# Gravitational Equivalent of Rindler Space

1. Mar 27, 2012

### Staff: Mentor

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!!!!

2. Mar 27, 2012

### Mentz114

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.

3. Mar 27, 2012

### PAllen

4. Mar 27, 2012

### Bill_K

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.

5. Mar 27, 2012

### Mentz114

The acceleration of a static observer in this space-time
$${ds}^{2}={dy}^{2}\,{\left( 1-3\,g\,z\right) }^{\frac{4}{3}}+{dx}^{2}\,{\left( 1-3\,g\,z\right) }^{\frac{4}{3}}-\frac{{dt}^{2}}{{\left( 1-3\,g\,z\right) }^{\frac{2}{3}}}+{dz}^{2}$$
is
$$-\frac{g}{3\,g\,z-1}\partial_z$$
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 $\nabla_\nu U_\mu V^\nu$ 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]

Last edited: Mar 27, 2012
6. Mar 28, 2012

### Staff: Mentor

7. Mar 28, 2012

### Mentz114

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
$$T_{xx}=T_{yy}=\frac{2\,{g}^{2}}{{\left( 3\,g\,z-1\right) }^{2}},\ \ T_{zz}=-\frac{4\,{g}^{2}}{{\left( 3\,g\,z-1\right) }^{2}}$$
So we have squishing in the z-direction and stretching in x and y-directions. The terms sum to zero so volume is preserved.

Last edited: Mar 28, 2012
8. Mar 28, 2012

### PAllen

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.

9. Mar 28, 2012

### Mentz114

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.

10. Mar 28, 2012

### PAllen

Yes, it implies that observers riding on a rotating disk, taking only local measurements, will detect non-Euclidean spatial geomoetry.

11. Mar 28, 2012

### Mentz114

Of course, thanks. I once knew that. D*rn memory.

12. Mar 28, 2012

### DrGreg

You may, perhaps, find the following useful, which I wrote 3½ 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)
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.

13. Mar 29, 2012

### Staff: Mentor

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.
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?

14. Mar 29, 2012

### PAllen

Yes. They simply measure that there are no tidal forces.

Last edited: Mar 29, 2012
15. Mar 29, 2012

### A.T.

16. Mar 29, 2012

### Staff: Mentor

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