Does gravitational time dilation imply spacetime curvature?

[PAllen]

If we parallel transport a vector in a small circle, and if the final vector is different (not parallel) to the starting vector, we attribute that effect to "space-time curvature". If I am not mistaken we can do the same in the accelerated elevator and the final vector remains parallel to the original, but is "time dilated". It's different terminology, but to attribute the effect of the second case to a physical process which is any different from the first case seems to me to be in contradiction to what Einstein says in his defense of the EP.

Nonsense. In the vicinity of a planet per GR, the vector parallel transported in a closed path changes. Far away from any bodies in an accelerating rocket, it does not. Time dilation has nothing to do with the parallel transport in a closed path in either case. Further, the EP specifically applies to a sufficiently small spacetime region that curvature can be ignored.

 

[A.T.]

If we parallel transport a vector in a small circle, and if the final vector is different (not parallel) to the starting vector, we attribute that effect to "space-time curvature".

Note that to test for space-time curvature you have to parallel transport a 4-vector, which has a time component.

If I am not mistaken we can do the same in the accelerated elevator and the final vector remains parallel to the original, but is "time dilated".

What is that supposed to mean? If the time-component would change, then the 4-vectors would not be parallel. But they actually are parallel, because there is no space-time curvature in the accelerated elevator.

 

[RockyMarciano]

you can try to model gravity as a spin-2 field on a flat spacetime background, but what you end up with is a curved spacetime--in other words, you contradict your starting point.

I don't see any contradiction or what exactly you refer to as my starting point, which was that Schild's argument is fine and is not contradicted by the Rindler coordinates examples(effects that are obtained using properties of noninertial observers/curved coordinates), this actually means that as you say "what you end up with is a curved spacetime" when trying to model effects like gravitational redshift. There has been an attempt in this thread(basically by PAllen) to mix velocity time dilation with gravitational time dilation when they are different.

 

[RockyMarciano]

In the vicinity of a planet per GR, the vector parallel transported in a closed path changes. Far away from any bodies in an accelerating rocket, it does not. Time dilation has nothing to do with the parallel transport in a closed path in either case.

But you are missing the difference between curvature at a point and path dependence of vector orientation along a curve in the presence of curvature. Gravitational redshift refers to the latter.

Further, the EP specifically applies to a sufficiently small spacetime region that curvature can be ignored.

Again you should be more specific, there is no sufficiently small region that curvature at a point can be ignored in principle, there is no region smaller than a point. There is indeed a local region where the higher than first order effects of curvature are not detected by certain measurements that apply to those sufficiently small spacetime region.
The EP raises rocket acceleration to a curvature phenomenon, it doesn't reduce gravitation to a flat spacetime effect as you seem to imply.

 

[RockyMarciano]

If the time-component would change, then the 4-vectors would not be parallel. But they actually are parallel, because there is no space-time curvature in the accelerated elevator.

Well, this is what Schild's argument shows. That in the gravitational redshift case they are not parallel.

 

[A.T.]

The EP raises rocket acceleration to a curvature phenomenon,

Not if "curvature" means "intrinsic space-time curvature" (case C in the below diagram). But you can describe the accelerated frame in intrinsically flat space-time using curvilinear coordinates (case B in the below diagram).

This is my own non-animated way of looking at it:

• A. Two inertial particles, at rest relative to each other, in flat spacetime (i.e. no gravity), shown with inertial coordinates. Drawn as a red distance-time graph on a flat piece of paper with blue gridlines.
  
• B₁. The same particles in the same flat spacetime, but shown with non-inertial coordinates. Drawn as the same distance-time graph on an identical flat piece of paper except it has different gridlines.
  
  B₂. Take the flat piece of paper depicted in B₁, cut out the grid with some scissors, and wrap it round a cone. Nothing within the intrinsic geometry of the paper has changed by doing this, so B₂ shows exactly the same thing as B₁, just presented in a different way, showing how the red lines could be perceived as looking "curved" against a "straight" grid.
  
• C. Two free-falling particles, initially at rest relative to each other, in curved spacetime (i.e. with gravity), shown with non-inertial coordinates. This cannot be drawn to scale on a flat piece of paper; you have to draw it on a curved surface instead. Note how C looks rather similar to B₂. This is the equivalence principle in action: if you zoomed in very close to B₂ and C, you wouldn't notice any difference between them.

Note the diagrams above aren't entirely accurate because they are drawn with a locally-Euclidean geometry, when really they ought to be drawn with a locally-Lorentzian geometry. I've drawn it this way as an analogy to help visualise the concepts.

 

[RockyMarciano]

Not if "curvature" means "intrinsic space-time curvature" (case C in the below diagram). But you can describe the accelerated frame in intrinsically flat space-time using curvilinear coordinates (case B in the below diagram).

The equivalence principle refers to what is measurable and predictable locally to first order, i.e. the description you mention. So it doesn't make the distinction you refer to about spacetime. It just makes sure that the local effects can be described, and they can as you admit both with B(effects based on the curved coordinates) and with C to first order.

 

[A.T.]

So it doesn't make the distinction you refer to about spacetime.

But you should make that distinction, if you don't want to be misunderstood. Because "spacetime curvature" in the title of this thread refers to intrinsic curvature, as shown in C.

 

[RockyMarciano]

But you should make that distinction, if you don't want to be misunderstood. Because "spacetime curvature" in the title of this thread refers to intrinsic curvature, as shown in C.

That intrinsic distinction can be made with second order effects(like pehihelion shift), but gravitational redshift, also in the thread title, is a first order effect of curvature.

 

[MikeGomez]

Nonsense. ...Further, the EP specifically applies to a sufficiently small spacetime region that curvature can be ignored.

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.

 

[A.T.]

but gravitational redshift, also in the thread title, is a first order effect of curvature.

Since redshift also happens in case B of the diagram, It doesn't imply the existence intrinsic space-time curvature, which is not present in B.

 

[RockyMarciano]

Since redshift also happens in case B of the diagram, It doesn't imply the existence intrinsic space-time curvature, which is not present in B.

Remember that we are talking about physical effects that we are assuming can only come from intrinsic curvature and not from curved coordinates alone(this is the premise that I think we are all accepting). The argument by Schild underlines that they could never come from flatness. The Rindler case that corresponds to B is obtained because of the curvature of the hyperbolic coordinates, but we know that coordinates can never give physical effects by themselves, even if we can compute a redshift from them. It is this fact that solves the issue in favor of Schild's argument that the physically observed gravitational redshift can never be due to flatness by itself, even if the first order approximation of intrinsic curvature can be modeled by Rindler curved coordinates in flat spacetime, and therefore the only possibility left is that the observed gravitational redshift implies intrinsic curvature.

So gravitational redshift wouldn't actually happen in the flat case B, even if we can compute it with B(since it is a first approximation to intrinsic curvature) given the fact that it actually happens, by using coordinates that are similar to the Kruskal ones used in the Schwarzschild solution. Again coordinates effects are never physical by themselves in the context of flatness, even if we can derive results from them.

 

[MikeGomez]

Remember that we are talking about physical effects that we are assuming can only come from intrinsic curvature and not from curved coordinates alone(this is the premise that I think we are all accepting). The argument by Schild underlines that they could never come from flatness. The Rindler case that corresponds to B is obtained because of the curvature of the hyperbolic coordinates, but we know that coordinates can never give physical effects by themselves, even if we can compute a redshift from them. It is this fact that solves the issue in favor of Schild's argument that the physically observed gravitational redshift can never be due to flatness by itself, even if the first order approximation of intrinsic curvature can be modeled by Rindler curved coordinates in flat spacetime, and therefore the only possibility left is that the observed gravitational redshift implies intrinsic curvature.

So gravitational redshift wouldn't actually happen in the flat case B, even if we can compute it with B(since it is a first approximation to intrinsic curvature) given the fact that it actually happens, by using coordinates that are similar to the Kruskal ones used in the Schwarzschild solution. Again coordinates effects are never physical by themselves in the context of flatness, even if we can derive results from them.

This.

 

[A.T.]

Remember that we are talking about physical effects that we are assuming can only come from intrinsic curvature...

No, it comes from the proper acceleration of the elevator, not from the intrinsic curvature of space-time. The intrinsic curvature of space-time near a mass just makes the red shift non-linear.

 

[MikeGomez]

No, it comes from the proper acceleration of the elevator, not from the intrinsic curvature of space-time. The intrinsic curvature of space-time near a mass just makes the red shift non-linear.

That's just a restatement of the idea that gravity which can be transformed away is different from the gravity which can't.

 

[PeterDonis]

So is the term "flat-spacetime" just an approximation?

It is if the spacetime is curved. Flat Minkowski spacetime is a valid solution of the Einstein Field Equation, so as far as the math is concerned a spacetime that was truly flat (not just an approximation) could exist. But it would have to have absolutely no stress-energy anywhere, ever.

 

[PeterDonis]

I don't see any contradiction or what exactly you refer to as my starting point

If you are claiming that the "spin-2 field in flat spacetime" model of GR is just as valid as the curved spacetime model, then your starting point is that spacetime is flat. But the end result of the "spin-2 field in flat spacetime" model is that spacetime is curved. So the model contradicts itself.

 

[PeterDonis]

this is what Schild's argument shows. That in the gravitational redshift case they are not parallel.

No, that's not what Schild's argument shows, or claims to show. It talks about opposite sides of a parallelogram having different lengths. It does not talk about 4-vectors not being parallel.

 

[PeterDonis]

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.

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.

 

[PeterDonis]

The argument by Schild underlines that they could never come from flatness.

That's what Schild appeared to be claiming, but I don't think that claim can be correct as it stands (that was my basic reason for starting this thread, since the Rindler case is an obvious counterexample). It's possible, as others have brought up in this thread, that Schild only meant to claim that gravitational redshift/time dilation implied spacetime curvature if the observers in question were at rest relative to an inertial observer at infinity (which brings up other issues, but those can be fixed in the case of a static body like the Earth by using the observer at the center of the body, who is also inertial and who is at a finite spatial location).

If that possibility I just mentioned is correct, then Schild's argument including that qualifier looks correct to me, but it focuses IMO on the wrong thing--the gravitational redshift/time dilation, instead of the fact that we have a family of accelerated observers who are all at rest relative to an inertial observer, something that obviously can't happen in flat spacetime, whereas the redshift/time dilation can.

gravitational redshift wouldn't actually happen in the flat case B

Redshift between Rindler observers is an obvious, easily computed fact about that congruence of observers. So this claim of yours makes no sense.

 

[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.