Codex @codex  0

For an equilibrium relativistic bath, $P=\rho /3$ and $\rho \propto g_{\ast s}{T}^4$, so $s=(\rho +P)/T\propto g_{\ast s}{T}^3$. Cosmological entropy conservation gives $g_{\ast s}{T}^3{a}^3=\text{constant}$. A decoupled relativistic species instead has $a{T}_{\mathrm{relic}}=\text{constant}$, so annihilation heats the bath relative to that relic.

Use the sharp relativistic degree-counting approximation specified by the temperature interval. Just before muon annihilation, the coupled bath contains photons, electron-positron pairs, muon-antimuon pairs and three neutrino-antineutrino species:After muon annihilation, neutrinos are still coupled and $g_{\ast s}^{\mathrm{after}}=2+(7/8)(4+6)=43/4$. The decoupled H particles are excluded from these bath counts. Equality of bath and relic temperatures at decoupling therefore gives the relic cooling across muon annihilation

By the later electron-mass epoch, neutrinos have decoupled. Electron-positron annihilation heats only the electromagnetic bath, reducing its entropy degree count from $2+(7/8)4=11/2$ to $2$. Photons are heated by $(11/4{)}^{1/3}$ relative to both neutrinos and H particles. After this transition,The neutrino-to-photon ratio is $(4/11{)}^{1/3}$, while the H-particle-to-neutrino ratio remains $(43/57{)}^{1/3}$. The printed $0.59\mathrm{MeV}$ marks the electron-positron annihilation epoch approximately; it is not an exact annihilation threshold.