Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 62 / 3 / b / Solution

Ignoring helium, charge neutrality gives $n_e=n_p$, while baryon conservation gives $n_B=n_p+n_H$. Defining the hydrogen ionization fraction $X_e=n_e/n_B$ yields $n_p=n_e=X_e{n}_B$ and $n_H=(1-X_e)n_B$. Substituting in the abundance relation givesUse $n_B=\eta n_{\gamma }$ and $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$ to obtain the Saha equation:If the positive right side is denoted by $A(T)$, the physical root isThis form avoids subtracting nearly equal numbers when $A$ is small, and makes the limits transparent: $X_e\simeq 1-A$ for $A\ll 1$, whereas $X_e\simeq A^{-1/2}$ for $A\gg 1$. It is an equilibrium prediction; a kinetic cosmological recombination treatment is needed after reactions cease to keep up with expansion.