<Detailed balance> says that each equilibrium transition is balanced by its time reverse:
$$
P_{\rm eq}(1)\mathbb P_F(1\to2)
=P_{\rm eq}(2)\mathbb P_B(2\to1).
$$
With the <Boltzmann distribution> $P_{\rm eq}\propto e^{-\beta H}$ and $\Delta H=H_2-H_1$, this requires
$$
\frac{\mathbb P_F}{\mathbb P_B}=e^{-\beta\Delta H}.
$$
Parts a and b instead give $\exp[-(2\zeta/\sigma^2)\Delta H]$. Equality for every pair of endpoints yields the <fluctuation-dissipation relation for a Langevin particle>
$$
\boxed{\sigma^2=\frac{2\zeta}{\beta}=2\zeta k_BT.}
$$

Solution

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

{title2=$\sigma^2=2\zeta k_BT$}

For a Langevin particle with drag coefficient $\zeta$ and white-force covariance amplitude $\sigma^2$, detailed balance at temperature $T$ requires $\sigma^2=2\zeta k_BT$.


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 Incoming links: Fluctuation-dissipation relation for a Langevin particle