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The entropy density of a photon gas is its entropy per physical volume, $s=S/V$. For homogeneous equilibrium vacuum radiation with two states of electromagnetic polarization and zero photon chemical potential,The first equality is the entropy density at zero chemical potential, obtained from the thermodynamic Euler relation $U=TS-pV+{\mu }_{\gamma }{N}_{\gamma }$ with internal energy $U=uV$. The second uses the photon gas equation of state, $u=a_R{T}^4$ and $p=u/3$. If an independently constrained photon number gives a nonzero chemical potential, the general entropy density is $(u+p-{\mu }_{\gamma }{n}_{\gamma })/T$ and $u=a_R{T}^4$ need not hold. This distinction explains why the absence of a conserved number matters for entropy as well as for the spectral shape. ^[ref]

A short first law of thermodynamics check, for neighboring equilibrium states with ${\mu }_{\gamma }=0$, isThe thermodynamic Euler relation fixes the additive normalization. Equivalently, this radiation entropy tends to zero with temperature at fixed volume; no contribution from the containing walls is included. Interpreting $TdS$ as transferred heat and $pdV$ as boundary work requires a reversible thermodynamic process. For a reversible thermodynamic process that is also an adiabatic process, there is no entropy production or entropy exchange, soAn irreversible adiabatic process can produce entropy even without heat exchange, so the word adiabatic alone is insufficient to infer $T^{3}V=\text{constant}$.

For an analytic example, expand an initially equilibrated photon gas reversibly from $V_0$ to $8V_0$, keeping only the same two photon states of electromagnetic polarization. Its temperature halves, its energy density becomes $u_0/16$, its entropy density of a photon gas becomes $s_0/8$, and its total entropy stays $S_0$. The internal energy halves rather than remaining fixed: the missing half leaves as work against radiation pressure,This is a stipulated reversible radiation model, not a report of a piston experiment. Maintaining a real cavity along this path requires controlling wall heat exchange and including the wall heat capacity in any actual apparatus balance.

For a computed scale at stipulated $T=2.725\mathrm{K}$, $s\simeq 2.04122\times 10^{-14}\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{m}}^{-3}$. Combining this with the photon number density givesThis is the total equilibrium radiation entropy divided by its mean photon number, not a separately assigned entropy carried by each labeled particle. During the ideal expansion, $n_{\gamma }\propto T^3$ and $T^{3}V$ is fixed, so the mean number $N_{\gamma }=n_{\gamma }V$ also stays constant. This follows from the path; it was not imposed as the independent number constraint used to establish chemical potential.

The beautiful connection is that expansion lowers each typical photon energy and dilutes the radiation in just the proportions needed to retain its equilibrium spectral shape and total entropy. For the cosmic microwave background, homogeneous expansion of the universe has physical volume proportional to the cube of the scale factor; fixed-species cosmological entropy conservation therefore gives $T\propto a^{-1}$. This describes preservation of an already established Planck photon distribution and connects the FIRAS spectral observation in photon gas to an entropy prediction; FIRAS did not perform a direct entropy measurement. When other equilibrated species disappear and transfer energy into radiation, use the effective number of relativistic entropy degrees of freedom, $g_{\ast s}$: the conserved quantity for the combined sector is $g_{\ast s}{T}^{3}V$, under the corresponding equilibrium and no-entropy production assumptions. A changing species population, separate decoupled sectors or entropy production requires an appropriate entropy balance rather than the pure photon gas formula. ^[ref]

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.

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.