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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?

Patrick.B
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  • The post for which this question has been assimilated is quite different and is not at all related to my question. Please re-open or give a correct reference that answers the question. – Patrick.B Dec 13 '19 at 15:47
  • How is it different? Yours looks like a special case of http://physics.stackexchange.com/questions/71476/gravity-as-a-gauge-theory . If you think it's not just a special case of that, please explain how it's different and what you want to know that isn't answered there. –  Dec 13 '19 at 21:29
  • The gauge is typically on the metric tensor which removes the scaling on the $dt^{2}$ term. Essential this is a coordinate transformation in GR. See Einstein's hole: "https://plato.stanford.edu/entries/spacetime-holearg/#ModSpaTheBegGui" and Google for the lapse function in FLRW metric. Roberston Walker is essentially the same metric - it has the same problems. – Cinaed Simson Dec 13 '19 at 23:28
  • I changed the question so that it ls hopefully clearer what I want and that the (both) given linked references do not help. – Patrick.B Dec 16 '19 at 15:12

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