Past exam of the kinetic theory course of the University of Oxford / 2024 / 3

For an integrable or approximately integrable background Hamiltonian, action-angle variables turn an otherwise complicated stellar orbit into $\dot{\mathbf{J}}=0$ and $\dot{\mathbf{\theta }}=\mathbf{\Omega }(\mathbf{J})$. This has two useful consequences. First, a phase-mixed equilibrium can be represented directly as $f_0(\mathbf{J})$: the action variables label the unperturbed orbits, while angle variables describe position along them. Second, the angles are periodic, so Fourier series reduce unperturbed transport to multiplication by $i\mathbf{n}\cdot \mathbf{\Omega }$ and make orbital resonances explicit. The action variables also separate rapid orbital motion from slow secular changes. These advantages apply on regular orbital regions; a globally chaotic background need not admit global action-angle variables.

The transformation is canonical, preserving the phase space measure and the stated total mass $M=\int f{d}^3\theta d^{3}J$. The collisionless Vlasov equation isHere $H$ is the single-star specific Hamiltonian; the normalization of $f$ by mass changes neither the Poisson bracket nor the transport equation.

Since $H_0$ and $f_0$ depend only on the action variables, $\{f_0,H_0\}=0$. Drop the quadratic term $\{\delta f,\delta {\Phi }^{\mathrm{tot}}\}$ from the Vlasov equation in a. Defining the orbital frequency vector by $\mathbf{\Omega }(\mathbf{J})=\partial H_0/\partial \mathbf{J}$, the linearized equation isThe sign of the driving term follows from the second term in the canonical Poisson bracket, because the background has no angle dependence.

Use the stated Fourier series normalization, so $g(\mathbf{\theta },\mathbf{J},t)={\sum }_{\mathbf{n}}{g}_{\mathbf{n}}(\mathbf{J},t)e^{i\mathbf{n}\cdot \mathbf{\theta }}$. Then b givesFor the Laplace transform with kernel $e^{i\omega t}$, integration by parts yieldsThe upper boundary vanishes for sufficiently large $\mathrm{Im}\omega >0$, and the initial term is zero by hypothesis. ThusorThe denominator identifies the orbital resonance between the perturbation and unperturbed orbital frequencies. The zero harmonic has no driving term and remains zero with these initial data.

In the periodic cube, $H_0=|\mathbf{v}{|}^2/2=(2\pi /L{)}^2|\mathbf{J}{|}^2/2$, so $\mathbf{\Omega }=(2\pi /L)\mathbf{v}$. The supplied action-angle variables are canonical: $d^3\theta d^{3}J=d^{3}x{d}^{3}v$. For $\mathbf{k}=2\pi \mathbf{n}/L$,Consequently c becomesFor each nonzero wavenumber, the Poisson equation for Newtonian gravity givesSubstitute the preceding response and add the externally imposed Newtonian gravitational potential:ThereforeThis gravitational dielectric function has the opposite susceptibility sign to the electron dielectric function, reflecting attractive rather than repulsive collective interactions. The homogeneous mean field is explicitly subtracted in this fictitious periodic model: the Poisson equation for Newtonian gravity acts on density fluctuations, and the $\mathbf{k}=0$ mode is omitted. Otherwise a nonzero uniform positive mass density would be incompatible with a periodic potential. This is the convention often called the Jeans swindle.

For analytic continuation, resolve velocity parallel to $\mathbf{k}$: let $u=\hat{\mathbf{k}}\cdot \mathbf{v}$ and $F(u)=\int f_0(u,{\mathbf{v}}_{\perp })d^2{v}_{\perp }$. ThenStart at $\mathrm{Im}\omega >0$ and deform the Landau contour continuously below the moving pole. At real $\omega$, the integral isFor $\mathrm{Im}\omega <0$, this causal continuation equals the undeformed real-axis integral plus $2\pi i{F}^{\prime }(\omega /k)$ wherever $F^{\prime }$ has the necessary analytic extension. It must not be replaced by the undeformed integral alone.

Zeros of ${ϵ}_{\mathbf{k}}$ are self-consistent collective Landau modes: with no external forcing, a nonzero Newtonian gravitational potential is possible at such a root. For time dependence $e^{-i\omega t}$, upper-half-plane roots grow, lower-half-plane roots describe Landau damping, and real roots are neutral. Damped Landau modes are poles of the continued response; they need not be regular discrete eigenfunctions of the original collisionless operator.

Regularity qualification. Continuation throughout the complex plane presupposes suitable analytic $F$. Arbitrary smooth $F$ need not allow it. For example, $F(u)=C\exp [-1/(1-u^2)]$ for $|u|<1$ and zero elsewhere is smooth but not analytic at $u=\pm 1$; its Cauchy integral cannot be holomorphically continued across those endpoints. The Landau contour gives the standard causal continuation on the domains where the equilibrium permits it, with possible singularities retained.

Represent the perturber by the periodic Dirac delta function at ${\mathbf{x}}_p+{\mathbf{v}}_{p}t$. With spatial Fourier series coefficient normalized by $L^{-3}$, its nonzero density coefficients areThe mean density is subtracted as in d. The Poisson equation for Newtonian gravity givesThe perturbation is zero before switch-on. For $\mathrm{Im}\omega >0$, its Laplace transform isHenceThe switch-on creates a forcing pole at the real frequency $a$, approached causally from the original upper inversion contour.

With $C_{\mathbf{k}}$ and $a$ from e, inversion givesFor $t>0$, deform the contour downwards. Closing below is clockwise, so the residue theorem multiplies the sum of enclosed residues by $-i$. The forcing pole has residue $i{C}_{\mathbf{k}}{e}^{-iat}/{ϵ}_{\mathbf{k}}(a+i{0}^{+})$ and therefore contributesA simple collective root ${\omega }_j$ of the continued gravitational dielectric function contributesThese are the Landau mode transients excited by introducing the mass abruptly. For a strictly damped background they decay, together with any remaining phase-mixing contour contribution under the usual regularity and decay assumptions. ThusThe causal boundary value is what the paper abbreviates by ${ϵ}_{\mathbf{k}}(\mathbf{k}\cdot {\mathbf{v}}_p)$. Physically, the moving mass polarizes the stellar population and carries a gravitational wake. Each spatial harmonic is changed in amplitude and phase by $1/{ϵ}_{\mathbf{k}}$ at the forcing frequency. Self-gravity can amplify the wake. The induced part alone is $(1/{ϵ}_{\mathbf{k}}-1)\delta {\Phi }_{\mathbf{k}}^{\mathrm{ext}}$; the wake's back-reaction on the perturber produces dynamical friction.

The waiting time is controlled by the slowest excited collective damping, not just an orbital or crossing time. If ${\omega }_j={\omega }_{rj}-i{\Gamma }_j$ with ${\Gamma }_j>0$, then a modal amplitude is suppressed by $e^{-{\Gamma }_{j}t}$, and one needs $t{\Gamma }_j\gg 1$ for every appreciably excited mode, together with decay of the phase-mixing remainder. Large transient prefactors can require additional time. Near marginal stability, a relevant ${\Gamma }_j$ tends to zero and the required waiting time diverges; the stationary wake approximation is then not uniform in the stability parameter.^[ref]

Qualification of stability. If stable means only non-growing, a persistent neutral mode invalidates the asserted asymptote. A concrete example is a homogeneous waterbag distribution with parallel marginal $F(u)=\rho 1_{|u|<V}/(2V)$. Its gravitational dielectric function isFor $k^2{V}^2>4\pi G\rho$, the roots $\omega =\pm \sqrt{k^2{V}^2-4\pi G\rho }$ are real, so the switch-on transients persist although that harmonic has no growing mode. Taking the smallest box wavenumber to satisfy the same inequality makes the whole waterbag model free of growing harmonics. The claimed late-time formula therefore requires damping of all excited modes and ${ϵ}_k(a)\ne 0$; exact forcing at an undamped root instead produces a secular response.