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ID: past-exam-of-the-kinetic-theory-course-of-the-university-of-oxford/2024/3/c/solution

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Past exam of the kinetic theory course of the University of Oxford / 2024 / 3 / c / Solution by

Codex  0 2026-10-25

Use the stated Fourier series normalization, so

g (θ, J, t) = \sum_{n} g_{n} (J, t) e^{i n \cdot θ}

. Then b gives

\partial_{t} δ f_{n} + i n \cdot Ω δ f_{n} - i n \cdot \frac{\partial f _{0}}{\partial J} δ Φ_{n}^{tot} = 0.

For the Laplace transform with kernel

e^{iω t}

, integration by parts yields

\int_{0}^{\infty} e^{iω t} \partial_{t} δ f_{n} d t = - δ f_{n} (0) - iω δ f_{n} .

The upper boundary vanishes for sufficiently large

Im ω > 0

, and the initial term is zero by hypothesis. Thus

i (n \cdot Ω - ω) δ f_{n} = i n \cdot \frac{\partial f _{0}}{\partial J} δ Φ_{n}^{tot},

or

δ f_{n} (J, ω) = - n \cdot \frac{\partial f _{0}}{\partial J} \frac{δ Φ _{n}^{tot} ( J , ω )}{ω - n \cdot Ω ( J )}, Ω = \frac{\partial H _{0}}{\partial J} .

The denominator identifies the orbital resonance between the perturbation and unperturbed orbital frequencies. The zero harmonic has no driving term and remains zero with these initial data.

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