Consider the Roberston Walker metric in 4D, $$ds^2 = -dt^2 + a^2(t) \left( \frac{dr^2}{1-kr^2} +r^2 d\Omega^2_2 \right)$$ Now, if we consider the collapse of a spherically symmetric fluid, spherically symmetrythen requires in general its 4-velocity to be as $$u^\mu = (u^t(t,r), u^r(t,r),0,0),$$ where I ignore normalization here for simplicity. In most papers, I see that they pick what looks like a special case of the above by taking $u^r = 0$ (which they then call co-moving coordinates). Is there a particular reason why we do not seem to mind that this is only a special case? Indeed, at first view, it would seem that it should be in principle possible to find other solutions for which $u^r \neq 0$ identically. How would these two different solutions then be related?
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