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

Table of contents
- Solution d

Solution

Initial equilibrium at

a_{I}

gives

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

The initial field and later noise are independent, so their cross terms vanish. Define

Δ t = t_{2} - t_{1} .

With the supplied Gaussian white noise covariance, the noise contribution is

2 k_{B} T Γ δ_{ij} \int_{t_{1}}^{t_{2}} d s e^{- 2 r (q) (t_{2} - s)} = δ_{ij} \frac{k _{B} T}{a _{F} + κ q ^{2}} (1 - e^{- 2 r (q) Δ t}),

Using the decay rate found in part (c), it follows that

⟨ p_{q, i} (t_{2}) p_{- q, j} (t_{2})⟩ = δ_{ij} [\frac{k _{B} T}{a _{I} + κ q ^{2}} e^{- 2 r (q) Δ t} + \frac{k _{B} T}{a _{F} + κ q ^{2}} (1 - e^{- 2 r (q) Δ t})] .

This is the covariance interpolation in a Gaussian Model A quench: each mode forgets its initial equilibrium with relaxation time

1/ r (q)

and approaches the final equilibrium variance. Larger-

q

modes relax faster.

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