How to compute the speed of sound in relativistic hydrodynamics? In Weinberg: Gravitation and Cosmology chapter 2.10 (Relativistic Hydrodynamics) the speed of sound is derived as
$v_s^2 = \left(\frac{\partial p}{\partial \rho}\right)_\sigma$
Here the derivative is taken isentropically, that is, at constant entropy $\sigma$ (this is because a sound wave travels so fast that its propagation can be approximated as an adiabatic process).
and the equation of state is given as
$p=(\gamma - 1) (\rho - nm)$
with $\gamma = 4/3$ for the case of a relativistic gas and $\gamma=5/3$ for the case of a nonrelativistic monoatomic gas ($nm$ is the rest mass density and $\rho$ the energy density).
Then Weinberg computes the sound speed for the relativistic case as
$v_s= \sqrt{\frac{1}{3}}$
in natural units and states that this is still safely less than unity (the speed of light). However what he doesn't show is the same calculation for the nonrelativistic case. This would yield
$v_s^2 = \frac{\partial ((5/3 - 1) (\rho - nm))}{\partial \rho} = 2/3$
and so the speed of sound in the non-relativistic case would actually be $v_s =\sqrt{\frac{2}{3}}$. So the speed of sound for a non-relativistic gas is actually higher than the speed of sound for an relativistic gas!? However one usually computes the non-relativistic speed of sound with another equation of state
$p=K\rho^{\gamma}$
which yields
$v_s^2=\frac{p}{\rho} = K \gamma \rho^{\gamma-1} = \gamma \frac{p}{\rho}$
This is a completely different result compared to the one we get from the equation of state given by Weinberg. So is the equation of state given by Weinberg wrong? If so, what is the correct equation of state for a relativistic gas and what is the real maximum speed of sound for a relativistic gas? If not, what is wrong with my calculation of the non-relativistic sound speed based on the equation of state given by Weinberg?
 A: Dear asmaier, sometimes it's useful to look into the real world to avoid some simple mistakes. The actual speed of sound in the air is 340 meters per second which is about one millionth of the speed of light. The squared speed of sound is one trillionth of the squared speed of light, so your claim that the non-relativistic speed of sound is comparable to the speed of light is surely incorrect, isn't it?
The first formula by Weinberg that you quoted is universally valid but you apply it incorrectly. Well, Weinberg doesn't make these errors so that his first $\sqrt{1/3}$ result for the relativistic gas is correct, too. However, in particular, $(\rho-nm)$ is supposed to measure the energy above the latent energy of $E=mc^2$, because only this "purely kinetic energy" contributes to the pressure; it must still be multiplied by $(\gamma-1)$.
Clearly, $(\rho-nm)$ is negligibly small in the non-relativistic limit. So your $c\sqrt{2/3}$ for the nonrelativistic speed of sound is clearly invalid. In fact, $(\rho-nm)$ measures the kinetic energy of the molecules, $mv^2/2$; the factor of $m$ cancels. Because it will get differentiated and multiplied by $(\gamma-1)$, and we take the square root at the end, you may see that the speed of sound comes out approximately equal to the speed of molecules in the air, just slightly smaller than that (mainly because of the factor of $1/2$ in the kinetic energy $mv^2/2$).
Recall that the (root mean square) speed of air molecules is 500 meters per second and the speed of sound is 340 meters per second.
There is no contradiction between the relativistic formulae and the non-relativistic ones, which are the limits of the former, and the non-relativistic materials' speed of sound is vastly smaller than the speed of light.
A: The speed of sound in the non-relativistic regime is much smaller then the speed of sound and agrees with your final formula. To see that you should remember that the particle density $n(\rho)$ is not a constant, but it is a function of the energy density. In fact, for the non-relativistic fluid
$$\rho(n)=nm c^2+\kappa n^{5/3}. $$
Now you can check that the energy dependence of the particle density will modify your relativistic calculation and the total result will agree with the correct final formula in your post.
