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For a relativistic fluid, the equation of state is given by:

$$ \rho = \rho_0 + \frac{3p}{c^2} \,.$$

The above expression is nicely derived in Weinberg (1972). Although I was told that for a compressible fluid that is relativistically hot (i.e. $p \gg \rho_0 c^2) $ under a constant acceleration, $g$, but absent of bulk flows in equilibrium, the following is the equation for motion and energy (cf. Allen & Hughes, 1984):

$$ \begin{align} \frac {\bf{V}}{c^2} \frac {\partial p}{\partial t} + \nabla p & ~~ = ~~ -\left(\frac {4p}{c^2} + \rho_0\right) \frac {\partial \bf{V}}{\partial t} - g \left(\rho_0 + \frac {4p}{c^2} \right) \\ \\ {\bf V} \cdot \frac {\nabla p}{c^2} & ~~ = ~~ \frac {3}{c^2} \frac {\partial p}{\partial t} + \nabla \cdot \left(\frac {4p}{c^2} \bf{V} \right) \end{align} $$

Are these equations relativistic and actually valid despite no Lorentz factor which may be necessary for thermal speeds approaching $c$? I tried checking by plugging the above equation of state into the non-relativistic hydrodynamic equations for momentum and energy (i.e. compressible Euler equations), and recover expressions very close to the above, but it is slightly off by coefficients of 3 or 4 which could either be due to errors or relativistic effects. Thus are these two equations correct? If so, when is the above generally valid?

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I'll sketch a derivation of the first equation, and show that it is an approximation for small speeds.

In GR if you start from the stress-energy tensor of a perfect fluid and assume a weak-field metric, you get the following equation for fluid particles: $$(\rho+p/c^2)(\partial_\beta u^\alpha +\Gamma^\alpha_{\lambda\beta} u^\lambda )u^{\beta}+\partial^\alpha p+\partial_\beta p\, u^\beta u^\alpha / c^2=0$$ In the Newtonian limit it reduces to the usual Euler equation. Next we substitute your equation of state and write $u^\alpha \approx (c, \mathbf v)$. For $\alpha = i$ we get: $$(\rho_0+4 p/c^2)\left(\frac{\partial \mathbf v}{\partial t}+\mathbf v \cdot \nabla \mathbf v +\Gamma^i_{\lambda\beta} u^{\beta} u^\lambda \right)+\nabla p+\left(\frac{\partial p}{\partial t}+\mathbf v \cdot \nabla p\right) \mathbf v / c^2=0$$ In the weak-field limit the only surviving Christoffel symbol is in this case $\Gamma^i_{00}\approx \mathbf g / c^2$, the gravitational potential. Ignoring terms $\mathcal{O}(\mathbf v^2)$: $$\nabla p+\frac{\mathbf v}{c^2} \frac{\partial p}{\partial t}=-(\rho_0+4 p/c^2)\left(\frac{\partial \mathbf v}{\partial t} +\mathbf g\right)$$ which is the first equation you wrote down.

It is therefore valid when: 1) the speeds involved are much less than the speed of light and the gravitational field is 2) weak and 3) static. The paper you quote (Allen & Hughes 1984) explicitly states that these conditions hold for the problem they're considering. For more on fluids in GR you can check the references they quote: Weinberg 1972 and Landau & Lifshitz 1963.

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  • $\begingroup$ To clarify: how do the assumptions you've taken differ from the classical Newtonian limit? Also, Allen & Hughes consider the situation of strong acceleration (i.e. $g \gg kc^2$)—is not the usage of weak gravity invalid in the above derivation? $\endgroup$
    – Mathews24
    Commented May 12, 2018 at 13:53
  • $\begingroup$ @Mathews24 In the Newtonian limit, there shouldn't be any factors of $c$, so this is not the usual Newtonian limit, but rather something ad hoc. As to the strong gravity thing, you can interpret that statement as saying something about $k$ rather than $g$: it would be the limit of long wavelengths $\endgroup$
    – John Donne
    Commented May 12, 2018 at 19:47
  • $\begingroup$ I agree; I suppose the assumptions you stated above appear almost exactly those assumed in the Newtonian limit except for the possibility of $p > \rho_o c^2$ and I was attempting to clarify any additional differences. I agree that "strong acceleration" can be viewed as a limit on wavelengths, but what does comparing $k$ to $g/c^2$ physically represent? Is the latter quantity of any significance in any contexts? $\endgroup$
    – Mathews24
    Commented May 12, 2018 at 19:52
  • $\begingroup$ I don't know exactly. I guess it's some sort of "size of the system". For a Newtonian body with zero energy $v^2/g\approx R$ $\endgroup$
    – John Donne
    Commented May 12, 2018 at 20:32
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    $\begingroup$ It appears to be a limit comparing the wavevector to the Christoffel symbol. $g/c^2 \gg 2/\lambda \implies g/(\lambda/\tau)^2 \gg 2/\lambda \implies \frac{1}{2}g\tau^2 \gg \lambda$ where $\tau$ is the period for a causal response over the perturbation wavelength, $\lambda$. The limit essentially indicates the displacement due to gravity is much greater than the perturbation (i.e. "strong acceleration"). $\endgroup$
    – Mathews24
    Commented May 12, 2018 at 22:24

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