Solution

ID: past-exam-of-the-kinetic-theory-course-of-the-university-of-oxford/2023/3/d/solution

Write , , and let the detuning from the rotating perturbation be
Substitute into the linear response in action-angle variables, then set :
At exact resonance the last quotient is interpreted by its continuous limit . Inserting this response in c gives
Pair with . Because , , and , the imaginary cosine terms cancel. The finite-time torque is
This is a leading, second-order response in the perturbation amplitude.
For the long-time limit, assume a continuum of occupied action variables, sufficiently smooth integrable weights, convergent harmonic sums, and regular resonance surfaces wherever a contributing surface intersects the distribution. Assume also that the perturbation is weak enough to neglect nonlinear trapping and secular changes in , and that the prescribed stellar bar pattern and its coefficients stay fixed. Then phase mixing removes nonresonant transients in the integrated response. The relevant long-time resonance kernel is the distributional convergence
It is not a pointwise limit: at the kernel equals . To justify its use under the integral over action variables, resolve the integral into detuning surfaces; the coarea formula gives
Smoothness and decay of this surface weight reduce the limit to the one-dimensional long-time resonance kernel. Equivalently, a causal convergence factor gives as .
Therefore the resonant torque from a rotating perturbation is
Only orbital resonances contribute to this steady integrated torque. The limit is a formal weak-perturbation limit, or an intermediate-time approximation after phase mixing but before substantial secular redistribution. It does not prove an unchanged halo can absorb angular momentum at this rate forever. A discrete set of orbital frequencies, singular resonance surfaces, or strong trapping can invalidate the stated steady limit.

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