Undergrad About global inertial frames in GR

[Demystifier]

Summary:: About the definition to use for a global inertial frame in GR

take a family of free-falling bodies (a geodesic congruence) foliating it: if we choose a frame (coordinate chart) in which those free-falling bodies are at rest

You can take such a family of bodies and define a basis (tetrad) based on that, but it will be a non-coordinate basis so you cannot chose a coordinate chart based on that. See e.g. the last appendix in the Carroll's book.

 

[PAllen]

You can take such a family of bodies and define a basis (tetrad) based on that, but it will be a non-coordinate basis so you cannot chose a coordinate chart based on that. See e.g. the last appendix in the Carroll's book.

I’m confused at what you are claiming. Lemaitre coordinates for Schwarzschild geometry are synchronous coordinates in which the congruence of radial infalling test bodies from infinity are at coordinate rest (similar to comoving observers in FLRW standard coordinates). The Lemaitre foliation of the infalling congruence is also hypersurface orthogonal.

 

[martinbn]

I’m confused at what you are claiming. Lemaitre coordinates for Schwarzschild geometry are synchronous coordinates in which the congruence of radial infalling test bodies from infinity are at coordinate rest (similar to comoving observers in FLRW standard coordinates). The Lemaitre foliation of the infalling congruence is also hypersurface orthogonal.

I think he means in general. In a spcific spacetime and a specific congruence it is possible.

 

[PAllen]

I think he means in general. In a spcific spacetime and a specific congruence it is possible.

Ok, but the OP specifically asked about the region around earth.

Also, you can construct Gaussian Normal (also called synchronous) coodinates in a finite region in any spacetime. As to lack of global coverage, I don't see how that is a differentiator, because in the general case, there is unlikely to be be any global coordinate charts.

Note the features of Gaussian Normal coordinates:

1) Lines of constant position are timelike geodesics
2) This timelike congruence admits foliation by spacelike surfaces everywhere orthogonal to the congruence
3) The proper time along any of the defining congruence lines between foliation surfaces is the same (this is why such coordinates are also called synchronous).

[add: Just to be clear, as noted by @PeterDonis and @Ibix much earlier in this thread, the conditions satisfied by a Gaussian normal chart do not define an inertial frame because the distances between the congruence lines change (in almost all cases). In curved spacetime it is impossible to add the constant distance requirement - there will be no solution to the combined constraints. As an aside, in Godel spacetime, maximum coverage of a Gaussian patch is small, being inversely proportional to the vorticity of the Godel spacetime. ]

 

[cianfa72]

I’m confused at what you are claiming. Lemaitre coordinates for Schwarzschild geometry are synchronous coordinates in which the congruence of radial infalling test bodies from infinity are at coordinate rest (similar to comoving observers in FLRW standard coordinates). The Lemaitre foliation of the infalling congruence is also hypersurface orthogonal.

So the geodesic congruence of radial infalling test bodies from infinity in Schwarzschild geometry basically define the Lemaitre coordinate chart: assign different spatial coordinate value to each infalling test body and take the proper time along any of the congruence worldline.

 

[PAllen]

So the congruence of radial infalling test bodies from infinity in Schwarzschild geometry basically define the Lemaitre coordinate chart: assign different spatial coordinate value to each infalling test bodiy and take the proper time along any of the congruence worldline.

Yes. And the result is a coordinate system that is completely regular at the horizon, and covers all of the physically interesting geometry - all that would exist in the case of BH that could form. The regions of Kruskal that it doesn’t cover only exist for eternal white hole/blackhole structures.

 

[pervect]

This thread seems to have gotten way too long for me to read all the responses.

What I'd actually use for a definition of a global inertial region is that the Riemann curvature tensor vanishes everywhere in that region, as that is a coordinate independent statement. At the moment I don't have a reference for this as a defintion, it's just what I'd use.

It may be sufficient to say that the geometry of some region of space is Euclidean, that if we omit a flash of light from one observer, the received frequency of the light by any observer is the same as the transmitted frequency (i.e. no doppler shift), and that the speed of light is constant for all observers. However, I don't have a reference for this.

Note that my original formulation omitted to specify the the geometry of space must be Euclidean, but I think we need that assumption to make the definitions equivalent.

What I'd try to do to prove this is to demonstrate that Bondi's k calculus derivation of the Lorentz transform follows from these assumptions.

 

[PeterDonis]

a global inertial region

The definition you give is for a flat region; "global inertial" is not a property of a region of spacetime, it's a property of a frame or coordinate chart on the spacetime. The connection between the two is that the only spacetime for which a global inertial chart is possible is flat Minkowski spacetime.

 

[cianfa72]

In fact, examples of timelike geodesic congruences in a curved spacetime in which the proper distance between worldlines is constant are very rare; the only one I can think of off the top of my head is the congruence of worldlines at rest in the standard coordinate chart on Godel spacetime.

Sorry, maybe I didn’t understand correctly, but - as said before - if a given spacetime admits a timelike geodesic congruence (which extends through the entire spacetime) with the property of constant proper distance between those geodesics, then it is necessarily the Minkowski spacetime and that timelike geodesic congruence actually defines a global inertial chart.

 

[PeterDonis]

if a given spacetime admits a timelike geodesic congruence (which extends through the entire spacetime) with the property of constant proper distance between those geodesics, then it is necessarily the Minkowski spacetime

No, this is not correct. I gave a counterexample in what you quoted: Godel spacetime. To restrict to Minkowski spacetime, you need to add the requirement that the congruence has zero vorticity. "Constant proper distance" already specifies zero expansion and shear, so those requirements together are sufficient to make the chart derived from the congruence inertial. But without the zero vorticity requirement, they're not--see below.

that timelike geodesic congruence actually defines a global inertial chart.

Not in Godel spacetime, no. The chart defined by the timelike geodesic congruence in Godel spacetime is not inertial, because that congruence has nonzero vorticity.

 

[cianfa72]

Not in Godel spacetime, no. The chart defined by the timelike geodesic congruence in Godel spacetime is not inertial, because that congruence has nonzero vorticity.

So, "constant proper distance" between worldlines of the timelike geodesic congruence is not the same as zero vorticity ? In other words zero geodesic deviation actually means "constant proper distance" between geodesics or zero vorticity ? Thank you.

 

[PeterDonis]

So, "constant proper distance" between worldlines of the timelike geodesic congruence is not the same as zero vorticity ?

Correct. Constant proper distance means zero expansion and shear, but does not require zero vorticity. Any rigid rotating congruence has zero expansion and shear, but is not inertial because it has nonzero vorticity. (The Godel spacetime congruence I described is an example of a rigid rotating congruence.)

In other words zero geodesic deviation actually means "constant proper distance" between geodesics or zero vorticity ?\

Zero geodesic deviation, by itself, just means constant proper distance (zero expansion and shear). Zero vorticity is an additional condition, not the same as zero geodesic deviation. As I noted, it is rare to find a timelike geodesic congruence that does not also have zero vorticity (the Godel spacetime one is the only example I can think of), which is why the additional requirement of zero vorticity for the coordinate chart derived from the congruence to be inertial is not often mentioned.

 

[PAllen]

Correct. Constant proper distance means zero expansion and shear, but does not require zero vorticity. Any rigid rotating congruence has zero expansion and shear, but is not inertial because it has nonzero vorticity. (The Godel spacetime congruence I described is an example of a rigid rotating congruence.)Zero geodesic deviation, by itself, just means constant proper distance (zero expansion and shear). Zero vorticity is an additional condition, not the same as zero geodesic deviation. As I noted, it is rare to find a timelike geodesic congruence that does not also have zero vorticity (the Godel spacetime one is the only example I can think of), which is why the additional requirement of zero vorticity for the coordinate chart derived from the congruence to be inertial is not often mentioned.

A relevant point is that the constant distance timelike geodesic congruence in Godel spacetime is not hypersurface orhtogonal. This is related to why Gaussian coordinate patches in Godel spacetime are small, with patch size inversely proportional to vorticity. Specifically, achieving the orthogonality condition requires use of a different timelike geodesic congruence, for which the geodesics intersect 'quickly' from an initial defining hypersurface.

Thus, I think it must be true that the following forces a flat spacetime region:

You can find a hypersurface orthogonal timelike geodesic congruence such that proper distance between the geodesics remains constant.

 

[cianfa72]

A relevant point is that the constant distance timelike geodesic congruence in Godel spacetime is not hypersurface orthogonal.

ok, so it actually relies on the Frobenius' s condition fulfillment for the tangent vector field to the given timelike geodesic congruence.

Thus, I think it must be true that the following forces a flat spacetime region:

You can find a hypersurface orthogonal timelike geodesic congruence such that proper distance between the geodesics remains constant.

So iff you can find a such timelike geodesic congruence that extends to the entire spacetime then the entire spacetime is flat (Minkowski) and the chart derived from that congruence (i.e. the coordinate chart in which the worldlines of the timelike geodesic congruence are at rest and the coordinate time is the proper time along each of them) is actually a global inertial chart.

 

[PeterDonis]

A relevant point is that the constant distance timelike geodesic congruence in Godel spacetime is not hypersurface orhtogonal.

Yes, this is equivalent to saying it has nonzero vorticity. The hypersurface orthogonality condition is just another way of stating the zero vorticity condition, since the two are equivalent by the Frobenius theorem.

 

[cianfa72]

So, do you think my post #44 makes sense ?

Btw it does mean the FLRW co-moving observers congruence is hypersurface orthogonal.

 

[PeterDonis]

do you think my post #44 makes sense ?

Yes.

Btw it does mean the FLRW co-moving observers congruence is hypersurface orthogonal.

If you mean your post #44 means that, no, your post #44 is not describing the FLRW spacetime comoving observer congruence, because that congruence has nonzero expansion (proper distance between geodesics is not constant). It does have zero vorticity (since it is hypersurface orthogonal), but it does not meet the "constant distance" condition for a congruence that can define a global inertial chart.

 

[cianfa72]

If you mean your post #44 means that, no, your post #44 is not describing the FLRW spacetime comoving observer congruence, because that congruence has nonzero expansion (proper distance between geodesics is not constant). It does have zero vorticity (since it is hypersurface orthogonal).

No, I was not claiming that. My point was related to what @Ibix said earlier in post#9: FLWR spacetime comoving observer congruence does not define a global inertial chart even if it is hypersurface orthogonal (since it has nonzero expansion).

 

[PeterDonis]

My point was related to what @Ibix said earlier in post#9: FLWR spacetime comoving observer congruence does not define a global inertial chart even if it is hypersurface orthogonal (since it has nonzero expansion).

Ah, ok. Yes, that's correct.

 

[PAllen]

So, I guess we have several sets of criteria for a inertial frame, that all are only satisfiable in flat spacetime:

1) There exists a timilike geodesic congruence with zero expansion, shear, or vorticity.

2) There exists a timelike geodesic congruence that is hypersurface orthogonal and has no expansion or shear.

3) There exists a timelike geodesic congruence that is hypersurface orthogonal and mutual proper distances between congruence world lines are all constant.

As stated, we can speak of satisfying these in a finite region of spacetime, meaning there there is zero curvature region of spacetime within an overall spacetime that is not necessarily all flat (e.g. within a spherical shell of matter). All of these criteria are well defined over a small region.

Further, we know that over some region, we can satisfy the "no vorticity", equivalently hypersurface orthogonal condition in any spacetime whatsoever. Then adding the no expansion or shear (equiv. constant distance) becomes impossible without zero curvature, because it rules out any type of geodesic deviation.

A remaining question for me, is whether you can find, in a small region of any spacetime, a timelike geodesic congruence with zero expansion or shear (but allowing vorticity / not hypersurface orthogonal)? I think a way to approach this is to write down these constraints formally and check if they are overdetermined in the general case.

 

[PeterDonis]

I guess we have several sets of criteria for a inertial frame, that all are only satisfiable in flat spacetime

All three sets of criteria are equivalent, so any congruence that does/does not satisfy one will/will not satisfy all three.

Specifically, the key equivalences are:

zero vorticity <-> hypersurface orthogonal
zero expansion and shear <-> Born rigid (see below) <-> constant proper distance

mutual proper distances between congruence world lines are all constant

Another more technical way of putting this would be that the congruence is Born rigid.

 

[PeterDonis]

A remaining question for me, is whether you can find, in a small region of any spacetime, a timelike geodesic congruence with zero expansion or shear (but allowing vorticity / not hypersurface orthogonal)?

The global timelike congruence in Godel spacetime that I have already described has these properties. It's the only example I'm aware of.

 

[PAllen]

The global timelike congruence in Godel spacetime that I have already described has these properties. It's the only example I'm aware of.

Right, and my question is whether something like that could be found in any small region of a general spacetime.

It comes down to counting constraints. The Gaussian normal coordinate condition imposes 4 constraints on the metric, and there are 4 degrees of freedom representing diffeomorphism invariance when solving the EFE. Thus it is not surprising that these constraints are generally satisfiable. So the question becomes how to translate existence of a timelike geodesic congruence of zero expansion and shear (with no other requirements) into conditions on the metric. If it can be expressed as 4 or fewer conditions, then it is likely generally satisfiable in some finite region.

Note, there is a lot of analysis behind showing that 4 simple conditions on metric expression are equivalent to the existence of a hypersurface orthogonal timelike geodesic congruence. So I don’t think my question is an easy one to answer - at least for me.

 

[PAllen]

Right, and my question is whether something like that could be found in any small region of a general spacetime.

It comes down to counting constraints. The Gaussian normal coordinate condition imposes 4 constraints on the metric, and there are 4 degrees of freedom representing diffeomorphism invariance when solving the EFE. Thus it is not surprising that these constraints are generally satisfiable. So the question becomes how to translate existence of a timelike geodesic congruence of zero expansion and shear (with no other requirements) into conditions on the metric. If it can be expressed as 4 or fewer conditions, then it is likely generally satisfiable in some finite region.

Note, there is a lot of analysis behind showing that 4 simple conditions on metric expression are equivalent to the existence of a hypersurface orthogonal timelike geodesic congruence. So I don’t think my question is an easy one to answer - at least for me.

Just to be specific, the constraint that g00 = 1, and g0i = 0 (for i = 1..3), can be shown to imply the existence of a hypersurface orthogonal timelike geodesic congruence - and vice versa, in that such a coordinate expression of the metric must be possible given the congruence existence. So, the question is whether there are similar statements that can be made for a timelike geodesic congruence with zero expansion and shear.

Also worth noting is that the Gaussian normal coordinate conditions actually have an infinite number of realizations in a sufficiently small neighborhood. However, in common exact EFE solutions, there is a unique maximal Gaussian normal chart, e.g. Lemaitre is the maximal Gaussian normal chart for the Kruskal geometry. There is a similar maximal chart for the Kerr geometry. Obviously, there is global maximal Gaussian chart for FLRW - the standard cosmological coordinates.

 

[cianfa72]

Just to be specific, the constraint that g00 = 1, and g0i = 0 (for i = 1..3), can be shown to imply the existence of a hypersurface orthogonal timelike geodesic congruence - and vice versa, in that such a coordinate expression of the metric must be possible given the congruence existence.

Yes, it should be the 'synchronous' reference frame as discussed so far in this thread Synchronous reference frame.

Also worth noting is that the Gaussian normal coordinate conditions actually have an infinite number of realizations in a sufficiently small neighborhood. However, in common exact EFE solutions, there is a unique maximal Gaussian normal chart, e.g. Lemaitre is the maximal Gaussian normal chart for the Kruskal geometry. There is a similar maximal chart for the Kerr geometry. Obviously, there is global maximal Gaussian chart for FLRW - the standard cosmological coordinates.

Does the attribute maximal in 'maximal Gaussian normal chart' actually mean that such a chart has the maximal possible extension in a given region of spacetime ? Anyhow that does not imply such maximal chart is actually global (i.e. it fills/extends to the entire spacetime).

The case of FLRW spacetime is, let me say, singular since that such unique maximal Gaussian normal chart turns out to be global.

 

[PAllen]

Yes, it should be the 'synchronous' reference frame as discussed so far in this thread Synchronous reference frame.Does the attribute maximal in 'maximal Gaussian normal chart' actually mean that such a chart has the maximal possible extension in a given region of spacetime ? Anyhow that does not imply such maximal chart is actually global (i.e. it fills/extends to the entire spacetime).

The case of FLRW spacetime is, let me say, singular since that such unique maximal Gaussian normal chart turns out to be global.

By maximal I simply mean the largest coverage of any of the infinite possible Gaussian coordinate patches in a given pseudo-Riemannian manifold. Two ends of the spectrum, due to the extreme symmetries of the spacetimes, are FLRW where coverage of the whole manifold is possible, and Godel spacetime, where all patches are the same "small size" limited by the vorticity of the Godel manifold.

 

[PAllen]

Just to be specific, the constraint that g00 = 1, and g0i = 0 (for i = 1..3), can be shown to imply the existence of a hypersurface orthogonal timelike geodesic congruence - and vice versa, in that such a coordinate expression of the metric must be possible given the congruence existence. So, the question is whether there are similar statements that can be made for a timelike geodesic congruence with zero expansion and shear.

I have a plausibility argument that this is generally impossible. I note that in the kinematic decomposition of a congruence, the vorticity tensor can actually be represented as a 3-vector, with 3 independent components. Meanwhile, the full expansion tensor (including both shear plus the scalar) irreducibly has 6 independent components. While the vanishing of one or the other are conditions on a congruence (rather than on a metric), it seems very suggestive that vanishing vorticity imposes a number of constraints less than the diffeomorphism degrees of freedom, while vanishing expansion tensor requires more.

 

[PAllen]

I have a plausibility argument that this is generally impossible. I note that in the kinematic decomposition of a congruence, the vorticity tensor can actually be represented as a 3-vector, with 3 independent components. Meanwhile, the full expansion tensor (including both shear plus the scalar) irreducibly has 6 independent components. While the vanishing of one or the other are conditions on a congruence (rather than on a metric), it seems very suggestive that vanishing vorticity imposes a number of constraints less than the diffeomorphism degrees of freedom, while vanishing expansion tensor requires more.

I see a flaw in this. The expansion tensor can, of course, be decomposed into a shear tensor and an expansion scalar. The shear tensor has only 3 independent components. So vanishing of both can be represented as 4 conditions, so the argument becomes unclear.

 

[PeterDonis]

The shear tensor has only 3 independent components.

No, it has 5. It's a traceless symmetric 3-tensor. Or, to put it another way, the expansion tensor as a whole is a symmetric 3-tensor, so it has 6 independent components. The expansion scalar is basically the trace, i.e., 1 of the 6 total, and the shear tensor is the other 5.

 

[PeterDonis]

the vorticity tensor can actually be represented as a 3-vector, with 3 independent components

That's because the vorticity tensor is an antisymmetric 3-tensor, which has 3 independent components, so it can be represented as a vector (actually a pseudovector). Of course this convenient representation only works in 3 dimensions.