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Comoving entropy is the entropy density multiplied by the physical volume factor $a^3$ for a fixed comoving region in an expanding homogeneous universe. In adiabatic expansion it is conserved. Separate decoupled radiation sectors conserve their own comoving entropy, so disappearance of massive species can heat the still-equilibrated sector relative to a decoupled species.

Let $s$ be proper entropy density in a homogeneous FLRW metric and $u^{\mu }$ the comoving fluid four-velocity. Reversible adiabatic expansion has a conserved entropy current,Since a fixed comoving volume has physical volume proportional to $a^3$, the covariant divergence givesThus the conserved quantity is comoving entropy; the proper entropy density falls as $a^{-3}$. This dilution does not require a decrease of the total entropy in the selected material volume.

For equilibrium relativistic matter at negligible chemical potentials, $s$ is proportional to $g_{\ast s}{T}^3$, where $g_{\ast s}$ is the effective number of relativistic entropy degrees of freedom. Conservation gives $aT{g}_{\ast s}^{1/3}=\text{constant}$, so $T\propto a^{-1}$ only while $g_{\ast s}$ is constant. Annihilation into a still-coupled bath can change its temperature relative to a decoupled species while conserving the bath's comoving entropy.^[ref]

Irreversible entropy production replaces the zero divergence by a nonnegative source. Separately decoupled sectors need their own entropy balances; imposing one common temperature after decoupling would add an assumption.

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.

For a relativistic species at zero chemical potential, $p=ϵ/3$ and $s=(ϵ+p)/T_i$, so its entropy density is proportional to $g_i{T}_i^3$. The Bose-Einstein distribution gives the bosonic normalization, while the fermionic thermal integral is $7/8$ of it. Species still in thermal equilibrium have $T_i=T$; freely streaming decoupled species can have another temperature and therefore contribute an extra factor $(T_i/T{)}^3$ when the total is expressed in units of $T^3$. This gives the two terms $N_{\ast }$ and $N_{SD}$ and their stated effective-degree-of-freedom weights.

After neutrino decoupling, the comoving entropy of the photon-electron-positron plasma is conserved separately from the neutrinos. Before electron-positron annihilation its effective entropy count is $2+(7/8)4=11/2$; afterwards it is $2$. Consequentlywhereas $a{T}_{\nu }$ stays equal to its pre-annihilation value. HenceThis is the instantaneous-decoupling, negligible residual-electron approximation used in the question. Heating of the still-coupled plasma, not cooling peculiar to neutrinos, creates the temperature difference.