 Solution

For a relativistic equilibrium species with temperature $T_i$, the entropy density is $s_i=({\rho }_i+P_i)/T_i=4{\rho }_i/(3T_i)$. In units $k_B=\hslash =c=1$, bosons have ${\rho }_i={\pi }^2{g}_i{T}_i^4/30$, while fermions have $7/8$ times this expression. Relative to a reference bath temperature $T$, define the effective number of relativistic degrees of freedom for entropy byThe sums here concern relativistic species, or their effective entropy contributions. Since a physical comoving patch has volume proportional to $a^3$, cosmological entropy conservation givesfor an isolated adiabatic bath. After thermal decoupling in cosmology, one must apply this law separately to sectors that no longer exchange entropy.

Before electron-positron annihilation in cosmology, the electromagnetic bath contains two photon polarizations and four electron/positron spin states. After neutrino decoupling, its entropy degrees of freedom are initiallyand after annihilation only the two photon polarizations remain. Electromagnetic entropy conservation therefore givesThe decoupled Cosmic neutrino background receives none of this entropy; its distribution temperature obeys $a{T}_{\nu }=\text{constant}$, and $T_{\nu }=T_{\gamma }$ before annihilation. ThusThis is the temperature of a decoupled relativistic relic in the instantaneous-decoupling approximation. The ratio remains the same afterwards in the absence of additional photon heating. At late times $T_{\nu }$ denotes the redshifted relic distribution temperature even when neutrino masses become relevant; it need not be a kinetic-equilibrium temperature.