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<p>If a particle with mass $m$ collides with a wall at right angles, and the collision is perfectly elastic. The particle hits the wall at $v\ ms^{-1}$. There is no friction or gravity.
So the particle will rebound at $-v\ ms^{-1}$?</p>
<p><img src="http://i.stack.imgur.com/Xp73i.png" alt="image description"></p>
<p>What will the change in momentum be?</p>
<p>I did:</p>
<p>$$initial\ momentum = final\ momentum$$
$$mv = m(-v)$$
$$mv = -mv$$</p>
<p>But this doesn't seem right because it's like saying $1=-1$?</p> | 7,406 |
<p>My science teacher used to say a lot of weird stuff, but I'm just making sure on this one.</p> | 7,407 |
<p>I know that half silvered mirrors are used in the production of laser emission. Can half silvered mirrors be used outside the instrument so the rays get reflected back on one side of the mirror and pass through the other side? </p>
<p>What will be the result if a half silvered mirror is kept rotating and laser is made to pass through it?</p>
<p>Similarly what will be the result if a prism is kept rotating and laser is made to pass through it? </p>
<p>Series of prisms and mirrors are used in the instrument to produce laser but can we obtain more patterns (movement of laser in entertainment lasers) when we keep a rotating half silvered mirror and prism (different cases) and laser is made to pass through it?</p> | 7,408 |
<p>Next semester, I am going to lecture about (the mathematics of) general relativity and I am still thinking hard how to organize and even more importantly how to motivate all the stuff.</p>
<p>I am wondering what minimal assumptions I have to make about the objects and their relations to be able to interpret the formulae and their relation to classical Newtonian physics. I should explain further:</p>
<p>I think the assumption that spacetime is modeled by a four-dimensional differentiable manifold M is fine and is easily to be motivated. I am also fine with assuming that we have an affine connection on the manifold because it can be measured by moving a (quantum) particle with spin along a closed loop and comparing spin direction (and relative position/phase for torsion) before and after going through the loop.</p>
<p>Then we may assume that the holonomy of the affine connection lies inside the Poincaré group (because we measure no other holonomy). Using this, we can parallel transport a chosen Lorentz metric in one tangent space to each other tangent space, so we get a Lorentz manifold. (Usually texts on general relativity start with a Lorentzian manifold, but they do not explain where the measuring of lengthes and angles should come from — a rod is itself a complicated physical object).</p>
<p>Now having such a manifold, we can write down the Riemannian curvature and the torsion tensor. For simplicity, let us assume that torsion vanishes for the moment. Given the Riemannian curvature, we can contract it and write down the Einstein tensor G. Now the Einstein field equations can be stated as a definition: "The Einstein tensor G is the stress-energy tensor", that is G tells us where we measure matter.</p>
<p>Mathematically this is fine (and actually of no content). From the viewpoint of physics, however, we want to be able to interpret the so <em>defined</em> matter (or stress-energy tensor to be more precise) as what is usually consider to be matter (or mass density or pressure or stress). What other inputs do I need to achieve this?</p>
<p>Do I have to add the geodesic equation for free-falling test particles, for example, or does this already follow from my definitions (that is the field equations) above (of course, one has to relate a test particle to the matter term)?</p>
<p>I am aware of the geometric interpretation of the Einstein field equation which relates the trace of the stress-energy tensor to the second derivative of the change of the volume of a ball of free-falling test particles. In order to use this, one has to know the equations of motions for free-falling test particles first. Further, one has to compare with the change of volume in the Newtonian limit. But how would we then get the pressure dependent parts in the trace of the stress-energy tensor, because Newtonian gravity depends only on the mass (the 00-part)?</p> | 7,409 |
<p>Yesterday my wife asked me that question, and I couldn't answer.</p>
<p>Consider a car, in a sunny day, and that is consumes x gallons per mile. Considering that everything is equal, except that it's traveling in a rainy day, but at the same temperature as the sunny day so that air density is the same.</p>
<p>Will the lower friction of the tires make it consume more or less fuel? And the fact that rain drops are falling over and in front of it ?</p>
<p>I answered that it'll consume more fuel, since friction is what makes car move and that the rain will act against it... but I'm not sure ?</p> | 7,410 |
<p>Since by definition we cannot observe black holes directly, how do astronomers determine the mass of a black hole?</p>
<p>What observational techniques are there that would allow us to determine a black hole's mass?</p> | 7,411 |
<p>How much of the Earth would a spoonful of the Sun scorch if held at ground level?</p>
<p>I basically would like to conceptualize the heat of the Sun on a smaller scale, please.</p> | 7,412 |
<p>The viscosity of water creates drag on swimmer's body so its effect is to slow down the swimmer. However the viscosity seems to be essential for pushing the water backwards by the swimmer's arms and legs. Would a human be able to swim in water with much lower viscosity? What standard stroke (front crawl, breaststroke, butterfly) would work best/worst? How would a lower viscosity affect fish motion in water?</p> | 7,413 |
<p>Water pressure is simply $pgh/2$ for a vertical wall parallel to the depth of the water at exactly the half the water's depth. Suppose the wall is tapered (angled) slightly at around 2 degrees from the y-axis (parallel to the wall). How would you measure the pressure at this point? Would it be the contributing pressure from both the vertical and horizontal pressure? As such: $pgh
\sin (\theta) + 1/2pgh\cos(\theta)$.</p>
<p>I'm assuming pressures can be summed as forces. This may be false.</p> | 7,414 |
<p>During a random reading through this site, I found this one: <a href="http://physics.stackexchange.com/questions/7131/origin-of-elements-heavier-than-iron-fe">Origin of elements heavier than Iron (Fe)</a>...</p>
<p>The answer was "The formation of many elements in earth was due to <a href="http://en.wikipedia.org/wiki/Supernova_nucleosynthesis" rel="nofollow">Supernova nucleosynthesis</a>" as told by some guy. Here, A question crosses my mind:</p>
<ul>
<li>If the elements were formed due to the explosion of a supernova, then there should be a <strong>remnant</strong> like a black-hole or a neutron star nearby... Were there any nearby? Or, the <em>famous</em> <strong>Big Bang</strong> is responsible for this?</li>
</ul>
<p>(<em>Down-voters - leave comments</em>)</p> | 7,415 |
<p>consider me sitting on the top of a train which is travelling close to the speed of light, will I be able to see my image on a mirror which I'm holding in my hand??</p> | 248 |
<p>I'm reading some chemistry-related papers that employ concepts of droplet evaporation. Since I am no chemist, I am wondering:</p>
<p>How can I estimate the actual size of a molecule, say <a href="http://en.wikipedia.org/wiki/Succinic_acid" rel="nofollow">succinic acid</a>?</p>
<p>An order of magnitude would suffice. I'm aware of the fact that most molecules can not be approximated by simple assuming that they are spherical.</p>
<p>Evidently, I don't want to do this bottom-up from quantum mechanics. A broad explanation or just a reference would suffice.</p> | 7,416 |
<p>I am rather a theoretician and looking for a formalism to represent biological clocks by Hermitian operators.</p>
<p>The simplest thought model I am looking for is a formal representation of a clock (for instance 12 hours clock) by applying a Hermitian operator (e.g. Hamiltonian)?</p>
<p><em>Note: a clock is a discrete modular representation</em></p>
<p>As questions came up, let me try to explain: </p>
<p>A clock has a modular arithmetic, that is a discrete oscillator, for instance:
$$a \equiv b \pmod n$$</p>
<p>Many biological clocks work like this, and our 12 hours clock is similar $\pmod {12}$.</p>
<p>It is possible as Raskolnikov indicated to apply an inverse Fourier type of approach and express such clocks, as discrete as they are, in the form of a sum such as (example):</p>
<p>$$\psi_q=\sum_{k=1}^q \frac{\omega_q}{4 \pi} \left( e^{-i(k-1)\;\omega_q x} + e^{i(k-1)\;\omega_q x}\right)$$</p>
<p>Both are the representation of the same type of clock.</p>
<p>My first question is <em>how does the parameters in the first equation relate to the second equation?</em></p>
<p>My second question is <em>then what is the Hermitian operator of the second equation?</em></p>
<p>My third question is <em>how can I find to a type of algorithm or general method to read the first equation directly into a Hermitian operator?</em></p>
<p>The fact that I use this for biological clocks is marginal. The $\pmod n$ operation is independent of where we use it.</p>
<p>I hope this helps, but if still questions, I am glad to explain.</p> | 7,417 |
<p>What is going on here <a href="http://youtu.be/LQFmKPA-iAg" rel="nofollow">http://youtu.be/LQFmKPA-iAg</a></p>
<p>Is ethanol leaving the liquid? If so, why?</p>
<p>Does the pressure add energy to the ethanol molecules causing them to gasify and move rapidly near border of gas/liquid and rapid depressurization allows it to escape the liquid and turn back liquid, this time as mist/vapor?</p>
<p>Is the added pressure needed? Would same thing happen if we'd rapidly depressurize it regardless what is starting pressure?</p> | 7,418 |
<p>Consider a big commercial airliner, like a 727, 747, or a 787. </p>
<p>At cruising altitude, under standard conditions, how much of the lift of the aircraft comes from the wings, and how much from the rest of the airframe -- the tube that holds all the people? </p>
<p>My general impression from watching airplanes in flight is that the velocity vector of the centre of mass of the airplane tends <em>not</em> to be pointing in the same direction as the nose of the aircraft. They're usually a little off, moreso during take-off and landing but it seems like they're almost always a little off. </p>
<p>When I look at a side-view of a 747, I see that the angle of attack of the airfoil doesn't quite match the line of the tube part of the hull. Moreover, the tube has a pretty large surface area relative to the wing area. </p>
<p><img src="http://i.stack.imgur.com/gg06c.jpg" alt="enter image description here"></p>
<p>My suspicion is that there's maybe stability reasons why you'd want the tube to not be flat in steady-state flight. And perhaps that's part of the reason why the bottom of the tube is more flat towards the front Is something like this what's going on? </p>
<p><img src="http://i.stack.imgur.com/BOqVu.jpg" alt="enter image description here"></p> | 7,419 |
<p>One way how to look at refraction by a dielectric medium like water or glass is that (phase) velocity of light decreases because it is the wavelength rather than the frequency of the light which changes.</p>
<p>I have read somewhere (but can't recall where) that the frequency must remain the same because otherwise principle of causality would be broken. Is that true?</p> | 7,420 |
<p>So I have faced a problem dealing with transient conduction and I need a little help with the problem solving concepts. I need to determine how long it would take to reach the final temperature but I need to know the convection coefficient for that.</p>
<p>I have done steady state and calculation of Reynolds'/Nusselt's Numbers and everything but that was when the film temperature, $\frac{T_{s}+T_{\infty}}{2}$ was constant throughout the process. In a transient problem, the surface temperature is constantly changing, causing the film temperature to constantly change and also the Reynolds'/Nusselt numbers to both constantly change with time because the film properties are changing. </p>
<p>How can I combat all of these changes? I could either:</p>
<ol>
<li><p>Calculate the convection coefficient using either the initial temperature or the final temperature when calculating the film temperature</p></li>
<li><p>Average the two temperatures in the beginning and then calculate the convection coefficient for that averaged temperature</p></li>
<li><p>Calculate the convection coefficients at the initial and final temperatures and then just average those numbers</p></li>
<li><p>Make a spreadsheet of the properties in the back of the book and then somehow write code on MATLAB that would continuously calculate the coefficient between different boundaries and then just add up the heat transfer, I would have to do that on MATLAB and it would take a long time to do.</p></li>
</ol>
<p>I am very open to any suggestions. I am having a lot of trouble on this because we did not do this in my class and my professor did not really give me an answer when I talked to him</p> | 7,421 |
<p>Why is the number of molecules in a standard volume of gas at a standard temperature and pressure a constant, regardless of the molecule's composition or weight?
Let's say I have a closed box full of a heavy gas, one meter on a side. It has a certain number of molecules inside. I want to be able to add a lighter gas to the box without changing its internal pressure (or temperature), so I connect a cylinder to the side of the box, which holds a frictionless piston for expansion (the piston has a constant force applied to it, to maintain a constant pressure inside the box and allow the volume of the gas to grow as new gas is introduced into the box).
Now I add Helium to the box. The piston moves back to maintain constant pressure, but why does the number of molecules in the box proper stay constant?
My mental image of this is that it would be like adding water to a bucket of marbles, and that, evidently, is wrong, but why is it wrong?</p> | 7,422 |
<p>In popular science books and articles, I keep running into the claim that the total energy of the Universe is zero, "because the positive energy of matter is cancelled out by the negative energy of the gravitational field".</p>
<p>But I can't find anything concrete to substantiate this claim. As a first check, I did a calculation to compute the gravitational potential energy of a sphere of uniform density of radius $R$ using Newton's Laws and threw in $E=mc^2$ for energy of the sphere, and it was by no means obvious that the answer is zero!</p>
<p>So, my questions:</p>
<ol>
<li><p>What is the basis for the claim - does one require General Relativity, or can one get it from Newtonian gravity?</p></li>
<li><p>What conditions do you require in the model, in order for this to work?</p></li>
<li><p>Could someone please refer me to a good paper about this?</p></li>
</ol> | 27 |
<p>Take an initial state and its environment, $E$, as follows,
$$
|\psi\rangle_i = |0\rangle |E\rangle + \sqrt{2}|1\rangle |E\rangle.
$$
Suppose that I've written it already in the basis in which the state de-coheres, such that after de-coherence, the wave-function is
$$
|\psi\rangle_f = |0\rangle |E_0\rangle + \sqrt{2}|1\rangle |E_1\rangle.
$$
where the environment states are orthogonal. In the many-worlds interpretation, if an observer has equal chance to be in each branch, he won't see a Born rule; each outcome is equiprobable.</p>
<p>But suppose that when the state decoheres, it decoheres such that,
$$
|\psi\rangle_f = |0\rangle |E_0\rangle + |1\rangle |E_2\rangle + |1\rangle |E_3\rangle.
$$
This can be obtained with a unitary transformation,
$$
U = 1 \otimes \left[\left(\frac{1}{\sqrt{2}}|E_2\rangle + |E_3\rangle\right)\langle E_1| + |E_0\rangle\langle E_0|\right]
$$</p>
<p>If the wave-function decoheres in the second way, as usual each branch is equi-probable, but this time it results in a Born rule! </p>
<p>What are the problems with positing that the wave-function always decoheres and branches in such a way that, if you assign equal probabilites to each branch, the results are what you would have got from the Born rule?</p> | 7,423 |
<p>If we observe light that originates X light years away, but it passes between black holes X/2 light years away, will it be normal or red shifted or blue shifted? What if the black holes were X/4 or 3X/4 light years away?</p>
<p>I have always wondered if the apparent red shift of distant galaxies must be due to relative velocity with us.</p> | 7,424 |
<p>There are 2 balls in a vacuum, next to each other but not touch. They are on the edge of a surface they will both leave the table at exactly the same time. One gets pushed harder than the other. The earth is completely flat. Apparently both balls will hit the floor at exactly the same time and even further be at the same vertical height from the floor at any point in time. How is this possible?</p>
<p>If you push the one ball much much much harder than the first so that it is going at a considerable speed 100m/s compared to 1m/s say. The balls can't possible hit the floor at the same time, right?</p>
<p>Now the earth is round. You hit the one ball so hard it goes into orbit the other ball just falls of the table straight down like before. As the one ball has gone into orbit it will never reach the ground. However the slow ball will reach the ground much faster. How can a rule be different? If it doesn't work at the extreme then it shouldnt work at all. </p> | 7,425 |
<p>I connected a bulb to a battery positive terminals with positive and negative terminals with negative .
It glows as it should but when i connect the positive terminal of the same bulb to the positive terminal of one battery and negative terminal of the same bulb to the negative terminal of another battery. The bulb does not glow. What I want to ask here is that electric potential difference is being maintained then why does the bulb does not glow.
NOTE : Batteries are not connected with each other. Both batteries are separately placed.</p> | 7,426 |
<p>I'm confused about the relation between the velocity expansion of a scattering cross section and the angular momentum (partial wave) expansion. For example, for dark matter annihilation, we write </p>
<p>$\sigma v = \sigma_s + \sigma_p v^2 + \cdots$, </p>
<p>where $v$ is the velocity and $\sigma_s$ corresponds to $s$-wave scattering (orbital angular momentum $\ell =0$) and $\sigma_p$ is the $p$-wave scattering ($\ell =1$). </p>
<p><strong>Question:</strong>: why does the $v^{2\ell}$ term correspond to initial states in an $\ell$-wave configuration?</p>
<hr>
<p>I understand that in these calculations we treat the scattering states as plane waves, and that these can be expanded as a series of partial waves ($\ell$ eigenstates). And I can see that $\langle \mathbf x | E,\ell,m\rangle \sim j_\ell(kr) \sim k^\ell$. Is there more to this story?</p> | 7,427 |
<p>The system is as follows -</p>
<p><img src="http://i.imgur.com/9hwHHIh.png" alt=""></p>
<p>Friction exists only between the 2 blocks.</p>
<p>I am trying to find out the accelerations of $m_1$ and $m_2$.</p>
<p>Let $a_2$ be acceleration of $m_2$, and $a_x$ and $a_y$ be the accelerations of $m_1$ in the respective directions.
Let $R$ be the normal reaction between the 2 blocks, and $N$ be the normal reaction between $m_2$ and floor.
Balancing components across the axes, I get the following equations -
$$N = m_2g + R\cos\theta \tag{1}$$
$$m_2a_2 = R\sin\theta \tag{2}$$
$$a_x = R(\sin\theta + \mu_s\cos\theta) \tag{3}$$
$$a_y = R(\cos\theta + \mu_s\sin\theta) – m_1g \tag{4}$$</p>
<p>I don’t think $(1)$ is necessary, since friction is not involved between the blocks and the ground.
Leaving that aside, I have 3 equations in 4 variables: $a_x, a_y, a_2, R$.</p>
<p>Is there are any way I could perhaps get a 4th equation so that the system of equations could be solved?
I can get $|a_1|$ in terms of $R$ from the expressions for $a_x$ and $a_y$, but I don’t think that would help.</p> | 7,428 |
<p>Given an alcohol solution and a mass of ice in whatever shape you like, is shaping it into a sphere the worst possible way to cool your drink without diluting it?</p>
<p>If the ice starts off at a sub-zero temperature, it is able to cool the drink without diluting it by first warming to 0 °C. After the periphery of the ice hits the melting point, it then waters down the drink in order to absorb more heat. However, if the conduction isn't perfect within the ice, melting could occur before the whole ice cube hits 0 °C, which seems sub-optimal.</p>
<p>I don't know if because it's not just water that's being cooled the thermo changes, or if there's some other important effect I'm neglecting</p>
<p>More important factors are probably the insulating properties of the container and ice, but all things equal, are all <a href="http://www.thewhiskeyball.com/" rel="nofollow">those</a> <a href="http://www.iceballmaker.com/" rel="nofollow">ice-ball</a> <a href="http://www.cirrusproducts.com/collections/ice-ball/products/2-75-cirrus-press" rel="nofollow">claims</a> categorically wrong?</p> | 7,429 |
<p>Thinking about the concept of symmetry breaking led me to the following question:
Let's say that I have a theory described by a Lorentz invariant Lagrangian, and the true vacuum of the theory is not invariant under Lorentz transformation, will there be massless Goldstone bosons (or fermions) similar to the breaking of a gauge symmetry? How would it be different from this kind of symmetry breaking? What would be some phenomenological consequences? Or would it be stupid (for some reasons) to look at this kind of vacuum from the beginning?</p>
<p>Thanks. </p> | 7,430 |
<p>I am reading a paper on quantum gravity (written circa 1988 but I don't think it's relevant to give a more precise reference) where I read the following statement: "universe loops will in general induce nonlocality, just as field theory loops induce nonlocality in relativistic-particle path integrals, and string loops induce nonlocality on the world-sheet". And to be honest, it's really unclear, I can't argue about the string theory case as I barely know anything about it, but regarding QFT, if those loops are the quantum corrections (appearing in an $\hbar$ expansion for example) I don't see how it is nonlocal, although it could be Wilson loops, then I'd like to know if the "path integrals" mentioned in the quote/article are the path ordered integral of the Wilson loops or the path integral of the partition function (or any correlation function). I hope that my question is clearer than my mind right now.</p>
<p>Thanks.</p> | 7,431 |
<p>I know that light does not interact with other light, but can interfere it, at least its amplitude. With that said, lights frequency can be changed via bouncing off matter, where matter might absorb some of that photons energy changing the frequency.. Does this imply that a high energy beam of light that intersects an <strike>non-</strike>visible beam of light, might be enough to bounce the light around at the intersection of the two laser beams to emit visible light? By changing its frequency? Could that also be controlled by the energy output of the other laser?</p> | 7,432 |
<p>Is there an easy way to understand and/or visualize the reciprocal lattice of a two or three dimensional solid-state lattice? What is the significance of the reciprocal lattice, and why do solid state physicists express things in terms of the reciprocal lattice rather than the real-space lattice?</p> | 7,433 |
<p>For two different chemical substances there are two open valves with <a href="http://en.wikipedia.org/wiki/Flow_rate" rel="nofollow">flow rates</a> $$Q_1=a\frac{m^3}{h}\ \text{ and }\ \ Q_2=b\frac{m^3}{h},$$ leading into seperate cables. </p>
<p>Next, the cables join, the substances mix perfectly and they flow along together. </p>
<p>Then during the route, there is an observational volume $V$ (e.g. a cylinder or length $d$ and cross section $A$).</p>
<blockquote>
<p>How long is the residence time $\Delta t$ of a particle of the mixture inside the volume?</p>
</blockquote> | 7,434 |
<p>I recently found <a href="http://www.mnn.com/food/beverages/stories/why-is-it-so-hard-to-walk-and-not-spill-your-coffee-physicists-have-the-answe" rel="nofollow">this article</a>, which describes how...</p>
<blockquote>
<p>It just so happens that the human stride has almost exactly the right frequency to drive the natural oscillations of coffee, when the fluid is in a typically sized coffee mug.</p>
</blockquote>
<p><em>Judging by appearance</em>, coffee and water's fluid dynamics seem rather similar. Unfortunately,they do not site a source, but they mention that the study was done by "a pair of fluid physicists at the University of California at Santa Barbara (UCSB)".</p>
<p>My question is <strong>how different are the properties of water and coffee? Is this difference significant enough to cause a dramatic change in how either fluid behaves in a mug (while walking)?</strong></p>
<p><strong>EDIT</strong>: <a href="http://pre.aps.org/pdf/PRE/v85/i4/e046117" rel="nofollow">Here</a> is the actual published article. </p> | 7,435 |
<p>Let us assume that there are two astronauts A and B who are floating in space. A sees B passing by and vice versa. A sends signals to B every minute. According to A since B is moving his clock will be slower. So B will receive the signals prior to the appointed minute. The same argument can be applied for B who will conclude A's clock is running slow. Who is right?</p> | 7,436 |
<p>I came across this proportionality statement in my quantum mechanics notebook:
$\psi(x,t)$ is proportional to </p>
<p>$$
\begin{align}
\cos(kx - wt) &= \exp(i(kx-wt)) + \exp(-i(kx-wt)) \\
&= \exp (i(kx-wt))
\end{align}
$$
I looked through most commonly used textbooks for quantum mechanics and I couldn't find this in them. Can you help me figure this out? Thanks.</p> | 7,437 |
<p>if a lightbeam is fired from space to earth is it blueshifted? I heard this happens with special relativity but not sure if it happens with genreal relativity</p> | 7,438 |
<p>QFT says that an unlimited number of bosons can occupy the same "state" (what I mean by that is that the whole system's wavefunction is composed of a product of many identical wavefunctions).</p>
<p>However, gravity increases monotonically with energy density. It seems that at some point, one additional boson would create a high enough energy density to create a black hole. Is this true? Could I calculate the number of bosons necessary to cause this?</p> | 7,439 |
<p>This question is closely related to <a href="http://physics.stackexchange.com/q/28834/7670"><em>Event horizons without singularities</em></a> from about a year ago (May 2012), which <a href="http://physics.stackexchange.com/users/1325/john-rennie">John Rennie</a> answered nicely and persuasively.</p>
<p>My variant of the question is this: Given an <em>existing</em> large-scale black hole and associated event horizon, how does matter manage to fall through the event horizon?</p>
<p>Here's the reference thought experiment I'm using:</p>
<p>Assume two clocks communicating in both directions via radio or laser, with the observer clock kept distant from the black hole, and the falling clock heading towards the event horizon of the black hole.</p>
<p>Each clock "sees" the time units (call them seconds) of the other clock through the radio link, and can express the size of those time units in terms of its own units of time. The observer clock watches the time units of the falling clock grow quickly in length, until at the event horizon the seconds of the falling clock become infinite in length. To the observer clock, it looks as though the falling clock has become suspended in time at the event horizon, since a clock with infinitely long seconds requires an infinite time to do anything.</p>
<p>Now the standard interpretation is that since the falling clock has its own time standard, it sees nothing amiss at the event horizon. In that interpretation, the apparent "freezing" of the falling clock is more-or-less an illusion caused by the falling clock red-shifting out of communication with the rest of the universe. By that interpretation, the observer clock is viewing what amounts to a massively slowed-down recording of the moments right up to the falling clock leaving the visible universe. The remaining, um, "singular" fate of the falling clock is simply hidden from view. It is an appealing scenario, one that "feels right" for interpreting the oddities of infinite time dilation.</p>
<p>My problem is with what the falling clock sees.</p>
<p>As best I can understand it -- and my question is what I'm seeing wrongly -- the falling clock will not see the event horizon as a "no big deal" event. Instead, it will see the time flow of the observer region accelerate very quickly, so that the falling clock can observe <em>and in principle exchange data with</em> events in the very distant future of the observer universe.</p>
<p>A second rather noticeable effect will be that unless the external universe stays very dark indeed, the falling clock will be incinerated by blue shifted radiation before hitting the event horizon proper. Why that is so is not hard to see: If at some point the falling clock measures the observer clock as having seconds that are one billionth the length of its own falling clock seconds, then the frequency of any electromagnetic radiation sent to it from the observer region will also be multiplied in frequency by a billion times. Or from the observer clock perspective, the falling clock has slowed down so severely in time that it begins accumulating energy over very long periods of time.</p>
<p>My biggest problem is that if the falling clock can <em>interact</em> with the future universe, no matter how painfully, its time dilation is necessarily real and observable, and not simply a left-over recording of the last moments of its fall out of this universe. So, if the falling clock is still available to interact with an observer clock a billion years from now, then it is not truly "in the black" yet, just very, very cold and slow -- and still perched very close to, but <em>still</em> not all the way through, the event horizon.</p>
<p>This would mean that regardless of how the black hole formed -- which is a separate question, and one that John Rennie addressed nicely last time -- then once it has formed, <em>external matter and light cannot penetrate its event horizon.</em></p>
<p>So what is the deal here? Is there something fundamentally wrong with my thought experiment? How exactly does a falling clock move through a region where the seconds are infinite in length? (And one more thought in passing: Does the observer clock also appear to become more distant in space? That might help... maybe?)</p>
<p><strong>Addendum 2013-10-27</strong></p>
<p>Here's the most succinct version of the question I can come up with:</p>
<blockquote>
<p>What is the mathematical procedure for calculating the last distant-clock time tag that the falling clock sees as it approaches the event horizon?</p>
</blockquote>
<p>The above version keeps the calculation firmly embedded in the time system of the falling clock, avoiding the dangers in statements such as "time flows normally for the falling clock." That assertion is patently true, but since it does not calculate the last time tag seen, it does not answer the question.</p>
<p><a href="http://physics.stackexchange.com/users/3099/twistor59">@twistor95</a> left a nice, highly relevant reference to an <a href="http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html">online article by John Baez</a> on why most physicists now think that a clock falling into a black hole will <em>not</em> see the end of the universe. <a href="http://physics.stackexchange.com/users/2451/qmechanic">@Qmechanic</a> noted an <a href="http://physics.stackexchange.com/q/21319/2451">an earlier Physics SE question</a> that to be honest I think is the same as mine. (I really did look, but Qmechanic is a lot better at such searches.) What's worrisome is that the answer to that question was the older physics view that the falling clock <em>would</em> see the end of the universe!</p>
<p>I must 'fess up that I exchanged a few emails with John Baez on this topic years before he wrote that piece. What left me baffled at that time was a subtle switch in whose time standard was used. So, in my current variant of this question, I tried as carefully as I could to ask the question in terms of predicting the last time stamp the falling clock would receive. This phrasing shifts focus from "Will he see the end of the universe?" (Baez: no) to what the math predicts. Prediction is, after all, the very essence of what good scientific theory is all about.</p>
<p>The old answer was that the cutoff occurs at infinity, that is, at the end of the external universe. If you accept that cutoff for a clock that is merely <em>approaching</em> the event horizon and has not yet fallen through it, then the idea that you can fall through an event horizon becomes very problematic indeed. See for example <a href="http://physics.stackexchange.com/users/1186/anixx">@Anixx</a>'s <a href="http://physics.stackexchange.com/a/21357/7670">accepted "collapsar" answer</a> for the older version of this question.</p>
<p>So, since the current answer is that the falling clock does <em>not</em> see the end of the universe, the visibility cutoff must necessarily occur for a tag that is well short of the end of the external universe. You cannot assert the one ("no end seen") without implying the other ("some tag will be the last one seen").</p>
<p>So again: If "no end is seen" is the answer, how is the implied final tag calculated?</p>
<p>I will be blunt on one point: As someone with an information technologies background, I see no strong reason to view either the old or the new answer as more persuasive. Untestable code, whether mathematical or programmatic, is always in danger of errors.</p>
<p><strong>Addendum 2013-10-30</strong></p>
<p>My question has been very nicely answered (no end-of-universe is seen!) in this new (2013-10-29) question asked by <a href="http://physics.stackexchange.com/users/1325/john-rennie">John Rennie</a>:</p>
<blockquote>
<p><a href="http://physics.stackexchange.com/q/82678/7670">Does someone falling into a black hole see the end of the universe?</a></p>
</blockquote>
<p><a href="http://physics.stackexchange.com/users/17320/michael-brown">Michael Brown</a> provided the answer, and John Rennie then iced the cake by providing an additional diagram that shows the actual intersection of the outside time stamp with the falling clock. Beautiful and elegant stuff!</p>
<p>Alas, though, it also means I don't have an answer to check here. @MichaelBrown, if you happen see this and don't mind adding in a link to your other answer as an answer here, I'd be happy to flag your link to close out this question.</p> | 7,440 |
<p>I understand that in general if we're adding more planes of atoms (increasing thickness of sample) then the intensity would increase because we have more constructive interference. But isn't there a breaking point for this? Shouldn't there be a finite thickness past which the intensity decreases? </p> | 7,441 |
<p>In the usual covariant approach to the development of surface charges of an asymptotic symmetry group, one works with the linearized theory as this ensures that the charges are integrable.</p>
<p>I also read </p>
<blockquote>
<p>In the full interacting theory with prescribed asymptotics, the expressions are one-forms on solution space indexed by asymptotic symmetries and one has to face the question whether these one-forms are integrable, i.e. whether one can construct suitable "Hamiltonians" for them.</p>
</blockquote>
<p>What is the meaning of integrable here? </p> | 7,442 |
<p>This question is motivated by <a href="http://physics.stackexchange.com/questions/71947/does-a-charged-particle-revolving-around-another-charged-particle-radiate-energy">similar one.</a>
If an accelerated point charge $q$ radiates with power $W$ then I assume the same particle with charge $-q$ will radiate with the same rate $W$. Now what if we make a dipole with these two charges and accelerate it with the same acceleration? What will be the radiation power? </p> | 7,443 |
<p>Well I imagine that there is a very fast rocket moving alongside the Earth and I know that people in the Earth regard the events on the rocket as happening more slowly as it is moving and time expands, and the same for people in the rocket as they could consider themselves at rest and the Earth moving so they would regard events in Earth as happening more slowly, my question arises in the moment they stop moving with respect to each other.</p>
<p>Lets say there were two identical twins, one on the rocket and one on Earth, for the twin in the rocket he would experience as if he was getting older quicker than his brother on earth as he regards the events there as happening more slowly and the same for the other twin, at the moment they stop moving, which one would be older? If no one, how is this time difference compensated just by stoping abruptly. I mean one clearly experienced how was he getting older quicker than his relative moving sibling, if we kept this motion long enough one could get to old age while seeing his brother stay younger, what happens <em>after</em> we stop this motion?</p> | 249 |
<p>I'm trying to get Ward-Takahashi identities using the approach used in <a href="http://books.google.com.br/books?id=nnuW_kVJ500C&pg=PA264&lpg=PA264&dq=%22leads%20to%20a%20differential%20equation%20for%20Z%22&source=bl&ots=vpxudGVM4Y&sig=b9ByfCjLwU9oT9wMXIInGj7z8vk&hl=pt-BR#v=onepage&q=%22leads%20to%20a%20differential%20equation%20for%20Z%22&f=false" rel="nofollow">Ryder's book (pages 263-266)</a>. I like that he starts from demanding gauge invariance of Z in a explicit way and them explores the consequences of that to functional generators of vertex functions. But the actual calculation is bugging me out. </p>
<p>The author seems oblivious to the fact that the fermionic fields and sources ($\psi$ and $\eta$) are Grassmann variables and keeps commuting them out with no regard for my sanity. For instance, equation (7.102) has a term:</p>
<p>$$i e (\bar{\eta}\psi-\bar{\psi}\eta) $$</p>
<p>that promply becomes (exchanging the fields by ${1\over i}$ times the derivatives on the sources, acting on Z to the right):</p>
<p>$$ e (\bar{\eta}{\delta \over \delta \bar{\eta} }-\eta{\delta \over \delta \eta }) $$</p>
<p>In my opinion that should be:</p>
<p>$$ e (\bar{\eta}{\delta \over \delta \bar{\eta} }+\eta{\delta \over \delta \eta }) $$</p>
<p>This one has no consequences because he commutes them again right after. But when I try to calculate it being careful with the Grassmann variables, I can never get the right signs in (7.111). I'm specially troubled by the derivatives below (this is what I'm getting, but one of them should have a different sign in order to get the right WT identities):</p>
<p>$${\delta \over \delta \bar{\psi}(x_1) } {\delta \over \delta \psi(y_1) }
\left[{\delta \Gamma \over \delta \psi(x) }\psi(x)\right]_{\psi=\bar{\psi}=0}=-\delta^4(x-y_1){\delta^2 \Gamma \over \delta \bar{\psi}(x_1) \delta\psi(x) }$$</p>
<p>$${\delta \over \delta \bar{\psi}(x_1) } {\delta \over \delta \psi(y_1) }
\left[\bar{\psi}(x) {\delta \Gamma \over \delta \bar{\psi}(x) } \right]_{\psi=\bar{\psi}=0}
=\\=-\delta^4(x-x_1){\delta^2 \Gamma \over \delta \psi(y_1) \delta\bar{\psi}(x) } = \delta^4(x-x_1){\delta^2 \Gamma \over \delta \bar{\psi}(x) \delta\psi(y_1) }$$</p>
<p>Does anybody ever did this calculation in detail and has some pointers? Are there any other references that follow this same approach?</p>
<p>EDIT: Just a shameless bump: I still looking for some light on this. Any reference on where this is done in detail would help.</p> | 7,444 |
<p>I guess that these questions were being asked by many people on the Northern Hemisphere during this summer (and other summers) and someone may give a nice, coherent answer. The general question is how many times more slowly one is getting suntan or damages his skin in the evening, relatively to the noon?</p>
<p>The Sun altitude (solar elevation angle) $\alpha_s$ apparently makes the ozone layer etc. look $1/\sin\alpha_s$ times thicker than when the Sun is directly above our head. Clearly, this makes the solar UV radiation weaker if we're further from the noon. So</p>
<ul>
<li>is that right that a particular spectral line gets weakened by the factor of $\exp(-C_\lambda/ \sin\alpha_s)$?</li>
<li>is that true that the changes to the UV-B radiation are the most important ones because UV-A is almost completely transmitted and UV-C is almost completely blocked?</li>
<li>because 98% of the UV radiation is said to be absorbed by the atmosphere, one would expect that the exponential reduction above will be dramatic. However, sources suggest that the "total amount" of UV radiation is only suppressed by a power law, probably by $1/\sin^2\alpha_s$ (10% thicker atmosphere implies 20% less radiation). Where does it come from? Is it from some integration of the exponentially suppressed function over frequencies? What is the approximate integrand and how does the decreasing exponential become a power law?</li>
<li>is it OK to assume that all transmitted UV rays cause suntan and potential diseases at the same rate, I mean that the ratio of "suntan vs harm" obtained by a photon is constant, or is it true that softer UV rays are giving us suntan with less harm to the skin? That would imply that it's healthier to get suntan in the evening</li>
<li>above, it was assumed that the solar photons travel straight from the Sun. But does Rayleigh scattering matter here? UV photons could go via the shortest path through the atmosphere (velocity orthogonal to the ozone layer) and then reflect to our skin via Rayleigh scattering – in this way, they would effectively see the minimally thin ozone layer. Rayleigh scattering is probably substantial for UV radiation, isn't it? In this way, one could explain why the Sun seems more powerful in the evening than the exponentially decreasing formula suggests. In the evening, one could still be getting suntanned from "all directions" of the sky (all places where it's "blue").</li>
</ul>
<p>Sorry for this mixture of facts, questions, and half-baked hypotheses. Please fix the claims that are incorrect. There seem to be many related questions above but I would really like to get some usable "rate of getting suntan" as a function of the Sun altitude.</p> | 7,445 |
<p>A very rough approximation from first principles, from the elementary charge and hbar, would suffice. But is there such an approximation at all? </p>
<p>(Alternatively, if water is too difficult: is there any other material or gas for which such a calculation is possible?)</p> | 7,446 |
<p>I have asked the question <a href="http://math.stackexchange.com/q/188313/28724">in math.stackexchange</a>, but perhaps it should be more relevant here. Hence I am re-posting it with necessary reediting.</p>
<p>Let $\mathcal{H}=\mathbb{C}^n$ be a Hilbert space. A state $\rho\in\mathcal{B(H)}$ is a positive semi-definite operator with unit trace. $\rho\in \mathcal{B(H)}$, where $\mathcal{H}=\mathcal{H}_1\otimes\mathcal{H}_2=\mathbb{C}^{n_1}\otimes\mathbb{C}^{n_2}$, is called entangled, if it can not be written as convex sum of one dimensional projections - like $P_x\otimes P_y$, where $|x\rangle\in\mathcal{H}_1$ and $|y\rangle\in\mathcal{H}_2$.</p>
<p>In a similar spirit can we define entanglement in the symmetric space $\mathcal{H}\bigvee\mathcal{H}$ and antisymmetric space $\mathcal{H}\bigwedge\mathcal{H}$? </p>
<p>As you have already understood, I am looking for entanglement in the indistinguishable particles. Hence the above definition for the entanglement of distinguishable particles does not work. A quick googling gave me a few papers, which refer to different definitions (for multipartite systems, and some are based on certain entropic conditions). Hence I want to know whether there is mathematical definitions for entanglement in such systems which is closest to the spirit of the definition mentioned above (for distinguishable particle). Advanced thanks for any help in this direction.</p> | 7,447 |
<p>On page 92, my still favorite supersymmetry <a href="http://rads.stackoverflow.com/amzn/click/0071636412">book</a> says, by making the global infinitisimal parameter of a SUSY tranformation spacetitime dependent (gauging) it forces one to introduce a new gauge field which turns out to have the properties of a graviton and one obtains supergravity. This is analogue to obtaining electromagnetism from gauging the global $U(1)$ symmetry of the Dirac Lagrangian.</p>
<p>This approach seems to me a nice way to learn more about supergravity, but unfortunately it was only mentioned as an aside comment in that book without explaining it any further. So can somebody explain (or outline) in a bit more detail how this works?</p> | 7,448 |
<p>I would like to understand better a phenomenon of a quantum heat bath. </p>
<p>Below I present one example, which seems quite clear to me. It would be great to see some less-discrete models, and more specific open quantum systems (with a given concrete Hilbert space). </p>
<p>(Discrete Picture)</p>
<p>Let $S$ be an open quantum system (e.g. an atom), $\mathcal{H}_{S}$ be a Hilbert space of $S$ and $\mathrm{H}_S$ be a Hamiltonian of $S$.
Let $\bigotimes_{k \geq 0} \mathbb{C}^{N}$ of copies of $\mathbb{C}^{N}$, where $N\geq 2$ is a fixed integer be the state space for the reservoir, that is a heat bath (e.g., radiation).
Let $\{e_i\}_{i = 0}^{N}$ be an orthonormal basis of $\mathbb{C}^{N}$ such that the countable tensor product is taken with respect to the ground state $e_0$. Single $\mathbb{C}^{N}$ is a state space for single photon, let $\mathrm{H}_{p}$ stands for a Hamiltonian of a single photon. When the system and a photon are interacting, we consider the state space $\mathcal{H}_{S} \otimes\mathbb{C}^{N} $ with a interaction Hamiltonian $\mathrm{H}_{I}$. </p>
<p>First question how can we determine the interaction Hamiltonian? How can we describe this interaction?</p>
<p>Now, consider the state of each photon to be given by a density matrix $\rho_{\beta}$ which
is a function of $\mathrm{H}_{p}$, e.g., a Gibbs state at inverse temperature $\beta= \frac{1}{T}$:
$$ \rho_{\beta}= \frac{1}{\mathrm{Tr}(e^{-\beta\mathrm{H}_{p}})}e^{-\beta\mathrm{H}_{p}}.$$
Denote the Total Hamiltonain by $\mathrm{H}_{T}$, that is, $\mathrm{H}_{T}(n)=\mathrm{H}_{S} \otimes I_{\mathbb{C}^{N}} + I_{\mathcal{H}_S} \otimes \mathrm{H}_{p} + \mathrm{H}_{I}(n) $.</p>
<p>The system $S$ is first in contact with the first photon only
and they interact together according to the above Hamiltonian $\mathrm{H}_{T}$. This lasts for
a time length $n$ . The system $S$ then stops interacting with the first photon and
starts interacting with the second photon only. This second interaction is directed
by the same Hamiltonian $\mathrm{H}_{T}$ on the corresponding spaces and it lasts for the same duration $n$ , and so on...</p>
<p>So we can see how the quantum system $S$ interacts with a heat bath. References - Attal and Joye (2007, <em>J. Funct. Anal.</em> <strong>247</strong>, 253--288).<a href="http://arxiv.org/pdf/math-ph/0612055v1.pdf" rel="nofollow">http://arxiv.org/pdf/math-ph/0612055v1.pdf</a></p>
<p>Informally speaking, $S$ is a `small' system and as we will go with time to $+\infty$ we should obtain a 'large' system after the interaction with a heat bath.</p>
<p>Some quantum noises should appear as well, if they were be `squeezed' how would we interpret it ? I meant by this that the driving noises form Araki-Woods representations of CCR in a squeezed (quasifree) state. I would like to know more about the physical interpretations of this phenomenon.
I would be also grateful if someone could show me some proper pictures presenting that or similar situation.</p>
<p>Thank you very much for any remarks and answers. </p> | 7,449 |
<p>We've got two long straight wires carrying current of 5A and placed along x and y axis respectively current flows in direction of positive axes we have to find magnetic field at a) (1 m,1 m)
b) (-1 m,1 m)
c) (-1 m,-1 m)
d) (1 m,-1 m)
now using ampere's law $$ \oint_{}^{}B.dl=\mu_oi $$ i found magnetic field for infinite wire be $$ B=\frac{\mu_oi}{2\pi r}$$ where r is distance from the wire but for evaluating the magnetic field we need to know direction of it and thus add it like vector but finding direction of magnetic feild at these point is confusing me , how do i find proper direction of magnetic field ?</p> | 7,450 |
<ol>
<li><p>If you were to draw a person on a Minkowski diagram, that was unmoving (though time was still passing) what would that look like?
(As a light cone and ignore the observer.)
<img src="http://i.stack.imgur.com/3y46m.jpg" alt="Minkowski diagram"></p></li>
<li><p>Also what would it look like if you drew a person looking into a black hole (at a safe distance) but NOT accelerating away? (also as a light cone)</p></li>
</ol> | 7,451 |
<p>I'm trying to understand Noether's theorem, and it's application to gauge symmetry. Below what I've done so far. </p>
<p>First, the global gauge symmetry. I'm starting with the Lagragian
$$L_{1}=\partial^{\mu}\Psi\partial_{\mu}\Psi^{\ast}-m^{2}\left|\Psi\right|^{2}$$
with classical complex fields. This Lagragian is invariant with respect to the global gauge symmetry $\Psi\rightarrow\tilde{\Psi}=e^{\mathbf{i}\theta}\Psi$, ... such that I end up with
$$\delta S=\int dv\left[\dfrac{\delta L_{1}}{\delta\Psi}\delta\Psi+\dfrac{\delta L_{1}}{\delta\Psi^{\ast}}\delta\Psi^{\ast}+\mathbf{i}\left(\Psi\partial^{\mu}\Psi^{\ast}-\Psi^{\ast}\partial^{\mu}\Psi\right)\partial_{\mu}\delta\theta\right]=\int dv\left[\partial_{\mu}j^{\mu}\right]\delta\theta$$
provided the equations of motion ($\delta L / \delta \Psi = 0$, ...) are valid. All along I'm using that
$$\dfrac{\delta L}{\delta\phi}=\dfrac{\partial L}{\partial\phi}-\partial_{\mu}\dfrac{\partial L}{\partial\left[\partial_{\mu}\phi\right]}$$
and that $\int dv=\int d^{3}xdt$ for short. The conserved current is of course
$$j_{1}^{\mu}=\mathbf{i}\left(\Psi^{\ast}\partial^{\mu}\Psi-\Psi\partial^{\mu}\Psi^{\ast}\right)$$
since $\delta S / \delta \theta =0 \Rightarrow\partial_{\mu}j_{1}^{\mu}=0$. </p>
<p><strong>Here is my first question:</strong> Is this really the demonstration for conservation of charge ? Up to now, it seems to me that I only demonstrated that the particle number is conserved, there is no charge for the moment... </p>
<p>Then, I switch to the local gauge symmetry. I'm starting with the following Lagrangian
$$L_{2}=\left(\partial^{\mu}+\mathbf{i}qA^{\mu}\right)\Psi\left(\partial_{\mu}-\mathbf{i}qA_{\mu}\right)\Psi^{\ast} -m^{2}\left|\Psi\right|^{2} -\dfrac{F_{\mu\nu}F^{\mu\nu}}{4}$$
with $F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$. This Lagrangian is invariant with respect to the local gauge transformation
$$L_{2}\left[\tilde{\Psi}=e^{\mathbf{i}q\varphi\left(x\right)}\Psi\left(x\right),\tilde{\Psi}^{\ast}=e^{-\mathbf{i}q\varphi\left(x\right)}\Psi^{\ast},\tilde{A}_{\mu}=A_{\mu}-\partial_{\mu}\varphi\right]=L_{2}\left[\Psi,\Psi^{\ast},A_{\mu}\right]$$</p>
<p>Then I have
$$\delta S=\int dv\left[\dfrac{\delta L_{2}}{\delta\Psi}\delta\Psi+\dfrac{\delta L_{2}}{\delta\Psi^{\ast}}\delta\Psi^{\ast}+\dfrac{\delta L_{2}}{\delta A_{\mu}}\delta A_{\mu}\right]$$
with $\delta\Psi=\mathbf{i}q\Psi\delta\varphi$, $\delta A_{\mu}=-\partial_{\mu}\delta\varphi$, ... such that I end up with
$$\dfrac{\delta S}{\delta\varphi}=\int dv\left[\mathbf{i}q\Psi\dfrac{\delta L_{2}}{\delta\Psi}+c.c.+\partial_{\mu}\left[j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu}\right]\right]$$
with $j_{2}^{\mu}=\partial L_{2}/\partial A_{\mu}$ and $F^{\nu\mu}=\partial L_{2}/\partial\left[\partial_{\nu}A_{\mu}\right]$</p>
<p>Then, by application of the equations of motion, I have
$$\partial_{\mu}\left[j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu}\right]=0\Rightarrow\partial_{\mu}j_{2}^{\mu}=0$$
since $\partial_{\mu}\partial_{\nu}F^{\nu\mu}=0$ by construction. Of course the new current is
$$j_{2}^{\mu}=\mathbf{i}q\left(\Psi^{\ast}\left(\partial^{\mu}+\mathbf{i}qA^{\mu}\right)\Psi-\Psi\left(\partial^{\mu}-\mathbf{i}qA^{\mu}\right)\Psi^{\ast}\right)$$
and is explicitly dependent on the charge. So it seems to me this one is a better candidate for the conservation of charge.</p>
<p>NB: As remarked in <a href="http://arxiv.org/abs/hep-th/0009058" rel="nofollow">http://arxiv.org/abs/hep-th/0009058</a>, Eq.(27) one can also suppose the Maxwell's equations to be valid ($j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu} = 0$, since they are also part of the equation of motion after all, I'll come later to this point, which sounds weird to me), and we end up with the same current, once again conserved.</p>
<p>Nevertheless, I still have some troubles. Indeed, if I abruptly calculate the equations of motions from the Lagrangian, I end up with (for the $A_{\mu}$ equation of motion)
$$j_{2}^{\mu}-\partial_{\nu}F^{\nu\mu}\Rightarrow\partial_{\mu}j_{2}^{\mu}=0$$
by definition of the $F^{\mu \nu}$ tensor. </p>
<p>So, my <strong>other questions</strong>: Is there a better way to show the conservation of EM charge ? Is there something wrong with what I did so far ? Why the Noether theorem does not seem to give me something which are not in the equations of motions ? said differently: Why should I use the Noether machinery for something which is intrinsically implemented in the Lagrangian, and thus in the equations of motion for the independent fields ? (Is it because my Lagrangian is too simple ? Is it due to the multiple boundary terms I cancel ?)</p>
<p>Thanks in advance.</p>
<p>PS: I've the feeling that part of the answer would be in the difference between what high-energy physicists call "on-shell" and "off-shell" structure. So far, I never understood the difference. That's should be my last question today :-)</p> | 7,452 |
<p>Assuming it's possible to vibrate a human at near light speed without harming him, would a few minutes of this from his point of view be much longer from a stationary observer's point of view?</p>
<p>In other words do vibrations work the same as normal movement with regards to time dilation?</p>
<p>So a person could walk into such a machine, and walk out hundreds of years in the future, even though a much smaller amount of time would have passed from their perspective?</p> | 7,453 |
<p>Question:-</p>
<blockquote>
<p>When a beam of $1.06eV$ photon of
intensity = $2.0 W/m^2$, falls on a platinum surface of area $1.0*10^{-4}m^2$, and work function $5.6eV$, $0.53$% of incident photons ejected photo electrons. Find the number of photoelectrons emitted per second and their minim and maximum energies</p>
</blockquote>
<p>the solution in my text book is as followed</p>
<blockquote>
<p>Number of photoelectron's emitted per second</p>
</blockquote>
<p>=$\frac{(2.0)(1.0*10^{-4})}{(10.6*1.6*10^{-19})}*\frac{0.53}{100} = 6.25*10^{11}$</p>
<p>but i am not able to understand what formula has been used here can any one please help me even the formula used here will also help me </p>
<p>Thank's </p>
<p>Akash</p> | 7,454 |
<p>What is <a href="http://en.wikipedia.org/wiki/Viscosity#Bulk_viscosity" rel="nofollow">bulk viscosity</a> and how does it affect the flow?</p>
<p>Explain the idea of introducing such a term in the Navier-Stokes equation. </p>
<p>What are the consequences if not taken into account?</p> | 7,455 |
<p>Can anyone clarify what does mean the angular frequency of a system in case of the vibrating membrane. </p>
<p>Angular frequency is measured in radians per second, what does this have with the vertical displacement in case of the wave equation.</p> | 7,456 |
<p>For example, if you have two hydrogen atoms and an oxygen atom, they are all electrically neutral and don't attract each other. But then if they manage to get "close enough" somehow they snap together releasing energy, and to get them apart again requires energy be input into the system. </p>
<p>This says that the bound state is a 'lower energy' state than the unbound state, and this is my question: </p>
<blockquote>
<p>From a Quantum Mechanical point of view, where does this energy come from that is released when the bond is formed? </p>
</blockquote>
<p>I've done some reading and it has something to do with filling electron shells and electron probability wavefunctions spreading between atoms, and something about the Virial Theorom where the kinetic energy of the electron is reduced as it's range of area is increased. </p>
<p>But that doesn't make sense to me, so I was wondering if someone could explain where the bond energy released comes from in terms of the electron wavefunctions? Assume high school physics/freshman maths.</p> | 7,457 |
<p>I use a bike to commute, so I spend a lot of time thinking about how to get the most bang out of my momentum.</p>
<p>Aside from the extra distance traveled in a wide turn, does making a sharp turn save you any energy? My guess is no, because these things tend to even out, but it definitely <em>feels</em> like I'm going much faster. I've even considered that maybe taking a turn sharp is worse, because the extra pressure will cause more friction in the bearings and the tires.</p>
<p>Either way it's more fun, though that probably doesn't get a term in the equations.</p> | 7,458 |
<p>Do quantum computer's tell us anything about the foundations of quantum theory? In particular Shor argued in the famous thread by 't Hooft </p>
<p><a href="http://physics.stackexchange.com/questions/34217/why-do-people-categorically-dismiss-some-simple-quantum-models">Why do people categorically dismiss some simple quantum models?</a></p>
<p>that quantum computation was at odds with 't Hooft's ideas. </p>
<p>Does quantum computation tell us anything new about hidden variables like <a href="http://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory" rel="nofollow">Bohmian mechanics</a> (which, at least so far, is 100% in agreement with everything we know about physics, contrary to what some people (e.g. Motl) claim)? </p> | 7,459 |
<p>According to numerous sources online, the percentage of oxygen is approximately the same at sea level and 10,000 meters. Since oxygen is heavier than nitrogen, shouldn't the percentage of oxygen decrease with altitude?</p> | 7,460 |
<p>If I had a 3V battery, and when no load connected it reads 3.2V, and with a load 2.8V (just a hypothetical example), what is the name for these two terms, with a load or no load?</p>
<p>I know the voltage drop occurs due to its internal resistance when a load is connected, however I still am not sure of the terms. Here's a few which may be of help:</p>
<ul>
<li><p>emf ($\mathcal{E}$)</p></li>
<li><p>Potential difference</p></li>
<li><p>Terminal voltage</p></li>
</ul>
<p>Any help would be appreciated in what these terms are called, this is one of those questions where it almost seems too simple to find this answer online.</p> | 7,461 |
<p>I remember reading about an experiment where fine rods of tungsten were super-heated with millions of amps of electricity, melting them into ionised gas and were then compressed (by magnetic fields?) into plasma.</p>
<p>The plasma heated up to temperatures never before reached. I can't remember exactly, but I believe it was a few billion degrees fahrenheit.</p>
<p>It was a number of years ago - and I can't find the report via a search engine.</p>
<p>Is there a limit to the temperature of plasma? What's the current highest recorded temperature of plasma? Is it hotter than nuclear reactions?</p> | 7,462 |
<p>What is the difference between air resistance/friction force and up thrust force?
I always think that air friction is the same as up thrust but it does not seem that way.</p> | 7,463 |
<p>I did see some posts on stackexchange on this matter, but I find them to be beyond my scope or not directly related to what I am looking for.</p>
<p>I am reading Feynman Lectures III, chapter 4. It talks about scattering of two particles $a$ and $b$ and it defines $f(\theta)$ to be the probability amplitude of particle $a$ scattering at angle $\theta$. Then it says </p>
<p>"You might also think that the amplitude for the second process (where particle $b$ instead enters the detector placed at angle $\theta$) is just $f(\pi-\theta)$. But that is not necessarily so, because there could be an arbitrary phase factor."</p>
<p>I'm not questioning why things behave the way they are in quantum mechanics. But, isn't it <strong><em>by definition of $f$</em></strong> that the probability amplitude of particle $a$ scattering at angle $\pi-\theta$ is $f(\pi-\theta)$? I don't see why there is a phase factor. If there's a phase factor, it should already be part of $f$.</p> | 7,464 |
<p>How can the transformer transfer exact power from primary winding to secondary winding if there are core losses, eddy current losses and hysteresis losses?</p> | 7,465 |
<p>So I've had a problem for a long time now understanding energy-mass equivalence, in particular I've had a lot of trouble understanding how something like interatomic potentials can be seen as mass. I've heard other people on the site say it depends on how you probe the system, which doesn't make too much sense to me, I don't see why mass should change based on 'what level you view the system at'. Can someone help me understand mass-energy equivalency? </p> | 7,466 |
<p>The Rarita-Schwinger action in curved $n$-dimensional spacetime is</p>
<p>$$
\int \sqrt{g} \overline{\psi}_a \gamma^{abc} D_b \psi_c
$$</p>
<p>Here $g = \det(g_{\mu \nu})$, and the indices $a, b \dots$ are 'internal' indices that transform under e.g. $\mathrm{SO} (3,1) $ in $3+1$ dimensions. $\gamma^{abc} = \gamma^{[a} \gamma^{b} \gamma^{c]}$ with the gamma matrices obeying $\gamma^a \gamma^b + \gamma^b \gamma^a = 2 \eta^{ab} $, and $\eta^{ab}=\mathrm{diag}(1,1 \ldots 1,-1,-1 \ldots -1)$ is the 'internal metric'. $\psi_{\mu} = \psi_{c} e^{c}_{\mu} $ is a spinor-valued one form. Spacetime indices $\mu, \nu$ can be 'converted' to internal indices using the frame field $e_a^{\mu}$, and vice versa. The covariant derivative is $D_{\mu} \psi_{\nu} =\partial_{\mu} \psi_{\nu} + \frac{1}{4} \omega_{\mu}^{ab} \gamma_{ab} \psi_{\nu} $. Here $\omega$ is taken to be the torsion free spin connection, and $\gamma^{ab} = \gamma^{[a} \gamma^{b]}$.</p>
<p>In flat space, the covariant derivative becomes a normal derivative, and the action then has a symmetry $\psi_c \rightarrow \psi_c + \partial_c \phi $, with $\phi$ an arbitrary function. This freedom can be used to eliminate some of the degrees of freedom from the field $\psi_c$ which correspond to lower spin. However, in curved space there is no corresponding symmetry under $\psi_c \rightarrow \psi_c + D_c \phi$. For this reason, it is said that the Rarita-Schwinger action in curved spacetime is inconsistent. My question is, what goes wrong when you don't have this extra symmetry? And do the problems manifest at the classical level or only at the quantum level?</p> | 7,467 |
<p>A moving star's relativistic mass is larger than its rest mass. Is its gravitational pull larger? </p>
<p>What about its inertial mass? Does it have larger inertial mass, keeping in mind that inertial mass should equal gravitational mass according to the equivalence principle.</p>
<p>I suppose this quote from wikipedia is relavant:</p>
<blockquote>
<p>However, it turns out that it is impossible to find an objective
general definition for the concept of invariant mass in general
relativity. At the core of the problem is the non-linearity of the
Einstein field equations, which makes it impossible to write the
gravitational field energy as part of the Stress–energy tensor in a
way that is invariant for all observers. For a given observer, this
can be achieved by the Stress–energy–momentum pseudotensor.[21]</p>
</blockquote> | 153 |
<p><strong>Note:</strong> Both diagrams are 2d(they might look 3d because I drew them manually.)
I have a prismatic frame that looks like this:<img src="http://i.stack.imgur.com/obZR5.png" alt="Frame"><br>
I made a section on the half of the first pillar(the red line represents the section), to study the internal forces in it(Normal force, shear force and moment.)
Here is a free body diagram at this section:<br>
<img src="http://i.stack.imgur.com/YtO2N.png" alt="Section"><br>
A is a fixed support so I calculated its reactions as follows:</p>
<ul>
<li> $X_A= 100 KN$
<li> $Y_A= 200 KN$
<li> $M_A= 300 KN$
</ul>
<p>Note: $AB=3m$<br>
I have also calculated the normal and the shear force at B, but I have a problem calculating the moment at B and here is why:<br>
now let's apply the equilibrium equation of the moments at B, as: </p>
<p>$$\Sigma M_B=0$$ </p>
<p>As you see there are the following moments: </p>
<ul>
<li> $X_A\times 3$
<li> $M_B$
<li> and at last $M_A$
</ul>
<p>I have done it like this:<br>
$$ 3X_A + M_B - M_A = 0 $$
I just need to know what to do with $M_A$ should I multiply it with the distance $AB$ or just leave it.(I tried both cases and I think both are wrong). now can somebody till me what to do with the $M_A$?</p>
<h2>Edit:</h2>
<p>I guess the diagrams are somehow misleading,(I am not good at drawing!).
The axis of moment B is the straight line perpendicular to the plane and passing through B and the same for A. Now please look at the first picture: imagine there is a force just above the red section, this force is causing the bending moment in B and the reaction moment in A. I want to calculate the internal moment B, and that's why I need the sum of moments at B.</p> | 7,468 |
<p>I read that as the air density is lower at high altitudes, the scattering is less, hence the darker blue but then other colors are scattering lesser than blue, why don't we see a darker shade of the other colors?</p> | 7,469 |
<p>Is there any open database of elemental sensitivity factors for Auger electrons?</p>
<p>I'm trying to figure out graphite oxidation degree from AES spectra, so relative sensitivity of O to C is required. Spectra were registered with the primary electron energy of 1500 eV. I found some reference data in Perkin Elmer AES handbook and this paper [1], but they are contradictive ( O/C = 0.37 @AES handbook; O/C = 0.26 @[1] with 3 keV) and presents quite distant points from my excitation conditions (1, 3, 5, 10 keV @handbook and 3, 5, 10 keV @[1] vs 1.5 keV of mine).</p>
<p>Experimental data would be preferable but if there aren't any, link to theoretical calculation software will be helpful too!</p>
<p>[1]: Calculated Auger yields and sensitivity factors for K L L–N O O transitions with 1–10 kV primary beams <a href="http://www.materialinterface.com/_files/Calculated%20AES%20yields%20Matl%20Interface.pdf" rel="nofollow">http://www.materialinterface.com/_files/Calculated%20AES%20yields%20Matl%20Interface.pdf</a></p> | 7,470 |
<p><a href="http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/Special_relativity_rel_sim/index.html" rel="nofollow">http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/Special_relativity_rel_sim/index.html</a></p>
<p>Talking about the situation of clocks shown on this page. Clocks A&B. Now suppose clock B is moving towards A with constant velocity. A sees B coming towards it, and figures out that the clock B is running slow, and by symmetry B says the same. Now, B & A meet at a space-time and actually compare readings. Now they will know one of the clocks ran slower. How is this possible</p> | 7,471 |
<p>I am looking at a manifold of dimension $n$ (And I am considering a local co-ordinates system $x_1,x_2,\ldots x_n$) and the metric defined by the components $g_{ij} = \frac{\delta_{ij}}{x_1^2}$. I'm wanting to find the components of the corresponding Riemann tensor. This should be a pretty straight forward task but I cannot seem to match my workings to the actual answer.</p>
<p>Of course $g^{ij} = x_1^2\delta^{ij}$ and I have that $R^l_{ijk} = \frac{\partial}{\partial x^j} \Gamma^l_{ik} - \frac{\partial}{\partial x^k} \Gamma^l_{ij} + \Gamma^s_{ik}\Gamma^l_{sj} - \Gamma^s_{ij}\Gamma^l_{sk}$</p>
<p>And I also have derived that $\Gamma^l_{ij} = \frac{1}{2}g^{lm}(g_{im,j}+g_{jk,i}-g_{ij,m})$</p>
<p>So I've done the hard stuff, and it should just be an easy calculation to get the the components of the Riemann tensor. But I can't get my answer to match.</p>
<p>Could someone please help me with the calculations please?</p> | 7,472 |
<p>Refraction gives rise to a momentum change orthogonal to the propagation direction. This must result in an equal and opposite change to the medium at the boundary. Entry and exit cancel, and the medium experiences no net force. The problem is that the forces on the medium do not in general line up, so we have a pair of forces, opposite but parallel - i.e. a couple. Can we therefore conclude that refraction imparts a couple to a refracting medium? (and if so, has the experiment been done?).</p>
<p>Assuming this couple indeed exists: Since the light loses no energy and loses no momentum, and yet the medium appears to have gained rotational energy and momentum, where does the medium's extra angular momentum come from?</p> | 7,473 |
<p>For a significant part of my life I have been taught that, if a photon of the "correct" energy meets an excited atom, the atom will then (with a certain probability) undergo transition to a lower energy state, emitting a photon, which
1)has the same energy as the incident photon,
2)propagates in the same direction,
3)and is in phase with that incident photon.</p>
<p>Unfortunately, this statement is rarely, almost never, followed by a discussion about reasons behind the stimulated emission being the way it is. After some searching I have even come up with a few texts claiming that the description above is "not exactly true", meaning that some variation in phase and energy of the stimulated photons is possible. Again followed by no explanation whatsoever. </p>
<p>Currently the most exhaustive discussion I have found is this: <a href="http://www.sjsu.edu/faculty/watkins/stimem.htm" rel="nofollow">http://www.sjsu.edu/faculty/watkins/stimem.htm</a> which is still far from perfect. At least, now I am somewhat convinced, that the two photons should propagate in the same direction. For those not willing to follow that link, the reasoning goes like this:
Take two principles (symmetry of physical phenomena under time inversion, and the idea that physical phenomena should be isotropic in an isotropic medium) and apply them to all the variations of the stimulated emission you can think of:
1)the stimulated photon propagates in a random direction. Under time inversion this becomes simple absorption, where one of the spatial directions (in which the former incident photon is propagating) is undesirable;
2)the stimulated photon propagates in the same direction as the incident photon. Under time inversion this becomes completely normal and isotropic absorption (as expected).</p>
<p>However, this argumentation is valid only if we presume that the two photons are completely indistinguishable (identical) in the first place. Therefore a "blue" photon (say, 400nm) stimulating emission of a "red" (600nm) photon in a random direction would still be decent physical process.
Another stackexchange question tackling this particular matter (can a photon stimulate emission of a different energy photon) <a href="http://physics.stackexchange.com/questions/119660/can-one-stimulate-emission-of-a-photon-with-an-energy-different-from-the-emitted">here</a> is apparently still waiting for a satisfactory answer.</p>
<p>Tl;dr / Long story short, why the photon produced in stimulated emission has the same (1)energy; (2)propagation direction and (3)phase as the incident photon?
Pointers to books/papers and decent web resources are much appreciated!</p> | 7,474 |
<p>I'm a physics tutor tutoring High School students. A question confused me a lot.</p>
<p>Question is:</p>
<blockquote>
<p>Suppose a mass less rod length $l$ has a particle of mass $m$ attached at its end and the rod is hinged at the other end in vertical plane. Another point object of mass $m$ is moving with velocity $v$ and hits the rod at its end and continues in its path with velocity $v/2$. The rod gets enough velocity that it can complete a vertical circular path. What would be the force due to hinge on the rod just after collision? Assume that time taken for collision is very small.</p>
</blockquote>
<p>From conservation of momentum, the particle attached to rod gets velocity $V' = v/2$ which is equal to $2\sqrt{gl}$ and therefore $$T-mg=\frac{mv'^2}{l}$$ The tension in the rod is the Normal reaction due to hinge on the other end of the rod. Hence, the force exerted by rod is $$\frac{mv'^2}{l}+mg$$</p>
<p>But other teacher claims that I'm wrong. He says there will be horizontal force due to hinge on the rod. I say, there won't be any horizontal force. He gives an example that suppose there is a rod and you flick it at one end with finger then won't the other end move (or tend to move)? Well it does.</p>
<p><strong>EDIT:</strong></p>
<p>I've understood your answers mathematically.I still don't know how to counter my friend's argument. I'm convinced by it. Because it sounds intuitive. He says, suppose you have a rod (massless or with mass) in space, and you flicked at one of its end, then the other end surely will have velocity.Similarly, in the above problem, the collision will impart velocity to one end, so other end also must move but it is hinged and can't move because of hinge. So, there is a horizontal force due to hinge on the rod. </p> | 7,475 |
<p>I am a <a href="http://en.wikipedia.org/wiki/Geographic_information_system" rel="nofollow">GIS</a> programmer implementing a visualization.</p>
<p>I am modeling the idealized trajectory of a particle ejected from a volcanic vent using:</p>
<p>$$\text{distance} = \frac{(v^2 \times sin(2\theta))}{g}.$$</p>
<p>Where $g = 1.62\:\mathrm{m/s^2}$, $v$ is velocity, and $\theta$ is ejection angle. $g$ is the lunar gravity constant I was supplied.</p>
<p>How can I incorporate the slope of the underlying surface assuming a single point of ejection? $$$$$$$$</p>
<p>EDIT:
My current workflow is to compute total travel distance, extract a topographic profile along the total theoretical travel distance and then check the height of the projectile to the height of the actual surface at 100m intervals. In this way I can compute the landing site for the projectile.</p>
<p>EDIT 2:
I updated the question with the correct formula. Apologies for the incorrect transposition. My implementation now assumes a completely flat surface. What happens when the ejection surface is sloped either uphill or downhill?</p> | 7,476 |
<p>If an equal torque of equal radius and size is produced on two bodies of the same weight and same center of gravity, but with different weight distribution (say one has a 1kg mass 1 meter above the center of mass, while the other has it 10 cm above). Will this affect the angular velocity of it?</p> | 7,477 |
<p>I am currently a high school student interested in a research career in physics. I have self taught myself single variable calculus and elementary physics upto the level of IPHO . And I am comfortable with euclidean geometry upto the point of some basic theorems like Pythagoras or Thales and the geometry in Cartesian coordinates and use of vectors. </p>
<p>I have read at a lot of places that somehow geometry is very much essential for doing physics? Now I don't know in what context, as in do you ever need the complex geometrical theorems that are upto the level of IMO, in physics or are you just good knowing simple theorems and their proofs and trigonometric relations and similarity etc.</p>
<p>Plus what other forms of geometry are essential for doing physics? </p>
<p>Is linear algebra a form of geometry? </p>
<p>What next step should I take to dive into geometry that will help me in my physics career for making an early start? Non-euclidean? And what is it about the geometry names, hyperbolic, elliptical? I can understand if someone tells me that plane geometry / 3d geometry, but what is the essence of hyperbolic/elliptical geometry?</p>
<p>And what are non euclidean geometries and topologies?</p>
<p>Is there a theoretical minimum that you need in all forms of geometries to do physics? I want to learn the math now properly so that I get used to its machinery and later on, don't have to struggle with it later, especially geometry, since the names of these geometries intrigue me .</p>
<p>Is there any need for learning euclidean geometry, upto the levels of <em>Geometry Revisited</em> by Coxeter?</p>
<p>Also I'd be grateful, if some book recommendations are also provided.</p> | 108 |
<p>I understand that quantum teleportation fidelity is the overlap of the initial quantum state with the teleported quantum state. If the teleportation is perfect, then the fidelity would equal 1 or 100% successful of retrieving the quantum state at the desired location.</p>
<p>In all real experiments I've glanced at and read thoroughly, the fidelities are never 1. Doesn't this suggest that what the experimenters "teleported" was really just a different state? Doesn't the fidelity have to be exactly 1 in order to really remove a quantum state from one location and make it appear at another location?</p>
<p>My concern is that either I've missed something, or that we aren't really teleporting at all because the final state achieved on the other side is not equal to the initial state!</p> | 7,478 |
<p>Looking at the Moody chart I think to myself, the <em>friction factor</em> doesn't decrease much at all with Reynolds number after a certain point. I wonder if laminar flow is more efficient in a sense, and what sense would that be?</p>
<p><img src="http://i.stack.imgur.com/1bIeg.jpg" alt="Moody diagram"></p>
<p>I understand that in laminar flow you have clear lines that the fluid doesn't cross, whereas that wouldn't be true in turbulent flow. We could imagine a pipe with a divider placed in the middle to keep the fluid from mixing in eddies, but that would just create more friction with the divider. There are thinkable cases, however, where you could introduce a divider that moves along with the fluid, in particular, the Taylor-Couette flow...</p>
<p><a href="http://en.wikipedia.org/wiki/Taylor%E2%80%93Couette_flow">http://en.wikipedia.org/wiki/Taylor%E2%80%93Couette_flow</a></p>
<p>This setup describes basically one cylinder rotating within another cylinder with a fluid in-between them.</p>
<p><img src="http://i.stack.imgur.com/zA1r3.png" alt="Concentric rotating cylinders"></p>
<p>Let's say that you kept the fluid the same, and the distance between the inner cylinder and outer cylinder the same. In that system, let's say you insert a divider at a radius in the middle of the annular area, and this divider was mostly buoyant in the fluid, so it's not experiencing friction on the edges, and it also is free to rotate with the fluid.</p>
<p>Would doing so actually reduce the frictional torque on the rotating inner cylinder? If you could introduce an infinite number of infinitely thin dividers is there a theoretical limit to how much you could reduce the retarding torque? Would that just make it laminar, or laminar-ish?</p> | 7,479 |
<p>I was told by a former physicist that he thinks that the now dormant S-Matrix Theory has the potential to lead to a breakthrough in formulating a Grand Unified Theory. He stated several reasons for this; all of which I do not remember except for one: The non-linearity of the differential equations that constitute it could mean that we could apply Dynamical Systems Theory to it, which could lead to a breakthrough in the unification of QM and GR. </p>
<p>Can anyone add this or elucidate on why this theory has this potential, and what would be the potential avenues for exploring it? </p> | 7,480 |
<p>Given that the boiling point of a liquid is the temperature at which the vapor pressure is equal to the ambient (surrounding) pressure, what significance does a liquid's vapor pressure have in the formation of bubbles that happens at and above the boiling point? </p>
<p>The definition of boiling point seems to imply that the pressure inside of the bubbles must be at least as great as the liquid's vapor pressure in order to balance the outside pressure, but is there any particular reason why the pressure inside of the bubbles is related to the vapor pressure? </p>
<p>The vapor pressure seems to be a measurement describing the tendency of the molecules to escape from the surface of the liquid, but I don't see how that relates to bubble formation within the liquid.</p>
<p>This question has bothered me for a while, so any help would be much appreciated.</p> | 7,481 |
<p>It is my understanding that in quantum mechanics (for 1/2 spin particles) the </p>
<p>probability function that describes the direction of a particle's spin state is </p>
<p>proportional to the overlap of the theoretical hemispheres formed by the motion of </p>
<p>the particle (in either spin state). However, with this theory, it is also </p>
<p>imperative to assume that the only eigenvalues to describe a particle's spin state </p>
<p>are ± ħ / 2. If we do assume that the spin state of a particle is well defined, </p>
<p>even if it is not observed,how would we formulate a function to determine the </p>
<p>expected value of a spin state for particles with more than two spin states (such </p>
<p>as the 3/2 delta baryons)? Are the results in any way linked to an overlapping of </p>
<p>the hemispheres in the various spin states? In addition, if it were ever </p>
<p>necessary, could we formulate such a function for particles with an even higher </p>
<p>number of spin states?</p>
<p><img src="http://i.stack.imgur.com/ySf3u.jpg" alt=" "></p>
<p>Here we see a visual of this theory. The spin components of particles (a) and (b) have an angle between their vectors, $\phi$. Using this, we can see that the probability of finding both particles in a certain spin state is proportional to the overlap of their hemispheres.</p>
<p>$\textbf{These calculations are mine, but I thought that it made sense intuitively }$</p>
<p>$\textbf{}$</p>
<p>This is the basic Idea behind the Hemisphere mathematics I described to you:</p>
<p>The expected value will be denoted as $\langle\sigma^{(ab)}\rangle $ and:</p>
<p>$\langle\sigma^{(ab)}\rangle $ = $P(u,u) + P(d,d) − P(u,d) − P(d,u)$, where $(d)$ donotes down and $(u)$ donotes up.</p>
<p>This spin values are relative to a set field direction that corresponds to a Stern-Gorlach apparatus. </p>
<p>Consequently, we can assume that if we observe the value $ +\cfrac\hslash2$, the spin of the particle will lie in the $"up"$ hemisphere, and contrarily, if we observe a value of $-\cfrac\hslash2$, our spin will reside in the $"down"$ hemisphere. </p>
<p>Now, if we denote angle between the spin states of 2 particles as $(\phi)$, the probability of observing both of the particles in a down state with be equivalent to $(P(u,u)=(P(d,d))=(\cfrac\phi\pi)$</p>
<p>Therefore, the proportionality coefficient has to equal (1) when $\phi=\pi$ and the probability of finding the particles in two different spin states will be $P(u,d) = x\phi + y$</p>
<p>This probability is equal to (1) when the two hemispheres coincide ($\phi$ = 0) and consequently $y = 1$ [note that this disappears if $\phi = \pi$]. Thus, $x = −\cfrac1π$ and $P(u,d) = P(d,u) = 1 − \phi$</p> | 7,482 |
<p>I ask <a href="http://math.stackexchange.com/q/884412/">this question</a> on math stackexchange but got no answer. Not sure how to move the post so I'm reposting it here.</p>
<p>I was arguing with my friend that <a href="http://en.wikipedia.org/wiki/Brownian_motion" rel="nofollow">Brownian motion</a>, in the sense of a pollen moving in the fluid, could be explained by physics laws (such as $F=ma$) and statistics laws. </p>
<p>To check it out I found Albert Einstein's paper, "Investigations on the theory of Brownian motion", however I found it a bit hard to understand. It starts with osmotic pressure which i have no idea, and the paper is 114 pages long, though in big font.</p>
<p>I'm wondering if there's an easier iteration? hopefully shorter?</p> | 7,483 |
<p>This is a continuation of <a href="http://physics.stackexchange.com/q/22183/2451">this question</a>.</p>
<p><a href="http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-1/" rel="nofollow">http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-1/</a> skip this lecture to around 25:50.</p>
<p>After doing dimensional analysis on $t\propto h^\alpha m^\beta g^\gamma$ Lewin concludes that:</p>
<p>$$\alpha = \frac{1}{2}, \gamma = -1/2, \beta = 0$$</p>
<p>This is all fully understood, but he then goes to conclude from this that:</p>
<p>$$t = C \sqrt{\frac{h}{g}}$$</p>
<p>How did he get to this? And why is he allowed to just assume that there is a constant, C, there when he doesn't even know its value or what it is?</p>
<p>Keep things as simple as possible please, I'm 16.</p> | 7,484 |
<p>I would like to know if there is an equation to predict the pressure drop in a <a href="http://en.wikipedia.org/wiki/Venturi_tube#Venturi_tubes" rel="nofollow">venturi</a> device using a compressible fluid as the working medium. In particular, I'd like to use this equation to predict the amount of vacuum created for a given $\Delta p$ across the device. Simplifying assumptions can be made, such as equivalent inlet and outlet areas.</p> | 7,485 |
<p>I am currently struggling with the formula for an exact current in QFT, a fermion with an upcoming momentum $p$ and an outgoing momentum $p'$. My problem is to show whether or not a term of the form $\bar{u}(p')F(q^2)i\sigma^{\mu\nu}(p+p')^{\nu}$ where $\sigma^{\mu\nu}=\frac{i}{2}[\gamma^{\mu},\gamma^{\nu}]$. I would like it to be forbidden by current conservation but I can't prove it, is my intuition wrong?</p> | 7,486 |
<p>For a centrifugal compressor, as found in most turbochargers on internal combustion engines, is there a noticeable change in flow rate versus a naturally aspirated flow rate? In other words, does the pump effectively increase the pressure of the gas only, or does it also increase the flow rate? Really what I'd like to know is how much of the work a turbocharger does goes into increasing the density of the fluid versus how much of it goes into raising its temperature, per the ideal gas law.</p> | 7,487 |
<p><strong>Note:</strong> This question may be difficult or impossible to answer within the rules of these forums due to its philosophical nature. I will delete the question if I am violating the rules.</p>
<p>Onto the question! Recently I have been curious whether classical electromagnetism is fully solved (up to the divergences). Specifically, can we completely mathematically describe and then interpret the classical world of charges and currents and them the associated fields. Let us assume that our world is classical and that electromagnetism is a complete theory even though there are certain inconsistencies (self-interactions, infinite energy of point charges, etc). A common description of $\textbf{E}(\textbf{r}, t)$ and $\textbf{B}(\textbf{r}, t)$ among physics textbooks is that changing electric fields induce magnetic fields and vice-versa. This is assuming that there are no external fields or sources of field present. In symbols,</p>
<p>$$
\partial_t\textbf{E} \neq \textbf{0} \implies \exists \textbf{B} \quad (1)
$$</p>
<p>$$
\partial_t\textbf{B} \neq \textbf{0} \implies \exists \textbf{E} \quad (2)
$$</p>
<p>Many physicists come to this conclusion from <a href="http://en.wikipedia.org/wiki/Maxwell's_equations" rel="nofollow">Maxwell's equations</a>. Specifically, they argue that Faraday's law,</p>
<p>$$
\nabla \times \textbf{E}(\textbf{r}, t) = -\partial_t\textbf{B}(\textbf{r},t),
$$</p>
<p>implies (1), and that Ampere's law (with Maxwell's correction term and no currents),</p>
<p>$$
\nabla \times \textbf{B}(\textbf{r}, t) = \partial_t \textbf{E}(\textbf{r},t),
$$</p>
<p>implies (2). Note that we are using natural units with $c = 1$. However, these equations do not have any obvious causal connection. While we may like to pretend that <em>right</em> implies <em>left</em>, this is purely notational convention. Who is to say from these equations alone that one field having a nonzero curl doesn't produce a changing dual field? One attempt at reconciling this problem seems to be in <a href="http://en.wikipedia.org/wiki/Jefimenko's_equations" rel="nofollow">Jefimenko's equations</a>. I will state the equations without derivation, but the fields can be solved for completely in terms of the source charges and currents (I'm lazy and the following equations are in mks units from Wikipedia):</p>
<p>$$
\textbf{E}(\textbf{r}, t) = \frac{1}{4 \pi \epsilon_0}\int [\frac{\rho(\textbf{r}', t_r)}{|\textbf{r} - \textbf{r}'|^3} + \frac{1}{c}\frac{\partial_t \rho(\textbf{r}', t_r)}{|\textbf{r} - \textbf{r}'|^2}] (\textbf{r} - \textbf{r}') - \frac{1}{c^2}\frac{\partial_t \textbf{J}(\textbf{r}', t_r)}{|\textbf{r} - \textbf{r}'|^2} d^3\textbf{r}',
$$</p>
<p>$$
\textbf{B}(\textbf{r}, t) = \frac{\mu_0}{4 \pi}\int [\frac{\textbf{J}(\textbf{r}', t_r)}{|\textbf{r} - \textbf{r}'|^3} + \frac{1}{c}\frac{\partial_t \textbf{J}(\textbf{r}', t_r)}{|\textbf{r} - \textbf{r}'|^2}] \times (\textbf{r} - \textbf{r}' )d^3\textbf{r}' ,
$$</p>
<p>where $t_r = t - |\textbf{r} - \textbf{r}'|/c$ is the retarded time. These equations seem to imply that neither of the fields "causes" the other. Instead, Jefimenko's equations imply that only the source charges and currents generate the fields (without the presence of external charges, currents, or fields). My question is related to this approach. Is it valid? What are the arguments for and against? Is the matter settled in the classical context of electromagnetism, or are there certain subtleties I've skipped over?</p>
<p>As an extra question, is it instead better to consider $F_{\mu \nu}$, and treat it as one object arising solely from $J^\mu = (\rho, \textbf{J})$, instead of looking at the individual fields?</p>
<p>Thanks in advance for any and all answers!</p> | 7,488 |
<p>What are the differences between comet versus asteroid versus meteoroid versus meteor versus meteorite in practical terms?</p> | 7,489 |
<p>As far as I know, particles vibrate with a frequency and wavelength determined by their energy level.</p>
<p>Is this vibration in 3D space?</p> | 7,490 |
<p>Right now, our resolution + light gathering power are still far too low to take direct images of exoplanets, so we're limited to subtracting the planet spectra from the parent star spectra when the planet undergoes a transit (and this isn't going to be possible for decades, according to Jim Kasting's latest book). So in this case, it seems that light gathering power is more important.</p>
<p>So, Is angular resolution more important when we want to measure the spectra of an Earth-like Exo?</p> | 7,491 |
<p>On the new <a href="http://astronomy.stackexchange.com/">Astronomy.SE</a> site, I was having a short discussion on one of my answers. The basic discrepancy was; can MACHOs like black holes/brown dwarfs/neutron stars be termed "dark matter"?</p>
<p>My reasoning is that these objects do not radiate EM radiation on their own but they do gravitate, and thus constitute a small part of the total dark matter in the universe. I agree that there is a lot of dark matter which doesn't</p>
<p>In other words, can the term "dark matter" be applied to nonradiating (or faintly radiating) bodies which still participate in the electromagnetic interaction (baryonic or otherwise)? Or is it necessary for all dark matter to not interact electromagnetically?</p> | 7,492 |
<p>The FRW is a nice isotropic and symmetric metric but I think its assumptions are too many. I was wondering about alternative models. Specifically are there any prominent alternatives that have more fleshed out local detail? For example, are there any models that can ascribe a more local interpretation to data which was considered as crucial support for the Big Bang. </p> | 7,493 |
<p><a href="http://space.stackexchange.com/questions/604/is-it-possible-to-get-pregnant-through-natural-means-in-space">http://space.stackexchange.com/questions/604/is-it-possible-to-get-pregnant-through-natural-means-in-space</a> indicates cosmonauts are at risk of irradiation whilst in space. </p>
<p>What little I know about space is that planets, and inhabitants are largely protected from deep cosmic rays by
- Sun's magnetic sheath
- Solar Wind
- Planet's own magnetic sheath</p>
<p>When in orbit, a cosmonaut and satellites are periodically in the Sun. At other times, they are in the dark. On the sunward side, any body is directly in the Solar wind, whereas in shadow the planet/body probably acts as a shield against the Solar Wind; I may be wrong!</p>
<ol>
<li><p>Where is the deep-space radiation higher - on the Sunward side of Earth, or on the side in Earth's shadow? I guess this may also be rephrased to read - what fraction of shielding from deep cosmic rays within the Solar System is provided by the Solar wind?</p></li>
<li><p>Is a cosmonaut at greater health risk on the Sunward Side, or in Earth's shadow? (<em>This second part may fit better on Biology SE but since it is contingent on the former I post it here</em>)</p></li>
</ol> | 7,494 |
<p>Related: <a href="http://physics.stackexchange.com/questions/72368/why-are-most-metals-gray-silver">Why are most metals gray/silver?</a></p>
<p>After reading Johannes’ impressive answer to Ali Abbasinasab question of why do most metals appear silver in color with the exception of gold (and copper), my lack of experience with quantum mechanics (I’m an undergraduate and I will take graduate quantum mechanics this coming quarter) is not enough for me to appreciate the relativistic quantum solution. What struck me most was Stefan Bischof’s statement: “Here physics has to explain more than just “is there a d valence electron”.” I took this as a challenge to see if I could explain the golden color of gold (and copper). Here is my humble attempt to answer this very question using scattering and EM approaches.</p> | 7,495 |
<p>Why is there an absolute entropy? Given any non-discrete probability distribution, we don't really have an absolute entropy because the entropy depends on the parametrization of the distribution (e.g. Beta vs. Beta-prime) which was arbitrarily chosen. Another way to put it is that instead of entropies we only have <a href="http://en.wikipedia.org/wiki/Kullback%E2%80%93Leibler_divergence" rel="nofollow">Kullback–Leibler (KL) divergences</a> aka relative entropies. So, why isn't there a physics analogue to KL-divergence? Just as we have relativistic velocity, which has some properties, why don't we also have relative entropies, which have some properties? Instead of saying the absolute entropy of the universe increases, why don't we say that the relative entropy given our prior belief of the universe increases?</p>
<p>Alternatively, what is the relation between entropy and number of microstates when the physical system is continuous, and how do we "count microstates"?</p> | 7,496 |
<p>I am working on a game involving flying and steering a paper airplane for WP7. I want the plane to fly just like how normal paper airplanes fly (see this game for an example <a href="http://armorgames.com/play/7598/flight" rel="nofollow">http://armorgames.com/play/7598/flight</a>) but I can't seem to find an equation for how paper airplanes fly.</p>
<p>Anyone have any experience with this? In my game now, it just follows the usual motion for an object in a vacuum, which makes for some flight, but it doesn't feel perfect, and traveling at a slight downward angle makes you lose speed, which isn't right.</p>
<p>Thoughts?</p> | 7,497 |
<p>Euler's laws of motion for a distributed mass are:</p>
<p>$$F = \frac{d}{dt} MV_{cm},\ N = \frac{d}{dt} L$$</p>
<p>$F$ are the sum of the external forces, $M$ the total mass, $V_{cm}$ the velocity of the centre of mass. $N$ are the sum of the moments of the external forces about some given point, L the total angular momentum about the same point.</p>
<p>If a gyroscope is supported at its base with its axis horizontal, it precesses at a constant angular velocity. Using the above equations, how does one show this?</p> | 7,498 |
<p>Clouds are made up of tiny water or ice droplets, depending on temperature. This implies that cloud density is greater than that of dry air. Why don't clouds sink through their surrounding atmosphere rather than float by in a variety of formations? </p> | 260 |
<p>Matt in <a href="http://photo.stackexchange.com/questions/9597/what-does-a-hexagonal-sun-tell-us-about-the-camera-lens-sensor/9601#9601">his answer</a> on <a href="http://photo.stackexchange.com/q/9597/64">What does a hexagonal sun tell us about the camera lens/sensor?</a> mentions</p>
<blockquote>
<p>Incidentally the number of (distinct)
points to the star is equal to double
the total number of unique
orientations* in the sides of the
aperture shape i.e. three blades would
be six points, six blades would also
be six points as a hexagon has only
three unique orientations in its
sides.</p>
<p>* a hexagonal aperture has six sides but only three unique orientations as
there are three pairs of parallel
sides.</p>
</blockquote>
<p>I think It makes sense to have three pointed star in case of three blades.</p>
<p>But how does a six blade aperture also result in a three pointed star?</p> | 7,499 |
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