The FRW metric is given by $$ds^2=dt^2-a^2(t)\Big[\frac{dr^2}{1-kr^2}+r^2(d\theta)^2+(r\sin\theta)^2(d\phi)^2\Big].$$ There is a time $t$ sitting in this metric. In which frame is this time measured?
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$\begingroup$ it's a coordinate time. It's the same as comoving time in comoving frame. $\endgroup$– KosmCommented Jun 11, 2017 at 15:17
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$\begingroup$ Is the time $t$ same for all comoving observers? $\endgroup$– SRSCommented Jun 12, 2017 at 6:12
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1$\begingroup$ yes, but there is no difference between any two comoving observers, they are at rest wrt each other $\endgroup$– KosmCommented Jun 12, 2017 at 6:24
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$\begingroup$ Well, I would not say that they are at rest which each other (even if in a sense it is true) as they experience for instance the redshift phenomenon from other co-moving observers... therefore there is relative motion also described as the expansion of space. $\endgroup$– Valter MorettiCommented Jun 12, 2017 at 12:03
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$\begingroup$ @Valter Moretti, I agree. They are at rest in a sense that they have 0 peculiar velocity, but redshift is still there $\endgroup$– KosmCommented Jun 12, 2017 at 18:50
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1 Answer
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It is nothing but the proper time measured at rest with one of the various galaxies where the expansion is isotropically observed.
answered Jun 11, 2017 at 17:21