- #1

karlzr

- 131

- 2

##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.