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:


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:


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: $$e=\int_{0}^{\infty} \epsilon(p) n_p(p) dp $$


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

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 $e(P,n)$?

  • $\begingroup$ Although relativistic fluids are sometimes within the field of astronomy, I think this question is better fit at the Physics SE. $\endgroup$ Apr 26, 2016 at 18:31
  • 2
    $\begingroup$ @Hohmannfan i tried asking there too, didn't get an answer there. it is defently related to astronomy/astrophysics. don't dismiss my question as off-topic just because you can't answer it. $\endgroup$
    – TensoR
    Apr 27, 2016 at 0:37
  • $\begingroup$ @RobJeffries i've been told there is a foruma derived by Taub to describe this intermediate case. i've tried looking it up but couldn't find anything useful. do you happen to know about this formula derived by Taub? or maybe you can give me the best known analytic approximation (for a non-degenerate gas)? $\endgroup$
    – TensoR
    Apr 27, 2016 at 22:41
  • $\begingroup$ This is standard (advanced) textbook stuff. Certainly deakt with in Clayton's Principles of stellar evolution.. $\endgroup$
    – ProfRob
    Apr 27, 2016 at 22:57
  • $\begingroup$ See demonstrations.wolfram.com/AModelForFermiDiracIntegrals $\endgroup$
    – ProfRob
    Apr 27, 2016 at 23:04

1 Answer 1

  • Your equations seem correct. Note, you can also get $v$ without having to differentiate, from $E=\gamma mc^2$ and $p=\gamma mv$.

Here are some notes on relativistic fluids as related to stellar interiors, including derivation of these equations and their solutions in the non- and ultra-relavistic limits http://www.jb.man.ac.uk/~smao/starHtml/equationState.pdf

  • Correct, the integrals cannot be solved analytically (only in the two limits) so there is no general analytic formula
  • $\begingroup$ if you put the kinetic energy you exclude the $nmc^2$ in e, so its only a matter of what you want e to represent - do you want to include the rest energy or not. so including the $mc^2$ isn't really a mistake. i suppose that if it is needed, those integrals can be solved numericly, but i thought that maybe one can write them using special function like the zeta function or gamma function, and therefore obtain a formula. $\endgroup$
    – TensoR
    Apr 27, 2016 at 22:58
  • $\begingroup$ The non-rel and ultra-rel limits do have integral solutions with special functions, but as far as I am aware, no analytic solution for the general case. Regards $mc^2$, true, it doesn't really matter $\endgroup$
    – cnosam
    Apr 27, 2016 at 23:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .