Before electron-positron annihilation, photons, electrons, and positrons share one temperature. Their effective entropy degrees of freedom are
$$
g_{*s}^{\rm before}=2+\frac78(2+2)=\frac{11}{2}.
$$
The neutrinos have already undergone <thermal decoupling in cosmology>, so $T_\nu a$ remains constant and they receive none of the electron-positron entropy. In the still-coupled electromagnetic plasma, <cosmological entropy conservation> gives
$$
\frac{11}{2}T_{\rm before}^3a_{\rm before}^3
=2T_\gamma^3a^3.
$$
The decoupled neutrino temperature at the same later time is $T_\nu=T_{\rm before}a_{\rm before}/a$. Dividing the two relations gives the <Cosmic neutrino background> temperature
$$
\boxed{\frac{T_\nu}{T_\gamma}=\left(\frac4{11}\right)^{1/3}}.
$$
This instantaneous-decoupling calculation neglects the small reheating correction from non-instantaneous neutrino decoupling.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 310 / 2 / a / Solution

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The cosmic neutrino background is the relic population of neutrinos that thermally decoupled before electron-positron annihilation. Entropy transfer to photons gives
$$
\frac{T_\nu}{T_\gamma}=\left(\frac4{11}\right)^{1/3}.
$$


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