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.
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:
$$n_p(p)= (2s+1)\frac{1}{ e^{({\epsilon(p)-\mu})/{k_B T}}+(-1)^{2s+1} } \frac{4\pi p^2}{h^3}$$
Here $s$ is the spin of the particles, for electrons $s=\frac{1}{2}$.
$$ \epsilon(p)=(m^2c^4+p^2c^2)^{\frac{1}{2}} $$
$$ v(p)= \frac{d\epsilon}{dp}=\frac{p}{m}\left(1+\left(\frac{p}{mc}\right)^2\right)^{-\frac{1}{2}} $$
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 $e(P,n)$?
