Put $E(t)=-\delta A/\delta\mathbf x(t)$. For a specified forward trajectory, the equation of motion determines
$$
\mathbf f_F=E+\zeta\dot{\mathbf x},
$$
whereas its time reverse requires
$$
\mathbf f_B=E-\zeta\dot{\mathbf x}.
$$
Assume the action is invariant under <time-reversal symmetry>, the position is even and velocity is odd under time reversal, and the trajectory-to-noise Jacobian is identical in the two directions. The <Onsager--Machlup path probability> is then
$$
\mathbb P_F[\mathbf x]
=\mathcal N\exp\left[-\frac1{2\sigma^2}
\int_{t_1}^{t_2}|E+\zeta\dot{\mathbf x}|^2dt\right],
$$
with $\zeta\dot{\mathbf x}$ replaced by $-\zeta\dot{\mathbf x}$ for $\mathbb P_B$. Since
$$
|E+\zeta\dot{\mathbf x}|^2-|E-\zeta\dot{\mathbf x}|^2
=4\zeta E\mathbin\cdot\dot{\mathbf x},
$$
their ratio is
$$
\boxed{\frac{\mathbb P_F}{\mathbb P_B}
=\exp\left[\frac{2\zeta}{\sigma^2}
\int_{t_1}^{t_2}
\dot{\mathbf x}\mathbin\cdot
\frac{\delta A}{\delta\mathbf x(t)}dt\right].}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 344 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-344/3/a/solution