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 Incoming links: Functional chain rule

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 344 / 3 / g / Solution 2026-09-28

For a specified field trajectory, the required normalized noise is

Λ_{F} = \frac{p ˙ + Γ δ F / δ p - Γ Y}{2 k _{B} T Γ},

and time reversal changes only

\dot{p}

to

- \dot{p}

. Assuming equal additive-noise Jacobians, the difference of the two Onsager--Machlup actions gives

lo g \frac{P _{F}}{P _{B}} = - β \int_{t_{1}}^{t_{2}} d t \int d r \dot{p} \cdot (\frac{δ F}{δ p} - Y) .

The functional chain rule identifies the first term as

- β Δ F

, so

\frac{P _{F} [ p ]}{P _{B} [ p ]} = exp [- β Δ F + β \int_{t_{1}}^{t_{2}} d t \int d r Y \cdot \dot{p}] .

The forcing performs generalized work

W_{Y} = \int Y \cdot \dot{p}

, and

W_{Y} - Δ F

is the heat dissipated into the bath. The formula is therefore the field-theory form of local detailed balance and quantifies nonequilibrium entropy production.

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Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 344 / 3 / b / Solution 2026-09-28

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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