 Solution

As the plasma cools through electron-positron annihilation, the pairs transfer their entropy to the coupled photons. Neutrinos have largely decoupled and do not share that heating in the instantaneous-decoupling approximation. Before annihilation the electromagnetic entropy count iswhile after it $g_{\ast s,\mathrm{after}}=2$. Conservation of electromagnetic comoving entropy, $g_{\ast s}{T}_{\gamma }^3{a}^3=\text{constant}$, gives the relative photon heating factorEquivalently $T_{\gamma }/T_{\nu }=(11/4{)}^{1/3}$ afterward. The exact $11/4$ count compares an asymptotic relativistic pair bath with the pair-free bath; at the supplied temperature some finite-electron-mass entropy loss has already begun.

The word “rises” requires care. The photons become hotter than an otherwise freely redshifting bath at the same scale factor; their absolute temperature need not rise during the expansion. In the smooth equilibrium process,since the pair contribution grows with temperature. This monotonic photon cooling during pair annihilation is slower than $a^{-1}$ while $a{T}_{\gamma }$ increases. The factor $1.401$ compares the photon temperature with the freely redshifting neutrino bath, not two consecutive absolute temperatures along the expanding history.