= Solution
Set the noise strengths to the <model A fluctuation-dissipation relation> and the <fluctuation-dissipation relation for a conserved flux>:
$$
\sigma^2=2\Gamma k_BT,\qquad\sigma_N^2=2Mk_BT.
$$
The preceding logarithmic ratio becomes
$$
\log\frac{P_F}{P_B}=-\frac1{k_BT}\int[\mu_j\dot p_j+\mu_j\partial_iW_{ij}+W_{ij}\partial_i\mu_j].
$$
The last two terms combine as $\partial_i(\mu_jW_{ij})$. They integrate to zero under the periodic <boundary conditions>, by <integration by parts>. The remaining term is $F_2-F_1$ by the <functional chain rule>, hence
$$
\boxed{\frac{P_F}{P_B}=e^{-(F_2-F_1)/(k_BT)}.}
$$
After including the equilibrium initial weights, the forward and reversed histories have equal probability. Thus <microscopic reversibility> is satisfied by the two channels together, with each channel's own thermal noise strength.
Back to article page
