# About global inertial frames in GR

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• cianfa72
In summary: Now consider a finite region of spacetime outside the Earth: 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 (i.e. having constant spatial coordinates in that frame) then it should match the above definition of (global) inertial frame -- global at least for that finite region of spacetime.This is just terminology. Call it what i want.You're using the word global when you mean local. If it is not the whole space-time, but just an open set, which is part of a coordinate chart, then it is local
From a mathematical point of view which are the definitions of expansion and shear for a congruence ?

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).

Last edited:
vanhees71
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.

vanhees71
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.

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.

cianfa72
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.

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.

cianfa72
@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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