category
stringclasses 107
values | title
stringlengths 15
179
| question_link
stringlengths 59
147
| question_body
stringlengths 53
33.8k
| answer_html
stringlengths 0
28.8k
| __index_level_0__
int64 0
1.58k
|
|---|---|---|---|---|---|
general relativity
|
Why is there a gravitational attraction between two objects at rest with respect to each other?
|
https://physics.stackexchange.com/questions/79676/why-is-there-a-gravitational-attraction-between-two-objects-at-rest-with-respect
|
<p>From my understanding of relativity, gravity is not a force, but a result of the curvature of spacetime. If Object1 moves past Object2, even though it's moving in a straight line, its direction may change due to the distortion caused by Object2's mass.</p>
<p>However, what about the situation where Object1 is not moving? How can there be an attraction between two objects that are at rest relative to each other? i.e. what makes them move towards each other?</p>
|
<p>Two objects that are initially at rest with respect to each other have initially parallel <a href="https://en.wikipedia.org/wiki/World_line" rel="nofollow noreferrer">world lines</a>.</p>
<p>However, the curvature of spacetime means that world lines that are initially parallel do not remain so. This is called geodesic deviation.</p>
<p><img src="https://i.sstatic.net/jDP8H.gif" alt="enter image description here"></p>
<p>(<a href="https://en.wikipedia.org/wiki/File:Earth_geo.gif" rel="nofollow noreferrer">Image credit</a>)</p>
<p>In the above image, the geodesic segments are parallel at the equator but, nonetheless, converge at the pole.</p>
<p>Imagine the time direction (<em>and remember, every object is relentlessly "moving" forward through the time direction even when "at rest" in space</em>) is along lines of longitude and the spatial direction is along lines of latitude.</p>
<p>If the surface were, instead, a plane, the two geodesics would remain parallel and the objects associated with those world lines would not move towards or away from each other.</p>
| 500
|
general relativity
|
what are the direct real life applications of general relativity and quantum physics
|
https://physics.stackexchange.com/questions/82121/what-are-the-direct-real-life-applications-of-general-relativity-and-quantum-phy
|
<p>What are the direct real life applications of general relativity other than nuclear technology? </p>
<p>What I meant was, was there any technology developed based on general relativity that can benefit mankind today? </p>
<p>Secondly, are there any adverse effects quantum technology today?</p>
| 501
|
|
general relativity
|
About divergence of a vector field and geodesic sphere
|
https://physics.stackexchange.com/questions/94459/about-divergence-of-a-vector-field-and-geodesic-sphere
|
<p>I have a question. I want to know the difference between the sphere and the geodesic sphere.
Another question:
given a vector field, $Y$, on a manifold $M$ defined by: $Y(p)=p$ for every point $p \in M$, I tried to calculate the divergence of $Y$. I think that it is equal to $n$, where $n$ is the dimension of $M$, but I don't know exactly how I can compute it.</p>
| 502
|
|
general relativity
|
What does adding a scalar field component to the Einstein field equations mean for black holes and string theory?
|
https://physics.stackexchange.com/questions/35863/what-does-adding-a-scalar-field-component-to-the-einstein-field-equations-mean-f
|
<p>If a scalar field component has to be added to the Einstein field equations (see below) to solve dark matter/energy, then how would string theory need to be modified and do black holes still exist?</p>
<p>The proposed modified equations are (ignoring physical constants)
$$ R_{ij} - \frac12 R g_{ij} = T_{ij} + \nabla_i\nabla_j \varphi $$
where $\varphi$ represents some kind of scalar potential. The conservation law for energy-momentum is proposed to be
$$ \nabla^i(T_{ij} + \nabla_i\nabla_j\varphi) = 0 $$</p>
<p>Refs: </p>
<ul>
<li><p>Dark Matter, Dark Energy, and the Fate of Einstein’s Theory of Gravity
<a href="http://mathgradblog.williams.edu/dark-matter-dark-energy-fate-einsteins-theory-gravity" rel="nofollow">http://mathgradblog.williams.edu/dark-matter-dark-energy-fate-einsteins-theory-gravity</a></p></li>
<li><p>UNIFIED THEORY OF DARK ENERGY AND DARK MATTER, TIAN MA, SHOUHONG WANG
<a href="http://www.indiana.edu/~fluid/paper/report.pdf" rel="nofollow">http://www.indiana.edu/~fluid/paper/report.pdf</a></p></li>
</ul>
|
<p>It does nothing more than saying that $T_{ij}$ can be decomposed into ''ordinary matter'' and ''scalar field.'' That equation is an assertion that the matter content of the universe contains a scalar field. </p>
| 503
|
general relativity
|
How to choose a solution from all possible solutions of general relativity
|
https://physics.stackexchange.com/questions/55435/how-to-choose-a-solution-from-all-possible-solutions-of-general-relativity
|
<p>So there are so many solutions for general relativity - then how does one "choose" the solution that is right one? By checking with observation? (though I also know that it is currently unknown which one is the correct solution.)</p>
|
<p>Generally speaking we start with a known stress-energy tensor and boundary conditions and look for solutions for the curvature. When doing this we're not usually overloaded with possible solutions, and it's normally pretty obvious which solutions are physically relevant.</p>
<p>Where multiple physically relevant solutions exist we select the appropriate one by comparing with experiment. For example we select the FLRW rather than the Gödel metric because experiment suggests the universe has no net rotation. Likewise the value of $\Lambda$ has to be fixed by experiment, as was done in 1998 by <a href="http://arxiv.org/abs/astro-ph/9812473" rel="nofollow">Perlmutter</a> and <a href="http://arxiv.org/abs/astroph/9805201" rel="nofollow">Riese's</a> groups.</p>
| 504
|
general relativity
|
Help me to understand this conversion (4-vectors)
|
https://physics.stackexchange.com/questions/60251/help-me-to-understand-this-conversion-4-vectors
|
<p>$u^{\mu}$ - 4-velocity</p>
<p>$b^{\mu}$ - 4-vector of magnetic field</p>
<p>$
u_{\mu}u^{\mu}=-1, \qquad u_{\mu}b^{\mu}=0
$</p>
<p>$$
u_{\beta}u^{\alpha}\nabla_{\alpha}b^{\beta}-u_{\beta}b^{\alpha}\nabla_{\alpha}u^{\beta}+\nabla_{\alpha}b^{\alpha}=0
$$
I don't understand why this equation gives this
$$
u^{\alpha}u^{\beta}\nabla_{\alpha}b^{\beta}+\nabla_{\alpha}b^{\alpha}=0
$$</p>
<p>Help me please!</p>
|
<p>This is because $u_\alpha \nabla_\beta u^\alpha =0$. To show this, just act with $\nabla_\beta$ on both sides of the equality $u_\alpha u^\alpha = -1$. You get
$$
u_\alpha \nabla_\beta u^\alpha + u^\alpha \nabla_\beta u_\alpha = 0
$$
and thus $u_\alpha \nabla_\beta u^\alpha =0$ as promised.</p>
<p>Cheers!</p>
| 505
|
general relativity
|
What is the physical meaning of charges at light-like infinity in asymptotically flat space-times?
|
https://physics.stackexchange.com/questions/60748/what-is-the-physical-meaning-of-charges-at-light-like-infinity-in-asymptotically
|
<p>In the case of charges defined at space-like infinity, I can understand the physical meaning of them because they can be related to measurements made by a physical observer (that is an observer whose wordline is time-like). For example in four dimensions, for the Schwarschild solution, the ADM mass coincide with the mass measured by an observer that is motionless (in Schwarschild coordinates) and far from the center of the solution.</p>
<p>My puzzle is then the following : since no physical observer can reach light-like infinity, how are physically interpreted the charges defined at light-like infinity? How are they related to any measurement?</p>
<p>Many thanks for your help.</p>
| 506
|
|
general relativity
|
Transforming an equation to the co-vector version
|
https://physics.stackexchange.com/questions/61641/transforming-an-equation-to-the-co-vector-version
|
<p>Ok, this question is more a result of my lack of knowledge of how to manipulate equations involving index notation rather than about physics...</p>
<p>I have the geodesic equation with $U^\lambda\equiv\dot{x}^\lambda$:-</p>
<p>$$
\dot{U^\lambda} + \Gamma^\lambda_{\mu\nu} U^\mu U^\nu
$$</p>
<p>And I want to transform to the co-vector $U_\mu=g_{\mu\lambda}U^\lambda$.</p>
<p>Can I simply multiply each vector by $g_{\mu\nu}$? Like so:-</p>
<p>$$
g_{\mu\lambda}\dot{U^\lambda} + \Gamma^\lambda_{\mu\nu}g_{\mu\alpha}U^\alpha g_{\nu\beta}U^\beta
$$</p>
<p>Or do I need to use $g^{\sigma\nu}g_{\nu\mu} = \delta^\sigma_\mu$ to rewrite $U_\mu=g_{\mu\lambda}U^\lambda$ and then sub it in?</p>
<p><strong>Edit: Here's my attempt at the sub in method</strong></p>
<p>So using $g^{\lambda\mu}g_{\mu\lambda} = \delta^\lambda_\lambda$ to rewrite $U_\mu=g_{\mu\lambda}U^\lambda$ as $U^\lambda=g^{\lambda\mu}U_\mu$. (Is this even correct?). Then differentiate:-
$$
\dot{U}^\lambda=\dot{g}^{\lambda\mu}U_\mu + g^{\lambda\mu}\dot{U}_\mu
$$
Can I assume that the differential of the metric wrt time is going to be zero? Obivously this is not going to be true in general since massive bodies move! But generally in simple problems would this be taken as true?</p>
|
<p>You generally need to do the second thing where you sub in. For an affinely parameterized geodesic $x(\lambda) = (x^\mu(\lambda))$ we have
$$
x_\mu(\lambda)= g_{\mu\nu}(x(\lambda))x^\nu(\lambda)
$$
Denoting derivatives with respect to affine parameter by overdots, it follows that
\begin{align}
\dot x_\mu(\lambda)
&= \frac{d}{d\lambda}\left[g_{\mu\nu}(x(\lambda))\right]x^\nu(\lambda) + g_{\mu\nu}(x(\lambda))\dot x^\nu(\lambda)
\end{align}
Therefore, notice that you can only "simply multiply each vector by $g_{\mu\nu}$" if the first term in this expression vanishes, which is only true then the metric components are constant along the curve. This is not in general the case, although it is the case, for example, in flat space coordinates where $g_{\mu\nu} = \delta_{\mu\nu}$. What is generally true, however, is that the directional <em>covariant</em> derivative of the metric is zero along a given curve;
$$
\frac{D}{d\lambda}\left[g_{\mu\nu}(x(\lambda))\right] = 0
$$</p>
| 507
|
general relativity
|
Interval and proper time
|
https://physics.stackexchange.com/questions/61353/interval-and-proper-time
|
<p>Is the definition of $$d s^2=-d \tau^2$$ assuming that $c=1$, so that we always have $$\left({ds\over d\tau}\right)^2=-1$$?
Is there a reason for this definition?
Don't we get an imaginary ${ds\over d\tau}$?</p>
|
<p>It depends on what convention you're using for the metric's signature. Some people use the metric signature (-+++), which is what you have there. The interval is then:</p>
<p>$$ds^2=-dt^2+d \mathbf{r}^2$$</p>
<p>On the other hand, some people use the (+---) convention:</p>
<p>$$ds^2=dt^2-d \mathbf{r}^2$$</p>
<p>In this signature $ds^2=d \tau^2$. Which one you use is a matter of preference. Special/General Relativity textbooks tend to prefer (-+++), though a lot of Quantum Mechanics textbooks prefer (+---).</p>
| 508
|
general relativity
|
A question about the relativity of time
|
https://physics.stackexchange.com/questions/8671/a-question-about-the-relativity-of-time
|
<blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://physics.stackexchange.com/questions/8659/invariant-spacetime-distance-circular-motion">Invariant spacetime - distance - Circular Motion</a> </p>
</blockquote>
<p>I understand that the closer something travels to the speed of light, that time will stretch by a factor, and distance will compress by the same factor.</p>
<p>My question is, if something travels in a circle, close to the speed of light, what does the distance of the journey look like to them? They measure that the trip took them 10 minutes. And an outside observer says that the journey took 20 minutes, and the outside observer measured that they did, for example, 1000 laps of a circle circumference 1000 km.</p>
<p>So if the plane had a distance trip counter, what would it read? And if they were looking out of the window, would the circle still look like it had a circumference of 1000km?</p>
| 509
|
|
general relativity
|
Calculating position in space assuming general relativity
|
https://physics.stackexchange.com/questions/10870/calculating-position-in-space-assuming-general-relativity
|
<p>Suppose two pointed masses are given in space. Suppose further that one of the masses has a given velocity at (local) time 0. Is there a way to compute its position in a future time?</p>
<p>Neglecting general relativity, I will simply compute an integral, but with general relativity, we see that the metric of the space changes with time, so I need to compute an integral with respect to a measure that changes along time.</p>
<p>Can this be done? If so, how?</p>
<p>Thank you!</p>
|
<p>If the moving mass is small enough, you can do this using the geodesic equation,</p>
<p>$$\frac{\mathrm{d}^2x^\lambda}{\mathrm{d}t^2} + \Gamma^{\lambda}_{\mu\nu}\frac{\mathrm{d}x^\mu}{\mathrm{d}t}\frac{\mathrm{d}x^\nu}{\mathrm{d}t} = 0$$</p>
<p>This is essentially the general relativistic equivalent of Newton's second law: it's a differential equation that governs how a test particle's position changes in time. The connection coefficients $\Gamma^{\lambda}_{\mu\nu}$ (also called Christoffel symbols) can be calculated from the metric. So if the metric is known, even if it's time-dependent, you can calculate the connection coefficients in your desired reference frame, plug them in, and find a solution to the geodesic equation that tells you how the particle will move. There may not be an analytic solution, but in all but the most extreme cases you can either solve the equation numerically, or make some approximation that might make it analytically solvable.</p>
<p>If the moving particle is not small enough to be considered a test particle, then the situation becomes more complicated because the metric, and thus the connection coefficients, will depend on the motion of the test particle itself. So you wind up with a coupled system of three equations: the geodesic equation for the moving particle, the geodesic equation for the other particle, and the Einstein equation</p>
<p>$$G^{\mu\nu} = 8\pi T^{\mu\nu}$$</p>
<p>which tells you how the metric changes in response to the motion of the two particles. In this case it's highly unlikely that you could find an analytic solution, but you could potentially still use a numeric differential equation solver, at least for some range of time. (All the equations are nonlinear so it's likely that your solution would lose accuracy quickly.)</p>
| 510
|
general relativity
|
General relativity at 0K
|
https://physics.stackexchange.com/questions/14338/general-relativity-at-0k
|
<p>Relativistic gravity affects particle in motion, does it affect particle that are resting too? How? and if not could one say that the matter at 0K is not affected by gravity?</p>
<p>I'm not a physicist; is just a thought and probably really naive.</p>
|
<p>Yes, gravity affects particles at rest, and particles at rest affect gravity.</p>
<p>In GR, the interaction between spacetime and matter and energy is described by <a href="http://en.wikipedia.org/wiki/Einstein_field_equations" rel="nofollow noreferrer">Einstein's equation</a>,
<img src="https://i.sstatic.net/gNZm5.png" alt="enter image description here">
(from Wikipedia).</p>
<p>The term $T_{\mu\nu}$ on the right is the <a href="http://en.wikipedia.org/wiki/Stress-energy_tensor" rel="nofollow noreferrer">stress-energy tensor</a>. Matter at rest contributes to the $T_{00}$ term, even if the matter has no other interactions than gravitational, so matter at rest contributes to the curvature of spacetime. If you have a lump of mass in space and you are not moving with respect to it, it will still attract you.</p>
<p>The motion of a particle acted on by gravity is described by a <a href="http://en.wikipedia.org/wiki/Geodesic_%28general_relativity%29" rel="nofollow noreferrer">geodesic</a>. This is true whether or not the particle is at rest. To test it, simply toss a ball up in the air. It comes to rest at the top of its arc, but gravity still pulls on it and it still falls.</p>
<p>More importantly, the <a href="http://en.wikipedia.org/wiki/Principle_of_relativity" rel="nofollow noreferrer">principle of relativity</a> states that frames moving with respect to each other have the same physics. (In GR, we are only going to compare such frames locally, but it doesn't impact the discussion much.) So if a particle is stationary in one frame, it is moving in other frames which are equally good descriptions of the universe. Thus, there is no such notion as "at rest", except relative to some particular frame with no special distinction. Every particle is always at rest in some frame, so if gravity has no impact on particles at rest, it has no impact on anything!</p>
| 511
|
general relativity
|
How do we resolve a flat spacetime and the cosmological principle?
|
https://physics.stackexchange.com/questions/16090/how-do-we-resolve-a-flat-spacetime-and-the-cosmological-principle
|
<p>As I've said elsewhere, I've not had the opportunity to take a class in general relativity. Nonetheless, I understand that two major aspects of the standard cosmological model are the <a href="http://en.wikipedia.org/wiki/Cosmological_principle" rel="nofollow">cosmological principle</a> and the observation of a <a href="http://en.wikipedia.org/wiki/Shape_of_the_Universe" rel="nofollow">flat space</a>. To get where I'm coming from, I'll try to give a brief description of my understanding of these concepts:</p>
<ul>
<li><p>cosmological principle - This principle states that there is no privileged position within the universe. In other words, wherever any observer is located, s/he will observe approximately the same thing. Obviously the specific celestial bodies observed will change, but the expansion of the universe will be judged the same and the universe will essentially appear isotropic.</p></li>
<li><p>flat space - This observation, tested and largely verified by the WMAP satellite, shows that the large scale universe is not curved.</p></li>
</ul>
<p>My natural inclination is that these two things cannot be simultaneously true. The reason it seems this way to me is that if any observer can see roughly the same amount of the universe in any direction, and the universe is of finite size, the observable portions must overlap somewhere. If the observable portions overlap, it must be possible to continue traveling in one direction and eventually end up where you started. To me, this seems to be what curvature is.</p>
<p>How do we reconcile these two concepts?</p>
<p>$\dagger$ I have read some of the articles on the subject such as those on <a href="http://en.wikipedia.org/wiki/Minkowski_space" rel="nofollow">Minkowski space</a> and <a href="http://en.wikipedia.org/wiki/Torus#n-dimensional_torus" rel="nofollow">multidimensional toruses</a>. I believe I can reconcile the two concepts and imagine a higher dimensional flat torus, but the concept is still a difficult one for me and I would love some clarification.</p>
|
<p>I'm not sure I understand what you don't understand. I am adding a answer since this would be too long for a comment... </p>
<p>Given the cosmological principle and the flat space time observation, the idea is that the flat spacetime is infinite, or at least very much larger than our horizon. So, yes, an observer that is, say 7 billion light years away from us would see parts of the universe that are beyond our horizon, and we would see parts of the universe beyond his horizon. There is no contradiction in this. This other observer at 7 billion years would, for example, also see a CMB with properties similar to ours but all the ripples would be different.</p>
<p>It is true that this could also be a flat spacetime that has the topological property of a torus but that is not at all required. This would give a flat spacetime that is finite. Do you think the universe cannot be infinite?</p>
<p>By the way, the cosmological principle is more than a principle, there is observational support for it. For example the CMB looks the same in all directions and the large scale structure of galaxy cluster seems uniform on the largest scales. </p>
<p>EDIT: Thinking about it more, you probably are thinking of the big bang as being one point (or very small region) in space at $t=0$ that then "explodes" into our universe. I had that same problem / misconception when I started learning about this. The problem is, we cannot say anything about t=0 of the big bang, because that would effectively be like a singularity filling all of space. Instead think of $t=\epsilon$, just very slightly after the big bang. At that time, you would have an infinite space filled with extremely high energy density at a very high temperature and the space would be expanding extremely rapidly. As time goes on from there the expansion rate slows, the density and temperature go down and you get the infinite universe we live in. Does that help?</p>
| 512
|
general relativity
|
Intuitively, what is the source term of the Einstein field equation?
|
https://physics.stackexchange.com/questions/16725/intuitively-what-is-the-source-term-of-the-einstein-field-equation
|
<p>My copy of Feynman's "Six Not-So-Easy Pieces" has an interesting introduction by Roger Penrose. In that introduction (copyright 1997 according to the copyright page), Penrose complains that Feynman's "simplified account of the Einstein field equation of general relativity did need a qualification that he did not quite give." Feynman's intuitive discussion rests on relating the "radius excess" to a constant times the gravitational mass $M$: for a sphere of measured radius $r_{\mathrm{meas}}$ and surface area $A$ enclosing matter with average mass density $\rho$ smoothly distributed throughout the sphere,</p>
<p>$$
\sqrt{\frac{A}{4\pi}} - r_{\mathrm{meas}} = \frac{G}{3c^2}\cdot M,
$$</p>
<p>wherein $G$ is Newton's gravitational constant, $c$ is the speed of light in vacuum, and $M=4\pi\rho r^3/3$. I don't know what $r$ is supposed to be, but it's presumably $\sqrt{\frac{A}{4\pi}}$. Feynman gratifyingly points out that $\frac{G}{3c^2} \approx 2.5\times 10^{-28}$ meters per kilogram (for Earth, this corresponds to a radius excess of about 1.5 mm). Feynman is also careful to point out that this is a statement about <em>average</em> curvature.</p>
<p>Penrose's criticism is: "the "active" mass which is the source of gravity is not simply the same as the energy (according to Einstein's $E=mc^2$); instead, this source is the energy <em>plus the sum of the pressures</em>". Damned if I know what that means -- whose pressure on what?</p>
<p>So my question is, taking into account Penrose's criticism but maintaining Feynman's intuitive style, what's $M$? I only care about average curvature, so please try not to muddy my head with the full Einstein equation. I have a copy of Wald too.</p>
| 513
|
|
general relativity
|
Local Charts in General Relativity
|
https://physics.stackexchange.com/questions/16951/local-charts-in-general-relativity
|
<p>We may consider a "local" region in curved spacetime (local in respect of the spatial and the temporal coordinates). A "local inertial frame" may be constructed by some transformation that produces flat spacetime locally. This transformation produces the diagonal [1,-1,-1,-1] in an approximate manner.</p>
<p>A physical point:</p>
<p>You are performing some experiment in a small laboratory room (stationary w.r.t. the Earth) for a small period of time. Are you in a local inertial frame? If there was a freely falling lift in front of you, it would have been a better approximation to the inertial frame concept. So your laboratory does not qualify to be an inertial frame
when compared with the falling lift. (The difference becomes even more conspicuous if you imagine "gravity" to be a 100 or a thousand times stronger.)</p>
<p>Now, let's move into the problem:</p>
<p>We take a small region of curved spacetime and use some suitable transformation to produce a "local inertial" frame. For this transformation we are <em>disregarding</em> all the anisotropies and inhomogeneities of the surrounding space in the original manifold.Incidentally the objective behind creating local inertial frames is to apply SR. The associated Lorentz transformations are supposed to hold good in the ideal conditions of isotropy and homogeneity of space</p>
<p>Our transformation transforms a small region of curved spacetime to a small region of flat spacetime. To this small/finite flat spacetime, we add the rest of it, ignoring all the anisotropies and inhomogeneities of the surroundings in the original curved spacetime [original manifold]. In effect we are <em>assuming</em> a "Global Transformation" without
working it out! (And we expect correct results when we go back to the original space by the reverse transformations)
We are in effect <em>assuming</em> a "Global Transformation" from curved spacetime to flat spacetime. Is it not as sacrilegious as "Flattening a Sphere"? </p>
<p>How much unreliability is the aforesaid error going to cast on calculations performed in the local inertial frames?To what extent will this error affect the Principle of Equivalence?</p>
|
<p>You cannot do a coordinate transformation in which a curved space becomes flat, and this is because the nonzero curvature is intrinsically meaningful.</p>
<p>The way you define a meaningful notion of curvature in two dimensions is by drawing triangles. In flat space, the sum of the angles in a triangle is 180 degrees, because when you walk around a triangle, turning to face the new walking direction as you hit each vertex, you make one full turn in orientation when you walk once around the triangle, so that the sum of the exterior angles is 360 degrees.</p>
<p>The curvature of a 2 dimensional plane is defined by the difference between the sum of the angles and 180 (or the sum of the exterior angles and 360). This is proportional to the area for small triangles, as you can see by dividing the triangle in two, and noting that the angle-deficit adds up over the two sub-triangles to the big triangle.</p>
<p>The amount of deficit angle per unit area for an infinitesimal triangle defines the intrinsic Gauss curvature of a 2d surface, and this is the content of the Riemann tensor. If a space is flat, this Riemann tensor vanishes indentically. You can compute the Riemann tensor from the metric tensor, and if it is nonzero in one coordinate system it is nonzero in all coordinate systems.</p>
<p>This proves that you can't choose a global frame where GR turns into SR.</p>
| 514
|
general relativity
|
The Light Ray Bends Round!
|
https://physics.stackexchange.com/questions/19429/the-light-ray-bends-round
|
<p>Let's consider the equation y=x in the x-y rectangular Cartesian frame in flat space time. We use the transformations in the first quadrant:
$$y=y'^2$$
$$x=x'$$
$$t=t'$$
For the first transformation we are taking the positive values of $y'$ only.
The equation of y=x in the transformed condition:</p>
<p>$$y'^2=x'$$
If the equation of the straight line:y=x in the original frame represents the path of a light ray in the flat space-time context , the corresponding path in the transformed frame is a parabola.</p>
<p>Options:
1.We may treat the second frame as just a mathematical workspace.
2. We can always think of some manifold in the physical sense where the null geodesic is parabolic!</p>
<p>It is also important to observe that the metrics in both the spaces represent orthogonal systems.One could of course think of going from the first space to the second via some intermediate non-orthogonal transformation.The value of the line element $ds^2$ does not change in the transformation used.</p>
<p>In view of the above considerations can we claim that we can pass from one manifold to another of a different type where the value of where $ds^2$ is preserved?</p>
<p>NB: It would be better to write $x=x’$ and $y=Ay’^2$: where A has the dimension $\frac{1}{Length}$
and y’ the dimension of length. The transformed eqn: $x’=Ay’^2$,y’ maintaining the dimension of length. Transformed metric is $ds^2=dt’^2-dx’^2-(2Ay’dy’)^2=dt’^2-dx’^2-4A^2y'^2dy’^2$ . Ay’ is a dimensionless quantity here.</p>
|
<p>Your coordinate transformation is double-valued, and the line y=0 is singular, the transformation turns around there. This is the reason that you see turning around. When we say coordinate transformations are allowed, they are restricted to be 1-1 and nonsingular Jacobian, so that they are differentiable and differentiably invertible.</p>
| 515
|
general relativity
|
Pseudo-Superluminal Motion and the Synchronization of Clocks
|
https://physics.stackexchange.com/questions/27908/pseudo-superluminal-motion-and-the-synchronization-of-clocks
|
<p>Let's consider two points A an B separated by a finite distance in curved space time. A light ray flashes across an infinitesimally small spatial interval at B. </p>
<p>Metric:
$$ds^2=g_{00}dt^2-g_{11}dx^2-g_{22}dy^2-g_{33}dz^2$$ ----------- (1)</p>
<p>We may write,</p>
<p>$$ds^2=dT^2-dL^2$$
Where,</p>
<p>$dT= \sqrt{g_{00}}dt$ and $dL=\sqrt{ g_{11}dx^2+g_{22}dy^2+g_{33}dz^2}$</p>
<p>dL is the physical infinitesimal(spatial) at B. Observer at A records the same value of dL
But for dT the observers at A and B record different values:
$$dT_{A}=\sqrt{g_{A(00)}}dt$$
$$dT_{B}=\sqrt{g_{B(00)}}dt$$</p>
<p>Local Speed of light[observation from B]$$=c=\frac{dL}{dT_B}=\frac{dL}{\sqrt{g_{B(00)}dt}}$$</p>
<p>Non-Local Speed of Light:[speed measured from A]</p>
<p>$$C_{non-local}=\frac{dL}{dT_A}=\frac{dL}{\sqrt{g_{A(00)}}dt}$$</p>
<p>Therefore :
$$c_{non-local}=c \times \sqrt{\frac{g_{B(00)}}{g_{A(00)}}}$$ --------- (2)</p>
<p>For a particle moving across an infinitesimally small spatial interval at B wehave in a similar fashion:
$$V_{non-local}=V_{local} \times \sqrt{\frac{g_{B(00)}}{g_{A(00)}}}$$ -----(3)</p>
<p>Since $V_{local}<c$
$$V_{non-local}<C_{non-local}$$ ----------- (4)</p>
<p>From (2) we find that the non local speed of light may exceed the local barrier. The non_local speed of a particle may also exceed the local speed barrier. </p>
<p><em>But the light ray is always ahead of the particle [in purely local or non-ocal observations]according to relation (4)</em>. There is no violation of relativity as such.</p>
<p><em>Average speed of light for a specified path</em>:</p>
<p>Let's consider a light ray is passing from P to Q along some path.Observation is being made from the point P.Time interval observed from P is given by:</p>
<p>$$dT=\frac{dL}{C_{non-local}}$$ ------(5)</p>
<p>$$=\frac{dL}{c\times \sqrt{\frac{g_{00}}{g_{P(00)}}}}$$</p>
<p>$g_{00}$ is the value of the metric coefficient at some arbitrary point of the path and $g_{P(00)}$ is the value of the metric coeff. at P.
The average speed of light for the path between P and Q obtained by integrating relation (5),to obtain the time of travelof the light ray, may be different from the constant standard value of “c”</p>
<p><em>Query</em>: Is it important to take care of such differences[differences in the local and non-local speed of light] in the synchronization of clocks in sensitive experiments?</p>
|
<p>This is more or less correct, the intention is clear, but you are saying it in a little bit of a clumsy way.</p>
<p>Your first statement is that if you have light crossing a certain distance, the distance is the same from the point of view of point A and point B, but the time taken to traverse the distance is different. This is correct in the way you set things up, but it isn't best to think of this as the local speed of light being different at A and at B, because the local speed of light is not defined by how far light goes by how much time it takes as measured from A. It is defined by the metric itself--- if the interval-length of the tangent vector to the path of the light ray is zero, then the light ray is <em>by definition</em> moving at the speed of light locally.</p>
<p>Your metric is purely diagonal. This means that little vectors in the direction defined by the t,x,y,z coordinates are always perpendicular to each other. The surface t=constant is therefore a kind of global notion of time, and when you say "a light beam crossed from A to B in a time $\Delta t_A$ as measured at A, you mean that the light crossed from point A at global time slice t, to point B at global time-slice $t+\Delta t$. The time this takes mismatches at A and B, but this is not an indication that the speed of light is different, because both A and B agree that the tangent vector to the light path is still null.</p>
<p>But A and B do not agree on the rate at which global slices separate from each other in time. This effect is extremely important, it is the main effect of general relativity. The first approximation to the effect of a gravitational field, is that there is a local variation in the clock rate. This is a metric of the form</p>
<p>$$ ds^2 = -(c^2 - 2\phi(x,y,z))dt^2 + dx^2 + dy^2 + dz^2$$</p>
<p>With explicit c's (usually you set these to 1), where $\phi(x)$ is the newtonian gravitational potential (divided by c^2). This approximation for the metric is not great--- there are spatial metric terms I ignored which are of the same order as the time metric term for the case of light (this is why Einstein's deflection of light is double Newton's--- this approximation reproduces Newton's value), but ignore these. The effect of the Newtonian potential is to make the clock rate incompatible at different positions, in exactly the way you describe.</p>
<p>For the earth's gravitational field $\phi(x,y,z)= gz$, where g is $9.8{m\over s^2}$the potential increases with height as g The difference in the transit time for different positions leads a person at a height h, who labels the time slices according to the faster clock at the higher location, considers the light travelling a shorter distance per time slice than a person who labels the time slices according to the slower clock at a lower position.</p>
<p>The sychronization failure means that clocks at different altitudes must be configured to run at different rates to stay synchronized, and this is the "general relativistic effect" that people talk about with regard to GPS satellites. In addition to their special relativistic time-dilation due to the orbital speed, which correction factor is proportional to $v^2\over 2c^2$ and makes them tick slower, they also have a General Relativistic correction factor proportional to $2\phi\over c^2$, which makes them tick faster than clocks close to the ground.</p>
<p>The virial theorem shows that in a Kepler orbit, the expected value of the Newtonian potential energy is proportional to the expected value of the Newtonian kinetic energy</p>
<p>$$ v^2/2 = GM/2r $$
$$ \langle {v^2\over 2}\rangle = -{1\over 2}\langle\phi\rangle $$</p>
<p>Where the brackets mean average over one orbit. You can see this directly for a circular orbit from setting the required centripetal equal the Newtonian gravitational force.</p>
<p>$$ {mv^2\over r} = {GmM\over r^2} $$</p>
<p>But the virial theorem says that it is true on average for elliptical orbits too. So the total slowing-down factor for a clock is</p>
<p>$$ {1\over c^2}({v^2\over 2} + \phi) = -{1\over 2} {v^2\over 2c^2}$$</p>
<p>And the GR effect is twich as big as the SR effect, and in the opposite direction--- so the orbiting satellite clock is faster than the stationary ground clock by the expected slowing down amount.</p>
<p>The General Relativistic slowing down of clocks near the Earth's surface is also measured by taking atomic clocks on a plane around the Earth, and this is a famous experiment from the 1970s. The effect is equivalent to the redshifting of light as they travel up a gravitational well (the light loses energy as it climbs up the gravitational well, each photon loses frequency). The lost frequency just means that the photon going from A to B <em>keeps its frequency fixed</em> in terms of the clock ticks of A (or in terms of the clock ticks of B) but since the clocks don't match, the frequency as measured locally is different.</p>
<p>These General Relativistic effects are generally known as weak-equivalence principle tests, since they only test that time dilates in the way described by the local metric above. This dilation was derived by Einstein by analyzing
a uniformly accelerated frame in special relativity, and deriving the relation between the tick-rate of synchronized clocks in the accelerating frame. He assumed that there is no local difference between the uniform Earth's gravity and constant acceleration, so he was able to find the relation between gravitational potential and the tick-rate for clocks that stay synchronized. The derivation is found in elementary GR books (Schutz is good and especially elementary), and it is also found on the <a href="http://en.wikipedia.org/wiki/Gravitational_redshift" rel="nofollow">Wikipedia article for gravitational redshift</a></p>
| 516
|
general relativity
|
How exactly does a real object (e.g. a triangle) behave considering the effects of non-euclidian geometry?
|
https://physics.stackexchange.com/questions/419477/how-exactly-does-a-real-object-e-g-a-triangle-behave-considering-the-effects
|
<p>The question is somewhat related to
<a href="https://physics.stackexchange.com/questions/220055/what-is-the-sum-of-the-angles-of-a-triangle-on-earth-orbit">What is the sum of the angles of a triangle on Earth orbit?</a> but still not quite what I think of.</p>
<p>Consider a real world triangle made of straight steel rods which are
connected and fixed with screws at the corners. When such a triangle is
constructed far away from any mass, i.e. in a euclidean geometry of
space-time, the sum of angles will be 180 degree.</p>
<p>Now bring this triangle in the vicinity of a massive object. The
geometry will become non-euclidean, the sum of angles will not be 180
degrees anymore. How exactly will this effect present itself on the
triangle? I can think of several possibilities:</p>
<ol>
<li>The sum of angles stays at 180 degrees but the rods will bend and/or
experience tension, or the rods will not bend and only experience tension.</li>
<li>The sum of angle deviates from 180 degree but the rods will still
look straight, or the rods will bend, with or without tension.</li>
<li>There is no visible effect because the angle measuring device is bend
exactly the same way as the triangle, with or without tension in the rods.</li>
<li>Any combination of angles, bending, tension, ...</li>
</ol>
<p>What happens when you loosen the screws?</p>
<p>Please note, that this just a warm up question. What I'm really interested in is the ratio of circumference and area of a circle, which happens <em>not</em> to be Pi in a non-euclidean geometry. How does the effect present itself when the circle is laid with tiny tiny tiles. These tiles can be used to approximate the radius and the area, so at least one of these numbers must change when the curvature becomes effective, don't they?</p>
<hr>
<p>Edited:
Clarification on the tiles. Consider a circle on the ground that was constructed in the usual way with a peg at the center, a string, and a pen to draw the circumference. When I lay tiles (that all have the same square shape and area) on the ground and count the number of tiles that are inside the circle or touch the border, and additionally count the tiles that make up the radius, I can get an approximation of $A/r^2$ which will be $\pi$ in a euclidian geometry. Thus, when I bring the circle together with the tiles into an non-euclidian geometry, something must happen to the tiles or the circle to compensate the change in the ratio $A/r^2$. I assume that the tiles will start spreading a bit and gaps will appear between them. Right?</p>
|
<p>I’m not sure if I understand you question good enough, so let me try.</p>
<p>There are several ways to look at the curvature in the vicinty of a central mass. As you talk about a triangle consider the bending of light in relation to the center of the mass. So heuristically it’s clear that the sum of the internal angles of a triangle made of intersecting null geodesics exceeds 180 degrees. </p>
<p>The rods of your steel triangle would feel some bending and would eventually break depending on the “strenght” of the bending. If you mean “tiny tiles” in the sense of freely falling “test particles” then again the sum of the angles of a triangle formed by these would exceed 180 degrees if released from the steel rods.</p>
<p>The circumference you are questioning is $2\pi{r}$, where $r$ is the coordinate radius. The relation between radial proper distance (the distance measured by rulers) and the r-coordinate follows from the Schwarzschild metric as
$d\sigma=\frac{dr}{\left(1-\frac{R_{S}}{r}\right)^{1/2}}$, with $R_S$ the Schwarzschild radius. By integrating one obtains a radial distance measured by rulers expressed by the corresponding difference of the r-coordinates. The above formula shows that the radial proper distance exceeds the coordinate difference and hence the circumference measured by rulers exceeds $2\pi{r}$ as required.</p>
| 517
|
general relativity
|
Angle of a hanging ball in system trying to approach speed of light
|
https://physics.stackexchange.com/questions/432698/angle-of-a-hanging-ball-in-system-trying-to-approach-speed-of-light
|
<p>A common example of acceleration is a ball hanging from the top of the car. The angle this hanging ball makes from zero is dependent on the acceleration of the car.</p>
<p>What happens as we <em>allow</em> the car to <em>attempt</em> to approach the speed of light at constant acceleration?</p>
<p>My expectation is that since the car cannot reach c, it's acceleration must begin to decrease at some point</p>
<p>From this expectation I would further assume that the angle the ball makes to an observer would decrease back to zero.</p>
<p>Is this expectation correct? Where did I go wrong?</p>
|
<p>You have come to correct conclusion, but the reasoning is not correct.</p>
<p>Indeed, from the point of view of observer who is staying still the acceleration of the car is decreasing to 0 as the speed of car approaches <span class="math-container">$c$</span>.</p>
<p>But you can't just use non-relativistic approach to calculate the angle of the thread the object hangs on. At least I do not know how to do it correctly.</p>
<p>Better approach would be first to calculate the position of thread in the frame of reference attached to the car. Looks like it's really simple(*): person inside the car "feels" constant acceleration, so the position of the thread will also be constant. Now we only need to change the frame of reference and find out how this picture looks like in frame of reference which is in rest. There is no need to calculate any forces to do that. Imagine that the body and thread are just painted on the wall of the car. <span class="math-container">$y_1 - y_0$</span> remains the same, <span class="math-container">$x_1 - x_0$</span> gets smaller because of relativistic length contraction, so the angle becomes smaller.</p>
<p>(*) I am not quite sure about this statement. Yes, the person inside the car would feel constant acceleration. But if the gravity field (which from the point of view of observer is now produced by some "earth" which flies around quite fast) - well, I do not know.</p>
| 518
|
general relativity
|
Mathematical question on Mathisson-Papapetrou-Dixon equations
|
https://physics.stackexchange.com/questions/443795/mathematical-question-on-mathisson-papapetrou-dixon-equations
|
<p>I am studying about Mathisson-Papapetrou-Dixon equations which govern the motion of a test particle around a central massive object in the pole-dipole approximation.</p>
<p>Given that <span class="math-container">$S_a=-\frac{1}{2}\epsilon_{abcd}V^bS^{cd}$</span> I want to prove that <span class="math-container">$S^{cd}=-\epsilon^{cdaj}S_aV_j$</span> where <span class="math-container">$\epsilon_{abcd}$</span> is the Levi-Civita symbol, <span class="math-container">$V^b$</span> is the four velocity of the particle and <span class="math-container">$S^{cd}$</span> is it's spin tensor.</p>
<p>Some useful relations I am about to use are <span class="math-container">$S^{cd}=-S^{dc}$</span>, <span class="math-container">$V_aV^a=-1$</span> and <span class="math-container">$V_aS^{ka}=0$</span>. The latter one is called spin supplementary condition and it is imposed in order to have a closed system of equations.</p>
<p>To prove the desirable result I contract <span class="math-container">$S_a$</span> with <span class="math-container">$\epsilon^{ajkl}$</span> so we have <span class="math-container">$\epsilon^{ajkl}S_a=-\frac{1}{2}\epsilon^{ajkl}\epsilon_{abcd}V^bS^{cd}$</span>.</p>
<p>But <span class="math-container">$\epsilon^{ajkl}\epsilon_{abcd}=-\delta_b^l\delta_c^k\delta_d^j+\delta_b^k\delta_c^l\delta_d^j+\delta_b^l\delta_c^j\delta_d^k-\delta_b^j\delta_c^l\delta_d^k-\delta_b^k\delta_c^j\delta_d^l+\delta_b^j\delta_c^k\delta_d^l$</span>. </p>
<p>That means that <span class="math-container">$\epsilon^{ajkl}S_a=-\frac{1}{2}(-V^lS^{kj}+V^kS^{lj}+V^lS^{jk}-V^jS^{lk}-V^kS^{jl}+V^jS^{kl})$</span> so <span class="math-container">$\epsilon^{ajkl}S_a=-(V^lS^{jk}+V^kS^{lj}+V^jS^{kl})$</span> which comes from the antisymmetry of the spin tensor. </p>
<p>Finally contracting with <span class="math-container">$V_j$</span> and using the fact that <span class="math-container">$V_aS^{ka}=0$</span> we get <span class="math-container">$\epsilon^{ajkl}V_jS_a=S^{kl}$</span> which means that <span class="math-container">$S^{kl}=\epsilon^{klaj}S_aV_j$</span> or even better <span class="math-container">$S^{cd}=\epsilon^{cdaj}S_aV_j$</span> which clearly differ with the relation I want to prove by one minus sign. </p>
<p>Can anybody help? Thanks in advance!</p>
|
<p>In Minkowski signature (-,+,+,+) we have
<span class="math-container">$$
\epsilon_{0123}= - \epsilon^{0123}
$$</span>
so I think your identity should read
<span class="math-container">$$
\epsilon_{abcd}\epsilon^{aijk}= -\delta^i_b\delta^j_c \delta^k_d\pm {\rm perms.}
$$</span>
which is minus what you have have. The same is true for the (+,-,-,-) signature.</p>
| 519
|
general relativity
|
Future pointing light cones in Black Hole in Schwarzschild Coordinates
|
https://physics.stackexchange.com/questions/448884/future-pointing-light-cones-in-black-hole-in-schwarzschild-coordinates
|
<p>In examining black holes in Schwarzchild Coords (ie without resorting to other coords) the r coord becomes timeline within the event horizon and the t coord spacelike.</p>
<p>Therefore the light cone is tilted by 90 degrees. However, how do we say which direction in r is future and which is past? (textbooks jump to the forward light cone being radially inward with minimal explanation). </p>
<p>This question is further complicated by the theoretical existence of white holes (even though they are not thought to be a physical reality) where the future light cone is pointing radially outwards. Is there a less hand wavy approach to explaining all of this? </p>
|
<p>You can't expect an explanation if you stick to singular coordinates,
where there is no smooth way to go from without to within.</p>
<p>Shift to well-behaved coordinates (e.g. Kruskal-Szekeres). Then
light-cones behaviour is quite simple and you may easily see which
their relationship must be wrt to Schwarzschild coordinates.</p>
<p>In K-S coordinates light-like geodetics are all at 45°, <span class="math-container">$t=$</span> const
lines are straight lines through the origin, <span class="math-container">$r=$</span> const lines are
equilateral hyperbolas with 45° asymptotes (see e.g. wikipedia article
<a href="https://en.wikipedia.org/wiki/Kruskal-Szekeres_coordinates" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Kruskal-Szekeres_coordinates</a> for
help).</p>
| 520
|
general relativity
|
Are the effects of general relativity accounted for in the calculation of the trajectory of the PARKER probe?
|
https://physics.stackexchange.com/questions/485500/are-the-effects-of-general-relativity-accounted-for-in-the-calculation-of-the-tr
|
<p>I think about the precession of the perihelion of the trajectory as the probe comes closer to the sun than Mercury, which was the first successful test for the general relativity.</p>
| 521
|
|
general relativity
|
Flat space limit of metric blows up in different coordinates
|
https://physics.stackexchange.com/questions/498407/flat-space-limit-of-metric-blows-up-in-different-coordinates
|
<p>Consider the metric </p>
<p><span class="math-container">$$ds^2=dv^2-dt^2+v^2 \mathcal{R}^2 d\Omega^2_3,$$</span></p>
<p>where <span class="math-container">$d\Omega^2_{3}$</span> is the metric of the <span class="math-container">$3$</span>-sphere, and <span class="math-container">$\mathcal{R}$</span> is the radius. I want to bring the metric in the form </p>
<p><span class="math-container">$$ds^2=\frac{du^2}{f(u)}+e^{2A(u)}(-f(u)dt^2+\mathcal{R}^2 d\Omega^2_3).$$</span></p>
<p>This can be done by substituting <span class="math-container">$v=e^u$</span>, and up to various rescalings of the coordinates we identify <span class="math-container">$f(u)=e^{-2u}$</span>, <span class="math-container">$A(u)=u$</span>. In this new coordinate system, the scalar curvature is </p>
<p><span class="math-container">$$R \propto e^{-2u},$$</span></p>
<p>whereas the Kretschmann invariant is </p>
<p><span class="math-container">$$R_{\mu \nu \kappa \lambda} R^{\mu \nu \kappa \lambda} \propto e^{-4u}.$$</span></p>
<p>Taking the limit <span class="math-container">$v \rightarrow 0$</span> corresponds to shrinking the radius of the <span class="math-container">$3$</span>-sphere to zero, and thus we are left with flat space. But the appropriate limit in the <span class="math-container">$u$</span>-coordinates is <span class="math-container">$u \rightarrow -\infty$</span>, where the curvature invariants blow up. This does not make sense, as in that limit the invariants must vanish. I thought I must have done a mistake in calculating the invariants, but I really can't find one. Can anyone think of what might by going wrong?</p>
| 522
|
|
general relativity
|
Consistent initial data for evolution in time with the Newman-Penrose formalism
|
https://physics.stackexchange.com/questions/531467/consistent-initial-data-for-evolution-in-time-with-the-newman-penrose-formalism
|
<p>In the <a href="http://www.scholarpedia.org/article/Spin-coefficient_formalism" rel="nofollow noreferrer">Newman-Penrose formalism</a>, one rewrites the Einstein equations in terms of a system of linear transport equations for the Newman-Penrose scalars. I am considering the initial value problem for this system of equations: I want to set initial data on a spacelike hypersurface, and then evolve (a subset of) the Newman-Penrose equations in time. </p>
<p>I do not know though how to set up initial data for all the Newman-Penrose scalars that are consistent with the constraints. For example, if I set
<span class="math-container">$\Psi_4$</span> and <span class="math-container">$\partial_t\Psi_4$</span> on the spacelike hypersurface (i.e. I treat their values as free initial data), presumably I cannot freely specify the rest of the Newman-Penrose scalars on the initial hypersurface without violating the constraint equations of the Einstein equations. (Note I want to evolve with the time function <span class="math-container">$t$</span>, and not use the Newman-Penrose equations in terms of the characteristic initial value problem). </p>
<p>Is there a general procedure for determining consistent initial data in the Newman-Penrose formalism?</p>
<p>EDIT: to clarify, I'm considering the initial value problem in vacuum (no matter fields)</p>
| 523
|
|
general relativity
|
Projection of 4D geodesic in spacetime onto 3D space is an ellipse
|
https://physics.stackexchange.com/questions/552566/projection-of-4d-geodesic-in-spacetime-onto-3d-space-is-an-ellipse
|
<p>In the Wikipedia article on "Geodesics in general relativity" we can find the following statement:</p>
<blockquote>
<p>In general relativity, gravity can be regarded as not a force but a
consequence of a curved spacetime geometry where the source of
curvature is the stress–energy tensor (representing matter, for
instance). <strong>Thus, for example, the path of a planet orbiting a star is
the projection of a geodesic of the curved four-dimensional (4-D)
spacetime geometry around the star onto three-dimensional (3-D) space.</strong></p>
</blockquote>
<p>Now, I take the last sentence to mean that, for example, the orbit of the earth around the sun appears to us as elliptical because with our eyes we can only perceive space as being three-dimensional. That is we can only perceive the curvature of spacetime indirectly. However, can this statement be made mathematically precise? Can we calculate a geodesic for this situation and show that its projection into three dimensions is an ellipse (maybe by choosing coordinates in such a way that the 3d orbit is embedded in a 2d plane)? When we want to project a great circle of the unit sphere <span class="math-container">$S^3 \subset \mathbb{R}^4$</span> onto a hyperplane <span class="math-container">$H \subset \mathbb{R}^3$</span> there's not really a natural choice for this hyperplane. So in this case we cannot speak of "the" projection of the geodesic onto three-dimensional space. Is the situation in general relativity somehow different? Is there a natural choice in this situation? If this can be found in a textbook I'm of course happy for a pointer. I'm a mathematics graduate student with a background in differential geometry but have not studied any GR yet.</p>
<p><strong>EDIT:</strong> In response to comments I feel the need to clarify the question maybe a bit more. The point of my question is not to which extent earths orbit is elliptical (which is an interesting question in itself, but not what I intended to ask). For all I care it can be perfectly circular. Above I highlighted a sentence from Wikipedia and I want to know if the statement can be made mathematically precise. Let me explain what I mean by this. Let's start with the geodesic equation from GR:
<span class="math-container">$$ {d^{2}x^{\mu } \over ds^{2}}+\Gamma ^{\mu }{}_{\alpha \beta }{dx^{\alpha } \over ds}{dx^{\beta } \over ds}=0\ $$</span>
A geodesic in spacetime <span class="math-container">$M$</span> (i.e. a four-dimensional, smooth, connected Lorentzian manifold) is a solution curve <span class="math-container">$\gamma:[a,b]\to M$</span> to this equation. As far as I understand it, the most basic model for a curved spacetime in the vicinity of a central fixed mass is given by the Schwarzschild metric. So we can probably neglect the mass of a planet placed nearby this central mass and just solve the geodesic equations using the Schwarzschild metric. As has been noted in the comments we would expect this planet to have a helical worldline (if its orbit in space is confined to a fixed 2d plane). I'd love to see a derivation of a geodesic from the equation above, such that we can see that the resulting curve is tracing out a 3d helix of some sort. So the final result of the calculation would be something like <span class="math-container">$$\gamma(t) = (t, \cos(t), \sin(t), 0).$$</span></p>
| 524
|
|
general relativity
|
Regarding nothingness and motion in the ocean of Matter Energy
|
https://physics.stackexchange.com/questions/658803/regarding-nothingness-and-motion-in-the-ocean-of-matter-energy
|
<p>I am but a humble college dropout with a keen interest in physics, but one way or another I feel I have discovered some key information that will lead to a solid theory of everything that should satisfy all parties it concerns, that is, if I can correctly format and word it. My hope here is simply some respectfully conversation regarding my outlook on things.</p>
<p>I have come to see energy and matter in a different way, in order to do. For any physical thing to move in any way there must be some kind of room to move within it, yes? In the fabric of space? This was one of the first thoughts I had regarding my physics research. How can things move without some kind of nothingness within or around them to allow them to move? Without some space within a particle, how could it even shape and form? It would be infinitely dense wouldn't it, it would have no energy and will have reached absolute zero and collapsed into a singularity. What is Nothing anyway? Nothing is infinite isn't it, if you think about it?</p>
<p>It helps me to build on this understanding picturing nothing (energy) as white, and matter (physical things) as black. Yet if you were to quantify it it would be 0, surely? Yet nothingness itself is infinite. Every physical thing must exist upon nothingness. It is the canvas for which all things lay – the very flow of time on which we exist.</p>
<p>Now I started to wonder how spacetime could bend, shape, and form, if it is truly the nothingness, background, that Einstein describes it as. Nothing cannot have shape? Also, seeing as bubbles and waves of matter and energy emerge from it, and from the singularity, in a big bang outlook, is it not feasible that spacetime is dark matter, dark energy respectively?</p>
<p>Lets talk about a singularity for a moment. When a star collapses due to every action having an equal and opposite reaction etc. The, its core collapses into a point of infinite density. It is pure matter (space) at that point, it has lost all its energy. The phenomena of dark energy is present as the singularity rips space and leaves a gap where the time, the energy, that flows though all is revealed and we are forever expanding into it, (visualize it) at an accelerated rate.</p>
<p>Dark matter is a phenomenon of space as it has properties where without relativity, which is only perception, space can be deeper at certain areas, or 'thinner' (for lack of a better word). Which would allow us to use time dilation for things if we could understand this relativity of space. Also it doesn't interact with the electromagnetic field – if it were space – that would explain why.</p>
<p>Look at the bells spaceship paradox for instance – two ships tied to each other with a string, identical, accelerating at the same rate. Eventually the string snaps. This is because of the property of space, in that there has literally been an increase in space between them. Because spacetime is matter and energy, is it not obvious?</p>
<p>Now if I may quickly talk about the 4 forces that we know of. I have been thinking about all the subatomic particles and their functions, it seems they would all share some similar traits, and this whole spin number etc. thing actually confused me. Here's how I see it: They spin to the left or vice versa, or in some cases, do not spin. This depends on the mass, motion and fields present. I think we can extrapolate a lot of info on why they interact that way on a small scale from simply looking at the field around a black hole.</p>
<p>Also this upstream contamination issue in physics, what's that about? Seems obvious how they could roll on up there.</p>
<p>Anyway I have a lot more to talk about but am a little distracted from other work at the moment, I can expand on/talk about this all day but I shall await responses before adding more. Seems we are on a wild goose chase for answers that may be simpler than we realize, the more you see the less you know, as they say.</p>
<p>Well I'll leave this here for now, see how the responses go though and add from there. Apologies for sloppy format and unclear explanations, there is a lot on my mind for now, I'm just looking for someone to try an envision these thought experiments that I have done in order to better explain my revelations to people – so, I'll leave it the for now and see where this goes.</p>
<p>I also recently wrote a second iteration of an outlook I hold that may allow to relations to an effect of a real TOE, on Reddit, however in my bad mood due to typical sheeple like (excuse the term please) responses, I also added another to sarcastically declaration that I am the crazy one.</p>
<p>Anyway, I'm just looking for potential conversation, video, or meet via video link etc. to talk to a physicist that may understand, it is of great importance.</p>
|
<p>If you don't know all the reasons why people think they are right, you'll never be able to convince them that they are wrong.</p>
<p>So I suggest first learning more about how most physicists see the Universe, if you want any hope of convincing them otherwise.</p>
<p>Theories of everything do attempt to connect empty space to matter and energy.</p>
<p>Many physicists do think dark energy is a property of empty space. But dark matter has to satisfy many properties (cosmological evolution, formation of structure, gravitational lensing of clusters, rotation curves) so theories for dark matter are very constrained.</p>
<p>The more you value your ideas, the more you ought to dress them up nicely. This means organizing your thoughts to make them as easy to understand as possible. Do your best to spend time formatting and making everything clear, you owe it to yourself.</p>
| 525
|
general relativity
|
Does a volume stay constant when freely falling?
|
https://physics.stackexchange.com/questions/709750/does-a-volume-stay-constant-when-freely-falling
|
<p>In general relativity, if a volume of particles moves unrestricted through spacetime, is their volume always conserved?</p>
<p>Say we let a collection of particles at rest wrt each other, fall freely in a gravitational field. Will tidal forces keep the volume they occupy constant? My intuition says yes, but how do we prove that? If the metric varies it seems reasonable that the number of small cubes stays the same (while the shape varies,).</p>
<p>Assume the small particles don't attract each other.</p>
|
<p>In general no, the volume is not conserved. To take an extreme example, consider a shell of particles surrounding the Earth. As they freely fall, their volume will definitely decrease!</p>
<p>EDIT: my example is extreme, of course. Baez at <a href="https://math.ucr.edu/home/baez/einstein/node5.html" rel="noreferrer">https://math.ucr.edu/home/baez/einstein/node5.html</a> points out that the volume of a "small" sphere of test particles in vacuum does not change, and I'm sure he's right about that. But in my example there's a mass at the center, so the whole volume is not in vacuum.</p>
<p>EDIT: to put some numbers on this, I'll calculate the volume of a 1 km thick shell falling near the Earth's surface (so the inner radius is about the same as Earth's radius, 6372000 meters, and the outer radius is 6373000 meters). Near Earth's surface Newton's equations are an excellent approximation for gravity, so the radius of a falling shell as a function of time is given by the differential equation <span class="math-container">$r'' = \frac{GM}{r^2}$</span>, where <span class="math-container">$GM$</span> is approximately <span class="math-container">$4 \times 10^{14} m^3 / s^2$</span> for Earth. Unfortunately there isn't a nice closed form solution for this equation, but we can solve it numerically using Wolfram Alpha with something like</p>
<blockquote>
<p>solve [y'' + (4*10^14) / (y-6372000)^2 = 0, y(0)=0, y'(0)=0] from 0 to 10 using r k f</p>
</blockquote>
<p>That is for the inner surface of the shell, and gives a delta of -492.569, which is reasonable (we know gravitational acceleration is about 9.8 m/s^2 near earth's surface, so we expect an answer near 490 m for a 10 second fall).</p>
<p>Doing the same for the outer surface gives a delta of -492.415, which again is reasonable (it falls slightly less because it starts further away).</p>
<p>Finally, calculating the volumes gives us a starting volume of
<span class="math-container">$4/3 * \pi * 1.21826269 * 10^{17}$</span> and a final volume (after 10 seconds) of <span class="math-container">$4/3 * \pi * 1.21826197641753715846882634 * 10^{17}$</span>, so the volume of the shell has indeed decreased as expected.</p>
| 526
|
general relativity
|
Correct my understanding of the logical flow of the formulation of General Relativity
|
https://physics.stackexchange.com/questions/731841/correct-my-understanding-of-the-logical-flow-of-the-formulation-of-general-relat
|
<p>Correct and direct me when I am wrong.
That is how I understand the logical flow of the formulation of the general theory of relativity:
First we have the equivalence principle, that one tells us that if want to see the effect of gravity just figure out the Physics in some frame accelerating with g in a semi flat space . We figured out the Physics and it looked in there like if we were doing Physics in some curved surface or something. So, we conclude that including the gravitational field in any metric makes the spacetime described by that metric curved. That is why gravity is the curvature of spacetime! So, we now throw away this accelerated frame and deal from now on with completely curved spacetime whose curvature is due the inclusion of the gravitational field in its metric and we are guaranteed to obtain the effects we hoped for ...Is anything of what I have said right?</p>
|
<blockquote>
<p>First we have the equivalence principle, that one tells us that if want to see the effect of gravity just figure out the Physics in some frame accelerating with g in a semi flat space</p>
</blockquote>
<p>This is ok, except it is important to emphasize that the equivalence principle is a local principle. It only describes a small neighborhood of spacetime around some given event.</p>
<p>Within the local frame it explains why an accelerometer does not detect gravity: locally gravity is a fictitious force and accelerometers do not detect fictitious forces. It also means that an object with an accelerometer that reads 0, an inertial object, forms a straight line in spacetime. The parabola that a projectile makes is due to using curved coordinates, the projectile itself forms a straight line in spacetime.</p>
<blockquote>
<p>We figured out the Physics and it looked in there like if we were doing Physics in some curved surface or something.</p>
</blockquote>
<p>Digging a bit deeper here is worthwhile. As mentioned above, an object with an accelerometer that reads 0 forms a straight line in spacetime. Furthermore, two objects that are at rest with respect to each other form parallel lines in spacetime.</p>
<p>Now, in the absence of gravity, two inertial objects that start out at rest with respect to each other remain mutually at rest. In spacetime this is the simple fact that two straight parallel lines keep the same distance and never intersect.</p>
<p>However, when there is tidal gravity, gravity that varies in space, you can have two objects that are initially at rest, have always 0 accelerometer readings, but eventually collide. In spacetime this implies two straight lines that are parallel at one point but intersect at another. This is impossible in a flat spacetime, but is possible in a curved spacetime.</p>
<p>This is why the equivalence principle implies that tidal gravity is represented by spacetime curvature.</p>
<blockquote>
<p>So, we now throw away this accelerated frame and deal from now on with completely curved spacetime whose curvature is due the inclusion of the gravitational field in its metric and we are guaranteed to obtain the effects we hoped for</p>
</blockquote>
<p>We don’t have to throw away the locally accelerated frame. We have to use a coordinate-independent formulation of the laws of physics. Once we have that we can use any coordinates we like just the same. Including the accelerated coordinates.</p>
| 527
|
general relativity
|
Relativity: A modification on Sea Tower experiment
|
https://physics.stackexchange.com/questions/223645/relativity-a-modification-on-sea-tower-experiment
|
<p>I first read about it on A Brief History of Time(Stephen Hawking). In 1962, a relativity experiment was executed: identical (classical) watches put on a water tower, one is on very high, other one is at the bottom. And of course, after a certain time, the one which is closer to ground was slower than the other.</p>
<p>What I want to learn is what happens if I add this:</p>
<p><a href="https://i.sstatic.net/1PUm5.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1PUm5.jpg" alt="sea tower experiment"></a></p>
<p>and re-do the experiment? The stick bounds the wheels of the identical watches externally. It has no mass, it is non-flexible, it has no friction, strong enough etc. So, think it as an ideal stick for phsic exam question.</p>
<p>What happens then? If without stick, lower clock is 30 minutes late than higher one, with stick, will it be 15 mins further from reference clocks? Or something entirely different will happen? Thanks.</p>
<p>P.S: All the mechanism that clocks use are fully Analog, not stepped! </p>
|
<p>The upper clock tries to force the lower clock to speed up, by exerting a force on the shaft, the force is produced by a motor.</p>
<p>The lower clock tries to force the upper clock to slow down, by exerting a force on the shaft, the force is produced by the mechanism that limits the clocks rate in normal use.</p>
<p>Which clock wins? The lower clock wins, except if the upper clock manages to break the lower clock's clock mechanism. </p>
<p>If the questioner does not like this kind of trivial answer, then the questioner should get rid of the clock mechanism of the lower clock.</p>
| 528
|
general relativity
|
Which Photon would win the race?
|
https://physics.stackexchange.com/questions/289025/which-photon-would-win-the-race
|
<p><a href="https://i.sstatic.net/VDHUg.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/VDHUg.jpg" alt="enter image description here"></a>Imagine that the Sun is not rotating. It also has a tunnel throughout its body exactly through the core. Please disregard any other effect then gravity.
From a far away point A (far from the Sun) I would shoot two entangled photons, one through the tunnel, and one a little bit next to the Sun (so just to pass next to the Sun). Point B is so that the line from point A to point B is on a straight (in 3D) line through the tunnel. </p>
<p>Now I am shooting the photons at an angle, such that both entangled photons will pass through both point A and B. (given GR/SR bending effects, the photon#2's path will be bent just right to pass through point B.)</p>
<p>Questions:</p>
<ol>
<li>Which photon would according to GR/SR arrive at point B first?</li>
<li>My question's answer could sound at first obvious, but considering that the photon travelling inside the tunnel would be affected by gravity too, the answer is not obvious.</li>
<li>What if I use neutrinos or electrons(or anything fast enough with some rest mass ) instead of photons?</li>
</ol>
|
<p>It seems to me that your central question is: <em>Does the particular lensed path light travels on affect its arrival time?</em> If so, the answer is a resounding <strong>yes</strong>, with evidence from astronomy! For example, the Twin Quasar (<a href="https://en.wikipedia.org/wiki/Twin_Quasar" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Twin_Quasar</a>) is a system consisting of two images of the same quasar, which traveled different paths to the observer and hence generate two different images. The interesting part of this is that <em>one of the images is 400 days behind the other</em>! Researchers identify this difference with a path length difference of 1.1 ly. So it seems that the photon with shorter path length (i.e. photon #1) would win.</p>
<p>To anticipate some attacks involving the fact that space is curved around a massive object and so maybe photon #2 wins, I say that the problem never provides dimensions, in particular the vertical separation between points A and B. Take those two points far enough away from each other, and it becomes extremely clear that, at least in this limit, photon 1 wins. At that point the total relative alteration to photon 1's path becomes miniscule, but photon 2 still has to travel in quite a large arc. My argument here is that since the Sun is (hopefully) not (yet) a compact object, but instead has a rather low density, significant curvature will not be seen within its body. Therefore, the same gentle-curvature limit should apply in all reasonable cases.</p>
| 529
|
general relativity
|
Motion of a planet around a spherically distributed mass
|
https://physics.stackexchange.com/questions/596430/motion-of-a-planet-around-a-spherically-distributed-mass
|
<p>I was studying a book which claims that, after a lot of math, the metric of the isotropic spacetime around a spherically symmetric is, approximately <span class="math-container">$$ds^2 = -A(r)dt^2 + B(r)dt^2 + r^2 d\Omega^2$$</span></p>
<p>So, the motion of a planet around this mass is given by the Lagrangian L</p>
<p><span class="math-container">$$L = [A(r)dt^2/d\tau^2 - B(r)dt^2/d\tau^2 - r^2 d\Omega^2/d\tau^2]^{1/2}$$</span></p>
<p>I am not sure how does we get this Lagrangian, probably i am forget something. Could you help me? I will not post any attempt to find it because, actually, i think this is derived by some definition of Lagrangian in spacetime that i am missing.</p>
|
<p>The Lagrangian of a single particle in General Relativity (see <a href="https://en.wikipedia.org/wiki/Relativistic_Lagrangian_mechanics#Lagrangian_formulation_in_general_relativity" rel="nofollow noreferrer">Wikipedia</a>, for example) is given by:</p>
<p><span class="math-container">$$\mathcal{L} = -mc^2\sqrt{g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu} + \mathcal{L}_I,$$</span> where <span class="math-container">$\mathcal{L}_I$</span> represents an interaction of some sort (not gravity). In this case, the particle is essentially "free" (i.e., in free fall in a gravitational field) and so <span class="math-container">$\mathcal{L}_I = 0$</span>. The form that you have therefore follows.</p>
| 530
|
general relativity
|
No mass in the Big Bang?
|
https://physics.stackexchange.com/questions/605193/no-mass-in-the-big-bang
|
<p>Roger Penrose says in the Big Bang there was no mass!!
<a href="https://www.youtube.com/watch?v=OFqjA5ekmoY" rel="nofollow noreferrer">https://www.youtube.com/watch?v=OFqjA5ekmoY</a>
because E=M.c2. ( minute ca. 6). So where came the mass from.</p>
|
<p>The big bang was before the spontaneous symmetry breaking, so there was no massive particles.</p>
| 531
|
general relativity
|
Where do these vectors come from?
|
https://physics.stackexchange.com/questions/607861/where-do-these-vectors-come-from
|
<p>I am reading <a href="http://www.fis.puc.cl/%7Erbenguri/3.pdf" rel="nofollow noreferrer">this paper</a> (link to pdf) by Benguria et al. <em>Aspects of the Hamiltonian dynamics of interacting grvitational gauge andd Higgs fields with application to spherical symmetry</em></p>
<p>At page 13 of the paper the authors set up to solve an equation of the form:
<span class="math-container">$$\mathcal{L}_{L_{a}}A^b=[L_a,A^b]=-\epsilon_{abc}A^c$$</span></p>
<p>where <span class="math-container">$A^b$</span> is a vector and <span class="math-container">$L_a$</span> are the angular momentum components and <span class="math-container">$\mathcal{L}_L$</span> is the Lie derivative. The authors say that the most general solution to this equation is of the form:
<span class="math-container">$$\alpha(r)L_a+\beta(r)M_a+\gamma(r)K_a$$</span>
where the vectors <span class="math-container">$M_a$</span> and <span class="math-container">$K_a$</span> are given by:
<span class="math-container">$$M_1=\cos\theta\cos\phi \partial_\theta-\dfrac{\sin\phi}{\sin\theta}\partial_\phi$$</span>
<span class="math-container">$$M_2=\cos\theta\sin\phi \partial_\theta+\dfrac{\cos\phi}{\sin\theta}\partial_\phi$$</span>
<span class="math-container">$$M_3=-\sin\theta \partial_\theta$$</span>
<span class="math-container">$$K_1=\sin\theta\cos\phi \partial_r$$</span>
<span class="math-container">$$K_2=\sin\theta \sin\phi\partial_r$$</span>
<span class="math-container">$$K_3=\cos\theta \partial_r$$</span></p>
<p>and satisfy the algebra:
<span class="math-container">$$[L_a,L_b]=-\epsilon_{abc}L_c \quad [L_a,K_b]=-\epsilon_{abc}K_c \quad [L_a,M_b]=-\epsilon_{abc}M_c$$</span></p>
<p>Where do these vectors come from?</p>
| 532
|
|
general relativity
|
If gravity impacts length measurements, and length measurements impact gravity, how do we resolve net gravity? (paradox I can't resolve)
|
https://physics.stackexchange.com/questions/608614/if-gravity-impacts-length-measurements-and-length-measurements-impact-gravity
|
<p>Here's the concept. We see a very dense 1.5 km radius asteroid, and my friend Charlie and I fly up in our spaceship to check it out. I fly close on my bathroom scale, equipped with rocket thruster, and hover above the non-rotating asteroid. I have 100 kg mass.</p>
<p>My scale reads 2.05864e15 Newtons, so gravitational acceleration g = 2.05864e13 m/s2. I then ping the asteroid with a radio signal, and the round trip takes 2 microseconds, by which I calculate the distance to the asteroid as 300 meters, and to the astroid's center as 1800 meters. I then use g = GM/r^2 to figure out that M = 1.00e30 kg. I radio back to Charlie, who has been watching from a million miles away.</p>
<p>Charlie says, "You are wrong about the mass. I've been watching via Zoom. Due to the high gravity, your time is running slowly by a factor of 1.41. The ping didn't take 2 microseconds, it took 2.82 microseconds. I figure your value of R to use in GM/R^2 is 1923.4 meters. That means the asteroid mass works out to 1.14e30 kg.</p>
<p>Who is correct? What is the mass of the asteroid? Things the two observers agree on: the asteroid is 1.5 km in radius as measured from far away, the gravitational acceleration is 2.05864e13 m/s2 at a point where the local observer measures the distance to the surface as 1 light-microsecond and the more distant observer measures the distance as 1.42 light-microseconds. What was the error of the incorrect observer?</p>
|
<p>There's an unfortunate tendency in descriptions of relativity (special especially) to say that observers "disagree" about measured quantities, as though they'd actually get into an argument about it, along the lines of "<span class="math-container">$Δx=2\text{m}$</span>!" "No, <span class="math-container">$Δx=3\text{m}!$</span>"</p>
<p>If you make the reasonable assumption that participants in thought experiments understand the physical laws that they're applying, then this will never happen, because they'll understand that they're describing the same physical situation in different ways, and the only reason they've quoted different values for a quantity named <span class="math-container">$Δx$</span> is that they're using the name <span class="math-container">$Δx$</span> to refer to two different things. Hopefully, one of them will agree to change the name of their thing to <span class="math-container">$Δx'$</span>, and then there will be no more "disagreement". Coordinates in relativity are like coordinates in Euclidean geometry, and there's nothing deeper in these "disagreements" than there would be in Euclidean geometry.</p>
<p>Given the extreme values involved here, you can't use a Newtonian approximation. Your proper acceleration is not <span class="math-container">$GM/r^2$</span> but <span class="math-container">$GM/(r^2\sqrt{1-r_s/r})$</span> (where <span class="math-container">$r_s=2GM/c^2$</span>), and the ping time is not <span class="math-container">$2(r_2-r_1)/c$</span> but <span class="math-container">$2(r_2-r_1+r_s\ln(r_2-r_s)/(r_1-r_s))\sqrt{1-r_s/r_2}/c$</span>. Solving these equations for <span class="math-container">$M$</span> and your radius is not entirely trivial but it's doable.</p>
<p>Charlie, knowing your proper acceleration, his own proper acceleration, and the ping time redshifted by a factor of <span class="math-container">$\sqrt{(1-r_s/r_3)/(1-r_s/r_2)}$</span>, can solve for M and the two unknown radii. Using the ping time he measures as though it was your measured ping time will produce the wrong answer because it's just the wrong calculation.</p>
<p>I should mention one other issue here: the notion of "radius" is not entirely unambiguous. Normally in Schwarzschild coordinates you use a so-called "reduced circumference" which is <span class="math-container">$1/2π$</span> times the circumference of a circle centered on the massive body. If you want to use these formulas then you need to use a reduced-circumference radius for the asteroid (<span class="math-container">$r_1$</span>). This is not "as measured from far away". In fact it makes no sense to talk about a radius measured from far away. You could measure the angular diameter of the asteroid from far away. Then you'd have to equate the measured angle to an angle calculated as a function of your position, the radius of the asteroid, and its mass (by calculating the geodesic path of a light beam that just grazes the asteroid). That would give you a system of one more equation in one more variable, which you could solve.</p>
| 533
|
general relativity
|
Using relativity to suppress classical inertial acceleration
|
https://physics.stackexchange.com/questions/611328/using-relativity-to-suppress-classical-inertial-acceleration
|
<p>This is a question from a mathematics student trying to visualize the fact that general relativity is based on a concept of, “identifying gravity with the curvature of spacetime” (sincere apologies for probable physics inaccuracies and useless details).</p>
<p>Suppose an absolute spacetime, a time-dependent mass distribution <span class="math-container">$p(t)$</span> and a time-dependent mass distribution <span class="math-container">$M(t)$</span>, both with smooth trajectories starting at <span class="math-container">$t=0$</span> (<span class="math-container">$t<0$</span> is not considered here) and momenta low enough so that Newtonian gravity approximates Einsteinian gravity.</p>
<p>Suppose also that total mass <span class="math-container">$m(p)$</span> doesn't change over time and that, at all times, <span class="math-container">$m(p)\ll M(t)$</span> and <span class="math-container">$p$</span> and <span class="math-container">$M$</span> are close enough so that the influence of <span class="math-container">$p$</span> over the gravitational field is neglected here.</p>
<p>Now denote <span class="math-container">$\vec{A}(t)$</span> the acceleration of <span class="math-container">$p$</span> at time <span class="math-container">$t$</span> and <span class="math-container">$\vec{A}_{G}(t)$</span> its component resulting from the gravitational influence of <span class="math-container">$M$</span>, the goal being to have a pretty arbitrary smooth vector field acting as inertial acceleration on <span class="math-container">$p$</span> trajectory.</p>
<p>Under those hypotheses, or similar or better-formulated ones, can we state, with an error controlled by scales of approximations made, that there exists an Einstein metric on spacetime such that the spatial acceleration <span class="math-container">$\vec{A^{*}}(t)$</span> of <span class="math-container">$p$</span> at time <span class="math-container">$t$</span> in the new manifold is the parallel transport of <span class="math-container">$\vec{A}(t)-\vec{A}_{G}(t)$</span> from <span class="math-container">$p(0)$</span> to <span class="math-container">$p(t)$</span> along the trajectory ?</p>
<p>In this case we could write (in a broad sense)<span class="math-container">$$\vec{A^{*}}=\vec{A}-\vec{A}_{G}$$</span>so the gravity component of acceleration woud have been “replaced” adequately.</p>
<p>Any explanation of any of my misconceptions would be greatly appreciated.</p>
|
<p>This is the simplest way I know to see the basis of General Relativity. It sacrifices math for simplicity.</p>
<p>See <a href="https://physics.stackexchange.com/q/178417/37364">Why can't I do this to get infinite energy?</a> for the basis of General Relativity. This tells you the physical reason why time runs slower near Earth than high above Earth.</p>
<hr />
<p>This shows that spacetime is curved.</p>
<p>Space is curved if you don't come back to your starting point when you walk around a square. Or equivalently, you wind up at different points if you walk east-then-north vs north-then-east.</p>
<p>The surface of the Earth is curved in this sense. It doesn't show for a small square. But try a really large square. Start on the equator.</p>
<ul>
<li><p>Walk 1/4 of the way around the world to the east. Turn left and walk 1/4 of the way around the world to the north. You are at the north pole.</p>
</li>
<li><p>Walk 1/4 of the way around the world to the north. Turn right, and walk 1/4 of the way around the world. (OK, it isn't east because coordinates are weird at the north pole.) But you are on the equator.</p>
</li>
</ul>
<p>Space-time is 4 dimensional, so you get an extra direction you can walk around the block. You can also wait a while.</p>
<p>So trace out this "square" where one side is distance, and the other time. Start at a point above the Earth.</p>
<ul>
<li><p>Have a person at the top wait a bit, then find the point/time a
distance X below him at that time.</p>
</li>
<li><p>Find the point/time a distance X below the top person right now. Have someone at that bottom point wait a bit.</p>
</li>
</ul>
<p>Time is slower at the bottom. In his travel through time, the bottom person passes through the the point/time the top person picks out. But when he does, he isn't done waiting yet.</p>
<hr />
<p>Finally, the curvature of spacetime causes gravity.</p>
<p>The easiest example is the deflection of starlight, the effect Einstein proposed as evidence for his theory.</p>
<p>Light from a distant star is a plane wave by the time it arrives at the solar system. If the geometry is just right, part of the wave will skim the surface of the Sun and continue onward to arrive at Earth. Normally light from the Sun would make the starlight invisible. But an eclipse is happening at the same time. The moon is positioned just right to block the sunlight, but not the starlight.</p>
<p>Part of the plane wave skims the Sun, where an observer sitting some distance above the Sun sees time running slowly. Part skims the distant observer, where he sees time running normally. He sees starlight near himself traveling at the speed of light. In time <span class="math-container">$t$</span>, light travels distance <span class="math-container">$x$</span>. Light near the surface also travels at the speed of light. But clocks run slower. In the same time <span class="math-container">$t$</span>, a shorter time will have passed at the surface. Light will travel a shorter distance.</p>
<p>This means that the wavefront will be tilted. The part near the Sun will fall behind. Since light travels perpendicular to the wavefront, light will be deflected.</p>
<p>On Earth, this means we will see the starlight traveling in a slightly different direction than it would have. We see this as an apparent shift in the position of the star.</p>
<hr />
<p>The experiment was done. The star did indeed shift, but the shift was twice as big as expected.</p>
<p>Einstein went back to the drawing board and this time showed that there would be distortions in distance as well as time. With this, one can derive a correct theory of gravity that predicts all the effects we measure.</p>
| 534
|
general relativity
|
How does the gravitational field of a massive body affect the orbit of that body around yet another massive body?
|
https://physics.stackexchange.com/questions/574129/how-does-the-gravitational-field-of-a-massive-body-affect-the-orbit-of-that-body
|
<p>If I understand the relativistic explanation of gravitation - that it is curvature of spacetime - then a particle left alone will travel along that curved spacetime in a path that depends on the local curvature at each instant in time (right?). If so, then the orbit of a small body without significant effects on that local curvature is something I can get my head around. However, if I have a body that non-negligibly curves spacetime (say...two black holes? But I expect that a planet around a star would fit my bill here), does the curvature caused by the smaller body affect its path? This sounds to me like it would be a weird self-interaction that wouldn't have any effect, but in my mind I am divorcing the curvature of spacetime from the force-interaction that bodies experience in that curved spacetime, and thus I am having trouble wrapping my head around it.</p>
|
<p>Yes, the gravitational field of the smaller body does have an impact on its own orbit. This effect is known as the “gravitational self force”. The gravitational field of the smaller body can be separated in two parts. The first is the “direct” field of the body. This is akin to the field it would produce in complete isolation. It is symmetrical and has no impact on the body’s orbit.</p>
<p>The second is the “tail” part and can be thought of as the difference to the body’s gravitational field due to sitting in the background curvature caused by the larger body as compared to sitting in a perfectly flat environment. It is this part that is responsible for the gravitational self force.</p>
| 535
|
general relativity
|
Misconception in the implications of the gravitational redshift experiments
|
https://physics.stackexchange.com/questions/575694/misconception-in-the-implications-of-the-gravitational-redshift-experiments
|
<p>In both Sean Carroll's book and Misner,Wheeler,Thorne, the authors take the gravitational redshift experiments to show that spacetime has a curved geometry.
There is a paper by Harvey R Brown(published in American Journal of Physics) that challenges this viewpoint by pointing out how this is a misconception. A single redshift experiment does not indicate curvature. Rather, different such experiments in different places is the main reason. (<a href="https://arxiv.org/abs/1512.09253" rel="nofollow noreferrer">https://arxiv.org/abs/1512.09253</a>) It presents 3 misconceptions, and the one I am asking about is the second one.</p>
<p>What I understood from the presentations of Carroll and Misner, Wheeler,Thorne book is as follows.</p>
<ol>
<li><p>Take two non inertial accelerated observers, they observe a difference in the emission and absorption interval ∆t0 and ∆t1. ∆t0 is the time interval two successive crests of the emitted light wave, ∆t1 is that for the absorption case.</p>
</li>
<li><p>The same experiment is done in the tower case, and we get the same result. This time we used a frame where the flat metric(in the coordinate system we use) must be diag(-1,1,1,1) had SR been strictly true, and we would then have ∆t0=∆t1. But in reality, we have ∆t0<∆t1. And we conclude that diag(-1,1,1,1) is not the true metric. Rather the true metric is like that of the accelerated case. But since this is the inertial frame and we use a Cartesian coordinate, here the metric being like that means the geometry is curved.</p>
</li>
</ol>
<p>This is where Dr Brown has his objection. He derives the relationship between ∆t0 and ∆t1 by using the accelerated frame metric and shows that the redshift relation is because of the fact that in the accelerated coordinate system, the minkowski metric does NOT have the simple diag(-1,1,1,1) form. And this is the main reason behind the relationship between ∆t0 and ∆t1. The same relationship between ∆t0 and ∆t1 holds in the gravitational case due to the Weak Equivalence Principle. And this implies that the metric in the gravitational case is also not diag(-1,1,1,1). Rather the metric is that of an accelerated frame in Minkowski spacetime. But this metric is a flat spacetime metric and the components of the Riemann tensor still vanishes even here. So no curvature is involved.</p>
<p>Question: Is my understanding of the argument correct?
And if it is correct, it makes me wonder why from Misner to Carroll so many prominent experts on GR simply missed this issue.</p>
|
<p>MTW certainly didn't miss this issue. Carroll may have. The paper you linked quotes Carroll's textbook as saying of the Pound-Rebka experiment:</p>
<blockquote>
<p>simple geometry seems to imply that the [emission and reception intervals] must be the same. But of course they are not; the gravitational redshift implies that the elevated experimenters observe fewer wavelengths per second [...] What went wrong? Simple geometry – the spacetime through which the photons traveled was curved.</p>
</blockquote>
<p>That does seem to be an error.</p>
<p>GR is widely misunderstood even by professional physicists, not because it's especially difficult but because it's almost useless, so few people are motivated to study it in any depth. But there always have been people who do understand it in depth. Nothing in Brown and Read's paper is new, and they aren't claiming that it is.</p>
<p>I don't think that it's necessary to do the Pound-Rebka experiment at multiple locations on the earth. It's enough to argue by symmetry that the result would be the same (or at least similar) elsewhere. It's hard to analyze, though, because I don't know of any rival to GR that predicts a different result for this experiment. Nordström's theory predicted no bending of starlight by the sun, and I think no precession of Mercury, but even it had a gravitational redshift.</p>
| 536
|
general relativity
|
Would gravity still act if all objects in a closed system somehow became stationary?
|
https://physics.stackexchange.com/questions/581982/would-gravity-still-act-if-all-objects-in-a-closed-system-somehow-became-station
|
<p>I have been trying to understand the implications of general relativity. I unfortunately don't have a good knowledge of advanced topics and I may have made some silly assumptions.</p>
<p>As far as I understand, spacetime dictates the trajectory of an object, and the object curves spacetime. Objects follow the shortest path, and it appears as if things are being pulled when instead they're just accelerating in a specific way due to the curvature. Gravity is a fictitious force.</p>
<p>I'm confused about what would happen if we imagine a universe with two identical stationary objects. I'm guessing that because it's not actually possible for anything to be completely stationary (because we cannot reach absolute zero (uncertainty principle?)), these objects will move along the curvature. But if we consider it was possible for objects to be completely stationary, does this mean that these objects won't follow the curvature since they're not moving to begin with and it would appear as if gravity has stopped working. The objects stay stationary instead of crashing into each other.</p>
<p>What would happen if there are two identical stationary objects and I apply a force to one of the objects, such that the direction of the force is perpendicular to the line connecting the two objects? I'm guessing that it should start to orbit the other object, but I also know that in this inertial frame of reference, since there are two objects, I shouldn't be able to tell who's moving. So the outcome of some force being applied should be symmetrical, so does this mean the objects would start to chase each other?</p>
<p>But then, in a situation where the objects are not identical, if I move the heavier object, would it still appear as if the smaller object has started orbiting due to a force pulling it? But this sounds like movement in one stationary object has induced movement in another stationary object (assuming stationary objects were possible)?</p>
|
<ol>
<li></li>
</ol>
<p>By 'stationary', you would mean 'stationary with respect to the spatial axes of a certain inertial frame of reference'. However, a stationary object would still 'move' along the time axis -- in fact, it will 'move' as fast as it can (=at the speed of light) along the time axis, if it is stationary spatially.</p>
<p>The thought is that your four-dimensional 'speed' is always constant. The only difference between a moving body and a stationary one is the composition of their four-dimensional 'speed'; if it's stationary, then all of its '4D speed' will come from its temporal 'motion'; if it's not, then the 'speed' will be a combination of its spatial motion and temporal 'motion'.</p>
<p>Of course, here the words 'move' and 'speed' should be interpreted as somewhat metaphorically, for you cannot really make sense of things moving along the temporal axis. (or maybe you could.)</p>
<p>Anyhow, from the four-dimensional point of view, you're always 'moving' at a constant 'speed', which is the speed of light, regardless of your spatial motion or temperature or whatever.</p>
<p>(By the way, all you need to articulate this is just special relativity:
Suppose that it took t seconds for an object S moved from a point A to a point B at a constant speed v in a certain inertial frame K. In the frame of reference of S (i.e., the frame of reference where s is always at the origin), let's say, the same journey took t' seconds.</p>
<p>Then we could think of <span class="math-container">$\frac{t'}{t}$</span>as the 'temporal speed' of the object S relative to the frame K. It tells you, so to speak, how slow S's clock ticks with respect to a clock in K.</p>
<p>Then there's the following special relativistic relation between S's 'temporal speed' and its 'spatial speed (v)':</p>
<p><span class="math-container">$(\frac{t'}{t})^2 + (\frac{v}{c})^2 = 1$</span></p>
<p>This is what I meant when I said your '4D speed' is always constant.</p>
<p>You always follow some spacetime trajectory regardless of your velocity with respect to a certain frame. So, yes, gravity still works even in the physically impossible hypothetical situation where things have zero spatial motion with respect to a certain inertial frame.</p>
<p>2.
Regarding the indiscernible objects: if you apply force to one of the objects but not to the other, then, of course, you can tell which one is which; one that experiences acceleration is the one that you pushed, and whether something is accelerated or not is not a relative matter in general relativity. What is completely relative is inertial motion (in SR) or geodesic motion (in GR). If a bucket of water is rotating, then everyone should agree that it's rotating, for angular motion is a form of acceleration and acceleration is not totally relative.</p>
| 537
|
general relativity
|
Coordinate Invariant Divergence
|
https://physics.stackexchange.com/questions/584772/coordinate-invariant-divergence
|
<p>I'm reading the book "Einstein Gravity in a nutshell" by Anothy Zee and I'm a bit stuck on one of the steps in the derivation for divergence in an arbitrary coordinate system. The proof goes as follows,</p>
<p>since we know
<span class="math-container">$$W^\mu\partial_\mu\phi$$</span>
where <span class="math-container">$W^\mu$</span> is a vector field, <span class="math-container">$\phi$</span> is a scalar field, and <span class="math-container">$\partial_\mu=\frac{\partial}{\partial x^\mu}$</span>
and
<span class="math-container">$$\int\sqrt{g}d^Dx$$</span>
where <span class="math-container">$g$</span> is the determinant of the metric <span class="math-container">$g_{\mu\nu}$</span>, and <span class="math-container">$d^Dx$</span> is the integral in D dimensions (e.g. <span class="math-container">$d^3x=dx^1dx^2dx^3$</span>), transform like scalars. We invoke the integral
<span class="math-container">$$I=\int W^\mu\partial_\mu\phi\cdot\sqrt{g}d^Dx$$</span>
which transforms like a scalar. Integrating by parts,
<span class="math-container">$$I=W^\mu\phi\sqrt{g}-\int\phi\cdot\partial_\mu\left(W^\mu\sqrt{g}\right)d^Dx$$</span>
However the book does not have the first term. Why is <span class="math-container">$W^\mu\phi\sqrt{g}=0$</span>?</p>
<p>One possible explanation that I have come up with is that it transforms like a vector, so in order for the LHS and RHS to be consistent (i.e. transform like a scalar), the first term can only equal zero, but I think this is really pushing it. What's a better explanation?</p>
|
<p>The integrated out term is a surface integral
<span class="math-container">$$
\int W^\mu \phi \sqrt g \,dS_\mu
$$</span>
at infinity (and not what you have written). <span class="math-container">$\phi$</span> is arbitrary, and as always in these types of arguments, can be taken to be zero at infinity. So the integrated out term vanishes.</p>
| 538
|
general relativity
|
Why are these vectors perpendicular?
|
https://physics.stackexchange.com/questions/591847/why-are-these-vectors-perpendicular
|
<p>I am reading <a href="https://doi.org/10.1016/S0370-2693(00)01125-4" rel="nofollow noreferrer">this paper</a> <em>The Bardeen model as a nonlinear magnetic monopole</em> by Eloy Ayón-Beato Alberto Garcı́a. A the end where the authors prove the that the weak energy condition is satisfied, they say that the vector
<span class="math-container">$$E_\lambda = F_{\lambda\mu}X^\mu$$</span>
is space like because it is by definition perpendicular to the time like vector <span class="math-container">$X^\mu$</span>.</p>
<p>Why are the perpendicular by definition?</p>
|
<p>Assuming that <span class="math-container">$\mathbf F$</span> is the Faraday tensor, then it is antisymmetric by definition. Therefore,</p>
<p><span class="math-container">$$E_\lambda X^\lambda = F_{\lambda \mu}X^\mu X^\lambda$$</span></p>
<p>is the contraction of an antisymmetric object <span class="math-container">$F_{\lambda \mu}$</span> with a symmetric object <span class="math-container">$X^\mu X^\lambda$</span>, which is equal to zero.</p>
| 539
|
general relativity
|
1) When can a time coordinate be separated in the interval (General relativity) ? 2) Unclear proper time expression
|
https://physics.stackexchange.com/questions/594680/1-when-can-a-time-coordinate-be-separated-in-the-interval-general-relativity
|
<ol>
<li><p>One has that <span class="math-container">$ds^{2} = g_{ij}(x)dx^{i}dx^{j}$</span>. I often see that the interval is re-expressed with a time "seperation" of the form:
<span class="math-container">$$
ds^{2} = g_{00}(x)dt^{2} + \tilde{g}_{ab}dx^{a}dx^{b} \;\; a,b = 1,2,3
$$</span>
When can this be done?</p>
</li>
<li><p>Why can the proper time infinitesimal always be written in the form (<em>according to Wiki "Proper time"</em>):
<span class="math-container">$
d\tau = \sqrt{g_{00}(x)}dt \; ?
$</span></p>
</li>
</ol>
<p>Thank you in advance for your answers</p>
|
<p><strong>1)</strong></p>
<p>I believe this can be done always, but I am not sure. You need to get rid of the cross terms <span class="math-container">$g_{0a}$</span> and you have 4 coordinate transformations at your disposal to get rid of the 3 metric functions, while keeping <span class="math-container">$g_{00}$</span> positive (assuming the signature is (+,-,-,-)). This looks like a problem that has a solution.</p>
<p><strong>2)</strong></p>
<p>It cannot. The proper time is always attached to some worldline. For any worldline whatsoever, the general formula is:
<span class="math-container">$$d\tau=c^{-1}ds=c^{-1}\sqrt{g_{\mu\nu}dx^\mu dx^\nu}$$</span>
This reduces to your formula only along worldlines that keep <span class="math-container">$x^1$</span>,<span class="math-container">$x^2$</span> and <span class="math-container">$x^3$</span> constant.</p>
| 540
|
general relativity
|
Gravity in compact space, like three-torus or a ball with ends identified
|
https://physics.stackexchange.com/questions/595287/gravity-in-compact-space-like-three-torus-or-a-ball-with-ends-identified
|
<p>What does gravity look like in a compact space, such as a universe with spatial periodic boundary conditions equivalent to a 3-torus, or a ball with opposite points on the surface of the ball identified? In particular, what is the equivalent to the Schwarzschild vacuum solution to the Einstein field equations? I have only a smattering of General Relativity, and so I am unsure if this is a meaningful question to ask.</p>
<hr />
<p><strong>Background:</strong></p>
<p>I know that to solve General-Relativity problems, we must solve the Einstein field equations <span class="math-container">$$G_{\mu \nu} = \kappa T_{\mu \nu}$$</span> along with the matter/field equations of motion.</p>
<p>The Schwarzschild metric describes a spherically-symmetric, static, vacuum (<span class="math-container">$T_{\mu \nu}=0$</span>) solution to Einstein's equations.</p>
<p><span class="math-container">$$ds^2 = -(1+GM/r)dt^2 + (1+GM/r)^{-1}dr^2 + r^2 (d\theta^2 + \sin^2\theta \ d\varphi^2)$$</span></p>
<p>According to my understanding, all that we need to do to find this metric is to enforce a spherically-symmetric form for the metric, yielding
<span class="math-container">$$ds^2 = -B(r)dt^2 + A(r)dr^2 + r^2 (d\theta^2 + \sin^2\theta \ d\varphi^2)$$</span>
and to specify <span class="math-container">$T_{\mu \nu} = 0$</span>.</p>
<p>The partial differential Einstein equations simplify to ordinary differential equations in <span class="math-container">$A(r)$</span> and <span class="math-container">$B(r)$</span>, and solving those yields the form of the Schwarzschild metric. <span class="math-container">$M$</span> is a constant of integration that is identified as the mass via the Newtonian far-field limit.</p>
<p>Can I enforce periodic boundary conditions, like <span class="math-container">$r=r+R$</span> for some constant <span class="math-container">$R$</span>? Would I find a different metric? Specifying periodic boundary conditions in space seems to me to overdetermine the ordinary differential equations in <span class="math-container">$A(r)$</span> and <span class="math-container">$B(r)$</span>, so I might be going the wrong way about this.</p>
| 541
|
|
general relativity
|
Make Newman-Penrose scalars definite spin weight
|
https://physics.stackexchange.com/questions/526164/make-newman-penrose-scalars-definite-spin-weight
|
<p>Reading through the original <a href="https://aip.scitation.org/doi/10.1063/1.1666410" rel="nofollow noreferrer">GHP paper</a>, I notice that some of the <a href="http://www.scholarpedia.org/article/Spin-coefficient_formalism" rel="nofollow noreferrer">Newman-Penrose scalars</a> do not have definite spin weight. For example,
<span class="math-container">\begin{equation}
\alpha\equiv\frac{1}{2}\left(n^il^k\nabla_kl_i+m^il^k\nabla_km^*_i\right),
\end{equation}</span>
does not have definite spin weight as, e.g. <span class="math-container">$n^il_i=1\neq0$</span> so that when
we transform <span class="math-container">$l_i\to\xi\xi^*l_i$</span> (where <span class="math-container">$\xi$</span> is a complex number), we also have terms like <span class="math-container">$l^k\nabla\left(\xi\xi^*\right)$</span> in the transformed <span class="math-container">$\alpha$</span> scalar.</p>
<p>Besides the approach taken in GHP, where new derivative operators (''thorn'' and ''edth'') are defined which incorporate <span class="math-container">$\alpha$</span>, is there another way to make <span class="math-container">$\alpha$</span> a scalar of definite spin weight (e.g. through the addition of another Newman-Penrose scalar, etc)?</p>
<p>EDIT:</p>
<p>I would be happy to have a way to construct definite spin weight objects from combinations of linear perturbations of the Newman-Penrose scalars about a type D background. </p>
<p>I've been reading through a recent paper <a href="https://inspirehep.net/record/1724549" rel="nofollow noreferrer">on the stability of Kerr spacetime</a>. There, the authors claim to define a properly spin-weighted version of a linear perturbation that is very closely related to a linear perturbation of <span class="math-container">$\alpha$</span> in Sec 3.1. Their last paragraph at the end of Sec 3.1 explains why their quantity has a definite spin weight, but I do not understand what they are saying.</p>
| 542
|
|
general relativity
|
Does curvature of spacetime changes the wavelength of light?
|
https://physics.stackexchange.com/questions/525469/does-curvature-of-spacetime-changes-the-wavelength-of-light
|
<p>Suppose we do an experiment on earth and light a monochromatic light near a highly dense object. Does it cause any change in the wavelength of light? </p>
| 543
|
|
general relativity
|
Rotating dust in Gödel universe
|
https://physics.stackexchange.com/questions/525772/rotating-dust-in-g%c3%b6del-universe
|
<p>I have a question regarding the Gödel metric. Supposedly the Gödel universe is filled with rotating pressure-less dust.
However, checking different sources, it seems like Einstein's field equations are satisfied in this case for a perfect fluid without pressure and 4-velocity <span class="math-container">$u^{\mu}=\left(1,0,0,0\right)$</span> , placing some restrictions on the cosmological constant and the matter density.
Since that 4-velocity represents a dust distribution that is at rest everywhere, I don't understand why it is said to be rotating. I guess it must be something very obvious but I can't see it.</p>
|
<p>Coordinate velocities in GR are generally meaningless. They have no special physical interpretation. To find out whether a particular spacetime has rotation, you need to consider some coordinate-independent criterion. <a href="https://en.wikipedia.org/wiki/G%C3%B6del_metric#Rigid_rotation" rel="nofollow noreferrer">WP has this</a>:</p>
<blockquote>
<p>The frame fields given above are both <em>inertial</em>, <span class="math-container">$\nabla_{\vec{e}_0} \vec{e}_0 = 0$</span>, but the <em>vorticity vector</em> of the timelike geodesic congruence defined by the timelike unit vectors is
<span class="math-container">$$-\omega \vec{e}_2$$</span>
This means that the world lines of nearby dust particles are twisting about one another. Furthermore, the <em>shear tensor</em> of the congruence <span class="math-container">$\vec{e}_0$</span> <em>vanishes</em>, so the dust particles exhibit <em>rigid rotation</em>.</p>
</blockquote>
| 544
|
general relativity
|
Gravitational motion of 2 point masses in free space
|
https://physics.stackexchange.com/questions/526989/gravitational-motion-of-2-point-masses-in-free-space
|
<p>I came across this question:</p>
<p><strong>"If there are two point masses in free space(i.e., there is no other mass/force/field acting in their vicinity), will those two point masses get closer to each other, or will they remain stationary as they are?"</strong></p>
<p>I approached this question through Newtonian Gravitation, and I thought they would come closer to each other. But the answer was that they would remain stationary, the way they are, and not move closer towards each other. It is a drawback of Newtonian theory, and I was supposed to approach it through General Relativity. </p>
<p>I have thought a lot about it, but I couldn't figure it out. </p>
<p>Any solution, approached through general relativity, helping me out with this will be appreciated. </p>
<p><strong>Note:</strong> There is a possibility that the answer is wrong. I'm not sure. Please help. </p>
|
<p>if the two masses do not circle with the right speed around their center of mass, the will move to get closer to each other, in no case they will stay at there initial distance. So the statement you heard is false.
trula</p>
| 545
|
general relativity
|
Does the Unruh Effect violate observer independence in General Relativity?
|
https://physics.stackexchange.com/questions/528820/does-the-unruh-effect-violate-observer-independence-in-general-relativity
|
<p>Observer independence means that the physic involved is independent of the reference frame of the observers but observers can't agree on the vacuum temperature due to the the Unruh Effect, does this not violate GR's principle? If we can't agree on the energy we can't agree on curvature, correct?</p>
|
<p>If you are at rest on the surface of the earth you receive the same Unruh temperature <span class="math-container">$T = \hbar g/(2\pi c k_b)$</span> as if you were in a rocket accelerating with <span class="math-container">$a=1g$</span> (see <a href="https://www.youtube.com/watch?v=qPKj0YnKANw&t=9m26s" rel="nofollow noreferrer">here</a>), so the equivalence principle is upheld. But it is not required that an accelerating and a nonaccelerating observer should see the same particle number, at least not from the quantum perspective. Relativity alone can't solve this problem.</p>
| 546
|
general relativity
|
If a photon does not experience time, how is it affected by gravitational pull of masses that might not yet be in its path?
|
https://physics.stackexchange.com/questions/532814/if-a-photon-does-not-experience-time-how-is-it-affected-by-gravitational-pull-o
|
<p>Ok, I apologize in advance if this has been asked before. I tried googling for it, but didnt find anything related. Im a comp sci guy, not a physics guy, so its possible Im not googling the right terms. </p>
<p>So my understanding is that: </p>
<p>A) Photons of light do not experience time in their reference frame. I.e., the time taken for a photon to travel from point A to point B is 0 (in the photons reference frame). </p>
<p>B) Light can and is bent by gravity. </p>
<p>So now lets say you shine a really bright flash light across the universe. Assume at the time you shine it, the light wont pass near any objects on its way across the universe. </p>
<p>In the lights reference frame, its just going to immediately end up on the other side of the universe (assuming that photon didn't hit anything). </p>
<p>However, since this example universe is so big, tons of objects had time to get near the path of the light without hitting it. Their gravitational pull altered the photons' path ever so slightly. </p>
<p>So, to my actual question: </p>
<p>In the example above, the moment the light photon is send flying across the universe, nothing is near its path of travel. However, in its reference frame, it immediately arrives at an altered location due to objects that got in its path. </p>
<p>How can this be possible? If light truly experiences no time, the time at point A and the time at point B should be equal. In either reference frame, the universe objects have not yet moved near its path when the light is at point A. </p>
|
<p>In the frame of an observer, a photon follows <a href="https://en.wikipedia.org/wiki/Geodesics_in_general_relativity" rel="nofollow noreferrer">a geodesic.</a> . In the observer's framemork the moving masses create moving geodesics for the incoming photon which is traveling with the speed of light in vacuum. In the photon's framework there is no time to be defined as "before" and "after" the motion of masses in the observer's framework . </p>
| 547
|
general relativity
|
Term for radius and gradient of spacetime distortion?
|
https://physics.stackexchange.com/questions/541651/term-for-radius-and-gradient-of-spacetime-distortion
|
<p>A black hole would distort spacetime time to a greater degree than planet Earth. That is, both the radius and the gradient of the distortion are greater. </p>
<p>Is there a term that combines "greater radius" and "greater gradient" into one?</p>
<p>In layman terms, a black hole distorts spacetime more "aggressively" than planet Earth.</p>
|
<p>Spacetime distortion is measured by the <a href="https://en.wikipedia.org/wiki/Riemann_curvature_tensor" rel="nofollow noreferrer">Riemann curvature tensor</a> <span class="math-container">$R_{\mu\nu\lambda\kappa}$</span>. This tensor has 256 components, but various symmetries reduce the number of independent components to 20. So, in general, it takes 20 numbers at each point in spacetime to fully describe how spacetime is distorted.</p>
<p>The simplest way to compare the spacetime curvature of a Schwarzschild black hole versus that of the Earth is to consider a curvature invariant like the <a href="https://en.wikipedia.org/wiki/Kretschmann_scalar" rel="nofollow noreferrer">Kretschmann scalar</a>. This is just a single number, and it is much larger at the event horizon of a stellar-mass black hole than it is at the surface of the Earth.</p>
| 548
|
general relativity
|
Do gravitational fields exist in vacuum region?
|
https://physics.stackexchange.com/questions/137153/do-gravitational-fields-exist-in-vacuum-region
|
<p>I was reading about "vacuum solution" in wiki, <a href="http://en.wikipedia.org/wiki/Vacuum_solution_(general_relativity)" rel="noreferrer">http://en.wikipedia.org/wiki/Vacuum_solution_(general_relativity)</a>. There are some words I'm confused.</p>
<blockquote>
<p>1.In general relativity, a vacuum solution is a Lorentzian manifold whose Einstein tensor vanishes identically. According to the Einstein field equation, this means that the stress–energy tensor also vanishes identically, so that no matter or non-gravitational fields are present.</p>
<p>2.Since <span class="math-container">$T^{ab} = 0$</span> in a vacuum region, it might seem that according to general relativity, vacuum regions must contain no energy. But the gravitational field can do work, so we must expect the gravitational field itself to possess energy, and it does.</p>
</blockquote>
<p>It seems to be conflicted about the existence of gravitational fields. So my question is , do <span class="math-container">$T^{ab}=0$</span> mean no matter and non-gravitational fields?</p>
|
<p>That article's choice of words could certainly be improved. Basically yes, $T$ captures all the non-gravitational "stuff."</p>
<hr>
<p>The idea of a "gravitational field" doesn't really fit in to GR. There is stress-energy $T$ everywhere, and there is a metric $g$ everywhere, and that's really all you need to define what exists.</p>
<p>The article is trying to say that even if spacetime is more or less empty of stress-energy, there is a potential for stuff to happen thanks to the metric being nontrivial. Really, this is nothing new -- in Newtonian gravity, two separated bodies have <em>as a system</em> a gravitational potential energy that doesn't really reside anywhere. Trying to localize this potential "energy" in space is more difficult/ill-defined in GR than in Newtonian gravity.</p>
<p>In GR you can have two isolated masses sitting in space initially at rest. The stress-energy tensor is nonzero only in the regions occupied by the masses. Exterior to the masses the homogeneous Einstein equation is the same as that for empty Minkowski space, but the <em>solution</em> depends on the boundary conditions imposed by the masses and is in fact not Minkowski but rather some nonlinear superposition of Schwarzschild solutions. The nontrivial nature of the metric (which one could perhaps misleadingly call a "gravitational field") means the masses will start to move toward one another.</p>
<p>Another example is gravitational waves. These are again solutions to the homogeneous Einstein equation (i.e. the equation in vacuum), but they are nontrivial solutions. The metric is not just Minkowski.</p>
<p>To see this more mathematically, one could turn to the <a href="http://en.wikipedia.org/wiki/ADM_formalism#Equations_of_motion" rel="nofollow">ADM equations of motion</a>. Take a hypersurface of constant timelike coordinate $t$. The induced $3$-metric on this surface has components $\gamma_{ij}$ and "conjugate momentum" $\pi^{ij}$. These can be written in terms of $g$ without reference to $T$. Then there are known equations for $\partial_t \gamma_{ij}$ and $\partial_t \pi^{ij}$. In particular, only specially contrived setups will have $\partial_t \gamma_{ij} = 0$.</p>
| 549
|
general relativity
|
Gravitational time dilation, does time of the observer at a lower gravitational potential looked slowed down in the frame of the higher one
|
https://physics.stackexchange.com/questions/154061/gravitational-time-dilation-does-time-of-the-observer-at-a-lower-gravitational
|
<p>This question is mainly inspired after watching the movie known as Interstellar</p>
<p>We knew that for time dilation caused by relativistic motion between A and B. A will measure <strong>B's clocks slowing down</strong>, and B will measure <strong>A's clock slowing down</strong> by the same rate, while they both measure <strong>their own clocks ticking normally</strong></p>
<p>In interstellar, there's a planet that experience severe gravitational time dilation relative to earth's frame due to being in close proximity to the black hole Gargantula, therefore for every hour spent there measured by the astronaut's frame, 7 years will have passed in Earth's frame</p>
<p>Thus this means to Earth's frame, they will measure the astronaut's clocks to be ticking slower relative to them</p>
<blockquote>
<p>But <strong>what about earth's clock as measured by the astronauts?</strong> Will they measure Earth's to be ticking faster than theirs (at the same magnitude it is slowed in earth's frame measuring theirs)?</p>
<p><strong>Because if they both measured the other's clocks as running slower, then how could the twin paradox like aging difference be possible?</strong></p>
</blockquote>
|
<p>Yes, the observers on Earth will measure the astronaut's clocks to be be running slow while the astronauts will measure the clocks on Earth to be running fast. So the situation is asymmetric.</p>
<p>The situation is asymmetric because the two sets of clocks are in different environments. Specifically there is a (gravitational) potential energy difference between them i.e. you get energy out going from Earth to the planet but you have to put energy in going from the planet to Earth. In fact you can directly relate the time dilation to the potential energy difference using the weak field equation:</p>
<p>$$ \frac{\Delta t_1}{\Delta t_2} = \sqrt{1 - \frac{2\Delta\Phi}{c^2}} \tag{1} $$</p>
<p>This asymmetry does not exist for observers moving at different speeds in flat spacetime.</p>
| 550
|
general relativity
|
How would Einstein calculate acceleration of a ball?
|
https://physics.stackexchange.com/questions/561508/how-would-einstein-calculate-acceleration-of-a-ball
|
<p>Let's say that we have two homogeneous spherical balls - one with mass <span class="math-container">$m_1=1000m$</span> and radius <span class="math-container">$r_1=1000r$</span> and second with mass <span class="math-container">$m$</span> and radius <span class="math-container">$r$</span>. Distance between centers of this balls is <span class="math-container">$R>1100r$</span>. The balls are not moving(Their relative velocities are equal to <span class="math-container">$0$</span>). Can equations of relativity theory be applied to problem of finding relative acceleration of a balls? If yes, then how exactly?</p>
| 551
|
|
general relativity
|
An easy way to determine the space-like and time-like paths on a spacetime manifold based on linear algebra?
|
https://physics.stackexchange.com/questions/565750/an-easy-way-to-determine-the-space-like-and-time-like-paths-on-a-spacetime-manif
|
<p>I think I have found a way to easily understand time-like and space-like paths with the contect of a little linear algebra. My question is: is my understanding, below, correct?</p>
<p>When I learned General Relativity, one source of some confusion was how to find the actual time or distance from a metric. In particular, suitable co-ordinates are just suitable co-ordinates and don't necessarily correspond to the actual physical time a clock, or a meter rule would actually measure. For that, you need to pick out the time-like, or space-like paths and integrate along them. I always found it a bit vague how to identify which was which.</p>
<p>Now however, based on some simple linear algebra, I think I have found way that makes it all very easy. I wanted to ask if it is correct or not.</p>
<p>I start with a metric in any suitable set of coordinates. In the analysis below I will eventually choose a time co-ordinate that measures time in meters (so that c is absorbed into the time and does not clutter the equations). The mathematical result is much clear this way. So I start with a metric that has been given to me by some GR calculation:</p>
<p><span class="math-container">$ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}$</span></p>
<p>In linear algebra, this is just a "quadratic form", i.e. it is a symmetric bi-linear map of the form</p>
<p><span class="math-container">$ds^{2}=dx^{T}Mdx$</span></p>
<p>where dx is a 4D column matrix, M is an invertable a symmetric 4 by 4 symmetric matrix, and the superscript T denotes transpose.</p>
<p>It is very easy to show that there is a linear transformation, P, of coordinates that will put M in "canonical" form, with all -1s or +1s on the diagonal. In the case of a metric tensor from General Relativity, the canonical form of M will be <span class="math-container">$\Lambda$</span> given by</p>
<p><span class="math-container">$\Lambda=\begin{bmatrix}
-1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}$</span></p>
<p>Thus, our initial metric can be written out in new coordinates as follows</p>
<p><span class="math-container">$ds^{2}=dq^{T}\Lambda dq$</span></p>
<p>where</p>
<p><span class="math-container">$dq=Pdx$</span></p>
<p><span class="math-container">$\Lambda=PMP^{T}$</span></p>
<p>and another property of symmetric matrices, of course, is that the vectors associate with the coordinates in <span class="math-container">$dq$</span> will be orthogonal.</p>
<p>So let <span class="math-container">$dq=\begin{bmatrix}
dt_{p}\\
dx_{p_1}\\
dx_{p_2}\\
dx_{p_3}
\end{bmatrix}$</span> and <span class="math-container">$dx=\begin{bmatrix}
dt\\
dx_{1}\\
dx_{2}\\
dx_{3}
\end{bmatrix}$</span></p>
<p>Then in these new coordinates, the metric becomes</p>
<p><span class="math-container">$ds^{2}=-dt_{p}^{2}+dx_{p_1}^{2}+dx_{p_2}^{2}+dx_{p_3}^{2}$</span></p>
<p>I have used of subscript p here, since, to me, these seem to be physical coordinates. These will not be the only choice of such coordinates that will have this property; the action of any member of the Poincare' group on these coordinates will of course bring about another set of coordinates with the same property.</p>
<p>The point is that I believe that when we have the metric in this form, then if we go from coordinates</p>
<p><span class="math-container">$\begin{bmatrix}
dt_{p}\\
dx_{p_1}\\
dx_{p_2}\\
dx_{p_3}
\end{bmatrix}=
\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}$</span> to <span class="math-container">$\begin{bmatrix}
dt_{p}\\
dx_{p_1}\\
dx_{p_2}\\
dx_{p_3}
\end{bmatrix}=
\begin{bmatrix}
0\\
1\\
0\\
0
\end{bmatrix}$</span></p>
<p>then I believe that we would have traveled through space an actual actual physical distance of 1m (without the passing of any time).</p>
<p>Similarly, if we traveled from</p>
<p><span class="math-container">$\begin{bmatrix}
dt_{p}\\
dx_{p_1}\\
dx_{p_2}\\
dx_{p_3}
\end{bmatrix}=
\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}$</span> to <span class="math-container">$\begin{bmatrix}
dt_{p}\\
dx_{p_1}\\
dx_{p_2}\\
dx_{p_3}
\end{bmatrix}=
\begin{bmatrix}
1\\
0\\
0\\
0
\end{bmatrix}$</span></p>
<p>I believe we will have advanced in time by 1m (in seconds this would be <span class="math-container">$(1m)/(3\times 10^{8}m)$</span> = 3.333 nanoseconds), but we would not have moved in space.</p>
<p>Thus, if I am right, a little elementary linear algebra makes it really easy, not only to pick out the time paths and space paths, but also to obtain coordinates consistent with the actual time or distance you would measure with a clock or meter rule. I was able to use this method to obtain the time dilation caused by Schwarzschild metric associated with a non-rotating black hole, so I think I am probably on the right track, but:</p>
<p>MY QUESTION IS THIS: is my analysis, above, correct?</p>
<p>I tend to find that if I can expressed the problem in the language of mathematics and see what is actually going on under the bonnet with the mathematics, it actually makes it much easier for me to understand the actual physics.</p>
|
<p>You say: "It is very easy to show that there is a linear transformation, P, of coordinates that will put M in "canonical" form, with all -1s or +1s on the diagonal." This is true at any one point. However, it's not true in a neighborhood. For any small 4-dimensional region of spacetime, <span class="math-container">$M$</span> can be put in canonical form everywhere in the region if and only if the spacetime is flat.</p>
<p>Another point: You write <span class="math-container">$\begin{bmatrix}
dt_{p}\\
dx_{p_1}\\
dx_{p_2}\\
dx_{p_3}
\end{bmatrix}=
\begin{bmatrix}
0\\
1\\
0\\
0
\end{bmatrix}$</span>, but you mean <span class="math-container">$\begin{bmatrix}
t_{p}\\
x_{p_1}\\
x_{p_2}\\
x_{p_3}
\end{bmatrix}=
\begin{bmatrix}
0\\
1\\
0\\
0
\end{bmatrix}$</span>. This distinction is important - if you're ok with <span class="math-container">$dq$</span> being an arbitrary differential form, then you can put <span class="math-container">$M$</span> in canonical form everywhere. The problem is that you want <span class="math-container">$dq$</span> to correspond to some choice of coordinates <span class="math-container">$q$</span>, so you need <span class="math-container">$dq$</span> to be closed.</p>
<p>Now, for any particular path, I think you can find a choice of <span class="math-container">$P$</span> that puts the metric in canonical form everywhere along the path. If you do this, you'll end up reinventing the usual approach, but it might make more sense to you once you've derived it from this perspective.</p>
| 552
|
general relativity
|
What will be the "Einstein field equations" for two or three bodies?
|
https://physics.stackexchange.com/questions/566281/what-will-be-the-einstein-field-equations-for-two-or-three-bodies
|
<p>In general theory of relativity the Einstein field equations e.g. relate the geometry of space-time with the distribution of one body within it.
<span class="math-container">$$R_{\mu\nu}-\dfrac{1}{2}g_{\mu\nu}R+g_{\mu\nu}\Lambda=\dfrac{8\pi G}{c^4}T_{\mu\nu}.$$</span></p>
<p>What will be the "Einstein field equations" for two or three bodies?</p>
|
<p>They are already included, that is a <strong>field</strong> equation. So the energy-momentum tensor on the right must include the distribution of energy (matter) and momentum in your model, all of it. Normally it is used the other way around, one takes a particular form of the energy momentum tensor to model a single "point"-like particle, by for example making the energy density into a delta function evaluated at the particles wordline.</p>
| 553
|
general relativity
|
When is a spacetime a black hole?
|
https://physics.stackexchange.com/questions/233034/when-is-a-spacetime-a-black-hole
|
<p>While reading $\textrm{Present status of the Penrose Inequality}$ by Marc Mars, 2009, I was confused with the following statement:</p>
<blockquote>
<p>... in order to determine whether a space-time is a black hole, detailed knowledge of its global future behaviour is required.</p>
</blockquote>
<p>Why is that ?</p>
|
<p>The most common definition of a black home is the portion of the spacetime manifold $\mathcal{M}$ that is $\mathcal{M} - J^-(ℐ^+)$, the manifold minus the causal past of null future infinity. That is, it's the region of spacetime where no signal can escape to infinity at some point in the future. This requires you to know the global structure of spacetime (such as the existence of null infinity and such).</p>
| 554
|
general relativity
|
Elementary question about non-Euclidean geometry in general relativity: "cannot move about without changing shape"
|
https://physics.stackexchange.com/questions/244842/elementary-question-about-non-euclidean-geometry-in-general-relativity-cannot
|
<p>One basic result of general geometry (from math) in curved spaces or on curved surfaces is that if you are in a surface of variable curvature, things like the Euclidean congruence postulates and theorems for triangles and other simple figures fail, and the reason this is is because those are actually strong statements about the homogeneity of the curvature of the space. Being non-homogeneous, they break down, and since as Euclid shows these are related to the notion of "moving something from place to place in space without altering its size or shape", it follows that in such a space, something CANNOT be moved about from place to place in the space without altering its size and shape.</p>
<p>It would therefore stand to reason this also holds in the variably-curved spaces of general relativity. So my question is, <em>physically</em>, what is the effect of being unable to move about from point to point without altering your size and shape?</p>
|
<p>It isn't necessarily objects that change. This is a 4 dimensional space with 3 "space" dimensions and 1 time dimension. </p>
<p>The definition of curvature can be stated as when you go around what should be a rectangle, you don't come back to the same place. Or equivalently, if you take the two different paths to the opposite corner, you wind up in two different places. </p>
<p>In a gravity well, time runs slower when you are deeper. So you can see curvature by considering a rectangle in the time-radial plane. Start at an event - a point in space-time. Wait 1 second and move 1 meter down, and note the event. Start over. Move one meter down and then wait 1 (slower) second. You arrive at the same place, but at a later event. </p>
| 555
|
general relativity
|
Can we distinguish between two mass distributions in spacetime having the same effect over a test partlicle
|
https://physics.stackexchange.com/questions/244884/can-we-distinguish-between-two-mass-distributions-in-spacetime-having-the-same-e
|
<p>Einstein's equation is</p>
<p>$$8πT_{ab}=G_{ab}$$</p>
<p>where the left side contains the stress-energy tensor and the right side contains the Einstein tensor. </p>
<p>Is there exactly one unique stress-energy tensor corresponding to a given spacetime curvature? Or is it possible for one curvature (i.e. one spacetime metric) to be produced by multiple different mass configurations</p>
| 556
|
|
general relativity
|
I can't understand where the minus sign in the second equation is coming from
|
https://physics.stackexchange.com/questions/696194/i-cant-understand-where-the-minus-sign-in-the-second-equation-is-coming-from
|
<p><span class="math-container">$R_{\mu v}=\frac{\partial}{\partial x^{\lambda}} \Gamma_{\mu v}^{\lambda}+\Gamma_{\mu \lambda}^{\eta} \Gamma_{v \eta}^{\lambda}$</span></p>
<p>equation (2) after the multiplication of the meteric tensor</p>
<p><span class="math-container">$g^{\nu\sigma}R_{\mu v}=\frac{\partial}{\partial x^{\lambda}} g^{\sigma\beta}\Gamma_{\mu \beta}^{\lambda}-g^{\alpha \beta} \Gamma_{\alpha \lambda}^{\sigma} \Gamma_{\beta \mu}^{\lambda}$</span></p>
|
<p>Metric compatibility
<span class="math-container">$$
\nabla_\lambda g_{\mu\nu} =\partial_\lambda g_{\mu\nu}+ g_{\alpha\nu}{\Gamma^\alpha}_{\mu\lambda}+ g_{\mu\alpha}{\Gamma^\alpha}_{\nu\lambda}=0. $$</span></p>
| 557
|
general relativity
|
Penrose diagram of a star in AdS?
|
https://physics.stackexchange.com/questions/527460/penrose-diagram-of-a-star-in-ads
|
<p>Often the Penrose diagram is drawn for pure AdS or AdS-Schwarzschild black hole. What is the Penrose diagram for a star that is not collapsing in AdS?</p>
| 558
|
|
general relativity
|
Does GR reflect Aristotelian time?
|
https://physics.stackexchange.com/questions/167044/does-gr-reflect-aristotelian-time
|
<p>This is really a philosophical question, which I've already asked at Phil.SE but I'm asking here with more physics detail.</p>
<p>In Newtonian Mechanics, space and time are independent of each other, and of motion; in that the geometry of space time is independent of the distribution of matter and energy and it's motion; does the same go for GR?</p>
<p>On the one hand, time is deliberately measured by motion, that of light; on the other hand there are solutions of GR that have <a href="https://en.wikipedia.org/wiki/Static_spacetime" rel="nofollow noreferrer">static spacetimes</a>.</p>
<p><strong>edit</strong></p>
<p>Aristotle, <em>Physics IV.14</em>:</p>
<blockquote>
<p>The reason, then, why people think of time as the change of the heavenly sphere is because all other changes are measured by this change, and time too is measured by this change.</p>
</blockquote>
<p>and Mach as quoted in Julian Barbours Paper <a href="https://arxiv.org/abs/0903.3489v1" rel="nofollow noreferrer">On the nature of time</a>:</p>
<blockquote>
<p>It is utterly beyond our power to measure the changes of things by time ... time is an abstraction at which we arrive by means of the changes of things; made because we are not restricted to any one definite measure, all being interconnected.</p>
</blockquote>
|
<p>There are a couple notions of time in GR. Typically when you write down a space you do so in some set of coordinates $(t,x,y,z)$. The $t$ there is the <em>coordinate time</em>. It marches forward forever, and doesn't care at all what the matter and energy content of the system may be. This seems very similar to the Aristotelian time.</p>
<p>However, the fundamental principle of general relativity is that your particular choice of coordinates doesn't matter. Your coordinate choice is really just a gauge freedom. Physical systems experience their own <em>proper time</em> $\tau$ which certainly can be different for different observers, even if the spacetime is static!</p>
<p>So the answer to your question seems to be: both yes and no :)</p>
| 559
|
general relativity
|
Derivation of Teukolsky equation doesn't match
|
https://physics.stackexchange.com/questions/713899/derivation-of-teukolsky-equation-doesnt-match
|
<p>I am trying to derive Teukolsky equation on Kerr spacetime using SageMath and</p>
<p>(1) [(Δ+3γ−γ∗+4μ+μ∗)(D+4ϵ−ρ)−(δ∗−τ∗+β∗+3α+4π)(δ−τ+4β)−3ψ2]ψ4=0.</p>
<p>Like he did in his <a href="https://articles.adsabs.harvard.edu/pdf/1973ApJ...185..635T" rel="nofollow noreferrer">paper</a>. But I did not have success, the expression is too big and doesn't match the one on his paper. I read about it and tried to use ψ4 = ψ*p^4, but still did not work. Can someone help me, please?</p>
<p>I need to obtain the teukolsky equation using (1), have ever someone used (1) to derivation of it and could give some hint? Thanks.</p>
| 560
|
|
general relativity
|
Does curvature of spacetime depend upon the "mass" or "density" of a object?
|
https://physics.stackexchange.com/questions/275562/does-curvature-of-spacetime-depend-upon-the-mass-or-density-of-a-object
|
<p>Suppose we have a object with mass "M" with small density and a object with same mass "M" but different density (like a large density). Does the curvature of spacetime same for the two object with same mass but different densities?</p>
|
<p>To start with, as pointed out in the comments, your question is not very precise. I will assume you are referring to spherically symmetric bodies. Then Birkhoff's theorem implies that the exterior is described by the Schwarzschild solution, so as long as we are in the exterior of both objects they are indeed equivalent. The interior will depend not only on the density but also on the internal flux of energy-momentum. There are several different known solutions that may describe the interior of a Schwarzschild star, at least according to general relativity, so spherical symmetry does not in any way imply interior equivalence. I can provide some links at a later time if you wish, otherwise I believe it is relatively easy to find for yourself. </p>
| 561
|
general relativity
|
Is the curvature of space-time a smooth function everywhere ? (except at black holes)
|
https://physics.stackexchange.com/questions/1628/is-the-curvature-of-space-time-a-smooth-function-everywhere-except-at-black-h
|
<p>Is the curvature of space-time a smooth function everywhere (except at black holes) in view of General relativity. By 'smooth' it is meant that it possesses derivatives of all order at a given point. </p>
|
<p>No, not at the boundary of a solid object like a planet. There's a step function in the stres-energy tensor, and so you'll have a step function in the Riemann tensor.</p>
| 562
|
general relativity
|
The Matter-Vacuum Boundary in General Relativity
|
https://physics.stackexchange.com/questions/3751/the-matter-vacuum-boundary-in-general-relativity
|
<p>A previous <a href="https://physics.stackexchange.com/questions/1628/is-the-curvature-of-space-time-a-smooth-function-everywhere-except-at-black-ho">Stack question</a> (before I joined) asking about continuity in GR received replies which suggested that Curvature would be discontinuous at say a planetary boundary (assume no atmosphere for simplicity). I will analyse some basics of this and then return to that question.</p>
<p>It is true that the Stress-Energy Tensor $T{_a}{_b}=0$ outside the body and is nonzero in the interior resulting in a discontinuity at the surface. This would imply that the Ricci Tensor $R{_a}{_b}$ is also discontinous at the boundary, and zero in the vacuum part as expected from the Einstein equations. However the Riemann Curvature Tensor $R{_a}{_b}{_c}{_d}$ (which generates the physically measurable accelerations) has contributions from the Weyl Curvature Tensor $C{_a}{_b}{_c}{_d}$ as well. In fact the Ricci Tensor "hands over" to the Weyl Tensor at the boundary: thus the Riemann Tensor stays non-zero there. However this "hand over" does not imply continuity, unless there is some GR Theorem which says that the Riemann Tensor stays continuous in this region.</p>
<p>Also in the Newtonian approximation the analogous role is played by the gravitational potential $\phi$ in the Poisson equation $\nabla^2 \phi = 4 \pi G\rho$. Clearly this shows a discontinuity too as the density $\rho$ suddenly drops off at the boundary. However the discontinuity is in the second derivative of the potential: the potential itself is continuous. This means that in exiting a planetary cave or mine one does not suddenly meet a change in Gravitational potential.</p>
<p>However I do not know any theorem in GR which guarantees such continuity. The applicable in-the-large scenario might be the surface of a neutron star; there may be in-the-small particle models too.</p>
|
<p>Roy, your wishful thinking is manifestly impossible. If the tensor $T_{\mu\nu}$ is discontinuous, and it surely is on the surface of a solid, then Einstein's equations guarantee that the Einstein tensor $G_{ab}$ is discontinuous as well - up to a normalization, it's the same tensor. It follows that the Ricci tensor and Riemann tensor, $R_{\mu\nu}$ and $R_{\kappa\lambda\mu\nu}$, must also be discontinuous because the Einstein tensor $G_{\mu\nu}$ can be easily calculated both from the Ricci tensor as well as from the Riemann tensor, so if the Ricci or Riemann tensor were continuous, the Einstein tensor would have to be continuous, which is an obvious contradiction.</p>
<p>I just proved the opposite theorem that the Riemann tensor is discontinuous.</p>
<p>You should realize that the Riemann tensor has a higher number of components than the Ricci (or Einstein) tensor, so its continuity - which means the continuity of all of its components - is an even stronger condition than the continuity of the Ricci (or Einstein) tensor. The argument above proves that none of these tensors is continuous in the presence of solids - which is why there can't be any theorem saying the opposite thing (it would be wrong). Another question is whether the Weyl tensor is continuous near such boundaries. I don't know the answer. The answer could be easily calculated from the very formula for the Weyl tensor. </p>
| 563
|
general relativity
|
Is there an energy density limit in GR?
|
https://physics.stackexchange.com/questions/7771/is-there-an-energy-density-limit-in-gr
|
<p>I am speaking about GR with classical fields and energy. One question, spread over three increasingly strict situations:</p>
<p>Is there an energy density limit in GR? (literally, can the energy density have an arbitrarily large value at some point in space at some point in time)</p>
<p>Is there an energy density limit beyond which a blackhole will always form?</p>
<p>Let's choose a small volume, for here I'll just choose the Planck volume. Is there an average energy density limit over this volume beyond which a blackhole will always form?</p>
<p><strong>Clarification:</strong></p>
<p>In light of <a href="http://en.wikipedia.org/wiki/Mass_in_general_relativity">http://en.wikipedia.org/wiki/Mass_in_general_relativity</a> , can those that are answering that the energy density is limited and referring to a mass $M$ in some equations please specifically state how you are defining the $M$ in terms of the energy density, or defining $M$ in terms of $T^{\mu\nu}$ the stress-energy tensor. Does your $M$ depend on coordinate system choice?</p>
<p>Also, reading some comments, it sounds like there is confusion on what energy density means. Based on wikipedia <a href="http://en.wikipedia.org/wiki/File:StressEnergyTensor.svg">http://en.wikipedia.org/wiki/File:StressEnergyTensor.svg</a> , it sounds like we can consider energy density = $T^{00}$ of the stress-energy tensor. If you feel this is not correct terminology, please explain and I'll edit the question if necessary.</p>
|
<p>The answer is NO. There is no energy density limit (for all three questions).</p>
<p>The easiest way to see this is that the energy density is just the $T^{00}$ component of the stress energy tensor. The solution in GR depends on the <em>full</em> stress energy tensor, so it is not enough to just talk about the energy density. Furthermore, because the energy density is just a component of a tensor, it is a coordinate system dependent quantity. So starting from a solution that doesn't become a blackhole, and has some energy somewhere, we can always choose the coordinate system to make the energy density arbitrarily large.</p>
<p>More clearly stated: Local Lorentz symmetry alone is enough to show that the energy density is not limited in GR. And furthermore since there exist non-zero energy solutions that don't become blackholes, this also answers your second question.</p>
<p>To make the answer to the third question more clear, let's discuss an exact solution. Consider the <a href="http://en.wikipedia.org/wiki/Friedmann-Lema%C3%AEtre-Robertson-Walker_metric">Robertson-Walker</a> solution with a perfect fluid. Here's an example stress energy tensor for a perfect fluid in the comoving frame:</p>
<p>$T^{ab} =\left( \begin{matrix} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{matrix} \right)$</p>
<p>Now if we change to a different coordinate system, using the coordinate transformation:
$\Lambda^{\mu}{}_{\nu} =\left( \begin{matrix} \gamma &-\beta \gamma & 0 & 0 \\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{matrix} \right)$</p>
<p>We see the energy density will transform as:
$\rho' = \gamma^2 \rho + p \beta^2 \gamma^2 = \gamma^2 (\rho + p \beta^2)$</p>
<p>So not only can the energy density be arbitrarily large, but even over a finite volume.</p>
| 564
|
general relativity
|
Decomposing geodetic/de Sitter effect into Thomas precession and spatial curvature
|
https://physics.stackexchange.com/questions/8043/decomposing-geodetic-de-sitter-effect-into-thomas-precession-and-spatial-curvatu
|
<p>According to Rindler the geodetic effect can be considered as consisting of Thomas precession combined with the effect of moving through curved space.</p>
<p>Wolfgang Rindler (2006) Relativity: special, general, and cosmological (2nd Ed.) p234</p>
<p>However according to Misner, Thorne, and Wheeler, Gravitation, p. 1118, Thomas precession does not come into play for a freely moving satellite.</p>
<p>See: <a href="http://en.wikipedia.org/wiki/Talk:Geodetic_effect" rel="nofollow">http://en.wikipedia.org/wiki/Talk:Geodetic_effect</a></p>
<p>I think that although a freely moving satellite doesn't feel gravity, it's relation to an observer is still subject to lorentz transformations and hence Thomas precesssion.</p>
<p>So who is right ? Rindler or Misner, Thorne, and Wheeler ?</p>
|
<p>The difference between Rindler's wording and the MTW wording is just a difference in the choice of coordinates.</p>
<p><strong>Thomas precession in STR</strong></p>
<p>First, what is the Thomas precession? It is a special relativistic effect so the original derivation of the Thomas precession only applies in flat spacetimes. In other words, Rindler's application of the Thomas precession in the context of general relativity requires one to specify how the flat special relativistic spacetime is identified with, or embedded into, the curved spacetime in general relativity.</p>
<p>The Thomas precession is a change of the angular momentum that a gyroscope undergoes when it is attached to another object whose velocity is changing and going along a curve in the space of velocities. Why does it occur?</p>
<p>Well, in special relativity, the velocity is a vector $u^\mu$ normalized so that $u^\mu u_\mu=1$. The space of such vectors is a two-part hyperboloid in a Minkowski space (its intrinsic signature is purely spacelike). Now, the angular momentum of a gyroscope moving together with an object is given by an antisymmetric tensor $J_{\mu\nu}$ that satisfies $J_{\mu\nu}u^\nu=0$. Now, if you try to parallel transport this tensor $J_{\mu\nu}$ along a path inside the hyperboloid, it will not return to itself because the hyperboloid is a curved submanifold. Instead, it will rotate by an angle around an axis (one that depends on the path in the space of velocities) - and this is what we mean by the Thomas precession.</p>
<p><strong>General relativity</strong></p>
<p>In the case of a gyroscope attached to a satellite in a gravitational field described by general relativity, we may describe the vicinity of the world line of the satellite as a piece of flat Minkowski spacetime. After a whole orbit, we return to the same place in space. However, the orbit won't be quite periodic in our "flat, thin, and long Minkowski cylinder" surrounding the world line. Instead, we will induce both a rotation of the velocity space as well as a rotation of the ordinary position space, caused by the actual curvature of the space. Rindler probably calculates the full effect by summing these two contributions - from the monodromy in the uniformly curved velocity space and from the monodromy in the actual spacetime curved by the presence of matter.</p>
<p>So I am pretty confident that Rindler did his job correctly and avoided any sign errors. Gravity Probe B has confirmed the result, after all.</p>
<p>MTW are arguably able to calculate the total result correctly, too. But they organize their calculation differently, attributing the whole effect to the curvature of space. Effectively, their coordinates for the "cylinder surrounding the satellite's world line" differ by a "twist" (a time-dependent rotation) from Rindler's cylinder - the two groups of relativists use different coordinates.</p>
<p>The MTW proposition seems to <em>directly</em> contradict Rindler's calculational strategy and I think that the MTW, in the very satellite context that is relevant and with the Rindler's choice of coordinates, is incorrect. A statement that would be true and similar to the MTW proposition is that if the satellite were freely moving in space and had "no intrinsic rotation" relatively to a chosen system of coordinates, then the velocity could be viewed as a "constant" and the Thomas effect would be exactly zero.</p>
<p>However, the coordinates chosen by Rindler contradict the assumption because the satellite itself - not just the gyroscope - is rotating in this system of coordinates (the gyroscope is also rotating, and differently). So the velocity itself is non-constant even in the flattened cylinder surrounding the world line, and the Thomas precession contribution is nonzero.</p>
<p>As you can see, the existence or absence of the Thomas precession depends on whether or not we consider the velocity of the satellite to be "constant" and this question depends on whether or not our "natural" (there is no natural one!) coordinate system is rotating relatively to other people's natural systems. If there is no Thomas precession in one system, it's because we embedded the special relativistic spacetime in such a way that the velocities stay "constant". However, if we choose a "twisted" coordinate system that is rotating with respect to the first one, the velocities will be non-constant and the precession will receive contributions from the Thomas precession.</p>
<p>By changing the coordinate systems, arbitrary parts of the overall precession after one period - that everyone has to agree with - may be moved from the Thomas precession contribution to the curvature-of-space contribution.</p>
| 565
|
general relativity
|
Perturbation of a Schwarzschild Black Hole
|
https://physics.stackexchange.com/questions/8307/perturbation-of-a-schwarzschild-black-hole
|
<p>If we have a perfect Schwarzschild black hole (uncharged and stationary), and we "perturb" the black hole by dropping in a some small object. For simplicity "dropping" means sending the object on straight inward trajectory near the speed of light.</p>
<p>Clearly the falling object will cause some small (time dependent) curvature of space due to its mass and trajectory, and in particular, once it passes the even horizon, the object will cause some perturbation to the null surface (horizon) surrounding the singularity (intuitively I would think they would resemble waves or ripples). Analogously to how a pebble dropped in a pond causes ripples along the surface.</p>
<p>Is there any way to calculate (i.e. approximate numerically) the effect of such a perturbation of the metric surrounding the black hole?, and specifically to calculate the "wobbling" of the null surface as a result of the perturbation,maybe something analogous to quantum perturbation theory?</p>
<p>Or more broadly, does anyone know of any papers or relevant articles about a problem such as this?</p>
|
<p>Your intuitive picture is basically correct. If you perturb a black hole it will respond by "ringing". However, due to the emission of gravitational waves and because you have to impose ingoing boundary conditions at the black hole horizon, the black hole will not ring with normal-modes, but with quasi-normal modes (QNMs), i.e., with damped oscillations. These oscillations depend on the black hole parameters (mass, charge, angular momentum), and are therefore a characteristic feature for a given black hole.</p>
<p>Historically, the field of black hole perturbations was pioneered by Regge and Wheeler in the 1950ies.</p>
<p><p>For a review article see <a href="http://arxiv.org/abs/gr-qc/9909058">gr-qc/9909058</a></p>
<p>For the specific case of the Schwarzschild black hole there is a very nice analytical calculation of the asymptotic QNM spectrum in the limit of high damping by Lubos Motl, see <a href="http://arxiv.org/abs/gr-qc/0212096">here</a>. See also his <a href="http://arxiv.org/abs/hep-th/0301173">paper with Andy Neitzke</a> for a generalization.</p>
<p>Otherwise usually you have to rely on numerical calculations to extract the QNMs.</p>
| 566
|
general relativity
|
A question on an assumption of space-time
|
https://physics.stackexchange.com/questions/10329/a-question-on-an-assumption-of-space-time
|
<blockquote>
<p>"A four-dimensional differentiable (Hausdorff and paracompact) manifold $M$ will be called a space time if it possesses a pseudo-Riemannian metric of hyperbolic normal signature $(+,-,-,-)$ and a time orientation. There will be no real loss of generality in physical applications if we assume that $M$ and its metric are both $\mathcal{C}^{\infty}$
."</p>
</blockquote>
<p>The above is an excerpt is taken from this <a href="http://rspa.royalsocietypublishing.org/content/314/1519/529.full.pdf+html" rel="nofollow">paper</a>. I'd like to know how the assumption that
$M$ and its metric are both $\mathcal{C}^{\infty}$ be made with out any real loss of generality in physical applications. Any intuitive answer is also appreciated.</p>
|
<p>Dear Rajesh, in reality, physics sometimes works with continuous functions that are not infinitely differentiable - for example look at the energy of the beam at atlas.ch (click at the "Status" button in the middle) when they ramp it up - there are all kinds of discontinuities.</p>
<p>But an arbitrary function that is smooth almost everywhere - and this is a description of functions that really covers everything that a physicist would use - may be approximated by infinitely differentiable functions with an arbitrary accuracy. So it doesn't really hurt if your theorems assume that all the spacetime fields including the metric tensor are infinitely differentiable; you may solve the situations involving functions that are not infinitely differentiable by taking a limit of the infinitely differentiable ones.</p>
<p>May we ask a physics question whether the fields in electromagnetism are infinitely differentiable? Well, we may but it is a meaningless question because the real world is <em>not</em> described by classical physics. So classical physics itself is just an approximation, so both "classical physics with all differentiable functions" and "classical physics with infinitely differentiable functions only" are just approximations of the reality, with none of them being more "physically real". </p>
<p>In quantum physics, we don't use classical fields but we may use wave functions. The time evolution is dictated by Schrödinger's equations - but may we ask whether $\psi(x,y,z)$ should be an infinitely differentiable function of $x,y,z$? Well, in this case, we usually don't make this assumption. Instead, we use all $L^2$ functions which are much more natural at the level of the Hilbert space, Fourier transforms, and so on. The $L^2$ condition surely doesn't require infinite differentiability.</p>
<p>On the other hand, a finite expectation value of the kinetic energy does force $\psi(x,y,z)$ to be a continuous function. Nevertheless, the infinite differentiability condition isn't ever natural in quantum mechanics.</p>
<p>I want to emphasize that all these extra conditions - whether something should be smooth or infinitely smooth etc. - are only a matter of mathematical taste. There can't exist an operational "physics" way to determine whether the world allows functions that are not infinitely differentiable. It's because the infinitely smooth and not-infinitely smooth functions can mimic each other with an arbitrary accuracy but the accuracy of any measurement in physics is always limited. So the restrictions on the "nice behavior" of the mathematical functions is always a matter of mathematical taste.</p>
<p>Exotic differentiable structures could be counterexamples to what I just wrote - because they're "qualitatively" different and their pathological behavior is the very point of their existence (so in some important sense, the ability of them to mimic the normal functions disappears) - but their role in physics remains very confusing and limited as of today. </p>
| 567
|
general relativity
|
Is Einstein's 1916 General Relativity paper a recommended way to start learning about the subject?
|
https://physics.stackexchange.com/questions/14241/is-einsteins-1916-general-relativity-paper-a-recommended-way-to-start-learning
|
<p>If a person has a good grounding in classical mechanics, electrodynamics and special relativity, is Einstein's 1916 paper a recommended way of learning about the subject?</p>
<p>After looking through it briefly, I like what I see because he explains all about tensors from first principles. On the other hand, I'm not too sure if his commentary on the following is outdated:</p>
<ul>
<li>Ehrenfest paradox </li>
<li>Mach's Principle</li>
<li>other sections</li>
</ul>
|
<p>No. It is not a good starting point. If nothing else, modern notation is very different from Einstein's original notation. Old notation left a lot to be desired about separating tensors from tensor components, if nothing else. </p>
<p>There has also been a lot of new insight into topology, surface charges, the action principle, the nature of black holes and exact solutions to Einstein's equations, and gravitational radiation, amongst many other things, over the past 100 years. If you would like, i could generate a list of better starting books, depending on your beginning fluency with math/physics. </p>
| 568
|
general relativity
|
Schwarzschild metric
|
https://physics.stackexchange.com/questions/15187/schwarzschild-metric
|
<p>Why, if the Schwarzschild metric is a vacuum solution ($T_{\mu\nu}=0$) , do textbooks state that $T=\rho c^{2}$ when approximating Poisson's Equation from the Einstein Field Equations? </p>
<p>Thank you.</p>
|
<p>There is a coordinate slicing known as the Kerr-Schild coordinate system where one can look at the Hamiltonian constraint $16\pi \rho = {}^{3}R - K^{ab}K_{ab} + K^{2}$, and find that the left hand side has the same singularity that you would find in $\nabla \cdot E = \rho$ when you put in the $E$ for a point charge.</p>
<p>So, you can interpret the Schwarzschild solution as having a delta function matter distribution. (though there are problems with this interpretation, too, and it's not something one should take <b>too</b> seriously)</p>
| 569
|
general relativity
|
Is 4-velocity normalized to -1 even for non-geodesic timelike curves?
|
https://physics.stackexchange.com/questions/16852/is-4-velocity-normalized-to-1-even-for-non-geodesic-timelike-curves
|
<p>In Hartle's General Relativity book ("Gravity"), one of the problems (chapter 8 problem 6) is to prove that $g_{\mu\nu}u^\mu u^\nu$ is conserved along geodesics (really not hard to show), where $u^\mu$ is the 4-velocity. My question is: Isn't it true that $g_{\mu\nu}u^\mu u^\nu$ is equal to $-1$ for <em>any</em> timelike curve whether it is a geodesic or not? This follows (I think) from</p>
<p>$$
g_{\mu\nu}u^\mu u^\nu = g_{\mu\nu}\frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} = \frac{g_{\mu\nu}dx^\mu dx^\nu}{d\tau^2}=\frac{ds^2}{d\tau^2} = \frac{-d\tau^2}{d\tau^2} = -1.
$$</p>
<p>Am I wrong about this? Why should we need the geodesic equation to prove this if it's true for any timelike curve?</p>
|
<p>I emailed my TA and here was his answer, which I think makes sense:</p>
<p>While it <em>is</em> true that a curve which is everywhere timelike can be parametrized so that its tangent vector has unit norm, it is also possible to draw a curve which starts out timelike and then becomes null or spacelike, so its norm won't be the same everywhere. The problem in Hartle simply shows that a <em>geodesic</em> which starts out timelike will always remain timelike (same holds for null or spacelike geodesics). </p>
| 570
|
general relativity
|
Proper distance and embedding diagrams?
|
https://physics.stackexchange.com/questions/18553/proper-distance-and-embedding-diagrams
|
<p>I'm trying to understand proper distance equation in Schwarzschild spacetime.</p>
<p>$d\sigma=\frac{dr}{\left(1-\frac{R_{S}}{r}\right)^{1/2}}$.</p>
<p>I'm sure I'm missing something really obvious here, but how do I use this to find the coordinate distance $r$
for a particular proper distance $\sigma$
. For example, if I found the proper circumference of circle going round the Sun that roughly coincides with the Earth's orbit. Then I move radially inwards one proper mile, how would I then find the circumference of the circle I now find myself on. This example is also in the context of trying to understand the spacing of concentric circles in embedding diagrams.</p>
<p>Thank you</p>
|
<p>You have the Schwarzschild metric</p>
<p>$ds^2=(1-R_s/r)c^2dt^2-(1-R_s/r)^{-1}dr^2-r^2(d\theta^2+sin^2\theta d\phi^2)$</p>
<p>For an equatorial orbit, put $\theta=\pi/2; d\theta=0$.
The proper distance between two events is defined as the integral of $ds$ along a spacelike path between them. I'm guessing the two events we're interested in are the (identical) start and end point of an elliptical path, where the path is traversed in zero coordinate time (so $dt=0$). The only things that vary on the path are $r$ and $\phi$. This path will be a function like</p>
<p>$r=a(1-e^2)/(1-e.cos\phi)$</p>
<p>where a and e are a couple of fixed parameters.</p>
<p>Since spacelike separations are negative in this signature, we apply an extra minus sign and get</p>
<p>$d\sigma = \sqrt{(1-R_s/r)^{-1}dr^2-r^2d\phi^2}$</p>
<p>Using the formula for the ellipse, you can get $dr$ in terms of $d\phi$ and take the square root, leaving just a $d\phi$ on the RHS. Integrating from 0 to 2$\pi$ should then give you your proper distance.</p>
<p>When you want to compare this with the value "one mile in", you need to decide what your measure of the radius of the elliptical path is (maybe average r), and adjust the parameters accordingly.</p>
<p>Edit: I just noticed you're using circular orbits. That simplifies it a bit ! In fact, doesn't it make it trivial, since $dr=0$ on your orbit ?</p>
| 571
|
general relativity
|
Projective Transformations in GR
|
https://physics.stackexchange.com/questions/18616/projective-transformations-in-gr
|
<p>A Thought Experiment:</p>
<p>We are in flat spaceime provided with a reference frame—a rectangular Cartesian frame. The coordinate labels[the spatial labels] are visible to us. Each spatial point is provided with a clock—and the different clocks are synchronized wrt to each other. Gravity is now turned on and made to vary upto some final state. During this process of experimentation the physical separations change while the coordinate labels remain fixed to their own positions. [Coordinate separations remain unchanged]. The length and the orientation of a vector changes in this process both in the 3D and in the 4D sense. We are passing through different/distinct manifolds in our thought experiment and if the 4D arc length does not change we are simply having a transition between manifolds for which ds^2 is not changing but the metric coefficients are changing.We consider the option of $ds^2$ changing in this posting.</p>
<p>Query:
Our experiment indicates at projective transformations operating in the physical sense[considering changes in the metric and in the value $ds^2$].A time dependent field is being observed where the metric coefficients are not being preserved. Is it important to include projective transformations[concerned with the non-preservation of the metric] in the mathematical framework of GR?</p>
|
<p>What you are talking about are not what are normally called projective transformations, but nonphysical coordinate changes, these are gauge changes in GR. Einstein was famously confused about this for years, beginning in 1913 when he wrote about the "Hole Argument" in General Relativity. He concluded that it is impossible to have generally covariant equations of motion, only changing his mind in 1915 when he realized the hole argument was completely bogus.</p>
<p>The hole argument is similar to your "projective transformations". He imagines a manifold where the physical situation stays the same, but the points get relabelled dynamically. This leads the metric coefficients to change in a certain way, which cannot be determined by the equation of motion, because it is totally arbitrary.</p>
<p>You didn't say it exactly the same--- you considered letting the metric coefficients change with time, so that the coordinate distance changes, but at the end of the day, the length of any two curves stays the same with the appropriate map, so that the transformation is just a coordinate change. So it's the same thing.</p>
<p>To get unconfused about this, it is best to think of the exactly analogous (but much simpler) gauge transformations in EM. Two vector potentials which have the same holonomy (the same E and B fields in spacetime of normal trivial topology) are identified. This identification means that oscillations of the electromagnetic potential of the form $A=\nabla \phi$ are unphysical, and they have no specific heat, they don't move charges, etc.</p>
<p>The coordinate transformations in GR are exactly analogous. They are more interesting, because in GR, unlike in EM, a gauge transformation, a coordinate change, can introduce a horizon where there was none before. This happens when you change coordinates to the frame of a constantly accelerated observer.</p>
| 572
|
general relativity
|
Modification of de Donder gauge
|
https://physics.stackexchange.com/questions/19684/modification-of-de-donder-gauge
|
<p>The de Donder gauge is often used to simplify the linearised equations of motion of general relativity. If the metric is linearised as $g_{ab} = \bar g_{ab} + \gamma_{ab}$, then the de Donder gauge reads<br>
$\nabla^a(\gamma_{ab} - \frac{1}{2}\bar g_{ab}\gamma) = 0$.</p>
<p>The partial differential equation for the gauge transformation vector $v^a$ is
$ \nabla^b\nabla_b v_a + R_a^b v_b = \nabla^a(\gamma_{ab} - \frac{1}{2}\bar g_{ab}\gamma)$. </p>
<p>In chapter 7.5 of Wald, I read that this equation can always be solved because it is of the form
$g^{ab}\nabla_a\nabla_b \phi_i + \sum_j (A_{ij})^a\nabla_a \phi_j + \sum_j B_{ij}\phi_j + C_i$.<br>
Theorem 10.1.2 of Wald says that in a globally hyperbolic spacetime this equation has a well posed initial value formulation on any spacelike Cauchy surface.</p>
<p>In stead of de Donder gauge, I want to use a similar gauge:<br>
$\nabla^a(\gamma_{ab} - n \bar g_{ab}\gamma) = 0$.<br>
The partial differential equation changes to<br>
$ \nabla^b\nabla_b v_a + (1 -2n)\nabla_a\nabla_b v^b + R_a^b v_b = \nabla^a(\gamma_{ab} - n\bar g_{ab}\gamma)$.</p>
<p>This equation is not covered by theorem 10.1.2 of Wald. My question is: is the existence of a solution for this equation guaranteed in an AdS background when $n=1$?</p>
|
<p>It may be far too late for this to be of any value to you, but in this paper by Mora et al: <a href="http://arxiv.org/abs/1205.4468" rel="nofollow">http://arxiv.org/abs/1205.4468</a>, it is shown that such a gauge choice does exist (for any $n$ in your notation).</p>
| 573
|
general relativity
|
What should be the equation satisfied by The Momentum commutators in a curved background?
|
https://physics.stackexchange.com/questions/21628/what-should-be-the-equation-satisfied-by-the-momentum-commutators-in-a-curved-ba
|
<p>This may be obvious but I have limited experience in physics , The generators of Spatial translation symmetry commutes with each other i.e [P(i),P(j)] = 0 but if Spacetime is a curved manifolds then the value of the commutator should not be zero but some invariant property related to curvature i.e a Function of the curvature tensor If this is false then what should the commutator be like e.g in the vicinity of a gravitational source according to GR , I'm sorry I do not know much in relativity nor differential geometry </p>
|
<p>The commutator you are interested in is non-trivial if you generalize the
translations to curvilinear coordinates. For a vector function $A^{\alpha
}\left( x\right) $, a «translation» along $dx^{\mu}$ is the following
transformation:
$$
A^{\alpha}\rightarrow A^{\alpha}-dx^{\mu}D_{\mu}A^{\alpha}
$$
(where $D_{\mu}$ is the so called covariant derivative) instead of $A^{\alpha
}\rightarrow A^{\alpha}-dx^{\mu}\partial_{\mu}A^{\alpha}$.</p>
<p>First of all, you should understand one basic fact from the group theory: a
commutator of generators corresponds to a generator of some transformation.
This transformation is a superposition of transformations which form a
infinitesimal closed contour in the parametric space of a group. It sounds
complicated but the idea is very simple. Let's imagine you have a group
element:
$$
T\left( \mathbf{a}\right) =\exp\left( i\mathbf{g}\cdot\mathbf{a}\right)
=1+i\mathbf{g}\cdot\mathbf{a+}\frac{1}{2}\left( i\mathbf{\mathbf{g}
\cdot\mathbf{a}}\right) ^{2}+\mathbf{\ldots,}
$$
where $a^{n}$ are parameters of the group, $g_{n}$ are generators of the group
and $\mathbf{g}\cdot\mathbf{a}=g_{n}a^{n}$. Let's now consider the following
sequence of transformations: $T\left( \mathbf{a}\right) $ then $T\left(
\mathbf{b}\right) $, so that $\mathbf{b}\neq\mathbf{a}$, then $T\left(
-\mathbf{a}\right) $, \ so that $T\left( -\mathbf{a}\right) T\left(
\mathbf{a}\right) =1,$ and finally $T\left( -\mathbf{b}\right) $. The
parameters of these transformations form a rectangle in the group parameter
space (see the picture below). Therefore, the total composite transformation
has the form:
\begin{align*}
& T\left( -\mathbf{b}\right) T\left( -\mathbf{a}\right) T\left(
\mathbf{b}\right) T\left( \mathbf{a}\right) =\\
& = \left( 1-i\mathbf{g}
\cdot\mathbf{b+\ldots}\right) \left( 1-i\mathbf{g}\cdot\mathbf{a+\ldots
}\right) \left( 1+i\mathbf{g}\cdot\mathbf{b+\ldots}\right) \left(
1+i\mathbf{g}\cdot\mathbf{a+\ldots}\right) .
\end{align*}
Let's now assume that $\mathbf{a}$ and $\mathbf{b}$ are infinitesimal small,
so that the expansion of the composite transformation has the form:
\begin{align*}
T\left( -\mathbf{b}\right) T\left( -\mathbf{a}\right) T\left(
\mathbf{b}\right) T\left( \mathbf{a}\right) & \approx1+\left(
\mathbf{g}\cdot\mathbf{a}\right) \left( \mathbf{g}\cdot\mathbf{b}\right)
-\left( \mathbf{g}\cdot\mathbf{b}\right) \left( \mathbf{g}\cdot
\mathbf{a}\right) \\
& =1+\left[ g_{m},g_{n}\right] a^{m}b^{n}=1+\frac{1}{2}\left[ g_{m}
,g_{n}\right] f^{mn},\qquad(1)
\end{align*}
where
$$
\left[ g_{m},g_{n}\right] =g_{m}g_{n}-g_{n}g_{m}
$$
and $f^{mn}$ is the so called oriented area element (or directed area
measure):
$$
f^{mn}=a^{m}b^{n}-a^{n}b^{m}.
$$
For example, if the parameter space of the group is three dimensional (as it
is for 3D translations) then the vector
$$
s^{k}=\frac{1}{2}\epsilon^{kmn}f^{mn}=\left[ \mathbf{a}\times\mathbf{b}
\right] ^{k}
$$
is transverse to $\mathbf{a}$ and $\mathbf{b}$, so that its length squared is
the area of the rectangle with the sides $\mathbf{a}$ and $\mathbf{b}$ (see
the figure below):
$$
s^{2}=\mathbf{a}^{2}\mathbf{b}^{2}-\left( \mathbf{a}\cdot\mathbf{b}\right)
^{2}.
$$</p>
<p><img src="https://i.sstatic.net/nUpg1.jpg" alt="enter image description here"></p>
<p>For a flat space, the parallel translation of a vector $\mathbf{v}$ along a
vector $\mathbf{a}$ doesn't change the direction of the vector $\mathbf{v}$.
Therefore, the parallel translation around the infinitesimal closed contour
$(\mathbf{a},\mathbf{b},-\mathbf{a},-\mathbf{b})$ equals to the identical
transformation, hence from the equation (1) we obtain $\left[ p_{i}
,p_{j}\right] =0$. For curved space, the parallel displacement along a small
4-vector $dx^{\nu}$ is non-trivial:
$$
\delta A^{\alpha}=-\Gamma_{\mu\nu}^{\alpha}A^{\mu}dx^{\nu},
$$
where $\Gamma_{\mu\nu}^{\alpha}$ are the so called <a href="http://en.wikipedia.org/wiki/Christoffel_symbols" rel="nofollow noreferrer">Christoffel symbols</a>.
Therefore the parallel translation around an infinitesimal closed contour $C$
is the contour integral:
$$
\Delta A^{\alpha}=-
{\displaystyle\oint\limits_{C}}
\Gamma_{\mu\nu}^{\alpha}A^{\mu}dx^{\nu}.
$$
Applying Stokes' theorem to this integral and assuming that the area enclosed
by the contour $C$ has the infinitesimal small value $\Delta f^{\mu\nu}$, one
can show that
$$
\Delta A^{\alpha}=-\frac{1}{2}R^{\alpha}{}_{\beta\mu\nu}A^{\beta}\Delta
f^{\mu\nu},
$$
where
$$
R^{\alpha}{}_{\beta\mu\nu}=\partial_{\mu}\Gamma_{\beta\nu}^{\alpha}
-\partial_{\nu}\Gamma_{\beta\mu}^{\alpha}+\Gamma_{\mu\rho}^{\alpha}
\Gamma_{\beta\nu}^{\rho}-\Gamma_{\nu\rho}^{\alpha}\Gamma_{\beta\mu}^{\rho},
$$
is the well known <a href="http://en.wikipedia.org/wiki/Riemann_tensor" rel="nofollow noreferrer">Riemann tensor</a>. Now you should remember that the
transformation around an infinitesimal closed contour is a commutator, see
(1). Hence for a vector function $A^{\alpha}\left( x\right) $ we have:
$$
\left[ D_{\mu},D_{\nu}\right] A^{\alpha}=R^{\alpha}{}_{\beta\mu\nu}A^{\beta
}.
$$</p>
<p>Therefore you are right saying that the commutator of translations in a curved
space is non zero, it is in fact Riemann tensor. Although the form of the
covariant derivative depends on the geometric type of the field it acts on,
e.g., for a co-variant tensor field:
$$
\left[ D_{\mu},D_{\nu}\right] A_{\alpha\beta}=A_{\alpha\rho}R^{\rho}
{}_{\beta\nu\mu}+A_{\rho\beta}R^{\rho}{}_{\alpha\nu\mu}.
$$</p>
| 574
|
general relativity
|
Tension on a cable in a gravitational field
|
https://physics.stackexchange.com/questions/22292/tension-on-a-cable-in-a-gravitational-field
|
<p>Consider a mass 'm' suspended in the gravitational field of a massive star. Assuming the Schwarzschild metric it is easy to calculate the gravitational acceleration at the location of the mass and thus the tension in the cable. The question is: how does this tension propagate up the cable?</p>
<p>I've tried to apply the stress-energy tensor, but I'm not convinced I know the correct principles to apply. In thinking about this I've come up with a thought experiment that gives a surprising result and would like some comments on it, and also ideas about the "correct" way to do this via the stress-energy tensor. My thought experiment:</p>
<p>-Consider a long loop of cable with pulleys at each end, one directly above the other. A generator is attached to the upper pulley, and the generator is operated to provide constant tension on the cable on one side of the pulley. An operator at the bottom pulley turns a crank and makes the pulley turn one full turn and stops.</p>
<p>-The work done by the lower operator is $2 \pi RT$. If we assume the tension is constant up the cable (i.e. the cable is massless and there is no transform of the tension by the metric) then the work received at the generator is the same.</p>
<p>-If we now convert this work into a photon and send it back down to the lower pulley, when it arrives there it will be blue shifted and have additional energy, proportional to the square root of the ratio of the $g_{00}$ components of the metric. This would violate conservation of energy and allow a perpetual motion machine, so we assume that this can't happen.</p>
<p>-My conclusion is that the tension on the cable must vary with the square root of the time metric component. I have not seen this described anywhere, however. Does someone know the correct answer, or see the fallacy in this thought experiment?</p>
|
<p>Your naïve interpretation of the work equation doesn't quite make sense in this context. Consider that the standard formula for work gives that $W=\int {\vec F}\cdot d{\vec x}$. At minimum, the presence of the dot product in the above equation should not be ignored, and we should interpret, for a radial force, the work to be equal to $\int \frac{F\,dr}{\sqrt{1-\frac{2M}{r}}}^{1}$. We should then note that this factor will exactly cancel the effect you cite.</p>
<p>${}^{1}$Note that, properly, we are defining the line integral in terms of three unit vectors normal to the line, so, the integral would have the form $\int \sqrt{|g|}\epsilon_{abcd}{\hat t^{a}}{\hat \theta^{b}}{\hat \phi^{c}}F^{d}$. When this is completely simplified, you'll find that the measure of the integral $\sqrt{|g|}$ has a factor of $\sqrt{1-\frac{2M}{r}}$ that cancels against the factor of $\frac{1}{1-\frac{2M}{r}}$ in $g_{rr}$, producing the term seen above. </p>
| 575
|
general relativity
|
Derivation of the Gauss-Codazzi equation
|
https://physics.stackexchange.com/questions/23749/derivation-of-the-gauss-codazzi-equation
|
<p>I'm interested in the derivation of the Gauss equation (Gauss-Codazzi). Usually we consider the definition of the Riemann tensor on the hypersurface.</p>
<p>$$^{(n-1)}R_{abc}^{~~~~~~~d}~w_d=[D_a,D_b]w_c$$</p>
<p>where $D$ is the connexion associated to the induced metric ($h$) on the hypersurface.</p>
<p>We have</p>
<p>\begin{align}
D_aD_bw_c=D_a\Bigl[h_b^{~d} h_c^{~e}\nabla_d w_e\Bigr]=h_a^{~f}h_b^{~g}h_c^{~e}\nabla_f\Bigl[h_g^{~d}h_k^{~e}\nabla_dw_e\Bigr]
\end{align}</p>
<p>where $\nabla$ is the connexion of the metric ($g$) in the full space. </p>
<p>Hence after some algebra we can arrive at the desired result. What I don't understand is why can we not do:</p>
<p>$$
D_aD_bw_c=D_a\Bigl[h_b^{~d} h_c^{~e}\nabla_d w_e\Bigr]=h_b^{~d} h_c^{~e}D_a\Bigl[\nabla_d w_e\Bigr]
$$</p>
<p>because $D_a h^b_c=0$.</p>
<p>But in that case I would have</p>
<p>$$
D_aD_bw_c=h_b^{~d} h_c^{~e}D_a\Bigl[\nabla_d w_e\Bigr]=h_b^{~d} h_c^{~e}h_a^{~\mu}h_d^{~\nu}h_e^{~\sigma}\nabla_\mu\Bigl[\nabla_\nu w_\sigma\Bigr]=h_a^{~\mu}h_b^{~\nu}h_c^{~\sigma} \nabla_\mu\nabla_\nu w_\sigma
$$</p>
<p>Therefore we would have</p>
<p>$$^{(n-1)}R_{abc}^{~~~~~~~d}~w_d=[D_a,D_b]w_c=h_a^{~\mu}h_b^{~\nu}h_c^{~\sigma} ~^{(n)}R_{\mu\nu\sigma}^{~~~~~~~~d}w_d$$</p>
<p>which is not the desired result.</p>
<p>So I understand the standard derivation, but I don't understand why in the way that I wrote, it doesn't work ?</p>
| 576
|
|
general relativity
|
The Light Cone in GR-----A Flickering One?
|
https://physics.stackexchange.com/questions/24514/the-light-cone-in-gr-a-flickering-one
|
<p>Events in relativity[SR or GR]are marked by coordinate values and not by physical values.We write a metric for motion along the x-axis:
$$ds^2=g_{00}dt^2-g_{11}dx^2$$ ----------- (1)</p>
<p>For physical values we may write:</p>
<p>$$ds^2=dT^2-dL^2$$ ------------ (2)</p>
<p>Where $dT^2=g_{00}dt^2$ and $dL^2=g_{11}dx^2$</p>
<p>For the null geodesic $ds^2=0$</p>
<p>From (1) we have for the null geodesic,</p>
<p>$$\frac{dx}{dt}=\sqrt{\frac{g_{00}}{g_{11}}}\ne1$$ [Generally speaking]-----(3)</p>
<p>Relation (2) provides a SR picture in the local context.
since,
$$\frac{dL}{dT}=1$$ ------------ (4)</p>
<p>[c=1 in the natural units]</p>
<p><em>Since events are marked by coordinate values</em> our light cone in GR should correspond to equation (3). It should be a flickering one in a time varying field[ and one with a distorted surface in a stationary field] since the metric coefficients go on changing in a time varying field.</p>
<p>As I advance long the time axis the distorted surface of my light cone goes on changing.Points which I expect to be at space-like separation in the future are now at a time-like separation[or vice versa]</p>
<p>Query:Does this relate to changes in the causal structure of the light cone?</p>
|
<p>The equation for determining the light-cone in the case where there are two varying metric coefficients is the one you wrote down---</p>
<p>$$ {dx\over dt} = \sqrt{g_{00}\over g_{11}} $$</p>
<p>And this does make a curved line in the coordinate-labelled spacetime. This is just stating that light will bend in response to gravity, so that the influence region is dynamical.</p>
<p>I am having a hard time understanding exactly what the question is--- you are saying correct things. The causal structure is dynamical in GR, and you need to understand how the future light-cones intersect each other to find the topological causal structure.</p>
<p>If all the light-cones are spreading outward in a curvy way (but not smooshing together too much, so gravity is not too strong at any point), the causal structure is qualitatively identical to ordinary SR. If the gravity is strong enough to bend the outgoing lightrays in some sphere so that they are really ingoing, you can tell this intrinsically by noticing that the area of the outgoing light-rays is shrinking instead of growing. When there is a sphere where the outgoing lightrays have shrinking area, you have a closed trapped surface, and the causal structure in the interior of the closed trapped surface can't reach infinity without passing through a singularity. The inevitable occurence of a singularity in the presence of a closed trapped surface is Penrose's celebrated theorem, and it is a consequence of the causal structure changing in the presence of enough matter to make a black hole.</p>
| 577
|
general relativity
|
what is relation between time and space in general relativity?
|
https://physics.stackexchange.com/questions/30877/what-is-relation-between-time-and-space-in-general-relativity
|
<p>there is a relation between time and space in special theory of relativity:
$$t^2c^2-L^2=\tau^2.c^2$$
what is relation between time and space in general relativity?</p>
|
<p>The remarkable property of spacetime in GR is that it is <em>locally</em> that of SR. Or, more technically, tangent to every event in the curved spacetime of GR is an SR spacetime. What this means is that, to first order, the line element at any event can be put into the (differential) form of SR in some coordinate system:</p>
<p>$c^2 dt^2 - dL^2 = c^2 d\tau ^2$</p>
<p>The departure from the flat SR spacetime shows up at 2nd order; curvature is characterized by the 2nd order derivatives of the metric.</p>
| 578
|
general relativity
|
Is a volumetric rate frame-invariant in general relativity?
|
https://physics.stackexchange.com/questions/38986/is-a-volumetric-rate-frame-invariant-in-general-relativity
|
<p>Imagine that I have a radioactive material with a long half life. The atoms in this material decay at a certain rate $R$. The rate is the decay constant times the number density $R = \lambda N $. It has dimensionality:</p>
<p>$$ \left( \frac{ \text{decays} }{m^3 s} \right) $$</p>
<p>Imagine that the material is on board a spaceship traveling at some significant fraction of the speed of light. Length is contracted and time is dilated.</p>
<p>$$ \Delta t' = \Delta t \gamma = \frac{\Delta t}{\sqrt{1-v^2/c^2}} $$</p>
<p>$$ L'=\frac{L}{\gamma}=L\sqrt{1-v^{2}/c^{2}} $$</p>
<p>The volumetric decay rate according to the lab reference frame is found by correcting for both the increased density (due to length contraction) and the decreased decay constant (due to time dilation).</p>
<p>$$ N' = L' A = \frac{N}{\gamma} $$</p>
<p>$$ R' = \frac{N'}{N} \frac{\Delta t'}{\Delta t} R = R$$</p>
<p>It's the same volumetric decay rate! Amusingly, the $Q$ value of the decay would be greater, but that's aside the point.</p>
<p>Question:</p>
<p>What if the material was put in a large gravity well? If you use the coordinates from outside the gravity well, would you obtain this same result?</p>
|
<p>I don't know whether it applies to all physically possible metrics, but the volumetric decay rate you define does stay constant in a Schwarzschild metric. Well, it does if the box is small compared to the curvature i.e. the time dilation etc is constant thoughout the box. I would need to think more about what happens if the box is very large.</p>
<p>Anyhow the Scharzschild metric is:</p>
<p>$$ ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1}dr^2 + r^2 d\Omega^2 $$</p>
<p>The time dilation is easy, as we see time moving more slowly for the box by a factor of $(1 - 2M/r)^{1/2}$. I had to think a bit about length contraction, but I think this is a sensible way to define it:</p>
<p>The Schwarzschild radial co-ordinate $r$ is defined as the radius of a circle with circumference $2\pi r$. So we can take a shell with circumference $2\pi r$ and another with circumference $2\pi (r + dr)$ and that defines our ruler of length $dr$. But the observer standing alongside the box would measure a different radial distance between the shells. Specifically they would measure the distance to be $dr/(1 - 2M/r)^{1/2}$. This distance is bigger than the observer at infinity measures, and therefore this means the shell observer's ruler is shorter than ours by a factor of $(1 - 2M/r)^{1/2}$. This factor is exactly the same as the time dilation factor, which means the time dilation and length contraction balance out, and the volumetric decay rate stays the same.</p>
| 579
|
general relativity
|
Do Christoffel symbols commute?
|
https://physics.stackexchange.com/questions/41437/do-christoffel-symbols-commute
|
<p>Do <a href="http://mathworld.wolfram.com/ChristoffelSymbol.html" rel="nofollow">Christoffel symbols</a> commute? For example, does $\Gamma^{e}_{db}\Gamma^{c}_{ea} = \Gamma^{c}_{ea}\Gamma^{e}_{db}$?</p>
|
<p>In classical theory, all observables commute. The components $\Gamma^a_{bc}$ are just real numbers so of course that they commute.</p>
<p>In quantum theory, they don't commute. It's probably a bit laborious to calculate the commutator.</p>
| 580
|
general relativity
|
How do the Einstein's differential equation of the curvature of spacetime come out of Einstein's field equation?
|
https://physics.stackexchange.com/questions/45307/how-do-the-einsteins-differential-equation-of-the-curvature-of-spacetime-come-o
|
<p>The classical theory of spacetime geometry that we call gravity consists of the Einstein equation, which relates the curvature of spacetime to the distribution of matter and energy in spacetime.
$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$
<strong>Mathematically</strong>, how do the Einstein's differential equation of the curvature of spacetime come out of the <a href="http://en.wikipedia.org/wiki/Einstein_field_equations" rel="nofollow">Einstein's field equations</a>, $$G_{\mu\nu}=8{\pi}T_{\mu\nu}$$. ?</p>
|
<p>$ds^{2} = g_{ab}dx^{a}dx^{b}$ isn't the Einstein equation. It's just the equation for what arc length is. It's the definition of the metric tensor, pretty much. You're implicitly using it if you've ever done calculus in three dimensions.</p>
<p>The Einstein equation and the Einstein field equations are the same thing and they are expressible as:</p>
<p>$$R_{ab} - \frac{1}{2}Rg_{ab} = 8\pi T_{ab}$$</p>
<p>Full stop.</p>
| 581
|
general relativity
|
How does one write the Einstein field equations in terms of Ricci tensor?
|
https://physics.stackexchange.com/questions/50141/how-does-one-write-the-einstein-field-equations-in-terms-of-ricci-tensor
|
<p>How can I go from the 'standard' Einstein equations
$R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R = \frac{8\pi G}{c^4}T_{\mu\nu}$ to these equations:
$R_{\mu\nu} = \frac{8\pi G}{c^4}(T_{\mu\nu} - \frac{1}{2}g_{\mu\nu}T)$?</p>
|
<p>Take the trace of the equation by contracting it with $g^{\mu\nu}$:</p>
<p>$$
g^{\mu\nu}R_{\mu\nu}-\dfrac{1}{2}g^{\mu\nu}g_{\mu\nu}R=\dfrac{8\pi G}{c^4}g^{\mu\nu}T_{\mu\nu}
$$</p>
<p>As $g^{\mu\nu}R_{\mu\nu} = R$, $g^{\mu\nu}T_{\mu\nu} \equiv T $ and $g^{\mu\nu}g_{\mu\nu} = 4$, the previous equation gives you $R = -\dfrac{8\pi G}{c^4}T$. Substituting this into Einstein's equation shall give you the result.</p>
| 582
|
general relativity
|
General Relativity Equivalence
|
https://physics.stackexchange.com/questions/55496/general-relativity-equivalence
|
<p>Is Einsteins Equivalence theorem in General Relativity correct? It seems to me that it neglects the fact that gravitational acceleration depends upon separation distance squared, thus neglecting the effect of tidal forces. </p>
<p>For example, as I sit on earth, I experience the affect of earth's gravity; Although the acceleration on my head is slightly less than the acceleration acting on my feet. If I make the claim that my frame is equivalent to me being in a space ship traveling at <code>g</code>, doesn't that mean my whole body is accelerating at <code>g</code> uniformly? This is contrary to the previous statement though.</p>
|
<p>The equivalence principle, as stated correctly by Einstein, says that these two situations are equivalent:</p>
<ul>
<li>An uniformly accelerating observer in the absence of a gravitational field</li>
<li>A free falling observer in an uniform gravitational field</li>
</ul>
<p>So, as you noted, this does not apply to the gravitational field of the Earth. Imagine you are in an elevator, free falling towards the Earth. You could let go of two pens - what you would see is that the two pens would come closer to each other, since each of them would be falling towards the centre of the Earth. You, as an observer, could then say with certainty that you are in a gravitational field.</p>
<p>In a hypothetical uniform gravitational field no experiment could reveal, whether you are in a gravitational field or not. Einstein took this thought experiment as a motivation for the development of General Relativity. In the mathematical construction of General Relativity, the equivalence principle does not play an important role.</p>
| 583
|
general relativity
|
General Relativity Paradox - Different local times of two frames a constant distance apart
|
https://physics.stackexchange.com/questions/56998/general-relativity-paradox-different-local-times-of-two-frames-a-constant-dist
|
<ul>
<li>Suppose there is a habitable star with a significantly large mass, and thus a huge gravitation field. It has a clock on it that ticks each local second. And it also has a mirror. This is Star A.</li>
<li>Suppose there is another habitable star with a much smaller mass, also with a clock, called Star B. </li>
<li>Finally, suppose that these two stars somehow maintain a fixed distance between them. (Eg: the two stars have perfectly calibrated rocket thrusters pointing toward one another).</li>
</ul>
<p>Please correct me if I'm wrong: An observer on Star A looking through a telescope at clock B would see it ticking quickly. Likewise, an observer on Star B looking at the clock on Star A would see it ticking slowly.</p>
<p>Now, suppose a person on A sends a light pulse towards B and starts a clock. They measure it takes 10 seconds for it to come back.</p>
<p>Now, a person on B sends a pulse to A, and measures how long it takes to get back. Does it also take 10 seconds? If so, there's a pretty clear paradox. If not, how could it take light different times to travel the same distance?</p>
<p>Thanks</p>
|
<p>Time is not the only measurement that is affected by a gravitational field. What makes you think that A and B measure the same distance between them? It helps to think about how you would actually measure the distance to a faraway object. If you are patient, you could do this measurement by bouncing a light beam off the object and seeing how long it takes to return, in exactly the manner you described already. (I have heard that this has actually been done to measure the distance to the Moon precisely.)</p>
<p>So let A bounce a light beam off B. It takes a time $2T$ to arrive back, therefore A thinks that B is at a distance $L = cT$, where $c$ is the speed of light.</p>
<p>Now let B do the same experiment. Of course, A measures the same time $2T$ for the beam to arrive and bounce back to B. But as you stated, if from A's point of view a time $2T$ has passed, from B's point of view a different time $2T^{\prime}$ has passed when the light returns. You know that the speed of light appears constant for all observers, so $B$ must measure a distance $L^{\prime} = cT^{\prime} \neq L$.</p>
<p>The physical meaning is that gravity is a warping of <strong>spacetime</strong>, not just space or time separately. The dilation of time measured by clocks due to any relativistic effect is exactly balanced by the contraction of metre rules (or yard sticks, or whatever) so that the speed of light is always constant. </p>
| 584
|
general relativity
|
Understanding Einstein's field equation
|
https://physics.stackexchange.com/questions/61179/understanding-einsteins-field-equation
|
<p><a href="http://en.wikipedia.org/wiki/Einstein_field_equations">Einstein's field equation</a>:</p>
<p>$$G_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu\nu} - g_{\mu\nu}\Lambda$$</p>
<p>I'm trying to understand each of the terms in this equation intuitively, but I'm struggling.</p>
<p>Basically, I want to understand how these equations allow me to predict the path of a particle, given the mass and energy distribution of a system.</p>
<p>I have some idea that $G_{\mu\nu}$ represents the curvature of spacetime, and that $T_{\mu\nu}$ represents the distribution of energy in the system, but it's not clear how.</p>
<p>Correct me if I'm wrong about any of this; I'm just starting.</p>
|
<p>$G_{\mu \nu}$ is the Einstein tensor, and is calculated by the following:</p>
<p>$$G_{\mu \nu} = R_{\mu \nu} -\frac{1}{2}g_{\mu \nu}R$$</p>
<p>where $R_{\mu \nu}$ is the Ricci curvature tensor, $g_{\mu \nu}$ the metric tensor and $R$ is the trace of the curvature tensor with respect to the metric, i.e. $g^{\mu \nu}R_{\mu\nu}$, where the metric tensor with indices raised is related to the other by taking the inverse. I have never used the Einstein field equations to calculate paths chosen by a particle - you use the geodesic equations for that. The E.F.E, as you said, equate physical quantities in a system to curvature. One can be given a metric tensor and compute those quantities (which is rather tedious, as you will see if you fully expand the expressions for the terms), or solve the equations, i.e. obtain the metric (which describes the resultant manifold) given a $T_{\mu\nu}$.</p>
<p>You can also think of the Einstein field equations as the equations of motion of the metric tensor which arise when applying the principle of least action. The Einstein-Hilbert action is:</p>
<p>$$S=\frac{c^4}{16\pi G}\int d^4x \, \ R \sqrt{-g}$$</p>
<p>If you have a matter field, you add a lagrangian into the expression, apply the principle of least action, i.e. $\delta S =0$ and you will obtain a term which is defined as the stress-energy tensor equated to $G_{\mu\nu}$. Also, note that the stress-energy tensor $T_{\mu\nu}$ can be obtained via Noether's theorem which states that a continuous symmetry that a Lagrangian exhibits implies a conservation law. See Peskin and Schroeder's book on quantum field theory for a discussion and demonstration of this.</p>
| 585
|
general relativity
|
Timelike/null generic condition in general relativity
|
https://physics.stackexchange.com/questions/70193/timelike-null-generic-condition-in-general-relativity
|
<p>My question concerns the following definition</p>
<blockquote>
<p><strong>Definition:</strong> The <em>timelike</em> (resp. null) <em>generic condition</em> in GR is fulfilled if
$$u_{[\alpha} R_{\rho]\mu \nu [\sigma}u_{\beta]}u^\mu u^\nu \ne 0$$
at some point of each timelike (resp. null) geodesic with tangent vector $\vec u$. ($R_{\rho \mu \nu \sigma}$ is the Riemann curvature tensor.)</p>
</blockquote>
<p>It is written in many places that this is the right condition to impose if one wants to assume that every freely falling (or light) particle encounters some form of matter or radiation in its history (or something to that effect).</p>
<p>But I don't understand, why the particular tensor $u_{[\alpha} R_{\rho]\mu \nu [\sigma}u_{\beta]}u^\mu u^\nu \ne 0$ is the right thing to look at in this context. For example, why don't we assume that $R_{\rho\mu\nu\sigma}u^\mu u^\nu \ne 0$ at some point? Or maybe that $R_{\mu \nu}u^\mu u^\nu \ne 0$, or perhaps that $G_{\mu \nu} u^\mu u^\nu \ne 0$?</p>
<p>I'm guessing that the last two condition could be too weak to derive the singularity theorems we want, so that something stronger must be assumed.
But the expression $u_{[\alpha} R_{\rho]\mu \nu [\sigma}u_{\beta]}u^\mu u^\nu \ne 0$ really looks a bit strange to me (i.e. I don't understand its significance). Could someone explain to me why this is the right condition to impose? </p>
<p>Thanks!</p>
|
<p>Carroll has this to say about this condition</p>
<blockquote>
<blockquote>
<p>These fancy conditions simply serve to exclude very special metrics for which the curvature consistently vanishes in some directions - Carroll P. 242-243</p>
</blockquote>
</blockquote>
<p>While that answers your question, it doesn't give much insight on what the curvature vanishing in some directions has to do with singularity theorems. (The generic condition on the metric is required to prove Hawking's and Penrose's singularity theorems) I hope someone else can give more insight on that.</p>
| 586
|
general relativity
|
Conservation of Energy and Birkhoff's theorem
|
https://physics.stackexchange.com/questions/71952/conservation-of-energy-and-birkhoffs-theorem
|
<p>I am reading the original paper by Bondi, van der Berg and Metzner (<a href="http://rspa.royalsocietypublishing.org/content/269/1336/21" rel="nofollow">link</a>) regarding gravitational waves in asymptotically flat axisymmetric spacetimes. In the introduction, he makes the following comment - </p>
<blockquote>
<p>The conservation of mass effectively prohibits purely spherically symmetric waves and similarly, conservation of momentum prohibits waves of dipole symmetry. </p>
</blockquote>
<p>I know that Birkhoff's theorem tells us that spherically symmetric asymptotically flat solution to GR is necessarily static, and therefore contains a timelike Killing vector, which implies conservation of mass (energy). Bondi et. al. seem to be stating the converse of this theorem, whose validity I do not immediately see. How do we show this?</p>
<p>Also, what is the corresponding proof of the second statement made above?</p>
|
<p>All you need to do is set up a multipole expansion of the gravitational waveform. You'll find that the monopole moment is proportional to the time derivative of the mass of the stress-energy tensor, and the dipole moment is proportional to the second time derivative of the momentum from the stress-energy tensor, both of which are conserved. Thus, the first nonzero moment comes from the quadrupole moment. This is worked out in great detail in MTW.</p>
| 587
|
general relativity
|
About an Einstein equation
|
https://physics.stackexchange.com/questions/72054/about-an-einstein-equation
|
<p>This is a question about an historical theory of gravitation, studied by Einstein quite a bit <em>before</em> he settled on General Relativity. At that time, Einstein did not know that gravity was a consequence of curved space-time. He identified the variations of gravity with the variations of light speed in a gravitational field. <br></p>
<p>In March 1912, Einstein postulated a first equation for static gravitational field, derived from the Poisson equation
$$\Delta c = kc\rho \tag{1}~,$$ where $c$ is light speed, $\rho$ is mass density and $\Delta$ is Laplacian.</p>
<p>Two weeks later, he modified this equation by adding a nonlinear term to satisfy energy-momentum conservation :
$$\Delta c = k\big(c\rho+\frac{1}{2kc} (\nabla c)^2\big)~. \tag{2}$$
Einstein's argument is the following:</p>
<p>The force per unit volume in terms of the mass density $\rho$ is $f_a$
$= \rho \nabla c$. Substituting for $\rho$ with $\frac{\Delta c}{kc}$ [equation (1)], we find $$f_a = \frac{\Delta c}{kc} \nabla c~.$$</p>
<p>This equation must be expressible as a total divergence (momentum conservation) otherwise the net force will not be zero (assuming $c$ is constant at infinity).
Einstein says: </p>
<blockquote>
<p>"In a straightforward calculation, the equation <strong>(1)</strong> must be
replaced by equation <strong>(2)</strong>."</p>
</blockquote>
<p>I never found the straightforward calculation. That's something that's actually hard for me!</p>
<p><b>addendum</b> <br>
The solution given and explained by @Gluoncito (see below) answers perfectly my question. However, it is likely that it is not the demonstration of Einstein for at least one reason : It is not a <em>straightforward calculation</em>.<br> Historically, Abraham, a german physicist, was the first to generalize the Poisson equation by adding a term for the energy density of the gravitational field (coming from $E=mc^2$). He published a paper in january 1912 containing a static field equation with the term :
$\frac{c^2}{\gamma}(\nabla c)^2 $ different but not far away from the Einstein term. After the publication of Einstein, Abraham claimed That Einstein copied his equation. I believe Einstein was at least inspired by Abraham. To what extent, I don't know. </p>
|
<p>There is a derivation of the equations above given by Giulini (may be more pedagogical?), You can look at it at :</p>
<p><a href="http://ae100prg.mff.cuni.cz/presentations/Giulini_Domenico.pdf" rel="nofollow">http://ae100prg.mff.cuni.cz/presentations/Giulini_Domenico.pdf</a></p>
<p>As you will see he arrives at the same equation (2) assuming the "variable speed of light" is actually proportional to the gravitational potential, as I first assumed, (no need of variable speed of light). Please note that in general relativity the speed of light is constant in local charts, and that`s enough for the theory.</p>
<p>ok, as asked by the operators I copy the main parts of the demonstration:
the field equation for the gravitational field in Newtonian mechanics is:
$$\Delta \phi = 4πG \rho$$,</p>
<p>the Newtonian force per unit volume (mass density x acceleration) is:
$$f = −\rho \nabla \phi$$.
Now, the work done against gravity to assemble a piece of matter $\delta \rho$ (along an incremental change $\delta \xi$ along the flow) is:
$$\delta A=-\int \delta \vec{\xi}.\vec{f} =\int \phi \delta \rho$$,</p>
<p>a small change in the density of matter produce a change in the gravitational potential:
$$\Delta \delta \phi = 4πG \delta \rho$$
with this the work can be written:
$$\delta A=\int \phi \delta \rho=\delta ( \frac{-1}{8 \pi G} \int (\nabla \phi)^2)$$
where the equality is given integrating by parts the lhs.
Thus it is possible to find the energy density of the gravitational field as:
$$\epsilon=\frac{-1}{8 \pi G} (\nabla \phi)^2$$
Now, the important point is that any source of gravitational field must be compatible with the principle $E=m c^2$.
Thus, (I jump to eq. 9) the mass equivalence of the gravitational field is:
$$\delta M_g=\frac{1}{4 \pi G} \int \Delta \delta \phi$$
*
Newtonian gravity fail this principle since the rhs is zero in absence of matter.
So Einstein added the energy of the gravitational field (the $\epsilon$ calculated before) as a source.
$$\Delta \phi=4 \pi G (\rho-\frac{1}{8 \pi G c^2} (\nabla \phi)^2)$$
Then computes the mass term (a complicated integral I`m a bit lost here, eq. 11), and redefines for consistency with the work $\delta A$ the field:
$$\phi \rightarrow \Phi=c^2 exp(\phi/c^2)$$
with this definition the equation becames:
$$\Delta \Phi=\frac{4 \pi G}{c^2} (\Phi \rho +\frac{c^2}{8 \pi G \Phi} (\nabla \Phi)^2)$$
The redistribution of the c-factors is due to the redefinition of $\phi$. The $\Phi$ multiplying $\rho$ cancels when you replace everything by $\Phi$ and gets the original eqs. * above.</p>
<p>Well, this was the derivation of Giulini, not very pedagogical because of eq. 11. If I understand it I tell you. It could be better to read the original Einstein paper but I have it only in German. I'm sure Einstein was clearer at the end...</p>
| 588
|
general relativity
|
Axial symmetry constraints on the metric
|
https://physics.stackexchange.com/questions/72074/axial-symmetry-constraints-on-the-metric
|
<p>I am reading the paper on Gravitational Waves in General Relativity. VII. Waves from Axi-Symmetric Isolated Systems by H. Bondi, M. G. J. van der Burg, A. W. K. Metzner. (<a href="http://rspa.royalsocietypublishing.org/content/269/1336/21" rel="nofollow noreferrer">link</a>) Here is a quote(s) from that paper</p>
<blockquote>
<p>Throughout this paper, we shall suppose that the 4-space is axially symmetric and reflexion symmetrical....</p>
<p>...... From the axial symmetry, the azimuth angle is readily defined. Suppose we now put a source of light at a point <span class="math-container">$O$</span> on the axis of symmetry and surround it by a small
sphere on which we can produce the azimuth co-ordinate <span class="math-container">$\phi$</span> together with a colatitude
<span class="math-container">$\theta$</span> and a time co-ordinate <span class="math-container">$u$</span>. We then define the <span class="math-container">$u$</span>, <span class="math-container">$\theta$</span>, <span class="math-container">$\phi$</span> co-ordinates of an arbitrary event <span class="math-container">$E$</span> to be the <span class="math-container">$u$</span>, <span class="math-container">$\theta$</span>, <span class="math-container">$\phi$</span> co-ordinates of the event at which the light ray <span class="math-container">$OE$</span> intersects the small sphere. In other words, along an outward radial light ray the three co-ordinates <span class="math-container">$u$</span>, <span class="math-container">$\theta$</span>, <span class="math-container">$\phi$</span> are constant. If we wish to write down the metric for such a system of co-ordinates (in which the part referring to the azimuth angle <span class="math-container">$\phi$</span> appears separately) then we know that since only the co-ordinate <span class="math-container">$r$</span> varies along a light ray, the term <span class="math-container">$g_{11}$</span> of the metric tensor must vanish, the four co-ordinates <span class="math-container">$u$</span>, <span class="math-container">$r$</span>, <span class="math-container">$\theta$</span>, <span class="math-container">$\phi$</span> being denoted by 0, 1, 2, 3 in that order. Moreover, we must have
<span class="math-container">$$\Gamma^0_{11} = \Gamma^2_{11}$$</span></p>
</blockquote>
<p><strong>QUESTION: Why do the above Christoffel symbols vanish?</strong></p>
<p>EDIT:
<strong>QUESTION 2: What does one mean by reflexion symmetry?</strong></p>
|
<p>I finally figured it out. It's simple really. He requires that we have radial null geodesics right? This implies that the geodesic equation
$$
\frac{d^2 x^\lambda}{d \tau^2} + \Gamma^\lambda_{\mu\nu} \frac{d x^\mu}{d \tau} \frac{d x^\nu}{d \tau} = 0
$$
should be solved by $(u,\theta,\phi)$ constant. Plugging this into the equation above, we immediately find the constraints
$$
\Gamma^0_{11} = \Gamma^2_{11} = \Gamma^3_{11} = 0
$$
He doesn't mention $\Gamma^3_{11}$ because it is trivially satisfied for the metric at hand, since
$$
\Gamma^3_{11} = \frac{1}{2} g^{3\rho} \left( 2 g_{\rho 1, 1} - g_{11,\rho} \right) = \frac{1}{2} g^{33} \left( 2 g_{3 1, 1} - g_{11,3} \right) = 0
$$
where we have used $g_{11} = 0$ ($(u,\theta,\phi)$ constant along null geodesics), $\partial_3 g_{\mu\nu} = 0$ (axial symmetry) and $g_{03}=g_{13}=g_{23} = 0$(reflexion symmetry)</p>
| 589
|
general relativity
|
Are orbits reversible in general relativity?
|
https://physics.stackexchange.com/questions/72359/are-orbits-reversible-in-general-relativity
|
<p>It seems if I reverse velocities then things begin orbiting backwards, at least in classical mechanics.</p>
<p>From <a href="https://en.wikipedia.org/wiki/Orbital_mechanics#Laws_of_astrodynamics" rel="nofollow">here</a>:</p>
<blockquote>
<p>Every orbit and trajectory outside atmospheres is in principle reversible, i.e., in the space-time function the time is reversed. The velocities are reversed and the accelerations are the same, including those due to rocket bursts. Thus if a rocket burst is in the direction of the velocity, in the reversed case it is opposite to the velocity. Of course in the case of rocket bursts there is no full reversal of events, both ways the same delta-v is used and the same mass ratio applies.</p>
</blockquote>
<p>What's up when I put relativistic effects into the mix?</p>
<p>So for example I watch a super light test particle orbiting a black hole in a highly precessing flower shaped orbit. Then I put a bouncy wall into it's path that's at rest from my viewpoint when the particle hits it, so the particle bounces back reversing it's velocity from my viewpoint.</p>
<p>Would it begin running its orbits backwards?</p>
<p>The actual reason I'm asking this, because I want to know whether I can use backwards ray-tracing to render a black hole.</p>
|
<p>Yes, the Schwarzchild space-time is reversible. Closed orbits and the like will stay closed in the time-reversed system. </p>
<p>There is, of course one obvious flaw: what about the horizon? Things go in, but they do not go out. Well, the answer to that is that the full spacetime doesn't JUST include a black hole, it includes a white hole/black hole pair. If you time reverse the spacetime, you tranform a particle falling into the black hole into a particle falling out of the white hole (and eventually, back into the black hole). This doesn't come up, because we typically don't consider spacetimes containing a white hole as real physical solutions. But in the idealized, mathematical case, you have to include it. </p>
| 590
|
general relativity
|
Is it possible that a matter field has a dependent on non-radial space-like coordinate in a spacetime with spherical symmetry?
|
https://physics.stackexchange.com/questions/78608/is-it-possible-that-a-matter-field-has-a-dependent-on-non-radial-space-like-coor
|
<p>After the work from Breitenlohner and Freedman, we know matter fields in asymptotically AdS spacetime can be stable out of the black hole under some special conditions.</p>
<p><strong>My question:</strong> In such a spherically symmetric spacetime, could the matter field has the non-radial variable of the coordinate? If it could be, what is the limit on it? </p>
<p>Suppose the metric tensor is
$$\mathrm{d}s^2 = -f(r)\mathrm{d}t^2 + \frac{\mathrm{d}r^2}{f(r)} + \frac{r^2}{L^2}\Omega_2(\theta, \phi)$$
$r \rightarrow +\infty$ is the infinite boundary, $f(r) \rightarrow \frac{r^2}{L^2}$ as $r \rightarrow +\infty$ and $L$ is the AdS radius, $\Omega_2$ is the 2-d spherical coordinate.</p>
<p>The matter field $\Phi(r,\theta)$ in question is a complex scalar field with a charge $q$
and the Maxwell field (1-form) reads:
$$A = A_t(r,\theta) \mathrm{d}t^2$$</p>
<p>the coordinate $r$ is the radial variable and $x$ is the non-radial. </p>
<p>Is it possible for the set-up above? </p>
<p>Thank you for your time.</p>
<p>-------------------<strong>Update</strong>---------------</p>
<p>To avoid more confusion, rather planar AdS black hole, we consider the metric as above showed. </p>
<p>Perhaps my question can be changed into this one:
Provided the stable spacetime is of spherically symmetric, if all matter fields coupling with the black hole in this spacetime should has the $SO(2)$ symmetry, and this property is equal that all matter fields haven't the non-radical space-like variable?</p>
| 591
|
|
general relativity
|
How strong is the spacetime curvature at distance $d$ for a nonmoving point mass?
|
https://physics.stackexchange.com/questions/92769/how-strong-is-the-spacetime-curvature-at-distance-d-for-a-nonmoving-point-mass
|
<p>Consider a point mass $A$ with mass $m$ in empty space. The point mass $A$ does not have a velocity and does not rotate.
Since gravity is symmetric for nonmoving objects, the spacetime curvature around $A$ is also symmetric.</p>
<p>So at a distance $d$ from the point mass $A$ how strong is the curvature $C$ ?</p>
<p>$$ C = f(d,m) $$ $$f = ???$$</p>
|
<p>It sounds as if you just want the acceleration given by the non-relativistic equation from Newton's law:</p>
<p>$$ a = \frac{GM}{r^2} $$</p>
<p>where $M$ is the mass of the object generating the gravitational field (strictly speaking this equation only applies when the mass of the accelerating object is much less than $M$).</p>
<p>For the GR version of this have a look at twistor59's answer to <a href="https://physics.stackexchange.com/questions/47379/what-is-the-weight-equation-through-general-relativity">What is the weight equation through general relativity?</a>:</p>
<p>$$ a = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}} $$</p>
<p>This is the simplest treatment of the problem I've seen, but even so I suspect you'll have problems with this unless your maths is reasonably advanced.</p>
| 592
|
general relativity
|
Is it possible to express "free"-ness of a time-like world line without referring to "tangent space" (but only directly to causal relations )?
|
https://physics.stackexchange.com/questions/93037/is-it-possible-to-express-free-ness-of-a-time-like-world-line-without-referrin
|
<p>I don't know much about <a href="http://ncatlab.org/nlab/show/tangent+bundle" rel="nofollow noreferrer"><i>tangent spaces</i>, or <i>tangent vectors</i>, "as such"</a>; nor about <a href="http://en.wikipedia.org/wiki/Schild%27s_ladder#affine_parametrization" rel="nofollow noreferrer"><i>affine parametrization</i></a> (which seems to be closely related to the notion of <i>tangent vectors</i>, as far as I understand for instance MTW, Box 10.2, section B).</p>
<p>Is it possible to explain/express whether some particular <i>identifiable point</i> (cmp. MTW, Box. 13.1) had been "<i>free</i>" (or had "<i>moved freely</i>"; or was "<i>in free fall</i>"; or was represented by a "<i>geodesic</i>"; etc.) without explicitly using the notion of <i>tangent vector</i> or <i>affine parametrization</i>?</p>
<p>I'd be especially looking for such a description being given explicitly and exclusively in terms of particular <i>identifiable points</i> and coincidence events in which they took part (or also, which <i>identifiable points</i> didn't take part in some particular coincidence event of other participants); or (equivalently, as far as I understand) in terms of whether coindicence events under consideration are <a href="http://en.wikipedia.org/wiki/Causal_structure" rel="nofollow noreferrer"><i>time-like</i> or <i>space-like</i> related to each other (or neither, i.e. <i>light-like</i>)</a>.</p>
<p>(I have already put forth some related attempts <a href="https://physics.stackexchange.com/questions/92606/a-question-about-a-relation-between-time-like-world-lines">here</a> or <a href="https://physics.stackexchange.com/questions/66636/which-causal-structures-are-absent-from-any-nice-patch-of-minkowski-space">there</a>. But perhaps this rephrased question helps to focus the effort ...) </p>
|
<p>Since I'm in a car right now I can't double check my answer, but I'll try my best not to lie to you. </p>
<p>To begin with I am a little unclear as to what you are asking. In most circumstances (i.e if the region under consideration is geodesically complete, which in physics is the case if there is no singularity present) any two points can be connected by a geodesic, and to any one point and direction with magnitude (tangent vector) there is a geodesic that passes through that point with the same direction and magnitude (velocity). </p>
<p>Now I believe that you are asking whether the motion of some body is that of a freely falling body, and specifically whether or not there is some description of the answer that does not specifically mention tangent vectors or affinely parametrized curves. The best I can do is that the body is in free fall if and only if the proper time between any two points on its world line is (locally) maximized. That is to say if any slightly different path that it would have followed would have produced a smaller proper time interval between the two points. The local part (slight variations) is there because there may be more than one geodesic connecting the same two points. </p>
<p>Of course, mathematically the proper time is defined through affinely parametrized curves, but at least it's a description of the physics behind the answer without explicit reference to the mathematics. I believe most of us can have a sort of intuitive understanding of proper time from introductory special relativity. </p>
| 593
|
general relativity
|
Sobolev norm for Schwarzschild metric
|
https://physics.stackexchange.com/questions/100996/sobolev-norm-for-schwarzschild-metric
|
<p>Considering a static spacetime of the metric form
\begin{equation}
\mathrm{d}s^{2}=-V^{2}\mathrm{d}t^{2}+h_{ij}\mathrm{d}x^{i}\mathrm{d}x^{j}
\end{equation}
with a timelike killing field $\xi^{\mu}=(\partial_{t})^{\mu}$ we
can choose a function space on each constant time hypersurface $\Sigma$
as$\mathcal{H}=\left\{ f\,\mid\,\parallel f\parallel<\infty\right\} $with
the Sobolev norm $\parallel f\parallel$being given by
\begin{eqnarray*}
\parallel f\parallel^{2} & = & \frac{q^{2}}{2}\underset{\Sigma}{\int}\mathrm{d}\Sigma V^{-1}f^{*}f+\frac{1}{2}\underset{\Sigma}{\int}\mathrm{d}\Sigma Vh^{ij}D_{i}f^{*}D_{j}f.
\end{eqnarray*}
In (negative mass) Schwarzschild coordinates this norm reads reads
\begin{equation}
\parallel f\parallel^{2}=\frac{q^{2}}{2}\underset{\Sigma}{\int}\mathrm{d}\mu\mathrm{d}rR^{n}V^{-2}\mathcal{\mid}f\mid^{2}+\frac{1}{2}\underset{\Sigma}{\int}\mathrm{d}\mu\mathrm{d}rR^{n}V^{2}\mid f'\mid^{2}
\end{equation}
with $\mathrm{d}\mu$ being the volume element of unit $n$-sphere
($\mathrm{d}\Sigma=\mathrm{d}\mu\mathrm{d}rV^{-1}R^{n}$).</p>
<p>My question is: How does this norm read for positive mass Schwarzschild spacetime? $V^2$ is assumed positive above (which is ok for $V=(1+2M/r)$ (M positive). But for $V=(1-2M/r)$ there will be the point of a change in sign so that the above norm cannot hold anymore.How does the norm then look like???</p>
<p>Thanks!</p>
|
<p>Notice that if $r<2M$, the coordinate $t$ turns out to be <strong>spacelike</strong>, so surfaces at $t$-constant are not appropriate for imposing, for instance, Cauchy data.
(I guess that your Sobolev norms are used to deal with the Cauchy problem). </p>
<p>If you are dealing with the whole Kruskal manifold you should fix a spacelike Cauchy surface instead, which for instance coincides to a $t$-constant surface in both the static wedges. </p>
<p>If you use the $t=0$-surface, the singularity region (where the measure you exploit to construct your norm is singular) is made of the <strong>bifurcation sub manifold</strong> $\cal B$ (a $2$-sphere) localized at $r=2M$ and you <strong>never</strong> encounter the region $r<2M$ on this spacelike Cauchy surface.
The submanifold $\cal B$ has obviously zero measure.</p>
<p>In
<a href="http://en.wikipedia.org/wiki/File:Kruksal_diagram.jpg" rel="nofollow">http://en.wikipedia.org/wiki/File:Kruksal_diagram.jpg</a>
the Cauchy surface I am mentioning is represented by the $x$ axis. $\cal B$ is
the intersection of the $x$ axes and the $y$ one. $r<2M$ holds in both the <strong>open</strong> regions indicated by II and IV, so you see that the Cauchy surface never meets them. </p>
<p>To construct a well defined Sobolev norm it is sufficient to change the measure in a neighbourhood of ${\cal B}$ to make it finite. I do not know if there is a canonical way to do it. </p>
<p>(Together with a pair of colleagues I used a similar technology to rigorously construct the so called Unruh vacuum for zero mass particles, proving that the state verifies the Hadamard condition at short distances so that it can be used for renormalization procedures (Adv. Theor. Math. Phys. 15, vol 2, 355-448 (2011) ))</p>
| 594
|
general relativity
|
The border of the System
|
https://physics.stackexchange.com/questions/111766/the-border-of-the-system
|
<p>In General Relativity, if the system accelerates, the inside of the system and the outside of the system will have different speed of time.</p>
<p>Where is the boundary of the system? </p>
<p>If a human accelerates closer to the speed of light, does that mean the human exist inside of their specific system? </p>
| 595
|
|
general relativity
|
Stationary and Static
|
https://physics.stackexchange.com/questions/112670/stationary-and-static
|
<p>I have some confusion about the concept of stationary and static.</p>
<p>A metric $g$ is called <strong>stationary</strong> if there is a time like Killing vector $K$.</p>
<p>$g$ is called <strong>static</strong> is further HSO (hyper surface orthogonal), i.e. there exists a foliation of hyper surface orthogonal to the integral curve of $K$.</p>
<p>The Schwarzschild solution requires $g$ to be static, and the Kerr solution only require $g$ to be stationary.</p>
<p>Here comes my question. To my understanding, requiring $K$ to be HSO is to enabling us to slice the spacetime into pieces with fixed time. The Kerr solution is for the metric outside a rotating body. In the note I am reading, it says </p>
<blockquote>
<p>Do not require $K$ to be HSO, after all we want a rotating body.</p>
</blockquote>
<p>Why a rotating body indicates that the metric cannot be HSO? I think I can also slice the spacetime into piece for every time.</p>
| 596
|
|
general relativity
|
What is the effective physical difference between a massive region of a polarized vacuum and a region of curved space-time?
|
https://physics.stackexchange.com/questions/114091/what-is-the-effective-physical-difference-between-a-massive-region-of-a-polarize
|
<p>What is the <em>effective</em> physical difference between a large region of curved space-time and an equally large region of a polarized vacuum? Consider the fact that vacuum polarization amounts to an effective deviation in speed of light in a local region due to the deviation in electric permittivity of free space.</p>
| 597
|
|
general relativity
|
Hyper surface orthogonal vector in Boyer-Lindquist coordinate
|
https://physics.stackexchange.com/questions/117325/hyper-surface-orthogonal-vector-in-boyer-lindquist-coordinate
|
<p>The Boyer-Lindquist coordinate coordinate of the Kerr Solution is
$$
ds^2=\left(1-\frac{2Mr}{\Sigma}\right)dt^2+\frac{4Mar\sin^2\theta}{\Sigma}dtd\phi - \frac{\Sigma}{\Delta}dr^2-\Sigma d\theta^2-\left(r^2+a^2+\frac{2Ma^2r\sin^2\theta}{\Sigma}\right)\sin^2\theta d\phi^2
$$</p>
<p>Let
$$
K=\frac{\partial}{\partial t}=(1,0,0,0)
$$</p>
<p>I am asking if it is possible for the vector field $K$ to be hypersurface orthogonal.</p>
<p>The answer is that it is possible if and only if $J=Ma=0$, i.e. the cross term in the metric, $g_{t\phi}$, vanishes.</p>
<p>I can figure it out only by the following (incorrect) argument.</p>
<p><strong>Assuming we want $K$ to be orthogonal to the surface $t=$ constant</strong>, the let $(t_0,r,\theta,\phi)$ be a curve in that surface, thus $V=(0,\dot r,\dot\theta,\dot\phi)$ is a tangent vector in that space. To make $K$ orthogonal, we must have
$$
\langle K,V\rangle=0
$$
which is
$$
\dot\phi g_{t\phi}=0
$$</p>
<p>so we must have $g_{t\phi}=0$.</p>
<p>But I think the above argument is incorrect, since we only proved that $K$ cannot be orthogonal to the surface $t=$ constant, there maybe other possibilities.</p>
<p>Then I tried to use the Frobenius Theorem, saying that $K$ is HSO if and only if
$$
K_{[a}\nabla_bK_{c]}=0
$$</p>
<p>We have
$$
K_a=K^bg_{ab}=g_{at}
$$</p>
<p>Then equivalently we must show
$$
g_{at}\nabla_bg_{ct}+g_{bt}\nabla_cg_{at}+g_{ct}\nabla_ag_{bt}=0
$$</p>
<p>But I don't know how to move on.</p>
<p>Furthermore, if $\nabla$ is Levi-Civita, i.e. it is metric compatible, the above equation holds naturally, no matter whether $g_{\phi t}=0$.</p>
<p>Could anyone point out to me how to deduce that $g_{t\phi}=0$ from this approach?</p>
|
<p>Your equivalence is not an equivalence. From the way you write, I think that your mistake comes from a problem with index notation. For the Levi-Civita connection, we have $$\nabla_a g_{\mu\nu} = 0. \tag{1}$$ But this does not mean that the $a$-derivative of the $\mu\nu$ component of $g$ is $0$. It means that the $a\mu\nu$ component of $\nabla g$ is $0$. Thus from (1) we cannot conclude that $$\nabla_b g_{ct} \overset{?}{=} 0 \tag{2}.$$
since in this equality $t$ is not an "index to be filled in", but is shorthand for $g_{c\mu} t^\mu$. We see that (2) holds iff $t^\mu$ is covariantly constant. Not surprising, since it's the definition of $t_c$. It is easy to become confused by this when doing or presenting concrete calculations.</p>
<p>Thus you really have to compute $$K_{[a} \nabla_b K_{c]}$$ and check when it vanishes. However since the Frobenius theorem has a formulation in terms of differential forms, one can use coordinate-independent, index-free differential forms language to solve your problem, and this could be faster. First, we have $$K_{[a} \nabla_b K_{c]} = (K \wedge dK)_{abc}.$$ (You may object that on the left we have the covariant derivative and on the right the exterior, and the former involves the metric but the latter does not. But it is easy to check that the Christoffel symbols cancel on the left, and we get the formula for $d$ on 1-forms. This and $\nabla_a f = (df)_a$ for scalars implies that for an arbitrary antisymmetric tensor $T_{\mu\nu\cdots}$, we have $\nabla_{[a} T_{\mu\nu\cdots]} = (dT)_{a\mu\nu\cdots}$. )</p>
<p>We have $$K_a = (1- \frac{2Mr}{\Sigma}) dt + \frac{2J\sin^2\theta}{\Sigma} d\phi.$$ I have not done the calculations myself from this point, but unless $J = 0$, $\Sigma$ depends on $\theta$, so probably we will get a $d\theta \wedge d\phi$ term in $dK$, that will not cancel.</p>
| 598
|
general relativity
|
Why do the space time get curved around a massive object?What problems do we face if we do not consider the space time to be curved?
|
https://physics.stackexchange.com/questions/132481/why-do-the-space-time-get-curved-around-a-massive-objectwhat-problems-do-we-fac
|
<p>As far as I have the knowledge of GTR that a mass bends the space time around it.But why does this bend occur?The example from real life that when a mass is placed on a net then the net bends but it us very difficult for me to visualise the situation of bending of spacetime due to a mass.What is actually happening?What is the physical basis behind this bend?I want to say that the books which I have studied till now just say that the space time "just bends" around an object and even my teacher told me that it just bends without any explanation so I just want to know the reason for the occurrence of this bend.What is the need for spacetime to bend around an object?What physical problem we can encounter if we do not the bending of space time.</p>
|
<p>As photons have energy, gravity affects light rays, turning their path from straight to curved, and changing their energy (frequency/color).</p>
<p>In classical relativity light always travels in a straight zero-lenght line, with phase speed = $c$.
If you include gravity in your relativistic model you this is not true anymore. As this is an important property of the model that helps verifying that the theory is sound, instead of using a model where you cannot trust light going straight, you change your definition of "straight" so that it stays coherent with the fact that light moves in a straight line. That's where the "curved spacetime" comes from.</p>
<p>So the logical process is not:</p>
<ul>
<li>I observe some strange physical phenomenon from which I conclude that the spacetime is curved</li>
<li>I know that light goes straight, but the space is curved</li>
<li>the light curves</li>
</ul>
<p>but instead:</p>
<ul>
<li>I observe light going around big masses</li>
<li>But I know that light goes in straight lines</li>
<li>I either have to change the idea that light goes in straight lines, or that the space is plain</li>
<li>I prefer keeping in my model the property that light goes in straight lines, and from my biased point of view the space is curved </li>
</ul>
<p>This is consistent to the idea beyond special relativity, in which Einstein decided that instead of preserving the rule that speed adds up while changing the system of reference preferred keeping the rule that the speed of light is the same in all (inertial) system of reference.</p>
| 599
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.