Gravitational time dilatation and curved spacetime - follow up

[cianfa72]

TL;DR Summary

gravitation time dilatation as proof of curved spacetime

Hi,
starting from this very interesting thread

In a recent thread, the question came up of whether the presence of gravitational time dilation implies spacetime curvature. My answer in that thread was no

I'm still a bit confused about the conclusions.
The main point, as far as I can understand, is all about conditions for a quadrilateral to be considered a parallelogram.

My first basic doubt is: the concept of 'parallel' applies just to geodesic paths or in another words: does make any sense to talk about 'parallelism' between two non geodesic paths ?

 

[jbriggs444]

The main point, as far as I can understand, is all about conditions for a quadrilateral to be considered a parallelogram.

My first basic doubt is: the concept of 'parallel' applies just to geodesic paths or in another words: does make any sense to talk about 'parallelism' between two non geodesic paths ?

You will have to focus that question a great deal more tightly.

It could make sense to ask whether a pair of tangent vectors, one from a specific event on one world-line and one from a different event on a different world line are or are not pointing in the same direction. In order to make that determination, one would "parallel-transport" the one vector to the other and then compare them locally. In flat space-time, parallel transport always gives the same result. The tangent vectors either will be unambiguously parallel or not. In curved space-time, parallel transport gives a result that can depend on the path taken. It is ambiguous whether the tangent vectors are parallel.

But, offhand, asking whether two non-geodesic paths are "parallel" sounds a lot like asking whether two squiggles on a piece of paper are "parallel". Not very meaningful without a definition for "parallel".

 

[PeterDonis]

The main point, as far as I can understand, is all about conditions for a quadrilateral to be considered a parallelogram.

That was one question that was debated in the thread, yes. It was not the only one. But it's a fair question for this thread to be focused on.

does make any sense to talk about 'parallelism' between two non geodesic paths ?

It depends on your definition of "parallel". Multiple possible definitions were proposed in the previous thread. I think the end result was that there is not a single unique definition of "parallel".

 

[cianfa72]

That was one question that was debated in the thread, yes. It was not the only one. But it's a fair question for this thread to be focused on.

Coming back to Schild's argument and taking in account what you said there in post#3 we can limit ourselves to just geodesic paths (observers paths at fixed height $z_1$ and $z_2$ and light rays paths are actually geodesics). The two light rays emitted from ground ($z_1$) can be obtained one from the other 'parallel transporting' one of them along the ground observer path: thus they form a pair of 'parallel' geodesics.

What about the paths of the observers staying at $z_1$ and $z_2$ ? Here $z_1$ and $z_2$ can be actually taken as Schwarzschild radius $r_1$ and $r_2$ I believe...

 

[PeterDonis]

The two light rays emitted from ground ($z_1$) can be obtained one from the other 'parallel transporting' one of them along the ground observer path: thus they form a pair of 'parallel' geodesics.

More precisely, you parallel transport the 4-momentum of the light ray at emission along the ground observer path from the first emission event to the second; then you construct the second geodesic from the parallel transported 4-momentum. Yes, this is one definition of "parallel".

What about the paths of the observers staying at $z_1$ and $z_2$ ?

Neither of them are geodesics, if that's what you're asking. If you're asking something else, what is it?

Here $z_1$ and $z_2$ can be actually taken as Schwarzschild radius $r_1$ and $r_2$ I believe

Yes, if you mean observers at constant height.

 

[PeterDonis]

taking in account what you said there in post#3 we can limit ourselves to just geodesic paths

If you do that, you can't use observers at fixed heights, since those worldlines are not geodesics, as I said in my previous post just now. What I described in post #3 in the original thread was an alternative formulation where both of the observers are free-falling--that means each one has to start and end at a fixed height, but in between needs to be in free fall, so they need to be moving upward at the start and downward at the end, i.e., each one follows a parabolic arc that starts and ends at their chosen height, $z_1$ or $z_2$. But in that case, I did not see any definition of "parallel" according to which those two worldlines would be parallel in Schwarzschild spacetime for the entire time of flight; at best they would be parallel only at the instant when each observer is at the top of his arc.

 

[cianfa72]

But in that case, I did not see any definition of "parallel" according to which those two worldlines would be parallel in Schwarzschild spacetime for the entire time of flight; at best they would be parallel only at the instant when each observer is at the top of his arc.

Thus, in any case -- in the Schild original one as in your alternative formulation, the observer paths are never 'parallel'. So basically we are 'making' a quadrilateral with for sure just two parallel geodesics (the two light ray paths)

 

[PeterDonis]

the observer paths are never 'parallel'

It depends on your definition of "parallel". There are ways to define "parallel" that can be applied to non-geodesic curves. Or, as it appears Schild did based on where we ended up in that previous thread discussion, one can redefine "geodesic" based on some non-local criterion (Schild appears to have used the criterion of worldlines of observers who were at rest relative to observers at infinity, based on exchanging round-trip light signals).

 

[cianfa72]

Or, as it appears Schild did based on where we ended up in that previous thread discussion, one can redefine "geodesic" based on some non-local criterion (Schild appears to have used the criterion of worldlines of observers who were at rest relative to observers at infinity, based on exchanging round-trip light signals).

Thus, in that case of redefined 'geodesics', the two observer paths at fixed Schwarzschild radius $r_1$ and $r_2$ (or simply $z_1$ and $z_2$) should be considered as 'parallel' ?

 

[PeroK]

Thus, in any case -- in the Schild original one as in your alternative formulation, the observer paths are never 'parallel'. So basically we are 'making' a quadrilateral with for sure just two parallel geodesics (the two light ray paths)

"Parallel", "quadrilateral" etc. are really terms from Euclidean geometry. In mathematics, of course, you can generalise concepts like those. But, the generalised mathematical objects may not resemble the originals in the way you would like to imagine.

If you read about the history of non-Euclidean geometry you see a common mistake, made by competent mathematical minds of the time, in trying to prove Euclid's parallel postulate. They assumed that because a "straight line" obeyed Euclid's first four postulates it must be intuitively what we think of as a straight line.

It's the same here. If you define "line" and "parallel" within the context of non-Euclidean spacetime, you can't rely on these objects necessarily to resemble their Euclidean counterparts.

 

[PeterDonis]

in that case of redefined 'geodesics', the two observer paths at fixed Schwarzschild radius $r_1$ and $r_2$ (or simply $z_1$ and $z_2$) should be considered as 'parallel' ?

That is what Schild did in his argument, yes.

 

[cianfa72]

That is what Schild did in his argument, yes.

So, just to go to the conclusions, can we assume gravitation time dilatation, as in Schild experiment and his arguments, is a proof of spacetime curvature ?

 

[PeterDonis]

can we assume gravitation time dilatation, as in Schild experiment and his arguments, is a proof of spacetime curvature ?

Not "gravitational time dilation" by itself, since, as I noted at the very start of that thread, "gravitational time dilation" exists for Rindler observers in flat spacetime.

If "gravitational time dilation" is supplemented by Schild's non-local definition of "geodesics" for Schwarzschild spacetime--that observers following these worldlines are at rest relative to observers at infinity--then gravitational time dilation between such observers does require spacetime curvature, but, as noted in the thread, there are other simpler ways of showing the existence of spacetime curvature if you have such a scenario.

 

[Nugatory]

can we assume gravitation time dilatation, as in Schild experiment and his arguments, is a proof of spacetime curvature ?

It depends on exactly what you mean by "gravitational time dilation" - that's a label that we apply to various invariant phenomena such as the redshift observed in the Pound-Rebka experiment. Are you considering the analogous redshift observed between tail and nose of an accelerating spaceship to be an example of "gravitational time dilation"? If so, the answer to your question is no.

 

[PAllen]

It depends on exactly what you mean by "gravitational time dilation" - that's a label that we apply to various invariant phenomena such as the redshift observed in the Pound-Rebka experiment. Are you considering the analogous redshift observed between tail and nose of an accelerating spaceship to be an example of "gravitational time dilation"? If so, the answer to your question is no.

Actually, the Pound-Rebka experiment did not detect curvature, per se. It is considered, in modern parlance (See, e.g. Clifford Will's Living Review article on testing GR), a test of Local Position invariance, specifically that a local accelerated frame near Earth behaves the same as a local accelerated frame 'far away from everything' i.e. an SR Rindler frame. To have tested curvature, the precision would have needed to be at least 8 orders of magnitude larger than it was. Alternatively, performing the test at two widely separated locations can be taken to establish curvature.

 

[Nugatory]

Actually, the Pound-Rebka experiment did not detect curvature, per se.

You’re right, and I chose a poor example.

 

[cianfa72]

On thinking this over I'm actually not positive that it is true. It is true that the figure described in Schild's argument, the original one that uses the worldlines of "hovering" observers rather than my modified one with geodesics, is not a "parallelogram", strictly speaking; it's a figure with two curved opposite sides and two straight opposite sides. But I think the following should still be true: if spacetime were flat, then the two curved opposite sides should be congruent, and therefore of equal length, since the two straight opposite sides are.

This argument does depend on the straight opposite sides being parallel, and the definition of "parallel" is something we haven't touched on in this discussion. The implicit argument, at least as I understand it from MTW, is that the two straight opposite sides are parallel because the spacetime is static--or, to put it in more precise technical language, because the two curved opposite sides are both integral curves of a Killing vector field, so the second straight opposite side can be viewed as simply the time translate along the Killing flow of the first one.

Quoting this argument discussed in that thread, can you elaborate the technical point supporting the statement the two straight opposite side (first and second light pulse paths) being parallel ?

 

[PAllen]

You might want to study post #177 in that thread for a summary of the accumulated understanding developed int the thread.

 

[PeterDonis]

can you elaborate the technical point supporting the statement the two straight opposite side (first and second light pulse paths) being parallel ?

Heuristically, in a geometry with a Killing vector field, if you take a curve and translate it along the flow of the Killing vector field, the "shape" of the curve remains unchanged. In that sense of "parallel", the two light pulse paths in Schild's argument are parallel. However, as @PAllen pointed out in the thread, a better term would probably be "congruent": both pulse worldlines have the same shape and the same geometrical relationship to the two other sides (the sides that are worldlines of stationary observers at the two different altitudes). The latter is actually the key point being used in the argument.

 

[cianfa72]

Heuristically, in a geometry with a Killing vector field, if you take a curve and translate it along the flow of the Killing vector field, the "shape" of the curve remains unchanged. In that sense of "parallel", the two light pulse paths in Schild's argument are parallel. However, as @PAllen pointed out in the thread, a better term would probably be "congruent"

sure, got it.

both pulse worldlines have the same shape and the same geometrical relationship to the two other sides (the sides that are worldlines of stationary observers at the two different altitudes).

sorry, I need some help about timelike KVF you are talking about here. Is it just $\partial_t$ (because of static feature of Schwarzschild spacetime) or does it exist a timelike KVF for which wordlines at fixed Schwarzschild radius $r$ (Schild's argument observer wordlines) are orbits of ?

 

[PeterDonis]

I need some help about timelike KVF you are talking about here. Is it just $\partial_t$ (because of static feature of Schwarzschild spacetime) or does it exist a timelike KVF for which wordlines at fixed Schwarzschild radius $r$ (Schild's argument observer wordlines) are orbits of ?

Both. $\partial_t$ (in Schwarzschild coordinates) is the timelike Killing vector field (at least it's timelike outside the horizon, which is the only region we're concerned with here), and worldlines at fixed $r$ are orbits of it.

 

[cianfa72]

Both. $\partial_t$ (in Schwarzschild coordinates) is the timelike Killing vector field (at least it's timelike outside the horizon, which is the only region we're concerned with here), and worldlines at fixed $r$ are orbits of it.

ok, thus as you said before in this thread at post #5, parallel transporting 4-momentum vector of the light pulse at emission along the ground observer path from the first emission event to the second, we find that at the event of second emission the 'angle/scalar product' between light pulse 4-momentum vector and tangent vector of the ground observer path (both belonging to tangent space at second emission event) is the same as that between the correspondent ones at the event of first emission, right ?

 

[PeterDonis]

parallel transporting 4-momentum vector of the light pulse at emission along the ground observer path from the first emission event to the second, we find that at the event of second emission the 'angle/scalar product' between light pulse 4-momentum vector and tangent vector of the ground observer path (both belonging to tangent space at second emission event) is the same as that between the correspondent ones at the event of first emission, right ?

Yes.

 

[pervect]

ok, thus as you said before in this thread at post #5, parallel transporting 4-momentum vector of the light pulse at emission along the ground observer path from the first emission event to the second, we find that at the event of second emission the 'angle/scalar product' between light pulse 4-momentum vector and tangent vector of the ground observer path (both belonging to tangent space at second emission event) is the same as that between the correspondent ones at the event of first emission, right ?

I haven't been following this thread. Is the "ground observer path" a geodesic? By which I mean a space-time geodesic.

 

[PeterDonis]

Is the "ground observer path" a geodesic?

No. It's the worldline of a stationary observer in the gravitational field of a planet like the Earth. Thus it's an accelerated worldline, not a geodesic.

 

[PeterDonis]

parallel transporting

I should clarify this to forestall a question that might be coming from @pervect in view of the post of his that I just responded to. Since the ground observer's worldline is not a geodesic, the correct term for how the 4-momentum of emission of the light pulse is transported along that worldline is Fermi-Walker transport, not parallel transport. Or, more relevant to the current discussion, Lie transport (i.e., transport by the Lie derivative along the worldline), since that is what preserves the inner product of the 4-momentum with the Killing vector field.

 

[cianfa72]

I should clarify this to forestall a question that might be coming from @pervect in view of the post of his that I just responded to. Since the ground observer's worldline is not a geodesic, the correct term for how the 4-momentum of emission of the light pulse is transported along that worldline is Fermi-Walker transport, not parallel transport. Or, more relevant to the current discussion, Lie transport (i.e., transport by the Lie derivative along the worldline), since that is what preserves the inner product of the 4-momentum with the Killing vector field.

Thus 4-momentum of light pulse emission transported by Levi-Civita affine connection along observer wordlines at fixed $r$ (orbits of the flow of the killing vector $\partial_t$ in Schwarzschild coordinates) could be different from that due to Lie transport along the same observer worldlines ?

 

[PeterDonis]

4-momentum of light pulse emission transported by Levi-Civita affine connection along observer wordlines at fixed $r$ (orbits of the flow of the killing vector $\partial_t$ in Schwarzschild coordinates) could be different from that due to Lie transport along the same observer worldlines ?

First, a clarification: if you have a non-geodesic worldline, "parallel transport" ("transport by the Levi-Civita connection") along it makes no sense. You have to take into account the path curvature of the worldline. Fermi-Walker transport is how you do that.

Second, "Lie transport" along an integral curve of a Killing vector field is the same as Fermi-Walker transport along that curve. The term "Lie transport" just makes clearer why it is relevant for this particular problem--heuristically, because it maintains the "geometric shape". Fermi-Walker transport in general makes no guarantee about that.

 

[PAllen]

First, a clarification: if you have a non-geodesic worldline, "parallel transport" ("transport by the Levi-Civita connection") along it makes no sense. You have to take into account the path curvature of the worldline. Fermi-Walker transport is how you do that.

Could you explain more? Parallel transport is well defined along any path. Specifically, transport around a closed loop (in the limit of ever smaller loops) is how curvature is defined.

 

[PeterDonis]

Parallel transport is well defined along any path.

Yes, I was sloppy in my wording. Let me try to make it more precise. I was thinking of the discussion of Fermi-Walker transport in Chapter 7 of MTW.

Suppose we have an orthonormal tetrad $e_0, e_1, e_2, e_3$ of basis vectors at some event, such that the timelike vector $e_0$ coincides with the 4-velocity of some observer whose worldline passes through that event. We want to figure out how to transport that tetrad along the observer's worldline so that the transported $e_0$ remains coincident with the observer's 4-velocity everywhere along the worldline. Or, equivalently, $e_0$ starts out tangent to the worldline and we want to transport the tetrad so $e_0$ remains tangent to the worldline and the tetrad remains orthonormal.

If the worldline is a geodesic, parallel transport does the job.

If the worldline is not a geodesic, however, parallel transport does not do the job. The parallel transported $e_0$ will not remain tangent to the worldline. So we need to correct for that. We do that, as MTW describes, by adding to our transport law a Lorentz boost at each event that rotates, in spacetime, the parallel transported $e_0$ so that it is tangent to the worldline at that event. (At the original event, where $e_0$ was already tangent to the worldline, this boost just becomes the identity.) This boost will also rotate the spacelike vectors of the tetrad so that they remain orthogonal to $e_0$--parallel transport preserves orthogonality, so the parallel transported spacelike vectors will be orthogonal to the parallel transported $e_0$, and the Lorentz boost also preserves orthogonality, so the boosted vectors will still be mutually orthogonal. The result of this entire process is termed Fermi-Walker transport.

If we are talking about transporting vectors along a Killing flow, as in this discussion, Fermi-Walker transport is the transport law that is equivalent to Lie transport along the Killing flow. So that's the one we are interested in for this discussion.

transport around a closed loop (in the limit of ever smaller loops) is how curvature is defined.

As I understand it, curvature is defined in terms of parallel transport around a closed geodesic loop. You can't use an arbitrary loop. Since parallel transport in curved spacetime is path dependent, it would not make sense to allow a loop composed of arbitrary curves, since that would make curvature non-unique.

 

[PAllen]

Yes, I was sloppy in my wording. Let me try to make it more precise. I was thinking of the discussion of Fermi-Walker transport in Chapter 7 of MTW.

Suppose we have an orthonormal tetrad $e_0, e_1, e_2, e_3$ of basis vectors at some event, such that the timelike vector $e_0$ coincides with the 4-velocity of some observer whose worldline passes through that event. We want to figure out how to transport that tetrad along the observer's worldline so that the transported $e_0$ remains coincident with the observer's 4-velocity everywhere along the worldline. Or, equivalently, $e_0$ starts out tangent to the worldline and we want to transport the tetrad so $e_0$ remains tangent to the worldline and the tetrad remains orthonormal.

If the worldline is a geodesic, parallel transport does the job.

If the worldline is not a geodesic, however, parallel transport does not do the job. The parallel transported $e_0$ will not remain tangent to the worldline. So we need to correct for that. We do that, as MTW describes, by adding to our transport law a Lorentz boost at each event that rotates, in spacetime, the parallel transported $e_0$ so that it is tangent to the worldline at that event. (At the original event, where $e_0$ was already tangent to the worldline, this boost just becomes the identity.) This boost will also rotate the spacelike vectors of the tetrad so that they remain orthogonal to $e_0$--parallel transport preserves orthogonality, so the parallel transported spacelike vectors will be orthogonal to the parallel transported $e_0$, and the Lorentz boost also preserves orthogonality, so the boosted vectors will still be mutually orthogonal. The result of this entire process is termed Fermi-Walker transport.

If we are talking about transporting vectors along a Killing flow, as in this discussion, Fermi-Walker transport is the transport law that is equivalent to Lie transport along the Killing flow. So that's the one we are interested in for this discussion.

Fine, I summarize my understanding as:

Fermi-Walker transport preserves angle between vector and curve, and defines 'non-rotation'.

Parallel transport preserves direction and magnitude of the vector (magnitude only relevant given metric), relative to itself.

Both rely on a connection to define what vectors are the same at 'nearby' points.

As I understand it, curvature is defined in terms of parallel transport around a closed geodesic loop. You can't use an arbitrary loop. Since parallel transport in curved spacetime is path dependent, it would not make sense to allow a loop composed of arbitrary curves, since that would make curvature non-unique.

There is no such thing as a geodesic loop, in general. Geodesics very rarely form loops. A sequence of segments of different geodesics forming a loop has no special properties. Especially, note that there uncountably infinite choices for such a construction. It is only the limiting process that makes the result unique.

 

[PeterDonis]

There is no such thing as a geodesic loop

To be clear, I didn't mean a closed loop formed by a single geodesic. I meant a closed loop formed by segments of multiple geodesics. The segments have to be geodesic segments because that is the only way to uniquely define a curve starting from a given event and with tangent vector at that event specified by a given vector. See, for example, MTW section 11.4.

 

[PeterDonis]

See, for example, MTW section 11.4.

Ah, I just looked at MTW Box 11.7, which generalizes the treatment of section 11.4 to small closed curves of arbitrary shape. So I was wrong that the general definition of curvature is limited to a loop formed of geodesic segments.

 

[PAllen]

Ah, I just looked at MTW Box 11.7, which generalizes the treatment of section 11.4 to small closed curves of arbitrary shape. So I was wrong that the general definition of curvature is limited to a loop formed of geodesic segments.

Saved me from summarizing the treatment in Synge and Schild.

 

[vanhees71]

As I understand it, curvature is defined in terms of parallel transport around a closed geodesic loop. You can't use an arbitrary loop. Since parallel transport in curved spacetime is path dependent, it would not make sense to allow a loop composed of arbitrary curves, since that would make curvature non-unique.

The Riemann curvature tensor is geometrically defined by parallel transport around an arbitrary "inifinitesimal" closed curve in your manifold. You don't need to make a closed loop of geodesics.

The construction is as follows: Consider covariant vector components. Then the change of this vector when parallel transporting it along an infinitesimal closed loop $\delta C$ is
$$\Delta A_{\mu} = \int_{\delta C} \mathrm{d} x^{\alpha} {\Gamma^{\beta}}_{\mu \alpha} A_{\beta}.$$
Now use Stoke's integral theorem. With the surface-elements $\delta f^{\mu \nu}$ of a surface whose boundary $\delta C$ you get
$$\Delta A_{\mu} = \frac{1}{2} \left [\partial_{\gamma} \left ({\Gamma^{\beta}}_{\mu \alpha} \right )-\partial_{\alpha} \left ({\Gamma^{\beta}}_{\mu \gamma} \right) \right] \delta f^{\gamma \alpha}.$$
Now up to higher-order corrections we have
$$\partial_{\alpha} A_{\beta}={\Gamma^{\gamma}}_{\alpha \beta} A_{\gamma},$$
and after some algebra you get
$$\Delta A_{\mu}=\frac{1}{2} {R^{\alpha}}_{\mu \beta \gamma} A_{\alpha} \delta f^{\beta \gamma}$$
with
$${R^{\alpha}}_{\mu \beta \gamma} = \partial_{\beta} {\Gamma^{\alpha}}_{\mu \gamma} - \partial_{\gamma} {\Gamma^{\alpha}}_{\mu \beta} + {\Gamma^{\alpha}}_{\delta \beta} {\Gamma^{\delta}}_{\mu \gamma} - {\Gamma^{\alpha}}_{\delta \gamma} {\Gamma^{\delta}}_{\mu \beta},$$
which are the components of the Riemann curvature tensor.