Does gravitational time dilation imply spacetime curvature?

[PeterDonis]

I don't know what Schild's argument is, and I don't really want to know.

It's referenced in MTW, as I explained in the OP to this thread. The specific reference is MTW section 7.3.

If it's famous, it's almost certainly right.

For the record, I do not agree with this general form of argument. But that's beyond the scope of this thread.

This thread is, however, about one specific instance that is an apparent counterexample to your claim, at least if we assume that "famous" includes any argument that the authors of a famous textbook like MTW thought worth including. The argument, as noted above, is given in section 7.3 of MTW; but as given, it appears to prove too much, namely, that flat Minkowski spacetime is curved, since gravitational time dilation exists between Rindler observers in that spacetime.

 

[PeterDonis]

it makes perfect sense in the Rindler spacetime

So how would you define it in Rindler spacetime? Specifically, consider the quadrilateral formed by the following points in Rindler coordinates:

##(-t, x_1)##

##(-t, x_2)##

##(t, x_2)##

##(t, x_1)##

where ##x_1 < x_2##. Is this quadrilateral a parallelogram?

 

[PeterDonis]

It's hopeless to define "parallel" in a curved spacetime

This would seem to indicate that the argument described in MTW section 7.3 (and ascribed there to Schild) is not well-defined, since it implicitly assumes a definite notion of "parallel" in a curved spacetime.

 

[PeterDonis]

Whatever definition of "parallelogram" you take

That's the problem; I don't know what definition I should take. More precisely, I don't know how to precisely formulate the definition that the argument given in MTW section 7.3 was implicitly assuming. That definition would (I think--see below) require that the quadrilateral in Rindler coordinates that I described in post #107 is a parallelogram; it also would require (this is explicit in MTW--see below) that a similar quadrilateral in Schwarzschild spacetime, described by the analogous set of 4 points in Schwarzschild coordinates (i.e., for "x" read "r"), is a parallelogram. The argument as presented in MTW specifically uses that term in reference to the Schwarzschild case. The extension to the Rindler case is mine, since the cases seem exactly analogous, at least as far as this discussion is concerned.

 

[PeterDonis]

So, there's no paradox. I think we need to agree on that before we going into deeper waters

If you mean, agree that Rindler spacetime (i.e., Minkowski spacetime in the Rindler chart) is flat, and Schwarzschild spacetime is curved, yes, of course I agree with that. I'm not saying any argument shows that Rindler spacetime is actually curved; of course it isn't.

 

[stevendaryl]

I'm thinking that Schild's argument is just wrong. For an empirical proof that spacetime is curved, it can't be enough to have a measurement involving [itex]g[/itex], the apparent acceleration due to gravity. It has to involve derivatives of [itex]g[/itex]. That's because [itex]g[/itex], the acceleration due to gravity in the x-direction, is basically the connection coefficient [itex]\Gamma^x_{tt}[/itex]. The curvature tensor is created from [itex]\Gamma^\mu_{\nu \lambda}[/itex] as follows:

[itex]R^a_{bcd} = \partial_c \Gamma^a_{bd} - \partial_d \Gamma^a_{bc} + \Gamma^a_{ce} \Gamma^e_{bd} - \Gamma^a_{de} \Gamma^e_{bc}[/itex]

So [itex]R[/itex] involves two types of terms: ones quadratic in [itex]\Gamma^a_{bc}[/itex], and ones involving derivatives of [itex]\Gamma^a_{bc}[/itex]. So no empirical measurement can prove curvature unless it involves derivatives of [itex]g[/itex] or is second-order in [itex]g[/itex]. Time dilation is first-order in [itex]g[/itex] and doesn't involve derivatives of [itex]g[/itex], so it can't demonstrate curvature.

So I'm not sure what Schild meant by his argument.

 

[martinbn]

Yes, this corresponds to using tidal gravity and its effects on geodesics to detect spacetime curvature. But that has nothing to do with Schild's argument, at least not directly, since he makes no attempt to argue that tidal gravity must be present if and only if gravitational time dilation is present (or even gravitational time dilation plus the additional feature that the observers are at rest relative to an observer at infinity).

I didn't mean tidal gravity. I was thinking of a simple geometry analogy. The sum of the angles in a triangle in the plane is 180. But if the sides are not geodesics, say a sector of a circle, then the sum is more than 180. Or if you want a parallelogram with unequal opposite sides, take two concentric circles and two radial sides. Of course this doesn't imply curvature. And it seems to me that this is analogous to Schilds argument.

 

[PeterDonis]

if you want a parallelogram with unequal opposite sides, take two concentric circles and two radial sides

I agree that this figure has two unequal opposite sides; the question is whether it is valid to call it a "parallelogram". And yes, similar remarks would apply to the corresponding figure in Schild's argument.

 

[MikeGomez]

I'm thinking that Schild's argument is just wrong. For an empirical proof that spacetime is curved, it can't be enough to have a measurement involving [itex]g[/itex], the apparent acceleration due to gravity. It has to involve derivatives of [itex]g[/itex]. That's because [itex]g[/itex], the acceleration due to gravity in the x-direction, is basically the connection coefficient [itex]\Gamma^x_{tt}[/itex]. The curvature tensor is created from [itex]\Gamma^\mu_{\nu \lambda}[/itex] as follows:

[itex]R^a_{bcd} = \partial_c \Gamma^a_{bd} - \partial_d \Gamma^a_{bc} + \Gamma^a_{ce} \Gamma^e_{bd} - \Gamma^a_{de} \Gamma^e_{bc}[/itex]

So [itex]R[/itex] involves two types of terms: ones quadratic in [itex]\Gamma^a_{bc}[/itex], and ones involving derivatives of [itex]\Gamma^a_{bc}[/itex]. So no empirical measurement can prove curvature unless it involves derivatives of [itex]g[/itex] or is second-order in [itex]g[/itex]. Time dilation is first-order in [itex]g[/itex] and doesn't involve derivatives of [itex]g[/itex], so it can't demonstrate curvature.

Yet we know that the term “flat space” in an approximation. Is there a method to add new Christoffel symbols for the non-inertial frame to account for the redshift effect?

 

[PeterDonis]

Is there a method to add new Christoffel symbols for the non-inertial frame

You don't "add" Christoffel symbols; they are determined by the metric. The frame @stevendaryl was implicitly assuming was a frame in which the source of gravity (e.g., the Earth) is at rest (so an observer "hovering" at a fixed altitude above the source is also at rest). At least, that's for the curved spacetime case--in the analogous flat spacetime (Rindler) case, the frame is just Rindler coordinates.

 

[phinds]

Or if you want a parallelogram with unequal opposite sides ...

By definition, that would not be a parallelogram. Am I missing something?

In Euclidean geometry, a parallelogram is a simple (non-self-intersecting) quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.

 

[PeterDonis]

By definition, that would not be a parallelogram.

Well, that's the question. Note that your definition says "in Euclidean geometry"--but we are dealing with arguments which assume the possibility of non-Euclidean geometry.

 

[phinds]

Well, that's the question. Note that your definition says "in Euclidean geometry"--but we are dealing with arguments which assume the possibility of non-Euclidean geometry.

Ah. THAT's what I missed. Thanks Peter.

 

[timmdeeg]

My impression following this discussion is that it might be difficult but not per se impossible to define a parallelogram in curved spacetime.
In Euclidean geometry all edges are in a plane. Taking this as a criterion wouldn't it require to define a parallelogram in curved spacetime in a plane of simultaneity? In which case the geodesics would be spacelike however.

 

[timmdeeg]

MTW write on page 35 "In Einstein's geometric theory of gravity, this equation of geodesic deviation summarizes the entire effect of geometry on matter."
From this I would expect that it doesn't make any sense to discuss parallelity of nearby geodesics in curved spacetime. Where am I wrong?

 

[PAllen]

My impression following this discussion is that it might be difficult but not per se impossible to define a parallelogram in curved spacetime.
In Euclidean geometry all edges are in a plane. Taking this as a criterion wouldn't it require to define a parallelogram in curved spacetime in a plane of simultaneity? In which case the geodesics would be spacelike however.

Of course you could do that, but that rules out Schild's parallelogram, which has two timelike and two lightlike sides. Further, it tells you nothing about curvature of the overall manifold because spacelike surfaces with intrinsic curvature are readily embedded in flat Minkowsli space (e.g. a 2 sphere), and Euclidean flat planes are embeddable in Schwazschild manifold which has intrinsic curvature. For polygon geometry to tell you anything about the overall manifold, all sides must be geodesics of the overall manifold rather than just geodesics of an embedded surface. This is a further problem for Schild's construction, because the two timelike sides are not geodesics.

 

[timmdeeg]

Of course you could do that, but that rules out Schild's parallelogram, which has two timelike and two lightlike sides.

Yes. I meant a parallelogram in curved spacetime in general not specifically Schild's.

Further, it tells you nothing about curvature of the overall manifold because spacelike surfaces with intrinsic curvature are readily embedded in flat Minkowsli space (e.g. a 2 sphere), and Euclidean flat planes are embeddable in Schwazschild manifold which has intrinsic curvature.

Thanks, so that can't be applied or at least makes no sense in the context we are discussing.

 

[PeterDonis]

From this I would expect that it doesn't make any sense to discuss parallelity of nearby geodesics in curved spacetime

Why not? Geodesic deviation is a well-defined concept, and "parallel" just means "zero geodesic deviation".

 

[timmdeeg]

Geodesic deviation is a well-defined concept, and "parallel" just means "zero geodesic deviation".

Do you say that "zero geodesic deviation" is possible in curved spacetime? Could you elaborate a bit on this or give an example? Thanks.

 

[PAllen]

Do you say that "zero geodesic deviation" is possible in curved spacetime? Could you elaborate a bit on this or give an example? Thanks.

A trivial case is a manifold with overall curvature but also regions of flatness. For an easy to visualize case, think of a flat plane with hills and depressions. Even without flat regions, there could be special pairs of geodesics that have zero deviation, perhaps only for part of their path.

 

[PeterDonis]

Do you say that "zero geodesic deviation" is possible in curved spacetime?

At a particular point, yes. It just won't stay zero as you move along the geodesics.

Could you elaborate a bit on this or give an example?

Consider two free-falling objects in the gravitational field of the Earth, moving purely radially. At some instant, they are both at rest relative to each other and the Earth, but at slightly different heights. At that instant, the two geodesics describing their worldlines are parallel. (At least, under the definition of "parallel" that appears to be used in, for example, MTW when discussing this kind of scenario.) The geodesic deviation between them is zero at that point. But then the objects will start falling again, and they will start to move apart, i.e., the geodesics describing their worldlines will diverge. So they won't stay parallel--the geodesic deviation doesn't stay zero.

 

[timmdeeg]

At a particular point, yes. It just won't stay zero as you move along the geodesics.

So it seems, if we talk about how to define a parallelogram in curved spacetime (in general, not specifically Schild's version) we don't have to respect an instant of time but instead the curves described by geodesics. These deviate however and thus the opposite sides of a "parallelogram" aren't parallel. But something must be wrong with this reasoning because otherwise a parallelogram couldn't be defined in curved spacetime per se and then it would be useless starting to talk about it, not even by accepting "deep waters" .

 

[PeterDonis]

These deviate however and thus the opposite sides of a "parallelogram" aren't parallel.

If you look back at my response to @DrGreg very early in this thread, where I proposed an alternative version of Schild's argument that uses geodesics, you will see, IIRC, that I said the resulting geodesics in the curved spacetime (Schwarzschild) case would indeed not be parallel. So yes, if we insist on using geodesics to form our quadrilateral, that quadrilateral will not be a parallelogram in the curved spacetime case.

 

[timmdeeg]

If you look back at my response to @DrGreg very early in this thread, where I proposed an alternative version of Schild's argument that uses geodesics, you will see, IIRC, that I said the resulting geodesics in the curved spacetime (Schwarzschild) case would indeed not be parallel. So yes, if we insist on using geodesics to form our quadrilateral, that quadrilateral will not be a parallelogram in the curved spacetime case.

Ah, I will look at that. Thanks for your answer.

 

[SlowThinker]

I'm having trouble seeing where is the problem...?
There are two physicists in a tall rocket accelerating upwards in a flat spacetime. One physicist, B, is near the bottom, the other, T, near the top.
B sends up a flash of light and jumps up. When he lands, he sends up another flash.
T jumps when he sees the first flash, in such a way as to land when the second flash arrives.
When viewed from inertial frame, the two flashes are parallel, but the paths of B and T are not, because T is moving at slower speed. So it's not surprising that the paths of B and T have different lengths. Further, it seems inevitable that the second flash travels a longer path than the first one, since the rocket is moving faster than during the first flash.
Now you can introduce Rindler coordinates, but then all 4 edges become curved (at least I think they do), so it's pointless to try to add angles at the corners. I think they do actually add to a full circle, but it's just a lucky coincidence.

As for the question how to define a parallelogram, you can have either geodesics for edges, or constant width, but not both (except in special cases).

 

[PeterDonis]

the paths of B and T are not, because T is moving at slower speed

Actually, T is moving faster while he's inertial, because he will keep accelerating while the first light flash is traveling from B to T, while B is moving inertially. But it's still true that B's and T's paths are not parallel, which is the key point.

it seems inevitable that the second flash travels a longer path than the first one

No, the paths of both flashes have zero length, because they are null paths (the paths of light rays). To fix this issue to be more intuitive, we would have to use timelike objects, like fast bullets or something, and impose some kind of condition on how they are launched.

It is true that the coordinate distance, in a fixed inertial frame, traversed by the second flash, will be longer than that traversed by the first flash. But that won't be true in non-inertial coordinates; in Rindler coordinates, for example, the coordinate distance traversed by both flashes will be the same.

 

[SlowThinker]

It is true that the coordinate distance, in a fixed inertial frame, traversed by the second flash, will be longer than that traversed by the first flash. But that won't be true in non-inertial coordinates; in Rindler coordinates, for example, the coordinate distance traversed by both flashes will be the same.

Hmm but... whatever definition of parallelogram you use, it can (should) depend on the metric, but not on the coordinates. If you want to analyze a "coordinate" parallelogram, you can get any result you desire, using the right coordinates.

I believe that a "constant width" (*) parallelogram has angles that add to a full circle in flat spacetime.
I'm not sure how to enforce parallelity in a "geodesic" parallelogram. Or how to hold the 4 edges in a plane. If you do know how, then the 4 angles should add to a circle as well.

(*) It might prove tricky to avoid twisting or curved edges but I think it can be done

 

[PeterDonis]

whatever definition of parallelogram you use, it can (should) depend on the metric, but not on the coordinates

Yes, but I was talking about your use of the term "distance", which is coordinate-dependent. To define an invariant figure of any sort, whether it's a "parallelogram" or not, you have to find a way of picking the four points that mark the corners in a way that's not coordinate-dependent. Your scenario with the light signals and the two observers jumping does that; but none of the sides of the figure obtained in that way are "distances" in the usual sense, since none of them are spacelike (two are timelike and two are null).

I'm not sure how to enforce parallelity in a "geodesic" parallelogram

You can't in a general curved manifold; you can only do it in a flat manifold, or a flat region of a manifold.

 

[timmdeeg]

If you accelerate a rocket and allow it to maintain its natural shape, then clocks at the bottom will run slower than clocks at the top. That's actually how gravitational time dilation was invented

If you assume the same proper distance between top and bottom in both cases wouldn't the difference of the clock rate be different? So that only in one the two cases the difference fits to Schwarzschild spacetime and thus justifies the term gravitational time dilation?

 

[PAllen]

If you assume the same proper distance between top and bottom in both cases wouldn't the difference of the clock rate be different? So that only in one the two cases the difference fits to Schwarzschild spacetime and thus justifies the term gravitational time dilation?

Proper distance between events is invariant, but between world lines it is frame dependent. In a local inertial (free fall) frame near Earth the proper distance between the top and bottom of building decreases with time, while in a noninertial frame where the building is stationary, the proper distance is constant. The difference, of course, is due to different choices about which pairs of events on the two world lines are considered simultaneous in order to compute proper distances. The situation is the same for the born rigid rocket.

 

[SlowThinker]

effects of the acceleration on the shape of the body will be quadratic in the acceleration and therefore will be ignored.

Yes the shape (contraction) of an accelerating rocket only depends on its speed, not acceleration.
But gravitational time dilation depends on acceleration, linearly.

Can we just rename gravitational time dilation to accelerational time dilation and move on?

 

[stevendaryl]

Once again, here are the facts:

For a rocket accelerating at constant proper acceleration:

• Take two clocks and bring them together at the rear of the rocket and synchronize them.
• Take one of the clocks to the front of the rocket. Wait.
• Take the front clock back to the rear, and compare. The relationship between the elapsed times on the two clocks will be, approximately: [itex]\tau_{rear}/\tau_{front} \approx 1 - \frac{gL}{c^2}[/itex], where [itex]g[/itex] is the acceleration of the rear of the rocket, and [itex]L[/itex] is the height of the rocket.

For a rocket at rest upright on the Earth:

• Take two clocks and bring them together at the rear of the rocket and synchronize them.
• Take one of the clocks to the front of the rocket. Wait.
• Take the front clock back to the rear, and compare. The relationship between the elapsed times on the two clocks will be, approximately: [itex]\tau_{rear}/\tau_{front} \approx 1 - \frac{gL}{c^2}[/itex], where [itex]g[/itex] is the acceleration of gravity at the rear rocket, and [itex]L[/itex] is the height of the rocket.

These numbers are only approximately true, in the limit of small [itex]L[/itex], and where the time the clocks spend in transit is negligible. It's the same effect; it's a reflection of the non-vanishing of the connection coefficients in the noninertial frame in which the rocket is at rest.

 

[timmdeeg]

Proper distance between events is invariant, but between world lines it is frame dependent. In a local inertial (free fall) frame near Earth the proper distance between the top and bottom of building decreases with time, while in a noninertial frame where the building is stationary, the proper distance is constant.

Yes, but I didn't consider an inertial frame. Is the proper distance between the top and the bottom of the rocket the same in both noninertial frames, hovering with g stationary above Earth and accelerating with g in flat spacetime?

Do we calculate the proper distance in the hovering case ##d\sigma=dr/\sqrt(1-2M/r)^{1/2}## but different when accelerating in flat space?

 

[PAllen]

Yes, but I didn't consider an inertial frame. Is the proper distance between the top and the bottom of the rocket the same in both noninertial frames, hovering with g stationary above Earth and accelerating with g in flat spacetime?

Do we calculate the proper distance in the hovering case ##d\sigma=dr/\sqrt(1-2M/r)^{1/2}## but different when accelerating in flat space?

We define it to be the same. We want to compare e.g. a 100 meter rocket or building in both cases, with non changing distance determined, e.g. by successive round trip radar measurements. Then the time dilation will be the same to high precision. There will be a small second order difference in the time dilation, but that can be made arbitrarily small by considering a hypothetical planet of very large mass and very low density (thus large radius) such that its surface gravity is g. Also, this second order difference plays no role in Schild's argument.

 

[stevendaryl]

We define it to be the same. We want to compare e.g. a 100 meter rocket or building in both cases, with non changing distance determined, e.g. by successive round trip radar measurements. Then the time dilation will be the same to high precision. There will be a small second order difference in the time dilation, but that can be made arbitrarily small by considering a hypothetical planet of very large mass and very low density (thus large radius) such that its surface gravity is g. Also, this second order difference plays no role in Schild's argument.

Well, are we any closer to figuring out what Schild's argument actually means? To first order, there is no observable difference between gravitational time dilation and acceleration-dependent time dilation. So how can gravitational time dilation prove that spacetime is curved?