Massive particles and antiparticles each with mass $m$ and respective number densities $n(t)$ and $\bar{n}(t)$ are present at time $t$ in the radiation era of an expanding universe with zero curvature and no cosmological constant. Assuming they interact with crosssection $\sigma$ at speed $v$, explain, by identifying the physical significance of each of the terms, why the evolution of $n(t)$ is described by

$$\frac{d n}{d t}=-3 \frac{\dot{a}}{a} n-\langle\sigma v\rangle n \bar{n}+P(t)$$

where the expansion scale factor of the universe is $a(t)$, and where the meaning of $P(t)$ should be briefly explained. Show that

$$(n-\bar{n}) a^{3}=\text { constant }$$

Assuming initial particle-antiparticle symmetry, show that

$$\frac{d\left(n a^{3}\right)}{d t}=\langle\sigma v\rangle\left(n_{\mathrm{eq}}^{2}-n^{2}\right) a^{3}$$

where $n_{\mathrm{eq}}$ is the equilibrium number density at temperature $T$.

Let $Y=n / T^{3}$ and $x=m / T$. Show that

$$\frac{d Y}{d x}=-\frac{\lambda}{x^{2}}\left(Y^{2}-Y_{\mathrm{eq}}^{2}\right)$$

where $\lambda=m^{3}\langle\sigma v\rangle / H_{m}$ and $H_{m}$ is the Hubble expansion rate when $T=m$.

When $x>x_{f} \simeq 10$, the number density $n$ can be assumed to be depleted only by annihilations. If $\lambda$ is constant, show that as $x \rightarrow \infty$ at late time, $Y$ approaches a constant value given by

$$Y=\frac{x_{f}}{\lambda}$$

Why do you expect weakly interacting particles to survive in greater numbers than strongly interacting particles?