Does gravitational time dilation imply spacetime curvature?

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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.]

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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).

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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.

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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
Staff Emeritus
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.

Mentor
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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
Staff Emeritus
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.

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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
Staff Emeritus
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
Staff Emeritus
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.

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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.

martinbn and PAllen
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
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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 arguement (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.

PeterDonis
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 arguement (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
Mentor
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a much better presentation would be...

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

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.)

MikeGomez
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.

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.

MikeGomez
PAllen