special relativity - What is the equation of state for a relativistic fluid/gas?

Say we have a relativistic fluid/gas, as we have in some astrophyical systems.

Now let us write:

• $e$ - energy density in the fluid's rest frame.




• $P$ - pressure in the fluid's rest frame.




• $n$ - number density in the fluid's rest frame.




• $m$ - mass of the particles.

I know that for the non-relativistic case we have:

$$e=nmc^2+frac{1}{hat{gamma}-1}P$$

where $hat{gamma}$ is the adiabatic index. $hat{gamma}=1+frac{2}{f}$ for a gas with $f$ degrees of freedom.

For the ultra-relativstic case we have:

$$e=3P$$

My question is what is $e(P,n)$ for a relativstic case (which is the general case of the 2 limits shown above)? I would also like to know how to derive it.

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Is the following way the correct way to do it ? :

The number density of particles is:
$$n=int_{0}^{infty} n_p(p) dp$$

The pressure is:
$$P=int_{0}^{infty} frac{1}{3} p v(p) n_p(p) dp$$

The energy density is:
$$e=int_{0}^{infty} epsilon(p) n_p(p) dp$$

where:

$$n_p(p)= (2s+1)frac{1}{ e^{({epsilon(p)-mu})/{k_B T}}+(-1)^{2s+1} } frac{4pi p^2}{h^3}$$

Here $s$ is the spin of the particles, for electrons $s=frac{1}{2}$.

$$epsilon(p)=(m^2c^4+p^2c^2)^{frac{1}{2}}$$

$$v(p)= frac{depsilon}{dp}=frac{p}{m}left(1+left(frac{p}{mc}right)^2right)^{-frac{1}{2}}$$

From calculating the three integrals above we can finally obtain $e(P,n)$.

• Can anyone confirm this is the proper way to do it, or am I missing something here?




• It seems as if those integrals cannot be solved analytically - is this
  true?




• Perhaps in this case there is no explicit formula for $e(P,n)$?

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