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ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2014/paper-1/1/c/solution

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Past exam of the theoretical physics II course of the University of Cambridge / 2014 / Paper 1 / 1 / c / Solution by

Codex  0 2026-10-20

Use normalized instantaneous modes

φ_{p} (x, t) = 2/ L sin (p π x / L)

and expand

Ψ = \sum_{p} a_{p} e^{- i ϑ_{p}} φ_{p}

. Differentiating at fixed

x

gives

\partial_{t} φ_{p} = - \frac{L ˙}{2 L} φ_{p} - \frac{2}{L} \frac{p π x L ˙}{L ^{2}} cos (p π x / L) .

For

n \neq = p

, integration uses

\int_{0}^{1} u sin (nπ u) cos (p π u) d u = - \frac{( - 1 ) ^{n + p} n}{π ( n ^{2} - p ^{2} )},

so the moving-wall derivative coupling is

⟨ φ_{n} ∣ \overset{φ}{˙}_{p} ⟩ = \frac{L ˙}{L} (- 1)^{n + p} \frac{2 n p}{n ^{2} - p ^{2}} .

For

n = p

the normalization derivative and the remaining integral cancel exactly, giving zero. The moving integration endpoint contributes no boundary term because every mode vanishes there. Therefore

\overset{a}{˙}_{n} = \frac{L ˙}{L} p \neq = n \sum (- 1)^{n + p} \frac{2 n p}{p ^{2} - n ^{2}} a_{p} e^{i (ϑ_{n} - ϑ_{p})} .

The real coupling matrix is antisymmetric, as required by conservation of total probability.

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