# About the Rindler metric and the Weyl tensor

1. Mar 14, 2010

### snoopies622

This question is a follow-up to the one I asked last week in the thread called, "about tidal forces". In that thread the question came up: what would happen to a sphere of free-falling particles (a "ball of coffee grounds") in a gravitational field described by the Rindler metric? After some analysis the conclusion was reached that - from the perspective of an observer at rest with respect to the Rindler coordinates - the overall shape of the set of particles would change because - as it approached the event horizon - the "top" and "bottom" of the sphere would move closer together.

My question is, since both the Ricci tensor and the Weyl tensor of the Rindler metric are zero, how is it that the size or shape of the sphere changes at all?

I know that the circumstances which create such a situation - being accelerated by, say, a rocket - are not what usually come to mind when one encounters the phrase "gravitational field", and that the spacetime involved is still flat, but my interpretation of the equivalence principle is that a "gravitational field" and an "accelerated frame of reference" should be thought of in the same way.

Last edited: Mar 15, 2010
2. Mar 15, 2010

### atyy

I haven't checked the analysis, but am not surprised that something like this can happen in Rindler coordinates. Expansion, contraction in this sense are coordinate dependent. Just like the expansion of the universe. I think even Minkowski spacetime (a part of it) can be made to look like expanding space.

Try George Jones's post #30 at https://www.physicsforums.com/showthread.php?t=234224&page=2

3. Mar 15, 2010

### bcrowell

Staff Emeritus
Although you're right that both the Ricci tensor and the Weyl tensor vanish in Rindler coordinates, what would actually be relevant here would be the Ricci tensor, not the Weyl tensor. The Weyl tensor describes tidal forces, which deform a cloud of test particles without changing its volume.

The Rindler coordinates describe a nonuniform gravitational field, in the sense that if an observer at rest with respect to the coordinate lattice releases a test particle at rest, the test particle's acceleration is dependent on height. Therefore I don't think it's surprising that a cloud of test particles appears to change its height; the particles at the top experience a different acceleration than the particles on the bottom.

If you view it all from a nonaccelerating frame, again it's not surprising. You explain the effect seen by the accelerating observer as arising from special-relativistic length contraction.

It is possible to think of the Ricci tensor as representing something like an "acceleration of volume," $d^2V/dt^2$. For geometrical visualization of this for a space with two spacelike dimensions, see this: http://www.lightandmatter.com/html_books/genrel/ch05/ch05.html#Section5.3 [Broken] However, writing it as $d^2V/dt^2$ clearly doesn't produce something that transforms properly as a tensor. For instance, I can do a coordinate transformation like $t'=t$, $x'=xe^t$. In the new (x',t') coordinates, all volumes are increasing exponentially with time. That doesn't mean I can infer anything physical about the existence of gravitational fields made by sources that are present in this region of space (which is what a nonzero Ricci tensor would indicate). The Ricci tensor is constructed so that if it's zero in (x,t) coordinates, it's also zero in (x',t') coordinates.

Last edited by a moderator: May 4, 2017
4. Mar 15, 2010

### Ben Niehoff

I think the phrase "at rest with respect to Rindler coordinates" is a bit misleading. An observer whose Rindler coordinates are not changing with time is NOT at rest--he can feel forces! This is analogous to keeping constant coordinates in a rotating reference frame: you can feel centrifugal forces. So the observer is not at rest, and he can measure the fact that he is not at rest.

Observers falling on geodesics in spacetime do not feel forces; however, observers maintaining their position in a curved spacetime do. In fact, one can identify the "acceleration due to gravity" with the amount of proper acceleration needed to stay in place at a given point in space around a gravitating object.

Obviously, since space is flat, the ball of coffee grounds will not change size or shape in its own reference frame. However, we already know about Lorentz contraction: that is, an observer moving at a constant velocity will see the ball compressed along his direction of motion. So naturally, an accelerating observer will see the ball becoming more and more compressed, and more and more redshifted, as it recedes away from him toward the Rindler horizon.

You could also ask what the ball actually looks like; i.e., trace rays of light to a given point in space time where they are observed. You will find that to a constant-velocity observer, the ball actually appears spherical, though it is rotated a bit. To an accelerated observer, I'm not sure if the ball appears spherical or not...I haven't tried to calculate.

5. Mar 15, 2010

### snoopies622

Thank you all - I'm finding these to be thought-provoking answers!

I'm just taking a stab in the dark here, but is the kind of volume change with respect to time that the Ricci tensor describes really the change in four-volume with respect to proper time? I can see how that would be a tensor, while a change in three-volume (space volume) with respect to either coordinate time or proper time would not be.

Last edited: Mar 15, 2010
6. Mar 16, 2010

### bcrowell

Staff Emeritus
I don't think that would make sense. You can't have time both as a variable of integration and as the variable you're differentiating with respect to.

7. Mar 17, 2010

### snoopies622

So in a "real" gravitational field - that is, one with a non-zero curvature tensor - the sphere's shape and/or size changes even in it's own frame of reference?

Hmm.. What if the variable of integration is coordinate time and the variable we're differentiating with respect to is proper time?

8. Mar 17, 2010

### atyy

No. Take a simplified universe with only the sun in it. The ball of matter that is the sun will be stationary in some reference frame, which you could call it's own. In this case, the sun is not free falling though. I think in relativity, these metrics are called "static", and they have a timelike Killing vector field orthogonal to spacelike hypersurfaces. In a general spacetime, the expansion of free falling balls is described by Raychaudri's equation applied to timelike geodesics. http://en.wikipedia.org/wiki/Raychaudhuri_equation

Last edited: Mar 17, 2010
9. Mar 17, 2010

### Ben Niehoff

I have to disagree, atyy. If you have a sphere of coffee grounds (i.e., a dilute dust with no internal forces) all following geodesics, then in curved spacetime they will distort and/or change size in their own reference frame! This is what curvature means after all. The Ricci tensor governs changes in volume with respect to proper time along a geodesic; the Weyl tensor governs changes in shape. You can derive this directly from the equation for geodesic deviation.

The reason the sun doesn't shrink is that it has internal forces causing it to accelerate outwards against the geodesic flow. The surface of the sun is not an inertial frame (and hence is not "at rest") precisely because it is not following geodesics.

Last edited: Mar 17, 2010
10. Mar 17, 2010

### atyy

Agreed - I think that is what I said - though your reply is more pertinent, since now that I read the OP, it was specified that the sphere consists of free falling test particles.

11. Mar 19, 2010

### snoopies622

Yet in spite of this height-based acceleration, or - if you will - height-based gravitational force, we still do not regard our sphere as experiencing a tidal force, because the spacetime is flat. Is this correct? That is - we define "tidal force" as a function of the spacetime curvature?

12. Mar 19, 2010

### bcrowell

Staff Emeritus
This sounds right to me, although I would say it in a different way. Some quantities are more useful to talk about in GR than others. Typically things that transform as tensors are more useful than things that aren't. A thing that doesn't transform as a tensor is not that interesting, because, e.g., you may be able to change it from a zero value to a nonzero value just by a change of coordinates. It's not really telling you anything, because it has this completely arbitrary coordinate dependence built into it.

When we describe the height-based acceleration, we're describing something that doesn't transform as a tensor. It's not as interesting as the Weyl tensor, which does transform as a tensor.