Entropy density at zero chemical potential ($s=(\rho +P)/T$)

For an extensive equilibrium system, the thermodynamic Euler relation gives $E=TS-PV+\mu N$. At zero chemical potential, divide by $V$ to obtain $s=(\rho +P)/T$. Equivalently, $dP=(\rho +P)dT/T$ makesThe first law of thermodynamics identifies this as $dS$; the thermodynamic Euler relation fixes the entropy normalization. For adiabatic expansion satisfying the cosmological perfect-fluid continuity equation, the numerator on the right vanishes, proving cosmological entropy conservation.

For a dilute nonrelativistic gas with $x=m/T\gg 1$ and zero chemical potential, the Bose-Einstein distribution approaches the Maxwell-Boltzmann distribution. Expanding $E=m+p^2/(2m)+\cdots$ in the number density integral gives $n\sim g(mT/(2\pi ){)}^{3/2}{e}^{-x}$. Since $\rho =mn+3nT/2+\cdots$ and $P=nT$, the entropy density at zero chemical potential gives $s=n(x+5/2+\cdots )$ and hence the displayed leading expression. This includes the internal multiplicity $g$; a single real scalar field has $g=1$.

The displayed asymptotic uses natural units $k_B=c=\hslash =1$. Thus $m$ and $T$ are both energy scales, and $x=m/T$ is dimensionless. Restoring SI units replaces this ratio by $m{c}^2/(k_{B}T)$ and includes $k_B$ in the physical entropy per particle.