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In many research papers about inhomogeneous cosmology, one often considers spherically symmetric (LTB) spacetimes where in the co-ordinate frame $(t,r,\theta,\varphi)$ wherein the metric assumes the form $$ds^2=-dt^2+\frac{(Y'(r,t)dr)^2}{1-k(r)}+Y^2(r,t)(d\theta^2+\sin^2\theta d\varphi^2)\qquad (1) $$ one solves for the Einstein equations sourced by a comoving dust: \begin{equation}T_{\mu\nu}(r,t) = \rho(r,t)dt^2 \qquad (2)\end{equation} How is that motivated?

(The objection against this choice is that in a metric in the above diagonal form (1) it does not seem that a dust $T_{\mu\nu}=\rho U_{\mu}U_\nu$, where $U$ has a non-vanishing radial component, can be brought in the form (2) by a $(r,t) \mapsto (r',t')$-diffeomorphism without destroying the form (1) of the metric)

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I may be misunderstanding the question, but: --

"Comoving" is normally a description of coordinates or an observer, not a matter field. For example, the FLRW coordinates are considered comoving because an observer at constant $(r,\theta,\phi)$ is at rest with respect to the local matter. This property of the coordinates is not preserved under a diffeomorphism. That's why some coordinates are comoving and some are not. So here, the dust isn't comoving, the coordinates are comoving because they're moving with the dust.

A perfect fluid can be defined as a matter field that is completely characterized by its pressure and density, so that a frame exists in which the stress-energy tensor has the form $\operatorname{diag}(\rho,P,P,P)$. A dust is one in which the pressure vanishes. So since the matter content of the LTB spacetime is dust, it is guaranteed that at every point in spacetime, there is a frame in which the stress-energy has the form (2) that you give. This is the comoving frame.

(The objection against this choice is that in a metric in the above diagonal form (1) it does not seem that a dust $T_{\mu\nu}=\rho U_{\mu}U_\nu$, where $U$ has a non-vanishing radial component, can be brought in the form (2) by a $(r,t) \mapsto (r',t')$-diffeomorphism without destroying the form (1) of the metric)

I'm not sure if I'm correctly understanding your point, but I think what you're saying just amounts to a statement that it's nontrivial to find solutions to the Einstein field equations. If you take an LTB solution and then alter it by changing the state of motion of the dust, it won't be a solution anymore, for the same form of the metric.

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  • $\begingroup$ A dust being "comoving" in a certain coordinate frame means that in the dust's contribution $T_{\mu\nu}=\rho U_\mu U_\nu$ to the energy-stress tensor, one has $U \propto (1,0,0,0)$. $\endgroup$ – Thibaut Demaerel Aug 7 '18 at 7:49
  • $\begingroup$ Let me clarify again my objection: in an arbitrary spherically symmetric spacetime sourced by dust only, one has insufficient "freedom of diffeomorphism" to choose co-ordinates wherein the metric assumes the stated diagonal form (moreover with $g_{00}=-1$) and wherein the dust is comoving $\endgroup$ – Thibaut Demaerel Aug 7 '18 at 7:53
  • $\begingroup$ So picking a comoving dust in this very co-ordinate frame seems to require physical rather than mathematical motivation. $\endgroup$ – Thibaut Demaerel Aug 7 '18 at 7:55
  • $\begingroup$ @ThibautDemaerel: A dust being "comoving" in a certain coordinate frame means that... Yes, I think we're in agreement on that, but it just seems odd and nonstandard to me that you refer to this as if it were a property of the dust. It's a relationship between the coordinates and the dust. So picking a comoving dust in this very co-ordinate frame seems to require physical rather than mathematical motivation. The choice of a matter field and the choice of coordinates are not independent choices. Without the dust, you would have some other spacetime, a vacuum spacetime. The coordinates [...] $\endgroup$ – user4552 Aug 7 '18 at 16:31
  • $\begingroup$ [...] of the LTB spacetime and the vacuum spacetime would not be relatable to one another. I don't know if it's true, as you claim, that one cannot in general find comoving coordinates for a dust spacetime with this particular set of symmetries. But whenever it is possible, it's obviously a mathematical advantage to do so. $\endgroup$ – user4552 Aug 7 '18 at 16:32

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