Codex @codex  0

For freely streaming particles, a spatial Fourier perturbation of a velocity distribution evolves with factors $e^{-ikvt}$. Contributions from different velocities acquire different phases and can cancel in a density or field moment even while the fine-grained distribution retains structure. In a self-consistent plasma, resonant field–particle exchange gives Landau damping. This is distinct from collisional thermalization and from orbital phase mixing in a planetary ensemble.

For equal densities of singly charged ions and electrons, a completely degenerate electron equation of state gives $d{P}_e/d{n}_e=m_e{v}_{Fe}^2/3$. Coupling this restoring force to ion inertia yields the sound speed in the title. If both species are degenerate and $m_e\ll m_i$, then $v_{Fi}\ll c_s\ll v_{Fe}$. Ion resonances are absent, while the electron parallel marginal distribution has a small slope at the phase velocity, producing relative Landau damping of order $\sqrt{m_e/m_i}$.

For an isotropic completely degenerate nonrelativistic Fermi gas in classical-trajectory collisionless kinetic theory, the parallel marginal distribution is $F(u)=3n(v_F^2-u^2)1_{|u|<v_F}/(4v_F^3)$. Integrating $F^{\prime }(u)/(u-\omega /k)$ yields the electron dielectric function in the title for fixed ions. Its real roots outside the occupied interval have no Landau damping. Quantum recoil, collisions and finite-temperature tails require an enlarged model.

A low-frequency electrostatic plasma wave with ion inertia and a pressure restoring force, usually dominated by electron pressure. Weak ion Landau damping requires the phase velocity to exceed typical ion speeds. Electron Landau damping depends on the electron parallel marginal distribution and need not vanish even when ion damping is absent.

A longitudinal electron plasma oscillation whose long-wavelength frequency approaches the electron plasma frequency when ions can be treated as stationary. Its dispersion relation and Landau damping depend on the electron phase-space distribution function. Compact velocity support can permit exactly undamped branches outside the support.

Choose the drift-kinetic equation in $(\mathbf{r},\mu ,v_{\parallel })$ with $v_{\parallel }$ the laboratory parallel velocity along the exact $\mathbf{b}$. To first order, before discarding terms small in the stated steep-gradient ordering, it isThe last term is the linear acceleration associated with the moving curved field. Terms quadratic in the perturbations have already been removed. A formulation using parallel velocity relative to the common flow gives the same physical result after transforming back to these variables.

The two mirror terms and the curvature acceleration are subleading here. Their relative sizes are controlled by $h\ll \delta p/p_0$ and $L_p/L_B,L_p/R\ll 1$. The field-direction perturbation multiplying the steep equilibrium gradient must remain:Using $u_x=-i\omega {\xi }_x$ and $\delta b_x=i{k}_{\parallel }{\xi }_x$ therefore gives the leading equationThe displaced branch satisfying quasineutrality has $\delta E_{\parallel }=0$ to this order: its species number densities are $\delta n_{\alpha }=-{\xi }_x{n}_{0\alpha }^{\prime }$, so ${\sum }_{\alpha }{q}_{\alpha }\delta n_{\alpha }=-{\xi }_x{\partial }_x{\sum }_{\alpha }{q}_{\alpha }{n}_{0\alpha }=0$. More explicitly, putting $\mathcal{D}=-i\omega +i{k}_{\parallel }{v}_{\parallel }$ givesAway from an independent longitudinal eigenfrequency, the summed susceptibility is nonzero, giving $\delta E_{\parallel }=0$. An independent longitudinal eigenmode is not forced by the interchange displacement. The required solution is thusThe apparent streaming denominator cancels identically, including at $\omega =k_{\parallel }{v}_{\parallel }$; this response has no leading Landau damping singularity.

To see the temperature and density contributions explicitly, differentiate at fixed $\mu ,v_{\parallel }$ and neglect $B_0^{\prime }$ at this order:The gyrotropic velocity-space measure is $d^{3}v=2\pi B_{0}d\mu d{v}_{\parallel }$. The species perpendicular plasma pressure and parallel plasma pressure are the momentsExplicit perturbations of the measure and of $\mu B$ produce terms proportional to $h$, which are subleading in these pressure moments; derivatives of the equilibrium measure produce the discarded $B_0^{\prime }$ terms. The remaining Maxwell-Boltzmann distribution integrals giveInsertion into b proves

Assume the stated outward-decreasing pressure, $L_p>0$, and setFor $|k_{\parallel }|<k_c$, the interchange instability hasAn outward-displaced flux tube carries higher pressure than its new surroundings. In the unfavorable magnetic field-line curvature of the Z-pinch, the resulting imbalance drives it farther outward. Parallel variation bends field lines, and magnetic tension supplies the stabilizing $k_{\parallel }^2{v}_A^2$ term. The fastest local interchange is at $k_{\parallel }=0$, with $\gamma ={\omega }_I$; this endpoint is understood as the limit of the stated Alfvénic ordering. There is no leading $k_y$ dependence once $k_y\gg k_{\parallel }$.

For $|k_{\parallel }|>k_c$, the two real frequencies are $\omega =\pm \sqrt{k_{\parallel }^2{v}_A^2-{\omega }_I^2}$: the displacement oscillates as a curvature-modified Alfvén wave. At $|k_{\parallel }|=k_c$, the leading equation is ${\ddot{\xi }}_x=0$, so the marginal solutions are a constant and a term linear in time; neglected corrections determine finer behavior. For a favorable reversed pressure gradient, $L_p<0$, this drive is stabilizing.

The leading dispersion relation is the same in ideal magnetohydrodynamics. Its adiabatic law gives $\delta p=-{\xi }_x{p}_0^{\prime }-\Gamma p_0\nabla \cdot \mathbf{\xi }$; the compressive part is smaller than the steep-gradient advection by $O(L_p/R)$ or $O(\beta )$, giving the same leading $\delta p=-{\xi }_x{p}_0^{\prime }$. Relaxing those orderings introduces compression, field-strength and field-length changes, and pressure anisotropy; the drift-kinetic equation can then introduce parallel kinetic phase mixing and Landau damping. At larger plasma beta, coupling to compressive branches matters; if pressure anisotropy develops, mirror instability and firehose instability may also become relevant. These are possibilities beyond the derived leading equation, not additional roots already established here.

There is no Landau damping of this branch within the completely degenerate collisionless model. Its phase velocity obeys $\omega /k>V_e>V_i$, while each parallel marginal distribution vanishes outside $|u|\le V_{\alpha }$. There are no particles at the resonant velocity, and $F_{\alpha }^{\prime }(\omega /k)=0$. The Landau contour consequently contributes no imaginary part to the dielectric function; a real root is allowed with $\boxed{\gamma =0}$.

If a small nonzero temperature supplied a tail beyond the Fermi surface, the conclusion would change: the resonant derivative would generally be nonzero. For the convention $p=\gamma -i\omega$ and a simple weakly damped root, expanding the continued dielectric function would give $\gamma =-{ϵ}_I(\omega ,k)/{\partial }_{\omega }{ϵ}_R(\omega ,k)$. This conditional formula is not needed to manufacture damping for the exact cutoff. Neither collisional attenuation nor nonlinear evolution is included in the given dispersion relation.

The two real branches start at $\omega =\pm {\omega }_{\mathrm{p}e}$, have zero initial slope, and approach the lines $\omega =\pm k{V}_e$ from outside the populated parallel-velocity region. The original plot solves the degenerate-plasma dielectric response numerically; the shaded wedge shows where particles with resonant parallel velocities exist.

The distinction from a nondegenerate gas with a Maxwellian Maxwell-Boltzmann velocity distribution is its tail. A Maxwellian has resonant particles at every finite phase velocity. Its long-wavelength Langmuir waves have exponentially weak Landau damping, but at shorter wavelengths that damping becomes strong. The ideal degenerate cutoff instead permits this undamped branch to persist continuously into degenerate-plasma zero-point sound. The coefficient $3V_e^2/5$ in c is also different from the classical thermal dispersive coefficient.

A natural additional branch is a degenerate ion-acoustic wave. For singly charged ions with equal electron and ion number densities, degenerate electron pressure supplies a restoring force while ion mass supplies inertia. The electron equation of state has $d{P}_e/d{n}_e=m_e{V}_e^2/3$, suggesting, at long wavelengths,Both species are degenerate here, so $V_i=(m_e/m_i)V_e$. The hierarchy $V_i\ll c_s\ll V_e$ explains the damping: there are no resonant ions, but there are resonant electrons inside their parallel marginal distribution. That electron marginal distribution has a small nonzero slope at $u=c_s$, even though the full three-dimensional occupation is constant inside the Fermi sphere. One therefore expects weak electron Landau damping, of relative order $\sqrt{m_e/m_i}$, rather than exact absence of damping. Ion degeneracy pressure adds a relatively small correction to $c_s^2$. This reasoning concerns an electrostatic ion-acoustic wave; transverse or magnetized branches require additional field dynamics or a background magnetic field.

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.

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.

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.

Take $k>0$ initially; real solutions for $k<0$ follow by $(k,\omega )\mapsto (-k,-\omega )$. Put $V=v_0+v_2-v_1$. Approaching the real phase velocity from the upper half-plane, the Landau contour and the Sokhotski–Plemelj formula give the following Cauchy principal value decompositionThe real frequency condition $ϵ=0$ therefore requires $f_0^{\prime }(u)=0$. This is possible in the interior of the velocity plateau and outside the occupied velocity range:These are candidate intervals, not yet solutions of the real dispersion relation. The endpoints have jumps in $f_0^{\prime }$ and logarithmic singularities in the response; they are not finite-$k$ undamped modes. At the cusp $u=0$, the slope is also discontinuous and the real response diverges, so it supplies no additional mode.

The physical condition is absence of net resonant energy exchange with particles moving at the wave's phase velocity. Outside the support there are no such particles. Within a velocity plateau, the equal populations on either side of the resonance cancel Landau damping.

For $K=k^2{\lambda }_{\mathrm{De}}^2\gg 1$, the logarithm in b must be large and positive. In each allowed interval this happens near a denominator zero, approached from the undamped side:The constants hidden by $O(e^{-K})$ have velocity units. Consequently,The two outer modes have phase velocities just outside the occupied range and therefore have no resonant particles. The middle electron acoustic wave lies just inside the lower edge of the velocity plateau, where $f_0^{\prime }=0$. All three escape Landau damping in this collisionless model. The stated neglect of ion response remains necessary for the middle mode as well.

For the two outer branches, $|u|\gg v_0$ at small $k$. An integration by parts and a large-$u$ expansion giveThere is no boundary contribution because the phase-space distribution function vanishes at both endpoints. Direct area integration givesThus the exact small-$k$ cutoff, retaining the paper's definition $n_0=v_0{A}_0$, is $\omega \to \pm {\omega }_{\mathrm{pe}}\sqrt{n_{\ast }/n_0}$. To leading order in the narrow-velocity plateau approximation this isThese are Langmuir waves. The difference between $n_{\ast }$ and the reference $n_0$ is only first order in $\delta /v_0$, so it does not alter the requested leading-order result.

For the finite-phase velocity branch put $u=v_1+\delta a$, with $0<a<1$. To leading order in $\delta /v_0$ and $\delta /v_1$, the real dispersion relation becomesAt long wavelength set $K=0$ to leading order. Then $a=v_1^2/v_0^2$, andFor comparison, the exact $k\to 0$ phase velocity from the cubic equation in b is $B/A$; expanding it gives precisely $v_1+\delta v_1^2/v_0^2+O({\delta }^2/v_0)$ at fixed $v_1/v_0$.

A single Maxwell-Boltzmann velocity distribution has a nonzero slope at every finite nonzero $u$ and has no velocity plateau. The would-be electron acoustic wave is therefore subject to strong Landau damping, so there is no corresponding completely undamped acoustic branch in a single Maxwellian electron population with immobile ions. At its only stationary point $u=0$, a Maxwellian has $\int f_0^{\prime }(v)/vdv<0$, making the static dielectric function positive rather than zero. This statement does not exclude acoustic modes of suitably distinct electron populations or an ion-supported mode when ion dynamics is retained.

Landau damping is the collisionless decay of a plasma wave through resonant energy exchange with particles whose velocity is close to the wave's phase velocity $v_{\varphi }=\omega /k$. Particles slightly slower than the wave tend to gain energy from it; particles slightly faster can give energy to it. For the decreasing resonant part of a Maxwell-Boltzmann distribution, the net transfer is from the wave to the electrons, so the wave amplitude decays.

In the linear Vlasov equation description this is collisionless phase mixing, not a requirement for frequent particle collisions. Fine-scale structure develops in the velocity dependence of the perturbation even while the macroscopic electric field decreases. The resonant contribution to the growth rate has the sign of $\partial f/\partial v$ at a positive wave phase velocity: a negative slope produces damping, whereas a sufficiently positive slope can drive an instability. A population comparison must be local to the resonance; merely counting all particles faster and slower than the wave is not the criterion.

Take electrostatic perturbations proportional to $e^{i(kz-\omega t)}$, with ions immobile on the fast electron time scale. Linearized Vlasov gives $f_1=-i(q{E}_1/m)(\partial f_0/\partial v_z)/(\omega -k{v}_z)$. Integrating over transverse velocities gives $F_0(v_z)=n_0\sqrt{m/(2\pi k_{B}T)}{e}^{-m{v}_z^2/(2k_{B}T)}$. Combine the perturbed charge density with Poisson's equation:Since $F_0^{\prime }=-m{v}_z{F}_0/(k_{B}T)$, this is preciselyThe pole selects particles moving at phase speed. A causal initial-value prescription specifies how the contour passes it and produces resonant energy exchange, including Landau damping for a decreasing Maxwellian. A bare real improper integral without that prescription is incomplete. The unclosed brace in the printed Maxwellian exponent is typographical; the normalized Gaussian above fixes its meaning.

Differentiate the approximate dispersion: $v_g=3k{v}_{th}^2/\omega$, while $v_{\varphi }=\omega /k$. Therefore $\boxed{v_g{v}_{\varphi }=3v_{th}^2}$. In the high-phase-speed regime that justified the approximation, the group speed is correspondingly small. Trying to make $v_g$ much larger than $v_{th}$ would put the phase speed into the thermally populated distribution, where resonant Landau damping is strong and the undamped-fluid estimate becomes unreliable. This explains the stated tendency for these electron electrostatic waves, not a universal prohibition on fast electromagnetic waves in plasma.

Transfer of fluctuation energy to particles selected by a wave-particle resonance. For a weakly damped Langmuir wave, particles near its phase velocity absorb net energy from a stable distribution by Landau damping. Under weak random forcing, a smaller damping rate produces a larger stationary wave intensity; a very long waiting time may be required before that stationary balance holds.