Past exam of the general relativity and cosmology course of the University of Oxford / 2012 / 4 / c / Solution

Assume an ideal nondegenerate hydrogen gas in thermal equilibrium and chemical equilibrium, with ground-state atoms and charge neutrality. Let $n_p=n_e=X{n}_B$ and $n_H=(1-X)n_B$. The ground-state Saha equation, with the usual electron/proton spin factors cancelling the atomic ground-state degeneracy, isInvert it and use $n_B=n_{\gamma }/s=(0.24/s)(k_{B}T/(\hslash c){)}^3$. This givessoHere $m_e$ is the electron mass denoted $M_e$ in the source, and $0.24(2\pi {)}^{3/2}=3.78$. At $X=1/2$, the left side is two. At $k_{B}T$ of order $B$, the right side is extremely small because $1/s\sim 10^{-9}$ and $B/(m_e{c}^2)\ll 1$. A large exponential is needed to compensate, requiring $B/(k_{B}T)\gg 1$ rather than a temperature of order the binding energy.

Physically, there are enormously many photons per baryon, so even a small high-energy tail of the thermal photon distribution can maintain ionization. Recombination is delayed until that tail is sufficiently depleted. This explains the low recombination temperature without solving the equilibrium equation; actual cosmological recombination also eventually requires rate equations once equilibrium fails.