Subtracting the two quadratic actions gives
$$
\log\frac{P_F}{P_B}
=-\frac1{2\sigma^2}
\int dt\,d\mathbf r
\left[
\left|\dot{\mathbf p}+\Gamma\frac{\delta F}{\delta\mathbf p}\right|^2
-\left|-\dot{\mathbf p}+\Gamma\frac{\delta F}{\delta\mathbf p}\right|^2
\right]
=-\frac{2\Gamma}{\sigma^2}
\int dt\,d\mathbf r\,
\dot{\mathbf p}\mathbin\cdot\frac{\delta F}{\delta\mathbf p}.
$$
The <functional chain rule> identifies the last integral as $F_2-F_1$, so
$$
\frac{P_F}{P_B}
=\exp\left[-\frac{2\Gamma}{\sigma^2}(F_2-F_1)\right].
$$

Microscopic time-reversal invariance implies <detailed balance>. The equilibrium probability density of a configuration with free energy $F$ obeys
$$
P_{\rm eq}\mathrel\propto e^{-F/(k_BT)}.
$$
Therefore
$$
\frac{P_F}{P_B}
=\frac{e^{-F_2/(k_BT)}}{e^{-F_1/(k_BT)}}
=e^{-(F_2-F_1)/(k_BT)}.
$$
Comparison for arbitrary endpoint free energies yields the <Model A fluctuation-dissipation relation>
$$
\sigma^2=2\Gamma k_BT.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 344 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-344/3/b/solution