Boltzmann equation collisional operator in thermal equilibrium

Edited after Thomas' answer

Question 30.6. "Detailed balance":

System is in thermal equilibrium, and the physics of the system is governed by the collision $1+2\leftrightarrow 3+4$

Show (assuming thermal equilibrium) that collision rates balance:

$f_1 f_2 (1 \mp f_3) (1 \mp f_4) = f_3 f_4 (1 \mp f_1) (1 \mp f_2)$

This comes from Boltzmann right hand side, equation 30.40.

Solution:

And we get to the solution by saying: $f_i = (e^{\frac{E_i-\mu_i}{T}} \pm 1)^{-1}$, where $i$ is a particle species

and then saying that in thermal equilibrium $E_1+E_2=E_3+E_4$, $\mu_1 + \mu_2 = \mu_3 + \mu_4$.

The problem with the solution

from what I know function $f \rightarrow f(\mathbf{p})$, and the $E_1+E_2=E_3+E_4$ in thermal equilibrium should apply only for total energies $E_i$, so e.g. for $E_1=\int dE_1'$.

Hence, $f$ is:

$f_i(\mathbf{p}) = (e^{\frac{E_i(\mathbf{p})-\mu_i}{T}} \pm 1)^{-1}$

In this case, we get $E_i=E_j$ which still leads to the desired answer.

Proposed solution to the problem

Should be:

$\int_0^{\infty} f_1 f_2 (1 \mp f_3) (1 \mp f_4) dE = \int_0^\infty f_3 f_4 (1 \mp f_1) (1 \mp f_2) dE$

Am I understanding something wrong?

Consider a particle with energy $E=p^2/(2m)+\epsilon_\alpha$, where $\alpha$ labels inernal states, such as the vibrational and rotational levels of an atom. The distribution function is $f_\alpha(p)$, and the phase space measure in the collision integral is $$\int d\Gamma = \int d^3p\sum_\alpha .$$ Consider $2\to 2$ scattering, with a cross section $\sigma_{\alpha_1,\alpha_2}^{\alpha_3,\alpha_4}(p_1,p_2;p_3,p_4)$. The cross section satifies the usual symmetries under $(1\leftrightarrow 2)$ etc. Then the equilibrium condition is $$f_1f_2(1\mp f_3)(1\mp f_4)=f_3f_4(1\mp f_1)(1\mp f_2),$$ where $f_1=f_{\alpha_1}(p_1)$ etc. This is satisfied by $$f_\alpha(p) = \frac{1}{\exp(\beta(E_\alpha(p)-\mu))\pm 1}$$ with $E_\alpha(p)=p^2/(2m)+\epsilon_\alpha$ as above and $E_1+E_2=E_3+E_4$.
• Thanks a lot for the reply Thomas, I have a question; in your notation is $E_1=\int dE_a$? – Otto Jun 22 '15 at 15:23