A Does gravitational time dilation imply spacetime curvature?

[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 (z,t) 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).

 

[PAllen]

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).

Reading the argument as presented in MTW a bit generously, I think the following can be said:

1) They posit a theory where the Minkowski metric is the observable metric for distance and time measurements.This is pretty clearly stated.

2) They also state free fall paths and light paths near a gravitating body are not geodesics. Gravity is governed by a field of unspecified nature that does not change observable geometry.

3) They posit it is possible to set up global Lorentz frame physically using a described procedure.

4) Though not clearly stated, the implication is that straight lines in the global Lorentz frame are Minkowski geodesics. I don't see any other reasonable way to read their argument.

5) The figure they set up has two parallel straight (geodesic) timelike sides and two congruent, possibly curved sides that need not be geometrically parallel.

6) They then note that allowing gravitational time dilation leads to a contradiction.

Thus the assumptions must be changed. With a fair amount of unstated reasoning, I think you could get to the conclusion that the observable geometry must be a curved pseudoriemannian manifold. Of course, I think there are other routes to this conclusion that are much more straightfowrd.

The interesting thing isn't the claimed proof of curvature (which is incomplete as given), but the concnclusion that gravity as a field theory of SR cannot accommodate gravitational time dilation between static observers (as long as the Minkowsi metric remains the observable metric).

[edit: a lot rests on the argumentation that the static observers would have to be straight lines in minkowski geometry if gravity is a field theory on SR (with SR geometry being the observable geometry). The argument stands or falls on how well the case for this is made.]

 

[PeterDonis]

Though not clearly stated, the implication is that straight lines in the global Lorentz frame are Minkowski geodesics. I don't see any other reasonable way to read there argument.

I agree.

The figure they set up has two parallel straight (geodesic) timelike sides and two congruent, possibly curved sides that need not be geometrically parallel.

I think that the two possibly curved sides do need to be geometrically parallel in order for the "parallelogram" argument, which is key to deriving the contradiction, to work. If the possibly curved sides are allowed not to be parallel, then there is no contradiction: the quadrilateral does not have to be a parallelogram, or the equivalent of one with curved sides, and therefore there is no issue with the two straight sides not being the same length.

If we assume an observable global Minkowski metric, then I don't think there's an issue with the two possibly curved sides being parallel, since one is the time translate of the other along a geodesic congruence. My real issue is with the assumption that there can be an observable global Minkowski metric which has geodesics that are not free-fall paths. In other words, I think a contradiction can be derived just from items 1 through 4 in your list, before we even get to constructing the specific quadrilateral that forms the basis of Schild's argument.

 

[PAllen]

I think that the two possibly curved sides do need to be geometrically parallel in order for the "parallelogram" argument, which is key to deriving the contradiction, to work. If the possibly curved sides are allowed not to be parallel, then there is no contradiction: the quadrilateral does not have to be a parallelogram, or the equivalent of one with curved sides, and therefore there is no issue with the two straight sides not being the same length.

You are missing a (subtle?) geometric point I made earlier. If you congruently translate a curve along a straight line in a flat plane, the two curves are generally not parallel, but that is not relevant; it is the congruence supports conclusion of the opposite straight side being equal. Requiring parallel curves would make them unequal.

I restate the example I gave earlier.

Translate a circular arc along a straight line. You find that mutually orthogonal distances between the two curves are not constant. What would have constant such distance and be parallel are concentric circular arcs. But concentric circular arcs would lead to the opposite parallel straight sides being unequal in length in most cases; while the congruent (translated) arcs would guarantee the opposite straight sides being equal. We all spent a lot of confusion assuming congruent meant parallel, when it is commonly mutually exclusive with parallel (for curves).

 

[PeterDonis]

Translate a circular arc along a straight line. You find that mutually orthogonal distances between the two curves are not constant.

If I'm imagining this right, the "mutually orthogonal distances" are not the ones that matter for Schild's argument. Schild's construction corresponds to this: take two parallel lines in a flat plane, and an arc that has one endpoint on each line. Translate the arc along the lines, maintaining the constraint that one endpoint is on each line. Compare the translated arc to the original arc.

It is true that, if you take any point on the translated arc, and draw a straight line through that point orthogonal to the arc at that point, when it intersects the original arc, it will in general not be orthogonal to the original arc. So the arcs will not be "parallel" in that sense. But that's not the sense that matters for Schild's argument.

The sense that matters for Schild's argument--more specifically for the "parallelogram" reasoning to hold--is this: affinely parameterize the original arc so that the endpoint on the lower curve is $\lambda = 0$ and the endpoint on the upper curve is $\lambda = 1$. Keep the parameterization fixed as you translate the arc. Then the distance between pairs of points on the two arcs that have the same value of $\lambda$ is constant along the arcs. This corresponds to drawing a congruence of parallel lines filling the space between the two original lines (the lower and upper observers' worldlines in Schild's construction) and seeing that the distance between the arcs along each such line is the same.

 

[PAllen]

If I'm imagining this right, the "mutually orthogonal distances" are not the ones that matter for Schild's argument. Schild's construction corresponds to this: take two parallel lines in a flat plane, and an arc that has one endpoint on each line. Translate the arc along the lines, maintaining the constraint that one endpoint is on each line. Compare the translated arc to the original arc.

It is true that, if you take any point on the translated arc, and draw a straight line through that point orthogonal to the arc at that point, when it intersects the original arc, it will in general not be orthogonal to the original arc. So the arcs will not be "parallel" in that sense. But that's not the sense that matters for Schild's argument.

The sense that matters for Schild's argument--more specifically for the "parallelogram" reasoning to hold--is this: affinely parameterize the original arc so that the endpoint on the lower curve is $\lambda = 0$ and the endpoint on the upper curve is $\lambda = 1$. Keep the parameterization fixed as you translate the arc. Then the distance between pairs of points on the two arcs that have the same value of $\lambda$ is constant along the arcs. This corresponds to drawing a congruence of parallel lines filling the space between the two original lines (the lower and upper observers' worldlines in Schild's construction) and seeing that the distance between the arcs along each such line is the same.

At this point, we are quibbling about words. Note that MTW does not use the world parallelogram, nor make any claim the light curves are parallel. They specifically use congruent and same shape. Since parallel curves have a well defined meaning, why insist on an alternate definition? What we need for Schild's argument to succeed is congruent translation not parallelism as it is normally applied to curves.

 

[PeterDonis]

What we need for Schild's argument to succeed is congruent translation not parallelism as it is normally applied to curves.

Hm, yes, I was using the word "parallel" in the wrong sense before. I agree that "congruent translation" is a better term.

 

[stevendaryl]

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).

I think the point may have been that if you considered gravity to be a force, rather than spacetime curvature, then gravity wouldn't affect whether or not someone was traveling a geodesic.

 

[PeterDonis]

I think the point may have been that if you considered gravity to be a force, rather than spacetime curvature, then gravity wouldn't affect whether or not someone was traveling a geodesic.

But that doesn't match the SR definition of "geodesic"--at least not the local one. That definition is "zero path curvature", which for a timelike worldline means "zero proper acceleration". That's one of the reasons I'm not sure there is a consistent theory that meets all the requirements for formulating Schild's argument.

 

[stevendaryl]

But that doesn't match the SR definition of "geodesic"--at least not the local one. That definition is "zero path curvature", which for a timelike worldline means "zero proper acceleration". That's one of the reasons I'm not sure there is a consistent theory that meets all the requirements for formulating Schild's argument.

How are you defining "zero proper acceleration"? If you're defining it to be freefall, then that's already assuming that gravity is not a force.

The equations of motion for a particle in a gravitational field can be written, in the nonrelativistic limit as:

m \frac{d^2 z}{dt^2} = -mg + F_{non-grav}

For zero acceleration, do you mean F_{non-grav} = 0, or do you mean -mg + F_{non-grav} = 0?

If gravity is an ordinary force, then you would mean the latter, and somebody at "rest" on the surface of a planet is non-accelerating.

 

[PeterDonis]

How are you defining "zero proper acceleration"? If you're defining it to be freefall

That's the only local way I know of to define it. Remember I'm specifically looking for a local definition (in response to @Denis who has claimed that there is one). I'm well aware that you can construct a non-local definition by just labeling any worldline that is at rest with respect to observers far away as having "zero acceleration" and therefore being a geodesic. That's my understanding of what Schild is doing.

 

[stevendaryl]

That's the only local way I know of to define it.

In light of the equivalence principle, it's the only sensible way to define it. But the whole point is whether gravity can be an ordinary force, as opposed to a manifestation of spacetime curvature. If it were an ordinary force, then the equivalence principle might be false, because there would not necessarily be any gravitational time dilation. I see it as: Gravitational time dilation is support for the equivalence principle, which is support for a geometric view of gravity.

 

[stevendaryl]

In light of the equivalence principle, it's the only sensible way to define it. But the whole point is whether gravity can be an ordinary force, as opposed to a manifestation of spacetime curvature. If it were an ordinary force, then the equivalence principle might be false, because there would not necessarily be any gravitational time dilation. I see it as: Gravitational time dilation is support for the equivalence principle, which is support for a geometric view of gravity.

Once you've heard the idea of the equivalence principle, it's hard to "unhear it" and it's hard to imagine that anyone would ever have thought that freefall was noninertial motion. But as far as I know, nobody actually thought of freefall as inertial motion until Einstein in the 20th century.

 

[PeterDonis]

In light of the equivalence principle, it's the only sensible way to define it.

The local definition, in itself, has nothing to do with the equivalence principle. It has to do with the physical meaning of the metric in SR. Locally, if I want to set up an SR inertial frame, the only way I have to figure out which curves are the "straight lines"--curves of constant $x, y, z$--in that frame is to use freely falling worldlines. Those are the only ones that are locally picked out physically. So if I am restricted to using local measurements only, using the freely falling worldlines as the geodesics of the metric is the only option.

So if I want to construct a theory that says "the metric is Minkowski", but picks out different curves as the "straight lines"--curves which are not freely falling worldlines--then the only way I can pick out which curves these are is to use some non-local criterion. In Schild's case, the criterion is to pick the worldlines that are "at rest" with respect to observers very far away, as verified by round-trip light signals. But there's no local way to tell which worldlines those are; there's no local way to say, the worldline with this particular proper acceleration is the "straight line" in this particular local region of spacetime. Only the nonlocal measurement can tell us that.

The equivalence principle amounts to the further claim, in the light of the above, that we should not use any such nonlocal criterion at all--we should insist on only using local measurements to pick out the "straight lines" (geodesics) of the metric. But I'm not saying that here. I'm only saying that, if we are going to say the metric is Minkowski but have some "straight lines" that are not freely falling worldlines--which we must do in the presence of gravity--then we have to use a nonlocal criterion to pick out which worldlines are the "straight lines", because there is no local way to do it.

 

[PAllen]

The local definition, in itself, has nothing to do with the equivalence principle. It has to do with the physical meaning of the metric in SR. Locally, if I want to set up an SR inertial frame, the only way I have to figure out which curves are the "straight lines"--curves of constant $x, y, z$--in that frame is to use freely falling worldlines. Those are the only ones that are locally picked out physically. So if I am restricted to using local measurements only, using the freely falling worldlines as the geodesics of the metric is the only option.

So if I want to construct a theory that says "the metric is Minkowski", but picks out different curves as the "straight lines"--curves which are not freely falling worldlines--then the only way I can pick out which curves these are is to use some non-local criterion. In Schild's case, the criterion is to pick the worldlines that are "at rest" with respect to observers very far away, as verified by round-trip light signals. But there's no local way to tell which worldlines those are; there's no local way to say, the worldline with this particular proper acceleration is the "straight line" in this particular local region of spacetime. Only the nonlocal measurement can tell us that.

The equivalence principle amounts to the further claim, in the light of the above, that we should not use any such nonlocal criterion at all--we should insist on only using local measurements to pick out the "straight lines" (geodesics) of the metric. But I'm not saying that here. I'm only saying that, if we are going to say the metric is Minkowski but have some "straight lines" that are not freely falling worldlines--which we must do in the presence of gravity--then we have to use a nonlocal criterion to pick out which worldlines are the "straight lines", because there is no local way to do it.

Now I will play a little devil's advocate. Suppose you want to locally know whether a charged body is following an inertial path (i.e. without knowing about distribution of all other charges and currents, and determining if they happen to cancel in a small region). You attach an accelerometer, or just compare to an uncharged body. Both depend on the existence of matter that doesn't couple to EM fields. While in GR as a geometric theory, you cannot define matter that doesn't couple to gravity, in a theory of gravity as a classical field in SR, there is no reason, in principle, you could not include weightless matter (totally violating the principle of uniform free fall, one aspect of the equivalence principle). That is, matter that simply doesn't couple to the gravitational field. Then, such matter readily allows determination of SR inertial paths locally in the presence of gravitating matter.

While the existence of such matter is optional, and not very plausible for such a theory intended to be serious (as were the many early attempts at accounting for gravity in SR), Schild's argument can be turned around to saying that all such theories must predict a violation of the principle of equivalence (between front to back redshift for a uniformly accelerating rocket vs no redshift for bottom to top of a tall building on earth). This is a more general critique of such theories than looking at specific predictions of specific such theories (of which there were both scalar and vector versions). MTW also explores the specific features of some of the proposed scalar and vector based SR gravity theories.

What Schild's argument doesn't do is measure curvature or even prove curvature must be present without additional argumentation that the only alternative to a pure SR based classical field theory is a metric theory with curvature.

 

[PeterDonis]

in SR, there is no reason, in principle, you could not include weightless matter (totally violating the principle of uniform free fall, one aspect of the equivalence principle). That is, matter that simply doesn't couple to the gravitational field. Then, such matter readily allows determination of SR inertial paths locally in the presence of gravitating matter.

Schild's argument can be turned around to saying that all such theories must predict a violation of the principle of equivalence (between front to back redshift for a uniformly accelerating rocket vs no redshift for bottom to top of a tall building on earth).

Hm, interesting. Just to explore this a bit further, presumably "weightless matter" in an SR gravity-as-a-field theory would not need to be held at a fixed altitude by some force (rocket, standing on the surface of a planet, etc.); it would just "float" at a fixed altitude, in free fall. So it could still be, logically speaking, that observers made of ordinary matter (which is not "weightless") could observe gravitational time dilation, because of their nonzero proper acceleration, while observers made of weightless matter would not. But that would violate the assumption of there being a spacetime metric in the first place (since the length of a timelike curve between two events would depend on whether the observer with that curve as his worldline was made of ordinary matter or weightless matter), which is required for an SR-based theory.

 

[PAllen]

Hm, interesting. Just to explore this a bit further, presumably "weightless matter" in an SR gravity-as-a-field theory would not need to be held at a fixed altitude by some force (rocket, standing on the surface of a planet, etc.); it would just "float" at a fixed altitude, in free fall. So it could still be, logically speaking, that observers made of ordinary matter (which is not "weightless") could observe gravitational time dilation, because of their nonzero proper acceleration, while observers made of weightless matter would not. But that would violate the assumption of there being a spacetime metric in the first place (since the length of a timelike curve between two events would depend on whether the observer with that curve as his worldline was made of ordinary matter or weightless matter), which is required for an SR-based theory.

No, that isn't what I was thinking. You've coupled my paragraphs in a way not intended. I brought up weightless matter only as a way to locally determine SR inertial paths. Just as with EM, you can still determine inertial paths without 'uncharged' matter using global operations, as we've discussed.

As an afterthought, I realized that (of course) weightless matter would be a direct violation of the principle of equivalence; but even without such matter, a different aspect of the POE must be violated in such a theory by Schild's argument. For this, all you need is some way to access SR inertial paths, it need not be local. That was the only point of my second paragraph - that I now best understand Schild's argument (and wish it were presented that way) as a proof that a pure SR classical field theory of gravity must violate the POE between acceleration far away from matter and being stationary in a gravitational field. This violation is first order, and local.

 

[PAllen]

No, that isn't what I was thinking. You've coupled my paragraphs in a way not intended. I brought up weightless matter only as a way to locally determine SR inertial paths. Just as with EM, you can still determine inertial paths without 'uncharged' matter using global operations, as we've discussed.

As an afterthought, I realized that (of course) weightless matter would be a direct violation of the principle of equivalence; but even without such matter, a different aspect of the POE must be violated in such a theory by Schild's argument. For this, all you need is some way to access SR inertial paths, it need not be local. That was the only point of my second paragraph - that I now best understand Schild's argument (and wish it were presented that way) as a proof that a pure SR classical field theory of gravity must violate the POE between acceleration far away from matter and being stationary in a gravitational field. This violation is first order, and local.

Yet a further thought on this is that there is funny tension between section 7.3 in MTW, which presents Schild's argument that gravitational time dilation implies curvature, and the very next section which discusses it in relation to the POE. There is some pussyfooting language about the seeming discrepancy (the POE deals all in locally flat spacetime physics).

IMO, with benefit of this very long thread, I would argue that a much better presentation would be to present gravitational time dilation via the POE. Then present Schild's argument recast in emphasis, to show how this rules out all pure SR theories of gravity (pure = the SR metric is the observable metric, there is no other) assuming the POE prediction is verified (as it was in the early 1960s).

 

[PeterDonis]

a much better presentation would be...

I don't suppose anyone on PF has Kip Thorne's contact information?

 

[Ibix]

I don't suppose anyone on PF has Kip Thorne's contact information?

https://www.its.caltech.edu/~kip/index.html/

His email address is mildly obfuscated so I won't post it.

 

[Adel Makram]

Once you've heard the idea of the equivalence principle, it's hard to "unhear it" and it's hard to imagine that anyone would ever have thought that freefall was noninertial motion. But as far as I know, nobody actually thought of freefall as inertial motion until Einstein in the 20th century.

Newton had already noticed this, and indeed he stated it, more or less, in Corollary VI to the laws of motion:
If bodies are moving in any way whatsoever with respect to one another and are urged by equal accelerative forces along parallel lines, they will all continue to move with respect to one another in the same way as they would if they were not acted on by those forces. (1726, p. 423.)

 

[PAllen]

Newton had already noticed this, and indeed he stated it, more or less, in Corollary VI to the laws of motion:
If bodies are moving in any way whatsoever with respect to one another and are urged by equal accelerative forces along parallel lines, they will all continue to move with respect to one another in the same way as they would if they were not acted on by those forces. (1726, p. 423.)

That in no way equates to the suggestion that such motion is inertial. In fact Newton argued for the notion of absolute rest, even though argued six ways from Sunday that you could never identify this state.

 

[Adel Makram]

That in no way equates to the suggestion that such motion is inertial. In fact Newton argued for the notion of absolute rest, even though argued six ways from Sunday that you could never identify this state.

When he mentioned " they will all continue to move with respect to one another in the same way as they would if they were not acted on by those forces., this means they move inertially.

 

[PAllen]

When he mentioned " they will all continue to move with respect to one another in the same way as they would if they were not acted on by those forces., this means they move inertially.

No it does not, as Newton meant inertial. You are looking back on it with today's understanding. If in a uniform elrctrric field, all objects had the same charge to mass ratio, would you say this makes their response inertial? Newton may have been intrigued but was very very far from considering this motion inertial.

 

[MikeGomez]

In fact Newton argued for the notion of absolute rest, even though argued six ways from Sunday that you could never identify this state.

You think that one of the great ones had a momentary lapse of reason here? Such is not the case. This is getting off topic from the OP, so I have created another thread here…

https://www.physicsforums.com/threads/Newton-on-absolute-motion.920783/

 

[stevendaryl]

So if I want to construct a theory that says "the metric is Minkowski", but picks out different curves as the "straight lines"--curves which are not freely falling worldlines--then the only way I can pick out which curves these are is to use some non-local criterion. In Schild's case, the criterion is to pick the worldlines that are "at rest" with respect to observers very far away, as verified by round-trip light signals. But there's no local way to tell which worldlines those are; there's no local way to say, the worldline with this particular proper acceleration is the "straight line" in this particular local region of spacetime. Only the nonlocal measurement can tell us that.

We have the claim:

Locally, there is no way to distinguish freefall from inertial motion

To me, that claim IS the equivalence principle. If the equivalence principle is false, then that claim is false.

For example, we have another criterion for inertial motion, which is that "An inertial path is one that maximizes proper time". It's conceivable that that would give a different answer as to what is an inertial path than freefall. GR says that freefall = inertial, but that's an empirical question. You can't assume it.

I'm only saying that, if we are going to say the metric is Minkowski but have some "straight lines" that are not freely falling worldlines--which we must do in the presence of gravity--then we have to use a nonlocal criterion to pick out which worldlines are the "straight lines", because there is no local way to do it.

Isn't maximizing proper time a local criterion?