H-theorem and Boltzmann equation applied to Boltzmann distribution

Question

Using the Boltzmann equation:

$$ \frac{dH}{dt} = \int_0^{\infty} dr \int_0^{\infty} ds W(r,s)[p_r - p_s][\ln{p_r} - \ln{p_s}],$$

and assuming $p_r = e^{-\beta r}$, the equation looks like

$$ \frac{dH}{dt} = \beta \int_0^{\infty} dr \int_0^{\infty} ds W(r,s)[e^{-\beta r} - e^{-\beta s}][s - r].$$

I would like to prove that as long as the transition rate satisfies detailed balance $$W(r,s) = W(s,r),$$ the system will be in equilibrium (meaning that the whole expression is equal to zero).

However, if I assume that $W(r,s)=1$, the integral doesn't seem to converge

Just for giggles I attempted $W(r,s)=e^{-(r-s)^2}$, which makes the integral converge, but it is clearly non-zero

Any idea what step is required here?

Answer 1

You are missing a condition that guarantees that the $W(r,s)$ function conserves probability:

$$ \frac{dP_i}{dt}=\sum_j{W_{ij} P_j} $$

$$ \frac{d}{dt}( \sum_i{ P_i}) = \sum_{ij}{W_{ij} P_j} = \sum_j{P_j (\sum_i{W_{ij}})}$$

The last expression must be zero for any distribution, which means that the $W_{ij}$ must satisfy

$$ \sum_i{W_{ij}} = 0$$

In your continuum limit, this means that

$$ \int_0^{\infty} ds W(r,s) = 0 $$

none of your sample transition functions satisfy this constraint, this is why you are getting absurd results