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

The action-angle variables $(\mathbf{\theta },\mathbf{J})$ are canonical coordinates, with $[{\theta }_a,J_b]={\delta }_{ab}$. In the convention $\dot{g}={\partial }_{t}g+[g,H]$, their Poisson bracket isIn particular, Hamilton's equations read $\dot{\mathbf{\theta }}={\partial }_{\mathbf{J}}H$ and $\dot{\mathbf{J}}=-{\partial }_{\mathbf{\theta }}H$, fixing the sign needed below.

Define the unperturbed orbital angular frequency vector byIt is the rate of advance of the angle variables along the unperturbed Hamiltonian flow; the action variables are constant. Expand the Collisionless Boltzmann equation to first order in the perturbations. Since $f_0$ and $H_0$ depend only on the action variables, $[f_0,H_0]=0$, andThe omitted Poisson bracket $[\delta f,\delta \Phi ]$ is quadratic in the perturbations. Use the Fourier series conventionFor each harmonic the linear ordinary differential equation isMultiplication by the integrating factor $e^{i\mathbf{k}\cdot \mathbf{\Omega }t}$ and use of $\delta f_{\mathbf{k}}(0)=0$ give the causal linear response in action-angle variables:The zero harmonic stays zero at this order. This formula retains the entire forcing history; replacing it immediately by a resonant Dirac delta function would discard the transient response.

Since $L_z=J_{\varphi }$, Hamilton's equations giveFor the spherical unperturbed Hamiltonian, $H_0=H_0(J_r,L)$; in these particular action-angle variables the third component ${\Omega }_{\varphi }=0$ describes the fixed orientation of the orbital plane. It should not be confused with the physical azimuthal angular velocity of a particle around its orbit.

The conserved normalization of a probability distribution $\int f{d}^3\theta d^{3}J=1$ makes the expected value of specific angular momentum equal to $\int L_{z}f{d}^3\theta d^{3}J$. Its time derivative is the torque per unit halo mass:This follows either by averaging the single-particle equation or by integrating the Collisionless Boltzmann equation against $L_z$, with vanishing boundary fluxes.

The term $f_0{\partial }_{{\theta }_{\varphi }}\delta \Phi$ integrates to zero over the periodic angle variable. Thus the first nonzero torque is quadratic in the perturbation. Use the Fourier series expansions of b; the source's reference here to part (a) is a numbering slip. Integrating the product over all three angle variables yieldsReality of the potential energy per unit mass gives $\delta {\Phi }_{-\mathbf{k}}=\delta {\Phi }_{\mathbf{k}}^{\ast }$; likewise $\delta f_{-\mathbf{k}}=\delta f_{\mathbf{k}}^{\ast }$. Conjugate harmonics make the total torque real. The sum is over all integer vectors, so no extra factor of two should be inserted.

Write $m=k_{\varphi }$, $B_{\mathbf{k}}=\mathbf{k}\cdot {\partial }_{\mathbf{J}}{f}_0$, and let the detuning from the rotating perturbation beSubstitute $\delta {\Phi }_{\mathbf{k}}={\Psi }_{\mathbf{k}}{e}^{-im{\Omega }_{\mathrm{p}}t}$ into the linear response in action-angle variables, then set $\tau =t-t^{\prime }$:At exact resonance the last quotient is interpreted by its continuous limit $it$. Inserting this response in c givesPair $\mathbf{k}$ with $-\mathbf{k}$. Because $B_{-\mathbf{k}}=-B_{\mathbf{k}}$, ${\Delta }_{-\mathbf{k}}=-{\Delta }_{\mathbf{k}}$, and $|{\Psi }_{-\mathbf{k}}{|}^2=|{\Psi }_{\mathbf{k}}{|}^2$, the imaginary cosine terms cancel. The finite-time torque isThis 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 ${\nabla }_{\mathbf{J}}{\Delta }_{\mathbf{k}}\ne 0$ wherever a contributing surface intersects the distribution. Assume also that the perturbation is weak enough to neglect nonlinear trapping and secular changes in $f_0$, 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 convergenceIt is not a pointwise limit: at $x=0$ the kernel equals $t$. To justify its use under the integral over action variables, resolve the integral into detuning surfaces; the coarea formula givesSmoothness and decay of this surface weight reduce the limit to the one-dimensional long-time resonance kernel. Equivalently, a causal convergence factor $e^{-\eta \tau }$ gives $\mathrm{Re}{\int }_0^{\infty }{e}^{-\eta \tau -i\Delta \tau }d\tau =\eta /({\eta }^2+{\Delta }^2)\to \pi \delta (\Delta )$ as $\eta \downarrow 0$.

Therefore the resonant torque from a rotating perturbation isOnly 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.

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.