# A Does gravitational time dilation imply spacetime curvature?

1. Jul 2, 2017

### Staff: Mentor

In a recent thread, the question came up of whether the presence of gravitational time dilation implies spacetime curvature. My answer in that thread was no:

This was based on the obvious counterexample of observers at rest in Rindler coordinates in flat Minkowski spacetime; two observers at different Rindler $x$ coordinates will be gravitationally time dilated relative to each other, yet the spacetime they are in is flat.

However, it occurred to me that there is an argument in the literature, originally due to Schild and described in MTW (which is where I encountered it), which purports to show that gravitational time dilation does imply spacetime curvature. I will give the argument here as it is given in section 7.3 of MTW:

Consider two observers at rest in the gravitational field of the Earth, one at height $z_1$ and the other at height $z_2 > z_1$. The lower observer sends two successive light pulses to the upper observer. This defines four events in spacetime as follows: E1 and E2 are the emissions of the two light pulses by the lower observer, and R1 and R2 are the receptions of the two light pulses by the upper observer. These four events form a parallelogram in spacetime--it must be a parallelogram because opposite sides are parallel. The lower and upper sides, E1-E2 and R1-R2, are parallel because the two observers are at constant heights; and the light pulse sides, E1-R1 and E2-R2, are parallel because the spacetime is static, so both light pulses follow exactly identical paths--the second is just the first translated in time, and time translation leaves the geometry of the path invariant.

However, the lower and upper sides of this parallelogram have unequal lengths! This is because of gravitational time dilation: the upper side, R1-R2, is longer than the lower side, E1-E2. This is impossible in a flat spacetime; therefore any spacetime in which gravitational time dilation is present in this way must be curved.

The problem is that the above argument would seem to apply equally well to a pair of Rindler observers in Minkowski spacetime! The worldlines of observers at rest in Rindler coordinates are orbits of a timelike Killing vector field, so two successive light pulses from a Rindler observer at $z_1$ in Rindler coordinates to a second observer at $z_2 > z_1$ should be parallel, and so should the worldlines of the observers themselves. So we should have a parallelogram in the same sense, but with two opposite sides unequal--which should imply that Minkowski spacetime must be curved!

So the question is: how do we reconcile these apparently contradictory statements?

2. Jul 2, 2017

### DrGreg

$E_1E_2$ and $R_1R_2$ look like parallel straight lines when drawn in Rindler coodinates, but they are not geodesics and are curved lines, of unequal length, when drawn in Minkowski coordinates.

A diagram drawn in Rindler coordinates does not have a uniform scale -- the Rindler components of the metric aren't all constant.

3. Jul 2, 2017

### Staff: Mentor

This is also true of the corresponding curves in Schwarzschild spacetime, which are used in Schild's argument. So if the Rindler argument is invalid on these grounds, so is Schild's argument.

This observation does lead to a suggestion, though: reformulate Schild's argument using only geodesic segments--for example, imagine two free-falling observers who launch themselves upward from $E_1$ and $R_1$, respectively, with just the right velocity so that they land again at $E_2$ and $R_2$. These two geodesics will, I think, have different lengths. However, the corresponding geodesics in Minkowski spacetime will also have different lengths! So I don't think this method answers the question I asked in the OP.

Neither are the corresponding metric coefficients in Schwarzschild spacetime.

4. Jul 3, 2017

### sweet springs

Hi.

$$g_{00}-1\neq 0$$ and $$G_{\mu\nu}=0$$ in vacuum space is common to the two systems.

The difference is $$G_{\mu\nu}=0$$ all over the region in Rindler system. But $$G_{\mu\nu}\neq 0$$ where energy-momentum exists in Earth gravity system.

When we look at vacuum region we find no difference, no curvature R=0, don't we?

Last edited: Jul 3, 2017
5. Jul 3, 2017

### Staff: Mentor

No. Schwarzschild spacetime is vacuum, but it is curved, not flat. It has zero Einstein tensor, but it does not have a zero Riemann tensor.

6. Jul 3, 2017

### Staff: Mentor

It is true that, as the scenario is formulated, it involves the Earth, and therefore involves the presence of a massive body. But we could just as easily formulate the scenario above a black hole, which is a vacuum solution--no matter or energy present. This would make no difference to the argument, so the presence of matter and energy cannot be the key difference between the two situations (gravity present vs. Rindler).

7. Jul 3, 2017

### Staff: Mentor

There is one key difference between this KVF and the corresponding one in Schwarzschild (curved) spacetime, though. The Schwarzschild KVF can be scaled so that its norm goes to $1$ at infinity (this scaling is in fact the standard one embodied in Schwarzschild coordinates). The Rindler KVF cannot (its norm increases linearly with height, so it increases without bound as height goes to infinity). This might be the key difference between the two situations; but if so, I would still like to understand exactly what role it plays in the argument.

8. Jul 3, 2017

### sweet springs

I feel confused with your posts #6 and #7. Depending on area, $R_{\mu\nu}=0$ thus curvature $R=0$ in vacuum
and $R_{\mu\nu}\neq 0$ in energy-momentum area, isn't it ?

Last edited: Jul 3, 2017
9. Jul 3, 2017

### Staff: Mentor

You might need to have more background in GR than you actually have. Remember that this is an "A" level thread; graduate level knowledge of the subject matter is assumed. I marked it "A" for that reason.

$R_{\mu \nu}$ is the Ricci tensor, not the Riemann tensor. A vacuum region of spacetime has zero Ricci curvature, but that does not necessarily mean it has zero curvature altogether. In the case of a vacuum region surrounding a spherically symmetric mass (or a black hole), the Riemann tensor is nonzero, so the spacetime is curved.

10. Jul 3, 2017

### martinbn

Isn't there a difference? A stretched rope between the two observers at different heights will remain as is, but in the Rindler case it will tear. The coordinates are chosen so that the observers stay with unchanging coordinates, but the "distance" between them does change. So it isn't really a parallelogram but more a trapezoid.

11. Jul 3, 2017

### Geometry_dude

Well, the issue is what do you mean with gravitational time dilation? If whatever you mean with that applies to observers in Minkowski spacetime, then it is really a misnomer IMHO - at least if you understand gravity as spacetime curvature.

It is important to keep in mind that the situation they appear to have in mind ("constant gravitational field") is actually adequately described by uniformly accelerated observers in Minkowski spacetime, so to speak of gravity in this setting means taking a Newtonian - not a relativistic - point of view.

12. Jul 3, 2017

### Staff: Mentor

No, it won't. You're confusing Rindler observers with observers in the Bell spaceship paradox. They are different scenarios. The distance between Rindler observers, as seen in the instantaneous rest frame of either observer, remains constant. Or, in more technical, invariant language, the expansion of the congruence of Rindler observers is zero. (The expansion of the congruence of Bell observers--the spaceships and string in the Bell spaceship paradox--is positive. That's why the string stretches and breaks in that scenario.)

13. Jul 3, 2017

### Staff: Mentor

The difference in rate of time flow between two observers with the same proper acceleration but slightly different heights (where "height" is "distance from some reference point along the direction of proper acceleration"). Or, equivalently, the redshift of light signals sent from the lower observer to the higher one.

Spacetime curvature is tidal gravity. But the term "gravity" is more general than that.

Is it? That's the question. Schild's argument appears to show that it can't be. So if it can, then something must be wrong with Schild's argument--or at least there must be some difference between the "gravitational field" case and the "uniform acceleration" case, that prevents Schild's argument from applying to the latter case even though it does apply to the former. So what is that difference?

14. Jul 4, 2017

### martinbn

Yes, I see it now.

15. Jul 5, 2017

### PAllen

Could it be as simple as a curved 2-surface can often be embedded in a flat manifold of higher dimension? That is, the Rindler x-t plane is curved, but is embedded in a flat 4-d Minkowski space. Equivalently, this argument fails show anything about curvature of a 4-manifold, because it only establishes curvature of a 2-d submanifold.

16. Jul 5, 2017

### Staff: Mentor

Hm, interesting. That would mean that just showing the presence of gravitational time dilation would not be enough; you would have to look at the details of how it varied with height and show that the resulting 2-d submanifold could not be embedded in flat 4-d Minkowski spacetime. I think this could be done for the r-t submanifold of Schwarzschild spacetime, but I admit I don't know how one would go about it in any detail.

17. Jul 6, 2017

### PAllen

Actually, I think my argument is simply wrong. The z-t plane in the Rindler case is the same 2-manifold with same induced metric as a standard Minkowski plane. It's just the coordinates imposed on it and the coordinate expression of the metric that are different.

I now think the Rindler case is just a counter example to the validity of the whole argument. To use arguments about polygonal geometry to establish curvature you must use geodesics of the the whole manifold. For example, you can embed a Euclidean flat plane in SC geometry [edit: in fact Gullstrand-Painleve coordinates show you can embed Euclidean flat 3 manifold in SC geometry ] and thus have triangles whose angles sum to 180 per geometry of that plane using its induced metric; but those triangle sides are not geodesics of the overall manifold, establishing the physical consequence that such a triangle would be under stress. Similarly, you can embed a 2-sphere in Euclidean 3-space, and using the induced geometry 'show' violations of sum of angles of a triangle; again, this establishes nothing about the overall manifold because these triangle sides are not geodesics of the overall manifold.

In this case, the 'static' observer lines in both Rindler and SC geometry are not geodesics of the overall geometry, so I just think the argument proves nothing. As compared to the more traditional (valid) argument that you can't have two geodesics (of the whole manifold) in the r-t plane of the SC geometry that maintain constant distance from each other, while you can trivially do this in the z-t plane of the Rindler case.

Last edited: Jul 6, 2017
18. Jul 6, 2017

### Staff: Mentor

Ah, that's right.

Yes, @DrGreg pointed that out earlier in this thread, and in response I suggested reformulating Schild's argument using only geodesic paths. See post #3.

Hm. Basically, what you're saying is that the upper and lower sides of the "parallelogram" in the gravity case, if the parallelogram is drawn correctly using geodesics (as in post #3), are not actually parallel. (This would be consistent with the fact that the velocities needed to launch the upper and lower observers in the right way as I described in post #3 would not be the same, so their initial 4-velocities would not be parallel--at least I don't think they would be.) So the fact that they are of different lengths doesn't tell us anything useful about curvature or lack thereof of the manifold. A correct argument would focus, instead, on the convergence/divergence of the geodesics (basically, that they are parallel at the instant halfway between the respective emission and reception events, and the distance between them at that instant is also smaller than the distance at the emission and reception events).

19. Jul 6, 2017

### PAllen

Right. Looking back at your amended procedure in post #3, you have established that trapezoids are possible in both Minkowsi manifold and SC manifold.

20. Jul 6, 2017

### Ibix

But if you are using only geodesics, you can't use Rindler observers in flat spacetime. Or was the reformulation intended to be of the Schwarzschild version only?