 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-344/2/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

\nabla^{2}

- q^{2}

. Each Cartesian component is an Ornstein-Uhlenbeck process:

\overset{p}{˙}_{q, i} = - r (q) p_{q, i} + f_{q, i}, r (q) = Γ (a + κ q^{2}) .

Since

r (q) > 0

, the transient from the remote past vanishes, and the integrating-factor solution is

p_{q} (t) = \int_{- \infty}^{t} f_{q} (s) e^{- r (q) (t - s)} d s .

Using the stated Gaussian white noise covariance and assuming

t_{1} \geq t_{2}

⟨ p_{q, i} (t_{1}) p_{- q, j} (t_{2})⟩ = 2Γ k_{B} T δ_{ij} \int_{- \infty}^{t_{2}} e^{- r (q) (t_{1} + t_{2} - 2 s)} d s = δ_{ij} \frac{Γ k _{B} T}{r ( q )} e^{- r (q) (t_{1} - t_{2})} .

Interchanging the times gives the stationary two-time correlation function

⟨ p_{q, i} (t_{1}) p_{- q, j} (t_{2})⟩ = δ_{ij} \frac{k _{B} T}{a + κ q ^{2}} e^{- Γ (a + κ q^{2}) ∣ t_{1} - t_{2} ∣} .

At equal times this is the equilibrium Gaussian field theory covariance, and the decay time is

r (q)^{- 1}

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