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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 $g=1$.

Consider a species in thermal equilibrium, with energy density and pressure depending on temperature and with vanishing chemical potential, as in the supplied thermodynamic relation. Put $U=\rho (T)V$. The first law of thermodynamics givesSince $dP/dT=(\rho +P)/T$,The differential is therefore exact:After fixing the irrelevant additive entropy zero, the entropy density at zero chemical potential isMore precisely the first law leaves an additive constant in $S$; extensivity fixes that constant for the usual entropy-density normalization. A nonzero chemical potential would instead require $Ts=\rho +P-\mu n$, so the stated formula is not a universal identity for a decoupled massive species with conserved particle number.

For a reversible thermodynamic process that is an adiabatic process in a comoving volume, $\dot{V}=3HV$ and the cosmological perfect-fluid continuity equation gives $\dot{\rho }=-3H(\rho +P)$. Differentiating the extensive expression explicitly, or using the exact differential above, givesThis is cosmological entropy conservation for the closed equilibrium gas with no entropy-producing energy injection.

For radiation in cosmology, $P=\rho /3$, so $s=4\rho /(3T)$. Comparing $s=c_1{g}_{\star }{T}^3$ and $\rho =c_2{g}_{\star }{T}^4$ yields the coefficient ratioWhen all relativistic species share the same temperature and count, the familiar normalizations are $c_1=2{\pi }^2/45$ and $c_2={\pi }^2/30$. More generally the energy and entropy effective counts can differ; the common $g_{\star }$ here uses the assumptions stated in the question.

Use the dark-sector temperature in every dimensionless integration variable: $x=m_{\varphi }/T_{\varphi }$ and $y=p/T_{\varphi }$. Assume a single real scalar field, $g=1$, and fast number-changing reactions keeping ${\mu }_{\varphi }=0$. For $x\gg 1$, the populated momenta satisfy $y=O(\sqrt{x})$, soThis is the nonrelativistic Maxwell-Boltzmann distribution limit of the Bose-Einstein distribution. The given Gaussian Gamma function integral givesSince ${\rho }_{\varphi }=m_{\varphi }n+3n{T}_{\varphi }/2+\cdots$ and $P_{\varphi }=n{T}_{\varphi }$, the entropy density at zero chemical potential is $s_{\varphi }=n(x+5/2+\cdots )$. Its leading nonrelativistic entropy at zero chemical potential is thereforeAn internal multiplicity $g$ would multiply this expression; the absence of $g$ in the printed result assumes a real one-state scalar.

Now use the constant separately conserved visible and dark entropy ratio and the relativistic visible-bath entropy law, $s_{\mathrm{SM}}=(2{\pi }^2/45)g_{\ast s}^{\mathrm{SM}}{T}_{\gamma }^3=\xi s_{\varphi }$. Since $T_{\varphi }=m_{\varphi }/x$,Thus the cannibal-sector temperature ratio hasThe numerical coefficient is approximately $0.525$. The visible-bath temperature is the photon temperature while that bath is internally thermalized; after distinct visible species acquire different temperatures, its total entropy must instead use the appropriate temperature-weighted $g_{\ast s}^{\mathrm{SM}}$.

The displayed relativistic scalar entropy formula must not be continued into this regime with $g_{\ast s}^{\varphi }=1$ held constant. If it is used as an effective definition at arbitrary temperature, thenand $T_{\gamma }/T_{\varphi }=(\xi g_{\ast s}^{\varphi }/g_{\ast s}^{\mathrm{SM}}{)}^{1/3}$ gives the same result. This specifies how the relativistic expression can legitimately be combined with the nonrelativistic one.