The reaction maintaining chemical equilibrium is
$$
\boxed{e^-+e^+\leftrightarrow\gamma+\gamma}.
$$
For $T<m_e$ and negligible <chemical potentials>, the <Electron> and <Positron> are dilute and obey the <nonrelativistic equilibrium number density>. Including two spin states for each particle,
$$
n_{e^-}+n_{e^+}=4\left(\frac{m_eT}{2\pi}\right)^{3/2}e^{-m_e/T}.
$$
The <photon number density> is $n_\gamma=2\zeta(3)T^3/\pi^2$, so the <electron-positron equilibrium pair abundance below the electron mass> is
$$
\boxed{
\frac{n_{e^-}+n_{e^+}}{n_\gamma}
=\frac{2\pi^2}{\zeta(3)}
\left(\frac{m_e}{2\pi T}\right)^{3/2}e^{-m_e/T}}.
$$
If $n_e$ denotes only one of the charge-conjugate species, this expression is divided by two.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 310 / 2 / b / Solution

For $T\ll m_e$ and negligible chemical potentials, the combined equilibrium density of electrons and positrons divided by the <photon number density> is
$$
\frac{n_{e^-}+n_{e^+}}{n_\gamma}
=\frac{2\pi^2}{\zeta(3)}
\left(\frac{m_e}{2\pi T}\right)^{3/2}e^{-m_e/T}.
$$
The exponential suppression eventually makes pair equilibrium incompatible with the residual electrons required by <charge neutrality>.

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