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.


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?


1 Answer 1


There are indeed internal states, but I believe that this does not affect the thermal equilibrium condition.

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 $E_1+E_2=E_3+E_4$.

  • $\begingroup$ Thanks a lot for the reply Thomas, I have a question; in your notation is $E_1=\int dE_a$? $\endgroup$
    – OTH
    Jun 22, 2015 at 15:23
  • $\begingroup$ Hey, I realized I had formulated the question in a slightly bad way but after reading your answer I noticed what my problem was. I had been thinking that the energy conservation applies only when integrating through the whole energy distribution. When I think of the collisional operator describing a collision (or annihilation) between particles I realize that the energy is conserved. Anyway, voted as an answer because this was easier for me to understand than the original solution. I'll also fix the question. $\endgroup$
    – OTH
    Jun 22, 2015 at 15:52

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