out what to do about that in a minute. And then, we do the formula to get the-- Gillespie's formula for how to sample from an exponential decay, as good as that is. So Gillespie thinks that the probability that something is going to happen, or the probability, the time, until the next thing happens is going as e to the negative a times t. So we're trying to sample from e to the negative, a, t. And that's what that crazy log of the random number is doing-- it's sampling from that distribution. And so we get a [INAUDIBLE] that how long we've waited from the time the last thing happened until the next thing happened. And now, we have to figure what happened. So there's four different processes that could have happened. A radical could have arrived. A radical could have left. A radical could have reacted with one of my hydrocarbon molecules, with one of my lipids, or the two radicals could have killed each other. So we list these guys out. A is the sum of all those processes. P1 is the probability that the first process happened, that a new radical arrives. So it's going to be the k for the first step, divided by a. That one was zero water, because it's subterfuging it from outside. So it's just a k of 1. Probably the second step is k2 times n rad. That's the rate of the second step divided by the total rate of all the processes. And I'm going to add that onto p1 to make a vector of possible things that could happen. So I'm picking a random number from zero to one. And I want to make a little bar. Here's the probability the first process happened. And here's the probability that the second process happened. And here's the probability the third process happened. And here's the probability the fourth process happened. I know that these guys have to add up to one, because I computed that something happened. And so I'm going to try to compare my random number from zero to one to these breaks and sort it out into these four possibilities, and then figure out which thing happened. OK, so this is the second step of Gillespie's algorithm. All right, so maybe you guys can help me finish the code here. If you want to go to Broadway in Chicago, you can go there. So I need to write the next one. So p3 is equal p2 plus what? So step three is-- sorry? Say again. AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: All right, that's enough, right? So now, I get to choose a random number, so, say, x is equal to rand. And then, I have to say if x is less than p1, than something. What's it gonna be? Then step number one happened. So that means that n rad is equal to n rad plus one, so one radical transported in. If x is less than p2-- [INAUDIBLE] Is that good? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: [INAUDIBLE] space. Thank you. What happens here? So this is b, n rad is equal to n rad minus one? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: I think they also take care of that, right? Is this right? X is less than p3, than nroh. Else and rand. Does that look right? Do I need an end? AUDIENCE: Semi-colon. WILLIAM GREEN JR: Semi-colon. Semi-colon, thank you. That would be really bad. All right, yes, Ziggy? AUDIENCE: So in the first problem, you have step size longer than that [INAUDIBLE] size. WILLIAM GREEN JR: So that's not good. It should be less than. Thank you. Any more bugs? AUDIENCE: I think you need spaces after the brackets. WILLIAM GREEN JR: Spaces after the brackets-- where? Is that here? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: [INAUDIBLE] separate line. AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: I can do that. It's easier to read anyway. So that's good. I definitely agree with that. [INTERPOSING VOICES] WILLIAM GREEN JR: Is that right? AUDIENCE: Yeah. WILLIAM GREEN JR: You like this? With all these [INAUDIBLE] sets, I don't need any n's, is that right? [INAUDIBLE] after the other like that, no problem. Just one end at the end. [INTERPOSING VOICES] WILLIAM GREEN JR: That's good. All right, so we're OK with this? We think this is going to work? AUDIENCE: [INAUDIBLE] [INTERPOSING VOICES] WILLIAM GREEN JR: All right, so now, we've computed what's going to happen. Now, we need to store results somehow. So I'm gonna say trajectory is equal to-- or trajectory of steps-- so step was zeros. Now, step is one. [INAUDIBLE] step is one, to start with. [INTERPOSING VOICES] WILLIAM GREEN JR: Does this look good? They had only commas, is that right? No commas? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: All right, one step is zero. I already have the first line filled in here. [INTERPOSING VOICES] WILLIAM GREEN JR: Is that good? We'll see. I don't know if it's OK or not, but I'll believe you. All right, so right now, we have a program that maybe or maybe not might actually compute something. Is that right? OK, now, we to figure out-- are we going to run this program? So we need an initial. So any suggestions? [INTERPOSING VOICES] WILLIAM GREEN JR: All right, so let's do an initial-- how about n rad is equal to 1 and nroh is equal to 0. Is that OK? We're cool with that? And how about k? So we need a vector of k's. Yup? AUDIENCE: For your line 18, I think that's just the [INAUDIBLE] WILLIAM GREEN JR: Yes, thank you. Yes? All right, so let's think about what's reasonable for the k's. So I think we can make the k for this stuff diffusing in very, very small. So I don't know-- [INAUDIBLE] make a small, 1e minus 5 something. Isn't that small-- only once an hour. You think it might be a little too small? And then, the next one is how fast do things diffuse out? Now, what's that going to physically depend on? It should be something like the diffusion. Diffusion [INAUDIBLE] get back to the wall, back to the outside. So it should be-- by one radical in here, what's the average time for it to diffuse back out? Any ideas? So if it does, say, just a random walk, maybe-- then you say that delta x squared is like d times tt. OK, so delta x squared, maybe make this like the radius of our lipid vesicle, so that delta t is like r squared over d. Seems like a reasonable scaling. And we actually want to rate, so we want the other way around. So k2 is going to be something like d over r squared. All right, so we need to pick a size of our LDL vesicles. I don't actually know what they are. I used to know this, but I don't remember. Anybody want to make a guess, how big is a vesicle inside your blood vessel? [INTERPOSING VOICES] AUDIENCE: Micron. WILLIAM GREEN JR: Micron, OK. So let's pick a mircron. So let's say r is equal to 10 to the minus 6 meters. What's a diffusion constant in liquid phase? [INAUDIBLE] AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: [INAUDIBLE] meter squared per second. OK, so feasible. [INTERPOSING VOICES] WILLIAM GREEN JR: Anybody know if it's meter squared or centimeter squared? What? [INTERPOSING VOICES] WILLIAM GREEN JR: You guys did the one of the drug. What was the drugs one? Was it 10 to the minus 6, meter squared or centimeter squared? [INTERPOSING VOICES] WILLIAM GREEN JR: Meters squared. OK, so for a light, smaller molecule, it's reasonable. So this number should be something like 10 to the minus 5, divided by 10 to the minus 12, so something like 10 to the 7. 1, 8, and 7. All right, now for the reaction, roo plus RH, we can go look it up on the [INAUDIBLE].. They'll have a list of reactions like that. For now, let's just guess. And so let's say that k3-- k of roo plus rh-- these guys typically have 8 factors of 10 to the eighth leaders per mole second. And they have ea's, about 15-- 15 k cals per mole over rt. OK so now, we need to figure out how to turn this into the k3 we want. Now, the k3 we want is really-- K3 is like this-- times the concentration of the rh, because we took the rh out of the problem, because we want this to have units of per second. Is that right? And this is in units of volume per meters cubed per mole second. And this would be moles per meter cubed. And so it's per second, which looks right. OK so now, we need to have-- I understand we have a calculator? We can try to calculate this number for room temperature. And this does not mean room temperature is r time t-- actually for body temperature. And the concentration of hydrocarbon-- it's actually the number of h's, because every h can be attacked in a hydrocarbon. So you typically have densities of 0.8 grams per centimeter cubed. It's for organic. And you typically have two h atoms per 14 grams of ch2 groups, because you only see h2 groups in there. S two h atoms. And then, if we're going to try to keep moles here, we need 6 times 10 to the 23rd atoms per mole. All right, and this is centimeters cubed, but this one has leaders, so I need 10 to the 3 centimeters cubed [INAUDIBLE].. So again, verifying what your high school teacher taught you, that the first thing is to learn units. [INAUDIBLE] So we take all these numbers. This is the RH, and this is the k. And we can multiply them all together, and we should get something reasonable, maybe. And since we have MATLAB we can make it do it for us. So let's just do that. So I think it's-- you have to help me, I can't read this very well. 1ea star exponential, [INAUDIBLE] 15,000, slash 1.987 times-- what's body temperature? 40ec you told me? So it's 310 Kelvin maybe-- times all these numbers, 0.8 times 2. Divide by 10 to the 23rd. More factors in there. OK, so 5 times 10 to the minus 25. We think we got this right? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: But I want to get moles, because I have this in moles-- it's [INAUDIBLE]. Three times mole [INAUDIBLE] This is really mole of [INAUDIBLE] atoms. OK, so we'll try and see what happens. So 5 times 7 minus 25. It does seem pretty small, so we have a problem. And then the last reaction-- those reactions are typically 10 to the fifth, liters per mole second. This is for a recombination of peroxy radicals. So now, we have to get rid of the volume unit. This is for roo-- sorry, [INAUDIBLE] Yeah? Yes, question? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: One mole per [INAUDIBLE].. Very good-- that's right, because [INAUDIBLE] that's just two atoms. It's actually two moles. Yes? That would change the [INAUDIBLE] Only a factor of 10 to the 23rd. So we'll get back. Thank you. Area 0.3, somewhere like it. OK, so now let's go do the same thing with this one, ro plus roo. So typically these reactions, the normal way they're written are numbers that are about 10 to 5th meters per mole second, for two radicals recombining. So now, I need to figure out how can I change the volume here. And the thing that is key is that I know the volume of this lipid particle. So what I really want is this. My k4 is going to be equal to normal k, divided by the volume of the reactor, because the rate we would write normally, which would be droodt-- it would be like negative 2k roo plus roo, times the moles, the concentration of roo squared. This how we would normally write it. And so we have to have the-- this unit is all right. So this is the volume of the reactor. So we were having nroo over v, so we might need Avogadro's number over here, because the rate here is in moles. But now, we are going to do single molecules. Yeah? AUDIENCE: Do you ever actually use k4 in calculations? WILLIAM GREEN JR: Yeah, for a. I think an a is there. Good question. All right, so we think this should be like a volume per molecule or something like that, for one molecule. So this should be k4, should be 10 to the 5th, meters per mole second, divided by 4/3 pi r cubed. And then, we need a mole-- [INAUDIBLE] 23rd molecules. And we're gonna have to make sure that the leaders and these guys match up, which is not going to be so easy. You might use 10 to the minus 6 meters. This is going to be meters cubed, so I need 10 to the 3 liters per liter cubed. And I think that will all work out to be per second. So let's try it. So this would be 10 to the 5th over 4/3 pi is about 4-- 10 to the negative 18, 6 times 10 to the 23rd. So all the big ones cancel each other out, is a 1 over 24. Oh, I lost it. That's bad. OK, so one over 24,000. It's a lot easier to do with a lot of people checking your work. All right, so four times n minus 5. So this is what we think k is. Anything else we need. We need a t-max, you want to guess the time? You want to wait? The time for the main reaction, the time constant is like seconds, right? So if we made 1,000 seconds, a lot of stuff is going to happen, right? So maybe 1,000 seconds will be OK for t-max. So let's try this all. Where's the main function. The main function-- [INAUDIBLE] my n initial was simple 1, 0. And k, which was decided was 1e minus 5, [INAUDIBLE],, 2.3, 4e, 25. Now, if you just look at these rates for a second, I think you can see a problem we're going to have. How many steps are we going to do? I don't know. 10,000. The time scales here-- this 27 is pretty fast. So the stuff is going to diffuse out of there, because it's going to disappear, and then nothing is gonna happen. Yeah? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: Centimeters squared, I thought so, yeah. Thank you, so we think this is 10 to the minus 9. So this is actually [INAUDIBLE] lower, is that right? We're still gonna have a problem. But it's not quite bad [INAUDIBLE] problem. So let's fix that. Um, so I think our problem is that the 1e minus 5 is actually too slow. So radicals are diffusing in from outside at some rate, and it's got to be a rate so that if there was no reaction, basically, the concentrations of the stuff here and here might be about the same, same order of magnitude, of the stuff that dissolved in the water, dissolved in the lipid. Maybe a couple of [INAUDIBLE] are different, but not a million times different. And so we need to have the rate of this stuff coming in to be something reasonable, that's going to be consistent with having a chance of having some radical in there. So now we to guess what we think the real outside world radical concentration is. So maybe-- I don't know, what? 1e minus 10 moles per liter? 1e minus 6 moles per liter? I don't know. Any biologists here? Do you know how many free radicals you have in your body? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: They can measure it, right? So it can't be zero. People talk about reactive oxygen species in your body, ROS. Yes, so I don't know what it is, 10 to the minus 6, maybe-- 10 to the minus 8. So we had the number for-- 1e3 e3 is for going out from the-- we have that rate leaving is 1e3 per second times the number in there divided by the volume, really. Well, it's times the number, that's the rate that's leaving. But we would think of this as a k. I don't know how to think of this, actually. But for sure, the volume matters. So in order for us to compute this, we use the r squared. Now, another way you look at this is say well, is diffusion times the gradient in the concentration? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: [INAUDIBLE] Well, it's OK. They would live in their little tiny thing for a millisecond, and then they diffuse out. I think that actually sounds very reasonable. It's only a micron. It's a very tiny, little thing. So the question is how do we complete the reverse? So if we have stuff outside, and suppose we didn't have anything inside, we'd have some rate of diffusion of the radical from the outside coming in. How do we think about it? It's really got to be the inverse of this. So if we think the time constant is 10 to the minus 3-- a millisecond to come out, it's probably about a millisecond to come in. I don't know. You think it's got to maintain whatever initial concentration we put there, one per volume, then it's got to be about the same. So this can't be as tiny as 10 to the minus 5. So if we take the average concentration as one, then this should 10 to the 3, I think. If we think the average concentration is 10, then it should be 10 to the 4. If we think the average concentration this 100, then it should be 10 to the five. So let's try 10 to the 3. So every millisecond, something comes in or goes out, some radical comes in and out. You guys buy this? Would you defend this to your boss? You can blame it on Professor Green. All right, here goes nothing. Do you think it's going to work? [LAUGHTER] Aww, aww, line 29. What's wrong? What's wrong with that? Parentheses off? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: Maybe I didn't save the latest version, let's try again. Ah, I didn't say the latest version. That's what it is. Now, here goes nothing. That [INAUDIBLE] help. Leave me alone. All right, trajectory, the first column is the times, so it uses the x-axis. And the last column is the oxidation, the products. So let's see if anything peroxidized or not. [LAUGHTER] [INTERPOSING VOICES] WILLIAM GREEN JR: Oh, it should have all the zeros, right? So how can we do it when we [INAUDIBLE] that way. That's really weird. Try it again. [INTERPOSING VOICES] WILLIAM GREEN JR: The time is 2 plus tau. AUDIENCE: [INAUDIBLE]. WILLIAM GREEN JR: I guess if you look at the [INAUDIBLE] So trajectory-- so it only [INAUDIBLE] 2 points. That's why it looks like that. I don't know why it doesn't show the zeros. Maybe they'll show on top of the origin there. So why did it only give us two points? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: Yeah, also-- seems like a 307 oxidations, but why 307 all of a sudden? That makes no sense, right-- because it should be one at a time. I should be seeing one at a time coming in. AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: Yes, it's very worrying. I agree, so we have a bug. So where is our bug. This is where I need your help. OK guys, let's figure out why didn't this work. So I'm storing-- by stepping the steps, they're going up. Ah, I know what's wrong. I need this step thing inside the loop here. So all of this-- so let's see if I can explain this. I'm storing them according to the step, but right now, the wild loop is just zipping around, and then, it ends at the step and just gets one. So I'd rather start again. It's gonna work this time? You guys don't have any faith. If people say that scientists and engineers don't have any faith, I think we have more faith than anybody else. We believe stuff like this is going to work. It's doing something now. Who knows what? [INTERPOSING VOICES] WILLIAM GREEN JR: Yup? AUDIENCE: [INAUDIBLE] time until your arrival time, 7:15 [INAUDIBLE] WILLIAM GREEN JR: [INAUDIBLE] is, not good, right? AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: Yeah, it's a very good point. I have a nested loop that should not be, right? AUDIENCE: [INAUDIBLE] You don't necessarily want to sample every arrival time. You might want to sample [INAUDIBLE] So you might want to have two. You don't necessarily want to [INAUDIBLE] every single-- WILLIAM GREEN JR: Yes, yes, I agree. AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: OK, so let's see how we can fix this. So we really want-- is it really a jump out? If the number of steps gets too large, we just want to jump out, so we want to go to? What do you think? [INTERPOSING VOICES] WILLIAM GREEN JR: Break, that's what we want. So while is not the correct thing to do here. AUDIENCE: Why would t [INAUDIBLE].. WILLIAM GREEN JR: So we're stepping time, every time we'll do the steps. And we just want a break. [INAUDIBLE] step greater than max steps. AUDIENCE: Why don't you do it with a while loop [INAUDIBLE] WILLIAM GREEN JR: What happens if you do a wild loop like that? Do you guys know? Does it work? [INTERPOSING VOICES] WILLIAM GREEN JR: It'll work. OK, that's easy. Double and-- I tired it with and this morning. It didn't work, so that's why I stopped doing it. Stuff like that will kill you, doesn't it? All right, we gotta get rid of one of the ends at the end. AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: I think it's likely to cause a problem, why it's taking 10 million years. So really, is there a reason it should be nested loop, right? Yes. AUDIENCE: [INAUDIBLE] WILLIAM GREEN JR: All right, try again. [INTERPOSING VOICES] WILLIAM GREEN JR: That's too fast. Try that again. That's not good. See we have a [INAUDIBLE] here. AUDIENCE: Oh no. WILLIAM GREEN JR: Oh, jumps to one. That doesn't look very good. Try that again. Boom. AUDIENCE: Oh! WILLIAM GREEN JR: Oh, that was pretty interesting, so much more like what you expect, so that at some time steps, I have one, two, three, or four, or five radicals in this vesicle. I only have one time separate because of the nine. So I think this actually looks like what I expect. I should see it jumping up and down, stuff goes in and out, sometimes it reacts, sometimes it goes away. So this one is correct. I don't know what's happened with the other one. So I'll have to figure out what's wrong with that equation. All right, we're almost done time. Well, I want to talk about for just one minute is-- what are we gonna do after we have this working? So we get this working, we have the trajectory, we really want to run 10,000 trajectories, because all we're doing is sampling from the probability, just the distribution of time. So we need to run zillions of them. So this whole thing we just wrote would be inside a loop that runs a lot of different cases, because every time you run this, because the random number, you're going to get a different trajectory. And what you care about is some kind of average behavior. Or you might want to know what's the percentage chance that I'm going to have [INAUDIBLE] oxidized and die? So you might say I want to count what fraction of the trajectories end up with number of peroxide greater than some number, that means I'm dead. And if that probability is too high, then I know I'm in big trouble. So that'd be one possible thing. You might have whatever objective function you want, but you need to run a lot to get good statistics for anything. So you're gonna have to embed this whole calculation inside a loop, and then, you're going to get a zillion of these trajectories out. And you figure out how are you going to analyze them in order to figure out what you want? So one way is if you could add the trajectories up-- so I have a trajectory-- suppose I have the number roh's versus time. And I do it once, and I have none. Then I have one, and then I have two, and then, I wait longer, and I get three. And then, I wait longer, I get four. Whatever, something like that. That's what it really looks like for one trajectory. And then, the next time I run it, it starts in a slightly different time. And this time it runs longer before something else happens. And then, it gets here, and then maybe, it goes over here. And I have a lot of them like that. I have 10,000 trajectories, all look like that. So if I could add them, then I can do an average, to get an average trajectory, so that would be one possible thing. Or I might want to histogram, so I might pick one time point. So like after 20 minutes, I want a histogram of what this looks like here or here. One of them has four peroxides, and one of them has five peroxides. And then the other 9,999-- some of the three, some of the whatever. Yes, question? AUDIENCE: [INAUDIBLE] you're not gonna get the same time points-- WILLIAM GREEN JR: You're not going to get the same time points, that's right. So see the first time it stopped, the first time point was here, the second time, the first time point was here. And they'll all be different. So one is I was asking a few of the students before the class started, there's got to be a program in MATLAB that will let you generate these kind of plots with the flat lines. Or even a linear interpolation would be OK too. But you need to have some way to add them up as continuous functions, because they all have different time points. So that's one practical issue about it, because you want to pick some special time you care about, and you want to know what does the trajectory say, what does this trajectory say? But you actually only have numbers here, just really you have that number and this number and this number. That's all you got. Yeah? AUDIENCE: You can plot like a-- an out of time to get to n amount of roh's and then, that would be [INAUDIBLE] WILLIAM GREEN JR: OK, so then you plot [INAUDIBLE] versus time instead, that's right. So that you just think of what you want. You're going to have all this trajectory information. And then you need to figure out what do you want, and then, what are you going to plot? And what are you trying to compute? And you might want to compute not only the number but the standard deviation of that number, because you don't know how many trajectories you have to run before the center of deviation is narrow enough that you'll be confident with it. So this is a general problem with this whole approach-- is that what you're getting out are just samples of things that happen. It's just like if you were doing a Monte Carlo, and you just got some of the energy values from the hydrogen peroxide calculation you guys did, you have 47 of those energy values. What are you going to do with that? So you have to figure out, how are you going to take the [INAUDIBLE] of this calculation. It's sampling from the real solution, so it should be OK. But what is it? How are you going to handle that and use it as a means-- it's sort of like as if you ran a zillion experiments, and then, what would you do with all that data? So that's like issue number one. Now alternatively, you can rewrite this as the master equation and solve it as an ODE, in which case, you'll get explicitly p of each of these ends. So n rad, nroh time-- it would be continuous in principle, but actually, the ODE solver gives you out random time points also. So you get a similar kind of thing out from the ODE solver, except it'll give you probability for every possible range of these guys. So if you have one radical, two radicals, three radicals, four radical, fie radicals, and number of peroxides from one to 1,000, or however many peroxides you get, you're going to get numbers. So you have all these numbers, all these probabilities, at different times. Again, what are you going to do with that? So one thing people do a lot is they would plot, say, the number of roh's versus time. So at different time points, compute the average over all your trajectories. Were with, if you had this solution, you can compute that as nrooh, p, nrooh. So like this, you're gonna have some kind of average if you do it that way. And either one are fine, but you have to think of what you want, and then in both cases, you'll probably be interested in the dispersion. So you [INAUDIBLE] want to worry about the n squared too. We'll talk more about this on Monday. All right and I posted a homework problem that was given last year about this, for Kinetic Monte Carlo for catalysis. It's used a lot in heterogeneous catalysis. And if you have extra time, feel free to do the problem. It's a really good problem. Those of you who are feeling like you don't have any extra time, and you're about to kill yourself, please don't kill yourself. And instead, just send an email to me or Professor Swan asking for an extension. And we can give you some more time to finish up the homeworks that's due tonight. All right, talk to you later. |
MIT_1034_Numerical_Methods_Applied_to_Chemical_Engineering_Fall_2015 | 5_Eigenvalues_and_Eigenvectors.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. JAMES W. SWAN: Let's go ahead and get started. I hope everybody saw the correction to a typo in homework 1 that was posted on Stellar last night and sent out to you. That's going to happen from time to time. We have four course staff that review all the problems. We try to look through it for any issues or ambiguities. But from time to time, we'll miss something and try to make a correction. The TAs gave a hint that would have let you solve the problem as written. But that's more difficult than what we had intended for you guys. We don't want to give you a homework assignment that's punishing. We want to give you an assignment that'll help you learn. Some people in this class that are very good at programming have apparently already completed that problem with the hint. But it's easier, as originally intended. And the correction resets that. So maybe you'll see the distinction between those things and understand why one version of the problem is much easier than another. But we try to respond as quickly as possible when we notice a typo like that so that we can set you guys on the right course. So we've got two lectures left discussing linear algebra before we move on to other topics. We're still going to talk about transformations of matrices. We looked at one type of transformation we could utilize for solving systems of equations. Today, we'll look at another one, the eigenvalue decomposition. And on Monday, we'll look at another one called the singular value decomposition. Before jumping right in, I want to take a minute and see if there are any questions that I can answer, anything that's been unclear so far that I can try to reemphasize or focus on for you. I was told the office hours are really well-attended. So hopefully, you're getting an opportunity to ask any pressing questions during the office hours or you're meeting with the instructors after class to ask anything that was unclear. We want to make sure that we're answering those questions in a timely fashion. This course moves at a pretty quick pace. We don't want anyone to get left behind. Speaking of getting left behind, we ran out of time a little bit at the end of lecture on Wednesday. That's OK. There were a lot of good questions that came up during class. And one topic that we didn't get to discuss is formal systems for doing reordering in systems of equations. We saw that reordering is important. In fact, it's essential for solving certain problems via Gaussian elimination. You won't be able to solve them. Either you'll incur a large numerical error because you didn't do pivoting-- you'd like to do pivoting in order to minimize the numerical error-- or you need to reorder in order to minimize fill-in. As an example, I've solved a research problem where there was something like 40 million equations and unknowns, a system of partial differential equations. And if you reorder those equations, then you can solve via Gaussian elimination pretty readily. But if you don't, well-- my PC had-- I don't know-- like, 192 gigabytes of RAM. The elimination on that matrix will fill the memory of that PC up in 20 minutes. And you'll be stuck. It won't proceed after that. So it's the difference between getting a solution and writing a publication about the research problem you're interested in and not. So how do you do reordering? Well, we use a process called permutation. There's a certain class of matrix called a permutation matrix that can-- its action, multiplying another matrix, can swap rows or columns. And here's an example of a permutation matrix whose intention is to swap row 1 and 2 of a matrix. So here, it looks like identity, except rather than having 1, 1 on the first two elements of the diagonal, I have 0, 1 and 1, 0. Here's an example where I take that sort of a matrix, which should swap rows 1 and 2, and I multiply it by a vector. If you do this matrix vector multiplication, you'll see initially, the vector was x1, x2, x3. But the product will be x2, x1, x3. It swapped two rows in that vector. Of course, a vector is just a matrix, right? It's an N by 1 matrix. So P times A is the same as a matrix whose columns are P times each of the columns of A. That's what this notation indicates here. And we know that P times a vector, which is the column from A, will swap two rows in A, right? So the product here will be all the rows of A, the different rows of AA superscript R, with row 1 and 2 swapped with each other. So permutation, multiplication by the special type of matrix, a permutation matrix, does reordering of rows. If I want to swap columns, I multiply my matrix from the right, IP transpose. So if I want to swap column 1 and 2, I multiply A from the right by P transpose. How can I show that that swaps columns? Well, A times P transpose is the same as P times A transpose transpose. P swaps rows. So it's swapping rows of A transpose, which is like swapping columns of A. So we had some identities associated with matrix-matrix multiplication and their transposes. And you can use that to work out how this permutation matrix will swap columns instead of rows if I multiply from the right instead of the left. Here's an important concept to know. Permutation matrices are-- would refer to as unitary matrices. They're transposed. It's also they're inverse. So P times P transpose is identity. If I swap the rows and then I swap them back, I get back what I had before. So there are lots of matrices that have this property that they're unitary. We'll see some today. But permutation matrices are one class, maybe the simplest class, of unitary matrices. They're just doing row or column swaps, right? That's their job. And so if I have some reordering of the equations or rows of my system of equations that I want, that's going to be indicated by a permutation matrix-- say, P1. And I would multiply my entire system of-- both sides of my system of equations by P1. That would reorder the rows. If I have some reordering of the columns or the unknowns in my problem, I would use a similar permutation matrix, P2. Of course, P2 transpose times P2 is identity. So this product here does nothing to the system of equations. It just swaps the unknown. So there's a formal system for doing this sort of swapping. There are a couple other slides that are in your notes from last time that you can look at and I'm happy to answer questions on. We don't have time to go into detail. It discusses the actual methodology, the simplest possible methodology, for doing this kind of reordering or swapping. So this is a form of preconditioning. If it's preconditioning for pivoting, it's designed to minimize numerical error. If it's preconditioning in order to minimize fill-in instead, that's meant to make the problem solvable on your computer. But it's a form of preconditioning a system of equations. And we discussed preconditioning before. So now we know how to solve systems of equations. It's always done via Gaussian elimination if we want an exact solution. There are lots of variants on Gaussian elimination that we can utilize. You're studying one of them in your homework assignment now, where you know the matrix is banded with some bandwidth. So you don't do elimination on an entire full matrix. You do it on a sparse matrix whose structure you understand. We discussed sparse matrices and a little bit about reordering and now permutation. I feel like my diffusion example last time wasn't especially clear. So let me give you a different example of diffusion. You guys know Plinko? Have you seen The Price Is Right? This is a game where you drop a chip into a board with pegs in it. It's a model of diffusion. The Plinko chip falls from level to level. It hits a peg. And it can go left or it can go right with equal probability. So the Plinko chip diffuses as it falls down. This guy's excited. [LAUGHTER] He just won $10,000. [LAUGHTER] There's a sparse matrix that describes how the probability of finding the Plinko chip in a certain cell evolves from level to level. It works the same way the cellular automata model I showed you last time works. If the chip is in a particular cell, then at the next level, there's a 50/50 chance that I'll go to the left or I'll go to the right. It looks like this, right? If the chip is here, there's a 50/50 chance I'll go here or I'll go there. So if the probability was 1 that I was in this cell, then at the next level, it'll be half and a half. And at the next level, those halves will split again. So the probability that I'm in a particular cell at level i is this Pi. And the probability that I'm in a particular cell level i plus 1 is this Pi plus one. And there's some sparse matrix A which spreads that probability out. It splits it into my neighbors 50/50. Here's a simulation of Plinko. So I started with the probability 1 in the center cell. And as I go through different levels, I get split 50/50. And you see a binomial or almost Gaussian distribution spread as I go through more and more levels until it's equally probable that I could wind up in any one of the cells. You can think about it this way, right? The probability at level i plus 1 that the chip is in cell N is inherited 50/50 from its two neighbors, right? There's some probability that was in these two neighbors. I would inherit half of that probability. It would be split by these pegs. The sparse matrix that represents this operation has two diagonals. And on each of those diagonals is a half. And you can build that matrix using the spdiags command. It says that there's going to be two diagonal components which are equal to a half. And their position is going to be one on either side of the central diagonal. That's going to indicate that I pass this probability, 50/50, to each of my neighbors. And then successive multiplications by A will split this probability. And we'll see the simulation that tells us how probable it is to find the Plinko chip in a particular column. Yes? AUDIENCE: [INAUDIBLE] JAMES W. SWAN: Yeah. So in diffusion in general? AUDIENCE: Well, in this instance in particular because [INAUDIBLE] JAMES W. SWAN: Well, OK. That's fair enough. This is one particular model of the Plinko board, which sort of imagines alternating cells that I'm falling through. We could construct an alternative model, if we wanted to, that didn't have that part of the picture, OK? So that's a matrix that looks like this, right? The central diagonal is 0. Most of the off-diagonal components here are 0 and 1 above and 1 below. I get a half and a half. And if I'm careful-- somebody mentioned I need boundary conditions. When the Plinko chip gets to the edge, it doesn't fall out of the game. It gets reflected back in. So maybe we have to choose some special values for a couple of elements of this matrix. But this is a sparse matrix. It has a sparse structure. It models a diffusion problem, just like we saw before. Most of physics is local, like this, right? I just need to know what's going on with my neighbors. And I spread the probability out. I get this nice diffusion problem. So it looks like this. Here's something to notice. After many levels or cycles, I multiply by A many, many times. This probability distribution always seems to flatten out. It becomes uniform. It turns out there are even special distributions for which A times A times that distribution is equal to that distribution. You can see it at the end here. This is one of those special distributions where the probability is equal in every other cell, right? And at the next level, it all gets passed down. That's one multiplication by-- it all gets spread by 50%. And the next multiplication, everything gets spread by 50% again. And I recover the same distribution that I had before, this uniform distribution. That's a special distribution for which A times A times P is equal to P. And this distribution is one of the eigenvectors of this matrix A times A. It's a particular vector that when I multiply it by this matrix AA, I get that vector back. It happens to be unstretched. So this vector points in some direction. I transform it by the matrix. And I get back something that points in the same direction. That's the definition of this thing called an eigenvector. And this will be the subject that we focus on today. So eigenvectors of a matrix-- they're special vectors that are stretched on multiplication by the matrix. So they're transformed. But they're only transformed into a stretched form of whatever they were before. They point in a direction. You transform them by the matrix. And you get something that points in the same direction, but is stretched. Before, we saw the amount of stretch. The previous example, we saw the amount of stretch was 1. It wasn't stretched at all. You just get back the same vector you had before. But in principle, it could come back with any length. For a real N-by-N matrix, there will be eigenvectors and eigenvalues, which are the amount of stretch, which are complex numbers. And finding eigenvector-eigenvalue pairs involves solving N equations. We'd like to know what these eigenvectors and eigenvalues are. They're non-linear because they depend on both the value and the vector, the product of the two, for N plus 1 unknowns. We don't know how to solve non-linear equations yet. So we're kind of-- might seem like we're in a rough spot. But I'll show you that we're not. But because there's N equations for N plus 1 unknowns, that means eigenvectors are not unique. If W is an eigenvector, than any other vector that points in that same direction is also an eigenvector, right? It also gets stretched by this factor lambda. So we can never say what an eigenvector is uniquely. We can only prescribe its direction. Whatever its magnitude is, we don't care. We just care about its direction. The amount of stretch, however, is unique. It's associated with that direction. So you have an amount of stretch. And you have a direction. And that describes the eigenvector-eigenvalue pair. Is this clear? You've heard of eigenvalues and eigenvectors before? Good. So how do you find eigenvalues? They seem like special sorts of solutions associated with a matrix. And if we understood them, then we can do a transformation. So I'll explain that in a minute. But how do you actually find these things, these eigenvalues? Well, I've got to solve an equation A times w equals lambda times w, which can be transformed into A minus lambda identity times w equals 0. And so the solution set to this equation is either w is equal to 0. That's one possible solution to this problem or the eigenvector w belongs to the null space of this matrix. It's one of those special vectors that when it multiplies this matrix gives back 0, right? It gets projected out on transformation by this matrix. Well, this solution doesn't seem very useful to us, right? It's trivial. So let's go with this idea that w belongs to the null space of A minus lambda I. That means A minus lambda I must be a singular matrix, whatever it is, right? And if it's singular, then the determinant of a minus lambda I must be equal to 0. So if this is true, and it should be true if we don't want a trivial solution, then the determinant of A minus lambda I is equal to 0. So if we can compute that determinant and solve for lambda, then we'll know the eigenvalue. Well, it turns out that the determinant of a matrix like A minus lambda I is a polynomial in terms of lambda. It's a polynomial of degree N called the characteristic polynomial. And the N roots of this characteristic polynomial are called the eigenvalues of the matrix. So there are N possible lambdas for which A minus lambda I become singular. It has a null space. And associated with those values are eigenvectors, vectors that live in that null space. So this polynomial-- we could compute it for any matrix. We could compute this thing in principle, right? And we might even be able to factor it into this form. And then lambda 1, lambda 2, lambda N in this factorized form are all the possible eigenvalues associated with our matrix A, right? There are all the possible amounts of stretch that can be imparted to particular eigenvectors. We don't know those vectors yet, right? We'll find them in a second. But we know the amounts of stretch that can be imparted by this matrix. OK? Any questions so far? No. Let's do an example. Here's a matrix, minus 2, 1, 3. And it's 0's everywhere else. And we'd like to find the eigenvalues of this matrix. So we need to know A minus lambda I and its determinant. So here's A minus lambda I. We just subtract lambda from each of the diagonals. And the determinant-- well, here, it's just the product of the diagonal elements. So that's the determinant of a diagonal matrix like this, the product of the diagonal elements. So it's minus 2 minus lambda times 1 minus lambda times 3 minus lambda. And the determent of this has to be equal to 0. So the amounts of stretch, the eigenvalues imparted by this matrix, are minus 2, 1, and 3. And we found the eigenvalues. Here's another matrix. Can you work out the eigenvalues of this matrix? Let's take 90 seconds. You can work with your neighbors. See if you can figure out the eigenvalues of that matrix. Nobody's collaborating today. I'm going to do it myself. AUDIENCE: [INAUDIBLE] JAMES W. SWAN: It's OK. OK. What are you finding? Anyone want to guess what are the eigenvalues? AUDIENCE: [INAUDIBLE] JAMES W. SWAN: Good. OK. So we need to compute the determinant of A minus lambda I. That'll be minus 2 minus lambda times minus 2 minus lambda minus 1. You can solve this to find that lambda equals minus 3 or minus 1. These little checks are useful. If you couldn't do this, that's OK. But you should try to practice this on your own to make sure you can. Here are some more examples. So the elements of a diagonal matrix are always the eigenvalues because the determinant of a diagonal matrix is the product of the diagonal elements. So these diagonal values here are the roots of the secular characteristic polynomial. They are the eigenvalues. It turns out the diagonal elements of a triangular matrix are eigenvalues, too. This should seem familiar to you. We talked about easy-to-solve systems of equations, right? Diagonal systems of equations are easy to solve, right? Triangular systems of equations are easy to solve. It's also easy to find their eigenvalues. So the diagonal elements here are the eigenvalues of the triangular matrix. And eigenvalues have certain properties that can be inferred from the properties of polynomials, right? Since they are the roots to a polynomial, if we know certain things that should be true of those polynomial of roots, that has to be true of the eigenvalues themselves. So if we have a matrix which is real-valued, then we know that we're going to have this polynomial of degree N which is also real-valued, OK? It can have no more than N roots, right? And so A can have no more than N distinct eigenvalues. The eigenvalues, like the factors of the polynomial, don't have to be distinct, though? You could have multiplicity in the roots of the polynomial. So it's possible that lambda 1 here is an eigenvalue twice. That's referred to as algebraic multiplicity. We'll come back to that idea in a second. Because the polynomial is real-valued, it means that the eigenvalues could be real or complex, just like the roots of a real-valued polynomial. But complex eigenvalues always appear as conjugate pairs. If there is a complex eigenvalue, then necessarily its complex conjugate is also an eigenvalue. And here's a couple other properties. So the determinant of a matrix is the product of the eigenvalues. We talked once about the trace of a matrix, which is the sum of its diagonal elements. The trace of a matrix is also the sum of the eigenvalues. These can sometimes come in handy-- not often, but sometimes. Here's an example I talked about before-- so a series of chemical reactions. So we have a batch, a batch reactor. We load some material in. And we want to know how the concentrations of A, B, C, and D vary as a function of time. And so A transforms into B. B and C are in equilibrium. C and D are in equilibrium. And our conservation equation for material is here. This is a rate matrix. We'd like to understand what the characteristic polynomial of that is. The eigenvalues of that matrix are going to tell us something about how different rate processes evolve in time. You can imagine just using units. On this side, we have concentration over time. On this side, we have concentration. And the rate matrix has units of rate, or 1 over time. So those eigenvalues also have units of rate. And they tell us the rate at which different transformations between these materials occur. And so if we want to find the characteristic polynomial of this matrix and we need to compute the determinant of this matrix minus lambda I-- so subtract lambda from each of the diagonals-- even though this is a four-by-four matrix, its determinant is easy to compute because it's full of zeros. I'm not going to compute it for you here. It'll turn out that the characteristic polynomial looks like this. You should actually try to do this determinant and show that the polynomial works out to be this. But knowing that this is the characteristic polynomial, what are the eigenvalues of the rate matrix? If that's the characteristic polynomial, what are the eigenvalues, or tell me some of the eigenvalues of the rate matrix? AUDIENCE: 0. JAMES W. SWAN: 0. 0's an eigenvalue. Lambda equals 0 is a solution. Minus k1 is another solution. What is this eigenvalue 0 correspond to? What's that? AUDIENCE: [INAUDIBLE] JAMES W. SWAN: OK. Physically, it's a rate process with 0 rate, steady state. So the 0 eigenvalue's going to correspond to the steady state. The eigenvector associated with that eigenvalue should correspond to the steady state solution. How about this eigenvalue minus k1? This is a rate process with rate k1. What physical process does that represent? It's something evolving in time now, right? So that's the transformation of A into B. And the eigenvector should reflect that transformation. We'll see what those eigenvectors are in a minute. But these eigenvalues can be interpreted in terms of physical processes. This quadratic solution here has some eigenvalue. I don't know what it is. You use the quadratic formula and you can find it. But it involves k2, k3, k4. And this is a typo. It should be k5. And so that says something about the interconversion between B, C, and D, and the rate processes that occur as we convert from B to C to D. Is that too fast? Do you want to write some more on this slide before I go on, or are you OK? Are there any questions about this? No. Given an eigenvalue, a particular eigenvalue, what's the corresponding eigenvector? We know the eigenvector isn't uniquely specified. It belongs to the null space of this matrix A minus lambda I times identity. Even though it's not unique, we might still try to find it using Gaussian elimination, right? So we may try to take-- we may try to solve the equation A minus lambda I times identity multiplied by w equals 0 using Gaussian elimination. But because it's not unique, at some point, we'll run out of rows to eliminate, right? There's a null space to this matrix, right? We won't be able to eliminate everything. We'd say it's rank deficient, right? So we'll be able to eliminate up to some R, the rank of this matrix. And then all the components below are essentially free or arbitrarily specified. There are no equations to say what those components of the eigenvector are. The number of all 0 rows-- it's called the geometric multiplicity of the eigenvalue. Sorry. Geometric is missing here. It's the number of components of the eigenvector that can be freely specified. The geometric multiplicity might be 1. That's like saying that the eigenvectors are all pointing in the same direction, but can have arbitrary magnitude, right? It might have geometric multiplicity 2, which means the eigenvectors associated with this eigenvalue live in some plane. And any vector from that plane is a corresponding eigenvector. It might have a higher geometric multiplicity associated with it. So let's try something here. Let's try to find the eigenvectors of this matrix. I told you what the eigenvalues were. They were the diagonal values here. So they're minus 2, 1, and 3. Let's look for the eigenvector corresponding to this eigenvalue. So I want to solve this equation A minus this particular lambda, which is minus 2, times identity equals 0. So I got to do Gaussian elimination on this matrix. It's already eliminated for me, right? I have one row which is all 0's, which says the first component of my eigenvector can be freely specified. The other two components have to be 0. 3 times the second component of my eigenvector is 0. 5 times the third component is 0. So the other two components have to be 0. But the first component is freely specified. So the eigenvector associated with this eigenvalue is 1, 0, 0. If I take a vector which points in the x-direction in R3 and I multiply it by this matrix, it gets stretched by minus 2. So I point in the other direction. And I stretch out by a factor of 2. You can guess then what the other eigenvectors are. What's the eigenvector associated with this eigenvalue here? 0, 1, 0, or anything proportional to that. What's the eigenvector associated with this eigenvalue? 0, 0, 1, or anything proportional to it. All these eigenvectors have a geometric multiplicity of 1, right? I can just specify some scalar variant on them. And they'll transform into themselves. Here's a problem you can try. Here's our series of chemical reactions again. And we want to know the eigenvector of the rate matrix having eigenvalue 0. This should correspond to the steady state solution of our ordinary differential equation here. So you've got to do elimination on this matrix. Can you do that? Can you find this eigenvector? Try it out with your neighbor. See if you can do it. And then we'll compare results. This will just be a quick test of understanding. Are you guys able to do this? Sort of, maybe? Here's the answer, or an answer, for the eigenvector. It's not unique, right? It's got some constant out in front of it. So you do Gaussian elimination here. So subtract or add the first row to the second row. You'll eliminate this 0, right? And then add the second row to the third row. You'll eliminate this k2. You have to do a little bit more work to do elimination of k4 here. But that's not a big deal. Again, you'll add the third row to the fourth row and eliminate that. And you'll also wind up eliminating this k5. So the last row here will be all 0's. And that means the last component of our eigenvector's freely specifiable. It can be anything we want. So I said it is 1. And then I did back substitution to determine all the other components, right? That's the way to do this. And here's what the eigenvector looks like when you're done. The steady state solution has no A in it. Of course, A is just eliminated by a forward reaction. So if we let this run out to infinity, there should be no A. And that's what happens. But there's equilibria between B, C, and D. And the steady state solution reflects that equilibria. We have to pick what this constant out in front is. And we discussed this before, actually, right? You would pick that based on how much material was initially in the reactor. We've got to have an overall mass balance. And that's missing from this system of equations, right? Mass conservation is what gave the null space for this rate matrix in the first place. Make sense? Try this example out. See if you can work through the details of it. I think it's useful to be able to do these sorts of things quickly. Here are some simpler problems. So here's a matrix. It's not a very good matrix. Matrices can't be good or bad. It's not particularly interesting. But it's all 0's. So what are its eigenvalues? It's just 0, right? The diagonal elements are the eigenvalues. And they're 0. That eigenvalue has algebraic multiplicity 2. It's a double root of the secular characteristic polynomial. Can you give me the eigenvectors? Can you give me eigenvectors of this matrix? Can you give me linearly independent-- yeah? AUDIENCE: [INAUDIBLE] JAMES W. SWAN: OK. AUDIENCE: [INAUDIBLE] JAMES W. SWAN: OK. Good. So this is a very ambiguous sort of problem or question, right? Any vector I multiply by A here is going to be stretched by 0 because A by its very nature is all 0's. All those vectors live in a plane. So any vector from that plane is going to be transformed in this way. The eigenvector corresponding to eigenvalue 0 has geometric multiplicity 2 because I can freely specify two of its components. Oh my goodness. I went so fast. We'll just do it this way. Algebraic multiplicity 2, geometric multiplicity 2-- I can pick two vectors. They can be any two I want in principle, right? It has geometric multiplicity 2. Here's another matrix. It's a little more interesting than the last one. I stuck a 1 in there instead. Again, the eigenvalues are 0. It's a double root. So it has algebraic multiplicity 2. But you can convince yourself that there's only one direction that transforms that squeeze down to 0, right? There's only one vector direction that lives in the null space of A minus lambda I-- lives in the null space of A. And that's vectors parallel to 1, 0. So the eigenvector associated with that eigenvalue 0 has geometric multiplicity 1 instead of geometric multiplicity 2. Now, here's an example for you to do. Can you find the eigenvalues and some linearly independent eigenvectors of this matrix, which looks like the one we just looked at. But now it's three-by-three instead of two-by-two. And if you find those eigenvalues and eigenvectors, what are the algebraic and geometric multiplicity? Well, you guys must had a rough week. You're usually much more talkative and energetic than this. [LAUGHTER] Well, what are the eigenvalues here? AUDIENCE: 0. JAMES W. SWAN: Yeah. They all turn out to be 0. So that's an algebraic multiplicity of 3. It'll turn out there are two vectors, two vector directions, that I can specify that will both be squeezed down to 0. In fact, any vector from the x-y plane will also be squeezed down to 0. So this has algebraic multiplicity 3 and geometric multiplicity 2. I'm going to explain why this is important in a second. But understanding that this can happen is going to be useful for you. So if an eigenvalue is distinct, then it has algebraic multiplicity 1. It's the only eigenvalue with that value. It's the only time that amount of stretch is imparted. And there will be only one corresponding eigenvector. There will be a direction and an amount of stretch. If an eigenvalue has a algebraic multiplicity M, well, you just saw that the geometric multiplicity, which is the dimension of the null space of A minus lambda I-- it's the dimension of the space spanned by no vectors of A minus lambda I-- it's going to be bigger than 1 or equal to 1. And it's going to be smaller or equal to M. And we saw different variants on values that sit in this range. So there could be as many as M linearly independent eigenvectors. And there may be fewer. So geometric multiplicity-- it's the number of linearly independent eigenvectors associated with an eigenvalue. It's the dimension of the null space of this matrix. Problems for which the geometric and algebraic multiplicity are the same for all the eigenvalues and eigenvectors, all those pairs, are nice because the matrix then is said to have a complete set of eigenvectors. There's enough eigenvectors in the problem that they describe the span of our vector space RN that our matrix is doing transformations between. If we have geometric multiplicity that's smaller than the algebraic multiplicity, then some of these stretched-- we can't stretch in all possible directions in RN. There's going to be a direction that might be left out. We want to be able to do a type of transformation called an eigendecomposition. I'm going to show you that in a second. It's useful for solving systems of equations or for transforming systems of ordinary differential equations, linear ordinary differential equations. But we're only going to be able to do that when we have this complete set of eigenvectors. When we don't have that complete set, we're going to have to do other sorts of transformations. You have a problem in your homework now, I think, that has this sort of a hang-up associated with it. It's the second problem in your homework set. That's something to think about. For a matrix with the complete set of eigenvectors, we can write the following. A times a matrix W is equal to W times the matrix lambda. Let me tell you what W and lambda are. So W's a matrix whose columns are made up of this-- all of these eigenvectors. And lambda's a matrix whose diagonal values are each of the corresponding eigenvalues associated with those eigenvectors. This is nothing more than a restatement of the original eigenvalue problem. AW is lambda W. But now each eigenvalue has a corresponding particular eigenvector. And we've stacked those equations up to make this statement about matrix-matrix multiplication. So we've taken each of these W's over here. And we've just made them the columns of a particular matrix. But it's nothing more than a restatement of the fundamental eigenvalue problem we posed at the beginning here. But what's nice is if I have this complete set of eigenvectors, then W has an inverse that I can write down. So another way to state this same equation is that lambda-- the eigenvalues can be found from this matrix product, W inverse times A times W. And under these circumstances, we say the matrix can be diagonalized. There's a transformation from A to a diagonal form. That's good for us, right? We know diagonal systems of equations are easy to solve, right? So if I knew what the eigenvectors were, then I can transform my equation to this diagonal form. I could solve systems of equations really easily. Of course, we just saw that knowing what those eigenvectors are requires solving systems of equations, anyway. So the problem of finding the eigenvectors is as hard as the problem of solving a system of equations. But in principle, I can do this sort of transformation. Equivalently, the matrix A can be written as W times lambda times W inverse. These are all equivalent ways of writing this fundamental relationship up here when the inverse of W exists. So this means that if I know the eigenvalues and eigenvectors, I can easily reconstruct my equation, right? If I know the eigenvectors in A, then I can easily diagonalize my system of equations, right? So this is a useful sort of transformation to do. We haven't talked about how it's done in the computer. We've talked about how you would do it by hand. These are ways you could do it by hand. The computer won't do Gaussian elimination for each of those eigenvectors independently, right? Each elimination procedure is order N cubed, right? And you got to do that for N eigenvectors. So that's N to the fourth operations. That's pretty slow. There's an alternative way of doing it that's beyond the scope of this class called-- it's called the Lanczos algorithm. And it's what's referred to as a Krylov subspace method, that sort of iterative method where you take products of your matrix with certain vectors and from those products, infer what the eigenvectors and eigenvalues are. So that's the way a computer's going to do it. That's going to be an order N cubed sort of calculation to find all the eigenvalues and eigenvectors [INAUDIBLE] solving a system of equations. But sometimes you want these things. Here's an example of how this eigendecomposition can be useful to you if you did it. So we know the matrix A can be represented as W lambda W inverse times x equals b. This is our transformed system of equations here. We've just substituted for A. If I multiply both sides of this equation by W inverse, then I've got lambda times the quantity W inverse x is equal to W inverse b. And if I call this quantity in parentheses y, then I have an easy-to-solve system of equations for y. y is equal to lambda inverse times c. But lambda inverse is just 1 over each of the diagonal components of lambda. Lambda's a diagonal matrix. Then all I need to do-- ooh, typo. There's an equal sign missing here. Sorry for that. Now all I need to do is substitute for what I called y and what I called c. So y was W inverse times x. That's equal to lambda inverse times W inverse times b. And so I multiply both sides of this equation by W. And I get x is W lambda inverse W inverse b. So if I knew the eigenvalues and eigenvectors, I can really easily solve the system of equations. If I did this decomposition, I could solve many systems of equations, right? They're simple to solve with just matrix-matrix multiplication. Now, how is W inverse computed? Well, W inverse transpose are actually the eigenvectors of A transpose. You may have to compute this matrix explicitly. But there are times when we deal with so-called symmetric matrices, ones for which they are equal to their transpose. And if that's the case, and if you take all of your eigenvectors and you normalize them so they're of length 1-- the Euclidean norm is 1-- then it'll turn out that W inverse is precisely equal to W transpose, right? And so the eigenvalue matrix will be unitary. It'll have this property where its transposes is its inverse, right? So this becomes trivial to do then, this process of W inverse. It's not always true that this is the case, right? It is true when we deal with problems that have symmetric matrices associated with them. That pops up in a lot of cases. You can prove-- I might ask you to show this some time-- that the eigenvectors of a symmetric matrix are orthogonal, that they satisfy this property that-- I take the dot product between two different eigenvectors and it'll be equal to 0 unless those are the same eigenvector. That's a property associated with symmetric matrices. They're also useful when analyzing systems of ordinary differential equations. So here, I've got a differential equation, a vector x dot. So the time derivative of x is equal to A times x. And if I substitute my eigendecomposition-- so W lambda W inverse-- and I define a new unknown y instead of x, then I can diagonalize that system of equations. So you see y dot is equal to lambda times y where each component of y is decoupled from all of the others. Each of them satisfies their own ordinary differential equation that's not coupled to any of the others, right? And it has a simple first-order rate constant, which is the eigenvalue associated with that particular eigendirection. So this system of ODEs is decoupled. And it's easy to solve. You know the solution, right? It's an exponential. And that can be quite handy when we're looking at different sorts of chemical rate processes that correspond to linear differential equations. We'll talk about nonlinear, systems of nonlinear, differential equations later in this term. And you'll find out that this same sort of analysis can be quite useful there. So we'll linearize those equations. And we'll ask is their linear-- in their linearized form, what are these different rate constants? How big are they? They might determine what we need to do in order to integrate those equations numerically because there are many times when there's not a complete set of eigenvectors. That happens. And then the matrix can't be diagonalized in this way. There are some components that can't be decoupled from each other. That's what this diagonalization does, right? It splits up these different stretching directions from each other. But there's some directions that can't be decoupled from each other anymore. And then there are other transformations one can do. So there's an almost diagonal form that you can transform into called the Jordan normal form. There are other transformations that one can do, like called, for example, Schur decomposition, which is a transformation into an upper triangular form for this matrix. We'll talk next time about the singular value decomposition, which is another sort of transformation one can do when we don't have these complete sets of eigenvectors. But this concludes our discussion of eigenvalues and eigenvectors. You'll get a chance to practice these things on your next two homework assignments, actually. So it'll come up in a couple of different circumstances. I would really encourage you to try to solve some of these example problems that were in here. Solving by hand can be useful. Make sure you can work through the steps and understand where these different concepts come into play in terms of determining what the eigenvalues and eigenvectors are. All right. Have a great weekend. See you on Monday. |
MIT_561_Physical_Chemistry_Fall_2017 | 31_TimeDependent_Perturbation_Theory_II_H_is_TimeDependent_TwoLevel_Problem.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: My spies tell me that there was a good mood after the exam. And it was an exam that was created to show the power of intuition and how you build intuition based on stuff that seems kind of ordinary. And all of a sudden you have the power to exercise chemical insight in ways which are more than just textbook ways of solving differential equations, memorizing pictures, and so on. So this is my purpose, and at the end of the course I'm going to try to really stretch your imagination. That's the probable schedule. And so I'm taking more time with the lecture on electronic spectroscopy and dynamics because this is really the core of what I do and what I care about. And so I don't know what I'm going to say in this lecture. I want to spend some time talking about Zewail's Nobel Prize work, and certainly that will be partly on Monday. Then, intermolecular interactions is basically why isn't everything in the gas phase? You know everything isn't in the gas phase, and this is the first step towards not being in the gas phase, kind of important. And then probably photochemistry, which is a little bit more of electronic spectroscopy but in a different framework. Delta functions is a way of dealing with complicated potentials without perturbation theory. And it's really neat and it's very abstract, and it's called the discrete variable representation, and it's a very powerful way of approaching complicated problems without having to do a lot of algebra because the computer does all of it, but it does it in a funny way. And I've been putting off this lecture on time-dependent Hamiltonians for a long time, and I hope that the last lecture will be an honest approach to time-dependent Hamiltonians. So this is kind of a motto for me. Small is a template for large. If you really understand everything a diatomic molecule can do, you're prepared to understand most of what big molecules do. And so this is sort of a bottom-up approach. And what I presented last time was sort of a template for everything a diatomic molecule can do. You have a ground state, and that dissociates the ground-state atoms. You have a repulsive state. Now it's not always just two states, but for H2+ or for H2 we sort of expect this sort of structure. And then there are excited states which are usually less bound than the ground states. So they have smaller rotational constants, larger internuclear distance. And if they don't, there's a good reason for it which you're supposed to be able to produce. Now the repulsive states often cross through a bound state, and this results in what we call predissociation. In other words, it happens before it was intended to. And it also is an example for all sorts of interesting effects associated with this curve crossing. And we'll talk more about that. And so predissociation and autoionization, these are dynamical effects that occur as you approach some special limit. Now this is you're above the dissociation limit, so it could happen. But as far as this data is concerned, you're not, and so it happens earlier. Now here you have ionization. And below the ionization there's a whole infinite manifold of Rydberg states which look almost identical to the potential energy curve for the ion. And there's all sorts of stuff that you can understand about the electron-ion interaction by studying Rydberg states. And that's been, for the last 20 years, roughly half of my research. Now encoded in Rydberg states is autoionization. If you have a vibrational level of a Rydberg state which lies above the ionization limit, well, then it can ionize, and that's interesting too. So this is a very quick summary of what I talked about before, and I want to add to that. So in order for a transition to occur, you either need a permanent dipole moment, a dipole moment derivative with respect to the displacement coordinate, or an electronic transition. So this is the rule. This is what you need for a pure rotation transition which is looked at in the microwave. This is what you need for a vibrational transition. The dipole moment has to depend on displacement from equilibrium for one of the coordinates. If it's a diatomic, there's only one. If it's a polyatomic, there's 3n minus 6. And so these two things are very important. And then the electronic transition moment, you can always have transitions where you promote an electron from the ground state to an excited state. But since the operator is a one-electron operator, the strong transitions are a very small subset of all the possible ones. And one of the important things for dealing with electronic transitions is the Franck-Condon principle. The Franck-Condon principle is basically a restatement of the Born-Oppenheimer approximation that the electrons move fast, nuclei move slowly. And as a result, delta R equals 0, delta P equals 0. Now I'm using capital letters just to mean all of the coordinates. And this is what we call a vertical transition. And delta P equals 0 is stationary phase. Now I like stationary phase. You have integrals of oscillating functions, and you're integrating usually over all space. But the integral accumulates to its final value in a very small region of space, and that region of space corresponds to where the momenta on the upper and lower states, or whatever you're looking at, where the momentum are the same. And this enables you to use classical insights because instead of having quantum mechanics where you're thinking about wave functions and thinking like I've got an integral to do and I can't say anything simple about it, but if you know that the integral accumulates in a particular geometry, then you can even estimate it. You know what determines how big that integral is. And this is almost always neglected in talking about the Franck-Condon principle, but it's a very important part, especially for me, because I want to find out simple things, and the simple things are always in simple places in state space or in coordinate space. And then there's the Franck-Condon factor. The Franck-Condon factor is something you calculate. If you know the potential or you have a reasonable estimate of the shape of a potential, you can calculate all the vibrational wave functions in that potential. Not a big deal. I haven't taught you how to do it. I will talk about it in 573, but that's next year, and that's a different life for most of you. But you can calculate the vibrational wave functions. And the Franck-Condon factor is the square of the overlap between the vibrational wave function for the initial state and the final state. So principle, factor, big difference. We want ideas. Franck-Condon factors are numerical. We want to know what's behind the Franck-Condon factors, and the Franck-Condon principle and stationary phase is what's behind it. And so you're now charged with being able to understand everything that happens in electronic transitions guided by this very important principle, which is really relegated only to calculating Franck-Condon factors, which is just something you do as opposed to something you think about. So that was kind of a summary of what we did last time. So we can have an electronic band system. So that's a transition between two electronic states, and these electronic states have vibrational levels, and the vibrational levels have rotation levels. And so it depends on how deep you want to go into understanding electronic transitions but what are the questions we want to answer? When we record a spectrum, we record a spectrum not just to publish a paper or to fulfill an assignment but to ask questions, and so what do we want to know? And there's how questions and why questions. How is a particular excited state different from the ground state? This means excited state. This means ground state. Spectroscopists use that notation. People who don't know anything about spectroscopy use whatever notation they feel like, and that's really stupid. But anyway, that's a different sort of thing. And so how are states different? Well, they have different equilibrium geometries. They have different vibrational frequencies. They have different energies. It's a transition between states pretty far apart in energy. There's permanent dipole moment. There's transition dipole moments. There's radiative lifetimes. There's all sorts of things that are distinct from one electronic state to another. And so this is how they're different, but then why? This is much more interesting, but you do this in order to start asking why. And so you already have qualitative electronic structure stuff like LCAO, MO. Now you could calculate things really accurately, but you wouldn't be using your brain, and this is MIT for God's sake. You do want to use your brain to explain why one state is different from another, and there's all sorts of qualitative theories that are your guide or your framework for doing this. And LCAO-MO, Huckel theory, these are the kinds of things that you can use. And we also have ideas about orbitals, bonding. So we have orbitals that are bonding, nonbinding, antibonding, and we know this by drawing a molecular orbital diagram. And we know the properties of what happens when you have bonding and antibonding orbitals. We have hybridization. So a tremendous amount is gained by asking for a carbon atom, and for many other atoms, what is the spn hybridization? We make a special set of orbitals which focus on the structure around each atom. And again, you can be guided by what you know from the spectrum or what you know from simple rules to saying, OK, if we have this hybridization, there is a certain geometry around that atom. Or if we have that geometry or if it's constrained to have that geometry, it will have this hybridization. And we haven't talked about this very much, but it's a very important part of understanding chemistry. And then we have these things where we-- so we look at families of molecules where you have the same number of electrons. And that was on the exam, and it's really beautiful how you can say, OK, even though we have the same number of electrons, if we change the difference in ionization energy between the two atoms, terrible things happen or wonderful, beautiful things happen. It depends on what you like. And homologous is just the same number of valence electrons but from different rows of the periodic table, and you had that on the exam too. These are the tricks we use. And then there's dynamics. And some people might say we study structure so that we can understand dynamics. I'm not quite there yet, but dynamics is interesting because it's harder than structure. You have a static structure. The spectrum is telling you what that is. But it's also telling you, if you look at it the right way, what the molecule can do when it gets excited. And I frequently start my talks with two slides, one called molecules at play where the molecules are in the ground state, and they're not doing anything interesting, and then molecules at work where molecules are highly excited and doing what they're supposed to do. And our job is really not to understand play. Let them do what they want, but we'd like to understand when they're working and how do we work with them? So dynamics is really that kind of a question. There's all sorts of kinds of dynamics. There's dissociation. There's ionization. There's Born-Oppenheimer breakdown. And you probably notice these letters, and that's what I do. I'm known for being a specialist in Born-Oppenheimer breakdown. It appears in many different ways. And so there's predissociation. There's autoionization. There's avoided crossings. And there's adiabtic versus diabatic. So these two things are really different sides of the same coin. Suppose you have two potential curves that cross. Well, this is what we think about as chemists where a potential curve corresponds to following the energy without really significantly changing the electronic structure. And so these crossing curves are what we call diabatic. But if you're a quantum chemistry, what you would do is you would calculate the adiabatic curves, and they would look like that. They don't cross. And so we have two ways of looking at the interaction between two electronic states, the adiabatic one and the diabatic one, and we use whichever one is simpler for the particular case. And at some point later in this lecture I hope to talk about a thing called Landau-Zener. And it basically is suppose you're driving at night on a very curvy road. And you're mostly awake, but you're driving too fast. And so if you have a situation like this where there's a sharp curve, and maybe if you're lucky there is another sharp curve on another road somewhere over here. You're going too fast. You're going to go off the road, and maybe you'll end up on this other road or you'll end up hitting a tree. So if you're going fast, you're going to jump off the curve. That's diabatic. And that's the point of view we take as chemists. And if you're going really slowly, you'll stay on the road. You won't even remember this intersection because you didn't die at it. And that's adiabatic. And depending on the gap and the curvature for a particular situation, the molecule knows how to either do this or do that. And it's not fatal for the molecule, but if the molecule is going through a curve-crossing region fast, it's going to jump the gap. And that's one of the things that Zewail looked at. If you can start a system at energy high above the energy where the curves cross, you're going through really fast. If you excite it really close to the energy of the curve crossing, you're going through slowly. And so by changing the conditions, you can understand what's going to happen. This is really powerful for insight. What I'm trying to convey is that there are very simple pictures to describe really complicated-looking things. And sometimes the data on these complicated things are very hard to either obtain or to interpret because it looks-- well, often it looks-- what would people say? It looks statistical. Now statistical is really a cop out. It says, I don't understand what's going on here, and so I'll approach it as if there isn't any law. And we just count the density of states, and we'd use simple formulas. And the shameful thing is that that usually works, but it only works so far. It doesn't lead to insight. It just leads to a representation, and our job is to understand, not just to represent. So back to this question of electronic transitions and the Franck-Condon principle. So if we have a transition between two potential curves that have the same geometry, we get a delta v equals 0 propensity rule. It has good and bad sides. One is the spectrum gets to be really simple because you only see transitions between the same vibrational quantum upstairs and downstairs. And this is true for polyatomic and diatomics as well. You get a simple spectrum and you say, oh crap, I can go home early. But you don't know anything about other vibrational levels, and if you're starting in the zero vibrational level of the ground state, well, you only know about the zero vibrational level of the excited state if the potential cures are the same, and there's all sorts of stuff you don't know. And if you're doing emission, well, maybe in an emission spectrum you excite more vibrational levels. But what's going to happen is if they're all have the same-- if the potential curves are the same, well, then all the vibrational transitions are going to be on top of each other in emission, and you might as well not have bothered. It's just a horrible situation. So delta v equals 0 depends on what your point of view is. If you want to understand things, this is bad. And then you can have transitions between states which have different shapes or displaced, and the Franck-Condon calculation tells you how this works. But here you have delta v equals many. But now for a polyatomic molecule where you have n atoms, you have 3n minus 6 normal modes. And some of them are going to be like this, and some of them are going to be like this. And that's good and bad because what it's saying is, yeah, you could have had a really complicated spectrum because there's so many vibrational-- I mean, for example, benzene has 30 vibrational levels, 30 vibrational modes-- not 1 but 30, and some of them have relatively low frequency. And benzene for most of you is a simple molecule. For me, it's really just beyond complex. And what happens when you have a lot of vibrational levels is the vibrational density states gets very high, very fast, so high that it's hopeless to be able to resolve the individual eigenstates. When that happens, frequency domain spectroscopy stops being so useful because a lot of the key details are hidden. So if it's not frequency domain, it wants to be time domain, and you can ask really good questions using time-domain techniques, which is what I'm going to talk about when I spend some time with Mr. Zewail. So you're going to have certain normal modes which are what we call Franck-Condon dark because they don't contribute to the spectrum. You only see delta v of 0 for modes. And the ones that lead to a change in geometry or a change in bond structure, those are the ones you see. Those are the Franck-Condon bright ones, and they're also much more interesting because why did the geometry change? What part of the molecule was responsible for the geometry change? This is what chemists want to ask. So that's the framework. Now, I've spent the last 30 years of my life looking at the acetylene molecule. Acetylene, for me, it's as big as I want to get, but for most of you that's really, really small. And acetylene has ground state which we can denote by a zero here, s0, lowest singlet. If we had a different number of electrons we might say-- if we had an odd number of electrons, the lowest state would be D0 or T0 if it had a triplet ground state like oxygen. The notation is simple. But we also use-- spectroscopists use notation where we use a Roman capital letter with a tilde over it for polyatomic molecules and not for diatomics, and the electronic state, the ground state is this. And have to be careful because the notation that organic chemists would use for acetylene is just this. And my point is I want to talk about where the hydrogens are. And so this is kind of a cheat because it sounds like I put a hydrogen on a butene because you have carbon here, these bonds. OK, you know the [INAUDIBLE]. And then there's an excited state which we could call S1. And the excited state looks like this. It's trans bent. This has a triple bond. This has a double bond. AUDIENCE: [INAUDIBLE] ROBERT FIELD: What? AUDIENCE: There's also [INAUDIBLE].. ROBERT FIELD: Yes, there's a-- you know about this paper here. But the lowest equilibrium geometry for the excited state is trans bent. You can have this structure too, and we have looked at the isomerization between these two, and how did we do that? How did we characterize the transition state for the isomerization? This is 20, 30 person years of work getting to the transition state, and it led to a paper in Science, my first in 40 years of trying. And so this is a molecule I really like because there are four modes which are Franck-Condon dark and two that are Franck-Condon bright. And we can do all sorts of really neat things. So in the ground state at high vibrational excitation, what do we expect? Well, one thing is nothing special. Well, the molecule could just go on doing nothing interesting, having four modes which are Franck-Condon dark and two modes that are Franck-Condon bright. It will remain boring. And another would be just increasing complexity because there can be anharmonic interactions between different normal modes. And so when that starts to happen people start to say, let me out of here. It gets complicated, and everything gets destroyed, and the spectrum just becomes uninterpretable, which is wrong. And then there is chaos which is another way-- or quantum chaos-- which is another way of waving your hands and saying it's statistical. And we know my opinion about that because bonds are really sacred in chemistry and if there are bonds, there isn't bag of atoms behavior. And then there's other possibility, isomerization. So this guy at high exaltation can go to that vinylidene. And I spent a hundred person years chasing after this. So there is a very shallow vinylidene well, and how is this kind of isomerization encoded in the spectrum? So in order to have a chance of understanding about vinylidene or things that happen in the ground state at very high excitation, I invented a fairly important method for looking at high vibrational levels. It's called stimulated emission pumping, and it looks like this. So we have two-- now this initially was demonstrated on a diatomic molecule, but its true importance is for something like acetylene. So what happens is you can excite to this distorted molecule from the ground state, and some vibrational levels are accessible because you have a little bit of room between two turning points. So you can excite a few vibrational levels by vertical transitions. But the important thing is that each vibrational level has an inner turning point, which is how you access it, and an outer turning point, which enables you to do-- so this is called the pump, and this is called the dump. Now I'm not sure whether I came up with pump and dump before it was talked about on Wall Street, but it was a very apt name for stimulated emission pumping. And so what we can do then is to access very high vibrational levels in the electronic ground state by pump and dump. And the way we detect it is we get fluorescence from this level. And when we hit the dump transition, the fluorescences decrease in intensity. And we can do a very significant map of what's going on at high excitation in the ground state. And when I invented SEP, I believed in quantum chaos. So I expected as we went up higher and higher we would see breaking of all the usual patterns. And I even wrote a paper saying we had seen quantum chaos when we excited to very high vibrational levels. And the reasoning for that is complicated, but the reasoning was correct except it didn't apply to chaos. It just applied to the spectrum getting complicated, really complicated. So what I really wanted to be able to do was to be able to somehow sample this transition state. And that experiment in its initial form was doomed to fail because this transition state involves a local bend. In other words, acetylene has normal modes, suspend, trans bend, CH symmetric stretch, CH antisymmetric stretch, and CC stretch. And this coordinate is mostly just that, a large-amplitude local band. And so when I started this, I didn't know that local bands would magically emerge from the spectrum, but they do and we saw them, and that led to a lot of good stuff. I should mention that I got tenure at MIT because of stimulated emission pumping. That was a long time ago. So we had not seen the isomerization in the ground state because we couldn't get high enough because Franck-Condon factors prevented us from getting high enough. And the spectrum also gets complicated, and there's all sorts of reasons why it didn't work. But as far as the cis-trans isomerization in the excited state, we killed that, and we killed that because we saw new patterns emerge, and that was more robust than normal modes. And then we saw those new patterns break, and the breaking is due to isomerization. And we worked out the theory, and that's what the Science paper was all about. So now, gas phase versus condensed phase. People who work in the gas phase don't talk to people who work in the condensed phase and vice versa because the spectra are profoundly different. So we have a ground state and we have some excited state. And whether it's a diatomic molecule or a polyatomic molecule, it doesn't matter that much. So what we have in the absorption spectrum is vertical transitions to a few vibrational levels, depending on the difference in geometry. And in the condensed phase, instead of staying in the vibrational level we populated, there is rapid transfer of energy from the molecule into its surroundings, and you end up in v equals 0 of the excited state. And then the spectrum is to a few vibrational levels of the ground state. And so you get something that looks like this. This is the sort of classic diagram for absorption versus emission spectra of a polyatomic molecule. So this is emission. This is absorption. And frequency is on the right. So what happens is in the absorption spectrum you observe transitions to higher vibrational levels-- to several vibrational levels, v prime equals 0 and higher. And so those are to the blue of the band origin. And in emission, you get emission only from v equals 0 in the excited state, and that's to the red of the origin. And so you get this classic double-hump picture. And there's all sorts of vibrational levels that are not resolved, and this is basically the kind of crappy spectrum you get when you look at a big molecule in the condensed phase. And so it's telling you frequency domain for this sort of thing is the wrong approach because there's nothing much here. So this thing, the expectation value, the Hamiltonian for the molecule as opposed to the overall system is time dependent. We're used to the Hamiltonian being time independent. There can be dynamics because you excite a coherent superposition of states. So for condensed-phase systems, the Hamiltonian or energy is not conserved in the molecule. But of course thermodynamics says if you have an isolated system it's conserved. That's irrelevant because we always want to look at something more interesting than a bulk sample. Now I want to talk about dynamics that is more subtle than the naive stuff-- so subtle dynamics. Well, let's have no collisions. Well, we can do that in the gas phase if the pressure's low enough. No breaking of molecules. So if there is no predissociation or autoionization-- so we can restrict ourselves to a low enough energy that things don't break. Now this is subtle. Molecules will fluoresce, and populations in excited states will decay. And as a rule, the fluorescence lifetime, if it's truly only fluorescence and not other dynamics, the lifetime is longer than 10 nanoseconds. So if we're looking at times shorter than 10 nanoseconds we can say there's no fluorescence. We're going to ignore that. That's OK because there still can be dynamics. So simplest case, two-level quantum beats, and you know about quantum beats. You've already looked at them in various problems. And so you have a system where you have a ground state and you have two excited states, and one of them is bright and one of them is dark with respect to transitions from this state. But there is an interaction between them, and so they're mixed character and with a short pulse. You prepare a coherent superposition of two eigenstates. The intensity of the fluorescence oscillates, and it isolates at the frequency difference of these two levels. And the modulation depth of the fluorescence tells you something about what is the coupling matrix element between these two states relative to the energy difference between them? So there's a lot of information in there, and so this is something that can happen faster than the spontaneous fluorescence. One is observing it in the spontaneous fluorescence, but the whole point is you're looking at some dynamics that is encoded in eigenstates. And that's one of my mottos. We know how to write all the theory for this. We have something that looks broad, but in reality it's a whole bunch of eigenstates. And the pattern of these eigenstates is telling you what the dynamics is. Now normally we think of dynamics as the width of something. And this collection of eigenstates behaving in some coherent way has a width, but that's only the beginning. What was the thing that gave intensity to these eigenstates? What was the zero-order state, and what was the coupling matrix element between the bright state and the dark state, and what are the rules for these? And so if you can resolve these eigenstates, you learn about much more detailed picture of dynamics than just saying, well, dynamics is a width. If you're below the energy where molecules can break, there is no width. But if you do a crappy experiment, you don't resolve stuff and it looks broad. Now, sometimes the molecule sets the standard for what's a real experiment by having its density of states so large you couldn't resolve it even if you had the best experiment in the world. But there's still the idea that inside this broad thing there is some interpretable dynamics. That's my goal. Well, one kind of interpretable dynamics is, OK, so you have some excited state and some ground state, and there's a range of vibrational levels that you can excite. And if you have a short pulse, you make a coherent superposition of all of those vibrational levels. And then what you have is some wave packet which has got an energy, sort of the average of these. And the wave packet starts out at a turning point because you started out with essentially zero momentum. And the wave packet moves, and it goes back and forth and back and forth. And it's telling you what the vibrational frequency of that mode is. And so you can then observe the fluorescence from this evolving wave packet, or you can monitor this evolving wave packet either by looking at fluorescence back to where you started. And so when it's over here, it can't fluoresce to where you started. When it's over here where it was born, it can. And so the fluorescence is going to be doing a kind of quantum beating. You could also use some kind of pump-probe experiment where you say, I'm going to look at the wave packet when it's here because I have a transition to another electronic state. And so the vertical excitation is at that energy, and so I wait for the wave packet to get to a point. And so you have a periodic motion. And you have dephasing because this wave packet is built out of vibrational levels which are not harmonic so that there is some dispersion of the vibrational frequencies around the average. As a result, that causes this thing to dephase. And its dephasing time could look like a width, but it's actually a particular mechanism. And then there are other things that can happen, and I love this. So we've got a few minutes. So suppose we have two electronic states that are crossing, and you're starting out in this level. You made a wave packet. So this is going back and forth, and every half oscillation it goes through this critical region. What's that region? What's special about that region? Come on. AUDIENCE: [INAUDIBLE] ROBERT FIELD: I'm sorry? AUDIENCE: [INAUDIBLE] ROBERT FIELD: Not if this is a bounce state. That's a good try. That is, in fact, if this were a repulsive state, then predissociation would occur when it cross through the region. This is the stationary phase region. This is where the momentum on the two potential curves is the same. And so the molecule can make up its mind which potential curve it wants to leave on because there's no impulse that causes a change in its momentum, and it just makes a decision which one. And so when that happens, if you're following the dynamics, you have a series of beats. And I'm just sketching them as a stick, but really, they have width. And then at some later type you start seeing a new family of oscillations because now you've got some amplitude on this other state, and it has a different frequency. But the important thing-- and I should stop here-- is that the only time the molecule has to change its mind between one state and another is when it's crossing through this stationary phase region. So it's a complicated thing, but it's isolating a particular region where the curves cross. And where the curves cross, instead of having crossing curves, you could have something like this and something like that. Now we're back to Mr. Landau and Mr. Zener. And so if there's a curve crossing, depending on how far above the curve crossing, it tells you how fast you're going through it. See, there's all sorts of wonderful stuff here, which is not the trivial dynamics that-- this is the only way you get broadening is the molecule breaks. And the fluorescence is usually so slow-- 10 nanoseconds is slow-- that you don't see any broadening because 10 nanoseconds corresponds to a pretty high resolution, and most people aren't looking at that kind of resolution. And so we have quantum beats which then can show beautiful dynamics in a way that if you're prepared to build a model and to say, well, we could have this case and this case-- this is the chemists' case. We have quantum chemists' case-- you can interpret everything. This is very much related to what Mr. Zewail does. And I will spend a significant amount of time on Monday talking about the Zewail experiments and how the whole point of his experiments are not just saying, OK, I excite the molecule and it breaks. It breaks in this much time. Well, that's a useful question, but what did it do before it broke? One particular bond breaks, but what about the motions of the other bonds? Or maybe you're talking about breaking in a particular normal mode. Does the molecule arrange itself to be broken? Or once it's excited it goes downhill to the graveyard in a particular path, but there are wonderful things you can do by asking, what is the mechanism by which the molecule breaks? And by varying how you excite the molecule, you're looking at different aspects of that mechanism. Now Zewail sold this really hard, and it worked. And so I'll leave that for Monday. |
MIT_561_Physical_Chemistry_Fall_2017 | 3_TwoSlit_Experiment_Quantum_Weirdness.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: All right. So last time, I said that atomic sizes are interesting or useful to keep in mind, because you want numbers for them, which are somewhere between 0.1 and 100, or something like that. Because then you have a sense for how big everything is, and you're in the right ballpark. If you have to remember a number and an exponent, it's a little trickier. And so I'm going to say something about atomic sizes in just a few minutes, but the main things from the previous lecture were this relationship between the wavelength and the momentum, which is true for both waves and particles, or things that we think are wave-like, like light, and things that we think are particle-like, like electrons-- and so this unifying principle. Then I have a little question. And that is suppose the wavelength for a particle is known, and suppose we have n particles. And so if we say each particle requires a volume of lambda cubed, and there are n particles, and we stick them into a volume smaller than n times lambda cubed, what's going to happen? Yes? AUDIENCE: They interfere with the structure? ROBERT FIELD: Their identities get corrupted, and the person who does that for the first time gets a Nobel Prize, and that was Wolfgang Ketterle-- and others. So when you have quantum mechanical particles that are too close together, they lose their individual identity. And so this is a very simple thing that anybody who was beginning to understand quantum mechanics in the 1920s would say, this is a puzzling thing, maybe we should think about that. And it's really hard, so it took a long time. Then a whole bunch of experiments led to the idea that we need to have a way for atoms to fill space and satisfy all of the other stuff, and that was Rutherford's planetary atom picture. The problem with that is it does fill space, but there's no way for the electron to continue orbiting around a nucleus, because it will radiate its energy, and fall into the nucleus, and game over. And so Bohr and De Broglie both had ways of fixing this. And Bohr's way was simply to say the angular momentum is not just conserved, but it has certain values-- an integer times h bar, h bar is h over 2 pi. And De Broglie said in order to keep the electron from annihilating itself, since it's going around in a circular orbit of known circumference, there must be an integer number of wavelengths around that orbit. Now, that's a much more physical and reasonable ad hoc explanation, but it's still ad hoc. It assumes that the particles are moving around circular orbits. I hinted that the way out of this is going to be that the particles aren't moving. Then we could still have angular momentum. We can still have all sorts of useful stuff, but they're not moving, and we don't radiate the energy of the particles. But that requires a completely new way of looking at particles in quantum mechanics. But the thing about both the Bohr hypothesis and the De Broglie hypothesis is that any sophomore who made such a proposal would be laughed at in the 1910-1920 period, because it's just ridiculous. But the reason these hypotheses were taken seriously is that this planetary model with those corrections explain the spectra of all one-electron systems-- hydrogen, helium plus, lithium two plus, uranium 91 plus, all of them, to better than measurement accuracy at the time, and better than measurement accuracy for a long time after this was proposed. So getting a whole bunch-- now, a whole bunch is an infinite number, actually-- of 10 digit numbers makes you think there's something here, but nobody knows where here is except the numbers. And the idea of the planetary model with the fixes doesn't explain anything other than the spectral lines, which is a lot, but it tells you there's something really good there. I guess one thing I forgot to mention in the summary is that the energy levels that you get from the Bohr model are going to explain spectra if you say a spectrum are transitions between these energy levels. And that was also a brilliant suggestion, and it was suggested by numerology. But let us go back to work now. We're going to talk about the two-slit experiment. And I have a personal thing to report about that. The first time I gave a talk, it was going to be a 15-minute talk at a spectroscopy conference, and so I did a practice talk, and it had to do with the two-slit experiment in relationship to spectroscopy. But I did a practice talk, and it took two hours. So I have a thing about the two-slit experiment. And I think this lecture is going to be not two hours. It's going to be on time. So we're going to talk about the two-slit experiment, and the important thing about the two-slit experiment is that it's mostly ordinary wave interference. There is no quantum mechanics, and so most of the hard stuff to analyze is classical physics. And I'm going to do the best I can with that. But after we get to understanding this problem, there will be a surprise at the end, which is a quantum surprise. And it's something that absolutely requires a postulate, the first postulate of quantum mechanics, some idea of what we are talking about here. What are we allowed to know about a system? And it tells you there's something there. And then I'm going to give you a little description of something which I think I have to call a semi-classical optics uncertainty principle. Semi-classical or semi-anything is usually quantum mechanics, just a little bit, mixed into something that was well understood before, or that is very convenient to use, and you only bring in quantum mechanics when you have to. And it's the easiest thing to understand. And that introduces the uncertainty principle, and we get a taste of the first of several quantum mechanical postulates. So let's start with the sizes. I should have given you these sizes in the previous lecture, but I was being mostly number-free. So the radius of a Bohr orbit is given by n over z squared, where z is the integer charge on the nucleus, times 0.5292 angstroms. So half an angstrom is the radius of, basically, any atom. And this charge on the nucleus, it gets smaller. And this is a very useful thing. The wavelength is given by n over z times 3.32 angstroms. And that's the situation where n times the wavelength is equal to 2 pi rn. So we have n wavelengths around an orbit, and this is really the De Broglie hypothesis. But again, something which is on the order of something that you can remember. And the energy levels-- these energy levels are z squared times 13.6 electron volts over n squared. If we were talking about energies in joules, or in any units you want, it's likely to have a big exponent. But this one is in electron volts, and that's actually what's happening. An electron is being attracted to a positive thing, and there's, basically, a voltage difference. And so that's another useful thing. Now this z-- I left out the Rydberg. This Rydberg constant-- there's a bunch of fundamental constants, and since I'm a spectroscopist, I think in terms of wave numbers, reciprocal centimeters. And for me, that's energy, it's frequency, it's everything, but anyway, for hydrogen, it's 1097677.581. For something with an infinite mass nucleus, it's 109737.3153 reciprocal centimeters. And this is the number that I have in my head, and I use it in all sorts of places, and you can imagine where. And to get to any particular nucleus, this is just r infinity mu nucleus, or mu atom, over the mass of the electron. And it turns out that almost everything except hydrogen is very close to this number. And so this hardly matters, but it does give you a little bit of dependence on the nuclear mass. So I said before that this Rydberg equation, or this equation, tells you nothing. It tells you where all the energy levels are, and anyone could tell you where the rest are. So it's a pattern, which is nice, but a pattern which says if things are well-behaved, like hydrogen atom, this is what the energy levels will be. But life is difficult. Life is not with everything well-behaved, and so this is a pattern which says I'm interested in how the real life is different from that pattern. It's a way of thinking about structure and how we learn about structure. Information about the details of a molecule or an atom is encoded in the spectrum, and this is the magic decoder, or one of the magic decoders, we use to begin to assemble the new insights. This doesn't appear in textbooks. In textbooks, you get the equations, you get the truth, and you don't get any strange interpretations. That's what you're getting from me. You're getting strange interpretations, and you'll have them throughout the course. So now what we want to do is talk about the two-split experiment. Let's just begin. So here's a diagram. And here we have a source of light. It's a light bulb in your notes. It's a candle here. And then we have two slits, and the slits are separated by distance d. And so this is the first slit, s1, and the second slit, s2. And the distance between them is much larger than the width of each slit. And now we go down to the screen. And the distance from the slits to the screen is l, and l is much, much larger than d. This means, of course, we're going to be using small-angle approximation and simple solutions, because everything is much larger than something else. And that's very convenient. Now I want to just put on axes. So this is the x-axis, and this is 0, and this is l. And now the screen-- we're going to see something on the screen that looks like this. I'm giving it away. And the distance here is 0, and the distance here is the z-axis. So the distance to this slit is on the x-axis, and the pattern, the diffraction pattern, is on the z-axis. And this 0 is right in the middle of the pattern. It would correspond to this point, the midpoint here. So what we want to do is calculate what's going to appear on the screen, and I've already given it away. What you see is a bunch of equally-spaced intensity maxima where we have constructive interference. And in between, we have less constructive interference-- or destructive interference, and we want to understand that. Now this is optics. This is no quantum mechanics at all. Let's look at this in more detail. So we have the z-axis, horizontal, and we have a path to the screen, and another path to the screen. So what we're interested in is here's one slit, here's the other slit, and we have two parallel lines that meet at infinity, which is where the screen is-- and what we're going to be interested in is what is the path difference between this one and this one. So we have an angle which is given by the perpendicular to this right. So this distance is d. This distance is L. And this angle is theta, as is this angle. And what we're interested in is this-- the extra path traveled by the lower slit. So we use trigonometry, and we can figure that out. And so delta is the path difference. Delta is equal to d sine theta. And in order for it to be constructive interference, we have to have this path difference to be an integer number of wavelengths. Now this is optics, and so that's something that we don't need quantum mechanics for. We know we're going to get interference. So we can now solve for where the constructive interference occurs. And so we have theta n is equal to the inverse sine of n lambda over d. But this is a small angle, even though I drew it not as a small angle. And so we can replace sine x by x or inverse of sine y by y. And anyway, we can say that the angles for constructive interference are given by n lambda over d. So we've derived the diffraction equation. We've solved. And now what we want to know is where do the spots occur. They occur at z equals 0, z equals plus or minus l sine theta, which is approximately equal to l over d n lambda. So we have l times lambda over d times an integer. So what we're going to see on the screen is a series of bright lines for constructive interference-- and they're not lines. It's a curve. But we can measure the maximum of the intensity, and we can say they're like this, and they're equally spaced. And they tell us things we knew. We knew l. We knew d. We knew lambda. So there's nothing surprising. This is just optics. So now suppose we go in and we cover one slit. What happens? Yes? AUDIENCE: The interference stops? ROBERT FIELD: The interference goes away. And you could imagine that there would be a little sign. If you covered the top slit, the pattern would be skewed a little bit in the direction of the bottom slit. And so there'll be a little bit of asymmetry, but you could actually know which slit your colleague covered. So if both slits are open, you have interference. If one slit is covered, you have no interference. We're getting into the realm of quantum mechanics. In quantum mechanics, one of the things we do is we say, suppose we did a perfect experiment. Maybe it's an experiment that's beyond what you're capable of doing with the present technology, but you can say, I could measure positions in time as accurately as I want, and one could also say that I could do this an infinite number of times. I could do the same experiment an infinite number of times. Without quantum mechanics, if you did the same experiment an infinite number of times, you'd get the same answer an infinite number of times. But with quantum mechanics, you're going to discover you don't. It's probabilistic, not deterministic. And under certain conditions, the range over which you have a finite probability is very small, and it looks deterministic, but it isn't. The perfect experiment business is an interesting hypothesis, but you can imagine defining what is perfect in terms of what is intrinsically possible to achieve, even if it's not currently possible. So what we want to do is decrease the intensity of the light that's going into the apparatus, so that there is never more than one photon in the apparatus. Never is a strong word, and we could never do that, and so that's not legal. But we can say, suppose we decrease the intensity so that for the time it takes for the photon to go from the slit to the screen, which we know because we know the speed of light, and for the intensity of the light, which we can measure with an energy meter, we can say the probability of there being more than one photon at a time in the apparatus is small, as small as we want, but not zero. And so then we do the experiment. And what we discover when we do the experiment is instead of having a uniform intensity or some kind of continuously varying intensity on the detector screen, we get a series of dots-- events. The photon went in, and the photon was a wave when it went in. There was interference, maybe, and the photon died on this detector screen. This is an example of destructive detection, which is something that is very important in quantum mechanics, because in quantum mechanics, most measurements destroy the system, or destroy the state that the system was in during the experiment. So this business of what is the state of the system is a really important quantum mechanical concept, which you don't normally encounter in classic mechanics. We send photons one at a time through the apparatus, and we get something like this. And we get something like this whether both slits are opened or one slit is covered. So we do this, and we do this for a long time, and what we see is we see a lot of events, and they're starting to arrange themselves where the interference fringes were supposed to be. So this pattern only emerges after you'd allow a large number of photons to go into the apparatus. There is no way classical optics gives you that. And so if one slit is covered, you get a uniform distribution of dots. If both slits are open, you get this kind of a distribution. I'm going to ask you to vote on this. So we do the experiment-- and I've sort of given away the answer, but I still want you to vote on this. What are the possible things? So we know there's only one photon in the apparatus at a time. Our concept of interference is light interference with itself. And this is a possibility that says, I think I know. And here is another, weak interference on top of constant background. This would reflect. Even though we decided that we would have one photon in the apparatus, occasionally there are two. And when there's two, there could be interference, and so we'd have some weak interference superimposed on the constant background. Now we get this 100% modulated interference structure. And the last thing is something else. You have to transport yourself back in time to around 1910. You haven't heard this lecture, but you do know what the experiment is. And so what would you expect? First, no interference, raise your hand. I've got one-- two-- I've got a few votes for no interference. In 1910, that's what you would have said. Weak interference on top of constant background-- that would be when you're hedging your bets and saying, the experiment, it isn't perfect, and this is really the right answer, but if someone was a little sloppy, I'd get this. Raise your hands for this one. What would you have said in 1910? I got nobody with the courage to say this. You would have gotten a Nobel Prize if you could have defended it. Something else-- maybe something else-- maybe the photons come in pairs or something ridiculous. So the correct answer is 100% modulated. What people would have said in 1910 is this. Some curmudgeons, like the climate change deniers, would have said there is a little bit of this or maybe that, but I can't possibly accept that, which is the truth. We won't belabor that anymore. This means that one photon can interfere with itself. It's a very disturbing idea, but it leads to a critical idea in quantum mechanics. In quantum mechanics, the state of the system is described by some state function, which is a function of position in time. And so what happens is you prepare the system in some state. You do something to it, like force it to go through two slits. Then we get some new state function. And then we detect it, and we get something else. So the actual experiment is a click, the preparation, and the click, detection. And somehow, what is the nature of the experiment is expressed on this initial state of the system. This is all very abstract, but it's about interference. So this guy had better have phase. So we have a wave that can constructively or destructively interfere with itself. And so we start talking about things like amplitude, but the crucial word is amplitude. And mostly, when we detect things, we're detecting probability. This is always positive. These guys can be positive and negative. This is essential for quantum mechanics. It's essential for understanding the two-slit experiment, but we have to do an awful lot more to make all this concrete. Why don't we look at the classical wave equation? Actually, we'll look at the classical wave equation next time, but I will say that the solution to the wave equation is some function of x and t, which has the form a sine kx minus omega t. So I'm doing this to introduce you to the crucial actors in this game, which is amplitude, wave number, and frequency. And this is a probability amplitude. It can have either sine, because you can see the sine function. So this is a wave of frequency omega propagating in the plus x direction. Now let's just identify the crucial quantity. Wavelength is the repeat distance. So that if we said we have u of x and t, it has to be equal to u of x plus lambda and t. So that's how we define the wavelength. And we discover that the wavelength has to be related to k times lambda is equal to 2 pi. Because this part has to change by 2 pi in order for there to be an exact replica of what we had before. So we know that the wavelength is the repeat distance, and k is 2 pi over lambda. It's called wave number, and it's the number of waves, complete waves, that occur in 2 pi times the unit length. Now, in 3D, we have a vector as opposed to a number. And that points in the direction of propagation of the wave. We know from quantum mechanics-- or from the experiments-- that the wavelength is related to the momentum. There's several reasons for this for waves. This could still be optics, but it could have been relativistic optics, because Einstein proposed that the momentum is e over c. So we put that together, and we get the relationship between k and momentum. Now h bar is h over 2 pi. And so the wave number is large if the momentum is large, and the wavelength is small if the momentum is large. Now what about the velocity? So we have a wave. Let's sit on this wave here and say, we're now sitting on some point where the phase is constant. I like to call this the stationary phase point, but that has other meanings, so you just have to be careful with this. So we want to know how fast this moves in space. And so we say, the phase is the phase of this function here, and so it's k x-- I'll put a little phi on this-- minus omega t. And we want to find how this moves in time, this stationary phase. We could choose this to be zero. We could choose a phase. This is zero, this is some maximum, but we can choose anything we want. So let's make it zero. And so then we can solve for x phi is a function of t, and that's omega t over k. We want the phase velocity, the velocity of this wave. So we take the derivative with respect to t. And so the velocity, which we'll call c, is equal to omega over k. That's true for light traveling in a vacuum. This is called the dispersion relation. Whenever you do a calculation of waves in material, the goal is to get the dispersion relation. This is the simplest possible one, because it says everybody at the same frequency-- I'm sorry-- that there is a relationship between omega and k so that everybody travels with the same speed. If that didn't happen, the waves would get out of phase with each other. That's what's dispersion is. So we have from simple optics, basically, everything we need. And now, the last thing is that the intensity of this wave is given by kkix minus omega i t. So this is the wave function. It's something that has sines. And this is the intensity, which is proportional to this quantity. We sum the amplitudes and then square. We do this in quantum mechanics all the time. And this is like quantum mechanics in the sense that we have our fundamental building block, which is something with phase. And the relationship of the thing with phase to the thing which is probability is sum of the square. Yes? AUDIENCE: What's the first character after the capital sigma? ROBERT FIELD: I'm sorry? AUDIENCE: The first character after the sum there? ROBERT FIELD: This? No, this. That's a constant. That's the amplitude of that particular frequency and wave vector. I'm sorry about the mess on the board. We have enough time. Oh, good. So this is sort of a taste of what you're going to be doing, but now let's produce a form of the uncertainty principle. You've heard about the uncertainty principle, and it's a very important part of quantum mechanics, but it's also something that you can have in optics. And so again, we resort to this simple idea of sending a particle through a slit and looking at what happens over here. Instead of having two slits, we just have one. But there are two important things. The two edges of this slit are special, because they're edges. And so we can ask, what about the interference between particles that was diffracted by this edge versus that edge. And the same thing goes. You get constructive interference when the difference in path length is an integer number of waves, and destructive interference when it's an odd integer number of half waves. We're interested in the destructive interference. So we analyze this in exactly the same way we did the two-slit experiment. And now this is the z direction. And this is the x direction. And what you end up finding is that the uncertainty in the z direction is going to be related 2 times lambda over delta s over l. Delta s is the width of the slit, so width of the image along the z-axis. So we have a maximum here, and we have minima here and here, and so this delta z is the distance between minima. So we can say, between minima, we have the particle localized. Its position in space-- or the photon localized-- its position in space is uncertain by this quantity. What about its momentum? And so now we just have to draw a little bit of conservation of momentum. So here we have the magnitude of the momentum starting from the middle of the slit, and this magnitude is constant along a circle-- the magnitude. So if we do draw a line here, this has got p. And if we ask the length here, that's also p. But now what we want to know is what is the uncertainty of the momentum in the z direction-- delta pz. This is the same sort of calculation we did before, and what you find is delta pz is approximately equal to magnitude of p lambda or delta s. And now we put in the Bohr relationship here, and we get that this is equal to h over lambda, lambda over the ds, which is equal to h over ps. Or ds in the z direction-- no, ds is the slit width. I got something wrong here in my notes. The final result of this calculation is dz is dpz approximately equal to h. Sorry about the glitch here. I don't know how to fix it right now, but if this were done correctly, we would have gotten this result. This is an uncertainty principle. If you tried to measure z and pz simultaneously with classical optics, you still would get something like this. Using the relationship for momentum that lambda is equal h over p, we've got to use that. But because of that relationship, we get this result. If you do a perfect experiment, and you make the slit smaller and smaller, you make the uncertainty in the momentum larger and larger. The best you can do is this. Now, this is actually a reasonable and rigorous derivation if I had done it a little better. But I've never liked this introduction to the uncertainty principle, because it says, for the kind of experiment we thought about, you can't do better than this. Maybe there's a different kind of experiment. It's sort of an artifactual as opposed to a physical derivation. We care. We will do a physical derivation of the uncertainty principle. And it will have to do with the ability of two operates to commute with each other. That's a purely mathematical definition, but this is the first sign of the uncertainty principle. At the end of your notes, there is a set of postulates from which, essentially, all quantum mechanics can be derived. Now, there are different sets of postulates proposed by different people, but these are things that can't be proven. They are things that you think is going to be true, and then you look at the consequences. The first postulate says, the state of a quantum mechanical system is completely specified by a function, psi of r and t, that depends on the coordinates of the particle and on time. This function, called the wave function or the state function, has the important property that this quantity times its complex conjugate integrated over the volume element is the probability that the particle lies in the volume element centered at r at time t. So we are saying there is a way, a complete way, of describing the state of the system. It's a function. It's not a bunch of discrete quantities, like velocity and position. And things are smeared out. And what we want to do when we want to calculate anything, we're going to be using this function. And this function comes from the Schrodinger equation. And we're going to get to the Schrodinger equation-- not in the next lecture. I'm going to spend the next lecture really laboring the wave equation, because the Schrodinger equation is just a tiny step beyond the wave equation. I've given you the five postulates. You have not a clue what any of them mean, except maybe a little bit about the first one. But what I don't want to do is give a lecture on the postulates. I want to bring them into action when you need them, because they'll mean much more. And so you won't be asked to memorize the postulates. You'll know when they are applicable and how to apply them. So it's a little premature to say you understand the first postulate, but that's what's at play here in the two-slit experiment and in this hand waving derivation of the uncertainty principle. So next time, we will look at the wave equation. Thank you. Hey, I finished on time this time. That's a bad sign. |
MIT_561_Physical_Chemistry_Fall_2017 | 36_Time_Dependence_of_TwoLevel_Systems_Density_Matrix_Rotating_Wave_Approximation.txt | The following content is provided under a Creative Commons license. Your support will help MIT Open Courseware continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: Today's lecture is one where-- it's a lecture I've never given before. And it's very much related to the experiments we are doing right now in my research group. And so basically, we have a chirped pulse of microwave radiation, which is propagating through a sample. It causes all of the molecules in the sample to be prepared in some way. We call it polarized. And this polarization relaxes by what we call free induction decay. And they produce a signal which-- so we have a pulse of radiation that propagates through the sample. The two-level systems in the sample all get polarized, which we'll talk about today. And they radiate that polarization. And we collect it in a detector here. And so the two important things are this is a time independent experiment, and that we have a whole bunch of molecules. And they're interacting with the radiation in a way which is complicated. Because this is not-- each one of them has quantum states, but all of the particles in this sample are somehow interacting with the radiation field in a way which is uncorrelated. So we could say all of these particles are either bosons or fermions. But we're not going to symmeterize or anti-symmeterize. Each of these particles is independent. And we need a way of describing the quantum mechanics for an ensemble of independent particles. So it's a big step towards useful quantum mechanics. And I'm not going to be able to finish the lecture as I planned it. So you should know where I'm going. And I'm going to be introducing a lot of interesting concepts. The first 2/3 of lectures votes are typed. And you could have seen them. And the rest of them will be typed later today. This is based on material in Mike Fayer's book, which is referenced in your notes. This book is really accessible. It's not nearly as elegant as some of the other treatments of interaction of radiation with two-level systems. Now I talked about interaction of radiation with two-level systems in lecture number 19. And this is a completely different topic from that, because in that, we were interested in many transitions. Let me just say the radiation field that interact with the molecule is weak. It interacts with all the molecules, and the theory is for a weak pulse-- and the important point in lecture 19 was resonance. And so we made the dipole approximation. And each two-level system is separately resonant and is weakly interacted with, and does something to the radiation field. Now here, we're going to be talking about a two-level system, only two levels. And the radiation field is really strong, or is as strong as you want. And it does something to the two-level system which results in a signal. And because the radiation field is strong, it's not just a matter of taking two levels and mixing them. The mixing coefficients are not small. It's not linear response. The mixing is sinusoidal. The stronger the radiation field, the mixing changes, and all sorts of interesting things happen. So this is a much harder problem than what was discussed in lecture number 19. And in order to discuss it, I'm going to use some important tricks and refer to something called the density matrix. The first trick is we have this equation which can easily be derived. And most of this lecture, I'm going to be skipping derivations. Some of the derivations are going to be in the notes. So we have some operator. And we want to know the time dependence of the expectation value of that operator. And it's possible to show that the expectation value of the operator a is given by the expectation value of the computator of a with the Hamiltonian plus the expectation value of the partial derivative of the operator a. So this is a general and useful equation for the time dependent of anything. And it's derived simply by taking the-- applying the chain rule to this sort of thing. So we have three terms. And when you do that, you end up getting this equation. So this is just this ordinary equation. And anyway, so this is what happens. So we're going to have some notation here. We have a wave function. And this is a capital psi, so this is a wave function, a time dependent wave function that satisfies the time dependent shorter equation. And we're going to replace that by just something called little t. And we can write this thing psi of x and t as the sum over n Cn psi n of x. And this becomes in a bracket notation, becomes C Cn n. So we have a complete ortho normal set of functions. And this thing is normalized to one. And now I'm going to introduce this thing called the density matrix. This is a very useful quantum mechanical quantity which replaces the wave function. It repackages everything we know from the time dependent shorter equation and the short in your picture of a wave function. It's equivalent. It's just arranging it in a different way. And this different way is extremely powerful, because what it does is it gets rid of a lot of complexity. I mean, when you have the time dependent wave functions, you have this e to the minus i E t over h-bar always kicking around. And we get rid of that for most everything. And it also enables us to do really, really beautiful, simple calculations of the time dependence of expectation values. It's also a quantity where, if you have a whole bunch of different molecules in the system, each one of them has a density matrix. And those density matrices add. And you have the density matrix for an ensemble. And so if the populations of different levels are different, the weights for each of the levels, or each of the systems, is taken care of. But we don't worry about coherences between particles unless we create coherence between particles. So this is a really powerful thing. And it's unlike the wave function, it's observable, because the diagonal elements of this matrix are populations. And the off diagonals elements, which we call coherences, are also observable. And if you look at the Fourier transform of the emission from this system, it will consist of several frequencies. And those frequencies are the off diagonal elements with the amplitude, the relative weights of those frequencies. And so one can determine everything in the density matrix experimentally. Now it's really-- it's still indirect because you're making experimental measurements. But we think about this thing in a way we don't think about the wave function. It's really important. And this is the gateway to almost all of modern quantum mechanics and statistical mechanics-- quantum statistical mechanics. And so this is a really important concept. And we've protected you from it until now. And since this is the last lecture both, in this course and in my teaching of this course forever, I want to talk about this gateway phenomena. So what is this? Well, we denote it by this, this strange notation. I mean, you're used to this kind of notation where we have the overlap of bra with a ket and or abroad with itself. But this is different. You know, this is a number and this is a matrix. And if we have a two-level system, then we can say that t is equal to C1 of t plus C2 of t. So state 1, state 2, and we have time dependence. Now those could be-- there's lots of stuff that could be in here. And this is going to be a solution of the time dependent shorter equation. So since it's an unfamiliar topic, I'm going to spend more time talking about the mechanics than how you use that to solve this problem. But let's just look at it. So we have for a two-level level system, we have-- it's a matrix of 1, 1; a 1, 2; a 2, 1; and a 2, 2 element. And so we want the rho 1, 1 matrix on there. And so it's going to be a 1 here, then we're going to have a 1 here. And then we have the C1, 1. Plus C2, 2. And then we have C1 star 1 plus C2 star 2. And so have I got-- am I doing it right now? So the first thing we do is we look at this inside part. And we have C1, C1 star. And we have 1, 1. And we have C2, C2 star. Now I'm getting in trouble, because I want this to come out to be only C1, C1 star. So what am I doing wrong? AUDIENCE: [INAUDIBLE] both have C2 halves the left hand and the right hand half are both [INAUDIBLE].. See, on the left side, you have 1 on 1 [INAUDIBLE].. ROBERT FIELD: So here-- And that's C2 C1 star, but that's 0. And so anyway, I'm not going to say more. But this combination is 1, this combination is 0, this combination is 0, this combination is 1. And we end up getting-- and then we end up just getting this. Rho 1, 2 is equal to C1 C2 star. Rho 2, 1 is equal to C2 C1 star. And rho 2, 2 is equal to C2, C2 star. So we have the elements of this matrix. And they are expressed in terms of these mixing coefficients for the states 1 and 2. Now, if we look at this, we can see that rho 1, 1 plus rho 2, 2 is equal to C1, C1 star plus C2, C2 star. And that's the normalization integral. That's 1. And we have 1, 1; 1, 2 is equal to 2, 1 star. And so each rho is formation. So the density matrix is normalized to 1. And it's Hermitian matrix. And we can use all sorts of tricks for Hermitian matrices. Now we're interested in the time dependence of rho. And so we're going to use this wonderful equation up here in order to get the time dependence of rho because rho like a, is Hermitian operator. And so we could do that. And so the time dependence of rho is going to be equal to the time dependence of t, t. Where we operating first here. And then t time dependent. And when we do this, what we end up getting-- well, so we have a time dependence of a wave function. So we use the time dependence shorter equation. And we insert that. And using the time dependence shorter equation we have things like-- so every time we take the wave function-- the derivative of a function, we get a Hamiltonian and so on. And so what we can express this time dependence of the density matrix by just using the time-- inserting the time dependent shorter equation repeatedly. This is why I say this is repackaging the Schrodinger picture, repackaging the wave function and writing everything in terms of these matrices. So that's the first-- that's what happens here. Sorry. That's what happens here. And then we write this one, and we get plus 1 over minus i h-bar t, t H of t. And we recognize that that is just one over i h-bar times H rho. So the time dependence of the density matrix is given by this computator. And the computators are kind of neat because usually what happens is these two-level things have very different structures, and you get rid of something you don't want to deal with anymore. And so now we actually evaluate these things. And we do a lot of algebra. And we get these equations of motion for the elements of the density matrix. And so we find the time dependence of the diagonal element for state 1 is opposite that for state 2. In other words, population from state 1 is being transferred into state 2. And that is equal to minus i over h-bar times H 1, 2 rho 2, 1 minus h 2, 1 rho 1, 2. And we have rho 1, 2 time dependence is equal to rho 2, 1 time dependent star. And that comes out to be minus i h-bar minus i over h-bar, H 1, 1 minus H 2, 2 rho 1, 2 rho 2, 2 minus rho 1, 1 times H 1, 2. This is very interesting, but now we have a couple differential equations and we can solve them. But we want to do a trick where we write the Hamiltonian as a sum of two terms. This is the time-- the independent part, and this is the time dependent part. This is the part that gives us trouble. This is the part that takes us into territory that I haven't talked about in time independent. But it's still-- it's perturbation theory. This is supposed to be something that is different from and usually smaller than H 0. And so we do this. So H 0, operating on any function gives En times n. And so we could call these E zeroes, but we don't need to do that anymore. And now we do a lot of algebra. We discover that the time dependence of the density matrix is given by minus i over h-bar times H 1 of t times the density matrix. So this is very much like what we did before, but now we have that the time dependence is entirely due to the time independent Hamiltonian. So everything associated with H 0 is gone from this equation of motion. So now let's just be specific. So here is a two-level level system. This is state 1. This is state 2. This difference is delta E. And we're going to call that h-bar omega 0. So this is the frequency difference between levels 1 and 2. H 0 is equal to minus h h-omega over 2 h-bar omega over 2 0, 0. We like that, right? It's diagonal. h1 is where all the trouble comes. And we're going to call that h-bar times e x 1, 2 E0 cosine omega t. This is not an energy. This is an electric field. So this is the strength of the perturbation. And this is the dipole matrix element between levels 1 and 2. So we have a dipole moment times an electric field multiplied by h-bar. So this quantity here, has units of angular frequency. And we call it omega 1, which is the Rabi frequency. It gets a special name because Rabi was special. And so we're going to be-- and this is-- expresses the strength of the interaction. So we have a molecular antenna mu 1, 2. And we have the external field. And they're interacting with each other. And so this is the strength of the badness, except its goodness. Because we want to see transitions. So now we do a little bit of playing with notation because there's just a lot of stuff that's going on. and we have to understand it. So we're going to call the state-- we're going to separate the time dependent-- the time independent part of the wave functions from the time dependent. And so state 1-- this is the full time dependent wave function. And it's going to be minus i omega 0 t over 2. in Other words, we should have had zeros here-- times 1, right. So this is the time independent part, and this is the time dependent part. And 2 is e to the minus i omega 0 t over 2, 2 prime. Notice these two guys have the same sign. This bothers me a lot. But it's true, because we have opposite signs here, and we have a bra and a ket. And they end up having the same signs. So that means that h 1 looks like this, 0, 0 omega 1 cosine omega t e to the minus i omega 0 t. And here we have omega 1 cosine omega t e to the pi omega 0 t. So this is a 2 by 2 matrix. Diagonal elements are 0. Off diagonal elements are this omega. The strength of the interaction times the frequency of the applied radiation times the oscillating factor. So now we go back and we calculate the equation of motion, bringing in this h 1 term. And so we have minus i over h-bar h 1 rho. And we get some complicated equations of motions. And I don't really want to write them out, because it takes a while, and they're in your notes. And I'm going to make the crucial approximation, the rotating wave approximation. Notice we have a cosine omega t. We can write that as e to the i omega t plus e to the minus i omega t. And so basically what we're doing is we're going to do a trick. We have the Hamiltonian, and we're going to go to a rotating coordinate system. And if we choose the rotational coordinate the rotation frequency right, we can almost exactly cancel omega 0 terms. And so we have two terms, one rotating like this, which is canceling or trying to cancel omega 0, and one rotating like this, which is adding to omega 0. And so what we end up getting is a slowly oscillating term, which we like, and a rapidly oscillating term, which we can throw away. That's the approximation. And this is commonly used. And I can write this in terms of transformations. And although we think about going to a rotating coordinate system, for each two-level system, we can rotate at a different frequency to cancel or make nearly canceling the off diagonal elements. So although the molecule doesn't rotate at different frequencies, our transformation attacks the coupling between states individually. And you can imply as many rotating wave core transformations as you want. But we have a two-level system. So we only have one. And so we do this. And we skip a lot of steps, because it's complicated and because we don't have a lot of time. We now have the time dependence of the 1, 1 element. And it's expressed as omega 1. I've skipped a lot of steps. But you can do those steps. The important thing is what we're going to see here. We have e to the i omega 0 minus omega t rho 1, 2. And we have a minus e to the minus i omega 0 minus omega t times rho 2, 1. And we have 2, 2 dot is equal to minus rho 1, 1 dot. And we have rho 1, 2 dot-- this is the important guy-- is equal to i omega 1 over 2 e to the minus i omega 0 minus omega t rho 1, 1 minus rho 2, 2. So we have a whole bunch of coupled differential equations, but each of them have these factors here where you have omega 0 minus omega. I've thrown away the omega 0 plus omega terms. And now it really starts to look good, because we can make these-- so when we make omega equal to omega 0, well, this is just 1. Everything is simple. We're on resonance. And so what we do is we create another symbol, delta omega, which is omega 0 minus omega. So this is the oscillating frequency applied. This is the intrinsic level spacing in the molecule. And so we can now write the solution to this differential equation for each of the elements of the density matrix. And we're going to actually define another symbol. We going to have the symbol omega sub e. This is not the vibrational frequency. This is just a symbol that is used a lot in literature, and that it comes out to be delta omega squared plus omega 1 squared. So in solving the density matrix equation, it turns out we care about this extra frequency. If delta omega is 0, well, then there's nothing surprising. Well, maybe e is just omega 1. But this allows for there to be an effect of the detuning. So basically what you're doing is when you go to the rotating coordinate system, you have an intrinsic frequency separation. And so in the rotating coordinate system, you have two levels that are different. And there's a stark effect between them. And you diagonalize this stark effect using second order perturbation theory or just the diagonizing the matrix. And so that gives rise to this extra term here, because you have the oscillation frequency and the Rabi frequency. And anyway, when you do the transformation, you get these terms. And so here is now the solution in the rotating wave approximation. Rho 1, 1 is equal to 1 minus omega 1 squared over omega e squared sine squared omega 0 t over 2. We have rho 2, 2 is equal to just omega 1 squared over the e squared sine squared omega e t over 2. We have omega 1, 2-- rho 1, 2, which is equal to something more complicated looking. Omega 1 over omega e squared times i omega 0-- omega e, sorry, or 2 sine omega e t minus delta omega sine squared omega e t over 2 times now e to the minus i delta omega t. It looks complicated. And we get a similar term for who 2, 1. It's just equal to rho 1, 2 complex conjugate. And so now what we see is these populations are oscillating not at-- It's a e, not a 0. They're oscillating at a slightly shifted frequency. But they're oscillating sinusoidally. And we have an amplitude term, which is omega 1 over omega e quantity squared. Omega e is a little bigger than omega 1. So this is less than 1. So it's just like the situation in the absence of this oscillating field, that you just get a slightly, slightly shifted oscillation frequency, and a slightly reduced co-factor. The coherence terms-- so these are populations, populations going back and forth between 1 and 2 at a slightly shifted frequency. And then we have this, which looks horrible. And now, for some more insights. If we make omega 1 much larger than delta omega-- in other words, the Rabi frequency much larger than the tuning, it might as well not be detuned. We get back the simple picture. We get rho 1 is equal to cosine squared omega 1 of t over 2, et cetera. So we have what we call free precession. Each of the elements of the density-- the density matrix is telling you that the system is going back and forth sinusoidally or co-sinusoidally cosine squared. And what happens to level 1 is the opposite of what happens at level 2. And everything is simple and the system just oscillates. Suppose we apply radiation or delta t a short time. And so what we're interested in-- here is t equals 0. This is time. And this is t equals 0. And before t equals 0, we do something. We apply the radiation. And we apply the radiation for a time, which gives rise to a certain flipping. And so what we choose. We have delta t is equal to theta over omega 1, or theta is equal to delta t omega 1, the Rabi frequency. And so if we choose a flip angle, which we call say a pi pulse, theta is a pi. And what ends up happening is that we transfer population entirely from level 1 to level 2. When we do that, we get no off diagonal elements of the density matrix. They are zero. So if at t equals 0, we have everything in level 1, and we have applied this 0 pulse or a pi pulse, we have no coherence. If we have a pi over 2 pulse, well then, we've equalized the two-level populations, and we created a maximum coherence. And this guy radiates. So now we have an oscillating dipole. And it's broadcasting radiation. And so all of the two-level systems, if you use a flip angle of pi over 2, you get a maximum polarization, they're radiating to my detector, which is up there. And I'm happy. I detect their resonance frequency. And so the experiments work. So we're pretty much done. So I mean, what we are doing is we're creating a time dependent dipole. And that dipole radiates something which we call-- if we have a sample like this, that sample-- all of the molecules in the sample are contributing to the radiation of this dipole. But they all have slightly different frequencies, because the field that polarized them wasn't uniform. In a perfect experiment it would be. And so they have different frequencies and they get out of phase. Or conservation of energy as the two-level system radiates from the situation where you have equal populations to everybody in the lowest state, there is a decay. So there's decays that causes the signal, which we call free induction decay, to dephase or decay. But the important thing is, you observe the signal and it tells you what you want to know about the level system, the two-level system, or the end level system. And it's a very powerful way of understanding the interaction of radiation with matter, because it focuses on near resonance. And near resonance for one two-level system is not near resonance. For another-- and so you're picking out one, and you get really good signals. And you can actually do-- by chirping the pulse-- you can have one two-level system, and a little bit later, another two-level-level system. They all radiate. They all get polarized. They all radiate at their own frequency. And you can detect the signal in the time domain and get everything you want in a simple experiment. This experiment has enabled us to do spectroscopy a million times faster than was possible before. A million is a big number. And so I think it's important. And I think that this sort of theory is germane, not just for high resolution frequency domain experiments, in fact, it's basically a time domain experiment. You're detecting something in the time domain, and Fourier transferring back to the frequency domain. So there are ultra fast experiments where you create polarizations and they-- it is what is modern experimental physical chemistry. And the notes that I will produce will be far clearer than these lectures, this lecture. But it really is a gateway. And I hope that some of you will walk through that gateway. And it's been a pleasure for me lecturing to you for the last time in 5.61. I really enjoyed doing this. Thanks. [APPLAUSE] Thank them. [APPLAUSE] Well, I got to take the hydrogen atom. [LAUGHTER] Thank you. |
MIT_561_Physical_Chemistry_Fall_2017 | 30_TimeDependent_Perturbation_Theory_I_H_is_TimeIndependent_Zewail_Wavepacket.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: This is the first of two lectures on spectroscopy and dynamics. Now, I'm a spectroscopist, and so this is the core of what I really love. And there are a lot of questions about, well, what are we trying to do? And you heard two lectures from Professor Van Voorhis, and he talked about ab initio calculations-- electronic structure calculations-- where you can get really close to the exact answer. And it's really a powerful tool. And it gets you the truth. But it gets you so much truth that you don't know what to do with it. And the same thing is true for spectroscopy. You get a spectrum. It contains a huge amount of information. And you can take lots and lots of spectra. But what are we trying to do? When I was a graduate student, we were doing a unique super high resolution spectroscopy. And so we thought what we were generating was excellent tests for quantitative theory. And if, in those days, I went to a lecture by a theorist, they were saying, we're generating stuff to check against experiment. And it's a circle, and that's not what we're trying to do. The theory and the experiment are just the beginning. And you can start from either extreme. What we want is, how does it all work? What is going on? Can we build a picture, which is intuitive and checkable and predictive, so that we can say, oh, yeah. If we want to know something, we can get to it this way. And the purpose of this lecture is to not provide a textbook view of spectra, but to give you a sense of how you can get to stuff that challenge your intuition. What we want to do as scientists is to be surprised. We want to do a good experiment or do a good calculation. And we want to find that the result is not what we expected. And we can figure out why it's not what we expected. And that's never conveyed in any textbooks. Now, this lecture is based on my little book of lecture notes. And I have a number of copies of it. And if people have a strong wish to have a copy, you can have one. I can give it to you. And a lot of this lecture is based on the first chapter of this book. But many of the topics are developed throughout the book. OK. So this is a two lecture sequence, and the first half will cover this stuff on this board. And it's in the notes, so you don't have to copy all this stuff. I just want you to see where we're going. So I've already talked a little bit about experiment versus theory. They complement each other. We can use theory to devise an experiment, which is path breaking. Or we can use an experiment to challenge the theorist to calculate a new kind of thing. And I hope that some of these things, those ideas, present themselves in this lecture. And it's really important that if there's something that you don't understand or don't capture the importance of, you should ask me a question. I really want to talk about what it's all for. OK. So I'm going to talk about spectra, and it will be what kinds-- rotation, vibration, electronic, and other ramifications. And going from atom to diatomic to polyatomic to condensed phase. Each step along this path leads to new complexities and new insights. Then I'll talk about, OK, we got a spectrum. What do we expect to be in the spectrum? Well, one of the things that's important is the transition selection rules or transition rules. Selection rules correspond to an operator-- eigenvalues of an operator-- that commutes with the exact Hamiltonian. And those correspond to symmetries. And there are propensity rules like, OK, which transitions are going to be strong and which are going to be weak? And a beautiful example of propensity rules are based on the Franck-Condon principle. And the Franck-Condon principle is one of the first keys you use to unlock what's in a spectrum, or what a molecule is doing. Because it's the first level of complexity that is presented to you in the spectrum. What are the vibrational bands, and how do we assign them, and what are they telling us? There is a very different kind of information in an absorption spectrum, because it's always from the lowest electronic state, lowest vibrational level. And so there's a simplicity, because of a kind of state selection. And in emission, it's a very different ballgame. Because in the gas phase, the emission is from many different levels. In the condensed phase, it's not. Why? OK. And then we get to dynamics. And the main thing I want to do in these two lectures is to whet your appetite for dynamics. And there are many kinds of dynamics ranging from a simple two level quantum beat to intramolecular vibrational redistribution. Which you can understand by perturbation theory to the strange behavior of electronically excited states losing their ability to fluoresce not because the molecule breaks, but because the bright state-- that's an important concept-- the bright state is something. It's not an eigenstate. It's a special state that we understand well. It's one of the things that we build perturbation theory around. The bright state mixes into an enormous number of dark states. And the molecule forgets that it knows how to fluoresce because the different components, different eigenstates dephase. And that is a beautiful theory. And when I was a graduate student, this theory of radiationless transitions-- Bixon-Jortner theory-- was just created and many people didn't believe it. They thought molecules, big molecules, they're really easy to be quenched by collision and the loss of the ability to fluoresce or the absence of fluorescence was somehow collision related as opposed to a physical process. So there's lots of good stuff. And some of the good stuff is Ahmed Zewail's Nobel Prize where he claims-- and that's why he got the Nobel Prize, because people believed that claim. Now, I'm not saying it's wrong. But part of getting famous is to have a package, which you can sell. And he sold the daylights out of it. And he calls it clocking real dynamics in real time. And it's basically wave packets. But they're wave packets doing neat stuff. And for example, one way a molecule can lose the ability to fluoresce is because the molecule breaks. And what is the mechanism by which a molecule breaks? Does the bond just simply break or is there some motion that precedes that? And what Zewail did was to show what are the motions that lead the molecule into the region of state space where the bond breaks. And, of course, if you want to manipulate molecules, you either want to get to those regions or avoid them. And so there's all sorts of insight there. OK. Now this is what I believe. That if you understand small molecules, you will see examples of everything you need to know to deal with almost any dynamical process in chemistry. Now, this is certainly an exaggeration, but this has been my motto for years. And so I really stress the small molecules. And it's not that small molecules are really hard. They're really beautiful, and they do enough so that you can anticipate what you need to deal with bigger molecules. So let's begin. OK. So what is a molecule? As chemists, we would never think of a molecule as a bag of nuclei and electrons. We wouldn't think of it as a bag of atoms either. We believe in chemical bonds. This is an important thing. It's not a conserved quantity. Bonds can break. But we believe that bonds tell an important story. And so almost all of our pictures for complicated phenomena are based on the-- I hesitate to use the word sanctity-- but the importance of bonds. OK. We start-- I'm going to erase this, because I've made my point, and it's embarrassing to keep emphasizing my secret motto. But it is true. OK. We have the Born-Oppenheimer approximation. And this is very important, because we can't solve a three body problem. We can solve a two body problem. But we have molecules, which are consisting of nuclei and electrons. And this Born-Oppenheimer approximation enables us to separate the nuclear part of the problem from the electronic part of the problem. Because these two things move at very different velocities. And so it's a profound simplification. We get potential energy curves or potential energy surfaces. And that is the repository of essentially everything we want to know. If we know the potential surface, we can begin to do almost anything. And certainly for a big molecule, it's not just a simple curve like this. If you have N atoms, there's 3N minus 6 vibrational modes. And well, that sounds terrible. But even for this, you have essentially an infinite number of vibrational levels and an infinite number of rotational levels. And so if you have a polyatomic molecule, you have 3N minus 6 infinities of infinities. So you're not wanting to get everything. You want to generate enough information to be able to calculate anything you want. And sometimes, you make approximations and you're not sure that those approximations are good. And you want them to break. You want to discover something new. So the Born-Oppenheimer approximation, we go from clamped nuclei calculation where the-- since the nuclei moves slow compared to the electrons, well, let's not let them move at all. And then we build a perturbation theory picture where we let them move. And we can deal with that because we understand vibrations and rotations. OK. So we have a potential energy surface. And there are things that we can anticipate about a potential energy surface. And LCAO-MO theory enables you to say a lot of important things about the potential energy surface. So it provides a qualitative framework. And so from molecular orbital theory-- and this is not what Professor Van Voorhis talked about. This is the baby stuff. And we don't expect to get the exact answer. But we do expect to be able to explain trends. Trends within a molecule and between related molecules. So this provides a framework for expectations. And there are things that we get like bond order. And we talk about orbitals that are bonding, non-bonding, and antibonding. And this comes directly out of the simple ideas. Recall when we had atom with a hydrogen, the hydrogen doesn't make pi bonds. And so they're pi orbitals for the atom A, which have nothing to interact with. And they're usually non-bonding. So we have these sorts of things. We have spN hybridization. And this is just telling you if a molecule wants to make the maximum number of bonds, you do something. And if it's sp cubed, you have four tetrahedrally arranged bonds. And if it's sp cubed, sp squared is planar with 120 degrees. These things tell you something about geometric expectations. Now, molecules don't follow the rules exactly, but they come pretty close. And so if you have some reason to believe that a particular hybridization is appropriate, then you have certain expectations for the geometry and how that's going to present itself in the spectrum. So bond order's related to internuclear distance and vibrational frequencies. Sp hybridization has to do with geometry, and all of these things are really important. So we have a potential surface. And let's say this is a boldface thing, implying that there are 3N minus 6 different displacement coordinates. This potential encodes the normal modes. What's a normal mode? Well, it's a classical mechanical concept. And it basically corresponds to situations where all of the atoms move at the same frequency in each normal mode. And these normal modes each has an expected frequency and expected geometry. Because if it's a polyatomic molecule, you have not just two things moving. They'll always be moving at the same frequency. But you have three or four or 100. And OK. So you learn about the shape of the potential and the force constants and so on. Now, we have the rotational structure. Now, molecules are not rigid rotors. But it's useful to think about molecules as rigid rotors to develop a basis set for describing rotation. And perturbation theory enables us to describe the energy levels of a non-rigid vibrating rotor. And it's straightforward. It may be ugly, but it tells you how you take information from the spectrum and learn about, say, the internuclear distance dependence of molecular constants like a spin orbit constant. Or some hyperfine constant. Or just the rotational constant. And it's a simple thing. And I'm toying with the idea of using this sort of problem on the exam. And I'm not sure whether I did on the second exam. But if I did, you'll have a chance to redeem yourself. OK. OK. Then in the gas phase, nothing much happens. You can have collisions, but the time between collisions can be controlled by what the pressure you use. And so you can sort of think about the gas phase as something where the molecules are isolated. And another way of saying that is the expectation value of the Hamiltonian in any state is time independent. The Hamiltonian is energy. Energy is conserved. And unless there are collisions, energy will be conserved. And so. But in the condensed phase, you have lot of collisions. They're very fast. And so one big difference between the gas phase and the condensed phase is energy is not conserved. And it's not conserved at different rates for different kinds of motions. And you want to understand that. OK. So now let's talk about the kinds of spectra. We have rotational spectra. And that usually is in the micro region of the spectrum. And it requires that the electric dipole moment be not equal to zero. I'll talk about this some more in a minute. We have vibration. And vibrational spectrum is in the infrared. And the requirement for vibration is that the dipole moment-- which is a vector quantity if you don't have a diatomic molecule-- changes with displacements, each of the normal modes. And so we have a molecule like CO2. CO2 does not have a dipole moment at equilibrium. It's a linear molecule, symmetric. But it can do this and that is a change in dipole moment. It can do this and that produces a new dipole moment. Or produces a dipole moment. And it can do this, which does not. So you have different modes, which are infrared active and infrared inactive. And it's related to this dipole moment. Now notice, I put a vector sign on it. It corresponds to motion, directions in the body frame where the dipole moment changes. When you do this, the dipole moment is perpendicular to the axis. When you do this, it's along the axis. And so there is stuff-- really a lot of stuff-- just looking at what vibrational modes are active. And then there's electronic. And that's mostly in the visible and UV. But the electronic spectrum could be in the X-ray region even. But mostly, molecules break when they get outside of the ordinary UV region. And so there's not much there. OK. What's needed. Well, what's needed is the electronic transition moment. Let's call this e1, e2. Going from two different electronic states, there is an electric dipole transition moment, which is not equal to zero. So H2, which has no vibrational spectrum and no rotational spectrum has an electronic spectrum. Everything has an electronic spectrum. OK. Now, a diatomic molecule. Here is sort of a template for everything a diatomic molecule can do. Now, they're cleverer than this. But this is sort of in preparation for dealing with greater complexity. And then we can have up here. OK. So this is the electronic ground state. And normally, if you have a diatomic molecule, you can predict what is the electronic ground state and how is it going to look-- how is its potential curve going to look relative to the excited states. That's something you should be able to do using LCAO-MO theory. OK. So this is the ground state. This is the repulsive state. And usually, the ground state correlates with ground state of the atoms. And so here we have a repulsive state, which also correlates with the ground state of the atom. So this is an excited state, and that correlates with an excited state of the atoms. And this is AB plus an electron. So that's one way the molecule breaks. And this dotted curve represents Rydberg states They're Rydberg states converging to every rotation vibration level of this excited state. And it can be complicated. But it's beautiful, because I know the magic decoder for how do you deal with Rydberg states. Because there's a lot of them, but they're closely related to each other. And we can exploit that relationship in guiding an experiment. And so here, now we have a curve crossing between a repulsive state and a bound state. And that leads to what we call predissociation. So the vibrational level of this state-- which would normally be bound above the curve crossing-- are not bound. And that's encoded in the spectrum too. And so for more complicated molecules, they're going to be these curve crossings or surface crossings. And we want to know how do we deal with them, and what is a diatomic-like way of dealing with them. And the important thing is at that internuclear distance where the curves cross, then you could be at a level-- starting at a level on this state-- the bound state. And it will have the same momentum at the crossing radius as the repulsive state. And that's where it goes. That's where it leaks out. I mean, we're normally used to thinking about processes-- non-radiative processes, all kinds of processes-- as an integral overall space. But because the momenta are the same on the two curves at this radius, they can go freely from one to the other. The molecules can go freely from one to the other. We get a tremendous simplification. And this is something that is always ignored in textbooks and is a profound insight. Because you know exactly where things happen and why they happen. And so you can arrange the information to describe what's going on at this point. And this is where semi-classical theory is really valuable. Because not only do you know that you have stationary phase at this point. But you know what the spatial oscillation frequency is. Because the spatial oscillation frequency is H-- or the wavelength-- is h over p. And you know what the momentum is at this point. And so you know where the nodes are. How far the nodes are apart and what the amplitude is here. And so it tells you exactly what you want to know in order to describe this non-radiative process. And my belief is that almost all of the complex things that molecules do happen at a predictable region in coordinated space. And you can get the information you need to understand them, because all of a sudden, the molecule is behaving in a kind of classical way. And we're entitled to think locally rather than globally. OK. I'm going very slowly. We may get through most of what I planned to talk about today. I have 11 pages of notes, and this is-- I'm on page three. So maybe we'll take off. OK. So how do you do spectra? Well, in the old days, you had some light source like a candle, and you had a lens, and you had an absorption cell. And there would be some sort of a spectrometer here. Could be a grading. It could be a prism. It could be anything. But it's something that says, OK, I looked at the selected wavelengths at which the gas in this cell removed light from the continuum. And then you have a detector, which in the old days was a photographic plate. But it could be a photo multiplier, and you're looking at the spectrum by scanning the grading or something like that. And so what you would get is some kind of a record where you have dark regions corresponding to where there has been no absorption and, well, actually it would be the other way around. There'd be bright regions, because there'd be no exposure of the emulsion on the plate and dark regions where there-- yes-- where the light hits. OK. So but we're much cleverer than this. And we can do all sorts of wonderful things. And again, I've been around for a long time and lasers were just beginning to be used when I was a graduate student. And I was one of the first small molecules spectroscopists to use lasers. But not as a graduate student. I wanted them, but we had such-- lasers were so terrible in the region between 1965 and 1971 when I was a graduate student. And so lasers were things to be admired, but hardly to be used. But one of the crucial things was dye lasers. Because these guys are monochromatic, and they can be tuned and tuned continuously over a wide region of the spectrum. And so that's way better than a candle or a light bulb. Because it's monochromatic, and you're asking one question at a time as you tune the laser. Now, lasers enable you to do many kinds of experiments. You can simply tune the laser through a series of transitions, and you get fluorescence every time the laser tunes through a transition. And if one laser is good, two lasers are better. And so you can do all sorts of things like suppose this spectrum is really complicated. And you want to be able to simplify it. And so you can do a double resonance experiment where you tune this laser to one line and then you tune this laser through a series of transition. That spectrum is going to be simple, and it's going to be telling you who this was. And there's just no end of tricks. And often, instead of detecting the fluorescence, you tune the laser. Let's do this. And so starting here is some kind of a continuum, ionization continuum. And so you have this photon being used twice. One here and one to take it above the ionization limit. And so you have an excitation, which you do want to know how strong it is. And so you might monitor the fluorescence. But you don't know who this is, and you find out by tuning this. But you would detect the excitation here by subsequent ionization. It's easy to collect ions. Every ion you produce, you can detect. Every photon you produce, you can't detect. Because you have a solid angle consideration and photo multipliers are not perfect. And so ionization detection is way more sensitive. So you can do that kind of thing. You can also do-- this is a kind of sequential excitation. You could imagine doing an experiment where you have an energy level here, and you have a laser, which is not on resonance. That's a coherent process. It uses the oscilltor strength at this level to get to here. And that's related to many other kinds of current experiments. And that's neat. Now, we're recording spectra, and we need to know what the rules are. And so there are certain transitions that are allowed and certain transitions that are forbidden. Now, I talked about the transition requirements for rotation, vibration, and electronic. But let's just talk about the electronic spectrum, because the other two are simple. The transition operator is equal to the sum over electrons of e times r sub i. It's a one electron operator. And that means if we have wave functions which are Slater determinates of spin orbitals like 2s alpha. This one electron operator can only have a non-zero matrix element if the two states differ by one spin orbital. That's a big simplification. And so one can actually use this to selectively access different kinds of states by designing an experiment. But the important thing is that for electronic transitions, we have the selection rule delta so is equal to 1. Not 2. Not 0. And now, the operator doesn't have any spin involved with it. And so that means delta s equals 0. You did not change, you did not go from a singlet state to a triplet state. And the only way you can get from a singlet state to a triplet state is if the triplet state is perturbed by a singlet state. So this picture I drew where I had a repulsive state crossing through a bound state, it might have been that one of those was a triplet. And as a result, and the other is a singlet, and you have spin orbit interactions, and you get extra states, extra lines appearing. But you get this wonderful selection rule. There is also for the electric dipole that we have plus to minus parity. Now what's parity? I don't like talking about parity, because the useful definition leads to complexity. But basically, parity corresponds to the symmetry, the inversion symmetry in the laboratory frame. Now, you say, well, a molecule doesn't have any inversion symmetry. But space is isotopic. And so you can go from a left handed to a right handed coordinate system. And that's what happens when you invert space. And so you can classify levels according to whether they're odd or even with respect to space inversion. This is close to the truth. This is close to all you need to know unless you're actually going to do stuff with the parity operator. But it's a useful way of saying, OK, I put parity labels on things. I learned how to do that, and that's enough. Now, you follow selection rules where good quantum numbers are conserved. Or they change in a way that you predict based on the way you did the experiment. A good quantum number, I remind you, is the eigenvalue of an operator that commutes with the exact Hamiltonian. There are very few rigorously good quantum numbers. But if a molecule has any symmetry, group theory tells you a bunch of things that commute with the Hamiltonian, and it gives you symmetry labels. And that's very important in inorganic chemistry where you have either molecules with symmetry or molecules with atoms with ligands in a symmetric arrangement. And since the transition is on the center atom usually that you can classify them using group theory as allowed or forbidden. So if we have an electronic transition, the easiest thing to observe is vibrational bands. If you have a relatively low resolution spectrum, you're going to see vibrational bands. You have to work harder to see the rotational transitions in each vibrational band, but you get an enormous amount of qualitative information just looking at the vibrational bands. Because the vibrational bands encode the difference between the ground state potential and the excited state potential. Now, this is a universal notation. Ground state is always double prime. Upper state is always single prime. Very strange, but that's the way it is. And so if these potentials are different, the vibrational bands encode the difference. And this comes from the Franck-Condon principle, which says nuclei move slowly, electrons move fast. The transition is an instantaneous process, as far as the nuclei are concerned. And so there's no change in nuclear coordinates, and there is no change in nuclear momentum. This is what's in all the textbooks, and nobody ever talks about this, because we don't really normally know or think about momentum. But we do know what it is. We do know what the operator is. And we know that kinetic energy is related to the momentum squared over 2 times the mass. So what is this? This means transitions are vertical. In other words, if we have a pair of electronic states, we draw these vertical lines. Not slanting lines. This means momentum is conserved. And this, here at this vertical point, we have this much momentum. And, well, there's the same amount here. Now usually, this just means-- the delta p is equal to 0-- means that of all the strong transitions, turning point to turning point transitions are the strongest. Because at a turning point, the vibrational amplitude is large. But there's more to it than that, because there are secondary maxima in the vibrational transition intensities. These correspond to stationary phase between the initial state and the final state. And so in addition to the strongest transitions, you get other transitions that you can explain by this delta p equals 0. Now, you don't know anything when you start. You know something maybe about the ground state. And this could be a polyatomic molecule if we're just looking at one mode. And suppose we have an excited state where the vibrational frequency is the same. In other words, there is no change in bonding character. And so what you end up getting is just the zero to zero transition in absorption. Or if you have many vibrational levels up here, you see v to v, delta v equals 0. Now, you could have a bound state, and it's usually true that the excited state is less bound. And so you would have-- the Franck-Condon active region corresponds to turning point to turning point in the lower state. So I shouldn't draw this. Let's draw a vertical transition. Well, I missed. Let me just start again. So we have an excited state, and we have a ground state. Ground state is bound. And this is v equal 0. And so we go from here to there. So if the excited state is less bound, it's a larger inner nucleus and smaller vibrational frequency. We have many vibrational levels accessed by the Franck-Condon principle. And if we have, in the other sense, we have an excited state, which is more bound than the ground state then the Franck-Condon region is narrower. Because this wall is nearly vertical. You have more transitions when it's displaced this way. And this branch of the potential is nearly much flatter, and you have fewer transitions here. So the vibrational pattern tells you qualitatively from the get go whether you're going to a more bound or a less bound state. That's very useful information. Now, you want to go further. You want to say, oh, well, I'm observing a bunch of vibrational levels in the excited state. What are their quantum numbers? Are we seeing the v equals zero level? How do we know what vibrational level we're observing? Or levels? And we can calculate Franck-Condon factors. But you can do many things. So I mean, if you have a situation like this, you might not get to the lowest vibrational level. But if you have a situation like this, you probably will get to the lowest vibrational level. So one way to get vibrational numbering is seeing the vibrational pattern terminate. That's a good sign it's v equals 0. If it terminates abruptly, it's v equal 0. But if it terminates slowly, you're not sure. But there are other really wonderful things about this. And you can do isotope separation. Isotope shifts. Since the vibrational frequency is the square root of k over mu, we can change and reduce mass. That changes which vibrational bands you observe, and it changes it in a quantitative way, and so you can often use that to tell. Another thing you can do is to now. If you have an excited state, the vibrational wave function is going to look like this. And there's a node here, and a node here, and node here, node here, node here. And so if we're looking at the vibrational progression observed from such a level, the intensities will have minima corresponding to how many nodes there are in the excited state. And the number of nodes is the vibrational quantum number. So there are lots of ways of doing it. And then there is something else. If you observe at moderately low resolution a vibrational band in an electronic spectrum, it will have the peculiar shape like that or like that. This is called a band head. And it corresponds to the fact that because the rotational constants in the upper and lower state are different, one of the branches-- you have delta J equals plus or minus 1, 0. And the delta J plus 1 is called the R branch and minus 1 is called the P branch. And delta J of 0 is called the Q branch. Now, these two guys are such that depending on the sign of the difference in B values, you get ahead on the high frequency side or ahead on the low frequency side. Well, actually, it's the other way around. But it tells you then the rotational constant, which is also a signal of how bound the state is. The rotational constant-- the shading of these band heads-- confirms your vibrational assignment. And typically, if you have a smaller vibrational frequency, you'll also have a smaller rotational constant. It's not always true. And it's always interesting when it doesn't happen. And so the vibrational bands have this asymmetric shape unless the rotational constant upstairs and downstairs is the same. And then you have a branch going this way and a branch going this way. And a gap in the middle that might be filled with some Q branch lines. And the Q branch lines tell you, oh, yeah, that was a transition where the lambda, the projection of L on the internuclear axis changed. And if it doesn't have this Q branch, it'll tell you this delta lambda equals 0. All sorts of stuff. And if you have no sign of band heads, but just this double hump structure, it tells you that the rotational constant is about the same as in the ground state. And it tells you also that the vibrational frequency is expected to be about the same. I better stop. So as soon as you go to polyatomic molecules. Instead of having just one vibration, you have 3n minus 6. And so 3n minus 6 downstairs, 3n minus 6 upstairs, that looks bad. But only some of the vibrational modes correspond to a distortion from-- a difference in geometry between the ground state and the excited state. And so most of the vibrational modes correspond to identical frequencies. And what we call that is Franck-Condon dark. Because the only transitions that are allowed are delta v equals 0 for that mode. And so only the modes that correspond to a change in structure appear as long progressions. And so there are only a few, and so the spectrum of a polyatomic molecule is not too much different from a diatomic molecule, because of the small number of Franck-Condon active modes. But you look closer and you see big differences. OK. So well, we'll talk more about this on Friday. And we might go to three lectures on this, because this is really-- you know the whole point of this course is to be able to understand how molecules talk to us. |
MIT_561_Physical_Chemistry_Fall_2017 | 32_Intermolecular_Interactions_by_NonDegenerate_Perturbation_Theory.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseware continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseware at ocw.mit.edu. PROFESSOR: Today, since the notes for this didn't exist until last night, I made copies for you. I will do a little more revisions on what appears on the web. But this is a really interesting and important topic. And so I think it's important that I give a little bit of extra guidance. So we're going to talk about what's in a spectrum. But in a very special sense, we're going to be talking about wave packet dynamics. And how do we understand what to expect in wave packet dynamics? And how do we approach it? And so let's begin talking about this. So you already know from my lectures earlier, if you have a time independent Hamiltonian, how do you create a wave packet? I mean, after this exam performance, I want some insights. AUDIENCE: [INAUDIBLE] the individual eigenstates. PROFESSOR: Yeah. So you use a short pulse. And you create some kind of linear combination. Usually, if you're not making eigenstates, that's a complication. And why would you want to do that? And the reason you do it is because these wave packets are going to move. And particles move. Your instincts are about particles moving. And there are rules for the particles moving, which are mostly what you have in your head. But there are some special stuff that has to do with quantum mechanics. And so I'd like to build from the bottom. And so I want you to be able to understand wave packets moving on a diatomic surface. And then we'll make a step into slightly more complicated situations. If we have a diatomic, we have a potential curve. And we have an initial state. And that initial state in V equals 0 gives us access to a range of vibrational levels in the upper state. And if we have a short pulse, we can create some kind of a wave packet involving a linear combination of these vibrational levels. So suppose you had a polyatomic, 3N minus 6 vibrational modes. So what would you do to go from a diatonic molecule to a polyatomic molecule? Initially, you want to do something simple that doesn't involve anything that's especially new. So what would you say about dynamic diatomic polyatomic surface where there's 3N minus 6 as opposed to 1 vibrational modes? What kind of a wave packet would you do, would you create? AUDIENCE: [INAUDIBLE] PROFESSOR: We had Franck-Condon bright modes and Franck-Condon dark modes. So it sounds like you're right on the verge of saying something profound. So say it. AUDIENCE: You're just going to have lot more coordinates. PROFESSOR: Yeah. You're going to have more coordinates. AUDIENCE: [INAUDIBLE] PROFESSOR: So you will create a wave packet on the Franck-Condon bright modes. And if there's more than one Franck-Condon bright mode, the wave packet will do just like what it would do on a diatomic molecule, except you might have two different frequencies. OK, well, there's additional complications. And that is, even a diatomic molecule, we don't have harmonic motion. So the wave packet, because of diagonal anharmonicity, will dephase. But the center of the wave packet will continue to move at the average of the vibrational frequencies for all these modes. You know that. And so in a polyatomic molecule, maybe there's two, maybe there's three Franck-Condon bright modes. And the wave packet will have several different components on the different Franck-Condon bright modes. And they'll be diagonal anharmonicities. And they will dephase and possibly do interesting things. So now, polyatomic molecules can do something else. Because there are, in fact, 3N minus 6 vibrational modes. And there's anharmonic couplings between the modes. So if you created some wave packet or set of wave packets in a polyatomic molecule, there will be anharmonic interactions between modes. And what you will have is the appearance of wave packets in the Franck-Condon dark modes with their own frequencies. So this is starting to get interesting. Because you hit this molecule. And it starts revealing its secrets. Where you had Franck-Condon dark modes, they're not going to talk to you. But now, because of the anharmonic couplings, they do. And then with a diatomic molecule, we might have an excited state. And it might be crossed by a repulsive state or crossed by another bound state. But if you have a wave packet operating on this bound state, it's going to be crossing through this region. What's so special about where curves cross? You know this, too. If you're in this state and at this point, you have this much momentum. And if you're on this curve at that point, you have the same. So you have two rapidly oscillating things with the same spatial oscillation frequency. And so there's a possibility that the wave packet can leak from this curve, from the bound curve, to the repulsive curve. And there are all sorts of things of that nature. I'm going to talk about all of that today. And one of the things-- the reason I made these notes is because I've always wanted to say something about the Landau-Zener model for crossing between different potential curves. And we talked a little bit before about driving too fast on a curvy road. Do you stay on it or do you hit the tree? And that's basically this picture of-- you have two curves that can cross or they could do this. This is the adiabatic representation, which is amenable to quantum chemical calculations. And this crossing curve is the diabatic representation, which chemists like because the electronic wave functions don't change as a function of internuclear distance. And so the issue is, when you have crossing curves or avoided crossing curves, how do you understand what the wave packet is going to do with those points? And that's the point of this lecture. And there are two experiments done by Ahmed Zewail's research group which illustrate many of the effects of wave packets and avoided crossings. And so the question is, how do we observe this kind of evolution of the wave packet? What kind of experiments do we do? If we're talking about motion, we're probably not interested in the frequency domain. Although, you can get a lot of information from the frequency domain that enables you to understand what you might see in the time domain. And one thing about frequency domain experiments-- when you have a molecule, which has 3N minus 6 vibrational modes, like benzene has 30, the vibrational density of states increases rapidly with excitation energy. So rapidly that, above the region of the fundamentals, the highest frequency fundamentals, they're no longer resolvable vibrational levels. The density of states is on the order of a billion per wave number. And you're going to be asking different kinds of experimental questions to say, do I understand what's going on in even a molecule as small, from your point of view, as benzene, big from my point of view. OK, so we're interested in mechanism. Why do things happen? What makes them happen? And how do we construct an experiment that reveals mechanism? And how do we interpret those sorts of experiments? And the Zewail experiments are beautiful examples of revealing mechanism, which is more complicated-- well, more revealing than what you might expect. For example, suppose you excite a molecule and it breaks. Well, how did it break? Did it just break? Or was there some motion preceding the breaking where the molecule arranged itself to receive this photon and do something with it? And it makes fragments. And what are the fragments? What state are the fragments in? There are all sorts of stuff there. And that's mechanism. And it's not just the breaking of a molecule and getting from the width of a spectral feature how fast that breaking occurred. There's things that happen in that time. OK. So we need to talk about diabatic versus adiabatic. And so if you did a quantum chemical calculation-- and you can. Because you still have access to the Athena cluster. Or is your membership expired? AUDIENCE: We should always have access to it. PROFESSOR: OK. Well, you do a calculation using the computer programs. And they're basically clamped nuclei. And that's just because you want to reduce the complexity of the calculation. It's an impossible calculation anyway to do exactly. And if we say, let's keep the heavy particles from moving around, we just solve the electronic Schrodinger equation, well, then we get what we call an adiabatic potential energy surface, or one for each electronic state. And well, if we clamp the nuclei, molecules don't hit the nuclei clamp. We have to unclamp them. And the effect unclamping leads to some perturbation term, H1. And so when we solve the clamped nuclei Schrodinger equation, we don't have vibrations. We don't have rotations. And so one of the things we don't know about is partial derivatives with respect to nuclear coordinates, which we have the nuclear kinetic energy, which is the second derivative. And when you have a second derivative, you can apply one derivative to the nuclear part of the wave function and one derivative to the electronic part. Or you can apply both. And so the electronic wave function you get is-- you get a wave function, which is an explicit function of the electron positions and parametrically dependent on internuclear distances or nuclear geometry. Well, that's there. And this nuclear kinetic energy term, which is present, is going to operate on these guys and lead to trouble. Or maybe not, depending on what the potential curves look like and what the their problem is. But certainly, this nuclear kinetic energy can operate on these functions and give something that we have to at least consider. So this guy has secrets embedded in it that we're going to have to look at. And they're going to be surprises. So let's think about diatomics. Because with diatomics, most of the things you're going to encounter in polyatomic molecules have examples that you can understand really clearly. So suppose there are two electronic states that cross or do something. And we can have a picture like this. So the calculation you do will give you these avoided crossing potential curves. Well, why do they avoid each other? If you have two states of the same symmetry, they can perturb each other. The Hamiltonian is totally symmetric. Any kind of term in the Hamiltonian that is totally symmetric can cause interactions between states of the same symmetry. Now, many of these terms of the same symmetry-- totally symmetric terms in the Hamiltonian-- are excluded when we do a calculation. Because we can. One is spin R.? But there are things that could affect the states belonging to these two potential curves. And so there are going to be interactions between states of the same symmetry. And this is the place. So you have the ab initio calculation. You get these curves. And you can see that there is some coordinate at which the difference between potential curves is a minimum. Nature tips its hand. These avoided crossings that are really important. And it tells you-- the quantum chemical calculations tell you two things. They tell you the internuclear distance at which the crossing occurs. And let's call this one-- let's have several names for these curves. OK, I have to-- so we can call this 2, 1, 1, 2 plus plus minus minus. I have two sets of labels. So let's say that the 2 and 1 describe the electronic character of this state, the kind of thing that the chemists care about. And plus and minus as upper and lower. And that's what you get from quantum chemistry. And so the quantum chemistry gives you the place at which these two curves have this minimum separation. And they have the plus at RC minus V minus at RC is equal to 2 H 12. You know from perturbation theory that, if you have two levels which are degenerate and they're interacting by some coupling term, the separation will end up being twice the matrix element. So nature gives you this and this from which you can construct the crossing curves, the curves that chemists like. So even though quantum chemistry doesn't know about diabatic curves, it tells you how you could construct them. And now, there's two limiting cases. We can have a very weak interaction. And so the curves get really close together. Or we can have a very strong interaction and get something like this. So these are the two limits. And it turns out that the way you would handle these two limits is experimentally and theoretically profoundly different. Why? This term, the nuclear kinetic energy term, does terrible things here. Because when it operates on the electronic wave function, it says, the electronic wave function is changing rapidly in this region. And so if we're going to try to set up a Hamiltonian that describes the energy levels of these two states, because of this term, there will be enormous couplings between the vibrational levels of the two states. And as a result, H0, the thing that ignores those coupling effects, will bear almost no resemblance to the observed energy level structure. To get from these kinds of curves to the observed energy levels is a lot of work. And it's work that people who use pre-written computer programs are ill prepared to do. The spectrum will not be predicted in any way by this, unless you do a huge matrix of interactions between vibrational levels of the different electronic states. And when you have a matrix element, which is large compared to the differences between levels, you can't use perturbation theory. You have to diagonalize a matrix. And usually, when you diagonalize a matrix, the eigenvalues are a bit of a surprise. Over here, the electronic wave function doesn't change very rapidly. And so we have no problem of the interactions among the vibrational levels of these two states. And so the adiabatic representation provides a good zero order picture of the energy levels. And there's a little bit of stuff that you can add to improve the fit. So when you have what's called a weakly avoided crossing, it tells you that the adiabatic representation is the wrong one. But when you have a strongly avoided crossing, well, it's the right one. Because all of the effects of this repulsion between the two states are included in your zero order Hamiltonian. Whereas here, none are. And you say, well, I could just go cruising through this region. And I go fast through it. And I don't notice the bad effects. Well, we'll see. OK. So now, if we had a diabatic representation like this, well, diabatic curves can cross. In fact, that's the whole point. We want them to cross. Because we'd like to keep track of electronic character 1 and electronic character 2. And so I'm just going to use this here. OK. But the really annoying thing is, when we did an initial calculation, we clamped the nuclei. We said there are certain terms in the exact Hamiltonian we're going to push aside and consider later. But for the diabatic picture, well, it's clearly there's some term in the electronic Hamiltonian that we'd like to turn off. But there isn't one. It's all or nothing. There is nothing you can identify in the electronic Hamiltonian that enables the two curves to cross. So it's not that it's a bad idea. It's just there is no simple way of dealing with this. Now sometimes, we have the Hamiltonian. And we have, say, the spin orbit Hamiltonian. And this is something we could turn off. But that's gilding the lily. I am a spin orbit aficionado. So I could deal with this. But if we don't say there are specific, named, small terms in the electronic Hamiltonian, there is no way we can get the diabatic limit directly. You have to use a trick. So let's draw a picture now, which illustrates the trick. And we're almost ready to start talking about Zewail's experiment. So here I'm going to draw. OK. So the adiabatic curves are the ones that don't cross. And The. Dash lines are the diabatic curves. And we know RC and H12. Or at least we know H12 at that internuclear distance. And it's a reasonable thing to say, well, let's let it not change with internuclear distance. And let's make some other approximation. We can say the diabatic curves are linear in the region of the curve crossing. We can always-- especially when the adiabatic picture is bad, that curve crossing is very, very compressed. And it's not a big step to say, OK, over this relatively small range of internuclear distance and energy, we're going to approximate the diabatic curves as linear. And so we know the exact Hamiltonian, whatever it is, be gotten from. OK. So here, we have what's available from quantum chemistry. These are the potential curves for the upper and lower adiabatic states. And here we have a potential curve for state 1 and a potential curve for state 2 at H12. And a unitary transformation of this matrix has to be equal that matrix. Well, we don't know these. But we're going to reduce them to one number, the slope at the crossing. And so we have one number here, one number here. And all of a sudden, we have enough to iteratively determine the difference between the slopes by fitting to the observed adiabatic potentials. And this is basically how it's done. Now, people who do these calculations for a living have a much encrusted picture of how they do this. But this is basically what's going on. Because people do want the diabatic picture. And a lot of insight is gained from the diabatic picture. OK. So if we're going to be approaching a spectrum where we have energy levels or an experiment where we learn something about the dynamics, we're going to want to think about this problem with two limits-- weakly avoided and strongly avoided. And completely different methods for dealing with both spectroscopic and dynamical information are appropriate for the two limits. OK. Landau-Zener-- this is a model for saying, OK, what is the probability of going for one curve to the other? And so we have a formula, which you can derive. I don't recommend it. Because derivations are things you do when you need it. And then you forget them. But I do want to give you a sense of what's in it. So we're going from one adiabatic state to another. And the probability is 1 minus e to the minus pi gamma. The important parameter is gamma, right? Everything in this. OK. And so P12 is small when gamma is-- well, we want this to be 1-- so when gamma is small. And P12 is large-- now, I just want to make sure that I have it-- yeah, when gamma is large. Now, what's gamma? Gamma is expressed-- and this is what you would derive if you were going to do this for real life. Because you would never accept somebody else's speculation, you derive it yourself. So V12 squared-- that's the matrix element. That's half the gap between the diabatic curves. And it's over H bar velocity S1 minus S2. This is the difference in slopes of the diabatic curves at the crossing point. This is the velocity of the particle going through the crossing point. And you can, of course, get the velocity from the momentum divided by the mass. And so all of this stuff is there. And so gamma is small when V is large and when S1 is approximately equal to S2. Well, if the slopes are nearly equal, that's a strongly avoided crossing, right? And V is large-- well, if you're going through a bend in the road and it's too fast, well, then you're not going to-- yes, you're not going to be able to make the curve. So anyway, this is the physical basis behind Landau-Zener. And it contains the connections to the stuff you have from your picture of the potentials. And this is basically how you organize a Zewail type experiment, which I've got to talk about now. You have those notes already on Zewail electronically. So I didn't make copies of them for you. So you're going to have to accept what I'm doing without looking at notes for this part, unless you have them already printed out. So Zewail had the 1999 Nobel Prize in chemistry. And it was really special. Because usually, these things are divided three ways. He got the whole damn thing. And this was because he offered something that we really want. We want to have an idea of, what is the mechanism by which dynamical processes occur? In other words, we're not just getting rate. But we're getting, what are the nuclei doing? And how does the motion of the nuclei affect the rate of the process? AUDIENCE: I think you want S1 is very different than S2. PROFESSOR: That's quite likely. And I'm very poor with logic. Let's see. So I want you to decide for yourself, OK? And let me just-- OK. So this is an example of a pump/probe experiment. So the pump pulse at t equals 0 starts things. And the probe pulse at t equals tau asks the question, has the wave packet gotten to what's called the OCR, the optically coupled region? In other words, you create a wave packet and you have a probe which can tell you the time at which the wave packet passes through where you're probing it. And this is really neat. And it is, I think, the essence of how Zewail created a really simple experiment using the crude technology that was available at the time to ask the kind of question he needed to ask. Now, I didn't like this whale experiment when it first came out. Well, because I'm a frequency domain spectroscopist. And I don't really care about dynamics, except how it's encoded in the frequency domain spectrum. So this is completely different. But anyway, so now, I want to describe the essence of what he did, and why it worked, and what it reveals. All right, and so I want to pick a blackboard. So there are two experiments that he did. One was dissociation of I-CN. And the other was dissociation of sodium iodide. Now, one thing to notice is we've got a big heavy atom in here. So that means the vibrational frequencies are low. That means that, with a not too short pulse, you could create a coherent superposition. When you think about what you would need to create a coherent superposition of vibrational levels, different by the canonical 1,000 wave numbers of a vibration, you realize that that experiment was only doable fairly recently when you have on the order of a few femtosecond time resolution. What Zewail did-- he had maybe a 100 femtosecond or maybe only a picosecond. And so he had to have a heavy atom. For I-CN, the beautiful thing about I-CN is there's CN here. CN is a diatomic molecule that has a very convenient spectrum. And so we're asking the question, how does the presence of the iodine affect the CN spectrum? Because the CN is going to give us the signal in this experiment. So here is-- it's really neat. This is a triatomic molecule. But we always use one dimensional pictures to describe everything that's going on. And you can do that when the one dimensional pictures are each applicable in separate times or at separate aspects of the experiment. But it's another beautiful example. If Zewail had been attempting to describe the full three dimensional potential , the pictures wouldn't have done anything for anybody. Because you would have to work too hard to understand what the pictures are telling you. OK. So we now have a repulsive curve. And we have another repulsive curve. OK. So we have V equals 0. And so we're exciting a wave packet. And because we have a few vibrational levels that are accessible with the Franck-Condon principle from V equals 0 and within the necessary Fourier transform of the post duration, you get a wave packet starting here. OK. So this wave packet is going to be toodling along at this energy. I'm sorry it's going to be moving on this potential. But it has a definite energy. OK. And then you can probe it this way, or this way, or this way. And the important thing is that these three places at which you probe correspond to different excitation energies. It's saying that, as the iodine atom is leaving-- these are transitions between CN. And in here, the iodine atom is close to the CN. And it is affecting the bonding in the CN. Over here, it's gone. And the CN is free. And what's happening is the transition frequency changes with time. And so if he has the probe pulse here, well, then when the wave packet reaches this point, you get a response. Before the wave packet gets there, there is no exaltation. So the OCR is defined-- the optically coupled region-- it's defined by the frequency, the center frequency, of the probe pulse. So if we then look at-- OK, now, what's happening here? When you excite to this state, you have CN electronically excited. When you look at this limit, you have CN not electronically excited. The CN A double pi x double sigma plus transition is one of the most studied things in the atomic molecules. And so it's known. And like all transitions, the intrinsic radiative lifetime is on the order of 10 nanoseconds. So if we excite to this state, we're going to get a photon out. But the lifetime of that photon is 10 nanoseconds. Are we going to measure dynamics? This is the beauty of the experiment right here. He catches the wave packet at a particular position. And he puts it in the bank. And then eventually, the CN flouresces. We don't care anything about when it flouresces or what frequency it flouresces at. All we cared about is that-- if you're looking at if the wave packet is in the optically coupled region, it gets excited to this surface. And then when it chooses to say, I got excited, here is my photon, then you can create a plot of-- so versus time. So you see nothing. And then you get something at the optically coupled region. And you do a sequence of experiments where you vary where the optically coupled region is centered by adjusting the center frequency of the probe laser. So sorry. I should really say, OK, here is the intensity of the fluorescence. And here is the 1 over the wavelength of the probe. So this is related to the energy of the probe. And so you do a series of experiments. And you discover that the wave packet reaches this. And then you move that. And you get another. And so you get a picture. So what do I want to say about this? So it basically is telling you that, when the molecule breaks, the frequency of the CN excitation is changing as a function of the distance of the iodine atom from the CN. And so that's mechanism. Now, it's really a very crude mechanism. But because it's reduced to one question, it's a function of one geometry. But nobody's ever be able to say they observed this motion of the wave packet in real time. And here is where people really get annoyed with Zewail. Because he's looking at real motion in real time. And it ain't real. He's probing one time at a time. But it's still mechanism. It's more than just looking at when does something break. What's the lifetime of something? There's more than one thing going on. Now, I really like the next experiment. And the next experiment involves sodium iodide. Now, if I were to ask you, what is the nature of sodium iodide at equilibrium? You'd say, it's probably ionic. But because iodine is at the bottom as opposed to the top of the periodic potential, it's maybe a little bit covalent. So here is the ground state potential. And I just have to make sure I understand what I'm trying to do. And here is an excited state potential. OK. So what I've tried is crossing curves. Those are the diabatic curves. Now, I've connected-- I should do it with dotted lines. Those are the adiabatic curves. And so the question is, suppose we create a wave packet on this potential. So it's created here. And this wave packet feels a force that way. Because it's the gradient of the potential. It's a particle. It's in a potential. And it's going this way. And then something happens when it goes through this curve crossing region. But what's going to happen? I should just say this is the ionic. This is the covalent. So this is NA plus plus I minus. And this is NA plus I. When you dissociate a neutral molecule, you always get neutral atoms. So there is a higher energy limit where you can get ions. And so even if you create this wave packet here, which is below the threshold for making ions, it only can leave by this path. So what happens is the wave packet is going back and forth, back and forth, back and forth. And each time it crosses through this region, it has to decide, am I ionic or covalent? And if I'm covalent, I can leave. And if I'm ionic, I'm stuck. I've got to go back and try again and again and again. So this is where Landau-Zener comes in. Because what's happening is the molecule is addressing this ionic covalent curve crossing. And it's deciding each time it goes through, what is the probability of being able to get out? OK. But then, there's something else. And that is, well, what are we probing? So we probe to an excited state. And so this is where the figures, even for this reduced representation, are confusing. But basically, you're probing by exciting this sodium atom to an excited state, which puts it in the bank. And it fluoresces when it chooses to, which is in 10 nanoseconds. Because sodium's radiative lifetime really is about 10 nanoseconds. And so how to describe this? So you have a choice of being able to excite at the free sodium transition frequency. And then what you're only seeing is these little packets of sodium that make it out. Or you can excite at a-- well, this is coming out wrong. But if you excite to the red of that, you will only see the sodium atoms while they are still close to the iodine. And so there are two sorts of signals you get. One is you get a series of pulses separated by the round trip time-- the round trip time is the period. And so the period is related the vibrational frequency. And so each time this packet goes through the optically coupled region-- I'm sorry-- goes through the curve crossing region, you get a little puff of sodium. If you're exciting so that you can see only the free sodium, well, then instead of this, you see the signal, another signal. So you get little steps. Each wave packet that makes it out dies at the turning point. And I'm way over time. But this is really the-- so there's two pictures. There is this sampling of the wave packet each time it goes through. And there is the delivery of the goods. And again, the signal is a slow signal. But the pump/probe delay is femtosecond time resolution. And so it can reveal this wave packet propagation. So that's all I have to say on the subject. I think it's really beautiful. I like it more and more the more I understand it. And the next lecture will be on why gases condense. |
MIT_561_Physical_Chemistry_Fall_2017 | 18_Rigid_Rotor_II_Derivation_by_Commutation_Rules.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: Now the main topic of this lecture is so important and so beautiful that I don't want to spend any time reviewing what I did last time. At the beginning when we talked about the rigid rotor, I said that this is not just a simple, exactly solved problem but it tells you about the angular part of every central force problem. And it's even more than that. It enables you to do a certain kind of algebra with operators which enables you to minimize the effort of calculating matrix elements and predicting selection rules simply by the basis of commutation rules of the operators without ever looking at wave functions, without ever looking at differential operators. This is a really beautiful thing about angular momentum that if we define the angular momentum in this abstract way-- and I'll describe what I mean by this epsilon ijk. If we say we have an operator which obeys this commutation rule, we will call it an angular momentum. And we go through some arguments and we discover the properties of all "angular momentum," in quotes. Now we define an angular momentum classically as r cross p. It's a vector. Now there are things, operators, where there is no r or p but we would like to describe them as angular momentum. One of them is electron spin, something that we sort of take for granted, and nuclear spin. NMR is based on these things we call angular momentum because they obey some rules. And so what I'm going to do is show the rules and show where all of this comes from and that this is an abstract and so kind of dry derivation, but it has astonishing consequences. And basically it means that if you've got angular momenta, if you know these rules, you're never going to evaluate another matrix element in your life. Now, it has another level of complexity. Sometimes you have operators that are made out of combinations of angular momenta, and you can use these sorts of arguments to derive the matrix elements of them. That's called the Wigner-Eckart theorem, and it means that the angular part of every operator is in your hands without ever looking at a wave function or a differential operator. Now we're not going to go there, but this is a very important area of quantum mechanics. And you've heard of three j symbols and Racah coefficients. Maybe you haven't. But there is just a rich literature of this sort of stuff. So today I'm going to talk briefly about rotational spectra because I'm a spectroscopist. And from rotational spectra we learn about molecular geometry. Now it's really strange because why don't we just look at a molecule and measure it? Well, we can't because it's smaller than the wavelength of light that we would use to illuminate our ruler, and so we get the structure of a molecule, the geometric structure, at least part of it, from the rotational spectrum. Now I'm only going to talk about the rotational spectrum of a diatomic molecule. You don't want me to go further because polyatomic molecules have extra complexity which you could understand, but I don't want to go there because we have a lot of other stuff to do. I also had promised to talk about visualization of wave functions, and I'll leave that to your previous experience. But I do want to comment. We often take a sum of wave functions for positive and negative projection quantum numbers to make them real or pure imaginary. When we do that, they are still eigenfunctions of the angular momentum of squared, but they're not eigenfunctions of a projection. And you could-- in fact, maybe you will on Thursday-- actually evaluate Lz times a symmetrized function. So by symmetrizing them, you get to see the nodal structure, which is nice, but you lose the fact that you have eigenfunctions of a projection of angular momentum, and that's kind of sad. OK, so spectra-- so we have a diatomic molecule, mA, mB, and we have a center mass. And so we're going to be interested in the energy levels for a rotation of the molecule around that axis which is perpendicular to the bond axis. When we do that, we discover that the energy levels are given by the rotational Hamiltonian. And for a rotation-- it's free rotation, so there's no potential. And the operator is L squared or J squared. That's another thing. You already have experienced my use of L and J and maybe some other things. They're all angular momentum. They're all the same sort of thing, although L is usually referring to electronic coordinates, and J is usually referring to nuclear coordinates. Big deal. But you get a sense that we're talking about a very rich idea where it doesn't matter what you name the things. They follow the same rules. So we have a single term in the Hamiltonian, mu r0 squared, or this might be the equilibrium instead of just the fixed internuclear distance. But we're talking about a rigid rotor, so r0 is the internuclear distance. And now I want to be able to write-- OK, so I want to have this quantity. I want to have this quantity in reciprocal centimeter units because that's what all spectroscopists do, or sometimes they use megahertz. In that case, the speed of light is gone. And when I evaluate the effect of this operator on a wave function, we get an h bar squared, which cancels that. We would like to have an energy level expression EJM is equal to hcBL L plus 1. So the units of B just accommodate the fact that we want it in wave numbers. But this is energy, So we need the hc. And when the operator operates, we get an h bar squared, and that's canceled by this factor here. And so the handy dandy expression for the rotational constant is 16.85673 times the reduced mass in AMU units times the internuclear distance in angstrom units squared reciprocal. So if you want to know the energy-- if you want to know the rotational constant in wave number units, this is the conversion. Big deal. So the energy levels are simply-- I'm going to stick with LM even though I'm hardwired to call it J. Now if I go back and forth between J and L, you'll have to forgive me because I just can't-- yes. All right, so we have the energy levels, hcB times L L plus 1. Now L is an integer, and for the simple diatomics that you're going to deal with, it's an integer. You can start at zero. And so the energy levels, the L L plus 1, the L L plus 1 is 2. I want to make this look right. B-- this is 1 times 2. This is 2 times 3. And the 3 times 4 is 12. And the important thing is that this energy differences is 2B. This energy difference is 4B. This energy difference is 6B. And so what happens in the spectrum-- here's energy. Here's zero. We have a line here at 2B, a line here at 4B, 6B, 8B. So if you were able to look at the rotational spectrum, the lines in the spectrum would be evenly spaced. The levels are not. That's very important, especially when you start doing perturbation theory because you're going to have energy denominators which are multiples of a common factor, but they're not equal to each other. But we have a spectrum, and it looks really, really trivial. And textbooks don't talk about this, but if you have a relatively light diatomic molecule and you have a laboratory which is equipped with a microwave spectrometer which is able to generate data that got you tenure and whatever, it's probably a spectrometer where the tuning range of the microwave oscillator is about 30%. That's a lot. If you think about NMR, the tuning range is-- 30% is huge. So, you see this and you say, oh yeah. I could assign that spectrum because an obvious pattern. But what happens in the spectrum is you get one line. And so you say, well, I need to know the internuclear distance of this molecule to 6 or 8 or 10 digits, but I get one line. There's no pattern. The textbooks are so full of formulas that they don't indicate that, in reality, you've got a problem. And, in fact, in reality you've got something that's also a gift. So there's two things that happen, isotopes and vibration. So we have this one line. We have a very, very strong, very narrow-- you can measure the daylights out of it if you wanted to. And then down here, there's going to be-- well, actually, sometimes like in chlorine and bromine, there's a heavy isotope and a light isotope, and they have similar abundances. And so you get isotope splittings, and that's expressed in the reduced mass mA mB over mA plus mB. Now the isotope splittings can be really, really small, but these lines have a width of a part in a million, maybe even narrower. And so you can see isotope stuff. That doesn't tell you anything at all that you didn't know except maybe that you were confused about what molecule it was because if you have a particular atom, it's always born with the normal isotope ratios. Except here we have a little problem where, in sulfur, if you look in minerals, the isotope ratios are not the naturally abundant of sulfur isotope. And this has to do with something really important that happened 2 and 1/2 billion years ago. Oxygen happened. And so isotope ratios are of some geological chemical significance, but here, if you know what the molecule is, there will be isotope lines. And they can be pretty strong depending on the relative abundance of the different isotopes, or they can be extremely weak. So there's stuff, so some grass to be mowed on the baseline. In addition-- and this is something that really surprises people. So here is v equals 0, and way up high is v equals 1. Typically, the vibrational intervals are on the order of a thousand times bigger than the rotational intervals. And typically, the rotational constant decreases in steps of about a tenth of a percent per vibration. Now we do care about how much it decreases because that allows us to know a whole bunch of stuff about how rotation and vibration interact. And I'm not probably going to do the lecture on the rotation and vibration interaction unless I have to give a lecture on something that I can't do and I'll slip in that one. So what happens is there are vibrational satellites. So here's v equals 0. It has rotational structure. And here is v equals 1. It has rotational structure. The v equals 1 stuff is typically a hundred to a thousand times weaker than the v equals 0 stuff. And that's basically telling you, how does a molecule changes its average 1 over r squared as it vibrates, and that's a useful thing. It may even be useful on Thursday. So in addition to hyperfine, there's other small stuff having to do with vibrations. And in some experiments that I do, we use UV light to break a molecule. And the fragments that we make are born vibrationally excited. And so by looking at the stuff near the v equals 0 frequency, you see a whole bunch of stuff which tells you the populations of the different vibrational levels. And that's strange because vibration is not part of the rotational spectrum. Vibration is big, but we get vibrational information from the rotational spectrum. And because the rotational spectrum is at such high resolution, it's trivial to resolve and to detect these weak other features. So as much as I'm going to talk about spectroscopy, it's a little bit more than I had originally planned. And now we're going to move to this topic which is dear to my heart, and it's an example of an abstract algebra that you use in quantum mechanics. And there are people who only do this kind of thing as opposed to solving Schrodinger equation or even just doing perturbation theory on matrices. So the rest of today's lecture is going to be an excursion through here as much as I can do. It's all clear in the notes, but I think it's a little bit strange. Oh, I want to say one more thing. How do we make assignments? You all took 5.111 or 5.112 or 3.091, and there are things that you learn about how big atoms are. And so you can sort of estimate what the internuclear distance is-- maybe to 10% or 20%. That's not of any chemical use, but it's enough to assign the spectrum. So what you do is you say OK, I guess the internuclear distance is this. That determines what rotational transition you were observing. And that has consequences of suppose you're observing L to L plus 1. Well, what about L plus to L plus 2 or L minus 1 to L? So if you make an assignment, you can predict where the other guys are. And that would require going to one of your friends who has a different spectrometer and getting him to record a spectrum for you, and that's good for human relations. And that then enables you to make assignments and know the rotational constants to as many digits as you possibly could want, includ-- all the way up to 10. It's just crazy. You really don't care about internuclear distances beyond about a thousandth of an angstrom, but you can have them. So first of all, you know we can define an angular momentum as r cross p, and we can write that as a matrix. Now I suspect you've all seen this. These are unit vectors along the x, y, and z directions. And this is a vector, so there are three components, and we get three components here. Now you do want to make sure you know this notation and know how to use it. So here is the magic equation. Li, Lj is equal to ih bar sum over k epsilon ijk Lk. Well, what is epsilon ijk? Well, it's got many names, but it's a really neat tool which is very wonderful in enabling you to derive new equations. So if i, j, and k correspond to xyz in cyclic order-- in other words, xyz, yzx, et cetera-- then this is plus 1. If it's in anticyclic order, it's minus 1. And if any index is repeated, it's 0. So it packs a real punch, but it enables you to do fantastic things. So if we have Lx, Ly, it's equal to ih bar plus 1 times Lz. And the point of this lecture is with this, you can derive all of the matrix elements of an angular momentum-- L squared, Lz, L plus minus, and anything else. But these are the important ones, and this is what we want to derive from our excursion in matrix element land. So the first thing we do is we extract some fundamental equations from this commutator. So the first equation is that L squared Lz is equal to Lx squared Lz plus Ly squared Lz plus Lz squared Lz. And we know this one is 0, right? This one, you have to do a little practice, but you can write this commutation rule as Lx times Lx comma Lz plus Lx comma Lz Lx. So if you have a square, you take it out the front side then the back side. And now we know what this. This is minus ih bar Ly. And this is minus ih bar Ly. And we do this one, and we discover we have the same thing except with the opposite sign. And so what we end up getting is that this is 0. Now I skipped some steps. I said them, but I want you to just go through that and see. So you know what this is. It's going to be Ly, and it's going to be minus Ly times ih bar. And you get the same thing here. But then you have an LxLy, and you have an LxLy. And when you do the same trick with this, you're going to get an Ly and an Lx again, and they'll be the opposite sign. So this one is really important because what it says is that you can take any projection quantum number and it will commute with the magnitude squared. The same argument works for Ly and Lz and Lx. So we have one really powerful commutator which is that L squared Li equals 0 for x, y, and z, which means since we like L squared and Lz-- we could add like Lx instead of Lz, but we tend to favor these-- that L squared and Lz are operators that can have a common set of eigenfunctions. If we have two operators that commute, the eigenfunctions of one can be the eigenfunctions of the other. Very convenient. Then there's another operator that we can derive, and that is-- let's define this thing, a step up or step down or our raising or lowering operator-- we don't know that yet-- Lx plus or minus iLy. So we might want to know the commutation rule of Lz with L plus minus. We know how to write this out because we have Lz with Lx, and we know that's going to be a minus ih bar Ly. And we have Lz with iLy, and that's going to be a minus Lx. Anyway, I'm going to just write down the final result, that this is equal to plus or minus h bar times L plus minus. The algebra of this operator enables you to slice through any derivation as fast as you can write once you've loaded this into your head. Yes? AUDIENCE: So for the epsilon, how do you [INAUDIBLE]?? Is it like xy becomes-- if it's cyclical it's positive? ROBERT FIELD: I'm sorry? AUDIENCE: When you say the epsilon thing, epsilon ijk, so you're saying that if it's in order, it's 1? ROBERT FIELD: Let's just do this a little bit. Let's say we have Lx and Ly. Well, we know that that's going to give Lz. And xyz, ijk, that's cyclic order. We say that's the home base. And if we have yxz, that would be anticyclic, and so that would be a minus sign. You know that just by looking at this, and you say if we switch this, the sign of the commutator has to switch. There's a lot of stuff loaded in there. And once you've sort of processed it, it becomes automatic. You forget the beauty of it. So are you satisfied? Everybody else? All right, so now let's do another one. Let's look at L squared L plus minus. Well, this one is super easy because we already know that L squared commutes with Lx, Ly, and Lz. So I just need to just write 0 here because this is Lx plus or minus iLy, and we know L squared commutes with both of them. Now comes the abstract and weird stuff. We're starting to use the commutators to derive the matrix elements and selection rules. So let us say that we have some function which is an eigenfunction of L squared and Lz. And so we're entitled to say that L squared operating on this function gives an eigenvalue we call lambda. And we can also say that Lz operating on the function gives a different concept mu. Now this lambda and mu have no significance. They're just numbers. There's not something that's going to pop up here that says, oh yeah, this means something. So now we're going to use the fact that this function, which we're allowed to have as a simultaneous eigenfunction of L squared and Lz with its own set of eigenvalues, this function, we are going to operate on it and derive some useful results that all are based on the commutation rules. So let us take L squared operating on L plus minus times f. And we know that L plus minus commutes with L squared. So we can write L plus minus times L squared f. But L squared operating on f gives lambda. We have L plus minus lambda f. Oh, isn't that interesting? We have-- I'll just write it-- lambda times L plus minus f-- L plus minus f. So it's saying that this thing is an eigenfunction of L squared with eigenvalue lambda. Well, we knew that. So L plus minus operating on f does not change lambda, the eigenvalue of L squared. Now let's use another one. Let's use Lz L plus minus. Well, I derived it. It's plus or minus h bar times L plus minus. And if I didn't derive it, I should have, but I'm pretty sure I did. And so now what we can do is write Lz L plus minus minus L plus minus Lz is equal to h plus or minus h bar L plus minus. Let's stick in a function on the right, f, f, f. So now we have these operators operating on the same function. Well, we don't yet know what L plus minus does to f, but we know what Lz does to it. And so what we can write immediately is that Lz operating on L plus minus f is equal to plus or minus h bar L plus minus f plus mu L plus minus f. Well, that's interesting. So we see that we can rearrange this, and we could write plus minus h bar L plus minus f is equal to mu L plus minus f plus Lz L plus minus f-- that's h bar, OK. Oh, I'm sorry, L plus minus f there. So what's this telling us? So we can simply combine these terms. We have the L plus minus f here. And so we can write mu plus h bar times L plus minus f. That's the point. So we have this operator operating-- AUDIENCE: I don't think you want the whole second-- ROBERT FIELD: I'm sorry? AUDIENCE: The first line goes straight to there. I think your second line's [INAUDIBLE].. ROBERT FIELD: I took this thing over to here. So let's just rewrite that again. We have Lz L plus minus f is equal to this. And so here we have mu, the eigenvalue, and it's been increased by h bar. And so what that tells us is that we have a manifold of levels-- mu, et cetera. So we get a manifold of levels that are equally spaced, spaced by h bar. AUDIENCE: I think it also should be plus or minus h bar, right? ROBERT FIELD: Plus minus h bar-- yeah. So we have this manifold of levels, and so what we can say is, well, this isn't going to go forever. This is a ladder of equally spaced levels, and it will have a highest and a lowest member. And so we can say, all right, well, suppose we have f max mu, and we have L plus operating on it. That's going to give 0. And at the same time we can say we have L minus min mu and that's going to give 0. We're going to use both of these. Now I'm just going to leave that there. Oh, I'm not. I'm going to say, all right, so since we have this arrangement-- all right, I am skipping something, and I don't want to skip it. So if we have L plus operating on the maximum value of mu, we get 0. And the next one down is down by an integer number of L, and so we can say that Lz operating on f max mu is equal to h bar L, some integer. Now this L is chosen with some prejudice. Yes? AUDIENCE: Why is there an f of x? ROBERT FIELD: Now I have to cheat. I'm going to apply an argument which is not based on just abstract vectors. We have an angular momentum. It has a certain length. We know the projection of that angular momentum on some axis cannot be longer than its length. I mean, I'm uncomfortable making that argument because I should be able to say it in a more abstract way, but this is, in fact-- we know there cannot be an infinite number of projection quantum numbers, values of the projection quantum number that aren't reached by applying L plus and L minus. It must be limited. And so we're going to call the maximum value of mu h bar L or L. Now I have to derive a new commutation rule based on the original one. No, let's not erase this. We might want to see it again. So let's ask, well, what does this combination of operators do? Well, this is surely equal to Lx squared, and we get a plus i and a minus i, and so it's going to be plus Ly squared. And then we get i times LyLx, and we get a minus i times LxLy. This is L squared minus Lx squared. We have the square root of 2 components of L squared, and so this is equal to the difference. And now we express this as i times LyLx. And what is this? This is plus ih bar Lx. AUDIENCE: I think you wrote an x [INAUDIBLE].. ROBERT FIELD: OK, this is, yes, x, and that's Lz. I didn't like what I wrote because I want to have everything but the z and the L squared disappearing, and so we get that we have L squared minus Lz squared. And then we have plus ih bar Lz. I lost the plus and minus. No I didn't. OK, that's it. And so we can rearrange this and say L squared is equal to Lz squared minus or plus h bar Lz plus L plus minus L minus plus. So we can use this equation-- OK, I'd better not-- to derive some good stuff. I better erase some stuff or access a board. We're actually pretty close to the end, so I might actually finish this. So we're going to use this equation to find-- so we want lambda, the value of lambda for the top rung of the manifold over here. So we apply L squared to f max mu. And we know we have an equation here which enables us to evaluate what the consequences of that will be, and it will be Lz squared f max mu minus and plus h bar Lz f max mu plus L plus L minus f max mu. So if we take the bottom sign, that L plus on f max is going to give 0. So we're looking at the bottom sign, and we have a 0 here, and so we have L squared f max mu is equal to Lz squared f max mu minus or plus h bar Lz f max mu plus 0. Isn't that interesting? So we know that Lz is going to give-- so we're going to get an h bar squared and mu max squared. We're going to get a minus h bar h bar mu max. So what this is telling us is that L squared operating on f max mu is given by l because we said that we're going to take the maximum value of mu to be h bar L. So I shouldn't have had an extra h bar here. So we get this result. So lambda-- so L squared operating on this gives the-- oh yeah, maximum mu. So it's telling us that L squared f max mu is equal to h bar squared l l plus 1. Now that is why we chose that constant to be l. And we do a similar argument for the lowest rung of the ladder. And for the lowest rung of the ladder, we know there must be a lowest rung, and so we will simply say, OK, for the lowest rung of the ladder we're going to get that mu is equal to h bar lambda bar mu min. And we do some stuff, and we discover that lambda has to be equal to h bar squared l l plus 1. And using this other relationship and the top sign, we get h bar squared l bar l bar minus 1. And there's two ways to solve this. One is that l is equal to minus l bar, and the other is that l bar is equal to l plus 1. Well, this is the lowest rung of the ladder. Wait a minute, let me just make sure I'm doing the logic correctly. It's this one, OK, here. So this is the lowest rung of the ladder, and l bar is supposedly larger than l. Can't be, so this is impossible. This is correct. And what we end up getting is this relationship, and so mu can be equal to-- and this is l l minus 1 minus l stepped to 1. This seems very weird and not very interesting until you say, well, how do I satisfy this? Well, if l is an integer, it's obvious. If l is a half integer, it shouldn't be quite so obvious, but it's true. So we can have integer l and half-integer l. That's weird. We can show no connection between the integer l's and the half-integer l's. They belong to completely different problems, but this abstract argument says, yeah, we can have integer l's and half-integer l's. And if we have electron-- well, we call it electron spin because we want it to be an angular momentum. Spin is sort of an angular momentum or nuclear spin. And we discover that there are patterns of energy levels which enable us to count the number of projection components. And if you have an integer l, you'll get 2l plus 1 components, which is an odd number. And if it's a half integer you get 2l plus 1 components, which is an even number. And so it turns out that our definition of an angular momentum is more general than we thought. It allows there to be both integer and half-integer angular momentum. And this means we can have angular momenta where we can't define it in terms of r cross b. It's defined by the commutation rule. It's more general. It's more abstract. It's beautiful. And I don't have time to finish the job, but in the notes you can see that we can derive the matrix elements for the raising and lowering operators too. And the angular momentum matrix elements are that L plus minus operating on a function gives this combination square root, and it raises or lowers m. So it's sort of like what we have for the a's and a daggers for the harmonic oscillator, but it's not as good because you can't generate all the L's. You can generate the m sub L's. And that's great, but there's still something that remains to be done to generate the different L's. That's not a problem. It's just there's not a simple way to do it, at least not simple to me. And so now anytime we're faced with a problem involving angular momenta, we have a prayer of writing down the matrix elements without ever looking at the wave function, without ever looking at a differential operator. And we can also say, well, let's suppose we had some operator that involves L and S, now that we know that we have these things, and L plus S can be called J. So now we have two different operators, two different angular momenta. We have S and we have the total of J. Well, they're all angular momenta. They're going to satisfy their selection rules and the matrix elements, and we can calculate all of these matrix elements, including things like L dot S and whatever. So it just opens up a huge area where before you would say, well, I got to look at the wave function. I've got to look at this integral. No more. But there is one thing, and that is these arguments do not determine-- I mean, when you take the square root of something you can have a positive value and a negative value. That corresponds to a phase ambiguity, and these arguments don't resolve that. At some point you have to decide on the phase and be consistent. And since you're never looking at wave functions, that actually is a frequent source of error. But that's the only defect in this whole thing. So that's it for the exam. I will talk about something that will make you a little bit more comfortable about some of the exam questions on Wednesday, but it's not going to be tested. |
MIT_561_Physical_Chemistry_Fall_2017 | 29_Modern_Electronic_Structure_Theory_Electronic_Correlation.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseware continue to offer high-quality, educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseware at ocw.mit.edu. TROY VAN VOORHIS: All right. Well, good morning, everyone. Hope you had some at least rest over the long weekend. Looking forward to getting back to talking about quantum. So last time, which was a whole week ago-- long time ago-- we started talking about electronic structure. We talked about the Born Oppenheimer approximation and how that gives us an electronic Schrodinger equation, how we look for-- what we're looking for are the eigenvalues and eigenvectors of that Schrodinger equation, how we can get lots of interesting information out of that. And then we broke it down so that we saw that there were going to be two main knobs that we were going to have to turn when we were making our approximations here. We were going out to look at our atomic orbital basis sets and how we compute the energy. And we spent the last half of class last time talking about choosing basis sets. We came up with some understanding of these strange names that are written at the bottom here. And then today, we're going to talk about some of these ways of computing the energy. And then I'm going to flip over, and we're going to talk about computational chemistry tools and how we sort of assemble that into doing a calculation. But again, the idea with basis sets was that we were sort of questing after this mythical complete basis set limit where the basis set was so big that it was infinitely big, that making it bigger didn't change your answer at all. And we never actually get there. But if we get close enough, we decide that's good. All right. So now we're going to talk today about computing the energy. And thankfully, most of the methods for computing the energy start from something that I believe you are familiar with already from 5.61. You guys talked about the Hartree-Fock approximation? Yeah? Yeah. OK. So-- or maybe we talked about-- did we talk about singled did we talk about determinants? Yes. All right. So there we go. So then we talked about a determinant. The idea of a determinant is that you have a set of orbitals, they're independent, and then you compute the average energy. And so, in one form or another, I would write the energy of a determinant something like this. I would say that there's a piece here which is the energy of each electron by itself. So these are one-electron energies. So for every electron in my system, if I have n electrons, I add up the energies of each of those electrons. But then, because my Hamiltonian has electron-electron repulsion terms, I end up with some additional contributions that can't be assigned to just one electron or another electron. And those terms are here in this sum. They're pairwise so for every mu and nu summing over independent pairs, I have a coulomb, term j, and an exchange, term k. And what these things show me is that there is an average repulsion here between the electrons. The electrons are independent. They can't avoid each other. But they still interact. There's still the electron-electron repulsion between electrons here. So that's an energy expression that, in one form or another, should be somewhat familiar, I think, from things you guys have done before. The idea of Hartree-Fock is relatively simple, which is we have an energy, then. This is true for any Slater determinant, any set of orbitals. So I have some orbitals here. And the idea of Hartree-Fock is to choose the orbitals to minimize the energy. This independent particle model energy, I want to minimize that. I'm going to choose the orbitals that minimize that. And the way I'm going to do that is I'm going to write my orbitals-- psi mu-- as a sum over some index, i, some coefficients c i mu, times my basis functions, phi i. And I'll just emphasize that what I have here are functions from my AO basis, the things that I talked about last time, like the actual 1s-like, 2s-like, 2p-like, 3p-like orbitals. And then on the other side, what I have here are my molecular-like orbitals. The molecular orbitals are linear combinations of my atomic orbital basis functions. So that's the general idea of Hartree-Fock there, is that we're just going to choose these orbitals to minimize the energy. And because the orbitals are defined by these coefficients, I can think of this energy as a function of the coefficients, c. So I can say, all right, that means that I have an independent particle model energy which depends on these coefficients, c. And if I want to minimize that energy, I then need to look for the places where the derivatives are equal to zero. So I take the derivative of that energy, set it equal to 0. Now as you might suspect, if I look at this expression, I have an orbital here, here, here, here, here, here, here, here, here, here. I have a lot of orbitals appearing in there. And then I'm going to have to use the chain rule, because every one of those orbitals depends on every one of the coefficients. So I'm going to have a chain rule expression. You might expect that there's going to be a lot of algebra involved in taking this derivative, and there is. This is where I get to use my favorite abbreviation, which is ASA. ASA stands for After Some Algebra. It means I'm not going to work through the algebra. There's some algebra. There's several pages of lecture notes where algebra is gone through. After you do the algebra, you end up with the left-hand equation reducing to an equation that looks like this. So there is a matrix, which I will call h, and then when I dot that matrix into a vector, c mu, the coefficient vector, I get some energy, e, mu, times the coefficient vector again. So that takes a lot of algebra to get to that, but you can get an equation that looks like this. And that's encouraging, because this is familiar to me. This is an eigenvalue equation. And that's nice because I remember eigenvalue equations from-- I mean, the Schrodinger equation is an eigenvalue equation. So it's nice that when I do this minimization, I get back an eigenvalue equation. That's nice. The one complication is that the Hamiltonian here depends on the coefficient, c. So it's some effective Hamiltonian. I'm intentionally collecting terms so that it looks like a Schrodinger equation, but it's a little bit funny because the Hamiltonian here depends on the coefficients that solve the equation. So this is sort of an eigenvalue equation. And so what we're seeing here is that H depends on the wave function somehow. It seems like, depending on what the wave function is, I have different Hamiltonians. And to understand why this is the case, I want us to think about the potential, v, that an electron feels in an atom or in a molecule. And there's generally going to be two pieces here. And they're actually-- I can color-code them with the terms that exist in the independent particle expression. So there is the one-electron terms, the terms in the energy that just come from one electron. And those are going to be things like, well, I might have a nucleus of charge plus z. And I might have my electron out here, the one that I'm interested in. And it's going to have some attraction to that nucleus. So that just involves one electron. It involves a nucleus, but just one electron. So there's some nuclear potential that each electron is going to feel. But then there's the term that correspond to these average repulsion terms. So there are other electrons in my system. And each of those electrons interacts with the electron that I'm looking at. So there's going to be some electron-electron repulsion term here. And that electron-electron repulsion depends on where the other electrons are. And those other electrons have some wave function. They're kind of spread out here, all over the place. And depending on how spread out those other electrons are, I might have more or less electron repulsion. So if they're very, very concentrated in the region that I'm looking at, the electron repulsion is going to be very large. If they're very spread out, and very diffuse, they're not going to have very much electron repulsion. But overall, the electron-electron repulsion depends on the wave functions that I've chosen. So the potential here is called a mean field, or average field. And it's because it's the potential that comes from the average repulsion. It's the effective potential due to the averaging, the smearing out, over all the other electrons. So what we do have, then, is that the Hamiltonian does into indeed depend on the wave functions, because the wave functions determine that average potential. But then if I know the Hamiltonian, I can solve for the wave functions, because the wave functions are the eigenfunctions of the Hamiltonian. And so I have this sort of chicken and egg situation, that the Hamiltonian defines the wave functions, and the wave functions define the Hamiltonian. And the solution to this is to do what are called self-consistent field iterations. And you can more or less think about these as the simple process of, guess some wave functions, build the Hamiltonian, get some new wave functions, build a new Hamiltonian, get some new wave functions, build a new Hamiltonian, and keep going until everything becomes consistent. And the idea, then, is that the wave function we're looking for in Hartree-Fock are the eigenfunctions of H of psi Hf, which we just defined to be equal to H Hartree-Fock. So there's some Hartree-Fock Hamiltonian. And the Hartree-Fock wave functions are the eigenfunctions of that Hamiltonian. Does that makes sense? Any questions about that? All right. So Hartree-Fock is the starting point. So we have that as an approximation. It's a useful starting point. In many cases, Hartree-Fock is not good enough. And so the real thing that we're trying to figure out is, how do we make something that uses what we've learned from Hartree-Fock about orbitals, and energies, and average repulsion-- how do we make something that's like Hartree-Fock, but better? So one way to do this is something that you've spent a lot of time on this semester, which is perturbation theory. You have something that's pretty good as a starting point, and you want to make it better, perturbation theory is the natural thing to turn to do that. And so the way we apply perturbation theory in Hartree-Fock is we say, well, we've got this Hartree-Fock Hamiltonian, which we can solve, and then we've got the actual Hamiltonian, H, which we can't solve. And so what we can do is then we say, aha, well that means that if I just take the difference between the Hamiltonian I can solve and the Hamiltonian I can't solve, and I treat that as a perturbation, then I can do a well-defined perturbation theory where I know the solutions to H0, and then I treat that additional term as a perturbation. And so then if I do that, I end up with a perturbation expansion where I can go to first, second, third, fourth, fifth order, and that additional term that's the difference between the two. And if I look at the first two terms added together, they give me the Hartree-Fock energy. So if I'm the one making up a perturbation expansion, I will always choose it so that the first order correction is 0. So that if otherwise I have chosen, I will just re-engineer my perturbation so that's true. So up through first order, I just get what I started with. I get Hartree-Fock. But then I have these additional terms. And so I can go to second order. And I will get a method that I call-- it's called MP2. So Hartree-Fock is named after Hartree and Fock. MP stands for Moller and Plesset, the names of the two people who wrote the paper in 1938, about this particular thing. And the 2 is second order. And then I could go on, and I could do MP3, MP4, dot dot dot dot dot. So in general, this is MP, and some number, n, telling me the order I go to in perturbation theory. So you could do this. In principle, this gets better and better answers. However, in practice, about 20 years ago, people discovered something that was rather disturbing. So if I take a look at the energy minus the [INAUDIBLE].. So let's say I have a case where I know the exact energy. And I look at the energy of these different approximate methods for, say, one particular simple thing, like a neon atom. So I take a neon atom, and I look at the energies of these MPn approximations, and I look at them as a function of n, so 1, 2, 3, 4, 5, 6, 7, 8. And I'll say that for neon-- neon is something for which Hartree-Fock is already pretty good. So the difference between Hartree-Fock and the exact answer is not too bad. So Hartree-Fock might be, let's say, there on this particular scale. And then if you do MP2, MP2 is better. So my MP2 is maybe there. MP3 is also a little bit better than that, but it overshoots a little bit. MP4 corrects back in the opposite direction. And then we've got 5, 6, 7, 8. And you can-- if I connect the dots, you can see what's starting to happen here. So that's a series. So mathematically, I've got something that there is a value for everything in n. That's a series. And the name for a series like this is one that is not converging. So there's an answer which is the infinite order answer, but these partial sums are not converging to the infinite answer. And very often this turns out to be the case with perturbation theory. There can be perturbation theories that are very useful at low orders for describing qualitatively how things shift. But if you go out to infinite order, they are series that, in principle, you can prove, give the correct, exact answer, because we've derived it as such. But just because a series in principle sums to something doesn't mean that any finite number of terms will give a convergent answer, or give smaller errors. And people find that for perturbation theory for electronic structure theory, this is-- not all of the time, but very often the case, that the perturbation expansion doesn't converge. And so this is one of those places where you make a nice empirical compromise, and you say, well, gee, every term-- it doesn't necessarily get better after MP2. MP2 usually improves on Hartree-Fock, but higher order might be better or worse. And certainly, higher order is more expensive. So MP2 is a good place to get off the elevator, because it's just going to get more expensive, and not necessarily more accurate. So this is a good place to stop. And so you'll see a lot of calculations out there that are variations on MP2, because it's the least expensive, and it's somewhat more accurate. OK so now we're going to take a pause, and we're going to talk-- and this is where I need audience participation. So we're going to discuss two different things that you might want to know. We've now got two methods, Hartree-Fock and MP2. And there are two things that you might want to know. One is, how accurate will one of these calculations be? If I do it, will it give me the right answer? And the other one is, how long will it take? In other words, will it finish by Friday when the P set is due? Will it finish by the end of the semester? Will it finish by the end of my time at MIT? What time scale are we looking at? So we'll start off talking about properties. So people have done many, many, many calculations using many different electronic structure methods, so that now there's sort of an empirical rule of thumb for just about any property you want to know about. There's a rule of thumb for how accurate one of these methods should be. So what's a property that you might want to know? This is where the audience participation comes in. I will only give you information about properties that you say are interesting. AUDIENCE: Time and space complexity. TROY VAN VOORHIS: Time and space complexity. All right. Could you be more specific. What type of time and space complexity? AUDIENCE: How well-- I mean, for a size of molecule, how long it takes-- TROY VAN VOORHIS: Ah, OK. AUDIENCE: --and also how much space it takes. TROY VAN VOORHIS: Yes, so that will be the second thing that we will talk about. Those are the two variables we'll talk about, in terms of how long and how fast. I'm interested here, where I'm saying, like, molecular properties-- things you might want to know about a molecule, or a reaction, or something like that. AUDIENCE: Like the energy of a molecule in transition state. TROY VAN VOORHIS: OK, so that would be sort of activation energy, right/ OK, what's something else? We only want to know about activation? Yeah. AUDIENCE: Charge distribution. TROY VAN VOORHIS: Right. OK, so I'll give one variable that probes that, which is a dipole moment, for example. You might want to know a dipole moment. AUDIENCE: What the MOs look like [INAUDIBLE] TROY VAN VOORHIS: OK, yes. So that is something we want to know about, but it's a qualitative one. So what I'm going to give is I'm going to give you things where I can say, oh, for a dipole moment, it's within this many debye, or this much percent. So MOs are qualitative, and we can get that, but I can't say, well, this MO is 75% right. AUDIENCE: Average distance of an electron from an atom. TROY VAN VOORHIS: OK, so that would be like size. AUDIENCE: Yeah. TROY VAN VOORHIS: Yeah, yeah. AUDIENCE: [INAUDIBLE]. TROY VAN VOORHIS: What was that? AUDIENCE: Color. TROY VAN VOORHIS: Yeah. I mean, so yeah, that's about an absorption-- like an electronic absorption spectrum. Other things people might want to know? AUDIENCE: Bond length. TROY VAN VOORHIS: Bond length. There you go. Anything else? We've got a good list. OK, we'll use that as our list. OK, so I will start off with bond lengths, because that's the one where Hartree-Fock does the best of these things. So it makes you feel like it's encouraging. So bond lengths-- Hartree-Fock usually predicts bond lengths to be a little bit too short. So the molecules are a little bit too compact. But it's not too bad. They're usually too short by about 1%, which is not too bad. It gets 99% percent of the bond length right. MP2 does better, so that it doesn't actually have a systematic error. It typically gets bond lengths correct, plus or minus one picometer. So a picometer is 10 to the minus 12th meters. So 0.01 angstroms, since a bond length is usually about an angstrom. So we'll go from that to size. So for the size of an atom or a molecule, this has to do with the Van der Waal's radius, basically. And actually, Hartree-Fock does a fairly good job of describing Van der Waal's radii. I would say that it more or less goes with the bond length prescription, so that the size is a little bit too small. So it might be minus 2% to 3% to small. And then there's very few people that actually do sizes with MP2, for various reasons. But I would say that it's-- I'll put "accurate" there. Because basically the size of an atom or a molecule is fuzzy. So I can talk about the radius, and it's a little bit fuzzy. The errors in MP2 would be smaller than whatever fuzziness I could have in my definition of size. For activation energies, Hartree-Fock is pretty poor, so the activation barriers here are too high, by 30% to 50%. So the barrier heights are very high in Hartree-Fock. MP2 has barrier heights that are still too high, but only by about 10% on average. Dipole moments in Hartree-Fock-- there are no systematic studies on that. I will just say that they are bad. They are bad enough that nobody does a systematic study. But for MP2, they're quite a bit better, and typically accurate say, 0.1 debye. And then for excitation absorption energy-- so this is going to be the difference between the lowest electronic state and the next highest electronic state-- so you can get these things through various means. The typical absorption-- and also, I'll say that doing that involves an extension beyond what I've talked about so far, because I've been focused on the ground state. But if you use the logical extensions of these things, the absorption energies in Hartree-Fock tend to be too big, and too big by, say, half an eV or so. And then for MP2, they're more accurate, but not that much. So plus or minus, say, 0.3 eV. Electronic inside states tend to be fairly difficult to get. But we see that MP2 is doing better on all accounts, than Hartree-Fock. So it's improving. That's good. But then, all right, what is this going to cost me? How long is this going to take? Well, there's two different ways that we can measure computational complexity. One is by the amount of storage that's required. It's the amount of information you have to store in order to do the calculation. The other way is by how long it's going to take, how many operations the computer has to do to solve the problem. And so for Hartree-Fock, the main memory piece that we need is actually the Hamiltonian. So we need that Hamiltonian matrix. We've got to store that in order to compute its eigenvalues. And it is an end-- it's a number of basis functions by number of basis functions. And so then now we have to say, all right-- well, first of all, we have to figure out how much RAM. So when I'm figuring out how much RAM is required-- does anybody know how much RAM they have on their laptop or desktop computer? Nobody knows. 16 gigs. All right, that's what's on my mine, too. 16 gigabytes, which is 16 times 10 to the 9th bytes, roughly. So now we have to say, all right, for storing n basis times n basis numbers, how much does that take? Well, let's say that we have a atoms. So we've got-- the number of atoms we have is a. Then the number of basis functions we have-- last time, we figured out that a DZP basis, which is a decent-size basis for carbon had around 15. It was 14 basis functions per carbon. I'm just going round to 15 because it's easier math. So that means that the number of basis functions is 15, roughly, times the number of atoms. So there's 15a times 15a things that I need to store to get this matrix. And that means I need about 225 a squared numbers. So these are how many numbers I have to store. And then I have to know that each number-- I'm usually storing these in double precision. So each number requires eight bytes in double precision. So that means that I have something like 1,700 a squared bytes to store that object. So now that says, all right, if I've only got 16 times 10 to the 9 bytes of storage space on my computer, and I need 1,700 a squared bytes for a molecule with a atoms, that just gives me a natural limit on the number of atoms. If you back it out, that implies that a is less than or about equal to 3,000, which is a big number. So your molecule could have up to about 3,000 atoms in it, and this calculation would run. And that rough calculation turns out to be about right. Now let's take a look at the CPU time required. So the CPU time, we're going to measure this in hours, because that's my human unit of time. So one hour is 3,600 seconds. And then what I need to know is well, what actually takes the computer time? What takes the computer time is doing mathematical operations-- so add, subtract, multiply, divide, if, things like that. And usually, we measure the speed of a CPU in terms of the number of operations you can do per second. And does anybody know the order of magnitude of operations per second the CPU on your computer can do? It's about a billion-- some number of billion operations per second, depending on how recent a model you have, and whether you play video games or not, you will do more or less than that. So that means that our computer, in one hour, can do about one-- and I'm going to round up, and say you've got a really good one. So we're going to say that your thing can do 10 billion operations per second. If you've got multiple cores, they can each go independently. So 1 times 10 to the 10 operations per second. And that means that you can do about 3 times 10 to 13 operations in an hour. I'm just going to set the hour as my patience limit. I don't want to wait longer than an hour. And so, then, for Hartree-Fock, the rate-determining step is computing the eigenvalues of the Hamiltonian. And that requires n cubed operations, where n is the dimension of the matrix. So if our matrix n basis on a side, it requires n cubed-- n basis functions cubed operations. And so then, again, using our translation of that into atoms, that's 15 times the number of atoms cubed. Which, if I did my math right back in my office, that's about 3,000 a cubed. And then if I back that out-- again, if I compare that to the number of operations, that gives me a limit on the number of atoms that I can handle in an hour, and that turns out to be about 1,000. So similar sizes. So I run out of storage about the same time as I run out of patience. If I was willing to be a little bit more patient, I might be able to do a few more. But order of magnitude, we're at 1,000. So I'll summarize. What we found here is that the storage requirements were something like a squared of order a squared. CPU time was of order a cubed. The maximum feasible number of basis functions that I could do here was something like-- so if I have 1,000 atoms, that was my limit. Something like 15,000 would be the maximum feasible number of basis functions. And maximum feasible number of atoms was about 1,000. So we won't go through the same exercise for MP2 in such detail. I will just tell you that everything is worse for MP2. Everything takes longer, it takes more storage, everything's worse. So it requires order a cubed storage. It requires order a to the fifth CPU time. The maximum feasible number of basis functions is therefore smaller, about 2,000. And in MP2, you actually require more basis functions per atom to get an accurate answer, so that the number of feasible atoms is more like 50 rather than 1,000. So that's the cumulative effect of larger basis sets and worse scaling with number of atoms that makes MP2 deal with much smaller systems. So questions about that. Yes. AUDIENCE: So what are some examples of [INAUDIBLE]?? TROY VAN VOORHIS: So C60 is 60 atoms. That's an easy one. There are a number of small dyes, for example, that we worked with in our group, where you might have seen absorption spectra, or emission spectra, or HOMOs, or LUMOs, or hole transport properties, or electron transport properties, that are around 50 atoms in size. And then the main thing that people get interested in that are on the 1,000-atom regimes are enzymes and peptides. Those are the sort of-- when you start saying, I want to do 1,000 atoms, it's usually because it's really some polymer or heteropolymer. Another case where you also are sometimes interested in things where you do simulations with 1,000 atoms, if you're interested in a surface. Because if you have a chunk of gold, for example, 1,000 atoms is just 10 gold atoms, by 10 gold atoms, by 10 gold atoms, which is just a little chunk-- a very, very small chunk of gold, but still a chunk. Other questions. OK, so I'm going to spend seven minutes talking about density functional theory. And then we're going to go over and do some examples. So the idea of density functional theory is that it'd be really nice if you had some magical way to do a Hartree-Fock calculation, but have a give you exactly the right answer. That basically would be the dream because we saw, well, for Hartree-Fock, we can do big systems, it's cheap, it has good scaling, all these kinds of things. It just doesn't give us very good answers. The results are pretty poor. So what if we had something that scaled like it had the computational cost of Hartree-Fock, but was, say, the accuracy of MP2, or even better than that? And so in density functional theory, what we use is the idea of looking at the electron density rho of r as the fundamental variable. So you can actually work out, for a determinant, what the electron density is. And it's just the sum of the squares of the orbital densities-- or the sum of the orbital densities. So you square each orbital, then you add that up, and that gives you the total electron density. It's basically just the probability of finding an electron at a point, r. OK, so that's a nice observable. The thing that's kind of amazing is that there is a theorem, a mathematical theorem that you can prove in about two pages using just sort of proof by contradiction. You can prove that there exists a functional, which is always given then in e sub v of rho, such that when you're given the ground state density, rho 0, if you plug that into this magical functional, it will give you the ground state energy. So if you found the ground state density lying around in the gutter, and you picked it up, and you put it into this functional, it would give you the ground state energy. And e0 is not just the approximate ground state energy. It is the exact ground state energy, the exact thing. Further, for any density, rho prime, that is not the ground state density, if you plug that density into eV, you get a higher energy. So what you could do is you could say, well, let me just search. Let me try density, see what energy it gives. Then I'll try another density. If it gives a lower energy, that's a better density. And I'll keep going, and going, and going, and going, until I find that ground state density. So the idea is that then you would say, aha, if I solve that equation-- so I search over densities, and that's not going to be a very hard search, because the density is just a three-dimensional object. It's not like the wave function that depends on a bunch of coordinates. It's just three dimensional. So I solve that equation, that will give me rho 0. And then I can take that rho 0, plug it back in there, and I will get the ground state energy. That gives me a closed set of conditions that lets me find the ground state energy, and then report back the energy of that density. AUDIENCE: Where do we get this [? EV ?] from? TROY VAN VOORHIS: That's a great question. The proof is a proof by contradiction. It's not constructive. It proves that it exists, but gives you no way of constructing it. So kind of the frustrating thing is like, oh, cookies exist, but we don't know how to make them. But the idea that such a thing exists gives you hope to say, well, maybe we can construct-- if we can't find the exact one, maybe we can find one that's very, very good. And that one that's very good, we would just use. We would pretend it was exact, minimize the energy, find the density, and then report back the ground state for that approximating. And that's actually what you do in density functional theory. You have approximate functionals. And they're all-- just like with basis sets-- named after the authors of the people who wrote the papers. Almost all of them are. So there's the Local Spin Density Approximation, LSDA. There's the Perdew-Burke-Ernzerhof functional. There's the Becke-Lee-Yang-Parr functional, the Perdew-Burke-Ernzerhof Zero functional, the Becke 3 Lee-Yang-Parr functional, and then on and on and on. Now these things have been sort of-- there's a mishmash of different exact conditions that were used to derive these functionals. And then, empirically, people have gone and shown that PBE is better than LSDA. It's about as good as BLYP. BLYP is not as good as PBE0, but it was just about as good as B3LYP. So we have, then, sort of an idea that, OK, if we go over towards the right-hand side of this, we're going to get better results out of DFT. But DFT, because it's based off of a Slater determinant, has exactly the same computational scaling, and storage, and everything, as Hartree-Fock. It's the same kind of expense. But if I go back to this accuracy thing, and I say, well, what would I get from that best density functional B3LYP, well what I find is that activation energies are still a little bit difficult. Now I underpredict activation energies. But my dipole moments are good. My sizes are again accurate. My absorption energies are just as accurate as MP2, and my bond lengths are just as accurate as MP2. There are other things, but basically B3LYP is as accurate or more accurate than MP2 for virtually every property you would want, but has Hartree-Fock-like cost. And so that makes density functional theory the workhorse of computational chemistry these days. Because it's inexpensive, you can do lots of calculations with it, but it's accurate enough for grunt work. And then there's a blank column there that, if I had time, I would talk about. There's a whole other category of things where you say, well, perturbation theory doesn't work, but are there more constructive ways that I could use wave functions to approximate the correlation energy? And the answer is yes, you can get good energies and good answers out. But then it comes at the cost of the calculation getting much more expensive. So there are methods that improve all these properties on the wave function side, that just require more. So if you want to know more about those, they're in the notes. And so now I think I will switch over and show you how we do some calculations with this. So now I will switch. There's two different tools that you have available to you for free on Athena that you can use to do electronic structure calculations. One is Gaussian, and the other one is Q-Chem. There may be even some other ones, but those are the two that I've collected notes about how to use. So I'm going to show you how we use Q-Chem. I'm not going to do it on Athena. I'm going to do it on my laptop, but it's basically the same. So there's this GUI called IQmol. And when you open it up, it gives you a window that looks like this. And there's a few things here. So there's, like, an open file, you can create something, you can open a file, you can save something, you can move something around after you've built it. And then these are the build tools over here. So that just turns it into build mode. If I click on this, I can choose any item in the periodic table. Right now, it's on carbon. This lets me select various fragments, and also what the attach point of the fragment's going to be in some cases. And I don't want to do it, so I'll select it for now, but then I'm going to go back to this one. This is the Add Hydrogens button. So if you have something that-- you just wrote your Lewis structure, it had no hydrogens in it, you click that button, it'll put all the hydrogens where they should be, or where it thinks they should be. This is a Minimize Energy thing, which basically if your structure looks really crazy, it'll kind of make it look less crazy. And then this is the Select button, which lets you pick certain atoms. This lets you delete certain atoms, take a picture, do a movie. And then that changes it over to fullscreen mode. And that's the life preserver for Help. But we'll do building a molecule. So does somebody want to tell me a molecule that has less than 15 atoms in it that they would like me to draw here, that I can actually draw? AUDIENCE: Diethyl ether. TROY VAN VOORHIS: Diethyl ether. Yes, I can do that. OK. All right, so put a carbon, carbon, then I got to switch to an oxygen, oxygen, carbon, carbon. And then I will click the Add Hydrogen button. [LAUGHTER] There we go. I've got my diethyl ether now. And while my geometry here is not perfect, it doesn't look totally crazy. So I'll just go with that. And then the thing we're going to be most interested in is this little tab here, called Calculation. So calculation here has a little button that says Q-Chem Setup. So I'll do Q-Chem Setup. And now it's got a bunch of things that I can specify. There are a few things that are most important. So the first thing is, what kind of calculation I want it to do. So I could have it just compute the energy of the molecule for the configuration I specified. That would be one useful thing. I could compute forces. In other words, the forces on the atoms, where the atoms want to move. I could optimize the geometry, which would then relax the thing down, and figure out what the best geometry is. I could do various things that scan across the potential energy surface. I can compute vibrational frequencies. I can compute NMR chemical shifts. I can do molecular dynamics. I can compute properties-- so lots of things. I'm just going to optimize the geometry. That's going to be what I'm going to choose for now. And then I get to choose a method. So I can do Hartree-Fock. I can do B3LYP. I can go down-- so all the things above this dotted line here are different density functions. And then, down here, we've got MP2, MP2 bracket v-- I'm not actually sure what that is. I've got various versions of MP2 that try to make it faster. And then I've got a couple of cluster methods that we didn't talk about yet, but that are in the notes. And then various methods for getting electronic excited states, down here at the bottom. So for now, let's just do B3LYP, and then I get to choose a basis. Because this is on my laptop, and we're in class, I'm going to choose a small basis. So I'm going to leave it at 321g, which is a small basis. And then I can also specify the charge, which is like the total charge of the molecule, and the spin multiplicity. So I want it to be a neutral molecule, spin 1-- or spin multiplicity 1, which is spin 0. Because it's 2s plus 1 is the spin multiplicity. And then there's various things about convergence control that you probably won't need to worry about. And then there's advanced things that you also probably won't need to worry about. So now I'm happy with that. I can submit my job. And I will call this diethyl ether, hit OK. And now it doesn't appear to be doing anything. In the background, it is doing something. I think this will finish in just a few seconds, but if you get concerned that the job might not be running or something might have happened, you can go over to the Job Monitor. Methane-- that was the job that I ran this morning to make sure that it works. Diethyl ether, we just submitted it at 10:50 and 21 seconds. Ah, there we go. Now it's been running for four seconds. Let's see how long it takes. OK, it's taking a lot longer than I think. All right. AUDIENCE: They close [? in 15 ?] [INAUDIBLE].. TROY VAN VOORHIS: That's all right. We still we've got a few minutes. So what will happen here in a minute, when we-- AUDIENCE: [INAUDIBLE]. TROY VAN VOORHIS: That's all right. It's going along. So when it's done, what will happen is, over here, it'll let us know. And over here will appear a thing where we can start looking at the results. It might have gone faster if I had minimized the energy to begin with, because then it would do less geometry steps. It's going along. I don't know. We might run out of class time before it finishes. It's 10:51. We've got four minutes. Let's see. Come on, laptop. I've got the MacBook Pro. Maybe I should have upgraded the processor, this kind of thing. AUDIENCE: How long do you usually let your group calculations go for? TROY VAN VOORHIS: Well, in my research group, we have 1,000 cores that we-- maybe around 1,200 cores, even, I think, actually. Oh, there we go. It finished. But we have around 1,200 cores for around 10 people. So that means that you can let a job run for a very, very, very long time, on multiple cores, and it doesn't get in anyone's way. So it's not unusual for us to put something in there that lasts a week. So now it's finished. We want to copy the results from the server. This is an arcane way of saying things. Because it's designed for supercomputers, where you are sitting at a terminal in your office, and then it sends all the results over to the supercomputer, it runs the calculation, and then you have to copy the results back to your little computer. I was actually running it on my computer. It's just the computer is both-- I'll just put it-- New Folder. OK. There we go. Now it has a little star next to it, which lets me know that it's finished. So let's see, Info-- so I can show the dipole moment. So there we go, that little blue arrow-- which, magnitude and direction is the dipole moment of diethyl ether. It does have a dipole moment, with its positive end here, pointing in that direction. That seems about right. Let's see. I can look at the files. These are only if you're really, really interested in details. You see these are very verbose. They have every bit of information you could possibly want in them. So if there's ever something that you can't get out of the GUI, if you just kind of go-- you can search. And I can look for converge. Oh, look, optimization converged, and then it tells me everything after that. I can look at the different atoms in the molecule, and it will highlight them for me. And then I can look at bonds, and it will tell me-- I don't if you guys can read, but down here, it has the bond length in the corner. So that C-H bond is 1.093. All these C-H bonds are going to be 1.09, plus or minus a bit. The carbon-oxygen bond there is 1.46, rounding. That one is also 1.46. And then this here-- so as it did the geometry optimization, it was adjusting the geometry trying to find the minimum. And these are the different total energies that it had as it went through the optimization. You'll see that eventually it sort of slows down. The energy is going down every time, and eventually it sort of slows down to where it's not changing any more. It says, all right, that's the minimum. You also notice that these are in atomic units. So that's minus 232 Hartrees. So that's like 4,000 kcals. No, that's more than 4,000 kcals. Anyway, it's tens of thousands of kcals. The reason it's such a low number is this is the energy it would take to rip every electron off of the molecule, including all of the 1s electrons. So the energies are typically going to be very, very large negative numbers. Don't worry about that. It's energy differences that matter. So you notice, even my bad geometry was already 232.1. The correct answer is 232.3. So the total energy change wasn't that much. And then the last cool thing that I want to show you is that we can plot orbitals. So I'm going to count some orbitals. There we go. So I just had to calculate the HOMO and the LUMO. I didn't go through all of it with you. But if you go there, it gives you some things-- I should actually go back and look, show you. So you get to choose a range of orbitals. The iso value is basically where it draws the surface. So it's inverse of what you expect. A larger iso value is-- it's an isosurface, so if you make it larger, the place where the iso value is larger is closer in. So it makes a smaller bubble. And a smaller iso value makes a bigger bubble. And you can change the number of points that it samples to make the surface, change the colors, change the opacity of it in terms of how you want it to look. But I already did this. So there is the HOMO. So the HOMO is something in the pi manifold. It's pi star here. And then I can take out the HOMO and show the LUMO. And the LUMO is, in this particular basis set, not very useful, because I don't have a lot of basis functions. So my basis functions, just there's not enough of them to really describe the LUMO, or the LUMO plus 1. If I want to get a LUMO, we'd need a bigger basis set than this. So that's the kind of thing that you can do with Q-Chem or with Gaussian. Gaussian has a GUI called GaussView. You can use either one. But they both work pretty well, and are pretty intuitive. And so I think, on the next problem set, you will have some homework problems that involve running some calculations with these things. And I hope you enjoy it. Well, I don't know if homework is ever really enjoyable. But I hope it's marginally enjoyable anyway. |
MIT_561_Physical_Chemistry_Fall_2017 | 1_Quantum_MechanicsHistorical_Background_Photoelectric_Effect_Compton_Scattering.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. BOB FIELD: I'm Bob Field. This is my 44th year at MIT, and I've taught this course roughly half the time, so it is my favorite course. And I'm a spectroscopist, and that means that I use quantum mechanics almost like breathing. And I'm going to try to convey some of the beauty and utility of quantum mechanics. This is not a typical undergraduate quantum mechanics class. It's not about history. It's not about philosophy. It's about use-- use for understanding complicated stuff, use for insight. And so a lot of the material is not in the assigned text, which is a very good book, but it's mostly a safety net. The printed notes, all of which are posted except for a few lectures that I'm still working on-- and that is the text. Everything in the notes, you're responsible for. There are sections of the notes which I call non-lecture. That's explanation of what's in the lecture or a little bit going beyond it. You're responsible for that. If I don't finish all the material in the notes, you're responsible for the material I didn't finish. You can ask me questions about it. You can ask your TAs questions about it. But there's a supplementary text, which is a book that I wrote which is intended to take undergraduates like you to the frontier of research in spectroscopy. This is the book that I used to use to teach graduate quantum mechanics. So I've written a lot of stuff which is already available on the web, but I just want to make sure that you know the structure of the course. There are going to be nine problem sets, always in weeks there is not an exam-- when there is not an exam. And the problem sets are worth 100 points. There will be three 50-minute exams, to which you will be allotted 90 minutes. And they'll be evening exams and on Thursday nights. OK, there will be a final during final exam period. And so the points for the course add up to 600. For each exam, you will be entitled to bring one 8 1/2 by 11 page that you've prepared with as dense or as coarse writing as you want-- a different one page, only one page, for each exam and for the final. One page-- the exam-- one page for the exam. One page for the final. Not four pages for the final, not two pages for the second exam, et cetera-- one page. I believe that writing these pages of notes or pages of things you should remember provide structure that you actually learn by preparing this, and so I stick to this. My job here is to force you to suspend what you expect about the way matter and light behave. You expect things based on your experience with macroscopic objects, and you're very smart. You're here at MIT because you are able to integrate those expectations and to express them in mathematical analysis of your observations. So you're pretty good at that. But quantum mechanics forces you to step away from what you think you understand perfectly because it doesn't apply in microscopic systems. And so I'm going to destroy your expectations about how things work at the beginning, and by the end, I will give them back to you in the form of things called wave packets. Wave packets are quantum mechanical particle-like objects. They obey the rules, and they do a lot of the things you expect. But initially, you have to suspend that belief in particles. OK, so I'm going to start by throwing a piece of chalk. I have to throw it at my TAs. So here is a piece of chalk, and the chalk followed a trajectory. If you're a major league outfielder, you look at the initial part of a trajectory, and you can pretty much figure out where you have to go. This sort of thing would be the subject of 801, but the outfielders don't know physics. They just have instincts, and they know that if they look at the beginning of a trajectory, they can predict where it will end and when it will end, and this is really important. That's the macroscopic view. We talk about trajectories. In quantum mechanics, there are no trajectories. We have possibly an observation of what was the initial condition and what was the detection event, but we can't describe what's going on by observing, point by point, the position and momentum. We're only allowed to do what we call click-click experiments. We start something in some kind of a well-prepared state, and then we detect what has happened at the end. We might do something to the system in between, but we cannot observe everything as the system evolves. And here, I will attempt to a little bit justify that you need to suspend what you believe. I threw a piece of chalk. It could have been a baseball. And suppose now I said, let's decrease the mass of the thrower, the catcher, and the object by 100. You pretty much know exactly what will happen. Now let's decrease it by a factor of 10 to the 20, which is like going from a baseball to an electron. You think you might know, but I guarantee you don't know. And you're going to be surprised, and it's because in the microscopic world, observation modifies the state of the system, and any sort of interaction of the evolving system is going to affect what you observe. And so we have to create a formal structure, which is a kind of measurement theory. What measurements are we allowed to make, and what do they tell us? Because you cannot observe the time dependence. You can calculate what the time dependence is if you know enough, but you cannot observe the thing evolving except by destroying it. OK, so quantum mechanics is beautiful because it describes the microscopic world, and it doesn't tell you you're wrong about the macroscopic world. It matches everything the macroscopic world does. OK, some of the key ideas of quantum mechanics-- so you do a series of identical experiments. You don't get the same result. It's probabilistic, and that should bother you because if you do an experiment carefully, you're trained to think that you do it carefully, you'll get the same result. But quantum mechanics says, tough luck. You can't do an experiment that carefully. There's also this wave-particle duality. You know what particles are and what the properties of particles are. You learn that in 8.01. You know what waves are, and you probably also learned that in 8.01 or 8.02. But you know instinctively that particles and waves behave differently. But in quantum mechanics, everything is both particle-like and wave-like, and this is also something that should bother you. And the third thing is energy quantization. Now, I said I'm a spectroscopist, so I live and die by these spectra, which consist of transitions between energy levels. And these transitions between energy levels encode everything we want to know about the mechanics of an object-- its structure, what it does. And so quantum mechanics is an elaborate encoder of information, and it's usually encoded in the form of these quantized energy levels. And so you're going to want to be able to calculate how the energy levels or the spectrum that you would observe for an object is related to the thing that describes what it can do. And that's the Hamiltonian, and the Hamiltonian we'll look at in many useful ways. Now, I just want to warn you that there are two ways of presenting quantum mechanics-- the Schrodinger picture, which is differential equations and wave functions, and the Heisenberg picture, which is matrix mechanics, where we have matrices and we have eigenvectors. And I am an advocate. I'm passionate about the matrix picture because what it does is presents everything in a way that you can then organize your insights, whereas the Schrodinger picture mostly involves solving one complicated differential equation after another, and the focus is on the mathematics, and it's much harder to see the big picture. Now, that means you need to know a little bit of linear algebra. Now, most of you haven't taken a course in linear algebra, but that doesn't matter because the amount of linear algebra you need to know for quantum mechanics is extremely small, and I will probably present all of it in lectures. But you will definitely have the TAs as a resource, and it's possible that they will give some formal lectures on linear algebra or some handouts, but it's not complicated. It's beautiful. OK, so light. Light is electromagnetic radiation. You've heard that. And it is both wave-like and particle-like. Now, the particle-like aspect of light is going to bother you, but it's very easy to show that it's necessary. OK, so first of all, what are some wave characteristics? If you have a wave, what kind of measurements are you going to make on it? OK, Sasha. AUDIENCE: Intensity at a point in space. BOB FIELD: There can be intensity at a point in space, but there could also be particles that are impinging on that point, so you need something a little bit more that is definitely wave-like, and you have-- yes? AUDIENCE: It has a frequency and a wavelength and an amplitude. BOB FIELD: Yes. I'm angling for something more, but it does have a frequency and a wavelength, and wavelengths are how you understand interference effects. And when you put light through a lens, there's refraction. When you put light on a grating, there's diffraction-- or through a pinhole. And there's the two-slit experiment, where you send light on an object which has two slits, and you can't tell which slit the light went through. This is a very beautiful thing which I will talk about in a coming lecture-- in fact, lecture number three. So the key properties of waves are refraction, diffraction, two-slit experiment. And behind that is interference effects. So we can have just a comic book picture or a cartoon. So here is a wave, and here is another wave. And this other wave has exactly the same frequency and phase. And if we add these two guys together, we get something that looks like that. And in fact, the addition is a little bit nonlinear. So we have constructive interference, or we can have destructive interference and anything in between. If we add these two guys, what you get is nothing. So here, you get an intensified wave-- constructive. And here, we get cancellation. Now, it's a little bit stressful to think that if particles have wave characteristics, they can annihilate each other, so you'll have to be prepared for that. So quantum mechanics exploits constructive and destructive interference. That's at the core, and you have to get used to that, and you have to get used to seeing particles do that. OK, so waves have a frequency and a velocity and a wavelength. And now here, what is nu? Nu is c over lambda. If someone on the telephone or on Skype constantly asks you "what is new," you can put an end to that behavior by saying c over lambda. And if you do it often enough, it'll stop. It may be that the question will be rephrased, but it's useful to remember this as a repellent and also as a crucial organizing principle. OK, waves are electromagnetic, and so that means we have transverse electromagnetic waves. And so here is one part of the wave, and here is another part of the wave. And this is the electric part of the wave. And so we have zero, and here is E0. And this is the magnetic part of the wave, zero and B0. And this is in the xz plane. This is the z direction. This is the z direction. And this is the yz plane. Now, this corresponds to a wave which is linearly polarized along the x-axis. And so this is-- it's kind of hard to draw a transverse electromagnetic wave, but there's an electric part and a magnetic part. And for the most part, I forget about this. The magnetic resonance-- this is the only thing you care about, but they're both part of light. And OK. So we have these waves, and now, the intensity of the light is proportional to the E0 squared-- the electric field squared. And it's also measured in watts per square centimeter. And this is in-- so this is in volts per centimeter squared. Now, I'm an old fashioned guy. I use centimeters instead of meters. You'll just have to forgive me for that, not using MKS units. But the important thing is that the intensity is related to the square of an electric field. So suppose we have light impinging on a hunk of metal. So a metal is something where the electrons are free to move around, and so we like a metal because of that. We could ask, well, what happens if we put light on salt or on some organic molecule? And in some sense, there is going to be a relationship, but the easy thing is if we put a beam of light onto a metal and it's pushing the electrons around. That's what an electric field does. It moves the electrons. And so you would expect classically that at high enough intensity, you're going to start ripping electrons out of the metal, and that would be wrong. And this is the photoelectric effect. This is what we observe. 0-- the current divided by the charge of the electron. So that's the number of electrons per second, and this is the intensity. And if we have infrared radiation, nothing happens, no matter how strong it is. If we have ultraviolet radiation, we have something happening. As the intensity of the light increases, the number of electrons per second ejected from the metal increases linearly. So why is that? We expect that it would be intensity that determines it, but it's the color of the light that determines. We can also do something like this, where we again ask, what is the current divided by the charge of the electron versus frequency? And again, we have 0 here. And what we see is nothing up until some critical point, and then this is-- so this is some special frequency, which is different for every metal. And then, all of a sudden, we start getting electrons. And this increases linearly, or at least for a while increases linearly with the frequency. So there's something about the frequency of light that rips out the electrons. It has to be above a certain minimum, and then the electrons increase with the frequency, but in a little bit more complicated way than linearly. And so I have something in the notes which is not quite right, but this onset is important. So these observations suggest that the electron is bound to the metal by some energy which we call the work function, and it's called phi. And so this is the energy that it takes to rip an electron out of a metal. And so it's called the work function, and work functions for metals range from a little over 1 electron volt to around 5 electron volts. I'm going to use those units. Well, you can forget that. But all metals are within a relatively narrow range of work functions, and this is somehow related to nu 0. So this is an energy, and this is a frequency. And so we expect that there is some relationship between nu 0 and this energy, and there is some proportionality concept, which I can call anything I want, but I'm going to call it h because it's going to become Planck's constant. That's the proportionality concept. And so the onset of ejection of electrons is when the frequency of light is greater than the work function. As I said, every metal has a different work function. OK, so this is looking like somehow, the light does not act in an additive way on the metal. It acts in a singular way somehow. There are particles of light that have definite energy. This is what it looks like, and we're going to call those particles of light photons. And so now we're trying to think of, all right, electromagnetic radiation comes as particles. What are the properties of particles? Well, particles-- well, I'm getting ahead of myself, but as long as I said this, particles have kinetic energy and momentum. Kinetic energy is a scalar quantity. Momentum is a vector quantity. And right now, what we want to do is measure the kinetic energy of the electrons that are produced by the annihilation of photons. OK, and so we can imagine an apparatus like this. Here is our metal, and here is the light impinging on the metal. And we have grids. We have a ground, and we have another grid and a detector. So we have ground voltage is 0. Here we have a voltage less than 0. Electrons don't like that. But these two grids are at the same potential. So the electrons don't know which way to go. And then there's another grid where we have the potential is V plus V stop. And so what happens is if the electron is ejected with enough kinetic energy to go uphill and cross through this potential, this grid, then it will make it to the detector and we'll count it. So this is just a crude apparatus. If you were going to do this experiment, you would probably design it better. But the idea is what we want to do is to measure what is the voltage that we apply that causes the electrons to stop reaching the detector. And that's the way we measure the kinetic energy of the electrons. When all of the electrons stop hitting the detector, we know the voltage, the stopping voltage. And we can now plot the stopping voltage. V stop. Versus frequency. And here is nu 0. And what we see is this. We see that below nu 0, there are no electrons ejected. Once we're above nu 0, electrons start being ejected. And they have a kinetic energy, which is measured by V stop. And it increases linearly with frequency. Now, when we do this experiment on different metals, the slope is constant, the same. It's universal. So the kinetic energy of the ejected electrons is equal to some constant. It goes as minus nu 0. And this is the same constant that we saw before. And this is Planck's constant. And now what we've seen is that these particles of light have definite energy. They have definite kinetic energy. And we can stop and they transfer that energy. And so we can draw an energy level diagram. So here's 0 and here. So this is energy. And this is minus the work function. And so we have a photon, which has this energy. And we have the kinetic energy of the electrons. And this is h nu minus h nu 0. OK. This story is complete. The photoelectron effect is the easiest thing to understand at the beginning of quantum mechanics. And it says that light comes in particles, which we call photons. And the energy of a photon is h nu. h is a fundamental constant. And we know what nu is. OK. I'm going to talk about Compton scattering. But Compton scattering is a little bit harder to understand than the photoelectron effect. And I'm going to put equations on the board which I'm not going to derive. They're easily derived. But this is just to complete the picture of the particle-like characteristic of Planck. So for Compton scattering, we have a beam of x-rays. X-ray is a form of light. It's a very high energy form of light. We have a block of paraffin. And what we observe is there are electrons kicked out. And there is x-ray scattered. So the experiment is we're looking for the particle characteristics. Particles have kinetic energy and they have vector momentum. And you have been trained painfully and completely in conservation of energy and momentum, because it enables you to solve all sorts of useful problems. So suppose the x-ray comes in, hits the block. And so we have the incident momentum and the x-ray is backscattered. And the scattering angle theta is measured this way. So the difference in momentum for this scattering is large for backscattering and much less for forward scattering. Now, this momentum has to be transferred to the electron. And so we can draw conservation diagrams. Or for this case, the backscattering. We have pn. We have the electron, p electron. And we have the momentum of the x-ray. OK, so this diagram determines the momentum transferred to the electron. And if the x-ray photon transfers energy to the-- it transfers momentum to the electron, it also transfers energy. So what we're going to see-- and I'm just going to wave my hands, because the mathematical analysis is something you can do, but I don't want to go through it. I just want to show the structure of the argument. What you measure is the wavelength of the x-ray out minus the wavelength of the x-ray in. And so if the x-ray has transferred energy to the electron, it has less energy when it leaves, and it has longer wavelength. So this is measurable, the change in the wavelength of the x-ray, and that's going to be-- so we'll call that delta lambda. And that's going to be a function of the scattering angle. So Compton scattering basically says, OK, light is a particle. Particles have kinetic energy and momentum. Conservation of energy and momentum predicts a difference, a redshift of the photon, which depends on the scattering angle. The rest is all 8.01. You can do that. OK. And so what you end up finding is that if the scattering is 0, then delta lambda is 0. There is no momentum. The photon just goes through. If the scattering is pi, in other words, it's perfectly backscattered, then delta lambda you can calculate is 2h over the mass of the electron times the speed of light. Now, this quantity, h over the mass of the electron times the speed of light, has dimensions of length. And it's 0.02 or 3 angstroms. And this is called the Compton wavelength of the electron. Because the photon is scattering an electron. The change in wavelength of the photon is determined by this. This is the momentum transfer. This gives you the momentum transferred. And so this is a universal constant. It says electrons have, when they're scattered out of any material, have this behavior. Again, universal and strange. The actual experiment, since what you actually want to measure is the change in lambda over lambda. The change in lambda doesn't depend on lambda but this does. And so what you want to do is make the observable thing large. And so you go to a short wavelength to make this a large fractional change that's easily measured. OK. The equations are derived in the lecture notes and better in texts. And what I want to do now is just tell you where we're going. We've talked about the wave and particle nature of photons, of light. And the particle nature is a surprise, but it's perfectly understandable and observable. The next thing we want to do is determine the wave nature of things that we call particles. And so we're going to do other kinds of simple scattering experiments where we discover that the electron-- so we have some solid. And we have UV light or x-rays. And we go through this solid. And it's mostly transparent. It's mostly free space. And so this light is interacting with the electrons. And maybe with the other stuff that is involved in matter, the nuclei. But the important thing is that the electrons that get scattered are in a kind of a diffuse state. This is mostly nothing. This material doesn't scatter x-rays. It's mostly transparent. And so an explanation for that is the Rutherford planetary model, where we have a nucleus with a charge and electrons orbiting that nucleus. And that's all very nice. It's a way of saying, well, OK, the electrons are forced to choose distances from the nucleus. And there's mostly empty space. But then you look more closely, and the electrons are doing this. They're oscillating in space, and they're going to radiate energy. And this radiation of energy will cause the electrons to spiral in and combine with the charge in the middle. And that's a problem. And so we have an explanation for why matter is mostly empty space and a conundrum. The electron should recombine with the nucleus. So how can matter be relatively non-compressible? And the best explanation is this planetary idea. But then we have this thing we have to explain. Now, there's two hypotheses for this, which are really just ad hoc things. One is the Bohr model where it says, in order for the electron to obey some laws of physics as it orbits the nucleus, it conserves angular momentum. Well, if it were rotating combined with the nucleus, it wouldn't do that. And so that's sort of OK. And then there's de Broglie, who was a very smart person. And he said that in order for the electron not to annihilate itself as it goes around the orbit, it has to have an integer number of half wavelengths along that trajectory. Then it won't annihilate itself. And that is much better than Bohr's hypothesis. But both of these hypotheses lead to line spectra hydrogen where you can build a really simple model and you can predict to eight decimal places all the absorption transitions of hydrogen atom. So a very simple model leads to an incredibly powerful and mysterious result. There is another experiment that I will talk about next lecture. And that is suppose we have a thin sheet of metal. We have photons, x-rays, or UV photons, impinging on this metal. There's diffraction because there is regular distances between the atoms in the particle in the foil. And so what we measure is diffraction rings for the electron and for the photon. And the diffraction rings are identical when you choose the wavelength of the particle to be equal to the wavelength of the photon. And so this is another way of showing wave particle duality. We can show that they have wavelengths and we can calculate to make them the same. So this is all very exciting. And this is about as much philosophy and history as you're going to get from me. Once we understand that there is something that we have to do, which is called quantum mechanics, we're going to start solving problems and then being able to understand incredibly complicated effects beyond the simple problems. And that should be exciting, but it should also be a little bit disturbing, because I will use a technique called perturbation theory, which for some reason all of the textbooks for a course at this level either ignore or treat in the most superficial way. But perturbation theory is a way of taking problems that we understand and can solve exactly. And exactly means an infinite number of states, all of which are given to us by solving one equation. And we can then use them to understand problems we can't solve exactly. And there'll be several ways in which I deal with that. OK, so it's time to stop. And I hope you're excited about what lies ahead, because it is strange and wonderful. And it says we can look inside a molecule. We can make measurements. We can understand what this molecule is going to do. But we have to develop our new way of doing that. OK, thank you. |
MIT_561_Physical_Chemistry_Fall_2017 | 25_Molecular_Orbital_Theory_II_H2_A2_AB_Diatomics.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: I like this lecture because the first half of my career was entirely devoted to the electronic structure of diatomic molecules. And this is how you make sense of diatomic molecules. Now in 5.111, 5.112, one of the most exciting thing that attracts people to become chemistry majors is the global interpretation of the periodic table in terms of some simple ideas so that those simple ideas enable you to predict the important properties of all of the atoms. And that's a pretty exciting thing to feel like, yes, it all makes some sense. And then we go on to other topics in chemistry, and we forget about the fact that we think we understand the periodic table completely. We don't use that information. And one thing that I've tried to do throughout my career is to take the simple insights into the periodic table, and explain molecular properties. And now, this is really a strange thing because we have computer programs that can calculate any of the properties-- any property you want-- of a small molecule. So why would we ever care about being able to predict the properties-- because we can calculate them? And the answer is we want to understand. We want to be able, without a computer, to say this is what we expect for a particular molecule not just a diatomic molecule. What are the things that control the bonding in molecules in general and the properties in molecules in general? And if you can understand what happens in diatomic molecules, there are very few cases which would leave you at a loss for what will be the properties of this new molecule that you're looking at. Or how would you manipulate a molecule so that you can force it to go into the states that will expose the critical properties that you want? So it's this sort of part of a toolkit that any practicing physical chemist who's worried about spectra, molecular properties, electronic properties would have to master. And this is really hard because, one, you have a computer-- you can generate the answers completely accurately. And two, you're smart MIT students. And you believe in the truth rather than estimating the truth. But I like to maintain that if you can estimate the truth and find out whether your estimates are wrong or right, that enables you to have confidence to deal with other properties, other problems. And so what I want to do today is to give a sense of how you begin to understand the electronic structure of simple molecules. And I want to tell you that a lot of the second-order or complicated patches you put on this simple picture are things that you develop and criticize throughout your career. You say, yeah, well, this works, but that doesn't. And why doesn't it work? And that's what we do as physical chemists. We look for when the simple picture breaks. And we have our own private simple pictures which we're constantly trying to make better than anybody else's. OK, so we have the periodic table. We know the properties of atoms. We know, especially, how the ionization energy is supposed to behave. We know how the low-lying electronic states are supposed to be ordered. Excuse me. We know about orbital sizes. Everything is related to this idea of atomic-orbital energies and shielding. And it's all related to the hydrogen atom. What we got from the hydrogen atom was a framework for understanding electronic properties of atoms. And what we know about electronic properties of atoms should be important in explaining electronic properties of molecules. So I want you to be able to make toy models, to draw pictures which are crude, but guiding, in your insights. And it's all based on what we saw. And you soon are going to have two lectures from Troy Van Voorhis, which will lead you to use some of the simple computer programs that enable you to calculate electronic property of anything. OK, so we want to extend what we had for H2+ to be able to understand H 2 AH, A 2, and AB molecules. And each of these steps involve extensions of relatively simple ideas. And so let's begin. OK, I told you that I believe the simplest explanation for bonding is when we have two atomic orbitals that overlap in the region where this overlap region can be attracted to both nuclei. Constructive interference causes the density in this special region-- the amplitude to be twice as large, the density to be four times as large. And that's a large, energetic effect. And the amount of overlap is crucial. You can have two little if the atoms are too far apart-- no bond. You could have too much if the atoms are too close together. OK, so we need to now look at cartoons of orbitals and learn about the names for these orbitals. So the simplest orbital is gerade-- sigma gerade NS orbital. OK, so "sigma" means there are no nodal planes containing the internuclear axis. "Gerade" means that in the body frame, it's even symmetry. So we have ns A plus ns B. And that looks like this. We can have an antibonding orbital. Again, we put the predominant atomic-orbital character in parentheses. And for heteronuclear molecules, we also put which atom is more important to that orbital. And this will be just ns A minus ns. And that will look like this. And there is no overlap here. In fact, there is negative overlap. And that's an antibonding orbital. And you can say, if there's no overlap, well, then the actual picture will sort of look like this, where the region in the binding region is actually depleted, and there's less there. And that's part of why it's antibonding. OK. Now we can also make sigma g np orbitals. And np z-- "z" is the internuclear axis-- A minus and np z, B-- well, that gives us something that looks like this-- plus, plus, minus, minus. And the minus sign is needed to turn this around. But this is overall gerade symmetry. And so everything is fine. And then we can have sigma u star np. And that would just be the plus sign here. And that would be an antibonding orbital. Now I put a star on these orbitals because when we don't have atomic symmetry, when we don't have g and u symmetry, we still use the star to imply antibonding. But in diatomic molecules, this is redundant. You should be able, just by looking at the orbital character or looking at a picture, to say, yes, it's a bonding orbital, and it's either g or u. Then we have pi u np. And that would be-- well, let's do it the opposite order. So we'd like to have something that's bonding. So we have to have the phases right. And that's ungerade because when you reflect through the center of the origin of coordinates, we go from plus to minus. So that's ungerade. But that's bonding. And that's a little trick because you start this think, from sigma, that ungerade is antibonding. And finally, we have the pi g star np. And that would be plus, minus, minus, plus. And that's gerade. OK, there are a couple other things that we know just from these pictures. And that is the p-sigma orbital is much more directional than the s-sigma orbital. That's all you get in sigma for s. And so the bonding interaction between p-sigma orbitals turns on at longer range than for s. So there's qualitative differences. And sometimes, you have a bond distance determined by favorable overlap between one pair of orbitals. And it makes things not so good for other orbitals. So usually, you have a ground state. You figure out what the ground state of molecule is. And that tells you something about what the internuclear distance is. And then you might grow to excited states of the molecule or excited configurations. And these excited configurations-- well, if you're going to learn about a molecule, you've got to do spectroscopy. So if you go from the ground state to an excited state, and the internuclear distance of the ground state is wrong for the excited state, well, then, you're going to have a big change in geometry. You'll have Frank-Condon factors, which-- I haven't talked about Frank-Condon factors-- I will-- that say, OK, the transition is going to have a lot of vibrational structure. And it'll tell you all sorts of good stuff about that. But the important thing is when you start looking at a molecule, you want to know what the ground state looks like and what transitions from the ground state ought to look like when you're ready to know how to use the simple structural information you're getting from these orbitals because there's no point doing an experiment if you don't have an idea about what the experiment will yield. That's what most people do when they do an experiment. They did the experiment because they thought it would yield something, and they wanted to be surprised because they'd like to show that maybe it didn't do what they were expecting. OK, so we have pictures. And we have notation. And this notation is what is used by professional spectroscopies, So we care about directionality. And we care about size of orbitals. And this is definitely something that you're empowered from 5.111, 5.112. I'll talk more about that too. How about right now? OK, in thermodynamics, when we're concerned about the enthalpy of formation of a molecule, we have to set a 0 of enthalpy. And we set it as separated atoms in their most-stable state and energy, or enthalpy, is something where there is no natural 0. We wanted a 0 which is convenient for all of our calculations and insights. And so what we do is we say, OK, every particle that we're going to be looking at-- we're going to make a molecule out of two atoms, like A and B and AB star. So we have the energy of A plus electron B plus electron, or AB plus electron. That's a common 0 of energy. So we talk about orbital energies relative to this common 0. Now in almost every textbook, the orbital diagrams are given to you. There's no discussion of, what's the 0 of energy? If you're going to know anything, you really want to be able to calculate or to say all of the actors in this game are acting in a way related to how far below this ionization energy they are because that determines their size. And the size determines the overlap. And bonding is related to overlap. OK, so we'll play that game. So for H 2+, it was very simple. We just had the 1s at r equals infinity, and over here, the 1s orbital at r equals infinity. And then we solved the minimal-basis variational problem. And we discovered that we had something like this, where this is the antibonding orbital. This is the binding orbital. The antibonding orbital is more antibonding than the bonding orbital is bonding. And, of course, this energy difference depends on internuclear distance, also not often talked about. And so we have the starting point at r equals infinity for the separated atoms. Now as we bring these atoms together, before there's any bonding at all, say, we have 1s on A. We make the internuclear distance smaller and smaller. And the bare nucleus of the other atom starts to penetrate into the-- so as you move the positively charged nucleus towards the static charge on the other atom, that's a favorable interaction. But there's a repulsion between the two nuclei. And when the bare nucleus of one atom penetrates inside the charge distribution of the other, it sees less attraction and more repulsion. And so this is why you get a potential curve that looks like this. The repulsion occurs at too-short internuclear distance. And from the variational calculation, you can determine the value of equilibrium internuclear distance. And so a lot of these diagrams are calculated at this particular internuclear distance. It's especially important because the repulsive states don't have a minimum. And so there is no particular internuclear distance that you care about so that if you want a simple picture, everything is calculated at that particular internuclear distance. OK, so sometimes, you want to have a little bit more insight in this picture. And instead of drawing the starting point at infinite separation, you could say, well, let's schematically suggest what happens as you-- so at shorter internuclear distance, the separated atom energy increases a little bit. And that could be useful. But it complicates things, so I won't do that. But often, we do like to know what happens as we bring particles together as opposed to starting at infinity and coming into the end point, which is the bond at the equilibrium internuclear distance. So you can put that insight in if you wish. So we know that this bonding interaction is smaller than the antibonding interaction. OK, we can now go from H 2+ to H 2. And the first thing we know is when we go from H 2+ to H 2, we can put an electron into the same orbital but with opposite spin. And as a result, we don't have to worry about mysterious things, like overlap repulsion or poly-repulsion because we're putting electrons into different spin orbitals. And that's all you have to do. If you try to put two electrons into the same spin orbital, that's really bad. And once we get beyond hydrogen-- hydrogen has this neat situation that there is no core. It's just the nuclei. But for anything other than hydrides or hydrogen, we have an inner core, which is filled with electrons of both spins. So anytime you have an orbital that gets too close to the other atom, there is going to be this mysterious overlap repulsion. And that makes the inner wall of most potentials other than hydrogen extremely vertical. OK, so let me just-- oh, I know what I wanted to say, but i will get to that When we go from H 2+ to H 2, we discover that the equilibrium in our internuclear distance decreases by approximately 30%. The vibrational frequency increases by 90%. The dissociation energy increases by about 70%. These are big effects. They're not a factor of 2. Naively, if one electron gives a bond or gives some sort of a bonding interaction, two should give twice that much. But this is pretty close to twice that much. Even though that's only 30%, it's a big effect. And it's because we're allowed to put an electron into an orbital which is bonding. And there is no magic-overlap repulsion. There is just the interelectronic-energy repulsion. So our picture is-- we're chemists. We believe in bonds. The most important thing is bonds. And in almost every situation, you want to conserve the number of bonds or know how much you have to pay for changing the number of bonds. That's the course. But we're more sophisticated. Now in our picture for H 2+, we had an effective Hamiltonian or a variational calculation. I like to talk about effective Hamiltonians. But the crucial actors in the H 2+ problem, or the overlap is a function R, the interaction energy as a function of R, and the orbital energy is a function of R. And these are the actors in almost all further calculations. You're going to want to either know these things or be able to estimate their sizes so you know the consequences. And we know much of the consequences from H 2+. We know that as R decreases, the overlap increases, the increases, and the interaction energy between the two atoms increases. So these are the factors that we put into a calculation. We know that there is a bonding region. And typically, we can say that the bonding region for H 2+ and for any molecule is going to be roughly between the atomic-orbital radius and twice the atomic-orbital radius because we've got two atoms. And so they want to see it at twice that radius. And then as we get closer, we get bonding. And as we get too close, we get antibonding. And so this is a critical region. Now we're ready to take the next few steps. So we want to do H 2+ to H 2 to A 2. And so the questions we ask is, how many electrons? That's easy. Feed the energy electrons into lowest MOs. Well, in order to do that, you need to know the energy order of the MOs. We get a configuration, which is just a list of the number of electrons in each of the molecular orbitals. And these configurations have particular states. And for a diatonic molecule, we know the projection of the electron orbital-angular momentum on the other nuclear axis, lambda, we know the spin. And sometimes, we know about this thing, omega, which is the projection of the spin, as well as the projection of the orbital-angular momentum. And so we have a notation for electronic states, which will be lambda 2s plus 1 omega. And so say we have a triplet. Well, this triplet will be 2s plus 1 is 3. And this is pi, say. So the correct way to say that is not "3 pi." but "triplet pi." And we normally don't talk about that in any special notation. So the splittings of the configurations for H2-- we have the ground state. And that's sigma g 1s squared, and the lowest excited state, triplet sigma u plus-- and that's sigma g 1s sigma u us. So we have these pictures. And this is bound. This has got a binding orbital and an antibonding orbital. This is a sigma u 1s. Sorry-- so antibonding. So this is repulsive. Or usually, it's repulsive at all internucleuses. Then there are higher-energy orbitals that come from 2p-- 2p pi and 2p sigma. And they give rise to many, many states. But we know, for hydrogen, when we go from 1s to 2p, we're at very, very high energy-- 3/4 of a Rydberg. And that's an energy that's larger than any chemical bond we know. So the excited states of H 2 are going to be a little bit weird because of this tremendous excitation energy. And because the excitation energy is so high, we're probably not prepared to understand what's going on up there. And in fact, terrible things happen up at high energy. We get there are some curves that are double minima. There are all sorts of things. And they have to do with-- you could have H+ H- ion-pair states. You can have H 2+ plus electron-- those are Rydberg states as well as the normal valence states. So normally, we don't want to worry about that. And there are two places we don't want to go without preparation. One is to very high excitation energy. And the other is to very large internuclear distance because things can happen. The weirdnesses are much more important when there is no core. So since everything is dependent on R, we need to know how to estimate the internuclear distance. And we know, for an atomic orbital, the ionization energy is hc times the Rydberg constant over n squared. And we know that the average r for an orbital is a0 times n squared over the charge. And so if we want to relate the size of an orbital to something that we can measure, well, we can simply put in this equation for n squared here. And we get that the average size of an orbital will be a 0, which is the Bohr radius of the hydrogen atom-- about half an angstrom-- times hcR. The Rydberg is 110,000 wave numbers-- and over z-- I'm sorry-- over I nl. So this ionization energy is measured. And so if you know the ionization energy, you know r. If you are talking about a molecular-orbital diagram, you have r and ionization energy on it. And so you can see how it would make everything relate to each other. OK, let's do an example. Let's look at the NH molecule. This is not a stable molecule, but it's a molecule that is important in some chemical reactions. It's easy to make an H by having, say, nitrogen-- N2-- and hydrogen and doing a discharge. And so in order to be able to understand H, you start out with all of the orbital energies. So let's make a little table. We have H 1s. And we have its ionization energy. And its 13.6 electoral volts. We have H 2s and 2p for hydrogen. They're degenerate. And that's 3.4 electron volts-- for nitrogen, 1s. This is an obscenely large energy. And so it's greater than 100 electron volts. We have N 2s, which is 18 electron volts. We have N 2p, which is 12 electron volts. And we're going to have, at the end, the ground state of NH, which turns out to be a triplet sigma, g minus state. And that's at 13.6 ev. So one of the things you want to do is ask, OK, all my actors-- where are they below the ionization energy? And the next thing you do is you arrange things in energy order. And so the lowest one is this. And it's out of the picture altogether. The next lowest one is this. And then we have two. And let's just say this is 3 to 4, and this is 4 to 3. They're roughly degenerate. That's what we're looking for because when the orbitals are degenerate, they have about the same energy. And we know from perturbation theory, if you want to have a mixture of two different things, you want the energy to not measure to be small. So all the actors-- the simple stuff-- we know what their energies are. We know that if we have two things that have the same, or nearly the same, energy, they're going to be mixed. And so now, with this sort of an annotated table-- the last one is this guy-- we can write down the low-lying configurations. And we can say something about the nature of the molecular-orbital diagram. [INAUDIBLE] But I stress the most important thing is something that's easily measured for atoms because what we want to do is build on what we think we know for the periodic table to have some sort of insight into molecular-periodic table, although it's a multidimensional periodic table. I know that's probably a stupid thing to say. OK, so here's the molecular orbital diagram for NH. So over on this side, we have H+. Over this side, we have N+. And we put the orbitals on-- OK, so here is minus 3.4, which is the hydrogen 2s or 2p. And down here, we have the hydrogen 1s. And that's at minus 13.6. And over here, we have the nitrogen 2p. And that's at minus 12. So what do we do? And then down much farther below way down here, we have the nitrogen 1s. The nitrogen 2s. OK, how many electrons do we have? Well, before I do that-- so how many electrons are on nitrogen? I mean, if you don't do that, you don't know the periodic table at all. How many? Did everybody see that? Well, then I can't ask you. Yes, it's seven electrons and plus 1 for the hydrogen, right? So we're dealing with eight electrons. And so we're going to put eight electrons into orbitals. And now on this diagram, basically, I have two orbitals that are roughly at the same energy. And so we know that all the action is going to be that. And so we can draw something like this and something like that. And now, we use our perturbation ideas. And this guy is higher in energy. And so it's closer to this other orbital. So this one is dominantly nitrogen, and this one is dominantly hydrogen. Make this dotted. OK, so we have two orbitals. This one is polarized towards nitrogen. And this one is polarized towards hydrogen. And then there is another orbital here. This is 2p pi. Well, what about 2p pi? There is no pi orbital here. The lowest pi orbital is way up here. So this guy is nonbinding. So we have a bonding orbital, a nonbonding orbital, and an antibonding orbital. And then up here, we have something that's so high energy, we might call it a Rydberg orbital or just something else. OK, so now we start putting electrons into orbitals in orbital-order energy. So we have sigma nitrogen 1s squared. That's the lowest one. That's way down here. I erased it. It's at minus 100 electron volts. And then we have nitrogen 2s. That's pretty far down too. And so these two-- we don't have any doubt. The electrons go into these sigma orbitals. And there's nobody nearby, and so these are localized on the nitrogen atom. We've got four more electrons to deal with. We have sigma NH sigma NH. Well, we'll call this sigma. We got two electrons there. And then we have-- so we got two electrons, two electrons in core orbitals. We have a bond. Then we have a nonbond. Yes. AUDIENCE: How far apart in energy do the orbitals have to be such that you consider them non-attracted? PROFESSOR: It's really up to you. I mean, at the lowest-- it's a question of how much detail you want to be honest about. There's always interaction. But the important thing is if the orbital energy is really low or stable, the orbitals are really compact. And so at equally internuclear distance, which is determined by a bonding orbital, they're just out of play. So it's not 0. But you really care about big effects-- things that are worth, roughly, the energy of a tenth of a bond. As you get more and more better at this, you could say, well, we'll worry about bonding effects that are worth less than a tenth of a bond or a typical bond. But you want to organize things in the order of how important they are, how careful you have to be, how many subtle effects you have to take into account. And so to avoid craziness, you want to say, I'm not going to worry about this because it's too small, or because their energy difference is too large. And it's very important to be able to shed unnecessary factors until you're ready to deal with them. Or maybe you never will. OK, so that was a really great question. All right, so we have the list of-- where did I put it? Oh, here it is. OK, so these are three filled orbitals. They give rise to a singlet sigma. But we have two electrons in a pi orbital. And if you care about this stuff, you'll learn how to figure out what two electrons in a pi orbital will give you. And they give you triplet sigma minus g-- I'm sorry-- no g because it's a heteronuclear-- triplet sigma minus singlet delta and singlet sigma plus. And this is actually much easier than for inorganic chemistry or for just figuring out the LS states of an atom because basically, we have sigma, pi, delta. And you can figure out, well, pi squared is going to give you this. Pi cubed is going to give you this again, except this-- it's going to give you this again. So you there's a finite number of things that you don't have to work out the energy levels by an elaborate procedure. And so you can sort of memorize them because there's very few that you actually are going to deal with. OK, so now Hund's rules apply to molecules. So what's the ground state of NH? Triplet sigma minus. So that's nice because you're going to be doing spectroscopy, and you're going to be starting with a molecule in a triplet state and you've got a pi bond. And so you know a typical internuclear distance for a pi bond, or you can figure it out. And we're only going to be seeing triplet states, and we're only going to be seeing triplet-sigma-minus states or triplet-pi states because of the selection rules. Now I'm telling you this. You could learn that. These are extra things. You've got to know the most-important stuff. So we have an intentionally naive primitive-molecular orbital diagram. OK, now-- my chance to say terrible things about textbooks. Every one of you has seen the molecular-orbital diagram for this mystical A2 molecule. Everybody's seen it. And it's presented either dishonestly or semihonestly with an asterisk. So let's just understand this. Again we have the ionization energy. We have A+ and A+. And we plot the energies. And naively, we say, OK, here's 2s-- same energy 2s. And here is 2p-- 2p-- for atoms other than hydrogen, 2p is above 2s. And so we know, well, we can draw some kind of a diagram like this. So these are the sigma star 2s and sigma 2s. No problem. And then we draw something like this for the p orbitals. And we have something like this. That makes sense. Now, the lowest-energy orbital has to be the more directional p orbital. So this is going to be the p-sigma orbital. And so we'll have a sigma 2p, a pi 2p, a pi star 2p, and a sigma star 2p. That's what you get using common sense. And it's wrong. And most of the textbooks say, well, if we have to correct this-- it turns out that the energy order of these two orbitals is frequently reversed. And most textbooks don't tell you why. They just say, well, this is what you would expect. And the canonical molecular-orbital diagram for homonuclear diatomics has the pi below sigma except for oxygen and fluorine. Well, that's not something that builds confidence. And that's something you learned in 5.111, 5.112. The energy gap between 2s and 2p depends on shielding. And so as you start out with lithium, 2s and 2p are-- lithium is more like hydrogen. So 2s and 2p are nearly degenerate. When you get to fluorine, the energy gap between 2s and 2p is enormous because 2s shields 2p. And as a result, the-- well, the shielding arguments explain this energy gap. And it gets large because 2s is unshielded, and it really gets very stable, whereas 2p is increasingly affected by the shielding by s, and as a charge, increases the orbital energy. OK, I thought I had a better explanation for it. But I do have one. It's just not handy. And so what happens? Well, what happens, as you go from lithium, where this and this are close together, you have interaction between orbitals of the same symmetry. Now the Hamiltonian is totally symmetric. Oh, my goodness. So because the Hamiltonian is totally symmetric, you can have interactions between orbitals of the same symmetry. And as a result, this guy gets pushed up. And this guy gets pushed up. And this guy gets pushed down. But these are so far out of the picture you don't care. So it's really just the relative energies of these two. And it's all due to the relative energies of 2s and 2p. And then you get the correct energy-level diagram. And you say, well, maybe as we went across the periodic table, there will be a switch. And there is a switch. And since I've lived most of my early life in CO and N 2 and O 2. I've encountered that. And I thought about it a lot. OK so that's pretty much all I'm going to say about the molecular-orbital diagram. I think next lecture is on Huckel theory. OK. |
MIT_561_Physical_Chemistry_Fall_2017 | 11_Wavepacket_Dynamics_for_Harmonic_Oscillator_and_PIB.txt | The following content is provided under a Creative Commons license. Your support will help MIT Open Courseware continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT Open Courseware at ocw.mit.edu. ROBERT FIELD: OK. So this is going to be a fun lecture because it's mostly pictures and language. And it's also an area of physical chemistry, which is pretty much-- I mean, there's three areas of physical chemistry where people have trouble talking to each other. There's statistical mechanics. There's quantum mechanics. And there's the time dependent, time independent forms of quantum mechanics. And these three communities have to learn how to talk to each other. And I'm hoping that I will help to bridge that gap. OK. So last time, we talked about the time dependent Schrodinger equation. It's a simple looking little thing. Remember though, it's an extra level of complexity on what you already understand. Now, in the special case where the Hamiltonian is time independent, then, if you know all of the energy levels and wave functions for the time independent Hamiltonian, you can immediately write down the solution to the time dependent Hamiltonian. And so you have a complete set of wave functions. And you have a complete set of energy levels. And with that, you can basically describe something that satisfies this thing. Another set of coefficients is often something that you arrange for simplicity or for insight. And I want to understand-- I want to talk about that. OK. And so in the previous lecture, we talked about the probability density, which is psi star psi. And that can move. And that can move in two ways. In the wave function, if there are two or more states belonging to two or more different energies, then you can have motion. And you can have motion like breathing, where probability moves towards the extremes, or motion where there is actually a wave packet going from one side to another. So for anything interesting to happen, this probability density has to include at least two different eigenstates belonging to two different energy levels. Now, the total probability is just the integral over all coordinates of the probability density. And probability is conserved. So one of the things that happens-- when you do an integral, the wave functions go away. And in this particular case, integral psi star psi, all you get is one. Or you get a constant, depending on how you normalize things. But it's not time dependent. Now, in understanding motion from classical mechanics, we know Newton's equations. We know how the coordinate and the momentum depend on time. And so if we calculate the expectation value of the coordinate or the expectation value of the momentum, we get some results. And it turns out that these things describe the motion of the center of a wave packet. A wave packet is anything that's not just a single eigenstate, so it moves. And so the motion of a single eigenstate-- the motion of a wave packet is described by Newton's laws. And this is Ehrenfest theorem. And it can easily be proven. But we're not going to prove it. We're just going to use it. So we are very interested in expectation values of the coordinate and the momentum. And for a harmonic oscillator, this is duck soup. Then we have nothing but trivial integrals involving the a's and a-daggers. And so even though in some ways the harmonic oscillator is a more complicated problem than the particle in a box, the harmonic oscillator is the problem of choice for dealing with motion and developing insight. There's another useful-- OK. We have this wave function here. There is a huge amount of information embedded in it, too much to just look at. We need to have ways of reducing that information. And one of the nice ways is the survival probability. The survival probability tells you how fast does the initial state move away from itself. And this is very revealing. Another very revealing thing is if this survival probability, which is a function of time and nothing else, because we've integrated over the wave functions-- if this survival probability goes up and down and up and down, there are recurrences. So at a maximum, which is usually a maximum at t equals 0 because the wave function is it at its birthplace, the survival probability will start high and go down, and come up, and go down. And there's all sorts of information about recurrences. As the wave function returns to its birthplace, it might not return completely. And so we get partial recurrences. And these tell us something. And the times at which the maxima in the survival probability occur tell us a lot about the potential. And there are some grand recurrences where, because all of the energy level differences are an integer multiple of a common factor, then you get a perfect recurrence. And these are wonderful for drawing pictures. Now, you really want to understand the concepts of time dependent quantum mechanics pictorially and to develop a language that describes it. And there are a lot of little pieces here that you need to convince yourself you understand so you can use the language. OK. One of the things I said at the beginning of the course is the central object in quantum mechanics is the wave function. Or we could generalize to this. This is actually the truth. This is a partial truth. This is everything that we can possibly know. But we can't know this. We can never observe the wave function. However, we do observations. And we construct a picture, which is what we call the effective Hamiltonian. This effective Hamiltonian describes everything we know, mostly energy levels. We have formulas for the energy levels. And we determine the constants. There's all sorts of stuff that we collect by experiment. And so we create an object which we think is like the true Hamiltonian. And by having the effective Hamiltonian, we can get these things. So we observe we make some observations. And then this is a reduced version of the truth. And we can get a pretty good representation of the thing that is supposedly hidden from us. OK. So let's start now at t equals 0. At t equals 0, which I like to call the pluck-- and it really makes it contact with those of you who are musicians, that you know how to create a particular kind of sound, which will evolve in time. And it depends on the details of how you handle the instrument. So at t equals 0, we have this thing which, if we have a complete set of eigenfunctions of the Hamiltonian, of the time independent Hamiltonian, we know we can write this as cj psi j of x. Now, often we don't need an infinite number of terms. But we know that there are lots of different kinds of plucks. And you can find out what these coefficients are by just calculating overlap integral of these functions with this initial state. OK. So this is telling you what happens at t equals 0. And if you can write the initial state as a superposition of eigenstates, then you can immediately write the time evolving object this way. Now, we have a minus i. Well, why do we need a minus i? Well, because we're going to take the time derivative of the Hamiltonian and multiply it by iH bar. We're going to want to get the energy, not minus the energy. And so we need a minus i here. And we need a divided by H bar here in order to satisfy the time independence or time dependence. So this always bothers people. Why should there be an i here? And why should it be minus i? And I recommend not trying to memorize it, but to just convince yourself. I need the minus i because I'm going to bring down a minus i over H bar times e. And we have this iH bar. The minus i times i gives plus 1. OK. So now, in understanding how the time dependent Hamiltonian can be sampled, we build the superposition states with a minimum number of terms, like two or three. Even though, in order to create a very, very sharply localized state, it needs a huge number of terms here. But you always build your insight with something you can do in your head. And so for some problems, you need only two states. And for others, you need three. And those are the ones you want to kill. You want to understand everything you can build with two and three state superpositions. OK. So there's two classes of problems. OK so for a half harmonic oscillator-- so for a half harmonic oscillator, we start with harmonic oscillator and divide it in half and say this goes to infinity. And so this side is not accessible. And so if the potential is half of a harmonic oscillator potential, then we know immediately what the eigenvalues and eigenfunctions are. They are the harmonic oscillator functions that have a node in the middle. So we know that we have v equals 1, v equals 3. No v equals 0 or 2. I mean, this corresponds to the lowest level, v equals 0 for the half oscillator. But this is basically-- now, what we're going to do is we create some state of a half oscillator. And then we take away the barrier. So that's a cheap way of creating localization. Of course, we can't do this experimentally. But you could, in principle, do it. And so basically, you want an initial state, which is a superposition of vibrational levels. OK. And this initial state needs to have the wave function be 0 at the barrier. So since we're going to be looking at the full potential, we're taking away the barrier. We're allowed to use the even and odd states. And so let's take the three lowest states. And so we have c0 psi 0 plus c1 psi 1 plus c2 psi 2. OK. This guy is OK. Yes? AUDIENCE: When you create the states of the half harmonic oscillator, do you have to re-scale them so that they normalize? ROBERT FIELD: Yes. We play fast and loose with normalization constants. We already always know that, whenever you want to calculate anything, you're going to divide by the normalization integral. However, it's OK. Because we're going to be using the full harmonic oscillator functions, which do exist over here. And it's fine once we remove the barrier. But we want to choose a problem which is as simple as possible. OK, now, this guy is not 0 at x equals 0. And this guy is not 0 at x equals 0. But you can say 0 is equal to c0 psi 0 of 0 plus c2 psi 2 of 0. We choose the coefficients so that these two guys together make 0 at the boundary. And why do we do this? We're going to be calculating expectation values of x and p. And if we only have half of the energy levels, all of the intricacies are going to be 0. So we have to have three consecutive levels to have anything interesting. OK, so the artifice of making these two-- arranging the coefficients so that you get a temporary node at x equals 0-- it won't stay a node. Because when we let things oscillate with time, these two will oscillate differently. And the node will go away. But that's how you do it. That's how you build a superposition. OK. OK. Now, what we want to do is to be able to draw pictures of what's happening and also to calculate what's going on. And one of the things we use for our pictures is the expectation value of x and the expectation value of p. So you know how to calculate the expectation value of x. We have this capital psi star x capital psi star capital psi integrate over x. And so we get to c0 c1 x01 cosine omega t plus 2 c1 c2 x12 cosine omega t. Now, how did I do that? Oh, come on. You can do it, too. So what is this x01? Well, x01 is the integral psi 0 star x psi 1 dx. And we know we can replace x by a plus a-dagger times the constant. We always like to forget that concept. We only bring it in at the end anyway. And so, well, this has a value-- it's a constant, the constant that we're forgetting-- times 1 square root, right? And x12 is the same sort of thing. It's going to be the same constant times 2 square root. This is easy. So getting from this symbol to this, that just requires a little practice, and then simplifying further to know what these x's are, and then the constraints on c2 and c0. We have all that stuff. OK. And now, we can draw a picture of what's going to happen. So here we have the full oscillator. And let's just draw some energy which is-- now that's complicated. We have a time dependent wave function, which is composed of several different energy eigenstates. So what is its energy? Well, you can evaluate what the energy is by taking the expectation value of the Hamiltonian. And so you can do that. So what we've made at t equals 0 is something that looks like this. It's localized on the left side. Or it's more localized on the left side. Now, sometimes, you're going to worry about phase. And so many times when you're working symbolically rather than actually evaluating integrals, there are symbolic phase choices that what you're using has made. And for example, for the harmonic oscillator, if you look in the book, you'll see that, for all of the harmonic oscillator functions, the outer lobe is always positive. And the inner lobe is alternating with the quantum number. So that's a phase convention that's implicit in everything that people have derived. And as long as the different things you combine in doing a calculation involve the same phase convention, which is implicit-- we don't want to look at what functions. We want to look at these xij's. OK. And so those things that we're manipulating have a phase implicit in how you define them. But that's gone in your manipulation. So you want to be a little careful. OK. So we start out and the expectation value is on this side. And the psi star psi is localized on this side mostly. And at a later time-- so this is a later time. I shouldn't make it bigger. At a later time, this thing has moved to the other turning point. Back and forth, back and forth. Now, since we have three energy levels-- and well, actually, we have a coherence term which involves the product of psi 0 and psi 1 and psi 1 and psi 2-- they differ in energy both by omega. So these two things have the same oscillation frequency. And so what's going to happen is the wave function, the wave packet, is going to move from this side to that side, back and forth always forever at the same frequency, no dephasing. In the middle, I can't tell you what it's going to look like. I don't want to tell you. Because you don't care. You care mostly about what's it going to look like at a turning point. Or what is it going to look like when it returns to home base? And all sorts of insights come from that. OK. Now, I said, well, if we want to know the energy of this wave packet, well, we take the expectation value of the Hamiltonian. And the expectation of the Hamiltonian is c0 squared e0 plus c1 squared e1 plus c2 squared e2. These things are easy once you've gone through it a couple of times. Because what's happening is you have these factors, e to the minus i ej t over H bar. And they cancel when you do a psi star psi. Or they generate a difference, omega H bar omega, when you have different values of the energy. So this is something that you can derive really quickly. And the fact that your eigenfunction-- the psi's in your linear combination up there are eigenfunctions of the Hamiltonian. So every time the Hamiltonian operates in a wave function, it gives the energy times that wave function. And then you're taking the expectation value. And so we have the wave function time itself integrated. And that goes away. So after doing this a few times, you don't need to write the intermediate steps. And you shouldn't. Because you'll get lost in the forest of notation. Because one of the things you probably noticed in the last lecture is the equations got really big. And then we calculate something else. And it gets twice as big. And then all of a sudden, it all goes away. And that's what you want to be able to anticipate. OK. So you can really kill this half oscillator problem. There's nothing much happening except you create a localization. And when you take away the other-- when you restore the full oscillator, everything oscillates at omega. And so you have a whole bunch of terms contributing to the motion of the wave packet. And they're all very simple. Because they're all oscillating at the same frequency. Now, as an aside, I want to say there's a huge number of stuff that's up in this lecture that's going to be on the exam, a huge amount. So if for example you wanted to calculate something like x squared, well, fine. You know what the selection rule for x squared is. It's delta v of 0 plus or minus 2. And so this thing is going to generate some constant terms and some terms at 2 omega. OK. So there are a lot of things about the harmonic oscillator that make it really wonderful to consider a problem, even a complicated problem, which is not explicitly a harmonic oscillator problem. Because you can get everything so quickly without any thought after a little bit of investment. OK. Now, we haven't talked about electronic transitions and potential energy curves. But we will. And I think you know about them. And so we have some electronic ground state. And we have some electronically excited state. And so each of these states has a vibrational coordinate. And we can pretend that it's harmonic even if it's not. Because we build a framework treating them as harmonic, and then discover that there's discrepancies which we can fit to a model. And we can determine from the time dependence what that model is. OK. So we start out with a molecule in v equals 0. And there is a much repeated truism. Electrons move fast, nuclei slow. Transition is instantaneous, or nearly instantaneous, because it involves the electrons. So what ends up happening is you draw a vertical line. And now, you transfer this wave function to the upper state. You just move-- this is the probability amplitude distribution of the vibration. We transfer that to the excited state. And that's not an eigenstate, the excited state. It's a localized state, localized at a turning point. Now, you want to know-- so there's the Franck-Condon principle, that is just another way of saying electrons move fast, nuclei slow. And there is delta x equals 0 delta p equals 0. If the nuclear state can't change while the electron is jumping, then the coordinate and the momentum are both constant. So you're creating a wave packet. And the best place to create it is near a turning point. Because then, you can match the momentum of this guy to that zero point momentum, which you can calculate. You know how to calculate this. Because the potential energy curve is 1/2 k x squared. And the momentum is given by this energy difference here. And so you can work it out. It's in the notes. I don't want to write it down. So you're going to create something which is vertical and is not quite exactly at the turning point. Because you have to have a little bit of momentum to match the zero point momentum here. So you know everything about the initial state. And so you can calculate what it is by taking the overlap of v equals 0 of the ground state with all of the vibrational levels of the excited state. And so we have the coefficients of each of those vibrational levels in the excited state that makes this wave packet. Now, of course, this wave packet is going to move back and forth, back and forth. And if this were a harmonic oscillator, it would move harmonically. And so the only thing that would appear in the expectation value of x is going to be this motion at omega. Now, this is usually a relatively high vibrational level. And the molecule is not being harmonic here. So there are correction terms called anharmonicity terms. So we have the energy level expression plus H bar omega e xe-- that's one number-- plus 1/2 squared. So we have a linear term and a quadratic term. And this omega e xe is on the order of 0.02 times omega e, 2%. So it's a small thing. But if you're going to be allowing a wave packet to be built out of many vibrational levels, this guy is going to de-phase a little bit. So you go around and come back. And you can't quite have everybody back where they started. And so you'll see a decreasing amplitude here. And that's best looked at by the survival probability. And you'll see characteristic behavior in the survival probably. OK. But let's go a little bit deeper before I move on to-- what time is it? Oh, I'm doing OK. So let's just say our superposition state, the electronically excited state, is a combination of v equals 10 and v equals 11. So we know immediately that psi star t psi t-- this probability amplitude-- probability, yes, this probability is c10 squared psi 10 squared plus c11 squared psi 11 squared plus 2 c10 c11 psi 10 psi 11 cosine omega t. That's not very legible. Now, I'm playing fast and loose here. Because I've done this before. I know what disappears. And I know, if it's harmonic, you just get this. Well, if there's two states, this thing is really just the frequency associated with these two levels. OK. Now, once we have this, we can also generate the survival probability. Well, the survival probability, capital P of t, that's defined as the square modulus of psi star xt psi x0 dx. OK. And you can immediately write what that is going to be. It's going to be c10 squared. And we've integrated so the wave functions go away. And so we have c10 e to the i H bar 10.5 omega t divided by H bar plus c11 squared e to the minus i plus H bar 11.5 omega e over H bar. OK. Why did I made the mistake here? Well, we have a psi star. And so we're going to get the complex conjugate of e to the minus i omega. So you end up getting this. Practice that. AUDIENCE: That whole thing is [? the ?] modulus squared, right? ROBERT FIELD: Yes. I have that right. Exactly. So and now, lo and behold, we know what to do with this too. And so we're going to get c1 0 to the 4 plus c-- not 1-0, 10, c11 to the 4 plus 2 c10 squared c11 squared cosine omega t. Isn't that neat? So we've got a whole bunch of constant terms. This is constant. Because it's square modulus and it's to the fourth power. And so all of these coefficients are positive. At t equals 0, this is 1. And so at t equals 0, the survival probability is at a maximum. And at some later time, that survival probability will be at a minimum. OK. And so you can say the maximum will occur at integer when omega t is equal to 2 n pi. So then the exponential factor is always 1. And we have a minimum when omega t is an odd multiple of pi. And all the exponential factors are minus 1. So we can also now look at the expectation value for x of t and p of t. And I'm going to just draw sketches. So for x of t, we have something that looks like this. This is at the left turning point. This is the right turning point, or near the right turning point. And this is at pi over 2 omega. This is at pi over omega. So that's the half oscillator point. Now, it starts out and the expectation value is at the left turning point. And it's not changing. The derivative of the expectation value with respect to t is 0. The momentum, which has a different phase-- the momentum starts out at 0 also. And at pi over 2 omega, it reaches a maximum. And at pi over omega, it reaches 9 again. But the important thing here is at t equals 0 the derivative of the momentum is as big as it can get. So what's that telling you? It's telling you this wave packet, as far as coordinate space is concerned, it's not moving at t equals 0. As far as momentum is concerned, it's moving like crazy. And so this survival probability is changing entirely dominated by the change in momentum, which is encoded in the wave function, which is neat. Because Newton's equation says that the time derivative of the momentum is equal to-- that's the mass times acceleration-- is equal to the force, which is minus the gradient of the potential-- the function of time. And for a harmonic oscillator, the gradient potential is x kx. And so it's telling you that we have this relationship between x and t and p. And we know what the momentum is. And I'm sorry. And so the change in the momentum, which is responsible for the change in the survival probability at t equals 0, is due to the gradient of the potential. So we're actually sampling what is the gradient of the potential at the left turning point. And often, for any kind of a problem, we want to know what kind of force is acting on the wave packet. There it is, classic mechanics embedded in quantum mechanics. It's really amazing. So if you have some way of measuring the survival probability near t equals 0, it's telling you what the slope of the potential is at the turning point. OK. Well, if we have an initial state which involves many eigenstates, the Hamiltonian, we still know that there's going to be some complicated behavior which is modulated by omega. So we have a cosine omega t always. And so no matter what kind of a wave function we make, if we start out at one turning point, it's going to go to the other turning point. And it'll keep coming back, and back, and back. And so one could imagine doing an experiment where-- let's just draw the excited state and some other repulsive state. So at this turning point, vertical transition to that repulsive state is within the range of your laser. And at this turning point, it's way high. And so what you can imagine doing is probing where this wave packet is as a function of time by having a probe pulse which creates dissociating fragments. And now, that is what Ahmed Zewail did. He talked about real dynamics in real time. So when the wave packet is here, it can't be dissociated. When it's here, it can. And you look at the fragments. Very simple. So there are lots of ways of taking these simple pictures of wave packets moving and saying, OK, I can use them to set up an experiment where I ask a very specific question. And I get answers. And I know what to do with the answers. Now, if it's not harmonic, you can still do Zewail's experiment or some other experiment. And you can ask, OK, the wave packet moved over here. And there's some dynamics that occurs. It dissociates or you do something. And when it comes back, it's not at the same amplitude. And so you can use the time history of these recurrences if it's harmonic. If it's not harmonic, then the recurrences-- even if nothing happens over here, the recurrences will be increasingly less perfect. And so you measure the anharmonicity. OK. Five minutes. Tunneling is a quantum mechanical phenomenon. And again, you want to use the simplest possible picture to understand the signature of tunneling. And so what do you do? Well, the simplest thing you can do is start with a harmonic oscillator. The next thing you do is you put a barrier in the middle. And you make it really thin. Thin is because you don't want to calculate the integral. You want to just say, oh yeah, I know that the wave function-- forget about the barrier. We know the magnitude of the wave function at the barrier, OK? And so if you have a state which has a maximum here, well, that means that it's feeling the barrier. And so you know what to do. If there is a node here, it doesn't know about the barrier. So you can do all sorts of things really fast. OK. So and it's harmonic because you're going to want to use a's and a-dagger's. So v equals 1, 3, 5. They have nodes here. They don't know about the barrier. Or if they do, it's a very modest effect. If you make this thin enough, you won't know about it all. And then v equals 0, 2, 2 4, et cetera-- well, they're maxima. Well, they have a local maximum here. And remember, this is a harmonic oscillator. So the big lobes are on the extremes. The smallest lobe is in the middle, but it's not 0. OK. There are also symmetry restrictions. This is a problem that has symmetry. We have even functions and odd functions. And for the even functions, d psi dx is equal to 0. And for the odd functions, psi of 0 is equal to 0. OK. So this is something that-- you choose the simplest problem. And you use symmetry. And bang, all of a sudden, you get fantastic insights. So let's look at what happens to the even functions. So let's just draw a picture of the harmonic oscillator energy levels. And we have a barrier, which we say maybe goes that high. And so we have 1. And we have 3. And we have 5. They're basically not affected by the barrier. Because they have a node at the barrier. And then we have v equals 0. What happens to v equals 0? Well, it hits this barrier. And it's got a large amplitude. And it can't accrue more phase as you go through the barrier. So when it hits the other turning point, it doesn't satisfy the boundary conditions. It'll go to infinity at x equals positive infinity. So it has to be shifted so that the boundary conditions at the turning points are met. And the only way that can happen is it gets shifted up in energy a lot. So v equals 0 is here. Now, it can't be shifted above v equals 1. Because that would violate the node rule. And if you draw a picture of the wave function for v equals 0 in the boundary region, the wave function has to look something-- well, it has to look something like that. So we have two lobes. And it's trying to make a node, but it can't. It's a decreasing exponential. I mean, if you have a wave going into this region of the barrier, you can have an increasing exponential or a decreasing exponential. You can never satisfy continuity of the wave function with the increasing exponential. But you can with the decreasing exponential. And so the height of the barrier determines how close this guy comes to having a node. It never will. And how does this compare to the wave function for v equals 1? Well, the wave function for v equals 1 looks like this. The amplitude in this lobe and the amplitude in that lobe are the same. But the sign is reversed. And so basically, this picture tells you that these two levels have almost exactly the same energy. So v equals 0 is shifted up. v equals 2 is shifted up, but not quite so much. And v 4 is shifted up hardly at all. And so you get what's called level staggering. And this level staggering is the signature of tunneling. This is how we know about tunneling, the only way we know about tunneling. So we measure the energy levels. And it tells us how are the wave functions sampling this barrier. Now, this is a childish barrier. And there is a real barrier-- molecules isomerize. And they isomerize over a barrier. And the barrier isn't necessarily at x equals 0. But if you understand this problem, you can deal with isomerization. Now, I'm an author of a paper that's just appearing in Science in the next few weeks on the isomerization from vinylidene to acetylene. There's a barrier involved. So these sorts of pictures are important for understanding those sorts of phenomena. Now, so I'm done really. So what we are now encountering with the time dependent Schrodinger equation is discovering how dynamics, like tunneling, is encoded in an eigenstate spectrum. Because the encoding is level staggering eigenstates. So people normally have this naive idea that eigenstate time independent Hamiltonian spectroscopy does not sample dynamics because it's only measuring energy levels. Real dynamical processes are time dependent. But the dynamics is encoded in energy level patterns. And that is actually my signature in experimental spectroscopy. I've looked for ways in which dynamics is encoded in eigenstate spectra. And chemists are interested in dynamics much more than in structure. So that's it for today. I will see you on Wednesday. |
MIT_561_Physical_Chemistry_Fall_2017 | 13_From_Hij_Integrals_to_H_Matrices_I.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: OK. So, today is the first of a pair of lectures taking us from the Schrodinger picture to the Heisenberg picture. Or the wave function picture to the matrix picture. Now almost everyone who does quantum mechanics is rooted in the Heisenberg picture. We use the Heisenberg picture because the structure of the problem is immediately evident when you write down what you know. And it's also mostly the way you program your computers to solve any problem. So I'm going to get you to the Heisenberg picture by dealing with the two-level problem, which is-- it should be called one of the exactly solved problems, except it's abstract as opposed to harmonic oscillator, particle in an infinite box, rigid rotor. It's a exactly-solved problem, and it leads us to-- or guides us to-- the new approach. I'm going to approach this problem the algebraic or Schrodinger way. And then I'm going to describe it in a matrix way and introduce the new language and the new notation. So I can say, at least at the end of this lecture, I'll be able to say the word matrix element and everybody will know what it is, and I'll stop talking about integrals. And one of the scary things is, when we go to the matrix picture, we stop looking at the wave function. We never think about the wave functions. And so there are all sorts of things like phases that we have to make sure we're not screwing up. Because we're playing fast and loose with symbols, and sometimes they can bite you if you don't know what you're dealing with. But mostly, we're going to get rid of wave functions, because we know all the solutions to standard problems, and we know a lot of integrals involving those solutions to standard problems. And so basically, all we need is an index saying this wave function or this state, this integral, and you have-- and then you can write down everything you need in matrix notation, and then you can tell your friendly computer, OK, solve the problem for me. OK. So once I give you the matrix picture, then I will talk about how you find the eigenvalues and eigenvectors of the matrix picture using a unitary transformation. And then I'll generalize from the two-level problem, which is exactly soluble. You don't need a computer for it, but it's convenient-- because you don't have to write anything down-- to N levels where N can be infinite. And the N-level problem is in principle difficult because even a computer can't diagonalize an infinite matrix. And all of your basis sets, all of your standard problems involve an infinite number of functions. And so I've introduced a notation and a concept of solving a matrix equation, but you can't do it unless you, say, let's make an approximation, and it's called nondegenerate perturbation theory. And this is the tool for learning almost everything you want to know about complicated problems. And it's not complicated to apply nondegenerate perturbation theory. It's just ugly. But it's really valuable because you don't have to remember how to solve a particular complicated differential equation, you just write down what you know and you do some simple stuff, and bang, you've got a solution to the problem. And this is really what I want you to come away from this course with-- the concept that anything that requires quantum mechanics you can solve using some form of perturbation theory. So hold onto your seats. So the Schrodinger picture is differential equations. And they're often coupled differential equations. And you don't want to go there, usually. And we're going to replace that with linear algebra. Now many of you have not had a course in linear algebra. And so that should make you say, can I do this? And the answer is, yeah, you can do this because the linear algebra you are going to need in this course is based on one concept, and that is that the solution of coupled linear homogeneous equations involves diagonalizing or solving a determinant. And when you solve this determinant of the equation, it's equivalent to diagonalizing a matrix. And that's the language we're going to be using all the time. And so the only thing I'm not going to do is prove this fundamental theorem of linear algebra, but the notation is easy to use and the tricks are very simple. So, we have with exactly-solved problems complete sets of the energy levels of the wave functions. And we know a lot of integrals. Psi i operator psi j d tau. We know these not by evaluating them, but because the functions are so simple that we can write these as simply a function of the initial and final quantum numbers. And so all of a sudden, we forget about the wave functions, because all the work has been done for us. We could do it too, but why? So we start with the two-level problem. And the two-level problem says we have two states, psi 1 and psi 2. Still in the Schrodinger picture. And this has an energy which we could we could call E1, and that would be H11, the diagonal integral of the Hamiltonian between the 1 function and the 1 function, and this is E2. Now these two states have an interaction. They're connected by an interaction term which causes them to repel each other equal and opposite amounts. And so this is H12. Again, an integral, which you usually don't have to evaluate, because it's basically done for you. And then we get E plus and E minus, and the corresponding eigenfunction. So this is the two-level problem, and it's expressed in terms of three parameters-- H11, H22, and H12. And we get the two energy levels and the two eigenfunctions. And we know how to solve this problem in the Schrodinger picture, and I'll do that first. And so, H11 is just this integral, psi 1 star H. And we put a hat on H for the time being-- 1 d tau. And H22 is a different integral of psi 2 star H psi 2 d tau. And H12 is psi 1 star H psi 2 tau. And we know that the Hamiltonian operator is Hermitian, and so we can write also that this is psi 2 star H star psi I d tau. Anyway, we call this thing V. And there's some subtleties. If the integral between these two things is imaginary, then the 1-2 and 2-1 integrals have opposite signs. Because the Hamiltonian is Hermitian. But let's just think of this as just one number. So we have E1, E2, and delta. So the two-level problem is going to let us find the eigenfunctions-- the plus and minus eigenfunctions. C plus 1. So we have eigenfunctions corresponding to-- which are eigenfunctions belonging to the eigenvalues, E plus and E minus. And there are a linear combination of the two states. Now, it's a two-level problem. This is a state space that contains only two states. It's not an approximation. And so completeness says we can write any function we want as a linear combination of the functions and the basis set. And so our job is going to be to find these coefficients and also to find the energy eigenvalues. OK. So let's start to do some work. We know this thing. And let's start some-- and we know that H psi plus-minus has to be equal to the eigenenergies psi plus-minus. Now, in order to get something useful out of this, we left multiply by psi 1. OK. And when you left multiply by psi 1-- well, let's just write it out. We're going to get psi 1 HC 1 plus. So let's just write this out. We have H11 C1 plus-minus. Because this is C plus-minus, and we have the integral of psi 1 with itself or the Hamiltonian is H11. And then we get another term. I'm doing this differently from my notes, so the other term is C plus-minus psi 2 V. OK. So plugging in what I have over here and knowing that H11-- we know that V is psi 1 H psi 2, and so this is what we get. We do the same thing, and we have the same left-hand side, but we can also now express this as that H-- that psi plus-minus is an eigenfunction. And so we have the same left-hand side, but on the right-hand side, we now get integral psi 1 star H psi plus-minus d tau. But that's equal to psi star E plus-minus C1 plus-minus psi 1 plus C2 plus-minus psi 2 d tau. OK. And so now we're talking about orthonormality of the wave function, and because we don't have any operators in here, then this thing becomes simply E plus-minus times C1 plus-minus plus 0 C2 plus-minus. There should be a 1 here. OK. Because psi 1, psi 2, that's orthogonal. So now we have an equation that's quite useful. So we're going to combine these two equations, and we get C1 plus H11 plus C2 plus-- I'm sorry, this is plus-minus, plus-minus. V is equal to E plus-minus C1 plus-minus. Now we need another equation. And so we do the same thing and we left multiply by psi 2 star and integrate. And we get another equation, and that is C1 plus-minus the plus C2 plus-minus H22 minus E plus-minus is equal to 0. So we combine the two equations that we have derived, solving-- because both equations can be solved for C1 plus-minus or C2 plus-minus. So we equate the C1 plus-minus over C2 plus-minus from the two equations. And we get this wonderful result, V over H11 minus E plus-minus is equal to H22 minus E plus-minus over V. Now you're not going to ever do this. So yes, you can attempt to reconstruct this from what I read on the board or what's in your notes, but the important point is that we're just using what we know. We say we have eigenfunctions, and we want to find something about the coefficients of the basis functions in each of the eigenfunctions. And we want to find the eigenvalues, and we get this equation. So, this is easy. We relate this to-- we just multiply through and we get V2, V squared is equal to H11 minus E plus-minus times H22 minus E plus-minus. Or we solve, and we know that E plus-minus is equal-- this is a quadratic equation. A quadratic and E plus-minus, and so we have this result, H11 plus H22 plus or minus this H11 squared-- H11 plus H22 squared minus 4 H11 H22 minus V squared over 2. This is just a quadratic equation. So we have the eigenenergies expressed in terms of the quantities we know. We simplify the notation. E bar is H11 plus H22 over 2. And delta is H11 minus H22 over 2. So when we do that, we simplify the algebra and we end up discovering that the eigenvalue equation is E plus-minus is equal to E bar plus or minus delta squared plus V squared square root. That's a simple result. This is something you should remember. So if you have a two-level problem, the energy eigenvalues are the average energy plus or minus this quantity. This is an exact solution for a two-level problem. For all two-level problems. OK. And we even simplify the notation more. We call X delta squared plus V squared. And so this becomes E bar plus-minus X squared. Pretty compact. Now, the reason for simplifying the notation is that we're going to derive the values for the eigenfunctions, and they involve a lot of symbols. And we want to make this as compact as possible, and so we'll do that. OK. So when we started out, it looked like we were going to solve for these quantities, but we took a detour and solved for the eigenenergies first. And this is one of the things that happens in linear algebra. You get something you can get easily-- quickly before you get the other stuff that you want. OK, but the second part of the job is to find these mixing coefficients. And so one thing you do is you say, well, we insist that the wave functions be orthog-- normalized. After that a lot of algebra ensues. And I'm not going to even attempt to work through it. You may want to work through it, but what I recommend doing is looking at the solution and then checking to see whether it does things that you expect it has to do. Because one of the things that kind of inspection leads you to is factors of two errors and sign errors. OK. But, C plus-minus-- C1 plus-minus is equal to 1/2 times 1 plus or minus delta over the square root of X. Two square roots. Whoops. OK, no square root in here because we get square root there. OK and C2 plus-minus is 1/2 1 minus or plus delta square root, x square root. OK? So these two things are kind of simple-looking, but there's an awful lot of compression and select and clever manipulation in order to get this. But the important thing to notice is that we have the energy difference divided by this thing, this delta squared plus V squared that expresses the importance of the two-level interaction. And these two things enter with opposite signs. OK. And so one of the checks that's really easy to do is let's let V go to 0, and let's let V go to infinity. OK, remember, our Hamiltonian-- the two-level problem-- I can't write it as a matrix yet, so remember that we had E1, E2, and V. And this was the higher energy, this was the lower energy, and so delta is E1 minus E2-- I better not write that. OK. So we normally write E1 over E2, and we have this V interaction between them. OK. And so if the interaction integral between the two functions is 0, then E1 is the higher energy, and it corresponds to-- it should correspond to psi 1 alone. Well, so the higher energy we're taking the plus combinations here, and if V is equal to 0, then this is delta over delta. And this is either 2 or 0. And when it's the higher-- the upper sine, it's 2 divided by 2 or 1, exactly what you expect. And this one is 1 minus 1. So these two are behaving, right? And the V to 0 rate. Now, in the V going to infinity, then this is infinite. So that just means that this term goes away, right? And so we have C1 plus and minus is equal to square root of 2-- 1 over the square root of 2. And same here. So what we get is what we call 50/50 mixing. OK, so this also works. So I don't say that this can-- this confirms that I have not made an algebraic mistake, I haven't, but it's a good test because it's really easy to do. And if you're not getting what you expect, you know you've either blown a sign or a factor or two, which are the two things that you fear most in a wave function free approach. Because you've got nothing to do except check your algebra, and usually, you made the mistake because it was subtle, and you'll make it again when you're checking it. So these kinds of checks are really valuable. OK. So once you know that this is likely to be correct, then you do some other things, and you check the wave functions you've derived with your C plus-minus 1 and C plus-minus two are both normalized and orthogonal. And that the energy you get is such that psi plus is E plus. And you know what E plus was. And so you just go in and you calculate what the energy should be giving using the values for PSI plus-minus And so we plug those into the original equations. And you can also, again, show that psi plus-minus a star H psi plus-minus is equal to 0 because the eigenfunctions are orthogonal. We already did that, but we plugged it into an equation here, and we got it a second time. So in the lecture notes, there is a lengthy algebraic proof or demonstration that E plus-minus is equal to plus or minus square of x, which are already derived, but then I just did it the long way. OK . So this is it-- we are about to move from the Schrodinger picture to the matrix picture. So the trick now is to go to linear algebra, go to the matrix picture and learn how to just write the equations and what the language is and show that it works. So suppose you have two square matrices, A and B, they're n by n squared matrices, you know the rules for matrix multiplication? So if you wanted, the mn element of this product or two matrices, you would go, n equals-- i equals 1. Let's use the same notation. j equals 1 to n of Am-- because I want the mn-- j Bjn. And so when you multiply two square matrices, you get back a square matrix. And this picture is not a bad one. So you can say, all right. So if you need a little q to remind yourself, you take this row and multiply term by term and add the results-- this row in that column, and you get a number here. And you just repeat that, and it's really easy to tell your computer to do this, and it's rather tedious to do it yourself if it's more than a two-by-two matrix. OK. So what about this thing C? Well, C is a N row column matrix. So it's N rows, 1 column. It looks like this-- C1, C2, Cn. And so if you want to multiply a square matrix by a vector, we know the rules too, OK? And again, this picture is a useful one. So let's just draw something like this. We do this and this, and that gives you one element in the column. OK. Now, the last thing that I want to remind you of-- I better use this board. If we have a two-state problem, then the vector c1 is 1, 0; the vector c2-- this should be a lower case c, because we tend to use lowercase letters for vectors and uppercase letters for matrices. And so if we do c1 dagger c2, or c1 dagger c1. Now this dagger means conjugate transpose, except, well, there isn't anything to do except convert a matrix-- a vector into-- so that becomes a row and this becomes a column. And so a row times the column gives a number. And that number is going to be-- well, let's do it. We have 1, 0; 0, 1. 1 times 0 is 0, 0 times 1 is 0, and so this is 0. Well that's orthogonality. This, we have 1 times 1 is 1. 0 times 0 is 0. So it's 1 plus 0. Right. OK, all right. OK. So the Schrodinger equation becomes in matrix language-- and now, one notation is-- I'm going to stop using the double underline. That means boldface. And we don't use hats anymore. Or at least if we were really consistent, when we go away from the Schrodinger picture, we don't put hats on operators, we make them boldface letters. OK. Now, the thing is we're so comfortable in matrix land, that we don't use either. OK. But remember, we're talking about different things. So the Schrodinger equation looks like this. And so we have a matrix, delta v, v delta times 1, 0. And that's delta, v; or delta times 1, 0 plus v times 0, 1. Isn't that interesting? Remember, the Hamiltonian or any operator operating on a function gives rise to a linear combination of the functions and the basis set. And so here's one of the functions, here's the other. OK. Nothing very mysterious has happened here. So this very equation is going to be HC is equal to EC. OK, so how do we approach this? Well I have to introduce a new symbol, and that's going to be this symbol T. It's a matrix. It's a unitary matrix. And we want it to have a special properties. Those special properties will be shown here. OK, so first of all, we have this matrix-- T11, T-- AUDIENCE: Your Hamiltonian matrix is incorrect. So I think you mean to say E bar plus-minus? ROBERT FIELD: I'm sorry? AUDIENCE: I think the diagonal elements of the Hamiltonian matrix should be-- ROBERT FIELD: Yeah, OK. If we wanted the eigenvalues, OK? This is-- AUDIENCE: So the infinite matrix in this case should be-- the diagonal elements should be E1 and E2. So I think you meant E bar plus delta, and then E bar minus delta. ROBERT FIELD: Yeah. OK. There is something in the notes and something in my notes which-- we can always write the Hamiltonian as E bar 0, 0, E bar plus delta v v minus delta. We can always take out this constant term. And this is the thing we're always working on. And so rather than-- and so we could call this H prime. Or we can simply say, oh, we have these two things-- this always gets added in, and it's not affected. If we take this diagonal constant matrix and apply a transformation to it, a unitary transformation, you get that matrix again, you get-- it's nothing. OK, so we wanted to describe some special matrix where the transpose of that matrix is equal to the inverse of that matrix. Now-- yes? AUDIENCE: So shouldn't there be a minus on there? On the prior equation? For the bottom right entry. The primary delta and the-- first line. First line. ROBERT FIELD: This is delta minus the-- AUDIENCE: Yes. But up there, you just had delta delta. ROBERT FIELD: Yep. Thank you. OK. Now, when you have a matrix, you really like to have T minus 1T is equal to the unit matrix. Getting the inverse of a matrix in a general problem is really awful. But for a unitary matrices, all you do to get the inverse is to take-- you flip it on the diagonal. Now strictly, the conjugate transpose of a unitary matrix is the inverse. But often we have real matrices. But the important thing is always this flipping along-- flipping the matrix on the diagonal. And that gives you the inverse unless there is stuff here that is complex. OK. So this conjugate transpose, it would look like-- OK? OK. Now, what we want to do is derive the matrix form of the Schrodinger equation using this unitary transformation. So we start out again with HC is equal to EC. And now, we insert TT dagger. And this is one of the things where you screw up. Whether you insert TT dagger or T dagger T, because they're both 1. And if you use the wrong one, all of your phases, everything is wrong. But it's still correct, the equations are correct, but the things you've memorized are no longer correct. OK, so we're going to insert this unit matrix between H and C. And of course, we have to-- we would have to insert this on the other side. T T dagger C and E. But that's one. OK, so we don't do anything on the left-hand side. And now we left multiply by T dagger. OK. And we call this H twiddle, and we call this C twiddle. So we have H twiddle, C twiddle is equal to EC twiddle. Now we're cooking. OK, because we construct this unitary matrix to cause this Hamiltonian, the transformed Hamiltonian to be in diagonal form. So we say that H twiddle, which is T dagger T, is equal to E plus, E minus, 0, 0 for the 2 by 2 problem. OK, that leads to some requirements. What are the elements of the T matrix? But the important thing is we say that T diagonalizes H. So this magic matrix gives you the energy eigenvalues. But we also have T dagger C is equal to C twiddle. And so this gives you the linear combination of the column vectors that correspond to the eigenvector. So what is it? We have here T dagger, which is TT, T12 T, T21 T, T22 T times C. And that's supposed to be equal to C twiddle or C1 twiddle, C2 twiddle. So we put everything together and we discover that OK, when we multiply this by that, we get the element that goes up here. And so the element on top is C1 T11 dagger plus C2 T12-- I'm using T's and daggers independently, and we do the same thing here. So we have a column vector, and it's composed of the original state. And so anyway, when we do everything, we discover that using the solution to the two-level problem from the Schrodinger picture, we know everything. So H twiddle, C twiddle plus is H plus C twiddle plus. And that's just E plus 1, 0. And H twiddle C twiddle minus is E minus times 0, 1. OK. This is all very confusing because the notation is unfamiliar. But the important thing is that everything you can do in the Schrodinger picture you can do in this matrix picture. And you can find these elements of this T matrix, and it turns out that what you-- the matrix that diagonalizes the Hamiltonian, the columns of T dagger are the eigenvectors. So you find this matrix, it diagonalizes H. The computer tells you the eigenenergies, and it also tells you T dagger or T plus-- TT. And so with that, you know how to write in the original basis set what the eigenbasis is for each eigenvalue. And so the next step, which will happen in the next lecture, is the general unitary transformation for two-level. Now, you expect that there is going to be a general solution for the two-level problem, because the two-level problem in the Schrodinger picture led to a quadratic equation. And that had an analytical solution. And so there is going to be a general and exact solution the two-level problem. And this unitary transformation is going to be written in terms of cosine theta sine theta minus sine theta cosine theta. This is a matrix which is unitary, and it, when applied to your basis set, conserves normalization and orthogonality. And so the trick is to be able to find the theta that causes the Hamiltonian in question to be diagonalized. And the algebra for that will happen in the next lecture. OK. Remember, you're not going to do this ever. You're going to use the idea of this unitary transformation, and you're going to use that to get-- to diagonalize the matrix. And this will lead to some formulas which you're going to like, because you can forget sines and cosines. Everything is going to be expressed in terms of things like this-- V over delta. So we have matrix element-- off diagonal matrix element over the energy difference. And that's the basic form of nondegenerate perturbation theory, which is applied not just in 2-by-2 problems, but to all problems. And so you want to remember this lecture as, this is how we kill the two-level problem. Then we can discover what we did and apply it to a general problem where it's not a two-level problem anymore, it's an infinite number of levels. And when certain approximations are met, it applies, and it gives you the most accurate energy levels and wave functions you could want. And you know how accurate they are going to be. And this is liberating, because now, you can take a not exactly-solved problem and you can solve it approximately. And you can use the solution to determine, OK, there's going to be some function of the quantum numbers which describes the energy levels. Well what is that function? And what are the coefficients in that function? And how do they relate to the things you know from the Hamiltonian? So it's incredibly powerful. And once you're given the formulas for nondegenerate perturbation theory, you can solve practically any problem in quantum mechanics. Not just numerically, but with insight. It tells you, if we make observations of some system, we determine a set of energy levels, which is called the spectrum of that operator. And the spectrum of the operator is an explicit function of the physical constants-- the unique interactions between states. And so you will then know how the experimental data determines the mechanism of all the interactions, and you can calculate the wave function. You can't observe the wave function, but you can discover its traces in the energy levels. And then you can reproduce the energy levels, and if you have the energy levels-- the eigenstates, you can also describe any dynamics using the same formulas in here. So this is an incredible enablement. And you don't have to look at my derivations. I'm not proud of the derivations, I'm proud of the results. And if you can handle these results that come from perturbation theory as well as you've done so far in the course, you're going to be at the research level very soon. OK, see you on Friday. |
MIT_561_Physical_Chemistry_Fall_2017 | 27_NonDegenerate_Perturbation_Theory_III.txt | ANNOUNCER: The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: When I was really young, I used to go to a television repair store as often as possible to take home one of the dead chassis. And then I would take it apart. And I don't know what I was looking for, but that was sort of the empirical stuff. I had no idea how a television worked, but I was really curious about maybe I could find it if I just did stuff. And what we've been talking about are ways in which we not just generate numbers, like parts of the television chassis, but insight. And there are we've talked about three ways so far. And one is Huckel theory, where Huckel theory is just a bunch of simple rules and simple ideas for how do you represent a large family of related molecules. And the Huckel theory is incredibly simple, but it enables you to make really sophisticated conclusions about how things work and what are the important factors. And so it's basically a procedure for distilling insight from random observations. Or maybe not random observations, but observations of many properties of many related molecules. Then you've seen LCAOMO. Based on the small variations all treatment of H2 plus, we got the idea of orbitals and what makes a bond. And then with the idea that we can describe the size of atomic orbitals by the energy below the ionization limit, we can make similar quantitative predictions based on this minimal basis set. So we draw molecular orbital diagrams. And these molecular orbital diagrams are especially valuable for isoelectronic and homologous comparisons. So isoelectronic would be, let's say, nitrogen, CO, BF. Molecules with the same number of electrons. And homologous would be something like CO, SIO, GEO, and so on. And with the concept of orbital size being related to the ionization energy of the atoms, we can make a lot of very useful comparisons. And so we develop insight. Now LCAOMO is a variation of method, and it can be a very large variation of method. And it could become of an issue. But it's usually not atomic orbitals. It's just Gaussian orbitals or something, which is computationally convenient. And so it can be a procedure to rationalize experience and make predictions as you go from the same number of electrons but increasing polarizability, polarity. And these-- this is where you develop chemical insight. And it's really exciting. And it's what was missing when I would take these chassis from the repair shop and chop them up into pieces. And oh yeah, there's a magnet in there and things like that. It's much deeper. And now perturbation theory is a kind of a different thing altogether. Perturbation theory says we do an experiment. We measure something, and we turn it into something we really wanted. The Rolling Stones are telling us that if you want to know how a particular property depends on internuclear distance, and you can't directly measure the property as a function of internuclear distance, well maybe you can measure it as a function of vibrational and rotational quantum numbers. And that's what perturbation theory does. It tells you how to get from what you observe to what you really want to know. And it can be horrible in terms of the algebraic exercises you have to go through in order to get what you really want or from what you-- yes. And today's lecture is going to be an example of not the worst thing you could ever do with perturbation theory, but pretty close to the worst. OK, but all three of the first things are related, are associated with getting insight from either a crude calculation refined against observations or just the observations and reducing those observations to something really neat about how things work. Then the next two lectures will be given by Professor Van Van Voorhis and it's-- they will be on ab initio theory, where basically you don't assume anything. You don't do anything except solve for the exact energy levels and wave functions. Now this is not possible. Directly you can do this by making approximations, doing an enormous variational calculation. Now many people think, well why bother with approximate methods when you can get the truth? And the answer is the truth is no more valuable than the parts of a disabled television. It's-- they're stuff. But there's no insight there. And quantum chemists who do these calculations are not just generating numbers. They're trying to explain how things work. And it's the same business, it's just the tools are different. The goals are not just getting-- in spectroscopy we can measure things to 10 or 11 digits. And nobody cares. I mean, how many digits? It's a little bit challenging to remember telephone numbers. And having big tables of 10-digit numbers, so what? But what do the numbers tell you? And this quest-- the same question is asked by good quantum chemists by doing a series of calculations where they turn on and off certain terms in the Hamiltonian. So it's the same thing. I'm an experimentalist, but many people who are experimentalists think I'm a theorist because I do weird stuff. But I'm not a theorist. Troy is a theorist. And he does weird stuff too, but he's not an experimentalist. And we're both after insight. And OK, so let's just-- I guess I'll just launch it through the perturbation theory. OK. But you do want to understand the difference in how these different methods, these approximate methods, work, and what they're good for, OK? So the goal of perturbation theory is to go from molecular constants to structural constants, or structural parameters. Molecular constants are like rotational constant, vibrational constant, stuff that you get by fitting the energy levels you observe to a dumb empirical expression, a power series in quantum numbers. Now there's lots of dumb empirical expressions you could use, and some are better than others. And perturbation theory will often tell you what is the right way to do it. So you go from molecular constants to things in the potential. So this is the displacement from equilibrium, and we would like to know something about how a molecule works. And there also might be other constants, like spin orbit, and hyperfine, dipole moment. And so let's just say we have some observable, and it is also a function of coordinate. And we'd like to know what that function is. But what we are able to do is measure energy levels as a function of quantum numbers. And so the information that we really want, the potential or the internuclear distance dependence of some electronic property, that's all gotten from what we can observe via perturbation theory. It's a very powerful tool. And it's not pretty, but it always works. And it's a good basis for insight. So perturbation theory is a fit model. The other methods, one doesn't generally do a least-squares fit to the-- of the adjustable parameters in Huckel theory to determine the properties of a molecule. One just says, OK, I'm going to try these because I think that attaching an electro-negative atom to a carbon atom is going to do something that I can predict. And maybe it's going to tell me some surprises about how its influence is not just where it's attached, but other places in the molecule. You know this from your first organic courses. You know all sorts of tricks to be able to predict reactivity and things that are related. But Huckel theory is not a fit model. LCAOMO theory is not a fit model. These are experience-based. And you integrate all sorts of stuff that you learn from comparing molecules. And the comparing of molecules is what makes up for the deteriable lists of the approximations. And your job is to hone your insight. And so you have nothing to protect you except the truth. You can observe molecules, but you can't be guided to something which is true, because there is no calculation. There's nothing exact and there's nothing complete. But in quantum chemistry you can get really close to the truth. But you don't know anything about why. And so you're doing the same thing, but from the opposite ends. OK, but perturbation theory is special because you take a huge amount of highly accurate experimental data of various types and you fit it. And so I want to talk about this. So you have the Hamiltonian. It's a matrix, and it's expressed in terms of parameters. Let's just use notation p, p of i. A set of parameters. So this is not a quantum chemical, Hamiltonian. It's a thing where we say there are certain degrees of freedom, and each one is controlled by a number of parameters. And we have the energy levels. And so what you do is you say let's choose a set of these parameters, and calculate the energy levels, and compare them to the observed energy levels. And it's not-- the first try, it's not going to be good. And then you say, OK, now I've got to do a least-squares fit. I have to vary the parameters in this matrix Hamiltonian to match the energy levels and maybe to match other things that you can observe in the experiment. But it's a complicated least-squares fit, because you have a matrix, and you want to diagonalize it. How are the parameters related to the eigenvalues and eigenvectors? You don't know. But when you're close, then you can start fitting these things. Then you can say, yeah, I know. And then once you've done that, you have not just the energy levels, but you have the wave functions. Now again, you don't know that there is something missing in your Hamiltonian. But you know that from the data you input you can match all the energy levels. But you didn't put it in all the levels. You put in just the ones that you measured. But you are arrogant. And you think, well, maybe I measured more than I-- I determined more than I measured, that I can extend the measurements to other things. And I told you at the beginning of the course you cannot observe the wave function. But if you do this, if you do a least-squares fit and match the energy levels, you have a pretty darn good representation of the wave function. And you can use that to calculate other stuff, especially dynamics. Remember dynamics, at least if the Hamiltonian is independent of time, the dynamics is going to be a function of x and t. But you start with the initial preparation. If you know that, then you, because you know the energy levels, you can calculate the full x of t for all time. So that's pretty powerful. If you fitted enough stuff, if your Hamiltonian has the important things in it, then you are basically able to do anything you want, whether it's static or dynamic. And if the Hamiltonian is time dependent, you can do that too, I just haven't showed you how yet. OK, so I really like perturbation theory, because it directly that deals with whatever data you have. And out from that, you get this thing which tells you everything, unless there's something that you didn't sample, that you didn't know you didn't sample. And then you discover that your data and the predictions are not in agreement. And then that's when you-- you don't go home and say, oh, I screwed up. I got to go take a nap. I don't-- you go home and you say, let's celebrate, because I discovered something that was really missing. And that's what we want to do. OK, so I have to mess around in the mud here. Because the perturbation theory you've done so far has been, perhaps, ugly, but it's been kind of simple because it's basically vibration. And now we've got vibration and rotation. And we have to combine the two, and we have to see whether there's something special that we can get from the combination. And you bet there is. OK, so we're going to look at the energy levels of a non-rigid, non-harmonic, or anharmonic oscillator. And we're going to find out how we generate from an expression of the potential energy surface or potential energy curve, because what I'm going to be talking about is diatomics. But it all extends to polyatomics. And we're going to calculate the relationships between what you observed and what you want to know. So for a diatonic molecule, we have a dumb representation of the quantum numbers. And we can write it. And I'm just really a stubborn person about spectroscopic notation, so I include this Hc convert wave number unit, the reciprocal centimeter units to energy. But you'll never catch me doing it in real life. But that leads to all sorts of algebraic errors that I don't know I'm making because I'm so unused to doing this. OK, so we have an expression for the vibrational energy levels. And I can't apolo-- I can't explain why we use four letters to represent one symbol, but that's the traditional thing. And for polyatomic molecules, which came after diatomic molecules, we start using only two letters. Instead of omega XE, we just use XE. But since I'm a diatomician, I'm going to do this sort of thing. And that's v plus 1/2 squared. And then the next term is omega YE v plus a 1/2 cubed. OK, and so then that's the vibrational part. And then we have the rotational part. The equilibrium internuclear distant rotational constant minus alpha e times v plus 1/2. Times JJ plus 1. And then minus DE, the centrifugal distortion constant, times JJ plus 1 quantity squared. And we can have more constants here. We can have additional constants there. But basically we have constants, which are known by multiplying a particular combination of vibrational and rotational constants, quantum numbers. And so these are the things we can determine, or sometimes we determine some of these and we want to get a prediction of the others because they're outside of our observation. And the potential is going to be dependent on the displacement from equilibrium and the rotational constant. And so there will be terms in the potential, like 1/2 KQ squared plus 1/6 AQ cubed, et cetera. So we have another dumb power series. And fortunately, we know that as we go up in the displace-- in this displacement coordinate, the coefficients get to be really small. And so we don't need to have a lot of these parameters. But we should have at least two, the harmonic and the cubic one. And we know for molecules the cubic parameter gives you a potential curve, which resembles reality. It's hard wall here, soft wall there. Molecules break. Now this has the unfortunate property of doing that. And so it's got death built into it. But it doesn't really matter, because you basically are looking at increasingly wide regions of the potential. And the fact that it does something terrible in your simple representation is almost irrelevant. But you have to be aware that you can tunnel through barriers. And when you start being able to tunnel through barriers, and it depends on how close you are to the top of the barrier and how wide it is, then you start seeing effects. And then this term is going to give nonsense. But there is a domain of goodness where you don't have to worry about that. So we have this, and then we have the rotational constant kind of contributing to the effect of potential, which is HCV of R JJ plus 1. Now here I've got-- here I have coordinates. Here I have quantum numbers. How can I do that? It's because central force problems can be represented in a universal form. And so we can just integrate over the angular part of the problem. And so now we have a mixed representation. But this guy is still a function of R or Q. And so this v of R is an operator. It's not just a constant. Some people call it the rotational constant operator. Kind of stupid. And so our job is now to figure out how to get from this parameter rise potential to the energy levels. Of course, we start with the energy levels. And we're going to want to determine the parameters in the potential. But you've got to have that connection. So this rotational operator has some constants in front of it. And this is in wave number of units, because I'm a spectroscopist. And we have the vibrational frequency, 1 over 2 pi C. K over mu square root. Also in wave number of units. Now the first problem is we don't like R, because harmonic oscillators are expressed in terms of displacement from equilibrium. And so we know that Q is equal to R minus RE, or R is equal to Q plus RE. So we want to replace this operator R by this, an operator plus a constant. So 1 over R squared can be expressed as 1 over RE squared, the equivalent internucleus, plus 1 over Q over RE. Plus [INAUDIBLE] squared. So this is the power series expansion, but we only keep the first term generally. And so now we're ready to go to work. And we can say we have B as a function of Q is equal to B of E, the equilibrium value 1 minus 2Q or RE plus 3Q squared over RE squared. And there's more terms, but this is enough. You bet it's enough. By the time this lecture is over, you're going to want not ever to see this stuff again. OK, so we have now a representation of the rotational operator in terms of a constant term, a linear term, and a squared term in Q. And we know how to do matrix elements of Qs, right? So we know we're in business. And we have the potential as another power series in Qs. So everything is going to come together. It's just going to be will the migraine wipe out your ability to pay any attention to this after we're done. OK, so I don't want to cover that up yet. So we know that this displacement operator can be replaced by the creation and annihilation operators. And so we have 4 pi C, mu omega E square root. A plus A dagger are friends. These guys, we hardly need to take a deep breath to know what to do with those. And so we're going to want to express all of the-- all of the actors here, these guys and those guys, in terms of expressions in the creation and annihilation operators. So for the rotational constant expansion, we're going to need something like 2Q over RE. And that, after a little bit of algebra, is 4BE over omega E square root times A plus A dagger. And we need 3Q squared over RE squared. And that's going to be 3BE over omega E, A plus A dagger squared. So all of the terms can be expressed, can be reduced to constants that we care about, or that are easily measured, and these powers of As and a daggers. Now this is-- you have notes, printed notes. And I'm following them pretty well, although there are a few typos, which I hope to correct. And so we do a similar thing for the terms here. And we're only going to keep-- we're only going to look at this. We're not going to go to Q 1/4. But this is bad enough. OK, and this is part of the zero order Hamiltonian. So we're going to have another term, and that's going to be 1/6 little a Q cubed. OK, so that's-- little a is the anharmonicity parameter. And what is its sign? Can we know its sign from experiment? You said yes. Why do you say yes? AUDIENCE: [INAUDIBLE] ROBERT FIELD: I'm sorry? AUDIENCE: It's cubed, so it's odd. So there's [INAUDIBLE]. ROBERT FIELD: The only way you know the sign of an off diagonal matrix element is-- I'm sorry, is if there is a diagonal element of it that can be put into the E, the first order correction to the energy. As soon as you have to square it when you do second order perturbation theory, you've lost the sign. But here we know the sign because of physical insight. We know that this kind of a potential would be nonsense. So we know that A has a sign and it's negative. OK, so we have a over 6 times our favorite parameter here to the 3/2 power times A plus A dagger cubed. Now this is something-- you do the operator algebra before you launch into a horrible calculation. And so we know how to do this in principle to reduce this to a simple expression. And now we take this part, this thing, and we're just going to call it capital A. Because we want it-- we don't want to clutter up what is going to be a terrible thing anyway. And we can always convert back to the little a at the end if we need to. OK, so now I've set the stage. I'll leave this for a while. That really is-- as you know, I really love revision theory, because it's the tool that I use all the time. And it's a psychological condition that is not curable, all right? So what do we want? We want E0 as a function of E and J. And that's just the J H0 VJ. And we know that HC will make EB plus 1/2 plus HCBEJJ plus 1. OK, well this goes in the energy denominators. And so that's good, because we need energy denominators as well as the zero order energies. Right now we need to know the bad stuff, the thing that's outside of the zero order Hamiltonian, which is everything. And so we have this is HC times BE JJ plus 1 times-- I'm sorry, I have to erase this. Minus BE over 4BE over a big E square root times A plus A dagger plus 3 BE. 3BE over omega E times A plus A dagger quality squared. So that's the rotational part. And then we have the vibrational power, which we have A, A plus A dagger cubed. OK, so now we look at this thing and we say, oh well, this has matrix elements delta V equals plus or minus 1. And this has delta V equals plus or minus 2 and 0. And this has delta V equals plus and minus 3 plus and minus 1. We always want to sort things according to the selection rules, because we always combine the things with the same selection rules. And it's best to do that at the beginning rather than somehow trying to do it at the end. Because if you have the same selection rule, you have cross terms. And that's really important. The cross terms are where a lot of good stuff happens. OK, so we know E0. And we would know E1 if there are any off the-- there are any diagonal matrix elements of this operator. This should be a plus as well. So this is H1. And this doesn't have any diagonal elements. And this doesn't have any diagonal elements. But this one does. And it's actually something you encountered, I think, on exam two. I'm not sure, but you've certainly encountered the diagonal element of this. And so if we look at A plus A dagger squared, we get A squared plus A dagger squared plus 2 number operator plus 1. And that's diagonal. So we have a diagonal element, and that gives us E1. So E1 of E and J is HC6BE squared over omega E times J, J plus 1, B plus 1/2. OK, well that looks pretty good, because this is the coefficient of-- we have an expression for the energy levels, which includes the E minus alpha E, B plus 1/2 of times JJ plus 1. So we have a term that involves B plus 1/2 and JJ plus 1. And it has a name. Alpha minus alpha. And here we have a term which has those, that quantum number dependents. And it has a value. And so it's telling us that alpha E, and I'll put on this harmonic oscillator, is equal to minus HC6BE v squared over omega E. One constant. Now it's also telling you that B of E increases with V. And for a harmonic oscillator, you have an equal lobe at each turning point, the largest lobe. But they're equal in magnitude. So one can ask, at the inner turning point is the change in the rotational operator, or is it larger-- is this change in BE larger relative to the equilibrium value, or is this larger? And the answer is we're talking about 1 over R squared. And 1 over R squared gets really large at small r as opposed to getting smaller at large R. And the amount of change is much greater at small r. And so that causes the effective rotational constant to increase. But we know for a harmonica-- for an anharmonic oscillator, we have a small lobe here and a big lobe here. And so we expect that there's going to be a battle between the harmonic contribution to alpha and the anharmonic contribution. And we expect that this is going to win. Why? Because every time anybody measures alpha, it's a positive number. And that's why it was expressed in the formula, which I've concealed, with a negative sign, to take into account that alpha is always-- the contribution is always negative, and so alpha is always positive. Yeah, and these things are-- it's historical. But anyway, so we have a contribution which has the wrong sign. And we know we're going to get another contribution from the interaction between the rotation and the vibration, and that this is going to make things right OK. And it comes from a term. OK. All right, there's no way I can make this simpler. Just have to bear with me and I will-- so we have delta V equals plus and minus 1 matrix elements from both the A plus A dagger and the A plus A dagger cubed terms. And we have-- we have this in both the anharmonic expression and in the rotational constant operator. We know what these-- how these things work out. And so delta V equals plus and minus 1 terms from the A plus A dagger cubed V plus 1 and V, V minus 1. So we can work these things out. And this one is 3V plus 1/2 to the 3/2. And this is 3V plus V to the 3/2. So all of this stuff, you want to simplify it as you go but it. And these come from the A term, the A, A plus A dagger cubed' term. We don't have a term that's linear at A plus A dagger from the anharmonicity, but we do-- and we do have a squared term, but that's the harmonic correction. So we start here, with A plus A dagger cubed. And then we're going to want to look at the delta V equals plus or minus 2, and delta V equals plus and minus 3 terms. And there's lots of algebra. You can do all these things. There's nothing challenging here. It's just how do you keep this stuff that you derived that you're going to need in a place that you can find it again, because the pages just get filled with garbage. OK. So the best way for me to do this is to write the results in the next-to-the-final step, and then in the final step, and show what goes to what. So the second order correction is a function of EJ. We're going to have terms that involve the delta V equals plus or minus 1. And we're going to get that from the cubic correction through the harmonic oscillator and the linear correction to the rotational concept. So we're going to get an expression that looks like this. B squared minus 2AB plus A squared. So this is a B term. This is the anharmonic term. We have a cross term. There are going to be three terms in this expression. OK, and so let's do that. How much time I have? Not very much time. Well that's good, because you won't have to watch much more of this. OK, so we have the HcBe JJ plus 1. And this guy is squared, because we're doing second-order perturbation theory. This is just the B term squared. And then we have 4BE be over omega E. And then we have the delta V of 1 and minus 1 matrix elements. V plus 1 over minus HC omega, and V plus V over HC omega. We have the energy denominators, two energy denominators of opposite side and similar identical magnitude. And we always do that because then the algebra is simple. If we do-- we fail to do that, you might as well just go to a booby hatch right away, because there's just no way you're going to retain your sanity if you don't combine the terms of delta V plus 1 and delta V minus 1. OK, and this is the first term. And there is the next term in this sum, which is the AB term. And so we have a minus 2 times HCV JJ plus 1. And it's not squared, because it's one B and one A term. And so we get 4BE over omega E square root. And then the A part. And we get 3B plus 1/2 to the 3/2 times V plus 1/2 to the 1/2, or minus HC omega E. And then the other term, which is 3V to the 3/2, V to 1/2, over HC omega E. We have to combine these two terms. And the combination is easy. And then there is-- so this term number two. And number three is the squared anharmonic term. So I'm going to skip it, since I'm basically done. There's just a lot of garbage like this. And at the end, we have an expression, term by term, which simplifies. And we get the second-order correction to the energy, VJ, is equal to minus HcBe cubed over HC omega. No, not HC. Just over omega E squared times JJ plus 1 squared. Well this is nice, because we have a term that just involves J, and it's negative, that says as a molecule rotates, the molecule stretches, the rotational constant decreases. This is called centrifugal distortion. And this is a very famous expression that DE is equal to 4BE cubed over omega squared. I left out a 4. So this is the Kratzer relationship. And that turns out to be of the most valuable ones, because the centrifugal distortion constant is extremely sensitive to contagion. If there is a perturbation, if there is something missing, this centrifugal distortion constant will have a crazy value. And if it doesn't ever-- if there's nothing wrong, then it will say, well, OK, I can-- if I know something about B and omega, I can determine how the rotational energy levels get closer together as you go up in V. And if you have only a pure rotation spectrum, then you actually-- where you're not sampling anything having to do with vibration, you actually can determine the vibrational constant from the centrifugal distortion. And I should tell you that spectroscopists come in families, or they used to. Pure rotation, infrared vibration rotation, electronic vibration rotation, electronic. And so the microwavers, who really have only rotation to look at, are kind of deprived. And they just don't know anything about electronic degrees of freedom. But they know how to get the vibrational frequency from the rotational spectrum, which is kind of nice. Gives them something to do. And there's more terms. And we're really done, but it's all in the notes. And so what we get is for the constant OAEXE, and center of distortion, and alpha E and beta, which is the vibrational distortion of the centrifugal distortion constant, we get all these in terms of omega and B. So it's a nice closed set where you get the-- you're able to predict how certain constants depend on the most fundamental ones, omega E and BE. That's kind of neat. And it's both a way of extending your observations and telling well, there's something missing, because these constants aren't coming out right. It's also a subject of great consternation. Because if you're a quantum chemist, what you do is you determine a lot of stuff by derivatives at equilibrium. The derivatives, the potential at equilibrium, are basically telling you the shape of the potential at moderately low energy. And it also turns out to be something with modern quantum chemistry. You can get all these derivatives pretty accurately. The trouble is measuring derivatives here and using the energy formulas gives you slightly different values from what you get from perturbation theory. So what we have is two communities who are very sophisticated using the same names for different quantities. It's dangerous. And it's also not understood by the two communities. And so people are struggling and struggling to get agreement between experiment and theory. And it's apples and oranges, but it's a subtle thing. OK, so that's basically all I want to say about perturbation theory. I will have one more lecture dealing with perturbation theory, and that lecture is on Van der Waals interactions between molecules and why we get liquids. OK. |
MIT_561_Physical_Chemistry_Fall_2017 | 19_Spectroscopy_Probing_Molecules_with_Light.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseware continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseware at ocw.mit.edu. ROBERT FIELD: This lecture is not relevant to this exam or any exam. It's time-dependent quantum mechanics, which you probably want to know about, but it's a lot to digest at the level of this course. So I'm going to introduce a lot of the tricks and terminology, and I hope that some of you will care about that and will go on to use this. But mostly, this is a first exposure, and there's a lot of derivation. And it's hard to see the forest for the trees. OK, so these are the important things that I'm going to cover in the lecture. First, the dipole approximation-- how can we simplify the interaction between molecules and electromagnetic radiation? This is the main simplification, and I'll explain where it comes from. Then we have transitions that occur. And they're caused by a time-dependent perturbation where the zero-order Hamiltonian is time-independent, but the perturbation term is time-dependent. And what does that cause? It causes transitions. We're going to express the problem in terms of the eigenstates of the time-independent Hamiltonian, the zero-order Hamiltonian, and we know that these always have this time-dependent factor if we're doing time-dependent quantum mechanics. The two crucial approximations are going to be the electromagnetic field is weak and continuous. Now many experiments involve short pulses and very intense pulses, and the time-dependent quantum mechanics for those problems is completely different, but you need to understand this in order to understand what's different about it. We also assume that we're starting the system in a single eigenstate, and that's pretty normal. But often, you're starting the system in many eigenstates that are uncorrelated. We don't talk about that. That's something that has to do with the density matrix, which is beyond the level of this course. And one of the things that happens is we get this thing called linear response. Now I went for years hearing the reverence that people apply to linear response, but I hadn't a clue what it was. So you can start out knowing something about linear response. Now this all leads up to Fermi's golden rule, which explains the rate at which transitions occur between some initial state and some final state. And there is a lot more complexity in Fermi's golden rule than what I'm going to present, but this is the first step in understanding it. Then I'm going to talk about where do pure rotation transitions come from and vibrational transitions. Then at the end, I'll show a movie which gives you a sense of what goes on in making a transition be strong and sharp. OK, I'm a spectroscopist, and I use spectroscopy to learn all sorts of secrets that molecules keep. And in order to do that, I need to record a spectrum, which basically is you have some radiation source. And you tune its frequency, and things happen. And why do the things happen? How do we understand the interaction of electromagnetic radiation and a molecule? And there's really two ways to understand it. We have one-way molecules as targets, photons as bullets, and it's a simple geometric picture. And the size of the target is related to the transition moments, and it works. It's very, very simple. There's no time-dependent quantum mechanics. It's probabilistic. And for the first 45 years of my career, this is the way I handled an understanding of transitions caused by electromagnetic radiation. This is wrong. It has a wide applicability. But if you try to take it too seriously, you will miss a lot of good stuff. The other way is to use the time-dependent in your equation, and it looks complicated because we're going to be combining the time-dependent Schrodinger equation and the time-independent Schrodinger equation. We're going to be thinking about the electromagnetic radiation as waves rather than photons, and that means there is constructive and destructive interference. There's phase information, which is not present in the molecules-as-targets, photons-as-bullets picture. Now I don't want you to say, well, I'm never going to think this way because it's so easy to think about trends. And, you know, the Beer-Lambert law, all these things that you use to describe the probability of an absorption or emission transition, this sort of thing is really useful. OK, so this is the right way, and the crucial step is the dipole approximation. So we have electromagnetic radiation being a combination of electric field and magnetic field, and we can describe the electric field in terms of-- OK. So this is a vector, and it's a function of a vector in time. And there is some magnitude, which is a vector. And its cosine of this thing, this is the wave vector, which is 2 pi over the wavelength, but it also has a direction. And it points in the direction that the radiation is propagating. And this is the position coordinate, and this is the frequency. So there's a similar expression for the magnetic part-- same thing. I should leave this exposed. So we know several things. We know for electromagnetic radiation that the electric field is always perpendicular to the magnetic field. We know a relationship between the constant term, and we have this k vector, which points in the propagation direction. Now the question is-- because we have-- we have a molecule and we have the electromagnetic radiation. And so the question is, what's a typical size for a molecule in the gas field? Well, anywhere? Pick a number. How big is a molecule? STUDENT: A couple angstroms? ROBERT FIELD: I like a couple angstroms. That's a diatomic molecule. They're going to be people who like proteins, and they're going to talk about 10 or 100 nanometers. But typically, you can say 1 nanometer or 2 angstroms or something like that. Now we're going to shine light at a molecule. What's the typical wavelength of light that we use to record a spectrum? Visible wavelength. What is that? STUDENT: 400 to 700 nanometers? ROBERT FIELD: Yeah, so the wavelength of light is on the order of, say, 500 nanometers. And if it's in the infrared, it might be 10,000 nanometers. If it were in the visible, it might be 100-- in the ultraviolet, it might be as short as 100 nanometers. But the point is that this wavelength is much, much larger than the size of a molecule, so this picture here is complete garbage. The picture for the ratio-- so even this is garbage. The electric field or the magnetic field that the molecule sees is constant over the length of the molecule to a very good approximation. So now we have this expression for the field, and this is a number which is a very, very small number times a still pretty small number. This k dot r is very small. It says we can expand this in a power series and throw away everything except the omega t. That's the dipole approximation. So all of a sudden, we have as our electric field just E0 cosine omega t. That's fantastic. So we've gotten rid of the spatial degree of freedom, and that enables us to do all sorts of things that would have required a lot more justification. Now sometimes we need to keep higher order terms in this expansion. We've kept none of them, just the zero order term. And so if we do, that's called quadrupole or octopole or hexadecapole, and there are transitions that are not dipole allowed but are quadrupole allowed. And they're incredibly weak because k dot r is really, really small. Now the intensity of quadrupole-allowed transitions is on the order of a million times smaller than dipole. So why go there? Well, sometimes the dipole transitions are forbidden. And so if you're going to get the molecule to talk to you, you're going to have to somehow make use of the quadrupole transitions. But it's a completely different kind of experiment because you have to have an incredibly long path length and a relatively high number density. And so you don't want to go there, and that's something that's beside-- aside from what we care about. So now many of you are going to be doing experiments involving light, and that will involve the electric field. Some of you will be doing magnetic resonance, and they will be thinking entirely about the magnetic field. The theory is the same. It's just the main actor is a little bit different. Now if we're dealing with an electric field, we are interested in the symmetry of this operator, which is the electric field dotted into the molecular dipole moment, and that operator has odd parity. And so now I'm not going to tell you what parity is. But because this has odd parity, there are only transitions between states of opposite parity, whereas this, the magnetic operator, has even parity. And so they only have transitions between states of the same parity. Now you want to be curious about what parity is, and I'm not going to tell you. OK, so the problem is tremendously simplified by the fact that now we just have a time-dependent field, which is constant over the molecule. So the molecule is seeing an oscillatory field, but the whole molecule is feeling that same field. OK, now we're ready to start doing quantum mechanics. So the interaction term, the thing that causes transitions to occur-- the electric interaction term, which we're going to call H1 because it's a perturbation. We're going to be doing something in perturbation theory, but it's time-dependent perturbation theory, which is a whole lot more complicated and rich than ordinary time-independent. Now many of you have found time-independent perturbation theory tedious and algebraically complicated. Time-dependent perturbation theory for these kinds of operators is not tedious. It's really beautiful. And there are many, many cases. It's not just having another variable. There's a lot of really neat stuff. And what I'm going to present today or I am presenting today is the theory for CW radiation-- that's continuous radiation-- really weak, interacting with a molecule or a system in a single quantum state initially. And it's important. The really weak and the CW are two really important features. And the single quantum state is just a convenience. We can deal with that. That's not a big deal, but it does involve using a different, more physical, or a more correct definition of what we mean by an average measurement on a system of many particles. And you'll hear the word "density matrix" if you go on in physical chemistry. But I'm not going to do anything about it, but that's how we deal with it. OK, so this is going to be minus mu-- it's a vector-- dot E of t, which is also a vector. Now a dot product, that looks really neat. However, this is a vector in the molecular frame, and this is a vector in the laboratory frame. So this dot product is a whole bunch more complicated than you would think. Now I do want to mention that when we talk about the rigid rotor, the rigid rotor is telling us what is the probability amplitude of the orientation of the molecular frame relative to the laboratory frame. So that is where all this information about these two different coordinate systems reside, and we'll see a little bit of that. OK, there's a similar expression for the magnetic term. I'm just not going to write it down because it's just too much stuff to write down. So the Hamiltonian, the time-independent Hamiltonian, can be expressed as H0 plus H1 of t. This looks exactly like time-independent perturbation theory, except this guy, which makes all the complications is time dependent. But this says, OK, we can find a whole set, a complete set of eignenenergies and eigenfunctions. And we know how to write the time-dependent Schrodinger-- the solutions of the time-dependent Schrodinger equation if this is the whole game. So we're going to use these as basis functions just as we did in ordinary perturbation theory. So H0 times some eigenfunction, which now I'm writing as explicitly time-dependent is En phi n t equals 0 e to the minus i En t over h-bar. So this is a solution. This thing is a solution to the time-dependent Schrodinger equation. And so when the external field is off, then the only states that we consider are eigenstates of the zero-order Hamiltonian, and they can be time dependent. But if we write psi n star of t times psi n of t, well, that's not time dependent if this is an eigenstate. So the only way we get time dependence is by having this time-dependent perturbation term. OK, so let's take some initial state. And let us call that initial state some arbitrary state. And we can always write this as a superposition of zero-order states. OK, and now, unfortunately, both the coefficients in this linear combination and the functions are time dependent. So this means when we're going to be applying the time-dependent Schrodinger equation, we take a partial derivative with respect to t, we get derivatives with this and this. So it's an extra level of complexity, but we can deal with it, because one of the things that we keep coming back to is that everything we talk about is expressed as a linear combination of t equals zero eigenstates of the zero-order Hamiltonian. OK, so the time-dependent Schrodinger equation-- i h-bar partial with respect to t of the-- yeah, of the wave function is equal to-- OK, that's our friend or our new friend because the old friend was too simple. And so, well, we can represent this partial derivative just using dots because the equations I'm going to be putting on the board are hideous, and so we want to use every abbreviation we can. This is written as a product of time-dependent coefficients and time-dependent functions. When we apply the derivative to it, we're going to get derivatives of each. And so that's the left-hand side. OK, and let's look at this left-hand side for a minute. OK, so we've got something that we don't really know what to do with, but this guy, we know that this is-- this time-dependent wave function is something that we can use the time-dependent Schrodinger equation on and get a simplification. So the left-hand side-- I haven't written the right-hand side. I'm just working on the left-hand side of what we get when we start to write this equation. And what we get is we know that the time dependence of this is equal to 1 over i h-bar times the Hamiltonian operating on phi n. Is that what I want? I can't read my notes so I have to-- I have to be-- yeah, so we've just taken that 1 over i h-bar. This is going to be the time-independent Hamiltonian, the zero-order Hamiltonian. And we know what we get here. Yes? STUDENT: So all your phi n's, those are the zero-order solutions? ROBERT FIELD: That's correct. STUDENT: So they're unperturbed states? ROBERT FIELD: They're unperturbed eigenstates of H0. And if it's psi n of t, it has the e to the i En of t-- En t over h-bar factor implicit, and we're going to be using that. All right, so what we get when we take that partial derivative, we get a simplification. OK, let me just write the right-hand side of this equation too. So we have the simplified left-hand side, which is psi n c-- I've never lectured on time-dependent perturbation theory before. And so although I think I understand it, it's not as available in core as it ought to be. OK, so we have this minus-- where did the wave function go? Well, there's got to be a phi in here and then minus i over h-bar En cn t over the times phi n of t. That's the left-hand side in the bracket here. OK, and the right-hand side of the original equation, that is just some n cn t En plus H1 of t phi n t. OK, it takes a little imagination, but this and the terms associated with that are the same. This happened when we did non-degenerate perturbation theory. We looked at the lambdas of one equation. There was a cancellation of two ugly terms. And so what ends up happening is we get a tremendous simplification of the problem. And so the left-hand side of the equation has the form, and the right-hand side has the form over here without the extra term-- sum over n, cn of t H1 of t psi n of t. OK, and now we have this equation. We have this simple thing here, and we have this ugly thing here. And we want to simplify this by multiplying on the left by psi F of t so-- and integrating with respect to tau. F is for final. So we're interested in the transition from some initial state to some final state. So we're going to massage this. And when we do that, we get-- I've clearly skipped a step, but it doesn't matter-- i h-bar cf dot of t is equal to this integral sum c n of t integral cf of-- phi f of t H1 f of t phi n of t, e tau. This is a very important equation because we have a simple derivative of the coefficient that we want, and it's expressed as an integral. And we have an integral between an eigenstate of the zero-order Hamiltonian and another eigenstate. And this is just H1 f n of t. OK, so we have these guys. So what we want to know is, all right, this is the thing that's making stuff happen. This is a matrix element of this term. Well, H1 of t, which is equal to v cosine omega t can be written as v times 1/2 e to the i omega t plus e to the minus i omega t. This is really neat because you notice we have these complex oscillating field terms, and we have on each of these wave functions a complex oscillating term. And what ends up happening is that we get this equation. i h-bar cf dot of t is equal to-- and this is-- you know, it's ugly. It gets big. A lot of stuff has to be written, and I have to transfer from my notes to here. And then you have to transfer to your paper. And there is going to be-- there will be printed lecture notes. And in fact, there may actually be printed lecture notes for this lecture. But if they're not, they will be soon. OK, and so we get this differential equation, which is the sum over n c n of t integral psi f star of t times 1/2 v, v to the i omega t plus e to the minus i omega t times psi n of t, e tau. Well, these guys have time dependence, and so we can put that in. And now this integral has the form psi f star 0 1/2 v, and we have e to the minus I omega and f minus omega t. Omega nf, the difference in-- so omega nf is En minus Ef over h-bar. And so we have minus this oscillating term, minus omega t, and then we have e to the minus i nft plus omega t. So here this isn't well, it's so ugly because of my stupidity here. But what we have here is a resonance integral. We have something that's oscillating fast minus something that's oscillating fast. And we have the same thing plus something that's oscillating fast. So those terms are zero because we have an integral that is oscillating. I'm sorry. It's oscillating between positive and negative, positive, negative. And as long as omega is different from omega nf, those integrals are zero because this integrand, as we integrate to t equals infinity or to any time, is oscillating about zero, and it's small. However, if omega is the same as minus omega nf or plus omega nf, well, then this thing is 1 times t. It gets really big. Now we're talking about coefficients, which are related to probabilities. And so these coefficients had better not go get really big because probability is always going to be less than 1. OK, so what we're going to do now is collect the rubble in a form that it turns out to be really useful. So we have an equation for the time dependence of a final state, and it's expressed as a sum over n. But if we say, oh, let's make our initial state just one of those. So our initial state is-- let's call it ci. And we say, well, the system is not in any other state other than the i state, and this is weak. So we can neglect all of the other states where n is not equal to i. And if they're not there, cn has to be 1, so we can forget about it. So we end up with this incredibly wonderful simple equation. So we make the two approximations. Single state, the perturbation is really weak, and we get cf of t is equal to the vfi-- the off-diagonal matrix element-- over 2i h-bar times the integral from 0 to t e to the minus i omega i f t minus omega times e to the i omega i f plus omega dt. Well, all complexity is gone. We have the amount of the final state, and it's expressed by a matrix element and some time dependence. And this is a resonant situation where if omega t, omega t-- if omega is equal to omega i f, fine. Then this is zero, the exponent is zero, we get t here from that. And we get zero from that one because that's oscillating so fast it doesn't do anything. But that's a problem because this c is a probability. And so the square of c had better not be larger than one, and this is cruising to be larger than 1. But we don't care about cw. What we really care about-- well, what is the rate as opposed to the probability? OK, because the rate of increase of state f is something that we can calculate from this integral simply by taking the-- we multiply the integral by 1 over T if the limit T goes to infinity. And now we get a new equation, which is called Fermi's golden rule. OK, so I'm skipping some steps, and I'm doing things in the wrong order. But so first of all, the probability of the transition from the i state to the f state as a function of time. So the probability is going to keep growing. That's why we want to do this trick with dividing by t. What time is it? OK. That's just cf of t squared, and that's just the fi over 4 h-bar squared times this integral 0 to t e to the plus and e to the minus term dt squared. OK, the integrals survive only if omega is equal to omega i f. And if we convert to a rate so that the rate is going to be Wfi, which is going to be Vfi over 4 h-bar squared times the sum of two delta functions Vi minus Ef minus omega plus sum of Ei minus Ef plus omega. So the rate is just this simple thing-- the square matrix element and a delta function-- saying either it's an absorption or emission transition on resonance, and we're cooked. OK, so now I want to show some pictures of a movie, which will make this whole thing make more sense. This is for a vibrational transition. So we have the electric field-- the dipole interacting with the electric field. And now let's just turn on the time dependence. OK, so this is the interaction term. We add that interaction term to the zero-order Hamiltonian, and so we end up getting a big effect of the potential. The potential's going like this, like that. And so the eigenfunctions of that potential are going to be profoundly affected, and so let's do that. Let's go to the next. All right, so here now we have a realistic small field, and now this is small. And you can hardly see this thing moving. OK, now what we have is the wave function of this. And what we see is if omega is 1/4 the energy, if omega is much smaller than the vibrational frequency or much larger, we get very little effect of the time dependent. You can see that the wave function is just moving a little bit. The potential is jiggling around, whether the perturbation is strong or weak. It's not on resonance. And now let's go to the resonance. Now what's happening is the potential is moving not too much, but the wave function is diving all over the place. And if we ask, well, what does that really look like as a sum of terms, the thing that's different from the zero-order wave function is this. So zero-order wave function is one nodeless thing. This is the time-dependent term, and it looks like V equals 1 So what this shows is, yes, there are-- if we have a time-dependent field, and it's resonant, then we get a very strong interaction even though the field is weak. And it causes the appearance of the other level but oscillating. And so resonance is really important, and selection rule is really important. The selection rule for the vibrational transitions has to do with the form. Oh, I shouldn't be rushing at all. OK, so let's draw a picture. And this is the part that has puzzled me for a long time, but I've got it now. So here we have a picture of the molecule. And this end is positive, and this end is negative. And we have a positive electrode, and we have a negative electrode. So that's an electric field. And so now the positive electrode is saying, you better go away, and you better come here. So it's trying to use compressed bond. And now this field oscillates, and so it's compressing and expanding. Now that's what's going on. But how does quantum mechanics account for it? Well, quantum mechanics says in order for the bond length to change, we have to mix in some other state. So we have the ground state, and we have an excited state that looks like that. And so the field is mixing these two. Now that means that the operator is mu 0 plus derivative of mu with respect to the electric field times q. So this is the thing that allows some mixing of an excited state into the ground state. This is our friend the harmonic oscillator-- operator, displacement operator. It has selection rules delta v equals plus or minus 1, and that's all. So a vibrational transition is caused by the derivative of the-- yeah, that's-- no, derivative of the dipole moment with respect to Q. So did I have it right? Yes. So this is something-- we can calculate how the dipole moment depends on the displacement from equilibrium, but this is the operator that causes the mixing of states. So one of the things I've loved to do over the years is to write a cumulative exam in which I ask, well, what is it that causes a vibrational transition? What does a molecule have to have in order to have a vibrational transition? And also what does a molecule have to have to have a rotational transition? Well, this is what causes the rotational transition because we can think of the dipole moment interacting with a field, which is going like that or like that. And so what that does is it causes a torque on the system. It doesn't change the dipole moment, doesn't stretch the molecule. It causes a transition, and this is expressed in terms of the interaction mu dot E. This dot product, this cosine theta, is the operator that causes this. We call the relationship between the laboratory and the body fixed coordinate system is determined by the cosine of some angle, and the cosine of the angle is what's responsible for a pure rotational transition. And we have vibrational transitions where they are derivative of the dipole with respect to the coordinate. Now let's say we have nitrogen-- no dipole moment, no derivative of the dipole moment. Suppose we have CO. CO has a very small dipole moment and a huge derivative of the dipole moment with respect to displacement. And so CO has really strong vibrational transitions and rather weak rotational transitions. So if it happened that CO had zero permanent dipole moment, it would have no rotational transition. But as you go up to higher V's, then it would not be zero. And you would see rotational transitions. And so there's all sorts of insights that come from this. And so now we know what causes transitions. There is some operator, which causes mixing of some wave functions. And the time-dependent perturbation theory when it's resonant mixes only one state. We have selection rules which we understand just by looking at the wave fun-- looking at the matrix elements, and now we have a big understanding of what is going to appear in a spectrum. What are the intensities in the spectrum? What are the transitions? Which transitions are going to be allowed? Which are going to be forbidden? And that's kind of useful. So there is this tremendously tedious algebra, which I didn't do a very good job displaying, but you don't need it because, at the end, you get Fermi's golden rule, which says transitions occur on resonance. Now if you're a little bit off resonance, well, then the stationary phase in the oscillating exponential persists for a while, and then it goes away. And so you get a little bit of slightly off-resonance transition probability, and you get other things too. But you already now have enough to understand basically everything you need to begin to make sense of the interaction of radiation with molecules correctly, and this isn't bullets and targets. This is waves with phases, and so there are all sorts of things you have to do to be honest about it. But you know what the actors are, and that's really a useful thing. And you're never going to be tested on this from me. OK, good luck on the exam tomorrow night. |
MIT_561_Physical_Chemistry_Fall_2017 | 26_Qualitative_MO_Theory_Hückel.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: Last time I talked about LCAO-MO for diatomic molecules. And I didn't finish, but the important point was that this is a toy model. This is based on a little bit of extension from something which is not really a toy model, H2 plus, to, basically, an interpretive framework that can be applied to, basically, all diatomic molecules. And the logic is relatively simple. We know this one. We get the molecular orbitals for H2 plus. There are basically two. There's a binding one and the anti-binding one. Of course, there's many more, but we don't care about them. And then we go to H2. And again, there's basically only two orbitals that we care about because the next higher principle quantum number has such high energy that we can just forget about them, because those states that derive from the higher principle quantum number are Rydberg states or complicated things, because they're at such high energy. And then it's a very small step to go from hydrogen to the AH molecules, because, well, we've got the electronic structure for the A atom, which is complicated-- more complicated than hydrogen. But because hydrogen only can make sigma bonds, because the p orbitals in hydrogen are so high that it's a simpler picture and can easily be understood. This is in all the textbooks, and it's more or less given to you as something to memorize. But there is a lot more to it than just memorization. And the important thing is that everything in LCAO-MO for diatomics is based on the periodic table. And the periodic table tells you about the ionization energies-- the periodicites of ionization energies. And so we can say for any non-integer-- Well, if we know what the ionization energy from a particular state is, we can use that ionization energy to derive a non-integer principle quantum number. This is all empirical. And then use that empirical principle quantum number to get the size of the orbital. And basically, the size is everything. Because everything is based on overlap, And the internuclear distance molecule is based on the relative sizes of the different atoms. And for different electronic states of the atoms, these sizes are different, because they have the ionization energy from that level implicitly expressed. So because you know about orbital sizes, and that the different atomic orbitals have different sizes, you can do an enormous amount as far as understanding the electronic structure of all diatomic molecules. Now, this-- when you have two states which have different energies-- We do something like this to describe the molecular orbitals that arise from them. That that's perturbation theory. So the solid line is the dominant character. The dotted line is the admix character of the other orbital. And this is all qualitative, but you have these different atomic orbital energies. You know them from the periodic table, basically. And so you can say, yes. There's going to be two orbitals arriving, derived from these two states. And one is polarized towards this atom. The other is polarized towards that atom. And so you get the shape of the orbital. And it's actually useful for saying, well, if I were to do chemistry, which end of this is electronegative, and which end is positive? Or which end of this would attach to a metal surface? Would it attach pointing into the surface or lying down? And there's all sorts of insights, if you can draw these sorts of pictures. And this is what we do as physical chemists. We take the crudest model, and we say, OK. We understand the important features. And as we discover new, important features, we build them in, or we forget about them because they're too subtle. And we're always looking for something where the crude picture doesn't work. And we then find the important thing that is needed. And for example, if you were to look at the molecular orbital diagram for C2, which is a perfectly legitimate gaseous molecule, you'll see that there's a little bit of ambiguity about which is the ground state. And this is an important thing. And there was a big controversy about this that was settled by spectroscopy. OK. So if you can build an intuitive picture for all diatomic molecules, you can also build an intuitive picture for what we could call chromophores. So there are a lot of molecules that are larger than diatomic molecules. But the electronic structure is mostly nothing, except a few atoms that are close to each other, where there's a double bond or there's something special. And so if you can do LCAO-MO for diatomic molecules, you can also address electronic structure in much larger molecules, because it's really due to a few important atoms or several different groups of important atoms. And so the idea is, in here, enable you to talk about the electronic structure of almost anything. And one of the things that we care about is spectroscopy. How do we learn about the structure of a molecule? And mostly, you learn about it from doing electronic spectroscopy. You do transitions between the ground state and some higher states. And you want to be able to predict what you are going to see. And for example-- and I think you may hear about this a little bit more-- if you know the spectrum of nitrogen, it doesn't start absorbing until far into the vacuum UV. And then go one atom over-- the spectrum of oxygen. Well, that defines the vacuum UV. It starts to absorb around 200 nanometers. And 2 and 1/2 billion years ago, oxygen started to enter into the atmosphere and profoundly changed life on Earth. Because with the vacuum UV radiation, nothing could live on the surface of land, because it would all be killed by this hard UV. And so everything was underwater at the bottom of the ocean. But all of a sudden, when oxygen appeared, there was protection. And then there's additional protection farther out towards the visible from ozone. It's not such a good shield, but it absorbs between 200 and 350 nanometers. And it protects us, too. But the big thing is oxygen. And why should oxygen be so different from nitrogen? And maybe you should think about that. OK. I should have mentioned the schedule up there. I'm going to give another lecture on perturbation theory-- one of my favorites, which I thought I wasn't going to get to give on Friday. And then Troy is going to give two lectures on quantum chemistry. And there's going to be significant lab experience. And you'll hear about that from the TAs. And so you will do real calculations. So LCAO-MO could be the organization structure for quantitative calculations. But quantitative calculations are usually huge basis sets, and you lose the LCAO completely. One could take the results of such a calculation, and then project it onto a LCAO-MO picture. And that's generally what we do. We reduce experiments, we reduce theory, to something that we can intuit. And it's really important to have intuition. OK. Huckel theory-- Huckel theory is another kind of non-rigorous theory. In fact, it's laughable in its simplicity. And the idea is that even such a crude theory can make sense out of huge families of molecules and enable you to be quantitative about things without having to do a huge calculation. Now, you almost always have to diagonize a matrix. And because usually, in Huckel theory, you're dealing with more than two or three relevant orbitals, it's a big deal. I don't know whether you have experience with matrix diagonalization routines, but there are many of them. And you're going to have experience with them. But the things you put into them have absolutely nothing to do with a real Hamiltonian. It's a make up picture. It's a picture based on, basically, two parameters-- alpha and beta. And I'll talk about this. And these are parameters that are agreed upon by international committee-- It's not a committee. But people say, OK. This is the value of alpha and beta that we're going to use to describe many problems. But I just want to make sure you understand that Huckel theory is only a little bit more ridiculous than LCAO-MO theory. Because in LCAO-MO, you really have some semi-empirical sense of how big orbitals are. And once you know how big orbitals are, you know what's going on. In this, you just have a couple of parameters that says, OK. These are the rules. Let's now apply that. OK. I'm carrying around too much stuff. So organic chemists are really wonderful, because they give you an abbreviated way of drawing a molecular structure. And almost everyone in this room is gifted in being able to visualize three-dimensional structures. Physical chemists tend not to be so gifted. But here is something where we don't have any hydrogens-- I mean, there are hydrogens, but they're implied. And we don't put carbons here. We just say, at every vertex, there is a carbon atom. And so we consider these things. And every organic chemist knows, if I draw something like this, that either I was stupid and I drew something that was impossible or what the structure is. OK. Now, this is a conjugated system. We have double bonds and single bonds. And that kind of thing is known to be unusually stable. And in order for it to work, it has to be planar. If you have a conjugated system which is not planar, then it's not as stable. So Huckel theory is based on the existence of unusual stability of conjugated systems. And it can be extended to non-conjugated systems-- to non-planar systems-- and we have rules for how to do that. But the simple rules are, you start with a picture, and you can write down a Hamiltonian. And it's a toy model, but it's still something that the computer has to diagonalize. So when we're doing a variational calculation expressed in matrix language-- We have a Hamiltonian matrix. We have eigenvectors, and we have eigenvalues. And we have the overlap matrix. And we have something where this is the alpha-- where, here, we can have non-orthonormal basis functions. And this makes them normalized. So these are the orthonormalized basis functions or basis vectors. And this is the kind of equation we have to solve. It's the generalized eigenvalue equation. And we don't like this, because it doesn't have the simplicity of just a Hamiltonian, where we have matrix elements along the diagonal and somewhere else. And the idea is, we want to take the secular determinate and make it 0 by adjusting values of the energy differences along the diagonal. But when we have this overlap matrix, it's complicated. Now, there's a way of dealing with the generalized eigenvalue equations, but one way to deal with it is to say s is equal to the unit matrix. You can do anything you want. And it's basically saying, there is no overlap between orbitals on adjacent atoms. We're going to neglect it, and then we're going to bring it back if we need it. But it's a wonderful simplification, because it enables you to write a simple, effective Hamiltonian, which looks just like h c alpha is equal to e alpha c alpha. OK. This, we know how to solve. And we can use the same procedure. OK. So these are the rules. If we have a planar molecule, we can say there are p orbitals-- one on each carbon atom or one on each atom that's not hydrogen, which are perpendicular to the plane of the molecule. So easy orbitals. And those are special. They give rise to pi bonds. Pi bonds are bonds where there is one plane of symmetry containing the bond, And then there are p x, and p y, and s. And these give rise to sigma bonds. So we have pi bonds and sigma bonds. Never the twain shall meet. And we don't care about the sigma bonds, because anybody can make sigma bonds. But only the special, perpendicular-to-the-plane guys, which are responsible for the fact that the molecule likes to be planar. And so we're only going to consider these p z orbitals-- one on each atom. So we don't care about no pi-sigma interactions. And we're going to neglect the sigma orbitals, because they take care of themselves. And we can do anything we want. It's just a question of, if we make too many ridiculous assumptions, we'll get ridiculous results. And this has been time tested, and it gives pretty useful stuff. And it provides a framework for making arguments about molecular structure and molecular reactivity. In organic chemistry you learn about resonance forms. And this is compatible with generating the resonance forms and saying, what is the relative importance? And what is the charge distribution, and bond strength, and everything like that? So it's really useful using the most primitive tools that organic chemists introduce at an early stage in your education. And that's one of the reasons why a lot of people become organic chemists, because it's so beautiful. OK. So our h i j matrix-- So we have a bunch of matrix elements, and we say, OK. h i i is equal to alpha. So every carbon atom has its alpha value. It's the same for all carbon atoms, regardless of who is nearby. And we have h I i plus or minus 1 adjacent atoms, and it's beta. That's it. That's the whole ball game. And it's really a simple-- There's far less here than in LCAO-MO. But it's still a toy. Both are toy models, and they're both very useful OK. And everything else which is not diagonal or off diagonal by 1-- 0. That's really convenient. And so you can draw the h matrix, regardless of what it looks like, as alpha, alpha, alpha, alpha, et cetera, along the diagonal, and beta, beta, beta, et cetera, beta, along the near diagonal. And if it's a ring, you have a beta here, here, and here. So that's pretty simple. So we have 0s, 0s-- so that it has a tri-diagonal structure with something up here and there. Never forget this thing here and there, which is present when you have a ring, and it's not present when you don't. That's it. That's Huckel theory. It's just there. OK. And so now, things can be more complicated. So if we're not content with what we get from the really primitive theory, we can do something like saying, well, we can make beta be dependent on internuclear distance. If the molecule, for some reason, is constrained to have not equal bonds lengths. So we can add an additional parameter-- some kind of reason for this beta to be dependent on r. But that's already getting sophisticated. And heteroatoms-- In other words, if you have a nitrogen instead of a carbon in a benzene-type ring, you can have-- So, well, nitrogen is different from carbon. It has a different-- In LCAO theory, the distance-- The ionization energy for nitrogen is different from carbon. It's larger. And so heteroatoms can be included if you use a different value of alpha. And now, alpha and beta are both negative. Now, this is a little bit fraudulent. Yeah. AUDIENCE: Do you also have to pick a new beta? Or-- ROBERT FIELD: Yes. But the main thing is alpha. AUDIENCE: I see. Why is that? ROBERT FIELD: Why is that? Because beta comes from overlap, even though we're neglecting overlap. And so the bond distances between normal and heteroatoms are not usually that different. But the thing is, you put what you need into the model. And the first thing you do is, you solve the most simple model. And you say, this is not quite what I wanted. And so I allow a couple of extra degrees of freedom. And it's really instructive how these things work. But the thing is, you're not calculating the matrix element of a real operator. It's all make believe. But it's really powerful, because what you're comparing is families of molecules. And the reality might be really complicated, but the complexity in each member of the family is pretty much the same. And what it's allowing you to see is, what are the differences? It allows you to see the big picture. I love this. And in fact, at an early stage of my education, I thought Huckel theory was wonderful. And it was what got me interested in quantum mechanics. Because you normally see Huckel theory before you know anything about quantum mechanics, because it's just a game. OK So heteroatoms-- You can fiddle with alpha. Now, the ionization energies for carbon and nitrogen are not that different. I'm sorry. They're very different, but the effect on the alpha value in Huckel theory is very small. Well, so it is. But the more electronegative or the higher the ionization energy, the alpha value becomes increasingly negative. Now, I was starting to say something is fraudulent, and I was distracted by a really good question. Alpha is on the diagonal. And we know we can determine the sign of a diagonal element. And we know we can't determine the sign of an off-diagonal element. But we say that alpha and beta are both less than 0. And the reason for this is that, when you saw the secular equation, especially-- You get two eigenvalues-- one where you have minus beta and one where you have plus beta. And so in a sense, beta is present, but it's just a sign choice. And since alpha and beta have to do with stability, we just say alpha is negative. We know that. And beta is chosen to be always negative. There's no, you could have had beta be positive. And you could do all the theory. It's just a lot more complicated explaining all the cases. All right. So let's continue with this. And so we can do heteroatoms. So if we have a non-planar system, we can say that beta is a function of the dihedral angle. We can put that in. We can do anything we want. We have a molecule. We're trying to describe its properties relative to the normal members of the group, which are planar and no heteroatoms. And we can do stuff that will accommodate these interesting differences, which you can impose by you putting the molecule in a constrained environment or doing stuff that distorts the geometry. So we do this. So when we do this, we get a Hamiltonian. We get the energies of the orbitals, and we get the eigenvector that corresponds to each energy. And the total energy involves the sum over the energies of each of the occupied orbitals-- the number of electrons in that. And we can also get bond order and charge. So sometimes, we want to know, what is the charge on each atom if the charge is going to be different from 0? Because that also controls chemistry. Negatively charged atoms are sought out by certain classes of reactants. Anyway, so you get all this stuff. And of course, some of it will require a little bit of patchwork, but you do this for your career. And you discover that there are certain things that I know how to handle. And I use my favorite parameters for it. And you get closer to the truth. And you always want to be surprised, because when the crude theory cannot be made consistent with observations, you know you did something special. You have a molecule, which has a property which is unexpected, and we like that. OK. So when you're doing quantum chemistry, there are five steps-- or when you're doing molecular orbital theory. And this is one of Troy's rules. We have a five-step procedure. And we define the atomic orbital basis set. And the basis set is one p z orbital per atom. And so we can have a molecular orbital, which is the sum i equals 1 to n c i u p z i. So this is the p z orbital on the i-th atom. This is the coefficient for this particular linear combination. OK. So we have a bunch of molecular orbitals. Next step-- compute h and s. But s is 0 because we made this ridiculous assumption. It's convenient. This is called the complete neglect of overlap. And we can do that. It's wrong. But when you include overlap, it leads to greater complexity of the calculation, and not much improvement in the results. And so we normally don't worry about the overlap. So all we care about is this. And as I said, h has this structure of alpha, beta, beta-- tri-diagonal structure and maybe something up here. That's it. Then we diagonalize the Hamiltonian. And so this gives us the eigenvectors or eigenvalues-- the energies and eigenvectors associated with each orbital energy. So we fill electrons into orbitals. Now, for benzene, what you would get-- Now, this is important. How many carbon atoms in benzene? Right. And how many orbitals will you get from six primitive orbitals? Right. And so there are six energy levels. And it happens that when you solve the secular equation, you get this pattern. This is a problem where the number of nodal planes determines the order of energy. And so if you have benzene, you can imagine that the orbitals that you can have will be no nodes-- nodal planes-- one nodal plane. And you could have the nodal planes between the atoms or through atoms-- and two nodal planes or three. You can't have any more with six atoms. And so you almost don't have to solve this secular equation at all. You can anticipate what the structure is going to be just by saying, no nodes, one node, two node, three node. And you can even anticipate double degeneracies, because if you have two nodal planes, you can put them through opposite bonds or through opposite atoms. And those are examples of the forms that you would deal with. The only thing you can't do is to know what the order the energies are. But there are tricks for that, too. The tricks for that include-- The sum of the eigenvalues is equal to sum of the diagonal nodes. And so we know that the six orbital energies will sum to 0-- will sum to 6 alpha. The betas go away. And there are other tricks that you can do, but generally, you solve the equation. You don't want to push your requirements for symmetry too far. So we end up doing that. And then stick diagrams-- We fill electrons into orbitals in energy order. And for benzene, there are six of them. And so this is, then, the lowest energy state of benzene. This is called the independent electron approximation. They don't know about each other. This is illegal. But it's legal in a sense, if you have two electrons in every orbital, you have nothing but singlet states. The ground state is always going to be a singlet state unless you have something really weird going on-- like in O2-- But that's not Huckel theory. But it's examinable. OK. Why does oxygen have a triplet ground state? That's something that every textbook says. You got to know that. And of course, they don't really tell you anything more except something to memorize. OK. So generally, you have two electrons in each orbital in singlets. Now, if you're going to do spectroscopy you would perhaps promote one of these guys to a higher state. And so you'll have singlets and triplets. But of course the only transition you would see would be a singlet-to-singlet transition. So you might as well forget about triplets. And you might as well forget about having to add [INAUDIBLE]. You can get away with murder, especially with Huckel theory. So we have a stick diagram. We fill electrons into the thing. And then we compute the energy of the many electron problem. And so let's just do this. And you've all seen this. So we number the atoms. We have our symmetric structure. And psi mu is going to be sum from i equals 1 to 6. C i mu p z i. And c mu is c 1 mu c 2 mu to c 6 mu. These are the mixing coefficients. Well, for benzene, it's pretty simple because you know that if you have no nodes and symmetry, all of these are going to be the same. And so if you put 1s here, you put a 1 over square root of 6 out in front for normalization. And you can figure out the eigenvectors for benzene-- all of them-- just by counting the number of nodes. And that's useful, because if you know the eigenvectors, then you can show what the eigenvalue is by multiplying the original Hamiltonian by an eigenvalue. So anyway, we have this. And then the Hamiltonian-- we have alpha, beta, and then 0s. And beta, alpha, beta, and then 0s, et cetera. And so we have this tri-diagonal structure. And because it's a ring, we have a beta here and here. OK. So then, we ask our computer to diagonalize this, or we use clever tricks from linear algebra to find the eigenvalues, but I don't recommend it. I mean, the general problem is going to be something where you have to use a computer. So don't develop tricks unless you want to check to see whether you programmed the computer correctly. And so when you do this, you get-- E 1, the lowest energy, is alpha plus 2 beta. E 2 is alpha plus beta. And E 3 is alpha minus beta. I'm sorry. E 2 and e 3 are both this. And E 4 and e 5 are this. And e 6 is alpha minus 2 beta. Now, we don't really care about these orbitals, because they don't put electrons in them. And so when you put the electrons in the orbitals, you get 2 for this, 2 for this, 2 for that. OK? And so we just calculate the sum of the energies. And we end up getting the energy levels for benzene. The ground state is going to be 2 alpha, plus 2 beta, plus 2 alpha, plus beta, plus 2 alpha, plus beta. Right? So this gives you 6 alpha plus 8 beta. Now, this-- You might have guessed it, but you didn't guess it. Your computer told you that. And the computer actually likes numbers rather than symbols. And so you actually obtain this simple structure from the computer-- requires a little bit of manipulation. But you still get 6 alpha plus 8 beta. And you also get the eigenvectors. And that's the stuff that you know. So c 1 is 1 over square root of 6. 1. 1. All of those. And c 2 is going to be something a little bit more complicated. We can have the nodal plane going through atoms. And so if it goes through atom one, it's also going to go through atom four, and so we have 0 and 0, 1, 1, 1, 1. And now, to figure out how to normalize that, we just-- 1 over square root of 4, and so on. We can figure these things out. C 3 is going to be an eigenvector. Now, instead of having the nodal plane going through atoms, it's going between atoms. And so instead of having any 0s, you're going to have something more complicated. And now I can see that there is something in my notes which is subject to how you'd actually impose the symmetry. But suppose you have something like this-- 2, 1, minus 1, minus 2, minus 1, 1. So why did I use these numbers? Well, I had to have a nodal plane here between atoms two and three. And the corresponding guy will be-- Well, that should have been 2 at the bottom. Let me just make sure I'm doing this right. No. I had a 2. OK. It's a 1. OK. And so the last 1 is here. So the sign change occurs twice. And those correspond to opposite bonds. Why were there 2s? Well, the 2s are between two atoms that have 1s, and so it's going to have a larger eigenvector. And you can figure it out lots of ways. You can also say, well, every atom has to be used up. Yes. AUDIENCE: So should there also be a sign change? ROBERT FIELD: I can't-- AUDIENCE: Should there also be a sign change in c 2? AUDIENCE: Yeah. In c 2 eigen c 2 6 should be [INAUDIBLE].. ROBERT FIELD: Did I screw up? AUDIENCE: Yeah. No, not-- in c 2. Not c 3. ROBERT FIELD: I'm sorry? AUDIENCE: The previous one. ROBERT FIELD: Oh, yeah. Yeah. So opposite sides. OK? All right. It doesn't matter. The computer tells you. But sometimes you can approach the problem very quickly, and just say, I know what the eigenvalues are going to be, and I have to normalize them. But one of the things that you often do if you're trying to be smart and skip steps is to say, OK. We have six eigenvectors. And each of the atomic orbitals gets used up completely among the six eigenvectors. And I do that all the time, because it's a very useful way of making sure I haven't screwed up. OK. So we get this, and that's perfectly OK. We don't know how significant it is, but say we had three of these-- three ethylenes-- I guess an organic chemist would just do that, right? But anyway, if we had three of these, when you do this you get 2 alpha plus 2 theta. OK? So we get 6 alpha plus 6 beta. And alpha and beta are both negative. And so the energy for three ethylenes-- did I screw up again? AUDIENCE: How are those ethylenes arranged with one another? ROBERT FIELD: They're separate. We could draw benzene. And so we could say, we have three isolated bonds. And since we don't care about the sigma structure, and no bonds are adjacent, you know that they're additive. And so we represent benzene. The primitive structure for benzene is three ethylenes. And benzene is 2 beta better. AUDIENCE: Yeah. ROBERT FIELD: That's the resonant stabilization. That's a great thing. OK. So we get a resonant stabilization. And now, the more subtle and wonderful stuff is, we have these eigenvectors. And we have the energies. And we can do stuff with them. And one thing is bond order. And so we have a formula. The bond order between atoms i and j is equal to the sum for mu equals 1, including only the occupied orbitals, of c i mu c j mu. So we have the new molecular orbitals, and we have the adjacent atoms. And this gives you the bond order. And so we can calculate the bond order. And we find that every bond is a pi bond order of 2/3, which is neat, because the non-resonant structure says, three of the bonds have a pi bond order of 0, and 3 have a bond order of 1. And so the average is 1/2. And this is bigger than 1/2. And it's uniform. So you can do this, and you can say, OK. The 1, 2 bond order-- the 2, 3 bond order-- you do the laborious calculation-- always get 2/3. Now, you could also calculate the charge on atom i. And that is, again, mu for the occupied orbitals. And that would be c i mu c i mu. And so for benzene, you would expect this to come out to be 0. And it does. So there is no charge. Let me just make sure that that-- Well, there is an equal charge on each atom, whether it's 0 or 1/6. That I haven't done. And I am a little uncertain about what this is going to come out to be. But they're going to be equal on every atom in benzene. And so now, suppose, instead of benzene, you have amylin. Well, you can do stuff. And you could say, well, the amylin out here is going to affect the alpha value here a lot, and here less and less. And then you do the calculation. And you find when you do this calculation, you get unequal charges. And you get the normal rules for ortho versus meta. And everything is great. And you also, when you do this, you can actually write resonance structures and you could calculate, well, what is the energy of that resonance structure-- in a Huckel-like theory. So there are lots of things you can do in order to say, OK. We do Huckel theory to get this 0 in our picture. And then-- I'm way over time. And then we can add special effects, and we know how to parameterize them. And if we're close to being OK, we'll get results that correspond to experiment. And this is so much better than you deserve, because it's all garbage. But it's empirically-calibrated garbage. And it it's calibrated over an enormous number of molecules and a huge amount of experience. And that's good. That's what we do. OK. So I'll see you on Friday with some perturbation theory. One of my favorite problems, too. OK. |
MIT_561_Physical_Chemistry_Fall_2017 | 22_Helium_Atom.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: I'm really sick today. And I hope I can make it through this lecture without disgracing myself. But this is a favorite topic of mine, too, so maybe it will be OK. So we're going to talk about the helium atom, which is really an amazing thing because hydrogen-- it was possible to solve it exactly. All sorts of fantastic things happened. And it seemed like just a small step from hydrogen to helium. And nobody can take that step. It's entirely approximate. And you can do better and better on the approximations, but we'd like to have some physical sense of what's going on, and what are the important terms, and can we steal things from hydrogen? The answer is, yeah, we can. And that was what last lecture was mostly about. The energies for hydrogen are minus z squared hc times the Rydberg over n squared. And we have the ionization energy from the nth level. And so this actually is the way we make a link to this concept of electronegativity because the dominant thing in electronegativity is the ionization energy. And this is the ionization energy. And when you put them together and solve for n, you get this arrangement hc z squared times the Rydberg over the ionization energy, square root. That's n, or the effective principle quantum number. And then we have all of these formulas that can be derived from hydrogenic wave functions, both for diagonal expectation values or off diagonal matrix elements. They're all closed form. And this is wonderful because it then enables you to use the formulas that are obtained exactly for hydrogenic wave functions. And so if we know the dependence of an electronic property on n and l-- well, n is experimentally determined-- distance real ionization. l is determined from angular momentum. Then we can determine the expected scaling of everything on the quantum numbers. And even if that's wrong, it tells you the qualitative effects when you get to more complicated systems in one electron atoms. And this is really what I meant by structure, where everything is related to basically one observation. There is one other thing that I didn't stress. And it's another concept from 5.112., and that's shielding. So suppose we have a sodium atom. Most of the electrons are in a compact core. The outer electron sees a charge of 1, even though the charge on the nucleus is larger than 1. And if you ask, well, suppose we look at this in a region, and we have an electron inside of it, the charge that that electron sees is the amount of electrons inside the sphere defined by this radius. And so the z, we can call it the z effective, is a function of r. And you have a different function every o. And so it's a way of putting in more insight. And I've made a lot of mileage in my career on these sorts of scaling arguments. OK, so helium-- the notes are beautiful. They're written by Troy. And they're very long. And they're very correct and incisive. And even if I weren't sick, I couldn't cover all those notes. So I'm going to make a rather express trip through the notes. And the critical thing is we have two electrons. And so we can't simply solve the problem by using hydrogenic functions, although we would like to. And the crucial thing that is missing is if we could ignore the interactions between the two electrons, which are enormous, then we would be using hydrogenic functions, and everything would be fine. So we can't. And we use what's called the independent electron approximation, where we're still using hydrogenic orbitals and energies, but we're using those to include the first order perturbation correction to the energies. So there's a term, the interelectron repulsion that leads to this extra correction energy. And that's most of the difference. But then, we have identical particles. And when we have identical particles, the permutation of those two particles has to commute with the Hamiltonian because the Hamiltonian only knows that we're dealing with all the electrons. It doesn't put names on the individual electrons. And so there's a permutation symmetry that has to be imposed. And this is very subtle because the permutation symmetry has nothing to do with the energy, but it has to do with the energy states. And so we'll explore that. Now, since we're going to be dealing with more than one electron, there's a lot of symbols that we're going to be carrying around. And so there's a useful thing called the atomic units, which sets most of the physical constants to 1. And that means all of the rubble that you have to carry around goes away. And it's not a big deal because if you use all of the things in your calculation in atomic units, namely value 1 or something like 1, then at the end, you're going to get the energy in atomic units. And we know what that is in conventional units. So it's a risky thing because you don't really know what you're doing. But at the end, you can probably convert your number, which is a pure number, into an actual energy. OK, so let me start with the good news. Well, maybe I should leave this exposed. The good news is we have two particles for helium-- vector r, momentum, momentum p vector for particle 1, vector, momentum. OK, so these are the actors. And when do we have to worry about commutation or non-commutation? Well, the answer is the coordinates associated with different particles, they contribute to the energy. But there is no problem of commutation. The momenta don't also commute. The coordinate for particle 1 and the momentum for particle 2, they commute. The only things that doesn't commute is r1x, p1x-- those kinds of things. They're commutators ih bar, and that means there's some problems. So most of the variables are commuting variables, but some are not. And you have to deal with that. OK. We know that if we have, as, for example, in a three-dimensional harmonic oscillator, if we have a Hamiltonian that doesn't commute-- if we can divide the system up into parts, then the Hamiltonian for each part commutes with the Hamiltonian for the other part, which is part of this. Then, you can write the Hamiltonian as a sum of individual system Hamiltonians that commute with each other, the energies as the sum, and the wave functions as the product. And so we like that. And we'd like to be able to build a picture for helium that is like that. And so if we write the Hamiltonian for helium, we have p squared for particle 1 over to me with p squared-- I don't know. Le me see. Let me organize better. Minus zb squared over 4 pi epsilon 0 r1. And then we have another term like this for particle 2. And we can call this Hamiltonian 1. I don't want to do that. Hamiltonian 1, and this Hamiltonian 2. And these are hydrogen atom Hamiltonians. And we know everything about them. So it's a fantastic way to build up a basis set consisting of energy, zero-order energies and zero-order wave functions, with which we can calculate anything. And so then we have this one other term, which plus e squared over 4 pi epsilon 0 over 1 minus r minus r2 vectors. That's the bad news. This says we have a Hamiltonian, where we can't use the eigenvalues and eigenfunctions of the hydrogenic things unless we do-- so we can say, well, these give the first-order energies for whatever state we want. But we can calculate matrix elements of this term in the basis set associated with hydrogen. And that's fine. That's a perfectly good way to calculate the first-order correction to the energy. But what about this? When we do perturbation theory and calculate the second-order corrections to the energy, what is that usually? Maybe I didn't ask the question clearly enough. What do you have to do to get these second-order energies in an infinite dimension problem? I just gave it away. Somebody has to have the courage to say something here. Remember, we have a sum of n not equal to n from 0 to infinity. AUDIENCE: Yeah. PROFESSOR: That's it. OK, I encouraged you. So the way we could use this basis set in a perturbation theory format is to calculate an infinite number of off-diagonal matrix elements of the hydrogenic wave functions. And that could yield results, but it's not the way it's done usually. There's a variational calculation that you use to get the better wave functions. But we're not going to do that here. We're going to go as far as we can without actually addressing what's wrong if we stop here. OK, so this guy is h1, the source of all trouble. And how we deal with it is there are three steps. One is ignore it. The other is use our hydrogenic functions to calculate matrix elements of this and get an h1, an e1. And the third step would be to get the exact answer by doing an infinitely difficult calculation, OK? All right. So this is going to be a notational voyage, which is horrible. Because we have many particles, and particles have coordinates and spins, we're going to have an explosion of stuff, even if we get rid of the extraneous units. But let's get rid of the extraneous units. And so atomic units-- we can choose any internally consistent set of units we want. And in atomic units, we say that the mass of the electron is 1. The charge of the electron is 1. The h bar is 1. 4 pi epsilon 0 is 1. That's good. A lot of stuff just going away. OK, and the energy in atomic units is not 1. The energy in atomic units is minus 2 times the energy of the hydrogenic 1s orbital. And that comes out to be 27.21 electron volts, in more convenient units. So you calculate a number, and then you multiply it by this to get the real energy. OK, and the unit of length is a0, which is the radius of the n equals 1 hydrogenic Bohr orbital, or Bohr orbit. And that's 0.529 angstroms, with a lot more digits if you need them, OK? And the speed of light is 137 in atomic units. So everything that you're going to need is either 1 or something that is related to-- OK, so you can do this, and I don't like it because it's so easy in a big calculation to get your units screwed up. But it's certainly great for lecturing and perhaps for writing computer programs. OK, so let's start with a non-interacting electron approximation. And in the notes, it's represented by these letters. I don't know whether that this is Troy notation or widely-adopted in quantum chemical circles. But it's easier to write this than that. And that just means we're going to ignore this. We just forget about it. Now, we know that's stupid because the electrons have charge of 1. And they repel each other. And the nucleus has charge of 2, in the case of helium. And so we know that this term is rather similar to h0. And we can't expect to get very far completely ignoring it. But we can do it, and we can do a calculation, and we can find out how well it works. So first of all, it's clear in the notes. I'm just going to skip to the answer. So the energy for helium in the 1s 1s orbital or 1s 1s configuration-- when we're going to use this word configuration, what that means is the list of orbitals that are occupied. Now, we're listing principal quantum number and orbital angular momentum. There is more stuff which doesn't get listed, and that leads to very interesting stuff. But if we have a list of the orbitals are occupied. That's called the electronic configuration. And I'm an electronic spectroscopist, so you can be sure this is something I care a lot about. So we can calculate this in the hydrogenic basis. And what you end up getting is-- where is it? I'm sorry. This is the experimental value. The calculated value is-- see, I'm trying to improve on my notes. And in my present state, I make a lot of mistakes. So that is 108.8 electron volts. So this is the energy of the 1s 1s state below the ionization limit. So that's what we predict. Except this is not the ionization limit. This is helium 2+ plus 2 electrons. Because we use the energy to ionize both hydrogen atoms, and so this is 108.8. Now, you can measure this experimentally. You can do helium to helium plus an electron, and then helium plus going to helium 2+ plus an electron. And the observed value is what I wrote before. Observed is equal to minus 79.0 electrons volts. Well, you might say, well, that's pretty good. It's the same order of magnitude. It's not even off by a factor of 2. But almost everybody in this room is a chemist. And this difference between the calculated value and the observed value is 30 electron volts. A chemical bond is typically 5 electron volts. So this is totally ridiculous for chemistry. We can't use it. We don't even want to think about it because it's an order of magnitude away from what we need. We improve on it by including e1 of n, which is related to h1, which is that simple-looking thing over there. So the interelectron repulsions can be dealt with, not exactly, but by calculating diagonal matrix elements of the interelectron repulsion. And that's what the independent electron approximation is. So that's all there is. But now we bring in something that we hadn't expected would be a problem. So we have this mystical, or mythical, or whatever operator-- p12 operating on some function of r1 and r2. And what it does is it permutes the electrons. We know that if something is operating on the names of the electrons, it has to commute with the Hamiltonian. So that means that this operator has to have eigenvalues if these states are eigenfunctions of the Hamiltonian. And there are two possible eigenvalues because we know p12 squared on psi is equal to plus psi. So the only way you can do that is having the eigenvalue of the permutation operator be either plus 1 or minus 1. And this is a fundamental symmetry. And I can't possibly tell you where it comes from or what I'm about to write on the board comes from. But so all half integer spins correspond to antisymmetry or belong to the eigenvalue minus 1. And they're called fermions. And integer spins are symmetric with permutations. And so integer or symmetric, and they're bosons. So we have to write wave functions which are wave functions of the electrons, which are antisymmetric with respect to the permutation of every pair of electrons. So helium, we only have two electrons. It's not so bad. But when you think about something like carbon monoxide, which I always think about, or some other molecule where there's more than two electrons, there's a God awful number of permutations. And you have to build them all into the wave function. And the normal feeling when you hear that is, oh, no, I can't possibly imagine how to do that. And if I could, there would be a ridiculous number of integrals that I would have to evaluate. And it's true. There are a ridiculous number of integrals. But there is a simple algebra that enables you how to deal with them. And it involves using what is called Slater determinantal wave functions. And I'll talk about that. Slater was an MIT professor. He wrote a lot of books. But the thing-- integer. Oh, they'res sort of a t. But the most important thing he did was to develop this way of dealing with an antisymmetric electronic wave functions. So the next thing we have to do is discover how do we build symmetric wave functions for the simplest possible case where there's two electrons? And already, we discover some surprising consequences for this. And so I'm going to go through this, but I'm going to skip a lot of steps because it's so clearly written in the notes and because we don't really have time to deal with 15 pages of single-spaced notes. And I'm not sufficiently at my best to get through maybe 12. And you don't want that, anyway. OK, so let us see what happens here. And so let's get to work. Suppose we have an electronic state 1s alpha spin. We know alpha is ms equals plus 1/2. And beta is ms equals minus 1/2. So this is the spatial coordinate. This is the spin coordinate. And, OK, so here is a simple product of two wave functions. And we know what they are. And so if we apply p12 to this, we get psi 1s alpha r2 sigma 2 1s beta r1 sigma 1. We don't even know what this is relative to that. So symbolically, these are two unrelated quantities. So weird, but they are eigenfunctions of the zero-order Hamiltonian. So we can take a linear combination of eigenfunctions that belong to the same eigenvalue, and it's still the same energy. So we do that. And so what we can do is write a combination of this function and that function. And if we do that, so now we can have psi 1s 2s. And what we get is 1 over the square root over 2 for normalization. And we have psi 1s alpha r1 sigma 1 minus psi 1s beta r1 r2 sigma 2. And this, in fact, is anti-asymmetric. If you apply the permutation operator to this, it is an antisymmetric function. So this does the job. Now, since the Hamiltonian doesn't depend on spin, the energy of this state and the energy of that state are the same. They are hydrogenic. And so using this wave function rather than the individual non-symmetrized functions doesn't change the energies. So are we wasting our time? AUDIENCE: I think you didn't [INAUDIBLE].. This should be the product of 1s alpha 1s beta. PROFESSOR: I'm sorry? AUDIENCE: So the [INAUDIBLE] must be [FINGER SNAPPING] PROFESSOR: Yes, yes, yes. AUDIENCE: Minus the other one. PROFESSOR: I just want to see what I wrote here. So yeah, what I wrote is wrong. 1s alpha, and we'll just put 1 in here. 1s beta 2 minus 1s alpha 2 1s beta 1. OK, this is what I should have written. And this is, in fact, antisymmetric. Thank you. I wrote what's in my notes, which is wrong. And I didn't realize I had done that because I knew the answer. All right. OK, so we can use this kind of function. It's legal because it's an eigenfunction of the permutation operator with negative symmetry, which we're told that's what we have to use for electrons. I can't tell you where it comes from, although it came from Pauli. And the exclusion principle really is this, rather than what you have memorized. Its consequence is that you can only put one electron in an orbital. But this is the exclusion principle, and it is Pauli. OK, so now we know what to do. And we can calculate the matrix elements. psi 1s alpha go in 1s beta. This is the antisymmetrized product of two electrons. Matrix element 1 for Hamiltonian 1 plus 1 over r1 minus r2 psi 1s alpha 1s beta. Though this is a perfectly legitimate thing to calculate and we know how to do it, or McQuarrie knows how to do it. And you don't really need to worry about doing these integrals because this is at such a low-level approximation, it does nothing for you except teach you how to approach these problems. But we can still do these integrals. And it's useful. And so what do you get? Well, you get minus 4y because these are hydrogenic with a nucleus of plus 2. And in the atomic units, the energy of a hydrogen atom charge of 1 is 1/2, not 1. And these go as z squared, so we get 4 over 2 plus 4 over 2. So we get this minus 4 or you get minus-- yes. And then we get something else. And that's this integral that we don't like, which is integral of 1s alpha 1s beta squared over r1 minus r2 dr1 dr2. So we can do this integral, or Mr. McQuarrie and many of his predecessors can do that integral, and you get a result. And this integral turns out to be 5z over 8. And converting that into units, and that's 5/4 au. And that's-- why is that? That doesn't seem right. Anyway, it's 34 electron volts or minus 34 electron-- plus. Plus 34 electron volts-- so the interelectronic repulsion for hydrogenic orbitals on the same nucleus is 34 electron volts. It's a big number. And so we can correct our the calculated versus observed energy. And instead of having-- well, the experimental value is 79. And now, the calculated value is minus 74.8 electron volts. So this is marginally relevant to chemistry because the error is about the typical bond energy. And so we can do stuff with this. We know that it's not accurate enough to do anything quantitative. But you can use it for qualitative arguments. And we can do much better than this using what's going to be called the variational principle, which is coming. OK, so now, we're going to be dealing with a problem that has lots of symbols and equations. And it's best to start using notation called stick diagrams because it's completely transparent what they mean. And so it's a shortcut for writing the correct equations. So we have for the ground state of hydrogen-- I'm sorry. We want to talk about excited states because 1s with 1s doesn't have enough complexity to attract our attention. And when we deal with excited states, 1s and 2s, we discover something important that's new. And it's the last thing we really need in order to understand electronic structure or how the electronic states that arise from a configuration are distinguished. And so we have for the 1s 2s configuration, there are four ways of writing this in stick diagrams. So we have 1s 2s, and we can have spins up, alpha alpha. And we can have 2s 1s. We can have spins up and down. 2s 1s, and spins down and up. And 2s 1s spins down. So it's much easier to write these things because they have an equation to associate with them, and you can play with these symbols. So the first thing you notice is this symbol can be written as an antisymmetrized product. And so this symbol-- we just take the alpha alpha. And so let me just write it out. 1 over square root of 2 times 1s r1 2s r2 minus psi 1s r2 psi 2s r1. And then times a spin part, alpha-- and we can just abbreviate the 1-- and alpha. OK, so we have factored the wave function to a spatial part and a spin part. Now, I'll make an outrageous statement. Doesn't matter how many electrons. You'll always be able to factor it into a spatial part and a spin part. Now, what is necessary is always this function has to be antisymmetric. And in this case, it's the spatial part that's antisymmetric and the spin part that's symmetric. Now, when we try to do these two stick diagrams. we cannot write a correct equation for this guy. We have to do both of them. We have to combine the two. And that will give us a separated spin part and spatial part. And it's antisymmetric. You can't make in an asymmetric function of this just switching the coordinates. And what you find when you do this is going to be that we get functions which look like alpha beta plus or minus beta alpha. OK, so spin 1 alpha spin 2 beta plus or minus spin 1 beta spin 2 alpha. And it turns out that there are three spin functions that we get, which are symmetric. And all three of these are associated with this antisymmetric spatial function. They have the same energy. And since they're three of them, we call them a triplet. And since we're talking about spins, this triplet corresponds to s equals 1. You use s equals 1 and creation and annihilation operators or raising and lowering operators to show that these are, in fact, the three eigenstates of s sub z. So we call it a triplet because they're three of them. And it also corresponds to s equals 1. And there's one, alpha beta minus beta alpha, which is antisymmetric and is associated with a different spatial function, which is symmetric. And it has to have a different energy because the Hamiltonian operates on the spatial part and not on the spin part. So antisymmetry has forced you to have two different spatial wave functions, and that guarantees that these guys will have different energies. And since there's only one of them, we call it a singlet, and it corresponds to s equals 0. So now, we want to know, which is higher in energy, the singlet or the triplet? And how much higher? And so, again, we take the wave functions that are correct as far as permutation is concerned, but only approximate as far as using hydrogenic functions, and we evaluate the relevant terms in the Hamiltonian. OK, so we can forget about the spin parts now. We know we're going to have two different spatial wave functions. And we're going to want to calculate-- yes? AUDIENCE: For the top state, [INAUDIBLE] both spins [INAUDIBLE]? PROFESSOR: That should be? AUDIENCE: Would that also be s equals 0? PROFESSOR: I can't hear. I understand the last word. AUDIENCE: With the statement, both spin be also s equals 0? PROFESSOR: No. Both spins up. The vector sum of 1/2 and 1/2 is 1. And so the total spin is going to be 1. And so we have 3-- I mean, if we're applying s minus-- suppose we start with-- AUDIENCE: I meant that the top spin [INAUDIBLE] both spins. PROFESSOR: Sorry. I can't hear properly. AUDIENCE: The topmost state with both spins on it? PROFESSOR: Yes. AUDIENCE: Would that have s equals 0 or s equals 1? PROFESSOR: It has s equals 1. OK. And now, I don't want to go into the angular momentum matrix element. That would be the last thing in this lecture. So yes, the projection quantum numbers are not uniquely associated with the total spin. But if you have alpha alpha, you can show that it's s equals 1. And alpha beta plus beta alpha is 1. And beta beta is 1. But alpha beta minus beta alpha is spin 0. You can do that using creation and annihilation operators. It's clear, OK? So you have to take it on faith until you prove what I've just said. And it takes about two minutes. OK, so we can evaluate the integrals. And something very beautiful occurs. And it's done very, very clearly in the notes. And it would take me more time than we have left to do it, and I didn't plan on doing it anyway. So we get the energies for the singlet and triplet states, which are E 1s plus E 2s. Now, remember, these are for a 1s orbital on helium, so it's going to be four times the energy of the 1s orbital on hydrogen. But we know what they are. Plus what we call J 1s2s plus or minus K 1s2s. So this is clear. And the integral associated with this is taking the square of the wave function. And basically, we're evaluating the interaction between an electron 1 in the 1s orbital and electron 2 in the 2s orbital. So that's a perfectly doable thing. It's a classical quantity. We have two charged distributions, and we're calculating the energy interaction between two electrons in these two charged distributions. So this is called the Coulomb integral. And that is an entirely classical quantity. And this is called the exchange integral. And the reason we call it an exchange integral is, basically, we're switching the electrons between the two orbitals. And this is something which is entirely quantum mechanical. It comes about because of the permutation symmetry that's required for fermions. And so this is the surprise. And it also depends on whether you have a symmetric, or an antisymmetric, or a single, or a triplet. And so you end up getting an energy level diagram for the state of a configuration, which looks like this. So we start out with E 0 1s2s. And then we allow the charged distributions to interact. And the electrons are going to repel each other. It's a positive energy. And so this is E 0 plus J. And then, we include the exchange interaction. And so this is the difference 2K. And this is E0 plus J plus K. And this is E0 plus J minus K. And this is the triplet state. And this is a singlet. And for helium, we call the singlet state singlet s because total angular momentum is 0. And spin, it has 0. And the triplet we call a triplet s. Now, if we did 1s 2p, we would get singlet and triplet p states. But the important thing is, how big is K? And remember, K is an approximation. It's a real quantity, but for hydrogenic orbitals, it's going to be incorrect. And so 2K 12 is 2.4 electron volts. And that's a perfectly doable calculation because you're dealing with simple functions. And you can do this integral. And you can actually do it in closed form. But it's 2.4 electron volts. And experiment, it's 0.8. And this tells you something really important. The repulsion between two electrons, when you have the correct wave function, is less than it would be if you're using the wrong wave functions. And basically, whenever you do anything using the wrong wave functions, your energy levels are too large or the energy differences are too large. And in this case, the electrons. And when you solve the exact Schrodinger equation, the electrons know about how to absorb it to avoid each other. It's called correlation energy. And this is what makes all the calculations hard because there is no closed form analytic expression for the correlation energy. And the way you optimize the correlation energy is throwing a huge basis set at the problem. And this is why it takes a long. And you never know that you've converged until you fill in another million functions. OK, but so there are several lessons. And that is we have something that's completely classical, and we understand what it is. And then, we have this extra thing that is a consequence of the permutation symmetry. And so this extra thing we calculated too big. But it says that if we have states that belong to the same electronic configuration-- in other words, if we specify the orbitals that are occupied, there's going to be more than one electronic state. And so configurations split into many different electronic states. And we know the machinery that enables us to calculate their relative energies. And some of it is simple Coulomb's law. And some of it is something that's caused by the required permutation symmetry. So I'm over time. But I can't believe that I got to this point because I delivered everything that was really important. When we have more than two electrons, everything I've said is true, it's just more complicated. And we have to evaluate matrix elements between determinantal wave functions, which looks like, if you have n electrons, there's n factorial terms in a determinate. And we have n factor l squared matrix elements. And you don't want to do that. But there turns out to be a simple algebra that makes it all pop out, almost with the same level of effort as this. And so we'll talk about many electron atoms, and then we'll go on to molecules after the many electron atoms lecture. OK. |
MIT_561_Physical_Chemistry_Fall_2017 | 7_Classical_Mechanical_Harmonic_Oscillator.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality, educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: Last time, I talked a lot about the semiclassical method, where we generalize on this wonderful relationship to say, well, if the potential is not constant, then we can say, well, the wavelength changes with position. And we can say that the momentum changes with position. But we're using this as the guide. And so the basis is really just saying, OK, we're going to take this kind of a relationship seriously. Because v is not constant. We have V of x. And we also know that the kinetic energy, which is an operator, is p squared over 2m. And the energy minus the potential is the kinetic energy. And so we can use this to get a classical, mechanical function that this p of x is going to be 2m E minus V of x. So why are we doing all this when we can just solve the differential equations? And the answer is we want insight. And we want to build our insight on what we know. And so we have this momentum function, which gives us a wavelength function, which tells us how far apart the waves are. But we also have another thing, which was demonstrated by my running across the room, and that the momentum is related to the velocity, which is related to the probability of finding the system at a particular place. And so we have, if we commit the travesty of saying, OK, we have a classical function-- it's not the momentum-- but it's going to somehow encode the classical behavior, we can determine without solving any differential equation what the spacing between nodes in the exact wave function is, and the amplitude in that region. And that's a lot because then we can use our knowledge of classical mechanics to say, oh, this is what we expect in quantum mechanics. And that's very powerful. And I keep stressing that you want to draw cartoons. And this is the way you get into those cartoons. OK, now, for the free particle, we have these kinds of wave functions, e to the ikx. And they're really great because often, we want to evaluate things like the integral of e to the ikx times e to the minus ikx dx. Well, that's one. And so instead of remembering that we have to evaluate trigonometric integrals, sine, sine theta, cosine, we can do this. And it really simplifies life. Now, using that insight, I'm asking you, OK, the expectation value for the momentum could be-- OK, if this is the expectation value for the momentum, what is psi of x? I promised I would ask you this question. I don't know if anybody really thought about it. But first of all, we're talking about the free particle. And this is some sort of eigenfunction of the Hamiltonian for the free particle. So what can we say about k? We have two parts to the wave function. We have an e to the ikx. And we have a e to the minus ikx. And the k is the same for both of them. So with all those hints, what is this? We have e to the ikx times something. And we have plus e to the minus ikx. So what are the somethings that go here? Yes? AUDIENCE: A and B? ROBERT FIELD: Right. And that didn't involve very much mental gymnastics if you really have done a little bit of practicing of integrals involving these sorts of things. Now there's two things. One is when you have the same k in e to the ikx and e to the minus ikx. The product is one. If you have different k's and you're integrating over all space, or over some cleverly chosen region, that integral is zero. Because these are eigenfunctions of-- these are different eigenfunctions of an operator belonging to different eigenvalues. And you always can count on those things being zero. Now, quantum mechanics is full of integrals. Basically, there's an infinite number of them. And most of them are zero. And you want to be able to look at an integral and say, oh, I don't need to evaluate that. And often, you want to look at an integral and say, I do know how to evaluate that. And I know an infinite number of those like that. And all of a sudden, it starts to be transparent again. Because the barrier between insight and quantum mechanics is usually a whole bunch of integrals. And they're all yours. And so we like problems where the wave functions have simple forms. And this is true for free particle. It's true for the particle in a box. We're going to start talking about their harmonic oscillator. And it seems like those integrals are not simple, but they are. I have to teach you why they're not simple. So today, we're starting on the classical mechanical treatment of the harmonic oscillator. Then we'll do the traditional quantum mechanical treatment. And then, we'll come back and use these creation and annililation operators, which are the magic decoders for essentially evaluating all the integrals trivially. And then, with all that in hand, we're going to make our first step into time-dependent quantum mechanics. And we're going to use time-dependent quantum mechanics. Well, we're going to use our facility with integrals to describe the properties of some particle-like state we can construct. And these constructions are really simple for the particle in a box or the harmonic oscillator, depending on which properties you want. So your ability to draw cartoons and to use classical insights for the particle in a box and the harmonic oscillator will be incredibly valuable once we take our first step into the reality of time-dependent quantum mechanics. Now, one of the things that's going to happen is that we're going to describe real situations, real situations that are not one of the standard solved problems. The standard solved problems are the particle in a box, or a particle in the infinite box, the harmonic oscillator, the hydrogen atom, and the rigid rotor. And the particle in a box-- so the potential for each of these is different. It looks like this. It looks like this. It looks like this. And it looks like that. So there is no stretching in a rigid rotor. And so all of the complexity is in the kinetic energy, not the potential. But each of these has a different potential energy. And the energy levels for the particle in a box are proportional to n squared. For the harmonic oscillator, they're proportional to n plus 1/2. For the hydrogen atom, they are proportional to 1 over n squared. And for the rigid rotor, they're proportional to n times n plus 1. So one of the valuable things you get from looking at these exactly soluble problems is that the energy level patterns for each of them are slightly different. And you can tell what you've got from the energy level pattern. And so when you take a spectrum, so often you want to know what kind of spectrum is this? Sometimes you can tell just by what frequency region it is. But usually, in a spectrum, there's a pattern of energy levels. And that gets you focused on well, what kind of problem-- what is the nature of the building blocks that we're going to use? So this is wonderful. Now, we're also going to find that when we solve these problems in quantum mechanics, we get an infinite number of eigenfunctions and eigenvalues. And often, we get presented to us a lot of integrals involving the operators between different wave functions. And one of the beautiful things is when the theory gives you an infinite number of those integrals. So now we would have a collection of all sorts of integrals that are evaluated for us in a simple dependence on quantum numbers. So what do we do with them? Well, there's two main things we do with these infinite numbers of integrals. One is we say, well, this problem is not the standard problem. There are defects. Instead of having the potential for the harmonic oscillator that looked like this, it might look like that. There might be some anharmonicity. Or there can be all sorts of defects. And these defects can be expressed by integrals of some other operator. And that goes into perturbation theory. Or if we're going to want to look at wavepackets, creating a particle-like state and asking how it propagates in time, those integrals are useful. So there's all sorts of fantastic stuff, once we kill the standard problems. It's not just some thing you have to study, and it's over with. You're going to use this. As long as you're involved in physical chemistry, you're going to be using perturbation theory to understand what's going on beyond the simple stuff. OK, so let's get started. So why is the harmonic oscillator so special? If we look at any potential energy curve, any one dimensional problem, it's typically harmonic at the bottom, no matter what it does. I mean, this is a typical molecular diatomic molecule potential. We have a hard inner wall. And we have bond breaking at the outer wall. And so this is an anharmonic potential. But the bottom part is harmonic. And so we can use everything we learn about the harmonic oscillator to begin to draw a picture of arbitrary potentials. So this is hard wall bond breaking. Now, we're mostly chemists. And breaking a bond is-- how much energy does it take to break a bond? Well, is that encoded in the spectrum? Yeah. So this is the sort of thing we would care about. And now let me put some notation here. So this is the potential. And this is the potential of-- the internuclear resistance, which is traditionally called R. We're going to switch notations really quickly from R to x. But this, the minimum of the potential, is the equilibrium internuclear distance R sub e. So for R near R sub e the potential V of R looks simply like k over 2, a forced constant, R minus R sub e squared. So this is harmonic oscillator. And now we're going to change notation a little bit. We're going to use lowercase x to be R minus Re. In other words, it's the distortion away from equilibrium. And we can take any potential and write it as a power series. Sorry. And so on. So if we know these derivatives, we know what to do with them. OK. So one standard potential is called the Morse potential. Because it looks like this. It looks like what you need for a diatomic molecule. And the Morse potential has an analytic form, where the potential v of x is equal to D sub e, the dissociation energy, 1 minus e to the minus ax. Well, that doesn't look like polynomial of x. But we power series expand this, and we get a polynomial x. OK, now, what is this D sub e? So if x is equal to 0-- this is 1 0. And so we get the energy. The potential at x equals 0 is 0. And if x is equal to infinity, v of x equals D sub e, so basically, this is the energy between the bottom of the well and where it breaks. This is D sub e. OK? So this is well interpreted. All the rest of the action is in here. AUDIENCE: [INAUDIBLE] ROBERT FIELD: I'm sorry? AUDIENCE: Do you just square the x? ROBERT FIELD: Yep! Thank you. It's an innocent factor. But it turns out to be very important. OK, so let's square this. We have v of x is equal to De times 1 minus 2e to the minus ax plus e to the minus 2ax. OK, so we're going to do power series expansions of these. And you can do that. And so we have plus some function of x plus some function of x squared, et cetera. Now, what is the function of x? We're taking our derivatives at the equilibrium, at x equals 0. And we have a potential, which has a minimum. So what's the derivative of the potential at equilibrium? Yes? AUDIENCE: Zero. ROBERT FIELD: Right. And so the x term goes away. And so we have a constant term, which we usually choose to be zero. And we have this quadratic term. Looks like a harmonic oscillator. And then there are other terms which express the personality of the potential. This is universal. And the rest becomes a special case. So we have v of x is De a squared x squared plus other stuff. OK, we're going to call this 1/2k. why? Because we'd like it to look like a harmonic oscillator. I mean, we know that the potential for a harmonic oscillator is described in this way. So let's just draw a picture. So we have a spring and a mass. And I should have drawn this up a little higher. OK, so at equilibrium there is no force. If the mass is down here, there is a force pulling it up. And if it's up here, there's a force pushing it down. Hooke's Law says that the force is equal to minus k x minus x 0. OK, and now we're going to switch to just the lowercase x in a second. But now, the force according to Newton is minus gradient of the potential. So the potential for this problem is v of little x is equal to 1/2k x squared. OK, so we have what we expect for a harmonic oscillator. And we're going to say the small displacement part of the Morse oscillator looks like 1/2k. x squared looks harmonic. And so what do we do now? Well, one of Newton's laws, force is equal to the mass times the acceleration. And so we can say, oh, well, the acceleration is the second derivative of x with respect to t. And the force is a minus gradient of the potential. So it's minus 1/2kx squared. Minus kx. OK, so this is the gradient of the potential. This is the mass-- well, I need the mass-- times the acceleration. And so this is the equation we have to solve. And so we want to find the solution x of t. And this is a second order differential equation because we have a second derivative. So there's going to be two terms. And they'll be-- we'll have some sine and some cosine function. And we're going to want a derivative, a second derivative, that brings down a constant k. And so we know that these are going to be things that have the form a sine kx k or m square root x plus b cosine k m x. So this is the solution. And we have to find A and B. Now, why do we have k over m? Well, you can look at this differential equation. And you can see that we have an m here. We have a k here. And that's what you need in order to solve it. So now our job is simply to find A and B. AUDIENCE: Should these be functions of t and not x? ROBERT FIELD: You know, sometimes I go onto automatic pilot because it's so familiar to me. I'm just writing what comes up in my subconscious. But yes, that's a good point. All right, so we want x of t. And it is this combination. OK, so now we put in some insights. We want to know what the period of oscillation is. And so x t plus tau has to be equal to x of t. And so when we do that, we discover-- and I'm just going to skip a lot of steps-- that k over m square root times tau has to be equal to 2 pi in order to satisfy that. We call k over m square root omega, just to simplify the equations. But we also discover that this actually is an angular frequency. So if we say omega tau is equal to 2 pi, then tau equals 2 pi over omega, which is equal to 1 over the frequency. Just exactly what we expect. So we have the beginning of a solution and omega, tau, and frequency make sense. Everything is what we sort of expect. OK, so now the next step is to determine the values of the constants. So normally, when we have a differential equation, after we find the general form, we apply boundary conditions. And so we're going to apply some boundary conditions. So here we have the potential. And-- so this is the potential as a function of coordinate. And this is the turning point, this is the inner turning point at energy E. This is the outer turning point. Well, what's true at the turning point? The turning point, the potential is equal to the energy at the two turning points. So 1/2k x plus or minus of E squared. So if we know E, we know where the turning points are. And so x of plus minus of E, we can solve this. And so we have 2E over k. Yeah. So this is the turning points. These are the turning points. Now, this is a fairly frequent exercise in quantum mechanics. You're going to want to know where are the turning points. Because this is how you impose boundary conditions easily. And so knowing that a turning point corresponds to where the potential is equal to the total energy is enough to be able to solve for this. OK, so suppose we start at x equals x plus. And so x of 0 is equal to x plus. And that determines one of the coefficients. So x of 0-- so we have at t equals 0 the sine term is 0 and the cosine term is 1. And so the first thing we get at t equals 0 is that B is equal to x plus. Right? The next step is to find a. There are several ways to do this. But it's useful to draw a little picture. So x plus is here. And that occurs at t equals 0. And x minus occurs at tau over 2. And in the middle we have tau over 4. So let's ask for, what is the value of the wave function when x is equal to 0? So how do we make the x be 0 at tau over 4? And that determines the value of B. I mean, the value of A. So we have everything we need. And now, before just rushing on, let me just say, well, this just gives A is equal to zero. OK, there's a different approach. When we have a sine plus a cosine term, we can always re-express it as x of t is equal to some other constant times sine omega t plus phi. And so the same sort of analysis gives c is equal to E over k square root. And phi is equal to minus pi over 2. OK, I'm just writing this. You want to do that. OK, so now we're going to be preparing to do quantum mechanics, the quantum mechanical solution of the harmonic oscillator. And so there are going to be other things that we care about, and one is the kinetic energy and one is the potential energy. And in particular, we'd like to know the kinetic energy, which we call t. And we'd like to know the expectation value of t as a function of time, or just T bar of t. And similarly, we'd like to know the expectation value of the potential energy as a function of time. And that's going to be V bar of t. And so from classical mechanics, we should be able to determine what these average values of the kinetic and potential energy. So what do we know? We know that the frequency is omega over 2 pi. We know the period is 1 over omega. OK, and so T of t is 1/2 mv squared of t, or p squared over 2m. But all right, v is the derivative of x with respect to t. And we have the solutions over here. And so we know that we can write x of t and v of t just by taking derivatives. And so we have x of t is 2e over k square root sine omega t plus phi. And v of t is omega times 2e over k square root cosine omega t plus phi. So we want v squared to be able to calculate the kinetic energy. And so we do that. And so the kinetic energy T of t is 1/2m m omega squared 2e over k cosine squared omega t plus phi. This here, omega is k over m, the square root of k over m. S this is m times k over m. So we just get k out there. And so now we would like to know the average value of the kinetic energy. OK, so we have m omega squared. And so that's k over k. And so we just get E integral from 0 to tau bt of cosine square omega t plus phi over tau. We want the time average. And so we calculate this integral and we divide by tau. That's how we take an average. And what we discover is that this integral is the numerator is 1/2 tau times E. No, I'm sorry. 1/2 tau, we have E times 1/2 tau divided by tau. And this becomes E over 2, an important result. And we're going to discover that the average value of the momentum for a harmonic oscillator is E over 2 in quantum mechanics. We do the same thing for the potential. And we discover that it is also E over 2. So we know that E is equal to T of t plus V of t. But it's also true that the average is equal-- so T bar is equal to V bar which is equal to E over 2. So now we have an important interpretation of the harmonic oscillator. The harmonic oscillator is moving from turning point to the middle to the other turning point. And what's happening is energy is being exchanged between all potential energy at a turning point to all kinetic energy in the middle. So energy is going back and forth between kinetic energy and potential energy. We can solve for the relationship between T of t and V of t. And we can find that V of t is equal to T minus pi over 4. tau over 4. So this is telling us just what I said before, that energy is being exchanged between potential and kinetic energy. And that the potential energy is lagging by tau over 4 behind the momentum. This is all very fast. But it's all classical mechanics, which you know. And we're going to be rediscovering all of this in quantum mechanics. And so we have to know what are we aiming for in quantum mechanics? So that we can completely say, yes, it's consistent with classical mechanics. And there's some really-- now, there's another really neat thing. So if we look at X of t, X of t is oscillating. And suppose we start out at a turning point, the outer turning point. So we can tell from this, the derivative of x with respect to t at x equals x plus is going to be 0. So at a turning point, the particle is hardly moving, not moving. And so what about the momentum? Well, the momentum is going to look like this. Yeah, it's going to look like that. And so at the time that we are starting at a turning point, the time derivative of the momentum is at its maximum value. So this is going to be really important when we start looking at properties of time-evolving wave functions. Because what we're going to discover is, suppose we start our system here, at a turning point. And that's actually something that we can do very easily in an experiment. Because when we excite from one electronic state to another, you automatically create a wave packet, which is localized at a particular internuclear distance. And so you go typically, to a turning point. So then, well, what's going to happen? Well, there is a thing in quantum mechanics, which you will become familiar with, called the autocorrelation function. No, it's not. It's called the survival probability. And that's going to be the product of the time-dependent wave function at x and T, the time-dependent wave function at x and 0. So this is expressing somehow how the wave function that is created at t equals 0 gets away from itself. It's a very important idea. Because you make something. And it evolves. And for a harmonic oscillator, if you make it at a turning point, this thing changes. Because the momentum changes. And the contribution of the coordinate to the decay of the survival probability is-- it's all due to the momentum, and not the coordinate change. That's actually, a very important insight. Because the momentum, the time rate of change of the momentum, is minus the gradient of the potential. So this is one way we learn about the potential simply by starting at a turning point and knowing that this thing, which we can measure, is measuring the thing we want to know. Now, we will get to this very soon. Because I haven't even told you about time-dependent quantum mechanics. But those are the things that we expect to encounter. OK, now I want to give you-- I'm going to throw at you-- some useful stuff, which turns out to be really easy. Suppose we want to know the expectation values, or the average values, of x, x squared, p, and p squared. OK, so we have a harmonic oscillator. And do we know what this is? Do we know the expectation value of the coordinate? AUDIENCE: 0. ROBERT FIELD: We have 0. Why is it 0? There's two ways of answering that question. But these are easy questions, which on an exam, you don't want to evaluate an integral. You want to know why is it 0. And there's two answers. You said 0, right? Why did you say 0? AUDIENCE: Because it's symmetric. ROBERT FIELD: Yes. OK, so there is a symmetry argument. Another is well, is the potential moving? The particle is in a potential. It's going back and forth. The potential is stationary. So there's no way that x could move. X could be time dependent. So we know this is 0. What about p? Same thing, whether you use symmetry or just physical insight. But what about x squared? Well, x squared is equal to V of x over k over 2. So expectation value of x squared is equal to the expectation value of V over k over 2. But we know the expectation value of V. It's E over 2. So we know without doing any integrals what the expectation value of x squared is, and similarly, for p squared. And that's just going to be m times E. OK, so why do I care about these things? Well, we have a little thing called the uncertainty principle. And we'd like to know the uncertainty in the coordinate and the momentum. And this is our classical view of it. But it's going to remind us of what we find quantum mechanically. So the uncertainty in x can be defined as the average value of x squared minus the average value of x squared square root. That's just the variance. And well, this is 0. And we know what this is. So we know that the uncertainty in x is E over k square root. And the uncertainty in p is going to be p squared average value minus p squared. This is still 0. And this one is then m times E square root. So delta x delta p, it looks like the uncertainty principle. This is classic mechanics, is just E over omega. But what's E over omega? AUDIENCE: h? h bar. ROBERT FIELD: Right. So it's all going to come around. This is related to the uncertainty principle in quantum mechanics. There is a minimum joint uncertainty between x and p. And it's just related to this constant. Now, this doesn't say the uncertainty grows as you go to higher and higher energy, as it will in quantum mechanics. But this is really a neat thing to see. OK, I've got two minutes left. So if we wanted to know the probability of finding x as a function of the coordinate, the turning points. So here are the turning points. And we can calculate what is going to be the probability of finding the particle at one turning point. Well, it comes down from infinity, goes back up to infinity. So here at the turning points, the particle is stopped. And so the probability of finding it at a turning point is infinite, times dx. So we don't have an infinite number. In quantum mechanics, we're going to discover that quantum mechanics is smarter than that. And what quantum mechanics does, suppose we have this is the energy. What quantum mechanics does is that the probability at a turning point is not 0. And there's something-- what happens is there's tunneling tails that instead of going to infinity, it has a finite value. And it reaches out into the forbidden region in an exponentially decreasing way. And that's basically the difference between classical mechanics and quantum mechanics. And there's an awful lot of important stuff that happens there. Now, I've been very fast through all of this because it's built on stuff that you're supposed to know. And it's built on stuff that we're going to work hard to understand from a quantum mechanical point of view. But this sets the stage for, what are the things we have to look for in quantum mechanics? So I do recommend that you look at the notes and make sure you can follow all the steps, which I went through incredibly rapidly. And it'll be really helpful. Because your job will be to draw cartoons. And these will guide you through it. OK, so I'll see you on Friday. |
MIT_561_Physical_Chemistry_Fall_2017 | 15_NonDegenerate_Perturbation_Theory_I.txt | The following content is provided under a Creative Commons license. Your support will help MIT open courseware continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: Now if you looked at today's notes, you'll notice that they're very long and very complicated. And that's not intended to be intimidating. It's just to show you the power of perturbation theory. So we have talked about the two-level problem. The two-level problem is one that's exactly solved. It's one of our favorite exactly solved problems, although it doesn't seem to have any physical relevance. It's just a nice numerical exercise. So we can take a two by two Hamiltonian and exactly diagonalize it. And that's done using a unitary-- or actually, in the case that we looked at, orthogonal-- transformation. And so it's a rotation in state space. And the rotation angle is an explicit function of the parameters in the Hamiltonian. Now all good things come to an end. We cannot do this for anything more than a two-level problem. But we can take the formalism for the two-level problem and say, oh, well, we can describe a transformation that diagonalizes the end level problem. And it has the same property of being unitary. But the good thing about it is it's solved by a computer. You tell a computer what the operator is, and it will diagonalize it. And not only will it diagonalize it, but it will give you the eigenvalues and the eigenvectors. It gets those eigenvectors by applying the same two level unitary transformation many, many times. So if you have a six-level problem or a hundred-level problem, the computer just cranks and cranks on each two by two and does transformations. And it keeps doing that until the off diagonal matrix element is small-- remaining off diagonal matrix element-- is small compared to the difference in energy between the eigenvalues it connects. And so you tell it, I want this convergence criterion to be a part in a hundred or a part in a million. And it just does this and does this. And eventually, it gives you the transformation and the eigenvalues. So that means that it doesn't matter how big the problem is. You just have to have a computer that's patient or fast. And it will crank out the results. Now you need to know how to use the t matrices, or the t dagger matrix, because these things enable you to solve basically any problem in time independent quantum mechanics and a lot of problems in time dependent quantum mechanics. So there is another way, and that's going to solve an n-level problem. And that's the subject of this lecture and the next lecture. It's called non-degenerate perturbation theory. Now our goal-- well, let's ask. I'm a spectroscopist. What do spectroscopists do? Now, there's a silly, stupid answer to that. I want something a little more profound. Yes? AUDIENCE: Find energy levels? ROBERT FIELD: Yeah, we record spectra. We get energy levels. And we get intensities. We get all sorts of things. We could also be working in the time domain, and we could we could be looking at some kind of quantum beating system or some decaying system. We make these measurements. But this is not why we do it. Remember, yes-- AUDIENCE: All spectra have buried information about the physical parameters of the system. ROBERT FIELD: Exactly, exactly-- that's what's kept me going for my entire career. I use the word encoded rather than buried. But I like that. I might start using buried as well. Yes, we're not allowed to look inside of small things. We're not allowed to determine the wave function by any experiment. But we are able to observe the energy levels and properties of the Hamiltonian. And we can regard the Hamiltonian as a fit model, a model where there are adjustable parameters which are the structural parameters, like the force constants and the reduced masses and whatever we need to describe everything in this problem. And today's lecture is mostly going to be on the interactions between normal modes of a polyatomic molecule. Now you might ask, why am I doing this rather than talking about just an ordinary, anharmonic oscillator. And the reason is, I'm going to do that in the next lecture. And it's in your problem sets. You're going to do that. So I'm going to show a little higher order of information. So we go from what we can observe to a representation of it, which we call the effective Hamiltonian, where in this effective Hamiltonian we have determined the values of all of the important structural parameters. And from that, we can calculate everything, everything that we could possibly observe, including things we didn't observe. So it's really powerful. It's a way of taking the totality of observations that you're going to make and saying, yes, I have looked inside this molecule. And I've determined everything that I'm allowed to determine. And I can calculate the wave function. The wave function is an experimentally determined wave function, but only indirect. This is amazingly powerful. And basically, everybody who deals with spectra is doing this whether they know it or not. And so I want to be able to give you the tools to be able to take any arbitrary spectrum and extract from it the crucial information. In the last lecture, we talked a little bit about matrix mechanics, and that involved linear algebra. And you have a wonderful handout on linear algebra, which will give you more than enough to be able to do any quantum mechanical problem we're going to be facing in this class. And there is notation. And the notation is unfamiliar, and you have to learn how to use it. And so we have for example-- the analogy to the Schrodinger equation in matrix language-- where we have the Hamiltonian, an eigenvector, an eigenvalue, and this eigenvector. So it's an eigenvalue or eigenvector equation. And this is the form of the Schrodinger equation and we can pretend that it is the real Schrodinger equation. And we can use the standard approaches, but it's useful to work in matrix notation. So we can solve for the energy levels, and we can solve for basically all of the eigenvectors. And our friend is this unitary transformation where t dagger is equal to t inverse. And t times t inverse is equal to this thing 1111100, or the unit matrix. I thought I saw a hand up. That was just-- that was not serious. And so these unitary transformations have this special convenient property. And we use these t daggers and ts to diagonalize the Hamiltonian. Now this is just a little bit of review. So how did we derive the thing that we are going to solve? Well, we took this equation and we inserted 1 between the Hamiltonian and the vector. And then this is 1. So we don't need to put it over here. And then we left multiply by t dagger, and we put parentheses around things. So this is now what we were calling h tilde. And this is c tilde. And this is now an equation that says, OK, we can transform the Hamiltonian into diagonal form-- E1, En, zeros. And when we have this in diagonal form, we can say, well, this equation is just E1, 100 et cetera. So for any eigenvalue, we have an eigenvector. Now what we'd really like to know is, well, how do we get these eigenvectors from the unitary transformation that diagonalizes H. We don't calculate this unitary-- yes? AUDIENCE: Is that i or--? ROBERT FIELD: This is a 1, this is a 1, this is a 1. AUDIENCE: I mean in your c tilde, third from the top. ROBERT FIELD: That's a j. This is a particular eigenvalue, and this is the eigenvector in that. Now it's easy for me to get screwed up, and it's easy for you to wonder, what the hell am I doing, until you've done it. And then it's completely transparent. It's really quite simple. But it's just this extra notation. And so it's possible to show-- and I don't want to do it. I did it last time. You can show-- and this is important-- that this transformation t dagger c equals c. Do I want?-- above, above. This is telling you that the columns of t dagger are the eigenvectors. They are linear combination of the basis vectors that correspond to each eigenvector. And you can look at this in more detail in the notes. Now this is great because mostly you want to know what the eigenvectors are for a particular eigenvalue. But sometimes, when you're doing dynamics, you prepare, not an eigenvector, but you prepare a t equals 0, a basis state. This is the most common, doable problem, and it's also something that one does in experiments. You can set up a problem so that, with a short pulse, you prepare the system at t equals 0 in something that's not an eigenstate. And usually it's a basis state-- it's one of the eigenvalues of an exactly solved problem. But your problems are not exactly solved ones, but your experiment selects that. And this, not being an eigenstate, needs to be expressed as a linear combination of eigenstates so that you can actually calculate this, which will describe how the system is behaving. And I like asking exam problems like this because it's easy to get hopelessly involved in ordinary algebra rather than just using linear algebra. So if this is the transformation to the eigenbasis, then the columns of t are the transformation back to the 0 order bases. And the columns of t are the rows of t dagger. Now why should you care? Because you don't know how to calculate the elements of the t matrices yet, but that's what perturbation theory is for. With perturbation theory, it doesn't matter that the computer could solve for the t and t dagger matrices. But there's no insight. You just get the numbers. I like to say that spectroscopy is not about creating archival tables of observed transitions and observed transition intensities. It's about understanding how things work. And so depending on whether you're doing a time domain experiment or frequency domain experiment, you're going to want to use either the columns of t dagger or the rows of t dagger. Now at some point in your life, you have to be exposed to non-degenerate perturbation theory because it is really powerful. But it's also incredibly ugly. So we have a whole bunch of problems-- I'll go over here-- that are related to the particle in a box or the particle in an infinite box or the harmonic oscillator. And so one of the things that you might do is round off the corners because physical systems don't have discontinuities. Well, that's going to be a very modest change to the energy levels and wave functions. Another thing you might do is have a barrier, and you could put the barrier anywhere. What does the barrier or an extra well do? Or you could do something like this. So the particle-in-a-box is a whole family of problems that you could solve using perturbation theory by saying, OK, the extra stuff is the perturbation. And we have to work out the matrix elements of the Hamiltonian that correspond to these extra things and then figure out what to do to get the eigenvalues and eigenfunctions. Now, for the harmonic oscillator-- again, you could put a barrier in the middle or you could make it an asymmetric like almost all molecular potentials are, where this is dissociation and this is two closed cells colliding with each other and doing very hard repulsion. And so that is harmonic near the bottom, but it's not harmonic elsewhere. And so how do you represent this? And the Morse oscillator is a cheap way of generating something with this shape and then doing the perturbation theory to understand how the nonharmonic aspect of the Morse can affect the energy levels. Or how did the energy levels determine, say, the association energy of this molecule? How is that dissociation energy encoded in the energy level pattern? Now for polyatomic molecules, if you have n atoms, there are 3n minus 6 vibrational normal modes. Well, how do I know that? n atoms, there's 3n degrees of freedom. There's 3 translations and 3 rotations. And so that leaves 3n minus 6. And all of that is vibrations. So we have many normal modes, and it's not too surprising that, if you stretch one normal mode, it'll affect the frequency of another. And we'd like to know that. And so perturbation theory is really valuable for polyatomic molecules. And that's the bulk of the examples that I worked in the non-lecture notes for this lecture. But there's also dynamics. So you prepare some initial state of t equals 0 and you want to know how it's evolving. But often that initial state is an eigenstate of one of the exactly solved problems. And you want to be able to re-express that in terms of the eigenstates of the real problem. And so you want to know C J-- I'm sorry. You want to know J, CJ eigenstates. So once you have this, then you know how to write the time dependent wave function because you are dealing with eigenstates and they have eigenenergies. And so you just write this thing out. It's just mechanical and boring. It's also true that molecules rotate. And when they rotate, there's centrifugal force and their internuclear distances change. And we can then calculate how that will affect the rotational energy levels using perturbation theory. And then there's the origin of life. Gases are not supportive of life. You need two particles to come together and start to condense into a liquid. That's the beginning. Perturbation theory explains the long range interactions by which all gas phase particles attract each other weakly. So that's important too. And so you'll be able to do all of this stuff. So here we have non-degenerate perturbation theory. And it is a mind numbing, formal derivation. So we start out with this rotary equation. And we say, well, let us expand the Hamiltonian. And let's put a little thing here. This is our friend. This is an exactly solved problem. This is what's new, and this is what's new and small. And we could probably neglect it. And we do the same thing to the energy levels. So these are the energy levels for the exactly solved problem. And these are the energy levels-- the first order corrected energy levels. And these are the second order corrected energy levels. And we do the same thing for the wave function. Now we write the Schrodinger equation with these three term expressions. Now I'm also going to say, we're never going to consider this one either. So life is simpler without them, but in the notes, I included them all. And so what we do is now we write the full equation and we sort it into sub equations corresponding to powers of lambda. So the lambda to the 0 equation is really easy. It's just H0 psi 0 is equal to E0 psi 0. We could put n's on this. And this is the exactly solved problem. It says, yeah, you build your foundation from the lambda to 0 equation, and it's just what you know already. And what you're going to do is now use the psi n 0 and en 0 to do everything else. The lambda to the one equa-- well, you might ask, well, what is lambda. It's just a mathematical trick. It has no significance whatsoever. It's a smallness parameter, but it's also something where you can say, these equations will be true for any value of lambda. And so it's just a way of separating the equations into things that have a structure that you can manipulate. People have waxed eloquent about the meaning of lambda, and it really doesn't have any meaning. So the lambda to the 1 equation-- well, we collect terms on the left hand side that have 1 lambda and on the right hand side. And so the lambda to the 1 equation is going to be, say, H0 psi 1 is equal to E0 psi 1. I'm sorry. There's more to it than that. There's four terms. We're just collecting the terms that have 1 lambda on the left-hand side and right-hand side. And we've got this equation. What are we going to do with it? Well, one thing we can do with it is multiply on the left. Let's just put some indices here. So we have n, n, n. We're going to multiply on the left and integrate by psi n 0. So we're going to get a bunch of terms. So we will have psi n 0 H 0 psi n 1 plus psi n 0 H 1 psi n 0. We have integral, integral. And on the left hand side, we have E n 0 integral psi n 0 psi n 1. And we have E 1 psi n 0 psi n 0. Well, this is kind of an ugly term. It's a psi 0 and a psi 1. But we know that H, when operating on psi 0, gives E 0. So we're going to get E n 0 integral psi n 0 psi n 1-- same two over here. Cancel them. And so we get a simple equation that is just H1 n n is equal to E1 n. So we've gotten now the diagonal matrix element of the perturbation term is equal to the first-order correction to the energy. And we can continue. The algebra isn't beautiful, but we end up getting the following equations. We have E n 1 is H 1 nm. We have psi n 1 is m not equal to n psi m 0 H nm. So what did I do here? I said, the wave function-- we have completeness. So if we want the first order corrections to the nth wave function, we can write this as a linear combination of the zero order wave functions. And when we do that, we end up with this formula. This is the mixing coefficient, and these are the state-- now why do I exclude n? Because we already have it. And then we get the second-order correction to the energy, which is m not equal to n h n m 1 H m n 1 over En 0. That's it. That's all we need. Now it does say non-degenerate perturbation theory. And so it's subject to the requirement that H1 n m over E n minus E m. So if the energy denominator is near 0, we know we're in trouble. But for the vast majority of energy levels, this term is much less than 1. And so that means we can deal with all of the interactions among the non-degenerate levels in one fell swoop. Now this is an infinite sum, and this is an infinite sum. So we don't like infinities. But we can say, all right, here's the Hamiltonian. It's an infinite Hamiltonian. And we're interested in this little corner of it. And so all of the interactions among these states with all of the infinite others get subsumed into this infinite sum. It's a small number. And then we're just interested in this little subspace of the energy levels that we're sampling in our experiment. So the molecule more or less tells you how to focus on the part that you care about and to get rid of the stuff that is of no trouble whatsoever. And it just contaminates the wave functions a little bit. And if you wanted to know what that contamination is, you could deal with it. So this is the tool that you can use to solve, basically, any problem involving molecules with a potential like a harmonic oscillator at the bottom. But it's usable for all problems, but there's a different basis set rather than the harmonic oscillator basis set. So this is your handy dandy key. And you don't need a computer, although when you see the horrible complexity that will result when you start dealing with these sums, you will say, well, I do want to use a computer. But now it's up to me to organize the program so that you can ask the computer to do what you need in a sensible way and you still get good answers. So in the notes, I'm dealing with a two-mode molecule. There are no two-mode molecules. There's one-mode molecules, and there might be three or four or six or whatever. But the complexity is the interaction between two modes. And so we're going to talk about that. So we have for a two-mode molecule-- the Hamiltonian will consist of Hamiltonian for mode 1, Hamiltonian for mode 2, and the Hamiltonian for modes 1 and 2, interacting with each other. Well, we know these. These are just ordinary harmonic oscillator. And we've got to do some work on this. And that's where perturbation theory comes in. So H12-- and we know, if this weren't here, we know that the energy levels are that some of the energy levels for the two independent oscillators and the wave functions are the products. So even though the energy levels have two quantum numbers, V1 and V2, and the wave functions have two quantum numbers, V1 and V2, if it were only this, we'd have completely solved that problem. But because of this, there's something else. So these then turn out to be the zero order states that you use to evaluate all the integrals here. And we have these and A and A dagger operators, which are enabling us to-- we have the operator for coordinate x is proportional to A plus A dagger. And this has a selection rule, delta V of plus and minus 1. So we like these things because we don't have to do any integrals. They're all done for you. And so we're then going to use these sorts of things to deal with the most important terms beyond harmonic. And so there's a cubic and there is a quartic. And although you haven't really explored this, what happens when you make dimensionless coordinates? You factor out something. And if you have cubic terms, they're 100 times smaller than quadratic terms. And the quartic terms are 100 times smaller than the cubic terms. And so you don't need to go much further. So the cubic terms for the two oscillators would be 1/2 K 122 Q1 Q2 squared plus 1/2 K 112 Q 1 squared Q2. And you could have Q1 cubed, but we already deal with that in the single-mode problem. So we won't worry about that. So these are the couplings between modes 1 and 2 that are cubic. And then we have 1/4 k 112 Q 1 squared Q 2 squared. There's also a Q1 Q2 cubed, and those terms usually are not important because they mostly are dealt with under here. So anyway, these are the parameters, and these are the things that you need in order to understand what this molecule is going to do when it's excited. Now when I was a graduate student, there was a great deal of excitement about doing what's called mode-specific chemistry. Ordinary compounds cost on the order of $1 a kilogram. But if you could do mode specific chemistry, you could make things that aren't makeable by ordinary, organic chemistry. And the organic chemists would love this because it's a load of garbage because these guys make the modes talk to each other. And so even if you could excite a pure overtone or combinational level, the energy moves around the molecule. And that's called intermolecular, vibrational redistribution. And those there are processes that you could talk about the primary paths for the energy to flow and the rates. And that was a major area of research for the last 30 years. And most people are tired of it. But it started out being IVR, yeah, anything can happen. It's statistic or whatever. But no, only specific things can happen, and they're controlled by these specific coupling terms. And you could calculate them. Now there's an interesting other thing. I told you that the first-order correction to the energy is equal to a diagonal matrix element of the correction term to the Hamiltonian. All of the second-order terms involve squares of matrix elements. The second-order terms you don't know the signs. If there is a first-order correction, you get the sine. And sometimes you know you want to know: is the perturbation like that or the perturbation like that? Is there a barrier or an extra well? And this enables you to know that in second-order perturbation theory when you have to square the matrix element, all that information is concealed. Well, anyway-- so there is a huge amount of algebra that's involved in using these equations. And I don't really want to go through that algebra. You can read my notes. I think the chances of you reading those notes are small. But if I don't lecture on them, it would be slightly greater. It's a complete treatment of a two-mode problem with all of the possible and harmonic terms cubic and quartic. And everything worked out. Now the trick is, you could easily say, well, the algebra is just so horrible. Why would I bother? But what you do is you take these terms and you sort according to selection rule. So for example, here we have a selection rule delta V1 is plus or minus 1 and delta V2 is plus or minus 2 and 0. And so there are then six possible selection rules associated with these sorts of terms. And what you want to do is do the algebra that combines all this horrible stuff according to selection rule that leads to simplification of the formulas. And then once you've got everything sorted according to selection rule, then you can calculate what happens. I can't promise that I will never give you a two-mode problem on an exam, but I can promise I will give you a one-mode problem. And so you want to really know how to do these sorts of things. So non-degenerate perturbation theory works when the energy denominator is large compared to the coupling matrix element. And accidents occur when you have, say, a situation where omega 1 is approximately equal to 2 omega 2. Now this isn't just blowing smoke. This happens an amazing number of times because stretches are higher frequency than bends. And it's very common for the bending modes to be roughly half or one third the frequency of a stretching mode. And so you get a resonance. So this is special because now it's violating the fundamental approximation of non-degenerate perturbation theory. But it's a two-level interaction. And so you can say, I know how to do a two-level problem. I can solve that. And these resonances have names. There is a Fermi and there is a Darling-Dennison. The Fermi resonance was discovered and understood by Fermi. And it has to do with CO2. Omega 1 in CO2 is approximately twice omega 2. The symmetric stretch and the bend are in Fermi resonance. So what happens then? Suppose we have a level that involves V1, V2, V3. And nearby there is a level V1 minus 1 V2 plus 2 V3. And because of the energy denominators, these two guys are nearly degenerate. So what happens is one gets shifted up, one gets shifted down a little bit. And they're out of the expectation of smooth behavior. And it might also be that this state is what we call bright and this is called dark. This state might be connected by an allowed transition from a lower-- an initial state and this might not. So the levels repel because they're interacting and they're out of position. And this guy is supposed to be bright. We're supposed to see a transition in the spectrum. And this one-- well, it should have been somewhere else. But we shouldn't see it because it's dark. It's forbidden. But because of the interaction between these two levels, the eigenstates have mixed character. And you get both level shifts and extra lines. This is called a spectroscopic perturbation. This is the core of everything I've done for the last 45 years-- spectroscopic perturbations. And so you can learn about some of these coupling terms because, instead of hiding in the forest of these small corrections, you get a big effect. And it's easy to observe and it's easy to determine each coupling term from these resonances. And the last thing I want to talk about today is a little bit of philosophy. This is why, among spectroscopists or physical chemists, there are two communities-- communities that like small molecules and communities that like big molecules, because they're really different. For the small molecule community, you have this accidental resonance and you get this sort of thing. For big molecules, you have a resonance with a dense manifold of levels. All of these levels share, say, the character of the dark state. And so they can all interact with this guy. And so in this case, you get the main transition in an extra line. In this case, if you have low resolution, you get a broadened line. And if you have high enough resolution, you see that there is a whole bunch of eigenstates under it. And if it's a large enough molecule, you couldn't resolve them anyway. And so some people are always dealing with fast IVR, fast dephasing of the bright transition into a dark manifold. And other people are always looking at these sorts of things. Now I like these because they really give you a lot of information. Now you can get the information you want from this because the width of this thing is related to the number of dark states and their average coupling matrix element. That's called Fermi's golden rule. We'll talk about that later. So I think there's a pretty good place to stop because what I try to do is to show you, yes, it can be really complicated. But it's something that you can do, and you can project out the coupling constants that you want in order to determine the stuff that is relevant to your experiment. But then there is this dichotomy between small molecules where the vibrational density of states is always smaller until you get to really high energy. And bigger molecules-- now they're not very big. Benzene is plenty big for this sort of thing. And there is the question. For example, in some big molecules, when you excite an electronic transition from the ground state to some excited state-- so here's S0, S1. Sometimes you can't see any fluorescence from S1 because the dephasing is so fast that there's nothing. And so it says, well, tough luck. You can't do spectroscopy in emission. But you can still see the absorption spectrum because then your signal is the removal of photons from your beam as opposed to fluorescence. So there's a huge amount of photochemistry and interesting stuff connected with a large density of states. And again, when I was a graduate student, there was a huge controversy about non-radiative transitions in medium sized molecules. And there was one community that says, the collisions which transfer population between levels are so fast that, in order to turn off this broadening, you had to go to really low pressure because then there wouldn't be collisions. And the answer, whenever somebody failed to see sharp spectra, was, well, you're not at low enough pressure. But this has nothing to do with collisions. And that got resolved by two gentlemen called Bixon and Jortner. Those are two names that any educated physical chemist will know to say, oh, that's the Bixon-Jortner. And they're still alive. They're still doing beautiful stuff. But anyway, that's all I want to say today. I will do details on one mode and Morse oscillator in other sorts of things next time. |
MIT_561_Physical_Chemistry_Fall_2017 | 12_Catch_Up_and_Review_Postulates.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at osu.mit.edu. ROBERT FIELD: You know that there is exam tomorrow night. It's very heavily on the particle in a box, the harmonic oscillator, and the time-independent Schrodinger equation. It's really quite an amazing amount of stuff. OK. So last time, we talked about some time-independent Hamiltonian examples. And two that I like are the half harmonic oscillator and the vertical excitation-- the Franck-Condon excitation. Now, when you're doing time-dependent Hamiltonians-- or when you're doing time-dependent problems for a time-independent Hamiltonian-- You always start with the form of the function at t equals zero and automatically extend it using the fact that you have a complete set of eigenfunctions of the time-independent Hamiltonian to the time-dependent wave function. And with the time-dependent wave function, you're able to calculate almost anything you want. And I did several examples of things that you could calculate. One is the probability density as a function of time. So the other is the survival probability. And the survival probability is a really neat thing, because it says, we've got some object which is particle-like. And the time evolution makes the wave function move away from its birthplace. And that's an easy thing to understand, but what's surprising is that we think about a wave packet as localized in position, but it also has encoded in it momentum information. And so when the wave packet moves away from its starting point-- if it starts at rest-- The initial fast changes are in momentum. And the momentum-- the change of the momentum-- is sampling the gradient of the potential. And that's something you usually want to know. The gradient of the potential at a turning point-- at an energy you know. And that is a measurable quantity. And it's really a beautiful example of how you can find easily observable, or easily calculable, quantum mechanical things that reflect classical mechanics. OK. I talked about grand rephasings for problems like the harmonic oscillator, and the rigid rotor, and the particle in a box-- You have this fantastic property that all of the energy level differences are integer multiples of a common factor. And that guarantees that you will have periodic rephasings at times related to that common factor. And so that enables you to observe the evolution of something for a very long time and to see whether the rephasings are perfect or not quite perfect. And the imperfections are telling you something beyond the simple model-- They're telling you about anharmonicity or some other thing that makes the energy levels not quite integer multiples of a common factor. Now, one of the nicest things is the illustration of tunneling. And we don't observe tunneling. Tunneling is a quantum mechanical thing. And it's encoded in what's most easily observed-- the energy level pattern. It's encoded as a level staggering. Now, we talked about this problem, where we have a barrier in the middle. And with a barrier in the middle, half of the energy levels are almost unaffected, and the other half are affected a lot. Now, here is a problem that you can understand-- In the same way, there's no tunneling here, but there is some extra stuff here. And the level diagram can tell you the difference between these two things. Does anybody want to tell me what's different? What is the qualitative signature of this? Yes. AUDIENCE: The spacings move up a little bit. The odd spacings are erased in energy, and the even spacings are just about unaffected. ROBERT FIELD: Even symmetry levels are shifted up. AUDIENCE: Right. Sorry. ROBERT FIELD: And so now you're ready to answer this one. If the even symmetry levels are shifted up, because they feel the barrier, what about the even symmetry levels here? Yes? AUDIENCE: They would be shifted down. ROBERT FIELD: Right. And so you get a level staggering where, in this case, the lowest level is close to the next higher one. And in this one, the lowest level is shifted way down, and the next one is not shifted. And then we get the doubling or the parent. So that's a kind of intuition that you get just by looking at these problems. Now, one thing that is really beautiful is, when you have a barrier like this, since this part of the potential problem is something that is exactly solved, you propagate the wave function in from the sides and they have the same phase. So there is no accumulation of phase under the barrier. And that means the levels that are trying to propagate under this barrier are shifted up, because they have to accumulate enough phase to satisfy the boundary conditions at the turning points. Here, you're going to accumulate more phase in this special region. Phase is really important. OK. So today we're going to talk-- And this this lecture is basically not on the exam, although it does connect with topics on the exam and makes it possible to understand them better. So instead of learning about the postulates in a great abstract way at the beginning, before you know what they're for, now we're going to review what we understand about them. And so one thing is, there is a wave function. And now we're considering not just one dimension, but any number of dimensions. And this is the state function that tells you everything you're allowed to know about the system. And if you have this, you can calculate everything. If you know how observables relate to this, well, you're fine. You can then use that to describe the Hamiltonian. Hermetian operators are the only kind of operators you can have in quantum mechanics. And they have the wonderful property that their eigenvalues-- all of them-- are real. Because they correspond to something that's observable. And when you observe something, it's a real number. It's not a complex number. It's not an imaginary number. So what is Hermetian? And why does that ensure that you always get a real number? OK. You all know, in quantum mechanics, that if you do an experiment, and the experiment can be represented by some operator, you get an eigenvalue of that operator-- nothing else-- one experiment, one eigenvalue. 100 experiments-- You might get several eigenvalues. And the relative probabilities of the different eigenvalues-- I tell you something about, what is this? What was the initial preparation? So I've already said something about this. The expectation value of the wave function for a particular operator is related to probabilities times eigenvalues. OK. The fifth postulate is the time-dependent Schrodinger equation. And I don't need to talk much about that, because it's all the way around us. And then I'm going to talk about some neat stuff where we start using words that are very instructive-- so completeness, orthogonality, commutators-- simultaneous eigenfunctions. And this is really important. Suppose we have an easy operator, which commutes with the Hamiltonian. And it's easy for us to find the eigenvalues of that operator-- eigenfunctions of that operator. Those eigenfunctions are automatically eigenfunctions. It's a hard operator. So we like them. And so we are interested in, can we have simultaneous eigenfunctions of several operators? And the answer is yes, if the operators can move. We use the term basis set to describe the complete set of eigenfunctions of some operator. And we use the words mixing coefficients and mixing fraction. So here we have a wave function that is expressed as a linear combination of basis functions. And the coefficient in front of each one is the mixing coefficient. Now, if we're talking about probabilities, we care about mixing fractions, which are basically the mixing coefficient squared-- or [INAUDIBLE] square modulus. So these are words that are part of the language of someone who uses quantum mechanics to understand stuff. OK. So I'm going to try to develop this in some useful tricks. So we have a state function. So we have a state function, which is a function of coordinates and time. And this thing is telling you, what is the probability of finding this system at that coordinate at the specified time? And the volume element is dx, dy, and dz. Now, often, we use an abbreviation-- d tau-- for the volume element, because we're going to be dealing with problems that are not just single particle, but many particles. And so we use this notation to say, for the differential associated with every coordinate associated with the problem, and we're going to integrate over those sorts of things-- or at least over some of them. So this is telling you about a probability within a volume element at a particular point in space and time. Now, one of the things that we're told is, if the wave functions are well behaved. And that says something about the wave functions and the derivatives of the wave functions. So what's well-behaved? Well, the wave function is normalize-able. Now, there are two kinds of normalization-- normalization to one, implying that the system is somewhere within a specified range of coordinates where there's one particle in the system. Whatever. And there's normalization to number density. When you have a free particle, the free particle is not confined. And so you can't say, I'm normalizing. So there's one particle somewhere, because that means there is no particle anywhere. And so we can extend the concept of normalization to say, there's one particle in a particular length of the system. And that was a problem that got removed from the draft of the exam. But one thing that happens is, if I write a beautiful problem, and it gets bumped from an exam, it might appear in the final. OK. So normalize-able is part of well-behaved continuous and single value-- we'll talk about all of these-- and square integrable OK. Continuous-- The wave function has to be continuous everywhere. The first derivative of the wave function, with respect to coordinate-- we already know from the particle in a box that that is not continuous at an infinite wall. So an infinite wall-- not just a vertical wall, but one that goes to infinity-- guarantees that this guy is not continuous. But that's a pretty dramatic thing. The second derivative is not continuous when you have a vertical step, which is not infinite. Now, when you have problems where you divide space up into regions, you're often trying to establish boundary conditions between the different regions or at the borders. And the boundary conditions are usually expressed in terms of continuity of the wave function and continuity of the first derivative. And we don't need this often. But you're entitled-- if the problem is sufficiently well-behaved-- All of these guys are continuous, and you can use them all. OK. So normalize-able-- Well, normalize-able means it's square integrable, and you don't get infinity unless we use this other definition of normalize-able. So one of the things that has to happen is, at the coordinate plus and minus infinity, the wave function has to go to zero. Now, the wave function can be infinite at a very small region of space. So there are singularities that can be dealt with. But normally, you say, the wave function is never going to be infinite, and it's never going to be anything but 0 at infinity, or you're in real trouble. Now, there is a wonderful example of a kind of a problem called the delta function. A delta function is basically an infinite spike-- infinitely thin, infinitely tall. And what it does is, it causes this to be discontinuous by an amount related to the value of the derivative at the-- by an amount determined by the value of the wave function at the delta function. And delta functions are computationally wonderful, because they enable you to treat certain kinds of problems in a trivial way. Like, if you have a barrier-- Yes. AUDIENCE: Does it relate at all to the integral of the spike? ROBERT FIELD: Yes. So we have an integral of the delta function at x i times the wave function at x, dx. And that gives you the wave function at x i. OK. I haven't really talked much about delta functions. But because it acts like a barrier and is a trivial barrier, it enables you to solve barrier problems or at least understand them in a very quick and easy way. And vertical steps are also not physical, but we like vertical steps because it's easy to apply boundary conditions. And so these are all just computationally tricky things that are wonderful. And we don't worry about, is there a real system that acts like a delta function or a vertical step? No. There isn't. But everything you get easily, mathematically, from these simple things is great. OK. Did I satisfy you on the-- AUDIENCE: Sure. I'll read into it. ROBERT FIELD: OK. And there's also a notation where you have x, xi, or x minus x i, and they're basically all the same sort of thing. If you have this-- the argument-- when the argument is 0, you get the infinite spike. And there's just lots of things. OK. So for every classical mechanical observable, there is a quantum mechanical operator, which is Hermetian. And the main point of Hermetian, as I said, is that its eigenvalues are real. And so what is the thing that assures that we get real eigenvalues? Well, here is the definition in a peculiar form. So we have an integral from minus infinity to infinity of some function-- complex conjugate-- a times some other function dx. So this could be a wave function and a different wave function. And the definition of Hermetian is this abstract definition. We have, say, interval from minus infinity to infinity g a-- Let's put a hat on it. Star, f-star, dx. Well, this is kind of fancy. So we have an operator. We can take the complex conjugate of the operator. We have functions. We can take the complex conjugates of the function. But here, what we're seeing is, the operator is operating on the g function-- the function that started out on the right. And here, what we have is this operator operating on the f function, which was initially on the left. And so this is prescription for operating on the left. And it's also an invitation to use a really convenient and compact notation. And that is this-- Put two subscripts on a. A subscript says the first guy is the function over here, which is complex conjugated. And the second one is the function over here, which is not complex conjugated. And so this equation reduces to a g f star, where, now, this is a wonderful shorthand. And this is another way of saying that A has to be equal to A dagger. Where now we're talking about operators, and matrix representations of operators. Because here we have a number with two indices, and that's how we represent elements of a matrix. And we're soon going to be playing with linear algebra, and talking about matrices. And so this is just saying, well, we can take one matrix, and it's equal to the complex conjugate of every term in the matrix. And the order switched. So this is a warning that we're going to be using a notation, which is way simpler than taking these integrals. And so once you recognize what this symbol means, you're never going to want to go back. OK. Why real? So let's look at a specific example, where instead of having two different functions, let's just look at one. So we have this integral f star A f d x. So that's a f f. And the definition says, well, we're going to get the-- So it says, replace the original thing by moving this-- anyway, yes. And this is a f f star. If you know how this notation translates into-- now you can see, oh, well, what is this? It's just taking the complex conjugate. And these two guys are equal. And so a number is real if it's equal to its complex conjugate. And this is just a special case. It's-- the Hermitian property is a little bit more powerful than that, but the important thing is that it guarantees that if you calculate the expectation value of an observable quantity, you're going to get a real number, if the operator is Hermitian. And it has to be Hermitian, if it's observable. Now, it's often useful if you have a classical mechanical operator, and you translate it into a quantum mechanical operator by doing the usual replacement of x with x, and p with i h bar, the derivative with respect to x. If you do all that sort of stuff, you might be unlucky and you might get a non-Hermitian operator. And so you can generate a Hermitian operator if you write-- that's guaranteed to be Hermitian. So you take classic mechanics, you generate something following some simple rules, and you have bad luck, it doesn't come out to be Hermitian. This is the way we make it Hermitian. So if this is not Hermitian, and this is not Hermitian, but that we're defining something that is Hermitian, so let's just put a little twiddle over A, that's Hermitian. OK then we're now talking about the third postulate, and each measurement of a gives an eigenvalue of a. We've talked about this enough. But your first encounter of this was the Two-Slit experiment. In the Two-Slit experiment, this experiment is an operator. And you have some initial photons entering this operator , and you get dots, on the screen, those are eigenvalues of this operator. Now it may be that the operator has continuous eigenvalues, but they don't have uniform probabilities. And so what you observe is a whole distribution of dots that doesn't look like anything special, and it's not reproducible from one experiment to the other, but you have this periodic structure that's appearing, which is related to the properties of the operator. And so there is not uniform probabilities of each eigenvalues, and so you get that. OK, these are simple, but really beautiful. OK, if we have a normalized state then we can say, OK, and we never use this notation. But whenever you see this symbol, it means the expectation value of an operator for some wave functions, so we could actually symbolically put that wave function down here, or some symbols saying, OK, which one? And that this is equal to psi star A psi d tau, or dx. And if the wave function is normalized, we don't need to divide by a normalization integral. If it's not normalized, like if it's a free particle, we divide by some kind of free particle integral. So now the next topic, which is related to this, is completeness and orthogonality. So we have a particular operator. There exists some complete set of eigenfunctions of that operator. Usually that complete set is infinite, but they're related to each other in a simple way. You have some class of functions, and you change an integer to get a new function. And orthogonality is, well, if you have all the eigenfunctions of an operator, if they belong to different eigenvalues, they're guaranteed to be orthogonal. Which is convenient, because that means you get lots of 0s from integrals, and we like that, because we don't have to worry about them. And you want to be able to recognize the zero integrals, so that you can move very quickly through a problem. Completeness means, take any function defined on the space of the operator you're interested in. You might have an operator that's only operating on a particular coordinate of a many electron atom or molecule. There's lots of ways of saying, it's not over all space, but for each operator you know what space the operator in question is dealing with. And then it's always possible to write. So this is some general function, defined in the space of the operator, and this is the equation that says, well, completeness tells us that we can take the sum over all of the eigenfunctions with mixing coefficients. c j. And this set of all of the eigenfunctions is called the basis set. It's a complete basis set. So it's always true, you can do this. So you know from other problems, that if you have a finite region of space, you can represent anything within that finite region via sum over Fourier components. That's a discrete sum. If you have an infinite space, you have to do a Fourier integral, but it's basically the same thing. You're expressing a function, anything you want, in terms of simple, manipulable objects. So sometimes the these things are Fourier components, and sometimes they're just simple wave functions. OK, now suppose you have two operators, a and b. If they operate over the same space, the question is, can we take eigenfunctions of one and be sure that they're eigenfunctions of the other? OK, but let's deal with something simpler. So suppose we have psi i and psi j, both belong to a sub i. So they both have the same eigenvalue. Well, in that case, we cannot be sure that these two functions are orthogonal. So there is a handy dandy procedure called Schmidt Orthogonalization that says, take any two functions, and let's construct an orthogonal pair. This is amazingly useful when you're trying to understand a problem using a complete orthonormal basis set. We know, I'm not going to prove it, because I don't know where it is in my notes, what the sequence is, I'm just going to forget it-- I think I'm going to do it. If you have functions belonging to different eigenvalues, they are automatically orthogonal. That's also really valuable, because you have to check. So you have harmonic oscillator functions, and they're all orthogonal to each other. You have, perhaps, one harmonic oscillator, and a different harmonic oscillator, and the functions for these two different harmonic oscillators don't know about each other. They're not guaranteed to be orthogonal. But any two eigenvalues of this guy are orthogonal, and any of those are orthogonal. But not here. OK, so let's just say we have now two eigenfunctions of an operator that belong to the same eigenvalues. And that happens. There are often very many, very high degeneracies. But we want to make sure that we've got two. So let's say, here is a number, the overlap integral between psi 1 and psi 2 dx. This is a calculable number. So you have two original functions, which are not guaranteed to be orthogonal, because they belong to the same eigenvalue, and you can calculate this number. And then we can say, let us now construct something up psi 2, which is guaranteed to be a psi 2 prime which is guaranteed to be orthogonal to psi 1. How do we do that? Well, we define psi 2 to be a normalization factor, times psi 2 the original, plus some constant times psi 1. And then we say, let us calculate the overlap integral between psi 1 and psi 2 prime. OK, so we do that. And so we have the integral, and we have psi 1 star times psi 2 plus a psi 1 dx. Psi 1 on psi 1 is 1. So we get an a. And psi 1 star times psi 2 gives s. So this integral, which is supposed to be 0, because we want psi 1 to be orthogonal to psi 2 prime, is going to be n times s plus a. Well, how do we satisfy this? We just make a to be minus s. Guaranteed. Now this is one of the tricks that I use the most when I do derivations. I want orthogonal functions, and this little Schmidt Orthogonalization enables me to take my simple idea and propagate it into a complicated problem. It's very valuable. You'll probably never use it, but I love it. So if a is equal to minus s, you've got orthogonality. And the general formula is psi 2 prime is equal to 1 minus x squared, the square root of that, times psi 2 minus s psi 1. So this is a normalized function which is orthogonal to psi 1. And it doesn't take much work. You calculate one interval, and it's done. Now later in the course, we're going to talk we're going to talk about a secular determinant that you use to solve for all the eigenvalues and eigenfunctions of a complicated problem. And often, when you do that, you use convenient functions which are not guaranteed to be orthogonal. And there is a procedure you apply to this secular matrix, which orthogonalizes it first. Then you diagonalize a simple thing, and then you undiagonalize, if you need to. Anyway, this is terribly valuable, especially when you're doing quantum chemistry, which you're going to see towards the end. OK. Often, we would like to express a particular eigenfunction expectation value as a sum over P i a i. So this is a probability. And so how do we do that? Certainly, the average of this operator over psi can be written as the eigenvalues of a times the probability of each operator. And what are they? Well, you can show that this is equal to c i-- I better be careful here. Well, let's just do it. Well, I'll just write it. Where this is the mixing coefficient of the eigenfunction of a in the original function. So we get probabilities mixing fractions, and we have mixing coefficients. And this is the language we're going to use to describe many things. AUDIENCE: [INAUDIBLE] ROBERT FIELD: I'm sorry? AUDIENCE: The probability doesn't have an a i, the average does. You just remove the a i. ROBERT FIELD: So the probability-- AUDIENCE: The probability-- ROBERT FIELD: Oh, yes. You see, when I start lecturing from what's in my addled brain-- OK, thank you. OK. Now let's do some really neat things. So we have a commutator of two operators. If that commutator is 0, then all non-degenerate eigenfunctions of A are eigenfunctions of B. If this is not equal to 0, then we can say something about the variances of A and B. So these quantities are greater than or equal to minus 1/4 times the integral-- I better write this on the board below. And this is greater than 0, and real. This is the uncertainty principle. So it's possible to prove this, and it's really strange, because we have a square of a number here. We think the square of a number is going to be real, but not if it's imaginary. Most non-zero commutators are imaginary. And so this thing is negative, and it's canceled. So the joint uncertainty is related to the expectation value of a commutator. And this is all traced back to x P x is equal to ih bar. And this commutator is imaginary. And everything that appears in here, it comes from the non-commutation of coordinate and momentum. And this is why this commutator is often regarded as the foundation of quantum mechanics. Because all of the strangeness comes from it. So yes, this is surprising. It's saying that, when we have a non-zero commutator, this is what determines the joint uncertainty of two properties. This commutator is always imaginary, that is a big, big surprise. And as a result the joint uncertainty is greater than 0, If the two operators don't commute, it's because x and P don't commute. It's really scary. OK, what time is it? We've got five minutes left, and I can do one more thing. Well, I guess I'm going to be just talking about-- So the uncertainty principle. If we know operator A and B, we can calculate their commutator. This is a property of the operators. It doesn't have to do with constructing some clever experiment, where you tried to measure two things simultaneously. It says, all of the problems with simultaneous observations of two operators, two things, comes from the structure of the operators. Comes from their commutation rule. Which traces all the way back to the commutator between x and P. So at the beginning, I said I don't like these experiments, where we try to confine the coordinate of the photon or the electron, and it results in uncertainty in the measurement of the conjugate property, the momentum, or something like that. These experiments depend on your cleverness, but this doesn't. This Is fundamental. So I like that a lot better. OK, the last thing I want to talk about, which I have just barely enough time to do, is suppose we have a wave function. Let's call it psi 2 in quotes of x for the particle in a box. Particle in a infinite box. This is a wave function, which is not the eigenfunction, but it is constructed to look like the eigenfunction. Mainly because it has the right number of nodes. And so suppose we call this thing some normalization factor x x minus a and x minus a over 2. This guarantees you have a node at x equals 0. This guarantees you have a node at x equals a. This guarantees you have a node at x equals a over 2. So this is that the generic property of the second eigenfunction for the particle in a box. And this is a very clever guess, and often you want to make a guess. And so, how well you do with a guess? And so this function, this part of it looks sort of like this between x and a, and this part of it looks like this. And we multiply these two together and we get something that looks like this. Which is, at least a sketch, looks like the n equals 2 eigenfunction with a particle in a box. So let's just go through and see how well we can do. So first of all, we have to determine n. So we do the normalization integral. Psi 2 star psi 2 dx. That has to be 1. So what we do, now this is kind of a cheat, because we do this because we don't know an eigenfunction. But we do know these eigenfunctions, so we can expand these functions, in terms of the particle in a box eigenfunctions. So we use these things, which you know very well. Now we have a sine function here. That's because I've chosen the box that doesn't have 0 left edge, it's symmetric about x equals 0. And that would be appropriate for this kind of function. So anyway, when we do this, we find the mixing coefficients. And I know I just said something wrong and that's, we don't have time to correct it. Because I said it the wave function is 0 at x equals 0, and at x equals a. And I'm now saying, all of a sudden I'm using symmetric box-- this does not matter, because the calculation is done correctly. And what we end up getting, is that the mixing coefficients for these functions of the general form, 800-- I'm sorry, 840-- this is algebra!-- Square root a to the minus 7/2 2 over n-- 2 over a. I think that's a. Well, I'm not sure whether that's a or n, but let's just say it's 2 over n square root-- oh, it's going to be 2 over a. And times the integral-- anyway, so we get c 2n is equal to 1680 square root over 6 over 2n pi cubed. And that becomes equal to 0.9914n to the minus 3. So this is a general formula which you can derive. I don't recommend it, and I don't think it's really important. The important thing is what I'm about to say. And I have no time. This is almost 1. So when we calculate the energy using these functions, we get that the energy of this 2n function is equal to 4E1 times 0.983 integral from n equals 1 to infinity times n to the minus 4. OK, the first term and this is 1. And so we have something that looks like 4 times E1. Now the sum is larger than one, and the product of these two things is larger than 1. And so what we get is that, E2n is larger, but only slightly larger, than the exact results. So this is sort of a taste of a variation calculation. We can solve for the form of a function by doing a minimization of the energy of that function. And that function will look like the true function, but its energy will always be larger than the true function. But it's great, because the bigger the calculation, the better you do. And that's how most of the money, the computer time in the world, is expended. Doing large variational calculations to find eigenfunctions of complicated problems. OK, good luck on the exam. I hope you find it fun, and I meant it to be fun. |
MIT_561_Physical_Chemistry_Fall_2017 | 6_3D_Box_and_QM_Separation_of_Variables.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: Well, we're now well underway into quantum mechanics. So a lot of the important stuff goes by very fast. So we represent a quantum mechanical operator with a little hat, and it means do something to the thing on its right. And it has to be a linear operator, and you want to be sure you know what a linear operator does and what is not a linear operator. This is an eigenvalue equation. So we have some function which, when the operator operates on it, gives back a constant times that function. The constant is the eigenvalue, and the function is an eigenfunction of the operator that belongs to this eigenvalue, and all of quantum mechanics can be expressed in terms of eigenvalue equations. It's very important, and you sort of take it for granted. Now, one of the important things about quantum mechanics is that we have to find a linear operator that corresponds to the classically observable quantities. And for x the linear operator is x, and for the momentum the linear operator is minus ih bar partial with respect to x. That should bother you two ways. One is the i, and the other is the partial derivative. But when you apply this operator to functions, you discover that out pops something that has the expected behavior of momentum, and so this is in fact the operator that we're going to use for momentum. And then there is a commutation rule. This commutation rule, xp minus px, is equal to this. This is really the foundation of quantum mechanics, and as I've said before, many people derive everything from a few commutation rules. It's really scary, but you should be able to work out this commutation rule by applying xp minus px to some arbitrary function. And going through the symbolics should take you about 30 seconds, or maybe it shouldn't. Maybe you can go faster. OK. We have an operator, and we often want to know what is the expectation value of the particular function, which we could symbolize here, but it's never done that way. So we have some function, and we want to calculate its expectation value of operator A, and this is it. And so this is a normalization integral, and this normalization integral is usually taken for granted, because we almost always work with sets of functions which are normalized. And so if you convince yourself that it is, in fact, they are normalized, fine, and then this is the thing that you normally would calculate. Then, we went to our first problem in quantum mechanics which is the free particle, and the free particle has some idiosyncrasies. The wave function for the free particle has the form e to the ikx plus e to the minus ikx, and the Hamiltonian is minus h bar squared over 2m second partial with respect to x plus v0. So there's no v0 here, and we have two different exponentials and so is this really going to be an eigenfunction of the Hamiltonian? This is really p squared, and so this is going to be an eigenfunction of p squared too. All right, so let's show a little picture here. Here is energy, and this is v0, and let's say this is the 0 of energy. What are the eigenvalues of this? What does the Hamiltonian do to this function? Well, in order to do that, you have to calculate something like where you have to calculate the second derivative of each of these terms, and the second derivative of this term brings down a minus k squared. Now, the second derivative of this term brings down a minus k squared, and so the energy eigenvalues are going to be given by h bar squared a squared over 2m plus v0. So these are the eigenvalues, eigenfunctions, the energies of a free particle, and they're not quantized. Now this v0 is something that will often trip you up, because it's hidden here. It's not in here. OK. I'm going to torture you with something. So why are these two k's the same? What would happen if the k for the positive term were different from the k for the negative term? Simple answer. Yeah? AUDIENCE: There'd be two eigenvalues? ROBERT FIELD: That's right. It wouldn't be an eigenvalue. It wouldn't be an eigenfunction of the Hamiltonian. It's a mixture of two eigenvalues, and so that's simple. But often we might be dealing with a potential that's not simple like this but has got complexity. So suppose we had a potential that did this. The potential is constant piecewise, and so what do we do? Yes? AUDIENCE: Break down the function into pieces for each in certain boundaries? ROBERT FIELD: Yes, and that's exactly right. You do want to break it up, but one of the things I'm stressing here is that you want to be able to draw cartoons. And so we know that if we choose an energy here, there is a certain momentum, or a certain kinetic energy here, and a different kinetic energy here, and so somehow, what you write for the wave function will reflect that. But now qualitatively, pictorially, if we have a wave function in this region which is oscillating like this, and it'll be oscillating at the same spatial frequency over here, well what's going to be happening here? Is it going to be oscillating faster or slower? AUDIENCE: Faster. Faster. ROBERT FIELD: Absolutely, and is it going to have amplitude smaller or larger than here? You're going to answer, yes? Well, let me do a thought experiment. So I'm going to walk from one side of the blackboard to the other, and I'm going to walk at a constant velocity. Then, I'm going to walk faster and then back to this original velocity. So what's the probability of seeing me in the middle region relative to the edge regions? Yes? AUDIENCE: It's less. ROBERT FIELD: That's right. The probability, local probability, is proportional to 1 over the velocity, and the wave function is proportional to 1 over the square root of the velocity. And the velocity is related to the momentum, and so we have everything. So we know that the wave here will be oscillating faster and with lower amplitude. This is what I want you to know, and you'll be able to use that cartoon to solve problems. If you understand what's going on here, these pictures will be equivalent to global understanding, and these pictures are also part of semi-classical quantum mechanics. I believe you all know classical mechanics at least a little, enough to be useful. And what we want to be able to do in order to draw pictures and to understand stuff is to insert just enough quantum mechanics into classical mechanics so that it's correct. Then, all of a sudden, it starts to make a lot more sense. OK. So the particle in a box, well, we have this sort of situation, and we have 0 and a. So the length of the box is a, and the bottom of the box v0 is 0 for this picture. Now, one of the things that I want you to think about is, OK, I understand. I've solved this problem. I know how to solve this problem. I know how to get the eigenvalues, and I know how to get the eigenfunctions, and I know how to normalize them. Well, suppose I move the box to the side. So I move it from say b to a plus b. So it's the same width, but it's just in a different place. Well, did anything change? The only thing that changes is the wave function, because you have to shift the coordinates. What happens if I raise the box or lower the box? Will anything change? AUDIENCE: [INAUDIBLE] ROBERT FIELD: You're hot. AUDIENCE: [INAUDIBLE] ROBERT FIELD: I'm sorry? AUDIENCE: [INAUDIBLE] ROBERT FIELD: Yeah. If I move the box so that v0 is not 0, but v0 is 10. AUDIENCE: Then, the weight function will oscillate slower. ROBERT FIELD: No. AUDIENCE: [INAUDIBLE] ROBERT FIELD: So if you move the box up in energy, the wave function is going to look exactly the same, but the energies are going to be different by the amount you move the box up or down, and this is really important. It may seem trivial to some of you and really obscure to others, but you really want to be able to take these things apart. Because that will enable you to understand them in a permanent way, and the cartoons are really important. So if you have the solution to the particle in a box, then it doesn't matter where the box is. You know the solution to any particle in a box. OK. There is something that I meant to talk about briefly, but when we write these solutions-- where did the other blackboard go? All right, well, I've hidden it-- so when we have solutions like e to the ikx and e to the minus ikx, so we have say a here and b here. When we go to normalize a function like this-- let's put the plus in here-- then we write psi star psi dx. So psi star would make this go a star and this go to e to the minus ikx, and this go to b star e to the plus ikx. So now, we multiply things together. We get an a, a star which is the square modulus of a, and we get e to the ikx and e to the minus ikx. It's 1. This is why we use this form. The integrals for things involving e to the ikx are either 1 or 0. So if you took e to the ikx, this term, and multiplied it by this term, you'd get an a, b star e to the 2 ikx integrated over a finite region. That's 0. So we really like this exponential notation, even if you've been brought up on sines and cosines, and you use the sines and cosines to impose the boundary conditions. OK, another challenge. So this is v0, and the only problem is this v0-- well, it looks like this. So this is v1. OK, so we have now a particle in this straight. It's a hybrid between the free particle and a particle in a box. So suppose we're at an energy like this. What's going to happen? Well, everything that's outside-- everything that's in the classically-allowed region, we understand. We know how to deal with it, but in here, well, that's OK too. But inside this classically-forbidden region, the wave function is going to behave differently. Now, I'm going to assert something. It doesn't have nodes. It doesn't oscillate. It's either exponentially decreasing or exponentially increasing, and it will never cross 0, never. OK. So now, if we're solving a problem involving any kind of 1D potential, number of nodes. So for 2D-bound problems, the number of nodes starts with 0, and it corresponds to the lowest energy state. The next state up has 1 node, and the next state has 2 nodes. So by counting the nodes, you would know what the energy order is of these eigenvalues which is also an extremely useful thing. If you're thinking about it or telling your computer to find the 33rd eigenvalue of something, because you just run a calculation that solves for an approximate wave function, and the 33rd, it needs 32 nodes. And so the computer says, oh, thank you, master, and here is your wave function, but you have to find the right thing. OK. Now, here is the picture that you use to remember everything about a particle in a box. And the wave function looks like this, and the next wave function looks like that, and the next wave function looks like this. And so no nodes, 1 node, 2 nodes, the nodes are symmetrically arranged in the space available. And the lobes on one side of the node and the other side have the same amplitude, different sine, and they're all normalized. And so the maximum value for each of these guys is 2 over a square root, where this is 0 to a. So that's a fantastic simplification, and it also reminds you of Mr. DeBroglie. He said, you have to have an integer number of half wavelengths-- well, for the hydrogen-- an integer number of wavelengths around a path. And for here, you need an integer number of wavelengths for that round trip which is the same thing or an integer number of half wavelengths. That's DeBroglie's idea, and it enables you to say, oh well, let's see if we can use this concept of wavelength to approach general problems. OK. Well, if you do something to the potential by putting a little thing in it, well, the wave function will oscillate more slowly in that region, and that causes it to be at a higher or lower energy? If it's oscillating more slowly here, it has to make it to an integer number of half wavelengths, and so that means it pushes it up. And if you do this, it'll push it down, and you can do terrible things. You can put a delta function there, and now you know everything qualitatively that can happen in a 1D box. OK. One of the things that bothers people a lot is, OK, so we have some wave function, it's got lots of nodes, and the particle starts out over here. How did it get across the node? How does it move across the node? Well, the answer is it's not moving. It's here. It's here. It's here. It's everywhere, and this is just the probability amplitude. There is no motion through a node, no motion at all. We are going to do time-dependent quantum mechanics before too long, and then there will be motion, but that motion is encoded in a different way. OK. Another thing, suppose you have a particle in a box, and it's in some state, and I'm going to draw something like this again. OK, first of all, one, two, three, four, five, which state is that? I got-- the hands are right, six, it's the sixth eigenstate. OK. Now, suppose-- nothing is moving. Right? This is a stationary state. How would you experimentally, in principle, determine that the particle is in this n equals 6 state? Now, this can be a completely fanciful experiment, which you would never do, but you could still describe how you would do it and what it would tell you. And so, yes. AUDIENCE: Try to find the n equals 6 to n equals 7 transition by irradiating it or something? ROBERT FIELD: OK. That's the quantum mechanical-- I agree, spectroscopy wins always. But if you want to observe the wave function or something related to the wave function, like the number of nodes, what would you do? And the reason I'm being very apologetic about this is because it's a crazy idea, But this is a one-dimensional system. Right? It's in the blackboard, and so you could stand out here and shoot particles at it from the perpendicular direction and collect the number of times you have a hit. And so you would discover that you would measure a probability distribution which had the form-- well, I can't draw this properly. It's going to have one, two, three, four, five, six, six regions separated by a gap, and what it's measuring is psi 6 squared. Well, you can't measure, you cannot observe a wave function, but you can observe a probability distribution wave function squared. You can also do a spectroscopic experiment and find out what is the nature of the Hamiltonian. And if you know the nature of the Hamiltonian, you can calculate the wave function, but you can't observe it. OK. Another thing, this harmonic oscillator-- this particle in a box has a minimum energy which is not at the bottom of the box. Well, we have something called the uncertainty principle. Now, I'm just pulling this out of my pocket, but I know that x, p is equal to i h bar not 0, and one can derive some uncertainty principle by doing a little bit more mathematics. But basically that uncertainty principle is where sigma x is expectation value of x squared minus expectation value of x squared square root. So if we can calculate this and calculate that, we can calculate the variance in x, and you can calculate the variance in p. That's exact, and that's what you can derive from the computation rule, but for our purposes, we can be really crude. And so if I'm in this state, what is delta x? What is the range of possibilities for x? AUDIENCE: The box link? ROBERT FIELD: Yeah, a. OK, and what is the possibility-- what is the uncertainty in p sub x? In an eigenstate, we've got equal amplitudes going this way and going that way. So we could just say p sub x positive minus p sub x negative which is 2p. That's the uncertainty. And if we know what quantum number we're in we know what the expectation value for p of the momentum is, and what we derive is that delta x delta P is equal to hn. You can do that, and maybe I should ask you to really be sure you can do that. In seconds, because you really know what the possible values of momentum or momentum squared are for a product of a box. OK. So why is there zero-point energy, because if you said, I had a level at the bottom of the box, we would have the momentum 0. The uncertainty and the momentum is 0, and the product of the uncertainty of the moment times the product of the uncertainty in the position has to be some finite number, and you can't do that here. And so this is a simple illustration of the uncertainty principle that you have to have a non-zero zero-point energy. That's true for all one-dimensional problems. OK. We've got lots of time. One of the beautiful things about quantum mechanics is that if you solved one problem, you could solve a whole bunch of problems, and so to illustrate that, let's consider the 3D particle in a box. So for the 3D particle in a box, the Hamiltonian can be written as a little Hamiltonian for the x degree of freedom y and z. OK. So we have three independent motions of the particle. They're not coupled. They could be, and we're interested in letting them be coupled. But that's where we start asking questions about reality, and that's where we bring in perturbation theory. But for this, oh, that's fantastic, because I know the eigenvalues of this operator and of this operator, eigenfunctions, and of this operator. So the problem is basically solved once you solve the 1D box. Now, one proviso, what you can do this separation completely formally as long as hx, hy commute, and basically we say the x, y, and z directions don't interact with each other. The particle is free inside the box. It's just encountering walls. There are no springs or anything expressing the number of degrees of freedom. OK. So we have now a wave function, which is a function of x, y, and z, but we can always write it as psi x of x, psi y of y, psi z of z. So it's a product of three wave functions that we know, and the energy is going to be expressed as a function of three quantum numbers, where the box is edge lengths a, b, and c. You didn't see me looking at my notes. I'm just taking the solution to 1D box, and I'm multiplying it. And so now we have the particle in a 3D box, and this is where the ideal gas law comes from, but not in this course. So anyway, this is a simple thing, and the wave functions are simple as well, and you can do all these fantastic things. So there are many problems like a polyatomic molecule. In a polyatomic molecule, if you have n atoms, you have 3n minus 6 vibrational modes. You might ask, what is a vibrational mode? Well, are they're independent motions of the atoms that satisfy the harmonic oscillator Hamiltonian, which we'll come to next time. And so we have 3n minus 6 exactly solved problems all cohabiting in one Hamiltonian. And then we can say, oh yeah, we got these oscillators, and if I stretch a particular bond, it might affect the force constant for the bending. So we can introduce couplings between the oscillators, and in fact, that's what we do with perturbation theory. That's the whole purpose. And with that, we can describe both the spectrum and how the spectrum encodes the couplings between the modes. And also, we can describe what's called intramolecular vibrational redistribution, which happens when you have a very high density of vibrational state. Energy moves around, because all the modes are coupled, and so even if you've plucked one, the excitation would go to others. And we can understand that all using the same formalism that we're about to develop. All right. I'm not using my notes this time, because I think there's just so much insight, so I have to keep checking to see what I've skipped. All right. So what I've been saying is whenever the Hamiltonian can be expressed as a sum of individual Hamiltonians, whenever we can write the Hamiltonian this way, we can write the wave function as a product of wave functions for coordinates, xi i1, 2 N. And the energies will be the sum Ein, i equals little ei n sub i. I equals 1 to N. So this is really easy. If we have simply a Hamiltonian, which is a sum of individual particle Hamiltonians, we don't even have to stop to think. We know the wave functions and the eigenvalues. OK. Now, suppose the Hamiltonian is this plus that. So here, we have a Hamiltonian, and this is this simply the uncoupled Hamiltonian. This is what we'd like nature to be, but nature isn't so kind, and there are some coupling terms. And so we know the eigenfunctions and eigenvalues for this Hamiltonian. We call them the basis functions and the zero-order energies, and then there is this thing that couples them and leads to complications, and that's perturbation theory. We're going to do that. OK. So now, let me just say, on page nine of your notes, there's the words next time, and those are going to be replaced by you should. There's a whole bunch of things that I want you to consider, and I was planning on talking about them, but they're all pretty trivial. And so there are a whole bunch of things you should study, because I will ask you questions about them. And of greatest importance is the ability to calculate things like that. OK. Now, I'm going to give you a whole bunch of facts which I may not have derived. But you're going to live with them, and you can ask me questions. Some of these things are theorems that we can prove, but the proof of the theorem is really boring. Understanding what it is is really wonderful. So all eigenfunctions that belong to different eigenvalues-- of whatever operator we want, the Hamiltonian, some other operator-- are orthogonal. That's a fantastic simplification. So if you have two eigenfunctions of the Hamiltonian, of the position operator, anything, those eigenfunctions are orthogonal. Their integral is 0, very, very useful. Then, one of the initial postulates about quantum mechanics is this idea that the wave functions are well-behaved. Well, if I were to state it at the beginning, you wouldn't know what's well-behaved and ill-behaved, but now I can tell you. One of the things is that the wave function is continuous, no matter what the potential does. The derivative is continuous, except at an infinite barrier. So you come along, and you hit an infinite barrier, and you've already seen that with the particle in a box. The wave function is continuous at the edge of the box, but the derivative is discontinuous, and it's because it's an infinite wall. That's a pretty violent thing to make the first derivative be discontinuous. The secondary derivative is continuous, except at any sudden change in the potential. So when you're solving 1D problems, and you've got a solution that works in the various regions, and you want to connect them together, these provide some rules about the boundary conditions. So now, most real systems don't have infinite walls or infinitely sharp steps. So for calculation of physically reasonable things, wave function for the first derivative, second derivative, they are continuous. But for solving a problem, we like these steps, because then we know how to impose boundary conditions, and that gets us a much easier thing to calculate. OK. Now-- oh, that's where it went, OK-- semi classical quantum mechanics. We know that the energy classically is p squared over 2m. Right? It's 1/2 mv squared, but that's the same thing as p squared over 2m. In quantum mechanics, when we talk about Hamiltonians, the variables are x and p not x and v. So that seems like a picky thing, but it turns out to be very important. And so we can say, well, the momentum can be a function of x, classically. So I just solved this problem, and if the potential is not constant, then the momentum, classical momentum, is not constant. But we know what it is everywhere, and we also know that the wavelength is equal to h over p. So we could make a step into the unknown saying, well, the wavelength for a non-constant potential is a function of coordinate, and it's going to be equal to h over p of x. That's semi-classical quantum mechanics, everything you would possibly want. Now, for one-dimensional problems, you can solve in terms of this coordinate dependent wavelength which is related to the local momentum. And so it doesn't matter how complicated the problem is, you know that you can calculate the spatial modulation frequency, you could calculate the amplitude, is it big or small, based on these ideas of the classical momentum function. So this demonstration of my walking across the room slow, fast, slow tells you about probability. So if you use this formula, you know the node spacings, and you know the amplitudes. Now, what you don't know is where are the nodes? You know how far they are apart, but I have to be humble about this. In order to calculate where the nodes are, I have to do a little bit more in order to pin them down. But mostly, when you're trying to understand how something works, you want to know the amplitude of the envelope, and that's a probability, and so it's related to 1 over the square root of the momentum. I'm sorry, the amplitude is 1 over the square root of the momentum, and the nodes spacings, those are the things you want to do, and you want to know them immediately. And a couple other facts that I told you earlier, but I think I want to emphasize them-- the energy order, number of nodes, number of internal nodes. For 1D problems, you never skip a number of-- so you can't say, there is no wave function with 13 nodes, even if you don't like being unlucky. And it's there, and so if you want the 13th energy level, you want something with 12 nodes, and that also focuses things. So these are amazingly wonderful things, because you can get them from what you know about classical mechanics, and it's easy to embed them into a kind of half quantum mechanics. And since, I told you, this course is for use and insight, not admiration of philosophy or historical development, and this is what you want to do. You look at the problem and sketch how is the wave function going to behave and perhaps how a particular thing at some place in the potential is going to affect the energy levels or any other observable property. And so the cartoons are really your guide to getting things right, but you really have to invest in developing the sense of how to build these cartoons. OK. I'm finished early again. Does anybody have any questions? OK. We start the harmonic oscillator next time. OK, so I can say a couple of things. The wave functions, the solution to the 1D box and the free particle, they're really simple. The solution to the harmonic oscillator involves a complicated differential equation which the mathematicians have solved and worked out all the properties. But there is a really important simplification that enables us to proceed with even greater velocity in the harmonic oscillator than we would in a particle in a box. And they are these things called a and a dagger, creation and annihilation operators, where when we operate on psi with this creation operator, it converts it to square root of v plus 1, psi v plus 1 further. These things, we don't ever have to do an integral. Once you're in harmonic oscillator land, everything you need comes from these wonderful operators. And so even though the differential equation is a little bit scary for chemists, these things make everything trivial. And so we use the harmonic oscillator, and the particle in a box to illustrate time-dependent quantum mechanics. They each have their own special advantages for simplifications, but it's wonderful, because we can use something we barely understand for the first time. And actually reach that level of, yeah, I can understand macroscopic behaviors too and how they relate to quantum mechanical behavior of simple systems. OK, so that's where we're going. We're going to have two or three lectures on harmonic oscillator. |
MIT_561_Physical_Chemistry_Fall_2017 | 4_Classical_Wave_Equation_and_Separation_of_Variables.txt | The following content is provided under a Creative Commons license. Your support will help MIT Open Courseware continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT Open Courseware at ocw.mit.edu. ROBERT FIELD: That's the outline of what we're going to cover. But before we get started on that, I want to talk about a couple of things. First of all, last time, we talked about the two slit experiment. And it's mostly classical. There is only a little bit of quantum in it where we talk about momentum as being determined by h over lambda. OK. But what were the two surprising things about the two slit experiment? There are two of them, two surprises. Yes? AUDIENCE: [INAUDIBLE] That a single particle can interfere with itself. ROBERT FIELD: Yes. That's the most surprising thing. And when you go to really low intensity-- so there's only one photon, which is quantum, in the apparatus, somehow, it knows enough to interfere with itself. And this is the most mysterious aspect. But then there's one other aspect, which is how it communicates that interference with itself. What is that? You're hot. You want to do another one? AUDIENCE: What do you mean by how it communicates? ROBERT FIELD: Well, here we have the screen on which the information is deposited. And you have possibly some sort of probability distribution, which is a kind of a continuous thing. But you don't observe that. What do you observe? So you could say the state of a particle has amplitude everywhere on this screen. But what do you see in the experiment? Yes? AUDIENCE: So you just see one [INAUDIBLE] point. [INAUDIBLE] thousands and thousands. And eventually, you see it mimic that probability [? sequence. ?] ROBERT FIELD: That's exactly right. So this state of the system which is distributed collapses to a single point. We say that what we are observing-- and this is mysterious. And it should bother you now. We're seeing an eigenvalue of the measurement operator. So we have this sort of thing. It goes into the measurement operator. And the measurement operator says, this is one of the answers I'm permitted to give you. And another aspect that's really disturbing or puzzling is that you do many identical experiments. And the answer is always different. There is no determinant. It's all probabilistic. So this really should bother you. And eventually, it won't. OK. Now, there's another question I have. When I described the two slit experiment, I intentionally put something up on the diagram that should bother you, that should have said, this is ridiculous. I used a candle in the lecture. And I used a light bulb in the notes. And why is that ridiculous? Yes? AUDIENCE: You said many frequencies. ROBERT FIELD: That's right. So the sources of light that I misled you with intentionally have a continuous frequency distribution. And the interference depends on the same frequency. So the only way you would see any kind of diffraction pattern, any kind of pattern on the two slit experiment, is if you had monochromatic light. It would all wash out. You'd still get dots. But the dots would never give you anything except perhaps a superposition of two or three or an infinite number of patterns. OK. This is, again, something that's really bothersome. Now, I also use the crude illustration of an uncertainty principle, that the uncertainty in position z-axis and the uncertainty in the momentum along the z-axis was greater than h. And I froze. And I didn't realize that I had delta s with the slit. But it really is the same thing as the z. Because the slit is how you define the position in the z-axis. And so this is the first taste of the uncertainty principle. And I said I didn't like it. OK. In quantum mechanics, you're not allowed to look inside small stuff. You're not allowed to see the microscopic structure. You're only able to do experiments, usually thought experiments, an infinite number of identical experiments. And they reveal the structure in some complicated, encoded way. And this is really not what the textbooks are about. Textbooks don't tell you how you actually think about a problem with quantum mechanics. They tell you, here are some exactly solved problems. Memorize them. And I don't want you-- I don't want to do that. OK. Last part of the introduction here-- suppose we have a circular drum, a square drum, and a rectangular drum. Have you ever seen a square drum or a rectangular drum? Do you have an idea how a square drum would sound? Yes. You do have an idea. It would sound terrible. Because the frequencies, you get are not integer multiples. It would just sound amazingly terrible. But if you had a square drum or a rectangular drum, you could do an experiment with some kind of acoustic instrument to find out what the frequency distribution is of the noise you make. And you would be able to tell. It's not round. It might be square. Or it might have a certain ratio of dimensions. This is what we're talking about as far as internal structure is concerned. And it's very much like what you would do as a musician. I mean, certainly, when a musical instrument is arranged correctly, it's not like a square drum. It sounds good. And that's because you get harmonics or you get integer multiples of some standard frequency. OK. So now, we're going to talk about the classical wave equation, which is not quantum. But it's a very similar sort of equation to the Schrodinger equation. And so the methods for solving this differential equation are on display. And so the trick is-- well, first of all, where does this equation come from? And it's always force is equal to mass times acceleration in disguise. OK. And then you have tricks for how you solve this. And one of the most frequently used and powerful tricks is separation of variables. You need to know how that works. Then once you solve the problem, you have the general solution. And you then say, well, OK, for the specific case we have, like a string tied down at both ends, we have boundary conditions. And we impose those boundary conditions. And then we have basically what we would call the normal modes of the problem. And then we would ask, OK, well, suppose we're doing a specific experiment or doing a specific preparation of the system. And we can call that the pluck of the system. And you might pluck several normal modes. You get a superposition state. And that superposition state behaves in a dynamic way. And you want to be able to understand that dynamics. And the most important thing that I want you to do is, instead of trying to draw the solutions to a differential equation, which is a mathematical equation, I want you to draw cartoons, cartoons that embody your understanding of the problem. And I'm going to be trying to do that today. OK. In this course, for the first half of the course, most of what we're going to be doing is solving for exactly soluble problems-- the particle in a box, the harmonic oscillator, the rigid rotor, and the hydrogen atom. With these four problems, most of the things that we will encounter in quantum mechanics are somehow related to these. And in the textbooks, they treat these things as sacred. And they say, OK, well, now that you've solved them, you understand quantum mechanics. But these are really tools for understanding more complicated situations. I mean, you might have a particle in a box. Instead of with a square bottom, it might have a tilted bottom. Or it might have a double minimum. But if you understand that, you then can begin to build an understanding of, what are the things in the experiment that tell you about these distortions of the standard problem? And the same thing for a harmonic oscillator. Almost everything that's vibrating is harmonic approximately. But there's a little bit of distortion as you stretch it more. And again, you can understand how to measure the distortions from harmonicity by understanding the harmonic oscillator. We did rotor, H atom. It's all the same. So I would like to tell you that these standard problems are really important. But nothing is like that. And what's important is how it's different from that. And this is my unique perspective. And you won't get that from McQuarrie or any textbook. But this is MIT. So there are templates for understanding real quantum mechanical system. And the big thing, the most important technique for doing that is perturbation theory. And so perturbation theory is just a way of building beyond the oversimplification. And it's mathematically really ugly. But it's tremendously powerful. And it's where you get insight. OK. Now, many people have complained that they found 5.61 hard. Because it's so mathematical. And maybe this is going to be the most mathematical lecture in the course. But I don't want it to be hard. Now, chemists usually derive insights from pictorial rather than mathematical views of a problem. So what are the pictures that describe these differential equations? How do we convert what seems to be just straight mathematics to pictures that mean something to us? And that's my goal, to get you to be drawing freehand pictures that embody the important features of the solutions to the problems. OK. So we're going to be looking at a differential equation. And one of the first questions you ask, well, where did that equation come from? And you're not going to derive a differential equation ever in this course. But you're going to want to think, well, I pretty much understand what's in this differential equation. And then we'll use standard methods for solving that equation. And one such differential equation is this-- second derivative of some function with respect to a variable is equal to a constant times that function. Now, that you know. You know sines and cosines are solutions to that. And you know that exponentials are solutions to that. Now, that pretty much takes you through a lot of problems in quantum mechanics. But now, one of the important things is this is a second-order differential equation. And that means that there are going to be two linearly independent solutions. And you need to know both of them. I'll talk about this some more later. Now, sometimes, the differential equations look much more complicated than this. And so the goal is usually to rewrite it in a form which corresponds to a differential equation that is well known and solved by mathematicians whose business is doing that. But we won't be doing that. OK. But usually, when you have a differential equation, the function is of more than one variable. And frequently, it's position and time. And so the first thing you do is you try to separate variables. And so that's what we're going to do. So we have a differential equation. And the first thing is a general solution. And one of the things that this solution will have is nodes. And the distance between nodes-- here's a node. Here's a node. That's half the wavelength. And we know that in quantum mechanics, if you know the wavelength, you know the momentum. So nodes are really important. Because it's telling you how fast things are moving. We can also look at the envelope. And this would be some kind of classical, as opposed to a quantum mechanical, probability distribution. And so it might look like this. But the important thing about the envelope is that it's always positive. Because it's probability, as opposed to a probability amplitude, which can be positive and negative. Interference is really important in quantum mechanics. But sometimes, the envelope tells you all you need to know. And the other thing is the velocity of a stationary phase. So you have a wave. And it's moving. And you sit at a point on that wave. And you ask, how fast does that point move? And I did that last time. And OK. So I've already talked a little bit about what we do next. But the important thing is always, at the end, you draw a cartoon. And you endow that cartoon with your insights. And that enables you to remember and to understand and to organize questions about the problem. OK. So let's get to work on a real problem. So we have a string that's tied down to two points. And so let's look at the distortion of that string. And so we chop this region of space up into regions. So this might be the region at x minus 1. And this might be the region at x0. And this might be the region of x1. And we're interested in-- OK, suppose we have the value of the displacement of the wave here at x minus 1 and here and here. OK, so these would be the amount that the wave is displaced from equilibrium. And we call those u of x. And so the first segment here, the minus 1 segment, this segment is pulling down on this segment of the string by this amount. And this one is pulling up on the segment by that amount. So we want to know, what is the force acting on each segment? And so we have the force constant times the displacement at x0 minus the displacement at x minus 1. So this is the difference between the displacements. And this is the force constant. We're talking about Hooke's law. Hooke's law is the force is equal to minus k times the displacement. And so we collect the forces felt by each particle. And the forces felt by each particle are, again, the force constant times the difference in u at 0 and minus 1 minus the difference-- plus 1 minus the difference in u at 0 and plus 1. And this is a second derivative. This is the second derivative of u with respect to x. So we've derived a wave equation. And we know it's going to involve a second derivative. So force is equal to mass times acceleration. Well, we already know the force is going to be related to the second derivative of u with respect to x. And now, this is something. And we know what this is. This is going to be the time derivative. OK. And this is just something that gets the units right. And it has physical significance. But in the case of this particle on a string-- this wave on a string, it's related to the mass of the string and the tension of the string. And it's also related to the velocity that things move. OK. So we have a differential equation that is related to forces equal to mass times acceleration. And the differential equation has the form second derivative of u with respect to x is equal to 1 over v squared times the second derivative of u with respect to t. That's the wave equation. So it is really f is equal to ma. But OK. And now, the units of this-- this is x. And this is t. In order to be dimensionally consistent, this has to be something that is x over t, OK? And so this may be a velocity. But it has units of velocity. That's the differential equation we want to solve. OK. Well, the original differential equation that I wrote-- but I'm getting ahead of myself. OK. So this U of x and t-- we'd like it to be X of x times T of t. We think we could separate the variables in this way. So we try it. If we fail, it says you can't do that. Failure is usually going to be a result that the solution to the differential equation in this form is nothing. It's a straight line. Nothing's happening. So failure is acceptable. But if we're successful, we're going to get two separate differential equations. OK. So what we do then is take this differential equation, substitute this in, and divide on the left. So we have 1 over X of x times T of t times the second derivative with respect to x of xt is equal to 1 over xt times the second derivative with respect to t of xt. OK. Well, on this side of the equation, the only thing that involves time is here. This doesn't operate on time. And so we can cancel the time dependence from this side. And over on this side, this derivative operates on t but not x. And so we can cancel the x part. And so what we have now is an equation X of x second derivative with respect to x squared of x is equal to this constant, 1 over v squared, times 1 over t times the derivative of T with respect to little t, OK? So this is interesting. We have a function of x on this side and a function of t on this side. They are independent variables. This can only be true if both sides are equal to the same constant. So now, we have to differential equations. We have 1 over x second derivative with respect to x of x is equal to a concept. And we have 1 over v squared 1 over t second derivative with respect to t of T is equal to K. So now, we're on firmer ground. We know about solutions to this kind of equation. And so there's two cases. One is this K is greater than 0. And the other is K less than 0. So let's look at this equation. If K is greater than 0, then if we plug in a sine or a cosine, we get something that's less than 0. Because the derivative of sine with respect to its variable is negative cosine. And then we do it again, we get back to sine. And so sines and cosines are no good for this equation if K is greater than 0. But exponentials-- so we can have e to some constant x or e to the minus some constant x. Or here, for the negative value of K, we could have sine some constant x and cosine of some constant x. We know that. So we have two cases. So to make life simple, we say K is going to be equal to lowercase k squared. Because we want to use this lowercase k in our solutions. All right. So I may have confused matters. But the solution for the time equation and the position equation are clear. And so depending on whether K is positive or negative, we're dealing with sines and cosines or exponentials. OK. So I don't want to belabor this, but the next stage is boundary conditions. We don't know whether K positive or K negative is possible. But we do know boundary conditions. And so if we have a string which is tied down at the end-- so this is x equals 0. And this is x equals L. Then we input impose the boundary conditions. Well, the boundary conditions are u of 0t is equal to 0. And u of Lt is equal to 0. So if we take the K greater than 0 case, u of 0t is equal to-- well, let's just do this again. 0t. We have x of 0 times T of t. OK, we don't really care about this. But x of 0 has to be 0-- I'm sorry. OK. So we have two solutions. If K is greater than 0, we have the exponential terms. So we have Ae to the 0 plus B to the minus 0. And this has to be equal to 0. That's x of 0. Well, e to the 0 is 1. E to the minus 0 is 1. And so this is good. A has to be equal to minus B. And then we have the other boundary condition, X of L. We have A e to the L plus B e to the minus L. Well, I'm sorry. Let's just put what we already know. Minus A. So we can write this as A e to the L minus e to the minus L. And that has to be equal to 0. Can't do it. This can never be 0. So that means the K greater than 0 solutions are illegal. Well, that's kind of bad news. Because it sounds like separation of variables is failing. But it doesn't. Because the K less than 0 solution works. So for K less than 0, X of 0 is equal to C sine 0 plus D cosine 0. And so that means D is equal to 0. Things are dying. And X of L, boundary condition, is C sine KL is equal to 0. And this we can solve. So sine is equal to 0 when KL is equal to 0, 0 pi, 2 pi, et cetera. And so we have KL is equal to n pi. So we can write this as Kn is equal to n pi over L. We get quantization. This isn't quantum mechanics. But there are certain allowed values of this K constant. And we have a bunch of solutions. And so what do they look like? So n equals 0, n equals 1, n equals 2. So what does the 0 solution look like? Yes? AUDIENCE: No node. ROBERT FIELD: Nothing. So we don't even think about this. We say, n equals 0 is not a solution. Because the wave isn't there. No nodes, one node. And if we look at this carefully, the node is always at a cemetery point. It's in the middle. If we have the next one, we'll have two nodes. And they'll be at the 1/3 2/3 point. And so we know where the nodes are. We also know that the amplitude of each loop of the wave function is the same. But it alternates in sine. So you can draw cartoons now at will. Because this spatial part of the solution to this wave is clear. Any value of the quantum number, or the n, gives you a picture that you can draw in seconds. And there are a lot of quantum mechanical problems like that. But sometimes, you have to keep in mind that the node separation, in other words-- well, let's just say node separation is lambda over 2. And lambda over 2 is equal to 1/2 h over p. So if we know what the momentum is, or we know what the kinetic energy is, we know what the momentum is. We know how momentum is encoded in node separations. So everything we want to know about a one-dimensional problem is expressed in the spacing of nodes and the amplitude between nodes. And the amplitude between nodes have something to do with the momentum, too. Because if you're going from here to here at some high velocity, there's not much amplitude. And at a lower velocity, you get more amplitude. And so the amplitude in each of these node to node separate sections is related to the average momentum of the classical particle in that section. So the classical mechanics is going to be extremely important in drawing cartoons for quantum mechanical systems. Not in the textbooks. We supposedly know classical mechanics pretty well, and especially here at MIT. So you might as well use that in order to get an idea of how all of the quantum mechanical problems you're facing will be behaving. OK. So the next thing we want to do is finish the job. And so I can simply write down the time-dependent solutions. They are E sine vkn t plus F cosine vkn t. And we can say that-- rather than carrying around all this stuff, we can say omega n is vkn. Isn't that nice? So we have a frequency for the time-dependent part, which is an integer multiple of this constant V times this [? vector. ?] Or this you can think of as just k sub n. So we can rewrite this in a frequency and phase form. We have now the full solution. We have A n sine n pi L over x. And then this e N sine n pi-- I'm so used to the pictures I don't even want to look at the equations anymore-- n omega t plus F n cosine n omega t. OK. So we can also take this and rewrite it in a simpler form, E n prime cosine n omega t plus phi n. So we can combine these two terms as a single cosine with a phase vector. OK. So now, we're ready to actually go to the specific thing that you do in a real experiment or a real musical instrument. We say, OK, here is the actual initial condition, the pluck of the system. And the pluck usually occurs at t equals 0. But I'll just specify it here. And what we have is now a sum over as many normal modes as you want. We have A n E n prime times sine n pi over L x times cosine N omega t plus phi N. So we have a bunch of terms like this, a spacial factor, and a temporal factor. And you can draw pictures of both. Now, but there is another simplification. From trigonometry, sine A cosine B-- we have sine and cosine-- can be written as 1/2 times sine n pi L x plus n omega t plus phi n plus sine n pi L x minus n omega t minus phi n. So these are the two possible solutions. And we can write them now in terms of position factor at a time factor in the same sine or cosine function. So these are the things. Now, we're ready to make a picture. OK. So these are the actual things that you make by exciting the system not in an eigenfunction. But it's a superposition of eigenfunctions. And again, there are certain things you learn. If you have a pure eigenfunction, you have standing waves. There's no left-right motion. There's no breathing motion. There's only up-down motion of each loop of the wave function. If you have a superposition of two or more functions, which are all of even n, then what happens is you have no motions, you just have breathing. In other words, you have a function that might look sort of like this at one time and like that at another time. So amplitude is moving. So it can be moving. And in between, it's sort of like this. Now, if you have a function which involves both even and odd n, you have left-right motion. This is true in quantum mechanics, too. So you only get motion if you're making superposition of eigen-- Yes? AUDIENCE: What is the difference between breathing and the standing wave with no nodes? ROBERT FIELD: Well, for this picture, this is over-simple. So I mean, you could have-- basically, what's happening is amplitude is moving from middle to the edges and back. And so yes. But you want to develop your own language, your own set of drawings so that you understand these things. And the important thing is the understanding, the ability to draw these pictures which contain the critical information about node spacings, amplitudes, shapes, and to anticipate when you're going to have left-right motion or when you're just going to have complicated up-down motions. Because there could be nodes. But there is no motion of the center of this wavepacket. Now, this is fantastic. Because I just said wavepacket. Quantum mechanics-- eigenfunctions don't move. Superpositions of eigenfunctions do move. If we make a superposition of many eigenfunctions, it is a particle-like state. What that particle-like state will do is exactly what you expect from 8.01. The particle-like states-- the center of the wavepacket for a particle-like state moves according to Newton's equations. So I'm saying I'm taking away your ability to look at microscopic stuff. And I'm going to give it back to you. By the end of the course, we're going to have the time-dependent Schrodinger equation. We're going to be able to see things move. And we're going to see why they move and how they encode that motion. Not in the textbooks, but I think it's something you really want to be able to do. If you're going to understand physical systems and use it to guide your understanding, you have to be able to draw these pictures and build a step at a time. And so this way the equation, the classical wave equation, gives you almost all of the tools for artistry as well as insight. OK. Now, in the notes, there is a time-lapse movie that shows what a two-state wave function-- what a two-state solution to the wave equation looks like if you have even and odd or only even and even terms. OK. Now, I'm going to make some assertions at the end. We're coming back to the drum problem. And suppose we have a rectangular drum. Well, solving the differential equation for this rectangular drum gives you a bunch of normal mode frequencies that depend on two indices. And so this is the geometric structure of the drum. And these are the quantum numbers. And these are the frequencies. And it's going to make a whole bunch of frequencies that are not integer multiples of each other or of any simpler thing. And that's why it sounds horrible. It's perhaps a little bit like playing a violin with a saw. It will sound terrible. You would never do it. But you would also never build a square or rectangular drum. But the noise that you make tells immediately not just what the shape of the instrument is but how it was played. For example, suppose you had an elliptical drum. That'll sound terrible, too. But here are the two foci. I'm not so sure it'll sound terrible if you hit it here or here. And certainly, if you are a circular drum, if you hit it in the middle as opposed to on the edges, it'll sound different. The spectral content will be the same. But the amplitudes of each component will be different. And so in quantum mechanics, you use the same sort of instinct as you develop as a musician in order to figure out how this system is going to respond to what you do to it. And that's pretty powerful. So many of you are musicians. And you know instinctively what's wrong when you do something that's not quite right, or your instrument is out of tune. But in quantum mechanics, all of those insights will come to bear. Not in the textbooks. Because the textbooks tell you about exactly solved problems. And then they tell you how to do spectra that are too perfect for anybody else to observe. And you won't see those spectra. And they don't tell you what the spectra you will observe tell you about the system in question. OK. So I should stop now. And I'm going to be generating Problem Set 2, which will be posted on Friday. Problem Set 1 is due this Friday. And-- Good. Thank you. |
MIT_561_Physical_Chemistry_Fall_2017 | 2_Wave_Nature_of_the_Electron_and_the_Internal_Structure_of_an_Atom.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation, or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: Last time, we talked about the photoelectric effect. What was that? And what were the important points? Yes? AUDIENCE: It's quantized and has energy associated with its frequency. ROBERT FIELD: Yes. OK, so, the idea of quantization of electromagnetic radiation and photons. And the photon has an energy h times nu. Nu is the frequency. And the evidence was mostly from a plot of what versus what. Somebody else? Yes? AUDIENCE: Frequency of the incoming photon versus-- or, that's the x-axis. So, kinetic energy of the ejected electron versus frequency. ROBERT FIELD: Exactly. And the slope of that plot, which was h, is universal. It doesn't matter where the electron came from. And this was really an amazing thing. And then, the other thing we talked about was Compton scattering. And what did Compton scattering tell us? Yes. AUDIENCE: Photon has a momentum-- ROBERT FIELD: Yes. AUDIENCE: --transfer. ROBERT FIELD: We're interested in the particle-like character of what we think of as waves. And we saw that the waves were particles. And particles-- or at least packets. And these packets had definite momentum. And that was a wonderful observation. So, today, this is the menu of what I'm going to talk about. And at the end, there's a magic word, "spectra." And I like that because what we're going to be discovering today is that the internal structure of atoms and molecules-- we are not allowed to observe it directly, but it's surprising. And it's encoded in something which we can observe, which is a spectrum. And the spectrum that I will show you at the end is one that is-- contains essentially no information, but acts as a template for what we really want to know about how things are different from hydrogen atom. And that's the beginning of our exploration of the structure of atoms and molecules. It's through the spectrum and it's how it's different from-- in subtle ways-- the spectrum of the hydrogen atom. OK, so, we're going to turn our focus today to the electron, as opposed to light. And we're going to play the same game. We know the electron is a particle. Does anyone want to tell me why we know that? Is there any-- anybody who's got a clue? You can-- oh, good. Yes. AUDIENCE: There's an experiment with the little oil drops, where they suspended them and found that-- ROBERT FIELD: Yes. AUDIENCE: --charge is quantized. ROBERT FIELD: I love that experiment. That's the Millikan experiment. One of the reasons I love it is because Millikan and Mulliken are two different people. So I find that it's really easy to come up with one of them. And I remember Mulliken is a spectroscopist and Millikan was a different kind of physicist. But they're both famous and they both have connections to important universities-- University of Chicago for Millikan and Mulliken, MIT and Caltech. So we're going to be looking at something that we know is a particle. And we're going to show that it has wave characteristics. We want to be able to show that the electron has a wavelength and it follows the same equation that we use to describe the behavior of electromagnetic radiation. So how would we control the momentum of an electron? Yes. AUDIENCE: So you can control the kinetic energy that it has by putting it through a certain potential. ROBERT FIELD: OK, so, we know we can easily measure the momentum of a particle. That's not a big challenge. But then how would we measure its wavelength? Yes. AUDIENCE: Some type of diffraction? ROBERT FIELD: Yes. We basically use some material, which acts a ruler. We have a thin metal foil and the distances between atoms in the foil are constant. So that's the ruler against which we measure the wavelength of light. And we'll talk about the Davisson-Germer experiment, where we measure the wavelength. And then we'll talk about the Geiger-Marsden experiment, where we say, well, atoms have electrons in them. And what is the structure of an atom? Remember, we can't look inside. So we have to use some kind of an experiment to be able to look inside the atom. Now, physicists have one trick they often use. To find out the internal structure of something, they shoot a particle, or a wave, at it that has a wavelength comparable to the distances you're hoping to measure. So if you have a very high energy probe particle, it will have a very short wavelength and it will look sort of like a bullet. And we know how bullets scatter off of targets or hit targets. We also, if we choose the wavelength to be comparable to the distances we're expecting to measure, then we're going to see diffraction. This is a kind of subject that lends itself to exam questions. So let's start out by talking about the Geiger-Marsden experiment-- I mean, the Davisson-Germer experiment. So we have a beam of X-rays or a beam of electrons-- either one. And we have an aluminum foil. And we have-- it's just an intense beam. We want to stop most of it before it hits a detector. And, so, this is some kind of a screen or-- So what we're looking for is, when the X-rays or the electrons scatter off of this ruler, we get something that appears on the screen as pairs of circles. This is the powder pattern because this is a multi-crystalline object. But in each object, we have a bunch of equally-spaced atoms, where this is the lattice constant and this is the square root of 2 times the lattice constant. So each atom has nearest neighbors and second-nearest neighbors. So we have two rulers going on. And one ruler will give one set of rings and the other ruler will give a different set of rings. And because these particles are randomly oriented, instead of having spots, you have circles. And there's all sorts of information in these powder patterns. But basically, they're saying, well, we're seeing a structure which is related to something we know. How would we know the distance between atoms in a foil? Using macroscopic measurements? Yes. You're hot today. AUDIENCE: You have access to density and you have access to the non-atomic weight. ROBERT FIELD: That's it, yes. So it's a simple matter to know at least what is the magnitude of the distance. There is an issue of what is the crystal structure. And there are different structures and that will give rise to different features in the powder pattern. But the important thing is we do this. We look at the pattern that emerges when we shoot X-rays at this screen. And we already know that X-rays have wavelengths. And we know that they have momentum. We know about the scattering of photons. And as a result, we know that we can predict exactly what the pattern associated with the X-ray scattering is going to be. And then we do the same thing with electrons. And now for the electrons, we can control the momentum. That's easy. And what we want to know is what is the wavelength. And we have the wavelength of the X-rays. And so what we do is we vary the momentum of the electrons until the powder pattern for the electron-- the diffraction pattern for the electron-- is exactly the same as the diffraction pattern for the X-ray. And we discover that, for the electron, we have the same result. OK, so, we have, now, photons. They have wavelengths. And the momentum was the surprise for the photons. And we have particles, which have wavelengths and momentum. And the wavelength was a surprise for the particle. So it doesn't matter. Everything follows this equation. And this equation was anticipated by de Broglie who, in his PhD thesis-- you know, he's a person about your age-- and he wrote his thesis in 1924. And, among other brilliant things, he said that everything should follow this simple equation. And that was a brave statement. And it predicted that de Broglie was going to make a lot of brilliant statements in his career. And this was just the first of-- and one of the nicest-- but we'll hear a little bit more about de Broglie by the time I'm finished with this lecture. So, we're now worried about atoms. And we already know that atoms have a diameter, roughly. We know that from the density, the typical size we like to have-- quantities that describe macroscopic objects. Which are not like 10 to the minus 20, but like 1 to 100. And so, the angstrom unit, which is 10 to the minus 10 meters, is a very useful thing for talking about sizes of atoms and molecules. So if we have an atom of a size about one angstrom, we can use this equation to say, well, what would it take for an electron to fit inside an atom? So we specify this-- we know this, we know that-- and that determines what the momentum would have to be. And these are all simple calculations. And since I don't like doing calculations on the board, and I don't really need to do this now-- you need to be able to do them, quickly, if I ask you on the exam. But basically, what we end up finding out is that the velocity of the electron would have to be 7.25 times 10 to the 6 meters per second. Which is OK, it's pretty fast. It's a few percent of the speed of light. But that would correspond to a kinetic energy, which is 2.4 times 10 to the minus 7 joules. Remember, I don't like these kinds of units. But it also corresponds to-- doing a unit conversion-- 149 electron volts. Electron volts are a good unit for energy for atoms because the ionization energy-- the energy it takes to pull an electron off of an atom-- is always somewhere between five and 15 electron volts. So you always want to calibrate yourself, your insight, in terms of numbers which are in the small scale, rather than having to remember the exponent. All right, 149 electron volts. Should that bother you? Well, it can't bother you yet because you don't know about what the ionization energy is. But I just told you. It's between five and 15. This is a factor of 10-- too big. So this is going to be a problem. How is it possible for things so small to have an electron fitting in that small size without it just leaving because it's just way too high energy? And so that leads us to ask questions about, well, what is the internal structure of an atom? How can an atom somehow accommodate this electron which needs to somehow fit inside? So that was the basis for the Geiger-Marsden experiment. Now, the Geiger-Marsden experiment looks very similar to the previous experiment. And here we have-- whoops-- alpha particles. Alpha particles are helium-2 plus ions. And they're produced by radioactive decay. And they have a tremendous amount of energy. More energy than was possible in the days these experiments we're doing to create for a particle. In fact, one of the earliest experiments, or apparatuses, capable of producing very high energy electrons was built by Robert Van de Graaff, here at MIT. And this was in the form of cylindrical towers, right near the parking garage. And it was there for the first 10 years I was at MIT. I'm very old, but that's still fairly recent. But anyway, Van de Graaff could make high energy particles and it was really neat. And what he could do was dwarfed by what can be done in electron accelerators, now. But in the days when the Geiger-Marsden experiment was done, which was 1911, there was no way of making and controlling the energy of an electron-- or of any particle. And here we have some particles which are produced by radioactive decay, which have tremendous energy. So they're heavy and they have high energy. And so that means the wavelength is very small. We want to use these helium ions. Yes? AUDIENCE: Could you not also control the energy of each particle as they're passing through, right? ROBERT FIELD: Yes. But you would need a very high voltage. And though that is something we could imagine doing now, but in 1911, the ability to do that sort of thing with control was not there. There needed to be advances in vacuum technology. There needed to be advances in power supplies. I mean, we're talking about very high voltages. And you wouldn't want to do that in your laboratory, even now with the capability. I remember when I was a graduate student, we had these things called Spellman power supplies. And they could produce 40 kilovolts. They were really scary. But that's nothing compared to what you need. OK, so, we want bullets. We want to have these helium particles interacting with a thin metal foil. We have a whole array of atoms here. And we have a little hole here. And what's going to happen is this radiation is going to hit these atoms. And there's going to be backscatter and forward scattering. And the crucial experiment was to measure the ratio of backscattering to forward scattering. Now, if we have a target that looks sort of like a smear, then the forward and backward scattering would be similar. If we have a target that looked like a bunch of tiny points, there would very rarely be backscattering. All of the scattering would be forward because most of the particles don't hit anything. And so what was found, and what was the surprise, is that there was very little backscattering. And that implied that the ratio of the size of the target to the size of the particle was enormous. The particles that scattered-- the alpha particles-- were tiny. They had a size-- something like 10 to the minus 4 times the typical dimension of an atom. So this is jellium. And this is a perfectly reasonable approach. That the positive and negative charges that make up an atom are distributed uniformly. This was the surprise. How do we explain, now, if atoms that are scattering the alpha particles are really small, even compared to the one angstrom? Well, how do they stick together? Why is matter not compressable? So what is going on here? Now, I'm not exactly sure of the genealogy here, but Geiger and Marsden were workers, or students, in the Rutherford lab. And the old man wanted to save face or to say, oh, here's an experiment. We learned something from this. You know, this is what we do-- this is my job. But anyway, Rutherford said, well, maybe it's like this. We have a nucleus and we have the electrons. So we have a nucleus where all the positive charge of the atom, and most of the mass, resides. And we have electrons in circular orbits. So maybe these circular orbits explain why you can't compress atoms to something that would be commensurate with the size of the nuclei. That the electrons cause a repulsion and the structure is stable. So this is a pretty reasonable hypothesis until one analyzes it. So what we have is a positive charge here, negative charge here. And so, there is Coulomb attraction and there's centrifugal force, or centripetal acceleration. And we have to have these two things match. So the inward force is minus the charge on the electron squared over 4 by epsilon 0 and 1 over r squared. And the centrifugal, it's-- OK, this is-- you know all this. You know to do this. Have known it since high school, probably. And so, you can combine all these things, say the inward and outward-- the inward force is exactly canceling the outward force. And you can solve for the velocity. And the velocity is q over the electron squared over 4 by epsilon 0 mass of the electron and the radius of the orbit, square root. This is a trivial derivation. I won't insult you by attempting to do it and try to increase your understanding of the equation, because you already understand it. So this is the requirement for the radius of the circular orbit. And it has the mass-- I mean, this is the requirement for the velocity. And this is the radius, here. So we know all that. There is nothing about quantization, yet. We know that for any radius, the electron will have a certain velocity. And we can choose whatever radius, we want whatever velocity we want. And that corresponds to whatever energy we get. What we're interested in is the frequency of the orbit. And so that will be 1 over the time it takes for the electron to go around. And 1 over the time it takes for electron to go around is 2 pi r, the circumference, divided by the velocity. So we have the velocity is equal to-- I mean, the frequency is equal to 1 over 2 pi r. And we can write an equation for this. And that's in the notes. In fact, there was a typo in the notes. Which has been corrected and read. But I don't need to tell you what it is. We can calculate this frequency. The reason we calculate the frequency is because we know if we have electrons moving back and forth at some frequency, they're going to radiate electromagnetic radiation at that frequency. Well, where did that energy come from? It came from the motion of the electron. So it has to give up kinetic energy. Now, the energy is the kinetic energy plus the potential energy. So if it gives up energy, some of these two things has to decrease. And what happens is this decreases faster that this increases. And what ends up happening is that the electron has a death spiral. What happens will be that the electron will go in a spiral, going faster and faster as it goes to smaller and smaller radius, and annihilate itself. So this is garbage. This can't be true. It violates laws that everybody knows are right. So one needs to find a way to live with this. Now, one really doesn't need to find a way to live with it until you realize what happens. Because this picture, subject to a couple of hypotheses, predicts an infinite number of 10-digit numbers. It's not an accident. Maybe one prediction would be fine. But all of the lines in the spectra of hydrogen atom, helium ion, lithium doubly-charged ion, all of those are predicted with no adjustable parameters to measurement accuracy. Now, at the time this was being done, the measurement accuracy might have been only a part in 1,000, or maybe a part in a million. But a part in 10 to the 10th, beyond that we're starting to get into fundamental physics. But this is an astonishing thing. And so I have to explain what the additional assumptions were because we've got something that predicts things we have no business knowing. And there was no explanation for the spectrum before these experiments, or this picture, was developed. So we have to first find a way, whether it's believable or not, of getting rid of the radiative collapse. So Bohr proposed that angular momentum, l vector r cross p is conserved. Well, we know that angular momentum is conserved. But for a microscopic system, what it means to be conserved may be a little bit more subtle. He proposed that angular momentum is conserved and that the angular momentum, the magnitude of the angular momentum, had to have a particular value. And that value was integer times h bar. h bar is the Planck's constant divided by 2 pi. Now, this is complete nonsense. Why should it be conserved? Why should it be restricted to this set of values? Well, the reason we accept that it's restricted is because it gives the energy levels that are observed. Now, before I get to energy level-- well, I do. OK, we have energy levels. We find that the energy is equal to minus some constant over n squared. Same n as in here. This is the Rydberg constant. And it's something that you can measure. It's basically a whole bunch of fundamental constants combined. And so, it has a value. This is one of the numbers in my permanent memory. And that's the value of the Rydberg constant in reciprocal centimeter units. And to get it into energy units you multiply by h times c, or to get it into frequency, you just multiply it by the speed of light. So, anyway, this is a number that is known to many decimal places. And it is generated by this idea that the angular momentum has to be certain values. Conservation is good, that's easy. This is weird. And it's also wrong because we find out later that the possible values of n include 0. Which would completely mess up the Bohr model. But, anyway, this, then, in combination with this amazing statement that-- OK, we have the nth energy and the n prime-th energy. And the spectrum corresponds to the frequency, corresponds to e n minus e n prime, over h. So everything we see in the emission spectrum of the hydrogen atom, or in a gas-- which is mostly H2-- there are transitions associated with the free atoms. And they're always around this simple equation, based on the Rydberg equation. This says the spectra are telling us about the energy level differences. And it's a simple equation. And it's true. It's true at incredibly high accuracy. And it tells you nothing except the mass of the particle. Because the mass of the particle-- for this Rydberg equation-- we have it expressed in terms of the mass of the electron. But it really should be the reduced mass of the mass of the nucleus times the mass of the electron, or the mass of the nucleus plus the mass of the electron. And this is something we know for all two-body interactions. That was known well before the time of these experiments. If you have two things interacting, the reduced mass is what you want rather than the individual masses. And so, the only information in these one-electron spectra is the mass of the nucleus. And there's not much difference in this reduced mass effect. But it's enough to say this is a spectrum of hydrogen, as opposed to lithium 2 plus. OK, but we still have a problem-- a very serious problem. Why is the angular momentum conserved? Why is the angular momentum forced to have a certain value? Well, I've really finished this lecture pretty fast. Let me just get to the end and I will go back. All right, so, the problem is this electron is assumed to be a particle and it's assumed to be moving. So the equation-- Maxwell's equations-- all of the equations about motion of charged particles say if it's moving it's going to radiate, whereas if it's oscillating, it's going to radiate. So maybe the problem is that it's not moving. Remember, the particles are both particles and waves. So we could imagine that, around this circular orbit, we have standing waves-- no motion. And this led to this Schrodinger equation, which talks about the states of the electrons that are allowed. It's basically the classical wave equation, with a couple little twists. And the thing about waves, you remember, we can have constructive and destructive interference. We can have standing waves. So there doesn't need to be a motion of our particle. There could be some static description of the probability of finding the electron everywhere around this orbit. And that's the Schrodinger equation. In the next lecture, I'm going to talk about the classical wave equation, which will be the warm-up for the Schrodinger equation. And the Schrodinger equation explains everything. People have made some really fantastic philosophical statements about the Schrodinger. It contains everything that we need to know. The problem is we can't solve the equation exactly. But it's true. And it contains everything that we'd ever want to know about the microscopic structure of atoms and molecules. So we've been led by these very simple experiments, which, now, you could do-- really trivially. You wouldn't have to be a smart student of a smart advisor, or stupid advisor. You would be able to do these experiments. And you could say, yeah, this is all very weird, but know now we know that spectra are everything. And I'm a spectroscopist. I'm very proud of this because the idea that you can make a few measurements and say something about the internal structure of an atom or molecule-- that's a fantastic thing. And what we've seen-- the spectrum of one-electron atoms-- is something which is really simple. It's the template for understanding all complexity. Because everything is different from hydrogen. Hydrogen has a point charge at the center. It's not quite a point charge, and that's actually a subject of even modern physics. And other atoms are not a point charge because they have electrons. And so there is a concentration of charge-- the thing to which the electron is attached is space filling. So that results in a shift at the energy levels. And the shift at the energy levels, and how that shift depends on the orbital angular momentum, tells you something about the shape of this charge distribution-- the radial shape. So everything we do in spectroscopy is somehow referenced to something we understand perfectly, but which is not of much interest. But it's a template for building up our understanding of everything. And this is a kind of a radical statement. And I get to say this because I'm up here and I do this for a living. I mean-- not teaching, but research. And I really believe that the things that we are enabled to observe about the microscopic structure of things are encoded in something completely unlike looking at it. And our job is to figure out how to break that code. And that's what I've done for the last 50 years and it's fun. OK, so, what more could I say to amuse you for five-- six minutes? Not much, because I've skipped a lot of really great stuff. But go back to de Broglie. De Broglie had the hypothesis that there is an integer number of wavelengths around the circular orbit. And that was telling you that it's stable because if it weren't an integer number of wavelengths, the electron would self annihilate. But that takes a valid point and moves it into something which is a little bit wrong. Because we're not trying to have a particle moving, we're having a distribution of probabilities. And there are still wavelengths and nodal structures. And the stable solutions do involve the particle not self annihilating. And so, de Broglie scores another triple. I mean, he didn't come up with the Schrodinger equation. So that's the home run that says, OK, we have the material, now to explain everything. The next lecture will be just introducing the mathematics of the wave equation and the crucial ideas. And that will lead into the following lecture, where we all talk about the Schrodinger equation. So I can stop now. Thanks. |
MIT_561_Physical_Chemistry_Fall_2017 | 14_From_Hij_Integrals_to_H_Matrices_II.txt | The following content is provided under a Creative Commons license. Your support will help MIT Open Courseware continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: So as I said at the beginning of the course, this is quantum mechanics for use, not admiration, and not historical. You're going to leave this course knowing how to solve a very large number of quantum mechanical problems. Or if not to solve it, to get insight into how you would solve it. And so I presented a couple of exactly solved problems, and that's not because they're historically important. It's because you're going to use them, and you're going to embed the results of those exactly solved problems in an approximate approach to almost any problem. And the vast majority of problems that you would face use the harmonic oscillator as the core of your understanding because almost all potentials have a minimum, and so that means the first derivative is zero at the minimum. So you don't care about the first derivative. You care about where the minimum is. And the second derivative is the dominant thing, and that's the harmonic aspect. And so using a harmonic oscillator basis set, you're going to be able to attack every problem. And another thing about the harmonic oscillator is that it uses these As and A daggers, which are magic. And what happens is you forget, totally, about the wave functions. The wave functions are there if you want to calculate some kind of probability amplitude distribution, but you never look at them when you're solving the problem, and that's a fantastic thing. Now, there is one more exactly solved problem that I want to talk about today, which is not usually included in the list of exactly solved problems because it's kind of special. The two-level problem. The two-level problem is exactly solved because there are only two levels, and the solution of that problem involves the quadratic equation. So there is an exact solution. And this is used a lot in introducing new techniques in quantum mechanics, and so you're going to see the two-level problem again and again as opening the door to being able to deal with much more complicated problems, and I'm going to try to refer to that a lot. OK, so we're about to go from the Schrodinger picture to matrix mechanics, and I'd like to have some comments about what are the elements of the Schrodinger picture. And there's no wrong answer here, but I'm looking for certain things. Anybody want to tell me? Yes? AUDIENCE: Is it based off wave equation or the Schrodinger equation, which is very similar to the wave equation? ROBERT FIELD: Yeah. So we have a differential equation, and the solutions are wave functions. More? AUDIENCE: [INAUDIBLE] ROBERT FIELD: Well, mathematics is challenging to some people. Some people really love it. But yes, it's much more mathematical because you're faced with solving differential equations, coupled differential equations, and you're challenged with calculating a lot of intervals, and sometimes the intervals are not over simple functions. They're complicated. But the main thing is you have this thing which you could never observe, but is somehow the core of everything that you can know, and that really-- as I get older, it bothers me more and more. It's not that I'm going to invent some way to do quantum mechanics totally without a wave function. The most important thing, I think, is that when you work in the Schrodinger picture, you get one wave function and one eigenvalue at a time. And so yes, you can solve these problems, but no, you don't get insight. You don't see the overall structure. You just see well, if you did this experiment, this is what you would observe. That's perfectly wonderful. And now, some anticipation. What's so special about the Heisenberg picture? Or what is the Heisenberg picture? Now, I haven't lectured on it so if you read my lecture notes, you know the answers to all of this, but can we speculate about what goes here? Yes? AUDIENCE: [INAUDIBLE] ROBERT FIELD: OK, so every operator can be represented by a matrix, and the elements of that matrix are calculated usually automatically. Often, they're an infinite number of basis states, so you sort of refer to the Schrodinger picture, but anyway, these matrices are arrays of numbers, and they can be infinite arrays of numbers, but you don't write an infinite array of numbers down. You write a few numbers and you recognize what the pattern is, and usually, you have some function of the quantum numbers, and that generates all the elements in this matrix. And the solutions are going to be certain eigenvectors, and instead of differential equations, it's linear algebra. Now, for differential equations, you'll learn a lot of tricks. For integrals, you learn a lot of tricks. For linear algebra, there aren't any tricks. It's all right there. And so it's a much more transparent-- now, it might be numerically demanding because you're dealing with very large arrays of numbers, and you're trying to get something from it, but the linear algebra is simple. OK, and we have these things called matrix elements. I can use this word now because the numbers in a matrix are called matrix elements. They're integrals, but they're integrals that are, more or less, given to you. And then we have infinite matrices. And in the Schrodinger picture, we might have infinite basis sets, but we don't really think too much about the problem of infinities because we're looking at things one at a time. Here, the operators are implicitly infinite, and we have to do something about that because no computer can find the eigenvalues and eigenvectors of an infinite matrix. So somehow, you have to truncate it, and you have to use some approximations, and the framework for that is perturbation theory. Now, I love perturbation theory. I've made my career out of perturbation theory and you're going to see a lot of it, but not today. The next exam is going to have a lot of perturbation theory. OK, so let's review where we've been. With a two-level problem, we have two energy levels, and there is some interaction between them, and we get E plus, psi plus, and E minus, psi minus. So we have these integrals, H2,2, H1,2, and because it's a Hermitian matrix, it's equal to H2,1 star. That's the definition of a Hermitian matrix, and that means if the H1,2 element is imaginary, the H2,1 matrix element is imaginary, but with the opposite sign. Now, almost all of the problems we're going to face are expressed in terms of real matrix elements, and so all you're really doing to go from the 1,2 to the 2,1 is just reversing the order of the index. Nothing else is happening, and that has also some great simplicity in the solutions we use to solve the two-level problems. OK, the solutions to the two-level problem are based on some simplifications. So what we do is we make the two by two matrix symmetric by subtracting out the average energy. We define then, two numbers, which are H1,1 minus H2,2 over two and this. So we have E bar and delta, and we have the interaction, which is H1,2, and we call it v. And we have to be a little careful because v could be complex or imaginary, but for now, we're just going to treat it always as real. OK, so for this two-level problem, the energy levels are given by E bar plus or minus delta squared plus v-squared square root. Or E bar plus or minus the square root of x, where that's x. And the eigenfunctions are expressed in terms of these coefficients-- one plus or minus is equal to 1/2 one plus or minus delta over x to the 1/2 square root. Square root is outside the bracket. And c2 plus or minus looks almost the same, except we have 1 minus or plus delta over square root of x. So in this reduced picture, there's not too much to remember. The difference between the eigenfunctions for the higher energy and the lower energy eigenstates differ only by these signs. OK, now, this is a lot. I derived it in lecture last time, and the algebra is horrible, and I'm not good at presenting algebra, but it's all in the notes. But if you take these formulas and you try them on for size, you will be able to verify that these give normalized and orthogonal eigenfunctions, which I recommend, mostly. OK, we take this expression for the eigenvalues, and let's just take one of the eigenfunctions and find out whether it belongs to the correct eigenvalue. If it doesn't, you've made an algebraic mistake. It's a very useful thing. OK, so now we're going to take the results for this two-level problem solved algebraically, and at the core of that was the quadratic energy formula. The quadratic formula for the equation that determines the energy. The quadratic formula is always applicable. It's exact. So what we end up getting is analytical expressions. It doesn't matter what the value of the two critical quantities, delta and x, are. You have solutions. So now we're going to take all this stuff and we're going to start over. We're going to rearrange it so that we have a different way of approaching the problem. So we're going to start talking about matrices rather than operators. And so I represent a matrix by a boldface-- this symbol is how you ask a computer to make it-- yes, it's how I represent a matrix, instead of a hat, which is the way you represent an operator. But now we're going to say every operator is represented by a matrix rather than a differential operator. And so this Hamiltonian is E bar, zero, E bar plus delta, v, v star, minus delta. And v is usually real. And another way we can say this is it's E bar times this fancy 1 with an under bar plus H bar. So this is a symmetric Hamiltonian matrix. This is the unit matrix. And now, since we're going to be doing matrix multiplications, let me just give you some mnemonics. So if we have a square matrix multiplying a square matrix, what we do is we multiply this row by that column, and we get one number, and you fill out the square matrix. And with a little practice, this will be permanently ingrained in your head. We also can have a matrix multiplying a vector. And so a matrix multiplying a vector gives a vector, and this product gives a number here. And you've probably all seen these sorts of things or could grasp them very quickly, but it's useful just to not get confused. We can also do something like this, and again, we use the usual vector and we get a vector. I'm sorry, we get a column. And this is a difficult symbol to make on a computer, but you get this first element here like that. And of course, you can do this times that, and you get a number. And you can also do it the other way around. You can do this times that-- I'm sorry, don't do that. And you get a square matrix. So those are the things that you have to practice, and it becomes second nature very quickly, and it's a lot easier than doing differential equations, or matrices, or integrals. OK, now, we use this superscript. This means transpose. This means complex conjugate and transpose. The theory deals with this, but when the Hamiltonian or the matrix you're interested in is real, the transformation that diagonalizes it is always just given. You need this to transpose. So these two symbols look similar and you won't have any trouble with that. And now, we have this kind of symbol. So this is a normalization. That should be equal to one, and it is because one times one is one plus zero times zero. And we can also look at this, and that's zero because zero times one is zero and one times zero is zero. So this is normalization, this is orthogonality. OK, we're playing with numbers, and we don't really look at the size, even though the numbers all are obtained by doing an integral over the wave functions and the operator, but that's something that you sort of do in first grade of quantum mechanics, and you forget how you did it, and you just know that they're there to be played with. OK, now, a unitary operator is one where the conjugate transpose-- the unitary matrix-- is equal to the inverse, which means t minus one times t equals one. So it's nice to be able to derive the inverse of a matrix. And for certain kinds of matrices, this is really easy because all you do is tip along the main diagonal. There are other matrices where you have to do a lot of work, but whenever you're dealing with matrices, you're not doing the work. The computer is doing the work. You teach the computer how to do matrix operations, and even if it's a hard one, the computer says OK, here it is. OK, so if you have a real symmetric matrix, then you can say OK, T transpose matrix T gives a1, a n, zero, zero. So you can diagonalize a real symmetric matrix by this kind of a transformation. That's called an orthogonal transformation. And if it's not real, then you use the conjugate transpose, use the dagger, and you still get the diagonalization. Now, in most of the books that you'll ever look at about unitary transformations, they actually are giving you what's called the orthogonal transformation, and it's what works for a real matrix, and I'm going to do that too. So when we have something like this, we say that this transformation diagonalizes A, or H, or whatever. And the word "diagonalizes" is really important because that's what you want because it presents all the eigenvalues. Remember, one of the things about quantum mechanics is you have an operator. You're going to observe something connected with an operator. The only things you can get are the eigenvalues. So here they are, all of them. That's kind of useful. OK, so in the Heisenberg picture, the key equation is the Hamiltonian as a matrix, some vector-- OK, this is the analog of the Schrodinger equation, but it's the Heisenberg equation. And mostly, it's just notation, and you have to get used to that. We want to find the vectors that are eigenvectors of this Hamiltonian with eigenvalue E. It's just like this the Schrodinger equation, except it's now looking for eigenvectors rather than eigenfunctions and eigenenergies. Now, in order to solve this problem, we exploit this kind of transformation, and we insert T T dagger between H and c, and that's just one. So we don't even have to worry about the other side. We're just playing with matrices, but they look like functions or variables, and everything is-- it's really neat how beautiful linear algebra is because you are now dealing with an infinite number of equations at once. You're dealing with these objects and you're using your insight from algebra as much as possible in order to figure out what's going on. Really beautiful. OK, and so now we must multiply this equation on the left by T dagger. OK, I'm dropping the under-bars now. OK, so now we say OK, here we have H twiddle, and here we have c twiddle, and here we have c twiddle. So this is now a different eigenvector equation, but we insist that this guy, H twiddle, looks like E1, E2, En, 0,0. A diagonal matrix, where all the eigenvalues are along the diagonal. And so this is what we want, and lo and behold, this is what we need in order to say, well, yeah, this thing has to be the eigenvector of this for one of the eigenvalues because this is an eigenvalue equation or an eigenvector equation. So if we can diagonalize the Hamiltonian, the transformation that diagonalizes the Hamiltonian gives you the linear combination of basis vectors that is the eigenvector, and we'll talk about this some more. So for the two-level problem, we want to find E plus, 0, 0, E minus. And usually, E plus is the higher energy eigenvalue than E minus. Always, when you do this stuff, you get eigenvalues and you get eigenvectors. And frequently, when you do the algebra as opposed to the computer during the algebra, you don't know that a particular eigenvector belongs to which eigenvalues. So it's useful to have a couple of things that you normally insist on. And so I like to label these things by plus and minus, corresponding to which is higher energy and which is lower. You could also say, well, the plus is going to correspond to a plus linear combination somewhere, but that's really dangerous. So now, let's just play a little bit. So we have simplified H magically so far to diagonal form. So H C twiddle-- I'm sorry, yes H C twiddle is going to be E c twiddle. So H c plus is going to be E plus 0, 0, E minus, one, zero because that gives us E plus times one and zero times one. So multiply these two things. We have a column vector, and that's the same thing as E plus times one, zero. And we do the same sort of thing to-- instead of using c plus, we use c minus. That's a zero, one, and that will give us E minus times zero, one. This is all just playing with notation and we're about to start doing some work. OK, so T dagger times c is supposedly equal to c plus. OK, and so well, we can write this formula in a schematic way, and so we have T dagger. Now, I always remember this because there used to be something analogous to Coke and Pepsi called Royal Crown Cola, and for Royal Crown, that just reminds me that row first, column second. I don't believe that any of you have ever heard of Royal Crown, but you could think of some other mnemonic. Now, it's really important to keep the rows and the columns straight. So we have 1, 1, and T dagger. Now, what goes here? This is in the first row, second column. So what do I put here? Yeah. You could even say it. OK, and here we have T dagger 2,1, and here we have T dagger 2,2. Now, if we multiply one, zero, because that's what we're supposed to do here, we'll get T 1,1 plus T 1,2 times zero, and then T 2,1 plus T 2,2 times zero. OK, and that's simply T 1,1 dagger times one, zero plus T 2,1 dagger times zero, one. So this thing gives the linear combination of the basis vectors that is equal to a particular eigenvector. So that means if we can find T, we can get T dagger, and we can get E plus and E minus, and c plus and c minus. So we have completely solved the problem if we know what T is. Well, with a two-level problem, we know algebraically that there is such a T, and that it's analytically determined. There is another way of approaching this, and that is to say the general orthogonal transformation, which we will call a unitary transformation, but it's missing a little bit of stuff if it really wants to be unitary. I'm going to call it-- so T dagger is cos theta, sin theta, minus sin theta, cos theta. So this is a matrix which is determined by one thing, theta. We want to find what theta needs to be in order to diagonalize the matrix. Now, since we know we're talking about sines and cosines, and that there is one theta, we abbreviate this to c, s, minus s, c because the algebra is heinous. Not as bad as in the Schrodinger picture, but it's terrible, and so you want to compress the symbols as much as possible. OK, so we want T dagger HT because that's H twiddle. That's the thing we want, and we want T dagger HT. And now since we've expressed the T in this form, we can multiply this out. And so we have c, c, minus s, c, delta, v, v, delta. Delta, v, v, minus delta, c, minus s, s, c. So we have three two by two matrices to multiply. Now, that's not something you do in your head. You could do two. So you multiply these two, and then you multiply by that, and the result-- I would be here for hours doing this, and you wouldn't learn anything except that I'm a real klutz. I should write this on its own board because it's really important. So that matrix becomes a big matrix, c squared minus s squared times delta plus 2sc times V, and c squared minus s squared times V minus 2cs delta. And we have the same thing down here, c squared minus s squared times V minus 2cs times delta. And the last one is minus c squared minus s squared times delta minus 2cs times V. So this is what we get when we take the general form for the unitary transformation, and transform the Hamiltonian with it. And the first thing we do is we say, well, we want this to be zero. If this is zero, then this is zero, right? So this turns out to be an equation that tells us what theta has to be. OK, and we also know from trigonometry, c squared minus s squared is what? I'm actually jumping ahead, but it's just cosine two theta, and 2sc is sine two theta. So we're going to get a simplification based on this, but now let's just say we want this to be zero. So that means that c squared minus s squared times V has to be equal to 2cs times delta. Which way am I going? 2cs over c squared minus s squared is V over delta. Well, that looks pretty good, especially because this is cosine-- this is sine two theta, and this is cosine two theta, which is tan theta is equal to this. So now we have a simple equation. We have the theta, and we have the V and a D-- a V over delta. I shouldn't do that. So now I can cover this again. So we can take this equation and solve it, and we have that theta is equal to 1/2 inverse tangent of V over delta. There it is. That's an analytic result. So for any value of V over delta, we know what theta is. That's not an iterative solution. That's complete analytical result, and that's fantastic, and it says, just like with the quadratic formula, which we used to look at the original Hamiltonian at the eigenvalues of the original-- well, yeah, it says no matter what, there is a solution, and you can express this solution as some combination of V and delta. OK, and so when you do that, you get that the energy levels are E bar plus or minus delta times cos two theta plus V times sin two theta. And there's no square root here. Why do you know that? Well, this is dimensionless. This is the units of energy, and so square roots keep coming popping up, but you don't want to put one here because that would be wrong. Even if you didn't do the derivation, if you saw a square root here, you'd know somebody is just writing down things from memory or copying badly, and making corrections. And that leads to E plus minus is equal to E bar plus or minus delta squared plus V squared, and there is a square root there. This is what we derived via the quadratic formula. We knew this, and it came out to be the same. Well, it better have. And we can also determine what T is. And I'm not going to write it, it's in the notes. It's a lot of symbols, but it's something-- it's so compressed that you can guess the form, and so you should look at that. We derived the eigenfunctions of the Hamiltonian, and they are exactly the same as what we get here. And remember that the column of T dagger or T transpose is eigenvector, and sometimes we want to know those eigenvectors. We're doing semi-OK. What happens if we go beyond the two-level problem? Well, you know from algebra that there is no general solution to a cubic equation. There are some limited range over which there is an analytic solution, but mostly, you don't use a cubic formula to solve the cubic equation. You do some kind of iteration. So for the number of levels greater than two, we know we're going to have a problem because just approaching it by transformation theory or linear algebra as opposed to the Schrodinger picture-- if you can't get a solution in one picture which requires solving an algebraic equation, you're not going to get it by playing with these unitary matrices. So we're going to be approaching a problem where we have to find the eigenvalues and the elements of the transformation matrix in some sort of computer-based way. We're not going to do it. It would be nuts, even for a three by three. Although, I will give you an exam problem which will be a three by three, and you're going to use perturbation theory to solve it. I haven't told you about perturbation theory. That's going to be next week, but we're leading up to it. OK, but now the point is we have the machinery in place. We have exactly solved problems, and we have the key parameters for exactly solved problems. So the structural-- So for the harmonic oscillator, we have the force constant and the reduced mass. For the particle in the box, we have the bottom of the box, v0, and we have the width of the box. For the rigid rotor, we're going to have the reduced mass and we're going to have the internuclear distance. There's things that determine all of the energy levels for exact solved problems, and they are basically the things that appear in the Hamiltonian, and we call them structural parameters. And we have energy levels. And often, these are some function of a quantum number. This is what we can observe. We observe the energy levels, and we represent them by some formula, and the coefficients of the quantum numbers relate to these things that we really what. So when you're not dealing with an exactly solved problem-- like instead of having a harmonic oscillator, you have a harmonic oscillator with something at the bottom, or you have a particle in a box with a slant bottom, or you have a rigid rotor where it's not rigid, you have additional terms in the Hamiltonian, and they are going to-- we are going to use perturbation theory to relate the numerical values of the things we want to know to the things we can observe, and perturbation theory gives us the formulas of the quantum numbers, and tells us the explicit relationships of the coefficients of each term in the quantum number expression to the things we want. This is how we learn everything. When we do spectroscopy, we measure these energy levels, and these energy levels encode all of the structure and all of the dynamics. It's really neat. OK, now, the last thing I want to-- do I have time? Yeah, maybe. Remember when we do time dependent quantum mechanics with a time independent Hamiltonian. We want to have psi of x and t. And usually, we're given psi of x, t equals zero. We're given the initial state, and that initial state, if this is an interesting problem, is not an eigenstate of the Hamiltonian. It's a linear combination of eigenstates of that Hamiltonian, and the kinds of flux we almost always use to test our insight, or actually, because they're feasible experimentally, is the initial state is one of the eigenstates of an exactly solved problem. It's some special combination of easy stuff. And we need to know how the coefficient-- this thing-- how that is expressed as j equals one to n of c j psi j. Because if we can express the t equals zero pluck as a linear combination of the eigenstates, then it's just a matter of mindless manipulation because we have c j, e to the minus i e j t over h bar times j. Bang, it's done. So what we want to know is how to go from a not-eigenstate to an eigenstate. And lo and behold, that's just the inverse of the transformation. So what we want to know is OK, since I don't have time to spell it out for you exactly, we have t dagger, which relates the zero order states to the eigenstates. We want to go in the opposite direction. We want the inverse transformation. So we want t, or we want to take instead of the columns of t dagger, we take the rows. And so if you have a machine or a brain-- and I'm not doubting this!-- that enables you to write down all of the elements in the t dagger matrix, you are armed to go both from zero order states to eigenstates, and from plucks to time evolving wave packets. It's really beautiful and simple. And normally, when it's presented, these are presented as separate project problems, and the whole point is you've got a unified picture that enables you to go get whatever you need in a simple way, as long as a computer is able to diagonalize your critical matrices. Well, I don't have time to talk about this in any detail, but if we look at the eigenfunctions or eigenvectors for the two-level problem, and we do power series expansions in theta, where theta is V over d-- theta is also called the mixing angle-- we discover that we have some formulas which says that the energy levels-- let's say the j-th energy level is equal to E bar plus-- now I'm doing this and I have to somehow grab something from in here. We have a sum of k not equal to j of V-- of H. This is the formula for the correction of the energy by a second order perturbation theory, and we can also write the formula for the corrected wave function. By doing a power series expansion in terms of powers of V/d or theta, we find what the structure has to be for the solutions to the general problem when n is not two. I'll develop the formal theory for non-degenerate perturbation theory in Monday's lecture, and that's really empowering. It's really ugly, but it gives you the answers to essentially every problem you will ever face in quantum mechanics. OK, have a nice weekend. |
MIT_561_Physical_Chemistry_Fall_2017 | 35_DeltaFunctions_EigenFunctions_of_X_Discrete_Variable_Representation.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: Last time, we talked about photochemistry. And the crucial thing in photochemistry is that the density of vibrational states increases extremely rapidly-- extremely, extremely-- and especially for larger molecules. And this enables you to understand intermolecular vibrational redistribution, intersystem crossing, internal conversion. The matrix elements are different in each of these cases. But what happens when the vibrational density of states gets large is that you have a special state, which is a localized state or something that you care about that you think you understand-- a bright state-- which is different from the masses. And because of the density of states, that bright state or that special state, gets diluted into an extremely dense manifold of uninteresting states. And as a result, the system forgets whatever the bright state wanted it to tell you. And so you get faster decay of fluorescence, not because the population goes away, but because the unique thing-- the bright state, the localized state-- the thing that you can prepare by excitation from the ground state, which is always localized, its character is dissipated. And so you get what's called statistical behavior. Now I have a very strong feeling about statistical behavior, and that it's mostly a fraud. But the current limit for statistical behavior is when the density of states is on the order of 100 per wave number. That's the usual threshold for statistical behavior. As we know more, as we develop better experimental techniques, we're going to push that curtain of statistical behavior back. And we'll understand more about the short-time dynamics and whatever we want to do about manipulating systems. So statistical just means we don't know, but it's not what we expected. And it's usually boring. OK. Today I'm going to talk about the discrete variable representation, which is really weird and wonderful thing, which is entirely inappropriate for a course at this level. But I think you'll like it. So in order to talk about the discrete variable representation, I'll introduce you to delta functions. And you sort of know what delta functions are. And then I'm going to say, well for any one-dimensional problem-- and remember, you can have a many-dimensional problem treated as a whole bunch of one-dimensional problems. And so it really is a general problem. But if you have a one-dimensional potential, you can obtain the energy levels and wave functions resulting from that one-dimensional potential. Regardless of how horrible it is, without using perturbation theory, which is should make you feel pretty good. Because perturbation theory is labor intensive before you use the computer. The computer helps you, but this is labor free because the computer does everything. At the beginning of the course, I said, you cannot experimentally measure a wave function. And that's true. but if you can deal with any potential and have a set of experimental observations, energy levels, which you fit to an effective Hamiltonian, you can generate the wave functions associated with the effective Hamiltonian directly from experimental observations. So it's not a direct observation, but you get wave functions. And you can have whatever complexity you want. OK, so let's begin. Delta functions. So this is not the Kronecker delta. This is the delta function that is a useful computational trick. And it's more than that. We write something like this. OK, this is a delta function. And it says that this thing is non-zero when x is equal to xi. And it's really big. And it's 0 everywhere else. And as a result, you can say that we get xi delta x xi. So that's an eigenvalue equation. So we have the operator, x, operating on this function. And this function has the magical properties that it returns an eigenvalue and the function. So this is more than a mathematical trick. It's an entry into a form of quantum mechanics that is truly wonderful. So this is part one. Part two will be DVR-- discrete variable representation. And the issue varies. Suppose we have a matrix representation of an operator. Well, suppose we wanted some function of that operator. How do we generate that? And there are lots of cases where you care about such a thing. For example, you might want to know about something like this-- e to the ih t over h bar. This would tell you something about how a system propagates under the influence of a Hamiltonian, which is not diagonal. This is kind of important. Almost all of NMR is based on this sort of thing. Also-- well, we'll get to it. So extend DVR to include rotation. And then the last part will be-- can be determined for even the most horrible situations, like a potential that does something like this, or something like that. Now we don't want to have a continuum. So I put a wall here. But this is a summarization from an unstable isomer to a stable isomer. This is multiple minima. Anything you want, you would never want to perturbation theory on it. And you can solve these problems automatically and wonderfully. OK, so let's just play with delta functions. And a very good section on delta functions is in Cohen-Tannoudji, and it's pages 1468 to 1472, right at the end of the book. But it's not because it's so hard, it's just because they decided to put it there. And so we have notation-- x xi. It's the same thing as x minus xi. So basically, if you see a variable at a specified value, it's equivalent to this, and this thing is 0 everywhere except when x is equal to xi. So the thing in parentheses is the critical thing. And it has the property that if we do an integral from minus infinity to infinity, some function of x-- dx-- delta x xi dx. We get f of xi. Isn't that wonderful? I mean, it's a very lovely mathematical trick. But it's more than that. It says this thing is big when x is equal to xi. It's 0 everywhere else. And it's normalized to 1 in the sense that, well you get back the function that you started with, but at a particular value. So it's infinite, but it's normalized. It should bother you, but it's fantastic and you can deal with this. It's an eigenvalue equation. I mean, you can say, all right, we have delta x xi xi delta x xi. And we can have delta functions in position, in momentum, in anything you want. And they're useful. So suppose we have some function, psi of x, and we want to find out something about how it's composed. And so we say, well, we have c of xi j delta x minus xi dx. This is the standard method for expanding a function. OK, what we want is these expansion coefficients. And you get them in the standard way. And that is by-- normally, when you have a function, you write it ck j the function-- I'm sorry. So here we have an expansion of members of a complete set of functions. And in order to get the expansion coefficients, you do the standard trick of integrating phi j star, psi j dx. OK, so this is familiar. This is not. But it's the same business. In fact, we've been using a notation incorrectly-- the Dirac notation. We normally think that if we have something like k and psi k of x, they're just different ways of writing the same thing. But the equivalent to the Schrodinger picture is really x psi. This is a vector, not a function. This is a set of vectors. And this is how you relate vectors to functions. We just have suppressed that. Because it's easier to think that the wave function is equivalent to the symbol here. It's not. And if you're going to be doing derivations where you flip back and forth between representations, you better remember this, otherwise it won't make sense. This is familiar stuff, you just didn't realize you understood it. OK, and so now, often, you want to have a mathematical representation of a delta function. And so this thing has to be localized and it has to be normalized. And there are a whole bunch of representations that work really well. One of them is we take the limit, epsilon goes to 0 from the positive side, and 1 over 2 epsilon e to the minus x over epsilon. Well this guy, as epsilon goes to 0, this becomes minus infinity. And it's 0 everywhere except for x is equal to 0. And so if we wanted to put x minus xi, then fine. Then it would be 0 everywhere except when x is equal to xi. So that's one. That's a simple one. Another one is limit as epsilon goes to 0 from the positive side of 1 over pi times epsilon over x squared plus epsilon squared. This also has the properties of being an infinite spike at x equals 0. And it's also Lorentzian. And it has the full width. It'd have maximum-- or a half width it would have-- a full width at half maximum of epsilon. If you make the width as narrow as you want, well that's what this is. OK, there are a whole bunch of representations. I'm going to stop giving them because there's a lot of stuff I want to say, and you can see them in the notes. OK, so what if we have something like xi, that just is the same thing as saying x minus xi is 0. You can produce a delta function localized at any point by using this trick. OK, there are some other tricks that you can do with delta functions, which is a little surprising. And that's really for the future. Delta of minus x is equal to delta of x. It's an even function. Derivative of a delta function is an odd function. Delta-- a constant times x is 1 over the absolute value of that constant, delta x. A little surprising, but if you have x, which-- I better not say that. OK, we have now another fantastic thing-- g of x. So what is the delta function of a function? Well it's a sum over j of the g ex and x of j times delta x minus xj. This is something where you're harvesting the zeros of this function. OK, when this is 0, we get a big thing. So the delta function of our function is a sum over the derivative-- the zeros of g of x. And at times, it's kind of neat. OK, there's also stuff in Cohen-Tannoudji on Fourier transforms of the delta functions. I'm not going to talk about that. But one of the things is, if you have a delta function and x, and you take the Fourier transformer with it, you get a delta function of p, momentum. Kind of useful. It enables you to do a transform between an x representation and a position representation and a momentum representation. It's kind of useful. OK, now we're going to get to the good stuff. So for every system that has a potential, and where the potential has minima, what is the minimum potential? What is the condition for a minimum of the potential? Yes? AUDIENCE: The first derivative has to be 0 with respect to coordinates. ROBERT FIELD: Right. And so, any time you have a minimum, the first term in the potential is the quadratic term. And that's the same thing as a harmonic oscillator. The rest is just excess baggage. I mean, that's what we do for perturbation theory. We say, OK, we're going to represent some arbitrary potential as a harmonic oscillator. And all of the bad stuff, all of the higher powers of the coordinate get treated by perturbation theory. And you know how to do it. And you're not excited about doing it because it's kind of algebraically horrible. And nobody is going to check your algebra. And almost guaranteed, you're going to make a mistake. (Exam 2!) So it would be nice to be able to deal with arbitrary potentials-- potentials that might have multiple minima, or might have all sorts of strange stuff without doing perturbation theory. And that's what DVR does. So for example, we want to know how to derive a matrix representation of a matrix. So often, we have operators like the overlap integral or the Hamiltonian. So we have the operator, the S matrix, or the Hamiltonian matrix. And often, we want to have something that is a matrix representation of a matrix. For example, if we're dealing with a problem in quantum chemistry, where our basis set is not orthonormal, there is a trick using the S matrix to orthonormalize everything. And that's useful because then your secular equation is the standard secular equation, which computers just love. And so you just have to tell the computer to do something special before, and it orthonormalizes stuff. And you can do this in matrix language. And if you're interested in time evolution, you often want to have e to the minus I h t over h bar. And that's horrible. But so we'd like to know how to obtain a matrix representation of a function of a matrix. Well, suppose we have some matrix, and we can transform it to be a1, a n, 0, 0. We diagnose it. And if a is real and symmetric, or Hermitian. We know that the transformation that diagonalized it has the property that t dagger is equal to the inverse. And we also know that if we diagonalize a matrix the eigenvectors that-- they say, if you want the first eigenvector, the thing that belongs to this eigenvalue, we want the first column of t dagger. And if you want to use perturbation theory instead of the computer to calculate t dagger, you can do that, and you have a good approximation. And you can write the vector, the linear combination of basis functions, that corresponds to t dagger as fast as you can write. And the computer can do it faster. So if the matrix is Hermitian, then all of these guys are real. And it's just real and symmetric. Well then these guys are still numbers that you could generate, but they might be imaginary or complex. Suppose we want some function of a matrix. So this is a matrix, this has to be a matrix too. It has to be the same dimension as the original matrix. So you can do this. Well we can call that f twiddle. But we don't want f twiddle. And so if we do this, we have the matrix representation of the function. So this is something that could be proven in linear algebra, but not in 5.61. You can do power series expansions, and you can show term by term that this is true for small matrices, but it's true. So if your computer can diagonalize this, then what happens is that we have this-- we can write that f twiddle is the-- So this is the crucial thing. We've diagonalized this, so we have a bunch of eigenvalues of a. So the f twiddle is just the values of the function at each of the eigenvalues. And we have zeros here. And now we don't like this because this isn't the matrix representation of f. It's f in a different representation. So we have to go back to the original representation. And so that's another transformation. But it uses the same matrix. So if you did the work to diagonalize a, well then you can go back and undiagonalize this f twiddle to make the representation of f. And so the only work involved is asking the computer to find t dagger for the a matrix. And then you get the true matrix representation of this function of a. Is it useful? You bet. Now suppose a is infinite dimension. And we know that this is a very common case. Because, even for the harmonic oscillator, we have an infinite number of basis functions. But what about using the delta function? We also have an infinite number of them. So what do we do? We can't diagonalize an infinite matrix. So what we do is we truncate it. Now the computer is quite happy to deal with matrices of dimension 1,000. Your computer can diagonalize a 1,000 by 1,000 matrix in a few minutes, maybe a few seconds depending on how up to date this thing is. And so what we do is we say, oh, well let's just take a 1,000. So here is an infinite matrix. And here is a 1,000 by 1,000 block. That's a million elements. The computer doesn't care. And we're just going to throw away everything else. We don't care. Now this is an approximation. Now you can truncate in clever ways, or just tell the computer to throw away everything above the 1,000th basis function. That's very convenient. The computer doesn't care. Now there are transformations that say, well you can fold in the effects of the remote basis states, and do an augmented representation. But usually, you just throw everything away. And now you look at this 1,000 by 1,000 matrix. And you get the eigenvalues. And so this might be the matrix of x, or q if we're talking in the usual notation. This is the displacement from equilibrium. Well, it seems a little-- so we would like to find this matrix. Well, we know what that is. Because we know the relationship between x and a plus a dagger are friends. And so we have a matrix, which is zeros along the diagonal, and numbers here and here, and zeros everywhere else. It doesn't take much to program a computer to fill in as many one-off the diagonal, especially because they're square roots of integers. So that's just a few lines of code, and you have the matrix representation of x. Now that is infinite. And you're going to say, well, I don't care. I'm going to just keep the first 1,000. We know that we can always write-- so we have v of x, and this is a matrix now. And it's an infinite matrix. But we say, oh, we just want v of x to the 1000th, and we'll get v of 1,000. We have a 1,000 by 1,000 v matrix. And we know how to do this. We diagonalize that. Then we write at each eigenvalue of x, what v of x is at that eigenvalue. And so now we have a v matrix, which is diagonal, but in the wrong representation for us. And then we transform back to the harmonic oscillator representation. And so everything is fine. We've done this. And we don't know how good it's going to be. But what we do is we do this problem. So we have the Hamiltonian, which is equal to the kinetic energy, I'm going to call it k, plus v. We know how to generate the representation of v in the harmonic oscillator basis by writing v of x, diagonlizing x, and then undiagonalizing-- or then writing v at each of the eigenvalues of x, and then going back to the dramatic oscillator basis. We know k in the harmonic oscillator basis. It's just tri-diagonal, and it has matrix elements delta v equals 0 plus minus two. So now we have a matrix representation of k, which is simple, add a v which is-- computer makes it simple. Add any v you want, there it is. So that's a matrix, the Hamiltonian. You solve the Schrodinger equation by diagonalizing this matrix. And so you have this h matrix, and it's for the 1,000-member x matrix, and you get a bunch of eigenvalues. And so then you do it again. And maybe use a 900 by 900, or maybe you use a 1,100-- you do it again. And then you look at the eigenenergies of the Hamiltonian, e1, say, up to e100. Now if you did a 1,000 by 1,000, you have reasonable expectation that the first 100 eigenvalues will be right. And so you compare the results you get for the 1,000 by 1,000 to the 900 by 900, or 1,100 by 1,100. And you see how accurate your representation is for the first 100. Normally, you don't even care about the first 100. You might care about 10 of them. So the computer is happy to deal with 1,000 by 1,000s. There's no least squares fitting, so you only do it once. And all of a sudden, you've got the eigenvalues, and you've demonstrated how accurate they are. And so depending on what precision you want-- you can trust this up to the 100th, or maybe the 73rd, or whatever, to a part in a million, or whatever. And so you know how it's going to work. And you have a check for convergence. So it doesn't matter. So the only thing that you want to do is you want to choose a basis set where we have-- x is the displacement from equilibrium. OK, so this is the equilibrium value. This is the definition of the displacement. And so you want to be able to choose your basis set, which is centered at the equilibrium value. You could do it somewhere else, it would be stupid. It wouldn't converge so well. And you want to use the harmonic oscillator, k over u. You have a couple of choices before you start telling the computer to go to work. And you tell it, well I think the best basis that will be what works at the equilibrium-- the lowest minimum of the potential, and matches the curvature there. You don't have to do that. But it would be a good idea to ask it to do a problem that's likely to be a good representation. And all this you've done. You get the energy level. So what you end up getting-- So you produced your Hamiltonian. And since it's not an infinite dimension Hamiltonian, we can call it an effective Hamiltonian. It contains everything that is going to generate the eigenvalues. And we get from that a set of energy levels-- e vi-- and a set of functions. So these guys are the true energy levels. And these are the linear combinations of harmonic oscillator functions that correspond to each of them. Who gives that to you? And so then you say, well, I want to represent the Hamiltonian by a traditional thing. Like, I want to say that we have omega v plus 1/2 minus omega x equals 1/2 squared, et cetera. Now we do a least squares fit of molecular constants to the energy levels of the Hamiltonian. And there's lots of other things we could do, but-- so we say, well in the spectrum, we would observe these things, but we're representing them as a power series in v plus 1/2. Or maybe in-- where we would have not just the energy-- the vibrational quantum number-- but the rotational constant. We could say the potential is v of 0 plus b x j j plus 1. Well, that means we could extend this to allow the molecule to rotate. And we just need to evaluate the rotational constant as a function of x, and just add that to what we have here. Another thing, you have t dagger and t for our problem. And you have, say, the 1,000 by 1,000, and maybe the 900 by 900 representations. You keep them. Because any problem you would have, you could use these for. So you still have to do a diagonalization of the Hamiltonian, but the other stuff you don't have to do anymore. Now maybe it's too bothersome to store a 1,000 by 1,000 t dagger matrix. It's a million elements. Maybe you don't, and you can just calculate it again. It takes 20 minutes or maybe less. And so this is a pretty good. OK I've skipped a lot of stuff in the notes, because I wanted to get to the end. But the end is really just a correction of what I said was impossible at the beginning of the course. We have in the Schrodinger picture H psi is equal to E psi, right? That's the Schrodinger equation. So this wave function is the essential thing in quantum mechanics. And I also told you, you can't observe this. So it's a very strange theory where the central quality in the theory is experimentally inaccessible. But the theory works. The theory gives you everything you need. It enables you to find the eigenfunctions, if you have the exact Hamiltonian. Or it says, well we can take a model problem and we can find the eigenvalues and wave functions for the model problem. We can generate an effective Hamiltonian, which is expressed in terms of molecular constants. We can then determine the potential. And we can also determine psi. All of them. So what I told you was true, but only a little bit of a lie in the sense that you can get as accurate as you want a representation of the wave function, if you want it. And DVR gives it to you without any effort. And so it doesn't matter how terrible the potential is, as long as it is more or less well behaved. I mean, if you had a potential, which-- Even if I had a v potential-- the discontinuity here-- you would still get a reasonable result from DVR. What it doesn't like is something like that, because then you have a continuum over here. And the continuum uses up your basis functions pretty fast. I mean, yeah, it will work for this. But you don't quite know how it's going to work, and you have to do very careful convergence tests, because this might be good. But up here, it's going to use a lot of basis functions. So for the first time in a long time, I'm finishing on time or even a little early. But DVR is really a powerful computational tool that, if you are doing any kind of theoretical calculation, you may very well much want to use something like this rather than an infinite set of basis functions and perturbation theory. It's something where you leave almost all of the work to the computer. You don't have to do much besides say, well what is the equilibrium value? And what vibrational frequency have I got to use for my basis set? And if you choose something that's appropriate for the curvature at the absolute minimum of the potential, you're likely to be doing very well. Now other choices might mean you don't get the first 100, you only get the first 50. But you might only care about the first 10. Or you could say, I'm going to choose something which is a compromise between two minima, and maybe I'll do better. But it doesn't cost you anything in the computer. Your computer is mostly sitting idly on your desk, and you could have it doing these calculations. And there is no problem where you can't use DVR. Because if you have a function of two variables, you do a two-variable DVR. It gets a little bit more complicated because, if you have two DVRs, now you're talking about a million-- 1,000 by 1,000-- two of them, and couplings between them. And so maybe you have to be a little bit more thoughtful about how you employ this trick. But it's a very powerful trick. And there are other powerful tricks that you can use in conjunction with quantum chemistry that enable you to deal with things like-- I've chosen a basis set, which is not orthonormal. And my computer only knows how to diagonalize an ordinary Hamiltonian without subtracting an overlap matrix from it. And so if I transform to diagonalize the overlap matrix, well then I can fix the problem. And so there is a way of using this kind of theory to fix the problem, which is based on choosing a convenient way of solving the problem, as opposed to the most rigorous way of doing it. OK, so I'm hoping that I will have a sensible lecture on the two-level problem for Wednesday. I've been struggling with this for a long time. And maybe I can do it. If not, I'll review the whole course. OK, see you on Wednesday. |
MIT_561_Physical_Chemistry_Fall_2017 | 33_Electronic_Spectroscopy_FranckCondon.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at OCW.MIT.edu. PROFESSOR: Today's lecture is on intermolecular interactions. And it's really one of these favorite topics where I can say, well, this is the origin of life. How do gases actually condense? And we can build a really, really simple theory for how gases condense based on what we know, based on second order perturbation theory, based on some hand-waving stuff. And it really is something from nothing. But it's also an example of what we do in physical chemistry where we explain one phenomenon by taking something that seems to be unrelated and building a model around it, and explaining this surprising thing. And that's what's going to happen today. We'll talk more about that. OK, now last time we talked about wavepacket dynamics. And so with wavepacket dynamics, there is first the pluck. How do you start the system in some superposition of eigenstates? And then there's the evolution of the pluck, which is usually some kind of particle-like state that follows Newton's laws. The center of the wavepacket follows Newton's laws for the position and the momentum. But if you make a wavepacket in a real molecule, the molecule is going to be not a harmonic oscillator. There will be anharmonic effects which causes the wavepacket to de-phase, but the center of the wavepacket continues to move as if it were in a simple potential. And it can de-phase, and rephase, and all sorts of complicated stuff. And that can be very interesting. Now, when you take this step from a diatonic molecule to a polyatomic molecule, you have 3n minus 6 oscillators cohabiting in the same house. And they're going to interact. And so if you start a wavepacket on one Franck-Condon bright mode because of anharmonic interactions handled by perturbation theory, between modes you start to get this wavepacket leaking out into other modes. And so the expectation value is going to exhibit an oscillating time dependence associated with the bright mode and some of the dark modes that are anharmonically coupled to the bright mode. And it can get pretty complicated. But you can understand all of it because you have perturbation theory. And so you can say, OK this time the wavepacket-- some amplitude of the initial wavepacket is transferred to a different mode. And when does that happen? Stationary phase. It happens at a particular geometry. And so you can be asking, OK, when is this excitation transferring? How strongly? This is a very, very complete picture. And it's very much like classical mechanics because it's localized. Then we talked about Landau-Zener. And the Landau-Zener is basically a simple idea that you also understand from real life. And it has to do with how does a wavepacket jump from one potential surface to another? And again, it's stationary phase. At the internuclear distance of the curve crossing, the wavepacket has to decide which curve it's going to follow. And it depends on the velocity of the packet at the curve crossing and the difference in slopes of the two paths it has to choose between. And all of this is very simple. Now, I am trying to create an exam problem which is related to Landau-Zener. And this is hard. But if I succeed, then you can expect to see it. The last thing I talked about the last lecture was the Zewail pump-probe experiments. And this is a beautifully graphical picture of how do you create something and then probe what's happening. And again, I hope that I can come up with something pictorial for the exam on the Zewail pump. OK, now before I talk about intermolecular interactions, I want to do a little bit of tying up loose ends. You know about non-degenerate perturbation theory. Now there's another kind of perturbation theory called degenerate perturbation theory. And this is when non-degenerate perturbation theory fails because H i j over E i 0 minus E j 0 is greater than one. Well then you've got a problem. You can't use ordinary perturbation theory. You have to diagonalize. You diagonalize it two by two or you diagonalize all of the little cases where this is violated. That's called degenerative perturbation theory. And there is a little bit of extra recipe for how do you put information about remote states into the part of the Hamiltonian you diagonalize? That's called a Van Vleck transformation, and I'm not going to talk about it. But basically you fix things up by diagonalizing two by two or three by three. And then do ordinary perturbation theory. And then there's variational. The variational model is something we've used mostly in a minimum basis set for talking about the atomic molecules. And in the variational model, we minimize energies by optimizing mixing coefficients. Now, perturbation theory is implicitly an infinite basis. Usually the infinity doesn't matter because the interesting stuff comes in from relatively low-lying levels. And the infinite number of very far away levels just modifies the energy levels a little bit, and you don't have to worry about it. The variational method, you choose your own basis set. Now, the choice of basis set is frequently dictated by convenience. And the basis set is often not very large or not orthonormal. And so there are problems with how do we deal with basis functions which are not orthonormal, especially not orthogonal to each other? And there is a transformation that you can do to make that work out. I mean, the problem is basically-- the secular equation is this for the variational method. And this is the overlap matrix. And the overlap matrix is not the unit matrix. We have what's called the generalized variational calculation. And there are methods for handling that. And I have often introduced those methods on a final exam. Those are just statements. OK, so variational is not guaranteed to give the exact answer. And it can't if you don't use an infinite basis. Perturbation theory implicitly uses this infinite basis. And you should feel confident that you're getting close to the right answer. And you can know what sort of errors you're making. OK, yes? AUDIENCE: When you go to very high orders in perturbation theory, your convergence is not necessarily better-- PROFESSOR: That's correct. AUDIENCE: --on the real solutions. PROFESSOR: OK, but that's-- AUDIENCE: [INAUDIBLE] when to cool it? PROFESSOR: This is-- the problem there is that you cannot include correlation in perturbation theory. And so there are things that are outside of the network of things that I've described. And so correlation effects where the particles can move in some way that they know about each other. That's very difficult to build in via perturbation theory. But when we don't have correlation, length perturbation theory makes an error. And you know how big that error is. OK, and then we know that we can diagonalize the matrix by some kind of unitary transformation. And the columns of T dagger are the eigenvectors. And the rows of T are the inverse-- rows of T dagger or the columns of T are the inverse. And so if you want to make a wavepacket where you have some additional state to equal 0, you want to express that as a linear combination of eigenstates. And this sort of transformation provides that information if you know how to use it. And you can get approximations to all of these elements of T dagger from non-degenerate perturbation theory. And on the second exam that was one of the things you didn't do very well on. And so you can see that again. All right. OK, so now we're going to talk about intermolecular interactions. And up until now, we've been talking about isolated molecules. And you have a lot of insights about how isolated molecules work, or at least what kind of stuff you need to be able to make predictions about isolated molecules. And perturbation theory is a very important tool in building that insight. In almost every problem which is not an exactly soluble problem, perturbation theory gives you at least a hand-waving argument for what's going on. Everything we did with LCAO-MO is really based on perturbation theory, or at least the interpretations, especially when you go from homonuclear to heteronuclear. And when we talk about what various orbitals look like and why they look that way. OK, so perturbation theory can also be used to talk about-- we have two particles, A and B. And we have the origin of coordinates. And so this is the coordinate for A, and this is the coordinate for B. And this is the interatomic coordinate. And so we're interested in what happens as a function of this interatomic distance. And can we use perturbation theory to be able to say something about that? And the answer is you betcha. Because basically we understand how these two molecules work. And so we have to somehow build in some insight into what is the interaction between these, and then use perturbation theory to quantify it. OK, now this is a situation where we're going to say atoms A and B are our field shell, or whatever you want to call it. They don't need to make another bond. They're not going to bond to each other. There's no covalent bond between A and B. There is no charge on A and B. There is no donor/acceptor. All of the things that we can understand by looking at individual systems, they are all turned off. These molecules do have energy levels. And so if we could find some way to talk about the weak interaction between them, well then we have the basis for perturbation theory. Because we have weak and we have energy levels that we know for the two molecules. And so if we're going to do perturbation theory it'll be second order perturbation theory because we are going to be interested in how interaction of A with B causes excited states of B to mix into B, and vice versa. And so one of the things you know about perturbation theory, if you're talking about the ground state of a system and you do perturbation theory, what happens? What happens to the energy levels? I mean, you have-- for second order perturbation theory-- I'm sorry? AUDIENCE: Everything gets-- things can increase or decrease. The H term gets squared. PROFESSOR: Yes. AUDIENCE: Some [INAUDIBLE] information. But depending on whether a level that's lower-- PROFESSOR: But if we're talking about the ground state-- AUDIENCE: If it's that ground state it has to go down. PROFESSOR: Right. And so that's the key. We know that if we're going to do second order perturbation theory and we're going to be asking questions about molecules in their ground state, they're going to be stabilized by interaction with other molecules. It doesn't matter, as long as there's no bonding interactions, or very strong short range interactions, this sort of a problem is guaranteed to give stabilization of the ground state of the system. And that's really a surprise, because-- Yes? AUDIENCE: You're treating the system as though it has like a ladder of energies and everything is already at its lowest? PROFESSOR: Yes. Well, OK-- you know, I'm not allowed to talk about statistics. I can't, actually. Because we do it-- we now do some statistical mechanics in 560. So you know a little bit. And you know that the molecules like to be in the lowest energy levels. And so we have these ground state molecules in a gas. And we understand they're isolated properties, or we could understand. And what happens here? Somehow there's interactions between them. And now first order perturbation theory, you're allowed to have a sign. But you're allowed to have a sign for diagonal elements of the perturbation operator. OK, so we'll just continue with this. So we want to talk about Hamiltonian for the system, which is the Hamiltonian for particle A, the Hamiltonian for particle B, plus some interaction term between A and B. OK, now the typical interaction between two particles is a dipole-dipole interaction. If molecules have dipole moments, then we already know something about what they want to do with each other. And from electricity and magnetism-- we know this isn't quantum mechanics-- we know that we can write a general formula for the dipole-dipole interaction. So that's the general formula for the dipole-dipole interaction. And it's a little more complicated than you would like because there's two terms. And this one is kind of a puzzling term because you're projecting a dipole on the interatomic axis. And so we'd like to be able to simplify this. And we can simplify it by simply drawing some vectors corresponding to the dipole moment. And so we have the dipoles oriented like that. Or we have the dipoles oriented like this, or like this and like that. Now, this is a reduction. You know, these are the simplified terms where these guys are irrelevant or at least you don't have to worry about the projections. And we know that this is an attractive interaction, this is an attractive interaction. This is repulsive and this is repulsive. And we know also by plugging into the formulas, we know that the energy for this one is minus mu-A mu-B over two pi epsilon 0 R cubed. And this is minus the same stuff, except a four in the denominator. So that means it's less attractive than this one. Minus means attractive. And then we have two terms that resemble these except they're repulsive. So there's four terms. So there's two terms that are attractive and two terms that are repulsive. And they come in pairs, one positive and one negative. And so you might naively say well, you can't win. If the particles are randomly distributed and randomly oriented, there will be no stabilization, even with dipoles. But that's not true because the molecules try to find an orientation which is more stable than the sum of the energies without the dipoles. And so maybe we could draw a picture and try to figure out, well, how that would be. So does anybody want to give me guidance to a picture showing, let's say, four dipoles oriented in the corner of a square? And what would be the stable arrangement for four dipoles oriented on the corners of a square? OK, right? So we have two attractive interactions, two more attractive interactions, and then there are repulsive interactions this way. But we have nearest neighbors. And they're stronger than the repulsive interactions. And so for one layer, that can win. Now, we have another layer of four dipoles. And if that layer is-- so we have another layer, and there will be four of them. And what would happen? What would be the stable arrangement if I have a dipole above this one? Is it going to be this way? Right. And so the next layer, all the dipoles are reversed. And so you can conceive of an arrangement where the dipoles would adopt a more stable arrangement. And statistical mechanics, or whatever you want to call it, guides that. And so we can say, all right, the ensemble of dipoles interacting with each other can arrange to be energetically stabilizing. And so we can reduce the problem to just one picture, say this type or that dipole-dipole arrangement, and do second order perturbation theory. And that will be an overestimate of the stabilization. But it will be proportional to the stabilization. And so we can get what we need for a very complicated infinite number of particle problem. OK, another thing that I want to ask you, suppose you have two particles and they come together. And so they come together with some kinetic energy. And they can't get rid of that kinetic energy, so they come apart. Now, if you have three particles, if something drives them together by mutual attraction, then one particle can take away the excess energy. You can end up leaving the two particles together. That's how you make liquids. So the driving force for the attraction between atoms or molecules is somehow related to what we're going to drive here. And that is the only way you make liquids from gases. And it's really strange because we're going to be talking about energy levels and interactions between molecules, and somehow coming up with a picture that leads to universal attraction between molecules that are not charged, not reactive, not anything. And they will always attract each other. And that's the first step towards making liquids. OK, now the important thing in all of this is that the interactions are small compared to the energy levels of the individual molecules. And so that's why we're doing perturbation theory. OK, so in perturbation theory we have an H 0. And the H 0 is-- I've got to make sure I use consistent notation. Well, I already see I have inconsistent notation in my notes. But anyway, OK, so we have the Hamiltonian is the individual system Hamiltonians, the sum of the two. And the first order term is VAB. And so whenever we can separate a system, we know that the zero order wave functions are products of the eigenfunctions of HA and HB. And the energy levels of 0 [INAUDIBLE] energy levels are sum of the energy levels for HA and HB. So we know that we can write H 0 psi A alpha psi B alpha. So this guy-- HA operates on psi A and treats this as a constant. HB operates on psi B and treats this as a constant. And when you do that, what you discover is that we get-- OK, so these are our energy levels. And these are our basis functions. And we are off to the races. So this is E 0. And this is psi 0 for the system. OK, now the first thing we do in perturbation theory is we say, let's look for the first order correction of the energy. That's the diagonal matrix element of the H1 term, this. So suppose we do have dipole moments on both particle A and particle B So that's the first order energy. Well, this is just the dipole-dipole interaction. And so we can take the reduced formula. We're just going to regard VAB as mu A dot mu B over R cubed. And so the energy is minus one over two pi epsilon 0 R cubed. I left out the two pi epsilon 0 on this formula. And so skipping a couple of steps, we have mu A mu B, the expectation value of these. So if there's a dipole-dipole interaction, there will be a first order correction term. And keep in mind that these guys can have signs. And that sign does appear in the first order correction to the energy. But that's basically these pictures we drew here. We have a dipole pointing this way, we have a dipole pointing that way. That's repulsive. And that's all-- so there's no surprises here. OK, but the important thing is that the energy-- If we have two molecules with dipoles, we get an interaction energy that we can calculate. But the interesting thing is going to be when we have molecules where at least one of the molecules doesn't have a dipole. Then we're getting into territory that we didn't think we understood. But we're continuing with the problem where we do have dipoles. I just want to make sure that-- yeah, OK. So for the second order correction to the lowest energy level, we have the sum over m, which is two quantum numbers. And m is not equal to the 0-0 state. Alpha is not 0, beta is not 0 sum over everything except the ground state. And what we have is an integral of psi m 0 star VAB psi 0 0 d tau squared over then E 0 0 minus E 0 m. Standard formula for second order variation theory. This is the ground state. This is some excited state. This is negative. This is squared. And this is squared. So this overall correction is negative. Now we can unpack all this and we can write the various terms in more instructive form. And what we'll have in the denominator will be E 0-- well, E A 0 minus E A alpha plus E B 0 E 0 E minus E B beta. Now we can rearrange that so we have E 0 A plus E 0 B minus E A alpha plus E B beta. So we have the ground state energy and the known excited state energies. And there is the difference between these. So that part is simple. And the matrix element also simplifies to the product of the dipole moment. OK, now suppose-- I just want to make sure that I'm doing this in the order it appears in the notes. Suppose mu A is not 0 and mu B is equal to 0. OK, well then all these formulas go out the window because nominally we think well, this B doesn't have a dipole moment. But suppose we have a particle B in an electric field. The electric field will polarize B. And if we have molecule A, which has a dipole moment, near B, it will also polarize B. And second order perturbation theory tells us how to handle that. And so we get a formula, a second order correction to the energy B. And we call this B comma induction. And that's going to be mu A squared over four pi squared epsilon 0 squared R to the six times sum beta not equal to 0 of the squared matrix element integral psi B star beta mu B psi B 0 d tau B squared over E B 0 minus E B beta. Now, I just told you that the dipole moment-- the permanent dipole moment for particle B is 0. But this is not the expectation value of the dipole moment. This is an off-diagonal matrix element of the dipole moment. This is the transition dipole. Transition between v equals 0-- well, between the ground state of B and some excited state, the beta, excited state of B. Well, we can write something like that because we know for all molecules there are transitions, which are electric dipole allowed. Now they could be between the ground state and an electronically excited state or the ground state and a vibrationally excited state. So we know something about these: we know the rules for when they're not 0. And this is a perfectly reasonable thing because it says this is related to the transition of-- the intensity of the observable transition. And this should really start to make you a little bit appreciative of perturbation theory. Because we're taking something which you can observe, a transition strength, and using it to describe an interaction between two molecules. So it's a relationship between two apparently unrelated phenomena. Physical chemistry is full of that. You build a model. And the model is telling you that certain things are related. And how are they related? Well, they're related by one over R to the sixth. They're related by the strength of the dipole on one, the permanent dipole on one, and some transition intensity-derived quantity on the other. Isn't that neat? And since the dipole-- we say well, this is a dipole-dipole interaction. And the mu A, the dipole on A, is inducing a dipole on B. It's not inducing anything. It's using transition dipoles. But it's creating something that looks like a permanent dipole. And you have dipole-dipole interaction. But the only difference is that it's one over R to the sixth because it's an induced dipole. OK, so we're using transition-- well, electric dipole transition moments. (There used to be two erasers!) So these electric dipole transition moments play the role of producing a dipole that can interact with the other dipole. And so-- well, do we need this? Well, we don't really. If you had dipoles, then the interaction would be one over R cubed. But if one of the dipoles is 0, then the interaction is one over R to the sixth. And it's one over R to the sixth times the permanent dipole of one and this thing from perturbation theory, which is the induced dipole of the other. Suppose mu A equals 0 and mu B equals 0. Will the particles interact with each other? Will they attract each other? And the answer is yes. And it's going to be an induced dipole-induced dipole interaction. Which, now since most people who write for general science have no clue about what quantum mechanics is, they will talk about all sorts of fanciful things-- you get an institute-- an instantaneous fluctuation on one particle which creates an instantaneous fluctuation on the other particle, and bam, you've got stabilization. Well, yeah, maybe. But you know, this is a simple perturbation theory. And so I don't really need to actually write out everything. Because one, you do that and you include the second order perturbation expression for particle A and particle B, you get a term that looks just like this. It has one over R to the sixth. But it now involves transition moments, particle A and particle B-- nothing new. You don't have to induce anything more. You don't get another factor of one over R cubed. You've already got the one over R to the sixth. So I'm going to skip a lot of steps because the notes are really clear in there. And we don't have a whole lot of time. And so I'm just going to write down the result. And now the name of this term is called dispersion, or London because a fellow by the name of London invented this. And the energy level formula for this correction term is one over four pi squared epsilon 0 squared R to the sixth. And then this sum of alpha equals one beta equals one to infinity. In other words, not alpha equals 0, not beta equals 0. And then-- Sorry about that. So we have a numerator, which is the squared transition moment between the 0 level of particle A and the alpha level and the 0 level of B and the beta level. We have the energy of the ground state and we have the energy of these excited states. All of this stuff, you know or at least you can parametrize it in terms of this strength of our transition. And there is an explicit formula that relates the dipole moment to the intensity depending on what kind of experiment you do. So I'm not going to take you through that. But you know you can measure the intensity, the molar absorptivity, or whatever you want that is related to this. So these are experimentally observable. These are experimentally observable. And usually this infinite sum is dominated by the lowest allowed transition from the ground state. And that can be an electronic transition or a vibrational transition. So for our homonuclear molecule, there isn't going to be an allowed vibrational transition. So nitrogen and oxygen cannot talk to each other through transition dipole-- vibrational transition dipoles because the dipole moment is not dependent on [INAUDIBLE] but they have electronic transitions. And so there is this electronic-- as opposed to a vibrational polarizability of the two. And that's what gives rise to attraction between N2 and O2, or N2 and N2. So everything is knowable and predictable. And even though the perturbation's sum is infinite, and even though we've made a gross approximation and say, well, we're just interested in this sort of an interaction or this sort of an interaction, and we know that the true interaction will be less because there will be an average over all of the random orientations. But there's going to be a lowest energy picture. And you can get that from statistical mechanics. And so you can do the configurational average over all possible orientations of all of the interacting particles. And you can calculate this. So that's beyond the level of this course. But the engine of the calculation is this formula. So molecules are polarizable. They react to an external electric field. And there are several ways you can describe the defined polarizability. One is to say well the polarizability-- let's call it alpha-- is equal to the second derivative of the energy with respect to the electric field. But the polarizability, you know, you also get from perturbation theory. And so each of these terms or the energy is associated with the interaction of a dipole with an electric field is the electric field times the dipole. And so we have electric field squared. We take the second derivative of that. And we've got back to this. So now here we come with intuition. If I asked you to tell me the van der Waals or the dispersion interaction between helium and helium, and compare that to xenon to xenon, would you be able to tell me? So which one is stronger? Which one-- which pair of inert gases is stickier? AUDIENCE: Xenon. PROFESSOR: Xenon, right. Why? Because the energy denominator for the first excited state is much lower for xenon than for helium. OK, what about large molecule versus small molecule? Do I need to provide a hint? 3n minus six. The more vibrational modes, the more transitions are available for polarization. And so again, now we're talking about molecules which can be floppy. And the floppier the molecule-- that means they have low frequency vibrations-- you can polarize the daylights out of them. And so large molecules are really sticky, stickier than inert gases. Because you have this dispersion interaction that brings them together. And once they're together, then they can do magic. But the thing that drives the condensation, or the formation of droplets, or whatever, is this dispersion interaction. And it's universal. Every pair of molecules or atoms is going to attract each other. And the strength is something that you can rate in terms of what you know about atomic and molecular properties. And isn't that neat? Because you're saying the viscosity, or whatever it is, the property of a gas, all of those things are related to the properties of the individual atoms or molecules embedded in perturbation theory. This is why I think perturbation theory is so wonderful, because it enables you to relate different things. And even if you're not going to evaluate the magnitudes of the terms, you can make comparisons which are always going to be right. And that's kind of wonderful. I mean, you don't have to use the equations. Or you just know that these particles will attract each other and the driving force is either a permanent or an induced dipole. And the induced dipole will be strong if you have a strong transition to a low-lying state, or a lot of strong transitions to a lot of low-lying states as you have in big molecules. OK, so I'm not positive what I'm going to talk about on Friday. It's probably going to be photochemistry. But I may panic and do something else. OK, I'll see you Friday. |
MIT_561_Physical_Chemistry_Fall_2017 | 9_The_Harmonic_Oscillator_Creation_and_Annihilation_Operators.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: Last lecture, I did a fairly standard treatment of the harmonic oscillator, which is not supposed to make you excited, but just to see that you can do this. And the path to the solution was to define dimensionless position coordinate, and then you get a dimensionless Schrodinger equation. And the solution of that involves two steps. One is to insist that the solutions to the Schrodinger equation have an exponentially damped form. And then the Schrodinger equation is transformed into a new equation called the Hermite equation. You shouldn't get all excited about that-- the mathematicians take care of it. And so the solutions to this Hermite differential equation gives you a set of orthogonal and normalized wave functions. They give you the energy levels expressed as a quantum number plus 1/2, times constant. The energy levels with the quantum number v are even or odd in v, and they're even or odd in psi. Even v corresponds to even psi. V is the number of internal nodes, and there are all sorts of things you can do to say, well, I expect if I know how to do certain things in classical mechanics, they're are going to come out pretty much the same way in quantum mechanics. So that's the standard structure, but more importantly, I want you to have in your head the pictures of the wave functions, the idea that there is a zero point energy, and that there's a reason for that-- that the wave functions have tails extending out into the non-classical, or the classically forbidden region. And this turns out to be the beginning of tunneling. You'll be looking at tunneling more specifically. I also want you to know how the spacing of nodes is, and that involves generalization of the bright idea that the wavelength is h over p. But if the potential is not constant, then p is a function of x. This is not the quantum mechanical operator, this is a function which provides you with a lot of intuition. And then if you know where the node spacings are, and you know the shape of the envelope, you have basically everything you need to have a classical sense of what's going on. And then-- I guess it's supposed to be hidden-- I did a little bit of semiclassical theory, and I showed that if you integrate from the left turning point to the right turning point at a given energy of this momentum function, you get h over 2 times the number of nodes. And this is the semiclassical quantization-- it's incredibly important, and it's useful either as an exact or approximate result for all one-dimensional problems. And so it tells you how to begin. Now, before I start talking about what we're going to do today, I want to stress where we're going. So we're going to be looking at some exactly solved problems. And so we have a particle in a box, the harmonic oscillator, the hydrogen atoms-- you have to count them-- and the rigid rotor. Now, all of these problems have an infinite number eigenfunctions, an infinite number of energy levels, and that's intimidating, but it's true. Now, these infinite number of functions are explicit functions of the quantum number. And so we have an infinite number, but in order to describe systems, we're going to be calculating integrals. We're going to be calculating a lot of integrals between these infinite number of functions. So we have an infinity squared of integrals. Well, that shouldn't scare you because what I'm going to show you is that all of the integrals that we are going to encounter are explicit functions of the quantum numbers, and they have relatively selection rules. In other words, which integrals are non-zero based on the difference in quantum numbers between the left-hand side and the right-hand side? So we're collecting these things in order to calculate a whole bunch of stuff. Now, I told you that this is a course for use rather than philosophy or history. And so if you encounter any quantum mechanical problem, you'd like to be able to describe what you could observe with it. And so if you're armed with the infinite number of energy levels and eigen solutions for our problem, you can calculate any property. So you define some property you're interested in-- there is a quantum mechanical operator that corresponds to that property. And in order to be able to describe observations of that property, you need to calculate integrals of that operator. Well, la dee dah. That should be intimidating, but it's not because almost all of these integrals can be expressed as a simple constant times a function of the quantum numbers or the difference of quantum numbers, and that's a fantastic thing. So we have any operator-- suppose the Hamiltonian is an exactly solved problem plus something else, which we'll call h1. And this is a complexity in the-- or it's the reality in the problem. And in order to deal with this, again, you're going to need to calculate integrals of this operator. And the last thing that's really going to be exciting is once we look at the time dependent Schrodinger equation, we're going to get wave packets. And these are functions of position and time, and these wave packets are classical-like, localized objects that move following the Newton's equations of motion with the center of the wave packet. And again, there are a whole bunch of integrals you're going to need in order to do these things. And so right now, we're starting with the best problem for these integrals, because a harmonic oscillator has some special properties. And the lecture notes are incredibly tedious, and they're mostly proofs. And I'm going to try to go fast over the tedious stuff, and give you the important ideas, but since there is some important logic, you should really look at these notes. So what we're going to be doing today is we start with the coordinate momentum operators, we're going to get these operators in dimensionless form, and then we're going to get these a and a-dagger guys. So this step is reminiscent of what I did at the beginning of the previous lecture, and then this is magic because this magic enables you to just look at integrals and say, I know that integral is zero, or I know that area is not zero. And with a little bit more effort-- maybe something that you'd put on the back of a postage stamp-- you can evaluate that integral, not by knowing integral tables, but by knowing the properties of these simple little a and a-dagger. And that's a fantastic thing. And it's so fantastic that this is one of the reasons why almost all problems in quantum mechanics start with a harmonic oscillator approximation, because there is so much you can do with this a and a-dagger formalism. Now, at the beginning I also told you that in quantum mechanics, the important thing that contains everything we're allowed to know about a system is the wave function. But I also told you we can never measure the wave function. We can never experimentally determine it, and so we need to be able to calculate what this wave function does as far as what we can observe, and these a's and a-daggers are really important in being able to do that. So I'm going to start with covering what I did in the notes, but I'm going to jump to final results at some point-- governed by the clock. And so the first thing we're going to do is these. And so what we do is we define the relationship between the ordinary position coordinate. And this little twiddle means it's dimensionless, and so we can write the inverse of that. And that's the one we are going to want to-- well, actually, we go both ways. And we do the same thing for the momentum-- p is equal to h-bar mu omega square root, p twiddle and the inverse which I don't need to write. And finally, we get the Hamiltonian, which is p squared over 2 mu plus 1/2 kx squared. And we'll put that into these new units. So we have h-bar mu omega over 2 mu, p twiddle squared plus k over 2. This is all very tedious, but the payoff is very soon. k over 2 times hr mu omega, x twiddle squared. Oh, isn't that interesting? We can combine-- we can absorb a k over mu in omega, and so we get, actually, a big simplification. We get h bar omega over 2 times p twiddle squared plus x twiddle squared. Well, that looks simpler. And so the next thing we do is-- it looks like a simple problem from algebra, let's factor this. Now, it's a little tricky because you know you can factor something in real terms if this is a minus sign. But we are allowed to talk about complex quantities, so we can factor that. And so this term, p twiddle squared plus x twiddle squared is equal to ip twiddle, plus x twiddle times minus ip twiddle, plus x. And you can work that out-- that ip times minus ip is p squared, and x times x is x squared. And then we have these cross-terms, ip times x and x times minus ip. Whoops, they don't commute. If this were algebra-- well, they would go away, but they don't. And so what you end up getting is p twiddle squared plus x twiddle squared, plus i times p-- I'm going to stop writing the twiddles. So we have this. I want to make sure that I haven't sabotaged myself-- that's going to be-- yeah, that's right. So we have something here that isn't zero. And it looks like i times the commutator of p twiddle with x twiddle. But we can work that out because we know the commutator of p ordinary with x ordinary. And so I did that. And so we have this commutator, p twiddle, x twiddle. After some algebra, we get plus 1. A number-- pure number-- no? I want you to check my algebra. So you just substitute in what this is in terms of ordinary p and the ordinary x. Use a commutator for ordinary xp, which is ih-bar, and magically, you get plus 1. So this very strange and boring derivation says, OK-- well, let's now give these two things names. This guy, we're going to call as the square root of 2 times a, and this one is going to be the square root of 2 times a-dagger. So H is going to be h-bar omega over 2, times square root of 2a-hat times the square root of 2a-dagger-hat minus 1. Remember, when we factored it, we got this extra term which was 1. And in order to make it correct, we have to subtract it away. And so this becomes h-bar omega, a-dagger-hat minus 1/2. Well, isn't that nice? Now, we have the Hamiltonian expressed as a constant, which we know is important because it's related to the energy levels, and times these two little things, which turn out to be the gift from God. It's an incredible thing, what these do. So we have gone through some algebra, and we know the relationship between a, and the x and p twiddles, and similarly for a-dagger. And we can go in the other direction, and we know the commutator, and now we're going to start doing some really great stuff. Well, one thing we're going to want to know about is a-hat. a-hat, a-dagger, that commutator-hat. And that turns out to be-- well, I already derived it-- it turns out to be plus 1. And as a result, we can say things like this-- a a-dagger-- So using this trick, we can show we can always replace something like a, a-dagger by a-dagger a plus this commutator, which is 1. And so we have this really neat way of reversing the order of the a's and a-daggers. So with this, we're going to soon discover the a operating on the eigenfunction gives square root of v times psi v minus 1. And a-dagger operating on this wave function gives v plus 1 square root of psi v plus 1. Which is the reason these things are valuable, because if you have any eigenfunction, you can get all the others. So suppose you have the lowest eigenfunction-- you apply a-dagger on it as many times as you need to get to, say vth function. So you don't actually-- you're not going to be evaluating integrals, you're going to be counting a's and a-daggers, and permuting them around, and getting 1's, and stuff like that. Yes? AUDIENCE: In this line with the commutator, you didn't move the dagger. ROBERT FIELD: I didn't what? AUDIENCE: For a, a times the square root of a a-dagger plus a-dagger a, it should be a a-dagger. And on the right-hand side, you need to move the dagger. ROBERT FIELD: OK, so this is to switch the order, and I've done that. And that then is-- no, I think-- wait a second. So we have a a-dagger-- so that's a a-dagger minus a-dagger a, and that's-- oh, yeah. Thank you. It's very, very easy to get lost, and once you're lost, you can never be found because you've made a mistake that's built into your logic, and you're never going to see it. You see it took me a couple of minutes to even accept-- that the insight from my TA who's sitting there calmly thinking, and I'm trying to do several things in addition to the thinking. So we can do things like this. Suppose we have psi-- I can't use this notation yet. So suppose we have psi-star v, and we have a-dagger, a-dagger, a-dagger, psi, v prime, dx. These are raising operators, so this is going to take v prime to v prime plus 3. That's the only integral that's not 0. And we get v prime plus 1, v prime plus 2, v prime plus 3, square what? It has the constants. And this would be v prime plus 3. So instead of evaluating an integral, looking at what the x's and p's are, we just have a little game we play. So now, we have to prove some of the things I've said. So we have h, and we're going to operate on a-dagger psi v. So what does the Hamiltonian do to this thing? So what we're going to want to do is to show that this thing is an eigenvalue-- eigenfunction of v plus 1, and that's what we are going to get. So let's just go through this. So we have h-bar omega, a-dagger a plus 1/2, times psi v. So what I did is-- where did I-- yeah, I showed that the Hamiltonian-- or did I not do that yet? Oh, yeah-- I did it right here. The Hamiltonian is h-bar omega, a a-dagger minus 1/2, or if we reverse these, it's equal to h-bar omega, a-dagger, a plus 1/2. So we can use either one, so I'm using that one-- except I wanted an a-dagger here, because we want to show what the Hamiltonian does to this. Now, we can pull in a-dagger out to the right, because this-- if it's 1/2 times a-dagger, well, that doesn't matter. This a-dagger, a, a-dagger-- well, we can pull this a-dagger out. So we have h-bar omega, a-dagger is equal to a a-dagger plus 1/2-- so iv. Now, we use our magic commutator trick to replace this by a-dagger a plus 1. So now, we have h-bar omega, a-dagger, and we have a-dagger a plus three halves is iv. Well, this is Ev plus h-bar omega. We've increased the number that started here. Here is Ev-- that was Ev plus 1. And so now, we have no operators in here, and we can stick the a-dagger back here. And so we have h-bar omega, Ev plus 1, a-dagger Well, what do we have here? We have an operator, we have this function, we have some constant times the same function. So what we've shown is that this thing is an eigenfunction of the Hamiltonian that belongs to the eigenvalue Ev plus 1. We've increased the energy by 1. So what we have-- so we can show that we apply a-dagger to any function-- we increase its energy, and we can do this forever. We could also do a similar thing if we apply a to psi v. We can go down, but at some point, we run out of steam because we've gone to the lowest energy, and if we go lower, we get 0. So a operating on psi min gives 0. So we have this stack of energy levels and wave functions, and we have the same stack being repeated as we go down, but this one has an end. We bring back what a is, and so a psi min is 0-- that's the equation. We bring in what a is, and it's ip twiddle x plus x twiddle. So we do some algebra, and what we end up with is a differential equation, psi min. dxx twiddle is equal to-- again, a little bit more algebra-- minus mu omega over h-bar times psi min. So what function gives-- there's an x in here too. So what function has a derivative, which is the function you had started with, times the variable, times a constant? And so the answer to that is that psi min has to have the form-- some normalization factor times e to the minus m omega-- or mu omega-- sorry-- over 2 h-bar x squared-- a Gaussian. Well, it had to be a Gaussian, right? We know when we did the algebra that we're going to get some function times a Gaussian. But for the lowest function, the Hermite polynomial is 1, and all there is is the Gaussian. And so we found the lowest level, and we can normalize it. So let's start over here. So what we have found is psi min of x is equal to mu omega over pi h-bar to the 1/4 power, e to the minus mu omega over 2 h-bar squared. Well, that's useful. We knew that, but this time we got it out of a completely different path. And now, we can get all higher v by a-dagger, a-dagger, et cetera. So remember, we don't care anything about what the function is, we just know that we can bring it in and get rid of it at will, because what we want is the values of integrals involving that function and some operator. So yeah, we can have all of those functions, and this is a way of generating all of the functions. And so if we wanted psi v, we would do a-dagger to the vth power divided by v-dagger-- v-- what do you call this with an exclamation point? Factorial-- ha! So we apply this operator that raises us to whatever level we want starting from this Gaussian at the bottom, and we have this normalization factor which cancels out the fact the stuff that you get by applying av. Now, there is some more logic in my notes, and I don't want to do that, but what we'd like to be able to show is that a-dagger on psi v gives some constant, and that this constant has some value-- we're going to evaluate what it is. And similarly, a psi v gives dv, and some constant v minus that. We can derive those things, and I'm not going to waste time deriving them-- I'm going to just give you the values. But we already know that cv is square of v plus 1/2-- e plus 1 and dv-- and you can see the derivation in my notes. I don't think going through them is going to be instructive. And that's just going to be v and 1/2. So now, we have something that's wonderful, because everything you need to know about getting numbers concerning harmonic oscillator is obtained from these five equations. a-dagger on psi v is v plus 1 square root psi v plus 1. a on psi v is v square root psi, v minus 1. I've said this before, but these are the most useful things you'll ever encounter. We have this thing called the number operator, and that number operator is a-dagger-hat, and the number operator operating on psi v gives v psi v. And so that's a kind of benign operator that can suck up all sorts of factors of a-dagger a, because it just gives a useful thing. And then we have a, a-dagger, and this is 1. Well, you sort of know it's going to be 1, because a a-dagger gives an increase-- it gives v plus 1, and a gives v minus 1. So it's plus 1, not minus 1-- you know that it's hardwired. Well, I did it-- I got to the point where it starts to get interesting. So we're going to be using this notation, a-dagger and a, for all sorts of stuff. And one sort of thing is transition intensities and selection rules. So you have a harmonic oscillator. A harmonic oscillator is, say a diatomic molecule which is heteronuclear. And so as the molecule vibrates, you have a dipole moment which is oscillating. And so any oscillating electric field will grab a hold of that dipole moment and stretch or compress it, especially if that field is in resonance with h-bar omega. And I've got some beautiful animation showing this, but we can't do that until we have time dependent quantum mechanics. So we have a time dependent radiation field, which is going to interact with the dipole associated with the vibrating molecule, and it's going to cause transitions. And so we can write the quantum mechanical operator that causes the transitions-- this is the electric dipole moment operator-- as a function of coordinate. And we can do a power series expansion of this, and we have-- so we have mu 0-- the constant term-- the first derivative of the dipole with respect to x, and the second derivative of the dipole with respect to x. And we have the x cofactor and the x squared cofactor. And so this guy doesn't have any x on it-- it's a constant. The only integrals involving-- the only integrals are delta v, v prime following the selection rule delta v v prime. So these integrals are 0 unless v and v prime are the same. And that says, well, an isolating field isn't going to do anything key to it, it's just going to leave it in the same vibration level. But it might have an electric Stark effect, but that's something else. So this term does nothing as far as vibration is concerned. This guy, which is a plus a-dagger, has a selection rule, delta v of plus and minus 1, and this guy has a selection rule, delta v of plus and minus 2 and 0. So if we're interested in the intensities of vibrational transitions, it says, well, this is the important term and it causes transitions, changing the vibrational quantum number by one, which is called the fundamental. This gives rise to overtones. So all of a sudden, we're in real problem land, where if we're looking at vibrational transitions in a molecule, that this enables us to calculate what's important, or to say these are the intensities I measure, and these are the first and second derivative of the dipole moment operator as a function of internuclear distance. Isn't that neat? I've gone so fast, I'm more or less at the end of my notes, but I can blather on for a while. So suppose you have some integral involving an operator and a vibrational wave function. So we have psi v star, some operator, psi v prime dx. And we'd like to know how to focus our energies. We're very busy people-- we don't want any value integrals that come out to be 0, we'd like to just know. So if this operator is some function of x or function of p, we'd have a power series expansion of the operator, and we then know what the selection rules are. So usually, you look at the operator and you find that it's a linear quadratic cubic function of x-- the leading term is usually linear. Bang-- you have a delta v of plus 1-- selection rule. Or if someone has bothered to actually convert the operator to some form-- oh, it's the operator-- this might have some form, a-dagger cubed times some constant. So if the operator looks like a-dagger cubed, we know that the selection rule is v to v plus 3, and we know that the matrix of the integral is v plus 1, times v plus 2, times v plus 3, square root of that, times the constant. So there is a huge number of problems that, instead of being pages and pages of algebra, are just reduced something that you can tell by inspection. So one of the tricks is we have an operator like x squared or x cubed-- what we want to do is write this in terms of a squared, a-dagger squared, and maybe some combination of a-dagger a. So we want to take the a-dagger a's with the a a-daggers and combine them using the commutation rule. And then we have expressed this in this maximally simple form, and then you just apply a squared, apply a-dagger squared, and you apply this, then you've got the value of the integral. So if you're a busy person and you want to actually calculate stuff, you want to know how to reduce operators-- usually expressed as some power of the coordinate or the momentum-- into a sum of terms involving these organized products of a's and a-daggers. And you're going to be absolutely shocked at how perturbation theory-- which leads to basically all of the formulas and spectroscopy-- it's an ugly theory, but it reduces everything to things that you can just write down at the speed of your pen or pencil, and that's a fantastic thing. So you can't do this with the particle in a box, you can't do this with a hydrogen atom, you can't do this with the rigid-- well, you can do some of this with a rigid rotor. But the harmonic oscillator is so ubiquitous because every one-dimensional problem is harmonic at the bottom. And so you can use it and then you can put in the corrections. But also because you want to describe dynamics, you almost always use the harmonic oscillator, because not only do you know the integrals, but you know there's only a few. Normally, you're going to be summing over an infinite number of quantum numbers, and that takes time, and it takes judgment to say, well, only certain of these are important. But for the harmonic oscillator, the sums are finite. All of these things are wonderful, and that's why whenever you look at a theory, you're going to discover that hidden in there is the harmonic oscillator approximation, because everything is doable in no effort. And sometimes when you look at a paper like that, it doesn't show you the intermediate steps, because everybody knows what a harmonic oscillator does. And there's also a lot of insight because something like this-- this is an odd symmetry term, this is an even symmetry term, and there are all sorts of things that have to do with, are you're conserving symmetry or changing symmetry? And sometimes, the issue is how does the molecule spontaneously change symmetry by doing something interacting with a field, or interacting with some feature of the potential surface. So this is a place where it's labor saving and insight generating, and it's really amazing. So maybe I've bored you with this, but this is the beginning of almost every theory that you encounter just because of the simplicity of the a's and a-daggers. OK, I'm done. Thank you. |
MIT_561_Physical_Chemistry_Fall_2017 | 17_Rigid_Rotor_I_Orbital_Angular_Momentum.txt | The following content is provided under a Creative Commons license. Your support will help MIT Open Courseware continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT Open Courseware at ocw.mit.edu. ROBERT FIELD: Today is the first of two lectures on the rigid rotor. And the rigid rotor is really our first glimpse of central force problems. And so although the rigid rotor is an exactly solved problem, it's also a problem that we-- it involves something that is universal. If we have a spherical object, then the angular part of the Hamiltonian for that spherical object is solved by the rigid rotor. And so if we understand the rigid rotor and know how to draw pictures, we're going to be able to understand the universal part of all central force problems. So it's not just a curiosity. It's a really major thing. OK. And I'm going to do a lot of unconventional stuff in this lecture and the next lecture, as far as the rigid rotor is concerned. Because it's covered to death, but with lots of equations in all textbooks. And there is very little effort to give you independent insights. And so my discussion of the vector model is something that I consider really special. And what I want you to be able to do is to draw cartoons that capture essentially all of the important points. And that you won't find anywhere except in my notes. But I think it's really important that you have your own point of view on these problems. So let me just do a quick review, with the exam in mind, of the perturbation theory stuff. So we had a harmonic oscillator. And it's perturbed by some anharmonicity terms. So the potential is going to 1/2 kQ squared plus bQ cubed plus cQ to the fourth. And we could go on. But this is enough. Because it gives a glimpse of all of the important stuff that emerges qualitatively as well as quantitatively from perturbation theory. So this is part of the H0 problem. And this is H1. And so you know now that doing second order perturbation theory is tedious and ugly. And it's easy to get overwhelmed by the equations. But what I want to do is to just review how you get the crucial structure of the problem. So the Q cubed' term-- there are no diagonal elements of the Q cubed term. And so we're always going to be talking about the second order corrections to the energy. The selection rules are delta v equals plus and minus 3 and plus n minus 1. And so what I told you is, if you're going to do the algebra, which you probably don't want to ever do, you're going to want to combine the delta v of plus 3 and minus 3 terms of the perturbation sum. Because the algebra simplifies. They have similar expressions. And you have a denominator, which is 3 Hc omega and minus 3 Hc omega. That factors out. And you just have a plus 3 and minus 3 in the denominator. And similarly here, so when you do this, you lose the highest order term. What's the highest order term? It's v plus 1/2 to the third power. Because we have delta v of 3. We have Q cubed. And the matrix elements of Q cubed are a plus a-dagger cubed. And the selection rule for them-- or the matrix-- must go as the square root of the power of the a's. And so all of the matrix elements have the leading term v plus 1/2 to the 3/2 power. But we're squaring. And we lose the highest order because of the subtraction. So this one gives rise to terms in v plus 1/2 squared. Q4 has selection rules delta v plus or minus 4, plus or minus 2, and 0. This is important. Because this one says, we have a contribution from the first order-- in the first order of energy. And so in the first order of energy, we're going to have v plus 1/2 to the fourth power-- I'm sorry-- to the 4/2 power. Because we only have the matrix element, we don't square the matrix element. So this gives rise to a term in v plus 1/2 squared. But it's sensitive to the sign of the Q to the fourth term in the Hamiltonian. It's the only thing that's sensitive to the sine. We also have terms that are v plus 1/2 to the 4/2 times 2. So we can have matrix elements to the four-- because it's a fourth power term, we get 4/2. And because it's squared, we multiply by 2. And then we lose the highest order. So we get, from this, v plus 1/2 to the third power. So what we're hoping to get from the perturbation theory is the highest order terms in the energy expression. And so Q to the fourth gives a signed v plus 1/2 squared term. And it gives an unsigned always positive v plus 1/2 cubed term. And this one gives v plus 1/2 squared term. OK. So this would tell you, when you're organizing your work, what to expect and how to organize it. And I'm really not expecting you to do very much with this level of complexity on an exam problem. But on homework, all bets are off. OK. So that's all I want to do as far as the review is concerned. So the rigid rotor-- we have a molecule. So we have a bond of length r sub 0. And we have two masses. And if the masses aren't equal, the center of mass isn't in the middle. The center of mass-- you need to know how to calculate where the center of mass is. There are all sorts of simple algebra-- but what is happening is this guy is rotating without stretching about the center of mass. Now, what we want to do is think about this problem as if it were-- here's our 0. And this is the reduced mass. This is a motion of a fictitious particle of mass mu. Mu is m1 m2 over m1 plus m2 on the surface of a sphere. They're all the same problem. But the question is, how do we interpret what we get from the solution to this problem? So what we care about is a description of, where is the molecular axis? The molecule is rotating. And we're solving the Schrodinger equation. And we get things like the expectation value of J squared and Jz and maybe some other things where these are the angular momenta. And how do the eigenfunctions for these states-- we have a state. And we have quantum numbers Jm and their probability amplitudes and theta and phi. That's a whole lot of stuff to digest. We want to go as quickly as we can to how this is related to the thing we really care about, which is, where is the bond axis? So the molecule is rotating. And so the bond axis is moving in laboratory frame. And we want to be able to take, from that, the minimal amount of information that we memorize or remember-- I don't like "memorize." Because "memorize" doesn't involve understanding. But "remembering" does. We want to be able to, at a drop of a hat, be able to say, yes, if we pick these two quantum numbers, which are related to the eigenvalues of these two operators, we can describe the spatial distribution of the molecular axis. There is an extra complication. So we live in the laboratory. And we have a coordinate system that we care about, the lab-fixed coordinates. And they're always going to be represented with capital letters. And there's also the body frame. And things are in the body frame. That's what the molecule cares about. And how do things in the body frame relate to the laboratory frame? It's not trivial. And that's where the real effort at understanding comes. And it's kind of trivial for a diatomic molecule. But it's far from trivial when you have a nonlinear molecule, a molecule with many atoms and properties of each of the atoms that somehow combine to give things that you observe. And so these two coordinate systems are very different. And actually, the solutions of the rigid rotor Hamiltonian gives you the relationship of the body coordinates to the laboratory coordinates. And that's kind of important. So we'll see how this develops as I go on. And I feel quite passionate about this. Because one of the things that I do as an experimentalist is I observe fully resolved spectra. And the big-- the most information-rich feature of the spectrum is the rotational spectrum. And so it contains a lot of really good stuff. OK. So first of all, the Hamiltonian is just the kinetic energy. Because it's a rigid rotor, it's free. There's no potential. But that's the last nice thing about it. Because we have to go from Cartesian to spherical polar coordinates. And there are a lot of unfamiliar things in this kinetic energy expression. And when I write it down, you're not going to like it. Unless you're a mathematician, and you say, oh, yeah, I want to be able to solve these kinds of equations. Because the Schrodinger equation is, considering the simplicity of the problem, terrible. And except for mathematicians who just love it because they know-- oh, I know that equation. I know how to write down everything that I care about. But we care about different things. So we want to know the energy levels of the rigid rotor. And we also want pictures. And remember, a picture is a reduced form of all the details that you have at your disposal. And you have to really sweat to make sure you understand every detail of the pictures and the stuff that you've averaged over, or you've concealed, because you don't need it, OK? All right. So I'm going to be generating a lot of pictures. And one of the things we want to understand is-- so we have this vector, J. The J is the angular momentum of the molecule. And it's a vector as opposed to a scalar. That means it has three components. So for this part of the problem that I've just hidden over here, that's a one dimensional problem. We have a momentum and a conjugate coordinate and it's 1D. This is 3D. And there's all sorts of subtle stuff that goes on. OK. So we would like to understand how this vector moves. Well, there is a question. I said moves. We're talking about the time independent Schrodinger equation. Nothing moves. We can use our concept of motion from classical mechanics to describe certain features of the average of some quantum mechanical system. And so there is kind of a motion. But we'll see what that is. OK. We have some operators that we like. OK. Total angular momentum, projection of the angular momentum on a laboratory z-axis-- it's kind of hard to draw capital Z and small z if you haven't got anybody nearby. But anyway, this is a capital Z. And this is the thing that's analogous to the creation and annihilation operators. We need them. And it gives us a lot of insight. And that will be the subject of Monday's lecture. And so we're going to have solutions, which are described by J and m quantum numbers. And we want to understand what these things look like. Where are the nodal surfaces? The nodal surfaces give us basically everything we want to know. There is the direct correlation to where the nodes are and how many of them there are with these quantum numbers. That's one of the important things you have to come away with. OK. In my next lecture on angular momentum, I'm going to introduce a commutation rule, Ji Jj equals I H bar sum over K epsilon I Jk Jk. Isn't that a strange looking thing? Because I haven't told you what this epsilon is. There's lots of names for it. It won't help. Because I'm not going to talk about it here. But this is actually the fundamental definition of an angular momentum. And from this commutator rule, we can generate all the matrix elements of angular momentum. And there's another family of commutation rules where we have a component of angular momentum and a component of what we call a spherical tensor. Any operator can be classified according to spherical tensor rank. And that then determines all the matrix elements of that operator. And so you can imagine that this is a really powerful way of approaching stuff. And what I said before about the commutation rule of x and P sub x is that many people regard that as the foundation of quantum mechanics. And this is just an extension of that. So there is a way of getting all quantum mechanics from a few well-chosen commutation rules. And that's really neat. OK. Let's get down to business. So I already drew this. But I'll draw it again. Because I'm going to drop a companion. OK. So we have here-- OK. So we have an angular momentum. And we have the bond axis. And this angular momentum is perpendicular to the bond axis. So if we know something about how the angular momentum is distributed in space, we know something about how the bond axis is distributed in space. But it's not trivial. But once you've learned how to make those connections, it's fine. OK. So now, let's draw-- So we have a right-handed coordinate system. And we have, say, the vector J. Now, this is the laboratory frame. We have J. And we have a molecule, which is perpendicular to J. And we have the projection of J on the body axis, on the z-axis, and that's M. OK. So when we talked about the quantum number J, which is the length of this vector, and m, which is the projection of that vector on this axis, that's beginning to be how we understand how this works. Because this is perpendicular to that. Now, here we have one of the sins. So we begin by saying, well, this is a picture. And J precesses, or moves, on a cone around z. How do we have motion? This is a time independent Schrodinger equation. It's a way of saying, well, it doesn't matter where J is. It's more or less equally distributed in probability on this cone. But not an amplitude. Remember, when we have something that's moving in the time independent Schrodinger equation, we get oscillations. But we need a complex function in order to have those oscillations. If we're going to take psi star psi, there is no complex part. And you do have uniform amplitude of J. And so it's sensible to say, well, it got that way because J precesses. Maybe. There is something if you say, well, maybe if I created a system at t equal 0 where J was at a particular position, and then I let that thing evolve-- this is not an eigenstate. It would be a complicated superposition. That thing would precess. And you would observe what we call polarization quantum beats. Now, that's getting way ahead of things. But it is helpful to think about J precessing on this cone. Because that gives you a sense that the length of J is conserved. And the orientation about the z-axis is not. And it's more or less uniform probability, but not uniform probability amplitude. But you don't need that for a lot of the things. If you're looking at the wave function, yeah, there's going to be some oscillation. The real part and the imaginary part will oscillate in such a way that the probability is constant about that. I finally realized that this morning. So it's not as much of a lie as you think. OK. So this is the picture. And this is what it looks like in the laboratory. And now, I'm going to prepare you for-- so suppose you had a molecule. And you have a little person standing on the molecule frame. And now, you have somebody out here observing as the molecule rotates. Well, and here's J. All information that you're allowed to know from outside the molecule comes from the projection of whatever is in the molecule frame on J. And that's what the vector model does. It tells you how to take stuff you know about the individual atoms and project it on the thing that communicates with the outside world. There are a couple of other things that I'm going to say about this. But now, let's get down to the business of actually doing a little bit of the solution of the Schrodinger equation. So for our free rotor, the potential is r equals r0 theta phi. And it's equal to 0 if r equals r0. And it's equal to infinity if it's not. So this is very much like a particle in an infinite box. In fact, you can do a really cheap solution if you say, well, let's consider a particle in an infinite circular box. Well, you can solve this problem just using the de Broglie wavelength. And you get that the energy levels for this circular box problem go as the quantum number squared. And that's almost exactly like the solutions to the free rotor. The difference is the energies for this go as the quantum number squared. And the energies for this go as the quantum number times the quantum number plus 1. It's almost the same. OK. So because the potential is constant as long as r is fixed, the Hamiltonian is just theta-- OK. And so we have to understand this kinetic energy. For 1D problems, the kinetic energy is P squared over 2 mu, the linear momentum. I should put a hat on this. Well, we have motion in three dimensions, or two dimensions, theta and phi. And so we want to do something. We want to generate a kinetic energy Hamiltonian, which is analogous to this, some kind of momentum squared over some kind of a mass. And so we can go by analogy. Or we can go back to classical mechanics. We know that the orbital angular momentum is r cross P. So these are two vectors. Cross product is a vector. We know all that stuff. We know how to write the cross product in terms of a matrix involving unit vectors and stuff like that. You've done that before. OK. So now, we have an angular motion, omega. And we would like to know what the velocity of the mass points are on a sphere. Or if we think of this as just a particle of mass mu rotating, then we want to know its velocity. And to get from the angular velocity to the linear velocity, the linear velocity is r omega. So we can say, all right, knowing-- well, for a vector cross product, if the two things-- the r and the P-- are always orthogonal, well, then we just do r times P. And in fact, for motion on this sphere, here's r. And the motion is always orthogonal to r. And so we can write that the angular momentum is m1 r1 squared omega plus m2 r2 squared times omega. Or it's just I omega. OK. And so we can write the kinetic energy term just goes as L squared over 2I where I is the sum of the individual masses M sub r I squared. OK. So we have an angular momentum operator, which is the guts of the kinetic energy. OK. Now, we're thinking Cartesian. And this is a spherical problem. And so we want to go from Cartesian coordinates to spherical polar coordinates. And that's a non-trivial problem. And it's also an extremely boring problem. And what you get is also something that's not terribly rewarding, except it's the differential equation you have solve. So T, when you go to spherical polar coordinates, is minus h squared over 2I 1 over sine theta partial with respect to theta times sine theta partial with respect to theta plus 1 over sine squared theta second partial with respect to phi. So that's the kinetic energy operator. So first of all, you say, what am I going to do? And the first thing you say is, separate the variables. So you do that. And you do the standard trick for separating variables. And you get two differential equations-- the theta equation and the phi equation. I'm sorry? AUDIENCE: H bar! ROBERT FIELD: When I wrote it, I said, why is that not H bar? OK. All right, so when you separate variables, we have this result-- 1 over phi of phi. This is the phi part of the equation. Whoops, yeah, that's right. m squared is the separation constant. So we arrange to have a differential equation that depends only on phi and another that depends only on theta. And they're equal to each other. And so we call the thing that they're equal to a constant. And we call it m squared. We like that because m can be positive or negative. And the sign of m determines whether you have an oscillating function or an exponential function. You're all familiar with that. And so this is the phi equation. And this one, you can solve easily. You know the solution to this. You could, with a little bit of thought, write down the solution and have normalized functions. So I'll just do that. So the phi part of the solution is-- it's not the order in which I wrote my notes-- 1 over the square root of 2 pi times e to the i m phi. And m is equal to 0 plus and minus 1 plus and minus 2 et cetera. Quantization comes from imposition of what we call periodic boundary conditions. The wave function has to be the same for phi and phi plus 2 pi and phi plus 4 pi. And so that gives quantization. So this is the phi part. That's simple. That's simple. It's wonderful. We understand that perfectly. And one of the nice things about the vector model is mostly that's what you're focused on. So immediately, we can draw some pictures. And we can look at the nodes in the xy plane. So m equals 0, there are no nodes. m equals 1, first of all, let's draw where phi is 0. So this is phi. And so there can be a nodal plane like that. And there could be a nodal plane like that. And already, you know this looks like an S orbital. And this looks like a Px orbital and a Py orbital. And then we could also have-- yeah, m equals 2, we could have this, and this, and that-- no node, one node, two nodes. I'm sorry, yeah. What am I doing here? This is one node. I have to be a little bit more careful here. So what I wrote down is not true here. There are going to be two nodes. So they're probably like this and like that. But it's clear that what I have in my handwritten notes-- I'm not sure what the printed notes say. But there are going to be two nodes. And so immediately, you know something about-- if you draw a reduced picture and you show the nodal structure in the xy plane, you can tell what the projection of the angular momentum quantum number is, which is really important. There's the other differential equation. And that's the theta differential equation. And that is complicated. It is an exactly solved differential equation. But one of the things it gives you, which is easily memorized-- remembered, I'm sorry-- is that the energy levels have-- where this is the overlying momentum. So you can solve this. And you can get the eigenenergies. And the eigenenergies do not depend on m. They only depend on the L quantum number. And they have this nice form. OK. Now, in my notes and in all the textbooks, there are these horrendously beautiful detailed expressions of the solution, the mathematical form to the solution of the theta equation. And it's the Legendre equation. And there's Legendre polynomials. And remember, with the harmonic oscillator, I told you, you don't ever want to look at these things. The computer will look at them. And if you have integrals, the computer will know how to do those integrals. Because you told it how. And you don't have to keep that in your head. You've got better things to do with your limited attention. So there is a solution. And it has a form, which I have decided that I won't talk about. Because you get much more insight into real problems from the vector model. Now, the vector model is really only marginally useful for a diatomic molecule in an electronic state, which you don't know about yet, which is called sigma sigma plus, where there is no angular momentum associated with the electron and when there is no angular momentum associated with the nuclei. So this is just practice for real life when you have to understand other things about what is living in the molecular frame. But this is hard enough to understand completely that it's a worthwhile investment. And you will be doing most of the understanding of more complicated problems on your own if you ever do anything with diatomic molecules or polyatomic molecules. And so, OK. So what is the vector model? Well, the things we want to know about the vector model is, how long is J? So if we have a state with J and m quantum numbers, how long is the vector associated with J? And what is the angle-- what is the projection of J on the laboratory z-axis? So we want to know the length of J. And we want to know the projection of J on z. OK. And we want to know the angle between the z-axis and J. That's the beginning of a picture. And again, we stick with this idea that J-- so here's the z-axis. And J precesses about z. And here is an angle we want to know. This theta is a different theta from the theta in-- in fact, I should call it alpha. OK. So some things we know-- we know that the eigenvalues of the rigid rotor equation are this. And we know that Jz operator operating on y Lm gives H bar m y Lm. And J squared operating on y Lm gives H bar squared J J plus 1. So what do we do with these two things? Well, one is to say, all right, the length of J is going to be the square root-- times y Lm. So the length of J is going to be the square root of H bar squared J J plus 1. So that becomes H bar and approximately J plus 1/2. Because the square root of J J plus 1 is almost exactly J plus 1/2. When you get really high J, it's exactly. At low J, It's a little bit less than exact. So we know the length of J is this. And we know the length of m is H bar m. The length of J z is H bar m. So now, we know that this is H bar m. And this is H bar times J plus 1/2. And so this is alpha. So we know how to calculate the cosine of alpha. So the cosine of alpha is equal to H bar m over H bar J plus 1/2, or m over J-- oh, right, plus 1/2, really small. So the angle, the cosine of the angle that j makes with the z-axis is m over J. And all of a sudden, now, we have lengths of things and angles of things. And we can do all sorts of stuff. Remember, what we care about is, if this is J, the body axis is like that. So we now know where the body axis is relative to the laboratory. So we have the body axis is perpendicular to J. And so we can then calculate where everything is. All right, so special cases-- if m is equal to plus or minus J, then what does that mean for this picture? Well, we have the z-axis. And we have J. And J is almost exactly along the z-axis. It would be exactly along if there wasn't this plus 1/2. And so what that means is the bond axis is almost exactly rotating in the xy plane. So if we choose m equals plus or minus J, the molecule is rotating in the xy plane. That's kind of nice to know. The other easy case is m equals 0. If m equals 0, J is perpendicular to the z-axis. And so the body axis is in the yz plane. Or if we're coming around here, it's in the xz plane. So we have xz yz. And together, what it says is the body axis is more along z than along anything else. That's a lot of insight. And so we can go to our friendly extreme cases, special cases. And we could say, where's the body axis? In the laboratory, which is where we're making the observations. That should make you happy. Because there's no equations here. There's just a lot of pictures that you can understand and develop your intuition. Now, I want a blackboard. OK. Suppose you have a laboratory in which you can make a molecular beam. And that's easy enough to do. You squirt molecules out of a pinhole. And you have some kind of an aperture. And so you have a directed flow of molecules in one direction. And you could also say, all right, I'm going to do something with my laser. And I'm going to make the molecules, which are like a helicopter, in other words, the molecular axis is rotating in a circle about the propagation axis of the beam, so a helicopter. And we can also do this. And so these are two ways that you can prepare molecules in this beam. And you could say, oh, I've got a whole bunch of molecules that are not in the beam. And these molecules are going to collide with the molecules outside the beam and get scattered. Well, which one is going to be scattered more? The one that sweeps out a big volume or the one that sweeps out something only essentially one dimension? That's insight. There's lots of other things, too. Now, let's do something which is actually slightly exam relevant. Suppose we have a diatomic molecule which is positive on one end and negative on the other, OK? And we're going to apply an electric field. And if the electric field is in this direction and the molecule is doing this, the molecule doesn't care. But if the electric field is in that direction and the molecule is doing this, the energy levels-- the energy goes up and down as the molecule rotates. Well, how is that? The molecule was free to precess. And we knew the amplitude along this axis. But if there's some angle dependence, what we're doing is we're mixing in a different J. And so that's telling you that, if you apply an electric field, there will be a Stark effect. And it will affect the m equals 0 levels differently from the m equals J levels. Now, I'm not going to tell you which one is more affected. But it's clear that one is hardly affected. And the other is profoundly affected. And that gives rise to a splitting of the energy levels in an electric field, which depends on J at m in a way that you would recognize. Time to quit. I did what I was hoping I could do. Now, there is some stuff at the end of the notes which is standard textbook material about polar plots of the wave functions. These polar plots are useful. But most of the time, when you first encounter them, you have no idea what they mean, except that they express where the nodes are. Now, you want to go back and you want to actually think about what these polar plots mean and whether you can use them as an auxiliary to this vector picture. OK. So I will then lecture, and it's exam relevant, on commutation rules on Monday. And then have a good weekend. |
MIT_561_Physical_Chemistry_Fall_2017 | 8_Quantum_Mechanical_Harmonic_Oscillator.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: So today we're going to go from the classical mechanical treatment of the harmonic oscillator to a quantum mechanical treatment. And I warn you that I intentionally am going to make this look bad, because the semi-classical approach at the end of this lecture will make it all really simple. And then on Monday I'll introduce creation and annihilation operators, which makes the harmonic oscillator simpler than the particle in a box. You can't believe that, but all right. So last time we treated the harmonic oscillator classically and so we derive the equation of motion from forces equal to mass times acceleration, and we solved it. And we saw that we have this quantity omega, which initially I just introduced as a constant, which was a way of combining the force constant and the mass. And then I showed that the period of oscillation is 1 over the frequency, which is 2 pi over omega. Now one of the things that people have trouble remembering under exam pressure is turning points. And this comes when the energy is equal to the potential at a turning point. And since the potential is 1/2 kx squared we get the equation for the turning point at a given energy, which is equal to plus or minus the square root of 2e/k. Now when you're drawing pictures there are certain things that anchor the pictures, like turning points. And now at the end of the previous lecture I calculated the classical mechanical average, and we use we use this kind of notation in classical mechanics. Sometimes we use this notation, too, but this is what we mean by the average value in quantum mechanics. And we found that this was the total energy divided by 2. And the average momentum is the energy divided by 2. And that's the basis for some insight, because as a harmonic oscillator moves it throws energy back and forth between kinetic energy and potential energy. And then one of my favorite things, and one of my favorite tortures on short answers, is the average values of-- And does anybody want to remind the class the easy way, or the two easy ways to know what this is? And the person who answered the question last time is disqualified. Yes. AUDIENCE: There is the symmetry method. ROBERT FIELD: What? AUDIENCE: There's symmetry. So if the system's symmetrical that the average value will actually be that symmetry point. ROBERT FIELD: Yes. So and the other is that the harmonic oscillator isn't moving, and so there is no way that the coordinate-- that the average value of either the coordinate or the momentum could be different from 0. However you do it you want to be doing it in seconds, not minutes. And certainly not by calculating integral. And now x squared and p squared, they're easy too. Especially if you know t and v. And so we then use those to calculate the variance. And that's defined as the average value of the square minus the square of the average square root. And what we find is that the variance of x times the variance of p is equal to e over omega. So as you go up in energy this joint uncertainty increases. And we'll find that that also is true for quantum mechanics. So this is sort of the kind of questions you want to be asking in quantum mechanics. And you want to be able to be guided by what you know from classical mechanics, and you want to be able to do it fast. Today's menu is what I would call-- This lecture is gratuitous complexity. Does everybody know what gratuitous means? This is one of my favorite Bobisms. And you'll hear other Bobisms during the course of this course. What I want you to be able to do for a lot of mechanical problems is to know the answer, or know what things look like without doing a calculation. In particular, not solving a differential equation or evaluating in the integrals. You want to be able to draw these pictures instantly. Now in the modern age everyone has a cell phone, and one could have a program in there to calculate what anything you wanted for harmonic oscillator. But chances are you won't be prepared for that. And if you want to have insights into how do various things you want to know about harmonic oscillators come about? You need the pictures, as opposed to the computer program. Now the pictures involve an advance investment of energy. You want to understand every detail of these little pictures. I'm going to write this runner equation, I'm going to clean it up to get rid of units which makes it universal. So it becomes a dimensionless equation. And the unit removal, or the thing that takes you from a specific problem where there's a particular force constant and a particular reduced mass, and makes it into a general problem. There is one or two constants that combine those things. And you've taken them out at the end of a calculation. If you need to have real units you can put those back in. And that's a very wonderful thing, and that enables us to draw pictures without thinking about, what is the problem? And then the solution of this differential equation-- which is actually quite an awful differential equation at least for people who are not mathematicians-- and the solution can be expressed as the product of a Gaussian function which goes to 0 at plus and minus infinity, so it makes the function well behaved. Times something that produces nodes, a polynomial. Does anyone want to give me a definition of a polynomial? Silence. OK, your turn. AUDIENCE: It's the linear combination of some numbers taken to different power ROBERT FIELD: Right. A sum of integer powers of a variable. And when we take a derivative of a polynomial we reduce the order of the polynomial. A little bit of thought, if you have a first order polynomial there'll be one node. If there is a second order there'll be two nodes. And nodes are very important. And so when we're going to be dealing with cartoons of the wave function, and then using semi-classical ideas to actually semi-calculate things that you'd want to know, the nodes are really important. And what's going to happen Monday is we'll throw away all this garbage and we will replace everything by these creation and annihilation operators. Which do have really simple properties, which you can use to do astonishingly complicated things without breaking a sweat. And the final exam is in this room on the first day of exam period, at least it's on a Monday, and it's in the afternoon. In the non-lecture part of the notes I replaced the mass for one mass on a spring by the reduced mass, which is m1 m2 over m1 plus m2 for two masses connected by a spring. And I go back and forth between using mu and m, and that's OK. All right so in the notes the differential equations in the first few pages are expressed as partial differential equations. They're total differential equations. That'll get changed. The Hamiltonian is t plus v, and in the usual form t is p squared over 2 mu. And so we get minus h-bar squared over 2 mu par-- not partial, I'm so used to writing partials that I can't stop. Second derivative with respect to x, plus 1/2 kx squared. So that's the Hamiltonian. Now that looks kind of innocent, but it isn't. And so the first thing we want to do is get rid of the dimensionality, the units. So this is xi and it's defined as square root of alpha times x. Where alpha is defined as k mu square root over h-bar. Now it would be a perfectly reasonable exam question for you to prove that if I take this combination of physical quantities this will have dimension of 1 over length. That makes xi a dimensionless quantity. I'm not even going to bother going through the derivation. The Hamiltonian becomes h-bar of omega times 2, minus second derivative with respect to xi plus xi squared. So this is now dimensionless. This has units. We divide by h-bar omega to make now everything dimensionless. And we get a differential equation that has the form minus the second derivative with respect xi plus xi squared, minus 2e over h-bar omega, times the wave function, expressive function of xi, not x. This is the differential equation we want to solve, and we don't do that in 5.61. You're never going to be asked to solve a differential equation like this. But you're certainly going to be asked to understand what the solution looks like, and perhaps that it is in fact a solution. But that's still pretty high value stuff, though you wouldn't really have to do that. This is the simplest way of writing the differential equation and it's dimensionless. The standard way of dealing with many differential equations is to say, OK, we have some function and it's going to be written as an exponential, a Gaussian, times some new function. And for quantum mechanics this is perfectly reasonable because we have a function in a well and the wave functions have to go to 0 at plus minus infinity. And this thing goes to zero at plus and minus infinity pretty strongly. So it's a good way of building in some of the expected behavior of the solution. And that's perfectly legal, and it just then defines what is the difference equation remaining for this? And it turns out, well we're going to get the Hermite equation. And this will be a Hermite polynomial, the solutions. Now one way of dealing with this is to simply say, well, we know the solution of this differential equation if this term weren't there. Because this is now the equation for a Gaussian. So building in a Gaussian as a factor in the solution, it's a perfectly reasonable thing. And then we have to say, what happens now when we put this term back in? And when we do we get this thing. Second derivative with respect to this polynomial-- I mean of this polynomial is equal to minus 2 xi times the hn d xi plus 2n hn. This is a famous differential equation, the Hermite equation, which is of no interest to us. And it generates the Hermite polynomials. These things are the Hermite polynomials. And they are treated in some kind of sacred manner in most of the textbooks, and I think that's really an offense because, well, we're not interested in mathematical functions. We're interested in insight and this is just putting up another barrier. Now with this equation, you can derive two things called recursion relations, and one of them is the derivative of this polynomial with respect to xi is equal to 2n times hn minus 1 of xi. Now that's not a surprise, because this is a polynomial. If you take a derivative of the variable you're going to reduce the power of each term by 1. Now it just happens to be lucky that when we reduce it to 1 you don't get a sum of many different lower order polynomials, you just get 1. And there is another one, another recursive relation, where it tells you if you want to increase the order you can do this, you can multiply hn by xi. And that's obviously-- it is going to increase the order, but it might not do it cleanly. And it doesn't. And so we get-- We have a relationship between these three different polynomials. Now it turns out that these two equations are going to reappear, or at least their progeny will reappear, on Monday in terms of raising and lowering operators. And what you intuit about what happens if you multiply polynomial by the variable? Or what happens if you take its derivative? And it is very simple and beautiful, but I don't think this is very beautiful for our purposes as chemists. And one of the things that these recursive relationships do, which also hints at what's to come on Monday, is that we can calculate integrals like this. That's not what I want. This is a quantum number, it's an integer. It's v, not nu, and multiply it by x to the n, p to the m, psi that should be complex conjugated, v plus l, dx. It turns out for almost everything we want to do with harmonic oscillators we're going to want to know a lot of integrals like this. And one of the things we like is when an integral is promised to be 0 so we don't ever have to look at it. And so there are selection rules. And the selection rules for this kind of integral is l is equal to m plus n, m plus n minus 2, down to minus m plus n. So the only possible non-zero integrals of this form are for the change in quantum number by this l, which goes from m plus n down to minus m plus n in steps of two. The two shouldn't be too surprising, because there is symmetry and we have odd functions for odd quantum numbers and even functions for even quantum numbers. And so something like this is going to have a definite symmetry and it's going to change things within a symmetry, and so it's going to change the selection rule in steps two. Now you don't know what selection rules are for, or why you should get excited about these sorts of things, but it's really nice to know that almost all the integrals you are ever going to face for a particular problem are zero. And you can focus on a small number of non-zero ones, and it just turns out that the non-zero ones have really simple value. There also exists what's called a generating function, which is the Rodriguez formula. And that is the hn of xi is equal to minus 1 to the n, e to the psi squared, the derivative with respect to xi, e to minus xi squared. We have one that has a positive exponent, and one has a negative exponent, and we have this. So we could calculate any Hermite polynomial using this formula, which you will never do. But it's treated with great fanfare in textbooks. Now the solution to the harmonic oscillator wave function in real units, as opposed to dimensionless quantities, is-- and I'm just writing this down because I never would ever think about it this way, but I have to at least provide you with guidance-- so we have a factor 2 to the v, again this is v. Now the reason I'm emphasizing this is that in all texts v quantum numbers are italicized. And if you've thought about it for a minute, an italic v-- for mortals-- looks like a nu. It isn't quite. I don't know what the difference is, but if you have them side by side they are different. And so a large number of people who should know better refer to the vibrational quantum number as nu, which marks that person as, well, I won't say but it's not complimentary! We have this factor to the square root, and that's a normalization-- oh we got another part of it. Alpha over pi to the 1/4 power. So that's normalization. Then we have the Hermite polynomial. And you notice I've got xi back in here, which is really a shame. And we have-- So this is the general solution, we have the exponentially damped function, we have polynomials. These are all the actors that we're going to have to deal with. And I promise you, you will never use this unless you want to program a computer to calculate the wave function for God knows what reason. The quantum numbers we are 0, 1, 2. And for a harmonic oscillator, which goes to infinity, v goes to infinity, too. There's an infinite number of eigenfunctions of the quantum mechanical Hamiltonian-- of quantum mechanical harmonic oscillator. We have-- and the functions are normalized, we have psi plus and minus infinity goes to 0, we have psi v of 0. So this for all the-- I put a v there. Psi v is 0 for odd v, derivative of psi with respect to x, at x equals 0 is 0 per even v. So we have symmetric functions. And we have the energy levels is equal to h-bar omega, v plus 1/2. Now this is h over 2 pi, and this is nu times 2 pi, the frequency times 2 pi. So it could also be h nu. I have trouble remembering when there is a 2 pi involved. And we have this wonderful thing. It's as if v prime is equal to v this integral is 1, it's normalized. And a v prime is not equal to v, it's zero. And that stems from a theorem I mentioned before, is if you have two eigenvalues of the same Hamiltonian, eigenfunctions of the same Hamiltonian, and they belong to different eigenvalues their overlap integral is 0. We like zeros. We like normalization because the integral is just 1, it goes away. Or the integral is 0, the whole thing goes away. So that's really good. So we call this set of v's is orthonormal. Orthogonal and normalized. And the orthonormal terminology is used a lot. And in almost all quantum mechanical problems we like using an orthonormal set of functions to solve everything. Sometimes we have to do a little work to establish that, and I'll show much later in the course how when you have functions that are not orthogonal, and not normalized, you can create a set of functions which are. And this is something that a computer will do without breaking a sweat. Now we're back to my favorite topic, semi-classical. Because it's really easy to understand. Not just to understand the harmonic oscillator, but to use it in many problems. So in classic mechanics the kinetic energy is e minus v of x, or p squared over 2 mu. And so we can derive an equation for p of x classical mechanically, which is 2 mu e minus v of x square root. This is an extremely useful function. It's not an operator. It's a thing that we're going to use to make sense of everything, but it's not an operator. And so this is classic mechanics, and then in quantum mechanics, we know that Mr. de Broglie told us that the wavelength is equal to h over p. And we can generalize and say, well, maybe the wavelength is a function of x for potential, which is not constant. And even though this is not an operator in quantum mechanics this is true. That you can say the distance between consecutive nodes is lambda over 2. We can use this node relationship to great advantage. For the pair of nodes closest to x we can use this to calculate the distance between them. Very valuable. Because I also want to mention something. If you have an integrand which is rapidly oscillating, or if you have two rapidly oscillating functions and you're multiplying them together, that integral will accumulate to its final value at the position where the two oscillating functions are oscillating at the same frequency. That's the stationary phase point. And this is also a wonderful thing, because if you can figure out where the things you're multiplying together are oscillating at the same frequency, your integral becomes a number. No work ever. And that's a useful thing. OK, so the stationary phase method enables you to use this in a really fantastic way. And it's a little bit like Feynman's path integral idea, that you can calculate a complicated thing by evaluating an integral over a convenient path as opposed to integrating overall space, because everything that you care about comes from a stationary phase. Quantum mechanics is full of oscillations, classical mechanics doesn't have oscillations, and the two meet at the stationary phase point. Now we're going to use these ideas to calculate useful stuff for quantum mechanical vibrational wave functions. The shapes of psi of x gets exponentially damped, but it extends into the classically forbidden e less than v of x regions. The wave function, if we have potential and we have a wave function, that wave function is going to not go to 0 at the edge but it's going to have a tail. And that tail goes to 0 at infinity. And so there is some amplitude where the particle isn't allowed to be, classically. And that's where tunneling comes in. But the important thing, the important insight is that there are no nodes in the classically forbidden region. There is only exponential decay towards 0, and if you've chosen the wrong value of the energy, in other words a place where there is no eigenfunction, the wave function in the classically forbidden region will usually go to infinity. Either over here or over here, and says, well, it's clearly not a good function. But there are no 0 crossings. It's oscillating in e greater than v of x, the classically allowed region. The number of nodes is v. So we can have a v equals 0 function that just goes up and goes down, no internal nodes. v equals one, it crosses zero right in the middle. And we have the even oddness. Even v, even function. Odd v, odd function. For an even function you have a relative maximum at x equals 0, and for an odd function you have a 0. And the opposite further derivatives. The outer lobes, the ones on the ends just before the particle encounters the classical wall, you get the maximum amplitude. And so you can draw cartoons which look sort of like that. Most of the valuable stuff is at the other turning point. And there's oscillations in between, but often you really care about these two outer lobes. That's a pretty good simplification. Now there's a nice picture in McQuarrie on page 226 which shows, especially for psi squared, that the nodes are pretty big. But they're not as big for relatively low quantum numbers as I've implied. But at really high quantum numbers we have a thing called the correspondence principle. And the corresponding principle says that quantum mechanics will do what classical mechanics does in the limit of high quantum numbers. And in the limit of high quantum numbers essentially all the amplitude is at the turning points. And in classic mechanics the particle is moving fast in the middle, and stops and turns around, and essentially all of the amplitude is at the turning point. So this is nice. Now we're getting into Bobism territory, because I'm really going to show you how to calculate whatever you need using these semi-classical ideas. We have the probability envelope. Psi star of x psi of x, and we're going to have both of these having the same quantum number, dx. So this is the probability of finding the particle near x in a region with dx. And this is the same thing as dx over v classical. It's not the same thing, there is a constant here, sorry. This probability density that you want is basically 1 over the classical velocity. And I demonstrated that when I walked across the room, when I walked fast in the middle and slow the outside. And you get the probability, you get this constant, by saying, OK, how long did it take for me to go from one end to the other? And comparing that, how long it took for me to go in some differential position. You get this constant in a simple way. v classical is equal to p classical over the mass, over the mu. But we know the function for p classical. We have 1 over mu times 2 mu e, minus v of x. So we know the velocity everywhere, and there's nothing terribly hard about figuring that out. And now we want to know what this proportionality constant is. And so for that we say the time to go from x to x plus dx, over the time to go from x minus to x plus. Because what's happening, the particle is going back and forth inside this well and so this is the time it takes to go one pass, and this is the time it takes to go through the region of interest. And so this ratio is the probability. And so we have the probability moving from left turning point the right turning point, and we want to know the probability in that interval. And so that's just dx over v classical at x, over tau, over 2. Because tau is the period, and we have half of the period, and so it's all together. I'm going to skip a little step, because it's taking too long. Psi star psi dx is equal to k over 2 pi squared, e minus v of x square of dx. Now so if we know the potential, and we know the energy, and we know the force constant we can say, well, this is the probability and-- but this is the probability of the semi-classical representation of psi star psi dx at x. Now this is oscillating, and this is not. So what we really want to know-- here's the-- if we have psi star whoops, slow down. This is oscillating and what we've calculated before is something that looks sort of like that. And if we multiply by 2 we have a curve that goes to the maximum of all these oscillations. The envelope psi star psi has the form. We've multiplied by 2, and so we end up getting 2k over pi squared, e minus v of x square root. Now you might say, well these are complicated functions, why should I bother with them? But if you wanted anything starting from the correct solution to the harmonic oscillator, using the Hermite polynomials, there's a whole lot more overhead. Notice also this is a function of x, not xi. This is the overlap function, it's the curve that touches the maximum of all these things and it's very useful if you want to know the probability of finding the system anywhere. And we get the node spacing from the equation h over p of x. And now here comes something really nice. It's called the semi-classical quantization interval. If we have any one dimensional potential, and we're at some energy, we'd like to be able to know how many levels are at that energy, or below. Or where are the energy levels? And we get that from this really incredible thing. We want to know that-- this is the difference between nodes. Right? And now if we would like to know x minus to x plus and some energy, we can replace lambda of x by h over p of x. And so we get p of x at that energy over h dx. pdx, that's called an action integral. Now I have to tell a little story. When I was a senior at Amherst College we had a oral exam for whether my thesis was going to be accepted or not. And one of my examiners asked me, what is the unit of h? Well, it's energy times time. And he wouldn't stop. He said no, I want something else. It's called action. Momentum times position. This is an action integral. And so anyway that's just a story. I spent a half an hour, and I was damn stubborn. I was not-- you know, it was energy times time. But that is much more insight here and that's maybe why I got so excited about this sort of an integral. If we want to know if we have an eigenvalue this integral has to be equal to h over 2 times the number of nodes. Well it's pretty simple. So we can adjust e to satisfy this. Or if we wanted to know how many energy levels are at an energy below the energy we've chosen we evaluate this integral, and we get a number like 13.5. Well, it means there are 13 energy levels below that. Now often you want to know the density of states, the number of energy levels per unit energy, because that turns out to be the critical quantity in calculating many things you want to know. And you can get that from the semi-classical quantization. We're close to the end, I just want to say where we're going. We have classical pictures and I really, really want you to think about these classical pictures and use them rather than thinking, well, I'm going to have my cell phone program to evaluate all the necessary stuff. And there are certain things you want to remember about this semi-classical picture. And now we have the ability to calculate an infinite number of integrals involving harmonic oscillator functions in certain operators. Well, la-di-da. Why do we want them? Well one of the things we want is to be able to calculate the probability of a vibrational transition. That's called a transition element, and that's an easy thing to calculate. Another thing we want to do is to say, well, nature screwed up. This oscillator isn't harmonic, there's anharmonic term, and I would like to know what is the contribution of a constant times x cubed in the potential to the energy levels. And that's called perturbation theory. Or I want to have many harmonic oscillators in a polyatomic molecule, they talk to each other, and I want to calculate the interactions between these harmonic oscillators affect the energy level. Remember, when we have a separable Hamiltonian we can just write the energy levels as the sum of the individual Hamilton. And then there's coupling terms, and we deal with those but perturbation theory. There's all sorts of wonderful things we do. But we're going to consider these magical operators, a creation and annihilation operator, where a star-- a dagger operating on psi v gives the square root of v plus one, times psi v plus 1. And a dagger operating on v gives the square root of v-- let's write it-- I mean a non-dagger gives v square root times psi, v minus 1. And that x is equal to a plus a dagger times a constant. And so all of a sudden we can evaluate all integrals involving x, or powers of x, or momenta, or powers of momenta without thinking. Without ever looking at a function. And I guarantee you that this is embodied in the incredible amount of work done wherever we pretend almost every problem is a harmonic oscillator in disguise. Because these a's and a daggers enable you to generate everything without ever converting from x to xi, without ever looking at an integral. It's all just a manipulation of algebra. And it's not just convenient, but there's insight and so this is what I want to convey. That you will get tremendous insight. Maybe, maybe I sold you on semi-classical, and I don't apologize for that because that's very useful. But the next lecture, when you get the a's and a daggers, it'll just knock your socks off. OK, that's it. |
MIT_561_Physical_Chemistry_Fall_2017 | 10_The_TimeDependent_Schrödinger_Equation.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: The outline for today is a little bit more review of feeling the power of the creation and annihilation operators. They enable you to do anything with harmonic oscillators, really fast. And so you don't spend time thinking about how to do the math. You think about the meaning of the problem and understanding how to manipulate, or how to understand harmonic oscillators. And then the exciting thing. The real Schrodinger equation is not the one we've been playing with. It's the time dependent Schrodinger equation. And I like to introduce the time independent one first, because it enables you to sort of develop some insight and assemble some of the tools before we hit the really serious stuff. And so I will talk about this and use the time dependent Schrodinger equation to show what it takes to get a motion of this product of the psi star psi. What does it take? It takes a superposition state consisting of at least two different energies. Then this normalization integral-- well, you'd expect normalization, or something like this to be preserved. And in fact, the normalization is time independent. Then, if we calculate-- now, this is a vector rather than just a simple coordinate. If we calculate the expectation value of the position, and the expectation value of the momentum, Ehrenfest's theorem tells you that this is going to describe the motion of the center of the wave packet. And the motion of the center of the wave packet is following Newton's laws. No surprise. And so we're seeing how classical mechanics is given back to you once we start making particle-like states. We can see how these particle-like states evolve. But there is a lot more than just particle-like behavior. You would, perhaps, challenge any outfielder for the Red Sox to go beyond calculating the center of the wave packet. That's what they know to do. But they don't know some other things like survival probability, how fast does the wave packet move away from its birthplace, or the wave packet does stuff and wanders around and comes back and rephases sometimes. And sometimes it just dephases. And these are words that you're going to want to put into your vocabulary. So there's a lot of really beautiful stuff when we start looking at the time dependent Schrodinger equation. And we are going to mostly consider the time dependent Schrodinger equation when the Hamiltonian is independent of time, because we can get our arms around that really easily. But when the Hamiltonian is dependent on time, then it opens up a world of complexity that is really best left to a more advanced course. But how do molecules get excited from one state to another? A time dependent Hamiltonian. So we are going to have to at least briefly talk about a time dependent Hamiltonian and what it does. And you'll see that eventually. I'm first going to do a little bit of review of the a's and a daggers. So the most important thing is, we're going to have some problem where we have this x operator to some integer power. And so we need to be able to relate that to the a's and a daggers. I'll drop the hats, because we know they're there. And one of the nice things is, if we have this operator to the nth power, we have this co-factor to the nth power. So we don't ever have to deal with it until the end of the problem when we say, oh, wow, we had x to the n. Well, we have then n h bar. I'm sorry. We have h bar over 2 mu n n over 2 power. So it really saves a lot of writing. And it means you solve one problem for a harmonic oscillator. And you've solved it for all harmonic oscillators. And this is just details about what is the mass and what is the force constant, which you need to know. And there is a similar thing for p. So that opens the door. It says, OK, we have some problem involving the coordinate and the momentum, some function of the coordinate and momentum, and we want to know things about the coordinate and momentum. And so we use this operator algebra. And so a operating on psi v gives v square root v minus 1. And a dagger, we know that. So that's hardwired. But suppose we wanted a to the 5th, operating on psi v. Well, what do we do? Well, we start, and we start operating on... You know, if this were a complicated operator, we would take the rightmost piece of that operator and operate on the wave function. And that will give us v and then v minus 1 and then v minus 2, v minus 3. I've got five of them. 1, 2, 3, 4, 5. v minus 4. Square root psi v minus 5. OK. It's mechanical. You don't remember this. You generate this one step at a time. And it's automatic. And so it doesn't stress your brain. You can be thinking about the next thing while you're writing that garbage. We have this number operator, which is a friend, because it enables you to just get rid of a bunch of terms. The number operator is a dagger a, and the number operator operating on psi v gives v psi v. If we want any harmonic oscillator function, we can operate on psi 0 with a dagger to the v power. And then we have to correct, because you know that this operating v times on this will give psi v. But it will also give a bunch of garbage, right? And you want to cancel that garbage. And so you write v factorial minus 1/2. And so that gives you the normalized function. So these are really, really simple things. And most of them, once you've thought about it a little bit, you can figure out yourself. OK. Now, many problems involve x, x cubed, x to the 4th. And we know that x is this. And x squared is this squared. And x cubed is this cubed. And so we have to do a little algebra to simplify. And what we want to do is simplify to sum of terms according to delta v selection rule. So there is some algebra that we do, when we have, say, a dagger, a dagger, a, a, a dagger. But we know that three a daggers and two a's means delta v of plus 1. And that came from x to the 5th power. But you get a lot of terms from x to the 5th power. And you have to simplify them. And in order to do that, you use this commutation rule-- a, a dagger is equal to 1. And that rearranges terms. So all of the work you do when you're faced with a problem involving integrals of integer powers of x and p for a harmonic oscillator is playing around with moving the a's and a daggers around, so that you have all the terms that have the same selection rule compressed into one term. That's the work. It's not much. And once you've done it for x squared, x cubed and x 4th, you've done it as much as you'll ever need to do. And that's it. That's the end. OK. Now here's an example of a problem that's a little bit tricky. And it's sort of right at the borderline of what I might use on the exam. So we have this, we have a dagger to the m power. A to the n power. Psi. OK. Now, here. What v is going to give a non-zero integral? We have a dagger to the m. And we have a to the n. And so that's going to be v minus n plus m, right? Because we lose n quanta, because of this. And we gain m quanta because of that. And now, well, that's good. You've used the selection rule. Now, how do we write out this interval? And there is a little bit of art there too. Because we have a whole bunch of terms. We have n plus m terms in the square root. And how do you generate them without getting lost? Yes? AUDIENCE: I think you have it backwards. Shouldn't it be plus n minus n? It means you're going to lower it n times. PROFESSOR: Yes. Yes. I wonder what I had in my notes. This is wrong. OK. So this has to withstand this and this. And so, yes. OK. However you remember it, you've got to do it right. And now we have the actual matrix element. And so the first term is going to be what does a n do to that? So you start on the right. And you start building up this way. And so the first term is going to be, what does a do to this? Well, it's going to leave it alone. But it's going to lower v. So we have v plus n minus m. And then we have v plus n minus m minus 1, et cetera, until we have n terms. And then we start going back up, because we're now dealing with this. And so I'm not going to write the rest of this. Maybe this is going to be a problem I start the exam with. So you don't want to get lost. And the main thing is, you have these operators. And you start operating on the right, and one step at a time. And this is a little tricky because you're changing the quantum number, and you're changing the wave function. And you have to keep both in mind, but you're only writing down this. OK? I want to save enough time so that we can actually do the time dependent Schrodinger equation. The time dependent Schrodinger equation. H psi. It still looks pretty simple. Instead of e psi here, we have this thing. This is the time dependent Schrodinger equation. This is it. This is quantum mechanics. Everything that comes from quantum mechanics starts with this. When we don't care about time, we can use the time independent Schrodinger equation. But when we do care about time, we have to be a little bit careful. So this is the real Schrodinger equation. And notice that I'm using a capital psi, rather than a lower case, or less decorated psi. And so this is usually used to indicate the time dependent Schrodinger equation. It's time dependent wave function. This is used to indicate the time independent equation. Now, if the Hamiltonian-- and this is wonderful-- if the Hamiltonian is independent of time, then if we know the solutions psi n en. If we know all of these solutions, then there's nothing new. We just are repackaging the stuff that we know from the time independent Schrodinger equation. So the first thing I want to do is to show you that if the Hamiltonian is independent, we can always write a solution to the time dependent Schrodinger equation-- e to the minus i e n t over h bar psi n x. So this, for a time independent Hamiltonian, is always the solution of the time dependent Schrodinger equation. So we're just using this stuff that we know, or at least we barely know, because we just started playing the game. But we can manipulate to see all sorts of useful stuff. OK. So let's show if this form satisfies the Schrodinger equation. So we have ih bar partial PSI with respect to t. So we get an ih bar. And then we take the partial with respect to t. This is independent of time. This has time [INAUDIBLE] And so we get a minus i e n over h bar. And then we get e to the minus i e n t over h bar psi n of x. Well. So, let's put this together. We have an i times i times minus 1. So that's plus 1. We have an h bar in the numerator and h bar in the denominator. That's 1. And so what we end up getting is e n e to the minus i e n t over h bar psi n. Well, this is psi. This function here is the solution to the time dependent Schrodinger equation. How do we know that? Well, we have this factor, h psi. Well, h doesn't operate on either the i e n t. So what we have is that we get E n e to the i omega t. Sorry. I'm jumping ahead. i e n t over h bar psi. All right. So, if we apply the Hamiltonian to this function, we get just e n times the function. And that's what we got when we did i h bar times a partial with respect to t. So what that shows is that this form always satisfies the time dependent Schrodinger equation, provided that the Hamiltonian is independent of time. Now, that's a large range of problems, things that we need to understand. But it's not the whole potato, because the Hamiltonian is often dependent on time. But it enables us to build up insight, and then treat the time dependence as a perturbation. And we're going to do perturbation theory in the time independent world. And then we're going to do perturbation theory a little bit in the time dependent world. OK. So what we always do here is we solve a familiar problem, and then we say, OK, well, there's something more to this familiar problem. And so we treat that as something extra. And we work out the formalism for dealing with that extra thing. But before we do the extra thing, we have to really kill the problem that is within our grasp. OK. So now, our job is to just explore what we've really got here. So the first problem is motion. So we have psi star x and t psi x and t. So we have this thing, which we're not going to integrate yet. Well, when is this thing going to move? Well, the only way this probability density is going to evolve in time is going to be if we have a wave function, psi, which is of x and t is equal to c1 psi 1, e to the minus i e 1, e over h bar, plus c2, psi 2, e to the minus i e 2, t over h bar, where e1 is not equal to e2. So this is the first, most elementary step. And remember, we have this notation e to the i something. And e to the minus i something. And when we take a plus 1 and a minus and put them together, we get 1. So this notation, this exponential notation, is really valuable. OK. So now let's just look at this quantity, psi star psi. OK. Psi star psi. Well, it's going to be c1 squared, psi 1 squared-- square modulus-- and we get c2 square modulus, psi 2 square modulus. Are we done? No. So, what we did is, I put in c1, psi 1 plus c2, psi 2. And I looked at the easy terms, the terms where the exponential factor goes away. And then there's two cross terms. Those two cross terms are c1 star-- whoops-- c2 e to the minus i e 2 minus e1 t over h bar psi 1 star, psi 2. And we have c1, c2 star e to the plus i e2 minus e1 t over h bar, psi 1 star. psi 1, psi 2 star. Now, this is just the automatic writing. OK. So we have two terms that are time independent. So, this is no big surprise. But then we have this stuff here. And if we say e2 minus e1 over h bar is omega 2, 1, everything becomes very transparent, because now we have something that looks like it looks like it's trying to be a cosine omega t. We have to be a little bit careful. These are the two time dependent terms. And they are the complex conjugate of each other. And so we know that if we have two complex numbers, c plus c star, we're going to get twice the real part of c. So this enables us to take these two terms and combine them. And we know it had better be real, because we're talking about a probability here. This probability has got to be real. And it's got to be positive. It's got to be real and positive everywhere and forever. Because there's no such thing as a negative probability. There's no such thing as a negative probability in a little region of space, and we say, well, we integrate over all space, and so that goes away. This psi star psi is going to be real everywhere for all time. And so that's a good thing, because we have a sum of a plus its complex conjugate. And so this is real. And we can simplify everything, and we can write it simply. But suppose we choose a particular case-- c1 star is equal to c1, which is equal to 1 over square root of 2. And we can say psi 1 and psi 2 are real. When we do that, then this complicated-looking thing, psi star psi, becomes 1/2 psi 1 squared, plus 1/2 psi 2 squared, plus cosine omega 1 2 t psi 1 psi 2. OK. Well, these two guys aren't moving. And they're real. And positive. Yeah. AUDIENCE: Wait, is c2 also 1 over square root of 2? PROFESSOR: I had 1 over square root of 2, one over square root of 2. That's 1/2. But then when you add the two terms, you get twice the real part. So we get a 1/2 times 2. And so this is not a mistake. It comes out this way, OK? AUDIENCE: I think you meant c2. You're only finding c1 star and c1 to be 2 over square root-- 1 over square root of 2. You didn't say anything about c2. PROFESSOR: Oh. OK. I want c1 and c2. That's what I wanted. OK. Thank you. That shows you're listening. And it shows that I'm sufficiently here to understand your questions, which is another wonderful thing. So this is now, we have something that's positive everywhere. It's time independent. And we have something that's oscillating. And so this term can be negative. But you can show-- I don't choose to do that. You can show that this term is never larger than psi 1 squared plus psi 2 squared. And so even though this term can be negative at some points, it never is a negative enough to make the evolving probability go negative. Now, you may want to play with that just to convince yourself. And it's an easy proof. And I'm just not going to do it. OK. So we have something that says, just like for the wave equation, what does it take to get motion? And to get motion you had to have a superposition of two waves of different energy or different wave vector. And so here we have, if e1 is not equal to e2, we have a non-zero omega 1, 2. It doesn't matter whether it's omega 1, 2, or omega 2, 1, because it's cosine. And so that's motion. So we get a standing wave sort of situation. And then we get this motion. OK, now, the fun begins. Suppose we want to calculate the expectation value of x and p. OK. And let's just take it as a 1D problem. And so x of t. No, there was no question. All right. We have-- OK. And again, we take this thing apart. And we say, all right, suppose we have the same superposition of psi is equal to c1. psi 1. OK. Actually, these should be the time independent. So what we get is, when we do this integral, we get c1 squared integral psi star x, psi 1 dx. And we get c2 squared integral psi 2 star x, psi 2 dx. And then we get cross terms. c1 star, c2. e to the minus i omega 2, 1 t times the integral psi 1 star x psi 2 dx. And we have c1, c2 star, e to the plus i omega 2 1 t integral psi 2 star x psi 1 dx. A lot of stuff here. Now, we're talking about the harmonic oscillator. This integral is 0. Because x is a plus a dagger. The selection rule is delta v of plus and minus 1. This one is 0. For the particle in a box, one can also ask, what about this integral? And for the particle in a box, it's a little bit more complicated. Because we've chosen a mathematically simple way to solve the particle in a box, with the box having a zero left edge. If we make the box symmetric, then we can make judgments and say, oh, yeah. This integral is also 0 for the particle in a box. And that's a little bit more complicated, the argument, than what I just did. So these two terms are 0. And now we have motion of the expectation value, which is described by these two terms. And again, they're the complex conjugate of each other. And so what we have is x of t is equal to twice the real part of c1 star, c2 e to the minus i omega 2, 1 t. Times x 1, 2. OK. This x 1, 2 is an integral. So x 1, 2 if these are the vibrational quantum numbers, well, then this is non-zero. This is just the square root of 2, or square root of 1. OK. So we have motion described by this. And so the only time we get motion is if v1 is equal to v2 plus or minus 1, for the harmonic oscillator. For the particle in a box, there's different rules. And often, for these simple problems, you want to go through in your head all of the simple cases. Now, we already can see that-- I don't want to talk about the particle in a box. So now let's just take another step. And we're going to have Ehrenfest's theorem, which you can prove, says that m times the derivative of vector p-- so this is a time dependent expectation value-- is equal to-- I'm sorry. This is not vector p. This is vector r-- is equal to-- So this is the vector p. So this is the derivative of the coordinate, with respect to time. That's velocity. Velocity times mass is momentum. And so we have a relationship between the expectation value of the position and the expectation value of the momentum. And that's for Newton's first law. And then there is another Newton's equation translated into quantum mechanics. So we have the expectation value of the momentum, dt, is equal to minus the expectation value of the potential. While this is acceleration times mass, and this is force-- minus the gradient of a potential is the force-- these are Newton's two equations. And what they're saying is, if we know these things, we know something about the center of the wave packet, and how the center of the wave packet moves. Now, the wave packet might be localized at one time. It might be mushed out at another time. But you can always calculate the center of the wave packet. It's just that there are only certain times that it looks like a particle. But this thing, these quantities, which you define by an integral, they evolve classically. So I told you at the beginning, you had to give up classical mechanics. It's all coming back. But it's coming back in a quantum mechanical framework, because we're talking now about wave functions, which have amplitudes and phases, and can do terrible things. But at certain limits, they're going to act like particles. But if you were to ask a question, well-- suppose we do this experiment. And so here we have an electronic ground state potential surface. Now, I'm jumping way ahead. But this is the wave function for that vibrational level. And you excite the molecule with a time dependent Hamiltonian, a time dependent radiation field light. And you vertically transport this wave function to the upper state, until you get something which is not an eigenstate. It's a pluck. This pluck is a superposition of eigenstates. And we can ask, how does this evolve? And what it's going to do is, it's going to start out localized. And at some point, it'll do terrible things. And then at some other time it'll be localized again at the other turning point. And it will come back and forth. And now, if it's not a harmonic oscillator, it won't quite relocalize. It'll mush out a little bit, and it'll come back and it'll mush out more. And so again, we can use the evolution of the wave packet to sample the shape of the potential. We can measure the anharmonicity. And so let's now talk about that. What are other quantities that we can calculate from the wave function of the time dependent Schrodinger equation? So let's talk about the survival probability. So this is a capital P. Survival probability. It's going to be the wave function. The time dependent wave function is created at t equals 0. It has some shape. Now, we always like to have a simple shape at t equals 0. And we'd like to know how fast that thing moves away from its birthplace. So we have this survival probability. It's a probability. So we're going to have integral psi star xt psi xt dx. And this integral is going to look like that. Now, whoops. I knew I was going to screw up. If I put a t here, then it would just be the normalization interval. And we already know what that is. But this is the birthplace. This is what was created at t equals 0. And this is time evolving thing. And so we can calculate how that behaves. And I'm just going to write the solution. So the result is, if we have the same kind of two state c1 psi 1, e to the minus i e1 t for h bar plus c2 psi 2 e to the minus i e2 e over h bar, well then, what we get is c1 to the 4th power plus c2 to the 4th power plus c1 squared, c2 squared, times e to the i omega 2, 1 t plus e to the minus i omega 2, 1 t. Now we're integrating. When we integrate, the wave functions go away. The wave functions become either 1 or 0. And so we're integrating. We're making the wave functions go away. And we have some amplitudes of the wave functions. And we have a time independent part and a time dependent part. And this is 2 cosine omega 2, 1 t. So what we see is this survival probability, the wave function, starts out at some place. And it goes away. And it comes back. And it goes away. And it comes back. And it does that. It completely rephases, because there's only two terms. But if there were three terms that are not satisfying a certain requirement, then when it comes back, it can't completely reconstruct itself. And this is the basis for doing experiments. One can observe the periodic rephasings of some initial pluck. And you can look at the decay of them, and that gives you something about the shape of the potential, the anharmonicity. And you can do all sorts of fantastic things. You can create a wave packet, and you can wait until it reaches the other turning point. And at the other turning point, it's possible that you could excite it to a different higher excited state. Ahmed Zewail got the Nobel Prize for that. So we're right at the frontier of what you can do, and what you can understand using this very simple problem of a time independent Hamiltonian and a 2 or a 3 term superposition. OK. Recurrence. This is a special property. When all of the wave functions, or all of the difference-- all of the energy levels, or differences in energy levels are integer multiples of a common factor, well then any coherent superposition state you make will rephase at a special time. And so what we can do is say, OK, for a particle in a box, the energy levels are e1 times n squared. For our harmonic oscillator, the energy level difference is ev plus n minus ev our n h bar omega. For a rigid rotor, which we haven't seen yet, the energy levels are a function of this quantum number j hcb. This is the rotational constant-- times j plus 1. So all of the energy levels are related. I'm jumping ahead. Sorry. So the energy levels are j times j plus 1. And the differences, ej plus 1 minus ej, are given by 2 hcb times j plus 1. So for these three problems, we have this perfect situation where one can have these oscillating terms all have a common factor. And at certain times, the oscillating terms are all 1. Or some are 1 and some are minus 1. And we get a really wonderful simplification. So at a time which we call, or I call, the grand recurrence time, where it's equal to h over e1 for this case of the particle in a box, or h over h bar omega, or h over 2 hcb, we get all of the phase factors becoming 1. And sometimes, something special happens at the grand recurrence time divided by 2. And that's a little bit like this. We have a wave function here. And at half the recurrence time, it's over here. And so you want to work your way through the algebra to convince yourself that that is in fact true. And in between, if you have enough terms in your superposition, which is usually the case, in between you get this. You get garbage looking. It's still moving. It's still satisfying the Ehrenfest theorem, but it doesn't look like a particle. It just looks like garbage. But you get these wonderful things happening. And now, if they're not perfectly satisfying this integer rule, then each time you get to a turning point, the amplitude has decreased and decreased and decreased. And at infinite time, it might recur. But nobody waits around for infinite time, because other things happen and destroy the coherence that you built. So this is a glimpse. Time independent quantum mechanics is complicated enough. Well, we can embed what we understand in the time dependent mechanics. And there are a lot of beautiful things that we can anticipate. Now, we can use these beautiful things to do experiments which measure stuff that is related to, how does energy move in the molecule? What is going on in the molecule? What are the mechanisms for stuff happening? And in magnetic resonance, there are all sorts of pulse sequences that interrogate distances and correlated motions. And it is really a laboratory for time dependent quantum mechanics with a time dependent Hamiltonian. But a lot of stuff that we do with ordinary laser experiments are usually understandable in a time independent way. Now, we want to get rid of things that are complicated. And so one of the things that we do is, we map a problem onto something which is time independent. And so one of the tricks that you will see is that if we go into what's called a rotating coordinate system, this rotating coordinate system is rotating at an energy level different divided by h bar. And that converts the problem into a time independent problem in the rotating coordinate system. So that you're in a rotating coordinate system and, basically, you use the ordinary perturbation theory. And then you go back to the non-rotating coordinate system. So there's all sorts of tricks where we build on the stuff that we understood. And we can have a picture which is intuitive, because we want to strip away a lot of the mathematics and see the universal stuff. And so I'm going to try to present as much of that as I can during this course. But a lot of the time independent stuff is heavy lifting, and I'm going to have to do a lot of that too. OK. Have a nice long weekend. I'll see you on Monday. |
MIT_561_Physical_Chemistry_Fall_2017 | 16_NonDegenerate_Perturbation_Theory_II_HO_using_aa.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: Today, I'm going to go over a lot of material that you already know and work an example with non-degenerate perturbation theory. This will give me an excuse to use interspersed matrix and operator notation, and one of the things that you will be doing is dealing with infinite matrices. And we don't diagonalize infinite matrices, but the picture of an infinite matrix is useful in guiding intuition. And so we have to know about what those pictures mean and how to construct them from things that seem to be obvious. OK, the problem I want to spend most of my time on is an example of a real molecular potential which is quadratic but with a cubic and a quartic perturbation term, and this is a simpler problem than what I talked about last time. The last time I really wanted to set the stage for what you can do with non-degenerate perturbation theory, and this is going to be more of dealing with a specific problem. Now, the algebra does not get simpler, and so it's easy to get lost in the algebra. And so I will give you some rules about what to expect, when you have a potential like this, and how does that come out of non-degenerate perturbation theory? Now, everything that we know about molecules comes from spectroscopy. That means transitions between energy levels, and so well, how do they happen? And we can lead into that by talking about the stark effect which is the interaction of an electric field with a dipole in the molecule. And so most transitions are caused by a time-dependent electric field which somehow interacts with the molecule through its dipole moment, and so you'll begin to see that today. Last time, I derived the formulas for non-degenerate perturbation theory. Now, I'm a little bit crazy, and I really believe in perturbation theory. Most of the textbooks say, well, let's just stop here at the first-order correction to the energy levels, and you get nothing from that except the Zeeman Effect which is kind of important for NMR. But in terms of what molecules do, you need this too, and this is equal to the zero-order energy. I'm sorry, the zero-order energy which comes from an exactly solved problem and then the diagonal matrix element of the everything that's bad in the world, the first-order Hamiltonian. And then, we have this complicated-looking thing, m not equal to n of H n m squared, E n 0 minus E m 0. Now, the only thing that you might forget about in this kind of a formula is is it n before m or the other way around? And what we know is, OK, here is the nth level, and here is the mth level, and the interaction causes this level to be pushed down. So if the mth level is above the nth level, this denominator will be negative, and that causes the pushing down. And so if you remember the idea of perturbation theory-- that things interact, and one level gets pushed down, the other level gets pushed up, equal and opposite-- you can correct your flaws of memory on which comes first. And we can write the wave function as an eigenfunction of the exactly solved problem plus a term, m not equal to n of H n m 1 E n 0 minus E m 0 times psi m 0. Now, the reason this sum does not include the nth level is, well, if you did, this denominator would blow up, and you don't need the nth level, because you've already got it here. So these are the formulas, and we had said, OK, we're going to write the Hamiltonian as H0 plus H1, and this is an exactly solved problem. This is everything else. Now, some people will sort this out into small everything else and big everything else. But it's foolish, because all that does is multiply the algebra in a horrible way, and you don't care, anyway. You just know that this stuff that's not in this exactly solved problem gets treated here, and you're going to know, in principle, how to deal with it, and there'll be some small problems that you know how to deal with. OK, and there is a rule that H n m 1 over E n 0 minus E m 0 absolute value is much less than 1. That's what we mean. Yes? AUDIENCE: Are you making a deliberate choice to put all the badness in the first-order perturbation? ROBERT FIELD: Yes. AUDIENCE: Then, what goes into higher order, if we make that deliberate? ROBERT FIELD: OK, you can say, well, we've got things that are important, that obviously affect the energy levels, and there's things that are smaller, like hyperfine structure. And when you use the word hyper, you generally mean it's small, and so what ends up happening is that you do the big picture, and then you see some more details. But it's only a very small number of people who actually segregate the badness into real bad and not so bad. OK, so this works only when this off-diagonal matrix element is smaller than the energy difference, and that's going to be true. Now, here is an infinite matrix, and we're interested in a certain space, a state space, which our experiments are designed to measure, and it depends on what experiment you do. This might be here, it might be somewhere in the middle, and the rest, we're not interested, because it's far away. Because the energy denominator is so large that the effect on the energy levels can be ignored, or you can say, well, we're not really ignoring it. We're folding the effect of all these levels through the matrix elements in here into here by second-order perturbation theory. We could in principle do that. We certainly are allowed to think that we are getting rid of this state space by, in principle, folding it into this block. Now, in this block of levels that we care about-- now I always like to draw these infinite matrices, or these matrices, by just saying, OK, here's the diagonal. And there might be a couple of levels within this state space we're interested in which, because of accident or because of something evil, their energy denominator is small compared to the zero-order energy differences. But that's usually true for only a few accidentally degenerate states, and we deal with them by using the machinery we obtained from the two-level problem, and so there, we are actually diagonalizing a small dimension matrix. We're not diagonalizing it, our computer is diagonalizing it. You really don't care how it's done, because you have a machine that will solve this like difficulty, and so it's OK for this rule to be violated. But so what happens, suppose we have two levels, and-- I just want to make sure I use the same notation as in the notes. So here, we have the two zero-order levels, and they interact and repel each other equal and opposite amounts. And these states, which are E2 and E1, are not pure state 2. They have a mixture of state 1 in them. Now, in spectroscopy, we always observe these levels by transitions. And so suppose we have state 0 down here, and let us say that this transition between the zero level and the zero-order level 1 is allowed. I'll symbolize that by mu 1,0 is not equal to 0. And the transition between this level and this zero-order level is forbidden, mu 2,0 is equal to 0. So we call this a dark state and a bright state, and now these two states interact, and we have mixed states. So the real eigenstates, you wouldn't have expected to see this level in the spectrum. But you do, because it's borrowed some bright character through the perturbation interaction, and so there's two surprises. One is everything is sort of describable by a simple set of equations, and then there's a couple of deviations, and there's some extra levels that appear. That's information very rich, because if it's dark, you can't see it. It's not a ghost. It's there, but you can't see it. Well, the perturbation somehow pulls-- sometimes-- pulls the curtain back and enables you to see stuff you need to know about. Now, you can imagine exciting this two-level system with a short pulse of light, where the uncertainty broadening of the short pulse of light covers these two eigenstates. What happens then is you get quantum beats, and that's going to be on the exam. OK. So this is a local perturbation, and it's just something in here that spoils the general rule but it's very information-rich, and it also is pedagogically fun. OK. So now, I'm going to talk about stuff that I've talked about before, but I'm going to go a little deeper and slower, and so let's do this problem. We have a potential, and for molecules, we tend to use Q rather than R or X as the displacement from equilibrium. The harmonic oscillator coordinate, and it's the same thing and so we have-- OK, so this is an exactly solved problem, and this is something extra. And so this is a cubic anharmonicity, and this is a quartic anharmonicity. In the previous lecture, I talked about anharmonic couplings between different modes of the same polyatomic molecule. Here, this is really a simpler problem, and I probably should have talked about it first. But I didn't, because I want to go deeper here than the big picture, which I described last time. So we're going to have two terms that we are going to treat by perturbation theory, and there are several important things here. Molecular potential looks like this, not like that. Right? This is bond breaking, and it breaks at large displacement. There is no such thing as a bond breaking as you squeeze the molecules together. So this is a crazy idea, and how do you get an asymmetric potential? Well, it's cubic, and which sign of b is going to lead to this? Yes. AUDIENCE: [INAUDIBLE] ROBERT FIELD: Right. OK. Now, if we're really naive, we say, yeah, negative is good, but boom, we ignore that. We don't worry that, if we took this seriously at large enough displacement, the potential will go to minus infinity. We don't worry about tunneling through this barrier. We just use this to give us something that has the right shape, and that we can apply perturbation theory to. And we're not going to worry about using perturbation theory to capture this tunneling, at least not now and not in this course. OK. So one thing is we have this term, and there is something you know immediately about odd powers. They never have delta v equals 0 matrix elements, they never have diagonal elements, and so they do not contribute to the energy in first order. The only way you know the sign of this perturbation is from a non-zero first-order contribution, because when you do second-order perturbation theory, the matrix element gets squared, and the sign information is lost. So there's nothing so far that tells you what the sign of b is, and the energy level pattern that you would obtain from either sign of b would be the same. It's just one of the signs is completely ridiculous. Now Q to the 4th has selection rules delta v of 4, 2, 0, minus 2, and minus 4. So that delta v of 0 term does enter in first-order perturbation theory, and so you can determine the sign of any even perturbation which is useful. Now, what does this do? Well, one thing that Q does is, if c is positive, instead of having a harmonic oscillator, it makes it steeper. And if Q is negative, it makes it flatter. This is typical of bending vibrations, and so bending vibrations tend to have flat bottom potentials, but there there's also something else. Suppose we have two electronic states. Now, I am cheating, because I'm assuming you'll accept the idea that there is something we haven't talked about yet. But there are different potential curves, and it's mostly true for atomic molecules. These two states can't talk to each other at equilibrium, because a symmetry exists that prevents that. And as you move away from equilibrium, there's a perturbation that gets larger and larger. And as a result, what happens is that this potential does something like that. So you get either a flattening or an actual extra pair of minima, and this one gets sharper. That's called a vibronic interaction, and that's fairly important in polyatomic molecules. But so we can begin to understand these things just by dealing with these terms in the potential. OK. So now, we're off to the races, and my goal here is to make you comfortable with either the operator or the matrix notation, and so I'm going to go back and forth between them in what might seem to be a random manner. OK. Now, you know that a dagger. Now, we can call it a dagger with a hat, or we can call it a dagger double underline-- bold, hat, matrix, operator. And you know that this operates on v to give v plus 1 square root psi v plus 1, or in bracket notation, we could write this as v plus 1 a dagger v. Now this is a shorthand. It's a wonderful shorthand. It's easier to draw this than wave functions. But we have to know, OK, what does this look like in the matrix for a dagger? So here is a dagger, and it's a matrix, and what goes in here? Well, the first thing you do, this is infinite, so you need some sort of a way of drawing something that's infinite so that you understand what it is. And so the first thing you do is you know that there is no-- all of the diagonal elements are 0. Now, most of the elements in this matrix are 0, and you don't want to draw them, because you'll just be spending all your time writing 0's. So we want a shorthand, and so now this is a matrix element. This is the row, and this is the column, and so where do I put this square root of E plus 1? AUDIENCE: [INAUDIBLE] ROBERT FIELD: So here, we have the row. I get confused about this, so before I accept your answer, which I want to reject, I have to think carefully about it. So let us say this is 0, and this is 1, so we have the 1-- you're right. OK, and so we have the square root of 1, the square root of 2, square root of n. And everything else is 0, so we can write big 0's. So that's what this matrix looks like. Now, if we're using computers, instead of multiplying these matrices to have say Q to the 13th, or a dagger to the 13th, you can just multiply these matrices. That's an easy request for the computer. It's not such an easy request for you, but you end up getting matrices. When you have integer powers of these, you get a matrix with a diagonal and then another diagonal separated by a diagonal of 0's, and so you know what these things are going to look like. So you might ask, OK, what does a look like? And there are several ways to say what a is going to look like, because it's the conjugate transpose of a dagger. Well, these are all real numbers, and so all you do is flip this on its diagonal, and now you get a. Then there's another actor in this game, and that's n, the number operator, and the number operator is going to be a, a dagger or is it going to be a dagger a? So which is it? AUDIENCE: [INAUDIBLE] ROBERT FIELD: This. AUDIENCE: [INAUDIBLE] ROBERT FIELD: Yes. OK, one way to remember this is when you operate with either a or a dagger, it connects two vibrational levels, and the thing you put in the square root is the larger of the two quantum numbers. OK. So what would the number operator matrix look like? So what do I put here? Yes. AUDIENCE: [INAUDIBLE] ROBERT FIELD: But what's the lowest vibrational quantum number? AUDIENCE: [INAUDIBLE] ROBERT FIELD: Right. OK, and so once you've practiced a little bit, you can write these things. And it's not just an arbitrary thing, because you want to be able to visualize what you're doing. Because you're dealing with multiple infinities of objects, and you want to focus only on the ones you care about. And with a little bit of guidance from these pictures, you can do what you need to do. OK. Well, we know what the operator Q is in terms of a dimensionalist version of Q, and that's h bar over-- now, I'm going to be using this. OK, this is the dimensionalist Q. That's what the twiddle means, and these are constants. And now I'm using omega twiddle, because spectroscopists always use omega in wave number units, reciprocal centimeter units. Which is a terrible thing, because first of all, wave number doesn't have a unit. It's a quantity. And centimeters are things that we're not supposed to use, because we use MKS. But spectroscopists are stubborn, and when we observe transitions, we always talk about wave numbers not energy. And so the difference between wave numbers and energies is the factor of hc, not h bar c. So anyway, this is the conversion factor, when omega is in wave number units. And we can go further and relate Q twiddle well, let's just not do that. We can relate this to h bar over 4 pi c u omega square root times a plus a dagger. OK? So this is something we did before. Omega twiddle is k over mu square root 1 over 2 pi c-- hc. So this what you're used to, and this is the extra stuff that we have to carry along in order to work in wave number units. OK. So we have operators that can be expressed like a, a, a dagger, a or a dagger, a, a, a or anywhere you put the dagger anywhere. You look at this, and you say immediately I know two things. I know the selection rule. The selection rule is count up the a's and count up the a daggers, and so this is delta v of minus 2. So is this. All of the positions of the a dagger are delta v of minus 2, but the numbers, the matrix elements, are different. You know this from the last exam. So we can have v minus 2 v and whatever comes in here in one of those forms. So those are the only non-zero elements, and you know how to mechanically figure out what is the value of the matrix element. Now, in order to simplify the algebra, which is not essential to the physics, you want to take all of the terms that results say from Q to the 4th and arrange them according to selection rule. And then take all of the terms that have the same selection rule and combining them to a single number. And you use the computation rule to be able to reverse the order of terms. Well, I don't want to do that. So suppose we have an a, a dagger, and we can write that as a, a dagger plus a dagger, a. And so if we want to convert something like this to something like that, we know that this has a value of plus 1, so we can do that. And that's tedious, and you have to do it. But one of the things that is kind of nice is when you have a problem that's cubic or quartic or quintic, you mess around with this operator algebra once in your life, and you put it on a sheet of paper, and you refer to it. It doesn't matter what the molecule is, what the constant in front of Q to the 3rd or 4th or 13th is. If you've done the operator algebra, you're fine. Now, you might say, well, I don't want to do that. I'm going to have the computer do that. Well, fine, you can have the computer do that, and then a computer will tell you what the matrix looks like, and you can do what you need to do. OK. So spectroscopists call the vibrational energy formula G of v. I don't know why the letter G is always use, but it is, and so this is the same thing as vibrational energy. And the vibrational energy, I'm going to put the tilde on it. You will never find a tilde in any spectroscopy note paper. We assume that you understand that the only units for spectroscopic quantities are reciprocal centimeters, but for this purposes, I have finally caved, and I said, OK, I'm going to put the tildes on. So the energy levels, now you might ask, why this? This is the first anharmonicity constant. It's not a product of two numbers. It's just what people wrote originally, because they sort of thought of it as a product of two numbers, but it's really only one. And that's times v plus 1/2 squared, and then the next term is omega e, ye, e plus 1/2 cubed. So this is a dumb power series in the vibrational quantum number, and so in the spectrum, one is able to fit the spectrum to these sorts of things. So that's what you get experimentally, but what you want to know is we want to know the force constant, the reduced mass, the cubic anharmonicity constant, the quartic anharmonicity constant, and whatever. So these are structural parameters, and these are molecular parameters. People like to call them spectroscopic parameters, but that implies something more fundamental. These are just what you measure in the spectrum. And so we want to know the relationship between the things we measure and the things we want to know, and so that's what perturbation theory is for. OK. Now, to risk boring you, I'm just going to go over material that we've done before. So if we have Q to the n, we have this constant out in front, h bar over 4 pi c mu omega twiddle to the n over 2, a plus a dagger to the n. And so we're going to be constantly dealing with terms like a plus a dagger squared and plus a dagger cubed and so on. These are the things I said you're going to work out once in your life and either remember or just become so practiced with it, you'll do it faster. And so these contain the values of the matrix elements and the selection rules. OK. So I'm going to do an example of the cube, and that's, of course, this thing in constant the 3/2 and then a cubed plus a dagger cubed plus a whole bunch of terms which have two a's and one a dagger. So let's just put it a squared and a dagger. This isn't a computation rule. This is just three terms that you have to deal with and three terms that have a dagger squared and a. So that's what you do your work on, because you don't want to be messing around once you start doing the perturbation sums, because it's ugly enough. So you want to simplify this as much as possible, and you do. And so the purpose is this is delta v of minus 3. This is delta v of plus 3. This is delta v of minus 1. This is delta v of plus 1. You arrange the terms according to selection rules, and so what you end up getting is Q cubed is equal to this thing, to the 3/2 times a cubed plus 3 a n plus 3 a dagger n plus 1 plus a. OK, the algebra is tedious, pretty simple. And what you usually want to do is have the thing that changes the vibrational quantum number after the thing that preserves it, because then it's easy just to write down these matrix elements just by inspection. OK? You could do it the other way around. It's more complicated if you put the n first and then the a, but it's up to you. So these then are what you work on, and now we start doing non-degenerate perturbation theory. The first thing you do is you want to know, well, what is the first-order correction to the energy? And if this is Q to the 3rd power, then this, that's 0. So you like 0's, but in this case, you're kind of disappointed, because you don't know what the sign of the coefficient of Q to the 3rd power is from any experiment or at least any experiment at the level we've described. When we introduce rotation as well as vibration, there will be something that reports the sign of the coefficient of Q 3rd. So now, we're stuck, and we have to start doing all of this second-order stuff. OK. Well, we have this bQ to the 3rd power, and so we get squares of the matrix element. So we get a b squared. We get this thing to the 3rd power, because we're squaring the matrix element, so this bunch of constants. And then, you get a matrix element v prime, some operator v, then we get an energy denominator. Now, there's a lot of symbols, but we want to simplify things as much as possible, hc omega times v minus v prime. So this is matrix element squared over an energy denominator, and what you really want to know is what is the quantum number dependence of everything? OK, so the operator here is either Q dagger cubed, Q dagger-- I'm sorry, a cubed, a times n, a dagger n plus 1, or a dagger cubed. And so we know how to write all of these matrix elements, trivially, no work. Once we've simplified here, it's really trivial to write the squared matrix element. OK, so we do it, and so we arrange things according to delta v and we have delta v of plus 3, plus 1, minus 1, and minus 3. And so what we get from the square of the matrix element a dagger cubed, we get v plus 1, v plus 2, v plus 3, and we also have an energy denominator. And we're going to get for this one 1 over minus 3, because the initial quantum number is v, and the second quantum number, v prime, is v plus 3, and so we get a minus 3 in the energy denominator. And then the plus 1, that comes out to be-- I've got it in a different order in my notes-- that comes out to be 9, v plus 1, v plus 2, v plus 1 squared, and the energy denominator for this is 1 over minus 1. OK. Then, we have the-- I'm going to skip this one-- we have the matrix element here, and that's going to be v minus 1, v minus 2. Remember, we're squaring these things. So that's why we don't have those square roots anymore, and we have an energy denominator 1 over 3. OK, now advice-- you don't like this, and you want to minimize your effort. And so it turns out that if you take the terms with the equal and opposite energy denominators and combine them, simplifications occur. One of the things you can see is you're going to have a v cubed here, and you're going to have a v cubed here. They're going to cancel, because we have a 1 over minus 3 and a 1 over plus 3. So you get an algebraic simplification when you take these terms pairwise and combine them. Now, you're never in your life going to do this, but if you did do it, this is how you end up with formulas which are simple. They're horrible getting there, unless you know how to get there. So now there's a rule. So if the perturbation is Q to the n, then the highest-order term involves v plus 1/2 to the n minus 1. The reason for that is in the matrix element the highest-order term is v plus 1/2 to the 3 over 2, and then you square it, you get to the 3rd power, and then the highest-order term cancels. So we know that if we're dealing with Q to the 3rd power, we're going to get a term v plus 1/2 squared. If we're dealing with Q to the 4th power, well, then we get something from second order from the first-order correction to the energy a delta v of one matrix element. What we've done is square it, and we get v plus 1/2 squared also. OK and the off-diagonal matrix element, we're going to get from the highest order from the off-diagonal is v plus 1/2 cubed. So one of the things that is great about the algebra is with a little bit of practice, you know how to organize things, and if you do the algebra, you collect the highest-order terms. Then, there's that kind of cascading result, and you get the lowest-order terms in simplified form. So this is irrelevant, but this is how you do it, if you're a professional. So with these results, we can determine the relationship between these molecular constants. Where did I put them? Oh, probably on this board. No. Oh yeah, it's right here. So we have omega e, omega e xe, omega e ye, and we have then the relationships between these things which you measure and these things that you want to know. So the stuff you want to know is encoded in the spectrum in a not particularly complicated way-- it's just not a very interesting way, but you have to do it in order to get it. OK, so I've got just a few minutes left, but I want to get to the really interesting stuff. So if you have some expression for the potential, it's easy to go and get the expression for the energy levels and the wave functions. And you can use the vector picture or the wave function picture. It doesn't matter, and so this is enough to do the spectroscopy. It tells you not only where are the energies, but because some transitions are supposed to be weak because of the dipole selection rule or whatever, it says, well, there are some transitions that have borrowed intensity. It also tells you-- suppose we make a coherent superposition state at T equals 0 using a short pulse, and suppose in the linear combination of zero-order states, there is only one that is bright. And so then it tells us, if we know how to go from basis states to eigenstates, we can go backwards, and we can write the expression for the T equals 0 superposition in terms of some sum over psi v0 cv. And if we have this, as long as we are writing a superposition in terms of eigenstates, we know immediately how to get to this, and then we've got all the dynamics. So the perturbation theory enables you to say, if I want to work in the time domain, I know what to do. You're going to get-- in the time domain-- you're going to get a signal that oscillates, and it oscillates at frequencies corresponding to energy level differences divided by h bar. And so what frequencies will appear in the Fourier transform of the spectrum, and what are the amplitudes of those Fourier components? You can calculate all of those stuff. It all comes from perturbation theory. These mixing coefficients you get by perturbation theory. And remember, if you have a transformation that diagonalizes the Hamiltonian that the eigenstates correspond to, the columns of T dagger, and the expression of the zero-order states, in terms of the eigenstates, corresponds to either the columns of T or the rows of T dagger. So once you do the perturbation theory, you can go to the frequency domain spectrum with intensities and frequency or the time domain spectrum with amplitude and frequencies. It's all there. This is a complete tool. It's the kind of tool that you can use for an enormous number of problems, and so you'd better get comfortable with perturbation theory, because the people who aren't comfortable can't do anything except talk about it. But if you want to actually solve problems, especially problems on an exam, you want to know how to use perturbation theory. And you also want to know how to read and construct the relevant notation in the vector picture, because a vector picture and the matrix picture is the one where you see the entire structure of the problem. And you can decide on how you're going to organize your time or what are the important things that I'm going to get from this analysis. And so it's much better than the Schrodinger picture, because with the Schrodinger picture, you're just solving a differential equation and one problem at a time, whereas with the matrix picture, you're solving all problems at once. This is really a wonderful thing, and so that's why I'm taking a very different path from what is in the textbooks. Your wonderful textbook McQuarrie does not do second-order perturbation theory. So nothing you want can be calculated, unless you're dealing with NMR, and you're dealing with magnetic dipole transitions. And then, you can get a lot of good stuff from first-order perturbation theory, and you can avoid second order until you grow up. OK, I'm done. I'll be talking about rigid rotor next time, and I will be talking about it also in an unconventional way. |
MIT_561_Physical_Chemistry_Fall_2017 | 34_Electronic_Spectroscopy_and_Photochemistry.txt | The following content is provided under a Creative Commons license. Your support will help MIT Open Courseware continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseware at ocw.mit.edu. PROFESSOR ROBERT FIELD: So last time we talked about intermolecular interactions. Intermolecular. It's complicated enough talking about individual molecules, what molecules can do, but we could also talk about interactions between molecules, and we can use perturbation theory, and what we discover is something universal about all pairs of molecules. Molecules which don't have any reason to interact with each other, like being charged or even having dipole moments, and we discover that all molecules are attracted to each other. And we know the force law, we know how everything scales, and we also know that we can look at molecules, and say this molecule is going to be a good molecule for intermolecular interactions because either it has a lot of low-lying transitions, or they have very strong transitions. And so you can compare different atoms or different molecules. And one of the important messages is that really big molecules are really sticky. They want to attract each other because they have a lot of vibrations, and each vibrational node can act as a dipole in this dipole-dipole interaction, whether it's induced dipole-induced dipole or whatever. So it's one of these things where something from one half of your brain is able to explain something from the other half, and that's really wonderful, and we're always aiming for that. Now, today we're talking about photochemistry. And this is an enormous topic. And I would say that there are vastly more physical chemists and biophysical chemists working on photochemistry than on spectroscopy. So what happens when a molecule gets excited by a photon? Does it just fluoresce or does it do interesting stuff? And there's a lot of really beautiful semi-empirical theory that enables you to estimate the rates of everything that a molecule can do, and the uses for what these things do. But in order to put this in perspective, I have to do a little bit of review of the different kinds of spectroscopy-- rotation, rotation-vibration, rotation-vibration-electronic in order to get the energy scales and to describe the actors in this problem of photochemistry. There's two sets of notes, what I'm going to be working from, which is because I'm a small molecule person, bottom-up, and I'm trying to take what I know and I've taught you about small molecules, and extend it to regions of incredible complexity, but using most of the same concepts. And Troy's notes, which are top-down, and he's basically interested in photochemistry, and condensed phases, and big molecules. And I'm afraid to go to condensed phases and big molecules, but the two shall meet, and many of the topics will be understandable with both approaches. OK, so let's just set the framework. We have three kinds of spectra, and the typical energy associated with that is one wave number. And we have rotation-vibration, and the typical energy scale is 1,000 wave numbers. And then there is the rotation-vibration-electronic, and that's, going to say 20,000 wave numbers and higher. Now, this is peanuts compared to the strength of molecular bonds and interactions between molecules, but this is big news because now the photon that excites the molecule is sort of acting like a reagent. This energy, it goes up as far as you want. These energies are comparable to the energies of chemical bonds, and the electronically excited molecules can do chemistry that the rotational or vibrationally excited molecules can't do. That's why it's so interesting, and that's why it's so complicated. OK, now, what can an excited molecule do? Well, maybe I should ask you. Yes? AUDIENCE: Rearrange. PROFESSOR ROBERT FIELD: OK, so we can call it isomerization. Anything else? AUDIENCE: Fragmentation. PROFESSOR ROBERT FIELD: I'm sorry? AUDIENCE: Fragmentation. PROFESSOR ROBERT FIELD: Yes. AUDIENCE: It can relax. PROFESSOR ROBERT FIELD: By? AUDIENCE: Emitting a photon. PROFESSOR ROBERT FIELD: Yes, fluorescence. That's what we do. There's lots of things that can happen and the kinds of things that can happen depends on what kind of exaltation we've produced, but the big photons are going to produce really complicated stuff. Now, one framework we can talk about is what is the rate of-- I can't imagine what I wrote there. So this is the total rate, and it can be composed of the radiative decay and the non-radiative decay. Now, there's a size scale here. Oh, I know what I wrote here. That's measured. So if you had a system and you measured the exponential decay of the population in that system, that's composed of the radiative part and everything else, and the radiative part corresponds to lifetimes no shorter than 10 nanoseconds. There is absolutely no way that molecules or atoms can fluoresce at a rate much faster than one over 10 to the minus eight seconds, and then there's other stuff. And so when the other stuff, the non-radiative decay is fast compared to the fastest the molecule could decay by emitting a photon, then we're getting into something complicated and interesting. When the molecule just radiates a photon, and this stuff is very small, well, that's nice, but it doesn't tell us very much except what is the radiative lifetime of the particular excited state. So there's various kinds of de-excitation, and there's collisional, and there's-- well, we can call it collision-free. Now, remember, if there's no collision, energy is conserved. We have a Hamiltonian. The expectation of the Hamiltonian for an isolated system is not time dependent. So we have energy conservation, yet in the absence of-- collisions are what change the energy. We remove energy from the system. We can have an incredibly complicated range of processes which are collision-free, and these collision-free processes have the signature of the molecule all over it. They are telling you something about what the molecule thinks it can do and does, whereas the collisional stuff is purely statistical. So the collisional processes, we can say-- we can estimate how fast they are. And first of all, there is this number that many physical chemists carry in their head, especially if we work in the gas phase. A collisional rate is on the order of 10 megahertz per torr. Now, I'm an old guy. I do two torr instead of bar or millibar, but a torr and a millibar are about the same thing. And at atmospheric pressure, that corresponds to 10 gigahertz per atmosphere, and so that means 10 to the minus 10 seconds. So the collisional rates can be competing with radiative rates at atmospheric pressure. Or in condensed phase, they really can compete. But this number, 10 megahertz per torr, is a good number to carry around for estimating the frequency of collisions. Now, there are different kinds of collisions. And one collision with this sort of rate is all it takes to relax rotation, and about 1,000 solutions are required to relax vibration. And now, it's very hard to estimate what it takes to relax electronic excitation because the electronic excitation is so big, there's so many different things that can happen. And so you could say, well, if nothing really chemical happened, it would take about a million collisions to relax electronic, but there are so many other things. I don't want to put a number here because this is really the interesting and complicated stuff. OK. In particular, suppose you have an electronically excited molecule, and it collides with an unexcited molecule, and you end up doing A plus BCD or all sorts of things. That's chemistry, right? But this wouldn't happen if this star wasn't there. And there's a tremendous complexity to chemistry. That's why we have-- it's a BSD subject. It's a major area of science, and so we're engaging with chemistry when we start talking about electronically excited atoms, and vibration and rotation excitation is largely irrelevant to chemistry. it's small potatoes. You can see chemical effects by rotation and vibrational excitation, but that's for the purists like me. OK, now collision-free effects are initially surprising. We can have several, and we've got some of those from the initial discussion, but I'm going to just list them again. We can have the molecule breaks, and that could be dissociation or ionization. This is something that all molecules do if you excite them enough. They either fall apart into smaller molecules or ionize. There's nothing surprising there. But then there's the unexpected processes, or at least they used to be unexpected. When I was a graduate student, there was still a lot of confusion about what happens in the absence of collisions. And there are several names for them. IVR-- intramolecular vibrational redistribution. This process, which you understand from perturbation theory-- I talked about this early in the semester. This put closed to the idea of vibration or mode-specific chemistry by putting an excitation into a molecule-- a high vibrational excitation into a particular molecule, activate a bond to chemistry. IVR is always faster than chemistry. Or faster than collisions until you get to the condensed phase. IVR, that's one. There was so much confusion about IVR in the old days that people didn't even know what it meant. It just meant I don't understand what's going on. But it is intramolecular vibrational redistribution. Then there's another process called internal conversion, and another process called intersystem crossing. So IVR is something that happens in one electronic state, internal conversion involves relaxation into high vibrational levels of a lower electronic state of the same spin, and intersystem crossing is a relaxation into a high vibrational levels of another electronic state with a different spin. These are really two sides of the same coin, except that they depend on different molecular properties, and one can make generalizations that are different for IC and ISC. So for a small molecule, we've looked at this before, but-- we can excite directly to an excited state, and that's direct association that comes like this, and there's all sorts of really neat things you can do, where you can calculate the Franck-Condon factors or the vibrational intensities for the different energies, here, by knowing that you have-- you can calculate the wave function for the continuum, and you know you have a big lobe here, and as it gets faster and faster, the lobes get smaller, and smaller, and closer, and closer together. And there's a thing called the reflection principle that describes the rate of excitation to a repulsive state. And by looking at the spectrum, you can learn the slope and position of this state. OK, there's also suppose you have a level here of this excited state. Well, that excited state has access to this crossing point, and so that can free-disassociate. So there's direct and free-dissociation. I like this stuff too because if I know the rate of free-disassociation, and what happens with free disassociation is you have a spectrum where you measure several vibrational levels in this state, and the vibrational lifetimes of these states are more or less the same, and all of a sudden the lifetime decreases. Or all of a sudden the spectrum goes from being sharp to broad. Or the fluorescence decreases. The intensity decreases because the quantum yield or the fluorescence quantum yield is one over tau radiated over one over tau radiated plus one over tau non-radiated. And the pre-disassociation rates can be much faster than the radiative rates. So this fluorescence quantum yield goes down. But the neat thing is that you generally know what the radiative component of the decay is, and so by measuring what the actual fluorescence quantum yield or the decrease in intensity, you learn something about the non-radiative decay. OK, I've also taught you we can also ionize. And we can have direct ionization, and it looks sort of like this, or we can have what's called auto-ionization where somehow, the molecule is able to convert rotation and vibration of the core, and transfer it into the electronic excitation, and the electron is free. So before you're actually accessing directly, an ionization continue, you can have this other process. These are things that I've studied for my whole career, and I know about it, and I love to talk about them, and we're not going to talk about them. OK, I also told you about quantum beats. So suppose we have a bright state and a dark state. In other words, we have two states, one of which has an allowed transition with the ground state, and one has a forbidden transition with the ground state. But these are zero order states, and they can interact to make mixed states because there's some term in the Hamiltonian connects the bright state to the dark state. So what you do with a short excitation pulse is you make some coherent superposition of these bright and dark states, which at t equals zero is the bright state, and you get quantum beats, and quantum beats can look like this. They start out phased up, and they go down like that, and so we have modulation of the fluorescence. Now, it doesn't have to be 100% modulation. It could be something really small, but it's always going to be phased up if we're looking at the bright state. So we understand quantum beats. Now, suppose instead of having one bright state and one dark state, light, dark, we have a whole bunch of dark states. Well, in that case, if you make this coherent superposition, you'll get many quantum beats, but they'll all have different frequencies, and the quantum beats will dephase, and it will look just like the fluorescence decays to zero faster than it's supposed to be able to. The ceiling on decay rate of one over 10 to the minus eight or 10 to the eight per second, that can be violated by mixing of the bright state into the dark state. So the ability of the molecule to fluoresce goes away fast, but the excitation doesn't decay. What you've done is you've mixed a bright state, a decaying zero order state, with states that don't decay, and the average lifetime is very long, but the experiment says the lifetime is really short. Because what's happening is not that the population is decaying, but the ability to decay has been spread among many eigenstates, and so none of them are very good at it. OK, so now you're starting to get the idea of how we can have fast decay without collisions. And all of these three processes, which I've hidden, IVR, internal conversion, intersystem crossing, are based on relaxation of one kind of state into many dark states. So nothing is happening without collisions. Energy is conserved without collisions, but the signal goes away, and it's tempting to say that either something magic has happened and the molecule has decayed, or you really haven't gone to low enough pressure, and there really are collisions. This was a very bitter battle in the mid 1960s. There was one faculty member who taught the quantum mechanics course, very much like the one I've been teaching for years, but he believed that there was no such thing as radiationless transitions. It's always that the fluorescence was quenched, but he's wrong. In the absence of collision, what decays? What decays is the ability to fluoresce or the ability to do other things like absorb another photon to some other excited state. You've produced the molecule not just in one eigenstate, but in a bunch of eigenstates, and this bunch of eigenstates, since the initial preparation, is the bright state that's in that bunch, and there's a couple among them. What happens is it's time dependent, and the evolution of this coherent superposition state is the key. Let's talk a little bit about IVR some more. When we have 3n minus six vibrational modes, there can be a very large number of dark states. States that couldn't be excited by whatever method you use to excite them, whether it be an ultraviolet photon, or a microwave-- anything. You have a bright state and a whole bunch of states that you are not supposed to be able to excite, an anharmonic interactions among them. And we know some things. So if we have a cubic interaction, where we have one normal mode. So this would be a term in the Hamiltonian. And again, I lectured on this before at a time when it was probably impossible to understand the significance of this, but you used perturbation theory to say OK, what's happening when you have an anharmonic coupling term that couples one mode to another? And we know how do all that sort of stuff. And the selection rules for that are delta vi equals plus and minus one, delta vj equals plus and minus two and zero. We also know when we go from the real coordinate to the dimensionless coordinate, we factor something out. And for each power of Q, that thing that we factor out is a factor of 100 smaller. We have cubic terms that we have quartic terms, and the quartic terms are a factor of roughly 100 smaller, but there's more of them. And so there is an order sorting thing, which we can call delta capital V, which is the sum from i equals one to 3n minus six of the absolute value of delta vi. So if the total change in number of vibrational quanta is three, well, we have a cubic term. If it's four, it's quartic term, and the quartic term has a coefficient, a factor of 100, roughly, smaller than the cubic one, but there's more possibilities. And so suppose we are going to talk about benzene. Well, it has 30 nodes. And so we could imagine delta v on the order o 30. Well, it's 10 to the 60 smaller than the delta v of three, except that there's a lot more possibilities. So we have to think about how big is the coupling matrix element, and how many states are being coupled? And so there's a thing that's really relevant called the vibrational density of states. So at a particular excitation energy, you want to be able to know how many vibrational states there are because that, more or less, tells you OK, we're going to have some kind of statistical coupling. And if we know the density of states and the average matrix element, we could do a Fermi golden rule of calculation, and we can predict what's going to happen. Now, there are ways of calculating the density of states, which are rigorous and beautiful, but there's also a very simple minded way of doing it, which I'm going to present here. So the vibrational density of states. This is one of the most primitive, crude estimates of density of states, and I'll explain how it works, but I'll explain it by going even cruder. And I also want to make sure that I have it-- OK, this frequency is in wave numbers, and this is in wave numbers. So we just changed both quantities. This is OK, and this will give me the number of states per wave number. OK, so let me do an example for benzene. We have 30 modes, and here's a big assumption, which is completely wrong, but it leads to an underestimate. Let us say that all of the modes have the same frequency-- 1000 wave numbers. That's a typical number for a vibrational frequency, but benzene has a lot of modes that have much lower frequencies, and only a few that have higher frequency, and so this is going to be a gross underestimate. And now let's say we have the vibrational energy divided by Hc, and that that is 10,000 wave number. So we have 10,000 wave numbers of vibrational excitation in benzene, and we have all the modes having the same frequency. So we need 10 quanta to make 10,000, and there's 30 choices. So we have 30 to the 10. So each quanta we pick, we have 30 choices. And we have to do 10 of them, and so it's 30 to the 10, but now we have 10 factorial. We have to divide that by that because the order of the 10 choices has to be corrected. OK, and so this number is 1.6 times 10 to the eighth. That's a big number. I'm used to talking about states one at a time, and it's very rare that I have two states within one wave number, and here we have 10 of the eight. So let's go down in energy a little bit. Let's say instead of 10,000, let's go to 3,000. And when we do the same calculation for 3,000, we get 4,500 states instead of 100 million states. So two messages. One is the number of states is really large, and it goes up really, really fast. And this takes us into a region of quantum mechanics that we hadn't thought about before because I'm doing bottom-up, and there is no place in what I do for a million states or even 100 states. OK. Now, let's talk about a typical energy level diagram. This is the ground state, This is the excited triplet state, which is almost always the first excited state, and then here is the first excited sigma state. And so for small polyphonic molecules, we have a very low density of states here, a somewhat higher density of vibrationally excited levels of the triplet here, and a very high density of states from the ground state. So you normally look at relatively low vibrational levels, and there's nothing much happening here, and maybe there's some perturbations by a triplet state. So we get local glitches in the energy levels and spectroscopic properties, and the density of high vibrational level in a ground state is so high that they might as well not be there. But now when we talk about these collisional processes of bigger molecules, we want to use Fermi's golden rule, and you've seen Fermi's golden rule. It tells you the rate of various processes. And so a rate, we can call gamma. Two pi over h bar times the vibrational density of states times H bright, dark squared. So that's the Fermi's golden rule rewritten in terms to describe radiationless processes. It's directly related to the first way we saw it. And so we have the vibrational density of states, which as you go up in excitation energy, increases really rapidly. You have the average of the matrix elements between the coupled states, squared. This is going to go down because remember when I talked about vibrational matrix elements, each additional number in the change in total vibrational quanta causes this to go down by roughly a factor of 100. We have a big increase here, and a big decrease here, but this guy wins. You can show that. So now, we can draw sort of a manifold of states. So what's happening is the density of triplet states is much higher than the density of singlet states. And so if you were just interested in IVR, well, you'd look at that. But this is a much faster process, and this is enormously faster because the energy gap between s1 and t1 is much smaller than in the energy gap between s1 and s0. And so the s1-s0 interaction is called internal conversion, and that's usually way bigger than the s1-t1 intersystem crossing, but not always. The intersystem crossing is due to spin orbit interactions. Spin orbit interactions are predictable for each atom. Extending what you got in five-eleven-one and five-eleven-two, you can predict what the spin orbit interaction is, and it's related to the amplitude of the wave function at the nucleus. But anyway, what happens is that carbon has a spin orbit coupling constant of 10 wave numbers, and oxygen has one of about 150 wave numbers. And since we're talking about squared matrix elements, oxygen is 225 times more effective than carbon. And so if we have a couple of oxygen atoms in the molecule-- or nitrogen atoms-- the spin orbit interactions are much bigger, and they can win over internal conversion sometimes. Depends on what the energy gap is. And now, molecules which have strong electronic transitions have chroma force. It's not the whole molecule. It's something about a part of the molecule. And if you have a strong transition at relatively low energy, there is always oxygen or nitrogen. And so the thing that you're exciting is going to have a fast intersystem crossing rate because it involves oxygen and nitrogen. But if you have just a pure hydrocarbon, intersystem crossing is probably not so important, but internal conversion is going to win. And for most hydrocarbons, you can have a nice, strong absorption spectrum, but very little fluorescence, and that's mostly because of internal conversion. But the crucial thing is the energy gap s1-t1, or s1-s0 controlling the relative importance of ISC versus IC. Now, you could also have several different excited states like s1 and s2, and they could be talking to each other through a small gap or to the ground state by an even larger gap. There's all sorts of stuff. Yes? AUDIENCE: The internal conversion process, you're going from the first singlet excited state to the lowest level singlet ground state. Where does the energy get redistributed to? PROFESSOR ROBERT FIELD: It gets excited into a vibrational excited level of the ground state, but the energy hasn't gone anywhere. It's just now these guys are not fluorescing, and if there are collisions, the collisions can cause removal of energy from the molecule. And since the vibrational energy gaps are really small, collisions start to be very effective in cooling. But if you have a laser, and the laser involves somehow exciting the s1 state, and the s1 state likes to fluoresce to high vibrational levels of the ground state because of Franck-Condon, so it has gain, but there is often a t1 to t2 excitation, which is in the region of the gain of the laser, and it quenches it. And so diolasers, which are a fantastic way of generating tunable radiation, are always worried about quenching by triplets, and you have to do something clever to quench the triplets to remove that. So there's all sorts of stuff about competing processes that you can use it as insights. OK, we don't have much more time, and the most important topic, I'm not going to talk about at all. So we can have isomerization, and isomerization takes many forms. I've spent the last 30 years looking at isomerization in a really simple polyatomic called acetylene. And in the ground state, you can isomerize from high vibrational levels of the ground state to another isomer called [? venility, ?] and this is a really interesting problem. And in the excited state, we have two conformers. The trans-bent conformer and the cis-bent conformer, and they're separated by a barrier, and we can understand both of those processes. But the important thing is if the molecule isomerizes, its chemistry changes, and so it will act differently. Now, there's all sorts of kinds of isomerizations. This is a 1,2 hydrogen shift, but there's also intramolecular proton transfer. There's torsions where a structure-- one isomer or conformer is locked in by a lot of hydrogen bonding, and the bigger the molecule, the larger number of isomerization structures you have to consider, and there's a lot of insight there. So isomerization is a very crucial thing in the behavior of large molecules. The topic that I don't really have time for is FRED-- fluorescence resonance energy transfer. Let's just draw a little picture of a molecule doing all sorts of neat things, and another confirmation of that molecule, and we have donor and acceptor. So at the opposite ends of this thing, clever chemists can put things that-- and so this is a molecule which has an electronically excited state. This fluoresces in the loop. We excited, say, in the UV, and it fluoresces to the red of the excitation because that would happen. In condensed phases, you remove some energy, and you fluoresce from v equals zero. And we have an acceptor which absorbs in the blue, and fluoresces in the red. And the donor-acceptor energy transfer is related to one over r to the six. Doesn't that look familiar? It's a dipole-dipole interaction, or an induced dipole-induced dipole interaction. It goes as one over r to the six, and so we might get a lot of red fluorescence from this molecule exciting in the UV, and much less red fluorescence or none because we've increased the distance. And this works in the range of I think one to 10 nanometers. Yeah. So it's a fantastic ruler. How much red shifted fluorescence do you get from donor to acceptor energy transfer? And it's a measure of the distance. And so you have a protein and a denatured protein. This distance changes, and one can use the way in which the red fluorescence appears as a measure of the structure. So this is not like small molecule spectroscopy where you calculate moments of inertia and vibrational frequency. This is really crude. This has nothing to do with quantum mechanics, except it is, and it's just a simple measurement of how far two things are apart. And there are all sorts of ways of using this, including time resolved waves, where you actually look at the rate at which the one kind of fluorescence goes away, and the other kind of fluorescence comes in, and many other things. OK, so that's it for me on photochemistry, and you should read Troy's notes because they are much more germane to what you might encounter in real life. But the bottom-up approach, I think, is useful too, especially in the importance of how the matrix elements and the density of states scale using simple ideas. So I'll see you on Monday and blow you away on Monday. |
MIT_561_Physical_Chemistry_Fall_2017 | 28_Modern_Electronic_Structure_Theory_Basis_Sets.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseware at ocw.mit.edu. TROY VAN VOORHIS: All right, well, good morning, everyone. I'm not Bob Field, but I'm Troy. Nice to see everyone. So I'm here. We're going to spend the next couple of lectures talking about electronic structure theory. And at this point, I'll give you the big reveal, which is that at this point, you have already encountered virtually every problem that you can solve by hand in electronic structure theory. So there's a handful of other things you could do if you really, like, knew your hypergeometric functions and some things like this, but basically, every electronic structure problem that you can solve by hand you've already encountered. And so if chemistry relied on us being able to solve things by hand, we wouldn't get very far. We'd be pretty limited in what we could do. So we end up using the fact that we have computers. And computers are far more good at doing tedious repetitive things that human beings are. They don't complain at all. And so we can use computers to do fairly neat things. So I'll show you just a couple of quick pictures here of things that we can do. So if you open up any research paper these days, even if it's just synthesis-- you know, we made this catalyst. We did this thing, whatever. There's always some part of it where they said, well, we did a calculation, and this is what the structure looks like. Or this is what the orbitals look like or something like this. So this was actually a calculation. We are computational chemists, I guess. I'll use this screen over here. Is this the one that you're-- or you're filming this one. This one? All right. so I'll film this one-- go to this one. Breaking the fourth wall there. So this is the picture of the HOMO, the reactive orbital water-splitting catalyst that we computed. So you can see that there's 50, 60 atoms here, probably several hundred electrons. We did not do this by hand. We didn't draw the surfaces by hand. The computer did all of this for us. And you can also do things like look at chemical reactions and run dynamics. So we can do-- there we go. So you can run dynamics of molecules. So this is a particular molecule in solution. And we can use the electronic energy in order to govern the dynamics of the molecule. And the computer can actually predict for us whether this molecule is going to react, what it's going to do, how it's going to behave in one solution, say, versus another or at one particular temperature versus another, those kinds of things. And all of these, again, are based off of the idea that we can use computers to solve problems in electronic structure, or at least approximately solve problems we couldn't do by hand. And so what the goal over the next couple lectures is for us to learn enough of the basics so that you can do some calculations like this yourselves and then see what you can do with that. So now I will switch over. You'll find where the other differences, which is that I don't like the feeling of chalk on my fingers, so I use my iPad as a chalkboard. So the rest of this will all be on the iPad. There we go. So for those of you who want to-- you may not have gotten in this time. But I've literally posted what I'm going to be writing on here, these blank notes, online. You can download them, print them off, if you want. If you have an iPad or a computer, and you like to take notes on that, you can download them to your iPad or computer and take notes that way. But I find it's good because there are some things that take a long time for me to write at the board and that it would take a long time for you to write in your notes. And we can just have them written down in advance. So the starting point for all of these things that you can do on a computer for electronic structure theory, we start with the Born-Oppenheimer approximations. So the Born-Oppenheimer approximation was the idea that because the nuclei are very heavy, you can clamp their positions down to some particular values. So in this particular case here, in this equation, I have it such that big R here are the nuclear positions. So there's going to be more than one nucleus generally. So there's going to be R1, R2, R3. All those crammed together, I'm just going to denote big R And that's where the nuclei are. And so in Born-Oppenheimer approximation, we clamp the nuclei down. And then what we're left with is the electrons. The electrons whiz around. They move in the field dictated by those nuclei. So we have a Schrodinger equation that describes the motion of the electrons in the presence of the nuclei. And so here I'm using the lower case letters, lowercase r. So they'll be, again, many electrons, so r1, r2, r3, so on and so forth. And it's the electrons that are really doing all the dirty work in this equation. They're the ones that are moving around. I've clamped the nuclei down. They're just parameters in this equation. And so what I have is I have an electronic wave function that describes the distribution of electrons. I have an electronic Hamiltonian that governs the motion of the electrons. Then I have an electronic energy after I've solved this Schrodinger equation. And the nice thing about this is that for an arbitrary number of electrons and an arbitrary number of nuclei, I can in one line, in about 10 seconds, write down the Hamiltonian. So the Hamiltonian is pretty easy to write down. I've even written it right here. So I did use the cheat of using atomic units. So the reason there's no H bars or mass of the electrons or any of that appearing in here is because I chose atomic units. But then the ingredients of my Hamiltonian are all pretty well-defined. So I'll have some kinetic energy. And that kinetic energy will just be for the electrons because the nuclei are fixed. I'll have an electron-electron repulsion term. So of course, the electrons repel each other. I'll have an electron nuclear attraction term because the nuclei are fixed in positions, and the electrons feel a potential due to that. And then I have a nuclear repulsion term. And in this case, because the nuclei are fixed, that nuclear repulsion is just a number because I know where all my nuclei are. They have some particular repulsion energy. That's just some number that I add onto my Hamiltonian there. But if I put all those things together, I get the Hamiltonian. So I can write down the Hamiltonian without too much difficulty. But of course, the interesting thing here is this Hamiltonian depends on these nuclear positions. So I have to know what-- depending on where the nuclei are, I'm parametrically changing this Hamiltonian. That changes H, and it also changes the eigenvalue, the electronic eigenvalue E. And so there's an obvious question of, well, what is this electronic eigenvalue? It turns out to be a very important thing. We will typically call this thing a potential energy surface. And it turns out to govern a whole host of chemical phenomena. So the first thing that it could generate that we can look at-- we've already actually seen potential energy surfaces. So on the left-hand side here, I have the picture from the case of H2-plus. So here we had the energy of the sigma orbital. And as a function of R, it forms some bonds. So as we made R shorter, the sigma orbital formed a bond. Then we had also this sigma star orbital as a function of R. When we brought the atoms together, in that one there was no bond, no swarming energy. It just went straight up. So we've got two of these things, and I should label my axes. So what I'm plotting here is the energy as a function of the bond distance R. So this was our first example of a potential energy surface, the electronic energy as a function of R. And we see that it taught us about the bond, the bond strength, so forth. There's one other thing that this picture reminds me of, and that's that for the same problem here, I actually had two different potential energy surfaces. I have the sigma potential energy surface and sigma star. And that's because I left something out in my Schrodinger equation up here. There's, of course, an index. I have different eigenfunctions of the Schrodinger equation. So I have an electronic Hamiltonian. It'll have many different eigenfunctions. So I have an index n for those eigenfunctions. And each of those eigenfunctions will have a different energy. And each of those energies is a different potential energy surface. So the sigma state was the lowest solution. The sigma start state was the next lowest solution. They had different potential surfaces. But I will often not refer to potential energy surfaces plural but instead to potential energy surface singular because most chemistry occurs on the lowest potential energy surface. So it involves just the lowest potential energy surface. And the reason for that is because electronic energies tend to be very big in chemical terms. So the difference between sigma and sigma star is several EV, typically. And chemical reactions on that same scale are usually tenths or hundreds of an EV. So the chemical reaction energy might be somewhere way down here, way way, way below the amount of energy you would need to get all the way up to sigma star. And so the sigma star, potential energy surface is just irrelevant. It's there, but it doesn't participate in the reactions at all. So in most cases, knowing about the lowest potential energy surface gets you most of chemistry. But then we can also, sort of, schematically talk about how these potential energy surfaces will look for a more complicated example. So let's say I had the water molecule, HOH. So this is a more general case. We have at least two coordinates here, R1 and R2. There's also going to be an angle, but I can't really plot three-dimensional hypersurfaces, even with my computer. So I'll stick with two. There's a third coordinate there, which is theta, which I'm not going to play around with. But we'll play around with just R1 and R2. So I can change individually the OH bond lengths in this molecule. So then this is going to give me the energy as a function of R1 and R2. So it'll be a surface. So it's a function of two variables. That's a surface. And that surface might look like what I have plotted there, where it would have various peaks and valleys on it. The very, very lowest valley, the very lowest minimum, just like we found for H2, that lowest minimum was the equilibrium bond-- bonding configuration. So for this two-dimensional potential surface, that lowest minimum's going to be the equilibrium configuration. But there's also the possibility of other minima on this potential energy surface. Those would be metastable intermediates, things that you could get trapped in, things that would live for a while, but eventually go down towards the lowest minimum. Then there'd also be reaction barriers. So these are the things that we learned about governing how fast things convert from these metastable states into the most stable states. So that's what you would get if you had just two coordinates. And so then you can sort of try to generalize in your head, well, what if we had 3, 4, 5, 6, 127, 368, lots and lots of coordinates? In some many dimensional space, it would still have the same qualities. There'd be minima and barriers and so forth. And we would get those out of the Schrodinger equation, the electronic Schrodinger equation. And so that is the electronic structure problem. How do we accurately solve for the electronic eigenvalue and the electronic wave function for an arbitrary molecule? And the key word here is accurately. I didn't just say how can we solve for electronic energy and the electronic wave function because, in general here, exact solutions are impossible. And I don't mean impossible for 5.61. I mean impossible for humans or computers. So in general, electronic structure theorists love approximations. We love to make approximations because that's the only way that you can make progress. And so what we're going to learn about today and next Monday are the, sort of, different categories of approximations that we make and what the pluses and minuses of those are. So I'll pause there and see if anybody has any questions. All right, so we'll move on then. So we have an out-- so there is actually an outline of how different modern electronic structure methods make approximations. And these follow the same kinds of steps that you would use in a molecular orbital calculation. So the first thing that you typically had to do in these calculations was to choose a set of basis functions. So you had to choose like 1s function over here or 1s function over there or a 2s function or 2p. You had to choose some set of atomic orbitals that you're going to use as the starting point of your calculation. Then you made linear combinations of those things to get better functions. And so then after you'd chosen that basis, you had to build some matrices. Then you had to solve the eigenvalues problem for that matrix. Then you had to pick out which orbitals you were actually going to occupy. Those were the next few steps. And then finally, you had to compute the energy. Now, that's already kind of a detailed outline. If it seemed like we had to understand the nuances of every one of those five steps, we'd be kind of-- well, you can do that. It would be an entire course unto itself. And that's not what we're going to be doing. The reason that we can avoid going over every single nuance is twofold. So first, we can avoid, more or less, steps two through four because they're done automatically. So once you have chosen your basis that you want to use, the computer knows how to build matrices. It knows how to diagonalize matrices and find eigenvalues. It knows how to pick out which orbitals to occupy. It knows how to do all of those things without you telling it to do anything. You don't really have to change anything. So the only things you have to worry about are choosing a good basis and then also telling the computer a good way to compute the energy. The other reason that we can actually do something useful here is because using a computer to solve an electronic structure problem is much like using an NMR to get a spectrum of a compound. I doubt that anybody in this room could build their own functioning NMR. I know generally the principles of how an NMR works. But building from the ground up-- oh, so we do have any takers? Anybody able to build and NMR, huh? AUDIENCE: [INAUDIBLE] [LAUGHTER] TROY VAN VOORHIS: From the ground up. I'm betting-- no. See? You have to actually, like, make your own superconducting magnet. You've got-- from the ground up, actually building your own NMR would be very, very difficult. But that doesn't preclude you from knowing how to use an NMR because you know, OK, well, this piece is-- this dial is roughly doing this, so I change this over here. I shim this. That's how I make the NMR work. It's the same thing with electronic structure code. You need to understand a little bit of how they work in order to use them effectively. But you don't have to understand every single line of code that went into building it, which is good because there's several hundred thousand lines of codes in many of these electronic structure packages. So we're going to try to understand the principles that go into this so that we know how to be good users of computational tools. And we can actually, then, organize. So roughly speaking, we'll be focusing on step one today and step five next Monday. But we can organize these approximations on a sort of two-dimensional plot, which just sort of summarizes the whole idea of what we're going to be getting at. So the idea here is that on one axis we have the step one thing, which is the basis, the atomic orbital basis. Then we can organize those basis, those choices of basis, from, roughly speaking, on the left hand, we have bad choices. And on the right hand, we have good choices. Now, you might say, well, why would we even include choices if we know they're bad? Because the bad ones also happen to be the fast basis. And the good ones tend to be slow. And so it's a time trade-off. So you have a lot of time, you might choose a really good basis. If you have not a lot of time, you might choose not a very good one. And then we can do the same thing with the energy. We have different choices of how we compute the energy. And we can roughly arrange those from bad choices to good ones. But again, these bad choices are fast, and the good ones are slow. And so the general thing that we want is we want to get up here into the upper right-hand corner, where the exact answer is. Somewhere up there, with a very good energy and a very good basis, we're going to get the exact or nearly the exact answer. So that's where we want to get. But moving from the lower left to the upper right. Every time we move in that direction we're making the calculation slower and slower and slower and slower and slower. Until eventually we just lose patience. And so then we cut it off and say, all right, this is as far as I'm willing to go. What's the best answer I can get? What's the most best-cost benefit analysis I can do? And so in terms of being good users of electronic computational chemistry, it's about knowing, OK, well, if I've only got three hours for this calculation to run or two days or however long I'm willing to let the computer run on this, what's the best combination of a basis and an energy to get me a decent result in that kind of time? So questions about that before I move on? OK, so we'll focus on choosing an atomic orbital basis. So we already have chosen an atomic orbital basis before. So when we did H2-plus, which is, sort of, a standard problem that you can actually solve by hand if you choose the right basis, when we did that calculation, we chose to have an atomic orbital on A, the 1s function on A, and the atomic orbital on B, 1s on B. And those two things together formed our basis. So we wrote our molecular orbitals as linear combinations of those two functions. We got sigma and sigma star out. And we were able to work out the energies. Now, this type of basis is a useful basis. And it has a name, and it's called a minimal basis. And it's minimal because you could not possibly do less. So if you want to form a bond between two atoms, you need at least one function on each atom. If you had only one function, period, you couldn't really form a bond. So it's as low as you can go, can't go lower than this. Now, you could think about the possibility of adding additional functions. So I could add in, say, the 2s function on A, the 2s function on B, dot, dot, dot, dot, dot. I could come up with various things. Now, you might ask, why would I want to do this? And there's a very good reason for doing this, which is one of the principles that we need to learn in constructing basis functions, choosing our basis, which is that adding basis functions always improves the calculation. So even though I think that adding 2s A and 2s B, why would those be important? I can't really say. Adding basis functions always make things better. And the reason for that is because we were doing a variation calculation. We're trying to approximate the lowest energy of the system. And just choosing the 1s A and 1s B functions doesn't give you the lowest energy. It may be close, but it's not exact. So then by adding the 2s functions, the result could get better by adding a little-- by making C3 and C4 not quite 0 here, maybe the answer gets a little bit better. Maybe it doesn't, but it can't get worse because the calculation could always choose C3 and C4 0. So adding basis functions always makes things either better, or at least not worse than they were. And so what you'll find is that when we're talking about AO basis, choosing an AO basis, we're going to be choosing bases that are much bigger than your chemical intuition would suggest. So for H2, I think, oh, 1s A, 1s B, that should more or less describe things. And if I was doing things by hand, that's what I would do. But if the computer's doing the work, well, no skin off my nose if this computer wastes some time doing some 2s or 3s or 4s intervals. I'll let the computer do that, as long as it gives me a better answer. So the bases we have will be much bigger because of this. But the other thing I want to note is that when I talk about these 1s functions, you probably all have in your head-- you know what 1s functions look like. They look like e to the minus a times r. So they just look like exponential decay functions. Turns out that for practical reasons, these are inconvenient things to use on a computer. And that is because integrals involving exponentials are not analytic in three dimensions primarily because of the cusp that occurs at r equal 0. So if you multiply two of these things times each other and try to do an integral, you have a cusp over here and a cusp over here. And the integral is just not something that can be worked out. And so when the going gets tough, the tough get empirical. And so instead of using these exponential functions, what we use in practice are Gaussians. So Gaussians are things instead of looking at like e to the minus ar, they look like e to the minus alpha r squared. So if I was to plot a Gaussian here on the same axes and choose the alpha value appropriately, I could get a Gaussian that might look like that. So it would be similar to that actual 1s function. But particularly near the origin, where it doesn't have a cusp, and particularly in the tails, where e to the minus alpha r squared decays very quickly, they don't actually look that similar. So whole reason to do this is not that we have some physical reason to think that Gaussians describe atoms is better than hydrogenic functions. They don't. It's just that we can do the integrals. There's easy integrals here. And if we really do want a hydrogen-like function, we can get that by just including more than one Gaussian. So one Gaussian doesn't look very much like an exponential. But if I take two Gaussians and choose their exponents and their coefficients appropriately, I can make a linear combination of two Gaussians. So this was the one Gaussian result. With two Gaussians I could make something that might look like that. And with three Gaussians I might be able do something that would look like-- and then four Gaussians and five Gaussians and six Gaussians. But it's clear, then, that by using a large number of Gaussians, I can get whatever I want just by brute force. And so those are the two things about building basis, choosing basis, AO bases. And at this point, you should start to feel a little bit intimidated because I've said that for atoms you're going to need more basis functions than you thought just to try to get a variation a lower energy. And those basis functions are likely to be constructed out of Gaussians, of which you'll need a bunch of Gaussians to even really approximate one hydrogen-like orbital. So you're going to have lots and lots and lots of Gaussians on every atom. And you're going to choose the exponents of every single one of those Gaussians yourself. And that would be just horrible to have to do. And the thing that comes to our rescue is a thing that's known as a basis set. So a basis set is constructed in the following way. A graduate student, probably long before you were bored, spent years of their life going through and figuring out, OK, for carbon, what is a good combination of Gaussian exponents? OK, they're 113.6 74.2, 11.3, and 1.6. They wrote that down. They said, OK, now, for nitrogen, what are a good set of Gaussian exponents? And then they went through, and they did this for every element, or at least many, many, many elements in the periodic table. Then they wrote a paper that said here is-- I did it. So I'm going to say here is the Troy basis set. And the Troy basis means when you say you're doing the Troy basis set means you're using those exponents that I wrote down for carbon or for nitrogen or for oxygen or for fluorine or for hydrogen. And it's all predefined. It's all laid out. And if I was really diligent, it's laid out for every element in the periodic table. If I was less diligent, maybe it's only the first two rows or something like that. But the result is that all you have to say is I want the Troy basis set because Troy makes really good basic sets. And so I'm going to use that. And then the computer can go and just say, OK, well, there's some file that has all those exponents. And the computer looks up the numbers and says, for carbon, you need this; for oxygen, this; for nitrogen, this. And so the key idea here is that these are predefined sets or AO basis functions. So somebody already defined these. You don't have to make a choice. Other than choosing the set, there's not other knobs that you have to turn. So you can obviously see the benefit of this, which is you don't have to put down hundreds and hundreds of numbers in order to get the calculation to run. So that's a big win, which hopefully you're remember for the next 25 minutes. Because the downside, then, is that you just noticed I called my basis set the Troy basis set, which would give you absolutely no idea of what was in that basis set, just that I made the basis set. And then somebody else might call it the Cambridge basis set because they did it in Cambridge. And somebody else might number their basis sets. You know, this is basis set number 17. And none of those things actually tell you what's actually in the basis set and how they designed it. They just give it a name. But then that name is-- you have to know the name in order to be able to specify the basis set. And you have to know what-- you have to sort of memorize, oh, well, this basis that does this. This basis set does this. Or at least remember where to look up what those basis sets do. So for the next 20 minutes or so, we're going to talk about some of those things. And you'll be annoyed at the fact that these basis sets have these funky, weird names and design principles, I think. But just remember, I don't have to put in 100 numbers. This is the price you pay for not having to put in 100 numbers. So the first thing that we'll talk about here-- so I'll say that basis sets are typically grouped by row. Of course, the basis sets are going to be different. So hydrogen's going to need a different number of basis functions than carbon. Argon's going to need a different number from carbon or hydrogen just because they have different numbers of valence functions and different numbers of core functions. And I'm also going to introduce a shorthand. So first of all, I'll note that we already have-- so actually, I'll just introduce the shorthand. So, say, for nitrogen and for the basis set we've talked about so far, the minimal basis set, the smallest basis set I can come up with for nitrogen would need to have the 1s function, the 2s function, and the 2p function. I need at least those functions to just have places to put all of my nitrogen electrons. Now, I'll get writer's cramp if I have to write all of these things out. So I'm going to develop a shorthand, which is when I have 1s and 2s, I'm just going to denote that as 2s. So I'm not indicating that 2s is the last function. It's Indicating that there are two of them. So I have two s-type functions, one that's sort of core like and one that's sort of valence like. But there's two of them. And then I have one p-like function. So I'll put 1p So that's just telling me how many of each of these things I have. So again, this is the number of s-type functions. This is the number of p-type function. And so then we can go, and we can talk about the one basis set we've dealt with so far, which is the minimal basis set. So for hydrogen and helium, it's just going to be that 1s function on each of those. And then when I go down to the next row, it looks just like nitrogen. They all are going to need in an s core function and an s valence function and a p valence function. So there'll be 2s, 1p-like. And then I go down one more row for sodium through argon, and I'll need another s function. So it'll be total of three s-type functions. And then I need two sets of p, a core p and a valence p. So there's 2p. So that's the minimal basis. That's the smallest basis that I could conceive of for any atom. And then we're going to have various ways of trying to make these things more elaborate, more broke. And the most common way to do this is to note that, well, when I bring atoms together, it's the valence functions that actually either contract or expand in order to describe the electrons moving around. The core functions don't really change very much. And so in order to give the valence functions flexibility to change, it makes sense to add more valence-like functions. So if the valence was s, it makes sense to add another s function. Or if the valence was p, it makes sense to add another p function. And so we get a name here, this concept, which is minimal basis is what's known as a single zeta basis set because it has just one set of valence functions. It might seem like would make better sense to call it a single valence basis set. But history made a different choice. It said that it's zeta. I don't know why they chose the name zeta, but they did. So the zeta just means that there's one set of valence functions. And then you can think about making a double zeta. So you'd take every valence function, just add another valence function that has a different exponent so it would allow more flexibility. So that'll be a Double Zeta basis set, DZ. So for hydrogen and helium, it's all valence. So we'd just get two s functions. For things in the same row as lithium and neon, the valence functions are in s and a p. And so when we double those, we're going to add another s-type function and another p-type function. So we would go 3s 2p here because we'd add one s function and one p function. And then for sodium through argon, we would have 4s 3p. And then we could go on and talk about, oh, well, what about a triple zeta? So triple zeta would be 3s for hydrogen helium. And then we add an s and a p for the next row. So it's 4s 3p. And then we add s and p for this, and we end up with 5s 4p. And so again, the idea here is that for carbon, we might have 1s that's the core. And then we'd have 2s and 2p as the first valence and then 3p. And these would respectively be-- so the single zeta or-- and then you include the next cell. And it becomes-- or sorry, single zeta would be all of this, sorry. Oh, it's the core. And then you could say, all right, well, but then I include another set. And this gives me a double zeta basis set. And then I could include another set, which would give me a triple zeta kind of basis set. Then you can go on and on and on. So you could just make a quadruple zeta and quintuple zeta basis set. And then this is where the names get annoying. So the name of the most common minimal basis set is STO-3G. And that stands for Slater-Type Orbitals 3 Gaussians, just in case you're wondering why it's STO-3G. And then for the double zeta basis set, there are ones that are called 3-21G, 6-31G, another one that is somewhat confusingly just called DZ, Double Zeta, as if there was only one such basis set. And then for a triple zeta, you can do 6-311G triple zeta valence. And there are others. So there are other abbreviations out there as well. But that's how you would do these kinds of-- how you change the zeta, the valence. So questions about those? Yeah? AUDIENCE: So is there [INAUDIBLE] in the example you're including as your additional zetas, like 3s 3p 4s. But would there be an advantage to including things that weren't atomic orbital-like things? TROY VAN VOORHIS: Yes. So there are occasionally times where people include basic functions that don't look like atomic orbitals. So the most common one is to make bond-centered functions. Like say, I want a bond here, and you could put basis functions there. There's a couple of reasons that becomes undesirable. One is that, then, you can't use a standard basis set because basis sets have to be sort of anchored to the atoms. That's how we catalog them. If you say, oh, I'm going to put one in the middle of the bond, you say, well, it would depend on both atoms involved in the bond and how long the bond was and things like that. So that's a little bit undesirable. The other one is that unless you pin that bond function down to a particular position, if you let the center of that bond function move, there's actually numerical instability that happens when two Gaussians come on top of each other. And that can sometimes cause calculations-- the computer to choke. So that in practice, we usually stick to increasing the atomic basis sets. Oh, but there is also one other case. In physics, they are much more fond of using solids. That's sort of one of the things that chemists tend to think about molecules as the example. Physicists tend to think about solids. And in solids, they're much more fond of using plane waves as their basis set. So they start off with the free-- instead of starting off with atoms as the reference, they start off with free electrons as their reference. And then say, oh, and then we introduce these potentials which are perturbations to the free electrons. So you get plane waves as your basis set. You make linear combinations of plane waves instead of atomic functions. Yeah? AUDIENCE: [INAUDIBLE] atomic orbitals do we need [INAUDIBLE]? TROY VAN VOORHIS: That's exactly the next-- you've hit on exactly the next thing, which is that none of those include any d functions. I mean, I can go up to like 18 zeta, and I would never have any d functions. And on hydrogen and helium, I can go [INAUDIBLE].. I wouldn't even have any p functions. And certainly there are situations where you would want these higher angular momenta. One of them is if you're in an electric field. So that polarizes electrons and tends to actually require slightly higher angular momenta functions in order to describe that polarization. And the other one is for just the directionality of bonding. That if you had, like, an SN2 reaction or something like this, where temporarily something was either actually hypervalent or sort of hypervalent, you would need these higher angular momentum functions to describe the hypervalency. And so those are called polarization functions. So here polarization functions are things that add higher angular momentum and valence. So you need this for things like polarizability, which is why they're called polarization functions. But you also need it for things like hypervalency. And so the nomenclature for polarization functions is also a bit weird. But the simplest one is that you just add the letter P, so P standing for Polarization to the basis set. And then the idea is that you're just going to add one function with that P. You when you add the P, that means you're adding one function that has one unit higher angular momentum. So if P was your highest angular momentum, you're going to add a d function. And if s was your highest angular momentum, you're going to add a p function. So for hydrogen and helium, that's going to add a P to lithium. To argon, it's going to add a D. If you did this also the transition metal atoms, you would note the transition metal atoms have d functions already. So the polarization would add an f function, just adding one higher angular momentum. So then going back to our table up here, let me use my magic, copy the double zeta basis results here-- copy. There's my double zeta. Now I'm just going to add the polarization to this, so double zeta plus polarization. Now I add my polarization functions. And so for hydrogen and helium, I'm going to add one p function to get a DZP basis set. Lithium to neon, I'm going to add 1 d function. For sodium to argon, I'm going to add 1 d function. And then I can do the same thing with triple zeta. So if I make a TZP basis set, then I'll add 1p, 1d, and 1d. And I'll note that you could also do a TZ 2p basis set. And then you're adding-- the 2p just indicates that I'm adding even more polarization functions. So I'm adding two sets of d functions instead of just one or two sets of p functions instead of just one. And then these things have names. So there's a 6-31G star, which is a very common DZP basis set. Or there's also that DZ basis that I mentioned above has a DZP generalization of it. Or you could have for the triple zeta, the TZVP basis, or the longest basis set name, 6-311G (d, p). This is where the annoyance started to come in. You're like this is annoying. But it's less annoying than the alternative. So we deal with it. So questions about that? Yep? AUDIENCE: [INAUDIBLE] 1p-- I'm sorry. That 1p, they're just like, we're going add DZ squared. Or [INAUDIBLE]? TROY VAN VOORHIS: It's one set of d functions. AUDIENCE: Oh, OK. TROY VAN VOORHIS: Yes. Yeah. Yeah? Yep? AUDIENCE: Why can't we have a single [INAUDIBLE]?? TROY VAN VOORHIS: Oh, we could. But, yeah, we could do that. So part of this has to do with-- and there are a couple of basis sets that do that. This mainly represents the hierarchy of empirically what people have found is most important. So if you just start with a minimal single zeta basis set, you say, well, what's the next most important thing? Usually almost always going to double zeta is more important than the polarization functions. So very few people bother making single zeta plus polarization. They'll say, OK, we'll do double zeta. And then after you do double zeta, then maybe you want to go to polarization. Or maybe you want to go to triple zeta, depending on what you're doing. Does that makes sense? AUDIENCE: [INAUDIBLE] TROY VAN VOORHIS: It would be 1s 1p for hydrogen, 2s 1d for carbon. It would just be to add that extra polarization function on top of the existing set. AUDIENCE: And would the polarization be the-- which AO did that come from? TROY VAN VOORHIS: What do you mean? AUDIENCE: So for hydrogen, would it be adding the 2p basis or-- TROY VAN VOORHIS: It would be like adding-- now, so for hydrogen, it would be like adding-- yeah. For hydrogen, it would be like the 2p function. But again, when we go back, we remember that we're not actually using the hydrogenic functions. We're using Gaussians to approximate them. And so all it really means is that it has the same angular distribution as a p function. But the radial part is whatever the person who made the basis set decided to make it. All right, then I have-- so then when I teach this class, I often have clicker questions. So this would have been the clicker question. So we'll do it by show of hands instead of by clicker, since we're small enough. So here's a test of whether we understood-- I can't show-- well, anyway. So I'll go back and put things up in just a second, if you need them. But my question is, how many basis functions will there be for C60 in a DZP basis set? I guess should say [INAUDIBLE]. So option A is there'll be 60 basis functions. So that would be how many basis functions per carbon atom? One. All right, or 180, that's three per carbon. 360, that's six per carbon. 840 is-- I can't-- anybody do the math better than me? Is that 14? No. No way. Yeah, it's 14. Yeah, 14-- 14 per carbon. This is-- huh? AUDIENCE: 18. TROY VAN VOORHIS: 18. And then this is 21, then, right? Yeah, so we've got, 1, 3, 6, 14, 18, 21 as the number of basis functions per. And I'll go back and throw this up for those who want the table up. We didn't write all that down. So here's our EZP basis set. Carbon's in this group here. All right, so then now I'll go back. Everybody feel like they got their answer? OK. So by a show of hands, how many people think the answer is A, 60? Nobody, all right. How about B? C? We go most on C, OK. D? I don't know, E? F? Everybody says C. C is not the correct answer. And every year people miss this question. So the key thing to get this right-- and we're going to re-poll in a minute here. But the key thing to get this right was actually the question that you asked, which is that you said, when I add a d function, do I add DZ squared, or do I-- and I said, no, you add a set of d functions, which is how many functions in d? 5. All right, so now we'll go back and look. All right, but don't feel bad. Every year, the majority of people choose that same answer. So now let's redo our math. AUDIENCE: I've never gotten it right. TROY VAN VOORHIS: Never gotten it right? OK. [INTERPOSING VOICES] All right, so now we'll go through. OK, how many people think A? How many B? Nobody's going to say C because I already told you that was wrong. How many people say D? All right, good, you're all right. Yes, the answer is D. In a DZP basis there are 14 basis functions per carbon atom. Now, you can already start to see even for this DZP basis, which isn't the biggest basis we've talked about, for C60, you've got 840 basis functions. That's a lot of basis functions. So that's why we're really glad that the computer's doing the work and not us because 840 integrals-- well, 840 by 840 matrices are hard to diagonalize by hand. Doing all those integrals is really, really a pain. So then going beyond those basis-- so those two things are the key basis set ideas. I'll just touch on a couple of other things before we finish up. So the first thing is diffuse functions. So occasionally, if you have anion, so you have an extra electron, that extra electron is not very well described by the valence because negative electrons tend to spread out a great deal. Bob is one of the world experts in Rydberg states, which is where these electrons spread out a really, really significant amount. And so these Gaussian functions that decay quickly don't describe those well. And so you have to add a Gaussian with a small value of alpha. And it's mostly useful for anions. So you add these in order to describe anions better. And the notation here is either you use the phrase aug or plus to the basis set. So if the basis set has the word aug in front of it, that means it has some diffuse functions on it in order to describe those weakly bound electrons. Or the plus indicates it has some of those functions in it. And again, those are added with the same angular momentum as the valence. They're not usually polarization and diffuse at the same time. And then the final thing is you can generalize these ideas to transition metals. It's a little bit hazy because a lot of this is predicated on us knowing what the valence of the element is. And for transition metals, it's like, well, is s in the valence? Yeah, probably. Is p in the valence? Well, maybe. I don't know. D is definitely in the valence. But depending on what you think the valence is, your definition of some of these things is a little bit fuzzier. Sometimes you add a p function. Sometimes you don't when you go from double zeta to triples for a transition metal. But the overall idea here is that we're trying to approach this particular limit, which is known as the complete basis set limit, which is the result that you would hypothetically get with an infinite number of atomic orbitals. So you just crank up, include all the atomic orbital in the universe. You would get to an answer that would be the right answer. And we're just trying to asymptotically approach that by making our basis sets bigger and bigger until we run out of steam. All right, so that's everything about basis sets. And tomorrow-- or not tomorrow, next Monday we will talk about how you compute the energy. All right, Happy Thanksgiving. |
MIT_561_Physical_Chemistry_Fall_2017 | 24_Molecular_Orbital_Theory_I_Variational_Principle_and_Matrix_Mechanics.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: So in the previous lecture, I told you about many electron atoms. And in order to describe their structure, we have to build a Slater determinant, because that builds in the antisymmetry between all pairs of orbitals. Now, Slater determinants are expanded into a very large number of additive terms. And there are algebraic methods, which I describe in a supplement, which enable you to reduce these algebraic terms, the sum over an enormous number of terms, to something almost identical to what you would get if you didn't antisymmetrize. They're just a few extra things. And so once you know the mechanics, it's not a big deal. It's a little bit more work than just calculating the individual orbital matrix elements, but it's nothing special. And Henry has produced a beautiful thing, a beautiful handout on Slater-Condon parameters, which is what you get when you evaluate these Slater determinal matrix elements. And they are the things that you use to describe electronic structure. So electronic structure is a really complicated many electron descriptive thing. But if you can reduce for every atom to a small number of parameters, most of which you understand intuitively how they depend on properties of the atom, like the ionization energy of the individual orbitals, you can say, yes, I understand all of the periodic table. That's where it comes from. The rules enable you to identify the lowest lying states and to calculate their properties and to be able to intuit their properties and to be aware of when something unexpected happens, which is what we're always looking for as physical chemists. We're always looking for a simple minded picture that breaks. Because then we can go deeper into our understanding of all the pieces. But we're always parameterizing. OK. So there's a lot of subtleties that I didn't have time to talk about in the evaluation of these Slater determinal matrix elements. And since whatever problems I give you connected with Slater determinal matrix elements or Slater determinants are going to be really simple. They're probably going to be 2 by 2 or 3 by 3. You don't really need to know and understand all the rules and all the shortcuts. But you've got the basic mechanism. So a Hamiltonian doesn't depend on spins. But because of antisymmetry, the spatial part of the wave function does depend on which spin state you've got. And so the Hamiltonian knows about spins in a peculiar, indirect way, which is beautiful. You've got to admire how clever this Hamiltonian is to be able to capture these spin effects. And of course, you have to capture them too. All right. So today we're finally getting to molecules. And the simplest molecule is the homonuclear molecule. Two hydrogens with one electron. That's the simplest problem. And because it is representative of all of the qualitative effects and most of the quantitative calculations you do, it's a really great introduction, because it captures most of the important stuff that molecules or small molecules do, that the electronic structure of small molecule. So the key to being able to describe small molecules using this LCAO, Linear Combination of Atomic Orbitals method, is the variational method. And this variational method says, we're going to throw a whole bunch of basis functions at the problem. A whole bunch of orbitals, which we choose because we like them or because the computer likes them. And we find that the linear combination of coefficients of these orbitals that minimizes the energy. And you can do this in wave mechanics or matrix mechanics. It's more elegant in matrix mechanics, and you should be comfortable going from one to the other. Now, I'm only going to be talking about a 2 by 2 problem. So you can go easily from one to the other without even taking a breath. But anyway, OK. In approaching this, you're going to see what's called the general eigenvalue problem where instead of diagonalizing a simple matrix, you have to deal with the Hamiltonian matrix and an overlap matrix, which makes the solution of the secular equation, finding the eigenvalues and eigenstates a little bit more complicated. So you're going to see it first appear here. And it can be solved by a simple matrix equation, which is in a supplement to a later set of lecture notes. But this is the thing that consumes the vast majority of computer time in maybe in the world, but certainly among chemists. So the crucial thing, and I wrote it all in caps, in dealing with this complicated many body problem, we make a crucial assumption. And that's called the Born-Oppenheimer approximation. It's based on the fact that electrons move really fast and nuclei move really slowly relative to electrons. And so what we do is we solve the problem with the nuclei clamped. Because that's a good approximation. And then after we solve the electronic problem, we have what we call a potential energy surface, which is the electronic energy as a function of nuclear coordinate. And it's parametric. But then once we have the potential energy surface, we can calculate the rotation and vibration. So basically, what we're doing in making the Born-Oppenheimer approximation is we're setting up to treat vibration and rotation by perturbation theory. Now, you never see it mentioned in that way, but that is, in fact, what we're doing. And the reason we can get away with it is the difference in velocity of electrons versus nuclei. But it's a fundamental philosophical point of view that we take about molecular structure that we can clamp the nuclei, get most of the insight, and then allow the nuclei to move and the electrons with them and calculate everything else. So this is central to chemistry, the Born-Oppenheimer approximation. Without that, we would have a bag of atoms. And that would be a very bad starting point, because we believe in bonds. And a bag of atoms is saying, well, yeah, we could be statistical. We could say every atom is capable of being anywhere and they don't interact in any rational way, and we would have no insight. So this is where most of the ideas that you learn to use as physical chemists and probably any kind of chemist comes from the Born-Oppenheimer approximation. And it's widely misunderstood and misrepresented. But we won't go into that. OK, we're going to derive two orbitals for hydrogen. H2 plus. The sigma bonding orbital and the sigma star anti-bonding orbital. This is enough to basically understand how you get bonding and anti-bonding in any molecule. And I like to say that we understand bonding in a very simple way. If you have wave functions for two adjacent atoms and they overlap in the region between the two nuclei, you get constructive interference, you get twice as much amplitude in the region where the electrons interact with both nuclei, which means four times the probability. So a little bit of overlap leads to a big energy effect because of this constructive interference. And that's really, really important. And all sorts of hand waving arguments about what makes a chemical bond have a lot of currency, even here at MIT, but that's what a bond is. It's constructive interference. All right, so let's get started. And the lecture notes are extremely long and extremely beautiful and clear. They're written by Troy. I have reduced them to a minimal picture, but I don't want to detract from them. The insight is very rich. OK, so what we have to do first is to look at a coordinate system. So we have one hydrogen nucleus. Another one. There's a center of mass. And so we say we have RA, this vector from the center mass to this RB like that. And up here is the electron. And it's at position R. This is at position RB. Position RA. Now, the center of mass can be anywhere in space, and we don't really care. So we tend to not talk about where the center of mass is. But there is another set of coordinates for it. OK. And now we have this, which is called R sub A. And it's a vector. And this is R sub B. So we have uppercase symbols for nuclear positions and lowercase for electrons, except they're between electrons and nuclei. OK. And once we have this picture, we can write what the Hamiltonian is, because we're basically looking at the interactions between these three particles. And so we have kinetic energy and we have potential energy. And the Hamiltonian-- excuse me. --in atomic units is this del squared, which is the second derivatives with respect to whatever we have here, which is little r. And we have 1/2. It's 1/2 because of atomic units and because the mass of the electron is treated as one. And then we have a similar sort of term for del for nucleus A. And we have a two, because 1/2 in atomic units. But we have the mass of A. Now, the minus signs, why do we have minus signs here? You know this. AUDIENCE: This is the Schrodinger. ROBERT FIELD: Yeah, right. Or we say it's p squared over 2m. And p is-- right. OK, but what we're trying to understand stuff, this is negative. That means it sort of leads to a stabilization. And many people talk about bonding in terms of velocity or momentum, but I don't. And then we have del squared B over 2 and B and then a repulsion term. A whole bunch of terms that we have. So these are the kinetic energy terms. And then we have the potential energy terms. 1 over RA vector minus little r vector minus 1 over RB minus little r. And then the repulsion between RA minus RB. OK, that's it. Now, that's a lot of stuff. And we'd like to make a lot of it go away. But again, we have potential. We have kinetic energy and we have potential energy. So as I've written it, the Hamiltonian, which is our big problem, is a function of ra rb r big RA big RB. That's just ridiculous. We don't want to look at a function of all these variables. We would like to be able to simplify it so that we can draw pictures on a two dimensional surface or just to reduce it now. We like to think in terms of orders of magnitude. And so it is a very simple matter to say, well, let's simplify this Hamiltonian by clamping the nuclei. They don't move. If they don't move, then the big Rs are no longer operators. They're just parameters. And the contributions to the kinetic energy of the heavy particles gone. So there's no rotation and no vibration. Now, it doesn't go away. We have to come back with it. But we come back with it when we have a good representation of the electronic structure for the nuclei clamped. And if we can do it for one internuclear separation, we can do it for all internuclear separations. And so if you have a grid of say 1,000 points, you do the same calculation 1,000 times, except you don't. Because you know how to extrapolate from one point to another. But basically, you are thinking about this as the problem of finding the electronic wave function at a grid of points. And so that turns all of the big R's into parameters, not operators, and simplifies the problem enormously. And so the reduced Hamiltonian becomes-- we'll call it the electronic Hamiltonian. R semicolon ra rb. So r is the variable, and these are just parameters that we sometimes include. And then the Hamiltonian is just del squared little r over 2 plus V effective. And this would be as a function of RA and RB. But it's because of R approximation, it's just a function of R. So these we pick. And this is the thing that we're worrying about. And then there's the 1 over RA minus RB, which is just a constant. This is the repulsion between two charged particles. So this is a lot nicer. We can also sort of forget about this. I mean, it's there, but it's a trivial thing that we can add in. So we have basically the kind of problem we like. A kinetic energy problem and a potential energy problem. So we're going to take this approach and we're going to learn how to apply it to the simplest system. And the simplest system is homonuclear. That's better than heteronuclear, because if we had two different atoms, we'd have to somehow represent their difference in properties. But because it's homonuclear, we just basically have an atom, the same atom, repeated, and there's a lot of symmetry that results. And the other is only two electrons. I mean, sorry, only one electron. Remember, we did hydrogen first, and that was wonderful. We got everything that was-- basically everything is related to anything else. Any property is related to any other property of the hydrogen atom through this orbital approximation. And so one electron makes us feel really comfortable, because we are going to expect that there is some structure, which means not just a random number or random collection of descriptive observations but some relationships. And we don't have to even antisymmetrize. So it's really something we can hit out of a park without getting too stressed. But there is a problem, and that is, how do we find the wave functions if we take a simple basis set? 1s a plus 1s b. So these are hydrogenic wave functions. We know them. Now, they may not be the most convenient thing when you're doing a large scale calculation. In fact, nobody uses hydrogenic functions, even though they make a lot of sense, because they're not computationally as easy as Gaussian functions. So if you need to describe a problem with, say, 10 hydrogenic functions and 1,000 Gaussians, you're way ahead of the game when you use 1,000 Gaussians, because the integrals are trivial. But in terms of understanding, it's really nice to take things we know and do everything in terms of them. OK. And so this is where we get the variational. So what we do is we say we're going to express the wave function as a linear combination of atomic orbitals. And that means we can adjust the coefficient of each atomic orbital to minimize the energy. And the variational principle or the variational theorem says, no matter what you do, you can't calculate an energy lower than the lowest energy of the system. Even if you don't know what it is, you know that anything you calculate is a little bit higher. And what you want to do, then, is to make the energy as small as possible by doing a min max problem on the coefficients. And so the variational method is basically that. OK. So basis set. I'm probably not going to finish my notes, even though these are a massively reduced version of what Troy had written, because I have some other things to say that aren't in the notes, and they're really important. When we do perturbation theory, we implicitly are dealing with an infinite basis set and an infinite set of known matrix elements. But we can't ask a computer to diagonalize an infinite matrix. So we have a tricky way of doing this so that we can use perturbation theory to get a good approximation of almost anything you want. But if your basis set is really terrible, then the perturbation theory is going to be terrible too. I mean, it's going to give correct answers. You're just going to have to do more calculations. The variational method says, OK, let's choose not an infinite basis set but a physically appropriate basis set or a computationally minimal basis set. And so we are making decisions about form and size. So we can use hydrogenic orbitals, because we know what they are. We know how to evaluate lots of integrals with them without too much trouble. And how many? Well, we've got two atoms. And so you sort of need one orbital on one atom and one orbital on another to even begin to capture the molecularness of the problem. So we've chosen the simplest possible form but not necessarily the most convenient form. And the size is two. Now, we could use four. We could use 100. And you could use a million. Most quantum chemical calculations, which you're going to be doing soon, involve not just millions but often billions of basis functions. And the computer does all the work and doesn't complain to you. It doesn't require a trip to the beach or anything like that. It just says, OK, here is what you wanted me to do. I don't know whether you-- it's a stupid question or a smart question, but I'm answering not what you thought you asked but what you did ask. OK. But we're going to show how it works with the simplest possible basis set. So our electronic wave function will be a function of electron position and parametrically nuclear position. C1, 1S A R R. I'm going to stop including this parametrically dependent on R. But eventually I'll do that. But for now. OK, so that's the wave function. The variational function. And we need to be able to calculate the average energy, because we know the average energy is going to be larger than the true energy. But if we do a good calculation, the energy difference between the truth and what you get is small as possible. And so you can say for a particular basis that we have, the best possible energies, if you want to do better, you have to use a bigger or smarter basis set. And so how do we calculate the average energy? Well, we do the usual thing where you say psi H psi d tau over psi star psi d tau. Now, I'm including this because we need it when we don't have orthonormal basis functions. And for this basis set, they're not orthonormal. They can't be. The wave functions for one atom are all mutually orthogonal. But because you have two atoms and there's overlap between them, they are not orthogonal, and they're not normalized. And so we have to do this. And so the problem is we want to minimize the energy with respect to C1 or C2 or C1 star or C2 star. Now, this is a subtle point. The complex conjugate of C is linearly independent of C. But they're related to each other. And so you can say that these derivatives are 0, and you can choose C1 and C2 or C1 star and C2 star, and you get the same result. And so it's important to understand this. So this is all by means of introduction. I'm going to start repeating myself. But let's draw some pictures. So the complicated thing is the potential. And so to illustrate the R dependence, I'm going to try to draw things. So we have r small and R large. This is the positions of the nuclei. And the potential will look like this. So the potential goes to minus infinity at the position of the nucleus, because the electron is attracted to the nucleus. And when the distance is 0, we have an-- it's a forgivable infinity because it-- anyway. So what I should really have done is exaggerated here. So when the particles are close together, the potential in the region between the nuclei doesn't go as close to 0 as it does when we do this. And so that's an important fact. So as the particles are brought together, something happens in this intermediate region. And when they're completely separated, it's as if they are completely separated. They're not talking to each other. And the wave functions look like this. I missed. And this is what I was telling you. This is a bond. This is why we have a bond. Because we have constructive interference in the spatial region where the electron can be attracted to both nuclei. Whereas here, there is no amplitude in between them or negligible amplitude, and so there isn't much of a contribution from the attraction to both nuclei. And so this is the separated atom or approaching separated atom limit. And this is the molecular limit. And now it's possible that you can squeeze the atoms too close together and bad things start to happen. Because then what you're doing is you're putting two electrons into the same spin orbital. And we know that that's overlap repulsion or it violates the exclusion principle and there is an energy penalty that you pay in order to get too close. But basically, this is what we're going to be explaining. And we're going to recover this from a real calculation. OK. So again, we want to minimize the average energy. And the average energy, as I've written somewhere, is given by psi star H psi d tau psi star psi d tau. And if we are writing these things as C1 1S A times C2 1S B, then we end up getting stuff that we know how to calculate now. 1S A. This orbital is not orthogonal to 1S B. It can't be. And so this is something that we had been able to avoid in all of our one dimensional problems. We could always think about a orthonormalized basis set. But the effect that we're really interested in is based on the fact that these guys aren't orthogonal. You can't just say we're going to write an orthonormal basis set. It's not possible. If you want to construct an orthonormal basis set, there is a transformation you can do. And it's a simple transformation, and it's done all the time. And you'll see it, but not today. But you still do the calculations on the non-orthogonal basis functions. And then you do a transformation that takes into account the lack of orthonormality. For now just accept it. And so we're going to have to worry about an integral like this one. SA 1 SB. And we call this the overlap integral. And this is something that we have to not ignore. So now we can approach this problem using wave function notation. And it's most familiar. But we can also use these things saying, OK, this is the complete set of the variational coefficients. And this is a step towards the matrix picture. Because it's basically a linear array of mixing coefficients. And you could think of that as a vector. And in fact, that's how you do the vector picture, but we'll use sort of a hybrid notation for the time being. But we can also say, OK, this means C1 C2 Cn. And this means C1 star Cn star. And here we are in the matrix picture. OK, but I will use sort of a hybrid picture until you're ready to take the complete jump. And so the average energy expressed as a function of this set of coefficients is going to be integral. Psi C star H psi C d tau over integral psi C star psi. The reason I'm being hesitant here is because I work in the matrix picture, and this hybrid picture is away from both comfort zones. But it is leading you towards what you need to see, and so we're going to do it. OK. So in the wave function picture, this thing becomes just C1 1S A plus C2 1S B, the whole thing conjugated, H C1 1S A C2 psi 1S B d tau. And the same sort of thing down here, except no Hamiltonian. And so we can now slavishly write down all of the terms that result. And what we get is from the Hamiltonian matrix element. So for H, we get C1 star C1 H11 plus C1 star C2 H1 H12 plus C2 star C1 H21 plus C2 star C2 H22. OK, we think we know some stuff. We think this is going to be the energy of the hydrogen atom. But it isn't. Because the orbital one sees both nuclei. And so this is something which we just call this guy. We're calling this epsilon. And that is a function of internuclear distance. But this guy is the same, because it's homonuclear. And this, well, we can call it V12 or V. And these two guys are the same thing. So we've reduced the problem to a simple equation. But I'm not going to go all the way there. But you know how to do this. You know how to do the overlap integral. You're gonna have a similar sort of thing. S11 is the overlap of the orbital 1 with itself. And that one is 1. But S12 is S. So anyway, what we end up doing is a lot of algebra. And what we end up getting when we impose the condition that the partial of everything with respect to C1 star is equal to 0, we get an equation for the average energy. And the average energy is-- oh, I'm sorry. C1. When we take the derivative with respect to C1, what we get is this equation. C1 star H11 plus C2 star H22 over C1 star S11 plus C2 star S21. And so we could also do the same thing, taking the partial derivative with respect to C2 and setting that equal to 0. But we get another equation. And so that other equation would be E average is equal to-- no, I'm not even going to write it, because I didn't write it in my notes, and I'm operating on batteries. You can write that. And so we have two equations for the average energy. And well, that's kind of good, because we have two unknowns. So we're going to be able to solve for them. OK. Now, we could also approach this problem using matrix methods. And since that's the way computers think about the problem and that's the way most people think about the problem, you really want to understand the matrix picture. And at risk of repeating myself, we know for this two level problem, we have H11 H12 H21 H22. And we know that the S matrix S11 S12 S21 S22. So we have an equation C1 star C2 star times the H matrix is equal to E average times the C1 star C2 star times the S matrix. So this is the H matrix and this is the S matrix. And you can write this down. And when you take the derivatives with respect to C, you get C dagger H is equal to E average times C dagger S. Now, if instead of taking derivatives with respect to the Cs we took the derivatives with respect to the C stars, we'd get a different equation. And that equation would be HC is equal to E average times S then C. This guy looks familiar. It looks like Hamiltonian times something is equal to the energy times-- uh oh. We have something else here. It's not just the energy. We have the overlap integral. So these two things are examples of the general eigenvalue equation. And so in almost all calculations, the fact that S is present leads to problems of algebra. But in this particular problem, we know S12 is equal to S21. And this epsilon parameter can be-- so we have H11 is equal to H22. They're both parametrically dependent on internuclear distance, and they're equal to epsilon of R. So what we end up with is an equation that looks really familiar. We have a 2 by 2 matrix, which we can diagonalize. We can find the energies. And so what you end up getting when you do that. And I want to spare you my attempting to do a derivation in real time. You can look at the notes and you can see that the results are what you care about and what we're going to interpret. So here we go. So what you find. The average energy for one of the orbitals that we get. Remember, we got a 2 by 2. We're going to have two eigenvalues and eigenstates. And so for one of those, we get epsilon plus V12 1 plus S1. So these are symbols that we've seen. And this tells us now this, this, and this are functions of R. So we're going to have an energy for the sigma orbital, which is a function of R. And we have another solution, which we're calling sigma star. And that one is epsilon minus V12 over 1 minus S12. So we'd like to be able to know which of these is the absolute-- one of these is-- so we have sigma and sigma star. Both as a function of R. And so if we draw a molecular orbital diagram and we say, OK, well here we have the energy of 1SA, which is epsilon as a function of R. And this is epsilon energy 1SB, which is equal to epsilon of R. Normally we choose R as equal to infinity for the separated atom. So these are actually just the energies of the individual orbitals. And what we end up getting is a molecular orbital diagram which, in an exaggerated form, is like this. We get bonding, because it's more stable than the separate atoms, and anti-bonding, which is less stable. And so one question we'd like to ask is, well, one question is why is it bonding? Why is this more stable? And the answer is constructive interference in the important energy region. But another question is, is this bigger? Is this a larger destabilization than this stabilization? And it's kind of hard to scope that out, because here we have two parameters, which are added, and here we have a difference of two parameters. And here we have a denominator, which is going to be larger for the sigma orbital than the sigma star. And so we don't know whether the plus versus minus V is dominant. But the energy denominator is saying, yeah, this one, the anti-bonding orbital is more anti-bonding than the bonding one. Now, you can actually do some algebra to calculate what is the difference in energy between the sigma star as a function of R. But evaluated at R sub E, which is the minimum-- the energy where the bonding orbital has its minimum energy. Then if you evaluate at that particular value of R and you make that comparison B 1S evaluated at R sub B, and you get epsilon times S12 minus-- so I'm doing sigma star, so that's this one minus B12 over 1 plus 1 minus S12. And for the sigma orbital, I have a similar equation, but I get epsilon S12 minus B12 over 1 minus S12. So we have the same numerator for these two cases but a different denominator. And so this is saying this simple variational calculation says you're going to get a pair of orbitals, one bonding, one anti-bonding. And the anti-bonding one is more destabilized relative to the separate atoms then the bonding one is stabilized. And this is a kind of general expectation that the anti-bonding orbitals are more destabilized than the bonding ones. But it's not a proof, because its just a proof for the simplest possible case. So I'm done. And in the next lecture, we're going to use molecular orbital diagrams like this to describe a series of examples. Homonuclear diatomic molecules, heteronuclear AH, and AB. H is special, and this is more general. And what we'd like to be able to do is without a great deal of thought construct the molecular orbital diagram for basically any diatomic molecule using some insight. Now, this also transfers to polyatomic molecules. And it's especially beautiful the way it transfers, because electronic structure is usually centered at a particular part of the molecule. And so, basically, you're reducing polyatomic molecules to a diatomic that you understanding the bonding and anti-bonding. OK, so I'll continue on Monday. And your problem set for this week is due on Monday. And the following set for the next week will also be due on Monday. Not that Monday. The following Monday. And that's the next to the last problem set. So you're inside of the home base. OK, have a good weekend. |
MIT_561_Physical_Chemistry_Fall_2017 | 21_Hydrogen_Atom_II_Rydberg_States.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: This is one of my favorite lectures for 5.61. And I hope you like it too. OK, last time, we talked about the hydrogen atom. And it's the last exactly solved problem. And you've probably noticed that like the way I treated each of the exactly solved problems, I avoided looking at the exact solutions. I said we should be able to build our own picture based on simplifications. And this is something that I really want to stress that, yes, the exact solutions are in textbooks. You can program them into your computers and ask your computer to calculate anything. But sometimes you want to know what it is you want to calculate and what is the right answer or approximately the right answer to build insights. Now, in 5.112 and possibly 5.111 and certainly not 3.091, you learn something about the periodic table. You learn how to make predictions about all properties of atoms based on some simple ideas. What are those simple ideas? I'm looking for three simple ideas. Yes. STUDENT: Things with similar valence configurations behave similarly. ROBERT FIELD: Yes, that's sort of a second order simple idea, that because-- I mean, it's using the idea that if you know that quantum numbers of the electrons and what orbitals are occupied, you can make connections between atomic and some molecular properties. Yes. STUDENT: If you understand the interplay between nuclear charge and electric charge, you can derive large physical magnitudes. ROBERT FIELD: Yes. I mean, there are things like electronegativity, shielding. And these are things where if we have some intuitive sense for how big orbitals are and what is the relative attractiveness to different nuclei of various orbitals, you build up everything. And most of it comes from this lecture. And so I'm going to talk about the hydrogen atom as a model for electronic structure. And in particular, I'm to talk about quantum number scaling effects. And I'm going to be using this semi-classical method a lot. So we know that for the hydrogen atom, you can factor the wave function into an angular part, which is universal. So if you understand it once for any problem, you understand it all the time. And so we can just put that aside, because that's a done deal. And then there's the radial part. And the radial part is different for every central force problem. And so we want to be able to know what goes into the radial part and how to use those ideas to be able to understand almost any problem, even problems where there is more than one electron. So I'm going to be making use of the classical, the semiclassical approach, where the classical momentum that's conjugate to the radius, in other words, the radial momentum, that we know what that is. It's related to the difference between the energy and the potential energy. And we get the momentum from that, the classical momentum function. And that gives us everything, because we use the DeBroglie relationship between the wavelength and the momentum. And I hope you believe what I'm going to try to do in the time we have for this. So there are several important things that I didn't get to have time to do in lecture. And I want to mention them, because I probably will write a question on the final connected to spin orbit and Stern-Gerlach. So you want to understand those. So the schedule for today is suppose you record a spectrum of the hydrogen atom. Now, that spectrum is going to reveal something that I regard as electronic structure, structure in this sense, not of bricks and mortar as you would talk about the structure of building, but what the architect had in mind and what the function of the building was. And so the structure is kind of a mystical thing, in which if you look at a little bit of it, you sort of get the idea of all of the rest. And so here, I'm going to show you how you can approach a spectrum and to know where you are, because the spectrum can be really complicated. And there are certain patterns that knowing about electronic structure enables you to say, yeah, I've got a Google map. I know exactly where I am. And it's based on some simple ideas. So this is first illustration of structure. Then I'm going to go and deal with the semiclassical methods for calculating all electronic properties of the hydrogen atom. And we get this business of scaling of properties with the principal quantum number. And that extends to Rydberg states of everything-- atoms and molecules-- not just the one electron spectrum. And the Rydberg states are special because one electron is special. It's outside of all of the others. And what we're going to learn about next time when we have many electrons or more than one, we don't have to think about antisymmetrization. We don't have-- we can still take our simple minded picture of an electron interacting with something. And so we can then take what we knew from hydrogen and describe everything about Rydberg states of molecules. And that's kind of exciting. And this will be followed by the bad news that when you have more than one electron, you have to do something else. You have to write antisymmetric wave functions, antisymmetric with respect to the exchange of all electrons. And that looks like it's going to lead to a tremendous headache. But it doesn't. But you do have to learn a new algebra. OK, so let's go on to this. OK, so for the hydrogen atom, it's easy to solve the time independent Schrodinger equation. And we get a complete set of these Rnl radial functions, complete set, an infinite number. We don't care. You can take this-- it's a simple one-dimensional differential equation. And you can solve it. You can tell your computer to solve it. You don't have to have any tricks at all. You can just use whatever numerical method to find the wave functions and calculate whatever you want. There's no insight there. It's the same thing as recording a spectrum, where you have tables of observed lines and maybe observed intensities. You don't know anything. It's a description. And you can have a mathematical description, or you can have an experimental description. You don't know anything. You want to have your insight telling you what's important and how to use what you know about part of the spectrum to determine other things. For example, we have two energy levels. Let's say n equals 1 and n equals 2. And there are several things that you would need to do in order to know what the selection rule for l is from this, because the l, the orbital angular momentum states of n equals 2 are degenerate. And so you could use the Zeeman effect. You can use radiative lifetimes. And what you would find is we have a p state and an s state. And the p state fluoresces to the s state and the s state doesn't, unless you apply an electric field and it mixes with a p state and it then fluoresces. So there are all sorts of things that will alert you to the fact that there's more going on here than just this simple level diagram. OK, so let's do some elementary stuff and then get to the Google map for the spectrum. We have this thing that makes the hydrogen atom interesting. The potential, the radial potential is a function of the orbital angular momentum. And the effective potential is given by h bar squared l, l plus 1, over 2 mu r squared-- and that's with a plus sign. And we have z e squared over 4 pi epsilon 0-- that's just stuff from the MKS units for the Coulomb equation-- and 1 over r. So we have an attractive thing that pulls the electron towards the nucleus. And we have a repulsion part, which keeps the electron away as long as l is not equal to 0. So this is the actual effective potential. This is the thing we would use for solving the Schrodinger equation for the Rnl of R functions exactly if you wanted to. And first surprise is that the energy levels come out to be-- for n equals 1, 2, 3, etc. So why are the energy levels not dependent on l? Because it takes more ink to write this than that. And this can be really important. But there is no l dependence. That's a surprise. It's not true for anything other than one-electron spectra. But if it's true and it has to be, you have to have something that says, OK, for something that's not hydrogen. Maybe there will be some l splitting. Why? Why is that? How does this effective potential have an effect for something other than hydrogen? And we'll understand that in a second. OK, now there's the question of degeneracy. So if we pick a value of n, we can have l equals 0, 1, up to n minus 1. That comes out of the mathematics. And I'm just assuming that you can accept that as a fact. We know from our study of the rigid rotor for the Ylm functions that the degeneracy of the l part of the wave function goes as 2l plus 1. If we have an orbital angular momentum of 0, there's a degeneracy of 1. Angular momentum of 1, it has a degeneracy of 3. We know that. So if we do a little game. And we look at the degeneracies for l equals 0, 1, 2, 3, the degeneracies-- so g sub l, is 1, 3, 5, 7. OK, now, the degeneracy for n is going to be all of the possible l's. And so if we have n equals 1, we only have one l. And so the degeneracy is 1. If we have n equals 2, we have p and s. And so we have 4. And for n equals 2, we have-- I'm sorry, for n equals 1, we have only 1. And then we have for n equals 2, 4. And for n equals 3, 9. So you see 1 plus 3 is 4. 1 plus 3 plus 5 is 9. And so the degeneracy for n is n squared. So as you go really high end, you get lots of states. And for hydrogen, they're all degenerate. But if they're not split, maybe you don't care. Of course, you do. I better leave this down. Now, here we are for-- suppose you take a spectrum. And suppose it's an emission spectrum. You run a discharge through gas of something that contains hydrogen. And you'll see a bunch of lines. And so one of the things you know is that the energy levels are separated-- they go as 1 over n squared. And so the energy levels-- well, actually, they're converging. And so let's just indicate that like this. So suppose you ask, well, suppose I'm at this level, there are going to be a series of transitions converging to a common limit from it. And that's a marker, a pattern in the spectrum. If you have essentially an infinite number of levels converging to a fixed point, that's easy to see. And so for each level, there will be convergence to the same fixed point. And so that tells you there's a kind of pattern recognition you would do. And you would begin to build up the energy level diagram and to know which levels you're looking at from these existence of convergent series. So if we're interested in the spectrum, we have the energy level differences, z squared Rydberg 1 over n squared minus 1 over n prime squared. That's the Rydberg equation. And so this also tells you something. So suppose you're in the nth level, well, you know this. And now this then converges to the ionization limit. But there's something else. And that is suppose we observe two consecutive members, n an plus 1. You don't know what n is, but because you see the pattern of this convergent series, you can see two levels that are obviously consecutive members of a series. And the question is what is n? And, again, you can solve that by knowing that the energy levels go as 1 over n squared. So the difference between energy levels goes as 2 z squared r over n cubed. This is just taking the derivative of 1 over n squared. We get 2 times 1 over n cubed. And so that kind of a pattern enables you to say, oh, yeah, I know what n is. So this is a kind of a hand waving, but this is what we do as spectroscopists. We're looking for a clue as to how to begin to put assignments on levels. And because there is this simple structure, which is represented by this Rydberg equation, we can say where we are. We know where we are. We have two choices. One is this 1 over n cubed scaling of the energy differences. And the other is the existence of a convergence of every initial level to a common final level. And that's very important. You can't assign a spectrum unless you know what the patterns are. And these are two things that come out of the fact that there is a structure associated with the hydrogen atom. OK, now, I'm going to have to put some numbers on the board. And it's another illustration of structure. Now, I'm just taking things from a table in McQuarrie. And this table is a table of expectation values of integer powers of r. And what is this? Every electronic property, anything you would measure is going to be a function of the-- is going to involve an integral involving a function of r. That's what we mean by electronic properties. And these electronic properties have a wonderful behavior. So let's make a little table. Here is the integer power. Here is the expectation value. And that's l. So we start with n equals 2. The result you get by actually accurately evaluating an integral-- and these are doable integrals. They are not numerically evaluated. And they have a simple formula-- a0 squared n to the fourth z squared times 1 plus 3/2 times 1 minus l, l plus 1 minus 1/3. These are done integrals. I can't do them. I wouldn't care to do them. So that's how any electronic property that goes as the square of r behaves, we get this. This is the Bohr radius. This is approximately half an angstrom. And it's the radius of the n equals 1 Bohr orbit, which doesn't even exist. Right? There are no Bohr orbits. But we have this Bohr model, which explains these energy levels. And so we use that. And there's nothing else here. That's the charge on the nucleus. Well, it's 1 for hydrogen. But there's nothing empirical here. And this is a sort of a fundamental constant now, but it's sort of empirical. And then the next one, it has a very interesting initial term, a0 not squared n squared over z. We've lost a power here. And it again is one of these complicated looking functions. OK, this r to the 1, that's how big an atom is, the radius of the atom. And again, we have a closed form. Notice that the l is present here, even though it's not present in the energies. 0, that just goes n to the 0 power. It's just a constant minus 1, minus 2, minus 3. Well, minus 1, that's what you have for the Coulomb attraction. And that goes as z over a0 n squared. And for minus 2, that goes as z squared a0 n cubed. And all that plus 1/2. And for minus 3, we get z cubed a0 cubed n cubed l, l plus 1/2, l plus 1. So there are all these formulas. Well, the important thing is the leading term. And what you discover is that right here something happens. The leading term goes as 1 over n cubed. It doesn't matter what the negative power of r is. The leading term goes as 1 over n cubed. Why's that? Well, that's really important, because this 1 over n cubed behavior is telling you something very important. So here's the nucleus. And here's the electron. We can think of it as a Bohr orbit. If we have a negative power of r, then as you get farther and farther away from r, the property gets small. And so what happens is for large enough negative powers of r, the only thing that matters is the first lobe, the innermost lobe of the wave function. And it turns out that's the only thing that matters for almost everything. And so one of the things I'm going to show you is why does the innermost lobe matter and how do we use that to understand everything, not just about hydrogen, but about Rydberg states of everything. So the scaling of electronic properties depending on the power of r is another example of structure. And if you know one thing about the hydrogen atom, if you make one measurement, if you know how to use it, you know everything. You don't have to do all this other stuff. That's a really beautiful example of structure. And this is insight too, because sometimes the structure that you have for hydrogen is only approximately valid for other things. And you're going to want to know what you can use and know how to modify it. And I guarantee you that nobody teaching a course like this would ever talk about that kind of stuff, because, you know, it's approximate. It's intuitive. And all the good stuff is tabulated in the textbooks and you have to memorize it. But you don't know how to use it. And this is how you use it. This is really important. So now let me draw some pictures. So this is hydrogen. This is sodium. This is carbon monoxide. So the electron sees a point charge in hydrogen. That's easy. Now in sodium-- we don't know yet why, but once I've talked about helium and many atoms, you will know why-- there are a whole bunch of electrons in the nucleus or in the core of sodium. And outside the core, you have something that looks hydrogenic. But there is this core. It's not a point. And that leads to the appearance of l dependence in the energy levels, because what you're going to find is that l equals 0 penetrates into the core. l equals 1 can't. And so as you have higher and higher l, you're seeing less and less of this. And so as a result, there is an l dependence. But it's an s emphatically decreasing one. As l increases, you get less and less sensitivity to this inner part. Now, CO isn't even round. But there are nuclei. But if you have an electron at a high enough end, it's outside of this core. But again, if you pick l-- now, you can't have l for something that's not round. But if you're far enough away from it, you can pretend it's round and use the not roundness as a perturbation. And so you can have Rydberg states of CO. And you can use the same ideas that we developed for the hydrogen atom to explain the Rydberg states of CO, anthracene-- I don't even know how to say the names of biological molecules. So just consider it. If you had it in the gas phase and you could make a high enough Rydberg, high enough end state, it would look following this kind of extrapolation with features common to hydrogen. OK, so now, semiclassical-- OK, remember, you can calculate this. If you want to use rigorous mathematics, you can find an analytic solution for this, for hydrogen. And that's good, but you haven't extracted any insight from it. You just have it. But the semiclassical theory extracts insight. So semiclassical theory starts out with writing the classical equation for the momentum. And the momentum as a function of r, that's a no-no from quantum mechanics. You can't ever say we know the momentum as a function of r. And it's also true in quantum mechanics, whenever you calculate an integral that you need to evaluate some property, you're evaluating it over all space. But what I'm going to show you is that many of these integrals accumulated a specific point in space, which is a reminder of the classical mechanics. So there are things where quantum mechanics says you have to do something, but there's insight waiting to be harvested because many of the things you have to calculate are trivially interpretable in terms of semiclassical ideas. And the semiclassical means classical. You can do classical mechanics. You can calculate the probability of finding a particle at a particular place. And you can use that information here. And so always the relationship between the classical momentum function and the potential is this equation. p squared over 2 mu is the momentum. And energy minus potential is the momentum, is the kinetic energy. And this is how you get there. OK, so this is the function. And so that's one of the important pieces. And the other important piece is really unexpected, but wonderful. And so we have this Vl of r. If l is 0, the inner wall of the potential is vertical. If l is not 0, the inner wall of the potential isn't quite vertical, but it might as well be because when r gets small, even the l l plus 1 over r squared term gets small. And so we get this vertical potential. And so what happens is we can use the idea from DeBroglie and say, OK, where is the first lobe? Where is the first node? Well, it's half a wavelength from the turning point. So if we know where the turning point is we can draw something that looks like that. Now, it's a wave function, so it's going to be oscillating. So the important thing is that the first lobe, the innermost lobe, of all n and l lines up because this is nearly vertical and the energy distance from here to the bottom of the potential at r not equal to 0 is really large. And so if you change n or l by 1 or 2, nothing much happens. And so if you just say, OK, for n greater than or equal to 6, 6, 36, the Rydberg constant divided by 36 is not a big number, 3,000, 3,000 wave numbers. The vertical distance is more than 100,000 wave numbers. So the correction from n equals 6-- so if we have n equals 6, n equals 7, there isn't much change. And so the first lobe is going to always be at the same place. And so one thing that happens is they line up. The first node for all n and l is at the same place, to a good approximation. And the amplitude in this first lobe scales as n-- so the amplitude scales n to the minus 3/2. I'm going to prove that. This is where all the insight comes from. If you have a property, which is essentially determined close to the nucleus, because you have a negative power of the r in the electronic property, then the amplitude of the wave function will scale as n to the minus 3/2. And we can use this in evaluating all integrals, because everybody has a lobe here, at the same point. And the only thing that's different from one state, at one n l and another is the amplitude of this lobe. And that goes as 1 over n to 3/2 power. OK, that's what I was afraid of. OK. So we're interested in turning points. And what's the definition of a turning point? Or what is the mathematical equation that tells you how to get the inner nuclear dis-- the r value for the turning point. Yes. STUDENT: Derivative equals 0. ROBERT FIELD: I'm sorry. STUDENT: When the derivative of V l d r equals 0. ROBERT FIELD: I'm not looking for that. I'm looking for-- so when the energy is equal to V l of r, plus or minus. That's the equation that tells you where the turning points are. And these are simple equations. So it's child's play, adult child's, to calculate what the inner and outer turning point is as a function of n and l. And so we know the effective potential. We pick an energy. We know what it is because it's the Rydberg over n squared. So we have enough to calculate this function, which is a0 n squared 1 plus or minus 1 minus l l plus 1 over n squared square root. Now I could derive this. You could derive this. It's a simple closed form equation. And that's the foundation of all of this. Now, the thing that's common for everything is the inner turning point, r minus, because what we're interested in is this. Where is the maximum or where is the first node and what is the amplitude? And so we're going to use this equation to get all that. All right, so the semiclassical theory lambda n l-- that's the wavelength-- is equal to h over Pr of r. That's DeBroglie. But it's generalized to a potential, which is dependent on r, or a momentum function, which is dependent on r. So now, we have to calculate several important things in order to build our model. One is the classical oscillation period. So if I told you we have a harmonic oscillator, we know the energy level spacing things are h bar omega. And they're constant. If we make a superposition state of a harmonic oscillator, there's going to be beat nodes at integer multiples of omega. So we can simply say, well, for Rydberg states or for anything, we can define the period of oscillation as h over, in this case, n plus 1/2 minus n Vn minus 1/2. This is the energetic separation between levels-- or even not levels. We have a formula that's giving the independence of the energy. So this is related to the period that we get from the harmonic oscillator. So this is a perfectly legitimate way of knowing the period. Well, when we do that, this is something I wrote down before, the energy separation of two levels differing in n by 1 centered at n is going to be the Rydberg divided by 2n-- well, I'll just write it down. So we have h. And now this hcR 2 over n cubed. And n cubed is really important. It came just from taking the derivative of the 1 over n squared energy dependent. So this is a formula that's perfectly legitimate. What n tells us is the period is proportional to n cubed. The higher you go, the slower you go. Second, node to node probability. So how long does it take-- you have this well. And I don't mean to draw it like a harmonic oscillator, but it's a well. And so you have a node here, and you have a node here. And you have a particle that's moving, classically. How long did it take for the center of the particle to go from here to here? That's an easy thing to calculate too, because we know classically what the momentum is at any point in space. The momentum is related to the energy difference between here and here. So what we're going for is the ratio of the time node to node, or to next node, to delta t turning to turning point. OK, well, this is 1/2 the period, right? So we know what 1/2 the period is. And the period goes as n cubed. So this is the probability of finding the particle within one lobe of the wave function. So we can calculate the probability of finding the particle within one lobe. And it's clearly going as 1 over n cubed. Now, we want an amplitude. What is the amplitude? It's square root of the probability. All of a sudden, we start to see that the amplitude in whatever lobe we want goes as 1 over n to the 3/2. Now, because we have this lining up of nodes, the first node and the first lobe being identical, what we have now is the scaling of the probability or the amplitude in the first lobe of the wave function. And the next thing is again something that's almost never talked about in textbooks, but it's a really fantastic tool for understanding stuff is suppose we want to know the value of an integral. So we have an integral. And that will be, say, the electronic wave function in this chi representation, as opposed to the Rnl representation. OK, so this is the integral we want. But what we really want is a way of saying, I can convert this integral to a number that I can figure out on the back of a postage stamp. No methods for integration. No numerical methods. Just insight. So we have some electronic property, which comes from this integral. But now, this is a rapidly oscillating function. This is another rapidly isolating function. And so if we were to ask, well, how does the integral accumulate? In other words, let's say we're integrating from r minus to r prime chi n l of r r of k chi n prime l prime of r Vr. So if we plotted the integral, the value of the integral, as it accumulates-- so we're not evaluating the whole integral. We're evaluating the integral up to the r point, the r prime point. And what we see that there is a stationary phase point, where at some point in space, the two functions are oscillating spatially at the same frequency. And we already saw this for Fermi's golden rule, right. You have a time integral where you have rapidly oscillating functions. And then when the applied frequency is resonant with the difference in frequency, we get the integral accumulating. So same thing. It's easier. And so we have this idea that there is a stationary phase point. And the integral accumulates from 0 to its final value there. And then there's just little dithering as you go out the rest of the way. You get in complications when they're two stationary phase points, because then there's destructive or constructive interference between them. But it's very rare that you have two. You only have one. And so if you know the amplitude of the wave functions at a stationary phase point, you can calculate the integral as a product of three things. No integration, just one number. And the thing that we care about is that almost all electronic properties accumulate in the innermost lobe, because they're the same. The wave functions are the same. And they get more and more different as you go farther out. And as a result, since the amplitude in this innermost lobe goes as 1 over n to the 3/2 and you've got two functions-- you've got the nl function and the n prime l prime function-- you have that the integral is proportional to 1 over n to the 3/2 1 over n prime to the 3/2. So not only do you get expectation values, you get off diagonal matrix elements. Now this is not exact. But it tells you this is the structure of the problem. And if I look at enough stuff, this is really important to say, how does everything scale? And if you know how things scale, you know that if you're combining things that aren't part of that scaling rule, that they're just not going to be relevant. This is a fantastic simplification, because, yes, you can do it exactly by programming your computer to calculate the integral numerically. And that's not a big deal. But you don't know anything. And here, you know what to expect. And this is a professional spectroscopist who survives by recognizing patterns and using patterns. This is beyond numerical integrals. This is actually understanding the structure of the problem. And the fact that the molecule gives you this beautiful-- or the atom gives you this beautiful lobe always at the same place, you know stationary phase. Now, there are other problems where you use the stationary phase approximation. You always want to be trying to use stationary phase. For example-- and I'm going to stop soon-- suppose you have two potential energy curves. Now, you don't know what a potential energy curve for a molecule is yet. But you can imagine what there is. And so they cross. And this is the place where at this energy, this curve is exactly the same as the momentum on that curve. And so the integral accumulates at this curve crossing point. Isn't that neat? If you know what the curve crossing point is, you know the amplitude of the wave function. And you can estimate any integral. So this is something you would use if you're actually creating new knowledge again and again, because it's deeper than the actual experimental observation. It's the explanation for it. And it's what you're looking for and how you use your knowledge of what you're looking for. That's why I like this lecture, because it really gives the approach that I take to all scientific problems. I look for an approximate way where I don't have to do any complicated mathematics or numerical integration. I can do that just sitting at my desk. OK, so on Friday, we'll hear about helium and why helium looks horrible, but isn't. |
MIT_561_Physical_Chemistry_Fall_2017 | 5_Quantum_Mechanics_Free_Particle_and_Particle_in_1D_Box.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: OK, let's get started. So last time we talked about the one-dimensional wave equation, which is a second-order partial differential equation. This is not a math course. If you have a second-order differential equation, there will be two linearly independent solutions to it. And that's important to remember. Now, there are three steps that we use in approaching a problem like this. Does anybody want to tell me what those three steps are? Yes? AUDIENCE: [INAUDIBLE] it has separate functions that take only one variable. ROBERT FIELD: OK, that's most of it. You want to solve the general equation, and one way to solve the general equation is to try to separate variables. Always you want to separate variables. Even if it's not quite legal, you want to find a way to do that because that breaks the problem down in a very useful way. So the first step is the general solution, and it involves trying something like this where we say u of x and t is going to be treated in the separable form. If it doesn't work, you're going to get 0. You're going to get the solution. The only solution with separated variables is nothing happening. And so that's unfortunate if you do work and you get nothing for it, but life is complicated. So then after you do the general solution, what's next? Yes? AUDIENCE: You need to set boundary conditions? ROBERT FIELD: Yes. Now if it's a second-order differential equation, you're going to need two boundary conditions. And when you impose two boundary conditions, the second one gives some sort of quantization. It makes the solutions discrete or that there is a discrete set rather than a continuous set. The general solution is more or less continuous, or continuous possibility. So we have now the boundary conditions, and that gives us something that we can start to visualize. So what are the important things that, if you're going to be drawing pictures rather than actually plotting some complicated mathematical function, what are the first questions you ask? Yes? AUDIENCE: Is it symmetric or asymmetric? ROBERT FIELD: Yes, if the problem has symmetry, the solutions will have symmetry. But there's more. I want another, my favorite. AUDIENCE: Where's it start? ROBERT FIELD: I'm sorry? AUDIENCE: Where does it start? So t equals 0 or x equals 0. ROBERT FIELD: The initial condition is really the next thing I'm going to ask you about, and that's called the pluck. You're right on target. But now if you're going to draw a picture, the best thing you want to do to draw a picture is have it not move. So you want to look at the thing in position, and what are the things about the position function that you can immediately figure out and use in drawing a cartoon? AUDIENCE: Nodes. ROBERT FIELD: Yep. So how many nodes and where are they? Are they equally spaced? And that's the most important thing in drawing a picture, the number of zero crossings that a wave function has, and how are they distributed? Their spacing of nodes is half the wavelength, and the wavelength is related to momentum. And so I'm jumping into quantum mechanics, but it's still valid for understanding the wave equation. So we want the number of nodes for each specific solution satisfying the boundary conditions and the spacing and the loops between nodes. Are they all identical? Is there some systematic variation in the magnitude of each of the loops between nodes? Because if you have just a qualitative sense for how this works, you can draw the wave function, and you can begin to make conclusions about it. But it all starts with nodes. Nodes are really important. And quantum mechanics, the wave functions have nodes. You can't do better than focus on the nodes. And then I like to call it pluck, but it's a superposition of eigenstates. And those superpositions for this problem can move. And so-- you're missing a great-- anyway. And so you want to know, what is the kind of motion that this thing can have? And so one of the things-- this is really the three steps you go through in order to make a picture in which you hang up your insights. And so if there's one state or two states, one state is just going to be standing waves. Two states is all the complexity you're ever going to need. And if two states have different frequencies, there will be motion, and the motion can be side-to-side motion or sort of breathing motion where amplitude moves in from the turning-point region to the middle and back out again. And so you can be able to classify what you can understand and to imagine doing experiments based on this simplified version of the flux. In my opinion, the most important thing you can do as a professional quantum machinist and in preparation for exams is to be able to draw these cartoons quickly, really quickly. That means you have to think about them in advance. And so this recipe is how you're going to understand quantum mechanical problems too. And this differential equation is actually a little more complicated than the first few that we're going to encounter because the first few problems we're going to face are not time dependent. There may still be a separation of variables situation and imposing boundary conditions and so on, but there is no motion. But eventually we'll get motion because our real world has motion, and quantum mechanics has to reproduce everything that our real world does. So, we're going to begin quantum mechanics, and first of all I will describe some of the rules we have to obey in building a quantum-mechanical picture. And then I'll approach two of the easiest problems, the free particle and the particle in an infinite box. So we have the one-dimensional Schrodinger equation, and the one-dimensional Schrodinger equation looks like the wave equation. And why? Because waves interfere with each other. We can have constructive and destructive interference. Almost everything that is wonderful about quantum mechanics is the solutions to this Schrodinger equation also exhibit constructive and destructive interference. And that's essential to our understanding of how quantum mechanics describes the world. The next thing I want to do is talk a little bit about postulates. Now I'm going to be introducing the quantum-mechanic postulates as we need them as opposed to just a dry lecture of these strange and wonderful things before we're ready for them. But a postulate is something that can't be proven right. It can be proven wrong. And we build a system of logic based on these postulates. Now one of the great experiences in my life was one time when I visited the Exploratorium in San Francisco where there are rather crude, or at least when I visited almost 50 years ago, there were rather crude interactive experiments where people can turn knobs and push buttons and make things happen. And the most wonderful thing was really young kids trying to break these exhibits. And what they did is by trying to break them they discovered patterns, some of them, and that's what we're going to do. We're going to try to break or think about breaking postulates and then see what we learn. So, let's begin. We have operators in quantum mechanics. And we denote them either with a hat or as a boldface object. We start using this kind of notation when we do matrix mechanics, which we will do, but this is just a general symbol for an operator. And an operator operates on a function and gives a different function. It operates to the right, or at least we like to think about it as operating to the right. If we let it operate to the left, we have to figure out what the rules are, and I'm not ready to tell you about that. So this operator has to be linear. And so if we have an operator operating on a function, A f plus bg, It has to do this. Now you'd think, well, that's pretty simple. Anything should do that. So taking the derivative does that. Doing an integral does that, but taking the square root doesn't. So taking the square root-- an operator says take the square root, well, that's not a linear operator. Now the only operators in quantum mechanics are linear operators. We have eigenvalue equations. So we have an operator operating in some function. It gives a number and the function back again. And this is called the eigenvalues, and this is the eigenfunction. AUDIENCE: Dr. Field? ROBERT FIELD: Yes? AUDIENCE: In the [INAUDIBLE] that should be b times A hat [INAUDIBLE].. ROBERT FIELD: What did I do? Well, it should just be-- sorry about that. I'm going to make mistakes like this. The TAs are going to catch me on it, and you're going to do it too. All right, so now here we have an operator operating on some function. And this function is special because when the operator operates on it, it returns the function times a number, the eigenvalue and the eigenfunction. We like these. Almost all quantum mechanics is expressed in terms of eigenvalue equations. Operators in quantum mechanics-- so for every physical quantity in non-quantum-mechanical life there corresponds an operator in quantum mechanics. So for the coordinate, the operator is just the coordinate. For the momentum, the operator is minus ih bar partial with respect to x or derivative with respect to x. Now this is not too surprising, but this is really puzzling because why is there an imaginary number? This is the square root of minus 1, which we call i. Why is that there and why is the operator a derivative rather than just some simple thing? Another operator is the kinetic energy, and the kinetic energy is p squared over 2m. And so that comes out to be minus h bar squared over 2m second partial with respect to x. Well, it's nice that I don't have to memorize this because I can just square this and this pops out, but you have to be aware of how to operate with complex and imaginary numbers. And there are so many exercises on the problem set, so you should be up to date on that. And now the potential is just the potential. And now the most important operator, at least when we start out, is the Hamiltonian, which is the operator that corresponds to energy, and that is kinetic energy plus potential energy. And this is called the Hamiltonian, and we're going to be focusing a lot on that. So these are the operators you're going to care about. The next thing we talk about is commutation rules or commutators. And one really important commutator is the commutator of the coordinate with the conjugate momentum, conjugate meaning in the same direction. And that is defined as xp minus px. And the obvious thing is that this commutator would be 0. Why does it matter which order you write things? But it does matter. And, in fact, one approach to quantum mechanics is to start not with the postulates that you normally deal with but a set of commutation rules, and everything can be derived from the commutation rules. It's a much more abstract approach, but it's a very powerful approach. So this commutator is not zero. And how do you find out what a commutator is? Well, you do xp minus px, operate on some function, and you find out. And you could do that. I could do that. But the commutator is going to be equal to ih bar. Now there is a little bit of trickiness because the commutator xp is ih bar and px is minus ih bar. And so I don't recommend memorizing it. I recommend being able to do this operation at the speed of light so you know whether it's plus ih bar or minus ih bar because you get into a whole lot of trouble if you get it wrong. So this is really where it all begins, and this is why you can't make simultaneously precise measurements of position and momentum, and lots of other good things. And then we have wave functions. So wave function for when the time independent Hamiltonian is a function of one variable, and it contains everything we could possibly know about the system. But this strange and wonderful thing, which leads to all sorts of philosophical debates, is that this guy, which contains everything that we can know, can never be directly measured. You can only measure what happens when you act on something with a given wave function. You cannot observe the wave function. And for a subject area where the central thing is unobservable is rather spooky. And a lot of people don't like that approach because it says we've got this thing that we're relying on, but we can't observe it. We can only observe what we do when we act on it. And usually the action is destructive. It's destructive of the state of the system. It causes the state of the system to give you a set of possible answers, and not the same thing each time. So it's really weird. So we have wave functions. And we can use the wave function to find the probability of the system at x with a range of x, x to x plus dx. And that's psi star x psi of x dx. So you notice we have two wave functions, the product of two, and this star means takes a complex conjugate. So if you have a complex number z is equal to x plus iy-- real part, imaginary part-- and if we take z star, that's x minus iy. So these wave functions are complex functions of a real variable. And so we do things like take the complex conjugate, and you have to become familiar with that. Now we have what we call the expectation value or the average value, and we denote this as A. So for the state function psi, we want the average value of the operator A. Now in most life, that symbol is not included just because people assume you know what you're doing. And this is psi star A hat psi dx over psi star psi dx. And this is integral from minus infinity to infinity. So this down here is a normalization integral. Now we normally deal with state functions which are normalized to 1, meaning the particle is somewhere. But if the particle can go anywhere, then normalization to 1 means it's approximately nowhere. And so we have to think a little bit about what do we mean by normalization, but this is how we define the average value or the expectation value of the quantity A for the state psi. So this is just a little bit of a warning that, yeah, you would think this is all you need, but you also need to at least think about this. That's great. I'm at the top of the board and we're now at the beginning. So the Schrodinger equation is the last thing, and that's the Hamiltonian operating on the function and gives an energy times that function. And if it's an eigenvalue, then we have this eigenvalue equation. We have these symbols here. So that's the energy associated with the psi n function. So now we're ready to start playing games with this strange new world. And so let's start out with the free particle. Now because the free particle has a complicated feature about how do we normalize it, it really shouldn't be the first thing we talk about. But it seems like the simplest problem, so we will. So what's the Hamiltonian? The Hamiltonian is the kinetic energy, minus h bar squared, or 2m second derivative with respect to x plus the potential energy, V0. Free particle, the potential is constant. We normally think of it as the potential is zero, but there is no absolute scale of a zero of energy, so we just need to specify this. And so we want to write the Schrodinger equation, and we want to arrange it in a form that is easy to solve. There is two steps to the rearrangement, and I'll just write the final thing. So the second derivative of psi is equal to minus 2m over h bar squared times E minus V0 psi. So this is the differential equation that we have to solve. So there was a little bit of rearrangement here, but you can do that. So the second derivative of some function is equal to some constant times that function. We've seen that problem before. It makes a lot of difference whether that constant is positive or negative, and it better, because if we have a potential V0 and we have an energy up here, well, that's perfectly reasonable. The particle can be there, classically. But suppose the energy is down here. If the zero of energy is here, you can't go below it. That's a classically forbidden situation. And so for the classically allowed situation, the quantity, this constant, is negative. For the classically forbidden situation, this constant is positive. You've already seen the big difference in the way a second derivative, this kind of equation, works when the constant is positive or negative. When this constant is negative, you get oscillation. When this constant is positive, you'll get exponential. Now we're interested in a free particle, so free particle can be anywhere. And we insist that the solution to our quantum-mechanical problem, the wave function is what we say well behaved. So well behaved has many meanings, but one of them is it never goes to infinity. Another is that when you go to infinity, the wave function should go to zero. But there's also things about continuity and continuity of first derivatives and continuity of second derivatives. We'll get into those, but you know immediately that if this constant is positive, you get an exponential behavior, and you get the e to the ikx and e to the-- not ik-- e to the kx and e to the minus kx. And one of those blows up at either positive infinity or negative infinity. So it's telling you that in agreement with what you expect for the classical world, an energy below the constant potential is illegal. It's illegal when this situation persists to infinity. But we'll discover that it is legal if the range of coordinate for which the energy is less than the potential is finite. And that's called tunneling, and tunneling is a quantum-mechanical phenomenon. We will encounter that. So we know from our experience with this kind of differential equation that the solutions will have the form sine kx and cosine kx. But we choose to use instead e to the ikx and e to the minus ikx because this cosine kx is 1/2 e to the ikx plus e to the minus ikx. And so we can use these functions because they're more convenient, more memorable. All the integrals and derivatives are trivial. And so we do that. So the differential equation-- and we saw before that we already have what k is. So minus k squared is minus 2m over h bar squared-- minus 2m over h bar squared E minus V0. We take the derivative of this function. This is the function, and this is the eigenvalue. We take the second derivative with respect to x. We get an ik from this term and then another ik, which makes minus k squared. And we get a minus ik and another minus ik, and that gives a minus k squared. And so, in fact, this is an eigenvalue equation. We have the form where this equation is an eigenfunction. With this, we have everything. So the energies for the free particle, h bar k over h bar k squared over 2m plus V0, so this is an eigenfunction, and this is the eigenvalue associated with that function. We're done. That was an easy problem. I skipped some steps because it's an easy problem and I want you to go over it and make sure that you understand the logic and can come to the same solution. Let's take a little side issue. Suppose we have psi of x is e to the ikx. Well, we're going to find that this is an eigenfunction of p, and the eigenvalue or the expectation value of p is h bar k. And if we had minus e to the minus ikx, then what we'd get is minus h bar k. So we have this relationship between p, expectation value, and h bar k. So this corresponds to the particle going in the positive x direction, and this corresponds to the particle going in the negative x direction. Everything is perfectly reasonable. We have solutions to the Schrodinger equation for the free particle. The solutions to the free particle are also solutions to the eigenvalue equation for momentum. And the two possible eigenvalues for a given k are plus h bar k minus h bar k. Now that's fine. So everything works out. We're getting things, although we have the definition of the momentum having a minus 1, an i factor, and a derivative factor. Everything works. Everything is as you would expect. And the general solution to the Schrodinger equation can have two different values, the superposition of these two. Right now, this wave function is the localized overall space. Now if we want to normalize it, we'd like to calculate integral minus infinity to infinity of psi star x psi of x dx. This is why we like this notation because suppose we have a function like this, psi star-- well, actually like this. Psi star is equal to a star e to the minus ikx plus b star e to the ikx, and psi is a e to the plus ikx. This would be e to the minus ikx. And so when we write this integral, what we get is integral of psi star psi dx is equal to a squared integral minus infinity to infinity of a squared plus b squared dx. So we have two constants which are real numbers because they're square modulus. They're additive. And we're integrating this constant from minus infinity to infinity. We'll get infinity. We can't make this equal to 1. So we have to put this in our head and say, well, there's a problem when you have a wave function that extends over all space. It can't be normalized to 1, but it can be normalized so that for a given distance in real space, it's got a probability of 1 in that distance. So we have a different form of normalization. But when we actually calculate expectation values, we can still use this naive idea of the normalization interval and we get the right answer, even though because both the numerator and denominator go to infinity and those infinities cancel and everything works out. This is why we don't do this first usually because there's all of these things that you have to convince yourself are OK. And they are and you should. But now let's go to the famous particle in a box. It's so famous that we always use this notation. This is particle in an infinite box, and that means the particle is in a box like this where the walls go to infinity. And so we normally locate this box at a place where this is the x coordinate and this is the potential energy, and the width of the box is a. And we normally put the left edge of the box at zero because that problem is a little easier to solve than the more logical thing where you say, OK, this box is centered about zero. And that should bother you because anytime you're interested in asking about the symmetry of things you'll want to choose a coordinate system which reflects that. Don't worry. I am going to ask you about symmetry, and it's a simple thing to take the solution for this problem and move it to the left by a over 2. So we have basically a problem where the potential is equal to 0 for 0 is less than or equal to x less than or equal to a, and it's equal to infinity when x is less than 0 or greater than a. So inside the box it looks like a free particle, but it can't be a free particle because there's got to be nodes at the walls. We know that outside the box, the wave function has to be 0 everywhere because it's classically forbidden, strongly forbidden. We know that the wave function psi is continuous. So if it's at 0 outside, it's going to be 0 at the wall. And so the wave functions have boundary conditions where, at the wall, the wave function goes to 0. So now we go and we solve this problem. And so the Schrodinger equation for the particle in the box where V of x is 0. Well, we don't need it. We just have the kinetic-energy term, h bar squared over 2m second derivative with respect to x. Psi is equal to e psi. Again, we rearrange it. And so we put the derivative outside, and we have minus 2me over h bar squared psi. And this is a number. And so we just call it minus k squared psi. We know what that k is as long as we know what the energy is, and k squared is equal to 2me over h bar squared. Now we have this thing which is equal to a negative number times a wave function, and we already know we have exponential behavior. But in this case, we use sines and cosines because it's more convenient. So psi of x is going to be written as A sine kx plus B cosine kx. This is the general solution for this differential equation where we have a negative constant times the function. So the boundary condition, psi of 0-- well, psi of 0, this is 0, but this part is 1. So that means that psi of 0 has to be 0, so B has to be equal to 0. And here, the other boundary condition, that also has to be equal to 0, and that has to be A sine ka. And ka has to be equal to n times pi in order to satisfy this equation. Sine is 0 at 0, pi, 2 pi, 3 pi, et cetera. So k is equal to n pi over a. And so we have the solutions. Psi of x is equal to A sine np a over x. And so we could put a little n here. And this is starting to make you feel really good because for all positive integers there is a solution. There's an infinite number of solutions. And their scaling with quantum number is trivial. And it's really great when you solve an equation and you are given an infinite number of solutions. Well, there's one thing more you have to do. You have to find out what the normalization constant is, so you do the normalization integral. And when you do that, you discover that this is equal to 2 over a square root sine n pi over a. So these are all the solutions for the particle in an infinite box, all of them. And the energies you can write as n squared times h squared over 8 m a squared or n squared times E1. There's another thing. n equals 0. If n is equal to 0, the wave function corresponding to n equals 0 is 0 everywhere. The particle isn't in the box. So n equals 0 is not a solution. So the solutions we have-- n equals 1 2, et cetera up to infinity, and the energies are integer square multiples of a common factor. This is wonderful because basically we have a problem. Maybe it's not that interesting now because why do we have infinite boxes and stuff like that? But if you ask, what about the ideal gas law? We have particles that don't interact with each other inside a container which has infinite walls. And I can tell you that in 5.62, there's a three-line derivation of the ideal gas law based on solutions to the particle in a box. Also, we often have situations where you have molecules where there's conjugation so that the molecule looks like a not quite flat bottom box with walls. And this equation enables you to learn something about the electronic energy levels for linear conjugated molecules. And this leads to a lot of qualitative insight into problems in photochemistry. Now the most important thing, in my opinion, is being able to draw cartoons, and these cartoons for the solutions the particle in a box look like this. So what you frequently do is draw the potential, and then you draw the energy levels and wave functions. I have to cheat a little bit. So number of nodes-- number of nodes, internal nodes. We don't count zeros at the walls. The number of nodes is n minus 1. The maximum of the wave function is always 2 over a. So here 2 over a, here 2 over a, here minus 2 over a i square root of 2 over a. This slope is identical to this slope. This slope is identical to this slope which is identical to that slope. So there's a tremendous amount that you can get by understanding how the wave function looks and drawing these cartoons. And so now if instead we were looking at problems where, instead of a particle in a box like this, we have a little dimple in the bottom of the box or we have something at the bottom or the bottom of the box is slanted. You should be able to intuit what these things do to the energy levels, at least have the beginning of an intuition. So, we have an infinite number of oscillating solutions. That means that we could solve the problem for any kind of a box as long as it has vertical walls, and that's called a Fourier series. So for a finite range, we can describe the solution to any quantum-mechanical infinite box with a terrible bottom, in principle, by a superposition of our basis functions. That's what we call them. Now, there are several methods for doing the solution of a problem like this efficiently. And you're going to see perturbation theory. And at the end of the course, you're going to see something that will really knock your socks off which is called the discrete variable representation. And that enables you to say, yeah, well, potential does terrible things. I solved the problem by not doing any calculation at all because it's already been done. So these things are fantastic that we have an infinite number of solutions to a simple problem. We're always looking for a way to describe a simple problem or maybe not so simple problem with an infinite number of solutions where the energies for the solutions and the wave functions behave in a simple n-scaled, quantum-number-scaled way. And this provides us with a way of looking at what these things do in real life. You do an experiment on an eigenfunction of a box like this, and it will have certain characteristics. And it tells you, oh, if I measured the energy levels of a pathological box, the quantum-number dependence of the energy levels has a certain form, and each of the constants in that special form sample a particular feature of the pathological potential. And that's what we do as spectroscopists. We find an efficient way to fit the observables in order to characterize what's going on inside something we can't see. And that's the game in quantum mechanics. We can't see the wave function ever. We know there are eigenstates. We can observe energy levels and transition probabilities, and between those two things, we can determine quantitatively all of the internal structure of objects that we can't see. And this is what I do as a spectroscopist. And I'm really a little bit crazy about it because most people instead of saying let's try to understand based on something simple, they will just solve the Schrodinger equations numerically and get a bunch of small results and no intuition, no cartoons, and no ability to do dynamics except another picture where you have to work things out in a complicated way. But I'm giving you the standard problems from which you can solve almost anything. And this should sound like fun, I hope. OK, so have a great weekend. |
MIT_561_Physical_Chemistry_Fall_2017 | 23_ManyElectron_Atoms.txt | The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high-quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. ROBERT FIELD: Let's get started. William Klemperer was my thesis advisor, and he died yesterday. It also happens that the subject of this lecture is really the core of what I got from him. He showed me how to evaluate matrix elements of many-electron operators, which is the key to being able to interpret-- not just tabulate-- electronic properties of atoms and molecules. Our goal is to be able to reduce the complexity of electronic structure, which is really complicated. The electrons interact with each other really strongly, and there are a lot of them. And it's very hard to separate the complexity of the many-body interactions into things that we can put in our head and interpret. And the whole goal of this course is to give you the tools to interpret complicated phenomena. We have the vibrational problem as a way of understanding internuclear interactions-- nuclear motions. We have electronic structure and the hydrogen atom as a way of understanding what electronic structure is, and to reduce it to, basically, the things we learn about hydrogen. When we go to molecular orbital theory, we take what we know about atoms, and build a minimally-complex interpretive picture, which is sort of a framework for understanding complicated molecular interactions. So one of the most important things about understanding electronic structure is, how do we deal with many-electron wave functions? And one of the terrible problems is that the electrons are indistinguishable. And so we have to ensure that the wave functions are anti-symmetric with respect to permutation of every pair of electrons, not just two. In helium we just dealt with two, and that wasn't so bad. But when we deal with n electrons, what we are going to discover is that in order to anti-symmetrize the wave function, we have to write a determinantal wave function, a determinant of orbitals. And when you expand an n by n determinant, you get n factorial times. And when you calculate matrix elements, you have n factorial squared integrals. So you're not going to be handling these one at a time, and looking at them lovingly. You're going to want to be able to take these things and extract what is the important thing about the electronic structure that you're going to need to know. And as a graduate student, I was collecting numbers. I was collecting numbers about spectroscopic perturbations, where non-degenerate perturbation 3 breaks down, and interesting things happen. But this was something that nobody in the world was interested in because it was the breaking of the usual patterns. And I was convinced that I had collected some stuff that told an interesting story. And I told Klemperer about this, and he said, well, have you thought about how to evaluate these integrals-- these numbers that you are extracting from the spectrum, by doing some tricks with the many-electron wave functions? And then he showed me, on a scrap of paper, how to do it. And I was launched. That was it. That has been the foundation of my career for the last 50 years. And I didn't think that Klemperer knew that. I didn't think anybody knew it, because I didn't think it was knowable. But he just gave it to me on a silver platter. And so I'm going to try to give you at least the rudiments of what it is you're up against, and how you reduce them to things that you care about, that you can think about. And you can understand the hydrogen atom in rather complete detail. Or at least you can understand how one observable relates to another. And so the relationship between the effective quantum number and the ionization energy of a state then provides a hydrogen-atom-based structural model for everything you can observe. Now spectroscopists have the unfortunate habit of saying we're interested in structure. Structure is static. Dynamics is magical, and special, and hard. But if you understand structure in a way which is not the exact eigenstates, not the exact wave functions, but something that the molecule was trying to do and sort of missed. And the dynamics is just what happens when this preparation isn't in eigenstate, which would be boring. And you get dynamics, which you can understand, as opposed to just saying, I'm going to tabulate the dynamics too. You don't know anything unless you have a reductionist picture of what's going on. And since the hardest part of dealing with molecules is the fact that they have a lot of electrons, this is really the core of being able to do important stuff. Now it's a horrendously complicated problem, and notationally awkward, too. And let me just try to explain it. And I'm going to try to do this without too much reliance on my notes, because they're terrible. We talked about helium. And helium has two electrons. And there's this 1 over r12 interaction between electrons, which looks innocent enough. You can write it down. You know, it's just a few symbols. And we can call it the first order perturbation. But that's really a lie, because it's as big as almost everything else. And so, yeah, we can, in fact, do a series of approximations. One is, ignore it, the non-interacting electron approximation. And that's basically repackaging hydrogen, and it's not quite enough. And then we can say, OK, let's calculate the first order energy by calculating expectation values of h12. So that's E1. And that's almost enough to give us a sense of what is going on. 1 over r12, commutate with any electron, that's not 0. 1 over r12, commutated with any orbital angular momentum-- any momentum is not 0. So that means that l and n are not good quantum numbers. What's a good quantum number? What's the definition of a good quantum number? Come on, this is an important question. Yes. AUDIENCE: [INAUDIBLE] count, and then you put them into some formula, and then you can read off eigenvalues. ROBERT FIELD: That's maybe 70% of what I want. You can put it into a formula. That means it's a rigorously good thing. It means it commutes with the Hamiltonian. A rigorously good quantum number corresponds to a eigenvalue of an operator that commutes with the Hamiltonian. So hydrogen, we rely on n and l to get almost everything. But here we find that, in addition to this being that small, it destroys the foundation of our picture. And so how do we think that we can make any sense of many-electron atoms and molecules? Well it turns out we can hide most of the complexity. And most of the complexity is just working out the rules for calculating these matrix elements. The matrix elements of operators that we care about, like transition moments, spin orbit, Zeeman effect, things that correspond to how we observe atomic and molecular structure. And so the main obstacle to being able to evaluate these matrix elements is the permutation requirement. And it turns out that there is a really simple way of dealing with the requirements for electron permutation, and that is to write the wave function as a determinant of one-electron orbitals. Because a determinant has three really important properties. One, it changes its sign when you permute any two columns. Two, it changes its sign when you permute any two rows. And three, if you have two identical columns or rows, it's 0. That's really fantastic. And that is the Pauli exclusion principle-- not what you learned in high school. What you learned is a small consequence of that. So if we can build anti-symmetric wave functions, we have aufbau, we can only put one electron in an orbital. We have all sorts of stuff, but it's too complicated to tell a student in high school that you can't-- just the question of indistinguishable electrons is such a subtle thing that you can't say, well, they have to be anti-symmetric. But it's easy to say, you can't put more than one electron in a spin orbital. But we don't talk about spin orbitals. We say, we can't put more than two electrons in an orbital, because we're protecting you from unnecessary knowledge. OK, well, I'm not going to protect you. [LAUGHTER] OK, so we know that the Hamiltonian has to commute with-- these capital letters mean, many electrons' angular momenta. And this is the spin, this is the projection of the spin. We know this is true because the Hamiltonian doesn't operate on spin. It's a trivial result, but it's a very important result. OK, so we have to worry about spin, and spin eigenstates, and other things like that. OK, so Slater determinants. J.C. Slater was an MIT professor in physics. He invented these things in 1929. I have a reference. I don't know if I've ever read this paper, but it's probably beautiful. So basically what Slater did is showed, yeah, you can do the necessary algebra to deal with any atom, and to be able to reduce an atom to a small number of integrals that you really care about. And there are two ways of doing this. One is the truth, and one is the fit model. Now the truth is really boring, because you lose all the insights, and the fit model gives you the things you have to think about and understand. And a fit model also tells you what are the import the actors. And maybe they're in costume, maybe they're not, but we can deal with them. But the truth is really very complicated. And as I said many times, when you go from hydrogen to helium, you can't solve the Schrodinger equation exactly. This was perhaps a little bit of a surprise, but I think it was only a surprise in newspapers. I think physicists knew immediately, when you go for a two-bodied problem to a three-bodied problem, there is no way you can have an exact solution. And that's the truth. You can't solve helium or any more-than-one-electron problem exactly. But you can do it really well, and it just costs computer time. And if the computer is doing the work, you don't really care. Because once you've told the computer the rules, then it's off to the races. You can go have lives or you can go have a life, and come back, and the computer will tell you whether you made a mistake and you're getting a nonsense result, or that you have the correct result. So what we know is this permutation operator, operating on any two-electron function, has to make-- OK, I'm skipping steps, and my notes are really kind of stupid sometimes. P 2, 1, which has to be equal to minus. And now if you apply the permutation operator twice, you get back to the same thing. So there's only two possible eigenvalues. You can have minus 1 or plus 1. And the minus 1 corresponds to fermions, things that have half-integer spin, like electrons. And the plus 1 corresponds to things that have integer spin, like photons, and vibrons, and other things. And actually, it's harder to construct a symmetric function than an anti-symmetric function. But the thing is, you've got lots of electrons, and you have very few quanta of vibrations in a single mode, and you have very few photons interacting with a molecule at once. And so the boson symmetry is less important in most applications. And so we just have to kill this one. OK, so suppose we want to talk about something like this, the 1s, 2s configuration. A configuration is a list of the occupied orbitals-- not the occupied spin orbitals, which is a spin associated with an orbital. The world of spin orbitals is where I live, but we do that for a reason. And so this two-electron thing can be expressed as a space part-- there are various conventions that-- times the spin part. And alpha 1, beta 2, and then we have minus or plus beta 1, alpha 2. I'm looking at my notes because some people always keep the electron in the first position, and some people keep always the orbital in the first position. And it doesn't matter, because you can permute rows or columns. But I just want to write what is in my notes. OK, so this thing, this two-electron function, has two anti-symmetrized possibilities. And one is a singlet, and one is a triplet. So s equals zero, s equals 1. We recognize this alpha beta minus beta alpha as the singlet spin state, and alpha beta plus beta alpha as the triplet spin state. So we have alpha beta plus beta alpha, and alpha alpha and beta beta, and we have alpha beta minus beta alpha. So we call s = 0 a singlet, and this a triplet, because of the number of states. And this wave function has the necessary spin symmetry and the necessary permutation symmetry. OK, so if, instead of two electrons, we have 1, 2 dot dot dot N, then Mr. Slater says we do this-- whoops. OK, N, 1, and then K1, and KN N. So that's a determinant-- an N by N determinant. And the rows correspond to the electrons, and the columns corresponding to the orbitals. Now this, because of the properties of the determinant, is anti-symmetric with respect to permutation of any two electrons or any two orbitals. But we don't really care about the permutation of the orbitals, because it's really the same thing is permuting the electrons. And so this N factorial is a consequence of normalization, because when you expand an N by N determinant, you get N factorial, additive products of N functions. It looks horrible. And because we're normalizing, we need this 1 over the square root of N factorial in order to have this thing come out to be 1 when you calculate the normalization integral. Now this notation is horrible because you've got too many symbols. And so depending on what you're trying to convey, you reduce the symbols, and you can reduce it simply by, instead of writing psi every time, just writing the state. Or you can-- since you don't need psi, you don't need the state letter, you can just have the state number. But the best way to do this is simply to say-- this is just the main diagonal of the determinant. It conveys everything you need. Again, if you permute any two of these guys, any adjacent pair, the sign changes. And it contains everything you need, and it doesn't require you to look at stuff you're not going to use. And your goal is going to be to take these things, and calculate matrix of them. And so you'll be dealing with the orbitals one or two at a time. And this is very convenient. And soon, you start to take this for granted. And it's a very simple thing, but it isn't, because you're doing a huge number of tricks. OK, I'm going to skip over what's in my notes. Demonstrating that for a two by two, that what I asserted is correct, you can do that very easily. OK, so we can count, and we have an atom, and we know how many electrons it has. And so we immediately know what our job is going to be. We're going to be having to write some Slater determinant of those number of electrons. And the goal is to be able to do the algebra in a way that maybe you can't describe to your friends because it's too complicated. I'm faced with the problem of trying to explain how to do this algebra. But it is something that you can learn, and you can ask a computer to do it, and there are all sorts of intuitive shortcuts where you can look at a problem, and you could say, I understand. OK, so you're used to orbitals. And that's perfectly reasonable, because for hydrogen, we have orbitals, and there's only one orbital, and it could have either spin. We don't mess with that. But now we're going to talk about spin orbitals. And that's just the combination of the name of the orbital with whether the spin is up or down. And the reason for this is it's easier to do the algebra. And the reason the algebra is-- it's initially harder to do the algebra, because there are certain selection rules, and stuff like that. But once you know how to do it you do the algebra. And then all of a sudden, everything pops out in a very useful form. So the stick diagrams are very important. But now I'm specifying the stick diagrams as spin orbitals rather than orbitals. Now another point, there are rules for how-- the number of spin orbitals is different between the left-hand side and the right-hand side of a matrix element. There are rules that are easily described-- and so for every kind of orbital, an orbital that is a scalar, that doesn't depend on quantum numbers, that has a selection rule delta SO of 0; and for something like a one-electron operator that has a selection rule delta spin orbital of 1 and 0; and then we have our friend 1 over rij, that has a selection rule delta spin orbital of 2, 1, and 0. And the algebra for each is something you work out, and then you know how to do it. And I'm going to try to give you just a little bit of a taste of this. So we already looked at something like this. But we use a slightly different notation. So I'm going to go back. And we have 1s alpha, 1s beta, 2s alpha, 2s beta. And so for the ground state of helium 1s squared, and we would do this. And the stick diagrams are great, because it's easier to see on a picture, who are the actors, and have I included all of them, or have I left something out? And so now we're interested in the stick diagram for the 1s 2s configuration. And there are several kinds of 1s/2s configurations, depending on what the alpha and beta are. So we have 1s alpha, 2s alpha. And we have 1s alpha, 2s beta. 1s beta, 2s alpha, 2s alpha, 1s beta, 2s beta. So there's four guys, and we can put our arrows on these things, and we know everything we need to know about these guys. It tells us what to do. Well, when we do this, the diagonal matrix elements of the 1 over rij Hamiltonian can be expressed. And we use this notation, J tilde minus K tilde. So for every two-electron thing, we're going to get this kind of-- now these are simple integrals, and some of them are 0. Because this doesn't operate on spins. And so if you had a 1s alpha, 2s beta, 1 over rij, 1s beta, 2s alpha, then the 1s alpha with the 1s beta is 0. The 1s alpha with the 2s alpha is not 0, et cetera. There are all sorts of stuff. But this tilde notation says, well, this is what we start with, and we have to convert it into things that really matter. So the operation of removing the tilde requires a little bit of work, a little bit of thought. And that's why my notes are crap, because I can't explain it well enough to really teach this. So when we do the 1s squared, the J 1s squared tilde, is equal to the J 1s squared, because the spins take care of themselves. But k tilde 1s squared is equal to 0. Because when we do 1s squared, we have an alpha with an alpha for the J term, and an alpha with the beta for the K term. And alpha with beta is 0, because the operator cannot change the alpha into beta. So this tilde notation is a convenient thing, because you can use any Slater determinant, and you can express it in terms of J's and K's. And the sign comes from switching the order of the orbitals. That's how the determinants work. And so you're going to see a whole bunch of stuff. But removing the tildes is the tricky business. OK, now when you get a problem where you have a configuration where a single Slater is not sufficient-- in other words, in order to make an eigenstate of s squared or sz, you sometimes need two or more Slaters, and you have to use a particular linear combination of them to get the right value of s and sz. And then what happens is you're looking at matrix elements of the 1 over rij operator, between Slaters. Now this is a headache. And I could talk until I'm blue in the face, and I cannot make it clear how to do this. Because it's just awful. But some things in life are worth suffering for. And so anyway, in the 1s, 2s situation, when you do everything right, you get-- this is just a general notation for a two-electron wave function-- 1 over r12, psi 2. So these guys are eigenfunctions of s squared and sz. And when you do that, you get 1/2 times 2 J, 1s, 2s, minus or plus 2K, 1s, 2s. Remember, when you have mismatched alpha and beta, the K's are 0. But when you have K, the 1 over rij matrix element between two Slaters, you can fix that. And so this is why it's so hard to explain, because-- yes, I'm not even going to apologize anymore. OK, so this is what you do. And the notes are pretty clear about how to do them, and what the problems are. But lecturing on it would be a little bit hard. OK, so now what are we going to do with it? Well, we'd like qualitative stuff and interpretive stuff. Qualitative is Hund's rules. Now if you looked at 100 textbooks, I think 95% of them will have Hund's rules wrong. You're never going to make that mistake. And interpretive-- well, we want to know the trends of things, and we want to be able to do something like what you did in freshman chemistry on shielding. Now you probably memorized some rules about what shields what. But I'm going to give you a little bit more insight into that. So we're going to talk about this for the rest of the lecture. OK, so you specify a configuration. And this configuration might be two electrons, two spin orbitals, two orbitals times e, or three, or 10. And often, when you specify the occupied orbitals, you neglect the field ones, which is nice, because you have fewer things to worry about. Because field orbitals have spin 0. And you don't have to do anything. They're automatically asymmetrized, and they basically act as a charged distribution in the core that is sampled by the electrons outside. And so you need some sort of a set of rules for how does that work. And that's shielding. So first, we specify a configuration. And you also learned-- in high school, probably-- how to determine the L, S, J terms that result from this configuration by some magical crossing-out of boxes. And if you didn't, I'm glad. Because it would have just clouded your mind, and caused earlier insanity than MIT causes. So anyway, so we have orbital angular momentum. And we can add the orbital angular momenta of the electrons following certain rules. And we have spin angular momentum. And J is equal to the vector sum of L and S. And we say we have an LS term-- like triplet P. And it can have J is equal to L plus S, L plus S minus 1, down to L minus S absolute value. These are the possible J's. And so Hund's rules is all about, of all of the states that belong to a particular configuration, which one is the lowest? One-- which one, not the second lowest. Which one is the lowest? And why do we care? Because in statistical mechanics everything is dominated by the lowest energy state. And so if you can figure out what is the lowest energy state, you've basically got as much as most people are going to want. So you want to know what are L, S, and J for the lowest energy state of a configuration. Configurations are typically far apart in energy. So if you know what the lowest energy configuration is, and the lowest energy state of it, as far as your friends-- the statistical machinations, you can tell them how to write their partition functions. And the rest is details. And mostly, you don't want details. If your friends tell you they want details, well, you tell them, this is what you have to do, but it's no simple three Hund's rules. OK, so Hund's rules-- you look at all of the L, S, J states that are possible for a particular configuration. And you can use the crossing out of ML/MS boxes if you want. And I could tell you why you would do that. But I don't want to cause insanity at this stage, either. But I'm an expert at that. And you can also use lowering operators to generate all the states, once you know stuff. OK, so once you know all the states, Hund says, which one of these has the largest S? which one? And that's easy to know. And for example, if you had 2p squared, you're going to get singlet D, triplet P, and singlet S. And well, here's the triplet. That has the largest S. So the triplet P is the lowest energy state. Now if there were multiple triplets, as there would be, say, for 2p3d, then you'd have to decide which of those triplets is the lowest. And all you care about, all you're allowed to say is which one is the lowest. And it's the one with the maximum L. And then the last step is, what is the lowest J for that LS state? And that's kind of cute. Because you have the P shell, there's-- for a P shell, you can have six P orbitals to fill the shells. 1 alpha, 1 beta, 0 alpha, 0 beta, et cetera. So the degeneracy of a P orbital is 6. If you have p to N, where N is less than 3, you have a less-than-half-filled shell. And then lowest is J equal L minus S, absolute value. And if N is greater than 3, you have the lowest being-- the highest possible value of J is equal to L plus S-- so for L, N greater than 3. And now when you have a half-filled shell, the lowest state is usually an S state with maximum spin. But it doesn't matter. When you have less-than or more-than-half-filled shell, you have generally a state with orbital angular momentum not equal to 0. And you have spin orbit splitting of that. And so you do want to care what is the lowest J. But when N is equal to 3, the lowest state is usually an S state. It doesn't have a spin orbit splitting, and it just has one value of J, which is whatever the spin is. So Hund's rules tell you how to identify, without knowing beans, what is the lowest energy state, and it's never wrong. Well, maybe sometimes wrong, but that's because of one of my things where you have a perturbation between states belonging to two configurations. But people get really excited when they discover a violation of Hund's rules. And it's just trivial. So there is this. What time is it? I have a few minutes to talk about shielding, and I will. OK, so we have a nucleus. And it has a charge of z. Bare nucleus-- there's no electrons, so the atomic number is the charge. And then we have a filled shell around it. It's spherically symmetric. And so if we penetrate inside of it, what we see-- suppose we penetrate inside, to this point, what we see is only the plus z, minus the number of electrons inside this sphere. Now if you took Electricity and Magnetism, you can prove this. If you didn't, you can accept it. And so outside the nucleus, the charge is plus one, because you have a neutral atom. And then when you penetrate inside this region of dense charge, and all of the spins are generally paired, this is spherical. So what you end up seeing is z effective, as opposed to z true. And let's say here here's r0-- or this is r0. And so beyond r0, the charge that you see is plus 1. Add 0, you see a charge of z. And so what ends up happening is you get z effective, which is dependent on distance from the nucleus. And it goes from the integer value that you know, from the atomic number, down to 1, because you've taken one electron away from a neutral atom, and taken it outside. And now we have this wonderful thing called the centrifugal barrier. So if we have a state that has a non-zero of l-- well, if we have a state with a zero value of l, it can penetrate all the way into the core, to the nucleus. And so that means that the shielding is less for s orbitals. And now if we have a non-zero l, it can't get in so far. And the larger the l is, the less it can see this extra charge. So high l's are very well shielded. Low l's are not so well shielded. And the shielding goes s least shielded, p, less, so on. Now there's some other interesting things. Which, you know, I hate to say this, but comparing 5.111 or 5.112 to 3.091, there is this business of what happens when you start to-- you start with potassium, and so you put an electron in the 4s, not in the 3d orbital. Right? Why is that? Well, the 4s sees the larger charge is less shielded. So it goes in. Then when you go from potassium to calcium, you put another electron in this. And that's true. So for calcium, you have a 4s squared. And 4s 3d is a higher-lying state. Now you take an electron-- I'm cooking my own goose. If you take one of these electrons away-- this is not the way I wanted it to come out-- you find yourself in a 3d state. Because 3D penetrates a little bit under 4s. I can't explain it in a way that's going to make sense. I really wanted to, because I care so much about these simple arguments. But I will just be wasting your time. So the order in which you feel orbitals comes out, naturally, different from the order in which you remove electrons from orbitals. And the shielding arguments are capable of explaining that. OK, so this is the end of atoms. And I've asked you to observe some complicated algebra which you're never going to do, or at least never going to do much of. Everything you need to know about atoms, you can tell a computer, and it can do it. Now molecules are much more complicated. And that's we're going to start on next time. We're going to start with molecular orbital theory. And I'm not going to be presenting the normal textbook approach. I'm going to present an interpretive approach, where you understand why things happen, as opposed to memorize just symmetries, and filling orders, and so on. OK, I'll see you on Wednesday. |
MIT_561_Physical_Chemistry_Fall_2017 | 20_Hydrogen_Atom_I.txt | FEMALE SPEAKER: The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So we're now starting to get into a really familiar territory, which is the hydrogen atom. And it's a short step from a hydrogen atom to molecules. And we're chemists. We make molecules. But one question that would be a legitimate question to ask is, what does a hydrogen atom have to do with molecules, because it's an atom and it's the simplest atom. And what I hope to show is that what we learn from looking at the hydrogen atom has all sorts of good non textbook stuff that prepares you for understanding stuff in molecules. And I am going to stress that because you're not going to see it in McQuarrie and you've probably never seen it elsewhere except in freshman chemistry when they expect that you understand the periodic table based on some simple concepts. And they come from the hydrogen atom. And so I'm going to attempt to make those connections. My last lecture was I told you something you're not going to be tested on. But it's about spectroscopy. That's how we learned almost everything we know about small molecules and a lot of stuff about big molecules too. The crucial approximations in that were the dipole approximation where the radiation field is such that the molecule field's a uniform but oscillating electromagnetic field. In developing the theory that I presented, we assume that only one of the initial eigenstates is populated at t equal zero. And only one final eigenstate gets tickled so that it gets populated. And so even though there is an infinity of states, the theory specifies we start with one definite one and only one final state gets selected because of resonance. And this is what we do all the time. We have an infinite dimension problem, and we discover that we only really need to worry about a very small number of states. And then we deal with that. There are a couple other assumptions that were essential for this first step into a time-dependent Hamiltonian. And that is the field is weak and so you get linear response. That means that the increase in the mixing coefficient for the final state is proportional to the coupling matrix element times time. Now mixing coefficients can't get larger than 1, and so clearly we can't use a simple theory without saying, OK, we don't care about the mixing coefficients. We care about the rate of increase. And so by going from amplitude to rates, we are able to have a theory that's generally applicable. Now what I did was to talk about a CW radiation field, continuous. Many experiments use pulsed fields. Many experiments use an extremely strong pulsed field. That I hope to revisit later in the course. But the CW probably is the best way to begin. So now let's talk about the hydrogen atom. We did the rigid rotor, and that led to some angular momenta. Well, let's just picture. So we have this rigid rotor, and it's rotating about the center of mass. And the Schrodinger equation tells us the probability amplitudes of the rotor axis relative to the laboratory frame. And we got angular momenta, which we can denote by a little l, big L, J, and many other things. And this is important because, if it's an angular momentum, you know it. You know everything about it. And you don't know yet that, if you have two angular momenta, you could know everything about the interactions between them. That's called the Wigner-Eckart theorem. That's not in this course. But the important thing is, if you've got angular momenta, you're on solid ground and you can do stuff that doesn't ever need to be repeated. And so one of the things that we learned about the angular momentum was that we have-- even though it's going to be L as the angular momentum mostly in the hydrogen atom, I'm going to switch to J, which is my favorite notation. So J squared operating on a state-- now I can denote it like this or I can denote it like this. It's the same general idea. These are not quite equivalent things. And you get H bar squared J A plus one JM. And we have JZ operating on JM. And you get H bar M JM. And we can have J plus minus operating on JM. And that's H bar times this more complicated looking thing, J plus 1 minus M, M plus or minus 1. So that's the only bit you have to remember. And you might say, well, is it plus or minus 1 or minus or plus 1? And so if you choose-- M is equal to J. Then you know that if you have M is equal to J, this is going to be 0 if we're doing a raising operator. And so there's nothing to remember if you are willing to go to an extreme situation and decide whether it's plus minus or minus or plus. Now the rigid rotor is a universal problem, which is solved. It deals with all central force problems. Everything that's round, the rigid rotor is a fantastic starting point. So one-electron atoms for sure-- many-electron atoms, maybe. We'll see. But anything that's round, the rigid rotor is a good zero point for dealing with the angular part of the problem. So for the hydrogen atom, we have a potential. It looks like this. And so the radial potential is new. For that rigid rotor, the radial potential was-- it's just 0 for the R is equal to R zero. And not zero-- then the potential is infinite. Now here we have potential, which doesn't go to infinity here and it does something strange here because you can't get to negative R in spherical polar coordinates. So is it a boundary condition or is it just an accident of the way we use coordinates? So we're going to apply what we know about angular momenta to the hydrogen atom. So our potential for the hydrogen atom is going to be expressed in terms of the distance of the electron from the nucleus and then the theta phi coordinates, which you already know. And the theta phi part is universal, and the R part is special to each problem that has physical symmetry. And there are different kinds of approximations you use to be able to deal with them. The nice thing about it is it's one dimensional, and we're very good at thinking about one dimensional problems. And even if the problem isn't exactly one dimensional or it has some hidden stuff to it, we can extend what we know from one dimensional problems and get a great deal of insight. And if you have a one dimensional problem, it's very easy to describe the potential in one dimension and the eigen functions in one dimension. And so we can begin to really understand everything. So we're going to have a wave function, which is going to have quantum numbers. And we're going to be able to write it as a product of two parts. Well, this is the same thing we had before for the rigid rotor. So we're going to be able to take this rotary equation and separate the wave function into two parts, a radial part and an angular part. And this is old and this is new. Because we've got this, we can use all of that stuff without taking a breath, except maybe getting the right letter L, J, S, whatever. So really the hydrogen atom is just one thing with some curve balls thrown at you in the latter stages. One of the things you also want to be able to do-- because nodes are so important in determining both the names of the states and how they behave in various situations, including external fields and excitation by electromagnetic radiation-- you want to be able to understand the nodal surfaces. And you already know this one. So how many nodal surfaces are there if you have a particular value of L? Yes. AUDIENCE: L. PROFESSOR: Right, and if you have a particular value of M, how many nodes are there in the xy plane? You're hot-- AUDIENCE: I mean, I know if M is 0, the entire angular momentum is tipped into the xy plane. So that means that the axis of the rotor has to be orthogonal to the angular momentum. So that means that the probability density is oriented along Z, I think. PROFESSOR: Now you're saying things that I have to stop and think about because you're not telling me what I expected to know. So if M is equal to 0, L is perpendicular to the quantization axis, and there are no nodes. AUDIENCE: Well, it depends on if you're talking about the nodes of the probability density of the rotor, where the axis is. PROFESSOR: That's true, but where the axis is determined by theta. And when theta is equal to pi over 2, you're in the xy plane. And the number of nodes in the xy plane is M, or absolute value of M. The phi part of the rigid rotor is simple. It's a differential equation that everybody can solve. We already know that one, and we know what the wave functions look like. We can write the hydrogen atom Schrodinger equation, and it's very quick to show separation of variables. And I'll do that in a minute. And then we have the pictures of the separated parts-- RNL of R and YLM of theta phi. And essential in these pictures is the number of nodes. In the spacing between nodes, remember the semiclassical approximation. We know that Mr. DeBroglie really hit it out of the park by saying that the wavelength is H over P. For every one dimensional problem, we know what to do with that. And we're going to discover that, for the radial problem, we have a very simple way of determining what the classical momentum is. And so we know everything about the nodes and the node spacing and the amplitudes between nodes and how to evaluate every integral of some power of R or Z. And so there's just an enormous amount gotten from the semiclassical picture once you are familiar with this. And so we can say that we have the classical wavelength, or the semiclassical wavelength. It has an index R. So we have a momentum, a linear momentum with a quantum number on it for L. See, that's a little strange. But that tells us what the potential is going to be and that tells us what to use in order to determine the wavelength. And this is really the core of how we can go way beyond textbooks. We can do-- in our heads or on a simple piece of paper, we can do this. And we can draw pictures, and we can evaluate matrix elements. Without any complicated integral tables, you can make estimates that are incredibly important. And from that you can get expectation values and also off-diagonal matrix elements of integer powers of the coordinate. That's an enormous amount of stuff that you can do. Now you can't do it yet. But after Wednesday's lecture, you will. And so this will be Wednesday. And then we'll have evidence of electron spin. And I will get to that today. So this is the menu. Now let's start delivering some of this stuff. I have to write some big equations. So the Hamiltonian is kinetic energy plus potential energy, and kinetic energy was-- I'm sorry. For the rigid rotor, V was zero. And everything was in the kinetic energy. Well, it's not quite true. And so, for hydrogen-- so we know this is P squared over 2 mu, and mu for the hydrogen atom, reduced mass. And we know this is just Coulomb's Law, minus e squared over 4 pi epsilon zero R, H bar squared. I'm sorry-- it was supposed to be e squared. So this is the classical-- So these are the parts. But the since it's spherical, we're not going to work in Cartesian coordinates. And in fact, that's a very strong statement. If you have a spherical problem, don't start in Cartesian coordinates because it's a horrible mess transforming to spherical polar coordinates. Just remember the spherical polar coordinates. So you know how spherical polar coordinates work. I'm not going to draw it. So the kinetic energy term is-- and this is the Laplacian, and that's a terrible thing. And Del squared-- it looks like it's going to be a real nightmare partial with respect to R, R squared partial with respect to R. And we have 1 over R squared sine squared theta partial with respect to theta, sine theta partial with respect to theta. And then a third term-- 1 over R squared sine squared. I got a square here. That's wrong. And this is sine squared theta. I heard a mumble over there. Yeah, but I'm just doing Del squared. AUDIENCE: With your first Del, it's H plus squared. PROFESSOR: My what? AUDIENCE: Your first time you used the Del operator, it's H plus squared. PROFESSOR: Oh, yes, yes. AUDIENCE: And then the second one, if you could clarify that that's a Del-- PROFESSOR: I'm sorry. AUDIENCE: The second Del squared, if you could clarify that's a Del squared. It looks like a-- PROFESSOR: It looks like a terrible thing. And the last part is a second derivative with respect to phi. This looks like a terrible thing to build on. But with a little bit of trickery-- and that is, suppose we multiply this equation by R squared-- then we have killed the R squared terms here and here. And we're going to be able to separate it. And so we are able to write an equation which has the separability built in-- and so partial with respect to R, R squared partial with respect to R plus L squared plus 2 mu H R squared V of R minus E psi. That's a Schrodinger equation. So we have an R dependent term and another R dependent term and a theta phi dependent term all in this one, nice operator that we've understood. So now there's one more trick. In order for the shorter equation to be separated into a theta phi part and an R part is we need a commutator. And so that commutator is this. What is a commutator between L squared and any function of R? Yes. AUDIENCE: Their operators depend on different variables? PROFESSOR: Absolutely, that's really important. We often encounter operators that depend on different variables. And when they do, they commute with each other, which is an incredibly convenient thing because that means we can set up the problem as a product of the eigenfunctions of the different operators. So this means that we can write a theta phi term, which we completely know, and an R term, which we don't know and contains all of the interesting stuff. So let's forget about theta and phi. Let's just look at the R part. Well, the way we separated variables, we had the Schrodinger equation and we divided by the wave function, which would be of R of R, Y L M of theta phi. And on one side of the equality, we have only the angle part. And so when we do that, when we divide by R, we kill the R part on this side. And on the other side, it's the opposite. And so we get two pieces-- one is only dependent on theta and phi and one is only dependent on R. So they both have to be a constant. So we get two separate differential equations. And we've already dealt with one of them. So what you end up getting is 1 over R of R times stuff times R of R is equal to 1 over YLM L squared YLM-- sorry. So this is the separation. This is all R stuff. This is all theta stuff and separation constant. Well, this one is inviting a separation concept because L squared operating on YLM is H bar squared, LL plus 1. That's the separation constant. So now let's look for the only time at the R part of the differential equation. And so this writing out all the pieces honestly-- we have-- there's one more. So that's the Schrodinger equation for the radial part. It looks a little bit annoying. The important trick is that we've taken the separation constant, and it has an R dependence but when we divide through by 2 mu HR squared. And we say, oh, well, let's call these two things together-- VL of R. This is the effect of potential, and it depends on the value of L. So this is just like an ordinary one dimensional problem except now, for every value of L, we have a different potential. And this potential is-- OK, when L is not equal to zero, this potential goes to infinity at R equals 0, which is bad, except it's good because it keeps the particle ever from getting close to the nucleus. And that's what Mr. Schrodinger-- I'm sorry what Mr. Bohr was thinking about, that we have only circular orbits or we have only orbits that are away from the place where the Coulomb interaction would be infinite. And I mean, there are several things that's wrong with Bohr's picture. One is that we have orbits. And the other is that we don't include L equals 0, but this is telling you a lot of really important stuff because the value of L determines the importance of this thing that goes to infinity at R equals 0. And that has very significant consequences as far as which orbital angular momentum states we're dealing with. So we're going to get the usual LML quantum numbers, and we're going to get another one from the radial part. And since this is a 1D equation, there's only one quantum number. And we're going to call it L. And now there's this word principal, and there's really two words-- principle with an LE and principal with an AL. Which one do you think is appropriate? I'm sorry. I can't hear. AUDIENCE: AL. PROFESSOR: AL is the appropriate one. Principle has to do with something fundamental. Principal has to do with something that's important. And I can't tell you how many times people who should know better use the wrong principal. You'll never do that now because it was like, what's nu? C over lambda. So this is the right principal. If we do something clever and we have the radial part and we say, let us replace it by 1 over R times this new function chi L of R, well, when we do that, this equation becomes really simple. So what we get when we make that substitution is H bar squared over 2 mu H second derivative with respect to R plus VL of R minus E chi L of R is equal to 0. That looks like a differential equation we've seen before. It's a simple one dimensional differential equation-- kinetic energy, potential energy. But it's not. There's some kinetic energy hidden in the potential energy. But this is simple, and we can deal with this a lot. I'm not going to. But if you're going to actually do stuff, you're going to be wanting to look at this differential equation. But I'm going to forego that pleasure. One of the problems with the radial equation is the fact that R is a special kind of coordinate. It can't go negative. And so treating the boundary condition for R is equal to 0 is a little subtle. And it turns out that when L is equal to 0, then R of 0 is not 0. But for all other values of L, R of 0 is 0. And all of NMR depends on L being 0 because the electron feels the nucleus. It doesn't experience an infinite singularity. It feels the nucleus. And when L is not equal to 0, then it's sensing the nucleus at a distance. And that leads to some very small splittings called hyperfine. And so there's different kinds of hyperfine structure. But anyway, this is a really subtle and important point. How much time left? Yes? No. So now pictures of orbitals-- So the problem with pictures of orbitals is now we have a function of three variables and it's equal to some complex number. So we need two degrees of freedom to present a complex number and we have three variables. And so representing that on a two dimensional sheet of paper is horrible. But we have this wonderful factorization where we have RNL of R. And we could draw that easily. We don't need any special skill. And we have YLM of theta phi. And we've got lots of practice with that, although what we've practiced with may not be completely understood yet. So we have ways of representing these. And so we go through the understanding of the hydrogen atom by looking at these two things separately. Now for the radial part, the energy levels-- where this is the Rydberg constant. It's a combination or fundamental constants. And for hydrogen, the Rydberg constant is equal to 109,737.319-- there's actually more digits, wave numbers-- times mu H over mu infinite. Well, this is actually the Rydberg constant for an infinite mass. And so mu H is equal to the mass of the electron times the mass of the proton over the mass of the electron plus the mass of the proton. Now the mass of the proton is much bigger than the mass of the electron. And so you can use this as a trick. You can say, well, what is it? Well, we know what this is. It's easy to calculate. But there are two limits. What is the smallest possible reduced mass? And that would be for-- if the mass of the proton is infinite, then we just get the mass of the electron. That's the biggest reduced mass. And the smallest is, if we have positronium, where we have an electron bound to a proton-- and then when we do that, we get half. I'm sorry. It's not a proton. It's a positively charged particle, which we call a positron. And so each of the terms here is the mass of the electron. And so the range is from 1/2 me to me depending on what particles you're dealing with. And so that's a useful thing. And so the Rydberg constant for hydrogen is smaller than the reduced mass of the infinite. I'm sorry. It's smaller than that for the infinitely mass nucleus. And it is the value that made Mr. Bohr very happy-- 679. So the important thing is this mass scaled, or reduced mass scaled, Rydberg constant explains to a part in 10 to the 10th all the energy levels of one-electron systems-- hydrogen, helium plus, lithium 2 plus. That's it. And that's fantastic. Now we have to talk to become really familiar with this Rnl of our function. And one of the questions is, how many radial nodes. And you know that for 1S there aren't any nodes. And you know for 2P there aren't any radial nodes. And so what we need is something that goes like n minus L minus 1. The number of radial nodes, which is all you need to know about the radial A function, is how many nodes are and how far are they apart and what's the amplitude of each loop between nodes. Semi classical theory gives you all of that. And often when you're calculating an integral, all you care about is the amplitude in the first loop. And so instead of having to evaluate an integral, you just figure out what is the envelope function based on the classical momentum function. Now the thing that everybody's been waiting for. Spin. So remember, we know an angular momentum is R cross P, and we know that there isn't any internal structure. Or at least we don't know about internal structure of the electron or a proton. And so we can't somehow say, well, it's R cross P. So we have the Zeeman effect. So we can look at the Zeeman effect for an atom in a magnetic field. And we have several things that we know. So we have the magnetic moment of the electron is equal to minus the charge on the electron times 2ME times L. Well, that's an angular momentum. So if we had circulating charge, well, that circulating charge will produce a magnetic moment, which has a magnitude which is related to the velocity. And what is L divided by mass? It's a velocity. And we have the charge here. So this is a perfectly reasonable thing. And that's good because we know that, if we have electrons with some kind of internal structure, we know what this is going to do. Now the magnetic potential is equal to minus the magnetic moment times the external magnetic field. And so what we have is E, BC, LZ over 2 NE. Well, this is another very good thing because, not only do we know what this is, we know that it has only diagonal elements. And so we can do first order perturbation theory. If this is so easy, if this is the Hamiltonian, we can just tack that on and it adds an extra splitting to our energy levels. The only tricky thing is, we don't know that it's LZ. We just know that it has a magnetic moment or it could have a magnetic moment. And so this wonderful experiment-- suppose we start with a 1S state, and you go to a 2P state. Now this implies that we know something about the selection rules for electromagnetic transitions. And so for an electric dipole transition, we go from 1S to 2P. Now this is not how it was done initially because the frequency of this transition is well into the vacuum ultraviolet. And in the days when quantum mechanics was being developed, that was a hard experiment. So one used an S to P transition on some other atom, like mercury, but let's pretend it was hydrogen. So this is an angular momentum of 0. So we would expect it would be ML equals 0. And here we have it could split into three components. And this is an L equals 1, 0 minus 1, because that's minus-- there with a minus sign somewhere. That's what we expect. And so being naive, you might expect transitions like this. Well, the transitions where you do not change the projection quantum number are done with Z polarized radiation. And so if it's Z polarized, you get only delta ML equals 0. And if you have x and y, you have delta ML equals plus and minus 1. So depending on how the experiment was done, you would expect to see one, two, or three Zeeman components. And they did the experiment, and they saw more than three components. They saw five components. And that's possibly for a number of reasons. But they saw five components. So they knew that there was something else going on. And so you have to try to collect enough information to have a simple minded picture that will explain it all. So one thing you might do is say, OK, suppose there is another quantum number, another thing. And we're going to call it spin. And we don't know whether spin is integer or half integer. We know from our exercise with computation rules that both half integer and integer angular momentum are possible. And so we can say we have a spin. And it could be 1/2. It could be 1. It could be anything. And then we start looking at the details of what we observe. And what we find is, all we need is spin 1/2 to account for almost everything. However, if it were spin 1/2, then you get two here. And you'd get two here, two, here and two here. And that's six. That's bigger than five. So you need to know something else. And one thing is quite reasonable-- you can say, well, the spin thing is mysterious, and electromagnetic radiation acts on the spatial coordinates. And there aren't any spatial coordinates of the internal structure of the electron. And so we have a selection rule delta MS equals 0. That still doesn't do it. You still need something else, and that is that the proportionality concept between the angular momentum and the energy, which is called the G factor. For orbital angular matter, the G factor is 1, or the proportionality constant is 1. So the G factor is L. And the G factor for the electron turns out to be 2-- not exactly 2, just a little more than 2-- Nobel Prize for that. And so with those extra little things, then every detail of the Zeeman effect is understood. And we say, oh, well, the electron has a spin of 1/2, and it like it acts like an angular momentum. It obeys the angular momentum computation rules. It also obeys the computation rules. Well, I won't say that. So I should stop-- there are other things that provide us information that there is something else. And one, I'm called by some people the spin orbit kid because I've made a whole lot of mileage on using spin orbits splittings. And the spin orbit Hamiltonian has the form L.S And this gives rise to splittings of a doublet P state, for example, as what you would see in the excited state of hydrogen into two components at zero field. And so there's all sorts of really wonderful stuff. And I'm going to cheat you out of almost all of it because we're going to go over to the semiclassical picture next time. And we'll understand everything about Rydberg states and about how do we estimate everything having to do with the radial part of the wave function. And there are some astonishing things. |
MIT_8962_General_Relativity_Spring_2020 | 20_Spherical_compact_sources_I.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: So we're switching gears. We're done with the large scale structure of the universe. Now we're done with cosmology. And I'm going to move into the topic that will dominate the last several lectures of 8.962. Let's look at how to construct the spacetime of a compact body. In particular, we're going to start by focusing on a compact spherical body. What I mean by compact-- so spherical, I hope, is clear-- but what I mean by compact is that the body occupies a finite spatial region. And it has a surface with an exterior that is vacuum. So what I'm going to do is design some kind of a spacetime. I'm imagining that there is a source that fills some compact region. I'm going to make it strictly symmetric. And the exterior of the source will be t mu nu equals 0. So if you think about this, this is telling you that if it's spherically symmetric, your metric can only depend on radius. And so the most general form, at least for a static spherically symmetric spacetime-- we'll talk a little bit about a few things related to non-static situations a little bit later. So the most general static form of a spherically symmetric compact body tells me that I can write my metric as some function of r, which for reasons of convenience I will write that is e to the 2 phi, t squared. Some other function of r, dr squared. And then some function of r times my angular sector. You might wonder why there's no cross-term. In particular, why can't I have a cross-term between my dt and my dr? I have a few lines on this in my notes. I won't go through them in detail here, but I'll just describe. Basically, what you find is that if there is a cross-term between, say, the t and the r pieces, you can get rid of it by a coordinate transformation. This way of doing things so that it is diagonal amounts to choosing a good time coordinate. We can simplify the angular sector by choosing a good radial coordinate. The word "good" is a little bit loaded here. I'm going to actually in a moment describe a slightly different choice of radial coordinate, which is also good, but for different purposes. The one I'm going to use is one where this function that gives length to the angular sector is just r squared. What this means is that the coordinate r, it labels nested spheres that have surface area 4 pi r squared. So if you imagine that in this spacetime you make a series of spheres, and each one of them has an area 4 pi r squared, that is the r that goes into this. Bear in mind-- [CLEARING THROAT] excuse me-- bear in mind, that does not mean-- let's say I've got one sphere of radius 4 pi r1 squared, and then outside of it a sphere of radius 4 pi r2 squared. That does not mean that the distance, the proper distance, between those two spheres is r2 minus r1. So this is unlikely to measure-- to label radial distance cleanly. In fact, it would only be the case that this cleanly labels radial distance if the function lambda were equal to 0. And as we'll see, that's generally not the case. So whenever we do things in a curved spacetime, things are always going to get a little bit messier. And what's nice about this is at least we do have a clean geometric meaning to the coordinate r. In fact, before I move on, this is called an areal radius, a radius simply related to surface area. So in the end, the spacetime that I will be working with in this lecture takes this form. And so this is a choice that is known as Schwarzschild coordinates. We will come back to the name Schwarzschild before too long. As I said, so this is a particularly good choice of radial coordinate. It'll be very convenient for what we do. It's not the only one. Other coordinate choices are possible. And another one that is quite useful for us-- we're not going to use it too much in this lecture-- is what are called isotropic coordinates. So this is a coordinate system in which you choose a radial coordinate which I will call r bar, such that your line element looks like this. What this does is this emphasizes the fact that we are working in a coordinate system-- or excuse me-- we're working in a spacetime in which all of your spatial slices are fundamentally isotropic. So get far enough away from this thing, and the three-- it's just emphasizing the fact that the three spatial directions are-- there is something special about it. There's presumably a body at r bar equals 0, in its vicinity, but other than that, there's nothing particularly special about-- if I go into a little freely falling frame, all three directions look the same. This helps emphasize this. We lose the areal interpretation of r here. I'll remind you that our weak field solution that we derived a few weeks ago, a few lectures ago rather, it looked like this. And voila, look at that. It's exactly the same form. So this is, in fact-- when we derive this, this is, in fact, actually in isotropic coordinates. What we found was that the mu that appears here is equal to minus phi with phi itself being very small. And the phi that came from this is Newton's gravitational-- the Newtonion gravitational potential. Let's switch back. We're going to use Schwarzschild coordinates. This is kind of an aside. We will use Schwarzschild shield coordinates for the bulk of our lecture. Actually, the last thing I'll say about this is that there is a problem. There's a problem on a problem set-- I might have made it optional. I don't quite recall-- in which for one of the spacetimes that you're going to be working with in about two more lectures, you convert between Schwarzschild coordinates and isotropic coordinates. I think it's an optional lec-- it's an optional problem at this point. But it's worth looking at. So let's build a bot. Let's build the spacetime of a compact spherical body. What we need are curvature tensors and matter terms. The curvature tensors, since I've given you-- whoops. Not there. No. The one at the top. I'm sorry. Since I've given you ds squared, you know all the metric elements. It's then just a matter of a little bit of having the algebraic stamina to run through and do the calculations. It's straightforward to construct all of the curvature tensor components. The GR tool Mathematica notebook that will soon be released on the 8.962 website, if it has not been released already, that will allow you to do this. I'm going to just sort of quote the results, what the results turn out to be. So for this spacetime, it's given in Carroll's textbook, a slightly different notation, but you can easily translate from the functions Carroll uses to the ones that I've written here. So we need to get those curvature tensor components. And we need matter. And so we will do what we generally always do in this class. We will treat our body as being made of a perfect fluid. So let's write out our curvature tensors that arise from this. I'm going to go straight to the Ricci tensor. If you want to work out all the complements to the Riemann tensor, knock yourself out. As I said, they are listed in Carroll's book. It's not terribly difficult to go and work all these guys out, though. Certainly not using Mathematica. So here's what you get for the rr complement of the Ricci. Notice we are seeing nonlinear terms in here. These are the kind of terms that in all of our calculations, when we were looking at bodies in a weak field spacetime, we set those to 0. So the form for the ttp. So notice it's quite similar to the form for rrp's. If you saw me hesitate while I was writing, I actually suddenly just got concerned I was writing the wrong line down, and was just verifying that I was writing the right thing. And two more non-trivial complements, or really one more non-trivial component, and then there's a fourth component that is simply related. So here is our Ricci tensor. We're going to solve it for a perfect fluid stress energy tensor. We will also solve it for-- sorry-- we will use this to make an Einstein tensor, equate it to a perfect fluid stress energy tensor, and use that to build the spacetime on the interior of this object. The exterior of the object is going to be treated as vacuum. That will have 0 stress energy tensor. And so it'll be a little bit easier when we construct the exterior solution just to set Ricci equal to 0 and solve it. Before I get into that, let's look at the source we're going to use to describe this compact spherical body. Like I said, we're going to treat this guy as a perfect fluid stress energy. So by now, you will have seen me written-- you'll have seen that I've written this form down multiple times. So these, the pressure and the density, are both functions only of radius, since this is spherically symmetric. And u denotes the four-velocity of fluid elements in this fluid-- in this body. I'm going to start calling this body a star. It doesn't necessarily have to be a star, but it's just a convenient shorthand. So I want this star's fluid to be static. So the four-velocity will have a timeline component, but there is no spatial motion. I want the star's fluid to be at spatial rest. I am going to require that this be properly normalized. u dot u equals minus 1. So if I do that in this spacetime, I'm led to the requirement that u upstairs t is e to the phi of r. Pardon me. e to the minus 5r. And u downstairs t is minus e to the 5r. Finally, we're going to assume-- so as I said, we're going to treat this as a compact body. So we're going to require that it have a surface. So we will assume rho of r equals 0, p of r equals 0 for r greater than or equal to r star. r star denotes the surface of this body. Let's attack. So it's a little bit easiest to do the exterior first. So we're going to consider the region outside of r star. t mu nu equals 0 there. And so my Einstein equations become Ricci equals 0. So I can just take every one of these things, set them equal to 0, and try to assemble solutions for phi and for lambda. Now a particularly convenient-- every one of these things has to equal 0 so I can make various linear combinations of them to make things that are particularly convenient to work with. And if you sort of stare at the tt and the rr equations, one thing you see is that if you multiply the tt by this combination of metric functions, add the r r, what that does is it allows you to cancel out those nonlinear terms. And this combination tells you that-- whoops-- this combination tells you that it's an overall factor of 2/r, the radial derivative of phi equals the radial derivative of lambda. And so this tells me that at least in the exterior phi equals minus lambda, perhaps up to some constant of integration. If we go when we plug this into our metric, what that k does is it just gives me a rescaling, depending upon whether I want to attach it to the function phi or attach it to the function lambda. What we see is that K just essentially rescales. But let's do the following. Let's sort of say that I'm going to plug this in and say that I'm going to plug it in for phi. So if I plug this into the metric, I can capture the impact of the constant of integration k here by just rescaling my time coordinate. So we are free just to say, OK, there's no physics in this thing. So let's just set k equals 0. So that's already great. We have found that at least in the exterior phi is equal to minus lambda. We've reduced this metric from having two free functions to one. So we're feeling spunky now. Let's take a look at a couple more of these equations, and see what we can learn from them. So we still have some of these ugly things involving nonlinear stuff in the rr and the tt, but it's all linear in the theta theta equation. So let's look at r theta theta equals 0. So I have e to the minus 2 lambda r dr of lambda minus r dr of phi minus 1 plus 1. All this equals 0. So now you can either substitute for lambda or substitute for phi. I'm going to follow my notes here and substitute in lambda equals minus phi. Stare at this for a second, and you'll realize that you can rewrite this term on the left-hand side as dr of r e to the 2 phi equals 1. And this can be easily integrated up. And the solution is phi of r equals 1/2 log 1 plus a/r, where a is some constant of integration. We're going to figure out a way to fix that constant of integration. And you can probably guess how I'm going to do that. But let's just hold that thought for now. So let's just take this solution and write down what our exterior spacetime looks like. So plugging this guy in, I get-- OK. So here's my line element. And we're going to do the usual thing. This is-- well, remember that this situation describes gravity. And gravity has a Newtonian limit, which these general relativistic solutions must capture. So let's make sure that this captures my weak field limit. So let's consider r gihugically greater than a. a must be some parameter with the dimensions of length. It might turn out to be negative. So let's take the absolute value of this thing. And let's consider a purely radial freefall. So I'm going to imagine-- when I work this guy out, I'm going to look at geodesics in this limit. I'm going to look at purely radial freefall. I'm going to consider the non-relativistic large r limit. So here I am imagining in my non-relativistic limit that dt d tau is much greater than the dr d tau. So when you work out all those components of the Ricci tensor, you will have computed all of the Christoffel symbols. And you go back to your handy dandy little table of these things that you will have computed. So here's what you get for that. Your phi is given by 1/2 log of 1 plus a/r. So-- taking the exponential and the proper derivatives, what this tells you is that this is negative a over 2r squared 1 plus a/r. And again, I'll emphasize, we are considering this sort of very large r limit. So this is approximately negative a over 2r squared. Dividing both sides by the factor of dt d tau squared, I finally get my equation of motion for non-relativistic weak field radial freefall to be that this thing's radial acceleration is a over 2r squared. Now, we want this to capture the Newtonian limit. Newtonian freefall is given by dt dr equals the gravitational acceleration minus gm over r squared. They're exactly the same, provided I select a equals minus 2gm. So what emerges out of this analysis is this truly lovely solution. This is known as the Schwarzschild metric. This is a good point to tell a brief story. Carl Schwarzschild derived this solution shortly after the publication of Einstein's field equation. So Einstein's field equations appeared in late November-- excuse me-- late in the year 1915. I believe in November of 1915, but we should double-check that. Schwarzschild published this exact solution, a few months later, I believe in early 1916. Students of history will note that if you were in Germany in 1915 and 1916, you were likely to be involved in some rather all encompassing, non-academic activities at that time. And in fact, Carl Schwarzschild did this calculation while he was serving as an artillery officer in the German Army on the Western-- excuse me-- on the Eastern Front. He, in fact, was quite ill at the time he did this. I believe it was a lung infection that he had accumulated while serving in the trenches. And so while recuperating from this disease, he received Einstein's manuscripts because he was a professor in his civilian life, and just wanted to keep up with the literature. He was inspired by this relativistic theory of gravity, and decided that it would be worth a little bit of his time while he was resting in the hospital to see if he could do something with these exciting new equations that his colleague, Herr Doktor Professor Einstein, had worked out. And he came up with this, the first exact solution to the Einstein field equations. And indeed, a solution that continues to be of astrophysical and observational importance today. He succumbed to the disease that he was suffering and died several weeks after coming up with this solution and publishing it. This was a solution which stunned Einstein, who did not think that such a simple solution would ever be found, such a simple exact solution would ever be found to these horrendous equations that he had developed. I always find this story to be somewhat-- well, I'm not quite sure what to make of it. In these days of the coronavirus, where we are all isolating, we are-- many of us now are probably beginning to hear stories of people we know who may have the disease. Sometimes it's hard for me to crawl out of bed in the morning. Somehow I do so. And I sometimes think to myself, what would Karl Schwarzschild do? He didn't let the fact that he was dying of a lung infection from-- he did not let the fact he was dying of a lung infection prevent him from leaving an indelible stamp on the history of science. I do not aspire to such greatness. I do not think I am capable of such greatness. But it at least helps me to get out of bed in the morning. So like I said, this is a solution that is of observational importance today. And one of the reasons it has such significance is due to a result known as Birkhoff's theorem. Birkhoff's theorem teaches us that the exterior vacuum of any spherically symmetric body is described by the Schwarzschild metric. This is true even if the source is time varying. So in doing this derivation, I focused on a situation in which my spacetime was static. There is no time dependence whatsoever. It turns out that if it is not static, if it is a time-varying spacetime, as long as the time variations preserve spherical symmetry, the exterior is still described by Schwarzschild. It's even stronger than if-- only if. As long as the time variations preserve spherical symmetry. So for instance, it could be radial pulsations. If it is not a variation that preserves spherical symmetry, you're very likely to have a time-varying quadrupole moment, and then you produce gravitational waves. But if you have time variations that preserve spherical symmetry, spherical symmetry does not allow us to have a time-varying quadrupole moment. So no gravitational waves are produced. And what this tells you is that as long as you're in this vacuum region on the outside of it, your thing could be down there going [VOCALIZING],, as long as it does so in such a way that leaves it spherically symmetric, then that is the spacetime that describes it. You are completely ignorant of what the radial extent of the matter is. It only matters that you have a mass of m, and you are distance r away from its center, at least in these coordinates. In that sense, it's kind of like Birkhoff's theorem plays a role similar to Gauss's law in elementary electricity and magnetism. If you have a spherically symmetric distribution of charges, you are completely agnostic as to what that-- what the radial distribution of that spherical symmetry is. And in fact-- although, this isn't usually discussed-- if that were to be sort of a time variable kind of thing, as long as it was just sort of homologous in moving these guys in and out like that, you would still just get the E-field associated with a point charge at the origin. This spacetime is kind of the gravitational equivalent of a point charge at the origin, a point I'm going to come back to in a future lecture. Let me just give you a brief proof of Birkhoff's theorem. So imagine I want to consider a spacetime in which I allow my metric functions to vary with time. So go ahead and work all of these guys. Use that as your metric. Work out all of your curvature tensors. We are focusing on the exterior vacuum region of the star. So what we're going to do is set those curvature tensors equal to 0. What you find when you do this-- so when you work out your curvature tensors, in addition to those four that we had on the center boards earlier, the four components of Ricci, you find a new component enters. What you find is that there is a component r sub tr, which looks like-- oops, pardon me-- which looks like 2/r times the time derivative of your lambda. This you are going to set equal to 0 in the exterior. So this tells you that the time derivative of lambda is 0. This means that your assumed lambda of t and r is just a lambda of r. So there is no time dependence in that lambda. So you just killed that. You can then look at your other curvature components. And in particular, one of the things that you'll see is several of those curvature components. You'll see that they pick up a couple of additional terms involving d by dt as the various of the lambdas and the phis. You can now get rid of those. Where I think it's particularly clean to look at this is if you look at r theta theta. That turns out to be completely unchanged. So your analysis of the r theta theta term, it could proceed exactly as before. But the other thing you can do is you can say, hey, guess what. I'm now going to imagine that my phi is time dependent. My lambda is not time dependent. We just proved that. But my phi could be. Let's look at the time derivative of this Ricci term. So if our r theta theta is 0, the time derivative of r theta theta must be 0. So when I compute this, this leads to the requirement that the double derivative, one derivative in r, one derivative in phi-- excuse me-- one derivative in r, one derivative in t of the potential phi must be equal to 0. And so this in turn tells us for this condition to be true, it must be the case-- pardon me. Just one second. How does this follow? Oh, no. Yeah, yeah. So this tells me it must be the case that phi can be separated into a function of r in a function of t. If I take the t derivative and then the r derivative, I get 0. If I take the r derivative and then the t derivative, I get 0. So my line element, my most general line element now appears to be something that looks like this. But once I have this, I can actually just redefine the time coordinate. I mean, if I look at this thing I've got here, I can just absorb a function of t into my time coordinate, just absorb it in a time coordinate like so. And what happens is that I then recover the Schwarzschild form I originally had. So that's really nice. What this tells us is that any spherically symmetric vacuum spacetime is going to have a time-like Killing vector. And if it has a time-like Killing vector, that means that I can define a notion of conserved energy for objects that are moving in that spacetime. It's going to prove very useful when we start discussing orbits. So this punch line is this totally describes the exterior of my compact spherical body. What about the interior? What we're going to need to do is-- the interior has non-zero stress energy. So we're going to need to look at components of the Einstein equation and equate them to the source, my perfect fluid stress energy tensor, and see what develops. So what I'm going to do is write my Einstein equation in the form of an upstairs downstairs index. And I'm going to do it that way, because t upstairs mu downstairs nu, it is represented in the coordinate system I'm working in here. But assuming it's a diagonal of minus rho ppp. And I had all the Riemann-- excuse me-- all the Ricci components on the blackboard a few moments ago. From them, I can construct my Einstein tensor. So there's a couple components that turn out to be quite nice. One of them is the tt component. This turns out to be minus 1 over r squared d by dr r 1 minus e to the minus 2 lambda. What's nice about this is notice it only depends on lambda. This is going to be equated to 8 pi g times t stress energy tensor tt, which is going to give me a minus rho there. So before I go and I do this, let me make a definition. I'm going to call e to the minus 2 lambda, I'm going to call that 1 minus 2gm of r over m. What I'm effectively doing is I am mapping my metric function lambda to a function m of r, which is a sort of mass function. Hopefully, you can kind of see that by doing this, remember, that this component of the metric, it's going to enter as e to the 2 lambda. And so when I match this, this is going to give me something that very nicely matches to my exterior spacetime. So plugging that in there, and valuing the derivatives, and equating this guy to the stress energy tensor is going to give me a very nice condition that the function m of r must obey. So plugging all this stuff in, I end up with minus 2g over r squared dm dr equals minus 8 pi g rho of r. Turning things around, I can turn this into an integral solution, which tells me that my mass function m of r is what I get when I integrate from 0 to r 4 pi rho of r prime. Two comments on this. You sort of look at this and go, ah, that makes perfect sense. This is what you would obviously get if you integrate up a density over a spherical volume. This is, hopefully, beautifully intuitive. And you kind of look at it and go, how could it be anything else? Well, it could be anything else. So two comments to make. First of all, I have set this up. I have imposed a boundary condition. I have imposed a boundary condition here that the mass enclosed at radius 0 is 0. You might look at that, and, well, duh. If I have a ball of radius 0, there can't be any mass in it. Of course, you're going to do that. Well, it's going to turn out that black holes are actually solutions that violate this boundary condition. Wah-wah. So there's that. The other thing to bear in mind is that 4 pi-- the reason why this probably makes intuitive sense is that in spherical symmetry in Euclidean space, 4 pi r squared dr is the proper volume element for a Euclidean three-volume. We are not in Euclidean space. And so this is indeed the definition of m of r, but it is not what we would get if we did a proper volume integration of rho over the volume that goes out to radius r. If I were to define a quantity that is the proper volume integrated rho, let's call this m sub b. Let's make it lowercase. I'll define why I'm calling that b in just a moment. I could define this as what I get when I do a proper volume integration of this thing in my curved spacetime. You get an additional factor. And what you find is that this in general-- so remember, this is going to go into your spacetime as a function. It's going to look like something like 1 over that factor there. This in general is greater than m of r. So there is a homework exercise where you guys are going to explore this a little bit. m sub b is what I get when I count up, essentially including a factor of the rest mass of every little-- let's call it baryon that goes into this star. This counts up the total number of baryons present in my star in something that's called the baryonic mass. This m that appears here defines the amount of mass that generates gravity. They are not the same. The gravitational mass is generally somewhat less than the baryonic mass. You might sort of think, well, where did it go? The missing mass can be regarded as gravitational binding energy. It's really not that it's missing. It's just that gravitational energy holds this thing together. And whenever you bind something, you do that by putting it in a lower energy state. Gravity is essentially this sort of mb with this-- the way to think of my mb here is bear in mind that we're including factors-- let's just put factors of c squared back in here. This is adding up the rest energy of every particle in this star. But when I put them all together, gravity binds it together. And that actually takes away some of that. It puts it into a bound state. And so the difference between the baryonic mass, or energy, and the gravitational mass, and energy, tells you something about how strongly bound this object is. So that's one important equation. It looks like I might finish this lecture up a little bit on the early side, which is fine, because I could use a break. Another one that is important for us will be the rr component of the stress energy tensor. So this guy, when you work it out, it looks like e to the minus 2 lambda 2 over r d phi dr plus 1 over r squared. That's 1 over r squared. So let's insert our definition. This prefactor is minus-- excuse me-- 1 minus 2gm of r over r 2 phi-- oh, shoot. Pardon me. Try that again-- 2 over r d phi dr plus 1 over p square 1 over r squared. And we equate this to the appropriate component of the stress energy tensor, which is just going to be the pressure. This can be rearranged and give us an equation for a differential equation governing the metric potential, the trigonometric function phi. So what results of that exercise something that looks like this over this. So I would like to do something with this in just a moment. But it's worth commenting on this. So when one looks at the gravitational potential in a spherical fluid body in Newtonian gravity, you get gm of r over r squared. So there's two interesting things going on here. First, your m of r, which appears in the numerator of this thing is corrected by a term that involves the pressure. What you are seeing here is the fact that pressure in some sense reflects work that is being done in this body. It's being squeezed. Gravity is sort of squeezing it down, or your hands are pushing it down. When you give this body some pressure, you're doing work on it. Work adds to the energy budget of that body. Energy gravitates. In the denominator, you're getting this r minus 2-- instead of just having an r squared, you get a correction that involves this 2gm term in here. And this is just correcting for the fact that you are not working in a Euclidean geometry. So now I want to do a little bit more with this. And so let me borrow a result from an old homework exercise. I believe this was on P set 3. Although I don't quite recall. So on this thing you guys took a look at the equation of local stress energy conservation for a perfect fluid and hydrostatic equilibrium. And what you found was that this turns into a condition relating the pressure, density, and what you can kind of think of as the four-acceleration of the fluid elements. So we see sort of gradients of the pressure entering here, the density entering into here, and some things related to the behavior of the fluid elements. The spacetime that we are working with is one where we have constrained the behavior of that fluid. So we are going to require that our fluid-- let's write it this way. So our fluid only has a-- it's only the time-like component of its four-velocity is 0. And so when I shoot that all through the various things involved in there-- oh, and, of course, phi only depends on r. So what this equation then boils down to is rho plus p times this Christoffel symbol equals minus dp dr. When you expand this out, this turns into d phi dr. So what this tells me is that d phi dr is simply related to the pressure gradient. When you put all of this together, what you get is the following equation. This is sometimes called the equation of relativistic hydrostatic equilibrium. My pressure gradient must satisfy this equation. Let me write down two more, because I think I'm going to actually begin with this board in my next lecture. So I'm going to just repeat this equation governing phi. And m is defined as this integral. So here are three coupled differential equations, which I can now use to build a model for the interior of this body. We integrate this thing out. Basically, the way it works is you need to choose an initial-- some parameter that characterizes the initial conditions. You typically choose a density at the center. You choose rho at r equals 0. You need an equation of state that relates the pressure and the density. And then you just integrate these guys up. There's a matching that must be done at the surface of the star. Then once you've done that, that'll end up giving you sort of a notice that by integrating up this equation, you determined the function phi up to some constant of integration. When you match at the surface of the star, that lets you fix that constant of integration. Boom! You've made yourself a spacetime. This is known as the Tolman-Oppenheimer-Volkoff equation, universally abbreviated TOV. These are the way in which we make spherically symmetric fluid bodies in general relativity, spherically symmetric static fluid bodies in general relativity. In the next lecture-- so we're going to end this one a tiny bit early. This is actually a very natural place to stop. I am going to start with this. And what we're going to do is talk a little bit about how one solves these. And we're going to look at a couple-- we're going to look at one particular special solution, which is not very physical, but it's illustrative. And I'm going to talk a little bit about how to solve this for more realistic setups. That is in preparation for what is one of my favorite homework exercises. It's one where I give you an equation of state. And you guys just take these equations. And using a numerical integrator-- I hope everyone has access to something like Mathematica, or WolframAlpha, or something like this in your dispersed lives. All MIT students should be able to access the student license for Mathematica. So as long as you have the hardware for that, you should be OK. And this is-- actually, it's pretty cool. You can make relativistic stellar models. I've actually had in the past students use this as the basis for some research projects, because when you do things like this, this is how professionals make models of stars. So we will stop there. And those of you who are watching at home, you can go on to the next video, where we'll be doing this. As for me, I'm going to take a bit of a break, and start recording again in about a half an hour. |
MIT_8962_General_Relativity_Spring_2020 | 19_Cosmology_II.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: So in the lectures I'm going to record today, we're going to conclude cosmology. And then we're going to begin talking about a another system in which we saw the Einstein field equations using asymmetry. So this is one where we will again consider systems that are spherically symmetric but we're going to consider them to be compact. In other words, things that are sort of, rather than filling the whole universe, the spacetime that arises from a source that's localized in some particular region in space. So let me just do a quick recap of what we did in the previous lecture that I recorded. So by arguing solely from the basis of how to make a spacetime that is as symmetric as possible in space but has a time asymmetry, so the past and the future look different, we came up with the Robertson-Walker metric, which I will call the RW metric, which has the form I've written there on the top one line. There were actually several forms for this. I've actually written down two variants of it on the board here. Key thing which I want to highlight is that there is, in the way I've written it on the first line, there's a hidden scale in there, r zero, which you can think of as, essentially, setting an overall length scale to everything that could be measured. There is a parameter, either k or kappa. k is either minus 1, 0, or 1. Kappa is just k with a factor of that length scale squared thrown in there. We define the overall scale factor to be-- it's a number. And we define it so that its value is 1 right now. And then everything is scaled to the way distances look at the present time. A second form of that-- excuse me-- which is essentially the change of variables is you change your radial coordinates to this parameter chi. And depending on the value of k, what you find then is that the relationship between the radius r and the coordinate chi, if k equals 1, you have what we call a closed universe. And the radius is equal to the sign of the parameter chi. If you have an open universe, k equals minus 1, then the radius is the sinh of that parameter chi. And for a flat universe, k equals 0, they are simply 1 is just the other modulo, a choice of units with the R of 0 there. All this is just geometry. OK. So when you write this down, you are agnostic about the form of a and you have no information about the value of k. So to get more insight into what's going on, you need to couple this to your source. And so we take these things. And using the Einstein field equations, you equate these to a perfect fluid stress energy tensor. What pops out of that are a pair of equations that arise in the Einstein field equations. I call these F1 and F2. These are the two Friedmann equations. F1 tells me about the velocity associated with that expansion premise. There's an a dot divided by a squared. We call that the Hubble parameter, h of a squared. And it's related to the density of the source of your universe as well as this kappa term. OK. And as we saw last time, we can get some information about kappa from this equation. The second Friedmann equation relates the acceleration of this expansion term, a double dot-- dot, by the way, d by dy. A double dot divided by a is simply related to a quantity that is the density plus 3 times the pressure of this perfect fluid that makes up our universe. We also find by requiring that local energy conservation be held, in other words that your stress energy tensor be divergence free, we have a constraint that relates the amount of energy-- the rate of change of energy in a fiducial volume-- to the negative pressure times the rate of change of that fiducial volume. And this, as I discussed in the last lecture, is essentially nothing more than the first law of thermodynamics. It's written up in fancy language appropriate to a cosmological spacetime. As we move forward, we find it useful to make a couple of definitions. So if you divide the Hubble parameter squared by Newton's gravitational constant, that's got the dimensions of density. And so we're going to define a critical density to be 3h squared over 8 pi g. And we're going to define density parameters, omega, as the actual physical density is normalized to that critical density. And when you do this, you find that the critical-- the first Friedmann equation can be written as omega-- oh, that's a typo-- omega plus omega curvature equals 1 where omega curvature-- pardon the typo here. Omega curvature is not actually related to identity but it sort of plays one in this equation. It is just a parameter that has the proper dimensions to do what is necessary to fit this equation. And it only depends on what the curvature parameter is. So remember that this kappa is essentially minus 1, 0, or 1 module of factor of my overall scale. So this is either a positive number, 0, or a negative number. All right. So let's carry things forward from here. Mute my computer so I'm not distracted by things coming in. We now have everything we need using this framework to build a universe. Let's write down the recipe to build a universe, or I should say to build a model of the universe. So first thing you do is pick your spatial curvature. So pick the parameter k to be minus 1, 0, or 1. Pick a mixture of species that contribute to the energy density budget of your model universe. So what you would say is that the total density of things in your universe is a sum over whatever mixture of stuff is in your universe. You will find it helpful to specify an equation of state for each species. Cosmologists typically choose equation of state that has the following form. OK. So you require your species-- if you follow this model that most cosmologists use, each species will have a pressure that is linear in the density. If you do choose that form, then when you enforce local conservation of energy, what you will then find is that for every one of your species, there is a simple relationship. There's a simple differential equation that governs how that species evolves as the scale factor changes. This can be immediately integrated up to find that the amount of, at some particular moment, the density of species i depends on how it looks. So 0 again denotes now. It is simply proportional to some power of the scale factor where the power that enters here can be simply calculated given that equation of state parameter. Once you have these things together, you're ready to roll. You've got your Friedmann equations. You've got all the constraints and information you need to dig into these equations. Sit down, make your models, have a party. Now what we really want to do-- OK. We are physicists. And our goal in doing these models is to come up with some kind of a description of the universe and compare it with our data so that we can see what is the nature of the universe that we actually live in. OK. So what is of interest to us is how does varying all these different terms change-- well, we're going to talk about various observables. But really, if you think about this model, how does it change the evolution of the scale factor? Everything is bound up in that. That is the key thing. OK. So if I can make a mathematical model that describes a universe with all these various different kinds of ingredients, yeah, I can sit down and make-- if I make a really complicated thing, I probably can't do this analytically so that's fine. I will make a differential equation integrator that solves all these coupled differential equations. And I will then make predictions for how AFT evolves depending upon what the different mixtures of species are, what the curvature term is equal to, all those together, and make my model. But my goal as a physicist is then to compare this to data. And so what I need to do is to come up with some kind of a way this will then be useful. We then need some kind of an observational surrogate for a of t. What I would like to be able to do is say, great, model A predicts the following evolution of the scale factor. Model B predicts this evolution of the scale factor. Can I look at the universe and deduce whether we are closer to model A or closer to Model B? And in order to do that, I need to know how do I measure a of t. And as we're going to see, this really boils down to two things. I need to be able to deduce if I look at some event in the universe, if I look at something, I want to know what scale factor to associate with that. I need to measure a. And I need to know what t to label that a with. So this kind of sounds like I'm reading from the journal of duh. But if I want to do this, what it basically boils down to is I need to know how to associate an a with things that I measure and how to associate the t with the things that I measure. That's what we're going to talk about today. What are the actual observational surrogates, the ways in which we can go out, point telescopes and other instruments at things in the sky, and deduce what a of t is for the kind of events, the kind of things that we are going to measure? Let's talk about how I can measure the scale factor first. OK. So let's ignore the fact there's a t on here. How can I measure a? We saw a hint of this in the previous lecture that I recorded. So recall-- pardon me. I've got a bit of extra junk here in my notes. Recall that in my previous lecture, we looked at the way different kinds of densities behaved. But I've got the results right here. For matter which had an equation state parameter of 0, what we found was that the density associated with that matter, it just fell as the scale factor to the inverse third power. That's essentially saying that the number of particles of stuff is constant. And so as the universe expands, it just goes with the volume of the universe. If it was radiation, we found it went as a scale factor to the inverse fourth power. And that's consistent with diluting the density as the volume gets larger provided we also decrease the energy per particle of radiation per photon, per graviton, per whateveron. If I require that the energy per quantum of radiation is redshifted with this thing, that explains the density flaw that we found for radiation. And so that sort of suggests that what we're going to find is that the scale factor is directly tied to a redshift measure. OK. I just realized what this page of notes was. My apologies. I'm getting myself organized here. Let's make that a little bit more rigorous now. OK. So that argument on the basis of how the density of radiation behaves. It's not a bad one as a first pass. It is quite indicative. But let's come at it from another point of view. And this allows me to introduce in a brief aside a topic that is quite useful here. So we talked about Killing vectors a couple of weeks ago. Let's now talk about a generalization of this known as Killing tensors. So recall that a Killing vector was defined as a particular vector in my space time manifold such that if I Lie transport the metric along that Killing vector, I get 0. This then leads to the statement that if I put together the symmetrized covariant gradient of the Killing vector, I get 0. Another way to write this is to use this notation. Whoops. OK. So these are equations that tell us about the way that Killing vectors behave. A Killing tensor is just a generalization of this idea to an object that has more than one index line, to a higher rank tensorial object. So we consider this to be a rank one Killing tensor. A rank n Killing tensor satisfies-- so let's say k is my Killing tensor. Imagine I have n indices here if I take the covariant gradient of that Killing tensor and I symmetrize over all n indices. That gives me 0. This defines a Killing tensor. Starting with this definition, it's not at all hard to show that if I define a parameter, k, which is what I get when I contract the Killing tensor, every one of its indices with the four-velocity of a geodesic. If my u satisfies the geodesic equation-- or this could be-- let's write this as a momentum. Which you say is the tangent to a world line. Could be either a velocity or a momentum. So if I define the scalar k by contracting my Killing tensor with n copies of tangent to the world line, and that thing satisfies the geodesic equation, then the following is true. You guys did this on a homework exercise for when we thought about a spacetime-- you did something similar to this, I should say, for a spacetime containing an electromagnetic field. We talked about how this works for the case of a Killing vector. Hopefully you can kind of see the way you would do this calculation at this point. Now the reason I'm doing this aside is that if you have a Friedmann-Robertson-Walker spacetime, search spacetimes actually have a very useful Killing tensor. So let's define k with two indices, mu nu. And this is just given by the scale factor. Multiplying the metric u mu, u mu, u nu. Where this u comes from the four-velocity of a co-moving fluid element. So this is the four-velocity that we use to construct the stress energy tensor that is the source of our Friedmann equations. So here's how we're going to use this. Let's look at what we get for this Killing vector. Excuse me, this Killing tensor when I consider it's a long a null geodesic. We're going to want to think about null geodesics a lot, because the way that we are going to probe our universe is with radiation. We're going to look at it with things like telescopes. These days people are starting to probe it with things like gravitational wave detectors. All things that involve radiation that moves on null geodesics. So let's examine the associated conserved quantity that is associated with a null geodesic. So let's say v-- let's make it a p, actually. So it's going to be a null geodesic, so we're going to imagine it's radiation that is following. It has a four-momentum, pu. And let's define k, case of ng, from my null geodesic. That is going to be k mu nu, p mu, p nu. Let's plug-in the definition of my Killing tensor. So this is a square root of t, g mu nu, p nu, p nu. This is zero. It's a null geodesic. Then I get u mu p nu, u nu p mu. Now remind you of something. Go back to a nice little Easter egg, an exercise you guys did a long time ago. If I look at the dot product of a four-momentum and a four-velocity, what I get is the energy associated with that four-momentum as measured by the observer whose four-velocity is u. So what we get here is two copies of the energy of that null geodesic measured by the observer who is co-moving. So what this null geodesic-- what this quantity associated with this null geodesic is two powers of the scale factor times the energy that would be measured by someone who is co-moving with the fluid that fills my universe. Energy of p mu, as measured by u mu. And remember, this is a constant. So as this radiation travels across the universe-- as this radiation travels across the universe, the product of the scale factor and the energy associated with that radiation as measured by co-moving observers is a constant. So this is telling us that the energy, as measured by a co-moving observer, let's say it is emitted at some time with a scale factor is a. When it propagates to us, we define our scale factor as 1, the energy will have fallen down by a factor of 1 over a. So this makes it a little bit more rigorous, this intuitive argument that we saw from considering how the density of radiation fell off. What we see is the energy is indeed redshifting with the scale factor. So if I use the fact that the energy that I observe-- if I'm measuring light, light has a frequency of omega-- what I see is the omega that I observe at my scale factor, which I define to be 1, normalized to that when it was emitted, it looks like the scale factor when it was admitted. Divided by a now, a observed. I call this 1. I can flip this over, another way of saying this is that, if I write it in terms of wavelengths of the radiation, the wavelength of the radiation and when it was emitted versus the wavelength that we observe it tells me about the scale factor when the radiation was emitted. Astronomers like to work with redshift. They like to work with wavelength when they study things like the spectra of distant astronomical objects. And they use it to define a notion of redshift. So we define the redshift z to be the wavelength that we observe, minus the wavelength at the radiation that when it is emitted divided by the wavelength when it was emitted. Put all of these definitions together, and what this tells me is that the scale factor at which the radiation was emitted is simply related to the redshift that we observe. So this at last gives us a direct and not terribly difficult to use observational proxy that directly encodes the scale factor of our universe. Suppose we measure the spectrum of radiation from some source, and we see the distinct fingerprint associated with emission from a particular set of atomic transitions. What we generally find is some well-known fingerprints of well-characterized transitions, but in general they are stretched by some factor that we call the redshift z. Actually, you usually stretch by-- when you go through this, you'll find that what you measure is actually stretched by 1 plus z. You measure that, you have measured the scale factor at which this radiation was measured-- was emitted. So this is beautiful. This is a way in which the universe hands us the tool by which we can directly characterize some of the geometry of the universe at which light has been emitted. This is actually one of the reasons why a lot of people who do observational cosmology also happen to be expert atomic spectroscopists. Because you want to know to very high precision what is the characteristics of the hydrogen Balmer lines. Some of the most important sources for doing these tend to be galaxies in which there's a lot of matter falling onto black holes, some of the topics we'll be talking about in an upcoming video. As that material falls in, it gets hot, it generates a lot of radiation, and you'll see things like transition lines associated with carbon and iron. But often all reddened by a factor of several. You sort of go, oh, look at that, carbon falling onto a black hole at redshift 4.8. That is happening at a time when the scale factor of the universe was 1 over 4.8-- or 1 over 5.8, forgot my factor of 1 plus there. So you measure the redshift, and you have measured the scale factor. But you don't know when that light was emitted. We need to connect the scale factor that we can measure so directly and so beautifully to the time at which it was emitted. We now have a way of determining a, but we need a as a function of t. And in truth, we do this kind of via a surrogate. Because we are using radiation as our tool for probing the scale factor, we really don't measure t directly. When we look at light and it's coming to us, it doesn't say I was emitted on March 27th of the year negative 6.8 billion BC, or something like that. We do know, though, that it traveled towards us at the speed of light on a null geodesic. And because it's a null geodesic, there's a very simple-- simple's a little bit of an overstatement, but there is at least a calculable connection. Because it's moving at the speed of light, we can simply-- I should stop using that word-- we can connect time to space. And so rather than directly determining the time at which the radiation was emitted, we want to calculate the distance of the source from us that emitted it. So rather than directly building up a of t, we're going to build up an a of d, where d is the distance of the source. And if you're used to working in Euclidean geometry, you sort of go, ah, OK, great. I know that light travels at the speed of light, so all I need to do is divide the distance by c, and I've got the time, and I build a of t. Conceptually, that is roughly right, and that gives at least a cartoon of the idea that's going on here. But we have to be a little bit careful. Because it turns out when you are making measurements in a curved space time, the notion of distance that you use depends on how you make the distance measurement. So this leads us now to our discussion of distance measures in cosmological spacetime. So just to give a little bit of intuition as to what's the kind of calculation we're going to need to do, let me describe one distance measure that is observation, which is about useless, but not a bad thing to at least begin to get a handle on, the way different parameters of the spacetime come in and influence what the distance measure is. So let's just think about the proper distance from us to a source. So let's imagine that-- well, let's just begin by first, let's write down my line element. And here's the form that I'm going to use. OK, so here's my line element. This is my differential connection between two events spaced between one another by dt, d chi, d theta, d phi, all hidden in that angular element, the omega. Let's imagine that we want to consider two sources that are separated purely in the radial direction. So my angular displacement between the two events is going to be 0. So the only thing I need to care about is d chi, and let's imagine that I determine the distance between these two at some instant-- So then you just get ds squared equals a squared or zero squared d chi squared, and you can integrate this up and you get our first distance measure, d sub p equals scale factor. Your overall distance scale are zero and chi. So Carroll's textbook calls this the instantaneous physical distance. Let's think about what this means if you do this. This is basically the distance you would get if you took a yardstick, you put one end at yourself, you're going to call yourself chi equals 0, you put the other end of the yardstick at the object in your universe at some distance chi in these coordinates, and that's the distance. d sub p is the distance that you measure. It is done, you're sort of imagining that both of the events at the end of this yardstick. You're sort of ascertaining their position at exactly the same instant, hence the term instantaneous, and you get something out of it that encodes some important aspects of how we think about distances in cosmology. So notice everything scales with the overall length scale that we associated with our spatial slices with our spatial sector of this metric, the r0. Notice that whatever is going on your scale factor, your a, your distance is going to track that. As a consequence of this, two objects that are sitting in what we call the Hubble flow, in other words, two objects that are co-moving with the fluid that makes up the source of our universe. They have an apparent motion with respect to each other. If I take the time derivative of this, the apparent is just a dot, or 0 chi, which is equal to the Hubble parameter times dp. Recall that the Hubble parameter is a dot over a, and if I'm doing this right now, that's the value of the Hubble parameter now. So this is the Hubble Expansion Law, the very famous Hubble Expansion Law. So we can see it hidden in this-- not even really hidden, it's quite apparent in this notion of an instantaneous physical distance. Let me just finally emphasize, though, that the instantanaeity that is part of this object's name, instantaneous measurements, are not done. As I said-- I mean, this is it sounds like I'm being slightly facetious, but it's not. The meaning of this distance measure is, like I said, it's a yardstick where I have an event at me, the other end of my yardstick is at my cosmological event. Those are typically separated by millions, billions of light years. Even if you could-- OK, the facetious bit was imagining it as a yardstick. But the non-facetious point I want to make is we do not make instantaneous measurements with that. When I measure an event that is billions of light years away, I am of course measuring it using light and I'm seeing light that was emitted billions of years ago. So we need to think a little bit more carefully about how to define distance in terms of quantities that really correspond to measurements we can make. And to get a little intuition, here are three ways where, if you were living in Euclidean space and you were looking at light from distant objects, here are three ways that you could define distance. So if spacetime were that of special relativity-- well, let's just say if space were purely Euclidean. Let's just leave it like that. Here are three notions that we could use. One, imagine there was some source of radiation in the universe that you understood so well that you knew its intrinsic luminosity. What you could do is compare the intrinsic luminosity of a source to its apparent brightness. So let's let f be the flux we measure from the source. This will be related to l, the luminosity, which-- suspend disbelief for a moment, we want to imagine that we know it for some reason. And if we imagine this is an isotropic emitter, this will be related by a factor of 4 pi, and the distance between us and that source. Let's call this d sub l. This is a luminosity distance. It is a distance that we measure by inferring the behavior of luminosity of a distant object. Now it turns out, and this is a subject for a different class, but nature actually gives us some objects whose luminosity is known or at least can be calibrated. In the case of much of what is done in cosmology today, we can take advantage of the behavior of certain stars whose luminosity is strongly correlated to the way that-- these are stars whose luminosity is variable, and we can use the fact that their variability is correlated to their luminosity to infer what their absolute luminosity actually is. There are other supernova events whose luminosity likewise appears to follow a universal law. It's related to the fact that the properties of those explosions are actually set by the microphysics of the stars that set them. More recently, we've been able to exploit the fact that gravitational wave sources have an intrinsic luminosity in gravitational waves, the dedt associated with the gravitational waves that they emit, which depends on the source gravitational physics in a very simple and predictable way that doesn't depend on very many parameters. I actually did a little bit of work on that over the course of my career, and it's a very exciting development that we can now use these as a way of setting the intrinsic luminosity of certain sources. At any rate, if you can take advantage of these objects that have a known luminosity and you can then measure the flux of radiation in your detector from these things, you have learned the distance. At least you have learned this particular measure of the distance. That's measure one. Measure two is imagine you have some object in the sky that has a particular intrinsic size associated with it. You can sort of think of the objects whose luminosity you know about as standard candles. Imagine if nature builds standard yardsticks, there's some object whose size you always know. Well, let's compare that physical size to the angular size that you measure. The angle that the object sub tends in the sky is going to be that intrinsic size, delta l divided by the distance. We'll call this distance d sub a, for the angular diameter distance. Believe it or not, nature, in fact, provides standard yardsticks type so that we can actually do this. Finally, at least as a matter of principle, imagine you had some object that's moving across the sky with a speed that you know. You could compare that transverse speed to an apparent angular speed. So the theta dot, the angular speed that you measure, that would be the velocity perpendicular to your line of sight divided by distance. We'll call this d sub m, the proper motion distance. So if our universe were Euclidean, not only would it be easy for us to use these three measures of distance, all three of them would give the same result. Because this is all just geometry. Turns out when you study these notions of distance, in an FRW spacetime, there's some variation that enters. Let me just emphasize here that there is an excellent summary on this stuff. I can't remember if I linked this to the course website or not. I should and I shall. An excellent summary of all this, really emphasizing observationally significant aspects of these things come from the article that is on the archive called Distance Measures in Cosmology by David Hogg, a colleague at New York University. You can find this on the Astro PH archive, 9905116. It's hard for me to believe this is almost 21 years old now. This is a gem of a paper. Hogg never submitted it to any journal, just posted it on the archive so that the community could take advantage of it. So the textbook by Carroll goes through the calculation of d sub l. On a problem set you will do d sub m. We're going to go through d sub a, just so you can see some of the way that this works. Let me emphasize one thing, all of these measures use the first Friedmann equation. So writing your Friedmann equation like so. i is a sum of all the different species of things that can contribute including curvature. So recall that even though curvature isn't really a density, you can combine enough factors to make it act as though it were a density. You assume a power law. So this n sub i is related to the equation of state parameter for each one of these species. And let's now take advantage of the fact that we know the scale factor directly ties to redshift. I can rewrite this as how the density evolves as a function of redshift. So when you put all of this together, this allows us to write h of a as h of z. This is given by the Hubble parameter now times some function e of z. And that is simply-- you divide everything out to normalize to the critical density, and that is a sum over all these densities with the redshift waiting like so. Let's really do an example, like I said. So if you read Carroll, you will see the calculation of the luminosity distance. If you do the cosmology problem set that will be-- I believe its p set 8, you will explore the proper motion distance. So let's do the angular diameter distance. Seeing someone work through this, I think, is probably useful for helping to solidify the way in which these distance measures work and how it is that one can tie together important observables. So let's consider some source that we observe that, according to us, subtends an angle, delta phi. Every spacial sector is spherically symmetric, and so we can orient our coordinate system so that this thing, it would be sort of a standard ruler-- what we're going to do is orient our coordinate system so that object lies in the theta equals pi over 2 plane. The proper size of that source-- so the thing is just sitting in the sky there-- the proper size of the source, you can get this in the line element. The delta l of the source will be the scale factor at the time at which it is emitted, r0. So this is using one of the forms of the FRW line elements I wrote down at the beginning of this lecture. And so the angular diameter distance, that's a quantity that I've defined over here, it's just the ratio of this length to that angle. Let's rewrite this using the redshift. Redshift is something that I actually directly observe, so there we go. This is not wrong, but it's flawed. So this is true. This is absolutely true. Here's the problem, I don't know the overall scale of my universe and this coordinate chi doesn't really have an observable meaning to it. It's how I label events, but I look at some quasar in the sky and I'm like, what's your chi? So what we need to do is reformulate the numerator of this expression in such a way as to get rid of that chi and then see what happens, see if we have a way to get rid of that r0. Let's worry about chi first. We're going to eliminate it by taking advantage of the fact that the radiation I am measuring comes to me on a null path. Not just any null path. I'm going imagine it's a radial one. We are allowed to be somewhat self-centered in defining FRW cosmology, we put ourselves at the origin. So any light that reaches us moves on a purely radial trajectory from its source to us. So looking at how the time and the radial coordinate chi are related for a radial null path we go into our FRW metric. I get this. So I can integrate this up to figure out what chi is. So this is right. Let's massage it a little bit to put it in a form that's a little bit more useful to us. Let's change our variable-- change our variable of integration from time to a. So this will be an integral from the scale factor at which the radiation is emitted to the scale factor which we observe it, i.e. now. And when you do that change of variables your integral changes like so. Let's rewrite this once more to insert my Hubble parameter. Now let's change variables once more. We're going to use the fact that our direct measurable is redshift. And so if we use a equals 1 over 1 plus z, I can further write this as an integral over redshift like so. And that h0 can come out of my integral. So this is in a form that is now finally formulated in terms of an observable redshift and my model dependent parameters. The various omegas that, when I construct my universe model, I am free to set. Or if I am a phenomenologist, that are going to be knobs that I turn to try to design a model universe that matches the data that I am measuring. r0, though, is still kind of annoying. We don't know what this guy is, so what we do is eliminate r0 in favor of a curvature density parameter. So using the fact that omega curvature-- go back to how this was originally defined-- it was negative kappa over h0 squared. That's negative k over r0 squared h0 squared. That tells me that r0 is the Hubble constant now, divided by the square root of the absolute value of the curvature, at least if k equals plus or minus 1. What happens when it's not plus or minus 1, if it's equal to 0? Well, hold that thought. So let's put all these pieces together. So assembling all the ingredients I have here, what we find is the angular diameter distance. There's a factor of 1 over 1 plus z, 1 over the Hubble constant. Remember, Hubble has units of 1 over length-- excuse me, 1 over time, and with the factor of the speed of light that is a 1 over length, so one over the Hubble parameter now is essentially a kind of fiducial overall distance scale. And then our solution breaks up into three branches, depending upon whether k equals minus 1, 0, or 1. So you get one term where it involves minus square root the absolute value of the curvature parameter times sine. That same absolute value of the curvature parameters square root. So here's your k equals plus 1 branch. For your k equals 0 branch, basically what you'll find when you plug-in your s of k is that r0 cancels out. So that ends up being a parameter you do not need to worry about, and I suggest you just work through the algebra and you'll find that for k equals 0, it simply looks like this. And then finally, if you are in an open universe-- that is supposed to be curvature-- what we get is this. So this is a distance measure that tells me how angular diameter distance depends on observable parameters. Hubble is something that we can measure. Redshift is something we can measure. And it depends on model parameters, the different densities that go into e of z, and-- which I have on the board right here-- the different densities that go into e of z. My apologies, I left out that h0 there-- and your choice of the curvature. When you analyze these three distances here is what you find. You find that the luminosity distance is related to the proper motion distance by a factor of 1 plus z, and that's related to the angular diameter distance by a factor of 1 plus z squared. So when you read Carroll, you will find that 1 plus z factor there-- excuse me, 1 plus z to the minus 1 power turns into a 1 plus z-- is the camera not looking at me? Hello? There we go. So that 1 over 1 plus z turns into a 1 plus z. When you do it on the p set, you do the proper motion distance, so it will just be no 1 plus z factor in front of everything. So the name of the game when one is doing cosmology as a physicist is to find quantities that you can measure that allow you to determine luminosity distances, angular diameter distances, proper motion distances. Now it turns out that the proper motion distance is not a very practical one for basically any cosmologically interesting source. They are simply so far away that even for a source moving essentially at the speed of light, the amount of angular motion that can be seen over essentially a human lifetime is negligible. So this turns into something that-- hi. MIT POLICE: [INAUDIBLE] SCOTT HUGHES: That's OK. Yeah, I'm doing some pre-recording of lectures. [LAUGHS] I was warned you guys might come by. I have my ID with me and things like that, so. MIT POLICE: That's fine. Take care. MIT POLICE: You look official. SCOTT HUGHES: [LAUGHS] I appreciate it. So those of you watching the video at home, as you can see, the MIT police is keeping us safe. Scared the crap out of me for a second there, but it's all good. All right, so let's go back to this for a second. So the proper motion distance is something that is not particularly practical, because as I said, even if you have an object that is moving close to the speed of light this is not something that even over the course of a human lifetime you are likely to see significant angular motion. So this is generally not used. But luminosity distances and angular diameter distances, that is, in fact, extremely important, and a lot of cosmology is based on looking for well understood objects where we can calibrate the physical size and infer the angular diameter distance, or we know the intrinsic brightness and we can determine the luminosity distance. So let me just give a quick snapshot of where the measurements come from in modern cosmology that are driving our cosmological model. One of the most important is the cosmic microwave background. So, vastly oversimplifying, when we look at the cosmic microwave background after removing things like the flow of our solar system with respect to the rest frame of-- excuse me, the co-moving reference frame of the cause of the fluid that makes up our universe, the size of hot and cold spots is a standard ruler. By looking at the distribution of sizes that we see from these things, we can determine the angular diameter distance to the cosmic microwave background with very high precision. This ends up being one of the most important constraints on determining what the curvature parameter actually is. And it is largely thanks to the cosmic microwave background that current prejudice, I would say, the current best wisdom-- choose your descriptor as you wish-- is that k equals 0 and our universe is in fact spatially flat. Second one is what are called type 1a supernova. These are essentially the thermonuclear detonations of white dwarfs. Not even really thermonuclear, it's just-- my apologies, I'm confusing a different event that involves white dwarfs. The type 1a's are not the thermonuclear explosions of these things. This is actually the core collapse of a white dwarf star. So this is what happens when a white dwarf it creates or in some way accumulates enough mass such that electron degeneracy pressure is no longer sufficient to hold it against gravitational collapse, and the whole thing basically collapses into a neutron star. That happens at a defined mass, the Chandrasekhar mass, named after one of my scientific heroes, Subramanian Chandrasekhar. And because it has a defined mass associated with it, every event basically has the same amount of matter participating. This is a standard candle. So there's a couple others that I'm not going to talk about in too much detail here. While I'm erasing the board I will just mention them. So by looking at things like the clustering of galaxies we can measure the distribution of mass in the universe that allows us to determine the omega m parameter. That's one of the bits of information that tells us that much of the universe is made of matter that apparently does not participate in standard model processes as we know them today-- the so-called dark matter problem. We can look at chemical abundances, which tells us about the behavior of nuclear processes in the very early universe. And the last one which I will mention here is we can look at nearby standard candles. And nearby standard candles allow us to probe the local Hubble law and determine h0. "Probble," that's not a word. If you combine "probe" and "Hubble" you get "probble." And when I say nearby, that usually means events that are merely a few tens of millions of light years away. It's worth noting that all of these various techniques, all of these different things, you can kind of even see it when you think about the mathematical form of everything that went into our distance measures, they're all highly entangled with each other. And so to do this kind of thing properly, you need to take just a crap load of data, combine all of your data sets, and do a joint analysis of everything, looking at the way varying the parameters and all the different models affects the outcome of your observables. You also have to carefully take into account the fact that when you measure something, you measure it with errors. And so many of these things are not known. Turns out you can usually measure redshift quite precisely, but these distances always come with some error bar associated with them. And so that means that the distance you associate with a particular redshift, which is equivalent to associating a time with a redshift, there's some error bar on that. And that can lead to significant skew in what you determine from things. There's a lot more we could say, but time is finite and we need to change topic, so I'm going to conclude this lecture by talking about two mysteries in the cosmological model that have been the focus of a lot of research attention over the past several decades. Two mysteries. One, why is it that our universe appears to be flat, spatially flat? So to frame why this is a bit of a mystery, you did just sort of go, eh, come on. You've got three choices for the parameter, you got 0. Let's begin by thinking about the first Friedmann equation. I can write this like so, or I can use this form, where I say omega plus omega curvature-- I'm going to call that omega c for now-- that equals 1. The expectation had long been that our universe would basically be dominated by various species of matter and radiation for much of its history, especially in the early universe. If it was radiation dominated, you'd expect the density to go as a to the minus 4. If it's matter dominated, you expect it to go as a to the minus 3. Now, your curvature density goes as a to the minus 2. And so what this means is that if you look at the ratio of omega curvature to omega, this will be proportional to a, for matter, a squared for radiation. If your universe-- in some sense, looking at these parameter k of the minus 1, 0, 1, that's a little bit misleading. It's probably a little bit more useful to think about things in terms of the kappa parameter. And when you look at that, your flat universe is a set of measure 0 in the set of all possible curvature parameters that you could have. And physicists tend to get suspicious when something that could take on any range of possible random values between minus infinity and infinity picks out zero. That tends to tell us that there may be some principle at play that actually derives things to being 0. Looking at it this way, imagine you have a universe that at early times is very close to being flat, but not quite. Any slight deviation from flatness grows as the universe expands. That's mystery one. Mystery two, why is the cosmic microwave background so homogeneous? So when we look at the cosmic microwave background, we see that it has the same properties. The light has the same brightness, it has the same temperature associated with it, to within in a part in 100,000. Now the standard model of our universe tells us that at very early times the universe was essentially a dense hot plasma. This thing cooled as the universe expanded, much the same way that if you have a bag of gas, you squeeze it very rapidly, it will get hot, you stretch it very rapidly, it will cool. There's a few more details of this in my notes, but when we look at this one the things that we see is that in a universe that is only driven by matter or by radiation-- so the matter dominated and radiation dominated picture-- it shows us that points on opposite sides of the sky were actually out of causal contact with each other in the earliest moments of the universe. In other words, I look at the sky, and the patch of sky over here was out of causal contact with the patch of sky over here in the earliest days of the universe. And yet they had the same temperature, which suggests that they were in thermal equilibrium. How can two disparate, unconnected patch of the sky have the same temperature if they cannot exchange information? You could imagine it being a coincidence if one little bit of the sky has the same temperature as a bit of another piece, but in fact, when you do this calculation, you find huge patch of the sky could not communicate with any other. And so how then is it that the entire sky that we can observe has the same temperature at the earliest times within a part in 100,000. You guys will explore this on-- I believe it's problem set eight. The solution to both of these problems that has been proposed is cosmic inflation. So what you do is imagine that at some earlier moment in the universe, our universe was filled with some strange field, and I'll describe the properties of that field in just a moment, such that it acted like it had a cosmological constant. In such an epic, the scale factor of the universe goes as exponentially with the square root of the size of the cosmological constant. What you find when you look at this is that this-- it still, of course, goes as a to the minus 2-- but a to the minus-- well, sorry. Let me back up for a second. So my scale factor in this case you'll find goes are a to the minus 2, and that's because the density associated with the cosmological constant remains constant. So even if you start in the early universe with some random value for the curvature, if you are in this epic of exponential inflation, just for-- you know, you have to worry about timescales a little bit, but if you do it long enough you can drive this very, very close to 0. So much so that when you then move forward, let's say you come out of this period of cosmic inflation and you enter a universe that is radiation dominated or matter dominated, it will then begin to grow, but if you drive it sufficiently close to 0 it doesn't matter. You're never going to catch up with what inflation did to you. On the P set you will also show that if you have a period of inflation like this, then that also cures the problem of piece of the sky being out of causal contact. So when you do that, what you find is that essentially everything is in causal contact early on. It may sort of come out of causal contact after inflation has ended, more on that in just a moment, and then things sort of change as the universe continues to evolve. OK so it looks like recording is back on. My apologies, everyone. So as I was in the middle of talk, I talked a little bit too long in this particular lecture so we're going to spill over into a little bit of an addendum, just a five-ish minute piece that goes a bit beyond this. Doing this by myself is a little bit weird, I'm tired, and I will confess, I got a little bit rattled when the police came in to check in on me. Let's back up for a second. So I was talking about two mysteries of the modern cosmological model. One of them is this question of why the universe is so apparently flat. The spatial sector of the universe appears to be flat. And we had this expectation that the universe is either radiation dominated or matter dominated, which would give us the density associated with radiation. If it was radiation dominated, then the density of stuff in our universe would fall off as the scale factor to the fourth power. If it's matter dominant, it's scale factor to the third power. When you define a density associated with the curvature, it falls off as scale factor to the second power. And so if we look at the ratio of the curvature density to any other kind of density, it grows as the universe expands. So any slight deviation from flatness we would expect to grow. And that's just confusing. Why is it when we make the various measurements that we have been making for the past several decades, all the evidence is pointing to a universe that has a flatness of 0? If you sort of imagine that the parameter kappa can be any number between minus infinity to infinity, why is nature picking out 0? Another mystery is why is the cosmic microwave background so homogeneous? We believe that the universe was a very hot dense plasma at very early times. It cooled as the universe expanded, and when we measured the radiation from that cooling expanding ball of plasma, what we find is that it has the same temperature at every point in the sky to within a part in 100,000. But when one looks at the behavior of how light moves around in the early universe, if the universe is matter dominated or radiation dominated, what you find is that a piece of the sky over here cannot communicate with a piece of sky over here. Or over here, or over here. You actually find that the size of the sky that, if I look at a piece of sky over here, how much of the universe could it talk to, it's surprisingly small. So how is it that the entire universe has the same temperature? How is that they are apparently in thermal equilibrium, even if they cannot exchange information? So I spent some while talking to myself after the cameras went out. So I'll just sketch what I wrote down. A proposed solution to this, to both of these mysteries, is what is known as cosmic inflation, something that our own Alan Guth shares a lot of the credit for helping to develop. So recall, if we have a cosmological constant, then the scale factor grows exponentially and the density of stuff associated with that cosmological constant is constant. As the universe expands, the energy density associated with that cosmological constant does not change. If our universe is dominated by such a constant, then the ratio of density is associated with curvature, the density associated with cosmological constant, actually falls off inversely with the curvature scale squared, and the curvature scale is growing exponentially, it means that omega c is being driven to zero relative to the density in cosmological constant, as e to a factor like this. It's being exponentially driven close to 0. You'd do do a little bit of work to figure out what the timescales associated with this are, but this suggests that if you can put the universe in a state where it looks like a cosmological constant, you can drive the density associated with curvature as close to 0 as you want. Recall that a cosmological constant is actually equivalent to there being a vacuum energy. If we think about the universe being filled with some kind of a scalar field at early times, it can play the role of such a vacuum energy. Without going into the details, one finds that in an expanding universe there is an equation of motion for that scalar field. How the field itself behaves is driven by a differential equation, looks like this. Here's your Hubble parameters, so this has to do with a scale factor in here. v is a potential for this scalar field, which I'm not going to say too much about. Take this guy, couple it to your Friedmann equations, and the one that's most important is the first Friedmann equation. What you see is that v of phi is playing the role of a cosmological constant. It's playing the role of a density. So if we can put the universe into a state where it is in fact being dominated by this scalar field, by the potential associated with the scalar field, it will inflate. A lot of the research in early universe physics that has gone on over the past couple decades has gone into understanding what are the consequences of such a such a potential. Can we make such a potential? Do the laws of physics permit something like this to exist? If the universe is in this state early on, what changed? How is it that this thing evolves? Is there is an equation of motion here so that scalar field is presumably evolving in some kind of a way? Is there a potential, does nature permit us to have a field of the sort with a potential that sort of goes away after some time? Once it goes away, what happens to that field? Is there any smoking gun associated with this? If we look at the universe and this is a plausible explanation for why the universe appears to be spatially flat and why the cosmic microwave background is so homogeneous. Is there anything else that we can look at that would basically say yes, the universe did in fact have this kind of an expansion? Without getting into the weeds too much, it turns out that if the universe expanded like this, we would expect there to be a primordial background of gravitational waves, very low frequency gravitational waves. Sort of a moaning background filling the universe. And so there's a lot of experiments looking for the imprints of such gravitational waves on our universe. If it is measured, it would allow us to directly probe what this inflationary potential actually is. I'm going to conclude our discussion of cosmology here. So some of the quantitative details, the way in which inflation can cure the flatness problem and cure the homogeneity problem you will explore on problem set seven. Going into the weeds of how one makes it a scalar field, designs a potential and does things like this beyond the scope of 8.962-- there are courses like this, and I would not be surprised if some of the people in this class spent a lot more time studying this in their futures than I have in my life. All right, so that is where we will conclude our discussion of cosmology. |
MIT_8962_General_Relativity_Spring_2020 | 2_Introduction_to_tensors.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: Good afternoon. So we spent our last lecture laying out some of the basic foundations, making a couple of definitions. I want to quickly recap the most important concepts and definitions. And then, let me be blunt, I kind of want to get through these definitions, which I think it's important to do them precisely, but there's nothing significantly challenging about them. We just need to make sure they are defined very precisely. So now that you've kind of seen the style of these things, I would like to sort of move through the next batch of these definitions quickly enough that we can start to move into more interesting material. So a quick recap-- and my apologies, my daughter has some kind of a virus that I am desperately trying to make sure I do not catch. And so I will be hydrating during this lecture. So I just want to recap some of the most important concepts we went over. So this whole class is essentially a study in spacetime. Later, we're going to connect spacetime to gravity. And general relativity is going to become the relativistic theory of gravity. So we began with a fairly mathematical definition. Spacetime is a manifold of events that is endowed with a metric. Manifold, for our purposes, is essentially just a set in which we understand how different members of the set are connected to each other. Events are really just when and where something happens. We haven't precisely defined metric yet. We will soon. But intuitively, just regard it as some kind of a mathematical object that gives me a notion of distance between these events. I tried and I will continue to try to be very careful to make a distinction between geometric objects that live in the manifold, the events themselves with things like a displacement vector between two events, other vectors, which I will recap the definition of in just a moment. They have an existence and sort of a reality to them that is deeper and more fundamental than the representation of that object. So when I say that the displacement vector is delta t, delta x, delta y, delta z, that is according to observer O. And when I write that down-- I will be very careful as much as possible. I will occasionally screw this up-- but I will try to write this down without using an equal sign. Equal sign implies a degree of reality that I do not want to impart to that representation. So an equal dot with a little o here, that's my personal notation for delta x is represented by these components according to observer O. And for shorthand, I was sometimes just write this by the collection of indices delta x alpha. A different observer O bar, they will represent this this factor using the exact same geometric object. They all agree that it's this displacement between two physical events in spacetime. But they assign different coordinates to it. They give a different representation to it. And we find that representation using a Lorenz transformation. And I'm not going to write out explicitly the Lorenz transformation matrix lambda. I gave it in the last lecture. And I'm assuming you're all experts in special relativity, and I don't need to go over that. So I will then using this as sort of the prototype, the general notion of a vector in spacetime, which we'll often call a 4-vector for the obvious reason that it has 4 components, I'm going to treat that as any quartet of numbers that has transformation properties just like the displacement factor I just went over. So if there's some quantity A that has time like an x or y and z component, as long as a different observer, for reasons having to do with the underlying physics or whatever the heck this A is, as long as that different observer relates their components to observer O's components via the Lorenz transformation, you got yourself a 4-vector. Any random set of four numbers, that ain't a 4-vector. You need to have some physics associated with it. And the physics has to tell it that it's a thing that's related by a Lorenz transformation. So I want to pick up this discussion by introducing four particularly important and special vectors, which to be honest, we're not going to use too much beyond some of the first couple of weeks or so of the course, but they're very useful. And one should often bear in mind, even later in the course when they've sort of disappeared, that they're coming along for the ride secretly. And these are basis vectors. So if I go into frame O-- I'm just going to pick some particular reference frame-- I can immediately write down four special vectors. So remember, this is a Cartesian type of coordinate system. So I'm going to introduce e0. I'm going to represent this by just the number 1 in the time slot and 0 everywhere else. e1, or ex if you prefer, I'm going to write that down like so. And you can kind of see where I'm going with this. The analogy, if you've all seen unit vectors in other classes, hopefully this is fairly obvious what I'm doing. I'm just picking out a set of, you know, little dimensionless simple quantities that point along the preferred directions that I've set up in this inertial reference frame. A compact way of writing this-- so notice, I have four of these vectors. And these vectors each have four components. And so what I can say is that the beta component of unit vector e alpha is representative, according to observer O, by delta alpha beta, the Kronecker delta. If you're not familiar with this one, I'm usually reluctant to send you to Wikipedia. But in this case, I'm going to send you to Wikipedia, Wiki via Kronecker delta. Yeah, so what that does is it just emphasizes that at least in-- by the way, I should put little o's underneath all of these things, because I have chosen this according to observer o's representation. So this is just saying that a coin to observer O, these are four very special vectors. The utility of these things-- up high for everyone in back to be able to see-- the utility of doing this is that if I now want to write the vector A as a geometric object, I can combine the components that observer O uses with the basis vectors that observer O uses. And I can sum them all together. I can put them together. And then I've got the-- it's not a representation. That's a damn vector. OK? I put it all together using sort of an internally consistent set of numbers and basis vectors. And so I am free to say this, where I actually use an actual equal sign. You might stop and think, well, but those aren't the components of the observer O bar would use. And you're right. Observer O bar would not use those components. They would also not use those basis factors. We're going to talk about how those things change. But observer O bar does agree that if they were handed O's components and O's basis vectors, this would give me a complete representation of what the vector is. OK? So again, I'm really harping on this sort of distinction between the geometric object and the representation. This is the geometric object. And we can take advantage of this. The fact that that combination of things is the geometric object is a tool that I'm going to now-- is a fact that I'm going to exploit in order to figure out how my basis vectors transform when I change reference frames. So let me just repeat what I wrote of there. But-- oops-- good point for just a slight editorial comment. When you're talking and writing and there are millions of little sub scripts and indices, sometimes the brain and the appendages get out of sync with one another. I caught that one. I don't always. OK? So if you see something like that and you kind of go, um, why did that alpha magically turn into a beta, it's probably a mistake. Please call it out. OK? All right, so this is how I build this geometric object, using the components and the basis vectors as measured by O, as used by observer O. Let's now write out what they would be if they were measured by a different observer, an O bar observer. I know how to get components, the barred components from the unbarred components. I don't yet know how to get the barred basis vectors from the unbarred basis vectors. But I know that they exist. And that once I know what they are, this equation is true. OK? These are just two different ways of writing this geometric object, which every observer agrees has its existence that transcends the representation. So let's rewrite what I've got on the right-hand side of the rightmost equal sign here. I'm going to write this as lambda mu bar beta A beta e mu bar. And then now I'm going to use a trick, which is what the-- it's not even so much a trick, but I'm just going to use a fact that is great when you're working in index notation. So often when students first encounter this kind of notation, your instinct is to try to write everything out using matrices and things like row vectors and column vectors. It's a natural thing to do. I urge you to get over that. If I get some bandwidth for that, I'm going to write up a set of notes this weekend showing how one can translate at least 2 by 2 objects-- two index objects and one index objects in a consistent way between matrices and row vectors and column vectors. But we're rapidly going to start getting into objects that are bigger than that, for which trying to represent them in matrix-like form gets untenable. In a little while we're going to have a three index object. And since we don't have three-dimensional chalkboards, making this sort of matrix representation of that is challenging. Soon after that, we'll have four index objects. And we'll occasionally need to take derivatives of that four index object, giving us a five index object. At that point, the ability to sort of treat them like matrices is hopeless. So really, you should just be thinking of this as ordinary multiplication of the numbers that are represented by these components as written out here. And an ordinary multiplication like this, I can just go ahead and swap the order of multiplication very easily here. So what I'm going to do is move the-- hang on a second. Yeah, yeah, yeah, now I see what I'm doing. Sorry about that. I misread my notes. I'm just going to move the A onto the other side of my lambda. And then I'm going to use the fact that beta on my right-hand side is a dummy index. So in that final expression I wrote down over there, beta is a dummy index. And I'm free to adjust it to put it into a form that's more convenient for me. So let's begin. Let me just write where I've got this equation right now over there. So A alpha E alpha equals-- and what I've got then is component A beta lambda mu bar beta-- keep mu bar. Sorry about that. So I'm going to remap my dummy index. The reason I did that is I can now move this to the other side and factor out the A alpha. OK, everyone can see that I hope. So see what I did was I arranged this so that I've isolated essentially only the transformation of the basis vector. So this equation has to hold no matter what vector I am working with. It's got to hold for an arbitrary A alpha. So the only way that can happen, this means that my transformation of basis vectors obeys a law that looks like this. Now, on first inspection, you're going to go, ah, that's exactly what we got for the components of the 4-vector. Caution-- I'm going to remind you, it's actually on the board over there. OK? The barred component is playing a different role than the barred basis vector here. If you want to get the barred basis vector from the unbarred basis vector, you need to work with the inverse of this matrix. That's what this is telling us here. OK? All that being said, if you're just working through this and you've got your components set up and you're sort of hacking through it, the algorithm for you to follow is actually simple. Really, all we're doing is lining up the indices. We're summing over the ones that are repeated and requiring that those that are on both the left-hand side and the right-hand side appear and equal one another. Or as an old professor of mine liked to say about 12 times a lecture, line up the indices. So that's essentially what we're doing. In this case, I line up-- if I have my Lorenz transformation matrix with the barred index up top and the unbarred down below. Boom, I line up the indices. And that tells me what the unbarred basis components are from the barred ones, and vice versa for the components. So what basically, as I just said, that tells me if I actually want to get this guy given this guy, I need to work with the inverse. Put this up high so that everyone in the back can see it. So this, again, is one of those places where you might be tempted to sort of write out a matrix and do a matrix inversion. But before you do that, remember this is physics. OK? The inverse is going to be-- the inverse matrix is going to be the one that does, at least for certain quantities-- what the Lorenz transformation does is that it relates objects, if I'm at rest, if I consider myself at rest, it tells me about things according to a frame that it's moving with respect to me. The inverse matrix does the opposite. It would say to that person being at rest, what is the matrix that tells them about things according to me, and they see me moving with the exact same velocity, but in the opposite direction. So to get the inverse matrix, to get the inverse Lorenz transformation, what we end up doing is we just reverse the velocity. So I've just wrote down for you-- and it's worth bearing in mind, every one of these lambdas that is there, they are really functions. And so my e alpha that I wrote down over there, I'm going to use an under tilde to denote a 3 vector. In one of your textbooks, it's written with a boldface. But that's hard to do on a blackboard. So if I want to go the other direction, well, I just need to have the inverse transformation. And I have that. Bear in mind, I mean, those are really the exact same matrices. OK? In terms of the function that I'm working with here, I'm just flipping around the direction in order to get these things out. OK? So, as I said, you might be tempted just to go ahead and do the matrix inversion. Let we just do a quick calculation to show you that that would work. And the reason I'm doing this is I just want to quickly step through a particular step, which is, again, sort of in the spirit of swatting mosquitoes with sledgehammers, it's the kind of analysis that you're going to sort of have to do off to the side now and again. So given that I know e alpha is-- let's see, let's use beta-- e beta bar be e beta bar. But I now know that I could write this guy as lambda-- let's use a gamma-- beta bar minus v in the gamma direction. Now, you look at that, and you go, ooh, look, I'm actually summing over the betas there. Let's gather my terms a little bit differently. So notice, I have unbarred basis vectors over here on the left, unbarred ones over here on the right, a bunch of junk in these braces here in the middle. The only way for that to work is if after summing over beta, that bunch of junk is, in matrix language, we'd say, it's the identity matrix. In component notation, we're going to call it the Kronecker delta. Just as little aside, if I started with the barred on the left side and the unbarred in the right-hand side and did a similar analysis, it would take you at this point now that you are all fully expert in this kind of index manipulation, it should take you no more than about a minute to demonstrate to yourself that you can get a Kronecker delta on the barred indices in a similar way, just by using them in a slightly different order. OK? All right, so far all still basically just formulas. So I'm going to start now doing a little bit of formalism that will very quickly segue into physics. We can all take a deep sigh of relief and go, ah, OK, something you can imagine measuring. So that's nice. So I have introduced 4-vectors. I haven't introduced the basis vectors and their components. I haven't really done anything with them yet. So before we start doing things with them, let's think about some operations that we can do with 4-vectors. So the first one which I'd like to introduce is a scalar product. To motivate the scalar product that I'm going to define in the same way that I defined 4-vectors as a quartet of numbers whose transformation properties are based on the transformation properties of the displacement, I'm going to motivate a general scalar product between 4-vectors by a similar kind of quantity that is constructed from the displacement. So let me recall a result that I hope is familiar from special relativity. So working in units where the speed of light is 1, you are hopefully all familiar with the fact that this quantity is something that is in variance to all observers. I did not use a represented by symbol there, because no matter whose delta t, delta x, delta y, delta z I use there, they will all agree on the delta s squared that comes out of this. If I want to-- actually, let me write of a few things done before I say anything more. So this is an invariant. This is the same-- let me actually write this in a slightly different way-- it is the same in all Lorenz frames. So pick some observer, get their delta t, delta x, et cetera, assemble delta s squared, pick a different observer, do the Lorenz transformation, assemble their delta t prime, delta x prime, et cetera, make that, boom, they all agree. So we're going to use this to say, you know what, I'm going to call that the inner product of the displacement vector with itself. So I'm going to call this delta x dotted into delta x. And so what this means is that built into my scalar product-- so if I write this as a particular observer would compute it-- this is the scalar product that I'm going to define with respect to the displacement vector. And this is usually the point where somebody in the class is thinking, why is there a minus sign in front of the timelike piece? I can't answer that. All I can say is that appears to be-- more than appears to be. There's a whole frickin' butt load of evidence-- that's not how nature is assembled. OK, it's connected to the fact-- so the fact that all of my spatial directions are sort of entered with the same sign, but my timelike direction, it has a different sign. It's connected to the fact that I can easily move left and right, front and back. It's a little bit of effort. I can move up and down. But I cannot say, oh, crap, I left my phone at home I'll go back 15 minutes and pick it up. You cannot move back and forth in time. Time actually, which is the timelike component of this thing, it enters into the geometry in a fundamentally different way from the spatial things. And that's reflected in that minus sign. Anything deeper than that, let's just say that we're probably not likely to get very far with that conversation. Depending on what kind of muscle relaxants you enjoy using on the weekend, you might have some fun conversations about it. But it is not something that you're really going to get very far with. You just kind of have to accept that it's part of the built-in geometry of nature. OK, so this I am defining as the inner products of the displacement vector with itself. I define vectors as having the same transformation properties as the displacement vector. We can similarly define an inner product of a 4-vector with itself. OK, now, I'll put this on another board. So A dot A, I'm going to define this-- or rather I will say it is represented according to observer O as minus A0 squared plus A1 squared plus A2 squared plus A3 squared. I realize there can be a little bit of ambiguity in the way I'm writing it here. You just have to-- if you're ever confused, just ask for clarification about whether I'm writing something to a power or whether it's an index label. Context usually makes it clear. Handwriting sometimes obscures context though. The reason why I'm doing this and a real benefit of this is that whatever this quantity is, because A has the same transformation properties of the displacement, this must be a Lorenz invariant as well. The underlying mathematics of Lorenz transformation doesn't care that I wrote a zero instead of delta x here. It doesn't care that I wrote A1 instead of delta x1, et cetera. It just knows that it's a thing that goes into that slot of the Lorenz transformation. So all observers-- this is how I represent it according to observer O. But all observers agree on that form. And as a consequence, this is going to be something that we exploit a lot. Even when we move beyond the simple geometry of special relativity, a generalization of this will be extremely important for, not an exaggeration to say, everything. So a little bit of terminology-- so if A dot A is negative, and depending upon which is bigger, A0 squared or the sum of the other ones-- it could very well be negative-- we say that A is a time like vector. This traces back to the fact that if A were the displacement vector, if the displacement, the invariant interval, were negative that would tell me that the two events, which are the beginning and the end of the interval, I could find a frame at which they're at the exact same location and are only separated in time. In the same way, this is basically saying that I can find a frame when this vector points parallel to some observer's time axis. If this is positive, A is spacelike. Everything I just said about timelike, lather, rinse, repeat, but replace time with space. OK? And if A dot A equals 0, we say A is-- there's two words that are commonly used-- lightlike or null. Null just traces back obviously to the zero. Lightlike is because this is a vector that could lie tangent to the trajectory that a light beam follows in spacetime. OK? OK, let me get some clean chalk. So, so far, I've only talked about this inner product, this scalar product. Oh, and, by the way, I'll use inner product and scalar products somewhat interchangeably. But this allows me to reiterate a point I made in Tuesday's lecture. When I say scalar, scalar refers to a quantity which is-- you know, it doesn't have any components associated with it. So in that sense, it's familiar from your eager intuition of, you know, not to vector. But it has a deeper meaning in this course, because I also want it to be something that is invariant between reference frames. So A dot A is the scalar product. It gives me a quantity that all observers agree on. Now, I've only done scalar products of vectors, A and the displacement vector, with themselves. So a more general notion, if I have vectors A and B, then I will define the inner product between them, as observed by O, as constructed by observer O, rather, like so. It's not hard to convince yourself, given everything we've done so far, that this quantity must also be invariant. I'll sketch a really quick proof. Let's define-- let's say we have two 4-vectors, A and B. Their sum, by the linearity rules that apply to these vectors, must also be a vector. And so if I compute, C dot C, this is an invariant. With a little bit of labor, that basically boils down to middle school algebra. You can show that C dot C is A dot A plus B dot B plus twice A dot B. This is invariant. This is invariant. This is invariant. And so this must be invariant. So this is really useful for us, because we now have a way-- I've introduced these objects, these geometric objects, these 4-vectors. We are going to use them in this class to describe quantities that are of interest to the physicist who wants to make measurements in spacetime. We've now learned one of the things when you're doing stuff in relativity is you have to be careful who is measuring what. What are the components that 4-vector as seen by this observer? What about their friend who's jogging through the room at 3/4 quarters speed of light? What about their friend who's driving 2/3 the speed of light in the other direction? You have all these really annoying calculations that you can and sometimes have to do. This gives us a way to get certain things that are invariant out of the situation that everyone is going to agree on. Invariants are our friends. So earlier today, earlier in today's lecture, I talked about how I can write my 4-vectors using the basis factors. So another way of writing this-- so what's sort of annoying is every time I've actually written out the inner product, I have used the represented by symbol. I don't want that. I want to have equal symbols in there, dammit. So let's take advantage of the fact that A dot B, I know how to expand A and B using components and basis vectors. And again, using the index notation, I can just pull everything out and rearrange this a little bit. Whenever you get down to a point like this, we now get to do what every mathematician loves to do-- give something a name. I'm going to define the inner product of basis vector A with basis vector B to be a two index tensor-- eta alpha beta. What's lovely about this, this is a totally frame invariant quantity. We know that. And so I've now found a way to write this using the components as something that gives me a result that is totally frame invariant. Now, when you hack through a little bit the algebra of this, what you'll find is that the components of this metric-- oh, shoot, I didn't want to actually say it out loud-- the components of this tensor, which pretend you didn't hear me say that-- the components of this tensor has the following components-- I just said something circular-- has the following values. This is, as I unfortunately gave away the plot, this is, in fact, the metric that I said at the beginning is the quantity that I must associate with spacetime in order for there to be a notion of distance between events. I haven't really said what a tensor is carefully yet. I'm going to make a more formal definition of this in just a moment. But this is your first example of one. And so the way in which this actually gives me a notion of distance is through this that I wrote down right here. If I have two events in spacetime that are separated by a displacement delta x, the delta s squared, which I obtained from this thing, is fundamentally the notion of distance between those two events that I use. And notice, it's a little less normal of a distance than you're used to when you do sort of ordinary Euclidean geometry. This is a distance whose square can be negative. What we like to say is that when you're working in special relativity, it's not necessarily positive-- the distance between two events is not necessarily-- the distance squared is not necessarily positive definite. If it's negative, though, that just means it's sort of dominated by the time interval between them. If it's positive, you know it's dominated by the space interval between them. If it's zero, well, you actually know-- it's actually a little bit confusing at that point. They could be, in fact, you know, very widely separated in both space and time, but in such a way that a light beam could connect them. So there's a lot of information encoded in that. Now, as we move forward-- hang on a second-- as we move forward, we're going to upgrade this. So right now, our metric is just this simple matrix of minus 1s, 0s, and 1s. One of things that we're going to do is sort of a warm-up exercise to the more complicated things we're going to do later is we're going to move away from special relativity and Cartesian coordinates. We're going to look at it in polar coordinates. That's going to be kind of like a warm-up zone. And so when we do that, we're always going to reserve eta for the metric of special relativity when I'm working in Cartesian coordinates. It's just a great symbol to have for that. And it's a useful thing to always have that definition in mind. I can continue to do special relativity, but then working in coordinates that are, you know, spherical-like or polar-like or something like that, then this is going to become a function. And what that is going to mean is that things like my little basis vector is going to have more complicated behavior. A little later in the course, we will then show that when gravity enters into the picture, essentially the essence of gravity is going to be encoded in this thing as well in a way where, again, it's going to be a function. It's going to vary as a function of space and time. And the dynamics of gravity will be buried in that. It's sort of funny that it really does just sort of start out-- I mean, if you take that thing and you set delta t equals 0, this is just the bloody Pythagorean theorem. That is all this is. Put time back in, and it sort of is the generalization of Pythagoras to spacetime. And, in fact, we're going to take advantage of that and sort of define a geometry that looks like this as being flat in the same way that a board is flat, and the Pythagorean theorem works perfectly on it. Then, we're going to start think about what happens when it becomes curved. And you start thinking about things like, what is the geometry on the surface of the sphere look like? That's just sort of pointing ahead. So I just throw that at you, so that you get ready for some of the concepts that we'll be talking about soon. So let me write this actually in terms of differentials. It's sort of useful for what I want to say next. So a little differential, if I have two events in spacetime that are very close to one another, I can write them like so. And what I've essentially done here is written dx as dx alpha e alpha. Before I get into some more sort of a couple of important, fairly important 4-vectors, the reason I did this is I want to make an important point about some notation and terminology that is used. If it is the case that the displacement vector is related to the differentials of your coordinates like so, we say that e alpha is a coordinate basis vector. What it does is it transforms a differential of your coordinate into a differential vector in spacetime. Now, you, may be thinking to yourself, OK, well, what other kind can there be? Well, this is where my little spiel there a second ago about how we're going to start looking at more complicated things, it's going to become important. So when we're working in a Cartesian-like coordinate system, the fact that this is what we call a coordinate basis vector isn't very interesting. Suppose I was working in some kind of a curvilinear coordinate system, OK? Spherical coordinates. Now, let's just focus on 3-space for a second. So if I write a sort of analogous equation in curvilinear coordinates-- OK, so here's the 3-space version of that. Now let's imagine that i equals 1 corresponds to radius, i equals 2 is theta, i equals 3 is phi. Then, this would be dr er plus d theta e theta plus d phi e phi. Does that disturb you at all? OK, this has dimensions of length. These have the dimensions of angle. In order for this to work, er must be dimensionless. e theta must have the dimensions of length. e phi must have the dimensions of length. This is what a coordinate basis looks like when I am dealing with-- well, we're going to use is a lot in this thing. I introduce this right now because you are all probably looking at that, and some small part of you inside is weeping, because what you want me to write down is this. Ah, isn't that better? OK, this looks like something you're used to. So I throw this out here right now just because I want to make sure you're aware that there are some equations and some foundational stuff you guys have been doing over the years, particularly, this shows up a lot when you've done E&M out of a textbook like Purcell or Griffiths or Jackson, because there's some derivative operators, which are assuming that your basis vectors are what we call orthonormal. So my e i hat here, it is an orthonormal basis. And orthonormal basis is defined such that the dot product of any two members of this thing gives me back the Kronecker delta. That is not necessarily the case when I work with a coordinate basis. Our basis has er dot er equals 1. Yay, that one's nice. But when I do e theta dot e theta, I get r squared. e phi dot e phi will be r squared sine squared theta. And what I'm going to do when I start generalizing these things, I'm going to change my-- this thing which I defined up here-- do I still have it on the board? Yeah, yeah, right here. So when I set eta alpha beta is e alpha dot e beta and I made it this thing, I'm going to generalize this and say at the dot product of any two basis vectors it gives me a more general notion of a metric tensor. And the values in the metric tensor maybe functions like this. Right now, throw that out there, you know, this might be sort of just like a peak of the horrors that lie ahead. OK, we're not going to worry about this too much just yet. But I want you to be prepared for this. In particular, it's really useful to have this notion of a coordinate basis versus an orthonormal basis in your head. We're going to start defining some derivative operations soon. In fact, probably won't get to them today, but they will be present when we start doing Lecture 3. And there's a couple of results that come up where everyone's sort of like, wait, I knew that the divergence had a factor of r on that derivative there. Where'd it go? It's because we're not working in an orthonormal basis. All right, I'm a little sick of math. So let's do a little physics. So, so far, the actual only physical 4-vector that I've introduced is the displacement vector. From the displacement vector, it's really easy to make the probably the first and simplest important 4-vector, which is known as the 4-velocity. This tells me the rate of displacement of an observer as this person moves through spacetime per unit-- and we're going to be careful about this in this class-- d tau is the time interval as measured along the trajectory of the observer with 4-velocity u. In other words-- that's a very long winded way of saying it-- it's an interval of proper time. In English, the word proper time sounds very like, woo, I don't want to use improper time. I better use that. But this actually I think it comes from French. It just refers to the fact that it's one's own time. Apparently in German people say eigenzeit. So, you know, there's a couple of different words for it. But proper time is what we use. In special relativity, if we see someone going by with constant velocity, a particular observer who sees, you know-- we're here in the room. Someone comes through. Their 4-velocity is u. We would see their 4-velocity to have the components gamma gamma v, where gamma, I'll remind you, is the special relativistic Lorenz factor. And I'll remind you again we've set speed of light to 1. A very useful thing, which we're actually going to take advantage of quite a bit, is that in the rest frame of u-- pardon me for a second-- in the rest frame of u, or I should say of the observer whose 4-velocity is u, they just have 1 in atomic direction, C, if you want to put your factors back into their. And that's basically just saying that the person is standing still, but moving through time, because you are always moving through time. All right, from the 4-velocity for an observer who has-- or for an object, I should say, who has some mass, we can easily define the 4-momentum, where this m is known as the rest mass of this object. It's worth a bit of description here. You will often see, particularly in some older textbooks that discuss special relativity, people like to talk about the relativistic mass. And that comes from the fact that if I write out what this thing looks like according to some particular observer, you have this gamma m entering into both of the components. And so older textbooks often called gamma m the relativistic mass. That's not really the way people have-- over the course of the past couple decades, they've moved away from that. And it's just more useful to focus on the rest mass as the only really meaningful mass, because, as we'll see in a moment, it's a Lorenz invariant. We'll see how that is in literally about 3 minutes. And so what we're instead going to say is that as seen by some particular observer, this has a timelike component that is the energy that that observer would measure and a set of spacelike components that are the momentum that that observer will measure. So where we get a bit of important physics out of all this stuff is by coupling these two 4-vectors to the scalar products that we made up. So the first one, if you do u dot u, according to any observer, that's just going to be minus gamma squared plus gamma squared v squared. And with about 20 seconds worth of analysis, you can find that this is always equal to minus 1. Actually, there's an even trickier way to do this. Suppose I evaluate this in the rest frame of the observer whose 4-velocity is u? Well, in the rest frame, v is 0, and gamma is 1, and I get minus 1. And this is an invariant. So whatever I get in that particular frame must be obtained in all frames. That's a trick we're going to use over and over and over again. Sometimes you can identify-- you know, you get some kind of God awful expression that just makes you want to vomit. But then you go, wait a minute, what would this look like in frame blah, blah, blah? And you sort of think about some particular frame. And in that frame, it may simplify. And if it does and it's a frame invariant quantity, mazel tov, you have just basically won the lottery. You've got this all taken care of. Go on with your life. So the 4-velocity has a scalar product of itself that is always minus 1. OK? How about the 4-momentum? Well, the 4-momentum is just 4-velocity times mass. So that's just minus m squared. But we also know it's related to these two other quantities, which are important in physics. This is related to the energy and to the momentum. So this is also equal to minus e squared plus-- so this is the ordinary 3-- the magnitude of the 3-vector part of this thing as measured by the observer who breaks up the 4-momentum in this way. So what this means is I can manipulate this guy around a little bit here. Anybody who works in particle physics is presumably familiar with this equation. Sometimes it appears with the p squared moved onto the other side. If it looks a little bit unfamiliar to you, let me put some factors of C back in this. So remember, we have set C equal to 1. When you put it back in, that's what this is. So it drops out of this in a very, very simple way. One of the uses of this-- and many of you have done exercises presumably in some previous study that does this. And there'll be one exercise on the p set that was just posted where you exploit this. So a key bit of physics, the reason why we care about 4-momentum is it's in one mathematical object allows us to combine conservation energy and conservation of momentum. So conservation of 4-momentum puts both conservation of energy and conservation of momentum into one mathematical object. So if I have n particles that interact, then the total 4-momentum is conserved in the interaction. OK? So, yeah, we talk a little bit more about this and then just sort of quickly move on. So combining this with the fact that we are free to change our reference frames often it gives us a trick that allows us to really simplify a lot of analysis. So if I have n particles that are sort of swarming around and doing some horrible bit of business that I need to study and I need to have a good understanding of, we can often vastly simplify our algebra by choosing a special and very convenient frame of reference in which to do our analysis-- choose the center of momentum frame. So this is the time frame in which that that p tote-- so C-O-M, center of momentum-- has zero spatial momentum. In that frame, you have just as much momentum going to the left as going to the right, just as much going forward as going backwards, as much going up as going down. And so this turns out to be-- so the classic example of where this is really useful is when you are studying particle collisions and you're looking at things like the production of new particles. So imagine you've got particle A with some 4-momentum PA coming in like this. Particle B's got some 4-momentum coming in like this. These guys collide. And they do so-- I work in the center of momentum frame. I might want to just calculate the energy at which they just happen to produce some new pair of particles at rest. I would like to find the threshold for this particular creation process. OK? So you're going to play with one problem on the p-set that's kind of like that. Let's see, what do I have time to do? I think I will do-- yeah, I think I can do two more things. So all the dot products that I have been talking about so far have been a dot product of a 4-vector with itself. I did u dotted into u. I did p dotted into p. I invented a frame in which p has a particularly simple form. And then when you actually do some of analysis, you would probably take that p tote and dot it into itself. I haven't done anything that looks at the crossing between these two things, dotting one into the other. So let me to go through a very useful result that follows by combining p with u. There's a very specific notion of p and a very specific notion of u. Let's let p be the 4-momentum of a particle. I call it a. Let's let u be the 4-velocity not of a, but the 4-velocity of observer O. So particle a might be a muon that was created the upper atmosphere and is crashing through our room right now. Observer O might be your hyperactive friend who is jogging through the room at half the speed of light. The question I want to ask is, what does O measure as the energy of particle a. So the naive way to do this, which I will emphasize is not wrong, what you might do is sort of go, OK, well, we're sitting here. This room is our laboratory. I've measure this thing in my lab. So I know p as I measure it. I can see O jogging by. So I know O's 4-velocity as I measure it. So what I should do is figure out the Lorenz transformation that takes me into the rest frame of O. Once I have that Lorenz transformation, I'll apply that Lorenz transformation to the 4-vector p. Boom, that will give me that energy. That will work. That will absolutely work. But there's an easier way to do it. So one thing you should note is that everybody represents that 4-velocity as an energy-- excuse me, they represent the 4-momentum as an energy and a 3-momentum. In particular, though, they represent it as the energy that they would measure and the 3-momentum that they would measure. So that means p as seen by O is e according to O and p according to O. That are acquainted o is what we want. And I just told you a moment ago, you know, if you have p in your own reference frame, and you have u in your own reference frame, you can do this whole math with Lorenz transformations and get it out. But you also know that in O's own reference frame, O represents their 4-velocity as 1 in the timelike direction, 0 in the spatial direction. So what this means is if I go into O's reference frame, if I go into their inertial reference frame, notice that if I take the dot product of p and u, I get e times 1 and p times 0. So that is just negative. It's exactly what I want-- modulo minus sign. And so you go, OK, well, I'll flip my minus sign around. And you think, OK, great, but I did that using those quantities as written down in O's reference frame. And then you go, holy crap, that's the invariant scalar product. I'm done. Mic drop. Leave the room. What this means is you start with p as you measure it, u as you measure it. Take the scalar product between the two of them. Boom, the answer you want pops out. No nonsense of Lorenz transformations. None of that garbage needs to happen. You just take that inner product and you've got it. So that sort of says in words, no matter what representation you choose to write down p and u in, take the dot product between the two of them, throw in a minus sign, you've got the energy of the particle with p as measured by the observer with u. It's sort of late in the hour, or the hour and a half I should say. So let me just sort of emphasize, there are occasional moments in this class where if you're dozing off a little bit, I suggest you pop up and tattoo this into a neuron somewhere. This is one of those moments. OK? This is a result that we're going to use over and over and over again, because this holds-- this isn't just in special relativity. When we start talking about the behavior of things near black holes, there's going to a place where I basically at that point to say, well, I'm going to use the fact that the observer measures an energy that is given by-- and I'm going to write down that. The dot product that's involved is a little bit more complicated, because my metric is hairier. But it's the exact same physical concept. OK? Let me just do one more. And then I'll go talk, without getting into the math, about what I will start with on Tuesday. Last 4-velocity, which is probably useful for us to quickly talk about is-- so we've talked a lot about 4-velocity. That is just one piece-- when we're talking about sort of the kinematics of bodies moving in spacetime, you need more information than just velocity. Sometimes things are moving around. There's additional forces acting on them. And so we also care about the 4-acceleration. And so this is what I get when I take the derivative with respect to proper time of the 4-velocity. So will there be some homework exercises that use this. The main thing which I want to emphasize to sort of conclude our calculations for today is that when I talk about a 4-velocity of acceleration, this has an extremely important property. It is always the case that a dotted into u equals 0. If you're used to sort of 3-dimensional intuition, that may seem weird. OK? Anytime you see a car accelerate from a stop, that's a case in which its acceleration is clearly not orthogonal to its velocity. But the issue here is, this is not a spatial dot product. This is a space time dot product. And some of your intuition has to go out the window because of that. It's very simple to prove this. Remember, u dot u equals minus 1. So d of u dot u, d tau, which is just 2, u dot a is the derivative of minus 1, which is 0. OK? So this is something that we will exploit. If you want to describe the relativistic kinetics of an accelerating body, this is a great thing that we can use to exploit. You often need a little bit more information. We have to give you as a bit of additional information some knowledge about what the orientation of the acceleration is and things like that. So whenever you are given-- I'll get to you in just a sec-- whenever you're given any kind of differential quantity like this, it's not enough to know-- it's like the acceleration of loss, you have to also have boundary conditions. And that sort of tells you what the initial direction is. Question? AUDIENCE: Is that time still the proper time? SCOTT HUGHES: That time is the proper time, yes. AUDIENCE: So it's not accelerating? SCOTT HUGHES: That's right. So you can still define a proper time for an accelerating observer. It will not relate-- hold that thought. You're going to play this a little bit more in a future problem set. I mean, the key thing is that the way-- if you have an d observer, an interval of proper time as compared to an interval of time for, you know, someone in a rest frame that sees this person accelerate away, the conversion between the intervals of time, the two measure, it evolves. So, you know, let's say you're in this room with me. In fact, it turns out that if you accelerate at 1G for a year, you get to very close to the speed of light. So let's say that you were in a rocket ship right now that launched with an acceleration of G. Initially, you and I synchronize our watches. And so an interval of a second to me is the same as a second to you. Half a year later, you're moving at something like half the speed of light. And I will see a noticeable time delay. An interval of a second as you measure it looks long compared to me. Six months later, you're actually quite close to the speed of light, and it gets dilated even more. So last thing, which I'm going to say, and I'm not going to get into too much detail with this yet, is we're going to begin next time by making a little bit more formal some of the notions that go around. So we've increased some physics and some vectors. I've given you guys one tensor so far, the metric tensor. And so I'm going to give you-- in fact, I will write down a very precise definition of this right now, and we'll pick it up from there on Tuesday. So the basic idea-- so you guys have see-- the one tensor you've seen so far is the metric tensor. And what the metric is is it's sort of a mathematical object that I put in a pair of 4-vectors. And it spits out a quantity that is a Lorenz invariant scalar that characterizes what we call the inner product of those two 4-vectors. More generally, I'm going to define a tensor of type 0 N as a function or mapping of N 4-vectors into Lorenz invariant scalars, which is linear in each of its N arguments. So I will pick it up here on Tuesday. And let me just say in words, the metric is a 0 2 tensor. I put in two 4-vectors. It spits out a Lorenz invariant scalar. We're going to before too long come up with a couple of things that involve three real vectors-- excuse me three 4-vectors, too many numbers here, a trio of 4-vectors, which it then maps to a Lorenz invariant scalar. Some of them will take in four 4-vectors and produce a Lorenz invariant scalar. Notice I wrote this in sort of funny way. The 0 N sort of begs for there to be sort of an N 0. OK? To do that I have to introduce an object that is sort of dual to a vector. We're going to talk about that. Those are objects called one-forms, which actually happen to be a species of vector. We're actually going to then learn that the vector is itself a tensor. And so we will make a very general classification of these things. And we'll see that vectors are just a subset of these tensors. And at last, we'll sort of have all the mathematics in place. We can sort of lose some of these distinctions and just life goes on, and we can start actually doing some physics with these. All right, I will pick it up there on Tuesday. |
MIT_8962_General_Relativity_Spring_2020 | 23_Black_holes_II.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: So in this final lecture, I want to think a bit sort of with an eye towards thinking about how one might actually make measurements that prove the nature of the black hole spacetime that was discussed the previous lecture. I'm going to discuss motion in a black hole spacetime. We touched on this a little bit in the previous lecture, where we discussed the motion of radial light rays, OK? We, in fact, used radial light rays as a critical tool for describing the properties of the spacetime. We use that to help us understand the location of events horizons. But I want to think a little bit more generally. What might it look like if I have material orbiting in the vicinity of one of these black holes? What if it's not light? What if it's made out of matter? And so what this is going to boil down to is understanding the behavior of geodesics in a black hole spacetime. And the naive approach to doing this is not wrong, but naive. What you do is you would just take the spacetime-- take your Schwarzschild or take your Kerr spacetime-- and turn a very large crank, grind out all of the connection coefficients, evaluate the geodesic equation, integrate it up. Solve for the geodesics. Boom, you got yourself your motion, OK? And that is absolutely correct. You can do that using your Kerr space time or your Schwarzschild spacetime. In fact, if you do that for-- you can get the connection coefficients describing this. Those are relatively easy to work out. I think they are listed in Carroll, in equation 5.53, according to my notes. That may just be for Schwarzschild. But at any rate, they're all listed there, and have a blast. This approach is not wrong, but-- my notes say, but it is not useful. That's not really true. I'll just say that there is a more useful approach to this. A more fruitful approach is to exploit the fact that these are highly symmetric spacetimes. Exploit the symmetries and the Killing vectors, and see how they can be used to reduce the number of degrees of freedom that you need to describe. In this lecture, I'm going to go through this in quite a bit of detail for Schwarzschild. The concepts that I'm going to apply work for Kerr and for Kerr-Newman as well, OK? Schwarzschild is just a little easier to work with. It's something that I can fit into a single lecture. In particular, one of the nice things about Schwarzschild-- so let's go ahead and write down that spacetime. So one of the nice things about Schwarzschild is it is spherically symmetric. This means I can always rotate coordinates such that-- well, when me think about it-- let's just back up for a second. Imagine that I have some kind of a body orbiting a Schwarzschild black hole. Spherical symmetry tells me that there must be some notion of a conserved angular momentum such that that orbit always lies within a given plane. Put it another way. Because it is fairly symmetric, there cannot exist a torque. The black hole cannot exert a torque that changes the orientation of the orbital plane. In particular, because it is fairly symmetric, there is no unique notion of an equator to this object. And so you might as well define any orbits to live in the theta equals pi over 2 plane. You can always rotate your coordinates to put any orbit in the theta equals pi over 2 plane. It's actually a pretty simple exercise. I won't do this, but if you start with an orbit that is in the theta equals pi over 2 plane, and it is moving such that its initial velocity would keep it in a theta equals pi over 2 plane, it's a very simple exercise using the geodesic equation to show that it will always be in the theta equals pi over 2 plane. So spherical symmetry says, you know what, let's just forget about the theta degree of freedom. I can always define my coordinates in such a way that it lives in the theta equals pi over 2 plane. Boom, I have reduced my motion from, in general, being three spatial dimensions to two spatial dimensions. That's another one of the reasons why, for pedagogical purposes, it's nice to start with Schwarzschild. For Kerr, this is not the case, OK? Kerr is a little bit more complicated. You have to treat the theta motion separately. It's not strictly symmetric, and so you can't do that. I have actually spent a tremendous amount of my career studying the orbits of objects around Kerr black holes. And I do have to say that the additional complications that arise from this lack of sphericity, they're really beautiful, OK? There's an amazing amount of fun stuff you can do with it. You know, there's a reason why I just keep coming back to this research problem, and part of it is it's just bloody fun. But if you're teaching this stuff for the first time, it's not where you want to begin. All right, so Schwarzschild allows us to reduce it from a three-dimensional problem to a two-dimensional problem. And Schwarzschild also has two Killing vectors. The time derivative of every metric component is equal to 0. That means p downstairs t is constant. So there is a timelike Killing vector. So p downstairs t is constant every level along the orbit. It means there exists a timelike Killing vector. And so what we do is we associate this with the energy of the orbit. We call this up to a minus sign. And we choose that minus sign because if we imagine orbits have very, very large radius, the spacetime is nearly flat, and we want to sort of clear out the minus sign associated with lowering our index here. We're going to call that constant negative of the energy of the orbit. The spacetime is also independent of the angle phi. And so p sub phi is a constant. This means that the spacetime has an axial Killing vector, something associated with motions around a symmetry axis. So I'm going to call this L-- well, actually, I will tend to call it L sub z. You can kind of think of this as, after I have put everything in the theta equals pi over 2 plane, this is like an angular momentum on the-- angular momentum parallel to the axis normal to that plane, and I call that the z-axis. It's worth noting that all three of these things are also true for Reissner-Nordstrom black holes. This one is not true for Kerr, but these two are, OK? So although the details change, many of the concepts will carry over when you look at different, more complicated classes of black holes. Let's now think about the forward momentum of a body moving in a Schwarzschild spacetime. So let's say it's got a rest mass m, then it will have three components. Remember, I have put this thing in a plane where there is no theta motion. It will have three theta motions describing its motion with respect to the time coordinate, the radial coordinate, and the actual coordinate phi, OK? So before I do anything with this, let's take advantage of the fact that we have these quantities that are constants. So using the fact that p downstairs mu is what I get when I hit this guy-- that be a mu, pardon me. This is what I get when I hit this guy with the metric. I can write out a p sub t and p sub phi. So let's do that on another board. Well, let's do a p sub t first. So p sub t is going to be gtt, p sub-- whoops. OK, this is minus. There's an m from that-- minus sign from my metric. And this whole thing I define as the negative energy, e. OK, so that combination-- we'll do something at the rest mass in just a moment. But you should basically look at this and saying that dt d tau, which tells me about how the body moves with respect to the Schwarzschild time coordinate-- that complement of the four velocity times 1 minus 2gm over r is a constant. Let's look at p downstairs phi. So this is m r squared sine squared theta. Ah, we've chosen our orbital plane, 1 times d phi d tau. This is equal to [INAUDIBLE] momentum L sub z. In my notes, I might flip around a little bit between L sub z and L. So let's massage these a little bit more. So I can take these two expressions and I can use them to write dt d tau and d phi d tau in terms of these conserved quantities. So dt d tau is equal to-- I'm going to call it e hat divided by 1 minus 2gm over r. d phi d tau is going to be Lz hat over our squared. And so these quantities with a hat are just the conserved values normalized to the rest mass, OK? They all are proportional to the rest mass, but it's actually if you want to know what the-- this is essentially telling me about how clocks on the orbit tick relative to clocks that are infinitely far away. And that can't depend on the mass of the orbiting object. This is telling me about how this small body is moving according to the clock of the orbiting observer. And again, that can't depend on the mass of the object. OK, so that's kind of cool. So I've now managed to relate three-- excuse me, two of the three components of the forward momentum to functions of r and quantities that are known to be constant. That's good. We have one more overall constraint. We know that if I take p dot p, I get negative of the rest mass squared. So this, when I write it out-- OK, notice every single term is proportional to m squared. So I can divide that out. I can insert dt d tau. I can replace for this e divided by this guy. I can replace d phi d tau by my Lz hat divided by r squared. Doing so, manipulating a little bit, I get-- we get something like this. And let me rearrange this a tiny bit. All right, I have managed to reduce this to a one-dimensional problem, OK? So going from that line over to there, basically all I did was insert the relationship between dt d tau and E, d phi d tau and L, cleared out some overall factors of things like 1 minus 2GM/r, manipulate, manipulate, manipulate, and what you finally get is this lovely equation here that tells you how the radial velocity, the radial velocity with respect to proper time, how it depends as a function of r given the energy and the angular momentum. I have written it in this form because the problem is very strikingly reminiscent to the Newtonian problem of understanding the motion of a particle in a 1/r potential, which we often describe as having an effective potential that has a gravitational term-- Newton's gravity, or if you're doing things like quantum mechanics, the Coulomb potential, and a Coulomb barrier associated with the angular momentum of an orbit. So one way to approach what we've got now, one thing that we could do, is essentially just pick your energy and your Lz-- pick your energy and your angular momentum-- pick an initial position-- let's imagine you synchronize the clocks at t equals 0-- and then just integrate. You've got your dr d tau given here. Don't forget, you also have d phi d tau and dt d tau related to the energy and your angular momentum, like so. Boom. It's a closed system. You can always do just sort of a little numerical integration of this. In a certain sense, this completely specifies the problem. But there's so much more we can do. In particular, what we see is that all of the interesting behavior associated with this orbit is bound up in this function V effective. So something that's really useful for us to do is to take a look at what this V effective looks like. So suppose you are given a particular value for E hat and Lz hat, and you plot V effective versus r. Well, what you typically find is that it's got a behavior that looks kind of like this, where this value right here is V effective equal to 1. Notice as r goes to infinity, you get 1 times 1. So this asymptotes to 1. Notice E hat squared has the same dimensions as V effective. In fact, in the unit choices I've used here, they are both dimensionless. So what we can do is plot-- let's imagine that we have-- you know what? I'm going to want to sketch this on a different board. Let me go over here. So I'm going to want to look at a couple of different values of V effective-- excuse me, a couple of different values of E hat. OK. Here's an example. This guy is asymptoting at 1. So since E hat, as I said, has the same units as V effective, let's plot them together. So example one-- imagine if E hat lies right here. OK. So let's call this E hat 1. So this is some value that is greater than 1. What is the point of doing this? Well, notice-- I'm going to flip back and forth between these two middle boards here. Let's look at the equation that governs the radial motion of this body. dr d tau has to be a real number. This has to be a real number. So we have to have E hat squared greater than or equal to V effective in order for dr d tau to have a meaningful solution. So let's look at my example here, E hat 1. E hat 1 is greater than my effective potential everywhere at all radii until I get down to here. Let's call that r1. So in this case, dr d tau, if I think about this thing-- so note that defines dr d tau squared. Let's suppose we take the negative square root. dr d tau is positive and inward-- or, well, it's negative-- negative and real, negative and real, negative and real, negative and real, negative and real-- boom. It's 0. Can it go into here? No. It cannot go into there, because there E hat squared is less-- sorry. That should have been squared. Inside here, E hat squared is less than the effective potential. dr d tau is imaginary. That doesn't make any sense. The only option is for this guy to change sign and trundle right back out. So E hat greater than 1 corresponds to a body that comes in from large radius, turns around at particular radius where E hat is the square root of the effective potential, and then goes back out to infinity or back out to-- let's not say infinity-- goes back out to large radius. OK? In Newtonian gravity, we would call this a hyperbolic orbit. This corresponds to-- so remember, when I did this I have not-- I'm not actually-- I'm only computing the radial motion. I'm not looking at the phi motion. So this actually, when you look at both the radial motion and the phi motion, what you see is that this is a body that comes in and then sort of whips around that small radius and goes right back out. Let's look at another example. Let's call this E2-- some value that is less than 1. OK. Well, for E hat 2, it's a potential. The potential is underneath E hat 2 squared only between these two radii, which I will call r sub p and r sub a. What we expect in this case is motion of this body essentially going back and forth and turning around at periastron and apoastron. This is a relativistic generalization of an elliptical orbit. In general relativity, they turn out generally not to be ellipses. So we call this an eccentric orbit. In the weak field limit, if you imagine r being very, very large, it's not hard to show that the motion is nearly an ellipse, but it's an ellipse whose long axis is slowly precessing. This actually leads to the famous perihelion precession of Mercury that Einstein first calculated. And this is an exercise that I am asking you to do on one of the final P-sets of this course. Using what I have set up here, it's really not that difficult to do. Let me go to another board. And note that one could imagine an energy such that dr d tau is exactly 0. So if you choose your energy so that you sit right here at the minimum of the potential-- I will label this as point s-- the energy that corresponds to exactly that point is what would be a-- that is, there is a single point at which dr d tau equals 0. Anywhere away from that, dr d tau would be imaginary, so that's not going to work. But right at that point, dr d tau equals 0, and you get a circular orbit. Notice there's a second point at which that can happen. Let's call this point u. Perhaps you can guess why I called these points s and u. If you imagine that you add a tiny amount of energy right here, well, what it will do is it will execute small oscillations around the point s. It will sort of move it up to something that's similar to what I drew up there as E2, but with a very small eccentricity. So if I slightly disturb a circular orbit down here at s, I essentially just oscillate in the vicinity of that circular orbit. S stands for stable. If I have an orbit up here at u and I very slightly perturb it, well, it'll either go in and eventually reach r equals 0, or it'll go out, and then it's completely unbound, and it will just keep trundling all the way out essentially forever. This guy is unstable. Stable orbits are particularly interesting and important. So let's look at these orbits with a little bit more care. So the very definition of a circular orbit is that dr d tau equals 0. Its radius does not change. If dr d tau equals 0, then it must have E hat equal the square root of the effective. Both of these orbits happen to live at either a minimum or a maximum of this potential curve. So I'm going to require that the partial derivative of that potential with respect r be equal to 0. So let's take a look at this condition. My effective potential is given up here. Do a little bit of algebra with this, set this guy equal to 0. What you'll find after your algebraic smoke clears is that you get this condition on the angular momentum. Notice as r gets really large-- oh, shoot. Try it again. Notice as r gets really large that this asymptotes to plus or minus the square root of GM r. That is indeed exactly what you get for the angular momentum of a circular Newtonian orbit. OK. So it's a nice sanity check. It appears to be somewhat pathological as r approaches 3GM, though. Hold that thought. OK. So now let's take that value of L, plug it back into the potential, and set E equal to the square root of that. A little bit of algebra ensues. And what you find is that this equals 1 minus 2GM/r over, again, that factor under a square root of 1 minus 3GM/r. Again, we sort of see something a little bit pathological happening as r goes to 3GM. Let me make two comments about this. So first of all, notice that this energy is smaller than 1. I can intuitively-- you can sort of imagine-- remember, E hat is the energy per unit rest mass. You can think of this as something like total energy over M-- so the energy associated with the orbiting body. It's got a rest energy, a kinetic energy, and a potential energy divided by m. For an orbit to be bound, the potential energy, which is negative, must have larger magnitude than the kinetic energy. So for a bound orbit, E kinetic plus E potential will be a negative quantity. So the numerator is going to be something that, when normalized to m, is less than 1. So this is exactly what we expect to describe a bound orbit. Notice, also-- so if you take this formula and look at it in the large r limit, it goes to 1 minus GM over 2r. This is in fact exactly what you get when you look at the energy per unit mass throwing in-- sort of by hand-- a rest mass. The minus GM/2r exactly corresponds to kinetic plus potential for a Newtonian circular orbit. So lots of stuff is hanging together nicely. So we've just learned that we can characterize the energy and angular momentum of circular orbits around my black hole. Let's look at a couple other things associated with this. So these plots where I look at the radial motion, this effective potential, as I mentioned a few moments ago, there's additional sort of degrees of freedom in the motion that are being suppressed here. So this thing is also moving in that plane. It's whirling around with respect to phi. We've lost that information in the way we've drawn this here. Let's define omega to be the angular velocity of this orbit as seen by a distant observer. OK. Why am I doing it as seen by a distant observer? Well, when things orbit, there tend to be periodicities that imprint themselves on observables. It could be the period associated with the gravitational wave that arises out of this. It could be the period associated with peaks and a light curve if this is a star orbiting around a black hole. It could be oscillations in the X-ray flux if this is some kind of a lump in an accretion disk of material orbiting a black hole. So if this is the angular velocity seen by distant observers-- remember, the time that distant observers use to-- the time in which the distant observers' clocks run is the Schwarzschild time t. So this will be d phi dt, which I can write as d phi d tau-- this is the angular velocity according to the orbit itself-- normalized to dt d tau. Now, these are both quantities that are simply related to constants of motion. What I've got in the numerator here is L hat over r squared. And what I've got in the denominator here is E hat over 1 minus-- I dropped my t. It would happen at some point-- 1 minus 2GM/r. So let's go ahead and take our solution here. My E hat is 1 minus 2GM/r divided by square root of blah, blah, blah. The 1 minus 2GM/r cancels. My L hat is square root GM r divided by, again, that square root 1 minus 3GM/r. Notice, the square root 1 minus 3GM/r factors all cancel. So this becomes plus or minus 1 over r squared square root GM r. Looks like this. Plus and minus basically just correspond to whether this motion sort of is going in the same sense as your phi coordinate or in the opposite sense. There's really no physics in that. It just comes along for the ride that both behave the same. If you guys get interested in this, and you do a similar calculation around a Kerr black hole, you'll find that your prograde solution gives you a different frequency than your retrograde solution because of the fact that the dragging of inertial frames due to the spin of the black hole kind of breaks that symmetry. Something which is interesting and-- well, I'll make a comment about this in just a second, which is kind of interesting. And here's what I'll say-- sometimes people think this is more profound than it should be-- is this is, in fact, exactly the same frequency law that you get using Newtonian gravity. This is actually exactly the same as Kepler's law. That seems really, really cool. And it is. It's actually really useful. It makes it very easy to remember this. But don't read too much into it. More than anything, it is a statement about a particular quality of this radial coordinate. So remember, in Newtonian gravity r tells me the distance between-- if I have an orbit at r1 and an orbit at r2, then I know that the distance between them is r2 minus r1. In the Schwarzschild spacetime, the distance between these two orbits is not r2 minus r1. However, r2 labels a sphere of surface area 4 pi r2 squared. And r1 labels a sphere of surface area 4 pi r1 squared. It's easy to also show that the circumference of the orbit at r2 is 2 pi r2, the circumference of the orbit at r1 is 2 pi r1. That, more than anything, is why we end up reproducing Kepler's law here, is that this areal coordinate is nicely amenable to this interpretation. So is this orbit-- so I described over here an orbit that is unstable and an orbit that is stable. I have described how to compute-- if I wanted to find a circular orbit at a given radius, those formulas that I derived over there on the right-most blackboards, they tell me what the energy and the angular momentum need to be as a function of r. Is that orbit stable? Well, if it is, I can check that by computing the second derivative of my effective potential. So my orbits are stable if the second derivative with respect to r is greater than 0, unstable if this turns out to be negative. Let's look at the crossover point from one to the other. What if there is a radius where, in fact, the stable and the unstable orbits coincide? In fact, what one finds, if you look at the effective potential-- you imagine just sort of playing with L sub z. So let's say we take that L sub z, and we just explore it for lots of different radii of the orbits. You find that as the orbits radius gets smaller and smaller, your stable orbit tends to go up, and this minimum sort of becomes flatter, and your unstable orbit just kind of comes down. They sort of approach one another. There is a point just when they coincide-- this should have an "effective" on it. My apologies. Right when they coincide, this defines what we call the marginally stable orbit. I may have put this one on a problem set. But it might be one of the ones I decided to drop. So I'm just going to go ahead and do the analysis. When you compute this, bearing in mind that your angular momentum is a constant-- so take this, substitute in now your solution for L sub z, which I've written down over there-- what you find is that the marginally stable orbit-- let's call it r sub ms-- it is located at a radius of 6GM. This is a profoundly new behavior that doesn't even come close to existing in Newtonian spacetime-- spacetime-- doesn't come close to existing in Newtonian gravity, excuse me. [SIGHS] I'm getting tired. The message I want you to understand is that, what this tells us is that no stable circular orbits exist inside r equals 6GM. So this is very, very different behavior. If I have-- let's just say I have a very compact body but Newtonian gravity rules. I can make circular orbits around it, basically go all the way down until they essentially touch the surface of that body. And you might think based on this that you would want to make orbits that basically go all the way down, that sort of kiss the edge of r equals 2GM. Well, this is telling you you can't do that. When you start trying to make orbits that go inside-- at least, circular orbits-- that go inside 6GM, they're not stable. If someone sneezes on them, they either are sort of blown out to infinity, or they fall into the event horizon. And in fact, one of the consequences of this is that, in astrophysical systems, we generically expect there to be kind of-- if you imagine material falling into a black hole, imagine that there's like a star or something that's just dumping gas into orbit around a black hole, well, it will tend to form a disk that orbits around this thing. And the elements of the disk are always rubbing against each other and radiating. That makes them get hot. They lose energy because of this radiation. And so they will very slowly sort of fall in. But they make this kind of-- it's thought that in most cases they'll make this kind of thick disk that fills much of the spacetime surrounding the black hole. But there will be a hole in the center. Not just because the thing is a black hole. I don't mean that kind of a hole. There'll actually be something surrounding the black hole's event horizon, because there are no stable circular orbits. Once the material comes in and hits this particular radius, 6GM in the Schwarzschild case, it very rapidly falls in, reduces the density of that material tremendously. And you get this much thinner region of the disk where essentially things fall in practically instantly. It should be noted that this 6GM is, of course, only for Schwarzschild. If you talk about Kerr, you actually have two different radii corresponding to material that goes around parallel to the black hole's spin and material that goes anti-parallel to the black hole's spin. And it complicates things somewhat. There's two different radii there. The one that goes parallel tends to get a little bit closer. The one that's anti-parallel tends to go out a little bit farther. But the general prediction that there's an innermost orbit beyond which stable orbits do not exist, that's robust and holds across the domain of black holes. So let me conclude by talking about one final category of orbits-- photon orbits. So let's recall that when we were talking about null geodesics, we parametrized them in such a way that d-- sort of the tangent to the world line has an affine parameter attached to it, such that we can write p equals dx d lambda. These guys are null. So p dot p equals 0. For the case of orbiting bodies, that was equal to minus mass squared. Mass is 0 here. So we're going to follow a very similar procedure. The spacetime is still time independent, so there is still a notion of a conserved energy. It is still actually symmetric, so there is still a notion of axial angular momentum. But because p dot p is 0 now, rather than minus m squared, when we go through the exercise of-- that sort of parallels what we did for our massive particle, we're going to derive a different potential. So let's go ahead and evaluate this guy again. And I get 0 equals minus 1 minus 2GM/r dt d lambda squared plus 1 minus 2GM/r dr d lambda squared. I'm still going to require the thing to be in the theta equals pi/2 plane so that we know theta term. And I have set theta to pi/2. So my sine squared theta is just one. So I get this. I'll remind you that I can relate-- so the relationship between the time-light component of momentum and energy, the axial component of momentum, and the angular momentum, it's exactly the same as before. There's no rest mass appearing. And so now I find-- so my energy, I don't put a hat on it. It's not energy bringing it mass, because there is no mass. That looks like so. And my L looks like so. Using them, I can now derive an equation governing the radial motion of my light ray. And sparing you the line or two of algebra, it looks like this. Pardon me just one moment. OK. OK. So kind of similar to what we had before, if you look at it you'll see the term involving the angular momentum is a little bit different. As you stare at this for a moment, something should be disturbing you. Notice that the equation of motion appears to depend on the energy. OK. That should really bug you. Why should gamma rays and infrared radiation follow different trajectories? As long as I am truly in a geometric optics limit, I shouldn't. Now, it is true that if you consider very long wavelength radiation, you might need to worry about things. You might need to solve the wave equation in the spacetime. But as long as the wave nature of this radiation is-- if the wavelength is small enough that it's negligible compared to 2GM-- I shouldn't care what the energy is. Energy should not be influencing this. So what's going on is I need to reparametrize this a little bit. What I'm going to do to wash away my dependence on-- wash away my apparent dependence on the energy here, is I'm going to redefine my affine parameter. So let's take lambda-- so L divided by lambda. And I am going to define b to be L divided by E. This is the quantity that I will call the impact parameter. And I'll describe why I am calling it that in just a few minutes. So I'm going to take this entire equation, divide both sides by L squared. And I get something that looks like this. What I'm going to do is say that the impact parameter, b, is what I can control. It's the parameter that-- all I can control that defines the photon that I am studying here. And everything after this, this is my photon potential. Notice that the photon potential has no free parameters in it. If I plot this guy as a function of r, it kind it just looks like this. Two aspects of it are worth calling out. This peak occurs at r equals 3GM. Remember the way in which our energy-- oh, still have it on the board here-- things like the energy per unit mass and the angular momentum per unit mass all blew up when r equal 3GM. 3GM is showing up now when I look at the motion of radiation, look at massless-- radiation corresponding to a massless particle, so to speak. If I look at the energy per unit mass, and the mass goes to 0, I get infinity. So the fact that I was actually seeing sort of things blowing up as r goes to 3GM was kind of like the equations hinting to me in advance that there was a hidden solution corresponding to radiation that could be regarded as a particular limit that they were sort of struggling to communicate to me. The other thing which I want to note is that the potential, the height of the potential right here, it has a peak value of 1 over 27 G squared M squared. Hold that thought for just a moment. Actually, let me write it in a slightly different way. This equals 1 over 3 root 3 GM squared. So now, to wrap this up I need to tell you what is really meant by this impact parameter. So go back to freshman mechanics. And impact parameter there is-- like, let's say I have a problem where I am looking at an object that is on an infall trajectory. There is a momentum p that is describing this. And it's moving in such a way-- it's not moving on a real trajectory, right? It's actually moving in a sense that is a little bit off of true to the radial direction. This guy actually has an angular momentum that is given by-- let's say that this is equal to px x hat. Let's say this is b y hat. This guy's got an angular momentum of b cross p. That's the impact parameter associated with this thing. It sort of tells me about the offset of this momentum from being directly radial towards the center of this object. Well, I'm going to use a similar notion to give me a geometric sense of what this impact parameter here means. Suppose here is my black hole. It's a little circle of radius 2GM. And I'm sitting way the heck out here, safely far away from this thing. And what I'm going to do is shoot light-- not quite radial. What I'm going to do is I'm going to have sort of an array of laser beams that kind of come up here-- an array of laser beams. And I'm going to shoot them towards this black hole. And what I'm going to do is I'm going to offset the laser beams by a distance b. And I'm going to fire it straight towards this thing. Go through sort of a careful definition of what angular momentum means, and you'll see that that definition of impact parameter gives you a very nice sense of the energy associated with the trajectory of this beam and a notion of angular momentum associated with this. So let's flip back and forth between a couple of different boards here. There are three cases that are interesting. Suppose b is small. In particular, suppose I have b less than 3 root 3 GM. So I start over here. Let's take the limit. What if b equals 0? Well, if b equals 0, this guy just goes, zoom, straight into the black hole. As long as it is anything less than 3 root 3 GM, what will happen is, when I shoot this thing in, it goes in, and it bends, perhaps, but it ends up going into the black hole. Let's look at this in the context of the equation of motion, the potential, and the impact parameter. If b is less than 1 over 3 root 3 GM, then 1 over b squared will be higher than this peak. And this thing is just going to be, whoo [fast motion sound effect], it's going to go right over the top of the peak. And it shoots into small radius. OK. Remember, L and E are constants of the motion. So that b is a parameter that defines this light for the this entire trajectory. 1 over b squared is less than V phot everywhere-- excuse me, greater than V phot everywhere. A rather crucial typo. And as such, it shoots the light ray, and it goes right over the peak into the black hole, eventually winds up at r equals 0. If b is greater than 3 root 3 GM, then 1 over b squared is less than V phot at the peak. This corresponds, in this drawing, to a beam of light that follows a trajectory that kind of comes in here at this point. dr d tau, if we were to continue to go into smaller radii, it would become imaginary. That's not allowed. So it switches direction and trundles right back out to infinity. On this diagram, that corresponds to a light ray that's perhaps out here. This guy comes in, and he gets bent by the gravity a little bit, and then just, shoo [light ray sound effect], shoots back off to infinity. The critical point, b equals 3 GM, is right where the impact parameter hits the peak. And so what happens in this plot is this guy comes in, hits the peak, and just sits there forever. You get a little bit more of a physical sense as to what's going on by thinking about it here. This guy comes in, [INAUDIBLE] this, and then just whirls around, and lies on what we call the photon orbit. What's the radius of that photon orbit? r equals 3 GM. Now, astrophysically, a more interesting situation is, imagine you had some source of light that dumps a lot of photons in the vicinity of a black hole. Some of those photons are going to tend to-- some of them are going to go into the black hole, some are going to scatter a little bit and shoot off to infinity. But if you imagine that there will be some population of them that get stuck right on the r equals 3GM orbit-- now, that is an unstable orbit. And in fact, if you look at it, what you find is that your typical photon is likely to whirl around a bunch of times, and then, you know, if it's not precisely at 3GM but it's 3.00000000000000001GM, it will whirl around maybe 10 or 15 times, and then it'll go off to infinity. So what we actually expect is if we have an object that is illuminated like this, then we will actually see these things come out here, and we will see-- so bear in mind, this is circularly-- this is symmetric. So take this and rotate around the symmetry axis. We expect to see a ring whose radius is twice the critical impact parameter. So it would have a diameter of 6 root 3 GM. It would be essentially a ring or a circle of radius 3 root 3 GM. This is, in fact, what the Event Horizon Telescope measured. So last year when I was lecturing this class, this was announced almost-- I mean, they timed it well. They basically timed it to about two or three lectures before I discussed this aspect of black holes. So that was-- thank you, Event Horizon Telescope. That was very nice of them. And of course, we don't expect-- so Schwarzschild black holes are probably a mathematical curiosity. Objects in the real universe all rotate. We expect Kerr to be the generic solution. And so there's a fair amount of work that goes into how you correct this to do-- so doing this for Schwarzschild is, because of spherical symmetry, it's beautiful and it's simple. Kerr is a little bit more complicated. But, you know, it's a problem that can be solved. And a lot of very smart people spend a lot of time doing this to sort of map out what the shadow of the black hole looks like, what this ring would look like in the case of light coming off of a spinning black hole. So that's it. This is all that I'm going to present for 8.962 in the spring of 2020 semester. So to everyone, as you are scattered around the world attempting to sort of stay connected to physics and your friends and your classwork, I wish you good health. And I hope to see you again at a time when the world is a little less crazy. In the meantime, enjoy our little beautiful black holes. |
MIT_8962_General_Relativity_Spring_2020 | 1_Introduction_and_the_geometric_viewpoint_on_physics.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: The key textbook for this class is Sean Carroll's textbook on general relativity. It's now almost 20 years old. I would say-- I think it's listed on the website as required. I would actually call it sort of semi-required. It is where I will tend to post most of the readings to the course. It's a really good, complete textbook for a one semester course, which is what we have, and I will not be going through the entire thing. You can't in a one semester course. And I will, from time to time, there will be a few topics that I cannot go as in into as deeply as I would like, and if we had two semesters maybe I'd be able to do a little bit more. And so for those who are interested, I will suggest readings in this. My personal favorite supplement to this is a textbook by Bernard Schutz , A First Course in General Relativity. OK, so these are all things where-- so the MIT bookstore, I'm not sure how much they carry. They're all available through Amazon, and you can definitely find these kind of things. If you get Schutz's textbook, definitely get the second edition. The first edition contains errors. There is actually a very important geometric object that we are going to introduce in a couple of weeks that his textbook has a really clever derivation of. I remember seeing that and thinking, wow, that's really clever. The reason he was able do it so simply is because it's wrong. AUDIENCE: [LAUGH] SCOTT HUGHES: It's corrected in the second edition. Another one is Gravitation by Misner, Thorne and Wheeler. This is sort of a bit of personal history for me. I was Thorne's graduate student, and I used this textbook when I learned this subject originally. I, frankly, do not recommend this textbook to somebody who is learning the subject for the first time. It's a good place for a reference for certain things. It's available in the reading room. Know where you can get a copy and pick it up. So first of all, picking it up is kind of a-- it's good exercise. It's actually like a-- you know, it's a huge book. It gravitates. So it's a great reference and has a couple of good sections in it for new students. But I will indicate from time to time, especially some of the stuff we do in the second half of the class, my lectures are kind of inspired by Misner, Thorne and Wheeler-- known as MTW. And for those of you who have sort of a more mathematically minded approach to things, General Relativity by Wald. In the course syllabus I call this the uber book. This really is sort of the self-contained book that really pins down the subject very, very well. It's quite terse and formal, but also very, very clear. If you are a mathematically minded thinker, you will find this to be a really good textbook to refer to. There is one particular derivation that I'm going to do in about a month and a half that I essentially take from Wald's textbook because it's just beautiful. Other things in it-- to me, it's a little too terse from my own personal taste, but others may find it to be pretty good. Two other quick things. So there are 11 problem sets and your grade is determined entirely on these problems sets. The syllabus gives the schedule for when they will be posted and when they are to be handed in. 11 does not divide evenly into 100, so what we do is we have 10 that are worth 9% of your grade and the 11th is worth 10% of your grade. Is that 11th one really 1% longer? I don't know. Again, I'm taking the viewpoint that you are graduate students, or at least you're playing one for the next hour and a half, and if you're going to sweat the difference of one percentage point on that thing, come on. This is about learning the material, don't worry about those little details that much. That is where all of your assessment is going to come from. The very first time I lectured this course I had a final exam, and it just turned to be a complete waste of my time and the students' time. You really cannot write-- you either write problems that are so easy that you can do them in your sleep, or they're so difficult you can't do them in the time period of an exam. So we're just going to stick with problem sets and that's fine. Let's move into, then, the way the course is going to be structured. So my presentation of this material-- the first half of the course basically up to spring break is essentially the mathematical foundations of general relativity. There are several choices that need to be made when you're doing this. This often, in some universities, if they have multiple semester sequences, what I'm going to cover in this first half of our semester in some places goes for a full semester. And what this means is there's a couple of things that I just cannot cover in quite as much depth. Those will be things where, for those of you who are interested in it, have that kind of a mathematical thinking of things, happy to push you to additional readings. We can dive in and look at it a bit more depth. My goal is to give you just enough stuff that we can do the most important applications of this subject. And I'm an astrophysicist, and to be blunt, most of my really interesting applications tend to be things that have to do with things like cosmology, black holes, dense stars, and things like that. And so I want to get enough formalism together that we can get to that part of things. And so the goal of this is that by the week right before spring break what we will do is "derive"-- I put that in quotes, and you'll see why a little bit later-- the Einstein field equations that govern gravity in general relativity. So there are several things that we could do that are not strictly necessary to get there, and just because of time limitations I'm going to choose to elide a few of these topics. The second half will then be applications. We will use everything we derived in the first half to see how general relativity gives us a relativistic theory of gravity. We will begin applying it. We'll see how Newton's law is encoded in these field equations. We'll see how we go beyond Newton's law, get some of the classic tests of general relativity, and then start looking at solving it for more interesting systems. Looking at the evolution of the universe as a whole, looking at the behavior of black holes, looking for gravitational waves, constructing the spacetime of neutron stars, things like that. So it's a fun semester. It sort of works well to fit these two things in like this. And for those of you who are interested in taking things more deeply, there's a lot of room to grow after this. And it does look like-- so people who particularly would like to go a little bit more detail on some of the math, if you've looked at the syllabus, I have one of my absolute favorite quotes from a course evaluation is put on there. Where a student in 2007 or so wrote, "The course was fine as it was, but Professor Hughes as an astrophysicist tended to focus on really mundane topics like cosmology and black holes." If you think those are mundane topics, what can I say, guilty as charged. But for those you who do want to take a different approach these sorts of things, we'll probably alternate lecturing this course in the future between someone from the CTP who works more in quantum gravity and things related to that. That'll be Netta Engelhardt in spring of 2021. We're now ready to start talking about, after doing all this sort of prep, we can actually talk about some of the foundations of the theory. So before I dive in, are there any questions? All right. What we're going to begin doing for the first couple of weeks-- well, not really first couple weeks. The first couple of lectures, is we're going to begin by discussing special relativity, but we're going to do special relativity using mathematical language that emphasizes the geometric nature of this form of relativity. What this does is it allows us to introduce basically the formalism, the notation, all the different tools that are important for when things get more complicated. When we apply a lot of these tools to special relativity, like we will be doing in the first three or so lectures of this course, it's kind of like swatting a mosquito with a sledgehammer. You really don't need that much mathematical structure to discuss special relativity. But you're going to be grateful for that sledgehammer when we start talking about strong field orbits of rotating black holes, right. And so the whole idea of this is to develop the framework in terms of a physical system where it's simple to understand what is going on. These are sort of a way to introduce the mathematical tools in a place where the physics is straightforward and then kind of carry forward from there. I will caution that as a consequence of this many students find these first three lectures to be a little on the dull side. So, sorry. It's just some stuff that we kind of have to get out there, and then as we generalize to more interesting mathematical objects, more interesting physical settings, it gets more interesting. All right, so let's dive in. So the setting for everything that we will be doing is a geometric concept known as spacetime. So we give you a precise mathematical definition of spacetime. A spacetime is a manifold of events that is endowed with a metric. So, a wonderful mathematical definition, and I've written it in a way that requires me to carefully define three additional terms. The concepts that I've underlined here, I've not defined them yet precisely exactly what I mean by them. So let's go over to the sideboard and talk about what exactly these are. So a manifold-- if you are a mathematician, you might twitch a little bit about the way that I am going to define this, and I will point you to a reading that does it a little bit more precisely. But for the purposes of our class, a manifold is essentially just a set or a collection of points with well understood connectedness properties. What I mean by that, is I'm going to talk about manifolds of space and time. I haven't defined an event yet, but I'm about to, but I'm going to say that there's a bunch of events that happen at this place and at this time and a bunch of events that happen at this place and at this time. And the manifold, the spacetime, gives me some notion of how I connect the events over here to the events over here. A manifold is a tautological concept. It's all about how one connects one region to another one. So if you're working on a manifold that lives on the surface of a donut, you have a particular topology associated with that. If it lives on the surface of a sphere you have a different topology associated with it. If you would like to see more careful and more rigorous discussion of this, this is one of the places where Carroll is very good. So go into Carroll, at least in the edition that I have it's on pages 54 to 62. He introduces a bit of additional mathematical machinery, discusses things with a little more rigor than I'm doing here-- this isn't rigorous at all, so significantly more rigor than I'm doing here. Those of you are into that, that should be something that you enjoy. So an event. This is when and where something happens. Could be anything. From our point of view, the event essentially is going to be the fundamental notion of a coordinate in space time. We will actually, in many cases-- actually, that's bad word choice, I should say. Coordinates are actually sort of labels that we attach to events. We are going to be free to adjust those labels, but the underlying geometric idea that the event is here, that's independent of the coordinates that we choose. So we will label these things with coordinates. But the event itself exists independent of these labels. Just to give an example, there might be one event which is I punch myself in the head. And so I'm very egotistic, so I will say this event happened at time 0-- x, y, and z equals 0. Because I define this corner of my skull as the origin the coordinate system, and I always think whatever's happening right now is the origin of time. Those of you out in the room are also egocentric and you would say whatever, I'm going to call that-- let's say you are at y of 3 meters and I'm going to put the floor as the origin of my z-axis, so z of 1.7 meters or whatever. You will come with your own independent labeling of these things. So you're all familiar with the idea that we can just change coordinate systems. I'm going to harp on this a bit, though, because there's going to be a really important distinction we make between geometrical objects that live in this manifold of spacetime and how we represent them using labels that might be attached to coordinate systems. And I'm going to come back to this when we start talking about some additional geometric objects in just a couple of minutes. So the last object that I have introduced into here is one that we will begin talking about in a lot more detail in the next lecture, but let me put it into here right away. And so that is the metric. Metric comes from a root meaning to measure, and what this is is it's something it gives me a notion of distance between events in the manifold. For physics to work, this has to exist, right? But it's worth knowing that the idea of a manifold is in some way more primitive than this. You can have a manifold without any notion of a metric attached to it. And if that's the case-- people like to joke that if you don't know what the difference is between a metric with a manifold and without a man-- excuse me, a manifold with a metric and without a metric, feel free to drink coffee out of a donut. Because topologically, those are the exact same thing, but their geometry, which is encoded in the metric, which tells me how the different points on that manifold are arranged, are rather different. What this basically does is it's going to give me a mathematical object that enforces or really conveys the idea that different events in this manifold have a particular distance between them. So without this, a manifold has no notion of distance encoded in it. So the two things together really make this concept come to life. You can see a lot more, like I said. You can get more information about many of these concepts from the readings in Carroll, Wald's textbook also goes into quite a lot of detail about this stuff. So this is the venue. This is the setting in which we are going to talk about things. And just cutting forward roughly 2 and 1/2 months' worth of lectures, what we are going to find is that part of Einstein's genius is that it turns out that this notion of the metric ends up encoding gravity. So that's kind of where we're going to end up going with things. The idea that the mathematical structure that tells me how far apart two events are is intimately connected to the properties of gravity. It's pretty cool, and it is something that-- physics is an experimental science. All of our measurements are consistent with it. So that's cool. So everything I have said so far, nothing but math. Nothing but definitions. So let's start working with a particular form of physics. So we are going to begin, as I said, with special relativity. This is the simplest theory of spacetime that is compatible with physics as we know it. Not fully compatible, but does a pretty good job. And we'll see that it turns out to correspond to general relativity when there is no gravity. So to set this up we need to have some kind of a way of labeling our events. And so I'm going to introduce kind of a conceptual-- you almost think of it as scaffolding, which we're going to use to build a lot of our concepts around. And in this one-- I mean that in a kind of an abstract sense. There's going to be sort of the edifice that we use to help us build the building that's going to be the mathematics of general relativity. But this one really is kind of like a scaffolding. Because what I want to introduce here is a notion of what is called an inertial reference frame. So I'm going to sketch this quickly, and I'm going to post to the course website a chapter from an early draft textbook by Roger Blandford and Kip Thorne. So when I talk about the inertial reference frame, I want you to sort of visualize in your head a lattice of clocks-- clocks and measuring rods-- that allows us to label, in other words, to assign coordinates any event that happens in spacetime. So just sort of, in your head, imagine that there's this grid of little clocks and measuring rods, and a mosquito lands on your head. It's right near a particular rod and a particular clock. It bites you, that's an event. You slap it, that's another event. And the measuring rods and events are what allows you to sort of keep track of the ordering of those events and where they happen in this four dimensional manifold of spacetime. So I'm going to require this lattice to have a certain set of properties. First, I'm going to say that this lattice moves freely through spacetime. What do I mean by moving freely? I mean no forces act on it, it does not rotate, it is inertial. Every clock and every measuring rod has no inertia, no force is acting on it at all. You look at that, you might think to yourself, well, why don't you make it at rest? Well, I did. It's at rest in respect to someone, but we might have a different observer who's coming along who has no forces acting on her, and she's moving relative to me at three quarters the speed of light. It's not at rest with respects to that observer. That's actually kind of the key here. So this inertial reference frame is at rest with respect to someone who feels no forces but not to all observers. I'm going to require that my measuring rods are orthogonal to each other. So they define an orthogonal coordinate system, and I am also going to require that the little markings on them that tell me where things happen are uniformly ticked. In other words, I'm going to just make sure that the spacing between tick marks here is exactly the same as the spacing between tick marks here. You may sort of think well, that's a result or an idea worthy of the journal Duh, but it's important to specify this. You want to make sure that the standard you are using to define length is the same in this region of spacetime as it is over in this region of spacetime. When we start getting into general relativity, we start to see there can be concerns about this coming about, so it's worth spelling it out and making it clear at the beginning. I'm also going to require that my clocks tick uniformly. We're going to make this lattice that fills all of spacetime using the best thing that Swiss engineers can make for us, we want to make sure that one second, an interval of one second, is the same here in this classroom as it is somewhere off in the Andromeda Galaxy. We want to make sure that there is no evolution to the time standard when we do this. Finally, I'm going to synchronize all these clocks with each other in the following way. This is going to use the Einstein synchronization procedure. This is the first place where a little bit of physics is actually beginning to finally enter our discussion. I'll comment that-- I'm going to go through what this procedure is in just a second. But an Easter egg here. Whenever you see a name that has Einstein in it, your ears should perk up a little. Because it probably means this is something important. This is, after all, a course in relativity, and that tends to be the things that end up mattering. Even when they end up being really easy to-- things we look at now and kind of see as fairly obvious, it's important, OK, and we often attach Einstein's name to this. So this Einstein synchronization procedure, this takes advantage of the fact-- and by the way, we're not going to teach special relativity in this course, 8.962. I assume you have already studied special relativity and you're all experts in this, and so I can freely borrow from its important results. This procedure takes advantage of the fact that the speed of light is the same to all observers, no matter what inertial reference frame they might be in. So the speed of light is a key invariant. It connects-- because it's a speed, it connects space and time, and because it is the same to all observers, it defines a particular standard for relating space and time that is going to have important invariant meaning associated with it. Just to remind you what this means-- let's see, do I have a laser pointer with me? I might, but don't worry about it. I have a pretend laser pointer. My chalk's a laser pointer. I point my laser pointer at the wall and you all see it dashing across the room at 300,000 kilometers per second. I then start jogging at half the speed of light, as is my wont, and continuing to point that, you guys measure the light going across the room. You still measure 300,000 kilometers per second. I, on the other hand, measure the light coming out of my laser pointer and I get 300,000 kilometers per second. So just because my laser pointer is moving at half the speed of light according to you doesn't mean the light that's coming out of it is boosted to a higher speed. If you studied special relativity you'll know that its energy is boosted to a higher energy level, but the speed of light is always just C. So we're going to take advantage of that to come up with a way of synchronizing our clocks. The way it works is this. So let's look at a two dimensional slice of my lattice here. Time, and let's make this be the x-axis. And so I will have, let's say, this is where clock one exists and this is where clock two exists. Let's go into the reference frame that is at rest with respect to this lattice. We want to synchronize clock one with clock two. So as time marches on, these guys stand still. So here's the path in spacetime-- the world line, traced out by clock one. Here's the clock path in spacetime, the world line, traced out by clock two. So let's say-- let's call this event-- let's say that this event happens at a time t1e. t1 is when clock one emits a pulse of light. So this light will just follow a little trajectory a bit through spacetime. This goes out and strikes clock two at which point it is bounced back to clock one. Let's ignore this point for just a second. So let's just say for the moment that this is then reflected back and then it is received back at clock one at a time t1r. Let's make this a little bit neater. t1e, t1r. Clock one receives the reflected pulse. So the moment at which it bounces, we'll call that t2b, that is the moment at which the light bounces off of clock two. And the way we synchronize our clocks is just by requiring that this be equal to the average of the emission and the reception time. Totally trivial idea, right? All I'm saying is I'm going to just require that in order for clock one and clock two to be synchronized to one another, let's make sure that when I bounce light between any two pairs of clocks they are set such that when the light bounces it's the midway point between the total light-- the halfway of the total light travel time that the light moves along. Very, very simple concept. So there's nothing particularly deep here, but notice Einstein's name is attached to this. And I don't say that to be-- I'm not being sarcastic. It really points to the fact that we are using one of the most fundamental results of special relativity in designing how these clocks work in this inertial reference frame. Believe it or not, this thing, this really simple concept, it comes back to sort of bite us on the butt a little bit later in this course. Because when we throw gravity into the mix, we're going to learn that gravity impacts the way light travels through spacetime. We're going to get to some objects where gravity is so strong that light cannot escape them. And we're going to find that our perhaps most naive ways of labeling time in the spacetime of such objects kind of goes completely haywire. Fundamentally, the reason why time is going haywire when you have really strong gravity is because we use light as our tool for synchronizing all of our clocks, and if the way that light moves in your space time is affected, the way you're going to label time is going to be affected. So this is a very simple concept. Again, I sort of emphasize these first couple lectures you're swatting a mosquito with a sledgehammer, but we're setting up this edifice because this will come back and it's important to bear this in mind when things get a little more interesting later in the course. So let's start setting up some geometrical objects here. Pardon me, let me do one other thing really quickly before I start setting up some geometrical objects. So when I sketched this thing called a spacetime diagram here, I should have talked a little bit about the units that I'm going to use to describe the ticking of my clocks and the spacing of tick marks on my measuring rods. What we will generally do in this course is choose the basic unit of length to be the distance light travels in your basic unit of time. While you parse that sentence, what that's basically saying is suppose I set my clock so that they tick every second. Well, if my clocks tick once per second, then my basic unit of length will be the light second. Do you want to put this into more familiar units? That's about 300,000 kilometers. One of my personal favorites-- if the time unit is 1 nanosecond, the length unit is, of course, one I will call it LNS-- Light Nanosecond. Students who are in 8.033 with me are not allowed to answer this question, does anyone know what one light nanosecond is? Actually, this is a little bit ridiculous. But to within far greater than a percent accuracy it is one foot. The English unit that comes on these asinine rulers that those of us educated the United States learned in all of our European friends sneer at us about. The speed of light is to incredible precision 1 foot per nanosecond. To be fair, let's make that a wiggly equal. So what this means is that in the units that I'm going to be working with, if I then want to express the speed of light in these units-- So C is one light time unit per time unit, which we are just going to call 1. So we will generally set the speed of light equal to 1. Just bear in mind what this essentially means is that you can think of, if you want to then convert to your favorite meters per second, furlongs per fortnight, whatever it is that you're most comfortable with, C is effectively a conversion factor then. And so what this means is that when we do this all velocities that we measure are going to be dimensionless. Really what we're doing is we're measuring them as fractions of the speed of light. Now, with this system of units defined, let's talk about a geometric object. So let's imagine that O is an observer in the inertial reference frame that I defined a few moments ago. This is a mouthful to say, it's even more of a mouthful to write, so I'm going to typically abbreviate this IRF. So o observes two events, which I will label P and Q. I'll just go to a clean board for this. So let's say here in spacetime-- so imagine I've got coordinate axes that have been drawn, it's going to be three dimensional, for simplicity I'm not going to actually write them out. Let's say I've got event P here and event Q over here. Now if we were just doing Euclidean geometry in three space, you guys have all known that once you've got two events written down on a plane or in a three dimensional space, something like that, you can define the displacement vector from one to the other. We're going to the same thing in spacetime. So let's call delta x-- and I'm going to make a comment on notation in just a moment-- this is the displacement in spacetime from P to Q. We're going to define the components of this displacement vector as seen by O. So when I write equals with a dot on it that means the geometric object that I've written on the left hand side is given according to the specified observer by the following set of complements, which I'm about to write down. So this looks like so. So, couple things that I want to emphasize that I'm introducing here, bits of notation that we're going to use over and over again in this term. Notice I am using an over arrow to denote a vector in spacetime. Different texts, different professors use slightly different notations for this. Those of you who took 8.033 at MIT with Sal Vitale he preferred to write a little under tilde when he wrote that. For us working in four dimensional space time, it is going to be of paramount importance to us, and so we're going to use this over arrow which you probably have all seen for ordinary three dimensional vectors. For us it's going to represent a four dimensional vector. Now in truth we're not going to use it all that much after the first couple of weeks of the class, our first couple lectures even. Occasionally we'll bust it out, but we will tend to use a more compact notation in which we say delta x, that displacement factor has the components delta x mu, where mu lies is in either t, x, y, and z, or 0, 1, 2, 3. When we set up a problem, we need to make a mapping to what the numerical correspondences between-- I need to tell you that mu equals 0 corresponds to time, and mu equals 1 corresponds to x. We'll switch to other coordinate systems and I'll have to be careful to say, almost always, mu equals 0 will be time. But what the other three correspond to, that depends on the coordinate system. It might be a radius, it might be a different angle, some things like that. Again, just sort of being a little overly cautious and careful defining these. I will note, though, that generally Greek indices in most textbooks, they tend to be used to label spacetime indices. And then there are times when you might want to just sort of imagine you've chosen a particular moment in time, and you want to look at what space looks like at that time. And so you might then go down to Latin indices to pick out spatial components at some moment in time. We'll see that come up from time to time, just want you to be aware there is this distinction. And as you read other textbooks, there are a few others that are used. Always just check, usually in some of the introductory chapters of the textbook, they will define these things very carefully. Wald is an example of someone who actually does anything a little bit different. He tends to use lower case Latin letters from the top of the alphabet to denote spacetime indices, and those from i, j, k, he uses them to denote Latin indices. If you're old enough to get this, this is often called by some of us who grew up in the Dark Ages it is sometimes called the Fortran convention. If you've ever programmed in Fortran, you know why that is. If you didn't, please don't bother learning it, it's really not worth the brain cells it would take. So we've got this geometric object that is viewed by observer o. Let's now think about what this looks like from the viewpoint of a different observer. A different inertial observer. Let's say somebody comes dashing through the room here, and observer o sees them running across the room at something like 87% of the speed of light. You know, since, as I have assumed, you are all experts in special relativity, that they will measure intervals of time and intervals of space differently than observer o does. So here's event P. Here's event Q. Here is delta x. This is all as measured by observer o bar. Something which I really want to strongly emphasize at this point is that this P, this Q, and this delta x, notice I haven't put bars on any of them. I haven't put primes or anything like that. It is the exact same P and Q and delta x as this over here. That is because P, Q, and delta x are geometric objects whose meaning transcends the particular inertial reference frame used to define the coordinates at which P exists, at which Q exists, and that then defined the delta x. These geometric objects exist independent of the representation. If I can use an intuitive example-- if I take and I hold-- let's be careful the pose I do with this-- let's say I stick my arm out. I say that my arm is pointing to the left, right? You guys will look at this and say your arm is pointing to the right, because you're using a slightly different system of coordinates to orient yourself in this room. We're both right. We have represented this geometric object, my arm, in different ways. But me calling this pointing to the left and you calling it pointing to the right doesn't change the basic nature of my arm. It doesn't mean that my blood cells changed because of something like this happening. This has an independent existence. In the same way, this delta x, it is the displacement, these two events. This might be mosquito lands on my head, this might be me smacking it with my hand, or flying near my head and me smocking with my hand. Those are events that exist independent of how we choose to represent them. So the key thing is we preserve that notion of the geometric object's independent existence. What does change is the representation that the two observers use. So I'm going to jump to this Greek index notation. And so what I'm going to say is that according to observer o bar, they are going to represent this object by a collection of components that are not the same as the complaints that are used by observer o. And to keep my notation consistent, let's play a little o underneath this arrow. So this is just sort of shorthand for delta x is represented according to o by those components. Delta x is represented according to o bar by these components. And again, since I'm assuming you all are experts in special relativity, we already know how to relate the barred components to the unbarred components. They're related by a Lorentz transformation. So what we would say is delta x is zero bar component, or t bar, if you prefer. It's given by gamma-- I will define gamma in just a moment, you can probably guess. So I'm imagining an observer that is just moving along the x-axis or the coordinate one axis. So what went on in that transformation I just wrote down is O bar moves with v along spatial axis 1 with speed v as seen by O. And of course gamma is 1 divided by square root of 1 minus v squared. Remember, speed of light is 1. We don't want to be writing this crap out every time we have to transform different representations, so we're going to introduce more compact notation for this. So we're going to say delta x mu bar-- this is what I get when I sum over index nu from 0 to 3 of lambda. Mu bar nu, delta x nu. So that's defining a matrix multiplication, , and you can read out the components of this lambda matrix from what I've got over there. And even better yet, delta x mu bar is lambda mu bar nu delta x nu. So in this last line, if you haven't seen this before, I am using Einstein's summation convention. If I have an index that appears in one geometric object in the downstairs position and an adjacent geometric objects in the upstairs position and it's repeated-- so repeated indices in the upstairs and downstairs position are assumed to be summed over their full range, from 0 to 3. We're going to talk about this a little bit more, what's going on with this, after I've built up a little bit more of the mathematical structure. In particular, what is the distinction between through the upstairs and downstairs positions. Some of you might be saying, well, isn't one way of writing it what we call a covariant component and wasn't one a contravariant component? If you know those terms, mazel tov. They're not actually really helpful, and so I kind of deliberately like to use this more primitive wording of calling it just upstairs and downstairs. Because what we're going to find, the goal of physics, is to understand the universe in a way that allows us to connect this understanding to measurements. And measurements don't care about contravariant versus covariant, and all these things are essentially just ways of representing objects with our mathematics that is sort of a go between from some of our physical ideas to what can eventually be measured. So covariant, contravariant, eh, whatever. At the end of the day, we're going to see as we put these sort of things together it's how these terms connect to one another that matters. The name is not that important. I do want to make one little point about this as I move forward. It is sort of worth noting that if I think of how I relate the displacement components according to my barred observer relative to those as measured by my unbarred observer, I can think of this Lorentz transformation matrix as what I get when I differentiate one representation's coordinates with respect to the other representation's coordinates. Kind of trivial in this case, and when we are doing special relativity there is a particular form of the Lorentz transformation that we tend to use, but I just want to highlight this because this relationship between two different representations of a reference frame is going to come up over and over and over again. This is a more general form, this idea that you are essentially taking the derivative. You're looking at how one representation varies according to the other representation. So using that to think about how to move between one inertial reference frame to another is going to be very important to us. This is a more general form. I want to make one further point, and then I will introduce a more careful definition of what is meant by a vector and I think that will be a good place for us to stop. So when I look at this particular form-- so let's look at the last thing I wrote there, where I use the Einstein summation convention. This is just a chance for me to introduce a little bit more terminology in notation. So when I wrote down the relationship delta x mu bar with lambda mu bar nu, delta x nu, how I labeled the index that I was summing over is kind of irrelevant. This is exactly the same as lambda mu bar alpha delta x alpha. I can switch to something else if you have enough fonts available. You can use smiley faces as your index, whatever. The key bit is that as long as you are summing over it, it's kind of irrelevant how you label it. I think that looks silly, so I'm going to go back to alpha. When I have an index like that that is being summed over, it's going to sort of disappear at the end of my analysis. Its only role is to serve as a place holder. It allows me to keep things lined up properly so that I can do a particular mathematical operation. So in this equation, nu or alpha is called a dummy index. Now, when I'm doing it with something like this, where I'm just relating one set of one indexed objects to another set of one indexed objects, it's kind of trivial to move these things around. We're going to make some much more complicated equations later in this class, and we'll find in those cases that sometimes it's actually really useful to have the freedom to relabel our dummy indices. It allows us to sort of pick out patterns that might exist among things and see how to simplify a relationship in a really useful way. On the other hand, I do not have the freedom to change that mu bar that appears there, right. I have to have that being the same on both sides of the equation. Because I'm free to mix around things where it's just going to be summed over and not play a role, but in this my mu bar-- pardon me for a second-- mu bar is not a dummy index, and so I do not have that freedom. We sometimes call this the free index. As I write that down, it seems like a bit of a strange name, actually, and it's really not free. You're actually constrained in what it can be. What can I say, probably there's some history there that I don't know about. So let's do our last concept for the day. So let's carefully define a spacetime vector. So a spacetime vector is going to be any quartet of numbers, those numbers we will call components, which transforms between inertial reference frames like the displacement vector. So if I represent some spacetime vector A as some collection of numbers that as observed by o has components A0, A1, A2, A3. If a second inertial observer relates their components to these by-- if that describes the components for the observer o bar, then it's a vector. If you have taken any mathematics, they carefully defined vector spaces, this should be familiar to you. It's a very similar operation to what is done in a lot of other kinds of analysis. The key to making this notion of a vector a sensible one is this transformation law. So if I had a quartet of numbers, which is say, the number of batteries in my pocket, the number of times my dog sneezed this morning, how many toes I have on my left foot, and-- I'm sick of counting so let's just say 0 for the fourth component-- that does not transform between reference frames by a Lorentz transformation. It's just a collection of random numbers. So not any old quartet of numbers will constitute the components of a vector. It has to be things that have a physical meaning that you connect to what you measure by a Lorentz transformation. For it to be a good vector, it also has to-- that A has to obey the various linearity laws that define a vector space. So if I had two vectors and I add them together, then their sum is a vector. If I have a vector and I multiply it by some scalar-- by the way, you have to be a little bit careful when I say scalar here, because you might think to yourself something like, you know, the mass of my shoe, that's a scalar. But you be careful when you're talking about things in relativity, whether the scale you're dealing with is actually a quantity that is Lorentz in variant. With quantity like mass, if you're talking about rest mass-- we'll get into the distinction among these things a bit later-- OK, you're good. You just have be careful to pick out something that actually is the same according to all observers. So when I say scalar, this means same to all observers. If this is the case, then I can define D to be that scalar times a vector and it is also a vector. Question? Oh, stretching, OK. I think, actually, I'm going to stop there. There is one topic that I just don't feel like I can get enough of it for it to be useful right now, so I think yes. I'm going to stop there for today. When we pick it up on Thursday we're going to wrap up this discussion of vectors-- again, I want to kind of emphasize, I can already see in some places you're getting the 1,000 yard stare. There's no question we're being excessively careful with some of these definitions. They are very straightforward, there's nothing challenging here. But there will be a payoff when we do get to times where the analyses, the geometries we're looking at, get pretty messed up. Having this formal foundation very carefully laid will help us significantly. So all right, I'm going to stop there for today, and we will pick it up on Thursday. |
MIT_8962_General_Relativity_Spring_2020 | 11_More_on_spacetime_curvature.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: So this is unfortunately a rather sad lecture, [LAUGHS] since this will be the last one that I'm giving to a live audience here, although any of you who are grad students, if you want to come and keep me company while I'm recording my next couple of lectures-- I'm actually not joking. It's really weird talking to an empty room, and so it might be kind of nice to at least have a face to react to. So I will be recording these, and making them available. And per the announcements that I sent out today, there are some additional assignments that I've begun posting. I've begun putting things like that up. But for the time being, ignore due dates. In fact, I would like to say, especially if you are getting ready to leave town, just ignore the assignments. There will be updates posted fairly soon. You know, [CHUCKLES] sometimes-- there's a saying that Martin Schmidt, our provost, said one time in respect to a particular initiative as it was unwinding, and I feel like it applies here: we're basically sewing together the parachute after we've jumped out of the airplane. [CHUCKLES] And so that's kind of where we are right now, and hopefully it'll be assembled before we hit the ground. But we're all kind of figuring out the way this goes. So there will be updates with respect to assignments and things like that, and you know, we're just trying to figure out-- I'll just be blunt-- how the hell we're going to actually pull this off. [LAUGHS] It's not curved enough. All right, so let me get back to where we were. I want to begin-- so what we worked with yesterday was we derived this mathematical object called the Riemann curvature tensor. And the way we discussed it was by thinking about some vector that I parallel transported around a closed loop. I chose a parallelogram because it allowed me to sort of nicely formulate the mathematics. So if I imagine a parallelogram who's got sides that are sort of close enough, that parallel is a well-defined term, and the sides are in two different directions, one's of length delta x in the lambda direction, one is a delta x in the sigma direction, when I take this vector all the way around the loop, I find the vector has changed. It's pointing in a different direction, and this describes the way in which it has sort of moved as I go around that loop. That object is-- even though it's created-- it's assembled from the Christoffel symbols, which are not tensor, but it's done in such a way that their nontensorialness cancels out, and this four-index object R is a tensor. As you can see, it involves terms like derivatives of the Christoffels and Christoffel times Christoffel. So that's where nonlinearity is going to enter. I want to sort of fairly quickly-- because I do want to get into some other stuff that we do with this-- talk a bit about the symmetry of this thing. So I discussed last time how you look at this, and naively, you think you've got-- in four dimensions, four dimensions, four [INAUDIBLE] states. It looks like you have 256 components. But in fact, there are quite a few symmetries associated with this which reduces it significantly. I kind of gave you the answer last time, but let me talk about where that comes from. I'm going to describe a way to see these symmetries, which there's a couple ways you can do it, and I'm going to pick the one that's essentially the simplest. It's perhaps not as-- there's a few ways to do it that a purist might prefer, but in the interest of time, I think this is sort of the best one. So the first thing is that just by inspecting this answer, you should see that this object is antisymmetric if you exchange the final two indices. OK, just look at that formula, switch the indices sigma and lambda, and you get a minus sign. Physically, that actually hopefully makes sense, because what that is basically telling you-- remember the way we did this, this operation of parallel transporting around a parallelogram. We sort of went along, say, the lambda direction, then the sigma, then the lambda, than the sigma. If I exchange those indices, it's kind of like I actually go in the opposite direction, OK? I go in the other direction first. I go on the sigma first, then the lambda. And so this corresponds to doing this operation, which I'll remind you is called a holonomy. It corresponds to reversing direction. OK? So that's a fairly easy one to say. The others are a little bit tougher, and like I said, what I'm going to do is follow kind of a shortcut to do this. One can do it using sort of the full glory of the Riemann tensor here, but life gets a little bit easier for us if we do the following. So what you want to do first is lower the first index. So what I'm going to do is look at R, alpha in the downstairs, mu lambda sigma. This is what I get when I contract, like so. And then what I'm going to do is I'm going to do all of my analysis that follows in a local Lorentz frame. When I go into the local Lorentz frame, the metric at a particular point is the metric of flat spacetime. It's the A to mu nu. My Christoffel symbols all vanish, but you have to be a little bit careful, because the derivatives of the Christoffel do not vanish, right? So one of the key points is that when you go into those local Lorentz frame, there is a little bit of curvature associated with it. So the derivatives do not-- OK, so the reason I'm doing this-- like I said, you actually could do the little counting exercise I'm about to go through with this whole thing. It's just messy, OK? And so this is sort of a quick way to see the way it all comes out. Because this is a tensor relationship, you are guaranteed that something you conclude in a convenient reference frame will hold in all of them, OK? So it's a nice way to do this. Those of you who are sort of, you know, particularly purist, knock yourselves out trying to figure out how to do this sort of in general. This is adequate for our purposes here. So when I go and I do this, here's what my all downstairs Riemann tensor turns into. I'm going to lose the Christoffel squared terms because Christoffel vanishes, and the only thing that is left are the terms that involve the derivative of the Christoffel, OK? Let's do something a little bit further. So let's now plug in the definition of the Christoffel, and we'll use the metric before we actually go into the frame, so that we can sort of get the form that we would get with this. I should have made myself be a little bit more careful here. So I'm going to put LLF on this, to make it clear that I'm doing this in a local Lorentz frame. So I do this in a local Lorentz frame, and what this turns into-- so there's a lot of algebra that I'm not writing out. This is an alpha, g sigma mu minus d sigma d mu g alpha lambda plus d sigma d alpha g lambda mu, OK? So notice, it is only the second derivative of the metric that is appearing here, OK? If I go into the local Lorentz frame, that's the one degree of freedom we were not able to transform away. Anything involving the metric in the first derivative dies in this particular frame. But this is good enough for us to begin to count up symmetries, and to see what things are going to look like. And so I don't have any great guidance for how to do this. This is one of those things where, literally, you just sort of stare at it for a little while, and you see what happens when you play around with exchanging various pairs of indices. So stare at this for a while. OK, so you can see right away-- just for counting purposes, let me write down the one that I argued was kind of obvious even in the full form. It's the one you get when you exchange the last pairs of indices. We'll call that symmetry one. Oh, bugger, thank you. [CHUCKLES] Yeah. [CHUCKLES] That's a good way to-- that's a very complicated way of writing zero, otherwise. [LAUGHING] All right, if you exchange the first ones, you also-- if you look at this formula, you can see when you exchange alpha and mu, you get a minus sign. So one that might be-- this one takes a little bit of creativity to sort of see it, where it comes out. Suppose what I do is I exchange the first two indices for the second two indices. Well, it turns out, if you do that, you just get the expression back. OK? So if I just swap alpha mu, make them my last two, make lambda sigma my first two, [CLICKS TONGUE] I get Riemann right back. So that's another symmetry. And then finally, someday, when I've got a little bit more bandwidth in my class to do this, I'd really like to sort of justify where you can derive this one a little bit more rigorously. But for now, it's sufficient just to sort of say, look at that formula, and you'll see that it's true. If you take-- and then you cyclically permute the second, third, and fourth indices, you get zero, OK? There is a variant of this particular symmetry, which is that if you take this and you antisymmetrize on these things, you get zero, OK? And I've got some words in the notes that, in the interest of time, I'm not going to go through them, but they are posted, demonstrating why those two are actually equivalent to one another. What it boils down to is, expand out this antisymmetrization-- and I'm going to do that again for a three-index object a little bit later in this lecture-- and then occasionally invoke one of these other ones, and you'll see that these two, four and four prime, are saying the same thing. So go through all of these different symmetries, and you'll find n to the 4 independent components goes over to n squared times n squared minus 1 over 12, OK? Not terribly hard to do that. It's a little exercise in combinatorics. And when you do that for n equals 20, as I mentioned, you end up with-- excuse me, not n equals 20. When you do this with n equals 4, you end up with 20. And just for completeness, let me actually write out that if you actually-- if you go into the local Lorentz frame, your spacetime metric is in fact minus 1, 1, 1, 1 on the diagonal. But you, of course, have these quadratic corrections, and one can in fact write them out explicitly. So that is what the time piece ends up looking like. There is an off-diagonal piece, which only enters at quadratic order. It looks like this, and move this down a little bit lower. That's what your space-based piece will look like. So this explicitly constructs the coordinate system used in a freely falling frame, including these second order corrections. So this particular form is known as Riemann normal coordinates. So if you are-- this is discussed in a little bit more detail in one of the optional readings in the textbook by Eric Poisson, A Relativist's Toolkit. Very nice discussion. All right, so now that we have the curvature tensor in hand, we really have essentially every tool that matters for 8.962, OK? There's a little bit more analysis we need to do with this guy, but we now have all the pieces. OK, there's no major new mathematical object or no object described in geometry that I need to build in order for us to make a relativistic theory of gravity. I do want to talk about the curvature tensor a little bit more, because there are a couple of properties about it that are very important for us. And I think we'll have just enough time today to sort of get to them. That will allow us to set up, and in the first lecture that I will record in an empty room, actually derive a relativistic gravity equation. So let me talk about a few variants of this curvature tensor. So first, suppose you take the trace on some pair of indices. So what I mean by the trace is, essentially, imagine all the indices are in the downstairs position, and I contract it with the metric, so that I'm summing over them. Well, first of all, you'll note that when you do this, if you were to take the trace on indices 1 and 2, they are antisymmetric. The metric is symmetric, so contracting metric with indices 1 and 2 is going to give me 0. If I can track symmetric with indices 3 and 4, I get 0, OK? So the only trace that makes sense is do it on either indices 1 and 3 or on indices 2 and 4. Let's do it on indices 1 and 3. So what I'm going to do is evaluate R, alpha mu alpha nu-- which if you like, you can write this as R alpha beta of R. Here, I'll turn it like this. Beta mu alpha nu-- we're going to define this as R with two indices, OK? Just little r and mu, nu. This actually shows up enough that it's given a name. This is called the Ricci curvature tensor. If you were to like, I said, if you do your trace on indices 1 and 2, you get 0. If you do it on 3 and 4, you get 0. If you do it on 2 and 4, it's exactly the same as doing it on 1 and 3, because of all these other symmetries. So in fact, this ends up, when you count up all your different symmetries, this is the only trace that is meaningful, OK? There are a few other pairs where you get a minus sign, but still the same thing. So this is the only trace that ends up being meaningful. This is actually, in fact, it's not too hard to show that this is symmetric. So even though, you know, Riemann had all these crazy antisymmetries and symmetries, this one is simpler, OK? One way to do this is just to take the exact expression that we wrote down for the Riemann tensor in terms of derivatives of Christoffel and Christoffel squared, and just write it out in this traced over form. So when you do this, you get this. This guy is symmetric on the bottom two indices. Now, this isn't obviously symmetric, but recall, a lecture or two ago, I worked through a couple of identities that involved the determinant of the metric. It turns out, if you go back and you look up these identities, you can rewrite this term as d nu of d mu of the log of j-- excuse me, log of square root of j. Symmetric when you exchange mu and nu. And then you get two terms that look like this. OK, this one, pretty obviously symmetric on exchange of mu and nu. When you do this one, you might think to yourself, hmm, that one doesn't quite look symmetric. But remember, alpha and beta are both dummy indices. And so in fact, when you exchange mu and nu, this last term just turns back into itself. So that's kind of interesting, because R mu nu-- the reason I went through that little exercise-- this is a symmetric tensor, and we're going to work in four dimensions of spacetime here. So symmetric 4 by 4 has 10 independent components. Riemann had 20. Somehow, this Ricci has-- in some sense, you can of it as having 10 of-- the 20 components associated with Riemann are encoded in this guy. Where are the other 10? I'm going to talk about that in just a moment or two. Before I do that, I will just note that we are going to want to also know about the trace of the Ricci tensor. So if you compute this, this is often just abbreviated R with no indices whatsoever. This is called the Ricci scalar or the curvature scalar. So there's another variant of curvature which we're not going to use very much, but I wanted to just talk about very, very briefly. The derivation of this is highly non-obvious, so let me just write it out, and I'm going to talk about it briefly. So suppose I define a four-index tensor, C alpha mu lambda sigma, to be the Riemann tensor-- minus-- so working in n dimensions-- 2 over n minus 2, g alpha-- I'm going to antisymmetrize here. So antisymmetrizing on lambda and sigma-- OK, so in the famous words of Rabi, who ordered that? So the way that this has been constructed, if you go through and you carefully look at the way this thing behaves under exchange of any pairs of indices, it has exactly the same symmetries as Riemann. But if you take the trace of this, turns out to be zero. In fact, when you go through, when you count up how many independent components it has, because it has no trace, it has 10 independent components. So this tensor has a name. It was first formulated by the mathematician Hermann Weyl. And so although it's a German name and it's not spelled that way, it is appropriately the "vile" tensor. It is a pretty vile thing to look at, but it plays an important role. So in keeping with the idea that it's got 10 independent components, Ricci has 10 independent components, Riemann has 20. Heuristically, you can sort of think and put big quotes around all these objects. This doesn't mean approximately. Let's just put it this way. This is, got all the same information as Ricci plus Weyl, OK? One can, in fact-- by bringing in appropriate powers in the metric and things like that, one can actually write down a real equation relating these. It's a bit of a mess. It's not that interesting. The key thing I want you to be aware of is that all of the curvature content of Riemann is sort of split into Ricci and Weyl. Looking ahead a little bit-- and because I'm going to be doing this in an empty room, I just want to make this point now-- we are soon going to see that Ricci is very closely related to sources of gravity. So we're going to-- when we formulate a theory of gravity, working with things, we're going to find that there's a tensor that is just a slight modification of the Ricci tensor, that is equal to the stress energy tensor that we use as our source. Stress energy tensor is a symmetric 4 by 4 object. Those 10 degrees of freedom in the stress energy tensor essentially determine the 10 curvature degrees of freedom encoded in Ricci. But what that tells you is that if you're in a region of vacuum, where there is no stress energy, there's no Ricci, OK? But there is curvature. We measure tides. We see gravitational effects. Weyl ends up being the quantity that describes behavior of gravity in vacuum regions. Ricci ends up very closely related to how it describes it in regions with matter. One reason I mentioned this-- so there's only-- [CHUCKLES] only one LIGO student here now. Hi, Sylvia. [CHUCKLES] But in fact, when one is describing gravitational radiation, the behavior of the Weyl tensor ends up being very important for characterizing the degrees of freedom associated with radiation and general relativity. So that's a little bit ahead. There's a few other things you can do with it which are related to what are called conformal transformations. I have a few notes on them, but they're not that important for our class. It's discussed a little bit in Carroll, so I'll leave that as a reading if you are interested, but we don't need to go through it here. So there are two other things that are much more important for us to discuss with respect to Riemann first, and I'm going to focus the time that we have on them. So one of the really important aspects of curvature that I've emphasized a few times now is the idea that initially parallel geodesics become no longer parallel when they're moving on a manifold that is curved. We're going to use the Riemann curvature tensor to quantify what the breakdown of parallelism actually means. There's a couple of different discussions of this that you'll see in various places. Carroll's discussion is very brief. It's rigorous. It's very brief. I'm doing something that's a little bit-- to my mind, it's a little bit more physically motivated OK so I'm going to do it in a slightly different way from the way it's done in Carroll. It's a little close to the way it's done in Schutz's textbook. So what we want to do is imagine we have initially parallel geodesics, and what we want to do is characterize how they become nonparallel. Well, we'll put it this way, how they deviate as one moves along their world lines-- around these world lines. OK, so here's what I want to do. That's the idea of the calculation that I want to do. So what I want to start out with is-- let's consider two nearby geodesics. And what I'm going to mean by nearby is that they are close enough that they're essentially in the same local Lorentz frame, OK? So they're going to have the same metric. I'm going to be able to choose coordinates such that the Christoffel symbol is zero for both of them. I will not be able to get rid of the second derivative, though, OK? So that will be where a bit of a difference begins to enter. Two nearby geodesics-- and I'm going to use just lambda as the affine parameter along them. So here's my first one. I'm going to make a few definitions on the next board. I'm going to call this gamma sub v. And here's my next one. I'm going to call it gamma sub u. Define two points here. OK, I'm going to make a couple of definitions on the next board. OK, so definitions-- I'm going to call u the tangent vector to the curve gamma sub u. OK? So this is equal to dx d lambda on that curve. v is the tangent vector to gamma sub v. The point A is at lambda 0 on curve gamma sub u. A prime is at lambda sub 0 on gamma v. So they're both parameterized by a parameter lambda, and I'm going to set the-- they're basically both synchronizing their clocks at the same time. Like, at those points, they're going to find their starting points as A and A prime. What I'm going to do now is I'm going to define what's called a geodesic displacement factor that points from lambda on the u curve to lambda on the v curve. And I'm going to use my favorite Greek letter, xi. So this points from lambda on gamma u to lambda that is basically to the same value. Let me make this a little bit more precise. Points from the event at lambda on gamma u to the event at lambda on gamma v-- OK? Apologies for being somewhat didactic there, but we need to define things carefully. So on this-- so here's my initial-- so that's what xi looks like at parameter lambda 0. And what we want to do is examine how xi evolves as one moves along these geodesics. That's going to be our goal. So let's make things a little bit more quantitative here. Xi is going to be equal to x gamma v at lambda minus x on gamma u at lambda. Finally, I'm going to assume that the curves begin parallel to each other. That's a statement that u of lambda 0 equals v of lambda 0. And it also tells me that I can use-- that this must equal 0 at the initial point. Not going to equal 0 everywhere. In fact, it won't. But I want to use this as a boundary condition in the calculation I'm about to do. OK. So what we're going to do is essentially say, you know, as we move along these two curves, these are geodesics. We're going to use the geodesic equation to slide along these two curves. We know what the equation is that governs u as it moves along gamma u. We know what the equation is that governs v it moves along gamma v. We're just going to take the difference between them, and we're going to kind of use that to develop an acceleration equation that governs that displacement xi. Here's the bit where I differ a little bit from Carroll. So Carroll, like I said, Carroll has a very brief discussion, which is-- it's absolutely right. But I want to give a little bit of physical insight, and I think you can do that by choosing to work in the local Lorentz frame. So we're going to work in the local Lorentz frame, and I'm going to center this local Lorentz frame on the event A. The reason why I'm doing that is this then allows me to say, g mu nu at the event is A mu nu. Pardon me a second. My Christoffel symbols at event A are all going to be zero. g mu nu at A prime is also going to be A mu nu. We have to be a little bit careful. Our Christoffel symbols do not vanish at point A prime because there's a little bit of curvature. They are in fact related to the fact that I'm sort of slightly far away, and I'm picking up that second order correction, OK? This is sort of where all the important bits of the analysis are going to come from. This is the fact that they're close to each other. I can set up a local Lorentz frame, but it's got that little bit of sort of schmutz at second order that's coming in, and kind of pushing me away from a simple local Lorentz frame there-- the simple mathematical form right there. OK, let's put that up. OK. So let's look at the equations that govern motion along these two curves. So the geodesic equation along curve gamma u-- and we'll just look at it at A. It's a second order-- I'm going to write it in terms of the coordinates. It looks like this. We're evaluating this at A. At A, my Christoffel symbols are all zero, so this is just zero. Let's look at it along the other curve, at A prime. OK? So in just a second, I am going to-- wait a second. I'm going to substitute in that derivative of Christoffel that's going to go in there. Before I do that-- so notice, this also depends on the velocity as I move along at that point, right? Typical geodesic-- So I'm going to substitute in the fact that both the x mu and the x nu velocity here, this is defined as v mu. But remember, these guys are defined as being initially parallel. OK? So what this tells me is-- OK, that's an alpha. So this is interesting. What I'm seeing is I end up with an equation where the derivative of the Christoffel symbol is coupling to sort of my motion along that world line and the displacement. Now, let me just remind you, what I really wanted to do was get an equation that governs how this guy changes, OK? But this guy is just defined as the difference between the position along curve v minus the position along curve u. So if I take the difference of these two things-- pardon me, forgot to label which curve this one is on. So I'm doing this at A prime, doing this one at A. This is an equation governing the acceleration of this displacement. OK? So this is interesting, OK? So what you're sort of seeing is that the geodesic displacement looks something like two derivatives of the metric coupling in the forward velocity and the displacement itself. Now, as written, this equation is fine if all you are doing is living life in the local Lorentz frame, OK? We want to do a little better than that, though. So I'm going to do a little bit more massaging of this, but I kind of want to emphasize that this already brings out the key physical point, OK? I can't yet-- you know, you're sort of looking at this, and you're thinking to yourself, that looks like a piece of Riemann, but it ain't Riemann, right? It's a derivative of a Christoffel symbol, and that's not a tensor. So this isn't quite the kind of thing we want yet. We need to do a little bit more work. And so what I would sort of say is, everything I did up to here, this is like the key important physics. Now I'm going to do a little bit of stuff to sort of put the suit and tie that a tensor is supposed to wear. I'm going to dress it up a little bit, so that it's wearing the clothes that all quantities in this class are supposed to wear. So we want to make this tensorial. And so for guidance of that, these time derivatives or derivatives with respect to the parameter, it's not really nicely formulated, OK? The thing which we should note is, d by d lambda, that's what I get when I contract the forward velocity with a partial derivative. What we should do is replace this with something like what I'm going to call capital D by d lambda, which is what I'm going to get when I do derivatives using-- when I use forward velocity contracted on a covariant derivative, OK? So-- Suppose I just-- I'm agnostic about what xi actually is. I just know it's a vector field, and I want to compute this thing. Well, we learned how to do that a couple lectures ago. I'm going to work this guy out. And-- [EXCLAIMS] I couple in a term that looks like this. You might be tempted at this point to go, oh, I'm in a local Lorentz frame. I can get rid of that Christoffel. Don't do that quite yet, because we're going to want to take one more derivative. Christoffel does vanish, but its derivative does not. Wait till you've done all of your derivatives before you insert that relationship. Just getting another piece of chalk. So my covariant derivative along the trajectory with my parameter is the usual total derivative plus-- that, OK? Now, I'm going to want to take one more derivative. OK? So there's a lot of junk in here. I'm going to get one term. It just looks like this. But now I'm going to get a whole bunch of other terms that involve-- so this is what I get when I expand this guy out, and I just have-- well, I should hang on a second. So first, I'm going to get one that involves essentially the first term, u on the partial, hitting both of these terms. OK, so when I do that-- pardon me just one moment. OK, never mind. Let me just write it down, and I'll describe where it comes from. OK, so this first term, this is basically the connection term coupling to this, OK? Associated with this covariant derivative. Now I'm going to have a whole bunch of terms that involve this guy operating on these terms over here. And my apologies that the board is not clearing as well as I would like today. And you know what? I'm going to just write it out like this, and move to a different board in a second. OK? All right, so I'm going to put this over on the other side here. What I have there-- so the first term, like I said, I'm basically just operating that thing on the two different terms. So it's a little easy when I first hit it on the d xi d lambda, and it's going to be messy when I expand out all the derivatives that operate on here, which is why I'm writing it with some care, because we're about to make a lot of mess on the board. OK. OK, so first, I'm going to do what happens when this guy hits the Christoffel symbol. Then I'm going to get a term that involves this combination of guys acting on the four velocity. And then I get these guys acting on my displacement. OK, [BLOWS AIR] now I'm ready to simplify. OK, so let me just emphasize, everything I have here, all I did was expand out those derivatives and write them in a form where I want to basically call out different terms and see how they behave. Two things to bear in mind, I am going to do this in the local Lorenz frame and in the vicinity of the point A. That's going to allow me to get rid of a lot of terms. I'll use the-- here's an open box of chalk. OK, so earlier, I didn't want to get rid of my Christoffel symbols because I was going to take another derivative. I'm done with that. So anything that's just a Christoffel on its own, I go into local Lorentz form-- Christoffel, you die! Die! Can't really do much with this. Let's set this aside. Let's look at this guy. What do we know about the forward vector u? It's a geodesic. It obeys the geodesic equation. The geodesic equation is, this thing in parentheses equals 0. You die! Finally, I have this condition here at the beginning. That is essentially the velocity at point A minus the velocity at point B. Our initial condition is that these things start out parallel to each other. These die because these are initially parallel. So initially parallel geodesic-- local Lorentz frame-- So the only derivative I'm going to need to expand out is this guy, and because I'm working in a local Lorentz frame, that's actually not that bad. So what I finally get after all the smoke clears is that my covariant acceleration of this displacement vector is related to my noncovariant acceleration, with a term that basically ends up just being-- oops-- a derivative of the Christoffel symbol. It couples in two powers of the four velocity and the displacement. But we actually already have-- we worked out earlier what this guy was, OK? So my noncovariant acceleration was a different derivative of the Christoffel symbol. So if I go and I plug this in-- My apologies for pausing here. There are a lot of indices on this page. Look at this, I've got a derivative of Christoffel minus a derivative of a Christoffel. That is exactly what the Riemann curvature tensor looks like in the local Lorentz frame. Now, if you actually go back and you look up your definitions, you'll see it's not quite right. There's a couple of indices that are sort of a little bit off, but they turn out to all be dummy indices. So what you should do at this point is just relabel a couple of your dummy indices. So if on the second term, you take beta to mu, mu to gamma, and nu to beta, what you finally wind up with is-- We'll write it first in the local Lorentz frame. This is nothing but the Riemann tensor in the local Lorentz frame. And from the principle that a tensor equation that holds in one frame must hold in all, we can deduce from this that this is the way the displacement behaves. So I start out with two geodesics that are perfectly parallel to one another. If they are in a spacetime that is curved, the Riemann curvature tensor tells how those initially parallel trajectories evolve as I move along those geodesics. So I want to make two remarks. One is just sort of a way of thinking about the calculation I just did. It may not-- you might be slightly dissatisfied with the fact that I decided to do the whole calculation in a whole special frame. If that's-- you would prefer to see something a little bit more rigorous, feel free to explore some of the other textbooks that we have for this class. They do treat this a little bit more rigorously than this. One reason why I wanted to do this is I wanted to really emphasize this idea that where this effect comes from is that if I think about the freely falling frame describing two nearby geodesics, it is that derivative, it's that secondary derivative of the metric, that really plays a fundamental role in driving these two things apart. What we get out of it is a completely frame-independent equation, and this ends up really being the key equation describing the behavior of tides in general relativity. We're going to-- when you go into a freely falling frame, we've basically at that point said there's no longer such a thing as gravitational acceleration. And if there's no gravitational acceleration that's coming, well, what the hell does gravity do? General relativity, tides become the key thing that really defines what the action of gravity is. One of the ones I am personally very interested in applying this to is that this equation gives you a very rigorous and frame-independent way to describe how a gravitational wave detector responds to the impact of a gravitational wave. If you imagine you have two test masses that are moving through spacetime, they just follow geodesics, OK? They feel gravity. They are free falling. They don't do anything. If you look at them, if you're falling with them, you don't see any effect whatsoever. But if they're sufficiently far away from one another that their displacement kind of takes the curvature to that freely falling frame, then this equation governs how the separation between the two of them actually evolves with time. And so for instance, if you're a student who works in LIGO, where some of the many important analyses describing how gravitational wave detectors respond to an [INAUDIBLE] radiation field, they are based on this equation. If you're interested in understanding how an extended object is tidally distorted due to the fact that it's sort of big enough that it doesn't all sort of live at a single point, but its shape spills across the local Lorentz frame, this equation governs-- it allows you to set up some of the stresses that act upon that body and change it from a simple point mass. So I'm waxing slightly rhapsodic here, because this is one of the more important physical equations that come out of the class at this point. All right, in our last 12 or 13 minutes, I would like to do one other little identity which is extremely important, but rather messy. In fact, I think I'm going to just-- if I have the time-- yeah, you know what? I think I will have time. I'm going to set up the first page of the next lecture. So let's move away from-- oh, by the way, I forgot to give the name of this. This is the equation of geodesic deviation. OK, so recall when we first talked about the Riemann tensor. There's a very mathematical action that the Riemann tensor does. You can think of it as what happens when you have a commutator of covariant derivatives acting on a vector. So if I evaluate the commentator, sigma lambda-- excuse me-- covariant of lambda covariant with sigma acting on v alpha, OK, it ends up looking like this. And I argue that this definition is kind of the differential version of that little holonomy operation by which I derived Riemann in the first place. A more generalized form of this is, suppose I act on some object with two indices, one in the upstairs, one in the downstairs. Oops. OK? So these are just some reminders of a couple definitions. I want to use this. I'm going to apply these things to two relationships that I'm going to write down, and I'm going to do something kind of crazy, then I'm going to do something a little bit crazier, and something kind of cool is going to emerge from this. And this is something that deserves a clean board. So I'm going to write down relationship A. That's what I get when I do the commutator of the alpha beta derivatives on the gamma derivative of some one form, OK? Now, if I apply those little definitions I worked out, this is Riemann mu gamma alpha beta-- OK? So this is relationship A. Relationship B, I'm going to take the covariant derivative along alpha, of the beta gamma commutator on p delta, OK? So notice the difference here. One, I'm doing the commutator acting on this particular derivative. Next one, I'm doing the derivative of commutator acting on this. And so I'm going to skip a line or two that are in my notes, but this ends up turning into minus p mu-- beta alpha delta minus R mu and delta gamma. OK? So the way I went to get this line, so the two lines that I skipped over, one is I just straightforwardly applied the commutator derivative, and then took a derivative. And then I took advantage of the fact that I can commute the metric with covariant derivatives, and sort of raising the next, move things-- move an index up on one side and down on the other, and take advantage of some symmetries of the Riemann tensor to slough my indices around. OK, so this board's kind of a disaster, so I'm going to go to a cleaner board here for what I want to do next. I'm just going to clean this so you're not distracted by its content. So two equations, what the hell do they have to do with each other? Here is where I'm going to do something which you might-- let's do the other board. I'm going through something that you might legitimately think is crazy. What I want to do is take equation A-- actually, I'm going to look at both of these equations. And I am going to antisymmetrize on the indices alpha, beta, and gamma. OK? So just bear with me a second. So the way I'm going to do this-- here's what I'm doing now. Here's my commutator. So if I look at equation A-- OK? So I need to expand this guy out. Let me just write out a step of this analysis. The way you do this is you add up every even permutation of those indices. That's a beta. And then you subtract off every odd permutation of these indices. Oh, and don't forget-- exactly on that one form, OK? So now we can-- if you expand these guys out, and gather terms exactly correctly, it's not hard to show that this turns out to be what you get if you just slide those commutators over by one. OK. You know, if you're really feeling motivated, just try it, OK? What this is is, the right hand side-- sorry, this is the left hand side-- we'll put this way. So when I apply this to the left hand side of equation A, what emerges is the antisymmetrized left hand side of equation B. So when I antisymmetrize on the indices alpha, beta, gamma, these two relationships become-- they say the same thing. So that means the right hand side must be the same as well. So let's apply this and see what happens if I require the antisymmetrized right hand side of A to equal the antisymmetrized right hand side of B. OK, so first, if I remind myself how I order these equations-- OK, yeah. So let's do-- here's A, and here is B. OK, so a few things to notice. This term, we actually showed when we looked at the different symmetries of the Riemann tensor. This is-- I didn't actually show this, but it's in the notes and I stated it. This is one of those symmetries. If I take this thing, and I just add up what I get when I permute the three final indices, I get zero. And that's equivalent to antisymmetrizing on these things. So this is zero by Riemann symmetry. This term here and this term here are exactly the same, OK? Because the only indices that are in a slightly different order are alpha, beta, and gamma. This one goes alpha, beta, gamma. Where I wrote it here, it just became beta, gamma, alpha. That's a cyclic permutation I've antisymmetrized. They are exactly the same. So these are on the same side-- or on opposite sides of the equation. So they cancel each other out. And so what arises from all this is p mu-- pardon me, I missed the derivative. I've set no properties on p, so the only way this can always hold is if in fact, this is equal to zero. This is a result known as the Bianchi identity. Let me just write another form of it, and then I'm fairly pressed for time, so I think I'm just going to write down the result of the next thing, and I will put the details of this into the first prerecorded lecture. If you expand out that antisymmetry and take advantage of the Riemann symmetry-- the various Riemann symmetry relationships, this is an equivalent form. So these two things both are an important geometric relationship the curvature tensor must go by. Now, in some notes that I am going to-- well, they're actually already scanned and on the web. What I do is take the Bianchi identity and contract on some of the indices to convert my Riemann tensor-- my Riemann curvature, into a Ricci tensor-- Ricci curvature. So in particular, if I use this second form of it, it's just covariant derivatives, right? And the metric commutes with covariant derivatives, so I can just sort of walk it through. What you find when you do this is that you can write this relationship in the following form, and the derivation of this will be scanned and posted, and I will step through it in the first recorded lecture. The divergence of a particular combination of the Ricci curvature and the Ricci scalar is equal to zero. This is a sufficiently interesting tensor that it is now given a name. We write this G, and we call this the Einstein curvature tensor. I sort of mentioned that one of our governing principles here is, we're going to change the source of gravitation from just matter density to the stress energy tensor. That is a two-index divergence-free tensor. We want our gravitational-- the left hand side of the equation to look like two derivatives in the metric, which is a curvature. So we need a divergence-free two-index curvature. Ta-da! It's got Einstein's name on it, so you know it's got to be good. All right, so that is where, unfortunately, the COVID virus is requiring us to stop for the semester, at least as far as in-person goes. So look for notes to be posted, videos to be posted, and I, of course, will be in contact, behind a sick wall or something. But anyway, [CHUCKLES] certainly, I am not going out of contact. And so those of you who are scattering to parts unknown-- known to you, that is, unknown to me-- good luck with your travels and getting yourselves settled. I really feel-- everyone feels terrible that you're going through this crap right now, but hopefully, this will flatten the curve, as they say, and it will be the right thing to do. But stay in touch, OK? This campus is going to be weird and sad without the students here, and so we want to hear from you. |
MIT_8962_General_Relativity_Spring_2020 | 8_Lie_transport_Killing_vectors_tensor_densities.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: Last Thursday, we began the work of moving from special relativity to general relativity, and we spent a lot of time unpacking two formulations of the principle of equivalence. So one, which goes under the name "weak equivalence principle"-- a simpler way of saying that is that, at least over a sufficiently small region, if there is nothing but gravity acting, I cannot distinguish between freefall under the influence of gravity or a uniform acceleration. The two things are equivalent to one another. Basically, this is a reflection of the fact that the gravitational charge and the inertial mass are the same thing. That is the main thing that really underlies the weak equivalence principle. When I gave my tenure talk a number of years ago here, I pointed out that there was this wonderful program called the Apollo program that was put together to test this. And the way it was done was that they put astronauts on the moon, and you actually show that, if you drop a hammer and a feather on the moon, they fall at the same rate. Of course, the Apollo program probably did a few other things as well. But I'm a general relativity theorist, so for me, that was the outcome of the Apollo program, was test of the equivalence principle. We also have a different variation of this we called the Einstein equivalence principle, which leads us to a calculation that we went through last time, which states that we can find a representation over a sufficiently small region of spacetime such that the laws of physics are reduced to those of special relativity. And we did a calculation to examine this, where we showed, given an arbitrary spacetime metric, I can find a coordinate system such that this can be written in the form metric of flat spacetime plus terms that are of order so we have coordinate distance squared. Let's put it this way-- so this, in the vicinity of a point pl-- I'll make that clear in just a moment. It ends up looking like of order of coordinate displacement squared with corrections that scale as 1 over second derivative of the metric. That sets the scale for what these end up looking like, or maybe it's actually times that. Sorry. It's times that. Why did I divide? I don't know. Oh, I know why, because I wanted to point out that that is what the scale of 1 over this thing-- for God's sake, Scott! Stop putting your square roots in there! So it looks like that. And so this is what I was saying. You have a curvature scale that is on the order of square root of 1 over the second derivative of the metric. Apologies for botching that as I was writing it up there quickly. OK, so we did a calculation that shows that. And indeed, what we did is we went through and we showed that a general coordinate transformation has more than enough degrees of freedom to make the metric flat to get the flat spacetime metric at a particular point. And in fact, there are six degrees of freedom left over, corresponding to six rotations and six boosts that are allowable at that particular point or event. Bear in mind we're working in spacetime. We have exactly enough degrees of freedom to cancel out the first order term, but we cannot cancel out the second order term. And in fact, we find there are 20 degrees of freedom left over. And in a future lecture, we will derive a geometric object that characterizes the curvature that has indeed 20 degrees of freedom in it or 20 independent components that come out of it. So this is the foundation of where we're moving forward. And so what this basically tells us is that, in a spacetime like this, we have what we call curvature. Trajectories that start out parallel to one another are not going to remain parallel as they move forward. And where we concluded last time, we were dealing with the problem that, if I want to take derivatives-- and we're going to start with vector field. If I want to differentiate a vector field on a curved manifold, it doesn't work. If I do the naive thing of just taking a partial derivative, it does not work. So I'll remind you where we left off. So we found partial derivatives of vectors. Let me put that in there. Partial derivatives of vectors, and it won't take much work to show it's true for one forms or any tensor, so let's say, partial derivatives of tensors do not yield tensors. We're familiar with this to some extent because we already encountered this when we began thinking about the behavior of, even in flat spacetime, flat spacetime and curvilinear coordinates. So that's the fact that the basis objects themselves have some functional dependence associated with them. We're interpreting it a little bit differently now. And so what we're doing is we're going to say that what's going on here is that, on my curved manifold-- and I want you to visualize something that's like a bumpy surface or, if you like, maybe a sphere. Think about if you are a two-dimensional being confined to the surface of a sphere. There's only two directions at any given point. You can go up, or you can go on the left-right axis or the north-south axis, right? And you would define unit vectors pointing along there. But as you move around that sphere, those of us who have a three-dimensional life, and can step back and see this, we see that these basis objects point in different directions at different locations on the sphere. The two-dimensional beings aren't aware of that. They just know that they're on a surface that's curved. And so they would say that the tangent space is different for all these objects. They've imagined that every one of these basis objects lives in a plane that is tangent to the sphere at any given point, and that plane is different at every point along the sphere. So we interpret this by saying that all of our basis objects live in this tangent space, and the tangent space is different at every point on the surface. So let me just write out one equation here. So when we looked at the transformation of a partial derivative of a vector, if we looked at just the components-- so this calculation is in the notes, so I would just write down the result. What we found was that, if I'm taking, say, the beta derivative of component A alpha, what I found when I want to go into a coordinate transformation is that there's an extra term that ruins the tensoriality of this. So this goes over to something that looks like-- So the first term is what we'd expect if this were a tensor relationship. That's exactly the matrix of the Jacobian between two different coordinate representations that we expect to describe how components change if they indeed obey a tensorial relationship. This extra thing here I wrote on the second line-- that is spoiling it for us. So I began to give you the physical notion of what we were going to do to fix this. So let me just reset that up again. So let's imagine I have a particular curve that goes along my manifold. I have a point P here on the curve, and I have a point Q over here. And let's say that P is at event x alpha. Q is at x alpha plus dx alpha. Here's my vector A at the event Q. And here's my vector A at the event P. So what we discussed last time is that, to get this thing, I'm just doing the normal partial derivative. I'm basically imagining that these are close to one another, that I can just subtract A at Q from A at P, divide by dx and take the limit. That's the definition of a derivative. And what this is telling us is, mathematically, yeah, it's a derivative, but it's not a derivative that yields a tensor quantity. And so we are beginning to discuss the fact that, to compare things that have different tangent spaces, that live in different points in my curve manifold, I need a notion of transport to take one from the other in order to compare them. So there are two notions of transport that we're going to talk about here. The first one is called parallel transport. So transport notion one-- we call this parallel transport. I'm going to actually focus a little bit on the math first and then come back to what is parallel about this afterwards. So what I essentially need to do is say, what I want to do is find some kind of a way of imagining that I take the vector at P and transport it over to the point Q, and I will compare the transported object rather than the object originally at point P. Abstractly, what I'm essentially going to do is I'm going to do what we always do in this. I'm going to imagine that there is some kind of an operation that is linear in the separation between the two of them that allows me to define this transport. So let's do the following. So I'm going to assume that we can define an object, which I will call Pi, capital Pi, alpha, beta, mu, which is going to do the following. So what I'm going to do is say that A alpha transported-- and to make it even clearer, how it's being transported. Let's say it's being transported from P to Q. This is given by alpha at P, and what I'm going to do is say that, whatever this object is, it is linear in both the coordinate separation of those two events and the vector field. So far, I've said nothing about physics, by the way. I'm just laying out some mathematical definitions. I'm going to bring it all together in a few moments. I'm then going to say, OK, I know that I had trouble with my standard partial derivative. Let's define a derivative operator in the following way. Define a derivative by comparing the transported vector to the field at Q. So what I'm going to do for the moment is just denote this notion of a new kind of derivative with a capital D. Right now, it's just another symbol that we would pronounce with a "duh" sound so that it sounds like derivative. So don't read too much into that for a moment. So I'm going to define this as A at Q minus A transported from P to Q and then divide by the separation. Take the limit-- usual thing. And when you do this, you're going to get something that looks like the partial derivative plus an additional term on here. It looks like this. So I've said nothing about what properties I'm going to demand of this thing. And in fact, there are many ways that one could define a transport of an object like this. In general, when you do this, this thing I'm calling Pi here is known as the connection. It is the object that connects point P to point Q. So let's make a couple demands on its properties. So now I'll start to put a bit of physics in this. So demand one is I'm going to require that, when I change my coordinate representation, when I evaluate this in my new coordinate system, that I get something that looks like this. If I do this, I'm going to find that, when I change coordinates and apply it to the entire derivative I've defined over here, that little extra bit of schmutz that's on the second line there is exactly what you need to cancel out this annoying bugger so that you have a nice tensor relationship left. So I'm going to demand that a key part of whatever this guy turns out to be is something that cancels out the irritating garbage that came along with the partial derivative. Combine that with what I'm about to say in just a moment-- that pins things down significantly. I'm going to make one more demand, and this demand is going to then be connected to the physical picture that I'm going to introduce in about five minutes. My final demand is I'm going to require that whatever this derivative is-- when I apply it to the metric, I get 0. If I do that, then it turns out that the connection is exactly the Christoffel symbol that we worked out earlier. If I put in this demand, the connection is the Christoffel, and this derivative is nothing more than the covariant derivative. Der-iv-a-tive. This is just the covariant derivative we worked out earlier. So much for mathematics. The key thing which I want to emphasize at this point in the conversation-- I hope we can see the logic behind the first demand. That's just something which I'm going to introduce in order to clean this guy up. This is my choice. Not just my choice-- it's the choice of a lot of people who have helped to develop this subject. But it's a good one, because, when I do this, there is a particularly good physical interpretation to what this notion of transport means. And bear with me for a second while I gather my notes together. So let's do the top one first. OK, so let's make that curve again. And actually, let's just go ahead and redraw my vector field. I do want to have a different copy of this. Let me introduce one other object. So I'm going to give this curve a name. I'm going to call this curve gamma. And this is a notion that I'm going to make much more precise in a future lecture, but imagine that there is some kind of a tape measure that reads out along the curve gamma that is uniformly ticked in a way that we will make precise in a future lecture. I will call the parameter that is uniform that denotes these uniform tick marks-- I will call that lambda. If you want to read a little bit about this, this is what's known as an affine parameter. We will make this a little bit more precise very soon. With this in mind, you should be able to convince yourself that this field U defines the tangent vector to this curve. So let's say that my original point here, that this is at lambda equals 2-- and let's say this is at lambda equals 7. And what I want to do is transport the vector from 2 to 7. Well, a way that I can do this is by saying, OK, I now know that the derivative that goes over here-- this is just the covariant derivative. So I forgot to write this down but from now on, this derivative I wrote down-- I'm going to go back to the gradient symbol we used for the covariant derivative. What I can do is take the covariant derivative of my vector field, contract it with this tangent vector. And what this does-- I'm going to define this as capital D alpha D lambda. This is a covariant derivative with respect to the parameter lambda as I move along this constrained trajectory. What this does is this tells me how A changes as it is transported along the curve. I'm now going to argue-- what parallel transport comes down to is when you require that, as you slide this thing along here, the covariant total derivative of this thing as you move along the curve, that it's equal to 0. OK, now let me motivate where that's coming from. Why is that? So let's put all of our definitions back in. Let's apply the definition of covariant derivative that we discussed in a previous lecture and we've all come to know and love. Now, this is the bit where we begin to introduce a little bit of physics. Let's imagine that points P and Q are sufficiently close to each other that they fit within a single, freely falling frame. They can go into the same local Lorentz frame. So remember, when I go into my local Lorentz frame, the metric becomes the metric of flat spacetime. G goes over to eta. There is a second order correction to that. So the second derivatives-- I cannot find a [INAUDIBLE] that gets rid of them. So there'll be a little bit of that there, which tells me how large this Lorentz frame can be. But I can get rid of all the first derivatives. And if you get rid of all the first derivatives, you zero out the Christoffel symbols in that frame. So in that frame, there are no Christoffel symbols. So what this means is that, in this frame, this just becomes the idea that a simple total derivative of this object-- you don't need to include all the garbage that comes along with the covariant derivative-- this is equal to 0. And that's equivalent to saying that, as I take this vector and transport it along, I hold all of its components constant as I slide it from one step to the other. So I start out with my A at point P here. And then I slide it over a little bit, holding all the components constant just like this-- slide over again, slide it over again. Da, da, da, da da. Da, da, da, da, da. Till finally, I get it over to there. What this is doing is that-- any two vectors in the middle of this transport process-- I am holding them as parallel as it is possible to hold them, given that, to be blunt, you can't even really define a notion of two objects being parallel on a curved surface if there's a macroscopic separation between them. But if you think about just a little region that's sufficiently small, that it's flat up to quadratic corrections, then a notion of these things being parallel to each other makes sense. And this idea, that I'm going to demand that, upon transport, the derivative of the metric equals 0, thereby yielding my connection being the Christoffel and this derivative being the covariant derivative-- it tells us that this notion of transport is one in which objects are just kept as parallel as possible as they slide along here. And that is why this is called "parallel transport." It's as parallel as it's possible to be, given the curvature. So as I switch gears, I just want to emphasize I went through the mathematics of that with a fair amount of care because it is important to keep this stuff as rigorous as possible here. This notion of parallel transport gets used a lot when we start talking about things moving around in a curved spacetime. In particular, if you think about an object that is freely falling, and it is experiencing no forces other than gravity, which we are going to no longer regard as a force before too long-- if you just go back to Newtonian intuition, what does it do? Well, you give it an initial velocity or initial momentum, and it maintains it. It just continues going in a straight line. In spacetime, going in a straight line basically means, at every step, I move and I take the tangent to my world line, my four-velocity, and I move it parallel to itself. So this notion of parallel transport is going to be the key thing that we use to actually define the kinetics of bodies in curved spacetime. There's a tremendous amount of work being done by all sorts of things these days that's based on studies of orbits in curved spacetime, and they all come back to this notion of parallel transport. All right. Bear with me a second. I just want to take a sip of water. And then I'm going to talk about the other notion of transport, which we are going to discuss-- I always say, briefly, and then I spend three pages on it, so we're going to discuss. All right. So parallel transport is extremely important, and there's a huge amount of physics that is tied up in this, but one thing which I really want to emphasize is that it is not unique. And there is one other one which we are going to really use to define one particularly important notion, instead of quantities, for our class. So suppose I've got my curve gamma, and I'm going to, again, take advantage of the fact that I can define a set of tick marks along it and make that vector U be the tangent to this curve. And I'm going to, again, have my favorite points, x alpha at point P plus dx alpha at point Q. There's another notion of transport that is-- basically what you do is you cheat, and you imagine that moving from point x alpha to x alpha plus dx alpha is a kind of coordinate transformation. So let's do the following. Let's say that x alpha plus dx alpha-- we're going to take advantage of the fact that, since we have this tangent notion built into the symbols we've defined, we'll just say that it's going to be the tangent times the interval of lambda. And what I'm going to do is define this as a new coordinate system, x prime. So that's the alpha component of coordinate system x prime. It's a little bit weird because your x prime has a differential built into it. Just bear with me. So what we're going to do is regard the shift, or the transport, if you prefer, from P to Q as a coordinate transformation. It's the best eraser, so I'll just keep using it. So what I mean by that is I'm going to regard x alpha, and I'm going to use a slightly different symbol. I will define what the L is. So this is transported, but I'm going to put an L in here for reasons that I will define in just a moment. This, from P to Q, is what I get if I regard the change from point P to point Q as a simple coordinate transformation and do my usual rule for changing coordinate representation. So expand what the definition of x prime is there, and what you'll see is that you get a term that's just basically dx alpha dx beta. Then you're going to get something that looks like the partial derivative of that tangent vector. And remember, this is being acted on. I should've said this is this thing evaluated at P. Great. So we fill this out. OK. So that's what I get when I use this notion to transport the field from P to Q. Let's think about it in another way. Now, these fields are all just functions. So I can also express the field at Q in terms of the field at P using a Taylor expansion. I'm assuming that these are close enough that everything is accurate to first order in small quantities, so nothing controversial about this. I'm assuming dx is small enough that I can do this. But now I'm going to get rid of my dx beta using the tangent field U. Now, before I move on, I just want to emphasize-- these two boards here, over the way the left-- we're talking about two rather different quantities. The one I just moved to the top-- that actually is the field-- if you were some kind of a gadget that managed your field A-- that would tell you what the value is that you measure at point Q. This would tell you-- what do we get if you picked P up and, via this transport mechanism, moved it over to Q? They are two potentially different things. So this motivates another kind of derivative. So suppose I look at A-- value it at a Q-- minus A transported-- whoops, that's supposed to be transported from P to Q-- defined by D lambda. I will expand this out in just a moment. Now I will, at last, give this a name. This is written with a script L. This is known as the Lie derivative of the vector A along U. Anyone heard of the Lie derivative before? Yeah. So at least in the context where we're going to be using it, this is a good way to understand what's going on with it. We'll see how it is used, at least in 8.962 in just a few moments. Filling in the details-- so plug in these definitions, subtract, take limits, blah, blah, blah. What you find is that this turns out to be U contracted on the partial derivative of A minus A contracted on the partial derivative of U. Exercise for the reader-- it is actually really easy to show that you can promote these partial derivatives to covariant derivatives. And what this means is that, when you evaluate the Lie derivative-- so notice, nowhere in here did I introduce anything with a covariant derivative. There was no connection, nothing going on there. If you just go ahead and work it out, basically, when you expand this guy out, you'll find you have connection coefficients or Christoffel symbols that are equal and opposite and so they cancel each other. So you can just go from partials to covariants. Give me just a second, Trey. And this is telling us that the Lie derivative is perfectly tensorial. So the Lie derivative of the vector field is also a tensor quantity. You were asking a question, Trey. AUDIENCE: In the second term, did you miss the D lambda? SCOTT HUGHES: I did. Yes, I did. Thank you. There should be a D lambda right here. Thank you. Yes. Yeah. If you don't have that, then you get what is technically called "crap." So thank you for pointing that out. For reasons that I hope you have probably seen before, you always compute the Lie derivative of some kind of an object along a vector field. So when you're computing the Lie derivative of a vector field along a vector, sometimes this is written using a commutator. I just throw that out there because you may encounter this in some of your readings. It looks like this. So let me just do a few more things that are essentially fleshing out the definition of this. So I'm not going to go through and apply this definition very carefully to higher order objects. What I will just say is that, if I repeat this exercise and, instead of having a vector field that I'm transporting from point to point, suppose I do it for a scalar field-- well, what you actually get-- pardon me for a second-- is this on the partial, but the partial derivative of a scalar is the covariant derivative because there's no Christoffel that couples in. If you do this for a one-form, where it's a 1 indexed object in the downstairs position, you get something that looks like this. And again, when you expand out your covariant derivatives, you find that your Christoffel symbols cancel each other out. And so, if you like, you can just go ahead and replace these with partials. And likewise, let me just write one more out for completeness. Apply this to a tensor. So it's a very similar kind of structure to what you saw when we did the covariant derivative in which every index essentially gets corrected by a factor that looks like the covariant derivative of the field that you are differentiating along. The signs are a little bit different. So it's a similar tune, but it's in a different key. OK, so that's great. And if you get your jollies just understanding different mathematical transport operations, maybe this is already fun enough. But we're in a physics class, and so the question that should be to your mind is, is there a point to all this analysis? So in fact, the most important application of the Lie derivative for our purposes-- in probably the last lecture or two, I will describe some stuff related to modern research that uses it quite heavily. But to begin with in our class, the most important application will be when we consider cases where, when I compute the Lie derivative of some tensor along a vector U and I get 0. I'm just going to leave it schematic like that. So L U of the tensor is equal to 0. If this is the case, we say that the tensor is Lie transported. This is, incidentally, just a brief aside. It shows up a lot in fluid dynamics. In that case, U often defines the flow lines associated with the velocity field of some kind of a fluid that is flowing through your physical situation. And you would be interested in the behavior of all sorts of quantities that are embedded in that fluid. And as we're going to see, when you find that those quantities are Lie transported in this way, there is a powerful physical outcome associated with that, which we are going to derive in just a moment. So suppose I, in fact, have a tensor that is Lie transported. So suppose I have some tensor that is Lie transported. If that's the case, what I can do is define a particular coordinate system centered on the curve for which U is the tangent. So what I'm going to do is I'm going to define this curve such that x0 is equal to lambda, that parameter that defines my length along the curve in a way that, I will admit I've not made very precise yet but will soon. And then I'm going to require that my other three coordinates are all constant on that curve. So if I do that, then my tangent vector is simply delta x0. In other words, it's only got one non-trivial component, and its value of that component is 1. And this is the constant. So the derivatives of the tangent field are all equal to 0. And when you trace this through all of our various definitions, what you find is that it boils down to just looking at how the tensor field varies with respect to that parameter along the curve itself. If it's Lie transported, then this is equal to 0. And so this means that, whatever x0 represents, it's going to be a constant along that curve with respect to this tensor field. Oh, excuse me. Screwed that up. The tensor does not vary with this parameter along the curve. This was a lot, so let's just step back and think about what this is saying. One of the most important things that we do in physics when we're trying to analyze systems is we try to identify quantities that are constants of the motion. This is really tricky in a curved spacetime because much of our intuition gets garbled by all of the facts that different points have different tangent spaces. You worry about whether something being true, and is it just a function of the coordinate system that I wrote this out in? What the hell is going on here? The Lie derivative is giving us a covariant, frame-independent way of identifying things that are constants in our spacetime. So we're going to wrap up this discussion. Let's suppose that the tensor that I'm looking at here is called the metric. Suppose there exists a vector C such that the metric is Lie transported along this thing. What does this tell us? So first, it means there exists some coordinate such that the metric does not vary. The metric is constant with respect to that coordinate. Essentially, if you go through what I sketched a moment ago, this is telling us that the existence of this kind of a vector, which I'm going to give a name to in just a moment-- the existence of this thing demands that my metric is constant with respect to some coordinate. I am not going to prove the following statement. I will just state it, because, in some ways, the converse of that statement is even more powerful. If there is a coordinate, such that dgd, whatever that coordinate is, is equal to 0, then a vector field of this type exists. So the second thing I want to do is expand the Lie derivative. So if I require that my metric be transported along the vector C, well, insert my definition of the Lie derivative. Now, what is the main defining characteristic of the covariant derivative? How did I get my connection in the first place? In other words, what is this going to be? OK, students who took undergraduate classes with me, I'll remind you of one of the key bits of wisdom I always tell people. If the professor asks you a question, 90% of the time, if you just shout out, "0," you are likely to be right. [LAUGHING] Usually, there's some kind of a symmetry that we want you to understand, which allows you to go, oh, it's equal to 0. By the way, whenever I point that out to a class, I then work really hard to make a non-zero answer for the next time I ask it. So the covariant derivative of g is 0, so this term dies. Because the covariant derivative of g is 0, I can always commute the metric with covariant derivatives. So I can take this, move it inside the derivative. I can take this, move it inside the derivative. So what this means is this Lie derivative equation, after all the smoke clears, can be written like this. Or, if I recall, there's this notation for symmetry of indices, which I introduced in a previous lecture. The symmetric covariant derivative of this C is equal to 0. This equation is known as Killing's equation, and C is a Killing vector. Now, this was a fair amount of formalism. I was really laying out a lot of the details to get this right. So to give you some context as to why this matters, there's a bit more that needs to come out of this, but we're going to get to it very soon. Suppose I have a body that is freely falling through some spacetime. And you know what? I'm going to leave this here. So this is a slightly advanced tangent, so I'll start a new board. So if I have some body that is freely falling, what we are going to show in, probably, Thursday's lecture is that the equation of motion that governs it is-- you can argue this on physical grounds, and that's all I will do for now-- it's a trajectory that parallel transports its own tangent factor. For intuition, go into the freely falling frame where it's just the trajectory from special relativity. It's a straight line in that frame, and parallel transporting its own tangent vector basically means it just moves on whatever course it is going. So this is a trajectory for which I demand that the four-velocity governing it parallel transports along itself. Now, suppose you are moving in a spacetime that has a Killing vector. So this will be Thursday's lecture. Suppose the spacetime has a Killing vector. Well, so there will be some goofy C that you know exists, and you know C obeys this equation. By combining these things, you can show that there is some quantity, C, which is given by taking the inner product of the four-velocity of this freely falling thing and the Killing vector. And you can prove that this is a constant of the motion. So let's think about where this goes with some of the physics that you presumably all know and love already. Suppose you look at a spacetime. So you climb a really high mountain. You discover that there's a spacetime metric carved into the stone into the top of it. You think, OK, this probably matters. You look at it and you notice it depends on, say, time, radius, and two angles. Suppose you have a metric that is time-independent. Hey, if it's time-independent, then I know that the derivative of that thing with respect to time is 0. There must exist a Killing factor that is related to the fact that there is no time dependence in this metric. So you go and you calculate it. So this thing, that C is a constant of the motion-- I believe that's P set 4. It's not hard. You combine that equation that we're going to derive, called the geodesic equation, with Killing's equation. Math happens. You got it. So suppose you've got a metric that's time-dependent and you know you've got this thing. So you know what? Let's work it out and look at it. It becomes clear, after studying this for a little bit, that the C for this Killing vector is energy. In the same way that, if you have a time-independent Lagrangian, your system has a conserved energy, if you have a time-independent metric, there is a Killing vector, which-- the language we like to use is-- we say the motion of that spacetime emits a conserved energy. Suppose you find that the metric is independent of some angle. We'll call it phi. Three guesses what's going to happen. In this case-- just one guess, actually. Conserved-- AUDIENCE: Angular momentum. SCOTT HUGHES: Angular momentum pops out in that case. So this ends up being the way in which we, essentially, make very rigorous and geometric the idea that conservation laws are put into general relativity, OK? So I realize there's a lot of abstraction here. So I want to go on a bit of an aside just to tie down where we are going with this and why this actually matters. OK. Let's see. So we got about 10 minutes left. So what we're going to do at the very end of today-- and we'll pick this up beginning of next time. So for the people who walked in a few minutes late, the stuff that I'm actually about to start talking about we need to get through before you can do one of the problems on the P set. So I'm probably going to take that problem and move it on to P set 4, but I'm going to start talking about it right now. So we've really focused a lot, so far, on tensors. We're going to now start talking about a related quantity called tensor densities. There's really only two that matter for our purposes, but I want to go through them carefully. So I will set up with one, and then we'll conclude the other one at the beginning of Thursday's lecture. So let me define this first. I'm going to give a definition that I like but that's actually kind of stupid. So these are quantities that transform almost like tensors-- a little bit lame, but, as you'll see in a moment, it's kind of accurate. What you'll find is that the transformation law is off by a factor that is the determinant of the coordinate transformation matrix. Take it to some power. So there's is an infinite number of tensor densities that one could define. Two are important for this class. So the two that are most important for us are the Levi-Civita symbol and the determinant of the metric. So we use Levi-Civita already to talk about volumes. And it was a tensor when we were working in rectilinear coordinates, where the underlying coordinate system was essentially Cartesian plus time. It's not in general, OK? And we'll go through why that is. That'll probably be the last thing we can fit in today. So let me remind you-- Levi-Civita-- I'm going to write it with a tilde on it to emphasize that it is not tensorial. So this is equal to plus 1 if the indices are 0, 1, 2, 3 and even permutations of that equals minus 1 for odd permutations of that. And it's 0 for any index repeated. Now, this symbol has a really nice property when you apply it to any matrix. In fact, this is a theorem. So I'm working in four-dimensional space. So let's say I've got a 4-by-4 matrix, which I will call m. Write a new [INAUDIBLE] notation, m alpha mu. If I evaluate Levi-Civita, contract it on these guys, I get Levi-Civita back, multiply it by the determinant of the matrix m. Now, suppose what I choose for my matrix m is my coordinate transformation matrix. So I'm just going to write down this result, and I'll leave it since we're running a little short on time. You can just double check that I've moved things from one side of the equation to the other, and you can just double-check I did that correctly. What that tells me is that Levi-Civita and a new set of prime coordinates is equal to this guy in the old, unprimed coordinates with all my usual factors of transformation matrices and then an extra bit that is the determinant of the coordinate transformation matrix. If it were just the top line, this is exactly what you would need for Levi-Civita to be a tensor in the way that we have defined tensors. It's not. So the extra factor pushes away from a tensor relationship. And so what we would say is, because this is off by a factor of what's sometimes called the Jacobian, we call this a tensor density of weight 1. So in order to do this properly-- I don't want to rush-- at the beginning of the next lecture, we're going to look at how the determinant of the metric behaves. And what we'll see is that, although the metric is a tensor, its determinant is a tensor density of weight negative 2. And so what that tells us is that I can actually put together a combination of the Levi-Civita and the determinant of the metric in such a way that their product is tensorial. And that turns out to be real useful because I can use this to define, in a curved spacetime, covariant volume elements, OK? With this as written, my volume elements-- if I just use this like I did when we're taught about special relativity, my volume elements won't be elements of a tensor, and a lot of the framework that we've developed goes to hell. So an extra factor of the determining of the metric will allow us to correct this. And this seems kind of abstract. So let me just, as a really brief aside, before we conclude today's class-- suppose I'm just in Euclidean three-space and I'm working in spherical coordinates. So here's my line element. My metric is the diagonal of 1r squared r squared sine squared theta. The determinant of the metric, which I will write g-- it's r to the fourth sine squared theta. What we're going to learn when we do this is that the metric is a tensor density of weight 2. And so to correct it to get something of weight 1, we take a square root. If you're working in circle coordinates, does that look familiar? This is, in fact, exactly what allows us to convert differentials of our coordinates. Remember, we're working in a coordinate basis. And so we think of our little element of just the coordinates. It's just dr, d theta, d phi. This ends up being the quantity that allows us to convert the little triple of our coordinates into something that has the proper dimensions and form to actually be a real volume element. And so dr, de theta, d phi-- that ain't enough volume. But r squared sine theta, dr, d theta, d phi-- that's a volume element, OK? So basically, that's all that we're doing right now, is we're making that precise and careful. And that's where I will pick things up. We'll finish that up on Thursday. |
MIT_8962_General_Relativity_Spring_2020 | 6_The_principle_of_equivalence.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: So we're just picking up where we stopped last time. So we are beginning to discuss how we are going to sort of do a geometrical approach to physics, using a more general set of coordinates now. So we began talking about how things change when I discuss special relativity, so for the moment keeping ourselves just at special relativity. We, by the way, are going to begin lifting our assumptions that it is simply special relativity fairly soon. But to set that up, I need to start thinking about how to work in more general coordinate systems. So we're going to do it in the simplest possible curvilinear coordinates. So it's basically just going from Cartesian coordinates in the spatial sector to plane polar coordinates. One of the things which I have emphasized a few times, and I'm going to continue to hammer on, is that these are a little bit different from the curvilinear coordinates that you are used to in your past life. In particular, if I write out the displacement, the little vector of the displacement element in the usual way, I am using what's called a "coordinate basis," which means that the vector dx is related to the displacement, the differential of the coordinates, by just that thing contracted with all the basis vectors. And so what that means is I have a little displacement in time, which looks normal. Displacement in radius, which looks normal. Displacement in the z direction, which looks normal, and a displacement in an angle, which does not. In order for this whole thing to be dimensionally consistent, that's telling me e phi has to have the dimensions of length. And that is a feature, not a bug. Last time, we introduced the matrix that allows me to convert between one coordinate system and another, so just basically the matrix-- it's sort of a Jacobi matrix. It's a matrix of partials between the two coordinate systems. And this idea that things are-- they look a little weird. So the way I did that was I didn't actually write it out, but I did the usual mapping between x, y and r and phi, worked out all of my derivatives. And sure enough, you've got something that looks very standard, with the possible exception of these r's that are appearing in here. So notice the elements of this matrix. These do not have consistent units-- again, feature, not bug. This guy is basically just the inverse of that. This is the matrix that affects the coordinates in the opposite direction. And notice in this case, you have some elements where their units are 1 over length. So let's just continue to sort of catalog what some of the things we are going to be working with look like in this new coordinate representation. And this will lead us to introduce one of the mathematical objects that we are going to use extensively as we move forward in studying this subject. So what I want to do next is look at what my basis vectors look like. So what I want to do is characterize what my e r and my e phi look like. And these are going to look very familiar from your intuition, from having studied things like E&M in non-Cartesian coordinates. So your e r is just related to the original Cartesian basis vectors, like so. And if you like, you can easily read that out by performing the following matrix multiplication on the original Cartesian basis vectors. Your e phi perhaps looks a little wacky. So you can see the length coming into play there. A good way to think about this is if your intuition about basis factors-- I have to be careful with this language myself-- your intuition about basis vectors is typically that they are unit vectors. These are not unit vectors. They do form a nice basis, but they are not unit vectors. In particular, the basic idea we're going to go with here is that e phi, it's always going to sort of point in the tangential direction. But no matter where I put it in radius, I want that vector to always sort of subtend the same amount of angle. In order to do that, its length needs to grow with r. So that's where that's a little bit different from your intuition. And there's a very good reason for this, which we will get to, hopefully, well before the end of today's class. So last time, when we first began to talk about tensors a couple of lectures ago, the first tensor I gave you-- so confining ourselves just to Cartesian coordinates-- was the metric, which was originally introduced as this mathematical object that came out of looking at dot products between basis vectors. It's essentially a tensor that allows me to feed in two displacements and get the invariant interval between those displacements that comes out of that. I am going to continue to call the dot product of two basis vectors the "metric." But I'm going to use a slightly different symbol for this. I'm going to call this g alpha beta. In the coordinate representation that we are using right now, so in plane polar coordinates, this becomes-- you can work it out from what I've written down right here. This is just the diagonal of minus 1, 1 r squared 1. So this equals dot here. This is-- I'll put PPC for plane polar coordinates under that. And then using that, you know that you can always work out the invariant displacement between two events. It's always going to be the metric contracted with the differential displacement element. And this is going to be minus dt squared plus dr squared plus r squared d phi squared plus dz squared. That, I hope, makes a lot of sense. This is exactly what you'd expect if I have two events that are separated in plane polar coordinates by dt, dr, d phi, dz. This is what the distance between them should be. So the fact that my basis vectors have this slightly annoying form associated with them, it all sort of comes out in the wash here. Remember at the end of the day, if we think about quantities that are representation independent-- and that's going to be the key thing. When you assemble scalars out of these things, the individual tensor components, they can be a little bit confusing sometimes. They are not things that we measure. They are not things that really characterize what we are going to be working with. And we really want to get into the physics, unless we're very careful about it. This is something you can measure. And so sure enough, it comes out, and it's got a good meaning to it. Let me just wrap up one last thing before I talk about sort of where we're going with this. So just for completeness, let me write down the basis one forms. Just as the basis vectors had a bit of a funny form associated with them, you're going to find the basis one forms likewise have a bit of a funny form associated with them. And the way I'm going to get these-- and so these are going to be the Cartesian basis one forms-- basically, I'm not carefully proving all these relations at this point, because you all know how to do that. I'm just using line up the indices rule. And when you do that, you get this. And likewise, your basis one form for the axial direction, I'll just write down the result. It's going to look like this. So the key place where all of this-- so right now, these are all just sort of definitions. Nothing I've done here should be anything that even approaches a surprise, I hope, just given the you guys have done-- the key thing that's probably new is all this garbage associated with coordinate bases, this extra factors of r and 1 over r that are popping up. But provided you're willing to sort of swallow your discomfort and go through the motions, these are not difficult calculations. The key place where all of this really matters is going to be when we calculate derivatives of things. It'll turn out there is an important rule when we talk about integrals as well a little bit later, but let's just say that. So for now, we'll focus on derivatives. So all the derivatives that we've been looking at so far, we have, indeed, done a couple of calculations where we've computed the derivatives of various vector valued and tensor valued quantities. And it was helped by the fact that all the bases, when I work in Cartesian coordinates, are constant. Well, that's not the case now. So now, we need to account for the fact that the bases all vary with our coordinates. So let me just quickly make a catalog of all the non-trivial-- there's basically four. In this one, where I'm just doing plane polar coordinates, there are four non-trivial derivatives we need to worry about. One of them actually turns out to be 0. So the radial derivative of the radial unit vector is 0. But the phi derivative of the phi unit vector is not. you go and take the phi derivative of this guy, and you basically get-- take the phi derivative of this, you're going to get this back, modulo factor of radius. So I can write d e r d phi as e phi over r. If I take the derivative of e phi with respect to r, I get e phi back, divided by r. So the simplest way to write this is like so. And finally, if I take the phi derivative of the phi unit vector, I get e r back, with an extra factor of r thrown in. And a minus sign. So we're going to see a way of doing this that's a little bit more systematic later, but I want to just keep the simple example, where you can just basically by hand calculate all the non-trivial derivatives easily. Of course, there's also a t unit vector and a z unit vector. But they're constants, so I'm not going to bother writing them out. All the derivatives associated with them are equal to 0. So let's imagine now that I have assembled some vector. So I have some vector field that lives in this spacetime. And I'm using this basis. And so I would write this vector with components v alpha. And let's let the-- so this is going to be a curvilinear coordinate system, so this will be plane polar coordinates being used here, plane polar coordinate basis vectors. And what I would like to do is assemble the tensor that you can think of essentially as the gradient of this vector. So let's begin by doing this in a sort of abstract notation. So the gradient of this guy-- this is sort of ugly notation, but live with it. Following what we have been doing all along, what you would want to do is just take the root of this whole thing. It's going to have a downstairs component on it. So attach to it the basis one form. If you prefer, you can write it using the d notation like I have there, but I just want to stick with the form I wrote in my notes. Looking at this the way I've sort of got this right now, I can think of, if I don't include the basis one forms here, this should be the components of a one form. So this should be a kind of object. So let's just expand out that derivative. Let's write it like this. So you just-- we haven't changed calculus. So when I do this, I'm going to basically use the old-fashioned Leibniz rule for expanding the derivative product of two things. Here's the key thing which I want to emphasize-- in order for this whole thing to be-- for this to be a tensorial object, something that I couple to this basis one form, the sum of these two objects must obey the rules for transforming tensors. But the two objects individually will not. So this is an important point which I'm going to emphasize in slightly different words in just a few moments again. This is one of the key things I want you to get out of this lecture, is that when I'm taking derivatives of things like this, you've got to be a little bit careful about what you consider to be components of tensors and what is not. Now as written like that, this is kind of annoying. So my first object has a nice basis vector attached to it. My second object involves a derivative of the basis vector. However, something we saw over here is that derivatives of basis vectors are themselves proportional to basis vectors. So what I'm going to do is introduce a bit of notation. So let me switch notation slightly here. So the beta derivative of e alpha can be written as-- in general, it can be written as a linear combination of basis vectors. So what we're going to do is define d-- I want to make sure my Greek letters are legible to everyone in the room here. So let me write this nice and clearly. d beta of e alpha, I'm going to write that as capital gamma mu beta alpha e mu. This gamma that I've just introduced here in this context is known as the Christoffel symbol. Fact I'm calling this a symbol, it's got three indices on it. You might look at it and go, ooh, smells like a tensor. Be a little bit careful. In much the same way that these two terms are not individually components of a tensor, but their sum is, this guy individually is actually not a component of a tensor, but when combined with other things, it allows us to assemble tensors. So for our plane polar coordinates, there are exactly three non-zero Christoffel symbols. So gamma phi r phi is equal to 1 over r, which is also equal to gamma phi phi r. Gamma r phi phi is minus r. And you can basically just read that out of that table that I wrote down over there. All the others will be equal to 0. Now from this example, this is what it makes it smell like every time you introduce a new coordinate representation. You're going to need to sit down for an hour and a half, or something like that, and just work out all the bloody derivatives, and then go, oh, crap, and read out all the different components of this thing, and assemble them together. There actually is an algorithm that we will get to at the end of this class that allows you to easily extract the Christoffel symbols provided you know the metric. But right now, I just want to illustrate this thing conceptually. The key thing which you should know about it is that it is essentially the-- I almost said the word "matrix," but it's got three indices. It's a table of functions that allows me to relate derivatives of basis vectors to the basis vectors. So before I go on to talk about some of that stuff, let's take a look at the derivative a little bit more carefully. So the derivative of the vector-- so let's basically take what I've written out up there. I'm going to write this as the beta derivative of vector v. And I can write that as the beta derivative of e of v alpha-- so the first term where the derivative hits the vector components. And then I've got a second term where the derivative hits the basis. I'm going to write this like so. This is sort of annoying. One term is proportional to e alpha, one is proportional to e mu. But notice, especially in the second term, both alpha and mu are dummy indices, so I'm free to relabel them. So what I'm going to do is relabel alpha and mu by exchanging them. As long as I do that consistently, that is totally kosher. And when I do that, I can factor out an overall factor of the basis object. This combination that pops up here-- so we give this a name. And this is a combination which, by the time you have finished this semester, if you don't have at least one nightmare in which this name appears, I will not have done my job properly. This shows up a lot at this point. This is called the "covariant derivative." And it shows up enough that we introduce a whole new notation for the derivative to take it into account. I'm going to call this combination of the partial derivative of v and v coupled to the Christoffel symbol-- I'm going to write this using the, if you're talking LaTeX, this would be the nabla operator. So I made a point earlier when we were talking about derivatives a couple of weeks ago that we were reserving the gradient symbol for a special purpose later. Here it is. So whenever I make a derivative that involves the gradient symbol like this, it is this covariant derivative. And the covariant derivative acting on vector components, it generates tensor components. Partial derivative does not. And what I'm going to do, just in the interest of time-- it's one of those calculations that's straightforward but fairly tedious-- I have a set of notes that I meant to make live before I headed over here, but I forgot. I have a set of notes that I'm going to show on the website by this evening which explicitly works out what happens when you apply the coordinate transformation using that-- it's been erased-- when you use that L matrix to construct the coordinate transformation between two representations. If you try to do it to partial derivatives of vector components, basically what you find is that there's an extra term that spoils your ability to call that-- it spoils the tensor transformation law, spoils your ability to call that a tensor component. So the partial on its own doesn't let you. You get some extra terms that come along and mess it all up. On next p set, you guys are going to show that if you then try to apply the tensor transformation law to the Christoffel symbols, you get something that looks tensorial, but with an extra term that spoils your ability to call it tensorial. There's a little bit of extra junk there. But the two terms exactly conspire to cancel each other out so that the sum is tensorial. So part one of this will be notes that I post to the website no later than this evening. Part two, you guys will do on the p set. So just saying in math what I just said in words, if I do this, like I said, you basically will eventually reach the point where what I am writing out right now will become so automatic it will haunt your dreams. Wait a minute, I screwed that up. It's so automatic I can't even write it properly. Anyhow, something like that will-- modulo my typo-- that should become automatic. And the key thing which I want to note is that if I take these guys, and I attach the appropriate basis objects to them, this is an honest-to-god tensor. And so this derivative is itself an honest-to-god tensor. A typical application of this, so one that will come up a fair bit, is how do you compute a spacetime divergence in each coordinate system? So suppose I take the divergence of some vector field v. So you're going to have four terms that are just the usual, like when you guys learned how to do divergence in freshman E&M in Cartesian coordinates. You get one term that's just dv x dx, dv y dy, et cetera. So you got one term that looks just like that, and you're going to have something that brings in all of Christoffel symbols. Notice the symmetry that we have on this one. Actually, there is Einstein summation convention being imposed here. But when we look at this, there's actually only one Christoffel symbol that has repeated indices in that first position. So when I put all this together, you wind up with something that looks like this. So go back and check your copy of Jackson, or Purcell, or Griffith, whatever your favorite E&M textbook is. And you'll see when you work in cylindrical coordinates, you indeed find that there's a correction to the radial term that involves 1 over r. That's popped out exactly like you think it should. You have a bit of a wacky looking thing with your phi component, of course. And let me just spend a second or two making sure. It's often, especially while we're developing intuition about working in a coordinate basis, it's not a bad idea to do a little sanity check. So here's a sanity check that I would do with this. If I take the divergence, I take a derivative of a vector field, the final object that comes out of that should have the dimensions of that vector divided by length. Remembering c equals 1, that will clearly have that vector divided by length. That will clearly have that vector divided by length, vector divided by length, explicitly vector divided by length. That's weird. But remember, the basis objects themselves are a little weird. One of the things we saw was that e phi has the dimensions of length. In order for the vector to itself be consistent, v phi must have the dimensions of v divided by length. So in fact, when I just take its phi derivative, I get something that looks exactly like it should if it is to be a divergence. Let's move on and think about how I take a covariant derivative of other kinds of tensorial objects. This is all you need to know if you are worried about taking derivatives of vectors. But we're going to work with a lot of different kinds of tensor objects. One of the most important lectures we're going to do in about a month actually involves looking at a bunch of covariant derivatives of some four-indexed objects, so it gets messy. Let's walk our way there. So suppose I want to take the derivative of a scalar. Scalar have no basis object attached to them. There's no basis object. When I take the derivative, I don't have to worry about anything wiggling around. No Christoffel symbols come in. If I want to take the covariant derivative of some field phi, it is nothing more than the partial derivative of that field phi-- boom. Happy days. How about a one form? The long way to do this would be to essentially say, well, the way I started this was by looking at how my basis vectors varied as I took their derivatives. Let's do the same thing for the basis one forms, assemble my table, do a lot of math, blah, blah, blah. Knock yourselves out if that's what you want to do. There's a shortcut. Let's use the fact that when I contract a one form on a vector, I get a scalar. So let's say I am looking at the beta covariant derivative of p alpha on a alpha. That's a scalar. So this is just the partial derivative. And a partial derivative of the product of something I can expand out really easily. So using the fact that this just becomes the partial, I can write this as a alpha d beta p alpha plus p downstairs alpha. So now what? Well, let's rewrite this using the covariant derivative. Pardon me a second while I get caught up in my notes. Here we are. I can write this as the covariant derivative minus the correction that comes from that Christoffel symbol. Pardon me just a second. There's a couple lines here I want to write out very neatly. So when I put this in-- oops typo. That last typo is important, because I'm now going to do the relabeling trick. So what I'm going to do is take advantage of the fact that in this last term, alpha and mu are both dummy indices. So on this last term that I have written down here, I'm going to swap out alpha and mu. When I do that, notice that the first term and the last term will both be proportional to the component a alpha. Now, let's require that the covariant derivative when it acts on two things that are multiplied together, it's a derivative, so it should do what derivatives ordinarily do. So what we're going to do is require that when I take this covariant derivative, I should be able to write the result like so. It's a healthy thing that any derivative should do. So comparing, I look at that, and go, oh, I've got the covariant derivative of my one form there. Just compare forms. Very, very similar, but notice the minus sign. There's a minus sign that's been introduced there, and that minus sign guarantees, if you actually expand out that combination of covariant derivatives I have on the previous line, there's a nice cancellation so that the scalar that I get when I contract p on a, in fact, doesn't have anything special going on when I do the covariant derivative. So I'm going to generalize this further, but let me just make a quick comment here. I began this little calculation by saying, given how we started our calculation of the covariant derivative of a vector, we could have begun by just taking lots of derivatives of the basis one forms, and assembling all these various tables, and things like that. If you had done this, it's simple to find, based on an analysis like this, that if you take a partial derivative of a one form, that you get sort of a linear combination of one forms back. Looks just like what you got when you took a partial derivative of the basis vector, but with a minus sign. And what that minus sign does is it enforces, if you go back to a lecture from ages ago, when I first introduced basis one forms, it enforces the idea that when I combine basis one forms with basis vectors, I get an identity object out of this, which is itself a constant. If you are the kind of person who likes that sort of mathematical rigor, some textbooks will start with this, and then derive other things from that-- sort of six of one, half a dozen of the other. So we could go on at this point. And I could say, how do I do this with a tensor that has two indices in the upstairs position? How do I do this with a tensor that has two indices in the downstairs position? How do I do it for a tensor that's got 17 indices in the upstairs position and 38 in the downstairs position? The answer is easily deduced from doing these kinds of rules, so I'm just going to write down a couple of examples and state what it turns out to be. So basically, imagine I want to take the covariant derivative-- let's do the stress energy tensor-- covariant derivative of T mu nu. So remember, the way that the Christoffel got into there is that when I looked at the derivative of a vector, I was looking at derivatives of basis objects. Well, now I'm going to look at derivatives of two different basis objects. So I'm going to wind up with two Christoffel symbols. You can kind of think of it as coming along and correcting each of these indices. I can do this with the indices in the downstairs position. Guess what? Comes along and corrects all them with minus signs. Just for completeness, let me just write down the general rule. If I am looking at the covariant derivative of a tensor with a gajillion upstairs indices and a gajillion downstairs indices, you get one term that's just a partial derivative of that guy, and you get a Christoffel coupling for every one of these. Plus sign for all the upstairs, minus sign for all the downstairs. That was a little tedious. You basically just, when I give you a tensor like that, you just kind of have to go through. And it becomes sort of almost monkey work. You just have to rotely go through and correct every one of the indices using an algorithm that kind of looks like this. Oh, jeez, there's absolutely a minus sign on the second one. Thank you. I appreciate that. So the way that we have done things so far, and I kind of emphasized, it sort of smells like the way to do this is you pick your new coordinate representation, you throw together all of your various basis objects, and then you just start going whee, let's start taking derivatives and see how all these things vary with respect to each other, assemble my table of the gammas, and then do my covariant derivative. If that were, in fact, the way we did it, I would not have chosen my research career to focus on this field. That would suck. Certainly prior to Odin providing us with Mathematica, it would have been absolutely undoable. Even with it, though, it would be incredibly tedious. So there is a better way to do this, and it comes via the metric. Before I derive what the algorithm actually is, I want to introduce an extremely important property of tensor relationships that we are going to come back to and use quite a bit in this course. So this is something that we have actually kind of alluded to repeatedly, but I want to make it a little more formal and just clearly state it. So this relationship that I'm going to use is some kind of a tensor equation, a tensorial equation that holds in one representation must hold in all representations. Come back to the intuition when I first began describing physics in terms of geometric objects in spacetime. One of the key points I tried to really emphasize the difference of is that I can have different-- let's say my arm is a particular vector in spacetime. Someone running through the room at three-quarters the speed of light will use different representations to describe my arm. They will see length contractions. They will see things sort of spanning different things. But the geometric object, the thing which goes between two events in spacetime, that does not change, even though the representation of those events might. This remains true not just for Lorentz transformations, but for all classes of transformations that we might care to use in our analysis. Changing the representation cannot change the equation. Written that way, it sounds like, well, duh, but as we'll see, it's got important consequences. So as a warm-up exercise of how we might want to use this, let's think about the double gradient of a scalar. So let's define-- let's just say that this is the object that I want to compute. Let's first do this in a Cartesian representation. In a Cartesian representation, I just take two partial derivatives. I've got a couple basis one forms for this. So I've got something like this. The thing which I want to emphasize is that as written, in Cartesian coordinates, d alpha d beta of phi-- those are the components of a tensor in this representation. And the key thing is that they are obviously symmetric on exchange of the indices alpha and beta. If I'm just taking partial derivatives, doesn't matter what order I take them in. That's got to be symmetric. Let's now look at the double gradient of a scalar in a more general representation. So in a general representation, I'm going to require these two derivatives to be covariant derivatives. Now, we know one of them can be very trivially replaced with a partial, but the other one cannot. Hold that thought for just a second. If this thing is symmetric in the Cartesian representation, I claim it must also be true in a general representation. In other words, exchanging the order of covariant derivatives when they act on a scalar should give me the same thing back. Let's see what this implies. So if I require the following to be true, that's saying-- oops. So let's expand this out one more level. So now, I'm correcting that downstairs index and over here. So the terms involving nothing but partials, they obviously cancel. I have a common factor of d mu of phi. So let's move one of these over to the other side. What we've learned is that this requirement, that this combination of derivatives be symmetric, tells me something about the symmetry of the Christoffel symbols itself. If you go back to that little table that I wrote down for plane polar coordinates, that was one where I just calculated only three non-trivial components, but there was a symmetry in there. And if you go and you check it, you will see it's consistent with what I just found right here. Pardon me for just a second. I want to organize a few of my notes. These have gotten all out of order. Here it is. So let me just use this as an opportunity to introduce a bit of notation. Whenever I give you a tensor that's got two indices, if I write parentheses around those indices, this is going to mean that I do what is called "symmetrization." We're going to use this from time to time. If I write square braces, this is what we call "anti-symmetrization." And so what we just learned is that gamma mu alpha beta is equal to gamma mu alpha beta with symmetrization on those last two indices. We have likewise learned that if I contract this against some object, if these were anti-symmetric, I must get a 0 out of it. So that's a brief aside, but these are important things, and I want to make sure you have a chance to see them. So trying to make a decision here about where we're going to want to carry things forward. We're approaching the end of one set of notes. There's still one more thing I want to do. So I set this whole thing up by saying that I wanted to give you guys an algorithm for how to generate the Christoffel symbols. The way I'm going to do this is by examining the gradient of the metric. So suppose I want to compute the following tensor quantity-- let's say is g the metric tensor, written here in the fairly abstract notation. And this is my full-on tensor gradient of this thing. So if you want to write this out in its full glory, I might write this as something like this. But if you stop and think about this for just a second, let's go back to this principle. An equation that is tensorial in one representation must be tensorial in all. Suppose I choose the Cartesian representation of this thing. Well, then here's what it looks like there. But this is a constant. So if I do this in Cartesian coordinates, it has to be 0. The only way that I can make this sort of comport with this principle that an equation that is tensorial in one representation holds in all representations-- this leads me to say, I need to require that the covariant derivative of the metric be equal to 0. We're going to use this. And I think this will be the last detailed calculation I do in today's lecture. We're going to use this to find a way to get the Christoffel symbol from partial derivatives of the metric. There's a lot of terms here and there's a lot of little indices. So I'm going to do my best to make my handwriting neat. I'm going to write down a relationship that I call "Roman numeral I." The covariant derivative in the gamma direction, G alpha beta-- you know what, let me put this down a little bit lower, so I can get these two terms on the same line. So I get this thing that involves two Christoffel symbols correcting those two indices. This is going to equal 0. I don't really seem to have gotten very far. This is true, but I now have two bloody Christoffel symbols that I've somehow managed to work into this. What I'm trying to do is find a way to get one, and equate it to things involving derivatives of the metric. So this is sort of a ruh-roh kind of moment. But there's nothing special about this order of the indices. So with the audacity that only comes from knowing the answer in advance, what I'm going to do is permute the indices. Then go, oh, let's permute the indices once more. So I'll give you guys a moment to catch up with me. Don't forget, these notes will be scanned and added to the web page. So if you don't want to follow along writing down every little detail, I understand, although personally, I find that these things gel a little bit better when you actually write them out yourself. So those are three ways that I can assert that the metric has no covariant derivative. They all are basically expressing that same physical fact. I'm just permuting the indices. Now there's no better way to describe this than you sort of just stare at this for a few moments, and then go, gee, I wonder what would happen if-- so stare at this for a little while. And then construct-- you know I have three things that are equal to 0. So I can add them together, I can subtract one from the other. I can add two and subtract one, whatever. They should all give me 0. And the particular combination I want to look at is what I get when I take relationship one and I subtract from it two and three. So I'm going to get one term that are just these three combinations of derivatives, gamma. And I get something that looks like-- let me write this out and then pause and make a comment. So I sort of made some lame jokes a few moments ago that essentially, the only reason I was able to get this was by knowing the answer in the back of the book, essentially. And to be perfectly blunt, for me personally, that's probably true. When I first wrote this down, I probably did need to follow an algorithm. But if I was doing this ab initio, if I was sitting down to first do this, what's really going on here is the reason I wrote out all these different combinations of things is that I was trying to gather terms together in such a way that I could take advantage of that symmetry. A few moments ago, we showed that the Christoffel symbols are symmetric on the lower two indices. And so by putting out all these different combinations of things, I was then able to combine them in such a way that certain terms-- look at this and go, ah, symmetry on alpha and gamma means this whole term dies. Symmetry on beta and gamma means this whole term dies. Symmetry on alpha and beta means these two guys combine, and I get a factor of 2. So let's clean up our algebra. Move a bunch of our terms to the other side equation, since it's a blah, blah, blah equals 0. And what we get when we do this is g mu downstairs gamma is equal to 1/2. What we're going to do now is we will define everything on the right-hand side-- I've kind of emphasized earlier that the Christoffels are not themselves tensors, but we're going to imagine that we can nonetheless-- we're not going to imagine, we're just going to define-- we're going to say that we're allowed to raise and lower their indices using the metric, in the same way you guys been doing with vectors and one forms and other kinds of tensors. So let's call everything on the right-hand side here gamma with all the indices in the downstairs position, gamma sub gamma alpha beta. And then this is simply what I get when I click all of these things together like so. If you go and you look up the formulas for this in various textbooks that give these different kinds of formulas, you will typically see it written as 1/2 g upstairs indices, and then all this stuff in parentheses after that. When you look things up, this is the typical formula that is given in these books. This is where it comes from. So I need to check one thing because it appears my notes are a little bit out of order here. But nonetheless, this is a good point, since we've just finished a pretty long calculation, this is a good point to introduce an important physical point. We're going to come back to this. We're going to start this on Thursday. But I want to begin making some physical points that are going to take us from special relativity to general relativity. So despite the fact that I've introduced this new mathematical framework, everything that I have done so far is in the context of special relativity. I'm going to make a more precise definition of special relativity right now. So special relativity-- we are going to think of this moving forward as the theory which allows us to cover the entire spacetime manifold using inertial reference frames. So we use inertial reference frames or essentially, Lorenz reference frames, and saying that Lorentz coordinates are good everywhere. We know we can go between different Lorentz reference frames using Lorentz transformations. But the key thing is that if special relativity were correct, the entire universe would be accurately described by any inertial reference frame you care to write down. And I will probably only be able to do about half of this right now. We'll pick it up next time, if I cannot finish this. The key thing which I want to emphasize is, gravity breaks this. As soon as you put gravity into your theory of relativity, you cannot have-- so we will call this a global inertial frame, an inertial frame that is good everywhere, so "global" in the mathematical sense, not "global" in the geographic sense-- not just earth, the whole universe, essentially. As soon as we put in gravity, we no longer have global reference frames and global inertial reference frames. That word "inertial" is important. But we are going to be allowed to have local inertial frames. I have not precisely defined the difference what "local" means in this case, and I won't for a few lectures. But to give you a preview as to what that means, it's essentially going to say that we can define an inertial coordinate system that is good over a particular region of spacetime. And we're going to have to discuss and come up with ways of understanding what the boundaries of that region are, and how to make this precise. So the statement that gravity breaks the existence of global Lorentz frames, like I said, it's a two-part thing. I'm going to use a very handwavy argument which can be made quite rigorous later, but I want to keep it to this handwaving level, because first of all, it actually was first done by a very high-level mathematical physicist named Alfred Schild, who worked in the early theory of relativity. It's sort of like he was so mathematical, if it was good enough for him, that's good enough for me. And I think even though it's a little handwavy, and kind of goofy in at least one place, it gives a good physical sense as to why it is gravity begins to mess things up. So part one is the fact that there exists a gravitational redshift. So here's where I'm going to be particularly silly, but I will back up my silliness by the fact that everything silly that I say here has actually been experimentally verified, or at least the key physical output of this. So imagine you are on top of a tower and you drop a rock of rest mass m off the top of this tower. So here you are. Here's your rock. The rock falls. There's a wonderful device down here which I label with a p. It's called a photonulater. And what the photonulater does, it converts the rock into a single photon, and it does so conserving energy. So when this rock falls, the instant before it goes into your photonulater, just use Newtonian physics plus the notion of rest energy. So it's got an energy of m-- mC squared, if you prefer, its rest energy-- plus what it acquired after falling-- pardon me, I forgot to give you a distance here-- after falling a distance h. So that means that the photon that I shoot up from this thing-- let me put a few things on this board. So the instant that I create this photon, this thing goes out, and it's going to have a frequency omega bottom, which is simply related to that energy. This photon immediately is shot back up to the top, where clever you, you happen to have in your hands a rerockulater. The rerockulater, as the name obviously implies, converts the photon back into a rock. Now, suppose it does so-- both the photonulater and the rerockulater are fine MIT engineering. There are no losses anywhere in this thing. So there's no friction. There's no extra heat generated. It does it conserving energy, 100%. What is the energy at the top? Well, you might naively say, ah, it's just going to go up to the top. It's going to have that same energy. It might just have it in the form of a photon and omega b. There will be some frequency at the top. And your initial guess might be it's going to be the same as the frequency at the bottom. But if you do that, you're going to suddenly discover that your rock has more energy than it started out with, and you can redirect it back down, send it back up. Next thing you know, you've got yourself a perpetual motion machine. So all you need to do is get your photonulater and your rerockulater going, and you've got yourself a perpetual motion machine here. I will grant that's probably not the weakest part of this argument. Suppose you had this. I mean, you look at this. If technology allowed you to make these goofy devices, you would instantly look at this and say, look, if I am not to have-- let's just say I live in a universe where I'm fine with photonulaters. I'm fine with rerockulaters, but damn it, energy has to be conserved. I am not fine with perpetual motion machines. If that's the case, we always fight perpetual motion. We must have that the energy at the top is equal to the energy this guy started with. When it sort of gets back into-- imagine that your rerockulater is shaped like a baseball catcher's mitt. You want that thing to just land gently in your mitt, and just be a perfectly gentle, little landing there. And when you put all this together, going through this, taking advantage of the fact that if you work in units where you've put your c's back in, there will be a factor of g h over c squared appearing in here, what you find is that the frequency at the top is less than the frequency at the bottom. In other words, the light has gotten a little bit redder. now I fully confess, I did this via the silliest argument possible. But I want to emphasize that this is one of the most precisely verified predictions of gravity and relativity theory. This was first done, actually, up the street at Harvard, by what's called the Pound-Rebka experiment. And the basic principles of what is going on with this right now-- I just took this out to make sure my alarm is not about to go off, but I want to emphasize it's actually built into the workings of the global positioning system. Because this fact that light signals may travel out of a gravitational potential, they get redshifted, needs to be taken into account in order to do the precise metrology that GPS allows. Now, this is part one, this idea that light gets redder as it climbs out of a gravitational field. Part two, which I will do on Thursday, is to show that if there is a global inertial frame, there is no way for light to get redder as it climbs out of a gravitational potential. You cannot have both gravity and a global inertial reference frame. That's where I will pick it up on Thursday. So we'll do that. And we will then begin talking about how we can try to put the principles of relativity and gravity together. And in some sense, this is when our study of general relativity will begin in earnest. All right, so let us stop there. |
MIT_8962_General_Relativity_Spring_2020 | 3_Tensors_continued.txt | [SQUEAKING][RUSTLING][CLICKING] SCOTT HUGHES: All right, welcome to Tuesday. So hopefully, you've all saw the brief announcement I send to the class. I have to introduce a colloquium speaker over in Astrophysics, basically at the second this class officially ends, so I will be wrapping things up a little bit early so that I can take into account the spatial separation and get back there in time to actually do the introduction. I've already posted the lecture notes of material I'm going to be covering today, and it'll probably spill-- I hope to wrap it all up today, but it's possible it'll spill a little bit into Thursday. So if you've already looked at those notes, today will essentially just be sort of my guided tour through that material. So I want to pick it up with where I left things last time. So we covered a bunch of material that, again, I kind of emphasize what we're doing right now is just laying the mathematical foundations in a very thorough, almost excessively thorough way in order that we have a very strong structure as we begin to move into more physically complicated situations in the special relativity that we're focusing on right now. So we talked about this definition of an inner product between two four-vectors, two vectors in space time. And it looks just like the inner product between two vectors that you are used to from your Euclidean three space intuition. It's just that we have an extra bit there that enters with a minus sign having to do with the time like components of those two four-vectors. And then, using the fact that I can write my four vector as components contracted onto elements of a set of basis vectors, I can use this to define a tensor, which I will get to the mathematical definition of more precisely in just a moment. The dot product of any two basis vectors, I will call that the tensor component, eta alpha beta. OK, and so this is the metric tensor of special relativity, at least in rectilinear coordinates. Rectilinear basically just means Cartesian but throwing time in there as well. OK, when we start talking about special relativity and curvilinear coordinates, it'll get a little bit more complicated than that, and I not use a symbol eta in that case. I am going to reserve eta for this particular form of the metric in this coordinate system. And of course, when you actually write out these components, a very compact way of writing this is it is just the matrix that has the elements minus one, one, one, one down the diagonal and zeros everywhere else. So this is a fairly common way of writing a diagonal matrix. This just takes into account the fact that there's zeros everywhere else. So I call this the metric tensor, which kind of begs the question, what's a tensor? So this is where we concluded things last time. I'm going to generally define a tensor of type 0n-- we're going to change that zero to something else by the end of today's lecture. You'll be able to see where I'm going probably from a mile away, but let's just leave it like this for now. So a tensor of type 0n is-- you can think of it as a function or a mapping of n vectors into Lorentz invariant scalars, which is linear in its n arguments. So the inner product clearly falls into this category. If I think of a dot b, let's say this inner product is some number, lowercase a. A will be a Lorentz invariant. I forgot to state this, but the reason we define the inner product in this way is that we are motivated by the invariant interval in space and time between two events. We wanted to find an inner product that duplicates its mathematical structure. If I take one of these vectors, multiply it by some scalar, linearity is going to hold. This will just be eta alpha beta. This is terrible notation. I'm using the [INAUDIBLE] symbol alpha for both a pre factor and an index. Minus five points to me. Let's call this gamma. You can quickly convince yourself. That just comes out, and you get a factor of gamma on this. You can do the same thing on the second slot. If I take the dot product of a with the sum of two vectors-- OK, et cetera. You can keep going. All the rules for linearity are going to hold. I'm not going to step through them all. You can see where they all go. So whenever I'm going to define a tensor, in my head, I'm imagining it's got properties like this that come along for the ride. Now, especially when you see this defined in certain textbooks, MTW is particularly fond of doing this. So we come back to this idea that it's sort of a function or a mapping. You can almost abstractly define the tensor as a mathematical machine that's got two slots in it. So several of the recommended textbooks will write down equations. I'm going to put two lines over the symbol to sort of-- if you actually read this, in for instance, MTW or Caroll or something like that, this will be written as a boldface symbol. It's hard to do on the blackboard, so I'm just going to write double bars over it. So if I imagine this with my slots filled with these, it's got two slots associated with it. I fill it with those two vectors. That is equivalent to saying a dot b, which is equivalent to it in this component notation, something like this. So we have repeatedly in the little bit of time we've spent together-- I should say I have repeatedly, in the little bit of time we've spent together, emphasized the distinction between frame independent geometric objects, things that reside in the manifold and have kind of an integral, physical, geometric sensibility of their own and their representations. I have emphasized quite strongly that you should think of the vectors a and b as being geometric objects. This is some thing that is pointing in space time. We all agree that this points, if that's a displacement vector, it points from event one to event two. OK, everyone agrees on that geometric reality of this. Different observers may represent it using different components. That's just because they're using different coordinate systems. So when I write down something like this-- so let's go back to where I wrote before. This is going to turn into some frame independent Lorentz scalar a. So this is what I like to call a frame independent object. Frame or Lorentz independent geometric object, as is this scalar. And so therefore, the tensor must be a frame independent geometric object as well. That's a lot of words around the blackboard, but I really want to nail that point home. So tensors just like vectors. Think of them as geometric objects that have an intrinsic geometric meaning associated with them that lives in spacetime. We will talk about certain examples of them. OK, you guys all have some intuition about vectors, because you've been doing vectors ever since you took your first kindergarten physics course, and so you know that there's some kind of an object that points in a certain direction. Tensors are a little bit more challenging in many cases to develop an intuition for. Some of them really do have a fairly simple geometric interpretation. You can kind of think of them as-- for instance, we're going to introduce one a little bit, which describes the flow of energy and momentum in space time. And so it'll have two indices associated with it, and those indices tell me about what component of energy or momentum is flowing in a particular direction. Really easy to interpret that. Some of the others, not so much. Nonetheless, they do have this geometric meaning underneath the hood. And that's bound up in the fact that if I put frame independent geometric objects into all the slots, I get a Lorentz invariant number out of it. It's the only way that that can sort of work. But one reason why I'm going through this is that just like with vectors, different observers, different frames will get different components in general. So there will be different representations according to different observers. I'm going to write down the same thing. Representations. So if you want to get a particular observer's components out of the tensor, there's actually a very simple recipe for this. All you do is, you take your tensor, and into its slots, you plug in the basis vectors that that observer uses. So if I want to get the components-- and this, unfortunately, is a fairly stupid example, but it's the only one we've got at the moment, so let's just work with it. If I take the tensor eta, the special relativity metric, I plug in some observer's basis factors into this thing, this by definition is eta alpha beta. Suppose I have a different observer who comes along, someone who-- we're doing special relativity, so someone who's dashing through the room at three quarters of the speed of light. I want to know what their components would be. Well, what I do is, I just plug into the slots-- let's put bars on the complements to denote this other observer. Do this operation for this other observers set of basis vectors, and you will get the components that they will measure. Now, one of the reasons why I'm going through this is that last time, we talked about how to transform basis vectors between different reference frames. We know that these guys are just related to one another by a Lorentz transformation matrix. So let's just take this a step further. So this is telling me that the components of the metric in this barred frame, they're going to be what I get when I put it into the slots. Those are the basis vectors in the barred frame using the Lorentz transformation matrix to go from the unbarred frame to the barred frame. Now, remember again-- this is one of those places where if you're sort of just becoming comfortable with the index notation, you're temptation at this stage is always to go, these are matrices. I should start doing matrix multiplication. If you set that urge within you aside, you go, no, no, no, no. Those are just a set of 16 numbers. For any particular set of complements, I can just pull them out. So because of the linearity of all these slots with this thing, this just becomes those two Lorentz transformation matrices acting on the abstract metric tensor with the unbarred basis vectors in the slots. And this we already know. This is, by definition, this is just eta mu nu. Repeat this exercise for any tensor you care to write down, any 0n tensor you care to write down. Go through all this manipulation, and you will always find that there's a very simple algorithm forgetting the components in, let's call it the barred observers frame, as converted from the unbarred observers frame. Essentially, you just hit it with a bunch of Lorentz transform matrices, and as an old professor of mine liked to say at this point, line up the indices. That's really all we're doing, is we're just going to line up the mus to convert them to alpha bars. Alpha bar here, mu there, put this guy here. I want to convert my new into a beta bar. I put my matrix there. Boom. Just line up the indices, and we're done. Now, this is, as I kind of emphasized, a fairly stupid example, because if you take the diagonal of minus one, one, one, one and you apply the most God awful immense Lorentz transformation you care to write down to it, you do all the matrix manipulation and you line it all up, what you'll end up finding is that this is the diagonal of minus one, one, one, one in all frames. That's actually one of the defining characteristics of the metric of special relativity. If you're working curvilinear coordinates, the metric is always minus one, one, one, one to all observers. So the recipe holds in general. This will hold whenever we are studying tensors from now and from henceforth. Just so happens that this first example we were given is kind of a dumb one. Nonetheless, learn the lesson and overlook the example, and wisdom shall be yours. I want to spend a few moments talking about a particular subset of tensors, of the 0n tensors, where n equals one. This is a subset of tensions in general that is known in many textbooks as one forms. For reasons that I will elaborate on in probably 10 or so minutes, these are also sometimes called dual vectors. And just if that's sort of making some neurons light up in your head, set that aside for a moment. I want to carefully go through them before I indicate the manner in which there is a duality that is being applied here. So if we go back to the definition of a tensor, this tells us that a one form is a mapping from a single vector to the Lorentz invariant scalar. So using some notation that I will probably only use in this lecture, because we're going to move past this notation pretty soon, let's say a one form-- I'm going to denote this with an over tilde on it. So let's say p is a one form. It will have in this sort of abstract notation a single slot, so I put the vector a into it. And this then gives me some scalar out. So in my notes, I go through some stuff indicating that this guy is-- it's a linear operation, but that's obvious, because it just inherits all the properties from tensors, so I'm not going to go through that. If you want to double check some of the details, they're in the notes that have been posted. Just like with the tensors, I extract components from this thing by putting my basis vectors inside. So if I take my one form p and I put it in my alpha basis vector, this gives me the alpha component of the one form. Notice, it's in the downstairs position. So one of the reasons why I want to go through this step is it gives me a way to think about what's going on at this scalar that I wrote down on the top board here. So what is this scalar that I get by putting a vector into my one form? So I take this, put my vector in here. I use the fact that I can write my vector using its components and the basis vector. I use linearity to pull this guy out. So the scalar that I get by doing this is just the scalar that one gets by contracting the upstairs components that I use to set up my vector with the downstairs components I use to set up my one form. This is an operation that is called contraction, for reasons that I hope are fairly obvious. So let me define a few other characteristics of this thing, and in just a few moments, we'll see what this is good for. OK, so one of things I'm going to want to do is change the representation of those things. So I'm going to want to know how these components transform between different frames of reference. But we've already done that. We did this using tensors, and this is just a tensor. So if I change reference frames, if I want to know what the components are according to some barred observer, I will step through the algebra in my notes, but I think you know where I'm going to go with this. You just take the components in the unbarred frame, line it up, contract it with the correct setup of my Lorentz transformation matrix, boom. Line at the indices, and we've got it there. So the last thing which I want to do with this before I talk a little bit about what this is really good for is say, you know, I've got these basis vectors that allow me to relate the components of my vectors to the geometric object in a way where I don't use a represented bi-symbol. I actually have an honest to God equal sign. Can I define a similar set of basis one forms? What I want to do is define a family of geometric objects, and I will denote them with an omega and a tilde such that any one form can be written as its components attached to these little basis vectors. Well, the way I'm going to do this is, I'm going to exploit the fact that I already know what basis vectors are. So I know, for instance, that if I take my one form and I plug in a basis factor, I get this. And so what I want to do is combine this thing which I would like to do with the defining operation of contractions. So I know p alpha a alpha is what I get when I've got my one form and I plug into its slot the vector a. OK, so let's insist that when I do this, I can write this as p beta omega beta. Now remember, these are the components. This is the actual basis one form. So I'm going to stick into my basis one form this form of the vector. I can then use the linearity of the tensor nature to pull out that component of a. So what this tells me is this is exactly what I want, provided whatever this geometric object is, it obeys the rule that when I plug basis vectors into it, I get the Kronecker delta back. Now, this may all seem really, really trivial right now. And indeed, if you think about this, in terms of just running down mathematical symbols, this is fairly trivial. One thing which I want to emphasize is that if you go through and say, well, if I'm working in a basis where this has a time-like component that I'll say is one, the time-like direction is zero everywhere else. Remember to set the points in the x direction, so it's zero, one along x, zero everywhere else. This then leads to-- so as an example, a set of basis objects that a particular observer would write just like so. I won't write out the two and three components. And again, you look at this and you think to yourself, dude, you're just repeating basis vectors. What's the big deal here? Now, I'm going to explain the fact that these are sometimes called dual vectors. So if we want to think about this in a language that is reminiscent of linear algebra, if you think of the basis vectors as column vectors, then my basis one forms are essentially row vectors. So these look a lot like my basis vectors. They enter in a dual way. And so they're going to play an important role in helping us to-- whenever I contract two objects together to make some kind of a Lorentz invariant scalar, I'm going to want to only combine objects that have a dual nature like. That's the only way I can get something sensible out of it. So let me give you an example. Mathematically, this is an equation that I can write down. No question. If I'm in a particular frame, I've got the components of vector a, I've got the complements of vector b, I can multiply their complements together, sum them, and square them. So this is mathematically well-defined but plays no role in the physics we are going to talk about this term, because this is not related to the underlying invariant structure of the manifold that we are working with. So remember I talked about how a manifold is essentially a sufficiently smooth set of points endowed with a metric? Well, the metric is what tells us that this is mathematically short. Write it down, but it means nothing. By contrast, of course this has frame independent meaning. This is important. So that's the sense in which these one forms are often called duels actors. It is when they are combined with vectors, they are duel to it in the sense that when they're combined in this appropriate way, we find that they describe the physics, they capture the invariant characteristics of the physics that are important in the theory that we are describing. If I may give one more example that is from a completely different field but I think helps to sort of illustrate a useful analogy to think about these things, suppose you are doing quantum mechanics and I give you two wave functions. So suppose you have a wave function psi of x and another wave function phi of x. If you wanted to, you could multiply them together and integrate overall space. I don't know what you would do with that, but you could. On the other hand, you could take the complex conjugate of one of them, multiply it by the other one, multiply it overall space, and in the notation that you learn about, this is, of course, just the inner product of wave function psi with wave function phi. Forming the one form, using a one form is akin to selecting an object that has-- it allows us to make a mathematical construction similar to the quantum mechanical enterprise that we use here. And as we'll see in just a couple of minutes, it's actually really easy to flip back and forth between one forms and vectors. OK, so a better way to move forward with this is rather than talking in terms of these more abstract things, let me you a good example. So imagine I have some trajectory through spacetime. So let's let the t axis go up. There's my x and y-axes. You all can imagine the z-axis. And some observer moves through spacetime like so. Last lecture, we talked about a couple of important examples of four-vectors. And so one which is germane to the situation is the four velocity of this observer. That just expresses the rate of change of its position in spacetime per unit proper time. And I'll remind you, tau is time as measured along this observer's trajectory. So roughly speaking, not even roughly speaking, exactly speaking, tau is the time that is measured by the by the watch of the person moving along there. So suppose in addition to this person sort of trundling along through spacetime here, suppose spacetime is filled with some field, phi, which depends on all of my spacetime coordinates. Question I want to ask is, what is the rate of change of phi along this observer's trajectory? So if you are working in ordinary Euclidean space, you would basically say, ah, this is easy. So you're three space intuition. You don't have this proper time to worry about. So you would just say that d phi dt along this trajectory is just what I get when I calculate dx dt. And then look at the x derivative of my field phi plus dy dt. This is one of those places where getting the difference between a partial and a total derivative right is important. So if you see me do that again, if I don't correct it, yell at me. OK, so you get this. And then you say, ah, this is nothing more than that particle's velocity dotted into the gradient of the field phi. It's a directional derivative along the velocity of this trajectory. So generalizing this to spacetime, you basically have the same thing going on. Only now, time is a coordinate. So we don't treat time as the independent parameter that describes the ticking of clocks as I move along it. I use the proper time of the observer as my independent parameter. So what I will say is, the rate at which the field changes per unit of this guy's proper time, every one of these is a component of the four velocity. So now, we introduce a little bit of notation. So this derivative is what I get when I contract the four velocity against a quantity that's defined by taking the derivative of this field. We're going to be taking derivatives like this a lot, so a little bit of notation being introduced to save us some writing. This is the directional derivative along the trajectory of this body. Now, this is a frame independent scalar. This is a quality that all observers will agree on. This is a four velocity. We know this is a four velocity. These are the components of a four-vector, so these are the components of a one form. So the generalization of a gradient is an example of a one form. So re-using that abstract notation that I gave earlier, I can write-- so the way you will sometimes see this is, the abstract gradient one form of the field phi is represented by the components, the alpha phi. You will also in some cases-- and I actually have a few lines about this in my notes-- sometimes, people will recycle the notation of a gradient that you guys learn about in undergraduate E&M. I urge a little bit of caution with this notation, because we are going to use this symbol for a derivative to mean something a little bit different in just a couple of lectures. It turns out that the something different reduces to this in the special relativity limit, so there's no harm being done. But just bear in mind this particular notion of a derivative here is going to change its meaning in a little bit. Another little bit of notation that is sometimes used here, and this is one more unfortunately, I think I'm stuck with this notation, one often says-- so this idea of taking a derivative along a particular four velocity, it comes up a lot. So sometimes, what people then do is, they define it as the directional gradient along u of the field phi. So don't worry about this too much. We'll come back to us a little bit more appropriate. I just want you to be aware, especially for those of you who might be reading ahead, when you see this. So just think of this as what you get when I am taking a gradient along a velocity u. It basically refers to the gradient one form contracted with the four-vector u. So the last thing I want to do as I talk about this is revisit this notion of one forms as being dual to vectors for just a moment. So we've just introduced the gradient as our first example of a one form. So the notion of the gradient as a one form, this gives us a nice way to think about what the basis one forms mean. So when I introduced basis one forms a few moments ago, it was a purely mathematical definition. I just wanted to have objects such that when I popped in the basis vectors, I got the Kronecker delta back. And after belaboring the obvious, perhaps for a little bit too long, we essentially got a bunch of ones and zeros out of it. Now, putting in math all of those words, I did a lot of junk to get that. But we also know that if I just take the derivative of my coordinate with my coordinate, I'm going to get the Kronecker deltas. This and this, these are the same thing. So I can think of this operation here as telling me if I regard this as this kind of abstract form of the gradient applied to the coordinate itself, this is nothing more than my basis one form. So my basis one forms are kind of like gradients of my coordinates. OK, you're sitting here thinking, OK, what the hell does this have to do with anything? So when we combine-- set that aside for just a second-- and remind you of some pretty important intuition that you probably learned the very first time you learned about the gradient. So imagine I just draw level surfaces of some function in-- I'll just do two dimensional space. OK, so I have some function h of x and y. This represents, like for instance, a height field. If I'm looking at a topographic map, this might tell me about where things are high and where things are low. So h of xy, there might be level surfaces on my map. It'd kind of look like this. And there'd be another one that kind of looks like this. And maybe it'd have something like this, and then something kind of right here. So we know looking at this thing that the gradient is very low here and very high here. Let's put this into the language of what we are looking at right now. OK, let's let delta x be a displacement vector in the xy plane, and dh is going to be my one form of my [] function. How do I get the change in height as I move along my displacement vector? Well, take my one form, plop into its slot delta x, I get something like this. The thing which I kind of want to emphasize here is, we have a lot of geometric intuition about vectors. So if I have a delta x-- let' say this is my delta x. It lasts for about this long over here. I take the exact same delta x, and I apply it over here. I get a very different result, because that goes through many more contours on the left side than it does on the right side. And the thing-- this is where this duality kind of comes in, and I'm going to put up a couple of graphics illustrating this here-- is that you should think of the one form as essentially that set of level surfaces. It's a little confusing. I'm not going to-- I mean, I can see a couple blank looks. Maybe even the majority of you have kind of blank looks on your faces here. And that's fine. So what I want you to regard is that when I'm talking about basis one forms and one forms of functions, they have a very different geometric interpretation, even though you're kind of used to gradient as telling you something about the direction along which something is changing. Actually, you define that direction. The thing that you're worried about is sort of how close the different level surfaces are of things. OK, so coming back to this idea that my basis one form that I use are essentially just the gradients of the coordinates. So I'm going to put some graphics up on the website, which I have actually scanned out of the textbook by Misner, Thorne, and Wheeler. And what they basically show is, let's say this is the time direction. Let's say this is my x-axis. And this is my y-axis. Your intuition is that the x basis vector will be a little arrow pointing along x. Well, what your intuition should be like for the x basis one form is a series of sheets normal to the x-axis that fill all of space. OK, spaced one unit apart, filling all of space kind of like that. So here's one example of one of those sheets. And notice, the x-axis pierces every one of those things. That's another way in which these are sort of a set of dual functions to the vectors themselves. This notion-- so I'm going to turn to something which is perhaps a little less weird in about 30 seconds. But the one thing which I kind of want to emphasize with this-- and again, I'm going to put a couple graphics up that help to illustrate this. And there's some really nice discussions of this. MTW is particularly good for this discussion. This ends up being a really useful notion for capturing how we are going to compute fluxes through particular directions. Because if I want to know the flux of something in the x direction, well, my x basis one form is actually like a sheet that captures everything that flows in the x direction. So it's sort of a mathematical object that's designed for catching fluxes. If isn't quite gelling with you, that's fine. This is without a doubt one of the goofier things we're going to talk about in this first introductory period of stuff. This is one of those places where I think it's sort of fair to say, if you're not quite getting what's going on, shut up and calculate works well enough. It's kind of like you know the Feynman's mantra on quantum mechanics. Sometimes, you've just got to say, OK, whatever. Learn the way it goes, and it's kind of like playing a musical instrument. You sort of strum it and practice it for a while, and it becomes second nature after a while. All right, so to wrap this up, what I want to do for the last major thing today is, I hope I can kind of put a bow on our discussion of tensors. Let's come back to the metric as our original example of a tensor here. So when I give you the metric as an abstract tensor, and I imagine I have filled both slots, I get a Lorentz invariant number. A dot b is what I get when I take this tensor and I put a and b into it slots. Suppose I only fill one of its slots. Well, if I take this, I plug in the vector a but I leave the other slot blank, well, what I've got is a mathematical object that will take a vector and produce a Lorentz invariance number. That's a one form. So let's do this very carefully and abstractly for a moment. But at this point, we basically have almost all the pieces in place. And so I'm going to kind of tone down some of the formality fairly soon. So let's define a one form, an object that takes a single vector inside of it as what I get when I take the metric and put that vector in there. If I want to get its components out, well, I know the way I do that is I put the basis vector in there. This guy is symmetric, so I can flip the order of the metric. It's symmetric, so I can flip the order of these guys. And what this tells me is that my one form component of a is just the vector component of a hit by the components of the metric. In other words, the metric converts vectors into one forms by lowering the indices. This is an invertebrate procedure as well. So this metric, I can define eta with indices in the upstairs position by requiring that eta alpha beta contracted with it in the downstairs position gives me the identity back. Incidentally, when you do this, you again find it's got exactly the same matrix representation. So this thing with its indices in the upstairs position is just minus one, one, one, one. It will not always be that way, though. Again, this is just because special relativity in rectilinear coordinates is simple. So I will often call that the inverse metric. And then, you shouldn't have a lot of trouble convincing yourself that if I've got a one form, I can make a vector out of it by a contraction operation. That now tells me that I have about 16 gagillian-- well, actually, it's a countable and finite thing. But I have many ways that I can write down the inner product between two vectors. This guy is-- if you like, you can now regard a vector as being a sort of a function that you put one forms into. These are all the same. These are all equivalents to one another. And actually making a distinction between vector in one form and all that, it's just kind of gotten stupid at this point. So the distinction among these different objects, the different names, kind of doesn't matter. And indeed, you sort of look at this. Up until now, we've regarded tensors as being these sort of things that operate on vectors. OK, but why not regard vectors as things that operate on one forms? What this sort of tells you is that this whole notion of tensors being separate from vectors that I talked about before is kind of silly. So I'm going to revisit the definition of a tensor that I started the lecture off with like so. So a new and more complete definition. A tensor of type mn is a linear mapping of m1 forms and n vectors to the Lorentz scalars. In this definition-- so we've already introduced zero, one tensors. Those are one forms. One zero tensors are vectors. Furthermore, as I kind of emphasized when I wrote this sentence here, the distinction between the slots that operate on vectors and the slots that operate on one forms, it's nice for getting some of the basic foundations laid. This is one of those places where now that the scaffolding is in place-- we've had the scaffolding in place for a while, but this wall of our edifice is pretty steady, so we can kick the scaffolding away. We can sort of lose this distinction. The metric lets us convert the nature of the slots on a tensor. So if I have an mn tensor and I use the metric to lower, that means I make it an m minus one n plus one tensor. So an example would be, if I have a tensor r mu beta gamma delta and I lower that first index, so this went from something that operates on one, one form and three vectors. Now, it's one it operates on four-vectors. Likewise using the inverse metric, you can raise-- and just for completeness, let's write that out. So an example of this would be if I have some tensor su beta gamma, and let's say I raised that first index to get something like this. OK, so let's see. In my last couple of minutes-- recall, I do have to leave a little bit early today, because I need to introduce a speaker. But I just want to wrap up one thing kind of quickly. I've spent a bunch of time talking about basis objects. And I'm going to go through this fairly quickly. The notes, if you want to see a few more details, you're welcome to download and look at them. They're not really tricky or super critical. We know we have basis objects for vectors, which hopefully you have pretty good intuition about. We have basis objects for one forms, where your intuition is perhaps a little bit more befuddled, but it'll come with time. So you might think, uh, now I've got two index tenors. I've got three index tensors. There's a four index tensor on the board. Scott's probably going to write 17 index tensor on the board at some point. Do I need a basis object for every one of those? So in other words, do we need-- glad that's caught on video-- do I need to be able to say something like the abstract metric tensor is these components times some kind of a two index basis object? Do I need to do something like this? Cutting to the chase, the answer is no. Basis one forms and vectors are sufficient. So what we're going to do is, abstractly just imagine that if I do have a tensor, I kind of have an outer product. I have both of the basis objects attached to this thing, and each one is just attached to those two slots. OK, so in this particular case, my two index basis two-form that would go with this thing here, the thing with two indices on it, I'm just going to regard this as-- I'll abstractly write this as just an outer product on the basis one forms. If I ever need a basis object for a tensor like this, I will just regard this as an outer product of these two things. So what's going on with that notation? What does this mean? Don't lose too much sleep about it. It's basically saying that I have separate objects attached to all of my different indices, and they're kind of coming along here and giving a sense of direction to these things. So for instance, if I have some kind of a tensor-- I mean, a great one, which we're going to talk about in just a little bit. It's a quantity known as the stress energy tensor, in which I can abstractly think of the tensor as having-- not just a [INAUDIBLE]. It will have two components. And I can think of it as essentially pointing in two different directions at once. Now, we're going to talk about this in a lot more detail in a couple of weeks. Actually, not even a couple of weeks. A couple of lectures. What I'm going to teach you is that the alpha beta component of the stress energy tensor tells me about the flux of form momentum component alpha in the beta direction. And so this is basically just saying, when I think of it as the actual object, not just the representation according to some observer, here is the thing that gives me the direction of my four momentum, and here is the direction in which it is flowing. And sometimes, we will make more complicated objects. And so you might need to imagine-- here's one which I actually wrote down in my notes here-- there will be times when we're going to care about a tensor which at least abstractly, we might need to regard as having this whole set of sort of vomitous basis vectors coming along for the ride here. And it is actually fairly important to have all these things that are in place. Where I will just conclude things for today is that the place where it is particularly important to remember that we kind of sometimes almost just implicitly have these things coming along for the ride here-- it's important when we calculate derivatives. So I gave you guys an example of a directional derivative for a scalar field that filled all its spacetime. I imagine that there was some trajectory of an observer moving through this. Now, imagine it isn't a scalar field that fills all of, spacetime but it's a tensor field. Here's t. We'll say this is the y direction. This is the x direction. Here's my observer moving through all this thing. And again, I'm going to say that this trajectory is characterized by a four velocity, dxt tau, and I'm going to imagine that there is some tensor field that fills all of spacetime when I go and calculate the derivative of this thing, when we're working in special relativity where we are right now, these guys are going to be constant, so it doesn't really matter. But soon, we're going to generalize to more complicated geometries, more complicated spacetimes, and the basis objects will themselves vary as I move along the trajectory. And I will need to-- in order to have a notion of a derivative that is a properly formed geometric object, I'm going to have to worry about how the basis objects change as I move along this trajectory as well. So that tends to just make the analysis a little bit more complicated. I have a few notes about this that I will put up on to the web page, but I don't want to go into too much detail beyond that until we actually get into some of the details of these derivatives. So I'm just going to leave it at that for now. So yeah, I'll just say, the derivative in principle and quite often in practice, it will depend on how these guys vary in space and time. And let me just say, you guys kind of already know that. Because when you studied E&M, you got these somewhat complicated formulas for things like the divergence and curl and stuff like that. And those are essential, because those are notions of derivative where you are taking into account the fact that when you're in a curvilinear coordinate system, your basis vectors are shifting as you move from point to point. The stuff we're going to get out this will look a little bit different, and it comes to the fact, as I emphasized in my last lecture, that we are going to tend to use what we call a coordinate basis, whereas when you guys learn stuff in E&M, you were using what's known as an orthonormal basis. And it does lead to slight differences. There's a mapping between them. Not that hard to figure it out, but we don't need to get into those weeds just now. All right, so I will pick it up from there on Thursday. So I'll begin with a brief recap of everything we did. The primary thing which I really want to emphasize more than anything is this board plus this idea that the metric can be used to raise and lower the indices of a tensor. At this point, talking about vectors, talking about one forms, many of you in a math class probably learned about contravariant vector components and covariant vector components. Once you've got a metric, it's kind of like, who cares? You can just go from one to the other. And that's why I tend to, almost religiously, I avoid the terms covariant and contravariant, and I just say, upstairs and downstairs. Because I can flip back and forth between them, and there's really no physical meaning in them. You have to think carefully about what is physically measurable, and it has nothing to do with whether it's covariant or convariant, upstairs or downstairs. All right, I will end there, since I have to go and introduce someone. |
MIT_8962_General_Relativity_Spring_2020 | 15_Linearized_gravity_II_Dynamic_sources.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: In the previous lecture, I began with the Einstein field equations in the most general form. And then I specialized to linearized gravity, to taking the spacetime metric, assuming that I can find a coordinate system in which it is close to the eta alpha beta form that we use for flat space time, plus a small perturbation, something that's defined. So I set whenever I have that h multiplied by itself I can neglect it. Doing so I was able to recast the Einstein field equation as a simple linear equation, which is the wave operator acting on a variant to that perturbation, the trace reverse version of that perturbation. It is minus 16 pi g times the stress energy tensor. So in doing this recasting, I have-- this is essentially just a trick of reorganizing the variables. I have taken my h alpha beta, which describes the way in which spacetime is shifted away from that of flat spacetime, the perturbation of flat spacetime, and I've just rearranged that to give it the trace with the opposite sign. And I've chosen a gauge such that the divergence of that trace reverse metric perturbation is zero. If we take this equation, which is completely general for linearized spacetime, the only thing which I had to do is choose a particular gauge. And there's no physics in that. That's essentially a coordinate choice. If I then imagine that my source is static, I will have static fields. And so my wave operator goes over to a simple Laplace operator. And that it's not too hard to show that the solution that emerges from this looks like this, where phi n is Newtonian gravitational potential, which arises from a distribution of mass like so. So this is the leading solution. And it's one that is very powerful and very useful. And it's actually using tremendous number of physics and astronomical applications. This restriction to a static source, though, is-- well, it's restrictive. What if I do not want to simply consider a static source? What if I am interested in sources of the gravitational interaction that are themselves dynamical? And so I emphasized last time that this wave equation, my box h bar alpha beta equals minus 16 pi GT alpha beta, this is something that I hope you have already seen in the context of electrodynamics, in particular whenever we have a wave equation, who's not even a wave equation, but any differential equation that is of the form linear differential operator on field is equal to a source. OK, so I'm separating out time and spatial behavior here. Whenever I have a situation in which I'm interested in studying a field in which some differential linear operator acting on that field is equal to a source, we can solve this using the technique of Green's functions. I am not going to be able to go through the technique of Green's functions in detail. I will quickly give you a synopsis of how it works and then I will look at the particular Green's function that applies to the wave operator. Students who would like to read more about this, I would point you to the textbook by Arfken. I believe it is called Mathematical Methods for Physicists. And in the third edition, Green's functions are described in Section 16.5 to 16.6. So in a nutshell, let me just give you a very brief synopsis of how this technique works. Suppose we take our source, which may be some complicated, very ornate entity. And what we're going to do is replace that source with a delta function. So imagine that I take s of t and x over 2a delta at some time t prime and a three-dimensional delta at some location x prime. What I'm basically saying here is I'm imagining I might have some source that's extended in time and extended in space, and I'm just going to look at the little blip at one particular event and see if I replace my source with this one single blip, what field emerges from that? What we're going to do is assert that a solution exists. And we will call the field that arises when my source is replaced with a delta function, we'll call it capital G. OK? This notation means that this is the amount of field at t, x, arising from my source at t prime x prime. So my differential equation for this solution G now looks like so. So whatever my linear operator is when it acts on this G, it gives me the source, which is now the delta function. Let me just introduce a little bit of notation here. So t and x, I call this the field point. This is the point at which the field G is being measured. t prime, x prime is my source point. This is the point at which the source is non-zero. In this particular case, it's actually a little delta function spike. The reason why we do this is pretty much by definition. Pardon me one second. Yes, pretty much by definition. If I take the source and integrate it against the delta function-- integrate over all time, integrate over all space-- I must get the source back. This is a tautology. All I'm doing here is really saying this is the properties of delta functions. However, what's sort of remarkable is if instead of integrating the source against the delta functions-- if instead of integrating the source against the delta functions-- I integrate the source against the Green's function, if I do this, then what I must get out of it is the solution f. This is actually very easy to prove. Simply take this equation-- let's call this equation A-- if you take a equation A and operate on it with your differential operator D, so if I hit equation A with D-- remember D, it's a differential operator that acts on only the unprimed coordinates. You know what? Let's just go and write this out. So D on this integral, when I hit D on this, it goes straight through the integral. This only acts on the unprimed coordinates. So it ignores the integrals. It ignores the s. The only thing that operator hits when I act on the integral is the Green's function itself. But I have asserted that the Green's function is such that when D acts on it, I get the delta functions out. And by the definition of delta functions, this just gives me my source back. So what this means is that if we can find the Green's function for our differential operator, all we need to do is integrate it against our source. And the result must be the solution to our field equation. It's a beautiful technique. So if you haven't seen this before, I highly recommend that you turn to that reading in Arfken. I'm going to take advantage of this. I'm going to use this technique now. In particular, I'm going to take advantage of the fact that if my differential operator is in fact the flat spacetime wave operator, there is a very well-known solution for the Green's functions. It's called the radiative Green's function in this case. I'm just going to write down what the solution is. So I'm going to talk about the properties of this in just a moment. So I'll just comment that this is something that you should be able to find developed in any good quality, advanced electrodynamics textbook. In my notes here I suggest looking at this place where I originally went through this carefully was in the second edition of Jackson. It's presumably also present in the latest edition. But it might be in a slightly different place. In the second edition, you can find this discussed in Section 6.6. When I have taught MIT's junior level electrodynamics course, I actually go through this. And by essentially introducing a Fourier transform, a Fourier decomposition, turning the wave operator into an algebraic operator, just rearrange a bunch of terms, then invert a Fourier transform. It's not too hard to show that this is the result that comes out of that. Notice the argument of that delta function. And please bear in mind, we are working in units in which we have set the speed of light equal to 1. So the delta function, it has support only at-- well, what's going on here is this is taking into account the amount of time it takes for information to travel from the source point to the field point. OK, if I can illustrate this with a little cartoon, imagine I have some dynamical source of mass and energy that's down here wiggling away. Here is my one example of a source point that is going to end up inside the integral when I integrate up the Green's function against my source. Here I am. I'm an observer making my measurement out here at position x. Whatever is happening at x prime takes time to be communicated to me making my measurements out here at distance x. The time lag is the distance that has to be traveled, x minus x prime. And, again, I'll remind you, we're working in units with a speed of light is equal to 1. We would divide by c if we were working in normal SI units. OK? So this factor in the Green's function is building in-- I'm going to do the integral in just a moment-- it is building in the fact that we need to account for the amount of time it takes for information about the source's dynamics to be radiatively communicated to observers far away from that source. All right, so we are now going to apply this to the linearized-- the Einstein field equation for linearized gravity. So here is our field equation. Sadly, the letter capital G is doing multiple duties in this lecture. So when we solve this, what we need to do is integrate t alpha beta regarded as a function of prime-- time t prime, position x prime against the radiative Green's function. Plug in the radiative Green's function, which we use. This is the Green's function that corresponds to that wave operator. We have a minus 1 over 4 pi hitting our minus 16 pi out front there. We're going to do the integral over the time variable. And we're going to leave our result expressed as an integral over space. This is an exact solution of the linearized field equations of general relativity. This is telling me that what happens-- the field that is measured, the trace reverse perturbation that is measured at time t and position x prime is given by integrating over my source, all the dynamics that happened earlier than that time t. What I need to do is take into account that over this source I have to fold in the amount of time it takes for information from source point x prime to go out to x. I integrate over the source. At every point in the integrand, I divide by the distance between the field point and the source point. Do that integral, multiply by 4G. There's my solution. This is wonderful. It's actually an exact solution. Like most exact solutions, it's not always the most useful thing in the world. There's one problem which we need to diagnose, which is that in the way that this has been formulated, it looks like every component of this trace reverse metric looks radiative. What do I mean by it looks radiative? Well, it came from a wave equation. They all depend on information that traveled out at the speed of light with a time lag appropriate to the speed with which radiation moves in flat spacetime. You know, you look at that, and you sort of say, wow, the entire metric has this sort of character that we associate with a propagating wave. All of spacetime is radiative. We need to be careful about that, because here is where we need to revisit and think carefully about what happens when we choose our gauge. What I'm going to show you is a little demonstration that in this case, the gauge we chose-- and let me emphasize that was a wonderful gauge for putting the horribly messy field equation into a form where we could actually solve it-- but the gauge we chose is unfortunately masking some of the physical character of the spacetime. So this Lorenz gauge was great for producing a set of field equations that we could bring to bear powerful mathematical techniques and get a closed form actually exact within the context of linearized theory, an exact solution. But it might be misleading us as to what the resulting solution means. Let me illustrate this with an example from electrodynamics. Suppose I gave you a vector potential that was of the form at a time-like component that was q over r minus q omega sine q dot r minus omega t capital R over r. Little r is square root of x squared plus y squared plus z squared. And capital R is some parameter with the dimensions of length. And my three spatial components of this thing, I'm going to write them as kqi sine k dot r minus omega t capital R over r minus qxi. OK, this is the potential that has a form that kind of looks like if I were to sort of give this to you on an oral exam and say without doing any calculation, what would you guess this is? And that would actually be, as you'll see in a moment, that'd be a very unfair question of me to ask. And I look at this and I say, you know what? That looks kind of like a Coulomb field. And then there is something radiative that it's embedded in, kind of a funny amplitude associated with that radiation. But maybe this is sort of like a dipole radiator with a net monopole moment going on it. You know, we've got things that involve fields that oscillate in space and time. They fall off as 1 over r's. A component falls off as 1 over r cubed. Yeah, that would be my guess-- a point charged with radiation. So your oral examiner, the next thing they say is, great, work out the electromagnetic field tensor corresponding to this. So you do this by taking some derivatives. And what results is this. This is the electromagnetic field tensor corresponding to a Coulomb electric field with no magnetic field. This potential is nothing more than a Coulomb point charge in a really dumb gauge. The way I actually generated this was I started with the q over r potential of a Coulomb point charge. And I just applied some crazy gauge generator to it to see what happened. So the parable of this story is whenever you are working with quantities that are subject to gauge transformations, you have to be careful about drawing conclusions about what those quantities mean in terms of the physics you want to get out of it. Gauge can obscure the physics if we are not careful. And in particular-- there's almost like a conservation of pain principle here-- it is often the case that the best gauge for formulating tractable equations for solving your problem turn out to be gauges that give you just kind of crappy results if you want to interpret what's going on physically. In the remainder of this lecture, I am going to introduce something that's a slightly advanced topic. And I pretty shamelessly stole this from a paper that I wrote, something that I did with a colleague of mine, Eanna Flanagan. And it's based on an idea that was developed in a different context by my MIT colleague Ed Bertschinger. So here is what we're going to do in the remainder of this lecture. What we're going to do is recast the metric and the source in a form, which allows us to categorize the radiative and non-radiative degrees of freedom of the spacetime. This will, in fact-- we're going to do this in a way where the results are, in fact, completely generic, provided we stick to linearized gravity, gravity linearized around flat spacetime, which will be sufficient for our discussion right now. I will give you the punch line upfront. What we will discover is that spacetime has four physical degrees of freedom that are non-radiative. And the hallmark of this is that they are going to be governed by differential equations that look like the Poisson equation. They will look like the Laplace operator on some potential is equal to a source. Let me rewrite that. This handwriting is terrible. Spacetime has two more degrees of freedom that are radiative. These are governed by wave equations. The metric tensor of spacetime has 10 components, 10 unique components. It's represented by about 4 by 4 symmetric matrix. OK? So it has 10 components in it. 4 plus 2 does not equal 10. You might look at this and think, what about the other four components of this thing? I've got 4 degrees of freedom. That must correspond in some sense to 4 of those components. I've got another 2 degrees of freedom. That must correspond to two of those components. What's going on the other four? Well, the other 4 degrees of freedom are essentially eaten up by our gauge freedom. OK, we are free to specify our coordinates via an infinitesimal coordinate shift. And so we have 4 degrees of freedom under our control. General relativity insists on giving physics to the other six. These four non-radiative degrees of freedom, the two radiative ones, and the four gauge degrees of freedom completely describe the spacetime metric. And just to use an analogy, these non-radiative of degrees of freedom are kind of like the non-radiative electric and magnetic fields that you get in electrodynamics. Your two radiative degrees of freedom are going to turn out to two polarization states associated with waves in the gravitational field in the same way that in electrodynamics, the electric and magnetic field also have two polarizations associated with them. So this is all worked out in great detail in a paper that Eanna Flanagan of Cornell University and I wrote about 16 years ago. And so I will make available through the 8.962 website a copy of that paper. And it was Flanagan who sort of got this started. And he and I then figured out all the details together. Ed Bertschinger is sort of the grandfather of this idea since he did something very similar in an analysis that characterized perturbations to cosmological spacetimes. All right, so here's how we proceed. Let's consider h mu nu as a tensor field on a flat background. In the previous lecture that I recorded, I described how when I am working in linearized theory, it's a useful fiction to think of h mu nu. Even though we know it's actually telling us about the curvature of spacetime, it's a useful fiction to think of it as a tensor field in a flat background. And we're going to take advantage of that. We are going to choose time and space coordinates. Once I have chosen time and space coordinates, I then think about how this 4 by 4 tensor, how its different components break up and how they behave with respect to rotations in the spatial coordinates. I can imagine going in and tweak my spatial coordinate system. And what I want to do is examine how the components break up into subgroups with respect to spatial only coordinate transformations. What we'll see is h mu nu is going to break up into-- you're going to get one piece. It's your time-time piece. OK? If I have chosen my time and my space coordinates and I imagine messing around with my spatial coordinate system, htt is unchanged. And so this-- I'm actually going to give it a name. I'm going to call it minus 2 phi-- this behaves as a scalar. My spacetime piece-- so components of this, of the form hti-- I'm going to characterize this in more detail a little bit later. But these three numbers are going to behave with respect to spatial only coordinate transformations like a vector. Likewise hij is going to behave like a 3 by 3 tensor. So let's break this down a little bit further. Whenever I am dealing with a function that is vector in nature-- and bear mind it's a field. So I'm dealing with a vector field-- I'm going to write this as a divergence-free function plus the gradient of some scalar. So what I'm going to do is say hti is equal to beta i plus the gradient of some gamma. And I'm going to require that the divergence of that beta be equal to 0. Notice, since all of my indices are spatial indices and I am working in nearly Lorenz coordinates-- the placement of the indices, whether it's upstairs or downstairs, is immaterial-- I will tend to write them all in the downstairs position. And so repeated indices-- we're going to sort of abuse the Einstein notation a little bit-- repeated indices will be summed over. Let's extend this logic to the tensor piece, to hij. This takes a little bit of thought. So let me just write out the answer and then describe the character of every piece that goes into this. So hij can be written as hij tt-- I will define that in a moment-- plus 1/3 hij plus d, sort of a symmetrized gradient of a vector epsilon, plus delta i delta j minus 1/3 delta ij plus operator on some function lambda. So let me go through and describe what each of these things mean. And while I'm at it, let me count up the number of degrees of freedom associated with them. Let's back up for just a second and do that over here. Gamma is a scalar. So there's 1 degree of freedom associated with gamma. Let me move this over to here. So that's 1 degree of freedom. Beta is a vector, but it's divergenceless. Because it's a vector, I have three components. But being divergence-free, those three components obey a constraint. So that's 3 minus 1. So this scalar and this divergence-free vector give me the 3 degrees of freedom I need to specify the 3 independent components of hti. Over here, as I look through all of these different functions, h is a scalar. I have defined it in such a way-- hold on just one second. I'm going to come back to that point in just a moment. Epsilon j is a vector. It is defined such that its divergence is 0. OK? So what is going on here is this contribution to hij, it gives me a piece of this thing that looks like the gradient of vector. This has 3 degrees of freedom minus 1 constraint. This is a scalar. It only has 1 degree of freedom associated with it. Lambda is another scalar. Notice whereas h feeds directly into hij its derivatives of lambda that feed into this. So this has one degree of freedom. But this is defined in such a way that if I take the trace of hij, this operator gives me 0. Right? So the trace-- what the trace will do is sum over elements when i and j are equal. So this will give me essentially the Laplace operator. But the trace of delta ij is 3. So I get Laplace operator minus Laplace operator, I get the 0 operator acting on lambda. So this gives me 1 degree of freedom. This gave me 1 degree of freedom. hij tt-- and you know what, I'm going to go to a separate board for this one. This is a tensor. tt stands for transverse and traceless. This is defined so that if I take its trace-- oops, I said this would all be downstairs-- if I evaluate this, I get 0. OK? So hij is a tensor but with no trace. And it's defined so that if I take it's divergence, I get 0. So this guy has 6 independent components. This is 1 constraint. This is 3 constraints, right? Because it holds for every value of j. So this ends up giving me 2 free functions. So as we sum all these up-- let's see, did I miss anyone here? Right. So I have totally characterized h mu nu by a set of scalar, vector, and tensor functions. I have a scalar phi, which I have over here with my time-time piece. I have a scalar gamma. I have a vector, beta i. I have a scalar h, a scalar lambda, a vector epsilon i, and finally, a tensor hij tt. OK? So what I've done is-- let me just step back for a second. What I've been doing in this exercise is trying to figure out with respect to its behavior under rotations and then just because it'll prove to be convenient in a moment its behavior when I sort of look at casting quantities in this irreducible form that involves divergence-free vectors and gradients of scalars. This is sort of the equivalent of doing that for a tensor function. The 10 independent components in h mu nu have now been encapsulated by-- I've got one function here 1 here, 2 here, 1 here, 1 here 2 here, 2 here. 1, 2, 4, 5, 6, 8 and 10. So all I've done is rewrite those 10 independent components of h mu nu into a form, that as we're going to see, is particularly convenient for allowing us to understand what the gauge invariant degrees of freedom in spacetime are, at least in linearized theory. All right, so, so far, all I've really done is rewrite my metric using a set of auxiliary variables that might look a little bit crazy. What we would like to do is examine what my linearized Einstein field equation looks like in terms of all of these new fields. So in terms of the phi gamma beta h lambda epsilon hij tt, I would like to run this through the mechanism, make my Einstein field equation, but express it in terms of these things. We'll see why that is in just a moment. But before I do this, we have to be a little bit careful. We have 10 functions here. We have 4 gauge degrees of freedom. And the whole point of this exercise is to try to understand how to deal with the fact that a gauge that is convenient for doing my calculation may leave me with a result that is confusing in terms of the physics I'm trying to understand. So what we're going to do is say that I can take my generator of a gauge transformation, xi alpha. I can write this since I have chosen time and space coordinates. I can break it into a time-like piece and a spatial piece. I'm going to say that the timeline piece is some scalar field A. And spatial piece looks like a vector field B sub i plus the gradient of some scalar c. And I'm going to require Di Bi to be equal to 0. The reason why I'm doing this is these functions that I came up with, my phis and gammas and betas and epsilons, whatever, they are going to change if I introduce a gauge transformation. So suppose I change gauge as we learned how to do in the previous lecture. When you do this, what you discover is the function phi changes to the original phi, so something that looks like the time derivative of that scalar A. The beta picks up a term that looks like the time derivative of the vector field B. Gamma goes to gamma minus A minus time derivative of c. H goes to H minus 2 nabla square root of c. So all of these functions that we sort of cleverly introduced trying to write these different pieces of the metric of spacetime in an irreducible form, when we change gauge, they change in this way. I left one out. It turns out when you change gauge, hij tt goes to hij tt. This piece is actually gauge invariant. That's really interesting, because that tells me that once I have worked out my field equations, the piece of it that describes this transverse and traceless piece of the metric perturbation, it has meaning in any representation that I write down. So the piece that looks radiative in one representation, in fact, is radiative in all representations. We still have to figure out what's going on with all this stuff. OK? So we have a little bit more work to do. But we've just learned something very interesting about this. So at this point, there's really no simple way to describe what you do next. Honest to god, what you essentially do is you just stare at these things for a while. And you start to notice that there are certain combinations of these different functions that, like hij tt, certain combinations of them are gauge invariant. So if I define capital Phi to be little phi plus the time derivative of my gamma minus 1/2 2 derivatives of lambda, if I define theta as 1/2 h minus Laplace operator on lambda. I define the vector field, psi i to be beta i minus 1/2 epsilon i-- note, this vector field is divergence-free-- and, of course, hij tt. Every one of these combinations is unchanged under a gauge transformation. But let's note something. Phi is a scalar field. Theta is a scalar field. So I have 1 plus 1 functions associated with them. This is a divergence-free vector. So it has 3 minus 1 degrees of freedom associated with it. And hij, this is traceless and divergence free. And as I already counted up, this guy has 6 minus 1 minus 3 degrees of freedom in it, also 2. So when I characterize the gauge invariant degrees of freedom in the spacetime, I've only got 6 functions left. But this is good, right? My original 10 degrees of freedom in the spacetime metric are characterized by these sort of 6 fundamental degrees of freedom in the gravitational field plus 4 gauge degrees of freedom that are bound up in my gauge generators. So what I would like to do is see if I can write the Einstein field equations in terms of these gauge invariant degrees of freedom. And you know what, I'm going to write it in its full form. We're, of course, going to linearize this. But what I'm going to do is take this thing, write it in linearized gravity, using my gauge invariant variables. Before I do this, it's really helpful to first decompose the stress energy tensor in a manner similar to how we decomposed our metric. So what I'm going to do is I've chosen time and space directions. So I'm going to call the time-time piece rho in energy density. The timespace piece, we know that this tells me something about the flow of energy or the density of momentum. And I'm going to write this as a divergence-free vector plus a gradient of some scalar. And I am going to describe the space-space piece as an isotropic pressure, an anisotropic term-- I'm going to give you a constraint on this in just a moment-- some kind of a gradient of a vector field, and then a second order trace free operator acting on a scalar. So, yeah, I'm going to want to introduce a couple of constraints here. I'm going to require-- I've already written out that the ISI is 0-- I'm also going to require that the divergence of sigma be equal to 0, the divergence of the sigma ij be equal to 0. And I'm going to require that the trace of this trace be equal to zero. Let me strongly emphasize that really all I'm doing is rearranging terms. I'm just trying to rewrite the components of my stress energy tensor using this decomposition under rotations and then looking at fields that can be written as divergence-free vectors plus gradients of scalars. And I have a typo. I'm just trying to do this in a way that parallels what I did for the metric. OK? Before I do this, don't forget that I am required to make sure that my stress energy tensor satisfies a law of local conservation of energy and momentum. When we do this, what we find is that these various things that I introduced here, some of them are related to each other. So in particular, what you find is the Laplace operator on s is equal to the time derivative of the density, the energy density. Laplace operator on the scalar sigma is related to the pressure and the time derivative of this s. Finally, Laplace operator on that sigma i is related to also the time derivative of this derivative of si. What this tells us is that-- so I really want to make sure people don't get too hung up on this. This is really just sort of the a convenient way of reorganizing the information in those terms. But if you do want to think about this, this is telling us that only rho, p, si, and sigma ij are freely specifiable. If you know these, there's a total of 6 functions here-- 1 scalar, 1 scalar, 3 vectors minus 1 constraint, 6 tensors minus 4 constraints, because this is divergenceless and trace-free. These are the only ones that are freely specifiable. They determine the other 4 fields. OK? So density, pressure, kind of an energy flow, and anisotropic stresses. OK, redemption is at hand. Take all of the framework that we developed and that we discussed in the previous lecture. And let's grind out the components of the Einstein tensors. Doing so to linear order in h, which is equivalent to saying linear order in all of these fields that we've introduced, what you find is Gtt minus Laplace operator times this field theta. Gti is minus 1/2 Laplace operator on this vector field psi minus the time derivative, the gradient of the time derivative of your theta. And Gij, it's like a wave operator on the tt piece of your metric. OK, that's a lot. Let's equate them to the source. And we'll take advantage of the fact that when you do this, you're always going to associate like with like, OK? Terms that are divergence-free, you're going to equate to a source that is divergence-free, things like that. This completely characterizes the Einstein field equations, solutions to the Einstein field equations in linearized gravity. Now, I want to make a couple of remarks about this. Part of what was the motivation for this entire lecture was this observation that when we solve the field equations using the radiative Green's function, by construction, the entire spacetime solution had this radiative character associated with it. Everything sort of fell off as 1 over r and had a time dependence that reflected a time delay, the time it takes for radiation to travel from the source to the point at which the field is being measured. But we saw via this electrodynamic example that we put up just for intuition's sake, that it's entirely possible to have a totally non-radiative field that looks radiative basically because we chose a gauge that masked its physical character. This was a lengthy and-- I'll be blunt-- a somewhat advanced calculation. I do not expect everyone to be able to follow this in great detail. But I want you to understand how we got to this final end result here. By decomposing the spacetime, the perturbation to spacetime in linearized theory, into sort of as irreducible as possible a set of functions, we found that there are exactly 6 degrees of freedom in that spacetime that are completely gauge invariant. These functions, they will have this form. They will obey these equations. In principle, you can solve these things and understand how these fields behave no matter what gauge you are working in. And what we see is 2 degrees of freedom-- remember, hij tt is a 3 by 3 tensor, but its traceless and its divergenceless. So there's actually only 2 degrees of freedom here-- it is indeed a wave equation. It is the only piece of the metric that is a wave equation in every gauge and in every representation. All other gravitational degrees of freedom obey something more like a Poisson equation. These are non-radiative. This is radiative. The way that we are going to use this-- the next thing which I want to talk about is gravitational radiation. And what this calculation told you is that if you are interested in gravitational radiation, this is the only piece of the spacetime and the source that you need to be concerned about. If you solve that equation, it will give you the gauge invariant radiative degrees of freedom in a spacetime no matter what representation you use. So we are going to-- for the next couple of lectures, we are going to focus on this. We're going to show how we can sort of solve the Einstein field equations in a particularly convenient gauge and then say, all right, I know I've now got these 10 functions. Four of them are just purely things I can get rid of by choosing my gauge. There's only six that are sort of truly physical. Of those six, four of them don't constitute true radiative degrees of freedom. There's only 2 degrees of freedom in my solution that described radiation. We're going to talk about how to pull them out, how to characterize them, and the important physical content that this gravitational radiation carries. So in a nutshell, you should take this lecture as demonstrating that in a very deep way spacetime always has some kind of a radiative component associated with it. But it's really only encapsulated in 2 degrees of freedom in the spacetime. It may look otherwise. But just be careful that because your gauge has confused you essentially. And with that, I will stop this lecture here. |
MIT_8962_General_Relativity_Spring_2020 | 16_Gravitational_radiation_I.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: OK. So let's just dive in. And I'll just quickly recap where we were last time. So what I did last time was, we are now beginning the adventure, so to speak, of looking at how one examines-- how one finds and then studies the properties of solutions to the Einstein field equations. In other words, given a particular source, what is the space-time that emerges? What is the space-time that is consistent with that source of gravity in Einstein's relativistic theory here? So we begin-- we're going to look at this in a couple of different ways. And the first one is where we're going to-- what we do-- linearize the Einstein equation. So we're going to imagine that space-time is a small deviation from the flat space-time that contains no gravity. And we are going to expand the Einstein field equations, discarding all terms that are non-linear in the perturbation away from flat space-time. So last time we did this, we developed our field equations, we recast them in a form that's particularly convenient for making solutions, and we noted that the coordinate degrees of freedom can be thought of as a kind of gauge freedom that, in some ways, it's very useful for us, because it allows us to write the equations in a form that is particularly amenable to solving, but can also be a little bit dangerous, because you can sometimes wind up with a solution whose physical character is masked by the nature of that gauge. I went through a somewhat advanced topic for the purposes of 8.962, and I do not expect students to follow this in detail. But it's good to at least survey this and understand what was going on with this. What I did was I decomposed the perturbation to flat space-time into irreducible degrees of freedom. I sort of chose a time coordinate. I then imagined the other three coordinates in our space, and I looked at how, having chosen time, the different degrees of freedom decompose in two scalars with respect to coordinate rotations, vectors with respect to coordinate rotations, and two-index tensors with respect to coordinate rotations. I then took those, particularly the vectors and the tensors, and I further broke them up into degrees of freedom that are either divergence-free vector fields or the gradients of scalar fields. And in doing this, what we found was that the 10 degrees of freedom that are specified-- excuse me, the 10 degrees of freedom that are present in the perturbation the space time, can be broken down into six functions that are gauge-invariant plus four gauge degrees of freedom. Applying a similar decomposition to the source, we found that those six gauge invariant degrees of freedom are governed by four Poisson type equations. And I will refer you to the previous lecture for exact definitions of these things, but there is a scalar field phi that is related to the mean density of energy in your space-time and pressures in your space-time-- sort of isotropic stresses. This DTS is an additional degree of freedom that is defined in that previous lecture. You find that a Poisson equation for a scalar field I called theta is directly related to the energy density. There are two degrees of freedom associated with sort of off diagonal terms in the metric-- the time-space terms in the metric. And they are related to the momentum flow. These are only two degrees of freedom, because the psi is a divergence-free field. And finally, what we found is if you take the spatial piece of the space-time metric, you project out the bits that are transverse-- in other words, it's divergence-free. We will come to a more physical meaning of this a little bit later in today's lecture. And you require that this be traceless-- then those components of the spatial space-time perturbation are already gauge invariant, and they're related to sort of anisotropic stresses in our source. They are governed by a wave equation. And as such, these represent a radiative degree of freedom. Because this is [coughs] (excuse me!)-- because this is trace-free and divergence-right, naively you would think that this has six numbers in it. A symmetric 3 by 3 matrix is how you would represent it. But because it is trace-free and because it is divergenceless, there are four constraints. And so those six independent numbers are reduced to two. And as we'll see in this lecture, those two degrees of freedom correspond to the polarizations of gravitational waves. Today's lecture is all about understanding these guys. Given our kind of funny schedule that we're on today, I got up early and have not had time to take a shower, so I am wearing my LIGO hat. And what we're going to be doing today is the foundations of how one calculates this and why one builds an observatory like LIGO to measure these things. All right. So let me just write that out. So our goal for today-- I should say our goal for this lecture, because this is the first of three lectures I'm going to record today. Our goal is to understand hij TT in terms of quantities that we can observe and also to understand how to compute this field-- this hij TT given us source. So let's begin by just assuming. Let's begin by focusing on how we can understand what hij TT actually means-- how we can go and measure this. So what I'm going to do is just we'll start out by let's just hand ourselves an hij TT and study its properties. So what I'm going to do is begin with a space-time perturbation that has everything except hij TT equal to 0. So this is something that has phi equals 0, theta equals 0, psi equal to 0. I am going to pick hij TT. If we have a form such that it only depends-- it's a function that depends only on the combination T minus z. So this represents something propagating-- as we see, it actually describes radiation. in the z-direction. So this assumed form can be represented by the following matrix. So if you take a look at this, I'm going to justify a few details of this later. There's a few constraints. So it looks like I have four numbers here. Because my space-time metric is always symmetric under-- we'll just say symmetry-- under exchange of indices, h xy is the same thing as h yx. So these two things are the same. And from the requirement that this construction be trace-free-- so from the way I've written this here, that tells me that h xx plus hyy h yy must equal 0. So these two quantities on the diagonal are not independent. They must be of opposite sign. So there are my two degrees of freedom. Why have I set the z part equal to 0? We'll look at that in a little bit more detail a little later in the lecture. But essentially, it comes from the fact that I'm making it propagate in the z-direction. What we're going to see is that if-- whatever direction the wave is propagating in, my field has to be orthogonal to it. And we'll prove that a little bit more carefully later. So, for now let's just take this as handed to us. We can see this is at least consistent with the properties I've written down here, from the fact that it depends on T minus z, it's not hard to show that this thing will satisfy the wave equation. So we have a good source for exploration in front of us-- excuse me-- a good space-time for us to explore the properties of. So let's begin by saying great. One of the most important things we've learned how to do is, given a space-time, I can compute how a freely falling body moves in that space-time. So let's consider what happens if I have freely falling bodies in a space-time, in a space-time such that this-- in a space-time that is this-- so I am not saying this very fluently. So this is what we want to do. We're going to consider freely falling bodies in a space-time that involves this h mu nu. And the way I'm going to do this, my first pass is to sort of say, OK, I know all about the motion of freely falling bodies in space-time. They are geodesics. Geodesics describe such bodies. So let's go ahead and look at the geodesic equation for this setup. Now I wanted to-- let's give ourselves a couple of conditions to try and understand this. Let's imagine that these are-- we're initially in a space-time that is just flat space-time. So we're going to imagine that h mu nu is 0 initially. And perhaps it then-- we're sort of imagining a space-time in which there's just you falling freely in empty space. Absolutely nothing going on here. And this h mu nu field passes over you. A wave packet passes over your freely falling particles. So h mu nu, not equal 0, comes along, and passes over your particles. I'm going to imagine that these bodies are initially at rest. So when I'm in this regime where h mu nu is equal to 0, these guys are just sitting stationary with respect to whoever is making these measurements. So we're going to say great. I've got these bodies sitting at rest in what is initially empty space-time. A wave passes along. What happens to these bodies? So if these guys are at rest, then I know that, at initial times-- so let's just say that this comes along, and it becomes h mu nu not equal to 0 at some time-- what is defined as the origin of time. So t equals 0 is when the non-zero wave comes along. So my space-time is just eta mu nu for all earlier times. My particles are sitting here at rest at all earlier times, then h mu nu comes along. So what we want to do, then, is look at how does this guy behave at t equals 0. Now, initially, the only component that will get picked out of this is the 00 component. And, working to linear order, I can write this. Just wanted to write one more step in here. Let's say that this, I'm going to write as eta alpha beta beta 00. Go back to your earlier lecture, when we defined the Christoffel symbol to look up the definition of this guy. So these are all the only components of the space-time that enter into this Christoffel symbol. Every single one-- let's go back and take a look at the space-time that we're looking down here. Notice this involves beta 0, 0 beta, and 00. That is the space-time parts, all of which are 0. The other space-time parts, all of which are 0. And the time-time part, which is 0. So the initial acceleration is 0. That means du alpha d tau is equal to 0. Which means that my particle begins on a geodesic, and it stays on the geodesic. In these coordinates, it does not move at all. This seems to suggest that gravitational waves have no effect. I have a small body here. Gravitational waves come along. It's unaccelerated. So what's this for? Well, if you think about this for a second, what you should convince yourself is that the calculation I just did, it's probably the first one you would think of doing when you're trying to ascertain how does a body move in a particular space-time. But, in some ways, it was kind of a dumb calculation to do. So let's bear in mind, what this is doing is tell me the geodesic equation describes motion with respect to some given coordinate system. Satisfying this equation is just saying, yup, you're following a geodesic. And guess what? These are-- this radiation is a form of gravity. Even if the body responds to it, it is still going to follow a geodesic. It's going to be in free fall. The gravitational field may be a little bit different from the ones that you're more used to from your previous experiences with gravity, but it is still just following the free fall trajectory that gravity demands. So free fall in these coordinates means the body is remaining fixed in these coordinates. This is one of the most important lessons in general relativity. When you ask a statement that depends on the coordinate system you've written down, you have to understand your coordinates. And I would submit we don't really understand the coordinates that we wrote down here with this. These coordinates essentially follows the body. Even if the body is moving, they essentially wiggle right along with the body itself. So if you want to understand what the impact of gravitational waves is on things, you have to try to formulate what you're able to measure using coordinate-independent language as much as possible. At the very least, you need to understand your coordinate system a lot better than I did in setting up this calculation over here. So let's reframe the way we do this analysis. Let's imagine-- so one of the things that we learned in thinking about how gravity works in general relativity is I can always, if I'm just focusing on a single body, a single point mass, I can always go into freely falling coordinates. And space-time is essentially the space-time of special relativity. I can find a representation so that everything looks just like special relativity near there. Gravity is basically completely thrown away, at least as far as the mathematical representation is concerned. I'm in free fall. I don't really see much. If I wanted to see the impact of these waves, what I need to do is think about how things behave over some finite region. I wanted to put together a setup such that I have two bodies, and the relative effect of things between those two bodies will allow me to see the effect of space-time curvature in action. So let's now-- with this in mind, let's consider two nearby bodies that each follow geodesics. So body A is right here. And I'm going to define body A as living at the origin of my coordinates. And we've got a body B over here. And we're going to say that this guy is displaced from the origin by a small amount, some epsilon, in what I will call the x direction. And what I want to do is say, OK, I've got these two guys. They are moving through a space-time. Let's make this be the space-time that we had just discussed a moment ago. So they are each moving on geodesics of this space time. We know that if I set up my coordinate system in the way that I've done it here, I am not going to see any acceleration on each of these bodies. Let's now try to focus on some kind of a quantity that tells me about the separation between these two. I'm going to do this in two ways. So the first thing which I'm going to do is say, let's imagine that I am bouncing a light pulse back and forth between A and B. What I'm going to do is write the 4-momentum associated with that light. So recall I can't really define a 4-velocity for light because proper time is not defined. But I can introduce some kind of an affine parameter such that d by d lambda tells me about motion on the world line that this light follows. And this is just bouncing back and forth between these two. So it moves in the x direction. Because this is light, p dot T equals 0. So these components, the 4-momentum associated with the light, the dt d lambda and the dx d lambda, they are related to each other like so. And look. The space-time metric, the gauge-invariant piece of the space-time, hxxtt, it is appearing in here. So let's say, OK, what I am interested now is computing what-- suppose I'm sitting at A, and I want to compute the time, according to A's clock, that it takes for a pulse of light to travel from A to B. Maybe I'll make it bounce back. Maybe I want to know how much time it takes for the light pulse to go from A to B and back to A. So let's turn this into an equation for the interval dt. So dt, I can do a little bit of this trick called division. Here is-- question is, did I do it correctly? Yes. So I'm doing the old physicist's trick. I'm just imagining that an interval of time-- an interval of affine parameter d lambda, the amount of dt that accumulates as it moves to an interval dx. They are related by so. And, remember, I am assuming that the perturbation from space-time, from flat space-time, is small enough that I can always linearize. So this can be-- I can take that square root using the binomial expansion. So suppose I want to integrate up the time it takes for the thing to go from A to B, and then back from B to A. Well, this is going to be given by integrating this guy's motion from 0 to epsilon, like so. And then integrating it back again. My tt is in here. It's going to switch direction when I do that, so I'm going to flip the sign on that dx. I'm leaving it written in this form, so you have to be a little bit careful. You guys will play with this a little bit more on a problem set. Bear in mind that as I move through x, the time argument of these guys is changing. And so that's where there's a little bit of work that you'll explore in a problem set. But I'm writing it in this form so that we get some-- naively, you might think, ah, I'll just flip the sign and I'll combine these. In a certain limit, that is indeed a good approximation. You got to be a little bit more careful in general though. This is the time it would take for light to go from A to B and back in the absence of this gravitational wave, this hxxtt. Notice there is an inference on the time of arrival of these pulses that depends on hxxtt. Time of arrival of the pulses depends on this particular piece of the space-time metric. One could imagine, if you had a precise clock, and you sat there and you timed the arrival of these things-- let's say you sent a pulse of light out-- bloop-- and you kept track of how long it took with a very precise clock, you could look for-- in the absence of a gravitational wave, that would be, if you imagine that your pulse was sent out perfectly periodically-- let's say every millisecond or something like that, you sent out a little pulse-- if it arrives back and your arrival pulses were just as spaced by 1 millisecond, you would say, great, I am sitting in empty space-time. But if you found that their arrivals varied, you would think, hmm, there's something else going on here. Gravitational waves is one of things that could lead to that variation. So this, at least, is kind of a proof-- this calculation I just did-- you guys are actually, on one of the upcoming problems sets-- I believe it's problem-- I can't remember if it's 7 or 8-- you do a little analysis where you will use this to understand how a detector like LIGO actually functions to measure out the effects due to a gravitational wave on the arms of a detector like this. Let me just do one more way of understanding how it is the gravitational wave leaves an imprint on these two separated particles. So when you look at that first naive calculation I did, the fact that I got no effect, I sort of joked that we did a dumb calculation. To be fair, it wasn't that dumb. What we're-- the first thing you usually think of doing when you want to understand the motion of a body in space-time, is you look at geodesics. Geodesics just tell me that this guy's acceleration with respect to the free fall frame is 0. And that's exactly what we ended up getting. It's basically just saying, ah, even in a gravitational wave, my two bodies are moving on geodesics. If I have two separate bodies, like I have in this calculation where I just calculated the variation in the time of arrival of light pulses, one of the tools we've learned is that I can have two geodesics, two free fall trajectories that deviate from one another. So let's look at the geodesic deviation of the free fall world lines of these two bodies. So I'm going to-- as I did when I did my geodesic calculation, I'm going to take them to initially be close enough that they have the same 4-velocity. And if you want to be careful about counting orders here, you can imagine, when the gravitation wave comes along, that one or both of them might pick up a correction of this that's of order h. Depends on what kind of coordinate system you pick up. But the key thing is that they start out like this. The gravitational waves may come along and it will move them around a little bit. But the amount it's going to move them, the amount of velocity that they're going to pick up, and the displacement they're going to get from their initial position if they're at rest, it can only be of order of the h that you're trying to measure there. So I'll remind you, the equation of deviation, equation of geodesic deviation, it defines a sort of proper acceleration acting on-- whoops, that's a typo. It is the proper acceleration on a vector xi that defines, in a precise geometric way, the separation of two geodesics. And what it looks like is the Riemann curvature of your space-time contracted on the 4-velocity of your bodies, and that separation vector xi. So, linearizing everything, bearing in mind that R will itself be of order h, and that any difference between the proper time along these things and the coordinate time of a particular clock is also going to be of order h, they are small enough that this equation becomes-- Something that looks like this. Notice I'm only looking at the spatial components of the separation vector xi. And that's because, since I'm working, since this is what my 4-velocity looks like, I'm going to pick out the 00 components of these things. And, don't forget, Riemann is antisymmetric when I exchange either indices 1 and 2 or indices 3 and 4. And because of that, the time-like component of this guy is going to be an uninteresting thing. I will just get that is equal to the acceleration of the time-like component is 0. So it doesn't really matter. So this is telling me that the geodesic separation of these two world lines is-- it's a second-order equation in time, and it's proportional to the separation itself. We need to work out all of these components. That is a fairly straightforward exercise, and I will just report to you the results. A brief aside. If you want to sort of verify the results I'm about to write out here, you could go into-- you could look up the definition of the Riemann tensor. Bear in mind that you are doing things at linear orders. You can throw away the terms of, like, connection times connection. It's just derivatives of the connection that are going to matter at linear order. But, even so, it's a bit of a grotty calculation. I am-- as I begin to reorganize the problem sets associated with this class under the new system that MIT is currently operating on, I will soon be releasing a few Mathematica notebooks that are very useful tools for doing tedious calculations, like computation of Riemann curvature tensor components. So I will put a few of those things up there. If you would like to validate some of the things that I am computing, that is where you will find a good way to test this. So, anyway, the non-zero Riemann components. So at linear order, you have R x0 x0. Bear in mind that, linear order, you are raising and lowering indices with the metric of flat space-time. So that's the same thing as this. And this turns out to be minus 1/2 two time derivatives of hxx. Y0 y0-- that's your yo-yo component-- pretty much looks exactly the same, but it's two time derivatives of the yy piece. Remember though-- I've erased it a while ago-- there are some constraints that we found due to the fact that my original h mu nu has to be-- its spatial piece has to be traceless. In particular, we found hyy was minus hxx. So I can remove that additional function, and just leave things in terms of hxx. And our final non-zero component looks like this. All others are either 0 or related by a symmetry. So notice, although I've written down three different complements here, there's really only two, because the symmetry that's-- it's not a symmetry of Riemann, but the symmetry associated with the trace-free nature of hijtt made these two equal to each other. So this and this are really the only independent degrees of freedom here. All right. Let's plug this into the equation of geodesic deviation, this guy, and see what we get. And what we get is an equation governing the acceleration of the x component of deviation. Here it is. Let's make a couple of assumptions and try to understand what this is telling us. So what I'm going to do is-- let's assume this is-- the first assumption is necessitated by the idea that I can do linearized gravity. So I'm going to assume that these guys are all very small. And I will imagine that my displacement vectors have some piece which-- basically, it's time constant. It does not change. And that there's a bit that will be of order h that describes how the separation of my two bodies responds to this incoming field. So what I'm going to do, when you go and you plug this in, assume that this guy is constant in time, and this one varies. What I'm going to do is imagine I don't just have two bodies. Imagine I have a ring of bodies. So suppose I have got my coordinate axes here. Horizontal is x, vertical on the board is y. And so imagine I have a ring of bodies, like so. And let's consider the following limit. Let's imagine that my hxx is some function that is sinusoidally varying in time, and my hxy is 0. I just want to call that the-- as we saw, there's really only two degrees of freedom in this field. And in the way that we formulated it, it's going to encapsulated by these two functions, xx and xy. So let's just look at one. Let's just look at the influence of the hxx. So when you do this, you essentially have the second derivative of xi x looks like two derivatives of your x field, your xx field times xi x. And two derivatives, your sort of acceleration in the y direction, is minus those two derivatives of your field times xi y. So what this is going to do, if you think about the way this is going to act on these things, you're going to get a sinusoidal acceleration associated with the displacement of this little sea of freely falling bodies as the gravitational wave comes along. And it's going to do so in such a way that the displacement in the x direction does the opposite of what the displacement in the y direction is doing. So as time passes, you'll find that-- so, initially, let's say it stretches along x, but then squeezes along y. Then it's a sinusoid, so, a quarter of a cycle later, it's back to being a circle. And, a quarter of a cycle later, it stretches along y and squeezes along x. Let's consider the limit now where hxx is equal to 0, and hxy is some sinusoid. So I start out with my circle. Look at your equations over here. Think about what they're going to do. Now we're going to find that the change in x and the change in y is correlated. So this guy stretches it out into an ellipse along a 45-degree line. A quarter of a cycle later, it's back to being a circle. And a quarter of a cycle later, it does the same thing but with opposite sign. These two fields, my hxx and my hxy, are what we call the fundamental polarizations of a gravitational wave. Polarization states are only defined with respect to a particular set of basis vectors. So if I make my basis vectors be this x and y, we call this one, for reasons that I hope are obvious, the plus polarization. The gravitational wave acts as a tidal stretch and squeeze. Notice what's going on here. I stretch along this axis, squeeze along this one. A quarter of a cycle later, I squeeze along this one, stretch along that one. So I get a tidal stretch and squeeze, where the main directions along which the stretch and squeeze are happening is plus shaped with respect to these axes. This one is known as the cross polarization. You also have a tidal stretch and squeeze, but the lines of force in this case are along axes that have an X shape. So at the level of 8.962, this is all we're going to say about the way that these things act on-- how gravitational waves act and what their observables are. And I deliberately chose to wear my LIGO today, because what LIGO does is look for these tidal stretches and squeezes. And the way it does so-- so, obviously, it's technologically not really feasible to set up a ring like this. But what we can do is sample the ring here and here. Actually, it's a little bit better to do on the plus polarization plot. So you make an experimental setup that samples the ring here and here. And if a gravitational wave comes along and it's lined up right, you see, ah, look at that. I'm getting that stretch and squeeze. The way you actually measure it out is essentially a timing experiment, exactly like this little calculation I did here. You bounce light between mirrors, and you very carefully time the round-trip travel time from one mirror to the other. And what's particularly cool is the experimental aspect of this. So how does one measure the time so precisely? Well, you use a laser beam as your clock. And what's beautiful about doing that is the way that you can actually sort of do the metrology at the level necessary to get the precision to measure your hxx's in here is by interference. So you can treat the incoming laser beam as defining your time standard, and then by interfering the light that has traveled down the arm and bounced back with that laser beam, basically the laws of nature allow you to automatically check and see whether there is a shift in the time of arrival associated with the action of the gravitational wave. So I imagine there will be some questions about this. There is a homework exercise that allows you to sort of test drive some of these concepts and allows you to see how it is that LIGO works. It really just sort of touches the-- it just scratches the surface of what an amazing experimental feat it is. But it should give you a good idea as to what goes into that. Now, there is an important part to all this that we've not talked about yet. I handed you hij. Really, I handed you h mu nu. The big thing that we care about next is, given a particular source, how do I compute this radiation field? So bearing in mind that hijtt, this is the gauge-invariant radiative degree of freedom in my space-time, how do I compute this guy? And what I'm going to do is borrow sort of one of the most important lessons of that gauge-invariant formalism. So we need to be cautious of the fact that, in some gauges-- for example the Lorentz gauge, which is so convenient for solving the Einstein equation-- components of the metric appear to be radiative whether they are or they aren't. This is a consequence of our gauge. But we also know only hijtt is radiation in all gauges. So that was what the previous lecture was about. As I've emphasized, it was a slightly advanced one. I don't expect everyone to follow all details. But if you can sort of grok these two main concepts, you're ready to now exploit this in the following way. What you do is, step a, compute-- find some metric perturbation in any convenient gauge by any convenient method. Once you have that, extract hijtt. You're done. So that's what we're going to do right now in the last 20 or so minutes of this recorded lecture. So the way we're going to do it is let's take the linearized Einstein equation in Lorentz gauge. As we showed in-- I believe it was two lectures ago, the Einstein field equations in this case reduce to a flat space-time wave operator on the trace-reversed metric perturbation, being up to a factor of minus 16 pi G of the stress-energy tensor. And this has an exact solution. So I'll remind you of this notation. So this is the field at time t and spatial location x. x and t-- well, let's write it like this. t and x are my field point. That is where I measure my field. x prime is my source point. That is where I am looking at my particular contribution to the integral. As we're doing this calculation, there was a t prime, but the radiative Green's function turned that t prime into this retarded time. So what we are seeing is that the field at t depends upon what's going on at the source at t minus the interval of time it takes for information to propagate from x prime to x. So what I'm going to do is, first, I'm going to introduce a couple of approximations that reflect the reality of the conditions on which we typically evaluate this integral. And then we're going to talk about how to solve this thing, and, given that we have now solved it, how to project out the gauge-invariant degrees of freedom associated with the radiation. Whoops, pardon me. Let's put this up. So first is I introduce this approximation. It's worth bearing in mind that when you are calculating this, x minus x prime, which is kind of like the distance from a particular point in the source to where you're making your measurement, it is generally a lot larger than the size of the source. So the kind of situation we're thinking about is, here we are, making our measurements at some point x here. Here's my source, and here is a point x prime inside the source. So the thing to bear in mind is that, for the kind of calculations where that-- when one is actually doing this, x minus x prime-- well, if I think about the kind of measurements that LIGO is doing, for example, that is typically hundreds of millions to billions of light years. The size of the source itself is of order tens to hundreds of kilometers. Bit of a separation of length scales there, you would say. So what I'm going to do is approximate the solution of that integral, and, for pedantic sake, I'm going to use an approximately equal. What I'm going to do is replace x minus x prime with r. Let's just say that this distance here is effectively the same for all points in the source. I can pull that out of the integral. If you like, you can refine this. There's a more careful expansion. And if you've done any electrodynamics, you should be familiar with this. And it uses the fact that 1 over x minus x prime can be expanded in Legendre polynomials, in a sum of Legendre polynomials. So if you do that, you can introduce refinements to what I'm about to derive right here. What this does, then, is it defines a multipolar expansion. I am not going to do that. I'm just going to look at the very first term. And you can sort of think of that as saying that I am just doing the leading, the most important multipole. Additional multipoles beyond that are certainly important, and people who work in the field of gravitational radiation, we certainly do not neglect them. In fact, much of my current research is based on sort of high-order calculations associated with looking at behavior of some of those multipoles. This will be fine for initial pedagogical purposes. Next, we're going to use the fact that only the spatial-- whoops. That's not supposed to be an h tilde. My apologies. That's supposed to be h bar, because this is the trace-reverse thing. Weird times, folks. Where was I at? Yes. So I am only going to care about the spatial pieces of this. You might protest, hey, isn't it the spatial part of the metric, not the trace-reverse metric, that matters? And you would be correct. If you do this carefully, to be perfectly blunt, you have to sort of finish the calculation before you can kind of justify the step I'm about to make. And we're not going to have time to do that very carefully, but I'll make a comment about it. For now, it suffices to say we only care about the spatial components. So I'm going to take my alpha beta over to an ij. So I now have something that involves the integral of the spatial pieces of the stress-energy tensor over my source. And now I'm going to take advantage of an Easter egg that was hidden in an early problem set. Back on problem set 2, I asked you to prove a result that I called the tensor virial theorem. One of the reasons why I assigned that was that I knew I was going to need it in this lecture. And what the tensor virial theorem told us is that if I integrate that, this ends up being equivalent to 1/2 two time derivatives. It's equal to the integral of two time derivatives-- excuse me. It's two time derivatives of the integral of the 00 piece of my stress-energy tensor. So it's the second moment of that, xi prime, xj prime. This is great, because what I've got in the left-hand side here, that's exactly what my integral is. Actually, you know what? This is going to give me a result that's sufficiently clean. I want a new board. So when that is done, what I finally see is that my hij, trace reversed, is 2 G over r two time derivatives of a tensor that I will call Iij. This is what I get when I integrate T 00, two moments of T 00. This is called the quadrupole moment of the source. The asterisk is because, as you guys will see on one of the problem sets, there's a slight tweak that goes in the exact definition of the quadrupole moment which has to do with the behavior of the trace of Iij. I will just leave this like so for now, and leave you guys to explore that on a problem set. So we're almost done. What I now need to do is pull out the transverse and traceless piece of this. Let's consider the transverse condition first. So, to do the transverse condition, I'm going to take advantage of the fact that, far from the source, box of h bar ij is equal to 0, because I'm in a region where the source is equal to 0 when I'm far away from it. This suggests that what we ought to do is expand our solution plane waves. So what I can do is write h bar. And I'm going to choose my indices carefully here, call it jl. It's some amplitude that does not depend on space or time, e to the i omega t minus a wave vector dotted with x. So this amplitude may depend on omega, may depend on k. It does not depend on t or on x. And the reason I changed my indices from i and j to j and l, and I did not use the index k, is just to avoid confusion between square root of minus 1 and the wave vector. To be absolutely careful, this is a single mode. If you like, maybe you should think about this as something that I have-- I'm doing a sum or an integral over omega and k in order to reconstruct the whole thing. This is just a single Fourier mode in the wave field. So if I want this to be transverse, that condition is that the divergence of this guy be equal to 0. But when I take the divergence of this, this becomes-- I pull down a factor of minus i, and I pull down the j-th component of my wave vector k. And I get my wave-- my Fourier term. That can be thought of as-- so we can clear out the factor of minus i. The transverse condition boils down to saying that kj-- whoops. No, that's right. That's right. kj contracted on the wave field is equal to 0. In other words, the wave field is orthogonal to the wave vector. If you have studied electrodynamics, this should be familiar to you. This is the same-- it's exactly the same as the idea that when I have an electromagnetic wave, my E field and my B field are orthogonal to the direction of propagation. Let's define a unit vector n sub j to be kj but just normalized. So my wave is propagating in the direction of the unit vector n. Go back to another Easter egg that was hidden on an earlier problem set. You guys studied projection tensors. If I introduce the projection tensor Pij, Pij will project any spatial vector. It will project out the components normal, orthogonal to the direction of propagation n. You can in fact think of Pij as defining the metric for the subspace that is orthogonal to n. So imagine that your wave is propagating in the n direction. There's a plane being carried along with it. The plane is orthogonal to n, and Pij is the metric you use to define spatial distances in that plane. Using Pij, we can now make my h trace-- excuse me. I can make it transverse. All I need to do-- I'll put one T on this to denote Transverse. Whoops. All I need to do is project each of its indices within that tensor. So we're almost there. It's transverse, but not yet trace free. So if I take the trace on this guy, taking the trace on this guy amounts to-- I am going to require that this be equal to 0. Now, suppose I have some tensor that is not trace free, and I want to remove its trace. I'm going to make this a two-index tensor. So I'm not going to prove the following result. You can easily verify for yourself. The formula for doing so is that aij trace free is aij minus 1 over N akl gkl gij. G is the metric that describes the manifold in which this tensor lives. N is its dimension. Now, for us, the tensor field whose trace I want to remove is hijT. So hij bar T is the field whose trace I wish to remove. This lives in the two-dimensional manifold whose metric is the projection tensor Pij. And we are working in a space where raising and lowering indices, there's really no important distinction between the two of them, because we're working in linearized theory. So I remove the trace by taking this guy and subtracting off-- pardon me just one moment. Yes, sorry. Subtracting off like so. Pardon me. It should have a T on it. Plugging in my various definitions, this turns into hlk Pli Pkj minus 1/2. So you might want to go and check some algebra there, but this is all-- I'm very confident in this. There's an overall factor of h bar lk. And the last thing which I will note is that, once I have removed the trace, this bar becomes meaningless. It's trace free, so the trace reversing of it has no action. Now, you should be a little bit careful about that. There's a few different notions of trace that went into this definition. Without proof, and just stating that by being a little bit more careful of some of the definitions we put down, the difference between-- when you count this up, you'll find that you may be making a small error, but it is on the order of terms that are quadratic in the perturbation. Putting all these pieces together, we finally obtain our transverse and traceless metric perturbation. Which looks like this. This is a result known as the quadrupole formula for gravitational waves. This is-- so given-- just take a stock of the final result. Suppose you have some dynamic gravitating source. This is telling me, what I do is I compute the quadrupole moment associated with that source. I take two time derivatives. I hit it with this combination of projection tensors, which has to do-- that tells me things about the geometry of the source. That's saying that my source, the waves that come from the source propagate along a particular direction to me. This picks out the components that are transverse with respect to that direction and are traceless, so that I don't have any gauge degrees of freedom in what results. Multiply by twice Newton's constant, divide by distance. And this is a wave field that our friends over in Building NW22 can now measure pretty much every day. You're going to do a few homework exercises associated with. One thing which is worth commenting on before I move on is, when you keep factors of c in this, that G goes over to G over c to the-- wait a minute. Let me double-check that. Is it G over c to the fourth or G over c squared? Anyhow, it's G divided by c to some power. Let's just say that it's missing factors of c. I may comment on that in some notes that I will add to the board. And so that kind of tells you that this is going to be a really small quantity. Makes it very hard for these things to be measured. Nonetheless, it turns out even though they have-- if you think about the impact they have on the detector, even though it's small, they carry a tremendous amount of energy. I'm going to take a break before I begin recording my next lecture. But the next lecture is actually going to lead us to an understanding of a different version of the quadrupole formula. It's based on this, but it describes the energy content of gravitational waves. Both of these formulas go by the name the quadrupole formula. It's a little bit confusing sometimes if both are used. And you're going to use both of them on upcoming homework assignments. All right. So I will end this recording here. |
MIT_8962_General_Relativity_Spring_2020 | 7_Principle_of_equivalence_continued_parallel_transport.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: All right, good afternoon. So today's lecture is a particularly important one. We are going to introduce the main physical principle that underlies general relativity. We'll be spending a bunch of time connecting that principle to the mathematics as the rest of the term unwinds. But today is where we're going to really lay out where the physics is in what is known as the principle of equivalence. So before I get into that, just a quick recap. So in terms of technical stuff we did last time, the most important thing we did was to introduce these mathematical objects called Christoffel symbols. Christoffel symbols are those capital gammas. We began by thinking of them as just what you get when you look at the derivative of your basis objects. Pardon me a second. There was a lot of chalk on the chalkboard, and I think I inhaled half of it while I was cleaning it. So take derivatives of your basis vectors. And there, you get something that's linearly related to your basis objects. And the gamma is the set of mathematical quantities that sets that linear relationship between the derivatives and the objects themselves. One can build under the demand-- so recall how we did this. We have a physical argument that tells us that a quantity we call the covariant derivative, which is a way of using these Christoffel symbols in order to get tensorial derivatives. So if I look at just the partial derivative of a tensor, if I'm working in a general curvilinear coordinate system, the set of partial derivatives do not constitute components of a tensor. OK? I posted some notes online, demonstrating explicitly why that is. But in a nutshell, the main reason is that, when you try to work this out or when you go and you work this out, you will find that there are additional terms. It was like the derivative of the transformation matrix that come in and spoil the tensoriality of this relationship. And what the Christoffel symbol does, as you guys are going to prove on P set three, it introduces terms that are exactly the same but with the opposite sign and cancel those things out. So that the covariant derivative is what I get when I correct each index with an appropriate factor of the Christoffel symbol. It is something that is a component of a tensor. In other words, it obeys the transformation laws that tensors must have. We have a physical argument that essentially is something that is tensorial, in other words, both sides of the equations are good tensor quantities in one representation, have to be tensorial in any other representation. And from that, we deduced that the covariant derivative of the metric must always be 0. And by taking appropriate covariant derivatives of the metric, sort of doing a bit of gymnastics with indices and sort of wiggling things around a little bit, we found that the Christoffel symbol can itself be built out of partial derivatives of the metric. And the result looks like this. So that's a little bit more of an in-depth recap than I typically do. But I really want to hammer that home because this is an important point. We then began switching gears. And we're going to do something. So I introduced, at the end of the previous lecture, a fairly silly argument, which, nonetheless, has an important physical concept to it. So where we were going with was the fact that, when we have gravity, we cannot cover all of spacetime with inertial frames, or I should say-- I should watch my wording here-- with an inertial frame. OK? We're, in fact, going to see that you can have a sequence of them. And we're going to find ways of linking them up. But for now, if you think of special relativity as the theory of physics in which we can have Lorentz coordinates that cover the entire universe, that's out the window. OK, and so I'm going to skip my silly derivation. Part one of it is essentially just a motivation that a gravitational redshift exists. So I gave a silly little demonstration where you imagined dropping a rock off of a tower. You had some kind of a magical device that converts it into a photon that shoots it back up and then converts it back into a rock. And you showed that, as it climbed out of the gravitational potential, it had to lose some of its energy. Or else, you would get more energy at the top than you put in when you initially dropped it. It's a way of preventing you from making perpetual motion machines. That is, without a doubt, a flaky argument. It can be made a little bit more rigorous. But I didn't want to. And I will justify the reason I can do this is that this is, in fact, a highly tested experimental fact. Every single one of you has probably used this if you've ever used GPS to find your way to some location where you're supposed to meet a friend. And what this basically tells me is that, if I imagine a tower of height h, if I shoot light with frequency at the bottom, omega b, when it reaches the top, it'll have a different frequency, omega t, where these are related by omega t equals, and I'm going to put my c squareds back in there, just to keep it all complete. OK? So this is, I emphasize, a very precisely calibrated experimental result. There are many ways of doing it, some silly, some serious. The silly one was sort of fun. But it does actually capture the experimental fact. OK? The key experimental fact is that it's necessary for light to lose energy as it climbs out of the gravitational field, in essence, because it's possible, thanks to quantum field theory, to convert photons into particles and particles into photons. This is a way of conserving energy. OK, so this was just part one of this argument that we cannot cover all of spacetime with an inertial frame. So here's part two of the argument. Suppose there was, in fact, a very large region of space-- now let's say it's near the Earth's surface-- that could be covered by a single Lorentz frame. OK? So I want you to imagine that you're sending your beam of light up. And we're going to take advantage of the fact that light is wave like. So what I want you to do is think of the spacetime trajectory of successive crests of this wave as it climbs out of the gravitational potential. So I"m going to make my spacetime diagram the way we like to do it with time running up. And so this direction represents height. So down here is the bottom of the tower. So let's just make the drawing clearer. Let's let this tick be the bottom of the tower. This tick over here will be the top. Now if we were just in special relativity, there was no gravity, we know light would just move on a 45 degree angle on this thing. We don't know yet what gravity is going to do to this light. OK? But you can imagine that whatever it's going to do, it's going to bend it away in some way from the trajectory that special relativity would predict. So let's just imagine that crest one of this light, it follows some trajectory in spacetime that maybe looks like this. So here is the world line of crest one. So what about the world line of credit two? So let's think about what the world line of credit two must be under this assumption that we can, in fact, cover all of spacetime with a single Lorentz frame. OK? If that is the case, so if that is the case, no position and no moment is special. OK? Everything is actually translation invariant in both time and space. So the second crest is going to be emitted one wave period later. But there's absolutely nothing special about spacetime one wave period later. And as it moves along, there can't be anything special about this. If it is to be a global Lorentz frame, whatever the trajectory is that the crest two follows through spacetime, it's got to be congruent with the trajectory of crest one. I've got to be able to just simply slide them on top of each other. OK? That is what this assumption demands. So this is going to look something like this. OK? Artwork a little bit off. The key thing which I want to emphasize, though, is that if this guy is congruent, the spacing between crest one and crest two at the bottom, let's call it delta tb, there'll be some spacing between the two at the top. We'll call it delta tt. They must be the same. But that's [INAUDIBLE]. That is just a period of the wave. And that's just up to a factor of 2 pi. That's 1 over the frequency. So this implies delta tb is not equal to delta t at the top. So a frequency at the top is smaller. So the period will actually be a little bit larger. So when you think about this argument carefully, the weak point is this assumption that I can actually cover the region with a large Lorentz frame. Remember what this is telling me is that, essentially, there is no special direction, right? I can do translation, variance in time and space, and they're all the same. But that's actually completely contradicted by common sense. If there were no special direction, why did it go down? There's obviously something special about down, right? So we conclude we cannot have a global Lorentz frame when we've got gravity. And in case you want to read a little bit more about it, I'll just give a highlight that this argument was originally developed by a gentleman named Alfred Shield. So that's unfortunate, OK, from sort of a philosophical perspective because the fact that we don't have a global Lorentz frame makes you think, oh, crap. What are we going to do with that framework we spent all this time developing? OK? The existence of a Lorentz frame and many of the mathematics we've been developing for the past few lectures, they all center around things you can do in Lorentz frames. So it's kind of like, OK, so do we just throw all that out? Well, here is where Einstein's key physical insight came in and gave us the physical tools that then needed to be coupled to some mathematics in order to turn it into something that can be worked with but, nonetheless, really save the day. So the key insight has there's a few steps in it. So first thing, I was very careful to use the word global when I ruled out the existence of Lorentz frames. But I'm allowed to have local Lorentz frames. Let's see. What does that actually mean? So I'll remind that a Lorentz frame is a tool that we use to describe inertial observers. In fact, we often call them inertial frames because they're sort of the constant coordinates. They represent the coordinates of an inertial observer who happens to be at rest in that Lorentz coordinate system. So let's switch over. Let's start calling them inertial frames for just a moment. So an inertial frame means that there is nothing accelerating, so no accelerations on observers or objects at rest in that frame. In other words, no forces are acting. Einstein's insight was to recognize that the next best thing is a freely falling frame. OK? If you are in a freely falling frame, to you, it sure as hell seems like there's a force acting on it and there's some accelerations acting upon it. But let's imagine you're in this freely falling frame. And you have a bunch of small objects that you release near you. Let's see if I can find some bits of broken chalk. So when you're in this really falling frame-- this was a lot easier when I was younger. All right, when you're in this really falling frame, OK, and these objects are all falling, there is no acceleration of these objects relative to you. The reason for this is that the gravitational force is proportional to the mass. And so the fact that we have f equals ma. And fg proportional to mass means that all objects in that freely falling frame-- this is a term that I'm going to start using a lot. So I'm just going to abbreviate it FFF. If I get really lazy, I might call it an F cubed. Anyway, l objects that are in that freely falling frame, they're experiencing the same acceleration. And so they experience 0 relative acceleration at least in the absence of other forces, OK? Perhaps one of them is charging. There's an electric force. Well, then it sort of suggests that the interesting force is the extra force provided to that relative to what we call the gravitational force that is driving this freefall. So I urge you to start getting comfortable with this idea because we're going to find this to be-- so one thing, which I'm going to do in a couple of minutes, is actually a little calculation that shows me I can, in fact, always find a Lorentz frame in the vicinity of any point in spacetime. OK? And so we're going to actually regard that Lorentz frame as being the preferred coordinate system of a freely falling observer, one who is not accelerated relative to freefall at that point. We are not in freefall right now. The damn floor is pushing us out of the way. But in this way of doing things, we would actually regard us as being the complicated people, OK? Someone who is merrily plummeting to their death, they're actually the simple observers, who are doing what they should be doing, in some sense. So the key thing, a way of saying this, so coming back to this and just one more point about that, because there are no relative accelerations, within this frame, objects maintain their relative velocities. That's basically the definition of something that is an inertial. If I have a frame where everything is-- two objects are moving with respect to each other at 1 meter per second, they're always moving at 1 meter per second, that's inertial. And so this sort of demands that we do all of our experiments in plummeting elevators or in space, right? OK, obviously, you can't do that. So there's some complications we're going to have to learn the math to describe. But this is the way we're going to from now on think about things is that the freely falling frame is the most natural generalization of an inertial frame that we describe using Lorentz coordinates. Now before I discuss a few details about this, an important thing that is worth noting is that a very important aspect of any realistic source of gravity is that it is not uniform. OK? Gravity here is a little bit stronger than gravity on the ceiling, OK? And it's a lot stronger than gravity in geostationary orbit. We call this variation in gravity tides. So tides break down the notion of uniform freely falling frames. What this is telling us is that this idea, if I'm going to create this inertial frame that is essentially a freely falling frame, it will have to be a finite size. Hence, it's local and not global. OK, physically, this is telling me that, if I have a really tall freely falling elevator, and I'm here, I have one friend here and one friend down there, even if we start absolutely at rest, let's say we're at rest. We measure reference back to the walls of this elevator. It's a very stiff elevator. We will see the three of us diverge away from each other with time, OK? Because this person, let's say earth is down here, this person is feeling the-- and using Newtonian intuition, gravity is a little stronger. Gravity is medium. Gravity is weak. And so as viewed in that freely falling frame, let's say we center it on me here in the middle, I will see the person in the top go up, the person at the bottom go down. Another way of saying this, so let me just say that, so a very tall elevator, we'll see a separation of freefall since gravity is not uniform. But there's another way of saying this. And what this is telling me is imagine I made a spacetime diagram that shows trajectories of the three of us, so the three of us that are plummeting in this elevator, OK? What we find, if we make our trajectories as we fall-- in fact, let's just go ahead and make a little sketch of this. OK, so let's say here is the middle. Here is the top. Here is the bottom. I'm going to draw this in coordinates that are fixed on the person in the middle. So the person the middle, in their freely falling frame, they think they're standing still. They see the person at the top moving up and away, the person at the bottom moving up and away. It's a very similar story to the little parable I sketched of that light pole moving up. These are not congruent trajectories. Another way to say this is that, in any moment, if I think of these as world lines, I can, at any moment along the world line, I can draw its tangent vector. The tangents do not remain parallel. Now I'm going to take you back to probably, I don't know when you guys all learned this, but when you first started learning about geometry. For me, it was in middle school. And when you first started learning about geometry, you were usually doing geometry on the plane. And they gave you various axioms about the way things behaved. One of them is now known as Euclid's parallelism axiom. And it states that if I have two lines that start parallel, they remain parallel. If you dig into the history of mathematics, this axiom bugged the crap out of people for many, many years, OK? Because when you look at pictures on a piece of paper or on a chalkboard or something like that, it seems right, OK? But it really can't be justified quite as rigorously as many other axioms that Euclid wrote down. And it turns out, there's an underlying assumption. The assumption is that you are drawing your lines on a flat manifold. There's a ready counterexample. Go to the Earth's equator. You start, say, somewhere in Brazil. Your friend starts somewhere in Africa. Both of you stand on the equator and go due north. You are moving on parallel trajectories. You will intersect at the North Pole. OK? Well, guess what? You ain't moving on a flat manifold. You're moving on the surface of a sphere. OK, curvature causes initially parallel trajectories to become non-parallel. What this is telling us is that the manifold I'm going to want to use to describe events when I have gravity cannot be flat. I'm going to have to introduce curvature into it. Take us a little while to unpack precisely mathematically what that means. But this is sort of a sign that things got more complicated. And tides are the way in which that complication is being introduced. So switch pages here. So what we have basically danced around and put together here is one formulation of what is known as the principle of equivalence. The principle of equivalence, or one formulation of it, tells me that over sufficiently small regions, the motion of freely falling particles due to gravity cannot be distinguished from uniform acceleration. The physical manifestation of this is that if you're an observer who is in that freely falling frame moving due to gravity, you are also experiencing that uniform acceleration. So you see no net acceleration. This particular formulation of it is known as the weak equivalence principle. I'm going to give you a couple variations on this in just a second. One thing that's important about it is, essentially, it's a statement that, if you think about Newton's laws, the idea that f equals ma and the force of gravity is proportional to m, it's a fairly precise statement that you think of that m as the gravitational charge. It's saying that the gravitational charge, the gravitational mass and the inertial mass are the same thing. That's actually a testable statement. What you can do is look at materials that have very different compositions. In particular, what you want to do is look at things that maybe have different ratios of neutrons to protons or have highly bound nuclei with lots of gluons in them or various kinds of fields that you can imagine perhaps couple to gravity differently than the quarks do. And so there are what are called free fall experiments to test this, which really basically just boils down to dropping lots of different elements of the periodic table and making sure they all fall at the same rate. And the answer is that they fall at exactly the same rate within experimental precision. And the last time I looked it up, well, what they do is they demonstrate the WEP, the Weak Equivalence Principle, is valid to about, I believe, it's about a part in 10 to the minus 13. Might be a little bit better now. So bear with me just one second here. OK, so good. We've got this notion here. And you might now think, OK, great. I have a way of generalizing what the notion of an inertial frame is. Is this enough for me to move forward? Can we now do all of physics by just applying the laws of special relativity in freely falling frames or, putting it this way, our new notion of an inertial frame? For two reasons that are closely related, this doesn't quite work. And it comes down to the fact that because of tides, we can't do that. What this basically means is that the transformation that puts you into a freely falling frame here is not the same as the transformation that puts you into a freely falling frame 10,000 miles over the Earth's surface, OK? There are different transformations at different locations in spacetime. This comes down the idea that it's not a global Lorentz frame, OK? We had a sequence of local Lorentz frames that we have to link up. So I'm going to do a calculation in just a moment where we explicitly show that we can always go into a frame that is locally Lorentz, but that there's a term that I'm going to very strongly argue is essentially the curvature associated with the spacetime metric that sets the size of what that region actually is. So what we are going to do, what is going to guide our physics as we move forward is, essentially, we're going to take advantage of a reformulation-- there we go-- of the equivalence principle, which I'm going to word as follows. In sufficiently small regions of spacetime, we can find a representation or a coordinate system such that-- I don't want to cram this into the margin. So I'll switch boards-- the laws of physics reduce to those of special relativity. This is known as the Einstein equivalence principle. At least, that's the name that Carroll uses in his textbook. A couple books use different ones. But this is a very nice one because it really is the principle by which Einstein guided us to rewriting the laws of physics. So we've got a weak equivalence principle, an Einstein equivalence principle. Just as an aside, there is something called a strong equivalence principle. We're not going to talk about this very much, but I'm going to give it to you because it's just kind of cool. It tells me that gravity falls in a gravitational field in a way that is indistinguishable from mass. That's a little weird. I'm not going to explain it very much right now. A way to think about it will become clearer in some future problem sets. Basically what it's telling us is that when you make very strong-- when you make any kind of a bound object, some of the mass of that object, in the sense of the mass that is measured by orbits, can be thought of as gravitational energy, OK? Energy and mass are equivalent to one another. So that gravitational energy should respond to gravity. And in fact, it falls just like any other mass, OK? This is another one. This is actually very-- well, prior to September 2015, this was very difficult to test. There were some very precise measurements of the moon's orbit that were done to test this. And there were observations of binary pulsar systems that were done to test this. Now every time LIGO measures a pair of binary black holes, which are nothing but gravitational energy, and our theoretical models match the wave forms, sort of like, boom, equivalence principle. Drop mic. Leave room. All right, so let's get precise. What I want to do is show that we can always find a local Lorentz frame. And what I mean by that is that I want to be able to show that I can always do a change of my coordinates, a change of my representation such that, at some point, I can convert the metric of spacetime into the metric we used in special relativity, at least over some finite region. So let me define a couple of quantities and then let me formulate the way this calculation is going to work. Let us let coordinates with unbarred Greek indices be the coordinate system that we start in. And let's say that, in this representation, the metric is G alpha beta. Let's demand that there exists a set of coordinates that I will represent with bars on the indices. When we transform to these coordinates, I want spacetime to be Lorentz at least in the vicinity of some point or some event. We'll call that event p. So we will assume that there is some mapping between these coordinates, so there's no nasty singularities there. And it's nothing that we can't deal with. So we can do it in either direction, but let's say x alpha is x alpha of p written as some function of these guys. And so there's a transformation matrix between the two of them. It takes the usual form, OK? So mathematically, the way I'm going to show this is I want to show that we can find a coordinate system such that, if I compute the spacetime metric and the barred representation, this is always going to be given by doing this coordinate transformation. Then I get the metric of flat spacetime over as large a region as possible. OK, we know we're not going to be able to do it everywhere. What I want to show is that I can make that happen at the point p and that the functional behavior of this thing is sufficiently flat that it remains at this over some region. So let me just sketch what the logic of this calculation is going to be. So what we're going to do is expand. So remember g is a function. All of these l's are functions. So what I'm going to do is I'm going to think of this whole thing as itself being some kind of a function. I'm going to expand in the Taylor series. There's something really cool that we're going to have to do with this. I mean, if I do this in general, you're going to get something pretty vomitous It's a giant, giant mess. OK? You might worry, what the hell are we going to do with this? OK, but there's something really cool that we can do. So we are given the metric g. But we are free to pick our coordinate transformation to be whatever we want it to be. So what we are going to do is compare the degrees of freedom offered by the coordinate transformation. Which again, I emphasize, we are free to make that whatever we need it to be. So the coordinate transformation and its derivative, right? We're free to play around with this thing. It's under our control. To the constraints that are imposed by the metric and its derivatives, which we are given. So what we're essentially going to do-- so in a moment or two, I'm going to write out very schematically what this coordinate transformation is going to look like when I do the Taylor expansion. And what I want to do is, essentially, just count up how many degrees of freedom the coordinate system offers me, count up the number of constraints that need to be matched in order to effect this transformation, see whether I've got enough. Assess what we learn from that, OK? We just set up with a few more things to help define the logic. So I'm going to write G alpha beta as G alpha beta at the point p plus x gamma. And you know what? Let's do the expansion in the bar coordinate. Minus x gamma bar at point p times the derivative at point p. And then I'm going to get something that involves a complicated form of x squared and second derivatives. OK, this can keep going. But this is going to be enough for our purposes. I am likewise going to expand my coordinate transformation matrix. Make sure I got all my bars in place. OK. OK. So let me go over to another board, just write down a few more important things. And then we'll start counting. So first key thing, which I want to note before I dig into the calculation, is the metric at this point, its derivative, its second derivative. These have been handed to us. OK? So we have no freedom to play with these. These are going to give us constraints that we need to satisfy. The coordinate transformation in its various derivatives, we are free to specify these. These are our degrees of freedom. OK, so let me just restate the calculation that we are going to do. Basically, we wanted to do this. Schematically, here's what this calculation is going to look like. And if you look at this right now, you might think this is going to be horrible. But we don't need to specify every step of this thing in absolutely gory detail. What we want to do is make this, which I can write as this guy at point p, this guy at point p, this guy at point p plus a term that is going to be linear in the distance from the spacetime displacement away from point p. If you want to go and multiply this out, knock yourself out. But it suffices to know that you are going to get some new things that involve first derivatives of all these quantities. So this will involve these two things. And this will involve second derivatives of all these things. Now what we're going to do is we're going to try to make what we would get multiplying all this out look as much like minus 1, 1, 1, 1 of the diagonal as possible. And like I said, it really just involves accounting argument. It's actually one of my favorite little calculations we do in this class because it's my favorite word for this is it really looks vomitous. But it's quite elegant. And I have an eight-year-old kid. And I'm constantly trying to teach her, math is just all about counting. And this is a perfect example. This is really just about counting. So let's look at this thing at 0th order. So what I'm trying to do is guarantee that I've got some l, which I can choose, which will make the components of the g at that point be minus 1, 1, 1, 1. So my constraints, the conditions I must satisfy involve things that hit these 4 by 4 symmetric tensor components. So this guy is something that is symmetric. Nature has just handed me a symmetric tensor with a symmetric 4 by 4 object. Those are 10 constraints that my transformation must satisfy. What I'm going to use to do this is this matrix at the point p. And I'll remind you, this matrix is just following set of partial derivatives at that point p. This is also 4 by 4. It is not symmetric. So this actually gives me 16 degrees of freedom to satisfy these constraints. Ta da! We can easily make this thing look Lorentzian at the point p. In fact, we can do so and have six degrees of freedom left over. Any idea what those six degrees of freedom are? Remember you're going into a Lorentz frame. Yeah? Three rotations, three boosts. Exactly right. Boom. So not only does it work mathematically, but there's a little bit of leftover stuff which hopefully hits our intuition as to how things should behave in special relativity. OK, that's not really good enough though, right? So I just showed that I can do it at this point. And if I move a centimeter away from that point, suppose, there's some really steep gradient, and then it goes completely to hell. OK? Then we're in trouble. So we have to keep going. We have to look at the additional terms in this expansion. So when we do things at 0th order, we have now-- so g has been handed to us. We have now specified behavior of l at that point. So l has been completely soaked up. I don't have any more freedom to mess around with it. When I go to the next order, OK, so the quantity that is setting my constraints is the first derivative of this. So these are four derivatives of my 10 metric functions. 4 by 4 symmetric by 4 components. So I have got 40 constraints I must satisfy. OK, well, the tool that I have available, I am free to specify the derivatives at this point. Again, remember this is itself a partial derivative. So I've got 4 components alpha, 4 derivatives with gamma bar, 4 derivatives with mu bar. Partial derivatives, it should not matter what the order is. So this needs to be symmetric on mu and gamma. So I, in fact, have 40 degrees of freedom at 1st order. Perfect match. OK? So I can make my coordinate system. Not only can I make it equal to the Lorentz form at-- I can make the spacetime metric Lorentz at the point p, I can also make it flat at that point p. All right, now we're feeling cocky. So let's move on. How far can I go? I lost the page I need. OK, so now you have a flavor for what we're doing. OK? I want to count up the number of degrees of-- sorry-- the number of constraints that are in the metric object, the new metric object I work with at this order and compare it to the degrees of freedom in the coordinate transformation at this order. So when I go to 2nd order, my new metric thing that I'm going to be messing around with is the second derivative. So I take two derivatives of g alpha beta at the point p. So this again needs to be symmetric in the derivative. So I have a symmetric 4 by 4 on these two guys. And the metric is itself 4 by 4, one alpha and beta. So there is 100 conditions, 100 constraints that we must satisfy. So let's look at the second derivative of this. OK, so I'm going to move this down a little bit. This is now going to look like the third derivative when I do this transformation. So it's going to be the third derivative of x alpha with respect to mu bar, delta bar, and gamma bar. So I've got 4 degrees of freedom for my alpha. And this must be perfectly symmetric under the any interchange of the indices mu bar, delta bar, and gamma bar. OK? This is a little exercise in combinatorics. So the number of equivalent ways of arranging this turns out to be n times m plus 1 times n plus 2 over 3 factorial. And I'm in 4 dimensions for n equals 4. Work that out, and you will find that you have 80 degrees of freedom. So what we can do is we can always find a coordinate transformation that makes it flat. It makes it Lorentzian at point p. It is flat in that region. In other words, there is no first derivative there. Sorry. Flat's not really the right word. There is no slope right at that point. But we cannot transfer away the second derivative. And so what this tells me is the coordinate freedom that we have means that we can always put our metric into the following form. I'm just going to write this schematically. Second derivative to the metric and sort of the quadratic separation in spacetime coordinate. So a couple things about this are pretty interesting. So this is basically telling me that we can make it Lorentz only up to terms that look, essentially, like the second derivative of the metric. Well, the second derivative of any function tells you about the curvature of that function. So this word curvature is going to be coming up over and over and over again. We are, in fact, because this is general relativity, we can't do everything simply. We're going to actually find that there is a rigorous way to define a notion of curvature that we're going to play with, which indeed looks like two derivatives of the spacetime metric. And we are going to find it has, I mean, it's actually going to end up looking like a 4 index tensor. OK? So it's going to be an object that's got 4 indices on it. Each of those indices goes over the 4 spacetime coordinates. And so naively, it looks like it's got 4 to the 4th power independent components. 256. It has certain symmetries we're going to see, though. And when you take into account those symmetries and you count up how many of those components are actually independent, any guesses what the number is going to turn out to be? 20. Yes, it exactly compensates for the number that cannot be zeroed out by this coordinate transformation. It's called the Riemann curvature tensor, and we will get to that fairly-- actually, really, just in a couple lectures. The other thing which this is useful for is in all of my discussion of the equivalence principle up to now, there's been a weasel word that I have inserted into much of the physics. I always said over sufficiently small regions. OK? I say that trajectories begin to deviate from one another when they get sufficiently far away from one another. OK? And you should be going, what the hell does sufficiently small, sufficiently far away, what do all these things mean? OK? And the issue is you need to have a scale. Well, this scale is going to be set by the second derivative of the metric. So the size of the region over which spacetime is inertial in this coordinate system, in this representation is approximately so imagine it's 1 over-- we can think of d2g as being 1 over a length squared. So 1 over the square root of that gives me a rough idea of how long the curvature scale associated with your spacetime actually is. And it tells you how big your inertial region actually is. So we're going to make this notion of curvature regress very, very soon. As a prelude to this, we now need to start thinking about how we do mathematics and physics on a curved manifold. So I'm going to start to set up some of the issues that we need to face. So we're going to need to define what I mean by a manifold that is curved. So a curved manifold is going to simply be one in which initially parallel trajectories do not remain parallel. So an example is the surface of a sphere, as I illustrated. You start somewhere on the equator in Brazil. Your friend starts somewhere on the equator in Africa. The two of you start walking north. You are exactly parallel when you take that first step of the equator. But your trajectories cross at the North Pole. Interestingly, an example that looks curved but is not, the surface of a cylinder. OK? If I take two lines, again, I need to imagine that this region continues all the way up here. I make two lines that are parallel to each other here and I have them extend around this thing, they would remain parallel the entire way out, OK? Another way of stating that is that you can always take a cylinder and, with an appropriate cut, but without tearing it, you can flatten it out and make it into a simple sheet, a perfectly flat sheet. You cannot do that with a sphere without tearing it in some places. So what is going to begin to complicate things is that we want to work with vectors and tensors that live in this curved manifold. We haven't really thought too carefully yet about the space. And let's just focus on vectors for now. We haven't thought too carefully about the space in which the vectors actually live. So implicit to everything we have talked about up until now is that we often regard vectors as objects that themselves reside in a tangent space. OK, if I'm working in a manifold that is flat, say it's the surface of this board, OK, and just two dimensions, every point on this board has the same tangent, all right? So if I draw a vector here and I draw another vector over here, it's really easy for me to compare them because they actually live in the same space that is tangent to this board. If I'm on the surface of a sphere, points that are tangent to the sphere at the North Pole are very different from points that are tangent to the sphere on the equator. So it becomes difficult for me to actually compare fields when they are defined on a curved surface like this. So this makes it a little bit-- and so I'm just going to set up this problem, and I will sketch the issue. And then we will resolve it in our lecture on Tuesday. This makes it a little bit complicated for me to take derivatives of things like vectors when I'm working in a curved manifold. So let's consider the following situation. So I'm going to define some curve, which I would just call gamma. It lives in a curved space of some sort. I'm going to draw it here on the board. But imagine that this is on the surface of a sphere. OK? So here's the curve gamma. And let's say this point p here has coordinates x alpha. And this point q here has coordinates x alpha plus dx alpha. Suppose there's some vector field that fills all this manifold. OK? And so let's say the vector a looks like this here at the point p. And it looks like this here at the point q. How do I take the derivative of the vector field as I go from point p to point q? Well, your first guess should basically just be do what you always do when you first learn how to take a derivative. OK? So that would be a notion of a derivative. There's nothing mathematically wrong with that. OK? But we're going to find that it gives us problems for the same reason that we ran into problems when we began working with vector spaces and curvilinear coordinates. OK? If I want this to be a tensorial object in the same way that we have been defining tensors all along here, I'm going to run into problems. So let me just actually demonstrate what happens if I try to do this. So the key bit, the way we want to think about it is that the points p and q don't have the same tangent space, which is a fancy way of saying that, as I move from point p to point q, the basis vectors are moving. They're starting to point in different directions. So if this were to be-- like I said, mathematically, if you just want to get that derivative, that is a quantity which has a mathematical meaning, OK? But it's not the component of a tensor, which we have called out as having a particularly important meaning in this geometric construction of physics that we are doing. And so if this were to be tensorial, then I should be able to switch to new coordinates, which I'll designate with primes, such that the following was true. The reason why this doesn't actually work is I'm going to demand that alpha-- excuse me-- that a is, in fact, actually already tensorial. OK? It's the compound of a vector, which is a particular kind of tensor. So I'm going to demand that the following be true. And I'm going to demand that my derivatives, they are just derivatives. They do the usual rule Jacobian rule when I switch coordinate systems. And so skipping a line of algebra, which you can get in my notes, this is actually very similar to the notes I've already posted. When you work this out, you're going to find you get one term that's correct, but you get another term that involves a derivative of your coordinate transformation matrix. And this is an actual term, spoils the tensorialness of this quantity. The way we are going to cure this is we are going to demand that if we want our derivatives to be derivatives that comport with a notion of taking tensor objects and getting other tensor objects out of them, before we take the derivative, we have to have some way of transporting the objects to the same location in the manifold, where I can then compare them. So here's one example of a notion of transport that we could use. So here's a at point q. There's a at point p. Notion one that we'll talk about is known as parallel transport. What that's essentially going to mean is I'm going to say, well, let's take one of these guys, either q or p. The way it's drawn in my notes, I've used q. But it could be either. And let's imagine I slide it parallel to itself until this gives me a alpha from q transported to p. And then I define a derivative with that transported notion of the vector. Now notice I've called this notion one. You can take from that this idea of how I transport the vector from one place to the other to do this comparison. I cannot uniquely define it. There are, in fact, multiple ways you can do this. We're going to talk about two that are useful at the level of 8.962. In principle, I imagine you could probably come up with a whole butt load of these things. These are two that the physics picks out as being particularly useful for us for the analysis that we are going to do. OK? So we'll pick it up there on Tuesday. We're going to start by coming back to this notion of I want to differentiate a vector field in a curved manifold. And let me just state before we conclude, now that I've transported a from q to p, they share the same tangent space. Since they share the same tangent space, I can compare them more easily. That allows me to make a sensible derivative that respects all the notions of what a tensor should be. We'll pick it up from there on Tuesday. |
MIT_8962_General_Relativity_Spring_2020 | 13_The_Einstein_field_equation_variant_derivation.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: All right, so we are ready to move to the next lecture. I'm going to record today. So if we just quickly recap the highlights of what we did in our previous lecture, the goal of this lecture was to take all the tools that we have built and develop a theory of gravity. So the two ingredients to this-- first we're going to assert that we take a valid law of physics, something that is well-understood in a local Lorentz frames, which we take corresponding to a freely falling frame when gravity is invoked, and we will carry it over into a general form by rewriting that law in a tensorial form. The example I went through was that if I assert that an object is unaccelerated, then I can normally write that by saying that the proper derivative of the velocity is zero-- or if you like the second derivative, positional on the world line is zero-- that carries over to saying that the parallel transport of the of the four velocity, along the four velocity, is equal to zero. If I require conservation, local conservation of energy and momentum to be asserted in my local Lorentz frame, my freely falling frame in the form that it has zero divergence with a partial derivative, that [? carries ?] over in general to zero derivative with a covariant derivative. Our second ingredient is to try to understand how it is that I can build my theory of spacetime-- excuse me. How I can build my spacetime given a distribution of energy, matter, and momentum. We are guided by the Newtonian field equation, which tells me the Laplace operator on Newtonian gravitational potential is 4 pi G rho. By studying the equation of motion for Newtonian gravity and comparing it to the geodesic equation in the slow motion limit for a spacetime that deviates just slightly from the flat space on a special relativity, we noted that this phi is closely related to G00. It's precisely equal to up to a factor of 2. It's a bit of an offset from the negative 1 that you get in the special relativity limit. We noted that this is not a healthy equation from the standpoint of being a ten-- of formulating things in terms of tensors, because rho is a particular component of the stress energy tensor. Doesn't make any sense to have a properly covariant eq-- if we want a properly covariant equation, it makes no sense to pick out one component of a tensor. So we upgraded rho to T mu nu. And we looked for, noting that two derivatives of the potential is something like two derivatives of the metric, which must be two derivatives of a curvature, we said, OK, it's got to be a curvature tensor on the left-hand side, and it's got to be a two index one, and it's got to be one that has no divergence so that this equation respects the local conservation of energy and momentum. And that led us by requiring that, in the appropriate limit of slow motion-- well, in this case, really it's just sort of a weak deviation from special relativity. By requiring this correspondence to hold, that force, the proportionality between my curvature tensor and my stress energy tensor will be 8 pi G. If you want to work in units where the speed of light is not set equal to unity, it's 8 pi G over C to the 4th. So that's fine. This is, in fact, exactly how Einstein originally derived this. And it's a very physical, very-- it's a very physicist kind of an argument. There is a another route to the field equations, though, which I would like to talk about in this next lecture. And what's interesting is that, historically, this route was actually developed at almost the exact same time that Einstein developed this way of coming up with the field equations that are going to describe relativistic gravity. This particular route helps us to-- I'd say it clarifies some of the choices that are made that lead to this particular framework for the field equations. And as we'll see as we get to the end of this lecture, it helps to illustrate how it might be-- if general relativity is incomplete in some way, how we might systematically add a little bit of additional framework, a little bit of an additional structure. It gives us, what is the most natural way to introduce additional degrees of freedom, additional ways of modifying things? So the framework I'm going to talk about today is something that allows us to much more naturally move beyond what this might be, just in case-- Einstein, at some point, we're going to likely find that general relativity is not quite adequate. It doesn't describe everything to the degree that we are able to measure everything. This will give us the tools to go beyond that. So the particular route I'm going to [? derive ?] is found by varying a particular Lagrangian. We're going to essentially define an action principle for the gravitational interaction. So I will call this lecture the field equation via the Einstein-Hilbert action. That is the same Hilbert of Hilbert Spaces, David Hilbert, mathematician who was active at about the same time that Einstein was at his peak research productivity. As I said, this way of thinking about things and this formulation that allows us to re-derive this, it was developed almost in parallel to Einstein's formulation of general relativity, so much so that they practically overlapped. But it must be said that Hilbert gives priority to Einstein on this. Hilbert gave Einstein the priority because he felt that Einstein really clarified the physics. And it was his insights that led to this way of thinking about gravity, even though in some ways he came up with a way of understanding where the field equations come from that is somewhat more elegant. So let me first sketch schematically how we're going to do this. Schematically, what we do is we define an action as the integral of a Lagrange density over all of spacetime. So we're going to write our action as S, and we're going to write it as an integral over our spacetime coordinates of some Lagrange density. So this is not a lot in field theories. This Lagrange density is something that depends on the fields you are studying. The action must itself be a Lorentz scalar. If I'm working in a freely falling frame, it cannot be something that depends upon the particular frame that one is using. And so one often writes this. You explicitly call out the square root of minus G. That is necessary in order to make your little coordinate interval here be a proper volume integral. And just to make a little bit of a notational difference, let's call L hat the version of the Lagrange density that has that weighted in there, like so. So this will, in fact, be-- so this entire integral, both of these entire integrals, are Lorentz scalars. This is a proper element of volume, and so this is a proper Lorentz covariant well-defined Lagrange density. So whenever you have an action principle, you proceed via an extremization procedure. So suppose your Lagrangian-- I'm going to write this very schematically. Suppose it depends on some set of fields. When you extremize this, that amounts to our requirement that the action be stationary with respect to a variation of the fields. So if I vary my fields, I'm going to vary my action. And let's just schematically write this using the first form of things here. I will then say, however, my Lagrangian varies, as I vary whatever fields it depends on, this must be zero. That is my stationarity requirement. For this to occur, for this to be valid for arbitrary field variations, I will then deduce from this that the variation of the Lagrange density itself with those fields must be equal to zero. When I do this, this will then lead to Euler Lagrange equations for my fields. Now, many students in the class may have already seen things like this. But for those who have not, let me work through a fairly simple example. So let's imagine that I have a Lagrangian that just depends on a scalar field phi and derivatives of that scalar field. So L is going to depend on some phi and derivatives of that phi. In my notes actually, imagine there might be multiple scalar fields. Let's just keep it simple here. We'll just have one scalar field and its derivatives. When we do our variation, it's going to be important that we treat variations of the derivative differently from variations of the field. At the moment, I'm going to focus on flat spacetime. We'll see what happens when the spacetime's a little bit more interesting in just a few moments. And what I'm going to do is imagine that phi varies like so, and its derivative varies like so, OK? So let's apply this. I want to look at a variation of this thing with respect to my fields. So when I do my variation of the action S, I get one term that will arise in the variation in the field phi, and I will get another term that arises due to the variation in the derivative. That term is potentially a little bit problematic for us. I'm going to gloss over a couple of details. But essentially what you want to do is, at this point, apply integration by parts. When you apply integration by parts, you can move this derivative onto this term here. And in so doing, your delta S becomes something like this. Caution-- I'm being a little bit glib here. That is only true if-- when you do this properly, there is kind of a boundary term that I'm assuming can be discarded. I don't want to get lost in the weeds associated with that right here. Our goal is just to sort of sketch the main physical ideas. But I will say that this is something that should be treated with a little bit of care. And I highlight this because there's going to be a similar step that I'm going to sort of go over with a little less care than I would like as we move forward, and I'll comment on that at that time. Having done this and assuming that the discarding of the boundary term is safe for me here-- actually, let me just make one further comment. This is being done at over all of spacetime. And so when I do that, I'm essentially assuming that the boundary is the boundary of infinity, which doesn't really exist, OK? So that's one reason why you can generally get rid of it in this case. And it turns out to be fairly safe. I'm highlighting this point because it should be stated a little bit more carefully than I have it in my notes. And when we apply this to gravity, there's a little bit of a subtlety there that I will comment on at that time. For now, I'm just going to say, great. So that sort of works good. That works out very good, very well. And in order for this to be equal to zero, the term inside my square braces must vanish, and so I wind up with Euler Lagrange equations, which my field, my field phi, must satisfy given the Lagrangian L, OK? Now let's just take it one step further. So I haven't said yet where the Lagrangian comes from, OK? As I sort of indicated, when I sketch this over here, it's going to depend on the nature of the field you study. So it's very common when you're studying, say, a scalar field and flat spacetime that you imagine that your Lagrangian has what we call a kinetic term, which is proportional to something that's quadratic-- whoops-- in the derivatives of your field. If it is massive, you may have another term which sort of looks almost like a potential energy in which you get something that looks like the field squared coupling to a parameter with the dimensions of mass. Take these derivatives, so taking the derivative of this Lagrangian with respect to the field, that's quite simple. You just get m squared phi. Take the derivative with respect to the field's derivative, what pops out of this is-- pardon me. [INAUDIBLE] looks like this. Now, take one more derivative this guy, pardon me. Just one second. Yeah. My apologies. I was just looking over my notes. You just get the wave operator, the flat spacetime wave operator, acting on the field phi. And the Euler Lagrange equations give us the following wave equation governing this field in flat spacetime. This is a result known as the massive Klein Gordon equation. So that's great. How do we apply this to our theory of gravity? So there's two things that we need to decide. One is, what fields am I going to build my Lagrangian out of? And so I'm going to come back to this point, because there actually is a choice to be made here, which is-- it's a little non-obvious. And I'm going to have a brief aside where I discuss for a moment a slightly different way of doing it than what I'm going to cover here. But we're going to vary the metric. We're going to treat the metric as our fundamental field. And so when I write down the Lagrangian for gravity, I'm going to imagine that it's a function of the metric, and that would be the thing that I vary. The other issue is, what should I use for my Lagrangian? So the principle which I'm going to use to make the choice here applies to the gravity. So my question is-- pardon me. My writing and my brain are getting out of sync here-- how do we choose our Lagrangian? How do we choose L? So it must be a scalar, or a scalar density if I want to take in bearing in mind that factor of square root G, must yield a scalar action. And here's why I'm going to make a real decision with consequences. It must be built from curvature tensors. The reason I choose to make it something that is constructed from curvature tensors is that I want this Lagrange density to be something that I cannot get rid of by changing my reference frame. If it's the metric, I can make the metric flat at any point simply by going into a freely falling frame. It has something to do with the Christoffel symbols, or rather with my connection. I can make that vanish by going into a freely falling frame. I cannot get rid of the curvature. So this point two here, this second point, this follows because this cannot be something that is eliminated by changing my frame of reference. I claim, based on that, that the simplest action of all, the one that you should at least start with, the one that is least ornate and that encompasses these two principles, is I'm going to choose the L hat. in other words, the one I'm going to separate out, the factor of square root minus G. I want this to be the Ricci scalar. It's the simplest possible thing that I can make out of curvature tensors that is a curvature and is going to give me a scalar. One could make other choices, OK? And in essence, that is the reason I am doing this lecture, is I want to highlight the fact that this particularly simple choice, we're going to see it reproduces the Einstein field equation. And so, in that sense, general relativity, we're going to see, is really the simplest geometric theory of gravitation. One could make more complicated ones, OK? But this is the simplest one. And in the last couple minutes of this lecture, we'll sort of explore how we can make it a little bit more ornate. So let's begin with this. So I'm going to write S equals the integral of all these things. And let's be blunt. Because I'm allowed to cheat because I'm the professor, I'm going to throw in a proof factor of 1 over 16 pi G, OK? Now, I am going to regard my curvature tensor. Like I said, I'm going to regard this as something that is built from the metric. So this form in which I've written here, I'm just saying schematically that Ricci itself is built from the metric. The Ricci scalar is the trace of the Ricci curvature. When I vary this thing, what I would like is to see what happens-- whoops, forgot my integral. And I'm going to vary it in the following way, OK? let's just clean up my indices a little bit here. I like to keep things consistent. I'm going to confuse myself. Let's leave them like that. This is fine, OK? So here's what I want to evaluate. I'm going to do a variation of these things. I'm going to assert the variation in the-- well, I'm not going to assert. What I'm going to do is do this variation. That will give me the variation in the action. For this to be stationary, I'm going to require that this equals zero. So what this tells me is I need to work this variation of what is in square braces out. I need to work this out right here. All right, so this needs be done with some care, so let's do it. First, the thing which I'm going to be looking at variations of, basically, I'm looking at a variation of-- I want to know how this changes when I vary the metric. This is going to break into three pieces. I'm going to want the variation of the square root of minus determinant of the metric. And then this will come along for the ride. I'm going to want to look at-- well, I'm going to get one piece that's just the variation of the metric itself. Don't need to do too much more with that. That actually already has the factor that I want to factor out there. And then I'm going to have a term that looks like what I get when I vary my Ricci curvature, OK? So these are the three pieces that I need to worry about, really just two. This one's kind of trivial. This one's a little more complicated. All right, so when I vary my square root of minus G term, we discuss this a little bit in a previous lecture where I introduce some, what I call, party tricks involving the determinant of the metric. You go back and take a look at this, you'll see that this can be turned into the following calculation, OK? When you vary your Ricci tensor, by something of a miracle, you will discover that this can be written as a term that looks like the derivative, should be the difference rather, of two covariance derivatives of variations in your Christoffel symbols, OK? Now, I like to work out what those variations in the Christoffel symbols are. These are a little bit lengthy. Just bear with me while I write these out. OK, as promised, it's a little bit lengthy. Using those two terms, OK, using that term, setting the indices as appropriate and taking those covariant derivatives, what you will then find when you evaluate the important combination to give me that term that involves my Ricci variation looks as follows. So I'm going to take those things, put them together to make my delta Ricci tensor, and then trace over it. And you get a lot of cancellations when you do that, such that the final result-- Notice no free indices on this, OK? So I've got derivatives hooking onto the Gs here. The mu nu is hooking up to that mu nu there. And before I manipulate this further, I'm just going to note that this is in the form of-- you can write this as the divergence of some vector field. So I'm going to say that this is the divergence of VA, OK? Excuse me. V alpha. V alpha is implicitly defined. It's basically this beta covariant derivative of all this junk here, OK? You might think, well, wait a minute. I'm taking derivatives of the metric here. Why doesn't my covariant derivative metric just give me zero? We need to be little bit careful about that because we are varying the metric, OK? And when we're varying the metric, once we have finished our variational principle, we will settle down to a metric where that is the case. But in the middle of the calculation, you just happy a little bit more careful than that. All right, so now let's put all these ingredients together. And what you wind up with, combining all of these pieces-- Oops, pardon me. Pardon me. Pardon me. Actually, let me move to another board. I've run out of room. This is the length the expression. My apologies. OK, so this becomes-- make sure I have enough room to get the whole thing on the line. OK, so fair amount of analysis goes into-- you put all those pieces together. You combine them all. And our goal is to reassemble what I've got up there on the top of this middle board here. So what this tells me is varying the action-- whoops-- leads me to an equation of this form. OK, let's stop and look at this for a second. So variation of the action gives me something it looks like the Einstein tensor contracted on my variation of my metric, plus this divergence of this goofy vector field. If that divergence of the goofy vector field were not there, we would be home free. Imagine we could ignore it. I would want to set delta S equal to zero, and that would give me G alpha beta equals 0. Now, I haven't said anything about a source, so that, in fact, gives me exactly what I would need to develop my Einstein field equations, at least in the case where there's no stress energy tensor. In a few moments, we're going to put a stress energy tensor back into this. But that's sort of the first step, that-- that first line, if I did not have that divergence of V, that would give me exactly what I needed there. So how do I get rid of that irritating divergence of a vector field? Well, in Carroll, it's pointed out that, because this is a perfect divergence and it is a proper volume integral, one can invoke the divergence theorem. If you invoke the divergence theorem, and then imagine this integral is being done over all of spacetime, and you basically say, well, I imagine there's no boundary the same because this covers all of time and all of space, you would then presumably lose contributions at the boundary. And then only the G alpha beta equals zero term would remain. That's not a bad physical argument. It's somewhat glib, however, OK? Carroll himself actually noticed this. So in Carroll's textbook, he himself says that this is not the rigorous way you get rid of this, OK? So for those of you who are more mathematically purist, if you're feeling dissatisfied-- but it's good on you. That is the way you should feel about it. I as a physicist am reasonably comfortable with this, but I would like to do this a little bit more rigorously. There are, in fact, two rigorous approaches to fixing this up. One of them is, in fact, quite beautiful. And it was my goal to overhaul this lecture and present method one that I'm about to write down in this lecture that I'm doing right here, right now. And among the various tragedies associated with the COVID-19 evacuation of our campus and everything moving into this online format, a very, very minor tragedy on the scale of everything that's going on, is that I just do not have the time to assemble the note to present what I was hoping to present to you. I hope by the end of the course to have enough bandwidth that I can maybe type some up and add them as supplemental notes to the course. But for now, they are not available. So let me just describe what I was going to present to you. Method one is you do what is called the Palatini variation. When you do a Palatini variation-- whoops-- you write the metric-- excuse me. You write the action like so, but you consider your Ricci curvature to be a function not of the metric but a function of the connection. Then when you do your variational principle, you don't just vary with respect to the metric. You vary it with both the metric and your connection coefficients. When you do this, if you do the calculation correctly, you, in fact, get both the Einstein equation, the Einstein-- so again, we've not introduced a source. I'm going to talk about source in just a moment. So the Einstein equation with no source emerges, and you find that your connection must be metric compatible. In other words, the connection is such that you are required to have covariant derivative of the metric equal to zero. It's actually a beautiful little calculation because we sort of argue that this was going to be the case-- when we were talking about how to define a covariant derivative, we noted that there are many ways to define a transport operation. This ended up being one that allowed us to define parallel transport. That became something that the physics drove us to selecting as a very natural choice for our connection. This variation doesn't necessarily know anything about that, but it shows that when you do your variational principle, treating metric and connection as separate degrees of freedom, that parallel transport-driven choice of the connection which gives us the Christoffel symbols emerges as the outcome of that. Now, this is not a terribly difficult calculation to do, but there's a lot of details you have to get right. And I've got about a fourth of it written up at this point and just haven't had a chance to get all the details completely straightened out. I really hope to have the bandwidth to finish it by the end of the term, in which case I will add it as supplemental notes to the course. The other thing you can do-- so there's absolutely nothing wrong with the approach that we started doing in which we vary the metric. And we just assume from the outset that my connection arises from the metric so that, the way we do it, the connection and the metric are not separate degrees of freedom, but the connection arises from the metric. And that is fine. But if we do that, then when you get down to this point, you just have to be a little bit more careful. You essentially take this argument that Carroll makes and you have to just do it with a lot more care. So what you need to do is define the boundary associated with the divergence integral very carefully-- whoops. Associated with divergence integral. In doing so, you have to very carefully treat the curvature on that boundary. So we are working in four dimensions. So it's a three dimensional boundary of our 4D spacetime. You can sort of think of the boundary as being a particular sort of slice through four dimensional space time. You pick a particular three dimensional slice, redefine that as the boundary region of this integral. And that requires us to introduce some additional concepts. In particular, once you pick out a particular slice of space time, some of the curvature associated with the metric on that slice isn't intrinsic to the spacetime itself, but it's imposed on the space time by the manner in which you have made that slice. Sort of in the same way that if I have a cylinder, the surface of a cylinder is intrinsically flat. If I put two geodesics on it and I parallel transport them around, they remain perfectly parallel forever and forever and forever. But it sure as hell looks curved. But that is because I have taken this flat surface and I have folded it up. I have embedded it in a higher dimensional space in such a way that there's extrinsic curvature that's been imposed on it. When you are carefully looking at the boundary of a four dimensional spacetime, you also need to very carefully treat the extrinsic curvature associated with how you put that boundary on top of things. When you do this correctly, you find you need to be a little bit more careful with how you define the Lagrangian. And when you, in fact, do that, it actually turns out you don't get that divergence term at all. You get sort of a counter term that cancels this guy away. So for students who are interested in seeing this explored and discussed in a lot more rigor than I am prepared to do, this is discussed in appendix E of Robert Wald's textbook, General Relativity. I will simply comment that Wald's textbook is one of the most mathematical textbooks on this topic. He does everything with great rigor and a lot of care. And the appendices is where he puts his particularly high-level, grotty details. So this really is something that is beyond the level of where we do a one semester course like 8.962. So it's unfortunate it sort of pops up here, which is why a textbook like Carroll does this in a somewhat glib way. I am personally a fan of doing this Palatini variation. And so, as I've emphasized, I'm going to try to clean that up and make that available. So what all of this ends up telling us-- so let's take the optimistic point of view here that we have either accepted the somewhat glib argument that we can discard that term, or we sat down and we slog through appendix E of Wald, or we wrote up the Palatini variation. Doing this variation in that case then leads us to G alpha beta equals zero. This is the Einstein equation when there is no source, when the stress energy tensor is zero. This is often called the vacuum Einstein equation. More generally, we should treat the action that we are varying as the Einstein-Hilbert action plus an action associated with matter. So a more general form would be to write S as the integral d4x square root minus G R over 16 pi G plus the Lagrangian for-- I'm just going to say m for matter. Could be matter and fields. If you do that, then when you look at your variation and enforce that the variation of the action equals zero, this leads you to an equation of the form-- that's given by this. And so up to a numerical factor, which I'll write out in just a second, you can basically see that this variation of my matter Lagrangian becomes the stress energy tensor. In particular, if you just look at this and say, oh, I am going to define T alpha beta as minus 2 divided by square root of minus G, like this, then the Einstein-Hilbert variation gives me exactly the same Einstein equation as we had before. I have a set of exercises in the notes, which I'm not going to go through because they're slightly tedious. And I think I'm going to actually end a little bit early here. But I want to just describe one of the ways in which this can be used. So suppose I pick as my matter Lagrangian minus one fourth f mu nu, f mu nu, where f mu nu is the Faraday tensor that describes electric and magnetic fields. So what I want to do is show that, by applying this variational principle that I've defined over there to this-- but this gives me exactly the stress energy tensor that one expects for electromagnetic fields. OK, so defining my electromagnetic action, pull out the overall factor of minus one fourth, I integrate over my spacetime coordinates. And then notice the way this is set up. So I have f mu nu upstairs, f mu nu downstairs. Let's write my f mu nu upstairs as f alpha beta downstairs but with enough prefactors of the metric to raise those indices. So I do a variation on this. And the way I'm going to do it is I will vary the metric. When I do this, I will not vary the fields, OK? If one does variations of those things, you can do that, and you can use that to do things like define your field equations. But what I want to do is just see how, by varying how the metric couples do these things, the stress energy tensor emerges. So in my notes, I go through this calculation with some care. And what emerges at the end is something that looks like this. And in fact, it's not too hard to show. Dig into something, a textbook like Jackson, what is in square braces there is, in fact, the standard stress energy tensor associated with the electromagnetic field. So this technique, this definition I've put in here, it does work in all the cases that we know about. And so this variational principal root to the field equations-- so punchline is selecting L equals R yields the Einstein field equations upon variation of the metric. So I'll take a sip of water here. And then the question I want you to sort of think of at this point is, so what? I mean, it's nice that there is a second way to do this. But did we need it? Is it really important that we have the second way of doing this? Well, here I want to come back to a comment I made as I began this lecture, which was that our guiding principle for formulating this Lagrangian was that it has to be built out of curvature tensors, and it has to be a scalar. I would argue that the Ricci scalar is the simplest such construction that does that. But it is not the only scalar made from curvature tensors that will do this. Imagine you wanted to modify gravity, or you wanted to come up with a framework that was consistent with general relativity. And remember, L equals R gives me general relativity, gives me Einstein's field equations. It gives me Einstein's theory of gravity. Suppose I want to modify it in a way such that things look just like Einstein's gravity for most of the universe, but perhaps small differences kick in when the curvature is small in some sense that will need to be characterized. Well, one way you could imagine doing this is by adding a correction. Suppose we try L hat equals R minus alpha over R. That sort of suggests that this is going to lead to corrections to the framework of gravity when the curvature scale is less than or of order square root of alpha. It's actually a work of a couple pages of algebra. I can take this modified Lagrangian. I can go through exactly this variational principles that we spent today doing. And when I do so, I find a new field equation coupling spacetime to my sources. So vary the metric, enforce stationarity of the action, what you get is a field equation that looks like this. Notice, set that coupling alpha to zero and it says G alpha beta equals 8 pi newtons GT alpha beta. But introduce this little scale here, suddenly you've got all this other stuff emerging. This is an idea that was proposed and explored in a paper, if you want to read a bit more about this, by Sean Carroll et al, published in Physical Review D. It's now almost coming up on its second decade. So the reason why Sean and his collaborators developed this, I mentioned in my previous lecture that there's observations that have driven us to consider whether there is, in fact, a cosmological constant contributing to the large scale structure of our universe. And like good scientists, you're always looking for alternate hypotheses that could be tested or falsified. And so what Carroll et al pointed out was the universe's accelerated expansion, so the behavior that is driving us to consider the possibility the cosmological constant, it's coming on very, very large scales where the curvature is quite small. Laboratory and solar system and many other astrophysical tests of the foundations of general relativity aren't as strong there. So why not try something a little bit out of the box? And they said, let's think about a theory of gravity in which there are potentially important modifications when the curvature is small. This is what resulted. And they showed, actually, that by appropriately choosing the parameter alpha, actually, this is a point that's worth making. Notice this is introducing a scale, OK? The theory doesn't tell us what alpha is. From the standpoint of the analysis that motivated it, it was a phenomenological knob that they sort of tweaked to see whether they could explain observations. And indeed, they did find that they could get pretty good explanation of the accelerating expansion of the universe with this modified theory of gravity. But all good theories are testable. They're falsifiable. They give rise to hypotheses that you can compare against data. It turns out, if you tune alpha to make the expansion of the universe work, you get the orbits of the planets in our solar system wrong. So this ended up not being a viable theory, but it was a pretty cool thing to explore, OK? And I think it serves as a really wonderful example of how, in a theory of gravity, one formulates a hypothesis and then tests it. And then giving Sean and his colleagues all their due, cheerfully accepting the falsification when it turned out it didn't work. You know, it's a very, very nice piece of work. And I think it just is a really good idea of the power of this action-based method for considering modifications to your theory of gravity, OK? And it really highlights the fact that just choosing your Lagrange density as the Ricci scalar, that's the simplest possible thing it can make. And that is a sense in which general relativity is the simplest of all possible relativistic theories of gravity. Let me just give you another example of-- actually, two more examples. I'm not going to go through them in great detail, but just sketching the way people sometimes imagine modifying gravity. So suppose I choose as my gravitational action my normal Einstein-Hilbert term plus a term that goes as the Ricci curvature squared. This is something that you might expect to become important when the curvature is large. OK, so when R is greater than R of order, 1 over beta, the second term, will become more important than the first term. Again, we've introduced a scale. There is some theoretical prejudice among the community that thinks about these kinds of models that perhaps this scale has something to do with where quantum gravity effects begin to kick in and classical general relativity must be modified. And so you can imagine, you might think that this beta is on the order of, when you work in units where G and C are equal to 1, H bar to the minus 1. Actions like this often emerge when you are studying-- they often show up when you are studying theories in which general relativity is itself an effective theory that comes about from averaging over small scale degrees of freedom. One more example. What if there were, in addition to the tensor, the metric tensor which describes the geometry of space time-- what if the manner in which the metric tensor coupled to its source varied depending on some additional field that filled all of spacetime? So you can imagine that your gravitational action took the usual form, but it was modified by some kind of a scalar field that might itself depend upon spacetime curvature, OK? And you can imagine then supplementing this with an action for the scalar field, which, up to an overall constant at least, I will write-- theories of this sort are called scalar tensor series. And what they essentially amount to is you imagine that there is some field in addition to the metric of spacetime which is affecting gravity. You can almost think of them as essentially saying that the effective gravitational constant varies depending upon the strength of the gravitational interaction. So perhaps down near a black hole the Newton constant is bigger or smaller or something than way out in empty space. Theories of this kind arise-- I should say terms of this kind and the action-- quite naturally in a lot of work in theoretical attempts to quantize gravity, things like that. And so people like to pay attention to these things and to see what kind of generic predictions they can make. These are known as scalar tensor theories. And if you're interested in seeing more about them, Sean Carroll's textbook has a full section describing them. A little bit out of date now. One thing that's kind of interesting as we sort of think about how the subject has evolved over the years that I've been teaching it is that when you have both a tensor component, in other words, a spacetime metric coupling to your stress energy tensor and you have a scalar field, you change the dynamics of the theory. And that ends up doing things like changing the character of gravitational radiation that is emitted from a system. And so, very recently, the most stringent limits on these possible modifications to general relativity have come about from things like LIGO observations which have looked at coalescences of black holes and neutron stars. And all of their measurements are consistent with there being no scalar modification to these things. And so it helps to set limits on many of these theoretical frameworks that introduce and look at modifications to GR which are motivated by certain considerations. This concludes the first half of 8.962. So if I'm just going to do a quick recap of where we are and where we are going, where I want you to sort of sit with things right now is that you have this arsenal which basically tells you that, given the metric of spacetime, you can compute the motion of bodies-- this is done using by geodesics-- you can characterize the curvature of that spacetime which tells you about tidal fields, and you now, as of these last two lectures, know how to get this by solving the Einstein field equation. What we will do in the remaining lectures that I'll be recording and distributing through this course's website is solving this equation and exploring the properties of these solutions. And so let me just sort of say this equation, if you go back to think about what goes into this-- at the beginning of my next lecture which I'll be recording tomorrow, we'll go through this in some care. But if you just think of this as a partial differential equation for the metric, it is a god-awful mess, OK? There's no nice way to put it. I mean, it's just a horrible mess of coupled, non-linear, partial differential equations. We are going to look at three ways of solving this. Method one, we will think about a kind of weak field expansion. I'm going to imagine that the spacetime corresponding to the sources we are studying looks like flat spacetime plus a small correction. That will allow me to take this horrible coupled, non-linear mess and linearize, OK? We will lay out the principles for doing that methodically in the beginning of the next lecture. Method two, we'll see what happens when we assert a symmetry. One of the reasons why those equations are such a mess is that, in general, there's just a tremendous number of terms that are coupling to one another, all sorts of things tying together, and it's just very, very complicated. But if we imagine that everything is, say, spherical symmetric, then that dramatically reduces the number of degrees of freedom that the spacetime could have even as a matter of principle. And so we will actually start by beginning to explore the most symmetric space times of all. And I will lay out exactly what that means. And that actually turns out, at least to within our ability to observationally probe this-- that appears to describe our universe on the largest scales. And so when we do the largest, the most symmetric things possible, that gives us spacetimes that worked very well for describing cosmology. Making things a little bit less symmetric but still highly, highly symmetric, that allows us to describe very strong field objects, things which I'm going to call-- I'm going to just call them stars. What I really mean by that are these are objects that are just self gravitating circle fluid balls, OK? And we will see how they can sort of hold themselves up against their own gravity, what their spacetime looks like. There's all sorts of interesting properties that emerge from them. And eventually we actually find that you reach a set of configurations where there is no-- no fluid can be allowed by the-- let me try that sentence again. You can find configurations such that no fluid allowable by the laws of physics can give you a stable object. They produce black holes. That will basically take us, essentially, to the end of the course, OK? These two things have a tremendous amount of content in them. And this is where a huge amount of the literature on general relativity has focused its attention over the past several decades. In my last lecture or so, I'm going to briefly describe general solutions. What do you do if there is no symmetry, there is no weak expansion, no weak parameter you can expand in? You just have to deal with the non-linear, bloody mess that these equations are and then try to solve it. And what this will do is allow me to explore and tell you a little bit about the field of numerical relativity, which is what you do when you just want to solve these things with no approximations, no assumptions that simplify it whatsoever. Just take Einstein's field equations, put it into a computer, write a code, hit Go, and see what happens. So under the normal schedule of things, where we would end today would've been right before spring break. And these lectures would start up right after spring break. Under the system that we are operating now, I do suggest you sort of save this for a little while. Make sure that you understand what is going on with these field equations. And we will be ready to explore the solutions, and not just compute these solutions but also explore things like motion in them and their properties. And we'll begin that process immediately afterwards. All right, and I will conclude this lecture here. |
MIT_8962_General_Relativity_Spring_2020 | 9_Geodesics.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: So let me just do a quick recap of what we did last time. So today, we're going to move into things that are a little bit more physics. Last time we were really doing some things that allows us to establish some of the critical mathematical concepts we need to study the tensors that are going to be used for physics on a curved manifold. So one of the things that we saw is that if I wanted to formulate differential equations on a curved manifold, if I just defined my derivative the most naive way you might think of, you end up with objects that are not tensorial. And so mathematically you might say, well, that's fine. It's just not a tensor anymore. But we really want tensors for our physics because we want to be working with quantities that have frame-independent geometric meaning to them. So that notion of a derivative-- if I just do it the naive way-- isn't the best. And so I argued that what we need to do is to find some kind of a transport operation in which there is a linear mapping between things like my vector field or my tensor field and the displacement, which allows me to cancel out the bits of the partial derivative transformation laws that are non tensorial. There's a lot of freedom to do that. One of the ways I suggest we do that is by demanding that when I do this, that derivative-- when applied to the metric-- give me 0. And if we do that, we see right away that the transport law that emerges gives me the covariant derivative as one of my examples. Now this shouldn't be a surprise. We introduced the covariant derivative by thinking about flat spacetime operations, but with all my basis objects being functionals. And this in some way is sort of a continuation of that notion. The other thing which I talked about is telling you we're not going to use a tremendous amount here except to motivate one very important result. And that is if I define transport by basically imagining that I slide my vectors in order to make the comparison-- along some specified vector field-- I get what's known as the Lie derivative. And so this is an example of the Lie derivative of a vector. And you get this form that looks like a commutator between the vector field you're sliding along and the vector field you are differentiating. Similar forms-- which are not really the form of a commutator-- but similar forms can be written down for general tensors. The key thing that you should be aware of is that it's got a similar form to the covariant derivative, in that you have one term-- let's focus on the top line for the second-- you have one term that looks just like the ordinary vector contracted onto a partial derivative of your field. And then you have terms which correct every free index of your field-- one free index if it's a vector, one free index if it's a one form, and corrections for an end index tensor-- with the sign doing something opposite to the sign that appears in the covariant derivative. What's interesting about this is that so defined, the Lie derivative is only written in terms of partial derivatives. But if you just imagine-- you promote those partial derivatives to covariant derivatives-- you find the exact same result holds because all of your Christoffel symbols-- or connections as we like to think of them when we're using parallel transport-- all the connective objects cancel each other out. And this is nice because this tells me that even though this object, strictly speaking, only involves partial derivatives, what emerges out of it is in fact tensorial. And it's an object that I can use for a lot of things I want to do in physics. In particular where we're going to use it the most-- and I said you're going to do this on the PSET, but I was wrong-- you're going to do something related to one of the PSETs-- but I'm going to actually-- if all goes well-- derive an important result involving these symmetries in today's lecture. We can use this to understand things that are related to conserved quantities in your space time. And where this comes from is that there is a definition of an object we call the Killing vector, which is an object where if your metric is Lie transported along some field C, we call C a Killing vector. And from the fact that the covariant derivative the metric is 0, you can turn the equation governing the Lie derivative along C into what I wrote down just there. And I should write its name. There's the result known as Killing's Equation. If a vector has a Killing vector-- if a metric has a Killing vector-- then you know that your metric is independent of some kind of a parameter that characterizes that spacetime. The converse also holds. If your metric is independent of something-- like say the time coordinate-- you know that there is a Killing vector corresponding to that independent thing. And so you'll often see this described as a differing amorphism of the spacetime-- if you want to dig into some of the more advanced textbooks on the subject. We'll come back to that in a few more details hopefully shortly before the end of today's lecture. So where I concluded last time, was we started talking about these quantities known as tensor densities, which are given the less-than-helpful definition-- quantities that are like tensors, but not quite. The example I gave of this-- where we were starting-- was the Levi-Civita symbol. So let me just write down again what resulted from that. So if I have Levi-Civita-- and the tilde here is going to reflect the fact that this is not really a tensor-- this guy in some prime coordinates is related to this guy in the unprime coordinates via the following-- let's get the primes in the right place-- the following mess of quantities. So I'm not going to go through this again. This is basically a theorem from linear algebra that relates the determinant of a matrix-- not metric, but matrix-- to what you get when you contract a bunch of matrices onto the Levi-Civita symbol. And so the key thing to note is that if this were not here, this would look just like a tensor transformation. But that is there. So it's not. And so we call this a tensor density of weight 1. So the other one-- which I hinted at the end of the last lecture, but did not have time to get into-- is suppose we look at the metric. Now, the metric-- no ifs, ands, or buts about it-- it's a tensor. And it's actually the first tensor we've started talking about back in our toddler years of studying flat spacetime, which by the way, was about three weeks ago. Obviously that's a tensor. It's a simple tensor relationship. Let's take the determinant of both sides of this. You might look at this and go, why do you want to do that? Well when I do this, I'm going to call the determinant of the metric in the primed representation. Let's call that G prime. I get 2 powers-- 2 powers of this Jacobian matrix is determinant. And I get the determinant in my original representation. Now I want to write this in a way that's similar to the way I wrote it over here. Notice I have all my primed objects over here on the left-hand side. And my factor of this determinant relates-- it's got primed indices in the upstairs position, unprimed in the downstairs. But the determinant of 1 over a metric is just 1 over the determinant of-- the determinant of the inverse of a matrix is just 1 over the determinant of that matrix. And so I can really simply just say this looks like so. So the determinant of the metric is a tensor density of weight minus 2. What this basically tells us is I now have two of these things. I've been arguing basically this entire course that we want to use tensors because of the fact that they give me a covariant way of encoding geometric concepts. I've got these two things that are not quite tensors. I can put them together and get a tensor out of this. So what this tells me now is I can convert any tensor density into a proper tensor. So suppose I have a tensor density of weight W. I can convert this into a proper tensor by multiplying by a power of that G. So multiply it by G to the W over 2. One slight subtlety here, when we work in spacetime-- let's just stop for a second and think about special relativity. In special relativity in an inertial reference frame, my metric is minus 1 1 1 1 on the diagonal. So its determinant is negative 1. And when I take negative 1 to some power that involves a square root, I get sad. We all know how to work with complex numbers. You might think that's all OK. It's not in this case. But the way I can fix that is that equation's still true if I multiply both sides by minus 1. I want this to be a positive number when I take the square root. So I'm allowed just to take the absolute value. So we take the absolute value to clear out the fact that in spacetime, we tend to have an indeterminate metric, where the sign depends on the interval. So remember the only reason we're doing this-- this is just a-- I don't want to say it's a trick. But it's not that far off from a trick. I'm just combining two tensor densities in order to get a tensor out of it. And minus a tensor density is still a tensor density. So I'm OK to do that. And I'm just doing this so that my square root doesn't go haywire on me. So a particular example-- in fact the one that in my career has come up the most-- is making a proper volume element converting my Levi-Civita symbol into a tensor that gives me a volume element. So my Levi-Civita symbol has a tensor density of weight 1. If I want to make that into a proper tensor, I multiply by the square root of the determinant of the metric. So now I will no longer have that tilde on there, which was meant to be a signpost that this as a quantity is a little bit goofy. You wind up with something like this. When you go-- and by the way, sometimes when you're working with this, you need to have this thing with indices in the upstairs position. You have to be a little bit careful. But I'll just give you one example. If you raise all four of the indices, what you find when everything goes through, this one is not that hard to see because you're basically playing with a similar relationship to the one that I wrote down over here-- just a short homework exercise to demonstrate this. And then you end up with the tensor density of the opposite sign. Weight minus 1, you wind up with a 1 over square root there. So like I said one of the reasons why this is an important example is that we use it to form covariant volume operators. So in four-dimensional space-- so imagine here's my basis direction for spatial direction 1, spatial direction 2, spatial direction 3-- you guys can figure out how to write spatial direction 0 on your own time-- I would define a covariant 4 volume-- 4 volume element from this. It'll look like this. And if this is an orthogonal basis, this simply turns into something like-- these are meant to be superscripts because these are coordinates. So it just turns into something like this if I'm working in an orthogonal basis. And again for intuition, I suggest go down to 3-dimensional spherical coordinates. And I wrote this last time. But let me just quickly write it up. I mean everything I did here, I tend to-- since this is a course on spacetime-- by default I write down all my formulas for three space dimensions, one time dimension. But it's perfectly good in 3 spatial dimensions, 2 spatial dimensions, 17 spatial dimensions-- whatever crazy spacetime your physics want you to put yourself in-- or space your physics wants to put you in. So I'll just remind you-- that when you do this, you've got yourself a metric across the diagonal of 1 r squared r squared sine squared theta. And just to be consistent-- I usually use Latin letters for only spatial things. So let's do that. This would be how I would then write my volume element. Did I miss something? AUDIENCE: Yeah. [INAUDIBLE] SCOTT HUGHES: Absolutely. Yeah. Thank you. I'm writing quickly. Yeah? AUDIENCE: Is there [INAUDIBLE]? SCOTT HUGHES: This is dx-- oh, bugger. Yep. I'm trying to get to something new. And I'm afraid I'm rushing a little bit. So thank you for catching this. And so with this, take the determiner of this thing. And sure enough you get r squared sine theta d r d theta d phi. So this is the main thing that we are going to use this result for-- this thing with tensor densities. I want to go on a brief aside, which is relevant to the problem that I delayed on this week's problem 7. So there are three parts of problem 7 that I moved from PSET 3 to PSET 4 because they rely on a result that I want to talk about now. So the main thing that we use the determinant of the metric for in a formal way is this-- that it's a tensor density of weight minus 2. And so it's a really useful quantity for converting tensor densities into proper tensors. And really the most common application of this tends to be to volume elements. But it turns out that it's actually really useful for what a former professor of mine used to like to call party tricks. There's some really-- it offers a really nice shortcut to computing certain Christoffel symbols. So in honor of Saul Teukolsky let's call this a party trick. So we're using the determinant of the metric to compute certain Christoffels. So this is going to rely on the following. So suppose I calculate the Christoffel symbol, but I'm going to sum on the raised index. And bearing in mind it's symmetric in the lower one, I'm going to do a contraction of the raised index with one of the lower indices. So let's just throw in a couple of definitions. This is equivalent to the following. And so throwing in the definition of the Christoffel with all the indices in the downstairs position-- this formula, by the way, is something that I've been writing down now for about 27 years. And I have to look it up every time. Usually by the end of a semester of teaching 8.962, I have it memorized. But it decays by then. So if you're wondering how to go from here to here-- this is the kind of thing-- just look it up. So let's pause for a second. Remember that the metric is-- it's itself symmetric. So in keeping with that, I'm going to flip the indices on this last term, which-- hang on a second. That was stupid. Wait. Pardon me. This is the term I want to switch indices on. My apologies. So the reason I did that is I want to have both of these guys ending with the alpha because notice this and this-- they're the same. But I have interchanged the beta and the mu. So these two terms-- the first term and the third term-- are anti symmetric upon exchange of beta and mu. They are contracted with the metric, which is symmetric upon exchange of beta and mu. Question? AUDIENCE: Does the metric have to be symmetric? SCOTT HUGHES: The metric has to be symmetric. [LAUGHS] I don't want to get into that right now, but yes [LAUGHS]. So these guys are anti symmetric. This guy is symmetric. And remember the rule. Whenever you contract some kind of a symmetric object with an anti symmetric m you get 0. So that means this term and this term die. And what we are left with is gamma mu mu alpha is 1/2 g u beta, and the alpha derivative of g u beta. There is the way it is contracting the indices in the 2 metric with the other one. Well here's a theorem that I'm going to prove in a second-- or at least motivate-- that it's going to rely on a result that I will pull out of thin air, but can be found in most linear algebra textbooks. It's not too hard to show that this can be further written as 1 over square root of the determinant times the partial derivative of the square root of the determinant, which is sometimes-- depending on your applications-- this can be written very nicely as the derivative of the logarithm of the absolute value of the-- the square root of the absolute value of the determinant. So before I go on and actually demonstrate this, you can see why this is actually a pretty-- so this actually comes up. I'm going to show a few applications as to why this particular combination of Christoffel symbols shows up more often than you might guess. It's really important for certain important calculations. And this is telling me that I can get it by just taking one partial derivative of a scalar function. And if you know your metric, that's easy. So this becomes really easy thing to calculate. So let's prove it. So the proof of this relies on a few results from linear algebra. So let's not think about tensors for a second. And let's just think about matrices. So imagine I've got some matrix m. I'm going to be agnostic about the dimensions of this thing. And suppose I look at the following variation of this matrix. So suppose I imagine doing a little variation. So suppose every element of m is a function. And I look at a little variation of the log of the determinant of that matrix. Well this can be written as log is basically a definition of this. Now, if I exploit properties of logarithms, this can be written as the log of the determinant-- m plus delta m-- divided by the determinant of m. Now I'm going to use the fact that 1 over the determinant of m is the determinant of the inverse of m. So taking advantage of that, I can further write this guy as something like this. Now I'm going to invoke an identity, which I believe you can find proven in many linear algebra textbooks. It just occurred to me as I'm thinking about this, I don't know if I've ever seen it explicitly proven myself. But it's something that's very easy to demonstrate with just a quick calculation. You can just do-- I'm a physicist. So for me I'll use Mathematica. I'll look at six or seven examples and go, it seems right. And so I've definitely done that. But I believe this is something that you can find proven explicitly-- like I said-- in most books. So remember these are all matrices. So this isn't the number 1. We want to think of this as the identity matrix. Oh and I'm also going to regard this variation as a small quantity. So if I regard epsilon as a small matrix-- this can be made formal by defining something like condition number associated with the matrix or something like that. But generally what I want to mean by that is if I take this epsilon and I add it to 1, all of this-- so my identity is 1 on the diagonal-- 0s everywhere else-- all the things that are put into the sum of 1 plus epsilon are much, much smaller than that 1 that's on the diagonal. That will be sufficient. So if epsilon is a small matrix, then the determinant of 1 plus epsilon is approximately equal to 1 plus the trace of epsilon. What that approximately refers to is-- of course you can take that further. And you'll get additional corrections that involve epsilon times epsilon, epsilon times epsilon, times epsilon. I believe when you do that, the coefficient is no longer universal. But it depends upon the dimensions of the matrix. But leading order it's independent of dimensions of the matrix. And that's something that you can you can play with a little bit yourself. Like I said, this is sufficient for what we want to do here. So I'm going to think of my small matrix as the matrix of inverse m times a variation of m. This is our epsilon. So we're going to apply it to the line that I have up here. And this tells me that my delta on the log of the derivative of m is the log of 1 plus the trace of m to the minus 1 on the matrix m. Log of 1 plus a small number is that small number. Now the application. So this is-- like I said, this the theorem that you can find in books that I don't know about but truly exist. This is something I've seen documented in a lot of places. Let's treat our m as the metric of spacetime. So my m will be g alpha beta. My m inverse will be g in the upstairs position. And I will write this something like so. And I'm going to apply this by looking at variations in my metric. So delta log-- I'm going to throw my absolute values in here. That's perfectly allowed to go ahead and put that into there. Applying this to what I've got, this is going to be the trace of g mu beta times the variation of g beta gamma. And I forgot to say, how do I take the trace of a matrix? So the trace that we're going to use-- we want it to be something that has geometric meaning and has a tensorial meaning to it. So we're going to call the trace of this thing g alpha beta epsilon alpha beta. If you think about what this is doing, you're essentially going to take your-- let's say I apply this to the metric itself. I put one index in the upstairs position, one the downstairs position, and then I am summing along the diagonal when I do this. You will sometimes see this written as something like that. So in this case, when I'm taking the trace of this guy here, that is going to force me to-- let's see. So this gives me a quantity where I'm summing over my betas. And then I'm just going to sum over the diagonal indices. I'm forcing my two remaining indices to be the same. So putting this together, this tells me-- so now what I'm going to do is say, I basically have part of a partial derivative here. All I need to do is now divide by a variation in my coordinate and take the limit. So it comes out of this as the partial derivative-- looks like this. Now let's trace it back to our Christoffel symbol. My Christoffel symbol-- the thing which I'm trying to compute-- is one half of this right-hand side. So it's one half of the left-hand side. And I can take that one half, march it through my derivative, and use the fact that 1/2 the log of x is the log of the square root of x. Check. So like I said, this is what an old mentor of mine used to like to call a party trick. It is a really useful party trick for certain calculations. So I want to make sure you saw where that comes from. This is something you will now use on the problem that I just moved from PSET 3 to PSET 4. It's useful for you to know where this comes from. You're certainly not going to need to go through this yourself. But this is a good type of calculation to be comfortable with. Those of you who are more rigorous in your math than me, you might want to run off and verify a couple of these identities that I used. But this is very nice for physics level rigor-- at least astrophysicists level rigor. So let me talk about one of the places where this shows up and is quite useful. So a place where I've seen this show up the most is when you're looking at the spacetime divergence of a vector field. So when you're calculating the covariant derivative of alpha contracting on the indices-- let's just throw in-- expand out the full definition of things-- all you've gotta do is correct one index. And voila, this is exactly the things that change-- where is it-- change my alpha to a mu-- that's exactly what I've got before. And so-- hang on just one moment. I know what I'm doing. These are all dummy indices. So in order to keep things from getting crossed, I'm going to relabel these over here. So I can take advantage of this identity and write this. So stare at this for a second. And you'll see that the whole thing can be rewritten in a very simple form. Ta-da. You haven't done as much work with covariant in your lives as I have. So let me just emphasize that ordinarily when you see an expression like you've got up there on the top line, you look at that, and you kind of go [GROANS] because you look at that, and the first thing that comes to your mind is you've got to work out every one of those Christoffel symbols and sum it up to get those things. And in a general spacetime, there will be 40 of them. And before Odin gave us Mathematica, that was a fair amount of labor. Even with Mathematica it's not necessarily trivial because it's really easy to screw things up. With this you calculate the determinant of the metric, you take its square root, you multiply your guy, and you take a partial derivative, and you divide. That is something that most of us learned how to do quite a long time ago. It cleans the hell out of this up. So the fact that this gives us something that only involves partial derivatives is awesome. This also-- it turns out-- so when you have things like this, it gives us a nice way to express Gauss's theorem in a curved manifold. So Gauss's theorem-- if I just look at the integrals that-- or rather the integral for Gauss's theorem-- let's put it that way. Let's say a Gauss's-type integral. So go back to when you're talking about conservation laws. If I imagine I'm integrating the divergence of some vector field over a four-dimensional volume, look at that, I get a nice cancellation. So this turns into an integral of that nice, clean derivative over my four coordinates. And then you can take advantage of the actual content of Gauss's Theorem to turn that into an integral over the three-dimensional surface that bounds that four volume. It's a good point-- so you're emboldened by this. You say, yay, look at that. We can do all this awesome stuff with this identity. It gives me a great way to express some of these conservation laws. You might think to yourself-- and I realized as I was looking over these notes-- I'm about to I think give away a part of one of the problems on the PSET-- but c'est la vie. It's an important point. Can we do something similar for tensors? So this is great that you have this form for vectors. The divergence of a vector is a mathematical notion that comes up in various contexts. So this is important. But we've already talked about the fact that things like energy and momentum are described by a stress energy tensor. So can we do this for tensors? Well the answer turns out to be no, except in a handful of cases. And I have a comment about those handful of cases. So suppose I take this-- and I'm taking the divergence on say the first index of this guy-- so there's the bit involves my partial derivative-- I'm going to have a bit that involves correcting the first index. So the first correction is it's of a form that does in fact involve this guy we just worked out this identity for. And in principle we could take advantage of that to massage this and use this identity. But the second one there's nothing to do with that. This you just have to go and work out all of your 40 different Christoffel symbols and sit down and slog through it. This spoils your ability to do anything with it, with one exception. What if a is an anti-symmetric tensor? If a is an anti-symmetric tensor, you've got symmetry, anti symmetry, and it dies. So that is one example of where you can actually apply it. And I had you guys play with that a little bit on the PSET. It's worth noting though that the main reason why one often finds this to be a useful thing to do is that when you take the divergence of something like a vector, you get a scalar out. You get a quantity that is-- really its transformation properties between different inertial frames or freely-falling frames is simple. So even when you can do this and take advantage of this thing, working with the divergence of a tensor-- exploiting a trick like this turns out to generally not be all that useful. And I'll use the example of the stress energy tensor. So conservation of stress energy in special relativity-- it was the partial derivative-- the divergence of the stress energy tensor expressed with the partial derivative was equal to 0. We're going to take this over to covariant derivative of the stress energy tensor being equal to 0. That's what the equivalence principle tells us that we can do. Now when I take the divergence of something like the stress energy tensor, I get a 4 vector. Every 4 vector always has implicitly a set of basis objects attached to it. When I've got basis objects attached to it, those are defined with respect to the tangent space at a particular point in the manifold where you are currently working. And so if I want to try to do something like an integral like this-- where I add up the four vector I get by taking the divergence of stress energy and integrate it over a volume-- I'm going to get nonsense because what's going on is I'm combining vector fields that are defined in different tangent spaces that can't be properly compared to one another. In order to do that kind of comparison, you have to introduce a transport law. And when you start doing transports over macroscopic regions, you run into trouble. They turn out to be path dependent. And this is where we run into ambiguities that have to do with the curvature content of your manifold. We'll discuss where that comes into our calculations a little bit later. But what it basically boils down to is if I use a stress energy tensor as an example, this equation tells me about local conservation of energy and momentum. In general relativity I cannot take the local conservation of energy and momentum and promote it to a global conservation of energy and momentum. It's ambiguous. We'll deal with that and the conceptual difficulties that that presents a little bit later in the course. But it's a good see the plant at this point. So let's switch gears. We have a new set of mathematical tools. I want to take a detour away from thinking about some more abstract mathematical notions and start thinking about how we actually do some physics. So what I want to do is talk today about how do we formulate the kinematics of a body moving in curved spacetime? So I've already hinted at this in some of my previous lectures. And what I want to do now is just basically fill in some of the gaps. The way that we do this really just builds on Einstein's insight about what the weak equivalence principle means. So go into a freely falling frame. Go in that freely-falling frame. Put things into locally Lorentz coordinates. In other words perform that little calculation that make spacetime look like the spacetime of special relativity of the curvature corrections. And to start with, let's consider what we always do in physics, is we'll look at the simplest body first. We're going to look at what we call a test body. So this is the body that has no charge, no spatial extent, it's of zero dimensional point, no spin-- nothing interesting, except a mass. So if you want to think about this-- I use a way that I find to think about this is all these various aspects to it, you're adding additional-- either charges to it or additional multipolar structure to this body. I'm thinking of this-- this is sort of like a pure monopole. It's nothing but mass concentrated in a single zero size point. Obviously it's an idealization. But you've got to start somewhere. So since it's got no charge, no spatial extent, it's got nothing but mass, nothing's going to couple to it. It's not going to basically do anything but freefall. In this frame the body moves on a purely inertial trajectory. And what does a purely inertial trajectory look like? Well you take whatever your initial conditions are. And you move in a straight line with respect to time as measured on your own clock. Simplest, stupidest possible motion that you can. So we would obviously call that a straight line with respect to the parameterization that's being used in this representation. So what does that mean in a more general sense of the representation? So if we think about this a little bit more geometrically, when a body is moving in a straight line, that basically means that whatever the tangent vector to its world line is, it's essentially moving such that the tangent vector at time T1 is parallel to the tangent vector at T1 plus delta T1, provided that's actually small enough that they're sort of within the same local Lorentz frame. So a more geometric way of thinking about this motion is that it's parallel transporting its tangent vector. Let's make this a little bit more rigorous. So let's imagine this body's moving on a particular trajectory through spacetime. So it's a trajectory parameterized. I will define its parameterization a little bit more carefully very soon. So for now, just think of lambda as some kind of a quantity. It's a scale that just accumulates uniformly as it moves along the world line. So I'm going to say the small body has a path through spacetime, given by u x of lambda. Its tangent is given by this. And if it is parallel transporting its own tangent vector, that is-- I'll remind you that the condition for parallel transport was that you take the covariant derivative your field. And as you are moving along, you contract it along the tangent vector of the trajectory you're moving on. And you get 0. So in my notes, there's a couple of equivalent ways of writing this. So you will sometimes see this written as the gradient along u of u. And you'll sometimes see this written as capital u u lambda equals 0. So these are just-- I just throw that out because these are different forms that are common in the notation that you will see. So let's expand this guy out. It's something like this. So what we're going to do-- so remember this is dx d lambda. This is d by dx. That's a total derivative with respect to the parameter lambda. So this becomes-- I'm going to write it in two forms. This is often written expanding out the u into a second order form. This is obvious but sufficiently important. It's worth calling it out. And this has earned itself a box. This result is known as the geodesic equation. The trajectories which solve these equations are known as geodesics. One of the reasons why I highlight this is it's-- I'm trying to keep a straight face with the comment I want to make. A tremendous amount of research in general relativity is based around doing solutions of this equation for various spacetimes that go in to make the Christoffel symbols. My career-- [LAUGHS] it's probably not false to say that about 65% of my papers have this equation at its centerpiece at some point with the thing that goes into making my gammas-- things related to black hole spacetimes. This is really important because this gives me the motion of a freely-falling frame. What does a freely-falling frame describe? Somebody who's moving under gravity. So when you're doing things like describing orbits, for example, this is your tool. A tremendous number of applications where if what you care about is the motion of a body due to relativistic gravity, this gives you a leading solution. Now bear in mind when I did this, this is the motion of a test body. This is an object with no charge, no spatial extent, no spin-- that describes no object. So it should be borne in mind that this is the leading solution to things. Suppose the body is charged. And there is an electromagnetic field that this body is interacting with. Then what you do is you are no longer going to be parallel transporting this tangent factor. It will be pushed away-- we like to say-- from the parallel transport. And you'll replace the 0 on the right hand side here with a properly-constructed force that describes the interactions of those charges with the fields. Suppose the body has some size. Well then what ends up happening is that the body actually doesn't just couple to a single-- remember what's going on here is that in the freely falling frame, I'm imagining that spacetime is flat at some point. And in a decent enough vicinity of that point, the first order corrections are 0. But there might be second order corrections. Well imagine a body is so big that it fills that freely-falling frame. And it actually tastes those second order corrections. Then what's going to happen is you're going to get additional terms on this equation, which have to do with the coupling of the spatial extent of that body to the curvature of the spacetime. That is where-- so for people who study astrophysical systems involving binaries, when you have spinning bodies, that ends up actually-- you cannot describe a body that's spinning without it having some spatial extent. And you find terms here that involve coupling of those spins to the curvature of the spacetime. So this is the leading piece of the motion of a body moving in the current spacetime. And it's enough to do a tremendous amount. Basically because gravity is just so bloody strong that all of these various things-- it's the weakest fundamental force. But it adds up because it's only got one sine. And when you're dealing with some of these things, it really ends up being the coupling to the monopole-- the most important thing. So all these other terms that come in and correct this are small enough that we can add them in. And that, to be blunt, is modern research. So let me make a couple of comments about this. A more general form-- this will help to clarify what the meaning of that lambda actually is. Suppose that as my vector is transported along itself-- so one way is recall how we derive parallel transport. We imagine going into a freely-falling frame and a Lorentz representation. And we said, in that frame, I'm going to imagine moving this thing along, holding all the components constants-- that defined parallel transport. Imagine that I don't keep the components constant, but I hold them all in a constant ratio with respect to each other, but I allow the overall magnitude to expand or contract. So suppose we allow the vector's normalization to change as it slides along. Well the way I would mathematically formulate this is I'm going to use a notation that looks like this. So recall this capital D-- it's a shorthand for this combination of the tangent and the covariant derivative. I'm going to call the parameterization I use when I set up like this lambda star, for reasons that I hope will be clear in just about two minutes. So what I'm basically saying is that as I move along, I don't keep the components constant. But I keep them proportional to where they were on the previous step. But I allow their magnitude to change by some function, which I'll call a kappa. So you might look at that and think, you know, that's a more general kind of transport law. It seems to describe physically a very similar situation here. It's kind of annoying that this normalization is changing. Is there anything going on with this? Well what you guys are going to do as a homework exercise, you're going to prove that if this is the situation you're in, you've chosen a dumb parameterization. And you can actually convert this to the normal geodesic parameterization by just relabeling your lambda. So we can always reparameterize this, such that the right-hand side is 0. And right-hand side being 0 corresponds to the transport vector remaining constant as it moves along. So I'll just quickly sketch-- so imagine there exists some different parameterization, which I will call lambda. So imagine something that gives me my normal parallel transport exists. And I have a different one that involves the star parameter. You can actually show that these two things describe exactly the same motion, but with lambda and the dumb parameterization, lambda star, related to each other by a particular integral. So what this shows us is we can always-- as long as I'm talking about motion where I'm in this regime-- where there's no forces acting-- it's not an extended body-- it's just a test body-- I can always put it into a regime where it'll [INAUDIBLE] geodesic and the right-hand side is equal to 0. If I'm finding that's not the case, I need to adjust my parameterization. When you are, in fact, in a prioritization such as the right-hand side is 0, you are using what is called an affine parameterization. That's a name that's worth knowing about. So your intuition is that the affine parameterization-- I described this in words last time. And this just helps to make it a little bit more mathematically precise what those words mean. Affine parameters correspond to the tick marks on the world line, being uniformly spaced in the local Lorentz frame. If you are working with time-like trajectories-- which if you're a physicist, you will be much of the time-- a really good choice of the affine parameter is the proper time of a body moving through the spacetime. That is something that is uniformly spaced, assuming that's-- you don't have to assume anything. Just by definition it's the thing that uniformly measures the time as experienced by that observer. So this is-- you guys are going to do on PSET 4-- this is the exercise you need to do to convert a nonaffine parameterized geodesic to an affine parameterized one. That kind of parameterization is not too hard to show that if we adjust the parameterization in a linear fashion-- so in other words, let's say I go from lambda to some lambda prime, which is equal to a lambda plus b, where and b are both constants-- we get a new affine parameterization. But that's the only class of reparamterizations that allows me to do that. And hopefully that makes sense. If you imagine that you're using proper time as your reparameterization, this is basically saying that you just chose a different origin for when you started your clock. And this means you changed the units in which you are measuring time. That's all. So I'm going to skip a bunch of the details. But I'm going to scan and put up the notes corresponding to one other route to getting to the geodesic equation, which I think it's definitely worth knowing about. It connects very nicely to other work in classical mechanics. So it's a bit of a shame we're going to need to skip over it. But we're a little bit behind pace. And this is straightforward enough that I feel OK posting the notes that you can read it. So there is a second path to geodesics. So recall the way that we argued how to get the geodesic equation, which we said we're going to go into-- it's actually in the board right above where I'm writing right now-- go into the freely-falling frame. I have a body that isn't coupling to anything but gravity. Therefore in the freely-falling frame, it just maintains its momentum. It's going to go in a straight line. Straight means parallel transporting tangent vector-- math, math, math-- and that's how we get all that. So what this boiled down to is I was trying to make rigorous in a geometric sense what straight meant. There's another notion of straight that one can imagine applying when you're working in a curved space. So your intuition for-- if you're talking about how do I make a straight line between two points on a globe-- your intuition is you say, oh, well the straightest line that I can make is the path that is shortest. We're going to formulate-- and I'll leave the details and the calculation to the notes-- we're going to formulate how one can apply a similar thing to the notion of geodesics. So imagine I've got an event p here and event q up here. And I ask myself, what is the accumulated proper time experienced by all possible paths that take me from event p to event q? I'm going to need to restrict myself. I want it to be something that an observer can physically ride-- so all the time-like trajectories that connect event p to event q. So I've got one a path that goes like this, got a path that goes like this, path that goes like this, path goes like this, path goes like-- some of them might have just become somewhat space like, so I should rule them out. But you get the idea. Imagine I take all the possible time-like paths that connect p and q. Some of those paths will involve strong accelerations. So they will not be the freefall path. Among them there will be one that corresponds exactly to freefall. So if I were talking about-- imagine I was trying to-- and this is something that Muslim astronomers worked out long, long ago-- they wanted to know the shortest path from some point on earth towards Mecca. And so you need to find what the shortest distance was for something like that. And when you're doing this on the surface of a sphere, that's complicated. And that's where the qibla arose from, was working out the mathematics to know how to do this. This is a similar kind of concept. I'm trying to define-- in this case, it's going to turn out it's not the shortest path, but it's the path on which an observer ages the most because as soon as you accelerate someone-- it's not hard. Go back to some of those problem sets you guys did where you look at accelerated observers. Acceleration tends to decrease the amount of aging you have as you move through some interval of spacetime. So the path that has no acceleration on it, this is going to be the one on which an observer is maximally aged. Why maximum instead of a minimum? Well it comes down to the bloody minus sign that enters into the timepiece of an interval that we have in relativity. And that's all I'll say about that, is just boils down to that. So what we want to do is say, well along all of these trajectories, the amount of proper time that's accumulated-- so let's just say that every one of these is parameterized by some lambda that describes the motion along these things. This is the amount of proper time that someone accumulates as they move from point p-- which is at-- let's say this is defined as lambda equals 0-- and it's indeterminate what that top lambda is actually going to be. It's whatever it takes when you get up to lambda of q. So what the notes I'm going to post do, is they define an action principle that can be applied to understand what the trajectory is that allows you to do this. So I'll just hit the highlights. So in notes to be posted, I show that this delta t-- this delta tau rather-- this can be used to define an action. It looks like this. And then if you vary the action-- or rather you do a variation of your trajectory-- where you require that the action remain stationary under that variation-- in other words I require delta i equals 0 as x goes over to such-- so-- what you wind up with-- is delta i equals-- Notice what I've got in here. This is just a Christoffel symbol. So when I do this variation, what I find-- and by the way going from essentially that board to that board, it's about 2/3 a page of algebra. Going down to this one, there's a bunch of straightforward but fairly tedious stuff. It's one reasons why I'm skipping over the details. We've got enough G mu nus on the board. So the key point is I am going to require that my action be stationary, independent of the nature of the variation that I make. For that to be true, the quantity in braces here must be equal to 0. Let me just write that down over here. This is a good place to conclude today's lecture. So we require this to be 0 for any variation. Yet the bracketed term being equal to 0, pull that out, and clear out that factor of the metric with an inverse, you've got your geodesic equation back. So we just quickly wrap this up. So it's worth looking over these notes. It's not worth going through them in gory detail on the board, which is why I'm skipping a few pages of these things. But what this demonstrates is that geodesics-- our original definition is that they carry the notion of a straight line in a straightforward way from where they are obvious in a locally Lorentz frame to a more covariant formulation of that-- so a generalized straight line to a curved spacetime. And they give the trajectory of extremal aging in other words a trajectory along which between two points in spacetime, an observer moving from one to the other will accumulate the most proper time. So I'm going to stop here. There's a bit more, which I would like to do, but I just don't have the time. But I'll tell you the key things that I want to say next. Everything that I've done here so far is I've really fixated on time-like trajectories. I've imagined there's a body with some finite rest mass where I can make a sensible notion of proper time. We are also going to want to talk about the behavior of light. Light moves on null trajectories. I cannot sensibly define proper time on long such a trajectory. They are massless. There's all sorts of properties associated with them that just make this analysis. The way I've done it so far, I'll need to tweak things a little bit in order for it to work. We will do that tweaking. It's actually quite straightforward and allows us to also bring in a bit more intuition about what affine parameters mean when we do that. So that'll be the one thing we do. The other-- it's unfortunate I wasn't able to get to it today-- but it's a straightforward saying, which I think I may include in the notes that I post-- is including what happens, if your spacetime-- so if the metric you use to generate these Christoffels has a Killing factor associated with it, you can combine Killing's equation with the geodesic equation to prove the existence of conserved quantities associated with that motion. And that's where we start to begin to see that if I have a spacetime that is independent of time, there's a notion of conserved energy associated with it. So we will do that on Tuesday. |
MIT_8962_General_Relativity_Spring_2020 | 4_Volumes_and_volume_elements_conservation_laws.txt | [SQUEAKING] [RUSTLING] [CLICKING] PROFESSOR: OK. Great. So at this point, we basically have all of the most important-- well, we now have the full understanding of how to work with tensors. I haven't really done too much physics with them at this point, but we've very carefully-- one might even argue excessively carefully-- laid out this mathematical structure that is going to be-- we're going to use it to contain the geometric objects that will describe the physics that we're going to study over the course of the entire semester. So there was a lot of twiddly detail in the preceding lecture, but I would say the two most important things I want you to take out of this is the idea that we now have tensors as a general class of geometric objects which map one-forms-- or dual vectors, if you prefer-- a combination of one-forms and vectors to the Lorentz invariant real numbers. And further, the distinction between one-forms, and vectors, and what is being mapped to what is not so important, because I can always use the metric to either raise or lower indices on tensors or vectors or one forms in order to convert from one to another. So if I raise an index-- if I have a tensor that's got m indices in the upstairs position and n in the downstairs, I raise an index, and I've got m plus 1 upstairs, n minus 1 downstairs. And likewise, I can lower index-- put one of those ones in the upstairs, down to the downstairs. We wanted to go through all that stuff with great care, because we're going to need the foundations of that to make a lot of what we talk about later rigorous. It is really overkill for where we're starting, but it's worthwhile to have that scaffolding in place. I'm sick of it, though, so today I'm going to try to do some things that are more physics, because, well, not all of you are necessarily physics students, but I'm a physics professor and I'm a little tired of doing math. So let's think about-- when we're studying physics, generally we are interested in the behavior of bodies and fields. Let's move this to the relativity framework. We're looking at the behavior of bodies and fields in spacetime. So far, if I think about the quantities that we have introduced-- and by the way, I apologize for the blackboards. A previous lecturer today used chalk that cannot be erased very well, so they're sort of grayish boards today, unfortunately. But we'll manage. Anyhow, so far, the quantities that we have introduced that have solid physics content are good for describing particles-- individual particles-- maybe a handful of them at once. So the two that I want to focus on in particular are the 4-velocity-- so if I have a body with a 4-velocity u, a particular Lorentz observer will describe that as having a timelike component gamma-- special relativistic gamma-- and a spacelike component gamma v. This is defined such that u dot u equals minus 1. It is great for describing-- it is the tool we use to describe the trajectory of a material body that follows a timelike trajectory through spacetime. Notice that its norm is minus 1. The fact that it's 1 means that it's normalized, and the minus tells us that it is timelike. By the way, because this is timelike, the 4-velocity is not going to be a useful tool for us when we want to describe the behavior of light. If we want to talk about the motion of a photon in our spacetime, we can't use a 4-velocity for it. The trajectory of a photon is null. Whatever quantity that is going to describe its trajectory-- I take its dot product with itself, I have to get 0. And you can intuitively get a sense what's going on with that. If I take the way a particular observer interprets these things and says, what would this turn into if v equals c, well, you have infinity and infinity, because your gammas are diverging there. So it's a singular limit. It doesn't behave well. We will overcome this difficulty. The other quantity that we have used-- let me see if I can clean this a little bit better. All right. The other quantity that we have defined which is good for particles is the 4-momentum, which is just that 4-velocity multiplied by a rest mass. So this is defined-- you can clearly, by trimming together a bunch of definitions-- you have p dot p equals minus m squared. We're actually going to interpret this, though. So if a particular Lorentz observer makes a measurement of this, they will call the timelike component E and the spatial component p. And so you put all these things together, and this tells me that E squared minus magnitude of p squared equals m squared. Notice this is perfectly well-behaved for a massless particle. This is actually going to be the trick by which, when I want to describe the trajectory of photons or of light in a spacetime-- I can't use 4-velocity. I can use 4-momentum. It works perfectly well. It's 0 times a 4-velocity, but if I use this intuition that this thing is diverging when I take the speed of light limit, I sort of-- I guess I cheat, but I'm basically getting 0 times infinity. And it behaves. We're going to do it a little bit more rigorously than that, but if you want a little bit of intuition, that's essentially why it works. So the last thing which I will say here is that this is good for m equals 0. OK. And when you do have m equals 0, it's often convenient to write this as h bar omega. So use the usual formula for the energy of a photon-- let me move this so that you can see-- and then I just put a unit vector into the spatial direction that defines the direction in which this is moving. OK. So this is all fine as far as it goes, but it's kind of restrictive. We want to a little bit more with our physics than just worry about the kinetics of individual particles. There's a lot we can do with that, but one of the things that we like to do when we're studying gravity is to have a description of continual matter, like things that gravitate. You want to be able to build a description of something like a star, or an exotic star-- a neutron star-- something like that. And so just having the behavior of individual particles is not good enough. I want to be able to describe things like fluids. I want to be able to have a continuum that describes things. So what I'm going to start doing today is begin to introduce the mathematical tools that we will use. Pardon me. I seem to have twisted this around myself in a crazy way. We're going to describe the mathematical tools that are useful for dealing with continuous matter, rather than just particles. So let's call this more interesting matter. So the simplest continuum form of matter which we're going to talk about is-- we're going to call it dust. OK. When you're here at the chalkboard, you can't get away with it, so you have an idea of what it means. Physically, what I mean by dust is that it is a collection of particles that do not interact with each other. So they pass through each other. There's no pressure that is generated. So they have energy density associated with them, but no interaction. So it's the most boring kind of matter that you like. You can think of it as essentially just particles, but it's a ton of them, and they're smearing out into a continuum. In particular, one thing I'm going to say is, so I can imagine, if I take my two erasers here, I have a nice dust field in front of me now, which I helpfully just inhaled. And you can think of this thing as a field of little dust elements that are all moving around. If I go in and I track an individual element of that dust-- I go in and I make a little cube that's a nanometer on each side-- every little element has its own rest frame. In that cloud-- I'm not going to do that again-- but as I made that cloud, different elements within the cloud have different rest frames, but for each individual element, I can define a rest frame. So we're going to use the properties of the dust in that rest frame as a way of beginning to normalize and get a grasp on how we're going to describe this. But I just want to emphasize again-- in a given cloud, different elements may have different rest frames. OK. But let's suppose that I clap my erasers, I make my little cloud, and I go and I zoom in on one particular nanometer by nanometer by nanometer chunk of that thing, and I go into the rest frame of that particular little bit of dust. So how am I going to characterize that? So presumably, each particle has its own mass associated with it. Each little bit of dust that's in that thing has its own mass. We're going to worry about how that comes into the picture a little bit later. Let's just begin by saying, one of the things it's going to be interested in knowing is, how much dust is there in that little cubic nanometer of dust in front of me there? So the first thing we're going to focus on is just counting how many bits of dust are in this element. In particular, I'm going to want to know, how many elements per unit volume? In other words, the number density. So I claim that's a good thing for us to get started on. And it looks like they haven't quite poisoned this board as badly. Good. So let's call n sub 0 the number density in the rest frame of that element. OK. So this will just be some number per unit volume. OK. So it's a quantity that has dimensions 1 over volume. Now, in the rest frame, like I said, we're going to talk about how I treat the mass associated with each element, and how much energy is in that thing, a little bit later. There's a few other tools I want to introduce first before I get to there. But in the rest frame, if I'm not worrying about the mass and the energy, this is it. This is really the only thing I can say about this. So what do I'm going to do now is say, OK, well, if I have a general cloud here-- as I described, when we made this little turbulent cloud of dust in front of us, it was all swirling around and doing its own thing. You're generally not in the rest frame. OK. So you also want to know how to characterize the dust out of the rest frame. So let's move out of the rest frame. When I do this, two things happen. So first of all, let's continue to focus on-- so we've got a particular nanometer cubed of dust that we are looking at here. So we're very attached to that particular thing. So I now boost into a frame where I'm going 3/4 of the speed of light or something like that, but I'm going to pay attention to that-- I'm very attached to the dust in that little cube, so I'm going to keep myself focused on that. Boosting into frame can't change the total number, so the amount of dust has to be the same. But of course, there's going to be a Lorentz contraction, and so the volume will get smaller. So the number in a particular volume stays the same while the volume Lorentz contracts. And so let's call n the number density in this new frame. This is just going to be the Lorentz gamma times n0. So again, I'm assuming you're all perfectly fluent with special relativity. So all I did there was say it's going to contract along that direction. The lengths are smaller by a factor of gamma, therefore the volume is larger by a factor of gamma. But there's a second thing that happens as well. When I'm in this new frame, there is now a flow associated with the dust. So this will now be flowing through space. In particular, what you can do is define a flux that describes the number of particles crossing unit area in unit time. So let's define n 3-vector to be this. And you use a little bit of intuition, and just think about the flow of these particles as they kind of go translating past you. This can only be the number density that you measure in that frame times the 3-velocity. OK? Think about that a little bit. It's very much like when you learn about current density in basic E&M. Notice it's n and not n0 here. So when you look at these two things for a moment or two-- let me just write down one more step. I can write this as gamma n0 v. These are screaming to be put together to make a 4-vector. They're looking at you and saying-- this bit's kind of like, I'm a timelike piece of a 4-vector. Spatial piece, I love you. Let's get together. They clearly are just screaming to go together. And when you do that, you say, hmm, let's see what happens here. Two great tastes that taste great together. OK. That's nice. Let's take advantage of the fact that both of these have a really simple form in terms of the density in the dust's own rest frame. And you look at this and go, holy moly. That is nothing more than the number density times a 4-velocity. So that's pretty cool. If I have this stuff in front of me-- you know, again, what the heck. This time I'm going to actually do my clap again. So I do this. Every little element-- I can think of it as having some trajectory through spacetime. I can attach a 4-velocity to that. And if I know what the density of dust is in that rest frame, I put it together-- now I've got a geometric object that describes this thing. And one of the reasons why this is powerful is, geometric objects-- as I have emphasized repeatedly-- have a geometric meaning that transcends a particular coordinate representation. Another way to think about this is that all observers-- I don't care whether you're at rest in this classroom, or you've had too much coffee and you're dashing through at 3/4 the speed of light. We all agree on N-- capital N vector-- but we choose different ways of splitting spacetime into space and time. And so I will have a different n from you, I will have a different nv from you. We all have different ways of splitting this spacetime into space and time, and so we have different ways of splitting these two things up. But this is a geometric object we all agree on. It's something that we can hang a lot of our physics on, and build frame-independent powerful geometric representations. So before going on to do a few things with this, let's just explore a couple of the properties of this. All right. So whenever you've got a 4-vector, at a certain point, you just sort of say to yourself, well, I've got this thing. Let's take the dot product with itself. Why not? So in this case, N dot N equals minus n0 squared. Or, turning this around, this tells you that if someone comes buy and goes, psst, here's a number 4-vector, you can figure out density in the rest frame associated with each element by the following operations. So that's nice. OK. In other words, the number density, in a sense, tells you what the normalization of this vector is. A few moments ago, I defined N as being kind of a flux of these things. It's a flux in the spatial directions. We want to think about flux in a more general kind of way. And so another thing which we're going to do-- and this is where 1-forms are going to play an important role for us-- is a nice systematic way to pick out a flux across a surface. So here's the bit where, when I was talking about it in Tuesday's lecture, I was getting the blank stares. And sorry, I'm coming right back to this again, so we're going to harp on this concept a little bit more. So recall we talked about the fact that if I have a particular coordinate system-- I have a particular set of coordinates x alpha that I use to represent spacetime. If I make 1-forms that are the gradients of those coordinates, I can represent them as essentially, level surfaces at unit ticks of the coordinate x alpha. OK. And basically, the way to think of it-- I kept emphasizing that the basis 1-forms are dual to basis vectors. And if you have this intuition that your basis vector is a little arrow pointing like so, then the thing that is dual to it is a surface that is everywhere but pointing along like so. That's kind of the way I want you to think about it. And those graphics that are scans from MTW I put on the web give better pictures of that than I am able to draw. So putting these concepts together, it basically tells me that I can use these basis 1-forms as a way of defining in an abstract form. So if I want to know the flux of N in the-- let's write it this way-- in the x alpha direction. So remember, the basis 1-forms-- the alpha in that case is not like a coordinate index. It's labeling a particular member of a set. And so what you would do is, you would construct this by saying, OK, let's take the beta component of a basis 1-form alpha. Contract it like so. And that tells me about the flux-- how much of this stuff is flowing in the direction associated with alpha. So this is a tool that we're going to use from time to time. One little thing which I want to emphasize-- actually, two things I want to emphasize. So if I'm working in an intelligent coordinate system, remember, this is basically just the identity matrix, right? And so this ends up being a very simple thing that I actually work out. And it says, if I want to know the timelike component, then this is 1 in the timelike direction, 0 everywhere else, and I just pick out N sub t-- in other words, the density itself in that frame. That is the other thing which I want to emphasize. So when you do this, if I pick out the timelike piece of this, this just gives me the 0 component of this thing, which is the density that I measure in this frame. This is very, very simple in terms of the calculation, but I want you to stop and think about what it means physically. So flux in the timelike direction-- what does it mean for a flux to be in the timelike direction, right? If i take my water-- actually, I'm kind of thirsty-- so I want a little bit of flux of water going down my throat. You have an intuition about what that means. The water's actually flowing in a particular direction. If it's flowing through time-- there it goes. That basically means it's just sitting there. It's not doing anything, but just moving in time, as we all are doing. One time I remember my wife saying about our daughter-- she's like, she's growing up so fast. I was like, eh, she's growing up at 1 second per second. And that's just the way things go. It's just sitting there and it's living its life at 1 second per second, as we all do. That is density. So we described the flux of a thing in time as just being its density. That's another way-- when we do a lot of our calculations, and we talk about the flow of things in the timelike direction-- that tends to be just the simple density associated with stuff. And then if I did this-- if I pulled out the x direction of this thing, I would pull out the x component of velocity times that. And that's sort of the flux of something in the x direction like you are probably used to from other classes. Now, a place where this turns out-- so doing it when I'm just using the basis 1-forms is kind of trivial, [INAUDIBLE] sucks. More generally, what you can do is define a surface as being the solution of some scalar function in spacetime. So let's say I do something like psi of t comma x comma y equals a constant. OK. This is very abstract, so let me just make it a little bit more concrete. Suppose my function psi were square root of x squared plus y squared plus z squared, and my constant was 5. OK. Well then, my scalar field would be picking out a sphere of radius 5. OK. You can make a little bit more complicated than that, and people often do. You can define the unit 1-form that is associated with the normal to this thing. Or rather, it's the 1-form associated with the surface. If you translate this to a vector-- you raise the index-- with the vector, it would be the vector normal to the surface. You might need to normalize it. Let's imagine that we normalize this thing. And then we would just need to contract it along this, and that tells me about the flux through this particular surface. OK? This is one of the things that-- this idea of 1-forms being level surfaces-- it tends to be useful for things like that. We're going to come back to a similar sort of picture in just a moment, because we'll have to start talking about integration. OK. In prep for that discussion, now that I'm talking about things a little bit more complicated than just the kinematics of simple particles, we're going to want to have some laws. And we want to have geometric ways of describing those laws that essentially act as constraints on what those particles can do. OK. So the form in which we're going to express them-- we're going to tend to put things into the form of conservation laws. So suppose, here's my little element that's got dust in it, and it's embedded in an environment with a bunch of dust around it. And over some time interval, some dust flows in, some does flows out. The density can go up, the density can go down. The total number-- it may vary depending on how things are flying. The spatial flux out of the sides must come-- and I'm going to say it's the flux out-- so let's say that that comes at the expense of the density of dust already there. OK. And so if you were to just to intuitively write down what kind of conservation law you would expect to see, you would write it as something like this, based on simple Euclidean intuition. What's kind of nice-- you look at that for a second. You go, ooh, if I think about this as being the time complement of my 4-vector. This is the space component of my 4-vector. This has a very obvious form when I write it in a geometric framework. This whole thing can be rewritten as a conservation law that looks like this. OK. So I'll remind you, d sub alpha equals d by dx alpha. I'm going to talk a little bit about some of the derivatives a little bit later, because there are a few subtle points that can get introduced. But for now, we just know that d downstairs t is d by dt. d downstairs x is d by dx. d by downstairs y, et cetera. So this is really nice. One thing which I want to emphasize-- again, coming back to what I said over there just a moment ago. When I write down this conservation law, I'm assuming that someone has defined what time means and someone has defined what space means. This is a form that's covariant, right? All observers agree that this goes together. When they actually make their own coordinate systems, they're going to have their own time coordinate and their own x-coordinate, but they're all going to be different flavors-- just different ways of instantiating what this is. Now, pardon me for just one moment. Depending on the pace of the course, we're about to switch over to a different set of lecture notes, and I want to make sure I smoothly go from one to the other. OK. So in many of your physics classes, you have learned, when you get a conservation law, both a differential form like this-- the rate of change of N in a box is related to the amount of N flowing out of that box and its sides. And you also learn an integral form of the conservation law. So without proof, let me just say that it should be intuitively clear that what I've written down over here is equivalent to-- OK, so it looks like this. You know what? Let me just fix up my notation a little bit. Let me call this lowercase n. OK. A few symbols that I've introduced here-- so V3 is some volume in three-dimensional space, and dV like this-- this is a symbol that means the boundary of that 3-volume. And that's basically-- it's a form of Gauss's law. That's what I've written down there. OK? So you've all seen things like that. Again, let me emphasize that when I write down a formula like that, I can only do that having assumed a particular Lorentz frame. That t is the t of some observer. That volume is the volume of that particular observer who is using t. You jump into a different Lorentz frame, their volumes will not be the same. Their times will not be the same. I'm going to make some coordinates up. So an integral form like this, as I've written it there, only works in one given Lorentz frame. Nonetheless, we are going to find it useful, even though in some sense, when you do it in an integral form like this, you're saying things in the framework of some particular observer. Sometimes you want to know in the framework of some particular observer. It could be you, right? And you might care about these sorts of things. And that's good, but the way I've written it here-- first of all, it's in language that-- it's in a mathematical formulation that's not easy to generalize as I take things up to higher dimensions. And so what I would like to do is think about how to step up a formulation like this in such a way that things are put into as frame-independent a language as is possible, and that will generalize forward when we start looking at more complicated geometries than just geometry of special relativity. So I want to spend the next roughly 10 or 15 minutes talking about volumes and volume integrals. And my goal here is to try to-- I'm going to start by just doing stuff that comes from the journal of duh. It's stuff you have seen over and over and over again, but I want to re-express it using mathematical formulation that maybe-- you have seen all the symbols, but perhaps not used in quite this way. And then it'll carry forward in a framework that generalizes in a very useful way for us. So let's begin with just simple 3D space. So I'm going to begin in 3D, and I'm going to consider a parallelepiped-- parallelepiped. Ha-ha. Got it right-- whose sides are a set of vectors. So there are three vectors, A, B, and C. OK. So here is a vector A. This one going into the board is vector B. And this one going up here is vector C. OK. So those are my three vectors. And if I go draw the ghost legs associated with these things-- OK. That's a little bit better. So these three vectors define a particular volume. And you guys have probably all seen-- you know you have three vectors. You can define a volume associated with this. A really easy way to get that volume given those three vectors is to take A and dot it into the cross product of B and C. This is a quantity which is cyclic, so if you prefer, you can write it as B dotted into C cross A, or C dotted into A cross B. That can be expressed as a determinant, or-- you guys can look at my notes to see the determinant written out if you like, but it's not that interesting, so I'm not going to use it very much. An equivalent way of writing all that is to use the Levi-Civita symbol. So that 3-volume is given by epsilon ijk Ai Bj Ck, where-- I'll remind you-- epsilon ijk equals plus 1 if i equals 1, j equals 2, k equals 3, and even permutations. Even permutations means I swap two pairs of indices. So 123, 231, 312-- those all give me plus 1. It gives me minus 1 for any odd permutations of those. So 132, 231, et cetera. Those will all give me minus 1. And it's 0 if any index is repeated. So you probably have all seen things like this before. This is fairly basic vector geometry. We are going to regard the Levi-Civita symbol as the components of a 0, 3 tensor. OK. Bear in mind for just a moment here I'm working only in-- sorry, just ran out of good chalk-- I'm only working in 3-space, so my tensor definition is slightly different. It's not going to be a set of things that maps to Lorentz invariants, but it's going to be invariant with respect to things like rotations and translations in three-dimensional space. So I'm going to regard these as the components of a 0, 3 tensor that basically takes in vectors and spits out the volume associated with the element whose edges are bounded by those vectors. OK. So I could say, in this abstract form I wrote down earlier, imagine a boldfaced epsilon which is my volume tensor. I put these slots into it, and voila. I get the volume out of it. Now, with that in mind, remember some of the games that we played with tensors in the previous lecture. So when I was talking about spacetime tensors, if I filled up all of their slots, I got out a Lorentz invariant number. In this case, I'm in 3-space, so I fill up all its slots, I get an invariant number in this 3-space. Suppose I only put in two vectors. So suppose I do something like, I plug in-- let's leave the first slot blank-- and I put in vectors B and C. OK, well, writing this out in component form, I know this is epsilon ijk Bj Ck. That's just B cross C, right? That's the area spanned by the side that is B cross C. And you guys have learned in other classes that you have an extra index left over, so it's a vector that has a direction associated to it. So it's sort of an oriented surface. We put the index in the downstairs position, so we're actually going to think about this as a 1-form corresponding to the side whose edges are B and C. So let's call this side 1-form sigma. This is a 1-form whose magnitude is the area of the side spanned by the vectors B and C. And although I can still tell it hasn't quite gelled yet, it's useful to think of 1-forms as being associated with surfaces. Guess what? This is the side of a parallelepiped. That's a surface. So it actually holds together. All right. So using all of this, if I wanted to write down how to do something like Gauss's theorem in this kind of geometric language-- and again, we emphasize this is very much in the spirit right now of mosquito with a sledgehammer. We don't need all this sort of stuff, but we're about to step up to something a little bit more complicated. So what you would do is say, OK, well, I know Gauss's theorem. I pick a particular 3-volume. I say the divergence of some vector field integrated over that volume is given by integrating the flux of that vector over the surface of this thing. So what you might want to do then at this point is say, oh, OK, well, what I'm going to do, then, is say that my volume-- oops, pardon me a second. First thing I'll do is define a differential triple. I'll define some x1 that points along one direction I care about, and x2, and an x3. And then I will say dV equals epsilon ijk dx1i dx2j dx3k. I can likewise define a 1-form associated with my area element, as I have done over here. I'm not going to actually write this out. The key thing which I want to say is you have all the pieces-- you put all these things together, and you can define this thing. It's now very easy for you to prove Gauss's theorem using this kind of ingredients. What I want to move onto-- there's a few more details in my notes. It's not super difficult or interesting to go through this. What I want to now start doing is generalize all of these ideas to the way we're going to approach them in spacetime. Basically, we're going to do exactly the same kind of operations that I just did in space-- three-dimensional space-- but I'm going to put an extra index on things, and I'm going to do all of my quantities in spacetime. OK. So imagine a parallelepiped with sides A, B, C, and D. Four dimensions, so it's going to point along four different-- these can be mutually orthogonal. I'm going to define the invariant 4-volume associated with these things like so. Where now my four-index Levi-Civita is defined such that epsilon 0123 equals plus 1. If I do an odd permutation of those-- I exchange one pair of them-- epsilon 1023-- this equals minus 1. If I repeat any index, I get 0. And likewise, all even permutations of this give me plus 1. All odd permutations of this give me minus 1. Or likewise, just do even permutations of this one. So that is how I'm going to generalize my Levi-Civita symbol. As they say on The Simpsons, it's a perfectly cromulent object. I'm going to need to talk about the area associated with the faces of each of these things. So what is the area of the face of a 4-volume? A 3-volume. So you can define a 1-form that tells me about the-- you can either call it the 3-volume or the 4-area. Knock yourselves out as to how you want to call it. And the obvious generalization-- let's say I leave off edge A. I want to get something like this. I do my similar exercise of defining a-- in this case, it'll be a differential quartet associated with directions 0, 1, 2, and 3. And so by going through a procedure very similar to this, you get a generalization of Gauss's theorem. It says that if I integrate the spacetime divergence of some 4-vector over a four-dimensional volume, it looks like what I get when I sum up the flux of that guy over all of the little faces. So I'm not going to step through the proof of that. It's fairly elementary, and basically it's just like proofs of Gauss's law that you have seen elsewhere, but there's an extra dimension attached to it. Really nothing new that's going on here. The thing which is new is, this is now being done in an additional dimension, and where this tends to be useful is when there is some kind of a conservation law that tells you something about this left-hand side here. So the whole starting point of this discussion was I wrote down, on intuitive grounds, that the rate of change of the total amount of-- so what I did was I had an integral of dust density over a 3-volume, and I said, d by dt of that was balanced by the flux through the surfaces on the edge of that 3-volume. As I argued, that's an observer-dependent statement, because you have chosen a particular time. You have chosen a particular space to make those volumes. This, on the other hand-- so let's switch my general vector field V to be N that we started this discussion off with. So this guy-- oops. Let me write it the way I wrote it over there. d4x over some 4-volume. This must be 0, because we are going to require that this thing be-- so if my number density is conserved, this must be 0. So this tells me that when I do this flux integral, it's going to have to be 0. Let me now break this integral up, and actually write this into a form that is a little closer to the way that you may have seen something like this before. So what I want to do is actually zoom in and think about the four-dimensional volume that I'm doing this integral over. So I'm just going to do a two-dimensional cut of it here on the blackboard. Let's let the time axis go up, and let's define the edges of my volumes to be t1 and t2. x-axis going across here. Boundaries are x1 and x2. OK. So here is my V4. And every face here is an example of my boundary of d4. OK. Of course there's one over here as well, and over here. So if I were to do the top line here, I know that's got to get me 0, so I'm going to take advantage of this and say, let's just look at what happens when I do the integral of the flux of this thing across the many different faces. So in a four-dimensional parallelepiped-- an n-dimensional parallelepiped has 2n phases, so there will be eight integrals we need to do. They're pretty obvious though, so I'm just going to write down a handful of them. OK. So over all those faces-- N alpha sigma alpha. So what I'm going to do now is say, OK, let's evaluate this on the face that is at t equals t2. So when I do this, I'm going to get N0 dx dy dz. Let's do next the contribution at moment-- actually, let me move this over, because I want a little bit more room. I also do one on the slice t equals t1. When I do this, though, in the same way that when you guys do fluxes in three-dimensional space, you get a sign associated with the orientation of these things, because the Levi-Civita symbol here has sort of a right-hand rule built into it. And so when I do it for the side that is on the future side of the box, I'm going to get a plus sign. Do all the analysis carefully on the side that's on the past side the box, you get a minus sign. Then I'm going to do it-- I'll pick out the component N1, and I'm going to do this along the face that's at x equals x1. Sorry. This is going to be x2. OK. You can write down four other integrals according to when you do the y side and the z side that I've not drawn. OK. What I'm going to do now is imagine that this box gets very small in the timelike direction. So let's let t2 go to t1 plus dt. And I'm going to do this and then rearrange things a little bit. Again, apologies. I'm eating chalk here. So I'm going to rearrange this integral so that it's of the form-- this side's being evaluated at t1 plus t2. So I'm going to rearrange stuff so that I can then write-- just one moment. Oh. Apologies. This whole thing, of course, equals 0. I'm just looking over my notes and went-- there was a magic sign flip. I was just looking at it and going, where the hell did that come from? OK. So I left off my equals 0 there. I'm basically moving a bunch of terms to the other side. So now I'm integrating this over the face at x2. And I do an integral over the face at x1. And then I'm not drawing in a bunch of faces along y2 and y1, z1 and z2. Divide both sides by dt. Take the limit of dt going to 0. What is this? This is the derivative of the volume integral of N0. So let's just write that out explicitly. So if I've got integral t1 plus dt N0 dx dy dz minus the integral at t1-- all this divided by dt. Take the limit as dt goes to 0. That becomes this. That i is a bad erasure. My apologies. What do I get for the other term? Well, when I divide out that dt, I am just left with the flux of the spatial component of the number vector N through all six sides at moment t, or t1. And at last-- so I will wrap things up in just a moment here. What this finally leads us to is that this fully covariant form that I have at the top there-- this becomes-- so in the language that you probably learned about in an earlier class, you can think of this as the area element of each side. These are completely equivalent to one another after you have chosen a particular Lorentz frame and invoked this four-dimensional analog of Gauss's law. So that was a lot to take, but I wanted to do it for this one particularly-- to be blunt, this was a particularly simple example, because-- so right now, the only matter that I've introduced beyond particle kinematics is dust. And dust is actually surprisingly important. When we actually get to cosmology, there are essentially two forms of matter that we consider when we study cosmology, and one of them is dust. In cosmological situations, a dust particle is basically a galaxy, or even a cluster of galaxies. We're thinking big. So it's not trivial to do this, but we are soon going to need to introduce mathematical tools that are a little better for describing things like fluids, and stuff like that. And when we do that, we are at last going to have things like conservation of energy and conservation of momentum. It will be really easy to write down a differential conservation law that describes conservation of both energy and momentum using that mathematical object. Via a process very similar to what I just did here, we can then turn this into integrals that describe how energy is conserved in a particular volume where things may be flowing into or flowing out of it, and how momentum is conserved per unit volume as things flow into it and flow out of it. I'm not going to go through that in detail on the board, but having done this-- I have some notes that I'm going to post to the website that describe this, and that will be the way we communicate this. And we're going to take advantage-- to be blunt, we will mostly just use this differential form. We'll use the fact there is a particular mathematical object whose divergence-like derivative there is equal to 0. OK. We have a few minutes left and we've ended early a few times, so I'd like to take advantage of this to switch gears a little bit and talk about another important 4-vector that plays some role in physics, and allows me to introduce a few other very useful tricks that we will often take advantage of at various points in this class. I forgot there's a straw. When you tip it, the straw doesn't work. Anyway, so the next example of stuff that we will occasionally talk about is an electric current. So switching gears very, very much now. So we will describe this as a 4-vector whose timelike component is the charge density as seen by some observer, and whose spatial component is the current density as seen by that observer. Bear in mind, you might look at this and twitch a little bit because the units look wrong. Don't forget c equals 1. So if you use this in other systems of units, sometimes we call this component charge density times c. So you guys have all-- you know what, let me just go ahead and write it out. So a couple properties about this are important and interesting. One is that the current and the charge density obey a continuity equation. We can think of it as a conservation of charge. It's expressed-- so in elementary E&M, you guys all presumably learned that the rate of change of charge density is related to the divergence of the current density. Well, this is exactly the same as saying that the spacetime divergence of the current 4-vector is equal to 0. This is really useful for us. And indeed, bearing this in mind, we find that if we want to express Maxwell's equations in a covariant way, we can do so such that this is built in automatically. So skipping over a few sets of things in my notes, we're going to find-- we're not going to find. We have found that electric fields and magnetic fields are inconvenient objects to describe using geometric objects that are appropriate for spacetime. If I want a geometric object that is appropriate for spacetime, the first thing you think of is a 4-vector. A 4-vector has four components. E-fields and B-fields have a total of six components among them. So what you going to do, have two 4-vectors and just ignore two of the components? That seems sketchy. So you think, eh, you know what? Why don't we make a tensor. Ah crap, a tensor has 16 components. That doesn't seem right. Then you go, ooh, I can make it symmetric. If you have a symmetric 4-by-4 tensor, well, that basically means that the number of independent numbers that go into this thing-- you have four down the diagonal. And then you count the number that are off the diagonal-- you have six. There's 16 in total, but the ones that are off the diagonal are equal to one another, so you have four on diagonal, six off. Too many. So you say, well, what about antisymmetric? If I have an antisymmetric object, that means that component F alpha beta is the negative of component F beta alpha. When you do that, that forces you the conclusion that there are, in fact, actually only six independent numbers in that thing. The diagonal has to be 0, because set beta equal to alpha here-- F alpha alpha equals minus F alpha alpha. That only works if that component is equal to 0. So the diagonal becomes zero, and only the six off-diagonals survive. And then you go, holy crap. Six. That's exactly what I need to have a geometric object that cleanly holds the three independent electric field components and the three independent magnetic field components. So many of you have seen all this. If this isn't familiar to you, take a look at a book like Griffiths or something like that. It goes through this. The punchline is that what you find is that the electric and magnetic field is very nicely represented by this antisymmetric 2-index object, whose components, in the units that we are using, are filled with the E and the B like so. OK. Last semester I had these memorized, but I have totally forgotten them. OK. So this is a geometric object that-- whoops. Anybody know why I just said whoops and had to put a dot on that there? The point is that electric and magnetic fields look different to different observers. This is its representation according to one particular Lorentz observer. So it's important to get that right. So in terms of this-- you're all familiar with the four Maxwell's equations. They turn out to be equivalent to-- so if you take a divergence of this F, it is-- depending on your units, so you might want to put u0's in there, and things like that. Basically, I'm setting everything that I can never remember to 1. The divergence of that thing-- by the way, actually divergence on the second index-- becomes the current density. This will actually only give you half of the Maxwell's equations. The other half-- what you do is you lower these two indices, and there's this cyclic permutation of derivatives that give you these. So you put these two things together, you apply it to this form I've written out here, and you will reproduce Maxwell's equations as they are presented in textbooks like Purcell and Griffiths. The thing which I want to emphasize here is, this form that we've got is written in such a way that the conservation of source is built into it. This geometric language requires that J mu have no divergence. And let me just show you can do this. All you need to know is that F is an antisymmetric tensor. You don't need to know anything about the properties of the E- and the B-field. So let's just try it. Let's look at the divergence of the current. So I'm going to do 4pi d mu J mu. So I'm going to take a mu derivative of this. Now bear in mind, mu and nu-- I'm using Einstein's summation convention-- they are dummy indices, so I can relabel them. I can change them. I can change mu to an alpha, nu to a beta, or I can just change mu to nu, nu to mu. So this-- as long as I do it consistently on all objects that have those things-- this is the exact same thing. But I also know that this tensor is antisymmetric. So if I switch these guys back, I get a minus sign. OK? Antisymmetry. What happens if I switch the order of the derivatives? Does it matter whether I take the x derivative first and then the y derivative, or the y derivative and then the x derivative? Partial derivatives commute with each other, right? They are perfectly symmetric. So this-- so what I've got is d mu d nu of F mu nu is minus d mu d nu of F mu nu, and that only works if the whole thing is 0. What I just did is, I actually just invoked a trick that we are going to use many times. And the one reason why-- this is a bit of a tangent. Or more than a tangent-- 150-degree turn from what I've been discussing before-- but I wanted to make sure you saw this little trick. Whenever I have an object that is antisymmetric in its indices and I contract it with an object that is symmetric in its indices, you just get 0. You can go through the little exercise if you don't feel fluent in this yet. You can do the exercise I just did up there again over. Take advantage of dummy indices, swap one. Work in the antisymmetry, swap the other. Work in the symmetry-- boom. You will necessarily prove that you've got 0. So I lay this out here because there's going to be several times later in the course, where I'm going to get to a particular calculation and there's going to be some godawful mess. We're going to look at it and go, oh, this is horrible, and then go, wait, symmetry-antisymmetry, boom. We just killed 13 terms. Tricks are fun. So I am going to stop there for today. There is another symmetry-antisymmetry thing which allows you to-- if you apply it to the equation of motion of a charge in an electromagnetic field, it just shows you that that equation of motion builds in the fact that in spacetime, the acceleration is always orthogonal to the 4-velocity. I won't do it in class. It is on page 6 of the notes that I'm about to post up. We will pick up next time. We'll begin by talking about the stress energy tensor. |
MIT_8962_General_Relativity_Spring_2020 | 10_Spacetime_curvature.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: We're in for a uncomfortable couple of weeks, but we will do our damnedest to make sure that-- I'm not going to say life won't be disrupted. Life is going to be bloody well disrupted. No question of that. But number one goal is making sure everyone remains healthy, both physically and mentally. When we're forced to isolate a little bit, we lose the social contact that makes life worth living, and so we're going to be working really hard. Look to your social group as well, and to your peers and and mentors and others to try to find a way to remain-- if you can't meet in person quite as much, there's the phone, there's Skype, there's FaceTime. Better than nothing, and that's sort of what we're looking at these days. And we're, of course, within the department, very committed to figuring out a way to make sure that the education that we're sort of here to do, we can deliver it to you in some form or another. There may be a few bumps in the road while we work this out, but we're getting there. Hopefully, this will-- if there is a disruption coming, it will be short-lived. If not, let's just let's focus on what the important things are. And today, the important things are the geodesic equation. At least that's what we're going to start things, and then we're going to take it into the next major concept that describes manifolds with curvature, I think we were going to take advantage of. So just a quick recap, where I ended things last time was we described geodesic trajectories. These are trajectories which parallel transport the tangent-- as they move along their world line, they parallel transport the tangent vector to that world line. And I forgot to put the physics up here. The key reason why this is interesting is that this corresponds to a free fall trajectory. Free fall basically means you are moving under the influence of nothing but gravity. And if you want to understand gravity in a relativistic theory, well, that's all you care about. So these are very important trajectories. And you know, it's not an exaggeration to say that solving this equation, OK-- so in this thing, I've kind of left agnostic what the spacetime is that you use to compute those covariant derivatives and to write your Christoffel symbols gamma, but it could be any spacetime that solves the relativistic field equations, which we haven't derived yet but we shall soon. This is what describes sort of small bodies moving through that kind of a spacetime. That is the starting point to a tremendous amount of analyses in general relativity. I made a crack last time that I think 65% of my published research work is essentially based on solving this equation. I actually went and checked. It's probably more like 75%, OK. That wasn't actually an exaggeration. It shows up a lot. All right, so this, as we went through, and it's basically just saying that I'm going to take the covariant derivative of the tangent vector u and contract it with u. OK, and there's a couple of other ways of writing this which I've written out here in just sort of notation. But if you expand it out, it looks like this. So what you're saying is that the vector u, the four-vector components u, are parameterized by some quantity which if it's a time-like trajectory, you can think of it as essentially the proper time as you move along that trajectory. This is just describing how this thing behaves as a function of that parameter, OK. So we're going to do some more with that. To begin with, there's a couple of cool results that we can derive. So there's a nice side note. We can rewrite this in terms of momentum. So if you imagine, let's focus on the version where I'm doing it per units proper time. And I'm going to take advantage of the fact that for a body moving on a time-like trajectory, so a body with a rest mass m, I just take this equation, and basically, you multiply by m twice, and it very clearly turns into something that is pretty much exactly the same, but I just replace my u's with p's. Two comments I want to make about this. So first, and you know what, let me actually expand this out. I mean, it's quite obvious. Why don't we write it in terms of the components in the Christoffel symbols. OK, so I'm going to write it like this. Let's do the following, OK. Recall that if I have-- so this trajectory, assuming, when you write like this, your parameter lambda is called an affine parameter. And that is a parameter such that the right hand side of [INAUDIBLE] equation is 0 on a free fall trajectory, OK. If it's something that's proportional to you, it is a valid parameter, but it's one that's sort of been defined in a bad way. One thing we showed last time is that I can shift lambda by any constant, OK, with the right units. That essentially amounts to just changing the origin of my clock. And I can multiply it by any scalar, which essentially amounts to changing the units in which I am measuring time. And it's still a good affine parameter. So here's an example of an affine parameter that I could use. Suppose that I defined this such that an interval of affine parameter delta lambda is an interval of delta tau divided by m. If I do this, well, then, p alpha is just d-- so remember this is now going to be-- let's write it like this. So this is my original definition of this thing. This then becomes dx alpha d lambda. So this is a way of choosing an affine parameter such that I'm essentially writing my tangent along the world line as the momentum rather than something like the four velocity, OK? But there's something cool about this, so let's now go and just write what my geodesy equation turns into. It's basically exactly the same. It's just that I'm going to absorb the m on the first term. Basically, it's exactly the same geodesy equation as I had before, but with u's promoted to p's. What's kind of cool about this is you can take a limit in which m goes to zero as long as your interval of proper time goes to zero at a rate such that delta t over m is constant. So what this allows us to do is just conceptually reformulate the geodesy equation, so that it's perfectly well behaved. Not just for time like trajectories, but for null or light like trajectories. OK, so that's very important for us. A lot of the most important tests of general relativity actually come down to looking at the behavior of light as it moves in some kind of a curved space time. And the geodesy equation, if you sort of interpret the way we thought about it before, you're kind of like, well, let's go back, and suppose I'm running it sort of in this one. And I'm thinking of my lambda as proper time. An interval with proper time is not defined along a light like trajectory, OK? So that just kind of makes it clear that that's fine. What we're going to do when we're talking about a light like trajectory is we're just going to find the parameter along the world line, such that the tangent vector is the momentum along that world line. Mass doesn't even make sense. So the p that goes into this-- so when I do this, this is going to be a p, such that p alpha p alpha equals minus m squared. We still have that rule that it's always going to be minus m squared, which is zero in this case. So this gives us a tool that we can use to study the motion of light as it reacts to gravity, for example. OK, we'll switch gears, so I want to do one other trick based on this momentum form of things. So I can rewrite the geodesy equation as follows, and it's going to start out where it just looks like I'm essentially doing what we call index gymnastics. I'm just sort of moving a few indices around. So let's write this as p alpha, and what I'm going to do is contract it on this. I'm putting the index and my momentum in the downstairs position now. First of all, you should stop and ask yourself, am I allowed to do that? If I do that, do I not generate some additional term that should then be moved to the right hand side? Well, think about what I'm doing. If I am lowering an index, that essentially means that I am-- let's do the following. Let's change this to a gamma for a second. What I have done here is I have essentially taken the equation I wrote over there, and I had hit it with g beta gamma, OK? I can always multiply by those things. The covariant derivative of the metric is zero. Because the covariant derivative of the metric is zero, it commutes with that derivative, so I can just walk it inside the derivative operator. So I wanted to go through that just a little bit carefully, because that's actually a trick that once you've seen it once, I want you to know it well. Because I'm just going to do it many, many times as we move forward. There's going to be a bunch of times, where I'm taking the partial-- it's going to be the covariant derivative is something. I know it's going be raising indices willy nilly on whatever it's operating on. But I'm taking it down to the fact that I'm effectively moving a metric inside and outside here. All right, so let's take that for the geodesy equation and expand it out. So I end up with mdp beta d tau, so I'm here using the fact that the p alpha that's on the outside, I'm writing that as mu alpha. And I'm using that to convert the derivative I get, expanding that into a d by d tau. And then I get a term that basically corrects the downstairs index. Because it's a downstairs index, it enters with a minus sign. Let's move this to the other side, and what I'm going to do is make all the indices be in the downstairs position. Let's see. Hang on a second. Did I do this right? Sorry. Yeah, I'm going make all the indices in the downstairs position on this capital gamma, so I'm going to write this as follows. OK, so I chose to write it this way as you'll see in just a moment. Because this is now symmetric on exchange of alpha and gamma. Let's expand that Christoffel symbol, so I'm going to have one term beta derivative alpha gamma. We'll put this up above. OK, so take a look at that last line of that expression. So as written here, I've got a term that is symmetric on exchange of alpha and gamma. But inside my parentheses, bearing in mind that my metric is itself a symmetric object, I've got two terms, this one and this one, where if I exchange alpha and gamma, I get a minus sign. So I got a symmetric contracted with anti-symmetric. Therefore, I can simplify this whole thing to something that only involves-- the only derivative I need to compute is one partial derivative of the metric. Now that's nice, but if you think about it, it might even be nicer than you realize. Suppose you're working in some coordinate system, such that for a particular derivative, for a particular, let's say, it's the derivative with respect to your time coordinate. Suppose the metric vanishes. Suppose that equals zero for some coordinate. Then you've just learned that a particular component of the four momentum, component of the downstairs four momentum mind you, that is a constant of the motion along the worldwide. I sort of said in words a few things about this a couple lectures ago when we were talking about Killing vectors. I'm going to actually tie this to that discussion in just a moment. This is often operationally the simplest way to deduce that you, in fact, have a constant motion, so there is some space times, very complicated ones that play huge roles in many of the kind of analysis we do. But you sort of look at them, and you kind of go, oh, thank god. It's time independent. All right, that means I know p downstairs t is constant. It's independent of the actual angle. P downstairs phi is a constant, and that ends up giving us some quantities that we can exploit. And later when we start talking about certain solutions of the field equations and looking at the behavior of these things, we're going to see how we can exploit them to understand the motion of bodies in very strong gravity. Before I do this, let me connect what you're saying here to-- what I'm saying right here to stuff that we did a lecture or two ago with the Killing vector. So we know that, if a metric-- so this is something that we wrote down a little bit before. We also know that, if the metric is independent of some particular coordinate, there exists a Killing field, or Killing vector, which I will call c beta. What I want to do now is say, OK, how does-- let's look at how. I'm going to define a particular scalar, so what do I get when I take-- oh, bugger. That made no sense. There we go. What would I get if I take that Killing vector? I contract it with my for momentum. How does this guy behave as I evolve along a trajectory? So the way we're going to solve this is we'll just look at the time of evolution as we move along the trajectory as we-- and the way we'll do that, I'll show you how to construct that time evolution in just a moment. We're going to assume p solves the geodesy equation, c solves Killing's equation. Let's see what happens. So what we're going to do is look at d by d tau, so that the proper covariant derivative along the trajectory of this guy. Well, this, if I take advantage of Leibniz's rule, so first of all, I can just write this as-- you know what? Let's throw an m into here just to make things nice and symmetric. The reason I did that is so that I can write that derivative in the following form, so this is what I want to evaluate. So one thing I do is expand out that derivative using Leibniz's rule, so this is-- let's first plot my Killing vector. That term is like this. Then I got a term, and it looks like this. Well, the first term, it's going to die. Because like I said, I'm going to assume p solves the geodesy equation, so p is a geodesy. I kill that. What about the second term? Well, for the second term, what I'm going to do is note that whenever you have some general two index object, so suppose I have some two index tensor, m alpha beta. I can always write this in the following way, right? Where remember, the parentheses denote the symmetric part, and the braces denote the anti-symmetric part. So this is just a theorem, right? You add it together. The 1/2s combine, and you get this thing back for the first term. And they combine the minus with the other one, OK? Very simple identity. So if I do that, applying it up here, this is symmetric under exchange of indices. This is anti-symmetric. It dies. The only thing that is left is this term. But if this is a Killing vector by Killing's equation, this equals zero by Killing's equation. So the importance of this, you've just shown that what you get when you contract for momentum with the Killing vector gives you a constant motion. You've also shown that the component of the four momentum, the downstairs component of the four momentum associated with whatever coordinate the metric happens to be independent of, is also a constant of the motion. The key thing to note is both are actually very powerful and important statements. One depends on the coordinate system and the representation you've chosen. The other does not, OK? So this is really true and useful, if you happen to have chosen the coordinate system such that this derivative is equal to zero. This is true, though, independent of your representation, so these are just two different ways of calling out constants of motion. And we actually find both of them to be very useful, so we're going to take advantage of them. There's a variation on this calculation that is on the next p set, and you will come back to this, again, when we start talking about motion in certain space times in the second half of this course. So let me just do a couple really quick examples. I've already kind of mentioned these, but let me give names to what these are. So if your space time has a time coordinate, such that the time derivative of any metric element is zero, then you know that a time like Killing vector, which I will call ct, and I'm leaving the vector sign on it. This thing exists, and you also know that p downstairs t is constant. Now the name that is given to this is negative energy. Why negative? Well, the main reason why it's negative is that we will often-- so let me just caution that this is not always an identity that we're going to use. We're going to use it in a huge number of problems that we care about. Many of the space times that we are going to work with are those that when you get really far away from whatever source is generating your gravity, it looks just like special relativity. We call such space times asymptotically flat. In other words, as you get asymptotically far away, it reduces to the space time that we studied in the first couple weeks of the class when we were doing geometric spatial relativity. And in that case, we knew the timelike component was energy. And in a flat space time, you lower that time like component's index. You get minus energy. It just so happens when you go through the math carefully that negative of the energy ends up defined in this way. It's going to be the quantity that is actually conserved everywhere. What's kind of cool is that we use this associated with asymptotic flatness to give you some intuition. But this is actually true, even if you're right outside of the vicinity of a rapidly rotating black hole. It is still the case that p downstairs t for the right choice of t is a constant. In my notes, I also show you that there is an example of an actual Killing vector that corresponds to angle momentum. Again, we'll come back to that a little bit later, so let me do one example geodesy before I sort of change topic a little bit. I'm going to write down a space time that we either in person or on video are going to derive basically right after Spring break. So suppose I hand you the following space time. This function phi, I'm not going to say too much about it quite yet. What I will say is that it is small in the sense that when you're doing various calculations with it, feel free to discard terms of order five squared or higher. And it only depends on the coordinates x, y, and z. No time dependence. I want to examine slow motion in the space time. So what I'm going to do is I'm going to imagine that my four momentum has the usual form. And I can think of it as having an energy, and sort of time like component, and momentum in the space like. The magnitude of the energy is always going to be much greater than the magnitude of the momentum, and in fact, will be approximately equal to the mass, where that is the mass of whatever body is actually undergoing this motion in this space that I've given you. OK, so the reason why I'm doing this is what we're going to do is you want to say, what does free fall look like in this space? Well, I look at geodesics, so there is my geodesy equation. This slow motion condition that I've applied over here, that tells me that when I expand all these terms out here, this is going to be dominated by the time time term, OK? So it'll be dominated because of the fact that when you just look at the numerical magnitude of the size of the components of the momentum, those are going to be the ones that dominate this calculation. Everything else, if you put your factors of c back in, they're going to be down by factors that look like v over c. So my geodesic equation turns into-- so all I did was say, it's dominated by this. I'm going to move it to the other side of the equation. So without giving away the plot, the beta equals zero component is going to turn out to be very uninteresting. Can you see why? Beta equals zero is energy. When I evaluate that Christoffel, I'm going to end up taking a bunch of-- you know, I'm going to have all these zero, zero, zero, time, time, time components. It's time independent, though, so all my time derivatives are going to be zero. It's going to vanish, but we expect that. Because it's a time independent metric, all the crap I went through a couple months ago guarantees that the time like components, energy's conserved, right? So that's kind of what we expect, so let's just focus on-- and if we had infinite time, which we clearly don't, it would be fun to talk about. Let's just move on and look at the spatial component of this. So as I look at the spatial component of this guy, let's focus on beta equals i. What you find when you actually evaluate this guy is this turns into-- OK, so go ahead and look up your Christoffel formulas. Again, these are one of those things that, eventually, you get this memorized by the end of the term, but don't feel bad if you keep forgetting it. No time derivative. No time derivative. The only thing that's left at the end of the day is essentially a gradient of the time time piece of the metric. So taking advantage of the fact gi alpha staying in the upstairs position, you can write this as-- it looks like this. Now, if you like-- remember, phi is a small value. You can do binomial expansion on that. Knock yourself out. It's not going to be important in just a second. Excuse me, minus one half one minus two, five to the minus one power. So let me talk a little bit about what I did here in this last line. My delta i alpha is coupling to a partial derivative, so that partial derivative, the zero component of that is the time derivative. Everything's time independent, so that's done. Let's just skip it, so what I did was I changed my alpha to a j. Because I only want spatial derivatives, so I'm allowed to do that. Because I'm just acknowledging the fact that all the time derivatives are uninteresting. When I do differentiate g00, I am differentiating negative quantity one plus two phi. The one doesn't contribute. All that is left is I took the minus on the inside, and there's my two phi. So putting all of these ingredients together, I at last get my Christoffel is delta ij. I'm saying it looks like spatial gradient of phi. And for keeping score, there are higher order terms, which we're going to collect under the assumption that this phi is small. Plug it back into my equation of motion. I end up with this. Now let's cancel out the m's that appeared in here. If we were not doing relativity, we would write this as-- ignore the fact that this is per unit proper time. This is dpdt is minus gradients of something that sure as hell looks like a potential. What we are going to do in the lecture rate after Spring break, so going into Spring break, depending on which-- as I said at the beginning, it's a little unclear how these lectures are going to be delivered. But bear with me. We're going to essentially put together-- we're going to take all the last ingredients and develop the field equations that describe relativistic gravity. The first thing we're going to do is solve this in a particular limit that describes a body that is weakly gravitating. This will emerge from this with phi being equal to Newtonian gravitational potential. What this is showing is that the geodesic equation, this equation that describes a trajectory that is as straight as possible in space time when it is given that particular space time, it does give you the Newtonian equation of motion, OK? Yeah? AUDIENCE: Is there still supposed to be a factor of m on the right-hand side? SCOTT HUGHES: No, so it's possible I dropped an m somewhere in there. Go through that just a little bit carefully here, but you know, it's meant to be-- actually, you know what? I take it back. Sorry, so I think I may have messed up. I remember what it was, I remember what it was. There's an m squared here. Thank you, Alex. There was an m squared here. There was an m and an m squared here. That's what I screwed up, and I think I have that wrong in my handwritten notes, which is probably why I messed that up. Yeah, so this equation was correct. And this should have been here like so. We'll clear this out and get this. Yeah, so module of that little bobble. I just want to show you this is essentially the Newtonian limit. Just give you a little look at where we go ahead, there are two ways that we are going to derive the field equations of general relativity. The first one essentially boils down to looking for certain tensors that have the right symmetries and allow us to have sort of a quantity that looks like derivatives on the field equaling the stress energy tensor as the source. That only works up to an overall constant, and this is actually the way that Einstein originally developed the field equations, was worked out all this stuff, and then by insisting that the solution that emerge from this reproduced the Newtonian equation and the Newtonian motion, he was able to fix what that constant actually is. There's a more sophisticated way of doing it, which I'm going to also go through. But it's worth noting that is the way Einstein originally did it. I had the privilege a couple of years ago-- I was at a conference in Jerusalem, where the Einstein Papers archive is located. And the guy who is the main curator of this was allowing those of us at the conference to look through them. And I actually found-- they had not yet quite categorized that, but it was the papers very much related to working with these peculiar equations. I don't think he was trying to fix the coefficient, but this spacetime can also be used to compute the perihelion precession of mercury. And so it actually showed Einstein working through that. And the thing which is really cool was he screwed up a lot. The page that I was looking at was full of errors. It would say things like-- big things crossed out and then "Nein nein nein!" written on the side. And so it made me feel better about myself. All right. So everything that we have done so far-- we've been dancing around this notion of what is called "curvature." So I have used this word several times, but I haven't made this precise yet. Curvature is going to be the precise idea of how two initially parallel trajectories cease to be parallel. So there's a couple of ways that we can quantify this. The one which I am going to use is one that's amendable to, with relative ease, developing a particularly important tensor, which characterizes curvature. And so what we're going to do is look at the behavior of a vector that is parallel transported in a non-infinitesimal region of a curved manifold. So for the purpose of this sketch, I'm going to make this closed figure be a triangle. When I actually do the calculation in just a moment, I'm going to use a little parallelogram. So around a closed figure, I want a curved manifold. So suppose my curvature's actually 0, and I do this for a triangle that is on the blackboard. So let's say I start out with a vector that points from A to B. And what I'm going to do is just parallel transport it. And in the case, goes, doo, doo, doo, doo, doo, doo, doo, doo, doo, doo, doo, doo. This is an experiment you can do at home. When it comes back, it's pointing exactly the way it was initially. Let me also just note that this triangle-- the sum of its internal angles is 180 degrees. Hopefully you all know that. Now, the next one-- if I'd had a little bit more time, I would have grabbed one of my daughter's balls to demonstrate this. But hopefully, if you guys have something like a soccer ball or a basketball, this is a little experiment you can do by yourself. Now imagine a triangle that is embedded on the surface of a sphere. So let's say this is the North Pole of my sphere. Here's the equator. So what I'm going to imagine is-- let's say I start up here. Let's make the North Pole be point A. I move on a trajectory that is as straight as I am allowed to be. And remember, if I'm a one-dimensional being living on the surface of this thing, that's a straight line. It only looks straight looks curved to us because we see a third dimension that this whole thing is embedded in. And this thing's going to come in, and it actually hits the equator at a right angle, OK? No ifs, ands, or buts about it. It's a bloody right angle. And then I'm going to-- let's call this point B-- walk back along the equator here till I reach a point which I will call C. And then I'm going to go straight north until I come back up to the North Pole. This is a triangle in which all three angles are 90 degrees. So here is a great little experiment that's very easy for you to do at home. Does anyone happen to have a ball with them? OK, never mind. Let me look at my notes for just a second. So let's say I start out here at point A, and I have my vector pointing in the south direction. So this guy goes down here. And what you'll see is it goes down to the equator, and it keeps pointing south. Then I bring it along over here, bring it back up to the north. The vector has been rotated by 90 degrees as it goes around that pass. It's a really, fun, exciting demo. If you've got a ball at home, you can do this over and over again. It's endless fun. I'm being slightly silly here, but there's an important point to be made. When you do this operation, parallel transport rotates the vector. It turns out that, if you are working on a two-dimensional manifold-- particularly, I think it's a two-dimensional manifold that is-- it may have to be of what's called constant curvature-- in other words, either a surface, a plane, or a hyperbola. It actually rotates by an angle of whatever is internal angle of the triangle minus 180 degrees. So in this case, it would rotate it by 90 degrees. If you took this thing and you actually opened it up all the way, you could basically, just by taking this leg and making it as long as you want, you can make it to 0. You can make it huge. And when you do so, you'll just rotate that vector all the more as it goes around. This operation, by the way, is called a holonomy. I throw that out there because, last time I looked, there was a decent Wikipedia page on this that has some cool animated graphics on it. Also, MathWorld.Wolfram.com had some good stuff. So this has good descriptions that you can find it all on Google. All right. What I want to do is take some of these somewhat vague notions-- so hopefully, I made it intuitively clear that there's something very interesting that happens when I parallel transport a vector around these figures, depending upon the underlying geometry of the manifold that they're embedded in. Let's try to make it more precise now. And I'm going to start all the way over here because I'm going to want big, clean boards to illustrate this. OK. So suppose I'm in some coordinate system, and this line I've written here represents a line of constant. So lambda is one particular member of your set of spacetime coordinates, so it might be time or radius or maybe you work in some crazy querying system. But lambda is meant to represent some particular member of your coordinate system. And then there's another track over here, which is displaced from it by delta x lambda, OK? So everywhere along here, one of your coordinates is equal to the value, x lambda. Everywhere along here wanted, that same coordinate's equal to x lambda plus dx lambda. Along this trajectory, there's a different coordinate that is kept constant. Lambda and sigma are not the same. So there is some coordinate whose value I will label as sigma that is constant along there. And along this one, it is also constant. Let me label the four vertices, A, B, C, and D. And let me number these four edges-- one, two, three, and four. What I am going to imagine doing is parallel transporting some vector, v alpha, around this loop. So what I'm going to do is generate the equations that describe how it changes as a transport. I'm going to start at A, so v is pointing along here. Transport it to B to C to D and then back to A. So let me very carefully do the first leg. Once you get the pattern, the others can be done a little bit more quickly. So the coordinate-- let's see. Hang on just one moment. Yeah, so I am going from A to B first. So as I move from A to B, x lambda remains constant, and the coordinate x sigma is increasing. So I am moving in a direction that points along the unit vector associated with the sigma coordinate. So I'm going to say that there's a basis vector. I shouldn't have said unit vector. I don't know its magnitude. I'm pointing along the direction in which coordinate sigma is increasing. And so parallel transporting this vector amounts to requiring that my covariant derivative along the sigma basis vector is 0. This can be written out. Turn this into index form. It looks like this. OK, no surprises. So now what I'm going to do is, essentially, I'm going to write down an integral that would describe how v alpha changes as I move from A to B. When I do this, I will then get the value of the vector at point B. So the way I'm going to write this is v alpha at B is equal to v alpha, the initial value of this thing, minus what I get when I integrate along leg one-- gamma alpha sigma mu phi mu dx sigma. Everyone happy with that? So everything I've done over here, so far, I think, is probably just fine. When you've got a differential equation, integrate it. Boom. You integrate it. You got your new thing. We're going to actually solve these integrals in a few moments, but we'll just leave it like this for now. So that's the first step. I got a couple more to do, but hopefully you can now see the pattern. If I go from B to C, I am now moving in the direction of lambda, and I'm holding the value of that coordinate constant at x sigma plus dx sigma. So the vector at C is going to be equal to this thing at B minus what I get when I integrate along path two, gamma alpha gamma mu. We got two more to go. So this one I am, again, integrating along the sigma direction. But notice, I switched the sign. I switched the sign because now my coordinate's going in the direction where it's decreasing rather than increasing. Get some fresh chalk. So we've taken it from A to B, B to C, C to D. Let's take it all the way around. So take it all the way around my second value at point A. This is going to be v alpha at D. And again, this guy is coming in the other direction. So I'll enter this one with a plus sign. And I get this. OK, so the way I'm going to quantify curvature is buried in all this stuff. Let's dig it out. So the first thing which you're going to do is I'm going to say, if I take v alpha final, basically what I want to do is write this guy out, substitute in for v alpha d, which requires me to substitute in for v alpha C, [INAUDIBLE]. So I'm going to get a big, old mess here. But in the end, the first term will be v alpha initial. So let's subtract that off. That is the change. When you actually work this out, it's going to involve four integrals. I have chosen to write this in a way that highlights a property I'm going to take advantage of in just a moment. OK, so the reason I wrote it in this way-- so I have the integral along four minus that along two plus integral along three minus that along one-- is that each one that I've written here-- they represent parts that are sort of parallel to each other on the figure, just offset from each other by a little bit, parallel but offset paths. Yeah, let's put this one high. Hang on just one moment. I have a thing in my notes that said I needed to fix something. Did I actually fix it? Yeah, I did. OK. So schematically, let's look at that first line. The integral along four of-- I have a something, dx lambda, minus the integral along path two-- of a something, dx lambda. So it's the same basic function inside each of these, but this one is being evaluated at x sigma. This one is being evaluated at x sigma plus dx sigma. I can combine them. So this becomes the integral-- let's say it's along two. Pardon me for a second. Make that a little bit bigger. So what I'm doing is I'm saying that I have a function evaluated at x sigma minus a function evaluated at x sigma plus dx sigma. Let's do a little binomial expansion. It's equivalent to an integral along a single path of essentially what I get, the first order Taylor term of that. Do the same thing for the other guy. Integral along three, I have a something, dx sigma minus integral along 1, same something, dx sigma. This guy is being evaled at x lambda x delta x lambda. This guy is being evaluated at x lambda. And so this whole thing is approximately equal to integral along one. So it looks like this, OK? So if you want to do this a little bit more carefully, knock yourself out. Part of that-- I probably should've said this explicitly, but hopefully the notation made it clear-- I'm treating these little deltas as small quantities, OK? So it makes sense that I can introduce a little first order expansion here. Let's leave the picture up, but I'm going to clear this board. With this way of doing things, let's rewrite my integrals. So what this gives me is delta vx alpha equals-- should really be an approximately equal because we're truncating this expansion. So the integral from x sigma-- x sigma equals dx sigma. Alpha x lambda [INAUDIBLE] x lambda of gamma alpha sigma mu phi mu dx sigma minus-- So this is just taking what I wrote there. Schematically, this is what you get when you actually expand all those guys out. All right. So I've got a couple derivatives here. And I'm doing an infinite test of a couple of infinitesimal integrals. When I'm doing infinitesimal integrals, they're very simple to evaluate. So let's just go ahead, evaluate them, and also expand out those derivatives. Doing so, this cleans up a fair bit. First one, I'm going to be able to finally get rid of those damn integral signs. So I'm going to wind up with something that is essentially quadratic in these things. It's going to look as the product of my little infinitesimal displacements. And I'm going to wind up with a term that involves a partial derivative of my connection here. So we've got one term looks like this. We've got another term that looks like this. So don't worry about the index gymnastics a little bit. If you go through it carefully, you'll see it. Pause here for a second. This expression sucks. The reason why it sucks-- it's not just because there's lots of terms. There's a bajillion indices on it. But it's because I've got one term that is linear in the vector and one that's linear in the derivative of vector. However, don't forget-- we get the derivatives of the vector by parallel transport. So we parallel transported this guy, which tells us that these derivatives are simply related to the vectors themselves. If I move that to the other side, that's equivalent to covariant derivative of v equals 0. So if I want to get rid of my derivative with respect to x lambda, here's I you'd write that. Likewise, if I want to get rid of-- and I do-- my derivative with respect to x sigma, just replace lambda with sigma. So now let's sub these in. Now, in 1980, the person who became my PhD supervisor wrote this giant review article on gravitational radiation. And either the last or second to last section of the paper-- I'm reminded of it right now. It begins with the sentences, "The end is near. Redemption is at hand. The end is near. We shall soon be redeemed." All right. Let's plug these in. So we plug these guys in here. What do we get? Delta v alpha equals these things. We're supposed to put this in a parenthesis. And now, what's going to happen when I get rid of all those derivatives is I'm going to have a bunch of terms that look like Christoffel squared. As Scooby Doo would say, ruh-roh, but that is just what we have to have. Incidentally, what you see when you do something like this is I now have terms entering into this whole thing that involved derivatives of the metric times derivative of the metric. Many of you may have heard sort of the slogan that general relativity is a nonlinear theory of gravity. There's where your nonlinearity is actually going to turn out to be entering, is the fact you add these squared terms in here that involve metric times itself entering in such a non-trivial and important way. What I'm going to do, finally, on this is-- so this is slightly annoying because I have one term in v mu, one term in v nu. But look. Both mu and nu are dummy indices. They're dummy indices, so what I'm going to do is-- on the last term or the last two terms, I'm just going to exchange mu for nu. And what I finally get is that the change in the vector v transported along a loop whose sides are delta x lambda and delta sigma-- it's a quantity that is linear in those two displacements. It's linear in the vector and involves this four index tensor whose value depends on derivatives of the connection and two nonlinear terms in the connection. This quantity is a mathematical entity known as the Riemann curvature tensor. Even though it involves connection coefficients, Christoffel symbols, and we argued before-- and you guys did a homework exercise where you show this-- that the connection, the Christoffel, is not tensorial, this combination of them, basically that the terms come together in such a way that, when you change your representation, the nontensorial bits cancel each other out from the terms that are being subtracted against one another. So this is, indeed, a true tensor. There's an equivalent definition, if you are reading Carroll-- so essentially, what I just walked through here is an integral equivalent of the following commutator being applied to the vector v. Some textbooks simply state the Riemann tensor is related to the commutator of partial derivatives acting upon a four-vector like so. With a little bit of effort, you can show that what this is is, essentially, a way of-- what I worked out over there is a geometric way of understanding what that commutator means. Incidentally, one thing, which I think is worth calling out-- when you apply this to a one-form or a downstairs component, you get this with a minus sign. If you are reading the textbook by Schutz, Schutz has this sign wrong in its first edition. Hopefully all copies of the first edition are rare enough now that, if you are looking at Schutz-- Schutz it's actually a wonderful textbook for an early introduction to this field, but if you happen to get a hold of the first edition, just be aware that there is-- I think it's on page 171 of the textbook, you'll see this written. So I've actually written a couple papers with Bernard Schutz, and so I'm allowed to tease him. Not only did he get it wrong. He actually came up with an intuitive argument that is wrong. Sometimes you just need to sit down and bloody well calculate something because you can almost always come up with an argument to convince you of something that's not true. And I'm afraid that's what he did in this particular case. So I have a couple notes in there about what is sometimes called curvature coupling, which essentially tells us-- I pointed out in the last lecture that when we're dealing with geodesics, strictly speaking they describe, completely point-like, almost just a monopole and no structure and no shape whatsoever moving through spacetime. If you have a larger body or a body that has any kind of multipolar structure associated with it, those multipoles-- you can think of that additional structure is essentially filling up part of the local Lorentz frame around the center of mass of that point, and they couple the spacetime and push it away from the geodesic. This Riemann tensor actually describes the way in which that body couples to the background spacetime, that it might be falling in. So this ends up playing a really important role. For instance, when you study the precession of equinoxes, we learn how to do this Newtonian theory using the action of tides from the Earth and the moon on a planet like the Earth. This ends up being the quantity that mathematically encapsulates tides and general relativity. So it enters into there. So I'm going to sketch through this very, very quickly, simply because we don't have a lot of time, and there's good discussion of this in various other places. But let me just point out that-- you look at this thing. It's a four-index tensor, and each index can take four values. That makes it look like it has 256 components. Now, I'm not going to step through this in detail right now. This will either be in the next lecture that I do this, or you'll watch me on a video once this gets recorded, depending upon how things unroll in the next 24 hours. Riemann has a lot of symmetries. I will go through those symmetries carefully, either in lecture or on a video. So symmetries-- Riemann-- and you know what? Let me write it out in n dimensions from n to the 4, which is what you'd expect for a four index object in n dimensions, down to n squared times m squared minus 1 over 12th. So where I want to conclude today is let's just take a look at what that turns into for a couple of different numbers and dimensions. So if you do n equals 1, you get 0. So the Riemann tensor has no components on a one-dimensional manifold. There's a simple reason for that. Remember the way we defined it, OK? We did this by parallel transporting around a particular figure. If you're in one dimension, this is all you can do. There's no holonomy operation in one dimension. You can't do that. So no curvature. If you want to be a real pedant and someone says, well, look at a curved line, you'll go, ah, but lines can't be curved. n equals 2. So you get 2 squared, 2 squared minus 1 over 12-- you get 1. So if you're working in two dimensions, there is a single number that characterizes the curvature at every point, and this is often thought of as just a radius of curvature. Simplest example is if you have a sphere. Sphere is completely characterized by its radius. But if you imagine that it's like a sphere that you squash, well, you can imagine at every point that there is a particular sphere that is tangent to that point, and the radius of curvature of the tangent sphere is the one that defines the curvature at that point. I'm going to skip three because it's not all that interesting. The one that is more important is, if you do n equals 4, you'll wind up with 16 times 15 over 12, which is 20. This is exactly the number of derivatives that we could not cancel out when we did the exercise a couple of lectures ago of assessing how well we can make spacetime have a flat representation in the vicinity of some point. It's the number of leftover constraints at second order in a freely falling frame. All right so I'm going to stop there for today. That's a nice place for us to stop. Keep watching your emails. We're in an interesting situation. Life at MIT is evolving. But when we pick it up in one form or another, what I'm going to do first is talk a little bit more about the symmetry of this object because there's a couple of explicit symmetries that lead to that reduction from n to the 4th to n squared n squared minus 1 over 12, and it's useful for us to go through them and see what they look like. And then I also want to talk about a couple of variants on this curvature tensor, OK? So just to give you a little bit of a preview-- the curvature tensor-- it's a four index object. We have argued already that we're going to end up doing things that look like looking at derivatives of the metric being equal to our source. Our source is a two-index object, the stress energy tensor. We've got gotta get rid of two indices. And so what we're going to do is we're going to essentially contract this guy with a couple of powers of the metric in order to trace over certain combinations of indices and make two index variants of the curvature tensor. And we're also going to look at derivatives of this, because it turns out that there is a particular combination of derivatives at the Riemann tensor that has an important geometrical meaning. What we're going to find is that, when we combine these two notions, there is a particular divergence of a particular variant of the curvature tensor that is 0. In other words, we can make a curvature tensor that is divergence-free. Our stress energy tensor is divergence-free. I wonder if one is related to the other. That, in a nutshell, is how Einstein came up with general relativity, by asking that question and then just seeing what happened. So that's what we're going to go through next. |
MIT_8962_General_Relativity_Spring_2020 | 12_The_Einstein_field_equation.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: All right. Good morning, 8.962. This is a very weird experience. I am standing in here talking to an empty classroom. I have some experience talking to myself, because like many of us, I am probably a little weirder than the average. But that does not change the fact that this is awkward and a little strange, and we already miss having you around here. So I hope we all get through MIT's current weirdness in a healthy and quick fashion so we can get back to doing this work we love with the people we love to have here. All right, so all that being said, it's time for us to get back to the business of 8.962, which is learning about general relativity. And today, the lecture that I am recording is one in which we will take all the tools that have been developed and we will turn this into a theory of gravity. Let me go over a quick recap of some of the things that we talked about in our previous lecture, and I want to emphasize this because this quantity that we derived about two lectures ago, the Riemann curvature tensor, is going to play an extremely important role in things that we do moving forward. So I'll just quickly remind you that in our previous lecture, we counted up the symmetries that this tensor has. And so the four most-- the four that are important for understanding its properties, its four main symmetries are first of all, if you exchange indices 3 and 4, it comes in with a minus sign, so it's anti-symmetric under exchange of indices 3 and 4. If you lower that first and next so that they are all in the downstairs position and you exchange indices 1 and 2, you likewise pick up a minus sign. Again, keeping everything in the downstairs position, if you just wholesale swap indices 1 and 2 for indices 3 and 4 like so, that's symmetric, and so you get the whole thing back with a plus sign. And finally, one that's a little bit non-obvious but can be seen if you're sort of pigheaded enough to sort of stare at this thing and kind of pound on the algebra a little bit, if you take the Riemann curvature tensor-- and this can be at the end of-- the first index can be either upstairs or downstairs, but if you cyclically permute indices 2, 3, and 4 and add them up, they sum to 0, OK? So that tells you that it's a constraint on this thing when I look at the behavior, this thing with respect to indices 2, 3, and 4. We introduced a variant of the Riemann curvature tensor called the Ricci curvature. So the way I do this is if I take the trace on the Riemann curvature tensor on indices 1 and 3, which is equivalent to taking it with the indices on the downstairs position and hitting it with the metric like so, I get this quantity, which I forgot to write down, is symmetric. One of the major important physical applications of the Riemann curvature tensor is that it allows us to describe the way in which two neighboring geodesics-- if I have two geodesics that are separated by a four vector c, and I look at how that separation evolves as they move forward along their geodesic paths, this differential equation describes how it behaves. And the key thing is that what we see is that the rate of separation is proportional to the Riemann curvature. It ends up playing the role-- when we think about-- what this tells us is this ends up, remember, geodesics describe free fall. And so what this is telling me is a way in which two nearby but somewhat separated-- separated by a distance c-- nearby but slightly separated geodesics-- both are in free fall, but their free fall trajectories are diverging from one other and perhaps being focused towards one another depending upon how R is actually behaving, this is the behavior of tides. Free fall is gravity, and this is saying that the free fall trajectory's change of separation is governed by the Riemann tensor, and that's telling me about the action of gravitational tides. The last thing that we did in Tuesday's-- excuse me-- in Thursday's lecture was I went through and I developed this proof of what is known as the Bianchi identity, which is an identity on the-- it's an identity on the covariant derivative of the Riemann tensor. And so notice what's going on here. I'm leaving indices 3 and 4 on the Riemann tensor-- oh shoot. I'm actually changing my notation halfway through, let me fix that. My apologies. OK. Apologies for that. Are you leave indices 3 and 4 unchanged, and what you do is you cyclically permute the index against-- in the direction which you're taking a derivative with indices 1 and 2. So my first term, it goes alpha, beta, gamma; then beta, gamma, alpha; gamma, alpha, beta. OK, notice the way they are cyclically permuting like that. Sum them up and you get 0. So let's take it from here. We're going to start with this Bianchi identity. What I want to do now is contract the Bianchi identity in the following way. So let's take this form that I've written out here-- and let me just make sure I've left it now in a form that comports with my notes. I did, good. So what I'm going to do is multiply the entire thing. Using the metric, I'm going to contract it on indices beta and mu. So remember, the metric commutes with the covariant derivative. So unless the derivative itself is with respect to either the beta or the mu index, that g just sort of waltzes right in top of there. So when I do this, it's going to beta-- g beta mu nu upstairs is going to walk right through this, it's going to raise the beta index, and what I wind up with here is this first term becomes the covariant derivative of the Ricci tensor, OK? I have contracted on indices beta and mu. When it hits this one, is just going to raise the index on that covariant derivative. So I've got a term here now that looks something like the divergence of the Riemann tensor, divergence with respect to index 3. When I hit this term, it walks through the covariant derivative again, and you see what I'm doing is a trace on indexes 2 and 3. Now I can take advantage of the anti-symmetry here-- let's reverse this. And so it's like doing a-- throwing in a minus sign and then doing a trace on indices 1 and 2-- excuse me, doing a trace on-- rewind, back up for just a second. I'm going to take advantage that anti-symmetry, I'll exchange indices 1 and 2, and then I am doing a trace on indices 1 and 3, which is going to give me the Ricci tensor. But because I have used that-- or that anti-symmetry, I will do so with a minus sign. So what I get here is this. I'm going to probably bobble more than once in this lecture, because again, doing this in an empty room is a little weird. All right. So I really want to get a relationship that simplifies the Riemann tensor, OK? A Riemann tensor's got four indices on it. I'm not scared, but I don't like it, OK? So we're going to do one more contraction operation to try to simplify this. Let's now contract once more using the metric on indices g and nu, OK? So when I do it with the first one, it walks right through-- right through that covariant derivative, and I get the trace of the Riemann tensor, the Ricci-- excuse me-- I get the trace of the Ricci tensor, the Ricci scalar. When I do it on the second term, OK? I am now tracing on indices 1 and 4. I will invoke anti-symmetry to change that into a trace on indices 1 in 3, and I get the Ricci tenser with a minus sign. And then the next one, I just trace on this and I wind up with something that looks like this. OK. Now these two terms are both divergences on the-- they are both divergences on the second index. The second index is a dummy index, so I can put these two together. So this is equivalent to-- or dividing out a factor of minus 2, the way this is more often written. You can also factor out that derivative. Let's write it like this. OK? So what I'm doing here is I divided by a minus 2 so that I can put this guy in front and I get a minus 1/2 in front of my Ricci term. And because I want to factor out my covariant derivative, I need to throw in a factor of the metric there so that the indices line up right. So what we do at this point is we do what-- whenever you reach a certain point in your calculation where you've got something good, you do what every mathematician or physicist would do, you give this guy a name. So switching my indices a tiny bit, we define g mu nu to be the Ricci tensor minus 1/2 metric Ricci scalar, and this is an entity known as the Einstein tensor. This is a course on Einstein's gravity, so the name alone should tell you, this guy is going to matter. One quick side note. So suppose I take the trace of the Einstein tensor. When we took the trace of the Ricci tensor, I didn't write it down, but if I take a trace of this guy, I just get the Ricci scalar R, which I used over here. So when I do this here-- oops. Suppose I just want to call this the Einstein scalar g. Well, applying that to its definition, this is going to be equal to the Ricci scalar minus 1/2 trace of the metric times R. And it's a general rule in any theory of spacetime that the trace of the metric is equal to the number of dimensions in your spacetime, OK? You can easily work it out in special relativity, you're basically just raising one index, and as we'll see, it holds completely generally. In fact, it follows directly from the fact that the upstairs metric is the matrix inverse of the downstairs metric. So this is equal to 4, so this whole thing is just the negative of the Ricci scalar. What this means is that the Einstein tensor is the trace-reversed Ricci tensor, OK? I just want to plant that for now. This is a fact that we're going to take advantage of a little bit later, but for now, it's just a mathematical fact that I want to point out, I want to set aside. We'll come back to it when it matters. OK. We now have everything that we need to take all of the framework that we have been developing all term and turn it into a theory of gravity. I just had a nightmare. Am I being recorded?-- yes, OK. Sorry, just suddenly thought I might have forgotten to turn my microphone on! So let's turn this into a theory of gravity. Ingredient 1 is something that we have discussed quite a bit before. I want to restate it and I want to sort of remind us. Several times over the past couple of lectures I have implicitly used this rule already, but I want to make it a little bit more explicit now. We're going to use the principle of equivalence. In particular, we're going to use what is known as the minimal coupling principle. So here's the way-- what this basically means. We're going to try to take laws of physics that are well-understood from laboratory experiments, from special relativity, everything that we have known and loved and tested for-- since we started studying physics, and we're going to try to see how that can be carried over to working in the curved spacetime that will describe gravity with as little additional sort of coupling to spacetime entities as is possible. So here's what we're going to do. Take a law of physics that is valid in inertial coordinates and flat spacetime, or equivalently, the local Lorentz frame, which corresponds to the local region of a freely-falling frame or a freely-falling observer, take that law of physics that is good in that form and rewrite it in a coordinate-invariant tensorial form. This is one the reasons why throughout this term, we have been brutally didactic about insisting on getting all of our laws of physics expressed using tensors, quantities which have exactly the transformation laws that we demand in order for them to be true tensors that live in the curve manifold that we use to describe spacetime. The last time I actually did something like this was when I derived-- I've erased it now, but when I derived the equation of geodesic deviation, OK? I first did it using very, very simple language, and then I sort of looked at it and said, well this is fine according to that local Lorentz frame, according to that freely-falling observer. But this is not tensorial, it's actually as written only good in that frame. And so what we did was we took another couple minutes and said, let's see how I can change this acceleration operator that describes my equation of geodesic deviation, put in the extra structure necessary so that the acceleration operator is tensorial, and when we did that, we saw that the result was actually exactly what the Riemann tensor looks like in the local Lorentz frame. We said, if it holds a local Lorentz frame, I'm going to assert it holds in all other frames. And that indeed is the final step in this procedure. We assert that the resulting law holds in curved spacetime. OK? So this is the procedure by which general relativity takes the laws of physics, good and flat spacetime, and rejiggers them so that they work in curved spacetime. Ultimately, this is physics, and so ultimately the test for these things are experiments. And I will simply say at this point that this procedure has passed all experimental tests that we have thrown at it so far, and so we're happy with it. So let me just describe one example of where we did this-- actually, I'm going to do two examples. So if I consider the force-free motion of an object in the freely-falling frame-- so recall, in the freely-falling frame, everything is being acted upon by gravity in an equal way. If I am in the local Lorentz frame, I can simply say that my object feels-- my freely-falling object, freely-flying observer-- feels no acceleration. That is a perfectly rigorous expression of the idea that this observer or object is undergoing force-free motion in this frame. This is not tensorial, though. And so we look at this and say, well, if I want to make this tensorial, what I'm going to do is note that the tensor operator that describes-- now let me keep my indices consistent here. My tensor operator that describes this equation is given by taking the covariant derivative of the four velocity and contracting it with the four velocity itself. These say the exact same thing in the local Lorentz frame. This one is tensorial, though, that one is not. And so we then say, OK, well this is the version that is tensorial, I'm going to assert that it holds in general. Another example in flat spacetime, local conservation of energy and momentum was expressed by the idea that my stress energy tensor had no divergence in the local Lorentz frame. Well, if I want to make this tensorial, all I do is I promote that partial derivative I used to define the divergence to a covariant derivative. This is how we are going to define conservation-- we're going to define local conservation of energy and momentum in a general spacetime theory. So that's step 1. We need-- or sorry, ingredient 1. Ingredient 2 is-- well, let's just step back for a second. We have done a lot of work to describe the behavior of curved spacetimes, OK? Spacetimes that are not just the spacetimes of special relativity, spacetimes when my basis objects are functional, where the Riemann curvature tensor is non-zero. We've done a lot to do that, but I haven't said anything about where that curved spacetime actually comes from. So the next thing which I need is a field equation which connects my spacetime to sources of matter and energy. That's a tall order. The way we're going to do this, we're actually going to do it two different ways. So in this current lecture, I'm going to do it using a method that parallels how Einstein originally did it when he derived-- what resulted out of this is what we call the field equation of general relativity or the Einstein field equation, and in this first presentation of this material, I'm going to do it the way Einstein did it. So what we are going to do is we will require that whatever emerges from this procedure, it must recover Newtonian gravity in an appropriate limit. This is a philosophical point about physics. When you come up with a new theory, you may conceptually overturn what came before. You may have an entirely new way of thinking about it. You may go from saying that there is a potential that is sourced by mass that fills all of space and that objects react to to saying something like, we now decide that the manifold of events has a curvature that is determined by the distribution of matter and energy in spacetime. It's very different philosophical and ultimately mathematical ways of formulating this, but they have to give consistent predictions, because at the end of the day, Newtonian gravity works pretty damn well, OK? We can't just throw that away. So what we're going to do is demand that in an appropriate limit, both the field equation for Newtonian gravity-- so this is the Laplace operator now, which I'm going to write in a semi-coordinate-invariant form, as the chronic or delta contracted-- basically it's the trace on a matrix of partial derivatives acting on a potential. This equals 4 pi rho, and I call this semi-coordinate invariant because part of what goes into this is this Newtonian limit only works if everything is sufficiently slowly varying in time, that things having to do with time derivatives can be neglected, OK? It's never really been-- prior to some of the more modern experiments that we've had to do, time-varying sources of gravity are very hard to work with. And so Newton was never really tested in that way. Nonetheless, whatever emerges from Einstein had best agree with this. And we are also going to require that the equation of motion in this framework agree with Newtonian gravity. We actually went through this-- this was a little bit of a preview of this lecture, we did this in our-- we concluded our discussion of geodesics. Let me just recap the result that came out of this. So our equation of motion was that-- you can write it as the acceleration of an observer is related to the gradient of the potential. All right. So let's follow in the footsteps of Einstein and do this. So what we're going to do-- let's do the equation of motion first. I've already gone through this briefly, but I want to go over it again and I want to update the notation slightly. So let's do the equation of motion by beginning with the geodesic equation. We will start with the acceleration coupled to the four velocity by the Christoffel symbols. All tests of Newtonian gravity, especially those that Einstein had available at the time that he was formulating this, were slow motion ones. We were considering objects moving at best in our solar system. And so things there on a human scale certainly move quickly, but they're slow compared to the speed of light. And so let's impose the slow motion limit, which tells us that the 0th component of the four velocity is much larger than the spatial components of the four velocity, OK? Remember working in units where the speed of light is equal to 1. And so if this is being measured in human units, kilometers per second, and things like that, this is on the order of the speed of light. So when we throw this in, we see that we expand this out, that the contributions from the dt d tau terms here are going to be vastly larger than any others. And so we can simplify our equation to a form that looks like this. OK? In the spirit of being uber complete, let's write out that Christoffel symbol. So dig back into your previous lectures' notes, remind yourself what the formula for the Christoffel symbol is. OK. Notice, two of the terms here are time derivatives. The Newtonian limit-- all the tests that were available when Einstein was formulating this, the limit that we care about here, the gravitational field, the gravitational potentials that he was studying, what the Newtonian limit emerges from, they are static. So we're going to do is neglect time derivatives to recover this limit. And when we do this, what we find is that the component-- the Christoffel component that we care about looks like one derivative of the 0 0 piece of the spacetime metric. It's not too hard to convince yourself that this, in fact, reduces-- oops, pardon me. I skipped a step. Pardon me just one moment. Just one moment, my apologies. I'm going to write the spacetime metric-- I'm going to work in a coordinate system such that spacetime looks like the flat space time of special relativity plus a little bit else, OK? This is consistent with the idea that every system we have studied in Newtonian gravity is one where the predictions of special relativity actually work really, really well, OK? Gravity is new, it's special, it's why we have a whole other course describing it. But clearly it can't be too far different from special relativity or we wouldn't have been able to formulate special relativity in the first place. So my apologies, I sort of jumped ahead here for a second. We're going to treat the g mu nu that goes into this as the metric of flat spacetime plus something else where I'm going to imagine that all the different components of this-- so a typical component of this h mu nu has an absolute value that is much smaller than 1. It's not too hard to prove that when you invert this, what you wind up with is a form that looks like so where this h mu nu with the indices in the upstairs position is given by raising h's indices using the metric of flat spacetime. We're going to talk about this in a little bit more detail in a future lecture, it doesn't really matter too much right now. I just want to point out that the inverse g, which we need to use here, also has this form that looks like flat spacetime metric and this h coupling into it. Now the reason I'm going through all this is that in order to work out this Christoffel symbol, I need to take a derivative. The derivative of eta is 0, OK? So the only thing that gets differentiated is h. So when you work out this Christoffel symbol, what you get is this. If you're being-- keeping score, there are corrections of order h squared, and pardon me, I should have actually noted, there are corrections of order h squared that go into this inverse. Let me move this over so I can fit that in a little bit better. But in keeping with the idea that-- in keeping with the idea that for the Newtonian limit, the h squared could-- h is small, we're going to treat the h squared corrections as negligible and we will drop them. OK. So let's look at what motion in this limit turns into, then. We now have enough pieces to compute all the bits of the equation of motion. So in keeping with the idea that I am going to neglect all time derivatives, this tells me that the gamma 0, 00 term is equal to 0. And from this, we find that there is a simple equation describing time. In our equation describing space, OK? So what I've done there is just taken this geodesic equation, plugged in that result for the Christoffel symbol, and expanded this guy out. So what results, I can divide both sides now by two powers of dt d tau. All right. If you take a look at what we've got here, this prediction of the-- no even a prediction, this result from the geodesic equation is identical to our Newtonian equation of motion provided we make the following identification. h00 must be minus 2 phi where phi was Newtonian gravitational potential. Or equivalently, g00 is the negative of 1 plus 2 phi. All right. So that's step 1. We have made for ourselves a correspondence between what the metric should be and the equation of motion. We still have to do the field equation, so let's talk about that. So very helpfully I've already got the Newtonian field equation right above me here. Let me rewrite it because I'm going to want to tweak my notation a tiny bit. I don't want to think about what this is telling me. So eta ij is the same thing as delta ij, I just want to put it in this form so that it looks like a piece of a spacetime tensor. This is manifestly not a tensorial equation. I have a bunch of derivatives on my potential being set equal to-- OK, there's a couple constants, but this, OK? When we learned about quantities like this in undergraduate physics, usually we're told that this is a-- excuse me, this is a scalar. But we now know, rho is not a scalar, it is the mass density, which up to a factor of c squared, is the same thing as the energy density. And when we examined how this behaves as we change between inertial reference frames, we found this transforms like a particular component of a tensor. And as I sort of emphasize, not that long ago we have been in something of a didactic fury insisting that everything be formulated in terms of tensors. Pulling out a particular component of a tensor is bad math and bad physics. So we want to promote this to something tensorial. So s on the right-hand side, we've got one component of the stress energy tensor. We would like whatever is going to be on the right-hand side of this equation to be the stress energy tensor, OK? We can sort of imagine that what's going on in Newton's gravity is that there is one particular-- maybe there's one component of this equation that in all the analyses that were done that led to our formation of Newtonian gravity, there may be one component that was dominant, which is how it was that Newton and everyone since then was able to sort of pick out a particular component of this equation as being important. Over here on the left-hand side, we saw earlier that the equation of motion we're going to look for corresponds to the Newtonian limit if the metric plays the same role-- up to factors of 2 and offsets by 1 and things like that-- the metric must play the same role as the Newtonian gravitational potential. So if I look at-- if I look at the Newtonian field equation, I see two derivatives acting on the potential. So I want my metric to stand in for the potential, we expect there to be two derivatives of metric entering this relationship. So now two derivatives of the metric is going to give me something that smells like a curvature. So we want to put a curvature tensor on the left-hand side of this equation. We have several to choose from, OK? It clearly can't be the Riemann tensor. There's too many indices, it just doesn't fit. It could be the Ricci curvature, OK? The Ricci curvature has two indices. That has two indices, that's a candidate. But it's worth stopping and reminding ourselves, wait a minute, this guy has some properties that I already know about. t mu nu tells me about the properties of energy and momentum in my spacetime, and as such, conservation-- local conservation of energy and momentum requires that it be divergence-free. So whatever this curvature tensor is here on the left-hand side, we need it to be a divergence-free 2-index mathematical object. At the beginning of today's lecture, I showed how by contracting on the Bianchi identity, you can, in fact, deduce that there exists exactly such a mathematical object. So let us suppose that our equation that relates the properties of the spacetime to the sources of energy and momentum of my spacetime is essentially that that Einstein tensor, g mu nu, be equal to the stress energy tensor. Now in fact, they don't have the same dimensions as each other, so let's throw in a kappa, some kind of a constant to make sure that we get the right units, the right dimensions, and that we recover the Newtonian limit. The way we're going to deduce how well this works is see whether an equation of this form gives me something that looks like the Newtonian limit when I go to what I'm going to call the weak gravity limit, and I'm going to then use, assuming it does work-- not to give away the plot, it does-- we'll use that to figure out what this constant kappa must be. So if we do, in fact, have a field equation of the form g mu nu is some constant t mu nu, it's not too hard to figure out that an equivalent form of this is to say that the Ricci tensor is k times t mu nu minus 1/2 g mu nu t where this t is just the trace of the stress energy tensor. Remember, I spent a few moments after we derived the Einstein tensor pointing out that it's essentially the same thing as Ricci but with the trace reversed. This is just a trace-reversed equivalent to that equation. This step that I'm introducing here, basically it just makes the algebra for the next calculation I'm going to do a little bit easier, OK? So I just want to emphasize that this and that are exactly the same content. All right. So to make some headway, we need to choose a form for a stress energy tensor. Our goal is to recover the Newtonian limit, and so what we want to do is make the stress energy tensor of a body that corresponds to the sort of sources of gravity that are used in studies of Newtonian gravity. So let's do something very simple for us. Let's pick a static-- in other words, no time variation, a static perfect fluid as our source of gravity. So I'm going to choose for my t mu nu-- dial yourself back to lectures where we talk about this, OK? So this is the perfect fluid stress energy tensor. We're working in the Newtonian limit, and we are working in units where the speed of light is equal to 1. If you put speed of light back into these things, you explicitly include it, this is actually a rho c squared that appears here. And so what this tells me is that if I'm studying sort of Newtonian limit problems, rho is much, much, much greater than P in the limit that we care about. Furthermore, I am treating this fluid as being static. So that means that my four velocity only has one component, OK? The fluid is not flowing. You might be tempted to say, oh, OK, I can just put a 1 in for this. Not so fast, OK? Let's be a little bit more careful about that. One of the key governing properties of a four velocity is that it is properly normalized. So this equals g mu nu u mu u nu is minus 1. We know that the only components of this that matter, so to speak, are the mu and nu equal 0. So this becomes g00 mu 0 squared equals minus 1. But g00 is-- well, let's write it this way-- Negative 1 plus h00. Go through this algebra, and what it tells you is u0 equals 1 plus 1/2 h00. Again, I'm doing my algebra at leading order in h here. We raise and lower indices. So in my calculation, I'm going to want to know the downstairs version of this. And if I, again, treat this thing-- treat this thing consistently, OK. What I'll find is I just pick up a minus sign there. OK. OK. Let's now put all the pieces together. The only component of my stress energy tensor that's going to now really matter is rho u0 u0, which, putting all these ingredients back together, is rho 1 plus h00. The trace of this guy, putting all these pieces together, is just equal to negative rho. Since I only have one component that's going to end up mattering, let's just focus on one component of my proposed field equation. OK? So this is the guy that I want to solve. I'll let you digest that and set up the calculation. We've got T00 minus 1/2 T00 T. This is going to be, plugging in these bits that I worked out on the other board, here's my T00. Just make sure I did that right earlier-- I did. OK. So this is my right-hand side of my field equation. It will actually be sufficient for our purposes to neglect this term, OK? We'll see why in just a moment. So plugging that in, I need to work out the 00 component of my Ricci. So I go back to its foundational definition. This is what I get when I take the trace on indices 1 and 3 of the Riemann tensor. I can simplify that to just doing the trace over the spatial indices, because the term I'm leaving out is the one that is of the form 00 here, which by the anti-symmetry, on exchange of those indices, must vanish. Plugging in my definition, what I find is it is going to look like this here. So I'm just going to neglect the order of gamma squared term because I'm working in a limit where I assume that all these h's are small. This is going to vanish because of my assumption of everything being static in this limit. So this, I then go and plug in my definitions. OK. Again, I'm going to lose these two derivatives by the assumption of things being static. And pardon me just a second-- yeah, so I'm going to lose these two because of the assumption of things being static. The only derivative-- the only term that's going to matter, the derivative here is h. And so when I hit it with the inverse metric, this becomes simply the derivative of the h00 piece, OK? I can go from g straight to eta because the correction to this is of order h squared, which as I've repeatedly emphasized, we're going to neglect. All right, we're almost there. Let me put this board up, I want to keep this. OK, where was I? So I've got it down to here, let me just simplify this one step more. Eta is-- if mu is not spatial, then this is just 0. So I can neatly change my mu derivative into a j. I can just focus on the spatial piece of it. So this tells me R00 is minus 1/2 Kronecker delta delta i delta j acting on h00. This operator is nothing more than-- it's a Laplace operator. So this is minus 1/2, our old-fashioned, happily, well-known from undergrad studies Laplace operator on h00. So putting all this together, my field equation, which I wrote in this form, reduces down to del squared h00 equals minus kappa rho. The Newtonian limit that we did for the equation of motion, the fact that we showed that geodesics correspond to this, that already led me to deduce that h00 was equal to minus 2 phi. My Newtonian field equation requires me to have the Laplace operator acting on the potential phi, giving me 4 pi g rho. Put all these pieces together, and what we see is this proposed field equation works perfectly provided we choose for that constant. Kappa equals 8 pi j. And so we finally get g mu nu equals 8 pi g t mu nu. This is known as the Einstein field equation. So before I do a few more things with it, let us pause and just sort of take stock of what went into this calculation. We have a ton of mathematical tools that we have developed that allow us to just to describe the behavior of curved manifolds and the motion of bodies in a moving curved manifolds. We didn't yet have a tool telling us how the spacetime metric can be specified, OK? We didn't have the equivalent of the Newtonian fuel equation that told me how gravity arises from a source. So what we did was we looked at the geodesic equation, we went into a limit where things deviated just a little bit from flat spacetime, and we required objects be moving non-relativistically so that their spatial four velocity components were all small. That told us that we were able to reproduce the Newtonian equation of motion if h00, the little deviation of spacetime from flat spacetime in the 00 piece, was equal to negative 2 times the Newtonian potential. We then said, well, the Newtonian field equation is sort of sick from a relativistic perspective because it is working with a particular component of a tensor rather than with a tensor. So let's just ask ourselves, how can we promote this to a properly constructed tensorial equation? So we insisted the right-hand side be t mu nu. And then we looked for something that looks like two derivatives of the potential, or, more properly, two derivatives of the metric which is going to give me a curvature tensor, and say, OK, I want a two-index curvature tensor on the left-hand side. Since stress energy tensor is divergence-free, I am forced to choose a character tensor that is divergence-free, and that's what leads me to this object, and there's the Einstein curvature. And then insisting that that procedure reproduce the Newtonian limit when things sort of deviate very slightly from flat spacetime, that insisted the constant proportionality between the two sides be 8 pi j. This, in a nutshell, is how Einstein derived this equation originally when it was published in 1915. When I first went through this exercise and really appreciated this, I was struck by what a clever guy he was. And it is worth noting that the mathematics for doing this was very foreign to Einstein at that time. There's a reason there's a 10-year gap between his papers on special relativity and his presentation of the field equations of general relativity. Special relativity was 1905, field equation was 1915. He was spending most of those intervening 10 years learning all the math that we have been studying for the past six or seven weeks, OK? So we kind of have the luxury of knowing what path to take. And so we were able to sort of pick out the most important bits so that we could sort of-- we knew where we wanted to go. He had to learn all this stuff from scratch, and he worked with quite a few mathematicians to learn all these pieces. Having said that, though, it did strike me this is a somewhat ad hoc kind of a derivation. When you look at this, you might sort of think, well, could we not-- might there not be other things I could put on either the left-hand side or the right-hand side that would still respect the Newtonian limit? And indeed, we can add any divergence-free tensor onto-- depending how you count it-- either the left-hand side or the right-hand side-- let's say the left-hand side-- and we would still have a good field equation. Einstein himself was the first one to note this. Here's an example of such a divergence-free tensor. The metric itself, OK? The metric is compatible with the covariant derivative. Any covariant derivative of the metric is 0. And so I can just put the metric over here, that's perfectly fine. Now the dimensions are a little bit off, so we have to insert a constant of proportionality to make everything come out right. This lambda is known as the cosmological constant. Now what's kind of interesting is that one can write down the Einstein field equations in this way, but you could just as easily take that lambda g mu nu and move it onto the right-hand side and think of this additional term as a particularly special source of stress energy. Let's do that. So let's define t mu nu lambda equal negative lambda over 8 pi g times g mu nu. If we do that, we then just have g mu nu equals 8 pi gt mu nu with a particular contribution to our t mu nu being this cosmological constant term. When we do this, what you see is that t mu nu is nothing more than a perfect fluid with rho equals 8 pi g in the freely-falling frame, pressure of negative lambda over 8 pi j. Such a stress energy tensor actually arises in quantum field theories. This represents a form of zero-point energy in the vacuum. You basically need to look for something that is a stress energy tensor that is isotopic and invariant to Lorentz transformations and the local Lorentz frame, and that uniquely picks out a stress energy tensor that is proportional to the metric in the freely-falling frame. So this is an argument that was originally noted by Yakov Zeldovich. Whoops. And much of this stuff was considered to be kind of a curiosity for years until cosmological observations-- we haven't done cosmology yet. We will do this in a couple of weeks-- a couple of lectures, I should say. And it turns out that the large-scale structure of our universe seems to support the existence potentially of there being a cosmological constant. So the behavior of all these things is a lot more relevant, it's been a lot more relevant over the past, say, 15 or 20 years than it was when I originally learned the subject in 1993. So I want to just conclude this lecture with a couple of remarks about things that are commonly set equal to 1 when we are doing calculations of this point. So one often sets G equal to 1 as well as c equal to 1. Carroll's textbook does not-- several other modern textbooks do not-- I personally like for pedagogical purposes leaving the G in there, because it is very useful for calling out-- helping to understand the way in which different terms sort of couple in. It can-- if nothing else, it serves as a very useful order counting parameter, something that we'll see in some of the future calculations that we do. But there's a reason why one often works with G equal to 1 in many relativity analyses. Fundamentally, this is because gravity is a very weak force. G is the most poorly known of all of the fundamental constants of nature. I think it's only known-- I forget the number right now, but it's known to about five or six digits. Contrast this with things like the intrinsic magnetic moment of the electron, which is known to something like 13 digits. What this sort of means is that because G is so poorly known-- well, let me just write that out in words first. So G is itself poorly known. And so when we measure the properties of various large objects using gravity, we typically find that something like G times an object's mass is measured much better than M alone. Basically, the observable that one is probing is G times M. To get M out of that, you take G times M and you divide by the value of G that you have determined independently. If you only know this guy to five or six digits, you're only going to know this guy to five or six digits. Whereas, for instance, for our sun, GM is known to about nine digits, maybe even 10 digits now. When you set both G and c to 1, what you find is that mass, time, and length all come out having the same dimension. And what that means is that certain factors of G and c can be combined to become convergent factors. So in particular, the combination over c squared, it converts a normal mass-- let's say an SI mass-- into a length. A very useful one is G times the mass of the sun over c squared is 1.47 kilometers. GM over c cubed, OK? You're going to divide by another factor of velocity. This takes mass to time. So a similar one, G mass sun over c cubed, this is 4.92 times 10 to minus 6 seconds. One more before I conclude this lecture. (I can't erase this equation, it's too beautiful!) If I do g over c to the fourth, this converts energy to length. I'm not going to give the numeric value of this, but I'm going to make a comment about this. So bear in mind that your typical component of T mu nu has the dimensional form energy per unit volume-- i.e., energy over length cubed. So if I take G over c to the fourth times T mu nu, that is going to give me a length over a length cubed-- in other words, 1 over length squared, which is exactly what you get for curvature. So when one writes the Einstein field equations, if you leave your G's and your c's in there, the correct coupling factor between your Einstein tensor and your stress energy tensor is actually 8 pi G over c to the fourth. And I just want to leave you with the observation that G is a pretty small constant, c to the fourth is a rather large constant, and so we are getting a tiny amount of curvature from a tremendous amount of stress energy. Spacetime is hard to bend. |
MIT_8962_General_Relativity_Spring_2020 | 18_Cosmology_I.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: All right. So at this point we're going to switch gears. Everything that we have done over the past several lectures has been in service of the approach to solving the Einstein field equations in which we assume a small perturbation around an exact background. Most of it was spent looking at perturbations around flat space time. A little bit in the last lecture we touched on some of the mathematics and some of the analysis when you curve-- you expand around some non-specified curved background. I didn't tell you where that curve background comes from. Today we'll be-- this lecture will be the first one in which we begin thinking about different forms of different kinds of solutions that arise from different principles. We're going to begin this by studying cosmology. It's the large-scale structure of the universe. So from the standpoint of-- from the standpoint of the calculational toolkit that we will be using, this is going to be the first example of a spacetime that we construct using a symmetry argument. We are not going to make any assumptions that anything is weak or small or any kind of approximation can be-- any kind of an approximation can be applied. What we're going to do is ask ourselves, suppose spacetime-- at least spacetime on some particular very large scales-- is restricted by various symmetries. So we will apply various restrictions to the equations and to the spacetime by the assumption that certain symmetric symmetries must hold. Let me reword this. By demanding that certain symmetries hold. Doing so will significantly reduce the complicated non-linear dynamics of the field equations of general relativity. This will allow us to reduce those complicated generic equations into something that is tractable. So let me describe-- whoops-- let's me give a little bit of background to this discussion. Let me get some better chalk, too. So as background, I'm going to give a little bit of a synopsis of some stuff that is described very nicely in the textbook by Carroll. So for background, part of what we're going to consider as we move into this is a notion of what are called maximally symmetric spaces. So I urge you to read Section 3.9 of Carroll for extensive discussion of this. But the key concept of this is that a maximally symmetric space is a space that has the largest number-- so let's say MSS, maximally symmetric space, has the largest number of allowed Killing vectors. If your space has n dimensions, it has n times n plus 1 over 2 such Killing vectors. And recall, if you do a lead derivative of the metric along the Killing vector, you get 0, OK? So it's that these n times m plus 1 Killing vectors all define ways in which, as you sort of flow along these vectors, spacetime is left unchanged. Intuitively, what these do is define a spacetime that is maximally homogeneous-- I shouldn't say spacetime yet, we haven't specified the nature of this manifold. So this defines a space that is maximally homogeneous. And homogeneous means that just-- it has uniform properties in all locations. And it is maximally isotopic. Which is a way of saying essentially that it looks the same in all directions. In the kind of spacetimes that we are familiar with, something-- a spacetime that is highly isotropic is one that is invariant with respect to rotations and boosts, and one that is homogeneous is something that is invariant with respect to translations. So let me give two examples. In Euclidean space, that is a maximally symmetric three-dimensional space, n times n plus 1 over 2 equals 6. And those 6 Killing vectors in Euclidean space correspond to 3 rotations and 3 translations. Minkowski flat spacetime: n equals 4. n times n plus 1 over 2 is equal to 10. I have 3 rotations, 3 translations, and 4 boosts, OK? The requirement that your space satisfy these properties, it leads to a condition that the Riemann tensor must be Lorentz-invariant within the local Lorentz frame. So for these two examples that I talked about, the Riemann tensor actually vanishes, and 0 is certainly Lorentz-invariant, so there's no problem there. But as I start thinking about more general classes of spacetimes, which I'm going to consider to be examples of massively symmetric spaces, they might not have vanishing Riemann tensors. But the Riemann tensors, in order to be massively symmetric, if I go into a freely-falling frame, that has to look-- everything has to look Lorentz-invariant. This leads to a condition that my Riemann tensor must take-- it is constrained to take one particular simple form. It must be R over n times n minus 1 times metric like so. This is-- so Carroll goes through this in some detail. Essentially what's going on here is this is the only way in which I am guaranteed to create a tensor that is Lorentz-invariant in a local Lorentz frame. So I go to my local Lorentz frame and I must have a form it looks like this, and this is a way of putting all my various quantities on my metric tensors together in such a way that I recover the symmetries of the Riemann tensor, and is my number of dimensions. Because it will prove useful, let me generate the Ricci tensor and the Ricci scalar from this. So R mu nu is going to be R over n times m minus 1. I'm taking the trace on indices alpha and beta. If I trace on this guy, I get n. The trace in the metric always just gives me back the number of dimensions. And when I trace on alpha and beta here, I basically just contract these two indices, and so I get the metric back over n times g mu nu. Take a further trace and you can see that that R that went into this thing is indeed nothing more than the Ricci curvature. Excuse me, the scalar Ricci curvature. OK? You can construct the Einstein tensor out of this, and what you see is that the Einstein tensor must be proportional to the metric. And in fact, there are-- the only solutions for the Einstein tensor that-- the only solution is the Einstein field equations in which the Einstein tensor is proportional to the metric are either flat spacetime or a cosmological constant. So empty space, empty flat spacetime and cosmological constant are the only maximally symmetric four-dimensional spacetimes. That does not necessarily describe our universe. So our universe, how am I going to tie all this together? We begin with the observation that our universe is, in fact, homogeneous and isotopic on large spatial scales. I emphasize spatial because the spacetime of our universe is not homogeneous, OK? In fact, the past of our universe is very different from the present. Because light travels at a finite time, when we observe two large distances, we are looking back into the distant past. And we see that the universe is a lot denser in the past than it is today. It remains the case, though, that it is still homogeneous and isotropic spatially, at least on large scales. So we are going to take advantage of these notions of maximally symmetric spaces to define a spacetime that is maximally symmetric spatially, but is not so symmetric with respect to time, OK? So we'll get to that in just a few moments. There is a wiggle word in here. I said that our universe, by observation, is homogeneous and isotropic on large spatial scales. What does large mean? Well the very largest scales that we can observe of all-- so that when we go back and we probe sort of the largest coherent structure that can be observed in our universe, we have to go all the way back to a time which is approximately 13-point-something or another-- I forget the exact number, but let's say about 13.7 billion years ago, and we see the cosmic microwave background. So the cosmic microwave background describes what our universe looked like 13.7 billion years ago or so, and what you see is that this guy is homogeneous and isotropic to about a part in 100,000, OK? With a lot of interesting physics in that deviation from-- that sort of part in 100,000 deviation, but that's a topic for a different class. We then sort of imagine you move forward in time, you look on-- so that tells you about the largest scales in the earliest times. Look at the universe on smaller scales. I mean, clearly you look in this room, OK? I'm standing here, there's a table over there, this is not homogeneous and not isotropic, things look quite different. What we start to see is things deviate from homogeneity and isotropy on scales that are on the order of several tens of megaparsecs in size, OK? Parsec, for those of you who are not astrophysicists, is a unit of measure. It's approximately 3.2 light years. So once you get down to boxes that are on the orders of 50 million light years or so on a side, you start to see deviations from homogeneity and isotropy. And this is caused by gravitational clumping. These are things like galaxy clusters. So when I talk about cosmology and I want to describe the universe as large-scale structure, I am going to be working on defining a description of spacetime that averages out over small things like clusters of galaxies, OK? So this is sort of a fun lecture in that sense, in that anything larger than an agglomeration of a couple dozen or a couple hundred galaxies, I'm going to treat that like a point. So here's what I am going to choose for my spacetime metric. This is where you start to see the power of assuming a given symmetry. So the line element, I'm going to write it as minus dt squared plus some function R squared of t gamma ij dx i dx j. The function R of t I've written down here. It's one variant of-- there's a couple functions that are going to get this name. We call this the scale factor. Caution, it's the same capital R we used for the Ricci scalar, it's not the Ricci scalar. Just a little bit of unfortunate notation, but it should be clear from context which is which when they come up. I have chosen gtt equals minus 1 and gti equals 0. Remember from our discussion of linearized theory around a flat background, that the spacetime-- the 10 independent functions of my spacetime metric, of those 10, four of them were things that I could specify by choosing a gauge. Well here, I have specified four functions pretty much by fiat. Think of this as defining the gauge that I am working in, OK? In a very similar way, I have chosen a coordinate system by specifying gtt to be minus 1 and gti to be 0. This means that I am working in what are called co-moving coordinates. So if I am an observer who is at rest in the spacetime so that I would define my four velocity like so, I will be essentially co moving with the spacetime. Whatever the spacetime is doing, I'm just going to sort of homologously track it. It's worth noting-- so those of you who think about astronomy, astrophysics, and observational cosmology, the earth is not co-moving, OK? We build our telescopes on the surface of the Earth which rotates. The Earth itself orbits around the Sun. The Sun is in a solar system. Or excuse me-- the Sun at the center of our solar system is itself orbiting our galaxy. And our galaxy is actually falling into a large cluster of galaxies called the Virgo Cluster. This basically means that when we are making cosmologically interesting measurements, we have to correct for the fact that we make measurements using a four velocity that is not a co-moving four velocity. This actually shows up in the fact that when one makes measurements, one of the most impressive places that shows up is that when you measure the cosmic microwave background, it has what we call a dipole isotropy. And that dipole is just essentially a Doppler shift that is due to the fact that when we make our measurements, we are moving with respect to the co-moving reference frame. All right. So that's the metric that we're going to use here. Setting gtt and gti like so means I have chosen these co-moving coordinate systems. I'm going to take gamma ij, I'm going to take this to be maximally symmetric, OK? So this is my statement that at any moment of time, space is maximally symmetric. So a few words on the units, a few things that I'm going to set up here. So my coordinate, I'm going to take my xi to be dimensionless, and all notions of length-- all length scales and the problem are going to be absorbed into this factor R of t. We're going to see that the overall scale of the universe is going to depend on the dynamics of that function R of t. So let's imagine that on a given-- at some given moment of time, you want to understand the curvature associated with that constant time slice. So the Riemann tensor that we build from our spatial metric, I'm going to write this as 3 R-- and I'm doing purely spatial things, so I'm going to use Latin letters for my indices. It's going to equal to some number k-- not to be confused with the index k. It's unfortunate, but there's only so many letters to work with. That looks like so. And if I take a trace to make my Ricci curvature, I get this, and your Ricci scalar will turn out to be equal to 6k. We won't actually need that, but just so you can establish what that k actually means. It's simply related to the Ricci curvature. Oops, that should have a 3 on it. Ricci curvature of that particular instant in time. Now I'm going to require my coordinate system to reflect the fact that space is isotopic. So if it's isotropic, it must look the same in all directions, and in a three-dimensional space, anything that is the same in all directions must be spherically symmetric. And so what this means is that when I compute gamma ij dx i dx j, it must be equal to some function of radius. The bar on that radius just reminds you this is meant to be a dimensionless notion of radius. Remember, all length scales are going to be absorbed into the function capital R. It's R squared d omega. And my angular sector is just related to circle coordinate angles the usual way. So it's convenient for us to put this f of R-- I can rewrite this as an exponential function, OK? Just think of this as the definition of the function beta. The reason why this is handy is that suppose I now take gamma ij, I compute the three-dimensional Christoffel symbols, I compute my three-dimensional Riemann, ignoring for a moment that my Riemann is meant to be maximally symmetric, OK? I'm just going to say, I know the recipe for how to make Riemann from a metric. I will do that. I will then make Ricci from that Riemann. When you do this, what you find is that-- let's just look at the Rr component of this thing. This turns out to be 2 over R times a regular derivative of beta, OK? It's just a little bit of tensor manipulation to do that using some of the tools, the mathematical tools I'm going to post in the 8.962 website, you can verify this yourself. If I compute Ricci from the maximally symmetric assumption, what I find is that this is equal to 2k times gamma, which is itself exponent of 2 beta. Let's equate these and solve for beta. So one side I've got 2 over r bar dr bar of eta. On the near side I have 2k and gamma r bar r bar is itself e to the 2 beta. Cancel, cancel. A little bit of algebra. First let's write it this way. Let's move this to the other side... e to the minus 2 beta is equal to this. Let us make the assumption-- so we have a choice of a boundary condition. Let's put beta equals 0 at r bar equals 0. This is basically saying that on my-- so r bar is sort of the origin. We're just sort of saying that things look like a flat spacetime in the vicinity of the origin of the coordinates we're using here, that's a fine assumption to make. Doing so, we can easily integrate this guy up, and here's what we get. So with this, we now have a full line element. So there's a few unknown quantities in here. What is k? What is R? So far I have only talked about the geometry of this spacetime. We haven't yet connected this-- any of the dynamics of the spacetime to a source. It is when we hook this up to a source that we're going to learn something about these two. So hold that thought for now. This essentially has just said that here is what my maximally spatially symmetric spacetime looks like, allowing for there to be a difference between the past and the present. Before I move on, so I can't tell you what k is yet, but I can make the following observation which allows me to restrict what values of k I need to worry about. Suppose I take k and I replace it with k prime equal alpha k. But in doing so, I define R tilde to be square root of alpha-- yeah. Square root of alpha times R bar. And I also require that my overall scale factor look like the original scale factor divided by square root of alpha. Rewriting my spacetime, my line element in terms of k prime and the tilde R into-- the two tilde R's, I get this. Basically, that transformation leaves the line element completely irrelevant to me. That was completely the wrong word. That re-prioritization of k and R and the two different R's here, that leaves the line element completely invariant. It is unchanged when I do this. So what this tells me is-- by the way, alpha has to be a positive number so that the square roots make sense there. It tells me that the normalization associated with k can be absorbed into my scale factor. And so what it suggests we ought to do is just-- you don't need to worry about whether k is equal to 15 or pi or negative the 38th root of e or anything silly like that. The only three values of k that matter for us are whether it is negative 1, 0, or 1. This stands for all negative values of k, 0 as a set onto itself, and all positive values of k, OK? So we will use this to say, great, the thing which I'm going to care about, once I start looking at the physics associated with this, is whether-- let's go back over to this version of it-- I'm going to care about whether k is negative 1, 0, 1, and I want to understand how my scale factor behaves. So before I start hooking this up to my source and doing a little bit of physics, many of you are going to do something involving cosmology at some point in your lives, and so it's useful to introduce a few other bits of notation that are commonly used here, as well as to describe some important terminology that comes up at this point. So here's some common notation and terminology. Let us define a radial coordinate chi via the following definition-- d chi will be equal to d R bar over square root of 1 minus kR bar squared. Now remember, we just decided that k can only take on one of three interesting values. I can immediately integrate this up, and I will find that my R bar is equal to sine of chi of k equals plus 1 is equal to chi of k equal 0. And it's the sinh of chi of k equals minus 1. So let's take a look at what this means with these sort of three possible choices. The three possible values that k can take. What is our line element looks like? So let's look at k equals plus 1 first. I get minus dt squared R square root of t d chi squared plus-- I'm going to use the fact that R bar, this describes a spacetime in which every spacial slice is what is called a 3-sphere, OK? You're all nicely familiar with the 2-sphere. So a 2-sphere is the three-dimensional surface in which you pick a point and every point that is, let's say, a unit radius away from that point that, defines a 2-sphere in three-dimensional space. So this defines the space-- the spatial characteristics of pick a point in four-dimensional space and ask for all of the points that are a unit distance away from it in three dimensions, that is a 3-sphere. Notice that my 3-sphere has a maximum-- there's a maximum distance associated with it, OK? So there's no bounds on chi, OK? Chi can go from 0 to infinity. But this one's periodic, isn't it? So as chi reaches pi over 2, the separation between any two points on that single slice, they've reached their maximum value, and as chi continues to increase, the distance gets smaller again, OK? And eventually, when chi gets up to pi, you come back to where you started. We call this a closed universe. This is something where if it were possible to step out of time and just run around on a spacial slice, you would find that it is a finite size. The best you could do is run around on that three-dimensional sphere in four-dimensional space. Let's do k equals 0 next. If I do k equals 0, there's my line element. Each spacial slice is simply Euclidean space. So this is often described as flat space. A significant word of caution. When you talk to a cosmologist, they will often talk about how the best-- we're going to talk about sort of the observational situation in the next lecture that I record a little bit. Our evidence actually suggests that this is what our universe looks like right now. We're in a k equal 0 universe in which space-- each spatial slice is flat. That does not mean spacetime is flat, OK? So when they say that it's flat, that is referring to the geometry only of the spatial slices in this co-moving coordinate system. k equals minus 1, you get a form that looks like this. This describes the geometry of a hyperbole. We call this an open spatial slice. So notice for both choices 2 and 3, if you could sort of step out of time and explore the full geometry of that spatial slice, it goes on forever, OK? Again, there's really no boundary on that chi as near as we can tell, and so that spatial slide can just kind of go, whee! And take off forever. This one sort of goes to large distances a little slower than this one does. This hyperbolic function means that this is really bloody large, OK? So both of these tend to imply a universe that is sort of spatially unbounded. The closed universe, because each slice is a 3-sphere, it's a different story. So another bit of notation which you should be aware of-- and I unfortunately am going to want to sort of flip back and forth between the notation I've been using so far and this one I'm about to introduce. It can be a little bit annoying when you're first learning it, but just keep track of context, it's not that hard. So what we're going to do is let's choose a particular value of the scale factor, and we will normalize things to that. So what I'm going to do is define some particular value of k such that the scale factor there I will call it R sub 0. And as we'll see, a particularly useful choice for this is to choose the value right now, OK? What we're doing, then, is we're kind of norm-- what we're going to see in a moment is this means we're normalizing all the scales associated with our universe to where they are right now. OK. Having done this, I'm going to define a of t to be R of t divided by this special value. For dimensional reasons, I'm going to need to put this into my radial coordinate. So notice, what's going on here is that my R will now have dimensions associated with it, and so essentially everything is just being scaled by that R0. And this is the bit where it gets a tiny bit unfortunate, you sort of lose the beauty of k only having three values. So I'm going to replace that with a kappa. This is unfortunately a little bit hard to read, so whenever I make it sort of with my messy cursive, it will be k; whenever it looks a little bit more like a printed thing, it will be kappa. And so kappa is k divided by R0 squared. And when you do that, your line element becomes this. OK? So that's a form that we're going to use a little bit. What's a little bit annoying about it is just that my-- the kappa that appears in there doesn't just come as a set of one of these parts of three, but basically if kappa is a negative number, then you know k must be minus 1; kappa equals 0 corresponds to k equals 0; if kappa is a positive number, then k equals plus 1. This form where we're using this sort of dimensionless scale factor a is particularly useful. If you look at this, this is telling you that with the choice that R0 defines a scale factor now, this means a now equals 1. And so this gives us a nice dimensionless factor by which we can compare all of our spatial scales at different moments in the universe to the size that they are now. OK. Everything I have said so far has really been just discussing the geometry that I'm going to use to describe the large-scale structure of spacetime. I haven't said anything about what happens when I solve the Einstein field equations and connect this geometry to physics. So what we need to do is choose a source. And so what we're going to do is we will do what is sort of the default choice in many analyses in general relativity. We will choose our source to be a perfect fluid. What's nice about this is that it automatically satisfies the requirements of isotropy and homogeneity. At least it does so if the fluid is at rest in co-moving coordinates. So let's fill this in: t mu nu with everything in the downstairs position looks like rho plus P mu nu mu nu plus Pg mu nu. And this becomes in my co-moving coordinate system. So then it looks like this, OK? A handy fact to have, this is going to be quite useful for a calculation or two that we do a little bit later-- actually, not just a minute later, almost right away. This looks like a diagonal of all this stuff, OK? All right. So what we want to do is use this stress energy tensor as the right-hand side. So we've worked out our Ricci tensor. With a little bit of work, we can make the Einstein tensor, couple it to this guy, we can set up our differential equations, and we can solve for the free functions that specify the spacetime. Before doing this, always a good sanity check, remind yourself, your fluid has to satisfy local energy conservation. Actually, let's just do the 0. So this is energy and momentum conservation, we set that equal to 0, this is local energy conservation. Expanding out these derivatives, what you find is that this turns into-- so it looks like this. And plugging in-- so using the spacetime-- by the way, I made a small mistake earlier. I should have told you that this spacetime, this is now called the Robertson-Walker spacetime. So this was actually first written down in the 1920s, and Robertson and Walker developed this basically just as I have done it here, just arguing on the basis of looking for something that is as symmetric as possible with respect to space if not time, and they came out with that line element. My apologies, I didn't mention that beforehand. This is my third lecture in a row, I'm getting a little bit tired. So if I take that Robertson-Walker metric, plug it into here to evaluate all these, this gives me a remarkably simple form. So rho is the pressure of my perfect fluid-- excuse me, the density of my perfect fluid, P is the pressure of my perfect fluid, a is my scale factor. If you like, you can put the factor of R0 back in there, and an equivalent way of writing this, which I think is very useful for giving some physical insight as to what this means-- so put that R back in. OK, so let's look at what this is saying. R cubed is modulo numerical factor, that is the volume of a spacial slice. And so this is saying, the rate of change of energy in a volume-- so a volume describing my spatial slice is equal to negative pressure times the rate of change of that volume. I hope this looks familiar. This, in somewhat more convoluted notation, is negative dp-- du equals negative P dv. It's just the first law of thermodynamics. All right. So this relationship, whether written in this form or in that form, is something that we will exploit moving forward. Let's now solve the Einstein field equations. So we'll begin with g mu nu equals 8 pi g t mu nu. The equations that are traditionally used to describe cosmology are a little bit more naturally written. If I change this into the form that uses the Ricci tensor-- so let me rewrite this as R mu nu equals 8 pi g t mu nu. OK? So this is equivalent where t is just the usual trace of the stress energy tensor. And what you find, there are two-- if you just look at the 0, 0 components of this equation, it tells you the acceleration of the scale factor a divided by a is-- it is simply related to the density and 3 times the pressure. If you evaluate Rii-- in other words, any-- this is-- there's no sum implied here, just take any spatial component of this guy, and add on R0,0 because it's a valid equation, it helps you to clear out some stuff, you get the following relationship between the rate of change to the scale factor, the density, and remember, kappa is your rescaled k. So I'm going to call this equation F1, I'm going to call this one F2. These are known as the Friedmann equations. When one uses them to solve to describe your line element, you get Friedmann-Robertson-Walker metrics. So just a little bit of nomenclature. Robertson-Walker tells you about the geometry, you then equate these guys to a source, and that gives you Friedmann-Robertson-Walker line elements. One other bit of information-- so let's introduce a little bit of terminology here. So a dot over a, this tells me how the overall length scale associated with my spatial slices is evolving as a function of time. This is denoted H and it's known as the Hubble parameter. H0 is the value of H that we measure in our universe corresponding to its expansion right now, OK? And the notes that I have scanned and placed online claim a best value for this of 73 plus or minus 3 kilometers per second per megaparsec. These notes were originally hand written about 11 or 12 years ago, that number is already out of date, OK? If you went to Adam Reese's colloquium shortly before MIT went into its COVID shutdown, you will have seen that there's actually little bit of controversy about this right now. So our best measurements of this thing, indeed, they are clustering around 72 or 73 in these units, but they're inconsistent with some other measures by which we can infer to be the-- what the Hubble parameter should be. And it's a very-- very interesting problems. Unclear sort of-- it sort of smells like something might be a little bit off in our cosmological models, but we're not quite there yet. Let's consider-- let's proceed with sort of the standard picture, and just bear in mind that this is an evolving field. The one thing I will note is that H has the dimensions of inverse time, OK? The way one actually measures it. So the dimensions in which most astronomers quote its value, it looks like a velocity over a length, which is, of course, also an inverse time. And that is because objects that are at rest with respect to the-- that are at rest in these co-moving coordinates, as this fluid is meant to be, if the universe is expanding, we see them moving away from us. OK, I'm going to make a few definitions. Let us define rho crit to be 3H squared over 8 pi j. So the way I got that was take F1-- imagine kappa is equal to 0, just ignore kappa for a second. Left-hand side as H squared, solve for rho, OK? Notice, rho quit-- rho crit is a parameter that you can measure. You can measure the Hubble parameter, I'll describe to you how that is done in my next lecture, but it's a number that can be measured. And then 3 and 8 pi are just exact numbers, g is a fundamental constant. So that's something that can be measured. Let's define omega to be any density divided by rho crit. Putting all these together, I can rewrite the first Friedmann equation. This guy can be written as omega minus 1 equals kappa over H squared a squared. Now notice, H and a, they are real numbers. H squared and a squared are positive definite. We at last can now see how the large-scale distribution of matter in our universe allows us to constrain one of the parameters that sets our Robertson-Walker line element. If omega is less than 1-- in other words, if rho is less than rho crit, then it must be the case that kappa is negative, k equals minus 1, and we have an open universe. If a omega equals 1 such that rho is exactly equal to rho crit, kappa must equal 0, k must equal 0, and we have a Euclidean spatially flat universe. If omega is greater than 1, kappa is greater than 1, k equals 1, and we have a closed universe. Clean up my handwriting a little bit here. OK, this is really interesting. This is telling us if we can determine whether the density of stuff in our universe exceeds, is equal to, or is less than that critical value, we know something pretty profound about the spatial geometry of our universe. Either it's finite, sort of simply infinite, or ridiculously infinite. Let me do a few more things before I conclude this lecture. First, this isn't that important for our purposes, but it's something that some of you students will see. A little bit of notational trickery. It's not uncommon in the literature to see people define what's called a curvature density. And what this is, is you just combine factors of kappa, g, and the scale factor in such a way that this has the dimensions of density. You can then find an omega associated with curvature to be rho with curvature over the critical density. And when you do this, F1, the first Friedmann equation, becomes simply omega plus omega curvature equals 1, OK? Just bear in mind, that is not really a density, it's just a concept-- it's a useful auxiliary concept. This is often for certain kinds of calculations, a nice constraint to bear in mind, OK? People are very interested in understanding the geometry of our universe, and this is a way of formulating it that sort of puts the term involving the k parameter or the kappa parameter on the same footing as other densities that contribute to the energy budget of our universe. OK. So the equations that we are working with here involve these-- it involves the pressure and density here. I haven't said too much about them so far. If I want to make further progress, I've got to know a little bit about the matter that fills my universe. So to make more progress, I need to choose what is called an equation of state that relates the pressure and the density to each other. What I really need is to know that my pressure is some function of the density, OK? This can be written down for just about all kinds of matter that we care about. In cosmology, one usually take and assumes that the pressure is a linear function-- it's just linearly related to the energy density. Let me emphasize that as a very restrictive form. When we finish cosmology, one of the next things we're going to talk about are spherically symmetric compact objects-- stars-- and we want to describe them as a fluid, and we'll need an equation of state to make progress there, we do not use a form like that for stars. As we'll see, though, for the kind of matter that dominates the behavior of our universe on the largest scales, this is actually a very reasonable form. So if I were to write down my thoughts on that, I would say restrictive but useful on the large scales appropriate to cosmology. Pardon me just one moment. Let's imagine that-- yeah, sorry. Let's imagine that I have a universe that's dominated by a single species of some kind of stuff, OK? So in reality, what you will generally have is a universe in which there are several different things present. So you might have a W corresponding to one form of matter, another W for a different form of matter, and you'll sort of have a superposition of them all present at one given moment. So to start, start by imagining a universe dominated by a particular what I will call a species rho i, and the pressure will be related to this by a particular W for whatever that rho happens to be. So before I even hook this up to the Friedmann equations, let's require that this form of matter respects stress energy conservation. OK, so the equation I just wrote, let me rewrite that in a slightly different form. I can divide both sides by R0 cubed. OK, that looks like so. Now it's not too hard to show using this assumed form here-- so if I plug in that my p is Wi rho i, in a line or two of algebra, you can turn this into-- and using-- well, I didn't even really do anything that sophisticated, I can just integrate up both sides. And what you see is that rho normalized to some initial time, some initial value. It is very simply related to the behavior of the scale factor. OK? But if you like, you can set a0 equal to 1, if you make that your stuff now, and this gives you a simple relationship that allows you to see how matter behaves as the large-scale structure of the universe changes. Let's look at a couple examples of how this behaves. So I'm going to call my first category matter. OK, that's pretty broad. When a cosmologist speaks of matter, generally what they are thinking of, this is stuff for which W equals 0. So this is something that is pressureless. And we talked about pressureless stuff very early in this class. This is what we call dust. So what we're talking about here is a universe that is filled with dust, which seems kind of stupid at first approximation, OK? Our universe sure as hell doesn't look like dust. But bear in mind, what we really mean about this is, go back to this pressureless. We're just referring to something that is sufficiently non-interactive, that when particles basically do not interact with each other. Our typical dust particle is going to actually be something on the scale of a galaxy. On cosmological scales, matter-matter interactions are, in fact, quite weak. So this is a very, very good description. Sort of imagine the universe is kind of a gas of galaxies and galaxy clusters, it's a pressureless gas of galaxy and galaxy clusters. When you put all this together-- so let's take a look at this form here. The density of matter looks like I'm going to set a0 equal to 1, it looks like the density now times a to the minus 3. OK? What I've done is I've just taken that evolution law there and I have plugged in Wi equals 0. What this is basically saying is that the conservation of stress energy demands that the-- excuse me-- that the density of this matter changes in such a way that the number of dust particles is constant, but their density varies as a to the minus 3. a sets all of my length scales. If I make the universe twice as big, the density will be 1/8 as large. Your second species of matter that your cosmologist often likes to worry-- or second species of stuff that your cosmologist likes to worry about is radiation. Here, just go back to Stat Mech. If you have a gas of photons, it exerts photon pressure, and that is of the form-- the radiation pressure is 1/3 of the energy density. Factors of speed of light are being omitted here. So this corresponds to a law in which-- here, let me put it this way. I should've make this an m. So this is my i equals m for matter. So my W for radiation is 1/3. And what you find in this case is that rho of radiation scales with the scale factor to the fourth power. What's going on here? Well let's imagine that the scale factor increases by a factor of 2, OK? Imagine that the number of photons is not changing. So what this is basically saying is, OK, I get my factor of 8 corresponding to the volume increasing by a factor of 8, but I have an additional factor of 2, where's that coming from? Well remember, that's an energy density. So this is saying that not only is the density being diluted by the volume growing, but each packet of energy is also getting smaller as the universe increases in size. Each quantum of radiation is redshifting. It's redshifting with a scale factor a. We are going to revisit that in the next lecture. That's an important point and we're going to re-derive that result somewhat more rigorously as we began exploring how it is that we can observational it probe the properties of our universe. Just for fun, there's another form of-- there's another kind of perfect fluid that cosmologists worry about, and that's the cosmological constant. So a cosmological constant has pressure equal to minus the density. This corresponds to an equation of state parameter W equal to minus 1. If I go to my form here, I plug in W equals minus 1, rho goes as a to the 0th power, a constant. Well, it is a cosmological constant, after all, so that shouldn't be too surprising. This is a very interesting one because it is basically telling us that the amount of energy in spacial slices-- the energy density does not change. The amount of energy appears to be Increasing. Now bear in mind, it's hard to define the total energy, we cannot really in a covariant way add up energy at various different kinds of points. Local energy is still being conserved, but there's no question that this guy is a little weird. So one of the next things that we want to do is take this stuff, run it through Einstein's equations. Einstein's equations, of course, give us the Friedmann equations. And solve to see what the expansion the universe looks like. We saw already that if the density of the universe relative the critical density is either higher, the same, or lower, that tells us about the value of this k parameter, or rather, the kappa parameter. We haven't yet seen how to solve for the scale factor a. However, we have the two Friedmann equations, and if nothing else, write them down, write out your stuff, you got yourself a system of equations, Odin gave us mathematica-- attack. To give you some intuition as to what you end up seeing when you look at these kind of solutions, let me look at the simplest kind of universes that we can solve this way. So let's examine what I will call a monospecies universe-- in other words, a universe that only contains one of these forms of matter that I have described here, one of these sources of stress energy that I described here. And for simplicity, I'm going to take it to be spatially flat. Neither of these two conditions are true in general, but they are-- they're fine for us to wrap our heads around what the characteristics of the solutions look like. So in this limit, Friedmann 1 becomes a dot over a squared equals 8 pi g rho over 3. I can borrow this form that I've got here to write this as 8 pi over 3 rho at some particular moment times scale factor to the minus n. So what I'm doing here is I'm assuming a0 now. My a right now is 1, and I'm defining n to be 3 times 1 plus W. This is easy enough to solve for. OK. Take the square root. What you find is-- I'm going to just sort of-- there's a constant you guys can work out on your own if you like. a dot must be proportional to a 1 minus n over 2, or a is proportional to t to the 2 over n. n equals 0 is a special case that we'll talk about in just a moment. If we are dealing with what we call a matter-dominated universe, well, in this sort of monospecies, spatially flat form, this would have W equal 0, n equals 3, and a scale factor that grows as t to the 2/3 power. A matter-dominated universe is one that expands, but it expands with this kind of a loss. So it slows down with time. A radiation-dominated universe, W equals 1/3, n equals 4, that is a universe that expands as the square root of t. What about my cosmological constant? Ah. OK, well that's a problem-- n equals 0, and my solution doesn't work for that one. So what you do is just-- let's just go back to our F or W equations. Or rather, our Friedmann equations. Let's write F1 down again. I have a dot over a equals 8 pi-- whoops-- 8 pi g rho over 3, and this is now a constant. Rho equals rho 0 because it's a constant. I can rewrite this in terms of the cosmological constant lambda. And so another way to write this is-- sorry about that-- a dot a-- a dot over a squared equals this, which equals lambda over 3. So this leads to an exponential solution. Now our real universe is not as simple as these three illustrative cases that I have put in here just to illustrate what the extremes look like, OK? We are not a monospecies universe, we have a mixture of matter, we have a mixture of radiation, we appear to have something that smells a lot like a cosmological constant, although the jury is still out if one is being perfectly fair. Work is ongoing. What you need to do in general is sort of model things. You try to make models of the universe that correspond to different mixtures of things that can go into it, and then you go through and ask yourself, do the observables that emerge in this universe match what we see? Generally you will see sort of trends that are similar to this that emerge, right? There might be a particular epoch where matter is more important, there might be an epoch where radiation is more important, there might be an epoch where cosmological constant is more important. And so you might see sort of you know transitions between these things where it's mostly a square root t expansion, and then something happens and the radiation becomes less important, there's an intermediate regime where both are playing a role, and then matter becomes important and it kicks over to a t to the 2/3 kind of expansion when it becomes matter-dominated. You don't want to assume the universe is flat, you need to do your analysis, including a non-zero flatness parameter in there, which makes things a little bit complicated. So in the next lecture, I am going to talk a little bit about how one extracts observables from these spacetimes. How is it that we are able to actually go into an FRW universe and measure things-- what can we measure, how can we use those measurements to learn about the energy budget of our universe and formulate cosmology as an observational and physical science? And with that, I will end this lecture. |
MIT_8962_General_Relativity_Spring_2020 | 17_Gravitational_radiation_II.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: OK. Let's move on to the next lecture. Might have gotten a little bit of me getting prepped here. I really have to give a lot of credit to the people in the Office of Digital Learning for making this entire thing possible. So I'm really grateful particularly to Elaine Mello, who is at home running the show remotely right now. This is-- a lot of people at MIT are doing a tremendous amount to keep this place functioning, and everyone really deserves a huge pat on the back. All right. So in the lecture that I just completed a few moments ago, we described-- I described how to characterize the gauge-invariant radiative degrees of freedom, which we call gravitational radiation. With gravitational radiation, we found at the end-- there's a formula that I will write down again a little bit later in this lecture, but we essentially found that it looks like certain projections of a term that involves two time derivatives of the quadrupole moment of a source. This has in recent years led to a-- well, we've now been able to directly measure this form of radiation. I'm wearing my LIGO hat in honor of the facility that first directly measured these things. And we have inaugurated an entirely new field of observational astronomy based on being able to measure those things. It's pretty exciting. I hope, though, that a few people noticed how restricted the analysis that I did is. I defined gravitational waves only in the context of linearized theory on a flat background. We haven't talked much about other solutions to the Einstein field equations yet, but you know it's coming. And I hope you can appreciate that, in fact, our universe is not described by a flat background. Many of the sources that we study, it is only weakly curved away from a flat background. So for instance, our solar system is accurately described as something that is weakly curved away from a flat background. And in that context, much of what I described carries over very cleanly. And so you essentially have-- in that case, you need to-- if you want to analyze things like gravitational waves propagating in our solar system or through a galaxy which has a spacetime metric, it's also accurately described as flat spacetime plus a weak deviation away from it. You can take advantage of the fact that your gravitational wave tends to be something that's rapidly varying and your solar system and your galaxy is something that has very slow time variation, or it's even static. So you can take advantage of the fact that there's these different time derivatives to separate terms. That actually leads me naturally to, what do we do if I want to describe the very realistic scenario of gravitational waves propagating not on a flat background, but on some kind of a curved background? Perhaps they propagate near a black hole where spacetime is very different from the spacetime of special relativity. Perhaps we need to describe them propagating across a large sector of our universe. We haven't talked about cosmology yet, but it's the subject of the next lecture I intend to record. And the spacetime that describes our universe on large scales does not look like the metric of special relativity. So how do I describe gravitational radiation, GWs, Gravitational Waves, on a non-flat background? Schematically, you can imagine that in this case, you're working with a spacetime that can be broken into a background. I'll put a little hat on it to say that this denotes my background spacetime. And then some kind of a wave field propagating on it. This background is no longer going to be that of special relativity. This could be something that varies with both space and time. This will be something that in general is small in the sense that in most of the cases, certainly all the cases we're going to study in this class, any term that involves an h times an h will be second order in a very small number, and so we can discard it. But it's still going to be challenging to figure out how I can even define what a wave means in this case. So to wrap your heads around that, think about what a local measurement might do, OK? So suppose I am in some region of spacetime. So first of all, we had a parable that we looked at in our previous lecture where we looked at geodesics in a spacetime that was flat plus a gravitational wave. And of course, geodesics are geodesics, and so we just found that they are unaccelerated relative to freefall frames. That sort of is-- you know, more or less defines what a geodesic is, so it was not too surprising. But we can then look at, for example, two relatively nearby geodesics, and we can look at how their geodesic separation changes. We can look at the behavior of light as it bounces back and forth. Something that's important to bear in mind is that when I make a measurement like that, I measure this or some quantity that is derived from this, the curvature tensor or an integrated light propagation time. We may know that the spacetime metric has two vary conceptually different contributions to it. It's got a background, and it's got a wave. How do we-- conceptually, we might see how to separate them, but how do we define some kind of a toolkit that allows us to separate? How do we define the separation between background and wave? Put it another way. Suppose what you are sensitive to is spacetime curvature. Suppose you are sensitive to things that depend on the Riemann tensor. How do you know from your measurement that you are measuring the Riemann tensor associated to the gravitational wave and not a Riemann tensor associated with a curvature-- excuse me-- associated with your background? That is an extremely important issue. If all I can measure is this, but I want to get this, what is the trick that I use to distinguish between these two? I call this a trick in my notes, but I don't like that term, actually. This is not a trick. This is a fundamentally important point when you are trying to measure some kind of a perturbation to a-- not just to spacetime, but to any kind of a field that you are studying, any kind of quantity you're studying which is itself spatially and temporally varying. What you need to do is take advantage of a separation of length and timescales. When I'm talking about general relativity and I'm talking about gravitational waves, I am going to use the fact that a gravitational wave is oscillatory. And as such, it varies on a much smaller length and timescales-- it varies on much smaller length and timescales than the background does. OK? It's useful to introduce an analogy here. Think about a water wave propagating on the ocean. In almost all cases when you look at a water wave propagating on the ocean, it's obvious what is the wave and what is the curvature in the ocean. It's actually associated with the fact that the Earth is round. So it is clear from the separation of both length scales and timescales-- not clean-- clear how to separate wave from the curvature of the Earth bending due to geological structures and things like that. A component of the curvature of the ocean that has a curvature scale of, say, 6,000 kilometers-- well, that's just the fact that the Earth is a sphere with a radius of 6,000 kilometers. OK? So you can kind of see that. But if you see something that is varying with a period of about a second that's got a wavelength of a meter or two, that has nothing to do with the curvature of the Earth. That's a wave. We want to introduce a similar concept here and apply it to our general relativistic calculation. So let's introduce two sets of length and timescales-- let's call them scales-- we will use in our analysis. So capital L and capital T, these are my long length in timescales. These describe the variation of my background. Lambda and tau are my short scales, and they describe the wavelength in the period of gravitational waves. So to put this in the context of the kind of measurements that LIGO makes, the LIGO interferometers make measurements of behavior of light moving in the spacetime near the Earth. Now, there is a component of that that varies on a scale that has to do with the curvature of spacetime near the surface of the Earth due to the Earth's mass distribution. That occurs on-- so first of all, that is practically static. It does vary somewhat because of the motion of the moon around the Earth and because of the behavior of the fluid in the Earth's core. There are small variations there which, interestingly enough, can be measured. But those tend to occur on timescales on the order of hours at the shortest. For the moon's orbit, the moon goes around in a time period of about a month. The Earth's rotating. It's turning on its axis. And so at a particular point, you're sort of casing the gravitational field of the moon as the Earth rotates under it, and so your time thing here is on the order of hours to about a day. The length scale associated with this-- you can calculate things like the Riemann curvature or tensor components near the surface of the Earth. That has the units of 1 over length squared. So take the square root and inverse it, and you're going to find that this varies on a length scale that is thousands or-- I should probably work this out before I try quoting numbers, but it's many, many, many, many, many kilometers, tens of thousands or millions of kilometers. Tens of thousands, probably. My gravitational wave, by contrast, it's got wavelengths that are on the order of-- they are also on the order of thousands of kilometers, but there is a time variation on them that, for the LIGO detectors, is on the order of seconds at most, really tenths of a second at the current sensitivity of the detectors, up to milliseconds. Way faster variation. So by looking for pieces of what I can measure that vary on these length scales and these time scales, I can separate it from effects that are varying on these long time scales and long length scales. So as I move forward in this lecture, let me emphasize that some of the things I'm going to introduce here are-- first of all, they're a little bit more advanced. You are certainly not responsible for knowing the gory details of how some of these tools I'm going to introduce are used. All right. So the point is, once I've introduced these long scales and these short scales, what we can then do is we can always remove the wave, remove the oscillation by introducing an averaging procedure. So if you imagine that you average over a timescale that is several times the wave-- excuse me-- several times the period of the wave and over a length scale that is several times the wavelength of the wave. These are taken to be shorter than the long scales, but longer than the short scales. And so when you do that-- let me put it this way. If I average-- so these angle brackets are what I'm going to define as my averaging procedure. I will make this a little bit more quantitative in a moment. On length scale l and timescale t, the background basically doesn't change. So when I average it, I just get the background back. On the other hand, my wave oscillates a couple of times. So if I average something that oscillates over a couple of cycles, I get 0. So given what I can actually measure, which is the spacetime or things related to the spacetime, this averaging procedure lets me pick out the background. The radiation is then what I get when I subtract that average bit from the field. So these are all things that, as a theorist thinking about how someone is going to go in and actually measure properties of the gravitational wave, these are tools I can use to separate one spatially and temporally varying quantity from another. Now, here I'm going to describe something that's a little bit more advanced than we need to get into, but it's important to discuss. How do you actually average a tensor like this? That's a little bit tricky. So the averaging was first made rigorous. And there have been various other mathematical formulations of this that are presented over the years, but I like this particular one because it's conceptually quite simple. It was first made rigorous by Dieter Brill and James Hartle-- Hartle is the author of a nice elementary textbook on general relativity-- in 1964. And the reference is given in the notes. And what you basically do is to define the average of a quantity like the spacetime metric, you integrate it over some proper volume in spacetime, proper 4 volume with a particular weighting function. So this is something that is defined such that if you integrate it over that same region, you get 1. And it's taken to be something that is-- you can sort of think of as a Gaussian in all four spacetime directions that is peaked around-- it's peaked at a particular location, and it has a width of l in all directions, l and t. We spent a lot of time talking about what makes a tensor a tensor. And one of the things that we discussed was the fact that when I am working in curved space time and my basis objects are functions, I can't really add a tensor here to a tensor there because they live in different tangent spaces. That's a poorly defined operation. An integral is nothing more than a sum on steroids. So how is it that I'm allowed to do this? Well, the idea here is this is not exact. OK? When one does this, you will find-- what comes out-- your average tensor modulo corrections or errors that are of order short length scale over long time length scale. This is the best you're going to be able to do. You're always going to be a little bit off when you define an averaging procedure like this because by definition, an average can't capture everything. That's why you're averaging. You're trying to actually throw away some piece of it. So this allows us to define our tensor calculus in this averaged way in a very useful framework provided with-- we just need to accept the fact that there's always going to be-- it really only makes sense in this regime where we can cleanly separate our length scales and our time scales. I emphasize this because sometimes one does an analysis, and one finds things are really weird going on. It's often useful to sort of step back and say, wait a minute. I'm trying to describe gravitational waves in a particular regime. Are these actually gravitational waves? And you sort of look at this, and you realize you've put yourself into a regime where your perturbation is varying on the same timescales and length scales as your background. And in fact, your amplitude is no longer really small. You sort of go, oh, crap. What I've done here is I've actually pushed this beyond the regime in which the radiative approximation is valid. So bear this in mind. You're always working-- you're always going to be working in this approximative form. Suppose we have now done this. We have separated length scales. So we now have cleanly and conceptually separated into background and perturbation. Now what I would like to do is just run through my exercise of computing all the quantities that I need to do to describe radiation-- really, all of my quantities that I need to do to develop and then solve the Einstein field equations. And I'm going to do so linearizing in h about this curved background rather than about the flat background. So let's develop all of our spacetime curvature tools, all the various quantities, linearizing in h around g hat. So I'm still doing linearized theory, but I am not linearizing around a flat background anymore. So to give you an idea, I'm going to do one term associated with this and a little bit of-- one term associated with this and a little bit of detail so you can just see what the complications end up looking like. So first thing you might want to do is compute your connection. Go back to your definition. That should be in your notes from a couple of lectures ago. So you insert this. Here is the definition of the connection. Let's now insert this split. So the inverse metric is going to have a form that looks very reminiscent to how we did the inverse metric in linearizing around a flat background, the exact same logic. I am raising indices using the background, so h alpha beta in the upstairs position is what I get when I raise using g hat. And you get a minus sign. Again, this is sort of like the tensor equivalent of binomial expansion of 1 over 1 plus epsilon. Pardon me just one moment. So I'm going to get two terms. So here's the first thing where I'm raising it to the inverse metric. This is going to split into two terms. Let me just write them all out for completeness here. So let's pause for just a second here. So when I multiply all of this out, I'm going to get one term that involves background inverse metric hitting all the derivatives of my background metric. I'm going to call it-- I'm going to call that Christoffel with a hat. OK? That is nothing more than the Christoffel symbol that you would have gotten if you were just working with the background metric. You're going to get another term that involves h alpha beta acting on the Christoffel symbol with all the indices in the downstairs position and constructed from the background. Then you're going to get another term that involves your background inverse hitting various derivatives of your wave or of your probation around the background. At this point-- oh, and there'll be a term of order h squared which I'm going to discard. At this point, you just have to kind of stare at this for a little bit. And when you do so, you find out something of a miracle occurs. You can write this as background plus a perturbation to the connection. I shift to the connection. And this shift-- when you stare at and you organize all the terms that appear here, this turns out-- whoops-- this turns out to look like something that is very reminiscent of the formula for the Christoffel symbols, but only acting on the h's and using covariant derivatives rather than partial derivatives. I don't have a good way to prove this other than to say, you know what? Expand out all of these covariant derivatives and then compare these terms. And this ends up being what you get when the smoke clears. I go through this one in some detail because simplifications of this sort happen at every level as we move on. What we're basically going to do is sort of brute force expand things. We'll find a term that looks like just a pure background piece, and then there'll be a whole bunch of little bits of crap that are linear in the field h. And you can gather them together and often write them in a form that's prettier, kind of like this, where it's just sort of a shift to the connection involving covariant derivatives. There's no great algorithm for actually working all these guys out. Basically, you just have a bit of labor to do to put them together. So let me write out a couple of other examples of what follows from this. Let's see. There's a point I want to get to in just a moment. I'll go through that in a bit more detail. OK. So when I take my Christoffel and I work out my Riemann tensor, I find a piece that looks just like Riemann computed solely from the background. And then I get a correction to it. And to linear order in h, this looks like when you work out-- OK. Hang on just a second. So at this point, we can now go ahead and start making things like the Ricci tensor and the Einstein tensor. I got slightly ahead of myself here because there's an important point to make at this point. So now, as we move forward, if we're assuming Einstein's general relativity, our curvature tensors end up coupling to the Einstein field equations, a couple things like the Ricci tensor and the Einstein tensor to the stress energy tensor. Accordingly, I am going to make the following simplification. Let's set the stress energy tensor equal to 0. So the analysis which I'm going to do in the remaining time in this lecture, it only describes vacuum regions of spacetime. One doesn't need to do this, and I strongly, strongly emphasize this. Doing so makes this lecture about one third as long as it would be if I did not assume this was equal to 0. So this is solely being introduced to simplify things. And what this means is that I am essentially describing gravitational waves as I-- I'm describing gravitational waves far from their source. Unfortunately, it does mean it complicates a little bit how they work in some cosmological spacetimes. But it is still sufficient for me to introduce some of the key concepts that I want to do. By assuming that t mu nu is equal to 0, what I'm going to do is I'm essentially, then, assuming that the background is equal to 0. The background spacetime arises from an Einstein tensor that is equal to 0. And it's not hard to show that this corresponds to Ricci being equal to 0, and of course, then, the Ricci scalar being equal to 0. So as I move forward here, my background Riemann tensor is nonzero, but my background Ricci tensor will be equal to 0. And I strongly emphasize that this is just a simplification that I am introducing in order to make today's analysis a little bit more tractable. So my background Ricci tensor is equal to 0. Let's compute the perturbation in my Ricci tensor. This is what I get when I trace over indices 1 and 3. I will skip over much of the algebra that is in this. But what one finds going through this sea of various definitions is that what results is a delta r mu nu. I will define this symbol in just a moment. OK. So a couple of definitions. This box operator that I'm introducing here, this is a covariance wave operator. And the trace is what I get when I contract my perturbation h with the background metric. OK? So I just want to quickly emphasize that there is nothing very profound about what has gone into this. This was all done by essentially just throwing together all the various definitions and linearizing in all the quantities that I care about. So what I would like to do now is take-- I'm going to take my Ricci tensor and my Ricci scalar. I will assemble them to make my Einstein tensor. And by requiring that that be equal to 0, I will get an equation that describes radiation as it's propagating in this curved background. But before I do this, I want to just pause for a second here and note that I'm going to get a mess when I do this. You can already sort of see I've got some interesting structures here. So here's a trace. Here's a trace. Here is a term that looks like a divergence. Here is a term that looks like a divergence. When we were doing linearize theory on a flat background spacetime, we got rid of those degrees of freedom. We sort of deduced that these were just kind of annoying for doing our analysis. And by changing gauge, we were able to rewrite the equations in a way that allowed us to get rid of them. Before I assemble this, I want to explore what changing gauge means now that I'm working on a curved background. So brief aside: let's generalize our notion of a gauge transformation. So what I'm going to do is just as when I was linearizing around a flat background spacetime, I began by introducing an infinitesimal coordinate displacement. And when I did this, I then had a matrix-- a matrix that affected this coordinate transformation, which looks like so. Just as when I-- whoops, typo. Just as when I was doing this around a flat background, I'm going to assume that these elements tend to be small. So let's apply this generalized gauge transformation to my metric. Something worth highlighting at this point-- my metric depends on space and time. It's not flat. So I have to build in the fact that it's a function of these coordinates. So this-- pardon me just one second. So now I need to find the inverse metric-- excuse me-- the inverse coordinate transformation that will essentially give me the same thing, flipping my indices around a little bit and introducing a minus sign. And when I expand this out-- again, bearing in mind I have to be careful about the coordinate that this is being evaluated at, OK?-- So I'm going to get one term here that involves my Kronecker deltas hitting this guy. Again, be careful with that argument. OK? So I'm going to get another term that involves-- I'm going to get two terms that involve my derivative of my infinitesimal displacement hitting the background. I'm going to get terms that involve the Kronecker hitting my metric perturbation. I'm going to discard terms that are quadratically small. Now, here is where we need to be a little bit careful. Notice I have shifted this guy here. I'm going to expand this with a Taylor expansion. Move this a little higher so that it's not blocking my view. OK. That looks pretty gross. We can clean it up, though. If I use the fact that-- for example, looking at one of these terms, this can be written as the covariant derivative of my infinitesimal displacement minus a connection coefficient. You can gather together a whole bunch of terms. And what you find-- something that looks remarkably similar to the form that we got. By the way, I may have forgotten to define that back here. My sincere apologies to those of you following along at home! My covariant derivatives with a hat mean this is a covariant derivative being taken with respect to the background spacetime. So just do your normal recipe for evaluating a covariant derivative, but compute all of your Christoffels using the background spacetime g hat. Carrying back over to here, what I find is that doing this infinitesimal coordinate transformation, it looks just like, why did you do all these steps properly? And this little bit of doing the Taylor expansion in the metric-- I speak from experience-- it's easy to overlook that bit. This looks just like what we did when we applied an infinitesimal coordinate transformation to a perturbation around flat spacetime. And it tells us that there is a kind of gauge that can be applied to my gravitational waves around a curved background provided I promote my derivatives of the infinitesimal coordinate generator from partials to covariance. So this defines-- if I introduce a generalized coordinate transformation, this generalizes the notion of a gauge transformation when I'm examining things like radiation on a curved background. This is my generalized gauge transformation. It will also prove useful for us to have a notion of a trace reversed perturbation. So I'm going to introduce an h bar mu nu. This will be my original h mu nu, the perturbation around my curved background. And I subtract off 1/2 of the trace like so. If you take the trace of this in the way that I defined earlier, this will give you minus little h back. So again, at this point, I'm going to skip over a couple of lines in my lecture notes because they're straightforward but a little bit tedious. What we're going to do is take my Ricci tensor, my Ricci scalar, my background spacetime. I'm going to assemble my Einstein tensor associated with this metric perturbation, but I'm going to write it in terms of the trace reverse [INAUDIBLE].. So let's write out delta g mu nu in terms of this guy. So assembling all the pieces, doing the algebra. OK. So pardon while I just write all that junk out. So this, again, I hope, reminds us of a step when we were doing perturbations around flat spacetime. We have all these terms that are sort of annoying divergences of that trace reverse metric perturbation. And so a line or two of algebra will justify this. I can change gauge. I can find a gauge in which my perturbation has no divergence if I choose these generators such that they satisfy a wave equation in which the source is the divergence of my old metric perturbation. Let's suppose we've done this. By the way, we will call this generalized Lorenz gauge. When you do this, what you find is that your metric perturbation-- oh, shoot-- is governed by the following Einstein tensor. And if you imagine that this guy is propagating in vacuum, what this tells you is that the wave satisfies something that's similar to the flat spacetime wave equation. Now, the operator is a little bit different because you have assembled it out of covariant derivatives and with a correction, which shows that your wave is actually now coupling to the curvature of spacetime. So I have a few more notes about this. It's not really important that we go through them in great detail here. Those of you who are interested, they will certainly be made available through the website. The key thing which I want to emphasize is essentially some of the key bits of-- an important piece of the technique, which is that what we are doing is using-- we are linearizing around this flat background. We have generalized our gauge transformation to allow us to simplify the mathematical structure of this equation. And then what results is an interesting correction to the usual wave equation that one sees. So I will also emphasize this is a somewhat more advanced topic. So one of the reasons why I'm eliding over some of these details is they are somewhat tedious. You don't need to know them in detail. It's good to be familiar with them and to be able to follow along here. I am going to post a couple of papers to the 8.962 website that lay out some of the foundations of this stuff. The key thing which I want to use this for is an important aspect of gravitational waves is that they carry energy. Electromagnetic waves carry energy. It shouldn't be a surprise if I have some kind of a source-- we're in a room right now. There's light shining down on me. The energy that is going on me, it heats me slightly. It is being drained from some power source somewhere else. We're all familiar with the story of electromagnetic radiation. For gravitational waves, ascertaining what the energy content of the wave actually is is a little bit more subtle. And indeed, understanding that there is an unambiguous energy carried by these waves is something that occupied a lot of the foundations of gravitational wave theory for a couple of decades. In part, this is driven by the fact that these things are so hard to measure that it was difficult to-- this is one of those things where if you could just go out and measure it, even if your theory was a little bit uncertain, you would say, well, goddammit it, it left an imprint on my detector here. It must have carried energy. And so you would know where you were going. But because there were no measurements of these things for years and years and years, no one quite knew which direction to step in. Part of what makes this complicated is the fact that when we are in general relativity, no matter what your spacetime is, you can always go into a local Lorenz frame. When you do that, spacetime is flat at a point. Furthermore, all the derivatives associated with it-- the first derivatives associated with it are 0. So it's sort of-- how the hell can I have a field h, a gravitational wave h that carries energy if I am free to change coordinates and make it equal to 0? Well, there's a couple of comments I'll make about that. So in sort of the same way-- well, let's back up for a moment. We made an analogy earlier to the electromagnetic potential when I was describing gauge transformations. So the metric is a quantity that is subject to gauge transformations, but the Riemann curvature tensor-- when I linearize around a flat background, I found that the Riemann curvature tensor was invariant to those transformations. It's a little bit more complicated when I linearize around a curved background, but it still remains the case that I can never make my curvature go away. It's reminiscent of the fact that I can choose a gauge that lets my electrostatic-- my electromagnetic potentials do any number of crazy-ass things. I must be getting tired because I'm swearing a little bit more. So I can let my electromagnetic potential do any sorts of crazy things, but my fields are the quantities that ultimately carry energy. They carry energy and momentum. So in the same way, the fact that I can get rid of the metric by going into a local Lorenz frame, that shouldn't bother us too much. That's kind of like going into a frame where-- or it's going into a gauge where I just make my potentials go away or make my potentials become static or something like that. This is basically telling me that the energy must be something that is bound up in the curvature. Furthermore, the fact that I can always make things look flat at a particular point-- I must use non-locality to actually pull out and understand the energy content. This sort of means that in a very fundamental sense, I am never going to be able to define, in a completely gauge-invariant way, the notion of energy in a gravitational wave at an event in spacetime. I may actually come up with gauges in which there is some notion of an energy like quantity that's defined at a particular point. It will not be gauge-invariant, though. I am going to need to-- if I want to really rigorously define what the energy content gravitation wave is, it's going to have to be based on something where I'm averaging over a region that is large compared to a wavelength, but small compared to the scales associated with my background. So let me sketch for you how we can understand this. And this will conclude with the derivation of a second quantity that is often called the quadrupole formula. Again turning to my electromagnetic analogy, the energy content of-- well, [INAUDIBLE] the content of the energy momentum carried by an electromagnetic wave is described by a pointing vector. And the pointing vector looks like the e field times the b field. It is quadratic in the field. In a similar way, we are going to expect the gravitational wave energy content to be something that is going to be quadratic in a field. So to do this properly, we're going to need to think about how to go in the second order in our theory, so second order in perturbations around the background. So what I'm going to do is imagine that my spacetime looks like some background plus some epsilon times h alpha beta. This h alpha beta is what we just spent the past hour or so computing. And imagine that there is some additional term j alpha beta. This epsilon that I've introduced here is just an order counting parameter. Its value is actually 1. OK? But what it does is it allows us to keep track of the order in perturbation theory to which I go. So if the gravitational waves are typically on the order of, say, 10 to the minus 22, here's my background curvature. These terms are all in the order of 1 or so. These are all in the order of 10 to the minus 22 or so. These are all in the order of 10 to the minus 44 or so. What I want to do is run this through-- and let's just focus on vacuum spacetime for now. Let's run this through the vacuum Einstein equation. So let's expand g alpha beta. Actually, let's not have too much crosstalk between indices. I'll call this g mu nu. And I'm going to require this to be equal to vacuum, so I'll set that equal to 0. What I want to do is expand this guy. That's a tremendous amount of work, but I'm going to sketch for you what the highlights of this look like. So when I expand my Einstein tensor, I'm going to get one term. That is basically just saying that my background spacetime is a vacuum solution. I'm going to get a term that looks like-- I'll call it g1 alpha beta. And this is going to depend on my first-order perturbation in the background. g1 is basically the delta g that I worked out in the first hour or so of this lecture. I will get another term that looks like the same g1, but in which my second-order term is coming along. But I'm going to get an additional term that results from non-linear coupling of h to h. I'll call that g2. So I emphasize again, this is just my ordinary Einstein equation. This is basically telling me my background satisfies the-- my background is a vacuum solution. This is the linear wave operator on a curved background that we just worked out in the first hour or so of this lecture. This is that same linear operator, but now applied to the second-order perturbation. And this is something new. It is very messy. It involves lots of terms that involve h hitting covariant derivatives of h, covariant derivatives of h hitting each other. I will post a paper by Richard Isaacson that steps through this in some detail. What we're going to now do is we require the Einstein equation to hold order by order. In other words, at every order in epsilon, this equation must work. So at order epsilon to the 0 or at order 1-- this is what I just said-- the background is a vacuum solution. Groovy. At order epsilon, this is my wave equation on my curved background. At order epsilon squared, now something new happens. Now, that is interesting. What we are seeing is that the terms which involve quadratic things with the linear perturbation, things that are quadratic in h are acting as the source to the wave equation that governs j. Let's try to make some headway on this. Let's go back to our separation of length in timescales. So let's define delta j mu nu to be j mu nu minus what I get when I average j mu nu on an intermediate length scale. So lambda-- I'll remind you that lambda and tau are sort of coming along for the ride here. This is my gw short scale. l is my background long scale. And so what I'm going to do is say, I'm going to take my j mu nu-- I don't really know too much about it quite yet. And I do know that by definition, my h only varies on the short scale. OK? So I'm going to take-- I'm going to do this averaging procedure here. Pardon me. That's an error. And so my l is my intermediate averaging scale. So I'm going to average things on the intermediate scale l here. And so this will be something that only varies on the short scale lambda. This, I will take out everything that varies in the short scale. This only varies on the long scale. So this is interesting. OK? By now doing this, let's regroup my metric. So this guy-- I'm going to first-- sorry, folks. Getting a little tired. Here are all the terms that vary on the long scale. And here are all the terms that vary on my short scale. With this idea than I am now-- so it's kind of interesting. What we see is that there is a piece of the second-order perturbation that has kind of become a correction to the background. So when I take my perturbation theory to the next order, there's no reason-- so at linear order, we define the linear order perturbation as being only the bit that varies on the short scale. But there is no reason why it has to remain the case that it only varies on the short scale when I go to the next order. And any bit that does, in fact, vary on the long scale, bearing in mind that the way we do our measurements-- we separate things and lengthen time scales-- it's going to look to us like a correction to the background. So with that in mind, let's revisit the second-order term in the Einstein equation. So here is my second-order Einstein equation. You should just assume that these also depend on the background. I'm not going to write that out explicitly. What I would like to do now is-- let's apply this averaging to this equation. So the average of this guy equals the average of this guy. Does this get us anywhere? Well, it definitely gets us somewhere with this, because this is a linear operator. OK? So when I go and-- I should be a little bit careful there. A useful trick which I forgot to state-- my apologies-- and this trick is worked out in the paper by Isaacson that I'm going to post-- is that whenever you take the average of second derivatives of various kinds of fields, it is equal to second derivatives of the average of that field plus corrections that scale as the square of the short length scale divided by the long length scale. So what that tells me is that at least up to these corrections, which the separation of length scales by default is assuming is a small number, I can take my averaging operator inside here. I can't do that trick on the right-hand side. I've not written out what this equation actually looks like here. It's a nonlinear operator, so it's going to be a lot messier. But this is something that's linear in j. And so as long as I bear in mind that I incur a small error by taking various-- by taking my averaging operation inside the differential operator, this is telling me that the wave equation-- sorry. My Einstein equation applied to this operator here is set by the quantity that I have there on the right-hand side. Now, remember, my background is itself a vacuum solution. Since my background is a vacuum solution, I can rewrite this whole thing as the Einstein tensor on my background metric plus my average quadratic correction. This is going to be equal to the average of that second-order piece. This is just the Einstein field equation with a strange source. If you look at this-- let's make the following definition. I'm going to define a stress energy tensor for gravitational waves as minus 1 over 8 pi g times the average of this operator. What this is now telling me is that as my gravitational wave propagates through spacetime, it carries, it generates a stress energy that changes the background in an amount that is quadratic in the wave's amplitude. Now, to get the details of what that operator actually is, I have a formula or two in my notes that will be scanned and made available. I want to cut to the punchline of this. The derivation of this is given in a paper by Richard Isaacson that I will be posting to the course website. So expand out this operator. Place in this the transverse traceless field. And what you find-- fill it with a transverse traceless field. That sort of simplifies a few things, and you get this remarkably simple result. The angle brackets, I'll remind you, they tell you that this quantity is only defined under the aegis of a particular averaging operation. That aside, what we now have is a quantity that tells us about how gravitational waves carry energy and momentum away from their sources. This is known as the Isaacson stress energy tensor. Let me comment, especially for students who work on things related to gravitational waves-- you will encounter many different notions of how to compute the energy, things like energy and angular momentum and things like that that are carried from a radiating source of gravitational waves. There is wonderful discussion of this point in a textbook by Poisson and Will. It's just entitled Gravity, I believe. The Isaacson stress tensor has the virtue that it is-- it makes it very clear that the waves are-- so it's a tensor. It makes it very clear that it arises out of quadratic derivatives of the wave field. And I really enjoy going through this derivation because you can see the way in which a term that is-- at least at the schematic level, you can see the way in which the second-order contribution to the Einstein tensor leads to a source that modifies your background spacetime. That's exactly what a stress energy tensor is supposed to do. For certain practical applications, it turns out to be a little bit hard to work with. And so you'll see this discussed in some detail in some of the more advanced textbooks. I'm happy to discuss this with students who are interested. For our purposes, this is not of concern. Let me just conclude this lecture by evaluating this for the solution that we worked out. So let us imagine that we go into a nearly flat region where those covariant derivatives become simple. And let's just compute one piece of the stress energy tensor. So let's look at the energy flux in a nearly flat region. So what I'm going to compute is the 0, 0 piece of this which defines how power flows through spacetime. So doing this in this region basically is going to look like the time derivative of my wave field contracted onto itself. Let's get the total amount of energy that is flowing through a large sphere by evaluating t0 0 integrated over a large sphere of radius r. And for my hij, I will plug in that quadrupole amplitude that I derived in the previous lecture. Over dots denote d by dt. So if you're interested in doing this integral yourself, it's actually not that hard, and that's a pretty-- it's a salubrious exercise. Let's put it that way. So I'll remind you that this projection tensor is built from-- built from the vectors that describe the direction of propagation. And a few useful tools to know about-- so should you choose to work this guy out, it's worth knowing that if I integrate ni nj over the sphere, this can be written in terms of just sines and cosines. These are basically direction sines-- or direction cosines. I get a simple result that depends only on the value of the indices i and j. If I integrate this over this sphere, I get 0. And any odd power of the ends multiplied together when integrated over the sphere will give me 0. And one more is necessary if you want to do this integral. If you integrate four of these buggers multiplied together, you get this object totally symmetrized on the indices i, j, k, and l. So throw it all together. Take the derivatives. Note that the only thing that depends on time is that quadrupole moment. And what you wind up with is this remarkably simple result. Whoops. This is a result that also is known as the quadrupole formula. It's worth comparing this to the dipole formula you get in electrodynamics. So the dipole formula that describes the leading electromagnetic-- the leading power in electromagnetic radiation that comes off of a source looks like what you get when you multiply two derivatives of the dipole moment with each other. The leading radiation in gravity is quadripolar. And so rather than being second derivatives, you have to take one more derivative out of this thing. There are three time derivatives of a quadrupole moment leading to this formula. So this result was actually originally derived by Albert Einstein. It's one of the first calculations that he did. He actually did it twice. He first derived something kind of like this in, I believe, 1916. And it was totally wrong. He just basically made a huge mistake. But then he did it more or less correctly in 1918. So he got this right. And it was a small-- a small error. He was off by a factor of 2, which is a simple algebra mistake. It's actually really enjoyable to count up the number of minor errors that Einstein made in some of these works. It just makes you feel a little bit better about yourself. So anyway, Einstein got this essentially right, and then it was hugely controversial for quite a few decades after that. Essentially, going back to these conceptual issues with which I began this lecture, people just grappled with the idea of, what does it mean for gravitational radiation to carry energy in the first place given that I can always go into a freely falling frame? It took a little while for those concepts to really solidify. Once they did, though, we were off to the races. And it's worth noting that if I take this formula and I apply it to a binary system-- I imagine I have two stars that are orbiting each other. I can compute the quadrupole moment and the time derivative of the quadrupole moment. I can compute the rate at which power is leaving these systems. Now, a somewhat more complicated variant of this calculation-- well, let me back up. So you go ahead, and you do that. You compute this for two stars that are orbiting each other. You will do this on a future problem set. What you find is that gravitational radiation carries energy and angular momentum away from the system, and it causes them to fall towards one another, which causes their orbital frequency to evolve in a very predictable fashion. That law that arises to very good approximation directly from this formula-- there are corrections that people have worked out over the years, but what you get just doing this quadrupole formula applied to two stars orbiting each other describes this chirp signal that I have on my hat to very good accuracy. And this is pretty much exactly what the LIDO detectors measure these days. They measure the evolution of systems that are evolving under the aegis of this formula. All right. So that is all we are going to say about gravitational radiation in 8.962 this term. As we figure out the way things are going to proceed moving forward, I will make myself available to answer some questions about this. I do want to emphasize as I conclude this lecture that many of these topics are fairly advanced. I don't expect you to be familiar with them. But I hope you understood the punchline of this. And I expect you to be able to apply some of these formulas that arise at the end. And I will conclude this lecture here. |
MIT_8962_General_Relativity_Spring_2020 | 14_Linearized_gravity_I_Principles_and_static_limit.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: All right, so in today's recorded lecture, I would like to pick up where we started-- excuse me. I'd like to pick up where we stopped last time. So I discussed the Einstein field equations in the previous two lectures. I derived them first from the method that was used by Einstein in his original work on the subject. And then I laid out the way of coming to the Einstein field equations using an action principle, using what we call the Einstein Hilbert action. Both of them lead us to this remarkably simple equation, if you think about it in terms simply of the curvature tensor. This is saying that a particular version of the curvature. You start with the Riemann tensor. You trace over two indices. You reverse the trace such that this whole thing has zero divergence. And you simply equate that to the stress energy tensor with a coupling factor with a complex constant proportionality that ensures that this recovers the Newtonian limit. The Einstein Hilbert exercise demonstrated that this is in a very quantifiable way, the simplest possible way of developing a theory of gravity in this framework. The remainder of this course is going to be dedicated to solving this equation, and exploring the properties of the solutions that arise from this. And so let me continue the discussion I began at the end of the previous lecture. We are going to find it very useful to regard this as a set of differential equations for the spacetime metric given a source. That, after all, is how we typically think of solving for fields given a particular source. And just pardon me while I make sure this is on. It is. I give you a distribution of mass. You compute the Newtonian gravitational potential for that. I give you a distribution of currents and fields. You calculate the electric and magnetic fields that arise from that. So I give you some distribution of mass and energy. You compute the spacetime that arises from that. But let's stop before we dig into this, and look at what this actually means given the mathematical equations that we have. So G alpha beta is the Einstein tensor. I construct it by taking several derivatives of the metric. I first make my Christoffel symbols. I combine those Christoffel symbols and derivatives of the Christoffel symbols to make the Riemann tensor. I hit it with another power of the metric in order to trace and get the Ricci tensor. I combine it with the trace of the Ricci tensor and the metric to get the Einstein. Schematically, I can think of G alpha beta as some very complicated linear-- excuse me, some very complicated nonlinear differential operator acting on the metric. So thinking about this is just a differential equation for the metric. The left hand side of this equation is a bit of a mess. Unfortunately, the right hand side can be a mess too. Let's think about this for a particular example. Suppose I choose as my force-- my source-- a perfect fluid. Well, then my right hand side is going to be something that involves the density and the pressure of that fluid, the fluid velocity, and then the metric. OK, so if I'm thinking about this as a differential equation for the metric, the metric is appearing under this differential operator on the left hand side, and explicitly in the source on the right hand side. Oh, and by the way, don't forget my fluid needs to be normalized. My fluid flow velocity needs to be normalized. So I have a further constraint, that the complement to the four velocity are related to each other by the spacetime metric. So if I am going to regard this as just a differential equation for the space time metric. In general, we're in for a world of pain. So as I described at the end of the previous lecture, we are going to examine how to solve this equation. In what's left of 8.962, we're going to look at three routes to solving this thing. The one that we will begin to talk about today is we solve this for what I will define a little bit more precisely in a moment as weak gravity. And what this is going to mean is that I only consider space times that are in a way that can be quantified close to flat space time. Method two will be to consider symmetric solutions. Part of the reason the general framework is so complicated is that there are in general, 10 of these coupled non-linear differential equations. When we introduce symmetries, or we consider things like static solutions or stationary solutions that don't have any time dynamics. That at least reduces the number of equations we need to worry about. They may still be coupled non-linear and complicated, but hopefully, maybe we can reduce those from 10 of these things we need to worry about to just a small number them, one, or two, or three. Makes it at least a little easier. In truth, symmetric solutions-- if you can then add techniques for perturbing around them, these turn out to be tremendously powerful. My own research career is-- uses this technique a tremendous amount. Finally, just basically say, you know what? Let's just dive in and solve this puppy. Just do a numerical solution of the whole monster, no simplifications. Eminent members of our field have dedicated entire careers to item three. I will give you an introduction to the main concepts in the last lecture. But it's not something we're going to be able to explore in much detail in this class. We're going to do one in a fair amount of detail. We will look at a couple of the most important symmetric solutions, so that you can see how these techniques work. And then in my last two lectures, I'm going to describe a little bit about what happens when you perturb some of the most interesting symmetric solutions. And we'll talk about numerical solutions for the general case. All right, so let's begin. We'll begin with choice one. Look at weak gravity, also known as linearized general relativity. So linearized general relativity is a situation in which we are only going to consider space times that are nearly flat. If I am in this situation, then I can choose coordinates, such that my space time metric is the metric of flat space time plus a tensor H-alpha beta, all of whose components-- so this notation that I'm sort of inventing here, double bars around H-alpha beta. This means the magnitude of H-alpha beta's components. These all must be much, much less than one. When you are in such a coordinate system you are in what we call nearly Lorenz coordinates. Such a coordinate system is as close to a globally inertial coordinate system as is possible to make. There are other coordinate choices we could make. So for instance, you're working in a system like that. This basically boils down to coordinates that are Cartesian like on their spatial slices. You could work in other ones. These are particularly convenient. Because for instance, if I work in a coordinate system whose spatial sector is spherical like, well, then there's going to be some components that grow very large as I go to large radius away from some-- the source of my gravitation. And this just makes my analysis quite convenient. In particular, where I'm going to take advantage of this. Whenever I come across a term that involves the perturbations squared or to a higher power, I'm just going approximate it as zero. I will always neglect terms beyond linear, hence the term linearized GR, in my analysis. Now, there are a couple of properties, before I get into how to develop weak gravity, linearized GR. I want to discuss a little bit some of the properties of spacetimes of this form. What are the particularly important properties of these? So let's consider coordinate transformations. My space time metric is a tensor, like any other. And so the usual rules pertain here, that I can change my coordinates using some matrix that relates my original coordinate system, which I've denoted without bars, to some new coordinate system that is barred. Now, recall that when we worked in flat space time, there was one category of coordinate transformations that was special. Those were the Lorentz transformations. So we are not working in flat space time. So on the face of it, we don't expect Lorentz transformations to play a particularly special role, except perhaps in the domain of a freely falling frame. But you know what? This is a nearly flat space time, so just for giggles, let's see what happens if you apply the Lorentz transformation to your nearly flat spacetime. So if I look at G mu bar nu bar as being a Lorentz transformation applied to my nearly flat metric. Well, what I get [INAUDIBLE] side is this. Now, one of the reasons why the Lorentz transformation was special in flat spacetime is that it leaves the metric of flat spacetime unchanged. And so this just maps to eta mu nu. I'm going to define what comes out here as h mu nu. I'm doing this in a fairly, I hope, obvious way. This is interesting. What this has told me is that when I apply the Lorentz transformation to my nearly flat space time, the background is unchanged. And the perturbation to the background transforms just like any tensor field would transform in flat space. Now, it should be emphasized, we are not working in flat space. We can compute curvature tensors. If we parallel transport-- If we consider two geodesics moving through the space time, we will see that parallel transport along those two geodesics, if they start out initially parallel, they do not remain parallel. So this is not flat space time. But in many ways, it's close enough that we can borrow many of the mathematical tools that were used in flat space time. In particular, we can introduce the following. Think of it as a useful fiction, which is that in this framework, we can regard the perturbation that pushes us away from flat spacetime as just an ordinary tensor field living in the manifold of special relativity, living in the eta alpha beta metric. It's worth bearing in mind in a fundamental sense, it's not. Space time is curved. h alpha beta is telling me about that curvature. But the mathematics works in such a way that you can borrow a lot of tools that we used in special relativity. And just imagine h as a tensor field living in that special relativity manifold. OK, so that's useful fact one that we want to bear in mind as we work in this nearly flat space time. Useful fact two is we want to think about what happens when we raise and lower indices. So suppose now that I'm going to regard h alpha beta as an ordinary tensor field living in this thing. I might want to know what it looks like with its indices raised. So I'm going to do what I usually do when I want to raise the indices on a tensor. I hit it with the metric. We're going to talk about what my upstairs metric looks like in just a moment. But clearly, it's going to be something that looks like the metric of flat space time plus terms that are on the order of h. Because I always drop terms of order h squared and higher, I can immediately say that this must simply be the metric of flat space time with the indices in the upstairs position acting on h. In other words, at least when I am acting on tensors that are built from the space time metric itself, I'm going to always want to raise and lower them using my flat space time, eta alpha beta. Bearing this in mind, let's carefully think about what the metric inverse actually looks like. I actually used this in one of my calculations in the previous lecture. And I said, I'm going to justify this in the next lecture. So here we are. Now we're going to justify it. So let's use this definition. So the upstairs metric is defined such that when it contracts with the downstairs metric, I get the identity back. This is the definition of the metric inverse. Working in linear theory, I know that this thing is going to be something like eta alpha beta plus a term of order h. I don't know what that term is yet, so let me just give it a new name. I'm going to call it m alpha beta. Whoops. Hopefully, the math will soon show me what this m actually is. It will be of order h, but as of yet, unknown. OK, so you know what? Let's rewrite that over here. So let's now multiply this guy out. Eta alpha beta hitting eta beta gamma. That gives me delta alpha gamma. m hitting eta. Now, remember what I just said over here. When I'm working with spacetime tensors, I always raise and lower indices using eta alpha beta. So when m alpha beta hits eta beta gamma, this is going to give me m alpha gamma. This eta hits that h. I get h, alpha upstairs, gamma downstairs. And then this guy is of order h, that guy is of order h. So additional terms of order h squared. So these guys cancel. And what I am left with is m alpha gamma equals negative h alpha gamma. I can raise my two indices, raise the gammas on both sides here. And we deduce from this that my inverse metric is eta alpha beta in the upstairs position, minus h alpha beta. At least two linear order in h. I'll just comment that what this is-- essentially, the matrix or tensor equivalent of a binomial expansion. 1 over 1 plus epsilon is approximately 1 minus epsilon. That's all this is. But this is important to do. In fact, I will just sort of remark somewhat anecdotally that when I work with graduate students on projects, where we have to do things in linearized theory, getting a sign wrong here is one of the most common mistakes that people make. All right, one final detail. So this detail, as I just labeled it, is a particularly important one. We talked about general coordinate transformations. And I immediately went in to discuss a Lorentz transformation. There's a different category of coordinate transformation that plays a very important role in understanding the physics of systems that we analyze in linearized theory. So let's consider a different kind of coordinate transformation. Let's consider a coordinate shift, which I'm going to define by x alpha prime. Really, they should be x prime alpha, but you'll see why I need to name it this way right now. So this is my original coordinate, x alpha, plus some little offset that's a function of the coordinates x beta. So think of this as suppose I have a coordinate grid that looks like this. And I have some function that says, I want to consider a different system of coordinates that maybe pushes me a little bit along away from each of these coordinate lines in a way that varies as a function of position here. This notation is kind of an abuse of the indices. I am really not trying to define a coordinate invariant relationship here. I am just trying to connect two quantities, and I'm trying to connect quantities in two specific coordinate systems. And as we'll see, even though this is a bit ugly. It works well for what we want to do. So my coordinate transformation matrix. OK, so I just take the matrix of-- I developed the Jacobian-- my matrix of partial derivatives-- of the new coordinate with respect to the old coordinate in the usual way. This is going to be my first term. It's just a Kronecker delta. And then I'm going to get a term that looks like matrix of derivatives of the function that defines my infinitesimal-- defines my shift. I just gave away what I was about to say. I'm going to require [in?] my work in these nearly Lorentz coordinates, all of these entries need to be small. These will all be much, much less than 1. And so we call this an infinitesimal coordinate transformation. We are going to need to use the inverse of this guy. And using the definition of the inverse of this, saying essentially that when I take this, and I contract-- Let me put it this way. I'll just write it out. So if I compute this. I get my Kronecker delta back. Taking advantage of the smallness of the transformation. It's not terribly hard to demonstrate that what comes out of it is this. The minus sign is the key thing which I want to emphasize here. That minus sign is very similar to the minus sign that I have here. What we're doing is, again, just kind of the matrix equivalent of expanding 1 over 1 plus epsilon for small epsilon. The reason that I am doing this is that I would now like to look at how the metric changes under this coordinate transformation. So what I'm going to do is define g mu nu in the new coordinate system. Usual operation. Let's now insert the many different definitions that we have introduced here. Notice that what I am using for my transformation matrix there is the inverse that I just wrote down. So let's fill that in. So I'm going to get a term that involves a Kronecker minus a matrix of partial derivatives. My other one gives me a nether Kronecker matrix of partial derivatives. And then finally, don't forget we are working in this nearly flat space time metric. And so I insert in my last term, eta alpha beta plus h alpha beta. So now, let's go and expand all of these terms out. My Kronecker, first, I get a term where both of the Kroneckers hit the metric of flat space time. So what I get is eta mu nu. Then I get a term which both the Kroneckers hit, the perturbation h alpha beta. Gives me h mu nu. Then, I'm going to get terms that involve these matrices of partial derivatives hitting the metric of flat space time. And what that's going to do is in keeping with our principle that when we're dealing with spacetime quantities, we raise and lower indices with eta. This is going to now give me-- pardon me just one moment-- a term that looks like partial-- everything in the downstairs position, d mu xi nu minus d nu xi mu. And then all the other terms are on the order of h times derivatives of the generators of my coordinate transmission. Small times small. These are infinitesimal squared. We are going to neglect them. Suppose that I insist that I have gone from one nearly flat spacetime to another. Bear in mind this picture. I'm just changing my representation a little bit. I've not changed the physics. So if I write this as eta mu prime nu prime, plus h mu prime nu prime. Well, I've got etas on both. The thing which is interesting is that I have generated a shift to my perturbation to the metric. Let's drop the primes for a second. And I'll just say that my nu, h mu nu is the old h mu nu minus the symmetrized combination of derivatives of-- the symmetrized combination of derivatives of infinitesimal coordinate transformation. Does this remind us of anything? This is starkly reminiscent of the way in which when we work with electromagnetic fields, I can take a potential, and shift it by the gradient of some scalar to generate a new potential. In so doing, what we find is that this leaves the fields unchanged. If you compute your Faraday tensor associated with this, it is unchanged. Similarly, we're going to write out the details of this in just a moment. When I generate the Riemann curvature from this, we find that although the metric has been tweaked a little bit by this coordinate transformation, Riemann is left unchanged. In acknowledgment of this, we call an infinitesimal coordinate transformation of this kind a gauge transformation. What the gauge transformation does is it allows us to change the metric, or change the way that we are representing our metric. And it's going to turn out to leave curvature tensors unchanged, in the same way that changing the potential and electrodynamics with a gauge transformation leaves our fields unchanged. And we're going to exploit this in exactly the same way that we exploit this in electrodynamics. We use this in electrodynamics in order to recast the equations governing our potentials into a form that is maximally convenient for whatever calculation we are doing right now. We're going to find-- and then we're going to derive this probably in about 20 minutes-- that the equations that govern h mu nu. If we leave things as general as possible, they're a bit of a mess. But by choosing the right gauge, we can simplify them, and wind up with a set of equations that are-- they cover all physical situations that matter, and that allow us to just cast things into a form that is much better for us to work with. All right. So we have now developed all of the sort of linguistics of linearized geometry that I want to use. Let's now go from linearized geometry to linearized gravity by running this through, and making some physics. What I want to do is look at the field equations in this framework. I am not going to run through every step of the next couple calculations. Doing so is a good illustration of the kind of calculation that a physicist likes to call straightforward but tedious. So I'm going to just write down what the results turned out to be. So let's run the metric through the machinery that we need to make all of our curvature tensors. OK, I'll remind you when we do this, we are linearizing. So anytime we see a term that looks like h squared, it dies. So we're only keeping things to linear order in h. So the first thing we find is the Riemann tensor turns into the following combination of partial derivatives of the metric perturbation h. In my notes, I have written out what happens when you switch from some original tensor h to a modified one using this gauge transformation. And what I show is that-- just a quick aside-- the gauge transformation generates a delta Riemann that looks like it's a whole bunch of-- let's see. Let's count them up. 1, 2, 3, 4, 5, 6, 7, 8. You have eight terms. Of course there's eight, because there's four terms here, and you get two more for each one. So you're going to wind up with eight additional terms that involve three partial derivatives of the gauge generator. So they're of the form d cubed on xi. And it's not hard to show. You just sort of look at them. They cancel in pairs. And so delta Riemann is zero. The Riemann tensor is invariant to the gauge transformation. All right, we want to take this Riemann and use it to build the Einstein tensor. Our goal here is to make the field equation in linearized coordinates. So let's start by making the Ricci tensor. So we're going to raise and lower indices in linearized theory with the flat spacetime. So when we make this guy, what we get is this. I've introduced a couple of definitions here. One of them, you've seen before. The box operator is just a flat spacetime wave operator. And h with no indices is what I get when I trace over h using the flat background spacetime. And let's do one more. Evaluating r, I get one further contraction. And this turns out to be d alpha, d mu, h alpha mu minus box of h. So we now have all the pieces we need to make the Einstein tensor. So I'm going to write out the result. And then we're going to stop and just look at it for a second. Einstein is Ricci minus 1/2 metric Ricci scalar. Keeping things to leading order in h. This becomes flat spacetime metric going into there. So when you put all these ingredients together, there's an overall prefactor of 1/2. And then there are 1, 2, 3, 4, 5, 6 terms. Let me write them out. OK. So recall at the beginning of the lecture, I pointed out that when one regards G alpha beta as just a differential operator on the spacetime metric, it's kind of a mess. Bearing in mind that what I have here is a simplified version of that, I have discarded all of the terms that are higher order in h than linear. This is already pretty much a bloody mess as it is. So you can sort of see my point there. If this were done in its full generality, it would be kind of a disaster. Now, in linearized theory, there is a bit of sleight of hand that lets us clean this up a little bit. Let me emphasize that the next few lines of calculation I'm going to write down, there's nothing profound. All I'm going to do is show a way of reorganizing the terms, which simplifies this in an important way. So what we're going to do is define the following tensor. h bar is h minus 1/2 eta alpha beta h. So this is a good point to go, well, who ordered that? Let's take the trace of this. Let's define h bar with no indices is what I get when I trace on this. That's going to be the trace of h. This would be the trace of h alpha beta, so I just get h back, minus 1/2 h times the trace of eta alpha beta. And the trace of eta alpha beta. This is what I get when I raise one index, and sum over the diagonal. That is 4. So the trace of h bar is negative the trace of h. We call h bar alpha beta the trace reversed metric perturbation. It's got exactly the same information as my original metric perturbation, but I've just redefined a couple terms in order to give it a trace that has the opposite sign of the original perturbation h. The reason why this is useful is recall the Einstein tensor is itself the trace reversed Ricci tensor. What we're going to see is that if we-- in acknowledgment that it's sort of a trace reverse thing, if I plug in a trace reverse metric perturbation, a couple of terms are going to get cleaned up. So here's how we do this. So let's now insert h bar. This guy is going to be equal to-- oops, pardon me. Insert h. This guy is h bar plus 1/2 eta alpha beta h. So just move that to the other side. All I'm doing is taking the definition, and I am moving part of it to the other side, so that I can substitute in for h. When you plug this into here, you'll see that there are certain cancellations. In particular, every term that involves the trace of h, h without any indices, is canceled out. And so what you find doing this algebra is that your Einstein tensor turns into-- that can't be right. OK. So now, my Einstein tensor has no trace of h in it. Every h that appears on its right hand side is the tensor with both of the indices. But now, it's the trace reverse version of that. This is still a bit of a mess. Now, we're going to do something that's got a little bit more-- it's not just sleight of hand. This is something that's got a little bit more of sort of the meaning of some of these manipulations that we've worked out. It's going to play a role in helping us to understand this. Notice this term involves delta mu on h mu. Excuse me, partial mu on h mu, partial mu on h mu. Partial mu and partial nu on h mu. This is the only term that does not look like a divergence. Three of the terms in my Einstein tensor look like divergences of the trace reverse metric. Wouldn't it be nice if we could eliminate them somehow? Well, if you studied gauge transformations and electrodynamics, you'll note that there's something similar that is done. You can choose a gauge, such the divergence of the vector for potential vanishes. Can we set the divergence of this guy equal to zero? So if you look at this, mu is a dummy index. This is four conditions that we are trying to set. This has to happen for mu-- well, we're going to sum over mu. Pardon me. It's going to happen for nu equal time, and for my three spaces. These are four conditions. My gauge generators, my xi nu are four free functions. That suggests that the gauge generators give me enough freedom that I can adjust my gauge such that if I start out with some original, I have an h old that is not divergence free. Perhaps I can make an h new that is. Well, let's try it. So remember, I just erased it. But in fact, I'll just write it down right now. The shift to the metric perturbation arising from the gauge transformation, it's on h. We need to look at how it affects the trace reverse stage. So if I start with my new perturbation is related to my old perturbation as follows. It's not too hard to show that your trace reversed metric perturbation. Pardon, pardon, pardon. My trace reverse perturbation transforms in almost the exact same way. I get one extra term. So now, what I want to do is look at how the divergence of this transforms. So I'm going to get one term here, d mu of this. It gives me a wave operator acting on my gauge generator. And then I get another term here that looks like-- remember, partial derivatives commute. So you can think of this as d nu of the divergence of xi, eta mu nu acting on this changes this into d nu on the convergence of xi. And I messed up the sign, my apologies. That plus sign should have come down here. These are equal but opposite. They cancel. So let me just highlight the result. So what this tells me is if I choose my gauge generators just right, I can adjust my trace reverse metric, so that it is divergence free. If I do that, then the first three terms in my Einstein tensor here vanish. And if I do that, then here is my Einstein tensor. So just as in e and m, all that you need to do is say, I'm going to change my gauge such that the following condition holds. The condition that describes going into this gauge such that the divergence of your trace reverse perturbation vanishes. This is a simple wave equation. So solutions to this are guaranteed to exist. If you sit down and you ask, can I come up with some kind of a pathological spacetime, or a pathological-- no. Imagine I'm in some original spacetime sufficiently pathological that doesn't allow me to do this. If you do that, you're going end up violating the conditions that define weak spacetime. You can't do that in linearized gravity anyway. So in practice, we can always choose the gauge that puts in linearized gravity my Einstein tensor in this form. This form is exactly analogous to the Lorentz gauge condition that is used in electrodynamics. And so we call this Lorentz gauge in linearized gravity. Once we've done that, here's what my Einstein field equations turn into. In the next lecture, we will solve this exactly. See, what you want to emphasize is this is one of those situations where the answer is so easy and simple for us all to work out, we don't actually really need to even do that much calculation. I'll remind you that in electrodynamics, if you work in Lorentz gauge of electrodynamics, the wave equation that governs the electromagnetic potential turns out to be-- could be factors of c and things like that, depending on which units you're working in. But we find an equation that has exactly the same mathematical structure. Possibly there's a plus sign. I should have looked that up. Wave operator on my vector potential is a source. And this is very easily solved using what's called a radiative Green's function. I will discuss this in the next lecture. You can look up the details in any advanced electrodynamics textbook. Jackson has very nice discussion of this. I have an extra index. I have a different coefficient. But the mathematical structure is identical. So as far as linearized theory is concerned, we're basically done. So I'm going to talk about the exact solution of this in the next lecture. To wrap up today's lecture, to wrap up this current lecture. Let me look at the solution of this in a particular limit. So I'm going to take my source to be a static, non-relativistic, perfect fluid. The fact that it is static means that all of my time derivatives will be zero. And if that's true for my source, it has to be true for the field that arises from it. Non-relativistic tells me that the fluid density greatly exceeds its pressure. And as a consequence, I can write my stress energy tensor as approximately density four velocity four velocity. And when you go and you look at the magnitude of these things, I sort of looked at this a little bit in the previous lecture. T00 is approximately rho. All others will be negligible. Probably there's a small correction to this, but we can neglect that on a first pass. So my field equation is dominated by the zero zero component. That's going to be the most important piece of this. Since this is static, I can immediately say-- I can change that wave operator into the plus operator. And we now notice this is exactly the equation governing the Newtonian gravitational potential, modulo a factor of four. Pardon me, factor of minus four. And so we see from this that h bar zero zero is just negative 4 times the Newtonian gravitational potential. At this order in the calculation, all other contributions to the trace reverse metric are zero. OK, so let's go from the trace reverse metric back to the metric. We use the fact that h mu nu is the trace reversed h mu nu. If we trace reverse it, we'll get the original metric back. Basically, we trace reverse twice. So the trace of this guy. OK, and so putting all these ingredients together, what we see are that the only non-zero contributions here are 8 0 0. Let's do this one carefully. This is minus 4 times Newtonian potential minus 1/2 times 8 0 0 and 4 times Newtonian potential. I have a c of minus sign here. This turns into minus 2 phi n. And h1 1 equals h2 2. h equals h3 3. This is going to be 0 minus 1/2 times 1 times 4 phi n. We put all these together. And what we get is-- And I'll remind you. This is a metric that I quoted in a previous lecture that I said we would prove in an upcoming one. Well, here it is. This is the Newtonian limit of general relativity. And it's worth remarking that this thing-- we are now in the very first lecture after having derived the Einstein field equations. 20 years ago, almost all laboratory tests, laboratory and astronomical observational tests of general relativity essentially came from this based on. This ends up being the foundation of gravitational lensing. This is used to look at post-Newtonian corrections in our solar system. To a good approximation, it describes a tremendous number of binary systems that we see in our galaxy and in a few other galaxies. You really need to look for a much more extreme systems before the way in which the analysis changes due to going beyond linear order starts to become important. There is an upcoming homework exercise. And for students taking this course in spring of 2020, it remains to be determined how we are going to do problem sets at this point. I will be making a decision on that in coming days. But I want to tell you about an exercise on P set number 7, in which you do a variation of this calculation. So instead of just having a static body with a body who has massive density rho, consider a rotating body. And the thing which is interesting here is that in general relativity, all forms, all fluxes of energy and momentum contribute to gravity through the stress energy tensor. So if I have a body that is rotating about an axis, there's a mass flow. There are mass currents that arise. And what you find if you do this calculation correctly is that there is a correction to the spacetime that enters into here, which reflects the fact that a rotating body generates a unique contribution to the gravity that is manifested in this space time. Now, when one looks at the behavior of a body in a spacetime like the one I've written down right here. It's very reminiscent of-- well, it's just like Newtonian gravity. It's the Newtonian limit. And Newtonian gravity looks a lot like the Coulomb electric attraction. So this is often called a gravito electric field. People use that term, particularly when they're talking about linearized general relativity. If I have a rotating body, I now have mass currents flowing in this thing. And the correction to the spacetime that arises from this, it's qualitatively quite different from us. It doesn't have that simple gravito electric Coulombic type of form. It, in fact, looks a lot more like a magnetic field. And in fact, when you ask, how does this new term that is generated affect the motion of bodies? You find something that looks a lot like the magnetic Lorentz force law describing its motion. So this is a very, very powerful tool. But it's already not enough. So we can go a lot further than this. We have done so far, the simplest possible thing that we can do with this toolkit that we have derived so far. In the next lecture that I will record, we're going to return to my linearized Einstein field equations. And I am going to explore general solutions of this. This is going to lead us into a discussion of how things behave when my gravitational source is dynamic. I do not want to lose the time derivatives that are present in that wave operator. And so this is going to lead us quite naturally, then, to a discussion of gravitational radiation. And so after the next lecture, we'll spend a lecture or two discussing the nature of gravitational radiation. And there will be an upcoming homework assignment or two in which you explore the properties of gravitational radiation. |
MIT_8962_General_Relativity_Spring_2020 | 21_Spherical_compact_sources_II.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: So we'll pick up where we ended last time. We're looking at the spacetime of a compact spherical body, working in what we call Schwarzschild coordinates. We deduce that the line element describing this body is of the form ds squared equals negative e to the 2 phi dt squared plus tr squared divided by 1 minus 2g m of r over r plus r squared d omega, where d omega is the usual solid element, line element. And let's see. Yes, the body is described, interior, as a perfect fluid with particular density profile rho of r, a pressure profile p of r. That describes this thing. Everywhere inside some radius r star, which gives its surface. Sorry, I just distracted myself. I don't know why I've been calling this d omega. Should be d omega squared. Whatever. Fluff in that notation. OK, so in the exterior of this thing-- so for everywhere for r greater than our star, it is vacuum. There is no density. There is no pressure. Once we're outside this thing, the only mass you see is the mass of the star. So another way of saying this is that when-- you know what? I'll get to that in just a moment. And in the exterior, the e to the 2, 5 becomes 1 minus 2g, that same mass over r. Everywhere in the interior, the pressure, the function phi, and the mass r are governed by these equations. So the mass, as I described last time, it looks like a deceptively simple spherical integral. But just be aware that when you do this, you are not integrating over a proper volume. If you were to integrate over a proper volume, you would get a larger mass, and such a mass, in fact, does have meaning to it. And the difference between that mass and this mass tells you something about how gravitationally bound this object is. And these two equations actually have a Newtonian limit associated with them. So this is a relativistic version of the equation of hydrostatic equilibrium, and this has a simple Newtonian analog. Did they drop the minus sign? No. This has a simple Newtonian analog, describing the gravitational potential inside a fluid object. These whole things taken together are called the Tolman-Oppenheimer-Volkoff equations, or the TOV equations. So the comment I was making was that we require m of r star to equal m, the total mass of this object. So that's another condition on this. Once you integrate up, then you switch over to the mass, the total mass, that is used in the exterior of the star. So if you wish to solve these things, what you basically do is just choose a central density. You have to choose an equation of state, which allows you to relate the pressure to the density. We'll talk about that a little bit more later in this class. And then you just start integrating. So you basically then integrate until you find that the pressure equals 0. The radius at which this occurs defines the star surface. As you integrate along, this allows you to build up. You then build the mass profile of the star, the pressure profile of the star. You are building up how the phi changes as you integrate along the star. So note, if you integrate this up, this is only defined up to a constant of integration. And so what you then need to do is once you have reach the surface of the star, you know m of r. And so what you're going to need to do is adjust the phi that you found in order to match with the exterior solution 1 minus 2gm over r. This is an exercise you will do on an upcoming homework assignment. This is one of my favorite assignments in the class. It's a really good chance to actually see the way we solve these equations. And this kind of an exercise, it's done all the time as part of modern research, and you will do this for a particularly simple kind of equation of state. So I want to look at what some examples of objects like this actually look like. So what we're going to do is consider an unrealistic but instructive idealized limit. So imagine a star that has rho equal constant. That is something where no matter how hard you squeeze it, you cannot change its density. The only way that that can happen is if you have an object that is infinitely stiff. This corresponds to a speed of sound, which is defined as dpd rho that is infinite. Now, of course, no speed of sound can actually exceed the speed of light. So this is a somewhat pathological object. Nonetheless, it's useful for us for the simple reason that the mass function that emerges for this is quite trivial. Rho is constant, so m of r just goes as the volume inside radius r times rho. The star then has a total mass of 4/3 pi r, to be pi rho r star cubed. So we know how its density behaves, it's a constant, we know how the mass function behaves. The challenge is solving for the pressure profile. So the pressure is governed by taking this differential equation and basically plug in that mass, plug in this mass function, and see what you get. So if I go and I throw this guy in here, this is what you end up getting. So bear in mind as you go through this-- yeah, I see what I did here. So what I do as I just pulled out my factor of 4/3, 4/3 pi rho r cubed-- excuse me, my 4/3 pi r cubed. I factor that out, and then cancel it an overall factor of r squared in the denominator. So this is simply what I have over here with this mass function defined. We want to solve this to find p. This is one of those rare moments where nature and analysis conspire and a miracle occurs. OK, it's a somewhat messy looking kind of solution, but nonetheless, it turns out that the pressure profile is determined by-- you can manipulate this equation. You can integrate it up. And this ends up describing what the pressure profile looks like, p sub c is the pressure at r equals 0. We define the surface of the star as being the location at which the pressure goes to 0. So by using that, so by exploiting that condition, we can use this equation to make a mapping between the radius of the star and the central pressure. So it's a spherically symmetric thing. Once we have chosen what this density is, then there is essentially a one parameter family of solutions. We-- if we choose a central pressure, that determines what the radius would be. Conversely, if we want to have a particular radius, that determines what the central pressure must be. Let's just do a little of analysis that follows from this. So p equals 0 at r equals r star. That defines the surface. So putting all that together, you can manipulate this equation to find that the radius of the star is determined from the central pressure, like so. If you prefer, you can write this as an equation for the central pressure in terms of the radius of the star. I've written this in terms of m tot but m tot is, of course, simply related to this radius-- excuse me, to this density and the star's radius cubed. So what's the importance of this? This is, as I've emphasized, a fairly idealized problem. Nature will never give us an object that has constant density. Any object, if you give it a little bit of a squeeze, the density will change. Causality requires that the speed of sound be less than the speed of light. So clearly, this is a somewhat fictional limit. But we can learn something very interesting about this. Notice that this formula for the central pressure, it diverges for a particular compactness of the star. So p usb c goes to infinity. Perhaps a little bit more easily to see, the denominator goes to 0, for a certain compactness, m over r. So let's look at the value at which the denominator goes to 0. So let's see. Move my 1 to the other side. Divide by 3. Square it. Rearrange terms. So if the ratio of this star's total mass to its radius is such that gm over r star exceeds 4/9 than the pressure diverges. What this tells me is that I cannot construct a physically allowable static object, even using this stuff as we can imagine fluid, a fluid that has an infinite sound speed. I cannot make a star more compact than that. Making it that compact requires infinite pressure at the core. This implies that stars have a maximum compactness. We cannot have physically realizable pressure profiles if-- and let's turn that around-- if the ratio of the radius to g times the total mass is smaller than 9/4. Now, this holds for the stiffest possible fluid that we can even imagine, one which the laws of physics actually do not permit. And so one infers from this, actually, that this bound basically tells me that given any physical fluid, any physically realizable star that I can construct, I must have a maximum allowed compactness. Putting this a little bit more precisely brings us to a result that is known as Buchdahl's theorem. Buchdahl's theorem tells me that there is no stable spherical fluid configuration in which the configuration's radius is smaller than 9/4 of gm total. For those of you who like to put factors of c in there, divide by c squared, and this tells you something that you can convert to SI units, which tells you how small you are allowed to make an object. So this is clear. It emerges, and very nicely, in this idealized but unphysical fluid limit. But it can be proven more generally. You just have to make a few assumptions about the way that the pressure profile is not singular in any place. Proof of this can be found in the beautiful, old textbook by Weinberg, section 11.6. Well, when you hear something like that, you gotta think to yourself, well, suppose I made a star with some kind of a fluid and I gave it a radius of 10/4 gm total over c squared. And I just came along and squeezed it. What would happen? Well, notice the word "stable" in this definition. You're free to do that, and should you do so, you would simply no longer have a stable object. So remember, part of what went into this analysis is we were assuming the spacetime and the fluid that is the source of the spacetime, we were assuming everything is static. OK, we're making sure everything just sits still. This is telling us you can't do that if you want to have a star as compact as that. So if you were to do this, it would collapse. It would become dynamical and the spacetime would transition into something else. What that something else might be will be a topic that we get into a little bit more after we have developed some additional material. So let's talk a little bit. This will help you with the homework assignment, where you guys are going to construct relativistic stellar models. Let's talk a little bit about how we describe real objects. They are not of constant density, and they instead have some p that is a function of the density. It's worth noting that in an even more general case-- this is actually worth a brief aside-- the equation of state relates the pressure to local density and s, where s is the entropy. For the kind of applications where general relativity tends to be important, for instance, when we're studying the stellar structure of a neutron star, the fluid ends up being so cold that you don't need to worry about the entropy. And where that comes from is that when I revisit my first law of thermodynamics and I include temperature and entropy effects, so the term that I left out in my earlier accounting, TDS. If t is small, I can ignore that term, and it ends up being something where my local energy, or energy density, ends up only depending on the pressure. But "cold" is a wiggle word. I have to define a scale to say whether an object is cold or hot. There's a few notes laying this out right in my notes. Let me just sketch the key idea. Cold depends on the fluids, and it's worth noting that the kind of fluid you're playing with here tend to be made out of fermions. And so it depends on the fluids' Fermi temperature. So the Fermi temperature is defined as the Fermi energy normalized to a Boltzmann factor. You get the Fermi energy by looking at how the energy levels are filled in your fermion fluid here. I have a few additional notes to lay out the way in which you can relate this to the local density, the mass of each particle that goes into this Fermi fluid. The punchline is that for neutron stars, one of the cases where we do, in fact, make dense general relativistic fluid stars, the Fermi temperature tends to be on the order of 10 to the 13 Kelvin, so about 10 trillion Kelvin. When we actually observe these objects, they are on the order of 10 to the 6 to 10 to the 9 Kelvin. So they're a factor of about 10 of the 7, 10 to the 4 to 10 to the 7 times colder than the Fermi temperature. Even though they may be a billion Kelvin, they are cold. So we're going to use what are called cold equations of state to describe these guys. So with that out of the way, let's talk a little bit about the kinds of equations of state that we will tend to use. So people who study the physics of dense matter, a lot of their lives is really down to understanding what the equation of state of that cold matter looks like. Some of them are concerned about hot matter as well, in which case they might be actually worrying about things at tens of trillions of Kelvin. But if you're looking at astrophysical applications, you're generally interested in the cold matter. And so they end up putting the other very complicated models using QCD and effective field theories to try to understand how it is that a particular fluid of dense matter, how its pressure and its density are related. And what I'm sort of wheeling around here is that you generally do not have a simple analytic form. You wind up with some kind of a fairly complicated function that emerges from a numerical calculation. It often ends up being-- if you are a user of this equation of state-- it ends up being in the form of a table. So they might actually just give you a file that's got a bunch of numbers, which says if the density is this, then the pressure is this. And you can fit little functions to that that allow you to look things up and do your calculations. But it's not in the form of a clean thing that you can write down on the blackboard. I want something clean I can write down the blackboard. So I'm going to introduce an approximation, which is useful for testing things out, test cases, and for pedagogy. What we do is we take the pressure to be a power law of the density. So what we do is we write p equals k rho 0 to the gamma, where k and gamma are constants. A form that looks like this, this is called a polytrope. Now, the thing which I particularly want to highlight, and for those of you who are going to do this highly recommended homework exercise, please pay attention at this point. This rho 0 is not-- oh, I erased it. This rho 0 is not the rho-- oh, there it is-- it's not the rho that appears in the equation of state. It's slightly different. Rho 0 is not the rho that appears, for instance, in the t of e equations. Rho 0 is what is called the rest mass density. It does not take into account the fact that if I take a big-- let's say I've got a big bucket of nuclear fluid. So I take my bucket here and I squeeze down on it. When I squeeze down, its density is going to increase, first of all, because I have decreased the volume. So the number of particles remains fixed, but I decrease the space in there. But I have also done work on it, because this thing exerts a pressure that opposes my squeezing. And I need to take into account the fact that the work I do in squeezing this fluid increases the density rho. So when you write out your t of e equations, rho is energy density. All forms of energy gravitate. This is just the way people traditionally write the equation of state. This is when one is doing nuclear physics. There's good reasons for doing this, but it's not the most convenient form for the kind of calculations that we want to do, and that you are going to want to do, in the problem set. Fortunately, it's not too difficult to convert, so let me describe to you how you do that. So we are going to use the first law of thermodynamics in a form in which I've written now a couple times. Interestingly, it showed up in our cosmology lecture. So my first law tells me du equals minus pressure dv. So this is my total energy in a fluid element and this is the work done on a fluid element. So rho is equal to the amount of energy in a fiducial volume. My rest energy-- excuse me, my rest density-- is the rest energy of every little body that goes into this per unit volume. This means that I can write du as d rho over rho 0, provided I throw in an extra factor of m rest to get the dimensions right. And I can write d volume as d1 over rho 0, provided I throw in that factor of m rest to get the dimensions right. I know this looks weird but, it's perfectly valid. So I'm going to rewrite my first law of thermodynamics as d rho over rho 0 equals minus pd 1 over r 0. Let's manipulate that right-hand side. So I'm going to assume this polytropic form. I'm going to use p equals k rho 0 to the gamma. But I'm going to switch that around. I'm going to write this as rho 0 equals p over k to the power of 1 over gamma. So when I do that, I get d rho over rho 0 equals kappa 1 over gamma over kappa pdp over p to the 1 plus 1 over gamma. Pardon me just one moment. I did something clever in my notes here and I'm just trying to make sure I understand what the hell I actually did. So I'm going to level with you. I've gone through this several times. There's a step in the calculation that at this point, I, for some stupid reason, didn't write down. I'm going to trust I knew what I was doing, though, because I know the final result was right. You can integrate up both sides here. Oh, I think I see what I did. OK. So you integrate up both sides here. And what you find is this becomes rho equals p over gamma minus 1 plus a constant. Yeah, not 100% sure how I actually did that, so my apologies on that. I'm going to assume I knew what I was doing. I will try to fix this and I may post an addendum here. The next step, actually, is you want to determine what that constant is. So the way you determine the constant is you take advantage of the fact that rho goes to rho 0. The energy density becomes the rest energy density if there is no pressure exerted. And so this gives us our final relationship here, which is that rho equals rho 0 plus p over gamma minus 1. OK, I will double check how I went from line 2 line 3 there, but the final thing that I have boxed online for is, indeed, exactly what you need to do in order to build a stellar model. I guess I've been emphasizing this is something you will do on an upcoming problem set. Let me just sketch the recipe. I've said this verbally, but let me just write it out explicitly here. So I will give you an equation of state. You then need to pick rho 0 at r equals 0. Using your equation of state and using that relationship between rho and rho 0, this will give you rho at the center, pressure at the center. Set m of r at the center to 0. I emphasize, again, that this may seem obvious, but it is somewhat important that you get it right. When you do this homework assignment, I'll give you a little hint as to how to build that in smoothly. It can be a little bit-- I don't want to say tricky, but it's worth thinking about a little bit. Then what you do is integrate your equations for the pressure in the mass from r equals 0. And this cannot be done analytically. You have to use a numerical integrator. If you have never used one of these before, I will give you a Mathematica notebook that demonstrates how to use it. This is a skill that is worth knowing. The plain truth of the matter is that the class of problems that are amenable to purely analytic solutions, those are interesting. They're illustrative. They're good to work with. But they tend to be unphysical and they're just not the ones that are of interest for many things that we study in science. So show while you're doing this-- this is not necessary to make your model, but it's very useful to do this. You can also integrate, whoops, d5 er from the center. So a caution is that you do not know phi at r equals 0. So what you should do is just temporarily set it equal to 0. And what you're going to be doing then when you integrate this up is you will calculate the delta phi that describes your model from the center to the surface, which brings me to step 4. When you find p equals 0, you've hit the surface. So what we do is we use the fact that p of r equals 0 defines the star's radius r star. Once you've done that, you now know the total mass and the radius. So you will find, when you're doing this, that your numerical integrator is not super well-behaved as you approach the surface. This is a feature, not a bug. What's going on is that as you begin to approach the surface, the gradient and the pressure gets quite steep. And so the way one numerically integrates a set of couple equations like this is by, essentially, if you take advantage of the fact that an integral, it's what you get by sort of dividing things up into tiny little pieces and add up like little rectangles. And when you're solving a differential equation like this, you're essentially taking the continuum solution-- that you guys have learned how to do in many cases-- and you're approximating it by a series of smaller and smaller finite steps. Because the gradient in the pressure gets large as you approach the surface, numerical integrators typically try taking an infinite number of infinitesimal steps, which makes the CPU sad, and so it's likely to exit with an error condition. Generally, when that has happened, you've gotten an answer that's probably good to within a part in a million, or something like that. Fine for our purposes. If you need to do something a little bit more careful, that's a subject for a numerical analysis class. For us, I will give you some hints on this when you begin exploring these solutions. So you have an additional boundary condition. You know by Birkhoff's theorem that the Schwarzschild metric describes the exterior. That means gtt is minus 1 minus, given by this for everywhere greater than r star. This gives us a boundary condition that phi of r star must be 1/2 log 1 minus 2m total over r star. By enforcing this boundary condition, you can go back to your solution for phi and you can figure out what the value at r equals 0 should have been to give you a continuous function that matches at the surface. So that's it for spherical stars. I look forward to you doing these exercises. My own biases are perhaps coming out, but these are a lot of fun. The one thing which I will do for you, and I regret that my notes didn't really have this, is I will try to figure out how on earth I went from line 2 to line 3 in this calculation over here, going from the rest mass density, the rest energy density to the energy density. My apologies that that's not there. All I can say is that there are many distractions these days and I overlooked that when I was reviewing my notes and preparing for today's lectures. I'd like to take this moment to take a little bit of a detour. Let's imagine that we have a spacetime that is Schwarzschild everywhere. In other words, it has this form for all r, not simply the exterior of some object. We already know that this spacetime is a vacuum solution. I know that t mu nu equals 0. Back up for a second. If I generate the Einstein tensor for this, I will get identically 0, which implies that this corresponds to a solution, which has t mu equals 0. I also know, though, that if I examine the behavior of radial geodesics in the weak field of this spacetime, I find that they fall towards this like an object that is falling towards a mass m. So this spacetime appears to be something that is everywhere vacuum. There is nothing in this spacetime, and that nothing has a mass of m. I hope that bothers you. That is among the sillier things that has been said in the name of physics. That sure sounds silly. But let me remind you that we, in fact, have seen something very similar in a much less complicated theory of physics. So if I look at the electric field of a point charge at the origin-- so that's the three vector e is just q displacement factor over r cubed. If I compute the divergence of this electric field-- the divergence, of course, tells me about the charge density-- and I get 0. So this is an electric field that has no charge density anywhere, but that lack of charge density has a total charge of q. This was something that we easily learn how to resolve. Usually at the MIT curriculum, this often shows up when you take a course like 8.07. What we do is we say, oh, all that's going on here is that I have a singular point charge at r equals 0. So yeah, I've got no charge density, but I have a total charge. Fine. We were happy with that. I want you to think of the Schwarzschild metric as doing something similar for gravity. There is no source anywhere, but there is mass. Maybe there's just something singular and a little funny going on at r equals 0. You might be concerned about what's happening there at r equal 0. When I say it plays a similar role, it plays a similar role to the pull on point charge. So there'll be nothing there, but perhaps there's something funny going on at r equals 0. And by the way, the field equations that govern gravity, my relativistic theory of gravity, they're non-linear. So when I say there's something funny going on at r equals 0, it could be really funny. So we're not going to get too worked up about that, but we're just going to bear in mind this is odd. t mu equals 0 but it has mass. So let's look at the spacetime itself. Just staring at this, we can see two radii where it appears something odd is going on. So you can see right away, lots of stuff kind of blows up and behaves badly at r equals 0. And you can also see that your gtt and your grr, they are behaving in a way that is potentially problematic when the radius is 2gm. So you look at that and think, yeah, there's two radii there that look sick. I am worried about this spacetime. Well, we should be cautious. One of the parables that we learned about when we studied linearized gravity is that we can sometimes put ourselves into a coordinate system that confuses us. When we study linearized gravity, we found a solution that looked everywhere. It looked like the entire spacetime metric was radiated. And it turned out only two of those 10 components were radiative. That turned out to be something that we were able to cure by introducing a gauge transformation. Doing that here's a little bit trickier, but we're going to need to think about, how can I more clearly call out the physical content of this spacetime? So one of the lessons that I hope has been imparted in this class so far is that if you really want understand the nature of gravity, you want to go from the metric to the curvature. So what I'm going to do is assemble an invariant scalar from my curvature. And I'm going to use the Riemann tensor because I know Ricci vanishes in this spacetime, so that wouldn't give me anything interesting. So what I'm going to do is assemble an object. I'm going to call it capital I. And that's just Riemann contracted into Riemann. This actually has a name. It is known as the Kretschmann scalar. And you can go in. You can work out all these components. The gr tool that is posted to the 8.962 website is something you can explore with us. And this is just a number. Turns out to be 48 g squared m squared over r to the sixth. What does this guy mean? Well, in an invariant way, it's kind of Riemann squared. Riemann tells me to go back and think about things like geodesic deviation. It tells me about the strength of tides. So roughly speaking, square root I is an invariant way of characterizing tidal forces. So if you're sitting around in the Schwarzschild spacetime and you want to give yourself an estimate of what kind of tidal forces are likely to act on you, compute the Kretschmann scalar, take its square root, and that'll give you an idea of how strong they typically tend to be. So notice, when we look at this, this tells us r equals 2gm. If you plug r equals 2gm in there, nothing special about it. It's a radius just like any other. As you go from 2.001gm gm to 1.99999gm, it increases a little bit. Of course, it's got the 1 over r to the sixth behavior, but it's not like there's a sudden transition, or anything particularly special happens right at that radius. But it is hella singular at r equals 0. So sure enough, r equals 0 is a place where tidal forces blow up. OK, fine. We're going to need to do a little bit more work then, because I still want understand, yeah, OK, r equals 2gm. There's no diverging tidal forces there, but that metric still looks wacky at that point. So what is going on there? So let's think about the geometry of the spacetime in the vicinity of 2gm. So let's imagine. Let's do the following exercise. Suppose I draw a circle at some radius r that's in the theta equals pi over 2 plane. So I'm just sweeping around in phi. I'm making this like so. So here is my r cosine phi axis. Here is my r sine phi axis. Here is my circle of radius r. And let's ask, what is the surface area that this guy sweeps out as an advance forward in time? So as this thing goes forward in time, it sort of sweeps out a cylinder in a spacetime diagram. Let's compute the proper area associated with this cylinder that this circle is sweeping out as it moves forward in time. So the surface area of my tube, I integrate from some start time to some end time. I'm going to integrate around in phi, and then the proper area element that I need to do this is going to be gtt g5 phi, the 1/2. There's actually a minus sign in there to get the sign right. Let's write it like this. To remind you how I do this, think of this area element as a 2 volume. Go back to some of our earlier discussion of defining integrals in spacetime, and this is the proper area associated with a figure that has some extent in time and extent in angle. So let's compute that guy. So I take my Schwarzschild metric. g5 phi in the theta equals pi over 2 plane is just r. gtt is the square root of 1 minus 2gm over r. So the area of my tube is going to be r, integrate from my start time to my end time, dt. So this is easy. So I get a 2 pi 2 pi r square root 1 minus 2g m over r, and let's just say my interval is delta t. Notice what happens as I take the radius of this thing down to 2gm. This goes to 0. This r goes to 2gm. If I go inside 2gm, I don't even want to compute that. Something has gone awry. But look, I can draw this thing just fine. Clearly, there's a surface there. It's got to have an area associated with it. Why are you telling me that the area of this thing is 0 in that limit and is a nonsense integral if I go inside this thing? Well, what's happening is we have uncovered a coordinate singularity. The time coordinate is badly behaved as we-- not well-- as we approach this radius, r equals 2gm. Let me give you an analogy that describes, essentially-- it's something that is very similar to that tube, that world tube that I just drew. But let me do it in a, perhaps, more familiar context. Suppose I want to draw a sphere, and all that I know about a sphere is that it has got two coordinates to cover it, an angle phi and an angle theta. And so I could say, OK, here is my sphere. Here is theta equals 0, phi equals 0. Here is phi equals pi over 2. Here's phi equals pi. Pi equals 3 pi over 2. Phi equals 2 pi. Theta equals pi over 2. Theta equals pi. There's my sphere. So this chart that I've just drawn here, it's true. This does represent the coordinate system, but it's a horrible rendering of a sphere's geometry. What I didn't realize when I wrote this down here is that, in fact, at theta equals 0 and theta equals pi, every phi value should be collapsed to a single point. This is reflecting the fact that if you look at a globe, all lines of longitude cross the north pole and the south pole. Every value of the azimuthal angle on the surface of the earth, they become singular at the north pole and the south pole. This drawing, well, it's like one of those, I forget the names of them, but the various renderings of a map that try to take the earth and write it on a flat space, and you wind up with Greenland being three times the size of Africa, or something like that. And it's because there should be zero area at the top here. As you approach the top the area, it should be getting much stronger. And when you do representation of your map like this, you're spreading everything way, way out. What is going on? And why this idea of drawing this world tube that is swept out by my circle of radius r as it advances forward in time? That drawing does not account for the fact that the Schwarzschild time coordinate is singular as you approach r equals 2gm. It's going to turn out all times t map to a single sphere, and r equals 2gm. To get some insight into what's going on here, let's do a little thought experiment. What I'm going to do is imagine I'm at rest in the Schwarzschild spacetime. So let's say that I'm at some finite radius r. I am not in a weak field. OK, maybe I am at something like r equals 4gm, or something like that. And what I'm going to do is drop a little rock, drop a particle. So I'm going to drop a particle from r equals r0. I'm going to integrate the geodesic equation, and I'm going to parameterize what its radial motion looks like as a function of "time." I put "time" in quotes here because you should be saying at this point, well, you just told me that time is doing something kind of funny here. What do you mean by that? I'm actually going to do this for two different notions of time. I'm going to do this for the coordinate time t, and I'm also going to do this for proper time, tau, as measured along that world, that infall. So I'm not going to go through the details of this calculation. It's a straightforward, moderately tedious exercise. I would just quote to you what the result ends up looking like. So let's first write down what the solution looks like, parameterized by the proper time. So this is most easily written as tau, proper time, and 2gm. Essentially, I'm just going to use it to set a system of units. I write this thing as a function of r. My solution turns out to be-- it looks like this. So if I were to make a plot of what this thing's motion looks like as a function of time, so here's our 0. Here is r of tau. And let's just put in, for fun, let's say this is 2gm. Zoom, fallen. You reach r equals 0 in finite proper time. The parable of the Kretschmann scalar is that as you do so, the tidal forces acting on you are diverging. So if you have any last wishes, send them out because you're not going to have a lot of time to tell people about them. Let's now write it as a function of coordinate time t. This ends up being-- bear with me while I write this out, this is slightly lengthy. OK, so what I mean on this last line is if you want to get the complete solution, just write both of these functions down. Again, subtract them off and place the r's with r0. When you look at this, here's what you see. The motion expressed in times of the coordinate time t asymptotically approaches the radius 2gm, but it never quite reaches it. As t goes to infinity, it eventually reaches-- so r, you can see it appearing in the behavior of this natural log. r gets to 2gm as t goes to infinity. So as measured by clocks on the infalling body, it rapidly reaches r equals 0. According to this coordinate time, it never even crosses r equals 2gm. What the hell is going on with that? Well, to give a little bit of insight into this, it's useful to stop for a second and ask ourselves, what is that coordinate time t actually measuring? So let me write down the Schwarzschild metric and let's think about this. So kind of hard to see what t means in this, but let's consider a limit. Suppose I consider observers who are very far away. If I look at people who are at r, much, much larger than 2gm. For such observers, spacetime looks like this, and this is nothing more than flat spacetime in spherical coordinates. This is what we call an asymptotically flat spacetime. As you get sufficiently far away from the source, it looks just like flat spacetime. essentially, special relativity rules apply. And that gives us some insight into what this coordinate t means. The t that we are using in the Schwarzschild coordinate system, this is time as measured by distant observers. Tau is time, as measured by this infalling observer. So what we are seeing here is the infalling observer crosses 2gm, reaches r equals 0, and has a very short life. But those who are using clocks, adapted to things very, very far away, never even see it cross 2gm. Why is that? Well, we will pick this up in the next lecture, but let me remind you that when we initially began working on this subject, one of the very first lectures, we talked about something called the Einstein synchronization procedure, where what we did was we imagined spacetime was filled with a conceptual lattice of measuring rods and clocks. And we synchronized all of those clocks by requiring that the time delay between different clocks is synchronized according to the time it takes for light to travel from one to the other. This is telling us we are actually working-- when we use Schwarzschild time, we are working in a system that reflects an underlying inheritance from special relativity. These are clocks that have been synchronized by the Einstein synchronization procedure. And so the pathological behavior that we see here, it must ultimately owe to the behavior of these clocks that we use to define our coordinate system, and the behavior of those clocks is linked to the behavior of light. So in order to get insight as to what is going on with this, why is it that if I use a clock adapted to the infalling body, I see painful death, but if I use a clock adopted to someone very far away, I don't even see it approach that dangerous r equals to a radius. In order to resolve that mystery, I'm going to need to examine what the motion of light looks like in this spacetime. We'll pick that up in the next lecture. |
MIT_8962_General_Relativity_Spring_2020 | 22_Black_holes_I.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: I'm going to record two lectures today. These are the final two lectures I will be recording for 8.962. I actually have notes on an additional two lectures. I have notes for an additional two lectures. But those additional two lectures are somewhat advanced material. It's sort of fun to go over them in the last week of the course, for certain students. It can be a great introduction to some of the most important topics in modern research. But I am not going to come into campus and present those lectures, OK, given everything that's going on right now. To be blunt, they are kind of bonus material. And this isn't the time. This is not the semester for us to go through our bonuses. I will make those notes available. I will be happy to discuss them in a saner moment with any student who is interested in them. But this semester, let's just focus on the core material. So that brings us to the topic of what we are studying today. Pardon me while I correct my handwriting. So if I can do a recap of what we discussed last time, we took a look at a spacetime that has this Schwarzschild solution written in the Schwarzschild coordinates everywhere. In our previous lecture, we looked at a spacetime that described a fluid body, a spherically symmetric fluid body. And it had a surface. This was the solution that we used for its exterior. This ends up describing a solution that has zero stress energy tensor. So it describes a vacuum situation. OK? So if I imagine a spacetime that looks like this everywhere, well, what I end up finding is that this is a vacuum solution. It has T-mu-nu equals 0 everywhere. However, it also has a mass m. So this is the vacuum solution with mass, which is, well, that's weird. We examined some of its curvature properties, and we found that at r equals 0, there is a tidal singularity, OK, in invariant quantity that we constructed from the tidal tensors, blows up at r equals 0. R equals 2GM looks a little bit funny. And it turns out tides are well behaved there. OK, there's nothing pathological in the spacetime there. But there is a coordinate singularity. Our coordinate t is behaving oddly. And where we left things last time is I did a little diagnosis of this by imagining a body that falls into spacetime from some starting radius R0. And I looked at the motion of this thing as a function of time. What we find is that if we look at the motion of this thing as a function of the proper time of that in-falling body, it crosses 2GM in finite proper time. And shortly afterwards, again, in finite proper time, it reaches the r equals 0 tidal singularity. If I look at that same motion as a function of the coordinate time t, it never even reaches, never even reaches r equals 2GM. We found a solution, what you can see in notes that I've put online and that are presented in the previous lecture. But we found a solution in which it just asymptotically approaches r equals 2GM, only reaching that radius in the t goes to infinity limit. These are two starkly different pictures of the kinematics of this in-falling body. So the question is, what is going on here? And as food for thought, I reminded us that these coordinates, if we sort of look at the way that spacetime behaves for a very, very, very large M-- Excuse me, very, very, very large R, not M, but for a very large R, R much, much greater than 2GM, this turns into the line element of flat spacetime. Flat spacetime is what we use in special relativity. And our time coordinate there, it is designed using this Einstein synchronization procedure, which means that the properties of light as it propagates through spacetime are built into the coordinate system. So that suggests what we might want to start doing in order to try to get some insight as to what is going on with this weird spacetime is to think about light as it propagates into spacetime. Let's think about what happens to radiation as it propagates. So let's imagine as the body falls that it emits a radio pulse with a frequency as measured according to the in falling. Let's say there's an observer who is falling in who's got a little radio transmitter that's beaming this message out. And according to that observer, this thing is emitted, the pulse has a frequency omega-- very far away, we can describe the momentum of this radio pulse like so. Imagine the thing is propagating out radially and very, very far away where the spacetime is approximately flat. It's simply the four-momentum that describes a null geodesic moving in the radial direction, OK? Minus sign here because with the index in the upstairs position and this asymptotically flat region, that would be the energy. So a couple of facts to bear in mind. The energy measured by an observer with four-velocity U is given by, let's call it E sub U, the energy measured by observer U. This is minus P-dot-U. We developed this in special relativity. But remember the way that we use the equivalence principle. We're gonna use the Einstein equivalence principle. And any law that holds in a freely falling frame, if I can write it in a tensorial way that works in that freely falling frame, it works in any frame. So this tensorial statement holds true, even though we are now working in the spacetime that is distinctly different from special relativity. We also know that in a time independent spacetime, the downstairs T component of four momentum is constant. So what this basically means is if I think about this radio pulse with its light propagating out through the spacetime, the value P0 associated with this thing's four momentum, it's the same everywhere along its trajectory. So let's consider a static observer sitting in the Schwarzschild spacetime. OK, I'm going to require G-alpha-beta U-alpha U-beta to be equal to minus 1. And I will require that this have some timelike piece, that this observer is static. So their spatial components of their four-velocity are equal to 0. It looks like, by the way-- just pause one moment here. It looks like the projector is on because of-- I should probably just leave it as is. All right, I'm not going to worry about that. Just leave that as it is. Sorry, let's go back to what I'm talking about here. So I have a static observer, so an observer who is not moving in space. They're only moving through time. And I need to normalize their four-velocity. Bear in mind, this is not a freely-falling observer. OK, this is an observer who must be accelerated, in some sense. And there must be some kind of a mechanism that is allowing this observer to hover at the fixed location in spacetime where they are at. So when I put these two constraints together and I tie it into that spacetime, what I find is that the timelike component is 1 square root of 1 minus 2GM over r. So let's compare the energy that is emitted at radius r to the energy that is absorbed at some radius big R. So I'm going to imagine-- And so the reason I formulate it in this way, what I want to do now is ask myself, well, what is the energy that an observer at little r would measure here? What is the energy that is observed by the person at radius capital R? And now, we know, if I imagine that these are both being measured by static observers, because P sub T, P sub 0 or P sub T, because it is a constant as the light propagates out, I can take the ratio of these two. And the energy observed at capital R versus the energy emitted at little r-- Hang on a second. As I was looking over my notes, there was a result that made no sense because I had a typo. So for those of you following along, the four-velocity component should have been a 1 over the square root of that quantity I had earlier. Suddenly, what I'm about to write down makes a lot more sense. So this becomes-- Let's imagine that the observer is so far away that they are effectively infinitely far away. So the question I'm asking is, if I imagine that the light pulse is emitted at some radius little r, what would a very distant observer measure the energy of that pulse to be? OK, so my light pulse or radio pulse is emitted at some finite radius. And we'll call e infinity, the value that is measured by a very distant observer. This becomes square root 1 minus 2GM over r. Notice that the energy, no matter how energetic the light is when you emit it, I described it as being a radio pulse, but it could be a laser pointer. It could be ultraviolet. It could be a gamma ray. You know, you could just hawk whatever massively-powerful source of photons you want and you're pointing it out in the radial direction and hoping that your distant friend can measure it. No matter what energy you emit as you are falling in, as you approach r equals 2GM, the amount of energy in that beam that reaches a distant observer goes to 0. No matter how energetic my pulse of light, no energy reaches distant observers as the emitter approaches R equals 2GM. In a similar way, imagine that as you are falling in and you've got this little beacon that you are sending messages out to your distant friends, imagine you send them out with pulses that are separated by delta T. So the person in the falling frame turns his beacon on for a moment every delta T. Let's say delta T is every second. The interval between pulses, as measured far away, one can show using a similar kind of calculation. You pick up a factor of 1 over square root 1 minus 2GM over r. This goes to infinity as r goes 2GM. So remember, the time coordinate was originally defined by imagining that I can synchronize all of my clocks using light pulses that are bouncing around. But what we can kind of see here is that light pulses that are emitted in the vicinity of r equals 2GM, they're kind of going to hell, OK? So if I imagine, let's just say for the sake of argument, I'm using a green laser pointer as the thing that I use for my Einstein synchronization procedure. And suppose that I communicate what time is on the clock by a modulation of the signal. So suppose that I modulate it by putting little spaces on it that tend to be about a second long. Or let's just say it's like a microsecond long so I can really pack some information into that. Well, the laser pointers that are emitting from r equals 2GM, their signal infinitely redshifts away. They lose all their energy. So they're not green. If they're really close to it, maybe they'll be red when they get to the next thing. Instead of being a pulse every microsecond, it'll be every two microseconds. And the closer we get to it, the more redshifted it becomes and the more extended the interval between pulses becomes. I'm going to post to the 8.962 website a set of notes that sort of cleans up this calculation a little bit, just goes through this in a little bit more detail. It was something I wrote in the Spring 2019 semester in response to a good question that a student had asked me about this. And I think they do a nice job of just going over this calculation. And they give a couple of examples of the way this is behaving. The key thing which I want to emphasize is that as r goes to 2GM, what we see is that-- well, let's just put it this way. In fact, let's remove the word "as." The surface r equals 2GM corresponds to infinite redshift. OK, we've talked already a little bit about, if I have light climbing out of a gravitational field, the light gets a little bit redder as it climbs out. Well, at this particular surface, the sphere of radius 2GM in this spacetime, if you can get down to there, that redshift then becomes infinite. All of the energy is drained out of it as it climbs out. So what this basically tells us is that this surface breaks the Einstein synchronization procedure and it renders that time coordinate bad, at least if we are concerned about understanding things right at this surface. OK, as long as we're concerned with the exterior of the surface, that's not a problem, OK? Everything works fine as long as we move away from this coordinate singularity, sort of in the same way that many of the pathologies associated with the spherical coordinate system here, the north and south pole, they are fine as long as you're not trying to do things like measure the longitude angle corresponding to the North Pole. So this is not even well-defined. In the same way, this is basically telling us that the coordinate T in which we wrote down this spacetime isn't really well-defined at r equals 2GM. So what we need to do, if we want to try to get some-- hey, I'm over here, camera!-- if we want to try to get a little bit of insight as to what is going on here, we need a better time coordinate. So I'm going to talk about a couple. And the way that we're going to formulate these is all the pathologies are revealed when we look at the behavior of light propagating in this spacetime. So let's play around with light. Let's look at null geodesics in this spacetime. So to begin with, let's stick, for just the next couple of moments, with the original Schwarzschild coordinates. OK, so I'm going to look at null geodesics. So I'm going to set 0, and I'm gonna ask myself, how does it move through an interval at dt and an interval of dr? OK, so we can solve this to find how, if an object is moving on a radial null geodesic, how do dt and dr behave? How does dt dr behave? Put it that way. OK, so that's what my dt dr looks like. Plus, the plus sign corresponds to a solution that is moving the outward in all direction. Minus sign is an inward directed null geodesic. These define what we consider to be the opening angle. So dt dr defines the opening angle of a light cone. So if we go to very, very, very large r, OK, we get dt dr equals 1. And this corresponds in units in which c equals 1 to light moving on a 45 degree angle in a spacetime diagram. This is familiar behavior from special relativity. But as r goes to 2GM, we see dt dr going to 0. Let's make a sketch and see what, sort of, dt dr, what the tangent to a null geodesic looks like in the t-r plane as a function of radius. So this little dash here means I'm sort of imagining that I'm going to stretch my r axis so that, out here, you're in the asymptotically flat region where things look like special relativity. So here we are out in this asymptotically flat region. My outward-going light ray goes off at 45 degree angle in the r-t plane. Inward one goes at a 45 degree angle pointing inside. Down here in the stronger field, it's going to be a little bit steeper than this. And so the opening angle of my light cone is closing up. Here, we'll have closed up a lot more. Here, it's closed up practically all the way. Now, as I approach r equals 2GM, both the inward and the outward direction in these coordinates go parallel to 2GM. So you can see the collapse of the light cone in these coordinates as you approach this coordinate singularity. So we need a healthier coordinate system. One thing that we can do is we can move the pathology out of our time coordinate and into our radial coordinate with the following definition. Suppose you choose a radial coordinate r-star such that dt equals plus or minus dr-star everywhere. OK? So if I replace my horizontal axis with-- pardon me-- if I replace my horizontal axis with the r-star, this will be 45 degree angles everywhere. OK? But to make this work, what you find is r-star must look like this. What you basically see is that this coordinate system takes r equals 2GM and it moves it to r-star equals minus infinity. So the way that this coordinate representation, that this different radial coordinate, the way that this makes the light cones always have 45 degree opening-- really 90 degrees opening-- it makes the light rays always go off at 45 degree angles is by essentially constantly stretching the radial axis so that this guy just gets stretched out so that it's opening at 90 degrees. This guy is stretched a little bit less. This guy is stretched a little bit less. Basically, not stretched when you're really far away. But as you approach r equals 2GM, you're infinitely stretching at these coordinates. These are known as tortoise coordinates, basically, because you start walking, taking ever slower and slower steps. Even steps, if you imagine you're stepping evenly in r-star, you're taking ever smaller and smaller steps in r, as you approach the infinite redshift surface. So with that tortoise coordinate defined, you use that as an intermediary to define a couple of new coordinates for your spacetime that are adapted to radiation. So we're going to define v to be t plus r-star. And the importance of this is that this is a coordinate that's not hard to convince yourself this is constant on an in-going radial null ray. I'm going to define a coordinate U to be t minus r-star. And this is constant on outgoing radial null rays. OK? So this basically means that if I'm working in this coordinate system, r-star, if I want to know the behavior of this guy, well, an outgoing radial null ray, you just might say, ah, that's the null ray that has U equals 17. OK? And that will then pick out, basically, a whole sequence of events along which that null ray has propagated. And, you know, as you can see, it vastly simplifies how we describe it. So once you've defined these two coordinates, you can rewrite the Schwarzschild spacetime in terms of them. It is generally best to choose-- So we're going to take this another step in just a moment, but we'll start-- you choose either V or U, and you replace the Schwarzschild time. So let's use V to replace time. So when you go and you look at what your new coordinate system looks like-- and remember, the way you do this is the usual thing, you're going to make your matrix of partial derivatives between your old coordinate system. So your old coordinates are t, r, theta, and phi. And then they go over to v, r, theta, and phi. OK? So you make your matrix of partial derivatives describing this. And here's what you find when you change to the metric in the new representation. Notice, there's no dr squared term at all. OK? We do still see something that's, you know, at least a coordinate singularity at r equals 2GM. We haven't quite gotten rid of it entirely here. But we've definitely mollified the impact of this coordinate singularity. So in this coordinate system, you can then set ds squared equal to 0, and you find two solutions describing radial null curves. So dv dr equals 0 for in-going. And, you know, by definition, these things are constant on an in-going radial null ray. And so as you move along it, V remains constant. OK, and you get something a little bit more complicated for the outgoing one. Let's redraw this using my new coordinates, OK? So I'm going to leave my horizontal axis as r. So I'm going to make my time axis be v. So out here, here's my in-going null ray. And here's my, eh-- let's make that a little closer to 45 degrees. Here's my outgoing null ray. OK? As I move in to smaller and smaller values of r, notice in the limit, as r goes to 2GM, the slope becomes infinite. OK, so this thing gets steeper. This one keeps pointing in, so this guy gets steeper, steeper. Right here, it lies pointing exactly straight up. What's kind of cool is that in these coordinates, I can actually look at what it looks like inside this thing. And so inside, this guy tips over and gets a negative slope and looks like this. Now, bear in mind, these two things, they denote the null rays. All radial timelike trajectories, and indeed all timelike trajectories, not just the radial ones, all timelike trajectories must follow a world line that is bounded by these two sides. OK, so if I'm out here, everything in here describes trajectories that a timelike observer can follow. Everything in here describes a trajectory a timelike observer can follow, everything in here, everything in here. Notice, when I am at this one here, right at r equals 2GM, every allowed timelike trajectory points towards r equals 0. At best, I can imagine an observer who's very close to the speed of light who sort of skims along inside of this thing. But they are timelike, so they will have a little bit of a slope that points them inward. OK. As you move inside r equals 2GM, it's even more so. OK, you can't even sort of go parallel to the r equals 2GM line. They all point towards this thing. Once you get to r equals 2GM, all trajectories, as they move to the future, must move to smaller radius. What this tells us, in particular, is that once you have reached r equals 2GM, you are never coming back. All allowed trajectories, everything that is permissible by the laws of physics, moves along a trajectory that points towards r equals 0 once you hit that r equals 2GM line. Because of this, this surface r, this surface of infinite redshift is given the name-- Let me come back to the point I was making. So the surface r equal 2GM, nothing that crosses it is ever going to come back. This surface of infinite redshift, we call an event horizon. OK? No events that are on the other side of r equals 2GM can have any causal influence on events on the outside. If you have a spacetime with an event horizon like this and this one that we are talking about right here, I'm going to talk at the end of my final lecture that I record about-- Maybe, actually, I'm gonna do it in this one. Yeah, I am. So at the end of this lecture that I'm recording, I'm going to go over a couple of other spacetimes that have this structure. Such spacetimes are called-- wait for it-- such spacetimes are called black holes. They are black because light cannot get out of them. And they are holes because you just jump into 'em, and you ain't never coming out. So let me pause for just one moment. I want to send a quick note to the ODL person who is helping me out here. I just want to let her know that this lecture may run a tiny bit long, since I spent a moment chatting with a police officer who checked in on me. OK, if you're watching, Elaine, hi! So I just sent you a quick note, letting you know that I'm likely to run a little bit long. All right, let's get back to black holes. So to sort of call out some of the structure of this spacetime, I want to spend just a few minutes talking about one final coordinate transformation that is very useful, but looks really weird. So just bear with me as I go through this-- a very useful, but unquestionably somewhat obtuse coordinate transformation. What I'm going to do is I'm going to define a coordinate v-prime. This is given by taking the exponential of the in-going coordinate time V, normalized for GM. U prime will be the exponent of U, a time that works well for the outgoing coordinate system divided by 4GM. I'm then going to define capital T to be 1/2 V-prime plus U-prime, capital R to be 1/2 V-prime minus U-prime. It's then simple to show, where simple is professor speak for "a student can probably do it in an hour or so." It's sort of tedious, but straightforward, just hooking together lots of definitions and slogging through a couple of identities. It's simple to show that capital T relates to Schwarzschild time T and Schwarzschild radius r like so. There's two branches. OK, so this is how one relates capital T and capital R to Schwarzschild t and Schwarzschild r in the region r greater than or equal to 2GM. You find a somewhat different solution in the region r less than 2GM. So if you're looking at this and kind of going, "what the hell are you talking about here," that's fine. Let me just write down two more relationships. And then I'll describe what this is good for. So a particularly clean inversion between TR and the original Schwarzschild tr, both the r greater than 2GM and r less than 2GM branches can be subsumed into this. And you find T over R looks like the hyperbolic tangent T over 4GM when you're in the exterior and the hyperbolic cotangent in the interior. So these rather bizarre-looking coordinates, these are known as Kruskal-Szekeres coordinates. I'll just leave that down like so. So when one goes into this, I'm not going to deny it, this is a bizarre looking coordinate system. OK, but it's got several features that make it very useful for understanding what is going on physically in this spacetime. So first, if you rewrite your metric in terms of capital T and capital R, what you get is a form that has no singularities. It's well-behaved everywhere, except at r equals 0. So you do get a singularity there, things blow up as r goes to 0. There's no other coordinate pathologies. And then you get sort of an angular sector. It's actually cleanest in terms of the Schwarzschild radius r, so we'll leave it in terms of that. One thing which is nice is notice that radial null geodesics, they simply obey dt equals plus or minus dr everywhere, OK? The only place where you run into a little bit of problem is as r goes to 0. And that's special, OK? So I got that just by setting ds squared equal to 0. It's radial, so my D-omega goes to 0. And then just dt equals plus or minus dr. So that's really nice, OK? I'm gonna make a sketch in just a moment. And the fact that I know light always moves along 45 degree lines in this coordinate system is going to help me to understand the causal structure of this spacetime. The causal structure is what I mean by which events can influence other events. What can exert a causal influence on what? So as I move on, I'm going to make a sketch in just a moment, I want to highlight a couple of behaviors that we see that are really sort of called out in this mapping between the two coordinate systems. So notice that a surface of constant Schwarzschild radius, constant r forms a hyperbola in the Kruskal-Szekeres coordinates. Notice that surfaces of constant time, they form lines in that they form lines of slope t over r equal to some constant. So they are lines with t over r equaling, on the exterior, let's just focus on the exterior, they have a slope that's given by the hyperbolic tangent of t over 4GM. On the interior-- replaced with cotangent, hyperbolic cotangent. The last thing which I'd like to note before I make a sketch here is let's look at the special surface of infinite redshift, this event horizon. So if I plug in r equals 2GM, plug this in over here, I get t squared minus r squared equals 0. This is the asymptotic limit to those hyperbolae. They just become lines t equals plus or minus r. So the event horizon in this coordinate representation is just going to be a pair of lines crossing in the origin of these coordinate systems, OK, a pair of 45 degree lines crossing into this coordinate system. Notice, also, that t equals plus or minus r, this corresponds to Schwarzschild t going to plus or minus infinity. So this, indeed, is a weird, singular limit of the Schwarzschild time coordinate. So you can find much prettier versions of this figure than the one I'm about to attempt to sketch. Let's see what I can do with this. So horizontal will be the Kruskal-Szekeres coordinate r, vertical will be the coordinate t. Here is the event horizon, r equals t or little r equals 2GM. Some different surface of r equal to some constant value greater than 2GM will live on a hyperbola like so. Some value of r equals constant, but less than 2GM, lies on a hyperbola like so. In particular, there is one special hyperbole corresponding to r equals 0. And this is where my artistry is going to truly fail me. This is an infinite tidal singularity. Now, the thing which is particularly useful about this particular coordinate system is, remember, light always moves in the capital R, capital T coordinates. It always moves on lines that go dt equals plus or minus dr. So what you can see is that imagine I start here and I send out a little light pulse, OK, a radially outgoing light pulse, it will always go away and go to larger and larger values of r, just sort of constantly moves along this particular trajectory. Let me write out again what I'm doing here. So a radial outgoing light ray, it will follow dt equals dr. But notice that this line goes parallel to the event horizon. If I am on the inside of this guy and I try to make a light pulse that goes outside, points in the radial direction, all it does is, in these coordinates, it moves parallel to the event horizon. It can never cross it. And in fact, because this is a hyperbola, one finds that even though you have tried to make this guy as outgoing as outgoing can be, it will eventually intersect the r equals 0 tidal singularity. Since I cannot have any event, here, that communicates with any event on the other side of this, this region, everything at r less than 2GM, these will be causally disconnected-- not "casually," pardon me. They are causally disconnected from the world that lies outside of r is 2GM. So in my notes and in Carroll's textbook, there is another coordinate system that you can do which essentially takes points that are infinitely far away and brings them into a finite coordinate location. And that final thing, it puts it in what are called Penrose coordinates and it allows you make what's called the Penrose diagram, which displays, in a very simple way, how different events are either causally connected or causally disconnected from the others. Fairly advanced stuff, not important, but many of you may find it interesting. Happy to talk further, once we all have the bandwidth to have those kinds of conversations. So let me summarize. So the summary is that this spacetime, so earlier we were looking at the spacetime of a spherically symmetric fluid object. We found a particularly clean form for the exterior of that object, where it was a vacuum solution. If we imagine a spacetime that has this everywhere, then we get this solution that we call a black hole. So I emphasize this is vacuum everywhere, but sort of the analysis kind of goes to hell at r equals 0. So there are some singular field equations there trying to describe the stress energy. Its behavior as you approach there, let's just say we're not quite sure what's going on. In this coordinate system, we see weird things happening as we approach r equals 2GM. This is simply a coordinate singularity. There's really nothing going bad with the physics here. But our attempts to use a time coordinate that's based on, essentially, the way light moves in empty space, it's failing in this region. And all of this lecture is about uncovering this and seeing that, in fact, this is a surface of infinite redshift beyond which things cannot communicate. So this is one of the big discoveries that came out of general relativity, OK, this creature we call the black hole. It is not the only solution of the Einstein field equations that we call a black hole. Let me talk briefly about two others. So another one has a spacetime that looks like this. So where did this come from? Well, suppose I bung this through the Einstein field equations, what I find is that it comes from a non-zero stress energy tensor. In fact, it's a stress energy tensor that looks like this. Pardon me. This is a stress energy tensor of a Coulomb electric field with total charge q. This represents a charged black hole. It turns out if you analyze this thing carefully, you find it has an event horizon. It turns out to be located at G quantity that involves the square root of m squared minus q squared. Now, if q-- don't even ask me what the units are that this is being measured in, they're pretty goofy units-- but if, in these units, the magnitude of q is greater than m, there is no horizon. There is still, however, an infinite tidal singularity at r equals 0. So such a solution would give us what is known as a naked singularity. I have a few comments and I have a couple notes on these. But they're not as important as other things I'd like to talk about. You might wonder, where does that r horizon actually come from? OK, that is what you get when you find the route, where you look for the place where the metric function vanishes. OK, you notice there's only one metric function that appears in there, 1 minus 2GM over r plus q squared over r squared. So in general, if you have what's known as a stationary spacetime, you can find coordinates such that surfaces of constant r are spacelike surfaces. If you look for the place where a surface of constant r makes a transition from being a spacelike surface to being a null surface, that tells you that you have located an event horizon. OK, this is discussed in a little bit more detail in my notes. There's also some very nice discussion in Carroll's textbook. What it boils down to is that if you can find a radial coordinate that allows you to do this, then the condition G upstairs r upstairs r equals 0 defines your event horizon, OK? It just so happens that it's also equal to G downstairs t downstairs t equals 0, in this case and in the Schwarzschild case. But it's not like that for all black holes that you can write down. In particular, let me write down the final black hole spacetime I want to discuss in this lecture. This is gonna take a minute, so bear with me. OK, so in this spacetime, the symbol delta I've written here, this is r-squared minus 2GM r plus a squared. Rho squared is r squared plus a squared cosine of the square root of theta. This thing turns out, if you compute the inverse metric components, you find that g upstairs r upstairs r is proportional to delta. And so there is a horizon where delta equals 0. Sorry. That turns out to be located at a radius that looks like this. This represents the spacetime of a spinning black hole. It was discovered by Roy Kerr, a mathematician from New Zealand. I think this was actually a big part of his PhD work. So it is known as a Kerr black hole. The parameter a is related to the angular momentum, the spin angular momentum of the black hole in the units that we measure these, normalized to the mass. So notice that in order for this to actually have a horizon, you need that a to be less than or equal to GM. If it saturates that bound, then you get what's known as a maximal black hole. So it has a couple of noteworthy features. First, it is not spherically symmetric. If it were spherically symmetric, we could write the d theta squared d phi squared piece, we could find some radial coordinate such that there was some simple radius such that G theta theta was simply sine squared G theta theta. This is the condition that defines spherical symmetry. And there is no angle-independent radial coordinate that allows you to do that. Notice also that there is a connection in this coordinate system between t and phi. Gt phi is equal to minus 2 GM a r sine squared theta over rho squared. Why minus 2 and not minus 4? Well, remember what I have there is Gt phi dt d phi plus G phi t d phi dt. One can show that this term, it reflects the kind of physics in which the spin of the black hole introduces a spinning, almost magnetic-like element to gravitation. If you guys do the homework assignment I have assigned in which you compute the linearized effect on a spacetime of a spinning body, you'll get a flavor of this, OK? This ends up giving you, that calculation gives you a similar term in the spacetime which further analysis shows leads to bodies. Essentially, what you find is that if you have an orbit that goes in the same sense as the body's spin versus an orbit that goes in the opposite sense of the body's spin, there's a splitting in the orbit's properties due to that. OK? So it breaks the symmetry between what we call a prograde orbit and a retrograde orbit. This is one of the most important solutions that we know of in general relativity because of a result that I'm going to discuss now. Oh, first of all, I should mention that, in fact, you can combine charge with spin. I'm not going to write down the result because it's just kind of messy. But it does exist in closed form. If you're interested in this, read about what is called the Kerr-Newman solution. So if you're keeping score, we have this spherically symmetric black hole, which only has a mass, you have the charged black hole whose name I forgot to list. Ah! Sorry about that. This guy is known as the Reissner-Nordstrom black hole. One of those O's, I believe, is supposed to have a stroke through it. Those of you who speak Scandinavian languages can probably spell it and pronounce it better than I can. So we have Schwarzschild, which is only mass, Reissner-Nordstrom, which is mass in charge, Kerr, which is mass and spin, and Kerr-Newman, which is mass, spin, and charge. You might start thinking, all right, well, does this keep going? Do I have a solution for a black hole that's got, you know, northern hemisphere bigger than the southern hemisphere? You know, every time you think about adding a bit of extra sort of schmutz to this thing, do I need another solution? Well, let me describe a remarkable theorem. The only stationary spacetimes in 3 plus 1 dimensions with event horizons are the Kerr-Newman black holes-- completely parameterized by mass, spin, and charge. If you take the Kerr-Newman solution, you set charge to 0, you get Kerr. If you take Kerr and you set spin to 0, you get Schwarzschild. So the Kerr-Newman solution gives me something that includes these other ones as sort of a subset. And what this theorem states is that the only-- so stationary means time-independent. So in other words, the only spacetimes that are not dynamical, but that have event horizons, at least with three space and one time dimension, are the Kerr-Newman black holes. Once you know these, you have characterized all black holes you can care about. And in fact, in any astrophysical context, any macroscopic object with charge is rapidly neutralized by ambient plasma that just sort of fills all of space. And so this Kerr solution, in fact, gives an exact mathematical description to every black hole that we observe in the universe. That is an amazing statement. OK? Of course, as a physicist, you want to test this. And this, in fact, is much of what my research and research of many of my colleagues is about. Can we actually formulate tests of this Kerr hypothesis? And many of us have spent our careers coming up with such things. Suffice it to say, in the roughly negative 5 minutes I have left in this lecture, that the Kerr metric has survived every test that we have thrown at it. So this metric, like I said, was essentially derived by the mathematician Roy Kerr as his PhD thesis. And it has really earned him a place in physicist Valhalla. Let me just conclude this lecture by making one comment here. An important word in this theorem is the statement that the only stationary spacetimes are the Kerr-Newman ones, stationary spacetimes with event horizons, so the Kerr-Newman black holes. What this means is that when a black hole forms, it may be dynamical, it may not yet be stationary. And so the way that this theorem, which is known as the No-Hair theorem, the way that it is enforced is that, imagine I have some kind of an object that due to physics that we don't have time to go into here, imagine that this thing, its physics changes in such a way that its fluid can no longer support its own mass against gravity, and it collapses to a black hole. OK? Initially, this could be a huge, complicated mess. So we have a mass, charge, spin, magnetic fields, who knows? The No-Hair theorem guarantees that after some period of time, it will be totally characterized by three numbers-- the mass, the spin parameter a and the charge q. What goes on is that during the collapse process, radiation is generated. What this radiation does is it carries away gravitational waves, carries away electromagnetic waves. Some of this is actually absorbed by this black hole. And it does so in such a way that it precisely cancels out everything in the spacetime that does not fit the Kerr-Newman form. You wind up-- so this is one of these things where we really can only probe this either with observations that sort of look at things like black holes and compact bodies colliding and forming black holes and looking at what the end state looks like. Or we can do this with numerical experiments where we simulate very complicated collapsing or colliding objects on a supercomputer and look at what the end result is. And what we always find is that the complex radiation that is generated in the collapse and the collision process always shaves away every bit of structure, except for exactly what is left to leave it in the Kerr-Newman form at the end. Really, when we do these calculations, we generally wind up with a Kerr black hole because we tend to study astrophysical problems that are electrically neutral. So this is a result that is sometimes called Price's theorem, based on sort of foundational calculations that were done by my friend Richard Price. He did much of this right around the time I was born in the days of being a PhD student and looking at the behavior of highly distorted black holes and seeing how the No-Hair theorem-- You can imagine making a spacetime that contains what should be a black hole, but you somehow distort it. What you find is it becomes dynamical and it vibrates in such a way as to get rid of that distortion. And you leave behind something that is precisely Kerr or Kerr-Newman if you have charge. Price's theorem is a semi-facetious statement that tells me everything in the spacetime that can be radiated, is radiated. In other words, any bit of structure, any bit of structure in the spacetime that does not comport with the Kerr-Newman solution radiates away and only Kerr-Newman is left. So that concludes this lecture. My final lecture, which I will record in about 15 minutes, is one in which we are going to look at one of the ways in which we test this spacetime, which is by studying the behavior of orbits going around a black hole. So I will stop here. |
MIT_8962_General_Relativity_Spring_2020 | 5_The_stress_energy_tensor_and_the_Christoffel_symbol.txt | [SQUEAKING] [RUSTLING] [CLICKING] SCOTT HUGHES: All right. So welcome back. I had a little bit of a break. Before I get into-- I go over my quick recap, you hopefully all have seen the announcements that we're going to delay the due date of the next problem set until Tuesday. That's in part because some of the material that appears on it is what we're going to talk about in today's lecture. So I want to make sure you've at least seen a lecture on all the topics before you try to do some of the problems on it. In truth, you could probably-- based on things I talked about in the previous lecture, you could probably deduce a lot of what you need to do, but still, it's good to go over it a little bit more methodically. And I'm trying to think. Is there anything else I want to say? Yeah. So then the next problem set will be posted a week from-- is that right? Yeah, I think we post a week from today. So let me just get back to what you were talking about last time. So I began introducing some geometric concepts and some quantities that we use to describe matter. We've done a lot of things so far that's appropriate for describing kind of the kinematics of particles, but that's rather restrictive. We want to talk about a broader class of things than that. And I began by introducing something which is, again, fairly simple, but it's a useful tool for beginning to think about how we are going to mathematically categorize certain important types of matter. Pardon me while I put some of my notes away. Also, pardon me. I'm recovering from a-- I caught a terrible cold over the long weekend, and I'm coughing incessantly. So I will be occasionally sucking on a cough drop. So I introduced the quantity called the number four-vector. And that is given by-- imagine you have sort of a-- it's best to think about this in the context of something like dust. So you have some kind of non-interacting little agglomeration of tiny particles. And in the rest frame of an element of this dust-- so imagine you go into-- well, I'm going to define the rest frame of that element by saying imagine you've got a cubic nanometer of it or something like that. And you go into the frame where everything in that cubic nanometer is on average at rest. We'll call n sub 0 the rest density. So I forgot to write that down. So that's the rest density. The rest-- pardon me-- number density. So I go into that rest frame. That tells me how many little dust particles there are per cubic volume, per element of volume. Multiply that by the four-vector that-- excuse me-- the four-velocity describing that element. And we're going to call that capital N. So that's a vector that describes-- its components describe the number density in some other frame of reference, in any frame of reference. And the flux, talking about how those dust particles flow from one element to another. So we spent a little bit of time talking about how to define volume elements in a covariant fashion. I'm not going to go through that, but part of the punchline of doing that is that we're going to define a sort of what I call a covariant formulation of conservation of number. It's going to be the spacetime divergence of that vector is equal to 0. That holds-- this equation holds in all frames of reference. If I choose a particular frame of reference, in other words, I define a particular time, I define a particular set of spatial coordinates, I can then break this up and say that this is equivalent to saying that the time derivative of the t component of that plus the spatial divergence, the spatial confluence is that they sum to 0. I can also take this thing and integrate over a-- oops-- it would help to have one element defined here. I integrate this over a four-dimensional volume. And by my conservation law, I must get 0 when I do that. Again, having chosen a particular frame, defined what time means, defined what space means, I can then split this out into a statement that the time derivative of the volume integral of the time component of this thing is balanced by the flux of the number across the boundaries of a particular three-volume. I remind you this notation, this sort of delta V3 means the boundaries of the V3 that's used under this integral. So we talked a little bit about a few other four-vectors that are used. In particular I introduced some stuff that we use in electricity and magnetism. But I want to switch gears. In particular, I want to introduce and discuss in some detail one of the most important tensors we are going to use all term. So I motivated this by saying imagine I have a cloud of dust. And I won't recreate one, because I am getting over a cold. And I don't want to breathe chalk dust. So I imagine I make a little cloud of these things here. So far we've characterized this thing by just counting the number of dust particles that are in there. But that's not all there is to it, right? That dust has other properties. And so the next thing which I would like to do is let's consider the energy and the momentum of every particle of dust in that cloud. So let's just imagine, again, we're going to start with something simple, and then we'll kind of walk up from there to a more generic situation. Let's imagine that my cloud consists of particles that are all identical, and each dust particle has the same rest mass. So suppose this guy has a rest mass m. So one of the things that I might want to do is in addition to characterizing this dust by saying that the number density-- [COUGHING] excuse me-- in addition to saying that the number density in a particular frame is n or n0, let's go into the rest frame of this thing. I might be interested in knowing about the rest energy density. So I'm going to denote the rest energy density by rho sub 0. Each particle has a rest energy of m. Remember, c is equal to 1. So mc squared, if you want to hold to the formula that we've all known and loved since we were babies. So that's the rest energy. And then how many-- if I want to get the rest energy density, I count up the number in each volume, or the number per unit volume. So that's m times n0. So that's the rest energy density of this thing. Great. Let's now ask, OK, so that's what it looks like in the rest frame of this dust element. Let's imagine that I now bop into a frame that is moving with some speed relative to this rest frame. In this frame, the energy density-- I'm not going to say rest energy density, because I'm no longer in the rest frame. But the energy density in this frame, which I will denote by rho without the subscript 0 is going to be the energy of each particle. So that's gamma times m. And the volume is going to be length-- it's going to be a little bit larger. Sorry. The volume is smaller because of a length contraction. And so this guy actually gets boosted up to gamma and 0. So in this frame, the energy density is larger than the rest energy-- excuse me-- the rest energy density with a gamma squared factor. Straightforward algebra. But I want you to stop and think about what that's telling us. If this rho were a component of a four-vector, is there any way I could get a gamma squared out of this thing by bopping between reference frames? No, right? When I do that, it's linear in the Lorentz transformation matrix. There's no way with a linear transformation that I would pick up two powers of gamma. This is not how a-- you guys are probably all used to from Newtonian physics of thinking of energy density as a scalar. This is not a Lorentz scalar. A scalar is the same in all frames of reference. So this is a different quantity. So it's neither a four-vector component nor a scalar. It's got two of those things, so it's not going to surprise you that what's actually going on here is we are picking out-- what we've done here is we've picked out a particular component of a tensor. Let's think a little bit more methodically about what tensor that must be. So when I originally wrote down my rho 0 here, I sort of just argued on physical grounds that it is the energy of every particle times the number of particles per unit volume of that thing. Well, the energy that I use in this thing doing it in both the rest frame and in the other frame, those are the time-like components of a particular four-vector. So we assembled rho by combining energy, which is the time-like component of the four-momentum with number density, which is the time-like component of the number vector that I just recapped a few minutes ago. So those are two time-like components of four-vectors here. So that rho is something like-- well, let's put it this way. I can write this as pt nt, which tells me that it belongs to a tensor. So let's define this as some tt of a tensor that I've not carefully introduced yet. There is some underlying tensor that we have built by looking at the-- there are various names for this. We'll call it the tensor product of two different four-vectors. So if I write this in the kind of abstract tensor notation that I've used occasionally-- in some of your textbooks this would be written with a bold faced capital T, I will use a double over line for this. We might say that this t is the tensor product of the number vector with the four-momentum. Now, the number vector is itself just n0 times the four-velocity. And my momentum, since all the particles, I'm assuming, are the same is just the rest mass times the four-momentum. This thing actually looks kind of like the four-velocity times four-velocity with a prefactor And that prefactor is nothing more than the rest density. So another way to say this is that if I make my cloud of dust, and I go in and I look at every little element in it, I take the rest energy density of every element in that dust. I construct the tensor that comes from making the tensor product of the four-velocity of that dust with itself. That is what this geometric object is equal to. If you want to write this in index notation, which is how we will write it 99.8% of the time, we would say that component alpha beta of this quantity t is that rest density times u alpha u beta. So this is-- in terms of the physics we're going to do this term, this is perhaps the most important quantity that we're going to talk about. So it's worth understanding what this thing is really telling us. So if I wanted to get these components, the alpha beta, out of this thing from this sort of abstract notation of the tensor, I'll remind you that we can do this by taking the tensor and plugging into its slots the basis 1 forms, which tell me something about-- basis 1 forms are really useful for sort of measuring fluxes across in a particular direction. And in fact, this quantity has the geometric interpretation-- think of this as the flux of momentum component alpha in the beta direction. So if we look at this component by component, it's worth doing that. So let's do the tt or 00 component. So this is the one we've already done. This is just rho 0 ut ut. It's just rho in that thing. According to the words I've written down here, this is the flux of pt in the t direction. So pt tells me about energy. The flux of energy in the energy direction refers to-- so here's my water flowing in the time-like direction, just sitting there apparently doing nothing, but no, it's moving through time. This is energy flowing through time. It's energy density, all sort of locked up in stable equilibrium here. But if somebody comes in someday with a nice cup of anti-water and combines it, we get to have all that energy released. And we can enjoy that for a few femtoseconds before we evaporate. Let's look at the other components. So T0i, you can take the definition and plug it in in terms of the components. It's obviously ut times ui, but still use the definition by words. So this is the flux of p sub t in the xi direction. This is talking about energy moving in a particular direction. This is nothing more than energy flux, energy sort of flux that we are used to think about, not flowing through time, but flowing through space. Now, T0i, this is the flux of momentum component i in the t direction. So this tells me about momentum density. So if I had this stuff flowing, there's some momentum associated with that flow. You can count up the amount of mass and every little volume. Divide by that volume. Take some ratios. That is the density of momentum. And finally, this last one, flux of p i in the xj direction, there's really no great wisdom for that. This is nothing more than momentum flux. If I have a bucket of water and things are sort of sloshing around, there's some momentum moving, and the whole assembly is moving in some direction, we can get a flow of momentum going in kind of a non-normal direction. As long as it goes in a normal direction, it might be moving along its flow. I'm going to talk about that in a few moments when I talk about a few different kinds of stress energy tensors. The key thing which I want you to be aware of is that under the hood of this thing, we're going to talk about different kinds of stress energy tensors for a little later in this class. But this basic interpretation of the way to think about the different components, it holds for all of them. So this is good intuition to have. One thing which is worth noting is if you look at-- if you actually-- let's take the form of the stress energy tensor that corresponds to the dust. So I'm going to write out all four of my components. I've already given you T00. T0i, I can write that as gamma squared rho 0 vi T0i equals gamma squared rho 0 vi. This guy is gamma squared rho 0 vi vj. There's two elements of symmetry here, which I want to emphasize. Notice energy density and momentum flux. Sorry. Energy flux and momentum density, those words are important. Energy flux and momentum density, they're exactly the same. That is actually true in all physics. It's clouded by the fact, though, that in most units that we measure things, we don't typically use speed of light equal to 1. So when you look at these kind of quantities in the physics that you are more used to, you would find your energy, your energy flux, your momentum density will have different units, because you'll have factors of c that come into there that convert from 1 momentum, and energy, and time flow, and space flow. Also, you don't usually include rest mass and rest energy in the energy flows and momentum densities that you've probably been familiar with in the past. So it's when we do relativity, and we set the speed of light equal to 1 that this symmetry between energy flux and momentum density becomes apparent. Question. AUDIENCE: The second one, do you mean rho 1 will be there? SCOTT HUGHES: Sorry. This one? AUDIENCE: Yeah, on the right-hand side. SCOTT HUGHES: On the right-hand side. This is-- AUDIENCE: Rho i. SCOTT HUGHES: No, this is definitely what I mean to write down here. So this is-- what I've done, I've taken advantage of the fact that I can write-- let's see. Do I have it down somewhere? Yeah. Yeah. So what I've done over here is I have written the fact-- so u sub t, ut is just the gamma factor. And I didn't explicitly write out what t-- the i components, but you get a similar factor that way. So no, that is correct that way. AUDIENCE: So that's why you mean the first and the second are the same. SCOTT HUGHES: Yeah. Yeah. And I just want to emphasize-- I mean, the key thing which I want to emphasize is when you do it in the physics that you guys have generally all seen up till now, it won't look that way. And there's two causes of that. One is you're not used to necessarily working in units where the speed of light is 1. And two, rest energy is built into these definitions. That shifts things a little bit. The other thing which I want to emphasize is notice this is symmetric under exchange of indices. Now, it's obvious for dust. It comes right back to the fact that these guys-- that those indices enter like so. It's an exterior product of the four-velocity with itself. I'm not going to go through in detail, but there is a physical motivation that I will sketch in a few moments that argues that it must be symmetric like this in all physical cases. I have a detailed sketch of this. I keep using the word sketch, because my mind wants to say proof, but it ain't a proof. It's sort of a motivation. And so I will post it for everyone to go through, but I will sort of illuminate it in a couple of moments. Basically, if this is not true, if it's not the case this is symmetric, a physical absurdity can be set up. I'll describe that in a few moments. So most of the world is not dust. So this is a decent example for helping to understand things. But we are going to want to do more complicated and interesting examples. Pardon me. Just one second. Let me just check here with this page. Yeah. So for us, the way that we are going to do this, there is a fairly general recipe that one can imagine applying to this. I'm going to save it for a little later in the class. It sort of borrows some techniques from field theory. Basically, if you can write down a Lagrangian density that describes the fifth system that you're under study, there's a particular variation you can do, that stress energy tensor emerges from. But for now, the key thing which I want to say is that we basically are going to deduce what the stress energy tensor looks like by essentially going into a particular frame-- we'll call it the rest frame. It's usually the rest frame of some element of the material being studied-- and thinking carefully about this physical definition of what the different components mean. This is not the most rigorous way to do it, but it's a good way to get started and develop some intuition. So let me give one example that for many of us in astrophysics, this is probably the one stress energy tensor that we write down and use over, and over, and over again in our career. And it's rare we do anything more than this in many, many cases. This is called a perfect fluid. So what is perfect about a perfect fluid? It begs the question here. What is perfect referring to here? So a perfect fluid is a fluid in which there is no energy flow in what I will call the "rest" frame. I put rest in quotes, because you have to sort of define it from the context of what your fluid is here. Basically, what it means is I can find a frame in which each fluid element there is no energy flow there. If a frame exists where that happens, there's a candidate to be a perfect fluid. And I also require there to be no lateral stresses. Lateral stresses refer to this Tij when i and j are not equal to one another. So this sort of refers to-- imagine that I have some-- let's say going into the board is the y direction. So if there's like some y stress that is somehow being transported in the x direction, that would be a lateral stress. Physically, this is actually-- that kind of a stress is hugely important when one is studying fluids. And one characterizes it by a quantity known as the viscosity. Viscosity tells me about how stress gets transported-- how momentum, rather, gets transported in a non-normal direction, against the direction in which the fluid is moving. So my perfect fluid has no viscosity. And I'll just conclude. I want to make-- there's a very important point here. A fluid that has no viscosity is a fluid that doesn't get anything wet. So this refers to when you pour water on yourself, the reason your hand gets wet is that there's some viscosity that actually prevents it-- causes there to be sort of a shearing force, which causes the water to stick to your skin. So a perfect fluid, this has been described as the physics of dry water. So it is-- anyone who's in applied math will know. They'll sort of roll their eyes and say, OK, fine, we'll do this sort of infant version of fluid first. And then a lot of the action and a lot of the fun comes from putting in viscosity and doing the real fluids. For the purposes of our class, what this boils down to is this tells me that the physics of this quantity is totally dominated by the fluid's energy density and its pressure. The pressure is an isotropic spatial stress. So in this particular frame, where I have no energy being transported, then the stress energy tensor can be represented as energy density as usual up in the-- whoops, over there now-- up in the tt component. There is no energy flux. By symmetry, if there's no energy flux, there is no momentum of density. And my spatial stresses are totally isotropic, and none of them are lateral. So it just looks like this, the diagonal of rho p, p, p, p. As I said, this is actually something that we're going to use over, and over, and over again. This is actually a-- so you can actually consider my dust stress energy tensor to be a perfect fluid with no pressure. So this actually subsumes this other one. I will-- just illustrative purposes, I'll show an example of a case that cannot be thought of as a perfect fluid. But we will tend to use this a lot in our class. And as I will demonstrate in just a moment, we're going to find that this ends up playing an important role in generating gravitational fields. What's interesting about this is you guys are probably used to the idea that mass generates gravity, and then throw in a c square that tells you energy to generating gravity. But it's also-- we're going to see pressure generates gravity. And it's connected to the mathematical structure of this guy here. So notice I've written an equal dot here. So this is just the way this is represented in this particular frame. I would like to write this in a more covariant form, something that does not rely on me going into a particular frame of reference. So the trick which I'm going to use for this is when I think about that form there-- so the rho piece of it, clearly, what I'm doing there is I'm picking out-- that is-- can be thought of-- as the energy density multiplied by the tensor product of the four-velocity that describes a particular element of this fluid. So I can write, again, using this sort of abstract notation. This piece of it looks like this. And so if I go into the frame, if I go into the rest frame of the fluid, that's just my u is 1 and 0. And the spatial components, that builds my upper left-hand corner of this tensor. How do I get the rest of this? Well, to get the rest of it, these are all sort of picked out of components of the tensor that are orthogonal to u. And I put an Easter egg in p set 1. You guys developed a tensor that allows me to build-- it's a geometric object that allows me to describe things that are orthogonal to a given four-vector. So if I take the projection tensor that you guys built one piece at 1, which looks like the metric plus the extern-- the tensor product of a four-vector with itself, that gives me the p's that go into that component. Or if I write this in index notation, I can do it in two ways. I will also emphasize that there are few moments in this class where I sort of urge you to take the sort of long-term memory synapses and switch them on. This is one of those moments. In a couple of lectures, we're going to introduce the principle of equivalence, which is the physical principle by which we're going to argue how we go from formulas that work in special relativity to formulas that work in general relativity. And by invoking the principle of equivalence, we're going to see that when we want to describe perfect fluids in a general spacetime, not just in special relativity and general relativity, it's exactly this formula. I just need to modify what the metric means, but that will allow me to carry that over. Let me see. So my notes are a little bit disorganized here. There's a piece that I-- every year I want to clean this bit up, and every year at this time of year, I have 70 gajillion administrative tasks. And I end up getting behind schedule. So I will clean this up on the fly. So there's a couple of points which I want to make about this. Let me do this point first. So by virtue of taking a graduate physics class, I'm confident you guys know about Newtonian gravity. One can write the field equation for Newtonian gravity as essentially a differential equation for the potential that governs Newtonian gravitational interaction. So let's call phi sub g the Newtonian gravitational interaction. Whoops. And I can write a field equation that's governing it as essentially its Poisson's equation. So the Laplace operator acting on that potential is up to a constant equal to the-- we usually learn it in terms of mass density. We're working in units where the speed of light is equal to 1. So it could just as well be the energy density. So you've all kind of seen that. Now, if we think about how we're going to carry this forward and make gravity a relativistic interaction, this equation should right away make us suspicious, because we spent several minutes earlier today talking about the fact that this is not a scalar. This is a component of a tensor. A physical theory which tries to pick out just one component of a geometric tensor is not a healthy theory. It would be like if you had sort of learned in E&M that there was a preferred direction of the electric field. If there's a particular set of Maxwell's equations for Ex, and a different set of Maxwell's equations for Ey. That would just be nonsense. Nature doesn't pick out spatial components of any one being particularly having some weight over the others, and the same thing holds in relativity for spacetime. So when we make this into a relativistic theory, we're going to say if this component of a tensor plays a large role in gravity to make this into a geometric object that makes a large role in gravity, I'm going to have to-- so let's call this my Newtonian equation. My Einsteinian equation-- I'm going to put an equal sign in quotes here, because there's a lot of details to fill in. But what's going to go on the right-hand side of this has to be something that involves the stress energy tensor. Newton picks out one component stress energy tensor. Relativity doesn't let me pick out particular components. So whatever I get when I do Einstein's gravity, the whole stress energy tensor is going to be important in setting the source of my gravity. That then sort of says, well, then what the heck do you do with this left-hand side? That is, in fact, going to be starting probably on Tuesday the subject of the next couple of lectures, basically going all the way up to spring break. The week before spring break is when we complete the story of what goes on the right-hand side-- excuse me-- the left-hand side of this equation. But what I will tell you is that it is indeed going to involve two derivatives of a potential-like object. And the potential-like object is actually going to turn out to be the metric of spacetime. So that's kind of where we're going with this. Let's do a little bit more physics with the stress energy tensor. So I have somewhat more detailed notes, which I will post online, which I am not going to go through in great detail here. But I'm going to kind of sketch this. So one is that I would like to prove-- again, that word is a little bit of an overstatement, but at least motivate the symmetry of this tensor. I'm just going to focus on the spatial bits of this. That'll be enough. Like I said, you can kind of see that T0i and ti0 are the same thing by thinking about the physical meaning of four-momentum and what a flux of four-momentum is. This one, there's kind of a cute calculation you can do. So imagine you have some little cube of stuff. It could be immersed in some field or fluid, something that is described by a field of stress energy. And so to start out with, let's look at how the flux-- remember what t alpha beta, or really Tij tells me about is the flux of momentum in a particular direction. It's telling me about the amount of momentum going into this box on one side and coming out on the other. So I'm going to look at the momentum flux into and out of this box. And so what I'm going to do is let's look at the momentum going into the sides that are-- so let's call this the top and the bottom here. What I'm going to do then is number the four sides, the other four sides of the box, not the top and the bottom-- so let's call this side, which is sort of facing away from us here, that side one. The one that is facing us that sort of points out in the x direction, I'll call that two. This side here, I will call three. And the one that is on the back I will call four. Apologies for the little bit of a busy picture here. But just what you want to do is sort of imagine there's a cube in front of you, and you go around and label the four sides. What we'd like to do is calculate what is the force that is flowing through each of these four sides, 1, 2, 3, and 4. Just ignore the top and the bottom for just a second. I'll describe why that is in just a moment. So if I look at the force on face one, well, face one, it is-- pardon me-- I mislabeled my sides. I want to be synced up with my notes, and so I realize it's annoying. And I'm very sorry about that. But if I mess this up, I will get out of sync with what I have written down. So the one that is facing us is side one. Two is on the right-hand side. Three is the back. Four is on the left-hand side. My apologies for that. But I think it's important we get that right so I don't get out of sync with myself. So the total force on face one is what I get by basically adding up all the flux of momentum flowing through face one. So force on face one, that is-- the i-th component of that is what I get when I integrate that over face one, which is normal to the x-axis. Each side of the cube is little l. So it's Tix l squared. Force on face two, now this is the one that is normal for the y-axis. And so this guy looks like Tiy l squared. And you continue this F3-- if you look at, and you if you assume that this thing is small, it's approximately the same as the force on F1. It'll become equal in the limit of the q becoming infinitesimally small. And this is approximately the same as the force on face two. So that tells us that there's no unbalanced force on this thing, which is great. Basically, it means that there's no unbalanced force causing this little element to accelerate away. One of the reasons why I am focusing on these four sides, though, is I would also like to consider torques that are acting on this thing. And here's where I'm going to skip over a couple of details and just leave a few notes for you guys to look at, because we're a little short on time with other things. This calculation is straightforward, but it gets-- I already screwed up a detail here. I don't want to sort of risk messing up a few other things. What I want to do is consider the torques about an axis that sort of runs through the middle of this thing that goes from the top and the bottom of this. So go through and just add up all the torques associated with these little forces about an axis in the center of-- that goes through the center of this cube. So I'm going to-- like I said, I'm going to leave out the details, but basically, you go through, and you do the usual r cross f to get the forces, the torques associated with each of these different faces. And what you'll find when you do this is that there is a net torque that looks like l cubed times Txy minus Tyx. Again, though, this scales with the size of the cube. So you look at this and kind of go, well, who cares? It's a little bit there. Maybe it spins up a little bit. But in the limit, the thing going to 0, there's no net effect. Well, let's be a little bit careful about this. What is the moment of inertia of this cube? I don't know exactly, but I know that it's going to be something like the mass of this cube. We'll call that l cubed times its mass density, or its energy density. And it's going to involve two powers of the only length scale characterizing this thing. And there will be some prefactor alpha. There will be some number that's related to the geometry of this. An integral or two will easily work out what the alpha is and make this more precise. But the key bit which I want to emphasize is that this is something that scales as the size of the fifth power. So yeah, the torque vanishes as l goes to 0, but it does so with the cubed power. It doesn't vanish as quickly as the moment of inertia vanishes. And I'll remind you, the angular acceleration of the cube, theta double dot, is the torque divided by the moment of inertia. So this is something that is proportional to Txy minus Tyx divided by l squared. So yeah, the torque does vanish as l goes to 0, but the moment of inertia vanishes more rapidly as l goes to 0. That sort of suggests that if we're in a screw universe, little microscopic vortices are just going to start randomly oscillating in any cup of water that you pour in front of you. Now, I don't have a proof that nature abhors that, but it seems pretty screwy. And so the case that most textbooks make at this point is to say, physics indicates we must have Txy strictly equal to Tyx in all cases, independent of what the size of this thing is. That had better just be a bloody 0 in the numerator in order to prevent this physical absurdity from being set up. Repeat the exercise by looking at torques around the other axes. And that drives you to the statement that this thing must, in general, be spatially symmetric. And again, physics of the way that energy and momentum behave in relativity makes our sort of timespace component symmetric as well. I emphasize this is not a proof. This is really just a physical motivation. This is the kind of thing that I'm a bloody astrophysicist. I like this kind of stuff. There is a different way of developing the stress energy tensor, as I said, that comes from a variational principle sort of based on sort of quantum field theoretic type of methods. And the symmetry in that case is manifest. It really does sort of come up. But this is a good way of just motivating the fact that you're not going to have any non-diagonal-- or excuse-- non-symmetric stress energy tensors. If you did, you would have really bizarre matter. Now, the stress energy tensor has this physical interpretation that it tells me about the flow of energy and the flow of momentum in spacetime. As such, it is the tool by which we are going to put conservation of energy and conservation of momentum into our theory. And the way we're going to do this is with a remarkably simple equation. The spacetime divergence of t alpha beta-- pardon me-- must equal 0. This is a covariant formulation of both conservation of energy and conservation of momentum. And if you want to say, well, which one? Is it energy or is it momentum? You can't say that in general. I can only say that after I have picked a particular reference frame, because it's only once I have defined time and I've defined space that I've actually defined energy and momentum. Prior to choosing a particular reference frame, all I have is four-momentum. One observer's energy is another observer's superposition of energy and momentum. Once I have picked a particular frame-- so once I have picked a particular frame, then if I evaluate d alpha T alpha either 0 or t, this is what conservation of energy looks like in that frame. So in that frame, here is conservation of energy. I can fit it. Hang on a second. No, no, no. I had it right the first time. Yeah. Sorry. I was just trying to make sure I got my indices lined up properly. So apologies you can't quite read the bottom one so well. The top one is conservation of energy in that particular frame. The second one is conservation of momentum in that particular frame. And again, the key thing which I want to emphasize is the covariant statement basically puts both of them together into a single equation. I can only sensibly state conservation of energy and conservation of momentum according to some particular observer. Now, I can repeat the game that I had done before with the number vector and turned these conservation laws into integral equations as well. Let me do it for energy. So if I take that integral equation and integrate over a three-volume, or if you prefer, I can take the original covariant formulation and integrate that over a four-volume. With a little bit of manipulation akin to the way I manipulated the conserv-- the integrals associated with the number vector in the previous lecture, you can write down a law that looks like this. And so this-- again, what I've done here is I've chosen a particular set of timeline coordinates. And the language that we often use in relativity is we basically say I'm going to take a single slice of time. And I would say that the rate of change of energy-- so integrate energy over a volume. It's the total energy in that V3. The rate of change of that thing is balanced by the amount of energy flowing into or out of through the boundaries of that volume. Add an extra index here. So make this-- make one of these be a j. Make this guy be a j. And you've got a similar statement for the conservation of momentum. This is a particular trick. I have a set of typed up notes that just sort of clean this up a little bit. They're basically exactly this point, but I'll put them online after today's lecture. You're going to want to use this on one of the p set problems this week. So you guys are going to do a couple exercises that involve integrating and actually finding essentially moments of the left-hand side of this thing. You're going to do a few exercise where you take advantage of this formulation of conservation of energy and momentum to derive a few identities involved in the stress energy tensor, one of which is another Easter egg that we're going to use quite a bit in a future lecture. So let me just wrap up our discussion of the stress energy tensor by just doing two more examples. And then I'm going to sort of begin switching gears a little bit. So far I have only described-- essentially, I've really only described perfect fluids. Dust is a perfect fluid with no pressure. There are a lot of other kinds of materials that we want to work with in the universe. One of which-- it's unlikely many of you are going to use this, but it's actually the stress energy tensor on which I have built-- well, I guess Alex is going to use it a little bit-- I have built a big chunk of my career. Suppose you have-- we kind of talked a little bit about how the four-velocity and the four-momentum are really only good for talking about like the kinematics of particles. Well, actually, particles aren't a bad thing to focus on in some of your studies. And much of my research is actually based on the idea of thinking about a binary system as one member being a black hole, and its companion being a particle-like object that is a particular limit of it, a particle-like object that orbits it. So a really simple stress energy tensor-- and I'm just throwing it out here, because I think it nicely illustrates the principle. A point particle with rest mass, we'll call it m0. And I'm going to say it's moving on a particular world line through spacetime. So the way we define a world line is we just say it's some four-vector that describes the place of this thing from some chosen origin. And it's generally most convenient to parameterize it by the proper time of whatever object or creature is moving on that world line. And so it's a lot like dust, only there's no volume. It's sort of like one particle of dust. And so the stress energy tensor we use for this-- let me back up for just a second here. When you guys learned about electricity and magnetism, one of the first things you learn about are point charges. And then a little bit later you learn about charge distributions. And you have charge densities. And then, usually, at some point in a class-- it's often like junior level E&M-- we say, how do you describe the density of a point charge? That's where you learn about the Dirac delta function. Well, if I have a point particle, I'm going to need to describe this thing's energy density as essentially a Dirac delta function. And so what we do is imagining that these things-- so this is the four-velocity of that body. They might be functions of time as this guy is moving along here. What we do is we introduce a Dirac delta function as this thing moves along through spacetime. What this does-- you can check the dimensions. This gives me exactly what we need in order to have something that's dimensionally correct, and has all the symmetries and all the properties that describe a particle moving at a particular four-velocity through spacetime. Now, you might want to just-- it's the thing which I kind of want to pause on for a second. You go, what the hell do you do with this, right? That's kind of inconvenient. Well, the trick we use to sort of clean up that Dirac delta function, it's very much like what you do when you encounter multidimensional delta functions in basic physics. You just build it out of a bunch of one-dimensional delta functions. Likewise, you'll have a term with a y component and the z component. Then you use the rule that if I integrate a function of x against a delta function whose argument is itself a function of x, I evaluate that function at the 0's of g. So let's say x0 is where g has a 0. Normalizing to the first derivative of g evaluated at that 0. When you put all of that together, that basically means you can do this somewhat abstract integral formally. I mean, exactly. You can just do it analytically. And what you do is you choose one of the delta functions to apply it to. Traditionally, people apply it to the timeline component. And what is the derivative of the world line z component with respect to proper time? It's the 0 component of the four-velocity. So you just divide by the 0 component of the four-velocity, and then you're left with a three-dimensional delta function for the spatial trajectory of this thing through all space. So this is an example of one that just, again, kind of the intuition is one of the things I want to emphasize. Notice we had-- this has the symmetries that we wanted to have. It's not hard to show that this. You can think of this as essentially being kind like a gamma factor. This all ends up giving me just what we need for this thing to have the right transformation properties. And it does, in fact, play a role in some-- well, I'll say in some research that's near and dear to my heart. Let me do another example. Suppose you want to know the stress energy tensor associated with a given electric and magnetic field. Well, first, let me just quote for you the exact answer, which is most compactly written if we use that Faraday tensor F, which describes electromagnetic fields in a frame-independent fashion, the way that I introduced it in the last lecture. So in units where basically everything but pi is set equal to 1, it ends up turning into this. So that's a bit of a mouthful. Let's go in and look at particular components of it, though. So let's say I go into a particular frame. I fill in my Faraday tensor with the form of electromagnetic field that I introduced last time. And I'll just go through, and I'll evaluate all the different components of this thing. So what you find when you fill this in is your T00 component, 1 over 8 pi e squared plus b squared. That's good. Hopefully, y'all remember from basic E&M, the energy density of an E-field is e squared over 8 pi and the right system of units. Energy density of a b field is b squared over 8 pi if you work in God's units. Let's do the timespace component. So again, hack through that mess there. This is going to be something that is a vector. In fact, it's the Poynting vector. Could it be anything else? If you use the recipe that I sort of suggested as the easiest way to approach this, this is kind of what you would have guessed for something like that. The bit that's actually kind of hard is then trying to get the spatial stresses of this thing. And here I don't have any great intuition for this one to convey to you. It's derived in textbooks like Griffiths and Jackson. So I'll just quote for you the result. So you get one term that looks like E squared plus b squared on the diagonals. Then there's a correction, which looks like this. I want to just quickly call out one example so you can see what the significance of this example is. So suppose you have something like a pair of capacitors. And there's just a uniform electric field between them. You want to evaluate the stress energy tensor between those pairs of capacitors. So my spatial electric field, let's say it just points in the Ex direction. And it's constant. So when you actually evaluate this guy, there's no energy flow. There's no magnetic field. So there's no Poynting vector. You, of course, have E squared over 8 pi for your energy density. Very different from a perfect fluid. And this kind of makes sense. That's sort of telling you that there is a stress that if you have a pair of plain parallel capacitors, you tend to attract the plates to each other. But there is a pressure associated with that electric field that actually goes in the other directions. This is your x direction. This is y and z. And as a consequence of this-- so there's some stuff which we're not going to do too much with in this class, but I may give you some pointers on this. Electric and magnetic fields, they generate pressures, at least in certain directions. They kind of generate like an anisotropic pressure. And when we start coupling this to gravity, you can get electric fields and magnetic fields that contribute non-negligibly to the gravity of their object. That is it for the stress energy tensor. As I said, we are going to use this guy over, and over, and over again. And the reason for this does come back to that little motivation that I gave probably about 45 minutes ago where we sort of looked at the Newtonian field equation, and then said, picking out a particular scalar as the source of gravity makes no sense in a relativistic covariant theory. It's got to be the whole tensor. And indeed, this is the one-- well, not the E&M one, but the general notion of a stress energy tensor is the one that we are going to use for that. So we have about 10 minutes left. And so I would like to start the process of switching gears. Before I do that, are there any questions? I will clean the board. So the reason we are switching gears is we now have probably the most important physical tools that we need in order to start thinking about making a relativistic theory of gravity. But we need a few more mathematical tools. In particular, I'm going to argue in a couple of lectures that flat spacetime is not sufficient for us to build a theory of gravity. We're going to need to-- first of all, you're going to have to understand what that means. And we're not quite ready to go there. So for now, it's just fancy words. But I'm going to have to introduce some kind of a notion of curvature into things. What does that even mean really? We need to have the tools to do that. And so as a prelude to going in that direction-- I will call this my prelude to curvature-- what we're going to start doing is flat spacetime in curvilinear coordinates. And the importance of doing this, why this is going to be useful to us is that it will introduce a-- it'll keep the physics simple. It's still going to be special relativity, but it's now going to special relativity using a mathematical structure in which the basis vectors are no longer constant. So that's going to allow us to begin making a couple of the mathematical tools that are necessary to build gravity into this theory. So we'll start by replacing this with just simple plain polar coordinates mapped in the usual way. So if we want to go back and forth, well, at least one way transformation, I build x and y from r and phi in that usual way. There's an inverse mapping as well, which involves trig functions. So I'm not going to write it down. One point which I really want to emphasize here is we are going to continue to use a coordinate basis. And to remind you what that means, so a coordinate basis means that the differential displacement vector from an event a to a nearby event b is simply related by differentials of my coordinate contracted onto my basis vectors. But when I'm working in a curvilinear coordinate system like this, that means one of them has a somewhat different form from what you are used to. When you guys talk about the differential of a displacement in just about every physics class up to now, if you have a differential angle, you'd want to throw an r. So this has the dimensions of length. We ain't going to do that. And so what this means, since this is an angle, every component of this has the dimensions of length. That means that my basis vector is going to be something that has the dimensions of length associated with it. This in turn means that this is not going to be a normal basis. That's unfortunately a somewhat loaded word. What I mean by that is that it has not been normalized. But given that you guys have spent all of your career thinking about the dot product of a unit vector with itself-- of a basis vector with itself being equal to 1, the other meaning of normal might be good for you, too. The key thing which I want to emphasize here is e phi dot e phi does not equal 1. One other little bit of notation which I would like to introduce-- so we are going to want to talk about transformations between different representations. We've done this-- so far, we have generally focused on moving between different reference frames. I want to generalize this notion. And I'm going to tweak my notation a little bit to indicate the difference. So I'm going to call capital L alpha mu bar what I get when I just look at the variation of the alpha coordinate system-- sorry-- the unbarred coordinate system with the barred one. So this is just-- there's no deep things here. I just want you to be familiar with the notation which I'm going to use. I will always reserve lambda for the Lorentz Transformation. So in the interest of time, I'll just write down one of these. So this sort of means like, for instance, suppose that I'm transforming. Let's let barred indicate my polar coordinates. Unbarred be Cartesian. These things are surprisingly sticky sometimes. So if I want to transform in one direction between my barred and my unbarred ones, this matrix will go through it. And it looks like this. So this is the thing which I will call l alpha mu bar. In my notes, I also give the inverse transformation. And I will actually write-- you know what? I have minute. I will write that one down as well. If you want to know how to go in the other direction, it's just the matrix that inverts this. The reason I decided to take the extra 30 seconds or so to write this down is I want to call out the fact that in this kind of transformation, because of the fact that this is a coordinate basis that has this somewhat unusual property, different elements of the matrix have different dimensions. It's a feature, not a bug. So this is going to show up. I'm going to do a couple more calculations with this a little bit tomorrow-- sorry-- not tomorrow, on Tuesday. This is going to show up in the way the metric looks. The metric is going to have a very different character in this coordinate system. And we're going to see the way in which-- it basically boils down to it's going to pick up a non-trivial functional form. It's not going to just be a constant. And that fundamentally reflects the fact that the basis vectors are not constant anymore. We'll end it there. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 253_Digraphs_Connected_Vertices_Video.txt | ALBERT MEYER: The topic for this video lecture is Connected Vertices, the kind of connections we can have between different vertices through edges. We're going to start by showing that the shortest walk between two vertices is a path. We're going to prove it by contradiction. Suppose we have some path from u to v and it crosses over itself. So here we have u and v. At some point, you get to c, and you go back to c, and from there you go to v. So you've gone through some vertice twice. But if you want to get the shortest path from u to v, why would you go through this loop? Why not just keep going straight from c to v? The path without that section that goes from c back to itself also for u to v and is shorter. So if we have any path that crosses over itself, we can just get rid of that part that loops around back into itself, and we still have a walk from u to v. So therefore the shortest walk from u to v is going to be a path. Now we're going to talk about the length and the walk relation. What this means is with two vertices, v and w, there is this Gn relation between v and w if there exists a length n walk from v to w. Gn is called the length n walk relation for G. So basically, if you can find a way to go from v to w in exactly n steps, then Gn applies from v to w. G itself, when you think about it, is a length 1 walk relation. The graphs define these relations. There is an edge from one vertice to another if there is a length 1 edge from one vertice to another. It is itself. Now, this lemma-- they say that Gm and relational composition with Gn equals Gm plus n. Let's explain with this means. So what that relational composition means is that the relational composition between the two applies from x to y if there is some vertice, z, such that there is a path, n, from x to z, and then a path, n, from z to y. Gm applies from x to z and Gn applies from z to y. And this is why this is the same thing as Gm plus n makes sense. If there is a path length n to z and a path length n from there to y, you just go from x to z in m steps, then z to y in n steps. And you have m plus n steps from x to y. The length 0 walk relation just make each vertice go back to itself. It points back to itself, the individual one. And the lemma is still true. G0 composes with Gn is just Gn, which makes sense. Everything itself plus Gn just gives you Gn. Let's talk about composition of matrices. So if we have some adjacency matrix for G and we do a composition with sum h, then we can get that by applying this Boolean and/or matrix multiplication. These adjacency matrices are ones and zeros. So we do matrix multiplication, but with Boolean operations instead of plusses and multiplications. So we can compute A G of n by fast matrix exponentiation. How do we do that? Well, basically you can do it for G n over 2 twice, and then Gn over f for each of those twice, and go doing those Boolean operations on each. So A G of n equals A G of n over 2, applying this operator to A G n over 2. So you can just break it down in 2 each time so you get logarithmic number of problems that we have to do. Now let's define another relation. G star is just called the walk relation of G. Basically, G star applies from u to v if there is a walk from u to v, no matter how long it is. If you can find some way to get from u to v, then it applies. If you want to get the walk relation, you just get everything inside the graph and apply self-loops. So add in an edge that points back to itself for every vertice. We called this G less than or equal to. It's basically G and then add in these G0 self edges. And G less than or equal to has a length n walk if G has a less than or equal to n walk. Now, think about that. If I can get from vertice x to vertice y in n minus some amount of steps in G, then I can get there in n steps in G less than or equal to because I can just loop around. If I want to get here from red to blue, I can get there in one step without those self-loops. But with the self-loops, I can just keep starting from red and go around to red as many times as I want, n minus 1 times, and then do this final step. So I can make a length n walk for any value of n greater than 1. Now let's compute the walk relation using what we've just defined. So G has n vertices. So the length of paths are going to be less than or equal to n. If you just go in a straight line from one thing to another, passing through each possible vertice, at most you're going to get n minus one length. Because you're going to pass through n vertices, so there's n minus 1 edge that's connected there. So G star, which is just the relation if there is a walk from u to v, is going to be this G less than or equal to n minus 1. So if we get G less than or equal to, add in all these self-loops, and then find all of the paths of length n minus 1 in there, which is basically all paths since G less than equal to n minus 1 is all paths less than or equal to n minus 1. But since G only has paths of less than or equal to n minus 1, that's just all the paths. It's just every single one of the vertices. And so we've defined G star and that's how we get all connected vertice pairs. And we can do this in n squared log n time using that composition of the adjacency matrices. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 343_Two_Pair_Poker_Hands_Video.txt | PROFESSOR: So let's do a basic example of counting that illustrates the use of these new generalized rules of the Division Rule and the Generalized Product Rule. And let's count some particular kind of poker hands called a 2-pair. So poker is a game where each player is dealt five cards from a deck of 52 cards. And the definition of a 2-pair hand is that there are 2 cards of some rank. The ranks are Ace, Deuce, up through King, so the ranks are 13 possible ranks. Ace is 1, 2, 3, up through 10, and then Jack, Queen, King is 11, 12, 13. So there are 13 possible ranks. We're going to choose 2 cards of some rank-- that's called a pair. Then we're going to choose 2 cards of a different rank-- a second rank. And finally, we're going to choose a card of still a third rank. So I get a pair, and another pair, and another card that does not match the ranks of either of the first two. And that is the definition of a hand that, in poker, is called 2-pair. Let's take an example. Here's a typical 2-pair hand. I've got 2 Kings. They both have rank 13. One is a King of Diamonds, the other is a King of Hearts. There are four of these suits, so-called-- Diamonds, Hearts, Spades, Clubs. There are 2 Aces, a pair of Aces. One is an Ace of Diamonds, the other's an Ace of Spades. And finally, there hanging loose, this third rank that doesn't match the Kings or the Aces-- namely a 3 of Clubs. Now, the way that I'm going to propose to count the number of 2-pair hands is to think about it this way. I'm going to begin by choosing the rank for the first pair, and there are 13 possible ranks that the first pair might have. Once I've fixed the rank for the first pair, the second pair has to have a different rank. So there are 12 ranks left. And once I've picked the ranks for the 2 pairs, then I have the rank of the last card, which is 11 possible choices. Then, in addition, once I've chosen the rank of the first pair, the rank of the second pair, and the rank of the loose card, the fifth card, I select a pair of suits to go for the first pair. So let's say if the first pair, I've figured out we're going to be 2 Aces. Which two aces should they be? Well, pick two of the four suits. And there are four choose two ways to choose the suits for the pair of aces. Likewise, there are four choose two ways to choose the two suits for the pair of kings. And finally, there are four possible suits I can choose for the rank of the last card. So that says that I might, for example, specify a two-pair hand by saying, OK, let's choose a pair of kings to come first and a pair of aces to be the second pair and a three to be the loose card. Let's choose the set of two elements diamonds and hearts for the kings, the two elements diamonds and spades for the aces, and a club for the three. This sequence of choices specifies exactly the two-pair hand that we illustrated on the previous slide, namely two kings, a diamond and a hearts; two aces, a diamond and a space; and the three of clubs. So I can count the number of two-pair her hands fairly straightforwardly. There were 13 choices for the rank of the first pair, 12 for the second, 11 for the rank of the third card, four choose way to choose the suits of the first pair, four choose way to choose two ways to choose the suits of the second pair, and four ways to choose the suits for the last pair. So this is the total-- 13 times 12 times 11 times 4 choose 2 twice times 4. And that's not right. There's a bug. What's the bug? Well, the problem is that what I've described in this number on the previous slide, that number, is exactly the set of six tuples, consisting of the first card ranks and the second card ranks and the last card rank and the first card suits and the second card suits and the last card suit. That is, if it's counting the number of possible ranks for a first choice, the number of possible ranks for a second choice, third, and so on, this set of six things are being counted correctly by the formula on the previous page. The difficulty is that counting these six tuples is not the same as counting the number of two-pair hands. We've counted the number of six tuples of this kind correctly, but not the number of two-pair, because this mapping from six tuples to two-pair hands is not a bijection. Namely, if I look at the six tuple, choose kings and then aces and a three with these suits and those suits and final suit for the three, which determines this hand-- the king of diamonds, king of hearts, ace of diamonds, ace of spades, three of clubs-- there's another six tuple that would also yield the same hand. Namely, what I can do is I'll keep the three of clubs specified. But instead of choosing the kings and their suits and the aces and their suits, I'll choose the aces and their suits and the kings and their suits. So I'm just switching these two entries and those two entries. And if I do that, here's a different six tuple that's specifying the same two-pair hand. That is, this tuple is specifying a pair of aces and a pair of kings, where the aces have suits diamonds spades and the kings have suits diamonds hearts and the three has suits clubs. So the bug in our reasoning was that when we were counting and we said there are 13 possible ranks for the first pair and there are 12 possible ranks for the second pair and we were distinguishing the first pair from the second pair, that was a mistake. There isn't any first pair and second pair. There are simply two pairs. And there's no way to tell which is first and which is second, which is why we got two different ways from our sextuples of mapping to the same two-pair, depending in the sextuple which of the two-pair I wanted to list first. So in fact, since either pair might be first what I get is this map, from six tuples to two-pair hands, is actually a two-to-one mapping. It's not a bijection, because there's no difference between the first pair and the second pair. There's just a couple of pair. If I do that, then I can fix this formula. Now that I realize that the mapping from these six tuples, which I've counted correctly to the things I want to count-- namely, the two-pair hands-- is two to one, then, by the generalized product rule, or by the division rule, all I need to do is divide this number by a half. And that is really the answer of the number of two-pair hands. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 444_Random_Variables_Uniform_Binomial_Video.txt | PROFESSOR: Certain kinds of random variables keep coming up, so let's look at two basic examples now, namely uniform random variables and binomial random variables. Let's begin with uniform, because we've seen those already. So a uniform random variable means that all the values that it takes, it takes with equal probability. So the threshold variable Z took all the values from 0 to 6 inclusive, each with probability 1/7. So it was a basic example of a uniform variable. And other examples that come up, if D is the outcome of a fair die-- dies are six-sided. Dice are six-sided. So the probability that it comes up 1 or 2 or 6 is 1/6 each. Another game is the four-digit lottery number where it's supposed to be the case that the four digits are each chosen at random, which means that the possibilities range from four 0's up through four 9's for 10,000 numbers. And they're supposed to be all equally likely. So the probability that the lottery winds up with 00 is the same as that it ends up with 1 is the same that it ends up with four 9's. It's 1/10,000. So that's another uniform random variable. Let's prove a little lemma that will be of use later. It's just some practice with uniformity. Suppose that I have R1, R2, R3 are three random variables. They're mutually independent. And R1 is uniform. I don't really care about the other two. I do care technically that they are only taking the values. They only take values that R1 can take as well. So I haven't said that on this slide, but that's what we're assuming. And then I claim is that each of the pairs, the probability that R1 equals R2-- the event that R1 is equal to R2 is independent of the event that R2 is equal to R3, which is independent of the event that R1 is equal to R3. Now, these events overlap. There's an R1 here and an R1 there and there's an R2 here and an R2 there. So even though the R1, R2, R3 are mutually independent, it's not completely clear. In fact, it isn't really clear that these events are mutually independent. But in fact, they're not mutually independent. In fact, they're pairwise independent. They're obviously not three-way independent-- that is, mutually independent-- because if I know that R1 is equal to R2 and I know that R2 is equal to R3, it follows that R1 is equal to R3. So given these two, the probability of this changes dramatically to certainty. So this is the useful lemma, which is that if I have the three variables and I look at the three possible pairs of values that might be equal that whether any two of them are equal is independent of each other. Now, let me give a handwaving argument. There's a more rigorous argument based on total probability that appears as a problem in the text. But the intuitive ideas, let's look at the case that R1 is the uniform variable, and R1 is independent of R2 and R3. So certainly, that implies that R1 is independent of the event that R2 is equal to R3, because R1 isn't mutually independent, both R1 and R2. Doesn't matter what they do, so it's independent of this event that R2 is equal to R3. Now, because R1 is uniform, it has probability p of equaling every possible value that it can take. And since R2 and R3 only take a value that R1 could take, the probability that R1 hits the value that R2 and R3 happens to have is still p. That's the informal argument. So in other words, the claim is that the probability that R1 is equal to R2 given that R2 is equal to R3 is simply the probability that R1 happens to hit R2, whatever values R2 has. This equation says that R1 equals R2 is independent of R2, R3. And in fact, in both cases, it's the same probability that R1 is equal to any given value, the probability of R being uniform has of equaling each of its possible values. You can think about that, see if it's persuasive. It's an OK argument, but I was bothered by it. I found that it took me-- I wasn't happy with it until I sat down and really worked it out formally to justify this somewhat handwavy proof of the lemma. All right. Let's turn from uniform random variables to binomial random variables. They're probably the most important single example of random variable that comes up all the time. So the simplest definition of a binomial random variable is the one that you get by flipping n mutually independent coins. Now, they have an order, so you can tell them apart. Or again, you can say that you flip one coin n times, but each of the flips is independent of all the others. Now, there's two parameters here, an n and a p, because we don't assume that the flips are fair. So there's one parameter is how many flips there are. The other parameter is the probability of a head, which might be biased that heads are more likely or less likely than tails. The fair case would be when p was 1/2. So for example, if n is 5 and p is 2/3, what's the probability that we consecutively flip head, head, tail, tail, head? Well, because they're independent, the probability of this is simply the product of the probability that I flip a head on the first toss, which is probability of H, which is p; probability of H on the second toss; probability of T on the third; T on the fourth; T on the fifth. So I can replace each of those by 2/3 is the probability of a head. 2/3, 1/3. 1 minus 2/3 is the probability of a tail. 1/3, 2/3. And I discover that the probability of HHTTH is 2/3 cubed and 1/3 squared. Or abstracting the probability of a sequence of n tosses in which there are i heads and the rest are tails, n minus i tails, is simply the probability of a head raised to the i-th power times the probability of a tail, namely 1 minus p, raised to the n minus i-th power. Given any particular sequence of H's and T's of length n, this is the probability that's assigned to that sequence. So all sequences with the same number of H's have the same probability. But of course, with different numbers of H's they have different probabilities. Well, what's the probability that you actually toss i heads and n minus i tails in the n tosses? That's going to be equal to the number of possible sequences that have this property of i heads and n minus i tails. Well, the number of such sequences is simply choose the i places for the n heads out of-- choose the i places for the heads out of the n tosses. So it's going to be n choose i. So we've just figured out that the probability of tossing i heads and n minus i tails is simply n choose i times p to the i, 1 minus p to the n minus i. In short, the probability that the number of heads is i is equal to this number. And this is the probability that's associated with whether the binomial variable with parameters n and p is equal to i is n choose i, p to the i, 1 minus p to the n minus i. This is a pretty basic formula. If you can't memorize it, then make sure it's written on any crib sheet you take to an exam. So the probability density function, it abstracts out some properties of random variables. Basically, it just tells you what's the probability that the random variable takes a given value for every possible value. So the probability density function, PDF of R, is a function on the real values. And it tells you for each a what's the probability that R is equal to a. So what we've just said is that the probability density function of the binomial random variable characterized by parameters n and p at i is n choose i, p to the i, 1 minus p to the n minus i, where we're assuming that i is an integer from 0 to n. If I look at the probability density function for a uniform variable, then it's constant. The probability density function on any possible value v that the uniform variable can take is the same. This applies for v in the range of U. So in fact, you could say exactly what it is. It's simply 1 over the size of the range of U, if U is uniform. A closely related function that describes a lot about the behavior of a random variable is the cumulative distribution function. It's simply the probability that R is less than or equal to a. So it's a function on the real numbers, from reals to reals, where CDF R of a is the probability that R is less than or equal to a. Clearly given the PDF, you can get the CDF. And given the CDF, you can get the PDF. But it's convenient to have both around. Now the key observation about these is that once we've abstracted out to the PDF and the CDF, we don't have to think about the sample space anymore. They do not depend on the sample space. All they're telling you is the probability that the random variable takes a given value, which is in some ways, the most important data about a random variable. You need to fall back on something more general than the PDF or the CDF when you start having dependent random variables, and you need to know how the probability that R takes a value changes, given that s has some property or takes some other value. But if you're just looking at the random variable alone, essentially everything you need to know about it is given by its density or distribution functions. And you don't have to worry about the sample space. And this has the advantage that both the uniform distributions and binomial distributions come up [AUDIO OUT] --and it means that all of these different random variables, based on different sample spaces, are going to share a whole lot of properties. Everything that I derive based on what the PDF is is going to apply to all of them. That's why this abstraction of a random variable in terms of a probability density function is so valuable and key. But remember, the definition of a random variable is not that it is a probability density function, rather it's a function from the sample space to values. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 113_Intro_to_Proofs_Part_2.txt | PROFESSOR: There's other kinds of bogus proofs that come up. Let's just run through this one quickly. Here's a fact that you about, roots of polynomials. Every polynomial has two roots, at least over the complex numbers, over c. And how do you prove that? Well, you just write down the formulas for the roots. You know the quadratic formula. One root is a minus b plus this square root of this quantity of 2a. And the other root is minus b minus this square root of b squared minus 4ac over 2a. And that's the end for the proof. You can just plug-in this formula for r1 for x into this polynomial and it would simplify to be equal to 0, which shows that this is a root. You could plug that one into this formula for x and simplify algebraically and discover it was 0, proving that r2 is a root. We've just proved that every polynomial has two roots. Well, that's not true. We haven't proved it. This is a proof by calculation that has problems. What's the problem? well let's look at a counter example. What about the polynomial 0x squared plus 0x plus 1? It doesn't have any roots. It's just a constant 1 which never crosses the origin. So it's got no roots. What about 0x squared plus 1x plus 1? Well that's 45 degree line, the y equals x line, and it only crosses the origin once. It has only one root. What happened to the two formulas, r1 and r2? And the answer was, in this case, we had to divide by 0 error. If you look at that formula, there's a quotient, there's a denominator of 2a. Divide by 0 and these formulas don't work right. They aren't the roots. And so implicitly, in order to have two roots, we need to assume that the denominator, a, the leading coefficient of the polynomial is not 0. So that fixes those two bugs. Is that all? Well, no, because look at this polynomial. 1x squared plus 0x plus 0 has one root. The only possible root of this is 0. Because if you look at this, the only way to get 1 times something plus 0 to equals 0 is if the something is 0. So there's only one root. And what's going on here? Well, what's happened is that in this case, the two formulas, r1 and r2, which were different formulas, evaluate to the same thing when b is 0 and c is 0 and a is 1. And that's why they look like different formulas but they evaluate to the same thing so there's only one root. The fix to that is to require the quantity by which the two root formulas, r1 and r2 differ to be non-zero. And that's the quantity that we were taking the square root of, the discriminant it's called. b squared minus 4ac needs to be 0 and then r1 and r2 will differ and we will get the two roots. Now, there's still a complication. It sounds like we've now verified that indeed our proof by calculation is correct now that we've put in these qualifiers, that a is positive and d is non-zero. But what happens if d is non-zero but negative? It's an interesting complication. And let's look at the formula, x squared plus 1, where b squared minus 4ac is going to be minus 3. And that will turn out to have two roots, namely i and minus i. And it's not possible to tell which is which. r1 is taking the square root of minus 1, and r2 is taking the square root of minus 1. One of them is adding the square root of minus 1. The other one's subtracting the square root of minus 1. But which square root of minus 1? There's no way to tell the difference between i and minus i, abstractly. They both behave the same way. And so we have an ambiguity about which one is r1 and which one is r2. It doesn't hurt at all for the theorem that r1 and r2 are different. And so there are two roots. But ambiguity can be problematic. And let me give you an illustration of that. When there's ambiguity, I can do things like proving easily that 1 is equal to minus 1. Here's the proof. And I will let you contemplate that and try to figure out just where in this reasoning that step by step seems pretty reasonable, but nevertheless I've concluded that 1 is equal to minus 1, which is absurd. It's taking advantage of the fact that you don't know whether the square root of minus 1 means i or minus i. So the moral of all of this is that, first of all, be sure that you are applying the rules properly. There's an assumption of an algebraic rule in there that isn't true. And again, that kind of mindless calculation is risky when you don't really understand what you're doing, you don't have a clear memory of what the exact rules are. So it's understanding that bails you out of this kind of blunder. Let's look at 1 equals minus 1 a little because it lets us wrap up with an amusing remark. What's terrible about 1 equals minus 1? Well, it's false, and you don't want to ever conclude something that's false. That's worrisome. It's disastrous when you conclude that something is false. Let me give you an illustration of why. Because let's suppose the 1 is equal to minus 1 and let's reason in a correct from that assumption that we have falsely proved. But let's assume that we start off with the assumption that 1 is minus 1. Well, if I multiply both sides of an equation by the same thing, it's equal. So I can multiply both sides by 1/2, and I get 1/2 is equal to minus 1/2. Now I can also add the same thing to both sides. That's a perfectly sound move for reasoning about equalities. If I add 3/2 to both sides, I've turn 1 equals minus 1 into 2 is equal to 1. Now I'm in great shape to prove all kinds of things. Here's a famous one. "Since I am the Pope are clearly 2, we conclude that I and the Pope are one. That is, I am the Pope." And I've just proved to you this absurd fact. This is a joke that's attributed-- a witty remark that's attributed to Bertrand Russell, the famous philosopher, logician, pacifist, Nobel Prize winner, who supposedly was approached by some socialite at a party who had heard that mathematicians thought that if 1 is equal to minus 1 then you could prove anything. And so she challenged him, prove that you're the Pope. And supposedly Russell, who was a notoriously quick wit, came up with this example. Who knows whether it's true. It's a good story. There's a picture of the great man. And you might care to learn more about this remarkable contributor to logic, and philosophy, and politics. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 261_DAGs_Video.txt | PROFESSOR: Directed acyclic graphs are a special class of graphs that really have and warrant a theory of their own. Of course, "directed acyclic graphs" is lot of syllables, so they're called "DAGs" for short. OK. So here's why they come up all the time. Let's look at a diagram that may be familiar to you. This shows the prerequisite structure of required courses in the 6-3 program of MIT Electrical Engineering and Computer Science department. There are similar charts for the other sub-majors of EECS and in other departments as well. So what does it mean? Well, let's look at this vertex corresponding to the first term calculus class 18.01. And there's an edge that points to 6.042. And that's because, if you look at the catalogue, 6.042 lists 18.01 as a prerequisite. If you look at the algorithms-- introductory algorithms-- class 6.006, you'll find, if you look in the catalog, that it has listed two prerequisites, 6.042 and 6.01. And the fact that they're explicitly listed in the catalog as prerequisites is why there's an arrow from 6.01 to 6.006 and from 6.042 to 6.006. Now when you're planning on when you want to take 6.006, of course, you have to attend to not just the fact that you have to take 6.042 first and 6.01 first, but you've got to take the prerequisites of those prerequisites first. So you really have to take 18.01 before you can take 6.006. And you need to take 8.02 before you can take 6.006. There are corequisites here. Let's just ignore those and pretend that they were prerequisites, because they're another kind of arrow that needn't distract us. OK. So that's what this diagram is telling us. And this is a DAG. It's simply a bunch of vertices, the course labels in rectangular boxes, and directed arrows showing catalog listings. And what I said was that when you're planning your course work, you're really interested in the indirect prerequisites. So "one class, u, is an indirect prerequisite of another class, v" means that there's a sequence of prerequisites starting from u and going to v. It means that you really have to have taken u some time before you took v. And that's a crucial fact and thing that you need to take account of when you're planning a course schedule. So in terms of graph digraph language, "u is an indirect prerequisite of v" just means that there's a positive length walk that goes from u to v in the digraph. In this case, we're talking about the 6-3 digraph of prerequisites. So "there's a positive length walk from 18.01 to 6.006" means that you really have to have taken 18.01 before you take 6.006. And of course, we're talking, then, about the positive length walk relation D plus of the digraph D. If D is the digraph shown in the prerequisite chart-- direct prerequisite chart-- then we're interested in D plus. And u D plus v just means there's a positive length walk-- that's what the plus is for-- going from to u to v. Now what happens if you have a closed walk? Well, a closed walk is a walk that starts and ends at the same vertex. And we can ask this question-- suppose there was a closed walk that started at 6.042 and ended at 6.042. How long does it take to graduate then? Well, it takes a long time. Because you can't take 6.042 until you've taken 6.042 and you're never going to be able to take it. That's a bad thing. We definitely don't want the prerequisite structure of courses in a department to have a closed walk of positive length. And in fact, there's a faculty committee that checks for this kind of thing. Bugs like this occasionally creep in when some busy curricular office of a department is planning a complicated program with dozens, if not 100 courses. And the Committee on Curricula's job is to check for that kind of thing. There's a whole staff that does it. I used to be the chair of that committee. And we did spend a lot of time with proposals from departments and making sure that those proposed course requirements satisfied faculty rules. OK. So a special case of a closed walk is a cycle. A cycle is a walk who's only repeat vertex is its start and end. Let me remark, because we keep talking about positive length cycles, that a single vertex all by itself is a length-0 cycle. So you're never going to be able to get rid of length-0 cycles, because they're the same as vertices. But positive length cycles, you can hope to ensure are not there. So if you're going to represent a cycle as a path, you'd show the sequence of vertices and edges, v0, v1, v2, where the understanding is that all of the vertices from v0 up to v n minus 1 are different-- that's what makes it a cycle-- except that the last vertex, v0, is a repeat of the first one. That's the one repeat that's allowed in a cycle. So it's natural to draw it in a circle like this where you start at v0, you follow the edges around from v1 to v i plus 1 all the way back around to v0. And that's kind of what a cycle is going to look like. So we have a very straightforward lemma about cycles and closed walks, namely that the shortest positive length closed walk from a vertex to itself-- "it's closed" means it starts and ends at v-- is a positive length cycle starting and ending at v. And the reasoning and proof is basically the same proof that said that the shortest walk between one place and another is a path from one place to the other. The logic is that if I have a closed walk from v to v and there was a repeat in it other than at v, I could clip out the piece of the walk between the repeat occurrences and I'd get a shorter walk. So the shortest closed walk can't have any repeats. It's got to be a positive length cycle. So a directed acyclic graph now is defined simply as a digraph that has no positive length cycles. It's acyclic, no positive length cycles. And of course, we can equally well define it, since cycles are a special case of closed walks and closed walks of positive length imply cycles, as a digraph that has no positive length closed walk. Some examples of DAGs that come up-- well, the prerequisite graph is going to be one. And in general, any kind of set of constraints on tasks, which ones you have to do before you do other ones, is going to be defining a DAG structure. One that you might not have thought of is, the successor function defines a relation on the integers, say, going from n to n plus 1. So I'm going to have an arrow that goes directly from n to n plus 1. And what's the walk relation then, the positive length walk relation, in this graph? Well, there's a positive length walk from n to m precisely when n is less than m. So the successor DAG, it's paths represent the less than relation. And of course, less than, it doesn't have any cycles. Because if a is less than b, you're never going to get around from b back to something that's less than it, like back to a. So there can't be any cycles in the successor DAG. And that's why it is a DAG. Another similar one is the proper subset relation between sets. So here I'm going to draw an arrow from this set to that set if this set is contained in that set but they're not equal. So {a, b} is a subset of {a, b, d}, but {a, b, d} has this extra element, d. So the left-hand set is a proper subset of the right-hand set. And I'm going to draw an arrow there. And by the same reasoning, there can't be any cycles in this graph-- a positive length cycle-- because if there was, it would mean that the set had to be a proper subset of itself, which doesn't happen. So this would be another basic example of a DAG. And I hope you begin to see, from these examples, why DAGs are really all-pervasive in mathematics and in other areas and why they merit attention. So when we're looking at a DAG though, we're basically usually interested in just the walk relation of the DAG. So if we're only interested in the walk relation of the DAG, then it would be typically the case that many different DAGs are going to have the same walk relation. And it's natural to ask, what's the most economical one. Is there a minimum, say, DAG that defines a given walk relation? So let's look at this example. Here's a simple DAG. And you can check that there are no cycles in it. What's the smallest DAG with the same relation as this one? And the way I can get it is by going through the edges one at a time and asking whether I can get rid of the edge because it's not contributing anything. So look here. There's a path from a to e that goes through b. Well, that tells me that having this direct edge from a to e is not contributing anything in terms of connectedness. And that means I could get rid of it and I'm still going to wind up with the same possibility of walking from one place to another. Because I can always walk from a to e going through b instead of going directly from a to e. I didn't need that edge. Another example is, here's a walk from a to d that goes through c. There's no need for me to walk directly from a to d. As long as I'm walking, I can take the longer walk and get rid of the short circuit from a to d. Likewise, if I look at this path from c to d to f, I don't need that edge from c to f. And as a matter of fact, now if I look at this length 3 path from a to c to d to f, there's no need for me-- in order to get from a to f, there's no need for me-- to take the direct edge. I can get rid of that too. It's kind of a redundant extra edge. Finally, if I look at the path from b to d to f, I can get rid of the direct edge from b to f. And at this point, I'm done. I'm left with a set of edges called covering edges, which have the property that the only way to get from one vertex to another is going to have to be to use a covering edge to the target for vertex. Or more precisely, the only way to get, say, from a to b is going to be to use that covering edge. If there was any other path that went from a elsewhere and got back to b without using this edge, then it wouldn't be a covering edge anymore. The fact that it's a covering edge means that if you broke it, there's no way anymore to get from a to b. So that's the definition of covering edges and you'll do a class problem about them, more precisely, in a minute. So the other edges are unneeded to define the walk relation. And all we need to keep are the covering relations to get the minimum representation of the walk relation in terms of a DAG. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 431_Independence_Video.txt | PROFESSOR: Independent events are events that have nothing to do with each other. And needless to say, it's a lot easier to work with them because you don't have to figure out some weird interaction between two events that do have something to do with each other. Typical case where independent events come up is, for example, you toss a coin five times, and then you're about to toss a coin the sixth time. Is there any reason to think that what the coins did the first five times is going to have any influence on the flip of the coin for the sixth time? Well, it's reasonable to assume not, which is to say that the outcome of the sixth toss is arguably independent of the outcome of all the previous tosses. OK. Let's look at a formal definition now in this short video of just what is the technical definition of independent events. So what we said is that they are independent if they have nothing to do with each other. What that means is that if I tell you that B happened, it doesn't have any effect on the probability of A. That is, the probability of A, given that B happened, doesn't change the probability of A at all. That's it. Now this is one definition. Maybe this is the more intuitive definition. But another definition that's equivalent and is standard is that two events are equivalent if the product of their probabilities is equal to the probability that they both happen, that is, the probability of their intersection. Now definition one and definition two are trivial equivalent, just using the definition of conditional probability. And if you don't see that, this would be a moment to stop, get a pencil and paper, and write down the three-line proof of the equivalence of these two equalities. In fact, the three-line proof has this as the first line and that as the second line. So you could argue it's really just the middle line that you're adding. OK. Definition two has the slight advantage that it always works, whereas definition one implicitly requires that the divisor-- remember probability of A given B is defined as the probability of the intersection divided by the probability B. It's only defined if the probability of B is positive. Whereas the second definition always works, so we don't have to put a proviso in about the probability of B being non-zero. So that's the definition of independence. Looking at this definition, what you can see immediately is that it's completely symmetric in A and B. Since multiplication is commutative and intersection is commutative, which is A and which is B doesn't matter. And what that implies then is that A is independent of B if and only if B is independent of A. Now another fact that holds is that if the probability of B happens to be zero, then vacuously B is independent of everything-- even itself. Which isn't important, but is a small technicality that's worth remembering by that definition. Now again, the intuitive idea that A and B have nothing to do with each other is that A is independent of B means that A is independent of whether or not B occurs. That is to say, if A is independent of B, it ought to be independent of the complement of B. And that's a lemma that's also easily proved. A is independent of B if and only if A is independent of the complement of B. It's a simple proof using the difference rule. And again, I encourage you to stop with a piece of paper and a pencil and convince yourself that that follows with a one-line proof. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 123_Proof_by_Cases.txt | PROFESSOR: Another basic proof technique is called Proof By Cases, which we prove something by breaking it up into pieces that are easy to prove but that together cover all possibilities. Let's look at an explicit, simple example from computer science. Here's a Java logical expression. The way to decipher this is that the double vertical bar means "or" in Java. And the double ampersand means "and" in Java. So this is a conditional test, an IF test, that is the guard on a bunch code to be executed if this test comes out to be true. Let's read the test. If x is greater than 0, or x is less than or equal to 0, and y is greater than 100, go ahead and do the code that's in there indicated by the vertical dots. We're going to assume here that x and y are variables that are declared to be of type floating point, [? or a ?] real number, or integers for that matter. OK. Now what I claim is this code can be improved if it's rewritten in the following way. Namely, if x is greater than 0 or y is greater than 100. So the claim is that these two hunks of codie, if I just replace this test, which has three components that require an extra step to evaluate some cases, by this code the programs are going to behave exactly the same way. And therefore it's just more efficient and easier to understand. One is one step faster if I replace this longer expression by this shorter expression. Now how do I argue that these two pieces of code are going to behave in exactly the same way, or come up with the same final output? They won't behave exactly the same because one will be faster than the other. But they're going to yield the same results. OK, let's consider how these two behave in two cases. The first case will be that the number x really is positive, that it's greater than 0. What happens then? Well, the first test above in the "or" comes out to be true. And that means that the whole "or" expression is true. Because when you have a true "or" anything at all, it comes out to be true. And you go ahead and execute code that follows. Likewise, the second expression starts with x greater than 0 "or." So it comes out to be true. So, in this case, if x is greater than 0, both conditional expressions will allow the code that follows them to be executed. Because they both evaluate to true. OK. The next case is that x is less or equal to 0. Let's see what happens then. Well, in the top expression, since x is less than or equal to 0, that first expression, x greater than 0, if evaluated, returns false. And same thing in the second expression. The initial test x greater than 0 returns false. Now one of the things that the ways "or" works is that if you have an "or" of a bunch of things, if the first thing is false, you ignore it and just proceed to the other things to see how they come out. So what that means is that in both of these expressions, since the first test in the sequence of things that are being "ored" together came out to be false, I can just ignore them. The code is going to behave as though, after the false was detected, it's just going to behave in the same way that the rest of the test says to behave. Well, in the top case, the expression to be check now is that x is less than or equal to 0, and y is greater than 100. But what do we know? Well, x less than or equal to 0 in this case. So this test comes out to be true. And we have something of the form true "and" something or other. That means that the net outcome of this expression, it depends entirely on the something or other. That is, it depends entirely on whether y is greater than 0. Because the x is less than or equal to 0. And so this expression can be simplified. It's going to behave exactly according to whether or not y is greater than 100. So look what I've just done. I've argued that in this case, both of these tests cards act like y is the test y greater than 100. Which is, they behave the same in this case as well. So what I just figured out was that, in both cases, these two expressions yield the same result. And the only possible cases are that x is greater than 0, or x is less than or equal to 0. So in all cases, they're the same, and we're done. That's why it's safe to replace the upper complicated expression by the lower, less complicated expression. So, in general as I said, reasoning by cases breaks complicated problems into easier sub problems. Which is what we just saw there. [? It ?] wouldn't be clear how to prove that these two things were equivalent, but I chose those cases and it made each case easy to figure out that the two things with the same. Now, the truth is that there are some philosophers who worry about reasoning by cases for kind of subtle reasons. They're called intuitionists. And here's what bothers them. Let me illustrate it. There's a million dollar Clay Institute question. One of the dozen or so questions that are considered to be the major open problems in various disciplines of mathematics. And one of the disciplines of mathematics is complexity theory in computer science, computational complexity theory. This question is known as the p equals np question. And we're actually going to talk about it a fair amount in just the coming few lectures. But for now, it doesn't matter what it means. Well, I'll tell you what it means. p stands for polynomial time, and np stands for nondeterministic polynomial time. I'm not going to define not deterministic polynomial time, but it would be momentous if those two things were equal. And everyone expects that they're not equal, but no one knows how to prove that. So the million dollar question is, is p equal to np, yes or no? And you get a million dollars for settling this question. Now, I claim that in fact the answer to this question is on my lecture table. And I will show it to you in class tomorrow. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 353_InclusionExclusion_Example_Video.txt | PROFESSOR: The final general counting rules that we'll examine is called inclusion-exclusion, and it is a straightforward generalization of the sum rule-- at least in the simple case of two sets that we'll look at first. So we're going to look at 6042 example of applying inclusion-exclusion, but let's begin by stating what it is. So the sum of the rule says that if you have two sets A and B that are disjoint-- no overlap between A and B-- then the size of A union B is equal to the size of A plus the size of the B. Well that's obvious. We took that as kind of basic axiom. But what if they're not disjoint? Suppose that A and B overlap, and there's some stuff in here that's the intersection of A and B where there are points in common, what then is the size of A union B in terms of simpler things that we can count? And the answer is that the size of A union B is the size of A plus the size of B minus the size of A intersection B. Now, the intuitive reason for that-- and it's not really very hard to make a precise argument-- is that when you're adding up the elements in A, you're counting all the elements of the intersection once. And then when you add in the elements in B here-- A plus B-- you're counting all the elements in the intersection a second time. The ones that are in A minus B get counted once. The ones that are in B minus A get counted once, but the ones that are in A intersection B get counted twice. So to get the right count, I have to subtract the size of A minus B so it's only counted once in the total formula, and that's an intuitive explanation of why inclusion-exclusion formula holds for two sets. Let's apply it. And I'm going to look at an example where we're looking at digit permutations, and I'm going to look at permutations of the 10 digits, 0 through 9 inclusive. There's a standard one where they're listed in order, and there is just a random seeming permutation of the digits 0 through 9. Notice that the 1 and 3-- the 2 is sort of out of order. The rest are in order. Now, what I'm going to be interested in is those permutations where certain patterns appear. So first of all, let's note it that the number of permutations we know how to count-- it's 10 factorial. I'm interested in how many permutations have a consecutive 6 and 0, a consecutive 0 and 4, or a consecutive 4 and 2. In other words, two of the consecutive numbers that appear in 6042. Well, the first one does not. There's no 60, 04, or 42 in this list. This one has a 42. So it would count as one of those permutations that has either a 60, a 04, or 42 because it's got the 42. Here's one where you've got a 2 and a 4, but that's not a 4 and a 2. And in fact, there is no pattern here of 60, 04, 42. So it's not one of the permutations that I'm interested in. On the other hand, here's one that's doubly good. This is a permutation that has both a 04 in it and a 42 in it. So it would be one of these permutations of the kind that I'm looking for that has at least one of the pattern 60, 04, or 42. Well, if I let P sub x be the permutations with the sub-sequence x, then what I'm really saying is that this one with a 42 in it is in P 42 because it's got the 42 pattern. This one with the 04 and a 42 in it is in the P 04 set of permutations with the pattern 04 intersected with the set of patterns that have a 42. So that's what that one illustrates. So what we're really asking for then is the union of three things-- the union of P 60, P 04, and P 42. How big is the set of sequences that have a 60 union, the set of things that have a 04 union, the set of things that have a 42? And as we saw illustrated in the previous slide, these are not disjoint. We'll I've been cheating a little because in order to figure out this one, I'm going to need inclusion-exclusion for three sets instead of two, which is slightly more complicated because I have a union of three things that overlap. And let's look at that. So what does inclusion-exclusion look like for three sets? If I want to know what's the size of A union B union C, here's a Venn diagram that shows a picture of A union B union C with all possible overlaps illustrated there. And the formula turns out to be-- you add up A, B, and C. You add up the size of A, the size of B, and the size of C. Now that has a consequence that just that sum of A, B, and C is counting this lens shaped region that is the intersection of A and C. It's counting it twice in the A plus C term. It's counting A intersection B twice, and it's counting this lens shape, which is C intersection B twice. So in order to get the sum right, I'm going to have to subtract one occurrence of A intersection B, one A intersection C, one B intersection C so that those items are only counted once in this sum. And then in fact, if you look at this region here, of the sort of rounded triangle region-- which is the intersection of A with B and C-- that one is actually getting counted three times. All three circles overlap it. So when I add in A and I add in B and I add in C, every one of those points there is being added three times. On the other hand, this rounded triangle shape, which is count of three times in the sum A plus B plus C, is being subtracted three times. Because when I look at A intersection B-- this region-- and I subtract it, I'm taking one away from the count on each point there. And likewise with A intersection C, takes one away. And B intersection C takes one away. Leaving the points in the rounded triangle in A intersect B intersection C not counted at all. So if I'm going to get the total count right so that every point discounted exactly once, I have to add back in the intersection of A and B and C. So that's an informal explanation of how the inclusion-exclusion formula works for three sets. We'll look at ways to rigorously prove inclusion-exclusion for an arbitrary number of sets shortly but not in this segment. Let's go on and apply the inclusion-exclusion rule for three sets to the example of digit permutations with the patterns 60, 04, and 42. And the way to remember this is that the intersections of an even number of sets occur negatively, the intersection of an odd number of sets occur positively, and of course, a single set can be thought of just am intersection of one set with itself. So it's also odd and occurs positively. All right. Well, now we can apply the formula and say that the set of permutations that have a 60, a 04, and a 42 is equal to the sum of the number that have a 60, the number that have a 04, and the number that have a 42 minus the numbers that have two of the patterns minus those that have all three patterns. At let's count these individual intersections and sets of permutations separately. It turns out that each one is easy to count, which is a typical situation which is why inclusion-exclusion is a valuable principle because this thing that is harder to count can be broken up into counting a bunch of other things-- intersections-- that are often easier to count. And they will be here. So let's begin by looking at P 60. P 60 is the set of permutations which have a 60 in them. Well, to count them, we can think about it this way. Think of the patterns with a 60 in them as a permutation of nine items-- the digits 1 through 5 and 7 through 9 and the object 60 that you can place anywhere, but it it's got to be lumped together. So there are really nine possible permutations of these things-- eight of them digits, and one of them is this pair of digits, 60. And the number of those permutations is equal to the number of permutations with the pattern 60. So the answer is there are 9 factorial permutations with the pattern 60. Same of course, applies to P 04, and P 42. The number of permutations with a given two digit pattern is 9 factorial. OK. What about P 60 intersection P 42? Well, you can think of this as the same way. You can think of this as saying, OK. I've got an object 60, I've got an object 42, and I've got the remaining digits 1, 3, 7, 8, 9 to permute. And the sequences of 10 digits that contain both a 60 and a 42 correspond exactly to permutations of the digits 1, 3, 5, 7, 8, 9, and the object 42, and the object 60. Now, there's eight of things. And so the number of permutations of these eight things is 8 factorial, which means the size of P 60 intersection P 42. The number of permutations of 10 digits that have both a 60 and a 42 pattern is 8 factorial. Now, that's the case of an intersection where these two things don't overlap. Let's look at the case of P 60 intersection P 04. Well, if it's got both a 60 and a 04, it actually is the same as having a 604. So the intersection of P 60 and P 04 is the set of sequences that have the pattern 604. And I count those in the same way. I say, OK. I've got an object 604 plus the remaining digits-- 1, 2, 3, 5, 7, 8, 9 for a total of eight objects, and the number of permutations of the 10 digits that half the pattern 604 corresponds to the number of permutations of these eight things. Again, 8 factorial. OK. Finally, how many permutations are there that have all three patterns-- 60, 04, and 42? That of course, is exactly the same as the set of sequences with the single pattern 6042, the four digit pattern. And again, we count that by saying that it's the number of permutations of the digits other than 6042-- six of them plus the 6042 object. There are seven of these , and so there are 7 factorial permutations that have all three patterns. So that means that I can go back to my inclusion-exclusion formula for the sequences that have one of the three patterns-- 60, 04, and 42-- and plug them in. So I get 3 9 factorials for the first sum of three terms. The intersections-- we all figured out each of them were-- I'm sorry it's 8 factorial. So I'm going to subtract 3 times 8 factorial. And this last term we figured out was 7 factorial. Well, I can think of 3 times 9 factorial as 9 times 8 times 3 times 7 factorial, and this is 3 times 8 times 7 factorial. And I wind up [NO AUDIO] PROFESSOR: 72,720. That's how many permutations of the digits 0 through 9 there are that have one or another of these three patterns. Turns out that's about 27% of the 10 factorial permutations of 0 through 9. So that's the significance of applying the disjunction of constraints, this union of having either 60, 04, or 42. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 459_Linearity_of_Expectation_Video.txt | PROFESSOR: So we've been saving for the last the property that makes expectation calculating really easy and short circuits a lot of the ingenious methods that we've used up until now-- namely, expectation is linear. So what that means is that if you have two random variables R and S and two constants a and b, the expectation function is linear. That is, you take a linear combination of R and S-- aR plus bS, and that's equal to the corresponding linear combination of the expectations. I'll read it again. Expectation of aR plus bS is equal to a times the expectation of R plus b times the expectation S. Expectation is linear. OK. That's an absolutely fundamental formula that you should be comfortable with and remember. It extends actually not only to any finite number of variables, but with some convergence conditions, it actually extends even to accountable sum of variables. But let's just settle for the two random variables case for today. Now, the crucial thing that makes it so powerful and useful is that this fact has nothing to do with independence. Whether R and S are independent or equal, it doesn't matter. This linearity holds. The proof is not terribly informative. It's just manipulating terms and rearranging terms into sum, but let's go through the exercise. Again, something I would never do in lecture. But in a video where you can skip it or fast forward or replay it, I think it might be worth doing, And it is a proof, by the way, that I think you should be responsible for. So let's go through it. OK. We're interested in the expectation of the random variable that you get by multiplying the random variable A by little a and the random variable B by little b. All right. One of the definitions of expectation is that you get this expectation by taking the sum over all the outcomes of the value of this linear combination at the outcome omega times the probability of omega. So what's the value of the linear combination aA plus bB? At omega, it's simply a times A of omega plus b times B of omega. OK. Now I've got a sum of these terms summing over omega. I can split them into two groups. I can take the sum over the aA's at omega times probability of omega and bB of omega times probability of omega. In other words, I'm multiplying by probability of omega here to get a sum of two terms, and then I'm rearranging all of the capital A terms first, followed by all the capital B terms. The result is that I wind up with the sum over omega of the A terms times the probability of omega-- and I factored out the little a-- plus b times the sum over all omega of B of omega times the probability of omega. It's just rearranging the terms in this sum after I've multiplied through by probability of omega. Well, of course, this is equal by definition to a times the expectation of A. Notice this is the expectation of A, and that's the expectation of B times b. And the proof is done. Not inspiring, but routine if you use the alternative definition of expectation in terms of summing over the outcomes. It's a messier proof if you have to use the definition of the expectation, being the value times the probability that the variable takes that value. And you wind up having to convert that formula into this formula in order to carry through the proof nicely. And we're done. OK. Let's make use of it. And in order to do that, let's make a really trivial but a very important remark about the expectation of an indicator variable. So remember I sub A is three random variables that's equal to 1 if the event A occurs and 0 if the even A doesn't occur. So what is the expectation of the indicator variable? Well, by definition, it's 1 times the probability that it equals 1 plus 0 times the probability that it equals 0. Those are the only two values it can take. Well, we can forget this term in 0 times something. But what is the probability that I A is equal to 1? That's exactly the probability of A, and that's the fundamental formula that we want to notice. The expectation of the indicator variable for the event A is nothing but the probability that A occurs. File that away. We're about to use it multiple times. So let's go back to the expected number of heads in n flips which we've now seen at least two ways to do-- one by generating function argument, another by a recursive argument using total expectation. Now where we're going to knock it off very elegantly using linearity because let Hi be the indicator variable for having a head on ith flip. So we look at the ith flip. Hi is 1 if the ith flip comes up head and Hi is 0 if the ith flip does not come up head. Then we can make the following crucial remark, and this is a trick that we'll use regularly by expressing some quantity that we're interested in as a sum of indicator variables. The total number of heads-- the random variable equal to the total number of heads in n flips-- is equal to the sum of the indicator variables for whether there's a head on the first flip plus whether there's a head on the second flip up through whether there's a head on the n flip. So suddenly the random variable that I want to compute is a sum of n random variables, in fact, n indicator variables. All right. Well, that tells me that the expectation of the number of heads is the expectation of the sum. After all, it's equal to the sum. But the expectation of the sum is going to be the sum of the expectations by linearity. So it's simply the expectation of H 1 plus the expectation of H 2 out through the expectation of H n. But what's the expectation of getting a head on ith flip? Well, the flips are independent. It's simply the expectation of a head. So what I have is-- each of these is equal to the probability of a head, and there's n of them. So the total is n times the probability of Head, or np, which is a formula that we had derived two other ways previously. And now it really falls out very elegantly with hardly any ingenuity other than the wonderful idea of expressing the number of heads as a sum of indicators. Let's look at one example and a very related example of asking about the probability of the expected number of hats that are returned. When n men check their hats at a hat check and then the hats get all scrambled up by incompetent staff and then they're given out again in such a way that the probability that the ith man gets his own hat back is 1 over n. What you could say is that all possible permutations of the n hats are equally likely. And we ask with all permutations equally likely, how many of them is it the case that the ith man gets his own hat back? And there's a 1 out of n chance that the ith man is going to get his own hat back because there's n hats, and he's equally likely to get all of them. OK. How many men do we expect will get their hat back in this setting? Well, let's let our Ri be the indicator variable for whether or not the ith man got his hat returned-- R i for hat returned to the ith man. Now, notice that Ri and Rj are not independent. In the previous case, those H's were independent because the coin flips were independent. But here, if I know, for example, that R 1 got his hat back, the probability that R 2 got his hat back has changed from 1 over n to 1 over n minus 1 because 1 is out of the picture, and R 2 is going to get his hat back among the remaining hats 2 through and n is n minus 1 of them. And he's got a 1 minus 1 over n minus 1 chance of getting his hat. His probability has changed given that R 1 got his hat back. So they're not independent. All right. Nevertheless, independence doesn't matter for linearity. So I can still say that the expected number of hats returned is equal to the expectation of the sum of the indicator variables for each man getting his hat back. And of course, the expectation of that sum is the sum of the expectations. And the expectation of each of these-- we figured out-- was 1 over n, and there's n of them. So it's n times 1 over n, or 1. I expect when all the hats are scrambled and all permutations of the hats are equally likely, that one man is going to get his hat back. None of the others will. OK. Now let's change the situation a little bit. And think instead of scrambling the hats in a way that all possible permutations of hats are equally likely, let's think about a Chinese banquet. So Chinese banquets are traditionally done with a table of nine in a circle, and there's a lazy Susan that spins around where there's dishes of food in front of each person. But let's generalize it to n. Suppose that n people are sitting around a spinning table-- a lazy-Susan) with n different dishes, one dish in front of each person. And now we spin the lazy Susan randomly, and we ask how many people do we expect will wind up with the same dish that they started with after the spin? Well, now we can let Ri indicate that the ith person got the same dish back. And now these Ris, which are different from the previous ones about hat returns, these are Ris are totally dependent-- much more so than the other ones work because they're all 1 or they're all 0. If one person gets their hat back, it means that the spinning table got back to where it used to be, and everybody has their hat back. And if one person doesn't have their hat back, then the table is shifted off where it was originally and nobody has the original dish. I said hats. I meant the dish that they started with. So everybody gets the same dish back, or nobody gets the same dish back. These variables are as dependent as they possibly could be, but it doesn't matter because linearity still holds. And that means that the previous argument about the expected number of dishes that get back to the person that they started with is still 1 even though all the Ris are equal. Well, that's so much for the lovely rule about linearity of expectation which holds regardless of assumptions about independence or not. There is a rule for products, but it requires independence. So the independent product rule says, sure enough that the expectation of a product of two random variables, x and y, is the product of their expectations, providing that they are independent. And this extends to many variables if they're mutually independent. Again, the proof by rearranging terms in the defining sum for the expectation of xy. Let's go through it. And again, you can fast forward or skip this part if you don't want to watch equations being manipulated. So by definition, the expectation of the product xy is the sum over all the possible values of x and y of the value of the product xy times the probability that the first variable, capital X, equals little x and the second variable Y is equal to little y. This is by definition, the expectation of the product-- by the first definition. Not the one in terms of outcomes. Now, using independence, this term here can be split into a product of X equals x and Y equals y. So let's do that. So this is the sum of xy times the probability that X equals x times the probability that Y equals y. Now I'm going to do a fairly standard trick, which is I'm going to organize this sum in a clean way. Right now it's an unordered sum over all possible pairs of x and y in the range of the variables x and y, but I'm going to do the sum so I first sum over all the y's, and then for each y, I'm going to sum over all the x's. This is an unordered some really. There's no order here, but now I'm giving you an arrangement which says that I'm going to lump together the sums over all the x's, and then for each of those I'm going to sum up over the y's. Well, when I do it this way, what I've got is an interesting thing here. I've got a sum over x, and there's some y terms here that don't depend on x. I can factor them out because they don't change with x. So if I factor out this y and probability of Y equals y, I wind up with the sum over y of this factored out term-- y times probability Y equals y-- times the sum over x's x times the probability that X equals x. Now, this is the same term that is the coefficient of every one of these y terms that depends on y, and this term does not depend on y. So I can factor it out. And if I do that, I wind up with the sum over x of x times the probability that X equals x times the sum over y of y times the probability that Y equals y. And guess what. This is by definition the expectation of X, and this is by definition the expectation of Y. And by that chain of equalities, I've proved, sure enough, that the expectation of XY is equal to the expectation of X times the expectation of Y QED. So the key step here was where independence was used at the very first step when I split up the probability that X equaled x and Y equaled y into the product of the corresponding probabilities. Now, let me just end with a warning about a couple of blunders that people make all the time. So first of all, don't forget independence as a crucial condition on the product rule for expectations. It can hold in some cases where the variables are dependent. Independent is not a necessary condition. It's efficient, but you need some kind of a condition in order for the product rule to hold. So if you're not careful, it won't if you forget to check for independence or something that is tantamount to it. So let's just take an easy example remember what happens if independence fails. Suppose I have a variable X-- a random variable X-- which takes positive and negative values with equal probability. So it takes 1 and minus 1 with equal probability. It takes pi and minus pi with equal probability. I don't really care what those values are as long as it's taking some positive and negative values, and it takes a positive value at the same probability that it takes that value negated. Well, that automatically means that the expectation of X is 0 because when I add up all these terms, the positive and negative terms cancel because they have the same probability. So any such X that's symmetric about 0 has expectations 0. On the other hand, if I square X, then all of those positive and negative values become positive. And so I'm taking the expectation of a variable that's strictly positive-- at least with nonzero probability at a bunch of outcomes. And therefore, the expectation of X squared is positive. So the expectation of X is 0, but the expectation of X squared is positive. Well, of course, if I multiply expectation of X times expectation of X, that's still 0. So here's a counter example. Expectation of X times expectation of X is equal to 0, which is less than the expectation of the square of X. Of course this is about as dependent as it could possibly be because it's the same random variable, but it illustrates the failure of the product rule if you don't have some kind of a condition like independence around. There's a second blunder that's more interesting and that people can fall in because there's a temptation to assume that if the product rule holds for independence, then so should the reciprocal rule. That is, you might think that the expectation of X over Y is equal to the expectation of X over the expectation of Y when X and Y are independent. But it's not true. Even when they're independent, the expectation of X divided by Y is in general, not equal to the expectation of X divided by the expectation of Y. In fact, the counterexample is if X is the constant 1, the expectation of 1 over Y [AUDIO OUT] PROFESSOR: [INAUDIBLE] complex instruction set code was better than risk. So I won't mention names, but prominent people have made this blunder. You shouldn't. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 457_Mean_Time_to_Failure_Video.txt | PROFESSOR: We're constantly asking how long we have to wait for things. And in the context of probability theory, it turns into the technical question called, the expected time to failure-- or the mean time to failure. Some examples might be that an insurance company wants to know, for a given policy holder, the expected time before that policy holder dies. A mechanical engineer wants to know the expected time before a button that's being pushed once per second is expected to fail. And I want to know when the part that my body shop has been waiting for is expected to come in. The mean time to failure problem, we can formalize in terms of flipping coins. So we're going to flip a coin until a head comes up. And we're going to think of a head as being a failure, and a tail as a success. So let's assume the probability of getting a head-- the probability of failure-- is p. Again, this is not a fair coin. It's a coin that may be biased in either direction. And let's let F be the number of flips until the first head comes up-- the number of flips until the first failure. And if we're counting as flips as time, it's the time to fail. So what we'd like to know is what's the expectation of F? What's the expected number of flips before a head comes up? Well, in order to calculate that expectation, we need to know some probabilities. So what's the probability that F equals 1? Well that's the same as saying that that's getting a head on the first flip-- it's the probability that you get just an H on the first flip, and that has probability, P. What's the probability that F equals 2? Well that's the probability that you flip a tail and then a head. And that has probability q times p, because we're assuming the flips are independent, so it's the probability of a tail-- which is q-- times the probability of a head-- which is p. Similarly, the probability of F equals 3 is the same as the probability of flipping tail, tail, head-- and it's q squared p. And of course, the probability density function of F-- the number of steps until you'd flip a head-- at n-- the probability that you have to flip n times before you get the first head-- is q at the n minus 1 p. By the way, a random variable whose probability density function is this value is called a geometric distribution. They come up all the time. All right, so what's the formula for the expectation of F? It's simply, of course, by definition, it's the sum over all the possible values of F-- which in this case are integers, n, greater than 0, of n times the probability that F equals n. We figured out that the probability that F equals n is q to n minus 1 times p. And now I'm going to observe that we really do know how to evaluate this sum easily enough. I'm going to factor out the p, and it becomes a sum over n greater than 0 of q to the n minus 1 times n. And then I can simplify matters if I replace n by n minus 1. And then I get a q to the n power. So this is equivalent to p times the sum over n greater than or equal to 0 of n plus 1 q to the n. Now this is a familiar generating function. It's simply equal to 1 over 1 minus q squared, as we've seen already. So in short, the expectation of F is p times 1 over the square of 1 minus q. So let's put them together, there. Of course 1 minus q is p, so it's p times 1 over p squared, or one over p. And we get this really very clean answer. The expected number of flips before you get a head is 1 over the probability of a head. So for example, with a fair coin, where p is 1/2, the expected number of flips until you get the first head is 2-- it's 1 over 1/2. If you had a biased coin, where the probability of getting a head was 1 in 3-- 1/3-- then, in fact, the expected number of flips until you got a head was 3. OK, that's a nice clean answer. But we got it in this way that doesn't really give much intuition. So let's look at another naive way to derive the expected time to the first head, without having to worry about generating functions, and all that sophisticated stuff about series-- which is easy to forget. So let's look at the outcome tree that describes this experiment of flipping until you get the first head. So starting at the root, with probability, p, you flip a head immediately and you stop. Or, with probability, q, you flip a tail, and then with probability, p, you finally flip the head and stop. If you haven't flipped the head by the end of the second step, then that is a possibility that, with probability, q, flipped a tail, and then there's a possibility that you stop after the third step, with a head, and so on. That's how this tree goes. Now, looking at the structure of this tree, it's an infinite tree, but it's very repetitive. In fact, if we call it the tree B, then what we're seeing is that this whole sub-tree is a copy of B. So now I have a nice recursive, but very finite, description of this whole infinite tree. B is a tree that has a left branch of p that ends in a leaf, or a right branch with probability, q, followed by a repeat of the tree, B. So now I can apply total expectation to find the expectation of F. F is the expected number of steps I make in this tree until I finally make the left branch to an H. How do I figure out what the expected time in the tree is until I make that left branch of flipping a head, finally? Well, using total expectation, what I can do is branch on whether or not the first flip is a head. So the expectation of F, according to the total expectation is going to be the expectation of F, given that the first flip is a head, times the probability, p, that it is a head. And the other term is that it's the expectation of F, given that the first flip is a tail times q, the probability that the first flip is a tail. Well, first of all, let's look at this term. What is the expected number of flips until you get a head, given that you got a head? Well, it's 1. So this term is easily taken care of. We understand that one. What about this term? This is the expected number of flips until you get the first tail. Sorry-- it's the expected number of flips until you get the first head, given that your first step was a tail. Well what that means is that we are here after the tail, and we're asking, what's the expectation of F-- the number of flips that you get starting at B-- when you do one flip that takes you to the start of B again? And the answer is, obviously, 1 plus the expected number of flips in B, which is expectation of F. In short, this term-- the expectation of F, given that the first flip is a tail, is simply 1 plus the expected number of flips that we're trying to figure out. Now look at this-- I have a very simple arithmetic formula now. Expectation of F is 1 times p plus 1 plus F times q. There it is. And now I can solve for E of F. Well, just taking a quick look at this, this is going to yield a q term, and p plus q is 1. And this is going to be q times E of F, and there's an E of F there. If I pull the E of F over, I'm going to do a little arithmetic-- I'm going to leave the rest to you, and realize, again, that the answer is what we had before, the expectation of F is equal to 1 over p. So let's do one more silly example for fun to remember what the significance of 1 over p is. [? Suppose we think ?] about the space station Mir. Now, it's spinning around, and there's a lot of garbage out there that it's likely to hit-- a lot of space junk. And suppose that, based on our previous statistics and estimations of the small stuff that has been hitting Mir that it could survive, that we estimate that there's about a 1 in 150,000 chance that in any given hour, it's going to run into some intolerable collision with space junk, or a meteor that's going to destroy. So suppose the space station Mir has a 1 in 150,000 chance of destruction in any given hour. So how many hours do we expect until destruction? Well, it's 1 of over 150,000, 150,000 hours, which [INAUDIBLE] So much for silly space station examples. Let's wrap up with an intuitive argument that could be made rigorous, but I'm not going to, because I think it's clearer just left in this informal way that makes it intuitive why you would expect that, of course, the expected time to failure is 1 over the probability of failing on a given flip. Well, how many failures do we expect in one try? Well, by definition, it's the expectation of getting a head on the first flip-- it's p. OK, now if you flip n times, you expect to get n times as many failures as you'd expect in one try. So the expected number of fails in n tries is n times p. That's an intuitive argument. In fact, it's the rigorously correct argument. Remember that if we flip n times, we're counting the number of heads and flips-- that's a binomial distribution we already figured out in a couple of ways-- that it's expectations is n p. But never mind that. I think it's intuitively clear that if you expect p heads in one try, and you do n tries that are all independent, you're going to expect to get n times p failures-- or heads. Now, what's the expected number of tries between failures? Well if you think about that, I've done n tries, and I've got n p failures, so if I divide the number of tries by the number of failures, that, by definition, is the average time between the failures. It's the expected time to a failure. So I divide the number of tries by the number of fails-- which, by definition, is the average number of tries between failures, and it's equal to n over n p, which is equal to 1 over p. And that's an argument that I hope you will remember. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 184_Strong_Induction_Video.txt | PROFESSOR: So now we come to an interesting variant of ordinary induction called strong induction, and here's how it works. With strong induction, just as with ordinary induction, you prove the base case P of 0. You're trying to prove for all nP of n, so you prove P of 0, but now in order to prove P of n plus 1 in the inductive step, assuming P of n with ordinary induction, with strong induction you can assume not just P of n but you can assume P of 0, P of 1. All the properties-- that all the numbers up through n have the property. And from this, of course, you could conclude that that everything has the property that for all mP of m. Now, an intuitive way to justify this is you think about it the way that induction works is you start off at 0, and then you make a step to 1, and you make another step to 2, and you make another step to 3, and the induction step going from n to n plus 1 justifies each of those steps. By the time you get to n and you have to prove you can get to n plus 1, you've already been through 0 1 up to n. You might as well take advantage of that fact. Instead of only remembering that you're at the n step, you might as well remember that you got there. That's an intuitive hand-wavy argument which can be justified formally in a way that will emerge in the next segment. So let's hold off on that and just bite the bullet and accept this as a basic principle of math that we're going to live with and use. As an application of it, let's prove something that we've already proved by well ordering and, in fact, strong induction and well ordering are closely related, as we'll also discuss later. So let's prove that using $0.03 and $0.05 stamps that you can get any amount of postage greater than or equal to $0.08 stamps, and I'm going to prove this by strong induction, with the induction hypothesis P of n that says I can form n plus $0.8 stamps. Clearly, if I can prove for all nP of n, then I've proved that I can get for every amount greater than or equal to $0.08 stamps. And let's do the base case. Well, the base case, P of 0. Can I make $0.8 stamps? Sure, $0.03 and $0.05. That's that one, and that's OK. For the inductive step, I have to get m-- I'm allowed to assume, rather, that I can get m plus $0.8 for any m from n down to 0, instead of just assuming that I can get n plus $0.8 to get n plus 1 plus $0.08. I can assume any amount less than what I'm aiming for, so I may as well assume that I can get any amount of postage from $0.08 up to n plus $0.08, and my objective then is to get n plus 1 plus $0.08, namely n plus $0.09. So I have to prove that for all n greater than or equal to 0, I can get n plus $0.09, assuming I can get from $0.08 to n plus $0.08. Well, that's not too hard to do. The inductive step is actually going to break up into a couple of cases, depending on the value of n. I have to prove n plus $0.09 for all n, so suppose n equals 0, I have to get $0.09. Well, three $0.03. If n is 1, I have to get 1 plus $0.09 or $0.10, two $0.05. So those cases are disposed of. So now my job is to get n plus $0.09, where n is greater than or equal to 2. Well, the nice thing about n being greater than or equal to 2 is that if I subtract 2 from it, it's a smaller number, and it's still not negative, and that means that I can get that amount plus $0.08 stamps. So I'm in this nice situation where I, by strong induction, I can get n minus 2 plus $0.08 stamps. There they are. And how do I get to n plus 9? Well, it's easy. I add a $0.03 stamp, and you have n plus $0.09, which completes the proof of the induction case, and the whole theorem is proved. We can conclude then that it works for all n and that you can indeed get n plus $0.08 using $0.03 and $0.05 stamps for all of them. So much for that example. All right, let's look at another example. This is a game that we used to play in class. You start off with a stack of blocks, say 10 blocks, and you're allowed to make a move that consists of splitting the stack into two smaller stacks. So if the stack has height a plus b, you can split it into a stack of height a and a stack of height b, and you get a score for that move. The score is a times b. And then you keep doing that until you can't make anymore moves. That is, when all you have left are stacks of height one, which you can't split anymore, and then your overall score is the total that you got for all the moves that you made until that point. Now, when we played this in class, we would have students competing, and they would try various strategies. So one strategy-- the simplest strategy, maybe not the best, but the simplest strategy, would be to start off with a stack of 10, so you take one of, and that leaves you with a stack of one and nine. Your score is nine. Then you take another one off of the stack of nine, and you're left with a one and an eight. Your score is eight, and so on, and you can see, in fact, if took one-at-a-time process then your score with a stack of height n would be n minus 1 plus n minus 2, down to 2 plus 1. Another strategy that might be sort of more in the spirit of computer science would be to keep splitting in two. So for example, if you had a stack of height 10, you could split it into two fives, and then you take one of the fives and split it into three and a two, and then you'd split the two into two ones, and so on, splitting as evenly as you can each time, and it seems like it might be a better strategy. And, again, we would have students try various strategies, and guess what? They all came in in a tie, and that's what we're going to prove now. Every way of unstacking n blocks gives the same score. Well, what score? Well, we know that the score for the simple strategy of taking one block off at a time is the sum from 1 to n minus 1, and that has a nice formula-- n times n minus 1 over 2, so we can formulate our claim that no matter how you play the unstacking game with the stack of size n, your final score will be n times n minus 1 over 2, and we're going to prove this by strong induction, with this statement-- call it claim of n-- is going to be the induction hypothesis. That's what we're trying to prove. Well, let's start in the usual way. The base case is n equals 0. Well, you might be bothered. That's no blocks. Well, let's see what happens. With no blocks, the score is 0 because there's nothing to do, and indeed the formula that is alleged to be your score comes out to be 0, so the base case n equals 0 works. Let's continue. For the inductive case, I have to assume that the given score formula holds for all stacks of height n or less, and I have to prove that it holds for a stack of height n plus 1. That is, that an n plus 1 stack score is n plus 1 times n over 2. Well, how shall I do that? Well, I'm going to split the inductive case into two cases. It turns out that I need to prove that c of n plus 1 holds, assuming c of n for n and less than n but, in particular, let's just deal with the case that n plus 1 is 1, the smallest value it could have and knock that one off separately. Namely, if the stack is of height one, again my score is 0, because there's no move to make, and the formula still evaluates to 0. So in the case that n plus 1 is 1, I've proved the claim at n plus 1, which I was obligated to do for the base case-- for the inductive step. Well, the other case in the inductive step is that n plus 1 is greater than 1. This is the interesting one, because now it's possible to make a move. So since n plus 1 is greater than 1, it's two or more blocks. I can make a move into two stacks that are both of positive. So suppose I do that? Suppose I split the stack of size n plus 1 into an a stack and a b stack, where a and b sum to n plus 1. And what's my score going to be then? Well, my score on that move that I make, where I split into the a stack and the b stack is a b, and the rest of the game consists of playing as well as I can on the a stack and as well as I can on the b stack, but a and b are smaller than n plus 1. They're less than or equal to n, which means that by strong induction, I know that no matter how I play on the a stack, I'm going to wind up with this score a times a minus 1 over 2. No matter how I play on the b stack, I'm going to wind up with b times b minus 1 over 2, so that means that my score on a plus b stack is going to be this formula, ab plus a times a minus 1 over 2 plus b times b minus 1 over 2. So you simplify that to organize it so it's a plus b times a plus b minus 1, which is exactly n plus 1 times n over 2, which is what we were trying to prove. We've proved C event plus 1. The inductive step is complete, and indeed we've proved that no matter how big the stack is, your score comes out the same. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 152_Predicate_Logic_2_Video.txt | PROFESSOR: So now we're going to talk about the concepts of validity and satisfiability, which take on some extra interest in the context of predicate calculus. So let's remember for propositional validity, if you have a propositional formula with variables taking the truth values ranging over true and false, then a formula is valid when it's true for all possible truth values. So here's an example that P implies Q or Q implies P. And you can check that for the four possible environments of P and Q-- true/false values of P and Q, this OR will come out to be true. Well, for predicate formulas there's a bunch of things that I need to give value to that are more complicated than just truth values. In particular, I will say that a predicate calculus formula is valid when it's true for all possible domains of discourse that the variables range over. There's a technicality that it has to be non-empty, but aside from that all possible domains. And whenever you have a predicate mentioned in the formula, in order to know whether the formula is true or not, I need to know what that predicate means. So a formula is valid if it comes out true no matter what the predicate means. Let's look at a concrete example to get a grip on this. Here is a valid formula of predicate calculus. It's mentioning predicates P and Q. It's of the form of a proposition because it's saying something about every possible z in the domain and every possible x and every possible y. The only thing that we need to know to make sense out of this formula, to figure out whether or not it's true, is what's the domain that x, y, and z range over and what exactly do P and Q mean? Well, I want to argue informally. Let's just look at what this is saying together. What this is saying is suppose that for everything in the domain, both property P of z and P of Q. In other words, everything in the domain have property P and property Q. Well, that certainly implies that everything in the domain has property P because they have both properties. And also, everything in the domain has property Q because everything has both properties. So when you say it that way, the sense that this is a fundamental logical fact that doesn't depend on what P and Q mean or what the domain is. It's just a fact about the nature of the meaning of the for all universal quantifier and the connectives [? AND then ?] implies. That's how we figure out that this is valid. Well, let me go one level in more detail to say again what I just said informally and try to be a little bit more precise and clear about a reason why this formula is valid. So suppose I wanted to prove that the formula is valid. Well, it's an implication. So the proof strategy-- there it is written again-- the proof strategy is I'm going to assume that the if part, the left-hand side of the implies or hypothesis, is true. That is, that for every z, P of z holds and Q of z holds. And then I'm going to try to prove, based on that, that the consequent holds, namely that the right-hand side for all x P of x and for all y Q of y holds. OK. How am I going to do that? Well, so here's the formula written just to fit on the line with the concise mathematical symbols. The upside down V means AND, and the arrow means implies. And we want to try to prove that this is valid a little bit more carefully. Well, the strategy, as I said, is to assume that the left-hand side holds. Well, what's the left-hand side say? It says that for every z, Q of z holds and P of z holds. That means that for every possible environment that assigns a value to z, Q of z and P of z both come out to be true. Well, suppose that the environment assigns the value c to z, where c is some element of the domain. Then what this means is that in that environment, Q of z and P of z is true, which means that Q of c and P of c holds. But Q of c holds and P of c holds, so Q of c certainly holds all by itself. All right. So now we're in an interesting situation because we just proved that Q of c holds. And we know nothing and have assumed nothing about c except that it's an element of the domain. c could have been any element of the domain, and we've managed to prove that Q of c holds. So it follows that, in fact, we have really proved that for every x Q of x holds. Now, that step of saying I proved it for Q of a given element without making any assumptions about the given element except that it's in the domain and therefore I can conclude that it holds for all domain elements, [? it's a ?] very natural and plausible and understandable rule. And it's a basic axiom of logic called UG-- Universal Generalization. We'll come back to it in a minute. Anyway, I've just proved that for all x Q of x holds. And by a completely symmetric argument, for all y P of y holds. And having proved both for all x Q of x and for all y P of y, clearly the AND holds. And I've just proved that the right-hand side of this implication is true given that the left-hand side is true. Now, having called this proving validity, let me immediately clarify that this is not fair to call a proof because the rules of the game are really murky here. This theorem, you could read it as saying that universal quantification distributes over AND is one of these basic valid formulas that is so fundamental and intelligible that it's hard to see what more basic things you are allowed to assume when you're proving it. And this proof really isn't anything more than translating upside down A and the AND symbol into English and using ordinary intuitive rules about for all and AND and using that in the proof. So this is a good way to think about the formula to get an understanding of it. But it's not right to say that it's a proof because we haven't been exactly clear about what the proof rules are. And with this kind of really fundamental valid fact, it becomes a quite technical problem to decide what a proof is going to be. What's fair to assume and what's fair not to assume? It would actually be perfectly plausible to take this as an axiom and then prove other things as a consequent of it. Anyway, going on, let's look at this just for cultural reasons. We're never going to actually ask you to do anything with it. But the universal generalization rule UG would be stated this way as a deduction rule in logic. The stuff over the bar means if you've proved this, then you can conclude you've proved the stuff below the bar. So what this is saying is if you've proved P of c for a constant c, then you can deduce that for every x P of x holds. And this is providing that c does not occur in any other part of the predicate P except where you're talking explicitly about it. It's hard to be more precise about that for now. Don't worry about it. But the idea is you're not supposed to assume anything about c other than it's in the domain and that it has property P, and you can then conclude that everything has property P. So let's look at a similar example where it is possible to prove something. Namely, I can prove that something's not valid. So here's a similar-looking formula. This one says that for every z if P of z holds OR Q of z holds, then for every x P of x holds OR for every y Q of y holds. And this one we're going to show is not valid. Let's think about it for a minute. What it's saying is if everything has either property P or property Q, that implies that everything has property P or everything has property Q. Well, when you say it that way, it's clearly not the case. But let's go one level more precise and lay that out. What I'm going to do is convince you that it's not valid by giving you a counter-model where I choose an interpretation. I choose a domain of discourse and predicates that P and Q are going to mean over that domain and that make the left-hand side of this implication true. And then I'm going to show you that the right-hand side is not true. And that means that in that domain with those interpretations of P and Q, this implication fails so it's not valid. So I need to make the left-hand side true and the right-hand side false. Well, I'm going to choose the domain of discourse to be the simplest one that will make this false, namely let's let the domain of discourse just be the numbers 1 and 2. And let Q of z be the predicate that says z is 1 and P of z be the predicate that says z is 2. Well, is the left-hand side true? Yeah, because the only things there are in the domain are 1 and 2, and so clearly everything in the domain is either 1 or 2. So the antecedent is true. On the other hand, is everything in the domain, does it satisfy P? Is everything in the domain equal to 2? No, 1's not equal to 2. What about, is everything in the domain equal to 1? Is it true that for all y Q of y holds? No, 2 is in the domain, and it's not equal to 1. And so we have found exactly what we wanted, a counter-model which makes the left-hand side of the implies true and the right-hand side of the implies false. Let me close with just one more example of a valid formula that we can talk through. This is the version of De Morgan's law that works for quantifiers. Remember De Morgan's law was the thing that said that the negation of P or Q was the same as not P and not Q. And remembering that the connection between universal quantification, [? an ?] AND, and existential quantification, [? an ?] OR, it turns out that by the same kind of reasoning, De Morgan's law comes out this way. It says that if it's not true that everything has property P, that's possible if and only if there's something that doesn't have property P. And so that's what De Morgan's law is. It's another thing you could take as an axiom, or you could try one of these hand-waving proofs about. But I think I've said enough to give you that example of another interesting valid formula, and we'll stop with that. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 2115_Optimal_Stable_Matching_Video.txt | PROFESSOR: So stability has some value, but it doesn't mean that everybody's happy. In fact, it just means that nobody can find anybody else who's equally unhappy that they would want to run off with. So let's examine the question of how well people do using the mating ritual, or in other possible ways of finding stable marriages. So basically, we want to begin with the question of who does better, the boys or the girls? Maybe it's a mixture. Maybe the boys do better, the girls do better, or maybe some boys do better than others, and some girls do better than others. One thing to notice is we know that the girl's suitors are getting better day by day, and that sounds like the mating ritual might be slanted towards them. Likewise the boy's sweethearts, the ones that they're serenading, are getting worse day by day, and that sounds like it might be an argument for the girls to do better, but that's not true. And the reason it's not is that, if you think about it, the boys are starting off with their first choice. They begin by serenading the girl at the top of their list, and it's true that day by day they keep going down, or staying the same or going down, but they're only sinking to, in fact, the best possible woman that they could be married to. Let's examine that. So I need a definition, which is that we'll say that a woman, Nicole, is called optimal for Keith when she is the highest ranked girl he can stably marry. So let's think about that for a minute. So Keith has his preference for different girls that he likes, to different degrees, and there may be some that he likes. Like Keith thinks that Angelina is terrific, but there's just no way that she's going to wind up with him, because she just ranks him very lowly, so there's no stable set of marriages in which Keith can wind up with this very desirable woman, Angelina. But if you look at all of the sets of marriages that are stable, that Keith can be involved in, among them Nicole is the woman that he most likes, so that's what we mean by Nicole is optimal for Keith. She's optimal among the feasible women that he could stably be married to. The claim that we're making is that the mating ritual yields a set of stable marriages, which is simultaneously optimal for Keith and all the other boys at once. Now, that's a kind of unusual thing. Usually when you're optimizing, you figure you're optimized for one boy, and it sacrifices the optimality for the other boys, but that's not what happens in the mating ritual. All of the boys get their absolutely optimal spouse in the mating ritual, and dually it turns out that all of the girls get the worst possible spouse that they can get, a pessimal spouse, among all possible stable marriages. Well, with that claim understood, let's go about proving it, and we're going to prove that the mating ritual leads to boy-optimal marriages by contradiction. So let's suppose that Nicole is optimal for Keith among all the women that Keith could possibly be married to in a stable way. Nicole is the best. Suppose that Keith does not wind up marrying Nicole in the mating ritual. So he doesn't marry Nicole in the mating ritual. That means that since the Nicole is optimal for Keith, he must be married to somebody that's less desirable to him than Nicole, so he must have crossed Nicole off on some day. Let's call that his bad day. So on his bad day, Keith is rejected by his optimal spouse. OK If this ever happens, there's going to be some boy who has the earliest bad day-- we may as well assume that it's Keith. So let's assume that Keith was the earliest among the boys to have a bad day, that is, a day on which he crosses off his optimal spouse, because he was rejected by her. Well, on this bad day when Keith crosses off Nicole, it's because Nicole rejected him, which meant that Nicole had a suitor that she liked better than Keith. Let's call that suitor Tom. So what we know is that Nicole prefers this guy Tom to Keith on the day that she rejected Keith, and he crossed her off, and we also know since this is the earliest bad day that anybody has, Tom has not yet crossed off his optimal girl. So what that means is that since he's serenading Nicole, and she's going to wind up rejecting Keith in favor of Tom, it must be the case that Nicole is at least as desirable to Tom as his optimal spouse, because he hasn't gotten to his optimal spouse yet. He's working his way down the list, and he hasn't had a bad day yet. So let's put these two pieces together. Nicole is at least as desirable to Tom as Tom's optimal spouse, and Nicole prefers Tom to Keith, but what that tells us is that if I had a set of stable marriages, with Nicole married to Keith, then in the stable set of marriages, of course, Tom is going to be married to somebody that's, at best, optimal for him, so he's married to somebody that he likes less than Nicole. And Nicole is married to Keith, and she likes Tom better than who she's married to. What that tells us is that Nicole and Tom are a rogue couple in any stable set of marriages where Nicole was married to Keith, but that contradicts the fact that Nicole is supposed to be optimal for Keith. There's supposed to be a stable set of marriages where Nicole is married to Keith. So a similar argument-- it's actually slightly easier-- is that the mating ritual yields a set of stable marriages in which all of the girls get the worst possible spouse that they can have in any set of stable marriages. So this leads to a whole bunch of questions, and it turns out that there's a very rich theory of stable marriages, as I mentioned. First question to ask is, well, are there other possible stable marriages? Well, one thing that you can obviously do is you could switch the roles of the boys and the girls. So if you switch the roles of the boys and the girls, you'll get a set of stable marriages that are optimal for the girls and pessimal for the boys. Maybe that's fair or you rather do that. So that's at least a possibility of using the mating ritual to get two different stable sets of marriages, unless the two happen to be the same. And the question arises, are there others that could exist that the mating ritual doesn't find, either by choosing the boys to act as boys or the boys to act as girls, and the answer is that, in general, there can be many. As a matter of fact, if there are N boys and girls, it's possible that there could be an exponential number of stable marriages in N, and that leads to the question of, well, which is one that might be a better one to choose compared to the one that completely favors the boys or completely favors girls. Another interesting question that comes up that's an issue that comes up with general protocols of negotiation and optimization among multiple parties is, does it serve anybody to lie? That is, instead of following the protocol and always going to the-- and the boys always serenading the girls that they like best, and the girls always rejecting anybody that's less desirable than their favorite suitor, suppose they violate the convention and lie. Can they do better? Well, it turns out that, of course, the boys in the mating ritual aren't doing optimal, so they don't gain anything by trying to lie, but the girls it turns out-- it's almost the case-- that girls can do better by lying. If they conspire together to lie, they can actually force the mating ritual to wind up turning into a stable set of marriages that's girl optimal. So that raises another issue about, are there protocols which are resistant to lying? We're not going to go into these questions. We mainly wanted to understand the stable marriage problem and its applications and how to find them. Again, if you want to learn more about this, you can look at the book by Gusfield and Irving that I mentioned in a previous video. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 274_Equivalence_Relations_Video.txt | Equivalence relations are another kind of binary relation on a set which play a crucial role in mathematics and in computer science in particular. And they can also be explained both in terms of digraphs and in terms of axioms. So let's begin with a digraph explanation of an equivalence relation. And the kind of relation that's an equivalence relation is the relation of there being a walk in both directions between two vertices. So if there's a walk between vertex u and vertex v and conversely there's a walk from vertex v back to vertex u, then u and v are said to be strongly connected, and strongly connected is going to be an example of an equivalence relation. So in terms of the walk relation, including 0-length walks, the relation we're talking about is u G* v and v G* u. Now, as a property of relations, this has a name. It's called symmetry. So a relation R on a set A is symmetric if and only if a R b implies b R a, and the first remark is that the strongly connected relation is symmetric. An equivalence relation is a symmetric relation that is transitive and reflexive. And again, we have immediately that the walk relation-- the mutual walk relation, the two-way walk relation or strongly connected relation in a digraph is an equivalence relation. Because clearly if there's two way paths between u and v and between and v and w, then there's one between u and w by going for u to v to w and back. Likewise, there is a length 0 walk from any element to itself. And by definition, strong connectedness is symmetric. So the strong connectedness relation in any digraph is an equivalence relation. And the theorem is, conversely, that any equivalence relation, anything that's an equivalence relation, is the strongly connected relation of some digraph. The proof is trivial. It's the strongly connected relation of itself. OK. Some examples of equivalence relations to see why they're so basic is that the most fundamental one is equality. Obviously, equality is symmetric and reflexive and transitive, and so it's an equivalence relation. Another one that we've seen is congruence mod n, which you could also check is symmetric and transitive and reflexive. And finally, another relation would be that two sets are the same size, providing they're finite sets. And another example would be a bunch of objects having the same color. Two objects have the same color is a relation among objects that have color that is symmetric and transitive and reflexive, so it's an equivalence relation. Let's illustrate some of these axioms that we have in terms of graphs. It can be helpful to remember them. So reflexive means that when you look at a digraph, it's reflexive when there's a little self loop from every vertex to itself. So there's a length 1 path or an edge from vertex to itself in reflexive graphs. Transitive means that whenever you have two edges connecting one vertex to another, there's a path of length 2 from one place to another that in fact is an edge from that place to its target. And of course as we said, once there is an edge wherever there's a path of length 2, it follows by induction that there's an edge wherever there's a path of any length, and that's what transitive means. Asymmetric means that whenever you have an edge from one vertex to another there is no edge back. So in particular, if I have an edge from this vertex to that vertex in blue, there is no edge that goes back in the other direction. Nor is there ever a self loop in an asymmetric graph. And finally, in a symmetric graph, wherever there's an edge, there's an edge that goes back the other way. So that can help you maybe remember what these properties mean. Now again, equivalence relations, besides being represented in terms of the strongly connected relation of a digraph, can be represented in two other very natural ways that really explains where they come from and what their properties are. So whenever you have a total function f on a set A, it defines an equivalence relation on the set A. Namely, if f is a total function from domain A to codomain B, then we can define a relation we can call equivalence sub f on the set A by the rule that two elements are equivalents of f if and only if they have the same image under f-- they hit the same thing. That is, A is equivalent sub f to A prime if and only if f of a is equal to f of a prime. And again, equivalence sub f immediately inherits the properties of equality, which makes it an equivalence relation. And the theorem that we have is that every relation R on a set A is an equivalence relation if and only if it in fact is equal to equivalence sub f for some function f. Let's illustrate that. We already remembered that congruence mod n can be understood as equivalence sub f, where the mapping is just map things to remainders. Two numbers are congruent mod n if and only if they have the same remainder on division by n. So map a number a to f of k, equal its remainder, and we have found the equivalence sub f representation of congruence, which is another way to verify that congruence is an equivalence relation. Finally, whenever you have a partition of a set, you can define an equivalence relation. So a partition of a set cuts up the set A into a bunch of blocks which are nonempty, and every element is a member of some block, and the blocks don't overlap. So in fact, every element is a member of a unique block. And that enables me to define an equivalence relation on A by the property that two elements are in the same block. In fact, that's the proof of the previous representation theorem in terms of a function that you can map an element to the block that it's in, in order to see that the block representation and the equivalence sub f representation are the same. The proof in the other direction, that every equivalence relation can be represented in this way, is an exercise in axiomatic reasoning, and elementary one that we're going to leave to a problem and not do in this presentation. So the theorem finally is that, again, a relation R on a set is an equivalence relation if and only if it is in fact the being in the same block relation for some partition. And that is the story and multiple ways of understanding what equivalence relations are. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 161_Sets_Definitions_Video.txt | PROFESSOR: We're going to look at the most fundamental of all mathematical data types, namely sets, and let's begin with the definitions. So informally, a set is a collection of mathematical objects, and the idea is that you treat the collection of objects as one new object. And that's a working definition. But of course, it might help a little bit, but it's a circular definition. This is not math yet because I haven't defined what a collection is, and a collection is no clearer or easier to define than a set is. So let's try to work up the idea of sets by looking at some examples. So we've already talked about some familiar ones. There's the real numbers for which we had this symbol R in a special font, and the complex numbers C, and the integers Z. And we might have mentioned, and you might have seen already, the idea of the empty set for which we use this symbol that looks like a zero with a line through it. Let's look at an example to pin things down. Let's look at this set of four things. Namely, it's got two numbers, pi/2 and 7, a character string in quotes, "Albert R," and the Boolean value true. So those are the four different things in it. They're of mixed type. And you might not like to have a mixed type like this in a programming language. But mathematicians don't worry about such things very much, rarely. Anyway, the first observation is that the order in which these elements are listed doesn't matter. This set-- the braces indicates that it's the set of these things-- is the same if I listed T first, then the string, and the two numbers last. There is no notion of order in a set. Now, to a computer scientist this is a little unnatural. The most natural thing to be would be to define a sequence of things, like the sequence that began with 7, then had the character string, then had the number, then had the Boolean. And you could get by with working with lists of things as long as they're finite. But they very quickly get out of hand when you have to talk about, say, a set of lists. Then it's not clear how to make a list out of those, and you wind up making sets again. So sets, in fact, are an unavoidable kind of idea. So another basic thing to understand about the notion of a set is that an element is either in a set or not in a set. So if I write down 7, pi/2, 7, this is the same description of the same set 7, pi/2. I'm just telling you the same thing twice here. That 7 is in the set, and the 7 is in the set again. So no notion of being in the set more than once. Now, sometimes, technically you want to add a notion of so-called multisets in which elements can be in a set a certain number of times, an integer number of times. But there's no real need for that. It's a secondary idea. And from our point of view, you're in or out of a set. If you repeat elements, it's the same as mentioning them once. So the most fundamental feature of a set is what's in it. And for that, there's a special notation. So we'll say that x is a member of A, where A is a set, and use this epsilon symbol to indicate membership. It's read x is a member of A. So for example, pi/2 is a member of that set that we saw before that had pi/2 in it. 14/2 is also a member of that set because 14/2 is just another description of 7. When I write 7 here, I don't mean the character 7. I mean the number 7. And so 14/2 is the description of the same number. It's in that set. On the other hand, pi/3 is a number that's simply not in that set. So I'm using the epsilon with a vertical bar through it, or some kind of a line through it, to mean not a member of. Membership is so basic that there's a lot of different ways to say it. Besides using the membership symbol x is a member of A, you can sometimes say x is an element of A, or x is in A, as well as x is a member of A. They're all synonyms. So for example, 7 is a member of the integers. Z is our symbol for the integers. 2/3 is not a member of the integers because it's a fraction that's not an integer. And on the other hand, the set Z of integers itself is a member of this three-element set consisting of the truth value T, the set of all integers, and the element 7. So here's an example where a set can contain sets, quite big ones even, and that's fine. That's not any problem mathematically. Related to membership is another fundamental notion of subset. So A is a subset of B, it's pronounced. So that horizontal U with a line under it is meant to resemble a less than or equal to symbol. So you can think of it as being A is less than or equal to B. But don't overload the symbols. Less than or equal to is used on numbers and other things that we know how to order. And this is a relation that's only allowed between sets. So A is a subset of B-- A synonym is that A is contained in B-- simply means that every element of A is also an element of B. If I wrote that out in predicate logic notation as a predicate formula, I'd say for every x, x is in A implies x is in B. If it's in A, then it's in B. Everything in A is in B. So some examples of the subset relation are that the integers are a kind of-- an integer is a special case of a real number. So the set of integers is a subset of the real numbers. A real number is a special case of a complex number, so the real numbers are a subset of the complex numbers. And here's a concrete example, where I have a set of three things, 5, 7, and 3, and this is the set with just the element 3 in it. Now, we sometimes are sloppy about distinguishing the element 3 from the set that's consisting of just 3 as its only element. But in fact, it's a pretty important distinction to keep track of. In this case, 3 is not a subset of this set on the right. But the set consisting of 3 is a subset of the set on the right because, after all, the only member of this set is 3, and that is a member of this set. A consequence of this general definition is that every set is a subset of itself because everything in A is in A . That's not really very interesting. Another important general observation is that the empty set is a subset of everything. The empty set is a subset of every set. Let's look at why that is in more detail. So the claim is that the empty set is a subset of everything. Let B be any old set, then the empty set is a subset of B. What exactly does that mean according to the definition of subset? Well, it says that everything that's in the empty set, if it's in the empty set, then it implies that it's in B. For every element, if it's in the empty set, then it's in B. Well, what do we know about this? The assertion that x is in the empty set is false. No matter what x is, there's nothing in the empty set. And now I have an implication that implies where the left-hand side, the hypothesis, is false. That means that the whole implication is true, and it doesn't depend on what B is. I'm not even going to look at B. I can see that x is in empty set is false, so the whole implication is true. And so what I'm saying is that for every x, something that's true has to be true. Well, it is. And that's why the empty set is a subset of B satisfies this definition in a formal way. And this is an example of why that convention that false implies anything is convenient and is made use here. So when you're defining sets, if they're small, you can just list the elements, as we did with that set with 7 and pi/2 and "Albert R." Sometimes we can even describe infinite sets as some kind of a list. Like I might describe the set of integers as saying, well, it's 0, 1, minus 1, 2, minus 2, and so on, and you would understand that. But in general, if I'm describing a set that is not so easy to list, say the real numbers, then what I'm going to do is define a set by a defining property of in the set. So I'm interested in a property P of elements, and I'm going to look at the set of elements x that are in some set A such that P of x is true, and that's the notation we use. So this would be read as the set of x in A such that P of x holds, that x has property P. So notice this vertical bar is read as "such that." It's just a mathematical abbreviation. This is those elements in A that have property P that P of x holds for, and that defines a set of those elements. Let's look at a simple example. The set E of even integers is simply the set of numbers n that are integers such that n is even. So in this case, the property P of n means that n is even. One last concept is the concept of the power set. So the power set of a set A is all of the subsets of A. So we could define it using set notation as it's the set of B such that B is a subset of A. An example would be-- let's take the power set of the two Boolean values true and false. So the power set of true and false, of that set consisting of two elements, is-- well, what are some of its subsets? The set consisting of just true is a subset of true, false. So is the set consisting of false, and so is the whole thing. It's a subset of itself. And one final element, the empty set, is a subset of the set of Boolean values true and false. So the power set of this two-element set is a set that has four things in it-- two elements of size 1, one element of size 2, one element of size 0. And that's going to be a general phenomenon that we'll examine more later. How big is the power set of a set? The even numbers, E that we just defined on the previous slide, is a member of the power set of Z because it's a subset of integers. Even integers are a special case of integers. And the integers are a member of the power set of R. That's just a synonym for saying that integers are a subset of reals. Every integer is a real, so the integers are a subset of reals, which means they're a member of the power set of reals. So the general property is that a set B is a member of the power set of A if and only if B is a subset of A. That was the defining condition for power set. And that's a fact to remember, and it may potentially confuse you. But it's a good exercise in keeping track of the difference between is a member of and is a subset of. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 317_Integral_Method_Video.txt | PROFESSOR: So we figured out that you can get the book stack, the overhang of the books, to be half the harmonic sum. With n books, you can get out Hn over 2 where Hn is this harmonic sum, or harmonic number. The question is, how are we going to estimate or calculate what this sum is? Now it turns out there is no simple formula for exactly what this sum is, but there is a simple formula that estimates it quite accurately. And we get the estimation by bounding the sum by integrals. And so let's look at this integral method for estimating sums. Remember what I wanted was the sum of 1 plus 1/2 plus 1/3 down to 1 over n. So let's form some unit width rectangles of heights equal to the amount that I want. So here's a rectangle of height 1, rectangle of height 1/2, rectangle of height 1/3. I'm going out here, if you actually count to 8, but let's propose that this is a height 1 over n. And what I know is that the total area of these rectangles is actually equal to the number that I want. The total area of these rectangles is the harmonic number. And I'm interested in a lower bound for Hn, because I want to know how far I can get out. I want a tight lower bound. It says Hn is larger than a certain amount. That's the amount that I'm sure that I can stack out in books. So the way I'm going to get a lower bound on this number Hn is by looking at this curve that goes through the corners of the rectangles. And if you check it, that curve is 1 over x plus 1. That is, the point here is-- when x is 0, I'm at 1 over 1. When x is 1, I'm at 1/2, the height of the second rectangle, and so on. So 1 over x plus 1 is a curve that is strictly below the boundaries of all these rectangles. That means that the area under 1 over x plus 1 going from 0 to n is a lower bound on Hn, because it's a lower bound on the area of the rectangles. So Hn equals the area of the rectangles. It's greater than the area of 1 over x plus 1, which of course is equal to the integral from 0 to n of 1 over x plus 1, which shifting variables is the same as the integral from 1 to n plus 1 of 1 over x dx, which of course we know is a natural logarithm of n plus 1. So there we have it. The overhang that you need for three books, which is Bn greater than or equal to 3, means that Hn has to be greater than or equal to 6. So by this estimate, I need log of n plus 1 greater than or equal to 6 in order to get three books out, that the back end of the top book is two books past the edge of the table and the right end of the furthest out book is three book lengths past the edge of the table. So my bound tells me that I need n books such that log of n plus 1 is greater than or equal to 6. Well, exponentiating both sides, the right-hand side becomes e to the 6th. And I figure out that as long as n is greater than or equal to e to the 6th minus 1 books-- rounded up, of course, because there's only-- you can't have fractions of a book-- you get an estimate that with 403 books, I can actually get my stack to stick out three book lengths past the edge of the table. Well if you do the actual calculation instead of the estimate, it turns out that 227 books are enough. So this estimate's a little off. But for our purposes, it tells us a dramatic fact, which is that we know that log of n plus 1 approaches infinity as n approaches infinity. And that means that, with enough books, I can get out as far as I want. You tell me how many book lengths you want to be out, I'll use the log n formula to calculate how many books I need to get that far out. So here's an example of some students in the class some years ago decided to do this as an experiment. Now when we used to do this in class, we first tried to do it with the big, heavy textbook that we used. And we kept trying to get them to balance and go out over the edge of the table, and they failed, because it turns out that textbooks are heavy and they compress. They're not the rigid rectangular cross sections that our model was based on. But CD cases work very well. They are more rigid. They don't compress easily, and they're very lightweight so that they don't cause problems with distortions because of the size of the stack. And so you can actually get CD cases to stick out pretty far. This is an example where it's 43 CD cases high, and the top four cases are completely past the edge of the table. The leftmost edge is about 1.81 or 1.91 case lengths past the table. There's another view of it from the guy who made the stack. And of course, they were right on the edge of stability in trying to get the CDs to stick out as far as possible. So if you notice these little spaces there, in terms of the balancing, it's really just on the brink of falling over. If you sneeze at it, it'll tip. But if you don't sneeze at it, it's stable, and you get the top CD out that far. So while we're at it, let's get an upper bound for Hn. We just got a lower bound of Hn, but the same kind of logic of using an integral will give you an upper bound for Hn. What I do now is I run a curve from the upper right corners of the rectangles. And that curve is simply 1 over x. So an upper bound for the harmonic number Hn is the area under 1 over x out to n plus this 1. And so I get an upper bound that says that the harmonic number Hn is less than the integral from 1 to n of 1 over x dx plus 1, or it's equal to 1 plus log of n. So combining those two bounds that I got by looking at a curve that's a lower bound on the area and a curve that's an upper bound on the area and integrating, I discover that Hn is bracketed between the natural log of n plus 1 and 1 plus the natural log of n. Now these two numbers, log of n plus 1 and 1 plus log of n, are very close, and they get closer and closer as n grows. So it turns out that what we can say pretty accurately is that Hn is asymptotically equal to log n. It's approximately equal to log n. And the precise definition of this tilde symbol that I've highlighted in magenta is called asymptotically equal. And the general definition of asymptotically equal, that Hn is asymptotically equal to log n, is that a function f of n is asymptotically equal to a function g of n when the limit of their quotient goes to 1. That is to say, as a multiplicative factor, each is within a constant one of the other in the limit. 1 plus epsilon of-- 1 plus or minus epsilon of gn is going to bracket fn, and vice versa. Asymptotic equivalents, or asymptotic equality. Let's do an example. So the remark would be, for example, that n squared plus n is asymptotically equal to n squared. Why? Well let's look at the limit as n approaches infinity of n squared plus n over n squared. It's the same as simplifying algebraically the limit of 1 over 1 plus n. As n approaches infinity, that term goes to zero. Sure enough, the limit is 1, and so these two terms are asymptotically equal. The idea of asymptotically equal is that all we care about is the high order term. The low order terms will disappear, and we're looking at the principal term that controls the growth rate of the functions when we look at them up to asymptotic equivalence. So let's step back for just a moment and generalize what we've done here with estimating the harmonic sum. There's a general method called the integral method where, in this particular case, suppose we have a function f, the positive real valued function that's weakly decreasing, like 1 over x. Then let's let S be the sum from i equals 1 to n of f of i. So I was interested where f of x was 1 over x, and I wanted the sum from 1 over-- 1 to n of 1 over i, which was the nth harmonic number. And I was comparing that to the integral from 1 to n of f of x, or 1 over x in our example. So if I is the integral and S is the sum that I want, what we can conclude really is that, in general, the sum is bracketed between the integral plus the first term of f of 1 in the sum, and the integral plus the last term of the sum. Remember, f is weakly decreasing, so f of n is smaller than f of 1. There's a similar theorem actually, just reverse f of 1 and f of n, to use an integral estimate to get a bound on a weakly increasing function. And that gives us a general tool for estimating the growth rate of sums. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 191_State_Machines_Invariants_Video.txt | PROFESSOR: So we're going to talk about state machines, which is a topic that you're going to see in many further courses, because state machines model step by step processes. And of course, if you think about a computation, normally the way you think about it is that it's doing instructions, step by step, one after another, until it finally reaches termination. Likewise, various digital circuits move through stages or states until they produce a final answer. So state machines come up in at least those circumstances and in many others. And the general model of state machine involves the idea that you can give it input and it responds to them, but we don't really need that for our purposes. So let's look at our example of a state machine. Here's maybe a particular simple one. This is a description of a state machine that counts to 99. So the circles are indicating what its states are, and we've named them from 0 through 99. And then there's a final state called overflow, and that funny arc is an indication that if you're in overflow mode and you make another step by following the arc, you get back to overflow mode. But if you're in 0, you can make a step to 1. And if you're in 1, you can make a step to 2, and so on. So starting off at the start state, which is generally indicated by that double mark. So to indicate where to start. Then the description of this machine, consistent in complete description is a set of states which are pictured as 0 through 99 plus overflow, a set of transitions which are indicated by the arrows, which is how one state can move to another state. And the transitions can be summarised by saying that there of the form of i to i plus 1 for i between 0 and 99. And then there's a 99 transition to overflow, and once you're in overflow, you stay in overflow. So the picture at the top is saying exactly the same thing as we've said here in mathematical notation, explicitly describing what the transitions are. So this is a machine that if you really build something to behave this way, it wouldn't be much use, because once it's overflowed, it's dead, because it stays there. Real machine to be useful would have a reset transition, which took overflow back to zero. But this illustrates the basic idea. So let's look at a fun example from a Die Hard movie. I've forgotten which one it was. But there was one with Bruce Willis and Samuel Jackson playing a detective and a friend that he meets who helps him deal with a bad man, as is the case in all these movies. This time, the bad man's name is Simon. And what Simon says to them as they stand behind the fountain in the park shown on the previous slide is that on the foundation, there should be two jugs, do you see them? A five gallon and a three gallon. Fill one of these jugs with exactly four gallons of water and place it on the scale, and the timer will stop. The timer and the scale are not shown in that picture, but there's a scale and a timer nearby. You must be precise, one ounce more or less will result in detonation. If you're still alive in five minutes, we'll speak. OK. So let's think about formalizing this as a state machine to see what's going on. So but first of all, to understand the specification informally. What there is a three gallon jug and a five gallon jug that's capable of holding water, and an unlimited supply of water that you can get from the fountain. And the basic moves that you can make-- So with this set up, the kind of moves that you can make would be, say you fill up the three gallon jug with water, and then you could pour the three gallon jug into the five gallon jugs. And the three gallon jug was empty and the five gallon jug you knew had exactly three gallons in it. And then you can do other things like empty a jug and fill a jug and empty them into each other. So let's model this as a state machine. And the first decision we need to make is what's the state going to be. Well, the state-- the obvious model for the state is the amount of water in each of the jugs. So b is the amount in the big jug and l is the amount the little jug. And what we know about b and l is that they're going to be some amount between 0 and 5 for b, and 0 and 3 for l. We're going to quickly realize that we need them to be integers, but off hand we can allow them to be real numbers. Because after all, you could just pour some arbitrary amount of water into the big jug, any amount that it'll hold between 0 and 5. Although, that'll be dangerous, because as soon as you do that, you're going to lose track of exactly how much is in there and you won't know when you have four gallons or not. So let's formalize the possible moves that we can have. So first of all, the start state is 0,0, because we start with both jugs empty. And then what are the possible transitions of how b and l moves? Well, let's see. The fill the little jug move amounts to saying that if you have an amount of b in big and l in little, then you can make a transition called fill the little jug into b, and big is still unchanged, and 3 in little for l less than 3. I'm going to forbid the vacuous move where the little jug is already full and you try to fill it. That doesn't count, so l has to be less than 3, you can make it 3 by filling the little jug. Similarly, you could fill the big jug if b is less than 5. Then you can turn it into 5 by filling it. And then you can empty the little jug, which is easy. If you go from b, l you go to b, 0. And likewise, you can empty the big jug. Those are the easy moves. A slightly more complicated move is pour the big jug into the little jug. Well if there's no overflow, what that means is that there's l in the little jug and b in the big job. And after you've poured the big jug into the little jug, there's b plus l in the little one, and nothing in the big one. But let's be careful here about what exactly-- we're doing math, we're not sort of-- we're trying to get away from the metaphor. So what is no overflow means? It simply means that b plus l will fit. b plus l is less than or equal to 3. All right. What's the other case of pouring the big jug into the little jug? Well that's when b plus l won't fit, in which case, you pour into the little jug. It's got l, so you pour in 3 minus l to fill it up. And then what's left in b is b minus the 3 minus little l that you poured. So that the other wise case, when there is overflow. And similarly, there are moves for pouring the little jug into the big jug. So that then is a formal specification of the Die Hard machine and the moves that we're going to allow. Now, you could allow other moves like randomly pouring out a little water, or randomly filling up a little water. But if you did that, again, you lose track of how much water is in the jug. So these are the only safe moves. And they're the only ones we're going to model. All right. So let's go back to Simon's challenge. He wants to disarm the bomb by getting exactly four gallons of water in the jug and measure it on the scale, or things will blow up. And how do you do it? Well, why don't you take a moment to think about it before I proceed to the next set of slides or before you let me proceed. But just to understand the rules again, watch the work here's how. We're going to start off with both jugs empty. So we start off in state 0,0, and the first move is going to be to fill the big jug, which takes us to state 5, 0. Where the big jug has 5 and the little jug is still empty. Then we're going to pour from the big into the little. So now, the little jug has 3. We're filling up the little one. That leaves two in the big jug. Now we're going to empty the little one, we still have 2 left in the big one. And now we're going to pour from the big one into the little one, so the little one has 2 gallons and the big one is empty. Now, we fill the big jug, and there's still 2 gallons in the little one and 5 gallons of the big one. Now we pour off from the five gallon jug until the one gallon jug is full, that's removing the 1 gallon that the 3 gallon jug still has capacity for. We reduced to full 2 gallons in the little jug, and four gallons, our target in the big jug. So we've accomplished it. And we're done. So the bomb doesn't go off. All right. So the point of this exercise is really just to practice how the moves work and what the states are, but the questions I want to raise is suppose that we want to have a Die Hard once and for all scenario, in which we're tired of the remakes of these movies. And we proposed that in the next movie, that Simon, if he's still around, offers an alternative challenge, where instead of a three gallon and a five gallon jug, there's a three gallon jug and a nine gallon jug. And now the question is, can you get four gallons into the big jug by pouring back and forth with rules like these, or can't you? And can you prove it? Well my guess is that you probably can figure out what's going on, because what's happening is if you start off with nothing in either jug, and you do these moves of filling up a jug and pouring one jug into another, you'll discover that the amount of water in any jug is always divisible by 3. This is a preserved invariant. Any state that you can get to, starting off from 0,0, 3 will divide the number of gallons in each jug. We could state it this way. There's a property of states, property of b and l, which is the state, which is that 3 divides b-- that vertical line is a shorthand for the symbol divides. So three divides b, or b is a multiple of 3. 3 divides l. Synonym, l is a multiple of 3. And the claim is that that's always going to be true. So in case that's not obvious, you might not have all the rules in your head. Let's just take a look at one of the more complicated rules. This was the rule where you're pouring the big jug into the little jug up until the little jug is full. And that transition is that if you're in state b, l, you move to b minus 3 minus l, and 3. And if you look at this now, clearly if 3 divides both b and l, both components of the state you're at, then in the new state, well 3 obviously divides the contents of the little jug, which is 3. But three also divides the contents of the big jug, which is a multiple of 3, namely b minus 3, which is a multiple of 3 minus w, which is a multiple of 3. When you take a linear combination of multiples of 3, you get a multiple of 3. And you look at all the other transitions, and they check equally well. If you're in a state b, l, and you move to a new state b prime, l prime, if 3 divides b and l, then 3 divides b prime and l prime. So this is what's called a preserved invariant. And of course the corollary is that in the Die Hard once and for all game with the 3 gallon jug and the 9 gallon jug, you can't get to any state of the form 4, x, because 4 is not divisible by 3, and therefore Bruce is going to die. Now what we've illustrated here is an argument that's known as Floyd's Invariant Principle, and it's really nothing but induction reformulated for state machines. The statement of what is invariant principle is that we're going to define a preserved invariant as a property of states. And a preserved invariants means it has the property that if you're in a state that has property p, and it's possible to make a single transition to state r, then r will also have property p. This is just like the induction step. We have to prove that p n implies p of n plus 1. So if you have a preserved invariant, then if the property holds at the start state, then it's obvious that the property will hold for all of the states that you can possibly get to. That p of r will hold for all reachable states. And you can prove this by induction on the number of states, but I think it's clear that if you have a property that you begin with, and it doesn't change making a step, it's never going to change. And that's all that Floyd's invariant principle states. So in particular, since the property p holds in all reachable states, if there is any final state which the machine reaches, then p is going to hold in that state. And what we saw was-- we're using the word preserved invariant to distinguish the definition here from another usage that's co-opted the word invariant to mean a property that's true in all states. And while it's nice to know that some property is true in all states, the way you prove that is by having a preserved invariance. You want to distinguish the two. Technically if you look at this, the predicate that's always false is a preserved invariant. Because of the condition, as usual the way implication works. If the antecedent is false, then the entire implication is true. So if you're always false, then it's always the case that if false held in a state, which it never does, then it has to hold in any state you can get to, so that implication is true. So just remember that preserved invariants that are constantly false exist, they are good preserved in variance. But they're not what other people would call an invariant. We use preserved invariance to prove that a property is true in all states. It's a methodology. So let's do one more example to wrap this up. Suppose I have a robot on a grid, the integer grid, and we can think then of the coordinates of the integer as a pair of-- the coordinates of the robot as the coordinates of the square that it's in, a pair of non-negative integers. Now the way that this robot can move is we can make a diagonal move in one step. So it could move one step to the northeast or southeast or northwest or southwest and that's it. And the question I want to ask is, suppose you start the robot off at the origin, at 0,0. Is there some way that it can wander around, following its moves, and get to a next state where it's moved 1 square over? That is, it gets to the square 0,1. The answer to that is settled again by a preserved invariant. I don't know whether it's obvious to you yet, but it will be in a moment. I'm claiming you can't get to the square 0,1, and the reason is that there's are preserved invariant of that set of robot moves, namely the sum of the coordinates is even is an invariant. If the sum of the coordinates is even, it stays even. And why is that? Well, any move adds plus or minus 1 to the coordinates of both x and y. Maybe x and y both go up by 1, in which case, the sum of x and y increases by 2. So if it was even, it stays even, or they both go down by 1, or maybe one goes up and the other goes down, in which case, the sum doesn't change in every case. If x plus y was even, it stays even. As a matter of fact, if it was odd, it stays odd. Moves actually preserve the parity of x plus y. But the invariant is that x plus y is even. Now, what else is the case. Well 0,0-- 0 plus 0 is 0, which is even. And so we're in Floyd invariant principal case, where all the positions you can get to from the origin 0,0, which has an even sum, have to have an even sum. And since 1 plus 0 is odd, you can't get to that square, 1,0. It's not reachable. So the parity invariant of the diagonally moving robot will set us up for an analysis of the 15 puzzle game. That's this logo that we've had on every slide in 6042 so far with 6 blank, 4, 2, on the diagonal. This is a game where there are these little numbered tiles that slide into the blank square, and the question is how to permute-- how to get from one permutations of the numbers to another. It turns out that the analysis of that game depends on a parity invariant reminiscent of a slightly more sophisticated than the diagonally moving robot. Let's look at one more example of using the invariant to verify a little algorithm that actually will be quite important as the course progresses, and that is fast exponentiation. So in this set up, a is a real number and b is a non-negative integer. I want to compute the b power of a. Let's say b was 128, and I want to compute the 128th power of some real number a. Well, I can multiply a by itself 127 times, that would work fine. But you think about it, suppose I square a and then I square it again, and I square it again, then in about eight squarings, instead of 99 multiplications, I'm going to get to 8 of the 128th. Now if b is not a power of two, you adjust it slightly, and using that idea, you can compute the bth power of a much more rapidly than if you just naively multiplied out b minus 1 times. So let's look at the pseudocode for this algorithm. Here, XYZ in our temporary registers y and z, hold-- y, z, and r all hold integers. And x holds this real number a. And you can read the code if you wish but in fact, I'm going to immediately jump to the slightly more abstract and easier to understand version of it as a state machine. So what matters about this fast exponentiation algorithm as a state machine is that first of all, there are three states to real numbers, and a non-negative integer. And the start state is going to be that the number a is in the first register, or in the first location, first coordinate of the states. 1 is the real number in the second coordinate, and b the target exponent, is the non-negative integer in the third component. The transitions are going to be as follows. Here's a simple one. If I have an amount x in the first location, y in the second location, z in the third, then if z is positive and even, then I'm going to square x, leave y alone, and divide z by 2. And that's the new state that I get. On the other hand, if z is odd and positive, then I'm going to square x, multiply y by x, and again take the quotient of z, divide z by 2. OK. Why does this state machine do fast exponentiation, why is it correct? And the insight is that there's a preserved invariant of this machine. And the preserved invariant is that y times x to the z is always a to the b. SO let's see how to verify that, that yx to the z is equal to a to the b. Let's just look at maybe the slightly more interesting of the two transition rules, which is when z is positive and odd, the xyz state moves to a new state, indicated in green. Where the new value of x is x squared, the new value of y is xy, and the new value of z is z minus 1 over 2. Remember, you went to the quotient of z divided by 2, and when z is odd, that means z minus-- it's literally z minus 1 over 2. Well, do the new values satisfy the invariant if I plug-in the green values of x squared for y and xy for x-- I'm sorry, x squared for x, xy for y, and z minus 1 over 2 for z? Well let's see what happens. If I take the new value of y, which is xy, and I multiply it by the new value of x, which is x squared, raised to the new value of z, which is z minus 1 over 2. Let's do a little algebraic simplification of that. Well, the x squared to the z minus 1 over 2 becomes x to the z minus 1. And I'm just carrying over the xy. And then that simplifies to simply y times x times x to the z minus 1, or yz to the z, which we assumed was equal to a to the b, so sure enough, the new values of x, y, and z satisfy the invariant. It's a preserved invariant, and an even simpler argument applies to the other transition, when z is positive and even. So we verified that this is a preserved invariant. Now at the start, remember, we start off with the real number a in register x, the real number b in z, and the real number 1 in y, which is the accumulator. And 1 times a to b is equal to a to b. So this is a-- this preserved invariant is true of the start state. That means by Floyd's Invariant Principle, that it is true at this the final state, if and when the thing stops. Well, when does this machine stop? As long as z is positive, it can keep moving. So it gets stuck when z is 0? What happens if it ever gets to z is 0? If it gets stuck, then the invariant says that yx to the 0 has to equal a to the b. But of course, yx to the 0 is nothing but y. And what we conclude is, that sure enough, this machine leaves the desired exponential value in the register y, which is where we get the answer. And that's why this algorithm is correct. Now another aspect of what's going on here is proving that the algorithm does terminate. So let me just say a word that Floyd distinguished sort of these two aspects of program correctness that typically come up. One is showing that if you get an answer, it's correct, and that's what we just did. If this machine stops, if it ever gets to the case where z is 0, then y has the right answer. But we haven't proved that it stops. So we've shown that it's partially correct like a partial function. It might not be defined everywhere, we haven't shown that yet, but when it is defined, if gives the right answer. The other aspect of correctness is termination, showing in effect, that the function is total, that the program always does stop. Well in this case, there's an easy way to see why it always stops. Because at every transition, z is being halved or more. z is a non-negative integer valued variable. And since we're halving it, or making it even smaller than half of it at every step, it means that since it starts with the value z, it can't get smaller, more than log to the base 2 of b times, because by then, it would have hit 0. And so we can be sure that this machine makes it most log to the base 2 of b transitions. And then it has to get stuck at the only place it can get stuck, which is when z equals 0. And there is a picture of my friend, an early colleague, Bob Floyd, whom I met at the very beginning of my career at Carnegie Mellon University. We worked together for about one year before he went off to Stanford. And you can read much more about his life in a warm and detailed eulogy written by his best friend, Don Knuth. Floyd won the Turing Award for his major contributions, both to program correctness and to programming language parsing. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 1101_Recursive_Data_Video.txt | PROFESSOR: Recursive data play a key role in programming, so let's take a mathematical look at what goes on. So the basic idea of recursive data is roughly that you're going to define a class of objects in terms of the simpler versions of the same object. With a little more precision, the idea is that you [? can ?] build up a recursive data type by starting with some stuff that you understand that's not recursive and you give me some base objects that I can begin with and declare that they belong to this recursive datum, and then you give me some rules called constructor rules which enable me to build new objects of the recursive type by applying these constructors to objects that I've already built up. There's nothing circular about it because I'm always building up new stuff from stuff I already have. Let's look at an example. I'm going to define a set E that's a subset of the integers, and I'm going to give you a recursive definition of E. The base case is that I'm going to tell you that 0 is an E and I'm going to give you two constructors. The first one says that if you have an n that's an E, you can add 2 to it and get a new element in E, providing that n is not negative. The second constructor is that if you have an n that's an E, you can negate it. You can take minus n, providing that n is positive, and those are the two constructor rules. Well, let's look at what goes on here. What is this telling us? Well, let's just use the first constructor rule and use it repeatedly. I can start off with 0. That's the base case, and then I can apply the constructor to add 2 to it. Then I can apply the constructor again to add 2 to 0 plus 2, and then I can apply the third time to get add 2 to 0 plus 2 plus 2. And it's clear what I'm getting is 0, 2, 4, 6, and so on, and I'm going to get all of the non-negative even numbers in that way. Now, I can apply to these, the positive numbers, I can apply the negation constructor. So I can get minus 2, minus 4, minus 6, and it becomes apparent then that I can get all of the even numbers. So we just figured out that E contains the even numbers. Is there anything else in E? And the answer is no, and the reason is that an implicit part of the understanding of a definition like this is that the only way that things can get into E is by being a base case or by being constructed from previously constructed elements by applying the constructor rules. In other words, there's an implicit clause here that says that's all. That implicit clause is called the extremal clause. And it's taken for granted and rarely mentioned explicitly as part of a recursive definition, but it's always to be understood. So what we can conclude from this is that E is exactly the even integers because there's nothing else there except those ones that were built up in the way indicated. So let's look at a slightly more interesting example now. I want to define the set of strings that consists only of left and right parentheses such that the left and right parentheses match up. Well, writing parentheses on the slide turns out to be confusing with parentheses that are actually used to delimit things, so I'm going to replace parentheses by brackets in blue-- a right bracket and a left bracket. This notation here, by the way, stands for the set of finite strings of right and left brackets. It's a general notation. If you have some collection of objects which you think of as letters and you write an asterisk as a superscript, that means the finite strings of those letters. So these are the finite strings of right and left brackets, and I want to give a recursive definition of a set M which I plan will be precisely those strings where the left and right brackets match up appropriately. The way to think about matching up is take let's say a standard arithmetic expression involving plus and times and so on, and make sure that it's fully parenthesized. So whenever you add two things, there's parentheses around that, and whenever you multiply two things, there's parentheses around that, meaning brackets. Then if you erased everything but the brackets, what you'd be left with would be a set of matched brackets. Actually, it would be a set of matched brackets or you could have several of them next to each other. Those would still could be considered matched, so that's the way our definition is going to behave. Let's give it. So the base case is about the simplest it could be. I'm going to start with the empty string. An empty string is this thing that acts like a zero under putting strings next to each other, or the concatenation operation. If you concatenate the empty string with any string, it doesn't change the string. And by definition, then, the empty string is a string with no characters, has length zero, and it acts as an identity under putting strings next to each other. There's going to be one constructor in M that's slightly ingenious. There's other maybe simpler or more natural ways to make up constructors that would define M, but this one is particularly nice because I can get away with just one, and it has some nice properties that we'll explore later. So here's the rule-- if I've built up two strings s and t of matched brackets that are in M, then I can build a new one by putting matched brackets around s and concatenating it with t-- that is, if s and t are strings of brackets in M, then if I start with a left bracket followed by the brackets in s followed by a right bracket followed by the brackets in t, that new string is yet another element that I've built up in M. Let's practice this to see how it works. So there's the constructor again. Well, how do I get started? To start, all I have is the base case. s and t have to both be the empty string because that's the only thing available to apply the constructor to. So if I do that, basically the s and the t disappear in this constructor expression and all I'm left with is a matching left and right bracket. But now I've got a matching left and right bracket, so I can use that to apply the constructor to, so I could let s be the matching brackets and t still be the empty string. Now when I plug into the constructor, the t still disappears, but I find brackets within brackets, and that's another string that I've built up in M. Now, being methodical, I could let s be empty and t be the brackets. And if I do that, then the s goes away and the t becomes the matched pair of brackets and I wind up with a matched pair next to a matched pair. Then, of course, I could let both of them be the matched brackets, and then I get a nested pair next to a matched pair. And now that I've got also going back to the very beginning the next most complicated string that I had was the nested pair of brackets, I could let s be that and t be empty, and then I would get brackets nested to depth three, and so on. That's the idea. Now, it may or may not be clear that you get exactly the strings of matched brackets in this way. That's taken up further in the notes and in some problems, but we're just trying to understand how this definition works and take it for granted that, in fact, it's right. Let's use that definition to prove some things about M, but I want to prove the things based on the definition of M not assuming that it works as intended. So I'm going to claim based on the definition that it's impossible to find a string in M that starts with a right bracket. Now, of course, since we're assuming M is the right definition of matched brackets, it's clear that a string that starts with a right bracket already has nothing to match-- no left bracket matching it-- so it shouldn't be in there, but let's just make sure that the definition behaves in the way that we intend or it might be wrong. So how do I prove that no string in M starts with a right bracket? Well, let's look at the definition. The base case doesn't have any brackets at all, so it certainly doesn't start with a right bracket. And looking at the constructor rule, all the strings that you can construct start with a left bracket. And so we're really appealing to the implicit that's all clause, the extremal clause that says that since the only way to get things in M is by applying the constructor, you're not going to be able to get anything that starts with a right bracket. One more example of a recursively-defined data type that's interesting, and we'll be doing some lovely class problems with, is the class that I call F18 functions. These are the functions from a first term of calculus, like as you study in 18.01, functions of a single real variable, and here's a recursive definition that I think covers all of the functions that are considered in 18.01. I'm going to start off with the identity function and any constant function and the function sine of x, and declare that those are the base cases. Those are the functions in F18. Then here are the constructor rules. If I have two functions that are in F18, I can add and multiply them or take 2 to the f where f is in there, and those will all also be functions in F18. So I can start building up a bunch of interesting stuff like polynomials and exponentials. In addition, if I have a function that's in F18, I can take it's inverse-- at least, insofar as the inverse is defined in the function-- and I can also compose two functions that are in F18 to get another one. Let's look at how this definition works. I claim that, in fact, the function minus x is in F18. How do I build up minus x from the rules? Well, minus 1 is a constant function, so I have that. And x is just the identity function, and I can multiply two functions that I have. So if I multiply minus 1 times x, guess what? I got minus x, so I've just figured out that that function is in F18. What about the square root of x? Well, if I multiply the identity by itself, I get the function x squared. And then if I take its inverse, that's square root of x. Well, I gave you sine x, but not cosine x or any other trig functions. Why not? Well, I want them all, but I can get them by the rules already. So how do you get cosine x? Well, cosine x is just sine of x plus pi. Well, why is that in there? Pi is a constant, x is the identity. So the sum is a function that's in F18. And then if I compose that function with sine, I get sine of x plus pi, which is cosine x. So cosine x is there. Now, this was actually pointed out to me by students, this simple way of getting cosine x. The original way that I thought, and I was using that square root operation where I was going to use the identity that cosine squared plus sine squared is equal to 1. So if I take 1 minus sine squared and then take the square root, that's another way to get cosine x, the point being that there's a lot of ways to derive the same function as being in F18 built up from the operations applied to other functions. What about log of x? Let's just close with that. How do I get log of x? Well, log of x is the inverse of e of the x. How do I get e to the x? Well, e to the x is what you get by taking 2 to the log to the base 2 of e, which is e, and then raising that to the power x. So if I take log e, which is a constant log to the base 2 of e, which is a constant, I multiply it by x the identity function and I take 2 to that power, I'm composing, in other words, x log x with the constructor 2 to the F, then I wind up with the function e to the x. And when I take its inverse, I get log of x, as claimed. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 331_Sum_And_Product_Rules_Video.txt | PROFESSOR: The topic of counting, or combinatorics, is an important one in a number of different disciplines, and notably in computer science. So the origins of the combinatorics and counting are a little bit disreputable. They come, historically, out of people studying gambling and trying to calculate odds and the fraction of times that various events occur to know what kind of bets to make on them. So a typical kind of question would be-- if you know how poker works, there are various classifications of five-card hands in poker. And you might ask, what fraction of all possible five-card poker hands translate into being a pair of jacks? And basically, this fraction of total poker hands which fit into the category "a pair of jacks" is the probability of a pair of jacks. So counting in gambling is one fundamental place where it really comes up. And historically, that's where a lot of combinatorics begins. Related to that is counting in games. When you're trying to write, for example, a computer program to play chess or checkers or so on, one of the aspects of it is getting a sense of how much searching you have to do in order to look ahead to find good moves. And you wind up counting, from a given chess position, how many possible further positions can you get to a given number of moves. A puzzle kind of question, in solving the Rubik's Cube toy, is how many different positions are there, and how many different positions can you get to from a given position? Concretely, in computer science, it comes up in algorithms. It's often the case that an essential question is, how many operations does it take to do a manipulation on a data structure and to update it from one to another? For example, how many comparisons does it take to sort n numerical items? And typically, the count is n log n, proved as a number of operations that are both achievable and a lower bound. One that we've seen when we looked at fast exponentiation is a question like, if you're trying to compute the nth power of a number d, how many multiplications does it take? It's roughly log n by using the iterated squaring method. And we want to be able to count that number of multiplications that a particular program uses to compute d to the n in the smallest number of multiplications you can get away with. And a place where, again, counting and combinatorics becomes critical is for security and the issue of cryptography. If you're going to have security from passwords, there needs to be too large a space of passwords for an adversary to search through exhaustively and check them likewise. If you're doing encryption with some kind of secret key that enables you to read messages, you want to be sure that the space of possible keys is also way too large to search exhaustively to see what keys work. So let's talk briefly now about the very basic counting methods, and two rules for counting things-- the most rudimentary of them, but in fact we'll get some mileage out of them. So the first rule is called the Sum Rule, and it's completely straightforward and obvious, which is that if I have two sets, A and B, that do not overlap, then the number of elements in A union B is simply the number of elements in A plus the number of elements in B. And there's no issue proving that-- it's self-evident-- but let's do an example. Suppose a class has 43 women and 54 men. How many people are in it? 43 plus 54 equals 97. This is implicitly assuming that there's no one whose sex is ambiguous and that there's no third sex, so that men and women are disjoint. The total number of students is the sum of the number of men and women. Another one is there are 26 lowercase Roman letters and 26 uppercase Roman letters and 10 digits. And so there are 26 plus 26 plus 10 equals 62 characters in that repertoire of symbols. The second rule is called the Product Rule, and just about as obvious. Suppose I have four boys and three girls. How many boy-girl couples can I assemble out of four boys and three girls? And the answer is, there are four ways to choose a boy, and for each of them, there are three ways to choose a girl. So there's 4 times 3, or 12, possible boy-girl girl couples in this setting. More generally, if I have a set A of size m and a set B of size n, then A cross B-- remember, that's the set of ordered pairs where the first element is from A and the second element is from B-- the size of A cross B is-- the vertical bars, remember, mean size-- is equal to m times n. So let's just do an example that illustrates it. Suppose that A is the set of four elements-- little a, b, c, and d-- and B is the set of three numbers, 1, 2, and 3. Then I can list A cross B in a nice orderly way, as a 4-by-3 matrix. But this is really meant to be just a list of elements, but I'm organizing this way so the pattern is more apparent. And for each element little a, I can pair it with each of the three elements in B. And for the second element, little b, in A, I can pair it with this three digits [? in A. ?] And c I can pair with three, and d I can pair with three. And that's where the 4 times 3 comes from, and more generally, the m times n comes from. A useful immediate application of this is, how many binary strings are there? How many strings of zeros and ones are there of length 4? Well, the length for binary strings, it can be explained as well as the product of B times B times B times B. We're not writing parentheses here. It's B times B-- cross B. So I'm thinking of a quadruple like this as being a pair whose first element is a triple. And a triple is a pair whose first element is a pair. And given that it doesn't really matter how you break it up, we just typically write it as B cross B cross B, and even abbreviate that as B to the fourth, where b is 0, 1. And the Product Rule says that the size of this is the size of B times the size of B times the size of B times the size of B, or 2 to the fourth. So in general, if I look at strings of length n, whose elements are from an alphabet of size m, the total number of such strings is m to the n. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 121_Proof_by_Contradiction.txt | PROFESSOR: It's doubtful if you really understand something if you can explain why it's true. That's what proofs are about in mathematics and in computer science. So we're going to be talking about proofs of lots of things that we're trying to understand. And in particular, we're going to look at a proof technique now called proof by contradiction, which is probably so familiar that you never noticed you were using it. And now we're going to call explicit attention to it, and think about it. So let's do an example first to see what's going on. Suppose that I wanted to prove that the cube root of 1,332 was less than or equal to 11. Or more precisely, suppose I didn't know and I'm asking this question, is the cube root of 1,332 less than or equal to 11? Well, one way to do it would be to simply compute the cube root of 1,332, which is a small bother, but manageable. But there's a simpler way than figuring out how to compute a cube root of a four-digit number. Let's just suppose that this inequality was true-- that is, that the cube root of 1,332 was less than or equal to 11. Well, if that was true, then what I could do is cube both sides. And I'll conclude that 1,332 is less than or equal to 11 cubed. Now, 11 cubed is a lot easier to compute than the cube root of 1,332. As a matter of fact, 11 cubed is 1,331. Wait a minute, I've just concluded that 1,332 is less than 1,331. That's obviously not true, which means that my assumption that this inequality held doesn't make sense. It leads to this immediate contradiction, which means that in fact, the inequality doesn't hold. And I have now precisely and unambiguously-- I hope clearly-- proved that the cube root of 1,332 is greater than 11, even though we never actually computed the cube root of 1,332. This is kind of a [? toy ?] and simple-minded example to illustrate proof by contradiction. So let's step back and explain, and say what it is in general. If an assertion implies something false, then the assertion itself must be false. That's what's going on here. If you're reasoning step by step, and at every step your reasoning is good-- which means that if you had something true and then you reached a conclusion from it in one step, the conclusion that you reached was also true-- then if you start off with some assumption, you keep proving things step by step in a way that preserves truth, and you arrive at something false, it's inevitable that what you started with must have been false. Or else the thing you finished with would have been true. OK, let's look at a real example of this-- an amazing fact that was known thousands of years ago to the ancient Greeks, which is that the square root of 2 is irrational. Now, let's remember that a rational number is a fraction. A rational number is a quotient of integers. And the way we're going to prove that the square root of 2 is not a quotient of integers is by assuming that it was. So let's assume that the square root of 2 was a rational number, which means that we've got integers n and d without common prime factors, such that the square root of 2 is equal to n over d. What I'm doing here is I'm saying squared of 2 as a fraction, n over d. And we know that you can always reduce a fraction to lowest terms, which means there are no common prime factors. So let's get that done. We have the square root of 2 is equal to n over d, with no prime that divides both n and d. From this assumption, I'm going to prove to you that both n and d are even. And that, of course, is an immediate contradiction, because then both n and d have the common factor 2. So all I've got to do in order to conclude that the square root of 2 is an irrational number-- it's not a fraction-- is prove to you that n and d are both even if the square root of 2 is equal to n over d. Let's do that. We'll start off with what I'm assuming-- square root of 2 is n over d. And let's get rid of the denominator. So let's multiply through both sides by d, and get that the square root of 2 times d is equal to n. Let's get rid of the square root of 2 now by squaring both sides. And I get 2d squared is n squared. Well, that's great, because look-- the left-hand side is divisible by 2. There it is. Which means that n squared is divisible by 2. The right-hand side is even. But if n squared is even, then n is even, and I'm halfway there. I've shown that the numerator is even. OK, let's keep going. Now, since n is even, it's equal to twice something. So n is 2k for some number k. I don't care what k is. Let's square both sides of that, and conclude that n squared is equal to 4k squared. Why did I square it? So that I could connect up here with the other question that I had about it n squared-- that n squared it was 2d squared. So combining these two, what I get is that 2d squared is equal to 4k squared. And of course, I can cancel 2, and get that d squared is equal to 2k squared. And again, I've got the right-hand side divisible by 2. So the left-hand side is divisible by 2. d squared is even, and therefore, d is even. And we've completed the proof as claimed. n and d both have 2 as a common factor, contradicting the fact that their in lowest terms. Now, I did assume something that is kind of obvious-- namely, that if n squared is even, then n is even. Why is this true? Well, you might think about it for a moment. There's a simple way to see it, and it's a proof by contradiction. We're going to use the fact that you can verify easily enough by doing a little arithmetic-- namely, the product of two odd numbers is odd. Let's assume that. So if the product of two numbers is odd, if I tell you that n squared is even, and suppose that n was not even, well, that means it's odd. But that would mean that n squared was odd, contradicting the fact that n is even. Therefore, it's a contradiction to assume that n is odd. It must be even That's an ad hoc proof that has to do with evenness and oddness. There's a more general way to understand this that actually will come in handy-- namely, that what I know is that numbers factor into primes in a unique way. So if I tell you that n squared is even, what I know about n squared is that all the primes that divide n squared come from n. So if there's a 2 among the primes that divide n squared, it has to be a 2 that is one of the prime divisors of n. And that would work even if I told you that n squared was divisible by 3. It would follow by that reasoning that n is divisible by 3. Now, that's a powerful fact. I'm assuming the prime factorization of integers. And it's not obvious at all that that's true, although it's very familiar. It's OK to assume. In a few weeks we'll actually look back at how to carefully prove that. But for now, it's OK to assume. And we also have the simple argument that worked based on properties of even and odd-- that if n squared is even, then n is even. That's the last gap in the proof, and so we're done. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 1111_Cardinality_Video.txt | PROFESSOR: Cardinality is the word that's used to refer to the size of infinite sets. And before we go further, let's take a quick look at why we're interested in infinite sets in a course that's Mathematics for Computer Scientists. Why does computer science care about infinite sets? Well, like every data structure that you'd examine in computer memory is finite and the integers individually are finite, you only calculate with finite things. But, the infinite abstraction happens right at the beginning. Although any given integer is finite, the set of all integers is infinite. And although any given matrix is finite, the set of all the matrices that might be represented in a computation are an infinite set. So, we take infinite sets for granted and reason about them all the time. The second, from a pedagogical point of view, introducing the concept of infinite sets and reasoning about them carefully forces you to go beyond your intuition and really follow the rules and reason in a careful mathematical way. Because although some properties that you're familiar with from finite sets carry over to infinite sets, others don't. And in order to know which is which, you have to be thinking carefully about the rules and properties that they have, as opposed to just going by intuition and familiar properties. And finally, the reasoning that goes into comparing the sizes of infinite sets, which is the topic of today's video, has profound implications in computer science because it leads to the insight about the logical limits of computation and examples of specific problems that computers can't solve, which we'll be taking up in a later video. But for now, let's go back to the topic of cardinality. So, there was this mathematician in the late 19th century named Cantor who was actually working on Fourier's series. And he discovered that the kind of series that he was working with diverged in infinitely many places, which sounds kind of bad. But he wanted to get across the idea that it didn't diverge in very many infinite places. And that led him to this idea of comparing the sizes of infinite sets. So, this is Cantor's idea. We know from the mapping dilemma that if you're looking at finite sets A and B, then the size of A is greater than or equal to the size of B if and only if A surj B, were surj is this technical relation which means there exists a surjection function from A to B that is a function with greater than or equal to one arrow into every element of B. And Cantor's idea was saying, well, it works fine for finite sets. Why don't we take this as the definition of what we mean by A is at least the size of B for infinite sets? So, we're going to think of A surj B now as saying, A is as big as B. And for finite sets, it's literally true that A surj B if and only if the size of A is greater or equal to the size of B. Now, let me take a moment to say that this notion of size or cardinality, when you're talking about infinite sets, it's kind of a no-no. There's an abstract concept of what cardinal numbers are, what these infinite numbers are. But the truth is, they're technical and not of very much use. So, we will never actually be talking about the cardinality or size of an infinite set. But what we will do is compare them. We're going to have a nice elementary theory of the idea that the cardinality of one set is greater than or equal to the cardinality of another set. And the basic definition is going to be based on surj. Similarly, bijection is even easier. A bij B means that there's a bijection from A to B. And we're going to interpret that as saying that A and B are the same size. That is, for finite sets, it literally means A and B have the same number of elements. We're going to adopt the notion of a bijective relation for infinite sets as meaning, I don't know what their size is, but I know it's the same because there's a bijection between them. There's a perfect one to one correspondence between As and Bs. Let's look at an example of where bijection comes up. The power set of N-- if N is the non-negative integers-- the power set of N is all the subsets of non-negative integers. And let me just remark that there is an obvious bijection between the subsets of integers and the infinite bit-strings, the infinite strings of zeroes and ones. So, N is the set of non-negative integers, 0, 1, 2. If you take any subset of N, here's one with has 0, missing 1, has 2 and 3, missing 4, 5, has 6, and so on, then what I can do is represent such a subset, possibly an infinite subset now, by an infinite sequence of ones and zeros. Put in ones in the position where elements in the subset occur and zeroes in positions where elements don't occur. This was exactly the same bijection that we had found between the non-negative-- the bit-strings and the finite subsets of the non-negative integers. But now, we're just extending it to arbitrary subsets of the non-negative integers. So, this defines a bijection between any subset of integers corresponds to an infinite bit-strings. And conversely, from any infinite bit-string, you can reconstruct what subset it refers to. So, we use this notation [? 0,1 ?] to the omega, meaning the infinite bit-strings that are infinite to the right. They have a beginning. In comparison to [? 0,1 ?] superscript star, which refers to the finite sets of bit-strings. So now, let's examine the standard size properties that you would expect if these relationships of surj and bij behaved like relationships between sizes. So, one basic property that finite sizes have is that if A is equal to B and B is equal to C in size, then the size of A and the size of C are the same. That's certainly true for finite sets. Does it hold for infinite sets, where now equality is going to be replaced by bij? Well, we have to check it. Is it true that if A bij B and B bij C implies A bij C, well, how do you prove that? Well, it's true, and here's how. By definition, since A bij B, that means that you have a bijection g from A to B. And since B bij C, you have a bijection f from B to C. Now, I need from these two bijections that I'm given, I need to find a bijection between A and C. Well, that's easy. What you do is you use g to go from A to B, and then you use f to go from B to C. And you compose them, and that gives you the needed bijection from A to C. So, define h to be the composition of f and g. And it's easy to check that if g and f are bijections, then their composition is a bijection. So, that's how I find the needed bijection from A to C. So, this property works out just fine. The similar property applies to at least as big as, greater than or equal to. For finite sets, if A is greater than or equal to B and B is greater than or equal to C in size, then A is greater than or equal to C. And actually, the same argument that worked for bij works for surj, because the composition of surjective functions is a surjective function. So, if A surj B and B surj C implies A surj C. Now again remember, although we're thinking of surj as meaning greater than or equal to in size, you cannot take these size properties for granted. They have to be proved. Surj has a technical definition having to do with surjective functions, functions with greater than or equal to one arrow in. That is not the same as talking about equality of some kind of sizes. Well, let's look at an example where the size properties hold but they're less obvious. Because here's another familiar size property. If A and B are each of size greater than or equal to the other one, then they're the same size. So, if the size of A is greater than or equal to the size of B and the size of B is greater than or equal to the size of A, then A and B are the same size. Now, this is certainly true for finite sets. It's kind of, you don't even think about that fact. And it holds for infinite sets. But, it's not so obvious. So, what we're saying is that if I have a surjective function from A to B and I have another surjective function from B to A, then there's a bijection between A and B. And the problem here is that this surjection from A to B might not be a bijection. And this surjection from B to A might also not be a bijection. So, where's the bijection going to come from? I have to build it. And so, this is not an obvious property, it's true. It's called the Schroeder-Bernstein Theorem. And the trick, basically, is you take the bijection from A to B and the bijection from B to A and you take parts of one and combine it with parts of the other. And in a slightly ingenious way that actually is contained in a problem in the text, you can find the bijection from A to B, but it does take a little bit of ingenuity. So this is a size property that works for surj and bij, but you can't say it's obvious. Well, let's look at an unfamiliar size property. Something that's not true of finite sets where we have to start being careful and not just hand wave and use our intuition about finite sets. Namely, for infinite sizes, size plus 1 is equal to size. Now, what exactly does that mean? Well, let's just illustrate it with an example. In fact, in some ways, you can the definition of an infinite set is that its size plus 1 is equal to its size. Let's look at a simple example. So, on the bottom, I have the non-negative integers. And on the top, I have the positive integers. So, I can get from the positive integers to the non-negative integers just by throwing in zero. So, that's where the plus one comes from. Here's a nice infinite set. I add another element to it, and I get another infinite set, but they are the same size. I have to show a bijection between them to show they're the set size. Well, you know what the bijection is. Map 0 to 1, 1 to 2, 2 to 3. This is a bijection which you know as the add one function. The add one function maps the non-negative integers to the positive integers, and it's a perfect bijection. Therefore, adding one element to the non-negative integers-- to the positive integers does not get me a larger set, it gets me another set of the same size. And this argument actually generalizes to any infinite set. If you throw in one extra element, you could still find a bijection between the original set and the set with one extra element. So, N is the same size as the positive integers. Well, in fact, let's look at this one. I can enumerate on the top all the integers, both positive and negative. 0, 1, minus 1, 2, minus 2, and so on. And that gives me the set consisting of all the integers. And over here, I can have 0, 1, 2, just the non-negative integers. And you can see the orderly way in which I've listed the integers at the top. That implicitly defines a bijection. I'm going to map zero to the zeroeth element of the sequence above. 1 to 1, 2 to minus 1, 3 to 2, 4 to minus 2. And in this way, I have actually defined a bijection between the non-negative integers and all the integers. In other words, you take half of the integers, namely the non-negative integers, and it's still the same size as all of them. There's a bijection between N and Z. Now, you could write a formula, actually, if you were trying to figure out what does the number N go to? What positive or negative integer? There's some not very hard formula involving dividing N by two and rounding. But, that doesn't matter. Once I've figured out some sensible way to list all the elements of the integers in a row, then I can wind them up against the non-negative integers. And that listing, in effect, defines the mapping in a perfectly clear way without necessarily having a formula. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 415_Sample_Spaces_Video.txt | PROFESSOR: So let's look now at the mathematical foundations of probability theory, the basic definitions of which we've just been doing examples up until now. So a key concept is a probability space. And that's what we're going to talk about in this segment. So the abstract setting of a probability space is the first thing you start off with is the set of outcomes, which is what we saw we were doing with the tree models in the previous videos. And the condition that we require is that there should be a countable set of outcomes. So there's something called the sample space. And the sample space is designed to model all the possible things that can happen as the result of your random experiment, all the possible outcomes. And we require that there be a countable number. Now, the examples that we've seen so far have only had a finite number. But we will shortly see a bunch of examples where we really need an infinite number. But only a countable infinite number. That's part of the definition of a probability space-- the set of outcomes. The next thing is a probability function whose task is to assign probabilities to the outcomes. So the condition is that the probability function, Pr, gives every element in S, every outcome, is going to be assigned a probability of between 0 and 1 inclusive. So every outcome gets a probability between 0 and 1. But the constraint on the probability function is that if I sum up the probabilities of all the outcomes-- omega is an outcome in the sample space S-- and I take the sum of all of those probabilities of omega, they have to sum to 1. That's the crucial condition that defines a probability function on a sample space. And the two together are what are called a probability space. A sample space with a probability function is a probability space. So the purpose of the tree model that we were using is really to figure out which probability space to use. And the mathematics doesn't really start until you have the probability space. Up until that, it's the modeling part that's very important mathematically. But you can't say that the model is right or wrong. It's a model, and its rightness or wrongness is a matter of judgment and comparison to how it stacks up against reality and things that we care about. When we're using the tree model, it's the leaves of the tree that correspond to the outcomes. And the outcome probabilities, which are crucial for having a probability space, we got by reasoning about the probabilities to assign to each possible branch of the tree as you worked your way from root to leaf. So the other key concept that we saw already is the idea of an event. An event, formally, is nothing but a subset of the sample space. An event is some set of outcomes. Presumably, the event is an event that you're interested in, like winning. And the definition of the probability of an event is simply the sum of the probabilities of all the outcomes in the event. And we were using this already for both Monty Hall and for the poker hands. But this is the official general definition-- that once we have a probability function that assigns probabilities to outcomes, then we can use that to define the probability of an event. This is the definition of the probability of an event-- simply the sum of the outcome probabilities. And as an immediate corollary of this definition, what we get is something that's central to probability theory. It's called the sum rule. And it says that if you have a bunch of events that are pairwise disjoint-- so there's no outcome in common to A0 or A1 or A1 or A2 and so on-- then the probability of the union of the A's, the probability that one of these events occurs, one or more of these events occurs, is simply the sum of the individual probabilities. And that is a rule that we'll be using all the time. It's very convenient for computing things. If you just break them up into separate cases, then you can handle the separate cases-- each A0, A1-- separately, and then add up the probabilities. And in some approaches to probability, more general ones, this is actually taken as an axiom. It's the axiom that defines a probability space, but where you start with an assignment of probabilities to events. But in the discrete case, we don't have to worry about that. It's a corollary of the way we define the probability. And that, of course, is a crucial rule-- sometimes called the countable sum rule. But we're just going to call it the sum rule. Expressed in concise notation, it's the probability of the union of the Ai's, as i ranges over the non-negative integers, is simply the sum of the individual probabilities of those events. Now, why it's called discrete probability that we're studying is because we have a countable sample space. And as we saw, that discrete combinatorics is the combinatorics of countable and even finite sets, really. The crucial reason why we're sticking to discrete probability is that allows us to work with sums instead of integrals. If you start allowing continuous intervals of time and the probability, say, of throwing a dart and it landing at a given interval on the line and a whole bunch of other situations where it's natural to want to use continuous probabilities, you're forced into defining a probability in terms of integrals, because every outcome has probability 0. And the theoretical basis of it is considerably more complicated. And we don't need it for, in fact, virtually any purposes that come up in computer science. And so we will, happily, not have to study integral calculus or measure theory, really, and just get by with sums. So let's quickly point out some rules that are now corollaries. They're really derived rules of probability theory that follow as a consequence of the countable sum rule. And the first one is the difference rule. The probability of A minus B is simply equal to the probability of A minus the probability of A intersection B. Now, notice how much this looks like the difference rule for cardinalities-- that the cardinality of the finite set A minus B is simply the cardinality of A minus the cardinality of A intersection B. And indeed, the proof of this is just like the proof of cardinality. It follows directly from the sum rule for probabilities, which corresponds, of course, to the sum rule for cardinalities. Namely, by the sum rule for probabilities, what we know is that A is equal set, theoretically, to A intersection B plus A minus b. That is, A breaks up into the points that it has in common with B and the points that it doesn't have in common with B. Since those are disjoint, you can add them. So the probability of A is equal to the probability of A intersection B plus probability of A minus B. And so I just transpose the A minus B to the left hand side. And I get the difference rule, which is a rule that's worth remembering. Similarly, we have inclusion-exclusion. If A and B are not disjoint, then the probability of A union B is equal to the probability of A plus the probability of B minus the probability of the intersection. And the proof is, in fact, exactly like the corresponding rule for cardinalities of finite sets. And of course, it generalizes to more sets. This is an example of the inclusion-exclusion for three sets in terms of probability. Another useful, it turns out, consequence is that the probability that A or B happens is guaranteed to be less than or equal to the probability that A happens plus the probability of B happens. And this follows as a trivial consequence of the inclusion-exclusion rule for two sets, because the probability of A union B is equal to this plus this minus some probability, namely, the probability of the intersection. So you're taking away something non-negative from these two in order to equal that. In particular, then, this must be less than or equal to that. And the closely related phenomenon is [? basi-- ?] [AUDIO OUT] The probability that A or B happens is greater than or equal to the probability that A happens. And finally, we can generalize that to a countable collection of sets. If I have a bunch of events-- A0, A1, and so on-- then the probability that at least one of them occurs, the probability of the union of the Ai's, is less than or equal to the sum of their probabilities. This is, again, another kind of obvious rule, not hard to prove. We're not going to bother proving it, because it really is obvious. But we will get some mileage out of it later on. So to summarize, the key concept here is a probability space. It consists of a countable set of outcomes, the sample space, and a probability function that assigns values between 0 and 1 to every outcome such that the sum of the probabilities is 1. And when we're using our tree model and so on, our objective is to construct one of these things. Usually, the hard part will be verifying that, in fact, the way we've assigned probabilities has the property that the sum of them is equal to 1. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 144_Truth_Tables_Video.txt | PROFESSOR: Just as truth tables provide a simple, direct way to define the meanings of the individual propositional connectives or propositional operators, they also provide a methodical way to understand the behavior of formulas, and in particular whether two formulas are equivalent or whether a formula is valid, meaning that it's always true. So let's take a look at that. To being with, if I'm thinking about a propositional formula that's a composite one that's built out of lots of more atomic, primitive ones, then in order to figure out the value of the whole thing, I need to know the values of the individual components. So if we think of a formula involving P's and Q's and R's that are true/false value propositional variables, then I need a truth assignment to know the values of those variables, in order to know whether or not the formula is true or false. So in computer science jargon, this process of assigning values to variables is called an environment, although logicians would call it a truth assignment. So an environment tells you, given a variable, whether or not it's true or false. Let's look at an example of three variables-- P, Q, and R. They are true/false valued, and I'm going to tell you that I've got an environment v in which P is true and Q is true and R is false. So v of P was T, et cetera. So I'm thinking of v as a function that maps a variable to its value. Now, let's see how I would use this particular environment to figure out the value of this composite formula whose atomic parts are P and Q and R, and Q again, I guess. Let's take a look at how we would go about figuring out the value of this whole formula given the values of P, Q, and R. It's pretty straightforward, but the methodical way to do it is sort of from inside out. Let's begin by attaching the truth values that we're given to those particular variables. Now, notice here that I've got both arms of the AND have been assigned truth values, and they're both true. That means I can assign true to the conjunction formula, the AND formula, and I do that by putting the T under that AND, which is the principal connective of this subformula. Now, looking back at this Q, I've got it's true, which means that NOT of Q is false, so I can put a false under the NOT. Now, notice I've got both arms of this XOR are defined. They're both false, and that means that the XOR is false because it's only supposed to be true if exactly one of them is true. And next, I had this true for AND. That means that NOT of that formula is false. And now that I have both arms of the OR, this one is false and that one is false, I can conclude that the entire expression is false by putting that final truth value under the principal connective of the whole formula. So that's actually an easily defined recursive process. I've described it from inside out, but if you are programming it, you would be doing it recursively top-down in order to evaluate the truth value of a formula given the truth values of its constituent variables given an environment. Now, a basic idea about propositional formulas is that two of them are equivalent if and only if they have the same truth values in all environments-- in all environments. No matter what the values of the P's and Q's and R's are, no matter what the truth value is, these two formulas come out to the same truth value, and that's what makes them equivalent. Let's look at an important example known as De Morgan's law. So De Morgan's law says that if I look at this formula, the NOT of P or Q, the negation of P or Q, I claim that that's equivalent to not P and not Q, and let's check that with a truth table. So here's going to be the truth table for the first formula, NOT P or Q, so let's write out the four possible values for P and Q. These are the four possible environments, one per row, and there is the formula whose value I'm trying to compute. So we do it in the usual way. I'm not going to repeat the truth values of P and Q because they're given here, but I can fill in the values of the OR because I know the truth table for OR and I discovered that the first three rows or subformula is true and the last place is false when both of them are false. And that means that I know the value of the whole formula, which is the negation. The NOT of all those values just flips the trues and falses, and that is the final truth values for this NOT P or Q in all possible environments. Let's do the same thing for NOT P and NOT Q. Well, this time, I'll fill in the values of NOT P and NOT Q because they're not just repeated there. So the NOT P is the flip of the P column and the NOT Q is the flip of the Q column, and now I can fill in the values at the AND. And of course, the AND is going to be true only when they're both true, and otherwise it's going to be false, and look what I've got. The possible truth values of the first formula, NOT P or Q, in all possible environments is exactly the same as the possible truth values of NOT P and NOT Q in those corresponding environments. The columns are the same, and that means these two formulas are equivalent, which is the proof by truth table. We've just examined all possible environments and verified that, in fact, they get the same truth value. This brings us to a useful other connective we haven't talked about yet called if and only if. So the value of the if and only if connective P if and only if Q is true, if and only if P and Q have the same truth value. Now, this if and only if is an English word that you're supposed to understand what it means, and that if and only if is an operator on truth values that we're defining. Since the English now can be confusing between that if and only if and this if and only if, let's disambiguate by having the truth table for if and only if. Here it is. What it says is that when both of them are true-- P if and only if Q is true-- and both of them are false-- P if and only if Q is true-- and otherwise, it's false. You can check that P if and only if is true exactly when the complement of P XOR Q is true. So now we come to two crucial properties of formulas called satisfiability and validity, and let's examine what those are. So a formula is satisfiable if and only if it's true in some environment. That is, it's satisfiable if there is some way to set the values of the variables P and Q to be truth values in such a way that the formula comes out to be true. And a related idea is that a formula is valid-- it's also called a tautology-- if and only if it's true in all environments. No matter what you set the variables to, it's going to come out to be true. Let's look at some examples to solidify those two concepts. So the formula P all by itself is satisfiable because it can be true if P is true, but it's not always true because P might be false. Symmetrically, NOT P is also satisfiable because it could be true if P is false, but it's not always true. It's not valid because P might be true, in which case not P would be false. A formula that's not satisfiable, a formula that means that there is no truth value that makes it true, which is the same as saying that it's always false, is the formula P and NOT P. It's probably the simplest not satisfiable formula or unsatisfiable formula. So this is clearly false because either P or NOT P has got to be false and the [? and ?] will then will definitely come out to be false. There is no value for P that makes this formula true. It's unsatisfiable. A valid formula, actually by De Morgan's law applied to the P and NOT P, is P or NOT P is going to be valid because no matter what truth value P has, it comes out to be true. That is, one of P and NOT P is true, and therefore the OR is going to get at least one true, exactly one true really, and going to come out to be true, so this is valid. Now we can connect up the pieces that we've just set up of relating validity and equivalence. Two formulas G and H are equivalent is the same as saying that G if and only if H is valid. So G if and only if H comes out to be true when G and H have the same truth value, and G and H are equivalent says that they do have the same truth value no matter what the environment is. So that's the same as saying that if G and H are equivalent no matter what the environment is, G if and only if H comes out to be true. So if G and H are equivalent, G if and only if H is valid, and the converse argument works the same way. A very important problem that comes up in multiple ways, and we're going to examine some of them later, is the problem of whether or not a formula is valid-- proving that it's valid-- and checking whether or not a formula is satisfiable. Now, there's a simple way to do that. The truth table tells it to you. If you want to know whether the formula is satisfiable, you just look at its truth table. Try every possible environment and see whether one of them yields the value of true. The problem with that approach, which theoretically is sound, but pragmatically the truth table size doubles with each additional variable. So with two variables, you got four rows. With three variables, you got eight rows. With four variables, you got 16 rows, and this very rapidly gets out of hand once the number of variables gets to be moderate sized. This is exponential growth. It's doubling each time you add a variable. So really when you start having hundreds of variables, truth tables are out of the question. You just can't write them down. They're too big, and in fact, in modern digital circuits when you think back about how we designed the [? adder, ?] if you look at what's going on in all those different wires corresponding to different variables, there are millions of those in a typical modern digital circuit. So truth table approach is just not going to be workable. It's hopeless. Now, one of the central problems in theoretical computer science is the question of whether or not there is a way to test for SAT that's more efficient than this impossible way of trying to come up with a truth table that's too big to fit in the universe when there are hundreds and thousands of variables. So we're interested in the question of is there some other way to check a formula for satisfiability than exhaustively trying to generate its whole truth table to see if some row yields true? This is the abstract version of the P equals NP problem. So we've talked about this before. P equals NP? is considered to be the most important open problem in theoretical computer science, in the theory of computation, and it's simply saying, is there some fast or efficient way to tell whether or not a formula is satisfiable that's more efficient than the truth table approach? Efficiency has a technical definition, which is that the number of steps grows much less than exponentially with the number of variables. Let's say it grows polynomial, like a quadratic or a cubic, but you're trying to beat exponential growth, which is what ruins things quickly. And this is an open problem. No one knows if there is some fast way to check for satisfiability or SAT. That's the SAT problem. By the way, very closely related to SAT is validity because if I wanted to know whether a formula G is valid-- well, valid means that it's always true. That means its complement, not G, is always false, which is the same as saying its complement is not satisfiable. So to check that G is valid, all I need to do is check that not G is satisfiable and vice versa, not G is not satisfiable, if and only if G is valid. So the point is that SAT and valid stand and fall together. If you had a fast way to do one, you would very quickly get a fast way to do the other one. So checking for one is just as difficult as checking for the other, and we're going to examine in subsequent lectures why it is that this problem is so important. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 323_Asymptotic_Properties_Video.txt | PROFESSOR: An advantage of expressing the asymptotic notations, in terms of limits, is that a bunch of their properties then become immediately obvious. Here's one. If f is little o of g, or f is asymptotically equal to g, then, in fact, f is big o of g. Or we can reason about this informally by saying that the first one means that f is much less than g, and the second one means that f is about the same as g, and the final one means that f is roughly less. So being about the same and definitely less, and certainly this implies roughly less. But we can in fact be entirely precise just using the definitions, because f equals o of g means the limit of f over g is 0. And f is asymptotically equal to g means that the limit of f over g is 1. And the definition of f equals big o of g is that the limit is finite. And clearly, if it's 0 or 1, then it's finite. Another such property is that if f is much less than g, then g is not roughly less than f. More precisely, if f is a little o of g, then g is not big o of f. The left hand side says that the limit of f over g is 0. But that implies that the limit of g over f is 1 over 0, or infinity, which means it's not finite, so g is not a big o of f. PROFESSOR: Now, the usual way that big o is defined in the literature doesn't mention limits at all. And, in fact, as I said, the definition really isn't a limit, it's a limsup. And let me show you the standard definition and then explain why the limsup soup captures it and is needed. So the official definition of f is big o of g is that there's some constant multiplier, c, that you can amplify g by, such that once g is amplified by the factor c, then, in fact, f is less than or equal to c times g. But this may not hold right at the beginning. There's a certain point, n 0, after which it holds forever. Let's try to illustrate this complicated alternation of quantifiers expression with a diagram that may make it clearer. So suppose that I want to express the fact that f is big o of g, where f is it a green line. So that green line is the graph of f of x, the function. And g in blue is shown-- and as a matter of fact, g of x is less than or equal to f of x. But nevertheless, f is going to be big o of g, because if you multiply g by a constant, it becomes sort of shifting it up to be this constant times g. It becomes this purple curve, and the purple curve, it gets to be above the green curve, from a certain point on. That's n 0. So by raising up the blue curve, g, by an amount c, to get it to be this purple curve, the purple curve gets above f from a certain point n 0 on. And that's why f is big o of g. Now, of course, multiplying the blue curve, g, by a constant doesn't raise it up a fixed amount. It alters it, but if we imagine that our curve was a log scale, than, in fact, multiplying g by c is the same as adding log c on a log scale. So the picture is actually accurate if the vertical scale is logarithmic. So using this standard definition, I can explain why in the equivalent definition of terms of limit, I couldn't say limit, I needed to say limsup. Here's what limsup does for us. Suppose I have a function, f, that's say, less than or equal to 2g. Which means that, surely, f is big o of g, according to the previous definition, because you amplify g by 2 and you get above f. The problem is that f of n over g of n may have no limits, so I can't simply say that f is o of g, because the limit of f over g is finite. Let's see how that could happen. Suppose that f is in fact equal to g times a number that varies between 1 and 2. That's an example where sin of n pi over 2 varies between 0, 1, and minus 1. And you square it, it becomes 0 or 1. And you add 1 to it, it becomes 1 or 2. So this is an expression, which as n grows, alternates between the values 1 and 2. And I'm multiplying g of n by this factor that's either 1 or 2. But the limit of f of n over g of n does not exist, it's alternating between 1 and 2. On the other hand, the limsup f of n over g is 2, which is finite, and therefore, according to the limsup definition, indeed f is o of g. Now, the technical definition of limsup is one that you can read in the text or find in a calculus book. It's basically the largest limit point of the fraction f of n over g of n. And if you don't know what a limit point is, it's stuff that we don't need to go into. But I did want you to understand why formally, we need limsup. In most cases, the limit exists, and we can use the simpler limit definition, rather than the exists a constant, such that for every number n greater than or equal to n 0, et cetera, which is a more complicated definition. OK. Let's collect a couple of more basic facts about little o and big o that we're going to need. Namely, that if a is less than b-- I know they can be negative numbers. I don't care, but real numbers. If a is less than b, then x to the a is little o of x to the b. The [? proof file ?] is almost immediately from the definitions, because to prove that x to the a is little o of x to the b, we want to look at the quotient of x to the a over x to the b. But, of course, the quotient of x to the a over x to the b is equal to 1 over x to the b minus a. And since a is less than b, b minus a is positive. So that means that as x approaches infinity, the denominator is x to a positive power also goes to infinity. And therefore, 1 over x to that positive power goes to 0, which is the definition of x to the a being little o of x to the b. Another crucial fact is that logarithms grow slower than roots. So if you think of epsilon as like a half or a third, saying that the log of x is less than or equal to the square root, it's less than equal to the cube root, it's less than or equal to the 50th root doesn't matter. OK. This is a proof that just falls back on elementary calculus. And I guess I've highlighted it, because it's definitely worth remembering. Logarithms grow slower than roots. The proof begins with the immediately obvious remark that 1 over y is less than or equal to y, because they're equal when y is greater to 1. 1 over y is equal to y when y is greater than or equal to 1. And as y increases, y gets bigger, and 1 over y gets smaller, so the inequality is preserved. That's easy, OK. Well that means that I can integrate both sides starting at 1. So if I take the integral of 1 over y from 1 to z, it's going to be less than or equal to the integral of y from 1 to z. Well, integral of 1 over y is log z, and the integral of y to z is z square over 2. So what we get is this new inequality, the log of z is less than or equal to z squared over 2, for z greater or equal to one. So we're on the way there. We've got log of z is less than z squared, but not z to any epsilon power. But we'll get that just by making a smart substitution for z, so that's the next step. We have that log of z is less than equal to z squared over 2 for any z greater than or equal to 1. Let's let z be the square root of x to the delta, where delta is simply some positive number. So in that case, what's the log of z? Well the log of the square root of x to the delta, the square root means it's half of log of x to the delta, which is delta log x. So log of z is delta log of x over 2. And, of course, z squared is just x to the delta, so z squared over 2 is x to the delta over 2. Now, I can just cancel the denominators too. And I get that log of x, and then transpose the delta log of x is less than or equal to x to the delta over delta. But as long as delta is less than epsilon, x to the delta is going to be a little o of x to the epsilon, which means that x to the delta times a constant, namely 1 over delta, is also going to be little o of x to the epsilon. And I've just figured out that I've shown that log of x is little o of x to the epsilon as required. One more crucial fact that I'm going to not prove, but I'll state is that polynomials grow slower than exponentials. This is closely related to the fact that logs grow slower than roots. But in particular, if c is any constant and a is greater than 1, then x to the c is little o of a to the x. And there's a bunch of ways to prove this using a L'Hopital's Rule, or McLaurin Series, and I'll leave it to you to look up your 1801 calculus text to find a proof of that fact. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 2119_Halls_Theorem.txt | PROFESSOR: So let's get set to state Hall's theorem in a way that doesn't mention boys and girls. But let's remember the girl-boy setup to start. So the general setup is a bipartite graph H. And a bipartite graph has two sets of vertices-- the girl vertices and the boy vertices. Formally, there's a set L of H, called the left vertices of H, and a set R of H, called the right vertices of H. The vertices of H altogether are L of H union R of H. Both of these are non-empty. And they don't overlap. Then the edges of H have the property that they only go between L of H and R of H. That is the definition of a bipartite graph. Now, we're interested in a matching in a bipartite graph. So let's be precise again of what's a matching without having to mention boys and girls and like. A match is a total injective function from the left vertices to the right vertices. So that means every L vertex, or girl, has a match m of L that is on the other side. And we need this total injective function, match function m, to follow the edges. What does follow the edges mean precisely? It simply means that the edge l m of l is a legitimate edge of H. Another way to say that is that the graph of this total injective function is a subset of the edges of H. So we can state the no-bottleneck condition, Hall's condition, as follows. If the size of S is less than or equal to its image under the edges, for every set of left vertices H, that's called Hall's condition, that the size of S is less than or equal to the image of S for every S, then there is a match. So this is a way of precisely stating Hall's theorem. No boys and girls mentioned. We'll be comfortable going back to the boy-girl language, because it's more memorable and easier to talk about. But just for the record, we have now formally stated, defined bipartite graphs, matches in bipartite graphs, and Hall's theorem really that there is a match when Hall's condition is satisfied without mentioning boys and girls. The puzzle is, how do you verify that there are no bottlenecks? It's pretty hard. The bottleneck condition involves checking that every subset S of L of H satisfies this cardinality condition. And there are a lot of subsets of L of H relative to the size of the graph, exponentially many. So we can ask, how do you verify that there are no bottlenecks? Well, it turns out in logarithms classes, you will learn a fairly efficient matching procedure, runs about quadratically, for finding perfect matches when they exist. But there is one particular special case that ensures no bottlenecks that we'll be making use of several times in the term. And that special situation is when it turns out that every girl likes at least d boys and every boy likes at most d girls. This is called a degree-constrained graph. If a graph is degree constrained, then there are no bottlenecks. So this condition that each girl likes at least d boys and every boy likes at most d girls could have been rephrased as saying that the minimum degree of the girls is greater than or equal to the maximum degree of the boys. But I think expressing it in terms of d maybe makes it a little clearer. Let's prove this. I'm going to prove that, if you have a degree-constrained bipartite graph, then it, in fact, satisfies Hall's condition. There are no bottlenecks. So we have that, if every girl likes at least d boys and every boy likes to most d girls, there are no bottlenecks. The proof goes as follows. Let's look at some arbitrary set S of girls. And suppose that the total number of edges that are incident to S has cardinality e. There are e edges that touch vertices in S. Well, that tells us that since every vertex in S has at least d edges out-- maybe more-- d times the size of S has to be less than or equal to the total number of edges coming out of S. Likewise, all of the edges that come out of S go, by definition, into E of S. And each vertex in E of S can only absorb d edges, d girls. So that means that the total number of edges absorbed by E of S-- that is, at most d times E of S-- had better be bigger than the number of edges that have to be absorbed e. So we get that d times the size of S is less than or equal to the total number of edges incident to S, which is less than or equal to d times the total number vertices in E of S. Well, I'll cancel the d's. And we have that the size of S is less than or equal to the size of its image. That's the definition of a bottleneck-- The violation of that would be a bottleneck. This says, there's no bottlenecks. And we have proved the degree-constrained condition is sufficient to verify Hall's condition. And by Hall's theorem, it's sufficient to guarantee that there is a match. Now, there's a lot of graphs with matches that are not degree-constrained. This is not an if and only if theorem. It's just a sufficient condition that comes up often enough that it's worth mentioning. Degree-constrained implies Hall's condition is satisfied, which implies that there's a perfect match. So let's turn now to the general case of Hall's theorem. And let's talk about proving Hall's theorem. Hall's theorem says that if there's no bottleneck, then there is a match. So let's begin by setting up a lemma that will play a crucial role. The strategy for proving it is basically going to be to try to break the problem of finding a match for a large set of vertices into sub problems of finding matches for smaller sets of vertices. And by strong induction, Hall's condition will apply to the subsets and will win. Let's look at that more carefully. So let's begin by supposing that there are no bottlenecks in some bipartite graph H. Well, in particular, if there's no bottlenecks anywhere, then if you restrict yourself to some set S of girls, no subset of that set S is going to have a bottleneck, obviously, because a bottleneck within S would be a bottleneck within the whole graph. So that part's trivial. What's not trivial and takes a little bit of arguing is that suppose you have a set of girls with the property that the number of boys that are compatible with that set of girls is exactly the same as the number of girls. So technically, the size of S is equal to the size of E of S. In that case, we can argue that there aren't any bottlenecks within the complement of S and the complement of E of S either. Well, let's look at a picture to illustrate what's going on. So here we have a bipartite graph. And there's S and its image on the right E of S. And we know that a bottleneck here would imply a bottleneck in the whole graph. What we want to argue is that a bottleneck in the complement of S would imply a bottleneck in the complement of E of S. So there's the complement of E of S. And there's the complement of S. Now, notice that some edges that come out of the complement of S may very well go into the E of S. That is, this would be a point that's both in E of S and in E of S complement. But we're not allowed to use that point, because we're trying to find a match only within S bar and E of S bar. And what we're really trying to argue is that a bottleneck here would imply a bottleneck in the whole graph. So let's see if we can argue this. We want to claim that there's no bottleneck between S bar any E of S bar given that there's no bottleneck anywhere and when S and E of S are the same size. So let's suppose that we had a set T over here that was a subset of S bar. And let's look at its image over here in orange. So we've got that the image of this set T, when we restrict it just to the part that's in the complement of S-- notice I'm leaving out this point here. I'm not taking the image of T. I'm taking the image of T restricted to the points that are not in the image of S. And what I want to know is, could that cause a bottleneck? After all, I've left out some points that used to be included in the image of T. And so the image of T might have been bigger than T because of those extra points that I'm leaving out, I have to be sure that that doesn't happen. Could there be a bottleneck that has been created by these points that I've left out? Could there be a bottleneck in E of T intersection, a complement to E of S. Could that orange guy be a bottleneck? Well, if it was a bottleneck, then I'd have a bottleneck in the whole graph. And that's because S and E of S are the same size, because all I do is, if I had a bottleneck there, I would take this set T along with S and this set orange along with E of S, and that would be a bottleneck. Because now, if you look at the size of T, I've added the same amount to the size of T as I've added to that orange set, because E of S and S are the same size. So if this orange part was smaller than T, then orange part along with E of S is smaller than T union S. And that defines a bottleneck in the whole graph. So again, if there was a bottleneck in here caused by restricting images to the complement of E of S, it means there was a bottleneck in the whole graph. If there's no bottleneck at all, then indeed, there's no bottleneck in this other part of the complement of S and the complement of E of S. And this gives me a hook on proving Hall's theorem, because that's basically the way I'm going to split the problem into two separate matching parts. So let's now proceed to prove that if there are no bottlenecks in a graph, then there's a perfect match. And it's going to be by strong induction on the number of girls. So case one is the case that we just examined, that there's a non-empty subset of girls-- a proper subset, not all the girls-- some subset of girls who-- another way to say it is there's a subset of girls that's not empty and its complement is not empty either. And the size of S is the same as the size of E of S. That is, the number of boys that are compatible with this set of girls is exactly the same size as this set of girls. Well, if there's going to be a match, there has to be one in that subset. And by the lemmas, if there's no bottlenecks anywhere, there's no bottlenecks when I restrict myself to just S and E of S. And also by the previous Lemma, there's no bottleneck in the complement of S restricted to the vertices in E of S. So this is a new bipartite graph, where I'm using S for the left vertices and E of S for the right vertices. And here, I'm using the complement of S for the left vertices and the complement of E of S for the right vertices. And the previous lemma says that if there's no bottlenecks overall, there's no bottlenecks in either of these two restricted graphs. So what I can do now is, by induction, since there's no bottlenecks in S, E of S, I can find a match there. So I can match up all of the girls in S with compatible boys in E of S. And I can match up all of the girls that are not in S, in the complement of S, with boys that I haven't used already. And there will be a perfect match there. And then these two separately will provide a match for the entire set of left vertices-- S union the complement of S will be all of the vertices. And these two matchings don't overlap. So combining them gives a complete matching. That knocks off this case. That's the nice case, where I can find some subset with the number of boys is exactly equal to the number of girls, which means that that has to work by itself. And I worked that one by itself and what's left over by itself. And that's how I can find an overall match. The second case is that there isn't any set of girls, which are compatible with exactly the same number of boys. Well, since Hall's condition holds, that means that every set of girls actually has to be compatible with a larger set of boys than the number of girls. That is, for every set S, the size of S is strictly less than the size of E of S for every non-empty and proper subset S. Well in that case, actually finding a match is easy, because now I've got slack. What I can do is just pick an arbitrary girl-- call her g-- and she's got to be compatible with at least one boy, because there's no bottlenecks. In fact, in this case, she'll actually be compatible with these two boys. But all we need is one boy. Since there's no bottlenecks, if I pick a girl, she's got to be compatible with some boy. And I'm just going to arbitrarily match her with that boy-- match g with b. Now, the consequence of that is that, if I now remove g and b from the graph, I am left with a graph in which at worst I've shrunk the set E of S by that girl that I've matched with B, which means that E of S, which used to be strictly greater than S has gone down by at most one. It's still greater than or equal to E of S. So I'm left that in the graph after I've removed this g paired with an arbitrary b that she's compatible with, I'm left with a graph that satisfies Hall's condition that has one fewer girl vertex. And so it has no bottlenecks. And therefore, it has a match. And that match in the sub graph, where I've taken out g and b, when I put back the match, when I add the part of g matching with b, it's a match for the whole graph. And that's how I get it. That completes the proof of Hall's theorem. |
MIT_6042J_Mathematics_for_Computer_Science_Spring_2015 | 1119_Russells_Paradox_Video.txt | PROFESSOR: So in this final segment today, we're going to talk about set theory just a little bit. Because if you're going to take a math class, if you're going to be exposed to math for computer science, it's useful to have at least a glimmering of what the foundations of math looks like, how it starts and how it gets justified. And that's what set theory does. In addition, we will see that the diagonal argument that we've already made much of played a crucial role in the development and understanding of set theory. So let's begin with an issue that plays an important role in set theory and in computer science, having to do with the idea of taking a function and applying it to itself, or having something refer to itself. And this is one of these things that's notoriously doubtful. There's all these paradoxes. But maybe the simplest one is when I assert this statement is false. And the question is, it true or false? Well, if it's true, then it's false. And if it's false, then it's true. And we get a kind of buzzer. It's not possible to figure out whether this statement is true or false. I think we would deny that it was a proposition. So that's a hint that there's something suspicious about self-reference, self-application, and so one. On the other hand, it's worth remembering that in computer science, we take this for granted. So let's look at an example. Here's a simple example of a list in Scheme Lisp notation, meaning it's a list of three things, 0, 1, and 2. And the black parens indicate that we're thinking of it as an ordered list. Now the way that I would represent a list like that in memory, typically, is by using these things are called cons cells. So a cons cell has these two parts. The left hand part points to the value in that location in the list. So this first cons cell points to 0, which is the first element in the list. The second component of the cons cell points to the next element to the list. And so here you see 1 pointing to the third element of the list. And there you see 2. And that little nil marker indicates that's the end of the list. So that's a simple representation of a list of length three with three con cells. One of the things that computer science lets you do and many languages let you do is you can manipulate these pointers. So using the language of Scheme, what I can do is I'll do an operation called setcar where I'm taking, in this case, I'm setting the second element of L, that is this cons cell, to L. And setcar is saying, let's change what the element in the left hand part of this cell is. This is where the value of the second element is. Let's change the value of the second element to be L. What does that mean as a pointer manipulation? Well, it's pretty simple. I just move this pointer to point to the beginning of the list L. And now I have an interesting situation, because this list now is a list it consists of 0 and L And 2. It's a list that has itself as a member. And it makes perfect sense. And if you sort of expand that out, L is this list that begins with 0. And then its second element is a list that begins with 0. And the second element of that list is a list that begins with 0, and so on. And then the third element of L is 2, and the third element of the second element of L is 2, and so on. It's an interesting infinite nested structure that's nicely represented by this finite circular list. Let's look at another example where, in computer science, we actually apply things to themselves. So let's define the composition operator. And again, I'm using notation from the language Scheme. I want to take two functions f and g that take one argument. I'm going to define their composition. The way that I compose f and g is I define a new function, h of x, which is going to be the composition of h and g. The way I defined h of x is I say apply f to g applied to x and return the value h. So this is a compose is a procedure that takes two procedures f and g and returns their composition, h. OK, let's practice. Suppose that I compose the square function that maps x to x squared, and the add1 function that maps x to x plus 1. Well, if I compose the square of adding 1, and I apply it to 3, what I'm saying is let's add 1 to 3, and then square it. And I get 3 plus 1 squared, or 16, because the add1 and then square it is the function that's the composition of square and add1. Now I can do the following. I could compose square with itself. If I take the function, square it, and square that, I'm really taking the fourth power. So if I apply the function composed of square square to 3, I get 3 square square, or 81, or 3 to the fourth. All makes perfect sense. Well now let's define a compose it with itself operation. I'm going to call it comp2. comp2 takes one function f. And the definition of comp2 is compose f with f. And if I then apply comp2 to square and 3, it's saying, OK, compose square and square. We just did that. That was the fourth power. Apply it 3. I get 81. And now we can do some weird stuff. Because suppose that I apply comp2 to comp2, and then apply that to add1, and apply that to 3. Well that one's a little hard to follow, and I'm going to let you think it through. But comp2 of comp2 is compose something four times. And when you do that with add1, what happens is that you're adding 1 four times to 3. If I comp2 of comp2 of square, I'm composing square with itself, and then composing that with itself. I'm really squaring four times. And I wind up with 2 to the fourth, or 16, is the power that I'm taking. And so compose2 of compose2 of square of 3 is this rather large number, 3 to the 16th. It could be a little bit tricky to think through, but it all makes perfect sense. And it works just fine in recursive programming languages that allow this kind of untyped or generically typed composition. So why is it that computer scientists are so daring, and mathematicians are so timid about self-reference? And the reason is that mathematicians have been traumatized by Bertrand Russell because of Russell's famous paradox, which we're now ready to look at. So what Russell was proposing, and it's going to look just like a diagonal argument is, Russell said, let's let W be the collection of sets s such that s is not a member of s. Now let's think about that for a little bit. Most sets are not members of themselves. Like the set of integers is not a member of itself because the only thing in it are integers. And the power set of integers is not a member of itself because every member of the power set of integers is a set of integers, whereas the power set of integers is a set of sets of those things. So those familiar sets are typically not members of themselves. But who knows, maybe there are these weird sets like the circular list, or a function that can compose with itself, that is a member of itself. Now let me step back for a moment and mention where did Russell get thinking about this. And it comes from the period in the late 19th century when mathematicians and logicians were trying to think about confirming and establishing the foundations of math. What was math absolutely about? What were the fundamental objects that mathematics could be built from, and what were the rules for understanding those objects? And it was pretty well agreed that sets were it. You could build everything out of sets. And you just needed to understand sets, and then you were in business. And there was a German mathematician named Frege who tried to demonstrate this by developing a set theory very carefully, giving careful axioms for what sets were. And he showed how you could build, out of sets, you could build the integers. And then you could build rationals, which are sort of just pairs of integers. And then you could build real numbers by taking collections of rationals that had at least upper bound. And then you keep going. You can build functions and continuous functions. And he showed how you could build up the basic structures of mathematical analysis and prove their basic theorems in his formal set theory. The problem was that Russell came along and looked at Frege's set theory, and came up with the following paradox. He defined W to be the collection of s in sets such that s is not a member of s. Frege would certainly have said that's a well defined set, and he will acknowledge the W is a set. And let's look at what this means. This is a diagonal argument. So let's remember, by this definition of W, what we have is that a set s is in W if and only if s is not a member of s. OK, that's fine. Then just let s be W. And we immediately get a contradiction that W is in W if and only if W is not in W. Poor Frege. His book was a disaster. Math is broken. You can prove that you're the pope. You could prove that pigs fly, verify programs crash. Math is just broken. It's not reliable. You can prove anything in Frege's system, because it reached a contradiction. And from something false, you can prove anything. Well Frege had to book. It was a vanity publication. And the preface of it had to be rewritten. And he said look, my system's broken. And I know that. And Russell showed that unambiguously. But I think that there's still something here that's salvageable. And so I'm going to publish a book. But I apologize for the fact that you can't rely on the conclusions. Poor Frege. That was his life work gone down the drain. How do we resolve this? What's wrong with this apparent paradox of Russell's? Well, the assumption was that W was a set. And that turns out to be what we can doubt. So the definition of W is that for all sets W, s is in W if and only if s is not in s. And we got a contradiction by saying OK, substitute W for s. But that's allowed only if you believe that W is a set. Now it looks like it ought to be, because it's certainly well defined by that formula. But it was well understood at the time that that was the fix to the paradox. You just can't allow W to be a set. The problem was that W was acknowledged by everybody to be absolutely clearly defined mathematically. It was this bunch of sets. And yet, we're going to say it's not a set. Well, it's OK. That will fix Russell's paradox. But it leaves us with a much bigger general philosophical question is, when it is a well defined mathematical object a set, and when isn't a set? And that's what you need sophisticated rules for. When is it that you're going to define some collection of elements, and you could be sure it's a set, as opposed to something else-- called a class by the way-- which is basically something that's too big to be a set, because if it was a set, it would contain itself and be circular and self-referential. Well, there's no simple answer to this question about what things are sets and what are not sets. But over time, by the 1930s, people had pretty much settled on a very economical and persuasive set of axioms for set theory called the Zermelo-Fraenkel set theory axiom. |
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