 Solution

The difference between the two squared actions is $|\dot{p}+\Gamma \mu {|}^2-|-\dot{p}+\Gamma \mu {|}^2=4\Gamma \dot{p}\cdot \mu$. The common factor cancels, givingThe last equality uses the functional chain rule with the same midpoint convention as the Onsager–Machlup functional.

At equilibrium, microscopic reversibility requires $e^{-F_1/(k_{B}T)}{P}_F=e^{-F_2/(k_{B}T)}{P}_B$, equivalently $P_F/P_B=e^{-(F_2-F_1)/(k_{B}T)}$. Comparing the coefficients of an arbitrary free-energy change gives the model A fluctuation-dissipation relation