How do We know that CMB is not moving itself, because the content of the universe was moving during decoupling?Dale said:They could compare each observer’s motion to the local inertial frame where the cosmic microwave background radiation is isotropic.
Because we arbitrarily choose the reference frame where it is isotropic. We do that because it makes the math easier and we arbitrarily prefer easy math over difficult math.exander said:How do We know that CMB is not moving itself, because the content of the universe was moving during decoupling?
Because seeing the CMB as isotropic is the meaning of "moving with space due to the expansion of the universe", which is what you asked about.exander said:How do We know that CMB is not moving itself
The basic point is that it is spacetime that underlies everything. There is no "space" or "time", just spacetime. If you want to talk about space or time separately you have to first slice 4d spacetime into a stack of 3d slices, each of which is "the universe at one time". But there isn't a unique way to do this. There are no gridlines built in to spacetime telling you how to slice (although there are limits about which direction you can sensibly call time).exander said:How can an observer distinguish two objects that are moving from each other in space compared to those that are moving with space due to expansion of the universe?
We know that the CMB has no net motion in space because we defined space to be the slicing of spacetime where the CMB has no net motion. It's true by definition.exander said:How do We know that CMB is not moving itself, because the content of the universe was moving during decoupling?
Claiming that CMB has not net motion is equivalent to say that spacetime's slicing it defines gives "space" slices that are all homogeneous and isotropic.Ibix said:We know that the CMB has no net motion in space because we defined space to be the slicing of spacetime where the CMB has no net motion. It's true by definition.
A non-tautological question would be: is the FLRW spacetime an accurate model of the real spacetime? Or, equivalently, are there observations which are inconsistent with us being in an FLRW spacetime (or at least one that looks FLRW on large scales)? The answer to those is a resounding maybe.
No. The Robertson-Walker spacetime is the most general spacetime satisfying spatial homogeneity and isotropy. It is basically just implementing those symmetries.cianfa72 said:Is possibile to conceive different spacetime models from FLRWs that are consistent with the homogeneity and isotropy features of CMB ?
Homogeneity no, not that I can think of. Isotropy is trivial; let the mass density vary with radial distance from the Earth. There are probably limits to how much variation you can have and be consistent with the observed existence of redshift of galaxies, and you probably get a complex relationship between redshift and intensity and time. It's also extremely contrived - of all thecianfa72 said:How the CMB is related to the FLRW spacetime models ? Is possibile to conceive different spacetime models from FLRWs that are consistent with the homogeneity and isotropy of CMB ?
Therefore CMB's both homogeneity and isotropic features in every point of the Universe force us to assume one of the 3 FLRW models (Spherical, Euclidean or Hyperbolic) ?Orodruin said:No. The Robertson-Walker spacetime is the most general spacetime satisfying spatial homogeneity and isotropy. It is basically just implementing those symmetries.
Then the model won't be homogeneous.Ibix said:Isotropy is trivial; let the mass density vary with radial distance from the Earth.
Sure - that's why I said I couldn't think of a homogenous model except for FLRW.PeterDonis said:Then the model won't be homogeneous.
I don't think we can conclude that the universe is homogeneous from the CMB alone. That only tells you that it's isotropic. You need to add the cosmological principle, or a study of other matter in the universe, or something of the sort to get homogeneity.cianfa72 said:Therefore CMB's both homogeneity and isotropic features in every point of the Universe force us to assume one of the 3 FLRW models (Spherical, Euclidean or Hyperbolic) ?
Yet, the only spacetime models that are both homogeneous and isotropic are the FLRWs models.Ibix said:I don't think we can conclude that the universe is homogeneous from the CMB alone. That only tells you that it's isotropic. You need to add the cosmological principle, or a study of other matter in the universe, or something of the sort to get homogeneity.
Yes. But the CMB is not the only piece of evidence supporting them.cianfa72 said:Yet, the only spacetime models that are both homogeneous and isotropic are the FLRWs models.
I see. I understood @cianfa72 to be asking if there were any other models that satisfied both properties, homogeneity and isotropy. If we only need to satisfy isotropy, then any spherically symmetric spacetime will do (for example, Schwarzschild).Ibix said:Sure - that's why I said I couldn't think of a homogenous model except for FLRW.
Depends how you read "consistent with the homogeneity and isotropy features of CMB". The CMB is isotropic, but without a few billion years' measurement I don't think we can say it's homogeneous with any confidence from CMB observation alone. Of course, we do have other lines of evidence pointing to homogeneity, like the large scale homogeneity of matter.PeterDonis said:I see. I understood @cianfa72 to be asking if there were any other models that satisfied both properties, homogeneity and isotropy.
Isotropy and a CMB. And we need to be at the middle, so it can't be pure Schwarzschild. That's actually why I was saying there were probably limits to the variation of density with radial distance that you can get away with. Presumably something that is not far from homogeneous would behave not that differently from an FLRW spacetime, would necessarily be isotropic to the special observer at the origin, but not be homogeneous.PeterDonis said:If we only need to satisfy isotropy, then any spherically symmetric spacetime will do (for example, Schwarzschild).
Sorry, when you talk of homogeneity and isotropy of spacetime you actually mean properties of the spacetime itself or properties of the spacelike hypersurfaces of specific foliations ?PeterDonis said:I see. I understood @cianfa72 to be asking if there were any other models that satisfied both properties, homogeneity and isotropy. If we only need to satisfy isotropy, then any spherically symmetric spacetime will do (for example, Schwarzschild).
When we say a spacetime is homogeneous and isotropic, what we mean is that there exists a foliation by spacelike hypersurfaces which have the required Killing vector fields.cianfa72 said:when you talk of homogeneity and isotropy of spacetime you actually mean properties of the spacetime itself or properties of the spacelike hypersurfaces of specific foliations ?
Ok, so assuming CMB isotropy does mean assuming isotropy (i.e. the existence of rotation KVFs about a point on foliation's spacelike hypersurfaces) at any point of any foliation's spacelike hypersurfaces ?PeterDonis said:When we say a spacetime is homogeneous and isotropic, what we mean is that there exists a foliation by spacelike hypersurfaces which have the required Killing vector fields.
Not necessarily, no. It is theoretically possible to have a model where the CMB is isotropic but something else is not.cianfa72 said:assuming CMB isotropy does mean assuming isotropy
Ok, claiming CMB is isotropic from measurement amounts to say it is isotropic just from the single point it is observed, I believe.PeterDonis said:Not necessarily, no. It is theoretically possible to have a model where the CMB is isotropic but something else is not.
Yes.cianfa72 said:claiming CMB is isotropic from measurement amounts to say it is isotropic just from the single point it is observed
Therefore we have not a direct evidence (by measurements) that CMB is actually isotropic w.r.t. other points in the Universe.cianfa72 said:Ok, claiming CMB is isotropic from measurement amounts to say it is isotropic just from the single point it is observed, I believe.
That's correct.cianfa72 said:we have not a direct evidence (by measurement) that CMB is actually isotropic w.r.t. other points in the Universe.
Well, the homogeneity of matter in the universe suggests that the universe is FLRW and hence the CMB is also homogeneous. But we haven't directly sampled the CMB elsewhere, no, due to not having been anywhere else.cianfa72 said:Therefore we have not a direct evidence (by measurements) that CMB is actually isotropic w.r.t. other points in the Universe.
But we don't actually know that by direct observation either. We assume it, and we have no observations so far that are inconsistent with that assumption (at least not when we specify that we only mean on large enough distance scales). But we have no direct confirmation of it either, for the same reason we have no direct confirmation of isotropy.Ibix said:the homogeneity of matter in the universe
Good point. I was thinking that we have a reasonable idea that the number density of galaxies is homogeneous, but really we only know its isotropic because we're seeing back into the past and the visible density does vary with distance. We attribute that to ##a## varying only with time, but the "only" is an assumption derived from the cosmological principle.PeterDonis said:But we don't actually know that by direct observation either. We assume it, and we have no observations so far that are inconsistent with that assumption (at least not when we specify that we only mean on large enough distance scales). But we have no direct confirmation of it either, for the same reason we have no direct confirmation of isotropy.
If the CMB is homogeneous everywhere and isotropic at a point then it is isotropic at any point (on foliation's spacelike hypersurfaces).Ibix said:Well, the homogeneity of matter in the universe suggests that the universe is FLRW and hence the CMB is also homogeneous.
Yes, exactly. In other words, our homogeneous model is consistent with the data we have (on large enough distance scales), but that in itself does not prove that the universe must be homogeneous (on large enough distance scales).Ibix said:I was thinking that we have a reasonable idea that the number density of galaxies is homogeneous, but really we only know its isotropic because we're seeing back into the past and the visible density does vary with distance. We attribute that to ##a## varying only with time, but the "only" is an assumption derived from the cosmological principle.
Yes.cianfa72 said:If the CMB is homogeneous everywhere and isotropic at a point then it is isotropic at any point (on foliation's spacelike hypersurfaces).
What could prove "that the universe must be homogeneous (on large enough distance scales)"?PeterDonis said:... but that in itself does not prove that the universe must be homogeneous (on large enough distance scales).
Spherically symmetric spacetime has only 3 linearly independent spacelike KVFs (Killing Lie algebra has dimension 3). 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).PeterDonis said:If we only need to satisfy isotropy, then any spherically symmetric spacetime will do (for example, Schwarzschild).
Strictly speaking, nothing. We don't prove physical theories - they are always provisional. However, going to a distant galaxy and checking that the universe appears isotropic from there too would spike Peter and my (pedantic and not terribly plausible) "it's just isotropic from our current location" objection. Or you could wait a few million years until you can see what the matter density in distant galaxies looks like it does now in our neighbourhood (although there may be model-dependant objections to that approach).timmdeeg said:What could prove "that the universe must be homogeneous (on large enough distance scales)"?
Same arguments apply, I'm afraid. It has to be said that the more independent bits of evidence you gather that are consistent with a homogenous model, the more contrived the isotropic-only model usually has to be. We'd usually prefer the less contrived model - but we don't ever strictly rule it out until it's directly falsified.timmdeeg said:Would the detection of primordial gravitational waves be a prove? If that proves the correctness of the inflationary theory then I think that even if the creation of matter particles at the end of inflation doesn't happen exactly silmultaneous this would cause slightly different matter densities only locally.
You would have to prove that there are no models that fit our current data that are not homogeneous.timmdeeg said:What could prove "that the universe must be homogeneous (on large enough distance scales)"?
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.timmdeeg said:But at least experimental support for inflation would support the L-CDM model
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.timmdeeg said:and thus postulates like homogeneity on which this model is based, right?
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.cianfa72 said:Spherically symmetric spacetime has only 3 linearly independent spacelike KVFs (Killing Lie algebra has dimension 3).
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.cianfa72 said: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).
Again, see the caveat I gave above.cianfa72 said: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).
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 said: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.
If you give it a small enough mass, yes--but then you're making the spacetime very close to just being flat.cianfa72 said:the volume of such "ordinary object" can be taken in principle as infinitesimal.
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.cianfa72 said:the 3 rotational KVFs are about that "location", I believe.
For a 2-surface the maximally symmetric condition requires ##2*3/2=3## independent isometries.PAllen said: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)
Yes, this is homogeneous but nowhere isotropic.cianfa72 said: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).
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.PAllen said:Yes, this is homogeneous but nowhere isotropic.
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.cianfa72 said: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, because the global topology of the cylinder is different from that of the Euclidean plane.cianfa72 said:that, however, cannot be globally extended as smooth KVF on the entire manifold.
In other words there is not any finite diffeomorphism/isometry that "rotate" points about any point on the cylinder.PeterDonis said:Yes, because the global topology of the cylinder is different from that of the Euclidean plane.
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 said:In other words there is not any finite diffeomorphism/isometry that "rotate" points about any point on the cylinder.
Ah ok, yes. Just take a finite open region and unroll it on the plane.PeterDonis said:But there are local isometries that do it for finite open regions centered on some chosen point.
Yes.cianfa72 said:Just take a finite open region and unroll it on the plane.
