= Solution
For an equilibrium relativistic bath, $P=\rho/3$ and $\rho\propto g_{*s}T^4$, so $s=(\rho+P)/T\propto g_{*s}T^3$. <Cosmological entropy conservation> gives $g_{*s}T^3a^3=\text{constant}$. A decoupled relativistic species instead has $aT_{\rm relic}=\text{constant}$, 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:
$$
g_{*s}^{\rm before}=2+\frac78(4+4+6)=\frac{57}{4}.
$$
After muon annihilation, neutrinos are still coupled and $g_{*s}^{\rm after}=2+(7/8)(4+6)=43/4$. 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>
$$
\boxed{\frac{T_{\rm relic}}{T_\gamma}=\left(\frac{43}{57}\right)^{1/3}\simeq0.910}.
$$
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,
$$
\boxed{\frac{T_{\rm relic}}{T_\gamma}=
\left(\frac{43}{57}\frac4{11}\right)^{1/3}
=\left(\frac{172}{627}\right)^{1/3}\simeq0.650}.
$$
The neutrino-to-photon ratio is $(4/11)^{1/3}$, while the H-particle-to-neutrino ratio remains $(43/57)^{1/3}$. The printed $0.59\,{\rm MeV}$ marks the electron-positron annihilation epoch approximately; it is not an exact annihilation threshold.
Back to article page
