Movement vs expansion in space

[timmdeeg]

But at least experimental support for inflation would support the L-CDM model and thus postulates like homogeneity on which this model is based, right?

 

[PeterDonis]

What could prove "that the universe must be homogeneous (on large enough distance scales)"?

You would have to prove that there are no models that fit our current data that are not homogeneous.

 

[PeterDonis]

But at least experimental support for inflation would support the L-CDM model

Not specifically. Experimental support for inflation by itself does not tell us anything about what happens after inflation ends. The Lambda-CDM model is about what happens after inflation ends.

and thus postulates like homogeneity on which this model is based, right?

As far as inflation is concerned, you have this backwards. Inflation models predict that the universe will be homogeneous to a very high degree of accuracy at the end of inflation, in which case we would expect it to be homegeneous on large enough distance scales today. But that in itself doesn't support the Lambda-CDM model specifically. It just tells us that the general class of FRW models is the right general class to be using.

 

[PeterDonis]

Spherically symmetric spacetime has only 3 linearly independent spacelike KVFs (Killing Lie algebra has dimension 3).

In general, yes. But note that FRW spacetimes that are homogeneous are spherically symmetric too. So a spherically symmetric spacetime can have more than just one set of 3 such KVFs.

They are rotational KVFs about a fixed point on any spacelike hypersurface of constant Schwarzschild coordinate time (contrast with a maximally symmetric spacelike hypersurface that has got $3*4/2=6$ linearly independent KVFs).

Note that this way of looking at it assumes that the vacuum Schwarzschild region ends at the surface of an ordinary object, i.e., that we are not talking about a black hole. For a Schwarzschild black hole, there is no "fixed point". But the rotational KVFs are still there and the spacetime is still spherically symmetric.

Therefore such spherically symmetric spacetime is isotropic only about a specific "spatial" location (i.e. about the points that a specific timelike worldline intersects any of the spacelike hypersurfaces of constant Schwarzschild coordinate time).

Again, see the caveat I gave above.

 

[cianfa72]

Note that this way of looking at it assumes that the vacuum Schwarzschild region ends at the surface of an ordinary object, i.e., that we are not talking about a black hole. For a Schwarzschild black hole, there is no "fixed point". But the rotational KVFs are still there and the spacetime is still spherically symmetric.

Yes, but the volume of such "ordinary object" can be taken in principle as infinitesimal. Therefore the 3 rotational KVFs are about that "location", I believe.

 

[PeterDonis]

the volume of such "ordinary object" can be taken in principle as infinitesimal.

If you give it a small enough mass, yes--but then you're making the spacetime very close to just being flat.

However, you don't need to do that. See below.

the 3 rotational KVFs are about that "location", I believe.

They are about the center of mass of the object. If the spacetime is spherically symmetric, the object must be, which means the rotational KVFs exist inside the object as well as outside. No need to make the object infinitesimal in size.

 

[PAllen]

Just to clarify relations between homogeneity and isotropy:

- isotropy about a point obviously does not imply homogeneity
- isotropy everywhere (about all point) implies homogeneity
- homogeneity does not imply isotropy (there can be the 'same' preferred direction at every point, e.g. a cylinder for a case of 2-surface geometry)
- homogeneity plus isotropy about one point implies isotropy everywhere.

 

[cianfa72]

homogeneity does not imply isotropy (there can be the 'same' preferred direction at every point, e.g. a cylinder for a case of 2-surface geometry)

For a 2-surface the maximally symmetric condition requires $2*3/2=3$ independent isometries.

In the case of the cylinder there are only 2 KVFs (the translation KVFs) and no global "rotation" KVF about any point (i.e. any linear combination with constant coefficients of the 2 independent KVFs doesn't result in a "rotation" KVF about any point).

 

[PAllen]

For a 2-surface the maximally symmetric condition requires $2*3/2=3$ independent isometries.

In the case of the cylinder there are only 2 KVFs (the translation KVFs) and no global "rotation" KVF about any point (i.e. any linear combination with constant coefficients of the 2 independent KVFs doesn't result in a "rotation" KVF about any point).

Yes, this is homogeneous but nowhere isotropic.

 

[cianfa72]

Yes, this is homogeneous but nowhere isotropic.

Nevertheless for the cylinder $\mathbb S^1 \times \mathbb R$, as explained in this video at minute 36:00, the Killing equation has a third local solution that, however, cannot be globally extended as smooth KVF on the entire manifold.

 

[PeterDonis]

for the cylinder $\mathbb S^1 \times \mathbb R$, as explained in this video at minute 36:00, the Killing equation has a third local solution

Yes, this is obvious when you consider that the metric of the cylinder is flat: it's the same as the metric of the Euclidean plane. So locally it must have the same solutions to Killing's equation as the flat Euclidean plane does.

that, however, cannot be globally extended as smooth KVF on the entire manifold.

Yes, because the global topology of the cylinder is different from that of the Euclidean plane.

An instructive exercise is to consider the spacetime version of this, where the metric on the cylinder is the 1x1 Minkowski metric.

 

[cianfa72]

Yes, because the global topology of the cylinder is different from that of the Euclidean plane.

In other words there is not any finite diffeomorphism/isometry that "rotate" points about any point on the cylinder.

 

[PeterDonis]

In other words there is not any finite diffeomorphism/isometry that "rotate" points about any point on the cylinder.

No, there is not any global isometry that does that for the entire cylinder. But there are local isometries that do it for finite open regions centered on some chosen point.

 

[cianfa72]

But there are local isometries that do it for finite open regions centered on some chosen point.

Ah ok, yes. Just take a finite open region and unroll it on the plane.

 

[PeterDonis]

Just take a finite open region and unroll it on the plane.

Yes.