Does gravitational time dilation imply spacetime curvature?

[PeterDonis]

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:

The difference in clock rates from bottom to top of the elevator does not, in and of itself, mean that spacetime is curved.

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?

 

[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.

 

[PeterDonis]

$E_1E_2$ 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.

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.

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

Neither are the corresponding metric coefficients in Schwarzschild spacetime.

 

[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?

 

[PeterDonis]

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

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.

 

[PeterDonis]

where energy-momentum exists in Earth gravity system.

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).

 

[PeterDonis]

The worldlines of observers at rest in Rindler coordinates are orbits of a timelike Killing vector field

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.

 

[sweet springs]

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.

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 ?

 

[PeterDonis]

I feel confused with your posts #6 and #7.

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.

curvature $R=0$ in vacuum

$R_{\mu \nu}$ 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.

 

[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.

 

[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.

 

[PeterDonis]

A stretched rope between the two observers at different heights will remain as is, but in the Rindler case it will tear.

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.)

 

[PeterDonis]

what do you mean with gravitational time dilation?

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.

at least if you understand gravity as spacetime curvature

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

the situation they appear to have in mind ("constant gravitational field") is actually adequately described by uniformly accelerated observers in Minkowski spacetime

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?

 

[martinbn]

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.)

Yes, I see it now.

 

[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.

 

[PeterDonis]

this argument fails show anything about curvature of a 4-manifold, because it only establishes curvature of a 2-d submanifold.

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.

 

[PAllen]

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.

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.

 

[PeterDonis]

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.

Ah, that's right.

To use arguments about polygonal geometry to establish curvature you must use geodesics of the the whole manifold.

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.

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

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).

 

[PAllen]

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).

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

 

[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?

 

[PAllen]

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?

Peter applied it in both, indeed not using Rindler observers in the Rindler case.

 

[PeterDonis]

if you are using only geodesics, you can't use Rindler observers in flat spacetime

You use the Rindler worldlines to determine the emission and reception events, but the paths you use in between those events are geodesics that connect them, not the Rindler worldlines that connect them.

 

[Ibix]

You use the Rindler worldlines to determine the emission and reception events, but the paths you use in between those events are geodesics that connect them, not the Rindler worldlines that connect them.

Ah - I see. I'll go back to thinking about it.

 

[PAllen]

I think this thread is a great case study on how even the best science texts should be read critically (but not crankily). It's a subtle distinction. I remember some saying about this from professor Carlip, but I can't seem to find it right now.

 

[PeterDonis]

To use arguments about polygonal geometry to establish curvature you must use geodesics of the the whole manifold.

On thinking this over I'm actually not positive that it is true. It is true that the figure described in Schild's argument, the original one that uses the worldlines of "hovering" observers rather than my modified one with geodesics, is not a "parallelogram", strictly speaking; it's a figure with two curved opposite sides and two straight opposite sides. But I think the following should still be true: if spacetime were flat, then the two curved opposite sides should be congruent, and therefore of equal length, since the two straight opposite sides are.

This argument does depend on the straight opposite sides being parallel, and the definition of "parallel" is something we haven't touched on in this discussion. The implicit argument, at least as I understand it from MTW, is that the two straight opposite sides are parallel because the spacetime is static--or, to put it in more precise technical language, because the two curved opposite sides are both integral curves of a Killing vector field, so the second straight opposite side can be viewed as simply the time translate along the Killing flow of the first one.

However, this gets weird when we compare it with the flat spacetime case, because if we look at the figure formed by the analogous curves for that case (using Rindler worldlines for the two curved sides), we find that the two straight sides are "obviously" not parallel--at least not if we look at them in inertial coordinates! But they are parallel (or at least they look that way) in Rindler coordinates. And since the Rindler worldlines are integral curves of a Killing vector field, we can't just say the inertial coordinate picture is "obviously" right based on what I've said so far.

But, as I noted in a previous post, there is a key property that the Killing vector field that generates inertial worldlines in Minkowski spacetime, and the timelike KVF in Schwarzschild spacetime, have, that the Rindler KVF in flat spacetime does not: being normalizable at infinity. So one way out of the apparent dilemma I described in the previous paragraph would be to say that, in order for the two straight sides of the figure in our thought experiment to be properly considered "parallel" on the time translation grounds described above (so that we can deduce curvature from the failure of the two curved sides to be the same length), the timelike KVF that generates the curved sides must be normalizable at infinity. (Note that MTW's description of the scenario seems to imply this requirement, since it talks about the two observers in the experiment being able to verify that they are "at rest" by exchanging light signals with an observer at rest at infinity.)

 

[PAllen]

@PeteDonis, I don't buy your latest argument because it seems to say we need global properties of a manifold to establish curvature. Yet curvature is uniquely determined via ininitesimal operations. I will think more, but my first reaction is poppycock.

 

[PeterDonis]

I don't buy your latest argument because it seems to say we need global properties of a manifold to establish curvature.

I don't think there is any dependence on global properties of the manifold in the part about still being able to argue that the two curved sides should be the same length in flat spacetime, even though those sides are not geodesics, as long as the two straight sides are parallel.

I agree that the part about the KVF having to be normalizable at infinity to define "parallel" seems to say that global properties are needed--but if that's the case, then the Schild argument as presented in MTW seems to say that too, since, as I noted, it includes the requirement that the observers verify that they are at rest by exchanging light signals with an observer at rest at infinity.

Perhaps an underlying issue here is pinning down exactly what a "static gravitational field" requires. The argument as presented in MTW seems to imply that part of the definition of that concept is that observers at rest in the field can verify that by exchanging light signals with observers at rest at infinity. (Making this precise would end up at something like the standard definition of asymptotic flatness.) But that would seem to make the concept of a "static gravitational field" inherently global.

The other possibility is that there is some local way to define "parallel" which makes it "obvious" that the straight sides of the figure in Schwarzschild spacetime are parallel while the straight sides of the figure in Rindler coordinates in flat spacetime are not.

(I also just realized that, because the straight sides are light signals, they actually are parallel, technically, in Minkowski coordinates in the flat spacetime case, because all light signals in the same direction are--at least with the intuitive notion of "parallel" that I've been using so far. But I think we could fix that, if necessary, by reformulating the argument using timelike geodesics for the straight sides. That would also make the straight sides have nonzero length, which is another aspect of using light signals that makes me wonder.)

 

[PAllen]

But the curved sides (non geodesic) in the Rindler case are parallel by two measurable criteria: exchange of successive light signals and born rigid distance. I reject out of hand that asymptotic conditions have anything to do with establishing curvature. For example, suppose infinity is not asymptotically flat. That should have no bearing on the curvature of a local region that has SC geometry, versus absence of curvature in a local Rindler region.

 

[PAllen]

Anyway, the whole notion that gravitational time dilation.implies curvature seems absurd to me. Pound Rebka only measured that a tall building was equivalent to similar height Rindler rocket. Its precision was not even close to distinguishing tidal gravity.

 

[PeterDonis]

the curved sides (non geodesic)

My comments were about the straight sides (the successive light signals, which are outgoing null geodesics). But light signals in the same direction in flat spacetime are always parallel, so I don't think that tells us anything useful.

However, in the Rindler case in flat spacetime, note that the spacelike geodesic segments that realize the "born rigid distance" you mention at different times are not parallel in Minkowski coordinates (since they are curves of constant coordinate time in different inertial frames), but they are all orthogonal to the Rindler observer worldlines, which are integral curves of a KVF, so they are "parallel" in that sense (which is the same sense in which the corresponding spacelike curves in Schwarzschild spacetime are parallel). But the latter sense of "parallel" is somewhat problematic, because in flat spacetime these "parallel" lines intersect (this isn't visible in Rindler coordinates because the intersection is on the horizon); and they also intersect in maximally extended Schwarzschild spacetime (at the origin of Kruskal coordinates).

 

[martinbn]

@PeteDonis, I don't buy your latest argument because it seems to say we need global properties of a manifold to establish curvature. Yet curvature is uniquely determined via ininitesimal operations. I will think more, but my first reaction is poppycock.

But, if the manifold is connected, a KVF is completely determined by its value and the value of its derivative at one point. To me it seems that global properties of a KVF are not really global, in some sense they are local.

 

[PeterDonis]

if the manifold is connected, a KVF is completely determined by its value and the value of its derivative at one point

By "value" and "derivative", do you mean "vector" (i.e., the vector at a point that is mapped to that point by the field) and "gradient" (i.e., the two-form $d \chi$, where $\chi$ is the KVF, at the chosen point--basically, the set of covariant derivatives in all possible directions)?

 

[martinbn]

By "value" and "derivative", do you mean "vector" (i.e., the vector at a point that is mapped to that point by the field) and "gradient" (i.e., the two-form $d \chi$, where $\chi$ is the KVF, at the chosen point--basically, the set of covariant derivatives in all possible directions)?

Yes.

 

[PAllen]

But, if the manifold is connected, a KVF is completely determined by its value and the value of its derivative at one point. To me it seems that global properties of a KVF are not really global, in some sense they are local.

But there is a kvf in both cases (Rindler and SC).

 

[PeterDonis]

there is a kvf in both cases (Rindler and SC)

Yes, but as I understand his point, the key difference between them that I pointed out--that the SC KVF is normalizable at infinity while the Rindler KVF is not--is fully determined by the local properties of the respective KVFs--in this case, the difference in their derivatives (the Rindler KVF's norm increases linearly with height, while the SC KVF's norm does not).