5
$\begingroup$

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?

$\endgroup$

1 Answer 1

3
+50
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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
    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
    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
    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
    May 12, 2018 at 20:32
  • 1
    $\begingroup$ 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
    May 12, 2018 at 22:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.