I About global inertial frames in GR

[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.

 

[PeterDonis]

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.

the vorticity tensor is an antisymmetric 3-tensor

Still another way of expressing all this is that, given the tangent vector field for the congruence, the orthogonal projection of its covariant derivative is a 3-D 2nd rank tensor, whose symmetric part is the expansion tensor and whose antisymmetric part is the vorticity tensor. A 3-D 2nd rank tensor has a total of 9 independent components, 6 in its symmetric part (or 1 trace + 5 symmetric traceless) and 3 in its antisymmetric part.

 

[PAllen]

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.

Ok, my momentary oversight was somehow thinking traceless implied zero diagonal. That’s just wrong.

So there probably is something to the idea that 6 constraints for Born rigidity versus 3 for vanishing vorticity is relevant to generally being able achieve a vanishing vorticity timelike geodesic congruence, while not generally being able to achieve a Born rigid timelike geodesic congruence

 

[cianfa72]

By maximal I simply meant the largest coverage of any of the infinite possible Gaussian coordinate patches in a given pseudo-Riemannian manifold.

Sorry, I've some problem with english in particular the meaning of 'any of'. Do you mean take the Gaussian normal coordinate chart that has got the largest coverage in the set of the infinite possible Gaussian normal coordinate patches in the given manifold ? Thank you.

 

[PAllen]

Sorry, I've some problem with english in particular the meaning of 'any of'. Do you mean take the Gaussian normal coordinate chart that has got the largest coverage in the set of the infinite possible Gaussian normal coordinate patches in the given manifold ? Thank you.

yes.

 

[cianfa72]

So in a given spacetime if we start to build Gaussian normal congruence (i.e. coordinate chart) from different spacelike hypersurfaces we end up in general with different coverages of the chart being built.

 

[PAllen]

So in a given spacetime if we start to build Gaussian normal congruence (i.e. coordinate chart) from different spacelike hypersurfaces we end up in general with different coverages of the chart being built.

yes. Often there is just one choice that leads to very large coverage.

 

[cianfa72]

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.

I was reading again this post: are expansion/shear and vorticity really two different things somehow related each other ?

As far as I understand, non-zero vorticity is the reason behind the impossibility to extend a Gaussian coordinate patch to cover the whole spacetime manifold (i.e. the exponential map results in geodesics crossing at some point).

Non-zero expansion and shear, instead, is the reason for non-constant proper distance between worldlines in the congruence (i.e. non-zero geodesic deviation actually prevents the possibility to build an inertial coordinate chart as discussed above in the thread).

 

[PAllen]

I was reading again this post: are expansion/shear and vorticity really two different things somehow related each other ?

They are separate features.

As far as I understand, non-zero vorticity is the reason behind the impossibility to extend a Gaussian coordinate patch to cover the whole spacetime manifold (i.e. the exponential map results in geodesics crossing at some point).

To construct a Gaussian patch of any size, you must have timelike geodesic congruence with zero vorticity. A spacetime with intrinsic vorticity (Godel, Kerr) makes it hard for such patch to be very large (though, with Kerr, there is a unique choice with substantial coverage).

Non-zero expansion and shear, instead, is the reason for non-constant proper distance between worldlines in the congruence (i.e. non-zero geodesic deviation actually prevents the possibility to build an inertial coordinate chart as discussed above in the thread).

In most any (all?) curved spacetimes, a congruence defining a Gaussian coordinate patch will have nonzero expansion and/or shear, so yes the proper distances between the world lines along the hypersurface orthogonal foliation will not be constant. As for how this relates to the rest of this thread, I have not read it all and do not plan to.

 

[cianfa72]

a congruence defining a Gaussian coordinate patch will have nonzero expansion and/or shear, so yes the proper distances between the world lines along the hypersurface orthogonal foliation will not be constant.

ok, you talk of 'hypersurface orthogonal foliation' since the Gaussian coordinate patch built starting from a given spacelike hypersurface has the property that timelike geodesics starting from it continue to remain orthogonal to the spacelike hypersurfaces of the foliation being built.

 

[PAllen]

ok, you talk of 'hypersurface orthogonal foliation' since the Gaussian coordinate patch built starting from a given spacelike hypersurface has the property that timelike geodesics starting from it continue to remain orthogonal to the spacelike hypersurfaces of the foliation being built.

but also any timelike geodesic congruence meeting the vanishing vorticity specified by the Frobenius condition, will have a foliation by surfaces each orthogonal to the congruence and also such that proper time difference between foliation surfaces along congruence lines will be the same for all congruence lines. But, in most spacetimes, you will only be able to satisfy these conditions [construction of timelike geodesic congruence with vanishing vorticity] in 'small' regions. If you further try to impose vanishing expansion and shear, there are likely no solutions for most spacetimes, even in a very small region.

 

[cianfa72]

From a mathematical point of view which are the definitions of expansion and shear for a congruence ?

 

[Ibix]

From a mathematical point of view which are the definitions of expansion and shear for a congruence ?

See this thread: https://www.physicsforums.com/threa...-for-rod-and-hole-relativity-paradox.1010444/

 

[cianfa72]

Sorry, just another piece of the puzzle. We said that the timelike geodesic congruence at rest in standard coordinates in Godel spacetime has zero shear and expansion, however it has not zero vorticity.

What about any two arbitrary timelike geodesics in Godel spacetime ? I believe they will not have costant proper distance (i.e. there will be a non-zero geodesic deviation between any two arbitrary timelike geodesics).

 

[PeterDonis]

Sorry, just another piece of the puzzle. We said that the timelike geodesic congruence at rest in standard coordinates in Godel spacetime has zero shear and expansion, however it has not zero vorticity.

Yes.

What about any two arbitrary timelike geodesics in Godel spacetime ? I believe they will not have costant proper distance (i.e. there will be a non-zero geodesic deviation between any two arbitrary timelike geodesics).

In general, yes, we would expect two randomly chosen timelike geodesics to have nonzero geodesic deviation. This will be true in any curved spacetime.

 

[cianfa72]

In general, yes, we would expect two randomly chosen timelike geodesics to have nonzero geodesic deviation. This will be true in any curved spacetime.

The point I would like to make is that if and only if (iff) there is a timelike geodesic congruence hypersurface orthogonal (i.e. zero vorticity) having zero expansion and shear then the underlying spacetime is flat (Minkowski) and the global chart in which the worldlines of the above congruence are at rest is a global inertial coordinate chart for the spacetime.

In the above case, any two randomly chosen timelike geodesics will always have zero geodesic deviation (i.e. constant proper distance) and each one zero coordinate acceleration in the aforementioned global chart.

 

[PeterDonis]

In the above case, any two randomly chosen timelike geodesics will always have zero geodesic deviation

Yes, but...

(i.e. constant proper distance)

...no. For example, consider the worldlines $x = 0$ and $x = 0.5t$. They are both timelike geodesics and their geodesic deviation is zero (their "relative velocity" is always the same), but the proper distance between them is not constant. This is why discussions of spacetime curvature normally focus on the case of initially parallel geodesics (which the above two are not).

each one zero coordinate acceleration

Yes.

 

[cianfa72]

For example, consider the worldlines $x = 0$ and $x = 0.5t$. They are both timelike geodesics and their geodesic deviation is zero (their "relative velocity" is always the same), but the proper distance between them is not constant.

ok, the above two are examples of non proper constant distance timelike geodesics in flat spacetime.

So zero expansion and shear for worldlines in a congruence (i.e. constant proper distance) implies their zero geodesic deviation however the reverse is not true in general.

 

[PeterDonis]

zero expansion and shear for worldlines in a congruence (i.e. constant proper distance) implies their zero geodesic deviation

"Geodesic deviation" is the wrong term here, it does not apply to worldlines in a congruence. It applies to the spacetime geometry. "Zero geodesic deviation" means flat geometry, independent of any properties that particular congruences of worldlines might or might not have. We have already discussed in this thread how the existence of a geodesic congruence with zero expansion and zero shear is not sufficient for a flat spacetime geometry.

The best term to use for the property of a congruence that you are describing is "Born rigid" (or "constant proper distance").

however the reverse is not true in general.

The two geodesics I gave cannot be part of the same congruence because they intersect. A congruence is a set of non-intersecting worldlines that fill a region of spacetime.

 

[PeterDonis]

@cianfa72 we are rehashing previous discussion in this thread. You re-started the thread after it had been dormant for three months; apparently you did not go back and review what had already been said carefully enough. Please do so.