Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 62 / 3 / ii / Solution

During radiation domination, the effective number of relativistic energy degrees of freedom, denoted $g_{\ast }$, gives the energy densityThe paper's ${\mathcal{N}}_{\mathrm{eff}}$ is this $g_{\ast }$; it is not the conventionally normalized effective number of neutrino species. With the specified reduced Planck mass, the Friedmann equation reads $H^2={\rho }_R/(3M_{\mathrm{Pl}}^2)$, whenceWrite the weak interaction conversion rate as $\Gamma =A{G}_F^2{T}^5$. The ratio $\Gamma /H$ scales as $T^3$, so it falls during cooling: conversion eventually becomes too slow to maintain neutron-proton chemical equilibrium. Equating the conversion and expansion rates givesOn suppressing the numerical coefficient, the cosmological weak freeze-out temperature is consequentlyThe one-sixth power reflects a balance between a conversion rate proportional to $T^5$ and a Hubble parameter proportional to $\sqrt{g_{\ast }}{T}^2$.