Does gravitational time dilation imply spacetime curvature?

[PAllen]

This verges on the absurd. If a experiment only detects only first order effects, like gravitational redshift, light deflection or GWs one cannot use this fact to imply that it discards spacetime curvature. It only means that it would take a different experiment to detect the higher order effects, like for instance Mercury's perihelion shift.

What's absurd is claiming that an experiment that does not distinguish curvature from absence thereof (experiment could be done in a regime with no curvature at all) can be used to prove the presence of curvature.

 

[PAllen]

I agree, I think the argument by Schild is misinterpreted in the OP.
Not necessarily, one just has to consider the point of view mentioned in MTW that's referenced above(and according to Thorne accepted by physicists because both approaches-flat and curved spacetime- give the same predictions) to make the Rindler case compatible with Schild's argument. Certainly the authors of MTW don't seem to find any problem or else they surely had noted it.

Gravitational field in flat spacetime, has the feature that the flat spacetime's geometry is not observable e.g. all lengths, angles, etc. measured are the same as for geometric interpretation of GR. You need not assume any geometry, only measurements, but as long as you do try to construct a geometry for the measurements, it must have cuvature.

 

[RockyMarciano]

What's absurd is claiming that an experiment that does not distinguish curvature from absence thereof (experiment could be done in a regime with no curvature at all) can be used to prove the presence of curvature.

Both claims are, but no one is making the latter, while you made the former several times.

 

[RockyMarciano]

Gravitational field in flat spacetime, has the feature that the flat spacetime's geometry is not observable e.g. all lengths, angles, etc. measured are the same as for geometric interpretation of GR. You need not assume any geometry, only measurements, but as long as you do try to construct a geometry for the measurements, it must have cuvature.

You just have to make a choice regarding the nature of the measurement tools.

 

[PAllen]

Both claims are, but no one is making the latter, while you made the former several times.

The argument described in the OP that is the topic of the thread effectively makes the latter claim. I also never made the former claim. Stating that an experiment doesn't demonstrate curvature is not the same as saying it disproves or is incompatible with curvature. You have misrepresented what I said.

 

[PAllen]

You just have to make a choice regarding the nature of the measurement tools.

The point remains the underlying flat minkowski geometry plays no role in observations. Its geometric invariants have no connection to observables.

 

[PeterDonis]

I really don't see why you would put the word gravitational in there

The word "gravitational" does not appear in what you quoted.

An important question would also be how you define that `distance'

The distance implied by the round trip light travel time is the obvious way to define it. You are right that this is not the only possible definition, but all of the possible definitions still have the property that the distance does not change with time between the two observers in question, which is the key point. The exact numerical value of the distance is not important.

I don't agree with this statement.

It's a statement about how the word "gravity" is used, not about physics. If you want to disagree with how words are used, you're going to have a tough time convincing everyone who uses them the way they want to, not the way you want them to.

 

[PeterDonis]

It's also known as "curvature without curvature" view.

It should be obvious that this is tongue in cheek, since it is immediately followed by

or--equally well--as "flat spacetime without flat spacetime"!

Physically, the key phrase is just before all this: the initial flat background space is no longer observable. In other words, 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. The spin-2 field view still works as an approximation in some scenarios (for example, in studying gravitational waves), but it is not a fully general model the way the curved spacetime model is. A proper reading of MTW (or even better, reading the actual primary source papers that MTW references) makes that clear.

 

[MikeGomez]

Physically, the key phrase is just before all this: the initial flat background space is no longer observable. In other words, 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. The spin-2 field view still works as an approximation in some scenarios (for example, in studying gravitational waves), but it is not a fully general model the way the curved spacetime model is. A proper reading of MTW (or even better, reading the actual primary source papers that MTW references) makes that clear.

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

 

[MikeGomez]

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.

 

[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.