Past exam of the kinetic theory course of the University of Oxford / 2024 / 2

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.