B Black hole at the beginning of time

[PeterDonis]

strictly speaking, there is a complete family of such charts since we can just continuously remap the values of spatial coordinates assigned to each of the worldlines orthogonal to the spacelike hypersurfaces of constant by adding the same constant to the old values assigned to them.

You can do that for the flat and open cases (i.e., zero and negative curvature), yes. For the closed case (positive curvature), the handling of the spatial part of the chart has to be somewhat different (strictly speaking, there can't be a single chart covering all of a spacelike 3-surface since an n-sphere can't be covered by a single chart).

More generally, there are of course a variety of transformations you can apply to the spatial part of the chart in any FRW spacetime, without changing the property that comoving worldlines have constant spatial coordinates in the chart. So yes, in that sense no chart is truly unique.

 

[cianfa72]

You can do that for the flat and open cases (i.e., zero and negative curvature), yes.

Ah ok, so for FRLW open cases there exist actually a global chart for the entire spacetime.

 

[PeterDonis]

for FRLW open cases there exist actually a global chart for the entire spacetime

For the flat and open cases (i.e., the ones with infinite spacelike 3-surfaces), yes.

 

[vanhees71]

A non-rigorous argument of the contrapositive (that inhomogenity implies anisotropy) is:

If points $A$ and $B$ are different, then we have anisotropy for a point midway between $A$ and $B$. That should at least help remember which way round the implication goes!

If I understand your arguments right the point is that you assume that the space is isotropic around any of its points. Only then follows homogeneity. If it's isotropic only around one point it's not necessarily homogeneous (e.g., "Schwarzschild spacetime").

 

[PeroK]

If I understand your arguments right the point is that you assume that the space is isotropic around any of its points. Only then follows homogeneity. If it's isotropic only around one point it's not necessarily homogeneous (e.g., "Schwarzschild spacetime").

The argument is that if it's not homogeneous then it can't be isotropic. This is logically equivalent to if it's isotropic then it must be homogeneous.

It's called the law of contraposition.

 

[cianfa72]

For the flat and open cases (i.e., the ones with infinite spacelike 3-surfaces), yes.

ok, so for the FLRW hyperbolic case (i.e. family of spacelike hypersurfaces homogeneous and isotropic each with a different constant negative curvature) the global chart described above is such that worldlines of comoving geodesic congruence are 'at rest' in it however the global chart itself is not inertial since the congruence is not 'rigid' (i.e. there is tidal gravity since the distance between comoving worldlines does change).

 

[vanhees71]

The argument is that if it's not homogeneous then it can't be isotropic. This is logically equivalent to if it's isotropic then it must be homogeneous.

It's called the law of contraposition.

This I don't understand intuitively. Situations, where the system is isotropic wrt. one point but not homogeneous just because of singling out this one point, are very common. Just take the situation of Schwarzschild spacetime. There you have a spherically symmetric star, and spacetime is isotropic around the center of the star but not homogeneous.

What's meant in cosmological principle is that space for a "standard observer" is isotropic around any point. Then space is necessarily both homogeneous and isotropic.

In Weinberg, Gravitation and Cosmology there's also a nice formal discussion analyzing symmetries of Riemannian spaces (Killing vectors) confirming this: If a space is istropic around all of its points it's also homogeneous.

 

[PeroK]

In Weinberg, Gravitation and Cosmology there's also a nice formal discussion analyzing symmetries of Riemannian spaces (Killing vectors) confirming this: If a space is istropic around all of its points it's also homogeneous.

Yes, a formal proof to show that for any manifold isotropy implies homogeneity is not going to be easy (and it's perhaps not even easy to see why this might be true). The logically equivalent contrapositive (non-homogeneity implies anisotropy) is, however, easier to justify. To construct a formal proof would still require precise definitions of a manifold and the properties in question.

 

[PeterDonis]

a formal proof to show that for any manifold isotropy implies homogeneity

Is impossible as you state it here since, as @vanhees71 has correctly pointed out, isotropy about only one point does not imply homogeneity. Off the top of my head I'm not sure what the minimum number of points is at which isotropy is required, in order for homogeneity to be necessarily implied.

 

[PeterDonis]

so for the FLRW hyperbolic case (i.e. family of spacelike hypersurfaces homogeneous and isotropic each with a different constant negative curvature) the global chart described above is such that worldlines of comoving geodesic congruence are 'at rest' in it

Yes.

however the global chart itself is not inertial since the congruence is not 'rigid' (i.e. there is tidal gravity since the distance between comoving worldlines does change).

No global chart in any curved spacetime can be inertial.

 

[PeroK]

Is impossible as you state it here since, as @vanhees71 has correctly pointed out, isotropy about only one point does not imply homogeneity.

It's clear from the context that "isotropic" means isotropic at every point. That must be the default terminology in any case.

 

[vanhees71]

Well, it's obvious that isotropy around only one point is not enough. It's however also pretty intuitive that isotropy around all points implies homogeneity (but not the other way, i.e., homogeneity does not imply isotropy around any point). Weinberg's definitions and proofs in Gravitation and Cosmology (Chpt. 13) is also not too difficult to follow.

 

[cianfa72]

No global chart in any curved spacetime can be inertial.

Yes that's is true in general, in the particular case of chart described in #56 the reason behind it should be that pointed out there, I believe.