Friedmann fluid equation in the non-relativistic case

Question

The Friedmann fluid equation I am referring to is: $$ a\frac{d\rho}{da} = -3(\rho+P) .$$

In the non-relativistic (low temperature) case for an ideal gas universe (representing matter), I know that the result is supposed to be that $\rho$ scales as $a^{-3}$. However, we know for the ideal non relativistic gas that: $$E=\frac{3}{2}kT.$$ Therefore, using $PV=kT$, we have $P=\frac{2}{3}\rho$. Substituting that to the Friedmann equation, the solution is that the density scales as the 5th power of $1/a$ instead of the 3rd. What's going on here?

Answer 1

If you correctly include the rest mass energy, then $$E=m+\frac{3}{2}kT$$ per particle on average (assuming only translational degrees of freedom), where $m$ is the particle mass. You could then multiply by the spatial number density of particles to obtain $$\rho_E=\rho_m+\frac{3}{2}p,$$ where $\rho_E$ is the energy density, $\rho_m$ is the mass density, and $p$ is the pressure. However, the nonrelativistic gas has by definition $kT\ll m$ and hence $p\ll\rho_E\simeq\rho_m$, so we normally just approximate $p=0$.