No. One could have a preferred direction (i.e., lack of isotropy) in a homogeneous spacetime, as long as the preferred direction was the same everywhere.cianfa72 said:Is the reverse also true ?
In this example the 'preferred/distinguishable' direction is that in which the curvature is null, I believe.PAllen said:The converse is false. Consider the simple example of a two cylinder - no point or small region is distinguishable from any other, so you have homogeneity. However, it is nowhere isotropic, as there is everywhere a distinguishable direction.
A simple example is the idealised uniform gravitational field. You have a homogeneous gravitational field (the same everyone), but a common direction of gravitational force.cianfa72 said:In this example the 'preferred/distinguishable' direction is that in which the curvature is null, I believe.
ok, so the condition of isotropy everywhere (that implies homogeneity everywhere) does mean the spatial curvature everywhere on each spacelike hypersurface point has to be constant. Hence the 3 allowed geometries with constant positive, null or negative curvature, namely spherical, flat or hyperbolic spatial geometry in each spatial slices.PeroK said:A geometric example is the surface of an infinite cylinder. Every point is the same, but there's always a circumferential direction and a longitudinal direction. So, homogeneous but not isotropic.
A non-rigorous argument of the contrapositive (that inhomogenity implies anisotropy) is:cianfa72 said:ok, so the condition of isotropy everywhere (that implies homogeneity everywhere)
No. Curvature is zero everywhere on a cylinder, and curvature doesn't have a "direction".cianfa72 said:In this example the 'preferred/distinguishable' direction is that in which the curvature is null, I believe.
Ah yes, the 'extrinsic' curvature of a cylinder does not matter.PeterDonis said:No. Curvature is zero everywhere on a cylinder, and curvature doesn't have a "direction".
So the 'preferred' direction on the cylinder is actually matter of topology.PeterDonis said:The "preferred" direction on the cylinder is the one in which the geodesics are closed circles.
No. Geodesics are not a matter of topology; you need to have a metric to know which curves are geodesics.cianfa72 said:So the 'preferred' direction on the cylinder is actually matter of topology.
Sure, however on the cylinder the metric is flat then from a metric point of view all geodesics are indistinguibile. It is the topology that picks a preferred direction.PeterDonis said:No. Geodesics are not a matter of topology; you need to have a metric to know which curves are geodesics.
No, they aren't. Geodesics go in different directions and can be distinguished that way. In fact, without a metric you don't even have a well-defined concept of "direction" to begin with (since you need a metric for angles as well as lengths), so it's not clear that you can define what isotropy means.cianfa72 said:on the cylinder the metric is flat then from a metric point of view all geodesics are indistinguibile.
The topology certainly is one factor, since it is what makes one particular set of geodesics closed curves. But you also need the metric to know that they are geodesics (as well as to define the concept of "direction", as above).cianfa72 said:It is the topology that picks a preferred direction.
ok, it is simple as defining a function of spacetime (i.e. the coordinate time function ##t##) such that its level sets are the given spacelike hypersurfaces (3-surfaces) foliating the spacetime.PeterDonis said:The assumption, stated in an invariant way, is that the spacetime can be foliated by spacelike 3-surfaces that are all homogeneous and isotropic. This of course implies that one can choose a coordinate chart in which each such 3-surface is a surface of constant coordinate time.
So, as far as I can understand, the above chart is the one that assigns constant spatial coordinates to each worldline in the family of timelike worldlines orthogonal to the given family of spacelike hypersurfaces and as timelike coordinate the coordinate time function ##t##.PeterDonis said:I believe that the existence of a family of timelike worldlines that is orthogonal to the above spacelike 3-surfaces can also be derived from the above assumption. This then implies that the metric in above coordinate chart
Yes.cianfa72 said:it is simple as defining a function of spacetime (i.e. the coordinate time function ) such that its level sets are the given spacelike hypersurfaces (3-surfaces) foliating the spacetime.
A chart whose coordinate time ##t## behaves as above does not have to assign constant spatial coordinates to each worldline that is orthogonal to the spacelike hypersurfaces of constant ##t##. There are an infinite number of possible charts with the same coordinate time ##t##. But there is a unique such chart that does assign constant spatial coordinates to each worldline that is orthogonal to the spacelike hypersurfaces of constant ##t##; that is the standard FRW chart used in cosmology.cianfa72 said:the above chart is the one that assigns constant spatial coordinates to each worldline in the family of timelike worldlines orthogonal to the given family of spacelike hypersurfaces and as timelike coordinate the coordinate time function ##t##.
Actually, strictly speaking, I believe 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 ##t## by adding the same constant to the old values assigned to them.PeterDonis said:There are an infinite number of possible charts with the same coordinate time ##t##. But there is a unique such chart that does assign constant spatial coordinates to each worldline that is orthogonal to the spacelike hypersurfaces of constant ##t##; that is the standard FRW chart used in cosmology.
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).cianfa72 said: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.
Ah ok, so for FRLW open cases there exist actually a global chart for the entire spacetime.PeterDonis said:You can do that for the flat and open cases (i.e., zero and negative curvature), yes.
For the flat and open cases (i.e., the ones with infinite spacelike 3-surfaces), yes.cianfa72 said:for FRLW open cases there exist actually a global chart for the entire spacetime
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 said: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!
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.vanhees71 said: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").
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).PeterDonis said:For the flat and open cases (i.e., the ones with infinite spacelike 3-surfaces), yes.
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.PeroK said: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.
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.vanhees71 said: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.
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.PeroK said:a formal proof to show that for any manifold isotropy implies homogeneity
Yes.cianfa72 said: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
No global chart in any curved spacetime can be inertial.cianfa72 said: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).
It's clear from the context that "isotropic" means isotropic at every point. That must be the default terminology in any case.PeterDonis said:Is impossible as you state it here since, as @vanhees71 has correctly pointed out, isotropy about only one point does not imply homogeneity.
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.PeterDonis said:No global chart in any curved spacetime can be inertial.