 Solution

The terms in the Boltzmann equation respectively describe dilution by expansion, pair annihilation, and pair production from the thermal bath. The annihilation rate per volume is $⟨\sigma v⟩n\bar{n}$; $P(t)$ is the corresponding production rate. Antiparticles satisfy the same equation with $n$ replaced by $\bar{n}$. Subtracting givesInitial particle-antiparticle symmetry therefore persists. Detailed balance in the thermal bath gives $P=⟨\sigma v⟩n_{\mathrm{eq}}^2$, soDuring radiation domination, with fixed effective relativistic degrees of freedom, $T\propto a^{-1}$ and $H=H_m/x^2$. Thus $\dot{x}=Hx$ and the dilution terms cancel in $Y=n/T^3$:After cosmological particle freeze-out, neglect $Y_{\mathrm{eq}}$. For constant $\lambda$ the exact reduced solution isFor the post-freeze-out relic abundance, the familiar answer $Y_{\infty }\simeq x_f/\lambda$ additionally neglects $1/Y_f$ compared with $\lambda /x_f$. It is an approximation, not an exact consequence of the post-freeze-out equation without initial data. A smaller annihilation cross-section means smaller $\lambda$, earlier decoupling and less depletion, so weakly interacting particles normally have a larger relic abundance.