Does gravitational time dilation imply spacetime curvature?

[timmdeeg]

For a rocket accelerating at constant proper acceleration:

• [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:

• [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.

So it seems rigorously applied the time dilation differs. Isn't this a reason to argue that if the time dilation for the rocket accelerating at constant proper acceleration and the gravitational time dilation for the rocket on Earth are unequal then the time dilation in the former case can't be named gravitational time dilation. It can't be named approximate gravitational time dilation either.

Please excuse should I've overlooked that this point was already clarified.

 

[stevendaryl]

So it seems rigorously applied the time dilation differs. Isn't this a reason to argue that if the time dilation for the rocket accelerating at constant proper acceleration and the gravitational time dilation for the rocket on Earth are unequal then the time dilation in the former case can't be named gravitational time dilation. It can't be named approximate gravitational time dilation either.

Please excuse should I've overlooked that this point was already clarified.

The difference between the two cases vanishes in the limit as [itex]gL/c^2 \rightarrow 0[/itex]. So first-order effects can't tell the difference.

 

[timmdeeg]

The difference between the two cases vanishes in the limit as [itex]gL/c^2 \rightarrow 0[/itex]. So first-order effects can't tell the difference.

So, which criterion allows one to say physical phenomena are the same and are named the same if they differ in second order effects though?

 

[stevendaryl]

So, what criterion allows one to say physical phenomena are the same and are named the same if they differ in second order effects though?

It's a matter of definition. You can define "gravitational time dilation" any way you want to. But the concept was invented in the context of the equivalence principle, and the equivalence between gravitational effects and acceleration effects is only valid to first order.

 

[timmdeeg]

Thanks

Also, this second order difference plays no role in Schild's argument.

From my layman point of view it seems weird to neglect second order effects if two well defined phenomena are compared. A difference in second order effects should inevitably be due to a different physical background. Am I wrong?

 

[stevendaryl]

From my layman point of view it seems weird to neglect second order effects if two well defined phenomena are compared. A difference in second order effects should inevitably be due to a different physical background. Am I wrong?

What do you mean by "neglect"? It's a matter of definition to call the first-order effect "gravitational time dilation". The second-order effect is called "tidal forces".

If I say that a rose and an apple are the same color, I'm not neglecting other differences between apples and roses. I'm just not mentioning them.

 

[PAllen]

Thanks

From my layman point of view it seems weird to neglect second order effects if two well defined phenomena are compared. A difference in second order effects should inevitably be due to a different physical background. Am I wrong?

The whole thread is about a particular argument for how curvature can be deduced. That argument makes no use of second order information.

A matter of definition rather than physics is whether there is one phenomenon or two. I believe most physicists would say there is one phenomenon, which includes second order curvature corrections when tidal gravity is present. The argument for one phenomenon is that in both cases the effect is primarily produced by setting up a noninertial frame.

 

[timmdeeg]

What do you mean by "neglect"? It's a matter of definition to call the first-order effect "gravitational time dilation". The second-order effect is called "tidal forces".

I see, "neglect" and "not mention" are different things.

 

[timmdeeg]

The whole thread is about a particular argument for how curvature can be deduced. That argument makes no use of second order information.

So as the thread shows it seems hard if not impossible to derive curvature from first order effects regarding the uniformly accelerating rocket. And I'm not knowledgeable enough to understand that attempting this makes sense given the flatness of the Rindler metric.

 

[stevendaryl]

I see, "neglect" and "not mention" are different things.

Yes. If I'm making a statement about the colors of objects, then not mentioning that one object is a fruit and the other object is a flower is not neglecting anything---what I'm leaving out is irrelevant to the statement I'm making.

On the other hand, if I say that French fries and apples are the same color, and I neglect to say that the French fries are covered in ketchup, then what I left out is relevant.

 

[PAllen]

So as the thread shows it seems hard if not impossible to derive curvature from first order effects regarding the uniformly accelerating rocket. And I'm not knowledgeable enough to understand that attempting this makes sense given the flatness of the Rindler metric.

Many of his here think it doesn't make sense, but are grappling with why Schild thought it did since he was a prominent GR physicist.

 

[PeterDonis]

I believe most physicists would say there is one phenomenon, which includes second order curvature corrections when tidal gravity is present. The argument for one phenomenon is that in both cases the effect is primarily produced by setting up a noninertial frame.

I don't think this is correct. Tidal gravity cannot be transformed away by changing coordinates, and is not "primarily produced by setting up a noninertial frame".

 

[stevendaryl]

I don't think this is correct. Tidal gravity cannot be transformed away by changing coordinates, and is not "primarily produced by setting up a noninertial frame".

But Schild's argument didn't seem to be about tidal forces.

 

[PAllen]

I don't think this is correct. Tidal gravity cannot be transformed away by changing coordinates, and is not "primarily produced by setting up a noninertial frame".

It depends on what observations you are talking about. For a local experiment, you are measuring exactly the same physics whether you do the experiment on Earth or in an accelerating rocket, and the primary effect is SR Doppler as it shows up in an accelerated frame. Tidal gravity introduces only a second order correction. This is noted in the reference I gave earlier in this thread to Clifford Will's living review article.

If you are talking about about measuring this at many points around the earth, the local physics is still the same. Tidal gravity contributes by allowing the existence of a configuration of local accelerated frames that could not exist in flat spacetime.

If you are talking about an inherently global measurement, then the tidal effects dominate. I'm not aware of good terminology for when curvature effects are large.

[edit: This seems related to absence of good terminology for Doppler in GR versus SR. Clearly locally, or with minimal tidal effects, claiming the Doppler is different phenomenon because of curvature elsewhere or overall curvature seems nonsensical. The change over to when curvature directly plays a substantial role is, of course, a continuum. Personally, in this case, my preference is to say the overall phenomenon is Doppler; that Doppler in GR includes a curvature contribution, and that SR doppler is GR doppler when the curvature contribution is absent or insignificant. Going from this, I might argue there is one phenomenon of gravitational time dilation, with the pure SR phenomenon being a special case. In all situations, a local measurement is primarily or exclusively (SR) due to setting up a non-inertial local frame. In the global case, you can't set up a global inertial frame, so you can't make it go away. Any global coordinates you try to set up are non-inertial.]

 

[PeterDonis]

Schild's argument didn't seem to be about tidal forces.

It wasn't. But in what you quoted, I was responding to a specific statement of PAllen's, not to Schild's argument.

 

[PeterDonis]

For a local experiment, you are measuring exactly the same physics whether you do the experiment on Earth or in an accelerating rocket, and the primary effect is SR Doppler as it shows up in an accelerated frame. Tidal gravity introduces only a second order correction. This is noted in the reference I gave earlier in this thread to Clifford Will's living review article.

Tidal gravity isn't observable in a local experiment that only probes first-order effects, yes.

If you are talking about about measuring this at many points around the earth, the local physics is still the same.

Yes.

Tidal gravity contributes by allowing the existence of a configuration of local accelerated frames that could not exist in flat spacetime.

Yes, and this cannot be transformed away by changing coordinates. Your statements about how local physics doesn't show this are irrelevant to that claim, because the local physics, which doesn't show any tidal gravity, is being described using coordinates that are only valid on a small patch--or, more precisely, that only take the Minkowski form to a good approximation on a small patch. If you try to extend those coordinates beyond the small patch, you will either find them telling you wrong information--giving you wrong spacetime intervals between events or wrong arc lengths along particular curves--or you will have to add back in the second order terms that were left out, and that show the existence of tidal gravity/spacetime curvature.

 

[stevendaryl]

Maybe the point of Schild's argument is just to prove, not that spacetime is curved, but that the 2-D coordinate system [itex](z,t)[/itex] is curvilinear (z is height above the Earth). This doesn't prove that spacetime is curved by itself, but the additional information that a person standing on the planet is not a Rindler observer presumably proves that.

 

[PeterDonis]

If the theory of gravity is based on SR (use the scalar theory of gravity of Poincare 1905 if you need an example) then straight lines of the Minkowski metric are the geodesics.

But how do you know, locally, that the worldlines of the observers in Schild's argument are such geodesics? I understand how we know it globally--exchange light signals with an observer at infinity. But how do you know it locally? Saying "they're just the straight lines of the Minkowski metric" doesn't help, because the Minkowski metric is not observable locally.

 

[SlowThinker]

[edit: I have also, several times, given a strictly physical definition for parallel, which you have ignored - mutual constancy of radar round trip times between observers. ]

So the two Rindler observers aren't parallel, since their radar distance is not equal both ways, right? Does that resolve this whole issue?

Anyway, can someone post the full citation of Schild? It looks like it needs to be read again. Otherwise, we can keep debating non-existing claims for 30 more pages.

BTW, "there are no parallels in a curved spacetime" is a fact, or just "I don't know a good definition"? I can see that null intervals cause trouble for most definitions, but the curvedness shouldn't be a problem.

 

[PeterDonis]

can someone post the full citation of Schild?

MTW gives three references, which they cite as Schild (1960, 1962, 1967). From the Bibliography, it looks like these are:

Schild, 1960, "Time", Texas Quarterly, 3, no. 3, 42-62.

Schild, 1962, "Gravitational theories of the Whitehead type and the principle of equivalence", in Moller 1962, Evidence for Gravitational Theories, Academic Press, New York.

Schild, 1967, "Lectures on General Relativity Theory", pp. 1-105 in Ehlers 1967 (ed.), Relativity Theory and Astrophysics: I, Relativity and Cosmology; II, Galactic Structure; III, Stellar Structure, American Mathematical Society, Providence, RI.

 

[PAllen]

So the two Rindler observers aren't parallel, since their radar distance is not equal both ways, right? Does that resolve this whole issue?

Anyway, can someone post the full citation of Schild? It looks like it needs to be read again. Otherwise, we can keep debating non-existing claims for 30 more pages.

BTW, "there are no parallels in a curved spacetime" is a fact, or just "I don't know a good definition"? I can see that null intervals cause trouble for most definitions, but the curvedness shouldn't be a problem.

No, the two Rindler observers are parallel curves by any reasonable definition (each finds the radar times to the other remain constant over time).

There are working definitions of parallel, and the ones discussed above are generally equivalent to e.g.:

https://en.wikipedia.org/wiki/Parallel_curve using the orthogonal distance definition. The treatment of lattidude lines as parallel on a 2 sphere is well established as well. These are curves (not geodesic).

Null paths are, indeed, the trickiest case, but I think one could come up with generalizations to cover these.

 

[PeterDonis]

"there are no parallels in a curved spacetime" is a fact, or just "I don't know a good definition"?

I would say the issue is that there is no general, unique definition in curved spacetime. You can construct special cases in which a particular definition looks reasonable, but they don't generalize, and they also don't nessarily have the same properties as parallel lines in flat spacetime.

 

[SlowThinker]

MTW gives three references, which they cite as Schild (1960, 1962, 1967). From the Bibliography, it looks like these are:

Well I meant the actual text. Without it, there's little to discuss.

 

[PAllen]

So the two Rindler observers aren't parallel, since their radar distance is not equal both ways, right?

Wait, I see what your are questioning. I see my wording was subject to ambiguity. I meant each finds radar times to the other remaining constant, not that this constant is necessarily the same for them. Using mutually orthogonal proper distance removes this ambiguity, and also says they are parallel, as does parallel transport of tangent vectors between events connected by mutually orthogonal geodesics.

 

[PeterDonis]

I meant the actual text

Unfortunately I have not been able to find any of the Schild references online. The text in MTW is a bit much to post here, but I tried to describe the argument as best I could in the OP of this thread.

To try to summarize the discussion, here are what I see as the main issues that have been raised:

(1) It's not clear whether the quadrilateral in spacetime formed by the four events described in the OP of this thread is actually a parallelogram, as the argument claims. If it isn't, then the fact that opposite sides of this quadrilateral are unequal does not necessarily imply spacetime curvature.

(2) The argument is formulated assuming a background flat spacetime (I didn't make this clear in the OP), and it is assumed that the worldlines of the two observers described in the OP of this thread are "at rest" in this background flat spacetime, in the sense that they can exchange light signals with observers at rest at infinity and verify that the round-trip light travel time is constant. However, it's not clear how to formulate this criterion in a local way, i.e., a way that doesn't involve hypothetical observers at infinity. Locally, as I described in the OP, the scenario can be duplicated in flat spacetime, including the gravitational time dilation, which would seem to indicate that gravitational time dilation can't require spacetime curvature.

(3) The assumption of a background flat spacetime would seem to be already contradicted by the fact that observers at rest in this background spacetime can have nonzero proper acceleration. To avoid bringing in hypothetical observers at infinity, one could imagine observers at different depths within the gravitating body itself (e.g., the Earth), and one can see that an observer at the center of the body would have zero proper acceleration, i.e., would be in free fall, but observers not at the center would have nonzero proper acceleration, even though all of these observers are at rest relative to each other. That in itself is impossible in a flat spacetime, and one doesn't even need to consider time dilation to see that.

 

[SlowThinker]

Wait, I see what your are questioning. I see my wording was subject to ambiguity. I meant each finds radar times to the other remaining constant, not that this constant is necessarily the same for them. Using mutually orthogonal proper distance removes this ambiguity, and also says they are parallel, as does parallel transport of tangent vectors between events connected by mutually orthogonal geodesics.

Now I see that, given a (infinitesimal) line (segment) we should be able to draw a parallel line through any point, and some pairs of points simply do not have the same radar distance both ways...
But it still seems strange that you can make 2 curves parallel just by changing coordinate system (from Minkowski to Rindler).
The last definition sounds good. I'd say 2 concentric circles aren't parallel but it eliminates other counterexamples I had in mind.

 

[PAllen]

Unfortunately I have not been able to find any of the Schild references online. The text in MTW is a bit much to post here, but I tried to describe the argument as best I could in the OP of this thread.

To try to summarize the discussion, here are what I see as the main issues that have been raised:

(1) It's not clear whether the quadrilateral in spacetime formed by the four events described in the OP of this thread is actually a parallelogram, as the argument claims. If it isn't, then the fact that opposite sides of this quadrilateral are unequal does not necessarily imply spacetime curvature.

I think John Baez first noted even in flat plane, a figure with opposite sides parallel but curved, need not have all sides the same length. I have thought of a concrete example. Consider concentric circles, connect a portion of them by parallel straight lines. You have a figure with two parallel curved sides and two parallel straight sides. Yet the length of curved sides will not be equal (for most choices of the straight lines). This is very similar to the rindler figure. Note also, in my example, the two straight sides are not generally the same length either.

This latter point is interesting, because even if we accept the view that Schild intended the two static observers to have SR geodesic world lines, he explicitly allowed the light paths to be curved in Minkowski space, in the presence of gravity. So if these are parallel curves, it then fails to follow that the parallel straight sides must have the same length!

So again, there seem to be holes, at least as presented by MTW.

[edit: reading MTW presentation again, there is a nuance. Schild isn't saying the light paths are parallel curves, instead that (on some physical arguments) one must be a time translation of the other. For curves, that would actually make them non-parallel, but then his argument holds. So the key to making the argument hold is noting that the static observers are required to be geodesics and the possibly curved sides are congruent in a specific sense, rather than parallel curves as geometers normally define them.]

 

[PAllen]

But it still seems strange that you can make 2 curves parallel just by changing coordinate system (from Minkowski to Rindler).
The last definition sounds good. I'd say 2 concentric circles aren't parallel but it eliminates other counterexamples I had in mind.

No, all the definitions of parallel curves I've given are coordinate independent. Constancy of radar times clearly is. But so is any definition besed on connecting curves by mutually orthogonal geodesics.

Well, if you don't think concentric circles are parallel curves, you disagree with all definitions I've found in the literature.

 

[PAllen]

Just calling attention to the edit in my last post, which now convinces me that if one accepts how I think Schild intended his 4 sided figure to be interpreted (two geodesic sides, two possibly curved sides which must be related by translation along the straight sides - not parallelism ) then his argument holds.

 

[PeterDonis]

Schild isn't saying the light paths are parallel curves, instead that (on some physical arguments) one must be a time translation of the other

Yes, he is basing this on the fact that the spacetime is static. Formalizing this would require transporting the light paths along integral curves of the timelike KVF.

the key to making the argument hold is noting that the static observers are required to be geodesics

Yes, but I don't think there is a local way to define this given that the worldlines of the static observers have nonzero proper acceleration. The only way I can see to define them as geodesics is as curves of constant spatial coordinates in the background Minkowski spacetime, but that requires exchanging light signals with observers at infinity, which you objected to before.

 

[PAllen]

Yes, but I don't think there is a local way to define this given that the worldlines of the static observers have nonzero proper acceleration. The only way I can see to define them as geodesics is as curves of constant spatial coordinates in the background Minkowski spacetime, but that requires exchanging light signals with observers at infinity, which you objected to before.

But after several readings of the MTW presentation, I see he is not attempting to make a local argument. He is trying to show by contradiction, that a certain broad class of SR based theories of gravity cannot exist if they must include gravitational time dilation.

Also, I see he is not requiring exchange of light signals with infinity, but only with some bodies placed sufficiently far from each other to have minimal gravitational influence, that can each verify mutual constancy of position. This sets up one instance of global Lorentz frame.

[edit: Perhaps calling the contradiction a proof of curvature is an over claim. All it really shows is a contradiction among all the requirements of the class of hypothetical SR based theory.]

 

[PeterDonis]

after several readings of the MTW presentation, I see he is not attempting to make a local argument.

I agree he isn't; but I also think your objection to that strategy, that you shouldn't have to make a global measurement to detect spacetime curvature, is valid.

Perhaps calling the contradiction a proof of curvature is an over claim. All it really shows is a contradiction among all the requirements of the class of hypothetical theory.

I agree.

 

[stevendaryl]

So Schilld's point is that someone stationary on the surface of a planet cannot be following a geodesic? So Rindler spacetime isn't a counterexample, because someone stationary in an accelerating rocket isn't following a geodesic, either.

 

[PAllen]

So Schilld's point is that someone stationary on the surface of a planet cannot be following a geodesic? So Rindler spacetime isn't a counterexample, because someone stationary in an accelerating rocket isn't following a geodesic, either.

I don't think his argument quite shows that. It shows that a certain set of assumptions leads to a contradiction. It doesn't say anything about which of the assumptions must be modified, or in what way.

Though there are many other paths to the same conclusion, I would sum up Schild's argument, tightened up as needed (perhaps done in the papers we can't access) as:

A theory of gravity that has Newtonian gravity as an appropriate limit, and SR as an appropriate limit, and includes gravitational time dilation between static observers near a gravitating body, must have as its accessible geometry, that of a curved pseudo-riemannian manifold.

 

[PeterDonis]

Schilld's point is that someone stationary on the surface of a planet cannot be following a geodesic?

In the argument as presented in MTW, the word "geodesic" is not mentioned at all. Without being able to see Schild's actual papers, I can't tell whether he used that term himself or not.

Even if he did, as I pointed out in response to PAllen, he must be using the term in a different sense from its usual one; if the point is that the timelike sides of the quadrilateral he describes must be geodesics in order for it to be properly termed a "parallelogram", then they can only be geodesics of the background flat Minkowski metric, which is not physically observable locally--that is to say, since these worldlines have nonzero proper acceleration, they are obviously not geodesics in the usual sense, so any definition of them as geodesics must be relying on some non-local measurement (such as exchanging light signals with observers who are very far away).