 Solution

ID: past-exam-of-the-theoretical-physics-major-option-course-of-the-university-of-oxford/2016/4/c/solution

Integrate the two velocity factors over the time square

0 \leq s, s^{'} \leq t

. The white noise contribution is

4 D t

. The orientation contribution depends only on the time separation, giving

M (t) \equiv ⟨ ∣ r (t) - r (0) ∣^{2} ⟩ = 4 D t + 2 v^{2} \int_{0}^{t} (t - s) e^{- D_{r} s} cos (ω s) d s .

This double-integral derivation fixes the delta contribution without ambiguity about a half-weight at an endpoint. Put

z = D_{r} - iω

. Elementary integration gives

M (t) = 4 D t + 2 v^{2} Re [\frac{t}{z} - \frac{1 - e ^{- z t}}{z ^{2}}] .

Using

Re (1/ z) = D_{r} / (D_{r}^{2} + ω^{2})

and multiplying out the second term yields

M (t) = 4 D t + \frac{2 v ^{2} D _{r} t}{D _{r}^{2} + ω ^{2}} + \frac{2 v ^{2} ( ω ^{2} - D _{r}^{2} )}{( D _{r}^{2} + ω ^{2} ) ^{2}} + \frac{2 v ^{2} e ^{- D_{r} t}}{( D _{r}^{2} + ω ^{2} ) ^{2}} [(D_{r}^{2} - ω^{2}) cos ω t - 2 D_{r} ω sin ω t] .

The constant and decaying terms cancel at

t = 0

, as required. The apparently singular case

D_{r} = ω = 0

is evaluated from the integral, giving

M = 4 D t + v^{2} t^{2}

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