Graduate Does gravitational time dilation imply spacetime curvature?

[martinbn]

@PeteDonis, I don't buy your latest argument because it seems to say we need global properties of a manifold to establish curvature. Yet curvature is uniquely determined via ininitesimal operations. I will think more, but my first reaction is poppycock.

But, if the manifold is connected, a KVF is completely determined by its value and the value of its derivative at one point. To me it seems that global properties of a KVF are not really global, in some sense they are local.

 

[PeterDonis]

if the manifold is connected, a KVF is completely determined by its value and the value of its derivative at one point

By "value" and "derivative", do you mean "vector" (i.e., the vector at a point that is mapped to that point by the field) and "gradient" (i.e., the two-form $d \chi$, where $\chi$ is the KVF, at the chosen point--basically, the set of covariant derivatives in all possible directions)?

 

[martinbn]

By "value" and "derivative", do you mean "vector" (i.e., the vector at a point that is mapped to that point by the field) and "gradient" (i.e., the two-form $d \chi$, where $\chi$ is the KVF, at the chosen point--basically, the set of covariant derivatives in all possible directions)?

Yes.

 

[PAllen]

But, if the manifold is connected, a KVF is completely determined by its value and the value of its derivative at one point. To me it seems that global properties of a KVF are not really global, in some sense they are local.

But there is a kvf in both cases (Rindler and SC).

 

[PeterDonis]

there is a kvf in both cases (Rindler and SC)

Yes, but as I understand his point, the key difference between them that I pointed out--that the SC KVF is normalizable at infinity while the Rindler KVF is not--is fully determined by the local properties of the respective KVFs--in this case, the difference in their derivatives (the Rindler KVF's norm increases linearly with height, while the SC KVF's norm does not).

 

[PAllen]

Yes, but as I understand his point, the key difference between them that I pointed out--that the SC KVF is normalizable at infinity while the Rindler KVF is not--is fully determined by the local properties of the respective KVFs--in this case, the difference in their derivatives (the Rindler KVF's norm increases linearly with height, while the SC KVF's norm does not).

But that has nothing to do with with the argument that gravitational dilation implies curvature. Again, curvature is local not global, and gravitational time dilation does not require curvature as proved by the fact that the defining experiment detecting it did not detect anything about curvature. The valid way if distinguishing Pound Rebka from an SR effect is to note that the world lines with proper acceleration maintain static distance from an inertial body (earth), which is not possible in SR. I remain convinced that your OP establishes that the claim that gravitational time dilation per se ( and the parallelogram argument) establishes curvature is simply false.

 

[martinbn]

It was a side comment that the use of a KVF is not a global but local argument. Now I have changed my mind. It actually is global not local, because it is uniquely determined globally.

 

[timmdeeg]

No, it won't. You're confusing Rindler observers with observers in the Bell spaceship paradox. They are different scenarios. The distance between Rindler observers, as seen in the instantaneous rest frame of either observer, remains constant.

If the rope which martinbn mentioned isn't stretched, doesn't that imply the absence of tidal gravity and hence that the difference in clockrates isn't due to curvature of spacetime?

 

[PAllen]

If the rope which martinbn mentioned isn't stretched, doesn't that imply the absence if tidal gravity and hence that the difference in clockrates isn't due to curvature of spacetime?

In the case of time dilation between the front and back of a rocket, indeed the effect is not due to curvature. In rocket centered coordinates you can say it is due to pseudogravity. In an inertial frame it is due to the fact that by the time a signal from the back of an accelerating rocket reaches the front, the front is moving faster than the back was at emission time, thus nothing but ordinary Doppler is involved.

Note that in a free fall frame, the time dilation from top to bottom of a tall building on Earth is primarily due to SR Doppler, with tidal gravity being only a second order correction. Pound Rebka only had the precision to measure the first order effect, thus it is considered a test of the principle of equivalence rather than test for tidal gravity.

 

[timmdeeg]

In the case of time dilation between the front and back of a rocket, indeed the effect is not due to curvature. In rocket centered coordinates you can say it is due to pseudogravity. In an inertial frame it is due to the fact that by the time a signal from the back of an accelerating rocket reaches the front, the front is moving faster than the back was at emission time, thus nothing but ordinary Doppler is involved.

Ok, thanks.

 

[RockyMarciano]

But why introduce the additional confusion of considering tidal gravity as the true trademark of GR's curved spacetime? Many solutions of the EFE have vanishing Weyl curvature and no tidal gravity.

On the other hand in the close neighborhood of a black hole, the geometry close to the event horizon can be described in Rindler coordinates so that should include tidal effects, right?

 

[PAllen]

But why introduce the additional confusion of considering tidal gravity as the true trademark of GR's curved spacetime? Many solutions of the EFE have vanishing Weyl curvature and no tidal gravity.

On the other hand in the close neighborhood of a black hole, the geometry close to the event horizon can be described in Rindler coordinates so that should include tidal effects, right?

The original question was whether gravitational time dilation implies curvature. Since this is a vacuum measurement, Weyl curvature is the only thing relevant. It is also trivially obvious that two measurements of such time dilation in different positions can measure curvature. The debate is over an argument that purports, from one measurement, to imply curvature. Peter presented that the argument would also work in flat spacetime for Rindler observers, so how could it be valid? I have argued that, indeed, the argument as given, is invalid, and cannot really be repaired.

The geometry near an event horizon is only described by Rindler coordinates and associated metric approximately. The degree of approximation is precisely the scale at which tidal gravity can be ignored, as there is NO tidal gravity in the Rindler metric. Its full curvature tensor is zero.

 

[RockyMarciano]

The original question was whether gravitational time dilation implies curvature. Since this is a vacuum measurement, Weyl curvature is the only thing relevant. It is also trivially obvious that two measurements of such time dilation in different positions can measure curvature. The debate is over an argument that purports, from one measurement, to imply curvature. Peter presented that the argument would also work in flat spacetime for Rindler observers, so how could it be valid? I have argued that, indeed, the argument as given, is invalid, and cannot really be repaired.

The geometry near an event horizon is only described by Rindler coordinates and associated metric approximately. The degree of approximation is precisely the scale at which tidal gravity can be ignored, as there is NO tidal gravity in the Rindler metric. Its full curvature tensor is zero.

My first point was that FRW solutions for instance have NO tidal gravity, and their curvature tensor isn't zero. So that's why I wanted to clarify that spacetime curvature is not the same as tidal gravity IN GENERAL, as implied in a post above just in case somebody might get that impression.

Now, as you say here we are restricting to the vacuum case so that's why I wrote the second part of my post. You are right that the coordinates only describe the geometry near the event horizon approximately, but that doesn't prevent one from performing higher order relativistic expansions using the mass source as a perturbative field.

 

[RockyMarciano]

By the way, one of the authors of the book MTW that originated this thread, Thorne, claims that there is no physical distinction between flat spacetime with a gravitational field and curved spacetime, just two representations of the same thing, so I guess for him this discussion would not make much sense.

 

[PeterDonis]

If the rope which martinbn mentioned isn't stretched, doesn't that imply the absence of tidal gravity

No, since you can have a Born rigid congruence of uniformly accelerated observers in both flat and curved spacetime.

 

[PeterDonis]

gravitational time dilation does not require curvature as proved by the fact that the defining experiment detecting it did not detect anything about curvature

This isn't a valid argument; an experiment detecting A but not B does not prove that A does not imply B. It just shows that, if A implies B, the implication must involve some condition that wasn't met in that particular experiment.

the world lines with proper acceleration maintain static distance from an inertial body (earth)

Hmm--yes, this argument could be made rigorous by observing that the center of the Earth is moving inertially, even though its surface is not--in fact, the existence of a gravitating body with this property would be sufficient to show spacetime curvature, even without looking at time dilation at all. Plus all this can be evaluated locally, or at least over a reasonably small finite region, without having to drag in any observers at rest at infinity or any normalizations of Killing vector fields.

I remain convinced that your OP establishes that the claim that gravitational time dilation per se ( and the parallelogram argument) establishes curvature is simply false.

I think I'm convinced as well, but since we're basically saying that MTW got something wrong, I want to make sure I've considered every possible avenue by which a counterargument might be made.

 

[PeterDonis]

It actually is global not local, because it is uniquely determined globally.

What does "uniquely determined globally" mean?

 

[PeterDonis]

Many solutions of the EFE have vanishing Weyl curvature and no tidal gravity.

Only if you define "tidal gravity" to only mean Weyl curvature. That is not the only possible definition. Many textbooks, including MTW, clearly define tidal gravity as spacetime curvature, period, i.e., a nonzero Riemann tensor. In other words, tidal gravity is defined as geodesic deviation, period--initially parallel geodesics don't stay parallel (and more generally geodesics don't maintain the same rate of convergence or divergence). That happens in FRW spacetime just as well as in a spacetime with Weyl curvature such as Schwarzschild spacetime.

in the close neighborhood of a black hole, the geometry close to the event horizon can be described in Rindler coordinates so that should include tidal effects, right?

No, because "close to the event horizon" means "close enough that tidal gravity is not observable". That's the necessary condition for the Rindler approximation to apply.

 

[martinbn]

This may be relevant. It makes the same point as PeterDonis and PAllen.

Does a gravitational red shift necessarily imply space‐time curvature?

Abstract: Schild’ has proposed a heuristic agrument which attempts, to show that any gravitational red shift requires that the geometry of space−time be curved. It is our intention to show that this argument is fallacious and we believe that no argument which attempts to infer space−time curvature solely from the gravitational red shift can be valid.
​

 

[Ibix]

I've been having a look at MTW and it's not entirely clear to me that the claim that Schild is making is quite as general as seems to be under discussion here. It seems to me that Schild isn't claiming that you can't have time dilation without curvature. Rather, he's claiming that you can't have a forcefield on a Minkowski background that induces time dilation without having curvature.

Basically I think he's excluding the Rindler case from consideration by having his observers under thrust at a fixed distance from a planet, which they can verify by radar with no knowledge of how gravity works beyond the time symmetry he assumes.

I can only see the abstract of the paper that @martinbn linked, so no comment.

 

[MikeGomez]

By the way, one of the authors of the book MTW that originated this thread, Thorne, claims that there is no physical distinction between flat spacetime with a gravitational field and curved spacetime, just two representations of the same thing, so I guess for him this discussion would not make much sense.

I believe that is in agreement with Einstein. Can you please provide the exact content of what Thorne said. MTW is not accessible to me.

 

[MikeGomez]

Defining space-time curvature as tidal gravity does seem to me to exclude Newtonian gravity (gravity in an accelerated elevator) as space-time curvature, if the rules for detecting tidal effects only permit comparing the paths of two or more objects (i.e. diverging or converging geodesics, or parallelogram tests). If, on the other hand, gravitational time dilation is recognized as a tidal effect that is detectable considering only the path of a single object, then there is consistency between the phrases “tidal gravity” and “space-time curvature” for all cases.

People seem to have a propensity for wanting to make a distinction between gravity which can be transformed away, and gravity which can not. Einstein defended the equivalence principle when Richenbacher tried to use that same (false) distinction to invalidate the EP.

In a letter Einstein wrote in reply to Reichenbacher http://einsteinpapers.press.princeton.edu/vol7-trans/220

“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. He (like Herr Kottler) raises the objection against the principle of equivalence that gravitational fields for finite space-time domains in general cannot be transformed away. He fails to see that this is of no importance whatsoever. What is important is only that one is justified at any instant and at will (depending upon the choice of a system of reference) to explain the mechanical behavior of a material point either by gravitation or by inertia. More is not needed; to achieve the essential equivalence of inertia and gravitation it is not necessary that the mechanical behavior of two or more masses must be explainable as a mere effect of inertia by the same choice of coordinates. After all, nobody denies, for example, that the theory of special relativity does justice to the nature of uniform motion, even though it cannot transform all acceleration-free bodies together to a state of rest by one and the same choice of coordinates.” - Albert Einstein

 

[PeterDonis]

one of the authors of the book MTW that originated this thread, Thorne, claims that there is no physical distinction between flat spacetime with a gravitational field and curved spacetime, just two representations of the same thing

Please give a reference. I strongly suspect you are getting this from a layman's book, not a textbook or peer-reviewed paper.

MTW is not accessible to me.

And it makes no claim of the sort RockyMarciano made anyway.

 

[PeterDonis]

I think he's excluding the Rindler case from consideration by having his observers under thrust at a fixed distance from a planet, which they can verify by radar

Yes, but in his formulation (at least as it's described in MTW), the radar signals are exchanged with an observer at rest at infinity, which is open to PAllen's objection that it should not require a global property to determine whether there is spacetime curvature or not.

 

[PeterDonis]

Defining space-time curvature as tidal gravity does seem to me to exclude Newtonian gravity (gravity in an accelerated elevator) as space-time curvature

You should look up the Cartan formulation of Newtonian gravity using curved geometry.

If, on the other hand, gravitational time dilation is recognized as a tidal effect

It isn't.

 

[PeterDonis]

This may be relevant.

Ah, so someone did spot this before. Unfortunately it's behind a paywall and it doesn't seem like a preprint is on arxiv.

 

[Geometry_dude]

The difference in rate of time flow between two observers with the same proper acceleration but slightly different heights (where "height" is "distance from some reference point along the direction of proper acceleration"). Or, equivalently, the redshift of light signals sent from the lower observer to the higher one.

Well, that's a very restricted definition - I really don't see why you would put the word gravitational in there. An important question would also be how you define that `distance' - even though there's a more or less straightforward answer in SR. As soon as you have curvature, however, this notion of `distance' gets physically problematic.

Spacetime curvature is tidal gravity. But the term "gravity" is more general than that.

No, I don't agree with this statement. By the Einstein equivalence principle, gravity is not detectable on sufficiently small spatio-temporal scales. The mathematical justification in GR for this is the existence of Lorentzian normal coordinates at each point. Minkowski spacetime is then viewed as the tangent space approximation to the curved spacetime at the point.
Yet, as in the case above, we'd be arguing about terminology and there is not much point in that.

It is important to keep in mind that the situation they appear to have in mind ("constant gravitational field") is actually adequately described by uniformly accelerated observers in Minkowski spacetime

Is it? That's the question.

Well, that's an excellent question to ask, because I don't think it has been adequately shown yet and I have looked quite deep into the literature. Actually, I have developed a mathematical formalism to give a clear-cut-mathematical answer to these issues in my master's thesis. It's not publicly available yet, because it's still being graded. But I can send it to you with a short explanation of how to prove or disprove it, if you pass me a PM with your email.

 

[RockyMarciano]

I believe that is in agreement with Einstein. Can you please provide the exact content of what Thorne said. MTW is not accessible to me.

If you get to consult MTW the relevant pages are 424-425, section 5 of box 17.2, and references therein. See also box 18.1.
It's also known as "curvature without curvature" view.

Please give a reference. I strongly suspect you are getting this from a layman's book, not a textbook or peer-reviewed paper..

There is indeed a layman's book(""Black holes and time warps") where he explains how the two views explained in MTW are routinely and interchangeably used by physicists(pages 400-402).
You'll see that some assertions from textbooks and peer-reviewed papers make it to general audience books, and that fact doesn't invalidate them as a matter of fact.
Besides the textbook references there are peer-reviewed papers by Thorne et al. such as: "Foundations for a theory of gravitation" 1973 Physical Review D7 3563-78

And it makes no claim of the sort RockyMarciano made anyway.

It does. See above.

 

[RockyMarciano]

I've been having a look at MTW and it's not entirely clear to me that the claim that Schild is making is quite as general as seems to be under discussion here. It seems to me that Schild isn't claiming that you can't have time dilation without curvature. Rather, he's claiming that you can't have a forcefield on a Minkowski background that induces time dilation without having curvature.

I agree, I think the argument by Schild is misinterpreted in the OP.

Basically I think he's excluding the Rindler case from consideration by having his observers under thrust at a fixed distance from a planet, which they can verify by radar with no knowledge of how gravity works beyond the time symmetry he assumes.

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.

 

[RockyMarciano]

But that has nothing to do with with the argument that gravitational dilation implies curvature. Again, curvature is local not global, and gravitational time dilation does not require curvature as proved by the fact that the defining experiment detecting it did not detect anything about curvature.

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.