I About global inertial frames in GR

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The discussion centers on the definition and existence of global inertial frames in General Relativity (GR). It clarifies that while a frame can be considered inertial in a finite region of spacetime, it cannot be classified as global due to the inherent curvature of spacetime. The participants emphasize that the definition of an inertial frame must include qualifiers regarding measurement accuracy and the constancy of proper distances between accelerometers. They also explore the conditions under which timelike geodesic congruences maintain constant proper distances, noting that such conditions are rare in curved spacetimes. Ultimately, the conversation highlights the complexities and nuances of defining inertial frames within the framework of GR.
  • #61
PeterDonis said:
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 said:
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.
 
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  • #62
PeterDonis said:
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
 
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  • #63
PAllen said:
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.
 
  • #64
cianfa72 said:
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.
 
  • #65
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.
 
  • #66
cianfa72 said:
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.
 
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  • #67
PAllen said:
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).
 
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  • #68
cianfa72 said:
I was reading again this post: are expansion/shear and vorticity really two different things somehow related each other ?
They are separate features.
cianfa72 said:
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).
cianfa72 said:
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.
 
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  • #69
PAllen said:
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.
 
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  • #70
cianfa72 said:
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.
 
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  • #71
From a mathematical point of view which are the definitions of expansion and shear for a congruence ?
 
  • #73
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).
 
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  • #74
cianfa72 said:
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.

cianfa72 said:
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.
 
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  • #75
PeterDonis said:
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.
 
  • #76
cianfa72 said:
In the above case, any two randomly chosen timelike geodesics will always have zero geodesic deviation
Yes, but...

cianfa72 said:
(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).

cianfa72 said:
each one zero coordinate acceleration
Yes.
 
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  • #77
PeterDonis said:
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.
 
  • #78
cianfa72 said:
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").

cianfa72 said:
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.
 
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  • #79
@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.
 

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