 Solution

ID: past-exam-of-the-kinetic-theory-course-of-the-university-of-oxford/2025/3/b/solution

\partial_{t} f + {f, H} = 0, {f, H} = \partial_{θ} f \cdot \partial_{J} H - \partial_{J} f \cdot \partial_{θ} H .

Ω (J) = \partial_{J} H_{0} = (Ω_{φ}, Ω_{R}) .

f_{0}

starts at second order in the perturbation. At first order, the Poisson bracket therefore gives

(\partial_{t} + Ω \cdot \partial_{θ}) δ f = \partial_{J} f_{0} \cdot \partial_{θ} δ Φ.

For the

k

th term in the Fourier series, this reads

(\partial_{t} + i k \cdot Ω) δ f_{k} = i k \cdot \partial_{J} f_{0} δ ϕ_{k} .

The lowercase

δ ϕ_{k}

here denotes the Fourier coefficient of the uppercase

δ Φ

in the Hamiltonian; these are the same perturbing potential. Multiply by the integrating factor

e^{i k \cdot Ω t}

and set the initial homogeneous response to zero. One obtains

δ f_{k} (J, t) = i \int_{0}^{t} d t^{'} k \cdot \frac{\partial f _{0} ( J , t ^{'} )}{\partial J} e^{- i k \cdot Ω (t - t^{'})} δ ϕ_{k} (J, t^{'}) .

Retaining the slow time dependence of

f_{0}

inside this integral is compatible with the subsequent second-order calculation; changes in

Ω

caused by that slow evolution affect only higher orders in this approximation.

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