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

Table of contents
- Solution b

Solution

Subtracting the two quadratic actions gives

lo g \frac{P _{F}}{P _{B}} = - \frac{1}{2 σ ^{2}} \int d t d r [\dot{p} + Γ \frac{δ F}{δ p}^{2} - - \dot{p} + Γ \frac{δ F}{δ p}^{2}] = - \frac{2Γ}{σ ^{2}} \int d t d r \dot{p} \cdot \frac{δ F}{δ p} .

The functional chain rule identifies the last integral as

F_{2} - F_{1}

, so

\frac{P _{F}}{P _{B}} = exp [- \frac{2Γ}{σ ^{2}} (F_{2} - F_{1})] .

Microscopic time-reversal invariance implies detailed balance. The equilibrium probability density of a configuration with free energy

F

obeys

P_{eq} \propto e^{- F / (k_{B} T)} .

Therefore

\frac{P _{F}}{P _{B}} = \frac{e ^{- F_{2} / (k_{B} T)}}{e ^{- F_{1} / (k_{B} T)}} = e^{- (F_{2} - F_{1}) / (k_{B} T)} .

Comparison for arbitrary endpoint free energies yields the Model A fluctuation-dissipation relation

σ^{2} = 2Γ k_{B} T .

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