In Dodelson's Book, chapter 3, we have the collision term in Boltzmann equation is written as

$$\int\frac{d^3p_1}{(2\pi)^32E_1}\int\frac{d^3p_2}{(2\pi)^32E_2}\int\frac{d^3p_3}{(2\pi)^32E_3}\int\frac{d^3p_4}{(2\pi)^32E_4}(2\pi)^4\delta^3(p_1+p_2-p_3-p_4)\delta(E_1+E_2-E_3-E_4)\{f_1f_2[1\pm f_3][1\pm f_4]-f_3f_4[1\pm f_1][1\pm f_2]\}$$

I want to understand why the distribution functions are related as $$f_1f_2(1\pm f_3)(1\pm f_4)-f_3f_4(1\pm f_1)(1\pm f_2)$$ and not as in the classical form $ f_1f_2 - f_3f_4$.


  • $\begingroup$ The $f_1f_2(1-f_3)(1-f_4)$ tells me that states 1 and 2 should be occupied and states 3 and 4 should be unoccupied so I would guess 1 and 2 are incoming and 3 and 4 are outgoing particles. The flipped expression is because the reverse process can also happen. I don not understand the $\pm$ so maybe provide some context. $\endgroup$ Jul 28 '16 at 17:09
  • $\begingroup$ Isn't it the same? The $f_1f_2f_3f_4$ term is just cancelled. $\endgroup$
    – svavil
    Jul 28 '16 at 20:19

This form of the distribution function accounts for Bose-Einstein (+) or Fermi-Dirac (-) statistics. For example, in fermions, if $f_3$ or $f_4$ is occupied (that is, equals one), that means a particle cannot scatter into that state due to Pauli exclusion, so the probability of that process vanishes. Note that for fermions $f$ is between zero and one, while for bosons it can be any non-negative number (which makes the notation of $f$ a poor choice, but ok).

As an example of where this comes from, recall that for bosons:

$$ a_1^\dagger |N_1\rangle=\sqrt{N_1+1}|N_1+1\rangle$$

so a transition probability to the state $|N_1+1\rangle$ is enhanced by a factor of the number of bosons in the state plus one, relative to the case where this state is unoccupied. Extending this to an average occupation number, and taking into account both the forward and reverse processes, gives you the case of $$f_1f_2(1+ f_3)(1+ f_4)-f_3f_4(1+ f_1)(1+ f_2)$$,

while a similar analysis for fermions gives you the case with minus signs.


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