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Integrated Boltzman equation for dark matter

  1. May 29, 2015 #1
    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. jcsd
  3. Jun 3, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Jun 6, 2015 #3
    Still need help!
     
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