 Solution

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

Let

⟨ \cdot ⟩

denote

(2 π)^{- 2} \int d^{2} θ

\partial_{t} f_{0} = - ⟨{δ f, δ Φ}⟩ .

⟨{δ f, δ Φ}⟩ = ⟨ \partial_{θ_{i}} δ f \partial_{J_{i}} δ Φ - \partial_{J_{i}} δ f \partial_{θ_{i}} δ Φ ⟩ = - \partial_{J_{i}} ⟨ δ f \partial_{θ_{i}} δ Φ ⟩ .

\partial_{t} f_{0} = - \partial_{J_{i}} q_{i}, q_{i} = - ⟨ δ f \partial_{θ_{i}} δ Φ ⟩ .

This sign also follows from Hamilton's equations,

\dot{J}_{i} = - \partial_{θ_{i}} δ Φ

: the flux is

⟨ δ f \dot{J}_{i} ⟩

δ ϕ_{- k} = δ ϕ_{k}^{*}

and

δ f_{- k} = δ f_{k}^{*}

⟨ δ f \partial_{θ_{i}} δ Φ ⟩ = - i \sum_{k} k_{i} δ f_{k} δ ϕ_{k}^{*} .

Insert b to obtain

q_{i} = - \sum_{k} k_{i} \int_{0}^{t} d t^{'} k \cdot \partial_{J} f_{0} (J, t^{'}) e^{- i k \cdot Ω (t - t^{'})} δ ϕ_{k} (J, t^{'}) δ ϕ_{k}^{*} (J, t) .

Although individual terms are complex, pairing

k

with

- k

makes the flux real.

Now take the marginalization specified by

F_{0} = 2 π \int_{0}^{\infty} f_{0} d J_{R}

. Assume vanishing radial-action flux at

J_{R} = 0

and

J_{R} \to \infty

: the first boundary does not permit negative radial action, and the second has no escaping flux in this calculation. Then

\partial_{t} F_{0} = - 2 π \partial_{J_{φ}} \int_{0}^{\infty} q_{φ} d J_{R} - 2 π [q_{R}]_{0}^{\infty} = - \partial_{J_{φ}} Q_{0},

with

Q_{0} = - 2 π k \sum k_{φ} \int_{0}^{\infty} d J_{R} \int_{0}^{t} d t^{'} k \cdot \frac{\partial f _{0} ( J , t ^{'} )}{\partial J} \times e^{- i k \cdot Ω (t - t^{'})} δ ϕ_{k} (J, t^{'}) δ ϕ_{k}^{*} (J, t) .

If radial-action boundary flux were present, its explicit boundary term would have to be retained.

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