Past exam of the kinetic theory course of the University of Oxford / 2024 (A15089W1)

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/a15089w1.pdf

Put $z_i=({\mathbf{x}}_i,{\mathbf{v}}_i)$ and $d{z}_i=d^3{x}_i{d}^3{v}_i$. Normalize the symmetric $N$-particle phase-space distribution function $D_N$ to one. Use factorial normalization for its one-particle marginal and two-particle distribution function:Thus $\int fdz=N$ and $\int f_2(z_1,z_2)d{z}_2=(N-1)f(z_1)$. For unit particle mass, $f$ also carries mass density. These conventions explain why no extra factor $N-1$ occurs in the result.

Separate the Hamiltonian as $H=H_1+H_{\mathrm{rest}}+H_{\mathrm{int}}$, whereand $H_{\mathrm{rest}}$ contains all terms involving only particles $2,\dots ,N$. The stated Poisson bracket gives Hamilton's equationsConservation of probability along the Hamiltonian flow gives the Hamiltonian Liouville equationAssume decay of the required velocity moments and either decay or periodic boundary conditions in position. Integrating over $z_2,\dots ,z_N$ eliminates the $H_{\mathrm{rest}}$ terms by integration by parts. The terms in $H_{\mathrm{int}}$ differentiating an integrated velocity also vanish. Indistinguishability makes the remaining $N-1$ terms equal, soRelabeling the two particles gives the first BBGKY hierarchy equation:This is exact for a differentiable pair potential; no independent-particle closure has been made.

Take the zeroth velocity moment of a. The right side is a velocity space divergence with vanishing boundary contribution, giving the continuity equationFor the first velocity moment, use integration by parts componentwise:Write $\mathbf{c}=\mathbf{v}-\mathbf{u}$; then $\int \mathbf{c}f{d}^{3}v=0$, and the momentum flux splits into bulk transport and the pressure tensor:Thereforewhere repeated Cartesian indices are summed and $n_2=\int f_2{d}^{3}v{d}^3{v}_{\star }$. The negative sign is the attractive or repulsive force $-\nabla \varphi$, rather than an additional sign convention for the pressure tensor.

For fixed $\mathbf{R}$ and $t$, define $h(\mathbf{x},s)=n_2(\mathbf{a},\mathbf{b},t)$ with $\mathbf{a}=\mathbf{x}+(s-1)\mathbf{R}$ and $\mathbf{b}=\mathbf{x}+s\mathbf{R}$. The chain rule givesso $\boxed{{\partial }_{s}h=\mathbf{R}\cdot {\nabla }_{\mathbf{x}}h}$. The derivative with respect to $\mathbf{x}$ here translates both arguments of the two-particle number density; it is not a derivative of its first argument alone.

Take the spatial divergence of the proposed interaction pressure tensor. Its pair potential factor is independent of $\mathbf{x}$, so the preceding identity givesExchange symmetry of the two-particle distribution function implies $n_2(\mathbf{x}-\mathbf{R},\mathbf{x})=n_2(\mathbf{x},\mathbf{x}-\mathbf{R})$. In the second term, substitute $\mathbf{R}\mapsto -\mathbf{R}$. Since $R_i{\varphi }^{\prime }(R)/R$ is odd, both terms become equal:On the other hand, at fixed ${\mathbf{x}}_{\star }=\mathbf{x}+\mathbf{R}$, the first-argument derivative of the pair potential is ${\nabla }_{\mathbf{x}}\varphi =-\mathbf{R}{\varphi }^{\prime }(R)/R$. Thus the interaction force density in b is $-{\partial }_{x_j}{P}_{ij}^{\varphi }$. Moving it to the left yields local conservation of momentum:The integral over $s$ distributes each pair's momentum flux along the straight segment joining the particles. The factor $1/2$ removes the double counting of ordered pairs. Central forces also make $P_{ij}^{\varphi }=P_{ji}^{\varphi }$.

Spatial homogeneity makes $\rho$ independent of position and makes the two-particle number density depend only on the separation. Isotropy removes its directional dependence. ConsequentlyHere $g_2$ is a radial distribution function, with the finite-$N$ normalization inherited from a; statistical independence is unnecessary. In particular, the translated pair in c has separation $\mathbf{R}$ for every $s$, so its $s$ integral is simply ${\rho }^2{g}_2(R,t)$.

Isotropy of the random velocity $\mathbf{c}=\mathbf{v}-\mathbf{u}$ makes its second velocity moment proportional to the identity. Taking its trace fixes the proportionality constant:Thus $\theta$ is the one-component random velocity variance; for unit mass in thermal equilibrium, $\theta =k_{B}T$. A Maxwell-Boltzmann velocity distribution is not required for the pressure tensor to be isotropic.

For the interaction contribution, the angular integral is $\int d\Omega {\hat{R}}_i{\hat{R}}_j=(4\pi /3){\delta }_{ij}$, by isotropy and its trace. With $d^{3}R=R^{2}dRd\Omega$, c becomesCombining the two pressure tensors gives the virial pressure from pair correlations:For a repulsive pair potential, the outward radial force is $-{\varphi }^{\prime }(R)\ge 0$. Since $g_2\ge 0$, the second term in $p$ is nonnegative, and positive when occupied separations feel nonzero repulsion. It expresses additional momentum transfer through interactions. The displayed integral assumes a differentiable pair potential and convergence; a hard core requires a contact formula or a limiting argument.

Use the electron electric charge $-e$, so $E=-{\partial }_x\phi$ gives acceleration $+(e/m_e){\partial }_x\phi$. The Vlasov equation, its average, and its linearized fluctuating part areThe quasilinear approximation retains the averaged quadratic term but drops products of fluctuations from the fluctuating equation. It freezes the slowly evolving $f_0$ during the fast response calculation. A real electrostatic potential has ${\phi }_{-k}={\phi }_k^{\ast }$, and the sum includes both signs of every nonzero wavenumber. Spatial averaging selects the opposite Fourier series harmonics; statistical homogeneity supplies the corresponding diagonal noise covariance.

Let $\hat{g}(p)={\int }_0^{\infty }{e}^{-pt}g(t)dt$. Zero initial $\delta f_k$ givesInsert this into the electrostatic self-consistency relation. Using ${\omega }_{pe}^2=4\pi e^2{n}_e/m_e$ and $p+ikv=ik(v-ip/k)$ givesThis derives the stated dielectric function with the Landau contour. A noise-free growing collective response is a zero of $ϵ$.

The Laplace transform of the white noise covariance is particularly simple. Writing $q=p^{\prime \ast }$,For clarity about contour signs, introduce the inverse-transform measuresThe first Bromwich inversion formula contour runs upward on $\mathrm{Re}p=\sigma$; its conjugate contour runs downward on $\mathrm{Re}q=\sigma$. Both measures therefore have the same positive orientation in their real frequency parameters. Combining the averaged equation with ${\widehat{\delta f}}_k$ yieldsAt a zero of $f_0^{\prime }$, the last expression means its continuous coefficient obtained before cancelling $f_0^{\prime }$. Substitution of the two inverse Laplace transforms and the transformed covariance givesThis is the paper's expression, with the orientation of $d{p}^{\prime \ast }$ made explicit. Reality follows by pairing $k$ and $-k$. The double integral retains switch-on transients; replacing it by a time-independent diffusion coefficient requires the additional long-time limit in b.

A linearly stable $f_0$ has no zeros of the causal dielectric function in $\mathrm{Re}p>0$. Hence the inverse Laplace transform contours can approach $\sigma =0^{+}$, retaining the causal boundary value at $p=-i\omega +0^{+}$. To obtain a stationary forced response, further assume that the excited collective perturbations actually damp. This is the decay condition in the question's long-time instruction.

A transparent evaluation of the double integral uses the stationary power spectral density. The white noise covariance $2A_k\delta (t-t^{\prime })$ has temporal spectrum $2A_k$. Linear filtering by $1/ϵ$ therefore givesThe causal solution for $\delta f_k$ in a gives, after the initial transients have decayed,The real part can be taken because the full $k$ sum pairs conjugates. Using the distribution theory identity $\mathrm{Re}{\int }_0^{\infty }{e}^{ia\tau }d\tau =\pi \delta (a)$, with a positive vanishing convergence factor, samples the temporal spectrum at $\omega =kv$:The factor $1/2$ from the one-sided time integral cancels the $2$ in the noise covariance. The result is nonnegative, as a stationary velocity-space diffusion coefficient should be.

Qualification of stability. Absence of growing modes alone does not guarantee this stationary limit. For example, the electron waterbag distribution $f_0(v)=n_e{1}_{|v|<V}/(2V)$ hasas undamped collective roots. White noise drives their field variance to grow instead of settling. Thus the formula requires decay of excited modes, not merely absence of exponentially growing ones. For weak Landau damping, the waiting time can be much longer than ${\omega }_{pe}^{-1}$, and must still be short compared with evolution of $f_0$ for the frozen-background quasilinear approximation to apply.

For a regular thermal $f_0$ and noise at $k{\lambda }_{\mathrm{De}}\gg 1$, the susceptibility $ϵ-1$ is small at thermal phase velocities: its scale is ${\omega }_{pe}^2/(k^2{v}_{\mathrm{the}}^2)$. To leading order the electrons cannot appreciably screen such short-scale fields, and b becomes independent of velocity:The electron kinetic energy per unit volume is $\mathcal{K}=(m_e/2)\int v^2{f}_{0}dv$. Assume $v^{2}D{f}_0^{\prime }$ and $v{f}_0$ vanish at infinity. One integration by parts gives the general heating formulaFor constant $D_0$, a second integration by parts and $\int f_{0}dv=n_e$ giveThis is the leading short-wavelength result. With vanishing boundary flux, the diffusion equation also conserves particle number; constant $D_0$ conserves the mean momentum while increasing the random kinetic energy.

Set $\kappa =|k|$ and first work with a representative positive $k=\kappa$. For $\kappa {\lambda }_{\mathrm{De}}\ll 1$, the thermal core is strongly screened. However, the causal dielectric function has weakly damped Langmuir wave roots ${\omega }_s+i{\gamma }_s$, with $s=\pm 1$, ${\omega }_s\simeq s{\omega }_{pe}$ and ${\gamma }_s<0$. Their phase velocities $v_s={\omega }_s/\kappa$ are in the tail. Both roots must be retained for each spatial Fourier series harmonic.

Write $ϵ(-i\omega +0^{+},\kappa )={ϵ}_R(\omega )+i{ϵ}_I(\omega )$. The supplied approximation givesExpanding $ϵ({\omega }_s+i{\gamma }_s)=0$ to first order gives the Landau damping rateIn particular, stable damping means ${\omega }_s{f}_0^{\prime }(v_s)<0$. Near either root,The narrow Lorentzian resonance approximation then gives the leading resonant contributionThis is a formula under the integral over velocity; it is not a pointwise approximation away from the resonances. Insert it into the heating formula in c. For one spatial harmonic, the contribution from one root isbecauseThus the tiny tail slope cancels the tiny Landau damping rate: weak absorption is compensated by a large noise-driven wave intensity. The two temporal roots supply two equal halves of the answer. Negative $k$ gives the same result by the conjugate response of a real electric field, without altering the original convention that the spatial sum includes both signs. ThereforeAs in c, equality means the leading result in the stated asymptotic limit. This calculation assumes nonzero damping, observation times $t\gg {\max }_s|{\gamma }_s{|}^{-1}$, and forcing weak enough that $f_0$ changes negligibly during that relaxation. Taking zero damping before the long-time limit invalidates the stationary calculation.

The long-wavelength limit in d is resonant heating. Most electrons screen the slowly varying spatial field. The noise nevertheless excites collective Langmuir waves, whose sharp temporal spectrum lies near $\omega =\pm {\omega }_{pe}$. Electrons with $v\simeq \omega /k$ remain in phase with a wave and exchange energy with it. A stable tail absorbs net wave energy through Landau damping. The heating is initially deposited in narrow tail regions of velocity space, even though the volume-integrated rate equals that in c.

The short-wavelength limit in c is stochastic heating in the terminology intended here. With $ϵ\simeq 1$, the imposed electric field is essentially unscreened. Its temporally uncorrelated kicks cause a random walk of electron velocity, so $⟨(\Delta v{)}^2⟩=2D_0\Delta t$. Electrons throughout the thermal population contribute to the resulting kinetic energy increase; no narrow collective wave peak selects a small tail population.

Both mechanisms are driven by white noise, and the formal condition $\omega =kv$ enters the stationary quasilinear diffusion calculation in both cases. The physical distinction is the sharp collective resonance in d versus the broad direct random forcing in c, rather than the mere presence of randomness. In collisionless theory this heating is an increase of the averaged kinetic energy; subsequent collisions can establish thermal equilibrium.

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.