Integrated Boltzmann equation for dark matter In Dodelson's book, the equation for a scattering process $ a + b \Leftrightarrow c + d $ is given as
\begin{align}
a^{-3} \frac{\mathrm d (n_a a^3)}{\mathrm d t}&=-n^{\text{eq}}_a n^{\text{eq}}_b\langle\sigma v\rangle\left(\frac{n_a n_b}{n^{\text{eq}}_a n^{\text{eq}}_b} - \frac{n_c n_d}{n^{\text{eq}}_c n^{\text{eq}}_d}\right)\\
 &= - \langle\sigma v\rangle\left(n_a n_b - \frac{n^{\text{eq}}_a n^{\text{eq}}_b}{n^{\text{eq}}_c n^{\text{eq}}_d}n_c n_d\right)
\end{align}
with $$n_i = g_i e^{\mu_i/T}\int \frac{\mathrm d^3p}{(2\pi)^3}e^{-E_i/T}$$ and the equilibrium number density $$n^{\text{eq}}_i = g_i \int \frac{\mathrm d^3p}{(2\pi)^3}e^{-E_i/T}.$$


*

*Why is this $n^{\text{eq}}_i$ the equilibrum number density since we don't necessarily have $\mu=0$ in equilibrium: $n_i=n^{\text{eq}}_i$ ?

*In papers about dark matter, it is very often to see $$ a^{-3} \frac{\mathrm d (n_a a^3)}{\mathrm d t}=-\langle\sigma v\rangle(n_a n_b - n^{\text{eq}}_a n^{\text{eq}}_b) .$$ This is true only if both $c$ and $d$ are in equilibrium, right? I have this question because in coannihilation dark matter hep-ph/9704361, the contribution from $\chi_a + X \Leftrightarrow \chi_b + Y$ is said to be $\propto (n_a n_X - n^{\text{eq}}_a n^{\text{eq}}_X)$ when both $\chi_a$ and $\chi_b$ are freezing out. I don't get it because $\chi_b$ isn't in equilibrium.
 A: In the reaction $A+B\leftrightharpoons C+D$
we can write the integrated Boltzmann equation as (Considering 
the FRW metric in flat space)
$$a^{-3}\frac{d(n_{A}a^{3})}{dt}=-\langle\sigma v\rangle\left(n_{A}n_{B}-\frac{n_{A}^{eq}n_{B}^{eq}}{n_{C}^{eq}n_{D}^{eq}}n_{C}n_{D}\right)$$
If $C$,$D$ are SM ( standard model ) particles $C$,$D$ through some 
sequence of intermediate reactions can be in thermal equilibrium 
with the photons. So,
$$\frac{n_{C}n_{D}}{n_{C}^{eq}n_{D}^{eq}}=\frac{n_{ph}n_{ph}}{n_{ph}^{eq}n_{ph}^{eq}}$$
But, $n_{ph}=n_{ph}^{eq}$ because $\mu_{ph}=0$.
Hence, $$a^{-3}\frac{d(n_{A}a^{3})}{dt}=-\langle\sigma v\rangle\left(n_{A}n_{B}-n_{A}^{eq}n_{B}^{eq}\right)$$
This equation is valid at every temperature either $A$,$B$ are in 
equilibrium or out of equilibrium.This equation gives the profile 
of $n_{A}$ as the universe expands.
If $A$,$B$ are dark matter particles and $C$,$D$ are SM particles. $C$,$D$ 
are in thermal equilibrium with the photons at all temperatures 
whereas $A$,$B$ are in equilibrium with photons only in the early 
universe where the temperature is high enough to maintain 
equilibrium. As the temperature drops below freeze-out $A$,$B$ go out 
of equilibrium and $A$ freezes out (Because rate of reaction becomes smaller than rate of expansion of universe).
