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So for a perfect fluid in General relativity, my impression is the average velocity only matters and the distribution of velocities does not in the fluid approximation (otherwise I would have some dependence on temperature). Can someone point me to a reference or explain on why this is the case?

It's not obvious to me why an average description can work in general relativity since it's not a linear theory.

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  • $\begingroup$ See (from page 505): archive.org/details/… , $\endgroup$
    – The Tiler
    Commented Jan 29, 2023 at 19:52
  • $\begingroup$ and page 1: .... Accordingly, when we speak of infinitely small elements of volume, we shall always mean those which are “physically” infinitely small, i.e. very small compared with the volume of the body under consideration, but large compared with the distances between the molecules. The expressions fluid particle and point in a fluid are to be understood in a similar sense. If, for example, we speak of the displacement of some fluid particle,.... $\endgroup$
    – The Tiler
    Commented Jan 29, 2023 at 20:08
  • $\begingroup$ "...If, for example, we speak of the displacement of some fluid particle, we mean not the displacement of an individual molecule, but that of a volume element containing many molecules, though still regarded as a point." L&L $\endgroup$
    – The Tiler
    Commented Jan 29, 2023 at 20:34
  • $\begingroup$ @TheTiler how does one decide/prove what a good volume is? $\endgroup$
    – More Anonymous
    Commented Jan 31, 2023 at 4:24
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    $\begingroup$ No problem, the important thing is to know how to transmit the idea and I gained to have deepened my own knowledge, thank you to you too with your well-posed questions :-) $\endgroup$
    – The Tiler
    Commented Feb 3, 2023 at 7:43

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A perfect fluid is in local thermal equilibrium, and the velocity distribution must be the one given by the equilibrium partition function. For a non-interacting gas this is the Maxwell-Juttner distribution. In thermal equilibrium the fluid is completely characterized by thermodynamic variables, for example energy density and pressure, and symmetries fix the form of the stress tensor, $T_{\mu\nu}=(e,p,p,p)$ in the local rest frame. The stress tensor is the only input required by GR.

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