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Gravitational Equivalent of Rindler Space

by Chestermiller
Tags: equivalent, gravitational, rindler, space
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Chestermiller
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Mar27-12, 08:29 AM
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In classical physics, the force on an object experiencing constant acceleration is equivalent to the "gravitational force" exerted on an object by a slab of material of uniform thickness and infinite lateral extent. The "gravitational field" is uniform above the slab, and the object's "acceleration" in free fall is constant.
Does this same type of equivalence apply in general relativity, in which the Rindler metric describes the geometry of an object's accelerated frame of reference (for constant acceleration)? IOW, for an observer at rest wrt a flat slab of infinite lateral extent, is the space above the slab described by the Rindler metric? I'm guessing that it is, since, when I evaluated the components of the Ricci-Einstein tensor for Rindler space, all the components came out to zero (if I did it right). Also, in Rindler space, the acceleration of the coordinate frame gradually decreases linearly in the direction of motion. Does the same type of effect happen in the region above a slab? I guess this would be interpreted as a tidal phenomenon that is not present in the classical analogy. Correct? Help!!!!
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Mentz114
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Mar27-12, 01:47 PM
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Rindler coordinates are transformations of Minkowski coordinates and so they will give flat-space numbers just like Minkowski. It's an interesting question, although the GR case will not represent the same physics, giving a constant coordinate acceleration as opposed to the proper acceleration of the Rindler frame.

I think the slab spacetime is the Taub metric. It is a vacuum solution but the Weyl tensor is not zero. I'm going to look at the EOMs later, if I don't fall asleep first.
PAllen
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Mar27-12, 01:55 PM
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This may be of interest (have no time right now to read this)

http://arxiv.org/abs/0709.3276

Bill_K
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Mar27-12, 02:24 PM
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Gravitational Equivalent of Rindler Space

Chestermiller, If you're interested in the General Relativistic analog of the Rindler metric, take a look at the so-called C metric, which is the field of a uniformly accelerating Schwarzschild mass. Varieties of the C-metric also exist for accelerating Kerr and Reissner-Nordstrom.

The acceleration must be supplied by something external, and in the C metric this is represented by a singular strut which is attached to the particle and extends along the axis to infinity.
Mentz114
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Mar27-12, 03:16 PM
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The acceleration of a static observer in this space-time
[tex]
{ds}^{2}={dy}^{2}\,{\left( 1-3\,g\,z\right) }^{\frac{4}{3}}+{dx}^{2}\,{\left( 1-3\,g\,z\right) }^{\frac{4}{3}}-\frac{{dt}^{2}}{{\left( 1-3\,g\,z\right) }^{\frac{2}{3}}}+{dz}^{2}
[/tex]
is
[tex]
-\frac{g}{3\,g\,z-1}\partial_z
[/tex]
I think g<0 so that is in the +ve z-direction. The metric is a vacuum solution of the region above ( z > 0 ) an infinite slab of constant density in the region z <= 0. I got the acceleration from [itex]\nabla_\nu U_\mu V^\nu[/itex] where U=(-1,0,0,0) and V=(1,0,0,0). This is the proper acceleration required to remain at a constant z-value.

The acceleration isn't constant as I was led to believe by the article (cited above by PAllen).
[correction: the article states the acceleration is constant for small z, which is true]
Chestermiller
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Mar28-12, 06:39 AM
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[QUOTE=Mentz114;3836727]Rindler coordinates are transformations of Minkowski coordinates and so they will give flat-space numbers just like Minkowski.

Thank you and the other responders so much for your replies. Some of the replies have been a little over my head (for now), but I will follow up.
The above quote was extremely enlightening, and has cleared up a source of confusion that I have been grappling with for months. As a retired engineer, I have become fascinated with relativity, and have been studying it on my own. One stumbling block I have encountered has been in trying to get an understanding of the relationship between accelerated reference frames in free space, and reference frames at rest in the vicinity of gravitationally massive bodies, within the framework of the equivalence principle. Somehow I got it in my head that, from the perspective of observers in an accelerated reference frame in free space, their space is curved. I studied accelerated reference frames in Chapter 6 of MTW, and found the development fascinating, but failed to recognize from the transformation equations and the development of the metric for the accelerated frame that flat space was also implied for the accelerated frame. In retrospect, I really should have, since, by applying the inverse transformation to the accelerated frame metric, you automatically obtain the Minkowski metric. However, MTW never articulated this point, which I regard as very important to my understanding. Your quote cleared all this up.
From this I conclude that the pseudo-gravity associated with an accelerated frame of reference in free space is not the result of curvature, but the true gravity in the vicinity of massive objects is the result of curvature. In this respect, the two situations are fundamentally different. Now I have to go back and think a little more about free fall in which, somehow, the curvature vanishes (at least locally).
Mentz114
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Mar28-12, 08:22 AM
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Quote Quote by Chestermiller View Post
From this I conclude that the pseudo-gravity associated with an accelerated frame of reference in free space is not the result of curvature, but the true gravity in the vicinity of massive objects is the result of curvature. In this respect, the two situations are fundamentally different. Now I have to go back and think a little more about free fall in which, somehow, the curvature vanishes (at least locally).
Yes, they are fundamentally different. But just to throw in another example, the Langevin frame in SR represents a rotating observer with an inward radial acceleration. Written in the Born chart this frame has non-trivial curvature! One component of the Riemann tensor is non-zero. This excellent article has the details. One needs to understand frame fields in GR to fully appreciate it, but well worth a read.

http://en.wikipedia.org/wiki/Born_coordinates

Speaking of curvature, the tidal forces seen by a static observer in the Taub spacetime above are given by
[tex]
T_{xx}=T_{yy}=\frac{2\,{g}^{2}}{{\left( 3\,g\,z-1\right) }^{2}},\ \ T_{zz}=-\frac{4\,{g}^{2}}{{\left( 3\,g\,z-1\right) }^{2}}
[/tex]
So we have squishing in the z-direction and stretching in x and y-directions. The terms sum to zero so volume is preserved.
PAllen
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Mar28-12, 08:36 AM
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Quote Quote by Mentz114 View Post
Yes, they are fundamentally different. But just to throw in another example, the Langevin frame in SR represents a rotating observer with an inward radial acceleration. Written in the Born chart this frame has non-trivial curvature! One component of the Riemann tensor is non-zero. This excellent article has the details. One needs to understand frame fields in GR to fully appreciate it, but well worth a read.

http://en.wikipedia.org/wiki/Born_coordinates
Note, it is not the 4-manifold that has curvature; that is impossible. It is quotient 3-manifold that is Riemanian (not semi-Riemannian) that has curvature. This is no more surprising than that curved 2-manifolds can be embedded in a flat 3-manifold.
Mentz114
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Mar28-12, 08:37 AM
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Quote Quote by PAllen View Post
Note, it is not the 4-manifold that has curvature; that is impossible. It is quotient 3-manifold that is Riemanian (not semi-Riemannian) that has curvature. This is no more surprising than that curved 2-manifolds can be embedded in a flat 3-manifold.
Thanks for the clarification. It would indeed be strange otherwise. Question - does this mean that the observer sees spatial curvature ? I suppose the spatial projection tensor hold the answer.
PAllen
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Mar28-12, 08:47 AM
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Quote Quote by Mentz114 View Post
Thanks for the clarification. It would indeed be strange otherwise. Question - does this mean that the observer sees spatial curvature ? I suppose the spatial projection tensor hold the answer.
Yes, it implies that observers riding on a rotating disk, taking only local measurements, will detect non-Euclidean spatial geomoetry.
Mentz114
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Mar28-12, 09:22 AM
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Quote Quote by PAllen View Post
Yes, it implies that observers riding on a rotating disk, taking only local measurements, will detect non-Euclidean spatial geomoetry.
Of course, thanks. I once knew that. D*rn memory.
DrGreg
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Mar28-12, 09:10 PM
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Quote Quote by Chestermiller View Post
One stumbling block I have encountered has been in trying to get an understanding of the relationship between accelerated reference frames in free space, and reference frames at rest in the vicinity of gravitationally massive bodies, within the framework of the equivalence principle. Somehow I got it in my head that, from the perspective of observers in an accelerated reference frame in free space, their space is curved.
You may, perhaps, find the following useful, which I wrote 3 years ago for someone else (and therefore it doesn't go into any mathematical detail): (you can click on the little blue arrowhead if you want to see my quote in context)
Quote Quote by DrGreg View Post
What we call the "curvature of spacetime" has a technical meaning; the equations that describe it are very similar to the equations that describe, say, the curvature of the earth's surface in terms of latitude and longitude coordinates, or any other pair of coordinates you might choose. This "curvature" need not manifest itself as a physical curve "in space".

For the rest of this post let's restrict our attention to 2D spacetime, i.e. 1 space dimension and 1 time dimension, i.e. motion along a straight line. This is adequate for many scenarios, even for black holes provided we are interested only in particles moving radially.

In the absence of gravitation, an inertial frame corresponds to a flat sheet of graph paper with a square grid. If we switch to a different inertial frame we "rotate" to a different square grid, but it is the same flat sheet of paper. (The words "rotation" and "square" here are relative to the Minkowski geometry of spacetime, which doesn't look quite like rotation to our Euclidean eyes, but nevertheless it preserves the Minkowski equivalents of "length" (spacetime interval) and "angle" (rapidity).)

If we switch to a non-inertial frame (but still in the absence of gravitation), we are now drawing a curved grid, but still on the same flat sheet of paper. Thus, relative to a non-inertial observer, an inertial object seems to follow a curved trajectory through spacetime, but this is due to the curvature of the grid lines, not the curvature of the paper which is still flat.

When we introduce gravitation, the paper itself becomes curved. (I am talking now of the sort of curvature that cannot be "flattened" without distortion. The curvature of a cylinder or cone doesn't count as "curvature" in this sense.) Now we find that it is impossible to draw a square grid to cover the whole of the curved surface. The best we can do is draw a grid that is approximately square over a small region, but which is forced to either curve or stretch or squash at larger distances. This grid defines a local inertial frame, where it is square, but that same frame cannot be inertial across the whole of spacetime.

So, to summarise, "spacetime curvature" refers to the curvature of the graph paper, regardless of observer, whereas visible curvature in space is related to the distorted, non-square grid lines drawn on the graph paper, and depends on the choice of observer.
I thin it is well worth getting to grips with Rindler coordinates as a bridge between SR and GR. Rindler coordinates have a horizon that behaves very similarly to the event horizon of a black hole, but you have the advantage that you can transform back to inertial coordinates to understand what is happening rather more easily than you can with a black hole.
Chestermiller
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Mar29-12, 07:28 AM
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Thanks very much DrGreg. You summarized the situation very well. Except for my conceptual "glitch" of failing to recognize that Rindler space is flat, I had already envisioned the remainder of what you so concisely and simply presented.
As someone studying relativity on my own, it has been hard to proceed without having someone to answer my questions. Discovering Physics Forums has been a real "find." I appreciate everyone's help.
Now for one additional question.
Since accelerated frames of reference are basically flat, and, in this respect, differ from reference frames at rest with respect to massive bodies, would it be at least conceptually possible for residents of an accelerated reference frame to experimentally measure the metrical properties of their space, and by doing so, ascertain that they are not experiencing gravitational curvature of spacetime?
PAllen
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Mar29-12, 07:31 AM
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Quote Quote by Chestermiller View Post
Thanks very much DrGreg. You summarized the situation very well. Except for my conceptual "glitch" of failing to recognize that Rindler space is flat, I had already envisioned the remainder of what you so concisely and simply presented.
As someone studying relativity on my own, it has been hard to proceed without having someone to answer my questions. Discovering Physics Forums has been a real "find." I appreciate everyone's help.
Now for one additional question.
Since accelerated frames of reference are basically flat, and, in this respect, differ from reference frames at rest with respect to massive bodies, would it be at least conceptually possible for residents of an accelerated reference frame to experimentally measure the metrical properties of their space, and by doing so, ascertain that they are not experiencing gravitational curvature of spacetime?
Yes. They simply measure that there are no tidal forces.
A.T.
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Mar29-12, 07:49 AM
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This might be relevant:
http://www.mathpages.com/home/kmath530/kmath530.htm
Chestermiller
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Mar29-12, 11:49 AM
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Quote Quote by A.T. View Post
Yes this seems to be very relevant. I will need some time to study it. Thank you so much. This write-up does some very nice modeling analysis.

One question: Is it correct to say that, statically, there is a unique solution for the metric in the region above the massive sheet?

Chet


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