Past exam of the kinetic theory course of the University of Oxford / 2023 / 3 / e / Solution

Apply the chain rule to $f_0=f_0(E)$ with $E=H_0(\mathbf{J})$:On the resonance selected by the Dirac delta function, $\mathbf{k}\cdot \mathbf{\Omega }=k_{\varphi }{\Omega }_{\mathrm{p}}$. ThusEvery weight is nonnegative, and $-f_0^{\prime }(E)>0$. Strict positivity additionally requires a populated nonaxisymmetric resonance with nonzero coupling on a set of positive surface measure. The unconditional statement is nonnegativity, rather than positive definiteness. As a concrete counterexample to strict positivity of this formula, retain one nonzero conjugate harmonic pair with $k_{\varphi }\ne 0$ and choose its smooth coefficient ${\Psi }_{\mathbf{k}}$ supported in an open region $|{\Delta }_{\mathbf{k}}|>d>0$. The perturbation is nonzero but has zero resonant coupling, so $\mathcal{T}=0$ even when $f_0^{\prime }(E)<0$. The even simpler zero-perturbation case also gives zero. Whether a particular physical stellar bar has nonzero resonant overlap is additional information.

For a coupled stellar bar and halo, the halo gains positive angular momentum and exerts the opposite torque on the stellar bar. If the halo mass is $M_h$ and the rigid stellar bar has constant positive moment of inertia $I_b$, thenThis bar-halo angular-momentum transfer is a form of dynamical friction: a resonantly coupled stellar bar tends to slow down while the dark matter halo spins up. Actual long-term evolution changes both the stellar bar pattern and the halo distribution function, so the fixed-pattern response must then be updated. If the stellar bar changes its moment of inertia, the general relation is $d(I_b{\Omega }_{\mathrm{p}})/dt=-M_h\mathcal{T}$; positive moment of inertia alone would not fix the sign of $d{\Omega }_{\mathrm{p}}/dt$ without that rigidity assumption.