# Integrated Boltzman equation for dark matter

1. May 29, 2015

### karlzr

In Dodelson's book, the equation for a scattering process $a + b \Leftrightarrow c + d$ is given as
$a^{-3} \frac{ d (n_a a^3)}{d t}=-n^{\text{eq}}_a n^{\text{eq}}_b<\sigma v>(\frac{n_a n_b}{n^{\text{eq}}_a n^{\text{eq}}_b} - \frac{n_c n_d}{n^{\text{eq}}_c n^{\text{eq}}_d}) = - <\sigma v>(n_a n_b - \frac{n^{\text{eq}}_a n^{\text{eq}}_b}{n^{\text{eq}}_c n^{\text{eq}}_d}n_c n_d)$
with $n_i = g_i e^{\mu_i/T}\int \frac{d^3p}{(2\pi)^3}e^{-E_i/T}$ and the equilibrium number density $n^{\text{eq}}_i = g_i \int \frac{d^3p}{(2\pi)^3}e^{-E_i/T}$.
(1)Why is this $n^{\text{eq}}_i$ the equilibrum number density since we don't necessarily have $\mu=0$ in equilibrium: $n_i=n^{\text{eq}}_i$ ?
(2) In papers about dark matter, it is very often to see $a^{-3} \frac{ d (n_a a^3)}{d t}=-<\sigma v>(n_a n_b - n^{\text{eq}}_a n^{\text{eq}}_b)$. 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 $\chi_i + X \Leftrightarrow \chi_j + Y$ is said to be $\propto (n_i n_X - n^{\text{eq}}_i n^{\text{eq}}_X)$ when both $\chi_i$ and $\chi_j$ are freezing out. I don't get it because $\chi_j$ isn't in equilibrium.

2. Jun 3, 2015