B Black hole at the beginning of time

[PAllen]

Actually, one can state the assumption in a way that does not depend on any choice of coordinates. 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.

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 has the general FRW form.

The most common derivations go about this in different order (which does not imply the above is incorrect). You first note or derive that any metric can be put in synchronous form (for some finite region). Then, argue that isotropy everywhere (which implied homogeneity) puts the space-space metric components in one of 3 forms, each diagonal only (for hyperbolic, flat, or positive curvature spatial slices). Then, argue that the coordinate coverage is global (also due to isotropy/homogeneity). This is basically how Weinberg goes about it, for example.

However, you are trying to get away from coordinates, as much as possible. Since the only assumption above that is unique to FRW is the isotropy/homogeneity, it suggests it should be possible to turn the arguments around as you suggest; I've just never seen it done 'rigorously' in that order.

 

[PeterDonis]

You first note or derive that any metric can be put in synchronous form (for some finite region).

This is equivalent to saying that there is a timelike congruence that is hypersurface orthogonal, correct? Then you can argue that that must be true for the entire spacetime, not just some finite region, because, as you say, that is implied by isotropy everywhere.

it suggests it should be possible to turn the arguments around as you suggest; I've just never seen it done 'rigorously' in that order.

My understanding has been that that is how the argument was originally framed by Friedmann et al., although I have rephrased it in more modern terminology.

 

[PAllen]

This is equivalent to saying that there is a timelike congruence that is hypersurface orthogonal, correct?

Yes. Actually, a timelike geodesic congruence that is hypersurface orthogonal.

Then you can argue that that must be true for the entire spacetime, not just some finite region, because, as you say, that is implied by isotropy everywhere.

Agree.

My understanding has been that that is how the argument was originally framed by Friedmann et al., although I have rephrased it in more modern terminology.

I didn't know that, interesting.

 

[cianfa72]

Then, argue that isotropy everywhere (which implied homogeneity)

So the point is that from isotropy everywhere follows homogeneity everywhere. Is the reverse also true ?

puts the space-space metric components in one of 3 forms, each diagonal only (for hyperbolic, flat, or positive curvature spatial slices).

As space-space metric components I believe you mean the metric tensor components $g_{\mu \nu}$ with both spatial indices. Are these 3 diagonal forms of spatial metric basically the 3 possibile homogeneous and isotropic FLRW models ?

 

[PAllen]

So the point is that from isotropy everywhere follows homogeneity everywhere. Is the reverse also true ?

I meant to type implies. Isotropy everywhere implies homogeneity. 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.

As space-space metric components I believe you mean the metric tensor components $g_{\mu \nu}$ with both spatial indices. Are these 3 diagonal forms of spatial metric basically the 3 possibile homogeneous and isotropic FLRW models ?

Yes.

 

[PeterDonis]

Is the reverse also true ?

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]

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.

In this example the 'preferred/distinguishable' direction is that in which the curvature is null, I believe.

 

[PeroK]

In this example the 'preferred/distinguishable' direction is that in which the curvature is null, I believe.

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.

A geometric example is the surface of an infinite cylinder. Every point is the same, but there's always a circumferencial direction and a longitudinal direction. So, homogeneous but not isotropic.

 

[cianfa72]

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.

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]

ok, so the condition of isotropy everywhere (that implies homogeneity everywhere)

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!

 

[PeterDonis]

In this example the 'preferred/distinguishable' direction is that in which the curvature is null, I believe.

No. Curvature is zero everywhere on a cylinder, and curvature doesn't have a "direction".

The "preferred" direction on the cylinder is the one in which the geodesics are closed circles.

 

[cianfa72]

No. Curvature is zero everywhere on a cylinder, and curvature doesn't have a "direction".

Ah yes, the 'extrinsic' curvature of a cylinder does not matter.

 

[cianfa72]

The "preferred" direction on the cylinder is the one in which the geodesics are closed circles.

So the 'preferred' direction on the cylinder is actually matter of topology.

 

[PeterDonis]

So the 'preferred' direction on the cylinder is actually matter of topology.

No. Geodesics are not a matter of topology; you need to have a metric to know which curves are geodesics.

 

[cianfa72]

No. Geodesics are not a matter of topology; you need to have a metric to know which curves are geodesics.

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]

on the cylinder the metric is flat then from a metric point of view all geodesics are indistinguibile.

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.

It is the topology that picks a preferred direction.

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]

So we can say both (metric and topology) are actually needed in order to define isotropy.

 

[cianfa72]

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.

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.

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

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]

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.

Yes.

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 $t$.

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]

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.

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]

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.