# Will the CMB rest frame always coincide with co-moving coordinates?

I understand that the CMB rest frame for a typical FLRW universe should coincide with co-moving coordinates, but under what conditions won't the two coincide?

For example if the universe had some net spin (an abhorrent thought I know!) then would the two frames still coincide?

EDIT: I need to explain further. Say the Universe is the closed FLRW model, EXCEPT that it has a net spin. In this case the rest frame of the CMB would necessarily be different from comoving coordinates. Wouldn't the frame in which space appeared isotropic and homogeneous not be comoving but rather moving with center of momentum frame? If the three-sphere's large enough, there appears no deviation from isotropy within an observer's hubble radius. Of course there's a curious situation at the "poles" of the rotation where the isotropic frame is a rotating one (making me think towards Mach's principle). anyway, this was my attempt at a an example for my question. thanks

## 1 Answer

By your assumption, we are talking about the FLRW Universe. Such a universe is by definition isotropic and homogeneous so there can't be any preferred direction at any point. If there were a difference between the CMB frame and the co-moving frame, it would indeed produce a preferred direction at some/most points (the direction of motion of one frame relatively to the other), so these two frames must indeed coincide in an FLRW Universe, pretty much by its definition.

To address the differences between such frames and effects, you have to go beyond the FLRW Ansatz. Once you do so, there are indeed various different, mutually inequivalent "generalizations of what used to be the co-moving frame" in the FLRW Ansatz.

• So a comoving observer is just a free falling observer in an isotropic and homogenous universe (which we have by definition in a FLRW-simplified universe) if I get that right? – Yukterez May 25 '16 at 7:21
• Yup, the comoving observer is surely a freely falling one. But note that "the" freely falling observer isn't unique. The free fall dictates the right acceleration but it doesn't dictate the right initial velocity. You must have the right initial velocity - one in which e.g. the stress-energy tensor (of CMB or anything else) has no mixed $T_{0i}$ components - to get the right comoving frame. – Luboš Motl May 25 '16 at 7:55
• Comoving observers are those who have constant spatial coordinates in standard coordinate chart. This (not very covariant statement, I know) will satisfy the demand that stress energy tensor has no mixed components, as Lubos Motl mentioned. However, I don't think this requirement fixes the comoving observer uniquely. For example, in Milne universe stress tensor vanishes (Milne universe is just part of Minkowski space in disguise of some non-standard coordinates) – Blazej May 25 '16 at 11:21