A Does gravitational time dilation imply spacetime curvature?

[PAllen]

Nonsense. Yes, for the purpose of showing the equivalence between gravity and inertia, the EP applies to a sufficiently small space-time region where curvature can be ignored. However, that does not equate the EP to making an argument for the absolute non-existence of curvature in that small region.

Additionally, Einstein explicitly states the validity of the EP for regions which are large enough that curvature does become a factor, when he says that it is of no importance whatsoever that gravitational fields for finite space-time domains in general cannot be transformed away.

You misrepresent what I said. I never claimed disproof of curvature, I only claimed that a test of the equivalence principle fails to establish curvature, and that gravitation time dilation is such a test.

Einstein used a fluid notion of equivalence principle which he did not precisely define. However I am using the modern formalization, which is inherently local. In particular, I use the well accepted formalization by Clifford Will, in the following classic living review, which also discusses gravitational redshift and time dilation experiments, classifying them as tests of "local position invariance" aspect of the equivalence principle.

See section 2.1 for modern formalization of equivalence principle, and 2.1.3 for discussion redshift and time dilation experiments in this context.

https://link.springer.com/article/10.12942/lrr-2014-4/fulltext.html

 

[MikeGomez]

@PAllen I didn't mean to misrepresent what you said. That is just what I thought you were implying.

Thank you for the link.

 

[MikeGomez]

... a spacetime that was truly flat (not just an approximation) could exist. But it would have to have absolutely no stress-energy anywhere, ever.

And there would be no gravity in that case, correct?

 

[MikeGomez]

The statement of Einstein's that you refer to was about general covariance, not the EP. General covariance is a much broader principle than the EP.

Ok. I need to study that. I thought he directed the subject matter specifically to the EP, since he directly prefaces the statements regarding gravity field that can be transformed away with...

"I now turn to the objections against the relativistic theory of the gravitational field. Here, Herr Reichenbacher first of all forgets the decisive argument, namely, that the numerical equality of inertial and gravitational mass must be traced to an equality of essence. It is well known that the principle of equivalence accomplishes just that."

 

[stevendaryl]

@PeterDonis, there is definitely a missing piece to the Schild argument for curved spacetime, and I don't know how to repair it. The idea is to create a "parallelogram" in spacetime. At point e_1, a light pulse is sent from the lower observer to the upper observer. e_3 is the event where the light pulse reaches the upper observer. Then a second light pulse is sent at point e_2, which reaches the upper observer at event e_4. Schild's argument is that since the "length" of the segment s_{12} is smaller than the length of s_{34}, spacetime must be curved.

But the issue is: how do we know that quadrilateral Q_{1243} is a parallelogram? In the case of Rindler spacetime, it clearly is not--it's a trapezoid. So what's the argument that it should be a parallelogram in the Schwarzschild spacetime case?

(Whether a figure is a parallelogram or a trapezoid is coordinate-independent. So we know that the figure is a trapezoid in Rindler spacetime because that's what it is in Minkowsky spacetime.)

 

[john baez]

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?

We reconcile them by noticing that one of your statements is erroneous, namely:

This is impossible in a flat spacetime.

It's impossible to have a parallelogram in flat space for which all four edges are geodesics and two opposite sides have unequal lengths. However, the 'parallelograms' you are talking about - in both the Schwarzschild and Rindler geometries - do not have all four edges being geodesics: two of the edges are paths traced out by accelerated observers. Such a 'parallelogram' can have opposite edges with unequal lengths in either flat or curved spacetime. Indeed, you have just observed this.

So, there's no contradiction.

 

[martinbn]

This may have already been said or I may misunderstand the problem, but here is an analogy from geometry. Two lines, in the plane, that start orthogonally from another line will remain the same distance apart (they are parallel) because the plane is flat. On the other hand two meridians starting orthogonally at the equator will converge and eventually intersect because the sphere has curvature. So one my use this as way of checking if there is curvature. But if the base line is not a geodesic it can be misleading. Two radii, in the plane, starting from a circle (towards the centre) are orthogonal to the circle and will converge, but that doesn't imply non-zero curvature.

 

[PeterDonis]

And there would be no gravity in that case, correct?

There would be no spacetime curvature. Whether that means no "gravity" depends on what you mean by "gravity".

 

[PeterDonis]

the numerical equality of inertial and gravitational mass must be traced to an equality of essence. It is well known that the principle of equivalence accomplishes just that

Yes, but this "equality of essence" still doesn't explain tidal gravity, i.e., spacetime curvature. It explains why all objects moving solely under gravity follow the same geodesics, but it doesn't explain why the geodesics are what they are. The spacetime curvature is what "cannot be transformed away"; but that fact doesn't affect the EP, it's just a separate aspect that needs to be explained in addition to the EP.

 

[PeterDonis]

how do we know that quadrilateral $Q_{1243}$ is a parallelogram? In the case of Rindler spacetime, it clearly is not--it's a trapezoid. So what's the argument that it should be a parallelogram in the Schwarzschild spacetime case?

I'm not sure, because I'm not sure how to rigorously define "parallel" in a curved spacetime. In fact I'm not even sure that the definition in flat spacetime, under which the figure $Q_{1243}$ is a trapezoid, is unique. Schild's argument, at least as it's presented in MTW, appears to be using the timelike KVF of the spacetime to define "parallel", but in flat Minkowski spacetime there are two sets of timelike KVFs (roughly the inertial and Rindler ones) which give different answers (and the Schwarzschild KVF corresponds to the Rindler one, since observers following its orbits have nonzero proper acceleration).

the 'parallelograms' you are talking about - in both the Schwarzschild and Rindler geometries - do not have all four edges being geodesics

@DrGreg brought up this objection early in the thread, and my suggestion to address it was to switch to a slightly different scenario where all four sides are geodesics. I described this variant in post #3; as far as I can see, the two timelike geodesic sides are still of unequal length. The question would be, do they still count as "parallel"? As I noted in response to @stevendaryl above, I don't know what definition of "parallel" is the right one to use in this context. I'm not even clear about this for the non-geodesic sides that appear in Schild's argument.

 

[PeterDonis]

one my use this as way of checking if there is curvature

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).

 

[PeterDonis]

I don't know what definition of "parallel" is the right one to use in this context.

To clarify this a bit more: I'm actually more concerned about the lightlike sides of the quadrilateral in the two cases. Schild's argument basically says that these sides are parallel because one is just the time translation of the other along the flow of a timelike KVF. But that is true of the two sides in the Rindler case as well--one is the time translate of the other along the flow of the Rindler KVF. But in Minkowski coordinates these two sides are clearly not parallel, which is why @stevendaryl said the figure is obviously a trapezoid. But if they're not parallel in the Rindler case, what justifies the claim that they are parallel in the Schwarzschild case?

 

[PeterDonis]

Such a 'parallelogram' can have opposite edges with unequal lengths in either flat or curved spacetime. Indeed, you have just observed this.

So, there's no contradiction.

Does this mean that Schild's argument is invalid?

 

[john baez]

I'm not sure, because I'm not sure how to rigorously define "parallel" in a curved spacetime.

It's hopeless to define "parallel" in a curved spacetime, but it makes perfect sense in the Rindler spacetime, and with that definition the opposite sides of a parallelogram in that spacetime have equal length.

I could try to explain all this, but that's unnecessary if all we want is to find the mistake in your "paradox". You claimed:

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.

Whatever definition of "parallelogram" you take, if it includes the figure in Rindler spacetime that you discussed later in your post, then it is possible to get such a thing with opposite sides having unequal lengths.

So, there's no paradox. I think we need to agree on that before we going into deeper waters, like "what's a parallelogram in the Rindler spacetime?"

 

[john baez]

Does this mean that Schild's argument is invalid?

I don't know what Schild's argument is, and I don't really want to know. If it's famous, it's almost certainly right. I just wanted to point out the hole in your apparent "paradox" - someone asked me to help.

 

[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 g, the apparent acceleration due to gravity. It has to involve derivatives of g. That's because g, the acceleration due to gravity in the x-direction, is basically the connection coefficient \Gamma^x_{tt}. The curvature tensor is created from \Gamma^\mu_{\nu \lambda} as follows:

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}

So R involves two types of terms: ones quadratic in \Gamma^a_{bc}, and ones involving derivatives of \Gamma^a_{bc}. So no empirical measurement can prove curvature unless it involves derivatives of g or is second-order in g. Time dilation is first-order in g and doesn't involve derivatives of g, 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 g, the apparent acceleration due to gravity. It has to involve derivatives of g. That's because g, the acceleration due to gravity in the x-direction, is basically the connection coefficient \Gamma^x_{tt}. The curvature tensor is created from \Gamma^\mu_{\nu \lambda} as follows:

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}

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