I've run into two conflicting derivations of the equation of state of a non-relativistic gas. However, the derivations of the relativistic equation of state of both sources agree. I think maybe the discrepancy is due to different definitions of energy (thermodynamics versus...?) I've summarized the arguments below.

$\textbf{Source 1}$: Stellar Structure & Evolution notes by O.R. Pols, page 24 http://www.astro.ru.nl/~onnop/education/stev_utrecht_notes/chapter1-4.pdf

Start with the pressure integral, $$ P = \frac{1}{3} \int_0^{\infty} \mathop{dp} n(p) p v_p = \frac{1}{3}n\langle pv_p\rangle, $$ where $n(p)$ is the distribution function in momentum, $p$ is the momentum, and $v_p$ is the velocity of a particle with momentum $p$. Also, recall that the "internal energy density" $u$ is $$ u = \int_0^{\infty}\mathop{dp} n(p) E_p = n \langle E_p \rangle, $$ where $E_p$ is the kinetic energy.

For the non-relativistic case, $v_p = \frac{p}{m}$, so $$ P = \frac{1}{3} n \langle pv_p \rangle = \frac{1}{3} n \langle \frac{p^2}{m} \rangle $$ $$ = \frac{1}{3} n \langle 2E_p \rangle = \frac{2}{3} u, $$ so the conclusion is that $$ P = \frac{2}{3} u.$$

$\textbf{Source 2}$: Ryden's $\textit{Introduction to Cosmology}$, page 55.

Equations of state here are denoted $$ P = w\varepsilon,$$ where $\textbf{I am assuming that Ryden's $\varepsilon$ is the same as Pols' $u$}$. Non-relativistic gas obeys the ideal gas law, $$ P = \frac{\rho}{\mu}kT,$$ where $\rho$ is the density and $\mu$ is the mean mass of the particles. For non-relativistic particles, $$ \varepsilon \approx \rho ,$$ so $$ P \approx \frac{kT}{\mu}\varepsilon $$

For a non-relativistic gas, $$ 3kT = \mu \langle v^2 \rangle $$ and thus $$ P = \frac{\langle v^2 \rangle}{3} \epsilon, $$ but where she claims $$\frac{\langle v^2 \rangle}{3} \ll 1,$$ which is clearly at odds with Pols' claim that that proportionality factor is $\frac{2}{3}$. In fact, Ryden takes the proportionality to be $0$, so that $P=0$ for a non-relativistic gas.

I am wondering how to reconcile these two arguments. Are Pols' $u$ and Ryden's $\varepsilon$ the same "type" of energy?

  • $\begingroup$ You are correct , I did the numbers and cannot make that term disappear, so I erased my answer. $\endgroup$ – Wolphram jonny Jun 20 '16 at 17:33

The derivation by Pols is correct. Ryden makes the strange decision to plug the relativistic rest energy $\varepsilon = \rho c^2$ into the classical ideal gas law. Surely it makes more sense to define a classical kinetic energy $$ u = \frac{1}{2}\rho\langle v^2\rangle $$ so that $$ P = \frac{2kTu}{\mu\langle v^2\rangle} = \frac{2}{3}u. $$

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  • $\begingroup$ Thanks for your answer! I actually found this derivation in a stat mech book later. I know, however, that it is common in cosmology to discuss "cold dust" equations of state, which are just as Ryden described, $P=0$. I'm still not quite sure how to reconcile that with the derivation you provided. Is it just that $u=0$ too, because maybe $u$ is just internal $\textit{kinetic}$ energy density (ie, not rest mass energy), and for cold non-relativistic matter, $\langle v^2 \rangle \approx 0$ to first approximation? $\endgroup$ – Physics_Plasma Jun 22 '16 at 1:54

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