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Abstract

These lecture notes were presented by Allan N. Kaufman in his graduate plasma theory course and a follow-on special topics course (Physics 242A, B, C and Physics 250 at the University of California Berkeley). The notes follow the order of the lectures. The equations and derivations are as Kaufman presented, but the text is a reconstruction of Kaufman’s discussion and commentary. The notes were transcribed by Bruce I. Cohen in 1971 and 1972, and word processed, edited and illustrations added by Cohen in 2017 and 2018. The series of lectures is divided into four major parts: (i) collisionless Vlasov plasmas (linear theory of waves and instabilities with and without an applied magnetic field, Vlasov–Poisson and Vlasov–Maxwell systems, Wentzel–Kramers–Brillouin–Jeffreys (WKBJ) eikonal theory of wave propagation); (ii) nonlinear Vlasov plasmas and miscellaneous topics (the plasma dispersion function, singular solutions of the Vlasov–Poisson system, pulse-response solutions for initial-value problems, Gardner’s stability theorem, gyroresonant effects, nonlinear waves, particle trapping in waves, quasilinear theory, nonlinear three-wave interactions); (iii) plasma collisional and discreteness phenomena (test-particle theory of dynamic friction and wave emission, classical resistivity, extension of test-particle theory to many-particle phenomena and the derivation of the Boltzmann and Lenard–Balescu equations, the Fokker–Planck collision operator, a general scattering theory, nonlinear Landau damping, radiation transport and Dupree’s theory of clumps); (iv) non-uniform plasmas (adiabatic invariance, guiding-centre drifts, hydromagnetic theory, introduction to drift-wave stability theory).

Lecture Notes for Physics 242A, B, C and Physics 250 1971–1972 Transcribed, edited, and graphics added by Bruce I. Cohen

Foreword

Allan Kaufman (b. 1927) grew up in the Hyde Park neighbourhood of Chicago not far from the University of Chicago. Allan attended the University of Chicago for both his undergraduate and doctoral degrees in physics. Chicago was replete with physics luminaries on its faculty and future luminaries among the doctoral students. Allan’s doctoral thesis advisor was Murph Goldberger who was relatively new to the faculty at Chicago and just five years older than Allan. Allan did a theoretical thesis on a strong-coupling theory of meson-nucleon scattering. Allan published an autobiographical article entitled ‘A half-century in plasma physics’ in A.N. Kaufman, Journal of Physics: Conference Series 169 (2009) 012002.

Allan worked at Lawrence Livermore Laboratory from June 1953 through 1963. While at Livermore Laboratory he taught the one-year graduate course in electricity and magnetism in 1959–1963 at UC Berkeley. In 1963 he first taught the first semester of the graduate course in Theoretical Plasma Physics 242A at Berkeley. He taught the plasma theory course at UCLA in the 1964–1965 school year while on leave from Livermore before joining the faculty at UC Berkeley in the 1965 school year. Allan frequently taught the graduate plasma theory course and the graduate statistical mechanics course until his retirement from teaching in 1998.

The lecture notes from Kaufman’s graduate plasma theory course and a follow-on special topics course presented here were from the 1971–1972 academic year and the first quarter of the 1972–1973 academic year. The notes follow the chronological order of the lectures as they were presented. The equations and derivations are as Kaufman presented, but the text is a reconstruction of Kaufman’s discussion and commentary. The content of Kaufman’s graduate plasma theory courses evolved over time motivated by new developments in plasma theory. Thus, the material reported here does not represent the totality of Kaufman’s lecture notes on plasma theory. Some of the graphics have been downloaded from material posted as open access on the Internet or from published material with full attributions to the sources.

I joined Kaufman’s research group during the 1971–1972 academic year. At that time Allan’s group included doctoral students Dwight Nicholson, Michael Mostrom, Gary Smith and myself. Claire Max was a post-doctoral research physicist associated with the group for part of this period. I graduated in August 1975. Harry Mynick, John Cary and Robert Littlejohn did their doctoral theses with Allan shortly thereafter. One can see the influence of Allan Kaufman’s formulation of plasma theory in the late Dwight Nicholson’s fine textbook Introduction to Plasma Theory (John Wiley & Sons, 1983).

I am very grateful to Allan Kaufman for his encouragement, interest, and feedback as I prepared these lecture notes and to Alain Brizard for reviewing the manuscript and making suggestions, and corrections. I also thank Gene Tracy, Robert Littlejohn and Jonathan Wurtele for their interest and encouragement. Lastly, I thank various authors for granting me permission to use their graphics.

Bruce I. Cohen

Part 1

1 Introduction to plasma dynamics

[Editor’s note: in the first lecture of Physics 242A Kaufman discussed the syllabus for Physics 242A, B and C. Kaufman used CGS units throughout his notes. The textbook used as a general resource for the class at that time was P.C. Clemmow and J.P. Dougherty, The Electrodynamics of Particles and Plasmas, Addison-Wesley (1969).]

1.1 Basic assumptions, definitions and restrictions on scope

Definition. An ideal plasma is a charged gas wherein no bound states exist (a ‘mythical beast’).

Postulate. We exclude the sufficiently dense plasma that requires quantum effects: $\hbar \rightarrow 0$ here.

Postulate. We further ignore special relativity: $\unicode[STIX]{x1D6FD}\equiv v/c\ll 1$ .

For purposes of an introductory study of plasma dynamics we initially assume no applied magnetic field $\boldsymbol{B}=0$ and dispense with the generality of Maxwell’s equations in favour of retaining only Coulomb interactions. We assume a gas of $N$ charged particles. Then the force on particle $i$ due to all the other particles is given by

(1.1) $$\begin{eqnarray}m_{i}\dot{\boldsymbol{v}}_{i}=e_{i}\mathop{\sum }_{j(\neq i)}^{N}\hat{\boldsymbol{r}}_{ij}\frac{e_{j}}{r_{ij}^{2}}\end{eqnarray}$$

where $m_{i}$ is mass; $\boldsymbol{v}_{i}$ is velocity; the dot indicates a time derivative; $e_{i}$ is the electric charge; and $r_{ij}$ is the distance from the $j$ th particle to the $i$ th particle and there are $N$ such equations; $\hat{\boldsymbol{r}}_{ij}$ points from particle $j$ to particle $i$ . We require an approximation method to solve this system of nonlinear equations. The charges and masses are parameters that have explicit dimensions. We also require initial conditions on particle positions and velocities, and need a statistical approach because $N$ is large.

Definition. $\ell _{0}$ is the average distance between nearest neighbours; $n\approx 1/\ell _{0}^{3}$ is the number density of particles; and $\bar{v}$ is an average velocity. These define the state of the plasma, statistically.

1.2 Definition of a plasma

Form the dimensionless quantity $e^{2}/m\ell _{0}\bar{v}^{2}$ . The classical electron radius is $r_{e}=e^{2}/\mathit{mc}^{2}$ ; so divide by another length $\ell _{0}$ to form a dimensionless quantity:

Definition.

Thus, we are comparing the interaction energy to the kinetic energy in the plasma; and we treat the interaction energy as a perturbation. The plasma is said to be weakly coupled.

Postulate. In our plasmas $N$ and $\unicode[STIX]{x1D6EC}^{\ast }\gg 1$ , equivalently $m\bar{v}^{2}\sim k_{B}T\gg e^{2}/\ell _{0}$ .

We are not assuming that the total kinetic energy $Nk_{B}T\gg N^{2}e^{2}/\ell _{0}$ . [Editor’s note: in what follows, units are employed for the temperature T such that $k_{B}\equiv 1$ .] There are some plasmas in which $\unicode[STIX]{x1D6EC}^{\ast }\leqslant O(1)$ , for instance in a metal where the interaction and the Fermi energies are comparable; and a quantum mechanical treatment is then necessary. In ionic crystals $\unicode[STIX]{x1D6EC}^{\ast }\ll 1$ is possible.

Exercise. (i) Find the region in temperature $T$ and density $n$ parameter space such that $\unicode[STIX]{x1D6EC}^{\ast }\gg 1$ . (ii) Impose the additional constraints $v/c\ll 1$ and $n\unicode[STIX]{x1D706}_{\text{de Broglie}}\text{}^{3}=n(h/mv)^{3}\ll 1$ .

Definition. The collision frequency is $\unicode[STIX]{x1D708}\sim n\unicode[STIX]{x1D70E}v$ where $\unicode[STIX]{x1D70E}\sim (e^{2}/mv^{2})^{2}$ , and the plasma frequency is $\unicode[STIX]{x1D714}_{\text{pe}}\sim (4\unicode[STIX]{x03C0}\mathit{ne}^{2}/m)^{1/2}$ . Then $\unicode[STIX]{x1D708}/\unicode[STIX]{x1D714}_{\text{pe}}\sim (\unicode[STIX]{x1D6EC}^{\ast })^{-3/2}\ll 1$ , i.e. the relative collisionality of the plasma is weak.

Definition. The Debye length $\unicode[STIX]{x1D706}_{D}\equiv \bar{v}/\unicode[STIX]{x1D714}_{\text{pe}}=(T/4\unicode[STIX]{x03C0}ne^{2})^{1/2}$ is the characteristic shielding length, i.e. the effective interaction distance. The shielded potential from a test particle is $V\sim (e/r)\exp (-r/\unicode[STIX]{x1D706}_{D})$ , and the number of particles in a region around a test particle of order the Debye length in dimension is then $\unicode[STIX]{x1D6EC}\sim n\unicode[STIX]{x1D706}_{D}^{3}$ . We must require that $\unicode[STIX]{x1D6EC}\gg 1$ for the validity of a statistical approach.

Theorem. $\unicode[STIX]{x1D6EC}\sim (\unicode[STIX]{x1D6EC}^{\ast })^{3/2}$ so that the conditions of weak collisionality and weak interaction energy are closely related. We will use $\unicode[STIX]{x1D6EC}\gg 1$ exclusively and call it the plasma parameter. [Note: sometimes the plasma parameter is defined as $\unicode[STIX]{x1D6EC}\equiv 4\unicode[STIX]{x03C0}n\unicode[STIX]{x1D706}_{D}^{3}$ .]

We note it is a very good assumption for most plasmas to assume that the Debye length $\unicode[STIX]{x1D706}_{D}$ is small compared to the plasma macroscopic dimension $L$ , so that $N\sim \mathit{nL}^{3}\gg \unicode[STIX]{x1D6EC}\sim n\unicode[STIX]{x1D706}_{D}^{3}$ .

2 Vlasov–Poisson equation formulation for a collisionless plasma

2.1 Equations of motion in phase space, Poisson equation and definition of distribution function

Consider the group collective, microscopic electric field $\boldsymbol{E}$ and the equations of motion

(2.1) $$\begin{eqnarray}m_{i}\dot{\boldsymbol{v}}_{i}=e_{i}\boldsymbol{E}^{i}=e_{i}\mathop{\sum }_{j(i\neq i)}\hat{\boldsymbol{r}}_{ij}\frac{e_{j}}{\boldsymbol{r}_{ij}^{2}},\end{eqnarray}$$

where $\boldsymbol{E}^{i}$ is the electric field on particle $i$ . We coarse-grain average the point charges to smear and smooth the collective electric field,

(2.2) $$\begin{eqnarray}m_{i}\dot{\boldsymbol{v}}_{i}=e_{i}\boldsymbol{E}^{i}(\boldsymbol{r}_{i})\rightarrow e_{i}\bar{\boldsymbol{E}}(\boldsymbol{r}_{i}).\end{eqnarray}$$

The six-dimensional phase-space equations of motion are then

(2.3) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}m_{s}\dot{\boldsymbol{v}}_{s}=e_{s}\bar{\boldsymbol{E}}(\boldsymbol{r})\\ \dot{\boldsymbol{r}}=\boldsymbol{ v}\end{array}\right\}\frac{\text{d}}{\text{d}t}(\boldsymbol{r},\boldsymbol{v})=\left(\boldsymbol{v},\frac{e_{s}}{m_{s}}\bar{\boldsymbol{E}}(\boldsymbol{r})\right).\end{eqnarray}$$

This phase space is not the same as the Gibbs phase space in statistical mechanics.

Figure 1. Flow in phase space (cartoon).

Theorem. Poisson’s equation is

(2.4) $$\begin{eqnarray}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{E}=4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}(\boldsymbol{r})\quad \unicode[STIX]{x1D735}\times \boldsymbol{E}(\boldsymbol{r})=0\quad \text{where }\unicode[STIX]{x1D70C}(\boldsymbol{r})=\mathop{\sum }_{i}e_{i}\unicode[STIX]{x1D6FF}(\boldsymbol{r}-\boldsymbol{r}_{i}),\end{eqnarray}$$

with the electrostatic constraint on $\boldsymbol{E}$ and the charge density $\unicode[STIX]{x1D70C}(\boldsymbol{r})$ needs to be smoothed.

Definition. $f_{s}(\boldsymbol{r},\boldsymbol{v})$ is the mean density of particles of a species $s$ in six-dimensional phase space; then

(2.5) $$\begin{eqnarray}\bar{\unicode[STIX]{x1D70C}}(\boldsymbol{r})\equiv \mathop{\sum }_{s}e_{s}\int \text{d}^{3}\boldsymbol{v}f_{s}(\boldsymbol{r},\boldsymbol{v}).\end{eqnarray}$$

The smoothed version of (2.4) becomes

(2.6) $$\begin{eqnarray}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\bar{\boldsymbol{E}}=4\unicode[STIX]{x03C0}\mathop{\sum }_{s}e_{s}\int \text{d}^{3}\boldsymbol{v}f_{s}(\boldsymbol{r},\boldsymbol{v})\quad \unicode[STIX]{x1D735}\times \bar{\boldsymbol{E}}=0.\end{eqnarray}$$

$f_{s}(\boldsymbol{r},\boldsymbol{v})$ evolves in time: what is the equation of evolution for $f_{s}(\boldsymbol{r},\boldsymbol{v};t)$ in time? Introduce $\boldsymbol{x}=(\boldsymbol{r},\boldsymbol{v})$ and $(\text{d}/\text{d}t)\boldsymbol{x}\equiv \boldsymbol{X}(\boldsymbol{x};t)$ ; then $f_{s}(\boldsymbol{r},\boldsymbol{v};t)\equiv f_{s}(\boldsymbol{x};t)$ .

Theorem. The number of particles $N_{V}$ for any species in a volume $V$ is

(2.7) $$\begin{eqnarray}N_{V}(t)=\int _{V}\text{d}^{6}\boldsymbol{x}f(\boldsymbol{x};t).\end{eqnarray}$$

Because the number of particles in the volume is conserved, except for net fluxes into or out of the surfaces bounding the volume, it follows that

(2.8) $$\begin{eqnarray}\frac{\text{d}N_{V}}{\text{d}t}=\int _{V}\text{d}^{6}\boldsymbol{x}\frac{\unicode[STIX]{x2202}f(\boldsymbol{x};t)}{\unicode[STIX]{x2202}t}=-\oint _{\text{surfaces}}\text{d}\hat{\unicode[STIX]{x1D70E}}\boldsymbol{\cdot }\boldsymbol{x}f=-\int _{V}\text{d}^{6}\boldsymbol{x}\unicode[STIX]{x1D735}\boldsymbol{\cdot }(\boldsymbol{X}f),\end{eqnarray}$$

where $\hat{\unicode[STIX]{x1D70E}}$ points out of the volume and the divergence theorem has been used. Given that the volume integrals in (2.8) are equal for whatever subdomain of phase space is enclosed in $V$ , the integrands must be equal; and we arrive at the Vlasov equation.

Theorem. Vlasov equation

(2.9) $$\begin{eqnarray}\displaystyle \frac{\unicode[STIX]{x2202}f_{s}(\boldsymbol{x};t)}{\unicode[STIX]{x2202}t} & = & \displaystyle -\unicode[STIX]{x1D735}\boldsymbol{\cdot }(\boldsymbol{X}f_{s})=-\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{r}}(\boldsymbol{v}f_{s})-\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{v}}\left(\frac{e_{s}}{m_{s}}\bar{\boldsymbol{E}}(\boldsymbol{r},t)f_{s}\right)\nonumber\\ \displaystyle & = & \displaystyle -\boldsymbol{v}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{r}}f_{s}-\frac{e_{s}}{m_{s}}\bar{\boldsymbol{E}}(\boldsymbol{r},t)\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{v}}f_{s},\end{eqnarray}$$

which can be rewritten as

(2.10) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}f_{s}(\boldsymbol{x};t)}{\unicode[STIX]{x2202}t}+\boldsymbol{v}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{r}}f_{s}+\frac{e_{s}}{m_{s}}\bar{\boldsymbol{E}}(\boldsymbol{r},t)\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{v}}f_{s}=0.\end{eqnarray}$$

In the presence of volumetric sources and sinks, e.g. ionization and recombination, and/or collisions, the right-hand side of (2.10) is no longer zero.

2.2 Continuity equation in phase space – Liouville theorem

A number of observations can be made immediately on inspecting the derivation of the Vlasov equation. From (2.7), (2.8) and $\text{d}\boldsymbol{x}/\text{d}t\equiv \boldsymbol{X}(\boldsymbol{x},t)$ we have

(2.11) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}f(\boldsymbol{x};t)}{\unicode[STIX]{x2202}t}=-\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{x}}\boldsymbol{\cdot }(\boldsymbol{X}f)=-\boldsymbol{X}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}f}{\unicode[STIX]{x2202}\boldsymbol{x}}-f\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{x}}\boldsymbol{\cdot }\boldsymbol{X}\end{eqnarray}$$

and hence,

(2.12) $$\begin{eqnarray}\left(\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}+\dot{\boldsymbol{x}}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{x}}\right)f(\boldsymbol{x};t)=-f(\boldsymbol{x};t)\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{x}}\boldsymbol{\cdot }\boldsymbol{X}=-f(\boldsymbol{x};t)\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{X}.\end{eqnarray}$$

Equation (2.12) is a phase-space continuity equation. The left-hand side of this equation is just a convective derivative, and the right-hand side allows for compressibility. If $\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{X}<0$ then $\text{D}f/\text{D}t>0$ , and $\text{D}f/\text{D}t<0$ if $\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{X}>0$ . We note that as an almost trivial consequence of the independent phase-space variables,

(2.13) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{r}}\boldsymbol{\cdot }\boldsymbol{v}=0,\quad \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{v}}\boldsymbol{\cdot }\left(\frac{e}{m}\left(\boldsymbol{E}+\frac{\boldsymbol{v}}{c}\times \boldsymbol{B}\right)\right)=0\Rightarrow \unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{X}=\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{r}}\boldsymbol{\cdot }\boldsymbol{v}+\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{v}}\boldsymbol{\cdot }\dot{\boldsymbol{v}}=0.\end{eqnarray}$$

Theorem (Liouville theorem). If $\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{X}=0$ , then the right-hand side of (2.12) is zero and (2.8.2) corresponds exactly to the Liouville theorem for Hamiltonian systems:

(2.14) $$\begin{eqnarray}\left(\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}+\dot{\boldsymbol{x}}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{x}}f(\boldsymbol{x};t)=0\right)\!.\end{eqnarray}$$

In this limit the phase-space flow is ‘incompressible’ and $\text{D}f/\text{D}t=0$ , i.e. $f$ is conserved along the phase-space trajectories. If the number of particles per unit volume $f$ is conserved then so is $1/f$ , which is the differential volume element per unit particle, i.e. the phase-space volume element is also conserved (although its shape may deform).

2.3 Nonlinear Vlasov equation with self-consistent fields

Theorem. Vlasov–Maxwell equations in a plasma with self-consistent fields

(2.15) $$\begin{eqnarray}\displaystyle & & \displaystyle \left(\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}+\boldsymbol{v}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{r}}+\frac{e_{s}}{m_{s}}\left(\boldsymbol{E}+\frac{\boldsymbol{v}}{c}\times \boldsymbol{B}\right)\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{v}}\right)f_{s}(\boldsymbol{r},\boldsymbol{v};t)=0\end{eqnarray}$$
(2.16) $$\begin{eqnarray}\displaystyle & & \displaystyle \qquad \unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{E}=4\unicode[STIX]{x03C0}\mathop{\sum }_{s}e_{s}\int \text{d}^{3}\boldsymbol{v}f_{s}(\boldsymbol{r},\boldsymbol{v};t)\end{eqnarray}$$
(2.17) $$\begin{eqnarray}\displaystyle & & \displaystyle \qquad \unicode[STIX]{x1D735}\times \boldsymbol{B}-\frac{1}{c}\frac{\unicode[STIX]{x2202}\boldsymbol{E}}{\unicode[STIX]{x2202}t}=\frac{4\unicode[STIX]{x03C0}}{c}\mathop{\sum }_{s}e_{s}\int \text{d}^{3}\boldsymbol{v}\boldsymbol{v}f_{s}(\boldsymbol{r},\boldsymbol{v};t)+\frac{4\unicode[STIX]{x03C0}}{c}\boldsymbol{j}_{\text{ext}}\end{eqnarray}$$
(2.18) $$\begin{eqnarray}\displaystyle & & \displaystyle \qquad \unicode[STIX]{x1D735}\times \boldsymbol{E}+\frac{1}{c}\frac{\unicode[STIX]{x2202}\boldsymbol{B}}{\unicode[STIX]{x2202}t}=0\end{eqnarray}$$
(2.19) $$\begin{eqnarray}\displaystyle & & \displaystyle \qquad \unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{B}=0\end{eqnarray}$$

and we could include the gravitational Poisson equation,

(2.20) $$\begin{eqnarray}\unicode[STIX]{x1D6FB}^{2}\unicode[STIX]{x1D719}_{g}=4\unicode[STIX]{x03C0}G\unicode[STIX]{x1D70C}_{m}=4\unicode[STIX]{x03C0}G\mathop{\sum }_{s}m_{s}\int \text{d}^{3}\boldsymbol{v}f_{s},\end{eqnarray}$$

where the gravitational field $\boldsymbol{g}=-\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}_{g}$ , $\unicode[STIX]{x1D719}_{g}$ is the gravitational potential, $\unicode[STIX]{x1D70C}_{m}$ is the mass density and $G$ is the universal gravitational constant. The gravitational field $\boldsymbol{g}$ could then be included in (2.15) as an additional acceleration term.

In a Hamiltonian system one can introduce the notation

(2.21) $$\begin{eqnarray}\boldsymbol{x}=(q_{i},p_{i})\quad \boldsymbol{X}=\left(\frac{\unicode[STIX]{x2202}H}{\unicode[STIX]{x2202}p_{i}},-\frac{\unicode[STIX]{x2202}H}{\unicode[STIX]{x2202}q_{i}}\right),\end{eqnarray}$$

where $H$ is the particle Hamiltonian and the $i$ index represents a phase-space degree of freedom. The Vlasov equation then can be written as

(2.22) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}f}{\unicode[STIX]{x2202}t}+\mathop{\sum }_{i}\left(\frac{\unicode[STIX]{x2202}f}{\unicode[STIX]{x2202}q_{i}}\frac{\unicode[STIX]{x2202}H}{\unicode[STIX]{x2202}p_{i}}-\frac{\unicode[STIX]{x2202}f}{\unicode[STIX]{x2202}p_{i}}\frac{\unicode[STIX]{x2202}H}{\unicode[STIX]{x2202}q_{i}}\right)=\frac{\unicode[STIX]{x2202}f}{\unicode[STIX]{x2202}t}+\{\,f,h\}=0,\end{eqnarray}$$

where $\{\,f,H\}$ denotes the Poisson bracket.

By simplifying the electromagnetic fields to be electrostatic, equations (2.15)–(2.19) become the Vlasov–Poisson equations, which are written as

(2.23) $$\begin{eqnarray}\displaystyle & & \displaystyle \boldsymbol{E}=-\unicode[STIX]{x1D735}\unicode[STIX]{x1D719},\quad \unicode[STIX]{x1D6FB}^{2}\unicode[STIX]{x1D719}=-4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}_{c},\quad \unicode[STIX]{x1D719}(\boldsymbol{r},t)=\int \text{d}^{3}\boldsymbol{r}^{\prime }\frac{\unicode[STIX]{x1D70C}_{c}(\boldsymbol{r}^{\prime },t)}{|\boldsymbol{r}-\boldsymbol{r}^{\prime }|},\nonumber\\ \displaystyle & & \displaystyle \frac{\unicode[STIX]{x2202}f_{s}}{\unicode[STIX]{x2202}t}+\boldsymbol{v}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}f_{s}}{\unicode[STIX]{x2202}\boldsymbol{r}}+\frac{e_{s}}{m_{s}}\frac{\unicode[STIX]{x2202}f_{s}}{\unicode[STIX]{x2202}\boldsymbol{v}}\boldsymbol{\cdot }\left(-\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{r}}\right)\int \text{d}^{3}\boldsymbol{r}^{\prime }\frac{4\unicode[STIX]{x03C0}\mathop{\sum }_{s^{\prime }}e_{s^{\prime }}\int \text{d}^{3}\boldsymbol{v}^{\prime }f_{s}(\boldsymbol{r}^{\prime },\boldsymbol{v}^{\prime },t)}{|\boldsymbol{r}-\boldsymbol{r}^{\prime }|}=0.\end{eqnarray}$$

We next consider the qualitative properties of the nonlinear self-consistent Vlasov equation in (2.23). The relative orders of the three terms are $f/\unicode[STIX]{x1D70F}:vf/\unicode[STIX]{x1D706}:f^{2}e^{2}\unicode[STIX]{x1D706}v^{2}/m$ , where $\unicode[STIX]{x1D70F}$ , $\unicode[STIX]{x1D706}$ and $v$ are the characteristic temporal, spatial and particle velocity scales. (i) Balancing the first two terms in the Vlasov equations yields $\unicode[STIX]{x1D70F}\sim \unicode[STIX]{x1D706}/v$ or $\unicode[STIX]{x1D714}/k\sim v$ , i.e. $v_{p}\sim v$ , where $v_{p}=\unicode[STIX]{x1D714}/k$ , $\unicode[STIX]{x1D714}$ is a characteristic frequency and $k$ is a wavenumber. (ii) Balancing the second and third terms yields $\unicode[STIX]{x1D714}\sim 4\unicode[STIX]{x03C0}(e^{2}/m)fv^{3}\unicode[STIX]{x1D706}/v\sim (4\unicode[STIX]{x03C0}\mathit{ne}^{2}/m)\unicode[STIX]{x1D706}/v$ which using $\unicode[STIX]{x1D714}\unicode[STIX]{x1D706}\sim v$ leads to $\unicode[STIX]{x1D714}^{2}\sim 4\unicode[STIX]{x03C0}\mathit{ne}^{2}/m$ , which is the plasma frequency squared. (iii) With $\unicode[STIX]{x1D706}\sim v/\unicode[STIX]{x1D714}$ and setting $v\sim v_{\text{th}}=(T/m)^{1/2}$ the electron thermal velocity, then $\unicode[STIX]{x1D706}\sim \unicode[STIX]{x1D706}_{D}\sim (T/4\unicode[STIX]{x03C0}\mathit{ne}^{2})^{1/2}$ where $\unicode[STIX]{x1D706}_{D}$ is the Debye length. We will see that many plasma phenomena can be characterized in terms of important dimensionless variables, for example, $\unicode[STIX]{x1D714}/\unicode[STIX]{x1D714}_{\text{pe}}$ , $\unicode[STIX]{x1D706}/\unicode[STIX]{x1D706}_{D}$ , $m_{e}/m_{i}$ , $T_{e}/T_{i}$ , $\unicode[STIX]{x1D714}/kv$ , $L/\unicode[STIX]{x1D706}_{D}$ , $\unicode[STIX]{x1D714}_{\text{ps}}/\unicode[STIX]{x1D714}_{\text{cs}}$ , $\unicode[STIX]{x0394}\unicode[STIX]{x1D714}/\unicode[STIX]{x1D714}$ , $v_{\text{th}}/c$ and the ratios of $E^{2}$ to $B^{2}$ and to $\mathit{nm}v^{2}$ . There are also plasma attributes and phenomena associated with non-uniformity and anisotropy.

2.4 Moment equations

2.4.1 Conservation of mass density, momentum density, energy density

Rewrite the collisionless Vlasov equation in the alternative form

(2.24) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}f_{s}+\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{r}}\boldsymbol{\cdot }(\boldsymbol{v}f_{s})+\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{v}}\boldsymbol{\cdot }(\boldsymbol{a}f_{s})=0,\end{eqnarray}$$

where the acceleration $\boldsymbol{a}$ is

(2.25) $$\begin{eqnarray}\boldsymbol{a}=\frac{e_{s}}{m_{s}}\left(\boldsymbol{E}+\frac{1}{c}\boldsymbol{v}\times \boldsymbol{B}\right)\end{eqnarray}$$

and define the number density $n_{s}$

(2.26) $$\begin{eqnarray}n_{s}(\boldsymbol{r},t)=\int \text{d}^{3}\boldsymbol{v}f_{s}(\boldsymbol{r},\boldsymbol{v};t).\end{eqnarray}$$

Definition. Moments of the velocity distribution are constructed from

(2.27) $$\begin{eqnarray}\langle A\rangle _{s}(\boldsymbol{r};t)\equiv \frac{\int \,\text{d}^{3}\boldsymbol{v}Af_{s}(\boldsymbol{r},\boldsymbol{v};t)}{\int \,\text{d}^{3}\boldsymbol{v}f_{s}(\boldsymbol{r},\boldsymbol{v};t)}.\end{eqnarray}$$

Examples.

  1. (i) $A=1\rightarrow$ identity operation.

  2. (ii) $A=\boldsymbol{v}\rightarrow \boldsymbol{u}\equiv \langle \boldsymbol{v}\rangle$ the average velocity.

  3. (iii) $A=(\boldsymbol{v}-\boldsymbol{u})(\boldsymbol{v}-\boldsymbol{u})\rightarrow nm\langle (\boldsymbol{v}-\boldsymbol{u})(\boldsymbol{v}-\boldsymbol{u})\rangle \equiv \unicode[STIX]{x1D64B}(r,t)$ the pressure tensor.

  4. (iv) $A=e$ then $\sum _{s}e_{s}n_{s}\equiv \unicode[STIX]{x1D70C}(\boldsymbol{r};t)$ the charge density using (2.26).

  5. (v) $A={\textstyle \frac{1}{2}}\,mv^{2}\rightarrow K(r;t)\equiv \int \text{d}3\boldsymbol{v}{\textstyle \frac{1}{2}}\,mv^{2}f=n\langle {\textstyle \frac{1}{2}}\,mv^{2}\rangle$ the kinetic energy density.

Theorem. A generalized moment equation can be derived directly from the Vlasov equation (the species index $s$ is understood),

(2.28) $$\begin{eqnarray}\displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}[n\langle A\rangle ] & = & \displaystyle \int \text{d}^{3}\boldsymbol{v}\left(\frac{\unicode[STIX]{x2202}f}{\unicode[STIX]{x2202}t}A+\frac{\unicode[STIX]{x2202}A}{\unicode[STIX]{x2202}t}f\right)=n\left\langle \frac{\unicode[STIX]{x2202}A}{\unicode[STIX]{x2202}t}\right\rangle +\int \text{d}^{3}\boldsymbol{v}A\left[-\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{r}}\boldsymbol{\cdot }(\boldsymbol{v}f)-\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{v}}\boldsymbol{\cdot }(\boldsymbol{a}f)\right]\nonumber\\ \displaystyle & = & \displaystyle n\left\langle \frac{\unicode[STIX]{x2202}A}{\unicode[STIX]{x2202}t}\right\rangle -\unicode[STIX]{x1D735}\boldsymbol{\cdot }[n\langle A\boldsymbol{v}\rangle ]+n\left\langle \boldsymbol{a}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}A}{\unicode[STIX]{x2202}\boldsymbol{v}}\right\rangle .\end{eqnarray}$$

Examples.

  1. (i) $A=1\rightarrow$ continuity equation

    (2.29) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}n(\boldsymbol{r};t)}{\unicode[STIX]{x2202}t}=-\unicode[STIX]{x1D735}\boldsymbol{\cdot }(n\boldsymbol{u})\end{eqnarray}$$

or

(2.30) $$\begin{eqnarray}\frac{\text{d}}{\text{d}t}n(\boldsymbol{r};t)=\left(\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}+\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\right)n=-n\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}.\end{eqnarray}$$
  1. (ii) $A=m\boldsymbol{v}\rightarrow$ momentum conservation

    (2.31) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\langle nm\boldsymbol{v}\rangle =\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}nm\boldsymbol{u}=-\unicode[STIX]{x1D735}\boldsymbol{\cdot }(nm\langle \boldsymbol{v}\boldsymbol{v}\rangle )+ne\left(\boldsymbol{E}+\frac{1}{c}\boldsymbol{u}\times \boldsymbol{B}\right).\end{eqnarray}$$

We can use the identity $\langle \boldsymbol{v}\boldsymbol{v}\rangle =\langle (\boldsymbol{v}-\boldsymbol{u})(\boldsymbol{v}-\boldsymbol{u})\rangle +\boldsymbol{u}\boldsymbol{u}$ , the continuity equation (2.30) and the definition of the pressure tensor $\unicode[STIX]{x1D64B}$ in conjunction with (2.31) to derive the fluid momentum balance equation,

(2.32) $$\begin{eqnarray}nm\frac{\text{d}}{\text{d}t}\boldsymbol{u}=-\unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D64B}+ne\left(\boldsymbol{E}+\frac{1}{c}\boldsymbol{u}\times \boldsymbol{B}\right).\end{eqnarray}$$

We can consider the Coulomb case, assume there is no magnetic field, sum over species, and integrate (2.31) over all space to demonstrate the total particle momentum is a constant,

(2.33) $$\begin{eqnarray}\displaystyle & & \displaystyle \int \text{d}^{3}\boldsymbol{r}\left[\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\mathop{\sum }_{s}\left\langle n_{s}m_{s}\boldsymbol{v}\right\rangle \right]=\int \text{d}^{3}\boldsymbol{r}\left[\mathop{\sum }_{x}\left(-\unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D64B}_{s}+n_{s}e_{s}\boldsymbol{E}\right)\right]\nonumber\\ \displaystyle & & \displaystyle \quad =\int \text{d}^{3}\boldsymbol{r}\left[\mathop{\sum }_{x}\left(-\unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D64B}_{s}\right)+\unicode[STIX]{x1D70C}\boldsymbol{E}\right]=\int \text{d}^{3}\boldsymbol{r}\left[\mathop{\sum }_{x}\left(-\unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D64B}_{s}\right)+\frac{1}{4\unicode[STIX]{x03C0}}\left(\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{E}\right)\boldsymbol{E}\right]\nonumber\\ \displaystyle & & \displaystyle \quad =\int \text{d}^{3}\boldsymbol{r}\left[\mathop{\sum }_{x}\left(-\unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D64B}_{s}\right)+\unicode[STIX]{x1D735}\boldsymbol{\cdot }\left(\frac{A\boldsymbol{A}}{4\unicode[STIX]{x03C0}}-\frac{E^{2}\unicode[STIX]{x1D644}}{8\unicode[STIX]{x03C0}}\right)\right]\nonumber\\ \displaystyle & & \displaystyle \quad =\int \text{d}^{3}\boldsymbol{r}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\left[\mathop{\sum }_{x}\left(-\unicode[STIX]{x1D64B}_{s}-\frac{\boldsymbol{E}\boldsymbol{E}}{4\unicode[STIX]{x03C0}}+\frac{E^{2}\unicode[STIX]{x1D644}}{8\unicode[STIX]{x03C0}}\right)\right]\nonumber\\ \displaystyle & & \displaystyle \quad =\oint \text{d}\boldsymbol{S}\boldsymbol{\cdot }\left(-\unicode[STIX]{x1D64B}_{s}-\frac{\boldsymbol{E}\boldsymbol{E}}{4\unicode[STIX]{x03C0}}+\frac{E^{2}\unicode[STIX]{x1D644}}{8\unicode[STIX]{x03C0}}\right),\end{eqnarray}$$

where $\text{d}\boldsymbol{S}$ is directed out of volume on its surface and we have used the divergence theorem and Gauss’ law, and assumed the fields and the velocity distributions vanish at infinity. We can include field stresses and field momentum using Maxwell’s equations as in § 6.7 of Jackson’s Classical Electrodynamics textbook to demonstrate that the total particle and field momentum is conserved.

  1. (iii) $A={\textstyle \frac{1}{2}}mv^{2}\rightarrow$ energy conservation

    (2.34) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}K_{s}=-\unicode[STIX]{x1D735}\boldsymbol{\cdot }n_{s}\left\langle \frac{1}{2}m_{s}v^{2}\boldsymbol{v}\right\rangle +n_{s}\langle e_{s}\boldsymbol{v}\rangle \boldsymbol{\cdot }\boldsymbol{E}=-\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{S}_{s}^{k}+\boldsymbol{j}_{s}\boldsymbol{\cdot }\boldsymbol{E}.\end{eqnarray}$$

We note that the magnetic field does no work and (2.34) can be summed over species to obtain the equation for energy conservation of all particle species. $K$ is an energy density. Equation (2.34) summed over species is extended to include the electromagnetic field energy density as follows:

(2.35) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\left(K+\frac{E^{2}+B^{2}}{8\unicode[STIX]{x03C0}}\right)=\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}(K+E)=-\unicode[STIX]{x1D735}\boldsymbol{\cdot }\left(\boldsymbol{S}^{K}+\frac{c}{4\unicode[STIX]{x03C0}}\boldsymbol{E}\times \boldsymbol{B}\right)=-\unicode[STIX]{x1D735}\boldsymbol{\cdot }(\boldsymbol{S}^{K}+\boldsymbol{S}^{EM}),\end{eqnarray}$$

where we recognize $\boldsymbol{S}^{EM}$ as the electromagnetic Poynting flux and note that $\boldsymbol{J}\boldsymbol{\cdot }\boldsymbol{E}$ terms identically cancel. By integrating (2.35) over volume, using the divergence theorem, and assuming all quantities vanish at infinity, we can demonstrate that total energy is conserved.

  1. (iv) $A=mr^{2}\rightarrow$ moment of inertia

    (2.36) $$\begin{eqnarray}\displaystyle I(t) & = & \displaystyle \frac{1}{2}\int \text{d}^{3}\boldsymbol{r}\mathop{\sum }_{s}m_{s}r^{2}n_{s}(\boldsymbol{r};t)\nonumber\\ \displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}I & = & \displaystyle -\frac{1}{2}\int \text{d}^{3}\boldsymbol{r}\mathop{\sum }_{s}m_{s}\boldsymbol{r}^{2}\unicode[STIX]{x1D735}\boldsymbol{\cdot }(n_{s}\boldsymbol{u}_{s})=\int \text{d}^{3}\boldsymbol{r}\mathop{\sum }_{s}m_{s}n_{s}\boldsymbol{u}_{s}\boldsymbol{\cdot }\boldsymbol{r}.\end{eqnarray}$$

Here, $I$ is the moment of inertia (not to be confused with the identity tensor $\unicode[STIX]{x1D644}$ ) and is a global scalar quantity that only has a time variation. Equation (2.36) is derived using the continuity equation (2.29), integrating by parts, and using $\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}r^{2}=2\boldsymbol{u}\boldsymbol{\cdot }\boldsymbol{r}$ .

2.4.2 Virial theorem

One can deduce a relation from the second time derivative (acceleration) of the moment of inertia relation in (2.36) that provides insight into how a plasma can radiate electromagnetic fields which allows the system to collapse. This is embodied in a virial theorem.

Viral Theorem. We begin by deriving the electromagnetic momentum conservation law from Maxwell’s equations,

(2.37) $$\begin{eqnarray}\displaystyle -\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\left(\frac{\boldsymbol{E}\times \boldsymbol{B}}{4\unicode[STIX]{x03C0}c}\right) & = & \displaystyle \unicode[STIX]{x1D70C}\boldsymbol{E}+\frac{1}{c}\boldsymbol{j}\times \boldsymbol{B}-\frac{1}{4\unicode[STIX]{x03C0}}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\left(\boldsymbol{E}\boldsymbol{E}-\frac{1}{2}E^{2}\unicode[STIX]{x1D644}+\boldsymbol{B}\boldsymbol{B}-\frac{1}{2}B^{2}\unicode[STIX]{x1D644}\right)\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D70C}\boldsymbol{E}+\frac{1}{c}\boldsymbol{j}\times \boldsymbol{B}-\frac{1}{4\unicode[STIX]{x03C0}}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D64B}^{\,EM}.\end{eqnarray}$$

We then take another time derivative in (2.36), use the momentum conservation equation (2.31) and include the $\boldsymbol{j}\times \boldsymbol{B}/c$ force in (2.33). We next eliminate $\unicode[STIX]{x1D70C}\boldsymbol{E}+\boldsymbol{j}\times \boldsymbol{B}/c$ using (2.37) to obtain

$$\begin{eqnarray}\ddot{I}=\int \text{d}^{3}\boldsymbol{r}\boldsymbol{r}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\mathop{\sum }_{s}n_{s}m_{s}\boldsymbol{u}_{s}=\int \text{d}^{3}\boldsymbol{r}\boldsymbol{r}\boldsymbol{\cdot }\left[-\unicode[STIX]{x1D735}\boldsymbol{\cdot }(\unicode[STIX]{x1D64B}^{\,K}+\unicode[STIX]{x1D64B}^{\,EM})-\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\left(\frac{\boldsymbol{E}\times \boldsymbol{B}}{4\unicode[STIX]{x03C0}c}\right)\right].\end{eqnarray}$$

We add the term involving the electromagnetic momentum to both sides of the equation to obtain

(2.38) $$\begin{eqnarray}\ddot{I}+\frac{\text{d}}{\text{d}t}\int \text{d}^{3}\boldsymbol{r}\boldsymbol{r}\boldsymbol{\cdot }\left(\frac{\boldsymbol{E}\times \boldsymbol{B}}{4\unicode[STIX]{x03C0}c}\right)=\int \text{d}^{3}\boldsymbol{r}\unicode[STIX]{x1D644}\boldsymbol{ : }\left(\unicode[STIX]{x1D64B}^{K}+\unicode[STIX]{x1D64B}^{EM}\right)=\int \text{d}^{3}\boldsymbol{r}\left(2K+E^{EM}\right)>0,\end{eqnarray}$$

where an integration by parts has been performed and $\unicode[STIX]{x1D644}$ : denotes the resulting double dot product of the identity tensor with the tensor(s) following it. In the absence of the radiation flux $\boldsymbol{S}^{EM}$ , one concludes that $\ddot{I}>0$ because the right-hand side of (2.37) is positive. In this limit the moment of inertia can only increase: in the absence of magnetic coils or gravity, the system cannot be contained by its own electromagnetic fields. With a finite radiation flux, the system can collapse and radiate energy away. An important caveat that limits these conclusions is that we did not include gravitation, which would introduce a negative term on the right-hand side.

Exercise. Verify (2.37) and the double dot product in (2.38).

2.5 Linear analysis of the Vlasov equation for small-amplitude disturbances in a uniform plasma

We can obtain exact solutions of the linearized Vlasov equation for infinitesimal amplitude perturbations.

Definition. Static solutions correspond to $\unicode[STIX]{x2202}f/\unicode[STIX]{x2202}t=0$ . This situation applies to a very small, but important, class of solutions. The time-independent Vlasov equation is

(2.39) $$\begin{eqnarray}\boldsymbol{v}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{r}}f+\frac{e}{m}\left(\boldsymbol{E}+\frac{1}{c}\boldsymbol{v}\times \boldsymbol{B}\right)\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{v}}f=0.\end{eqnarray}$$

Uniform solutions correspond to $\unicode[STIX]{x2202}f/\unicode[STIX]{x2202}\boldsymbol{r}=0$ .

In the absence of electric and magnetic fields there exists a solution for a spatially uniform $f(\boldsymbol{v})$ that can be an arbitrary function of velocity. For this simple case the solution of the time-dependent Vlasov equation is that $f$ is a constant along the phase-space trajectories and remains fixed at its initial, arbitrary function of velocity,

(2.40) $$\begin{eqnarray}\frac{\text{d}}{\text{d}t}f(\boldsymbol{v})=\frac{\text{d}\boldsymbol{v}}{\text{d}t}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{v}}f(\boldsymbol{v})=0.\end{eqnarray}$$

This almost trivial result has utility in that we now add a small-amplitude perturbation that can depend on time and space. For simplicity we restrict consideration to the case of a Coulomb model. The equation set for a small-amplitude, linear expansion of the distribution function and the self-consistent electric fields is as follows:

(2.41a ) $$\begin{eqnarray}\displaystyle & & \displaystyle f_{s}(\boldsymbol{r},\boldsymbol{v};t)=f_{0s}(\boldsymbol{v})+\unicode[STIX]{x1D6FF}f_{s}(\boldsymbol{r},\boldsymbol{v}\boldsymbol{ : }t)\end{eqnarray}$$
(2.41b ) $$\begin{eqnarray}\displaystyle & & \displaystyle \boldsymbol{E}(\boldsymbol{r};t)=0+\unicode[STIX]{x1D6FF}\boldsymbol{E}(\boldsymbol{r};t)\quad \unicode[STIX]{x1D735}\times \boldsymbol{E}=0\end{eqnarray}$$
(2.41c ) $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{E}=4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}(\boldsymbol{r};t)=4\unicode[STIX]{x03C0}\mathop{\sum }_{s}e_{s}\int \text{d}^{3}\boldsymbol{v}f_{s}\end{eqnarray}$$
(2.41d ) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FF}f_{s}}{\unicode[STIX]{x2202}t}+\boldsymbol{v}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FF}f_{s}}{\unicode[STIX]{x2202}\boldsymbol{r}}+\frac{e_{s}}{m_{s}}\unicode[STIX]{x1D6FF}\boldsymbol{E}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{v}}\left(f_{0s}+\unicode[STIX]{x1D6FF}f_{s}\right)=0\rightarrow \nonumber\\ \displaystyle & & \displaystyle \qquad \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FF}f_{s}}{\unicode[STIX]{x2202}t}+\boldsymbol{v}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FF}f_{s}}{\unicode[STIX]{x2202}\boldsymbol{r}}=-\frac{e_{s}}{m_{s}}\unicode[STIX]{x1D6FF}\boldsymbol{E}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}f_{0s}}{\unicode[STIX]{x2202}\boldsymbol{v}}.\end{eqnarray}$$
The term on the right-hand side of (2.41d ) that is nonlinear in the product of $\unicode[STIX]{x1D6FF}\boldsymbol{E}$ and $\unicode[STIX]{x1D6FF}f_{s}$ , is dropped because of the linearization, i.e. only first-order terms in a Taylor-series expansion are retained. One must be careful with the vector calculus in (2.41d ) when using non-Cartesian coordinates, and canonical coordinates can prove useful. We introduce the perturbed electric potential such that $\unicode[STIX]{x1D6FF}\boldsymbol{E}=-\unicode[STIX]{x1D735}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}$ and follow the prescription: (i) Solve for $\unicode[STIX]{x1D6FF}f_{s}$ in terms of $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}$ using (2.41d ). (ii) Construct the linearly perturbed charge density $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}$ from $\unicode[STIX]{x1D6FF}f_{s}$ using (2.41c ). (iii) Solve for $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}$ using Poisson’s equation derived from (2.41b ) using suitable boundary conditions.

2.5.1 Causality, stationarity and uniformity in the dielectric kernel

We can deduce a linear relation of $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}$ on $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}$ for the linearized system. The charges in the plasma respond to the small-amplitude field produced by the perturbed electric potential. The most general linear relation can be represented as

(2.42) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}(\boldsymbol{r};t)=\int \text{d}^{3}\boldsymbol{r}^{\prime }\int \text{d}t^{\prime }\unicode[STIX]{x1D712}(\boldsymbol{r},\boldsymbol{r}^{\prime };t,t^{\prime })\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r}^{\prime };t^{\prime }).\end{eqnarray}$$

The representation allows $\unicode[STIX]{x1D712}$ to be a generalized function. In fact, it can have some unusual properties, viz., including being a derivative of a delta function. However, $\unicode[STIX]{x1D712}$ is subject to at least three important constraints:

  1. (i) Causality: a perturbation in $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}$ will cause a later perturbation in $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}$ .

  2. (ii) Stationarity: the effect of $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}$ on $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}$ can depend only on the time interval between cause and effect $(t-t^{\prime })>0$ , and cannot depend on absolute time. This is a consequence of the underlying unperturbed system being stationary, i.e. time independent.

  3. (iii) Uniformity: $\unicode[STIX]{x1D712}$ can only depend spatially on $\boldsymbol{r}-\boldsymbol{r}^{\prime }$ (isotropy would imply $|\boldsymbol{r}-\boldsymbol{r}^{\prime }|$ ) because the underlying unperturbed system has no dependence on spatial coordinate.

Theorem. We introduce $\unicode[STIX]{x1D70F}\equiv t-t^{\prime }$ and $\boldsymbol{s}\equiv \boldsymbol{r}-\boldsymbol{r}^{\prime }$ , and express $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}$ in terms convolution integrals.

(2.43) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}(\boldsymbol{r};t)=\int \text{d}^{3}\boldsymbol{s}\int _{0}^{\infty }\text{d}\unicode[STIX]{x1D70F}\unicode[STIX]{x1D712}(\boldsymbol{s};\unicode[STIX]{x1D70F})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r}-\boldsymbol{s};t-\unicode[STIX]{x1D70F}).\end{eqnarray}$$

2.5.2 Solution of the dielectric function via Fourier transform in time

We introduce the Fourier transform in time.

Definition. The Fourier transform of $g(t)$ is

(2.44) $$\begin{eqnarray}g(\unicode[STIX]{x1D714})=\int _{-\infty }^{\infty }\text{d}t\,g(t)\exp (\text{i}\unicode[STIX]{x1D714}t).\end{eqnarray}$$

We need to impose initial conditions on the linear perturbations in the electric potential, velocity distribution function and an externally imposed free charge density $\{\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719},\unicode[STIX]{x1D6FF}f,\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}^{\text{ext}}$ ) to calculate the Fourier transform: $g(t)=0$ for $t<0$ and $g(\unicode[STIX]{x1D714})=\int _{0}^{\infty }\text{d}tg(t)\exp (\text{i}\unicode[STIX]{x1D714}t)$ . The externally imposed free charge density is internal to the plasma domain. $g(t)$ must die out with time in order that $g$ ( $\unicode[STIX]{x1D714}$ ) converges. However, in general $g(t)$ does not die out and may even grow. If $g(t)$ does not die, then $g$ ( $\unicode[STIX]{x1D714}$ ) can be made to converge if $\unicode[STIX]{x1D714}$ is complex, i.e. $\unicode[STIX]{x1D714}=\unicode[STIX]{x1D714}^{\prime }+\text{i}\unicode[STIX]{x1D714}^{\prime \prime }$ with $\unicode[STIX]{x1D714}^{\prime \prime }>0$ . $g(\unicode[STIX]{x1D714})$ will converge even if $g(t)$ is growing exponentially. For exponential growth $\unicode[STIX]{x1D714}^{\prime \prime }\geqslant 1/(\text{growth time})$ . If $g(t)$ grows faster than exponentially, no convergence is possible. The integration contour for the Fourier transform is shown in figure 2.

Figure 2. Fourier transform integration contour in the complex $\unicode[STIX]{x1D714}$ plane.

We next calculate the Fourier transform of (2.43) and use the convolution theorem for Fourier transforms to obtain,

(2.45) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}(\boldsymbol{r},\unicode[STIX]{x1D714})=\int \text{d}^{3}\boldsymbol{s}\unicode[STIX]{x1D712}(\boldsymbol{s},\unicode[STIX]{x1D714})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r}-\boldsymbol{s},\unicode[STIX]{x1D714})\Rightarrow \unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}(\boldsymbol{k},\unicode[STIX]{x1D714})=\unicode[STIX]{x1D712}(\boldsymbol{k},\unicode[STIX]{x1D714})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{k},\unicode[STIX]{x1D714}),\end{eqnarray}$$

where we have also Fourier transformed in space: $g(k)=\int \text{d}^{3}\boldsymbol{r}g(\boldsymbol{r})\exp (-\text{i}\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{r})$ .

Theorem. Inverse Fourier transforms

(2.46) $$\begin{eqnarray}g(\boldsymbol{r})=\frac{1}{(2\unicode[STIX]{x03C0})^{3}}\int \text{d}^{3}\boldsymbol{k}\exp (\text{i}\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{r})g(\boldsymbol{k})\quad g(t)=\frac{1}{(2\unicode[STIX]{x03C0})}\int \text{d}\unicode[STIX]{x1D714}\exp (-\text{i}\unicode[STIX]{x1D714}t)g(\unicode[STIX]{x1D714}).\end{eqnarray}$$

The Fourier transformed Poisson equation is

(2.47) $$\begin{eqnarray}\displaystyle -\unicode[STIX]{x1D735}^{2}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r},t)=4\unicode[STIX]{x03C0}\left[\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}(\boldsymbol{r},t)+\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}^{\text{ext}}(\boldsymbol{r},t)\right]\Rightarrow k^{2}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{k},\unicode[STIX]{x1D714})=4\unicode[STIX]{x03C0}\left[\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}(\boldsymbol{k},\unicode[STIX]{x1D714})+\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}^{\text{ext}}(\boldsymbol{k},\unicode[STIX]{x1D714})\right] & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

and we then use (2.45) to obtain an equation for $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\unicode[STIX]{x1D714},\boldsymbol{k})$ ,

(2.48) $$\begin{eqnarray}k^{2}\left[1-\frac{4\unicode[STIX]{x03C0}}{k^{2}}\unicode[STIX]{x1D712}(\boldsymbol{k},\unicode[STIX]{x1D714})\right]\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{k},\unicode[STIX]{x1D714})=4\unicode[STIX]{x03C0}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}^{\text{ext}}(\boldsymbol{k},\unicode[STIX]{x1D714})=k^{2}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{k},\unicode[STIX]{x1D714}),\end{eqnarray}$$

where we introduce the relation $4\unicode[STIX]{x03C0}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}^{\text{ext}}(\boldsymbol{k},\unicode[STIX]{x1D714})=k^{2}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{k},\unicode[STIX]{x1D714})$ to define $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\unicode[STIX]{x1D714},\boldsymbol{k})$ .

Recall the conventional formulation of Gauss’ law in a dielectric medium with free (external) charges present,

(2.49) $$\begin{eqnarray}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D63F}=4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}^{\text{ext}}\quad \unicode[STIX]{x1D63F}=\boldsymbol{E}+4\unicode[STIX]{x03C0}\unicode[STIX]{x1D64B}=\unicode[STIX]{x1D700}\boldsymbol{E}\quad \Rightarrow -\unicode[STIX]{x1D735}\boldsymbol{\cdot }\left(\unicode[STIX]{x1D700}\unicode[STIX]{x1D735}\unicode[STIX]{x1D719}\right)=4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}^{\text{ext}}.\end{eqnarray}$$

In a uniform plasma $\unicode[STIX]{x1D700}$ is spatially uniform and (2.49) yields $-\unicode[STIX]{x1D700}\unicode[STIX]{x1D735}^{2}\unicode[STIX]{x1D719}=4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}^{\text{ext}}$ which combined with (2.48) leads to the following result.

Theorem. Poisson’s equation and plasma dielectric response

(2.50) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D700}k^{2}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{k},\unicode[STIX]{x1D714})=4\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}^{\text{ext}}(\boldsymbol{k},\unicode[STIX]{x1D714})\rightarrow \left[1-\frac{4\unicode[STIX]{x03C0}}{k^{2}}\unicode[STIX]{x1D712}(\boldsymbol{k},\unicode[STIX]{x1D714})\right]\equiv \unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})\rightarrow \unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{k},\unicode[STIX]{x1D714})=\frac{\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{k},\unicode[STIX]{x1D714})}{\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})}.\hspace{-10.00002pt} & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Corollary.

(2.51) $$\begin{eqnarray}\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})=\frac{k^{2}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}}{k^{2}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}}=\frac{\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}^{\text{ext}}}{\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}^{\text{tot}}}=1-\frac{\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}(\boldsymbol{k},\unicode[STIX]{x1D714})}{\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}^{\text{tot}}(\boldsymbol{k},\unicode[STIX]{x1D714})},\quad \unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}^{\text{tot}}=\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}+\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}^{\text{ext}}.\end{eqnarray}$$

Fourier transform the linearized Vlasov equation,

(2.52) $$\begin{eqnarray}\displaystyle \displaystyle \left(\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}+\boldsymbol{v}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\boldsymbol{r}}\right)\unicode[STIX]{x1D6FF}f(\boldsymbol{r},\boldsymbol{v};t) & = & \displaystyle \frac{e}{m}(\unicode[STIX]{x1D735}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719})\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}f_{0}}{\unicode[STIX]{x2202}\boldsymbol{v}}\rightarrow \left(-\text{i}\unicode[STIX]{x1D714}+\text{i}\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{v}\right)\unicode[STIX]{x1D6FF}f(\boldsymbol{k},\unicode[STIX]{x1D714};\boldsymbol{v})\nonumber\\ \displaystyle & = & \displaystyle \frac{e}{m}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{k},\unicode[STIX]{x1D714})\text{i}\boldsymbol{k}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}f_{0}}{\unicode[STIX]{x2202}\boldsymbol{v}}\nonumber\\ \displaystyle \displaystyle \unicode[STIX]{x1D6FF}f(\boldsymbol{k},\unicode[STIX]{x1D714};\boldsymbol{v}) & = & \displaystyle \frac{\displaystyle \boldsymbol{k}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}f_{0}}{\unicode[STIX]{x2202}\boldsymbol{v}}}{(\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{v}-\unicode[STIX]{x1D714})}\frac{e}{m}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{k},\unicode[STIX]{x1D714}).\end{eqnarray}$$

Theorem. Linear susceptibility

(2.53) $$\begin{eqnarray}\unicode[STIX]{x1D712}(\boldsymbol{k},\unicode[STIX]{x1D714})=\frac{\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}(\boldsymbol{k},\unicode[STIX]{x1D714})}{\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{k},\unicode[STIX]{x1D714})}=\mathop{\sum }_{s}\frac{e_{s}^{2}}{m_{s}}\int \text{d}^{3}\boldsymbol{v}\frac{\displaystyle \boldsymbol{k}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}f_{0}^{s}(\boldsymbol{v})}{\unicode[STIX]{x2202}\boldsymbol{v}}}{\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{v}-\unicode[STIX]{x1D714}}.\end{eqnarray}$$

Definitions. $n_{0}^{s}=\int \text{d}^{3}\boldsymbol{v}f_{0}^{s}(\boldsymbol{v}),g^{s}(\boldsymbol{v})\equiv f_{0}^{s}(\boldsymbol{v})/n_{0}^{s},\int \text{d}^{3}\boldsymbol{v}g^{s}(\boldsymbol{v})=1$ .

Theorem. From (2.50)–(2.53) we obtain the linear dielectric function

(2.54) $$\begin{eqnarray}\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})=1-\frac{4\unicode[STIX]{x03C0}}{k^{2}}\unicode[STIX]{x1D712}(\boldsymbol{k},\unicode[STIX]{x1D714})=1-\mathop{\sum }_{s}\frac{\unicode[STIX]{x1D714}_{s}^{2}}{k^{2}}\int \text{d}^{3}\boldsymbol{v}\frac{\displaystyle \boldsymbol{k}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}g^{s}}{\unicode[STIX]{x2202}\boldsymbol{v}}}{\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{v}-\unicode[STIX]{x1D714}},\quad \operatorname{Im}\unicode[STIX]{x1D714}>0,\end{eqnarray}$$

where $\unicode[STIX]{x1D714}_{s}^{2}\equiv (4\unicode[STIX]{x03C0}n_{0}^{s}e_{s}^{2})/m_{s}$ .

From (2.50) one solves for $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{k},\unicode[STIX]{x1D714})=\unicode[STIX]{x1D700}^{-1}(\boldsymbol{k},\unicode[STIX]{x1D714})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{k},\unicode[STIX]{x1D714})$ from which we obtain the following using the convolution theorem:

(2.55) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{k},t)=\int _{-\infty }^{\infty }\text{d}\unicode[STIX]{x1D70F}\unicode[STIX]{x1D700}^{-1}(\boldsymbol{k},\unicode[STIX]{x1D70F})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{k},t-\unicode[STIX]{x1D70F})\text{ where }\unicode[STIX]{x1D700}^{-1}(\boldsymbol{k},\unicode[STIX]{x1D70F})=\int \text{d}\unicode[STIX]{x1D714}\exp (-\text{i}\unicode[STIX]{x1D714}\unicode[STIX]{x1D70F})\unicode[STIX]{x1D700}^{-1}(\boldsymbol{k},\unicode[STIX]{x1D714}).\end{eqnarray}$$

2.5.3 Stable and unstable waves/disturbances

Suppose $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{k},t)=a(\boldsymbol{k})\unicode[STIX]{x1D6FF}(t)$ for a pulse-type forcing function without specifying $a(\boldsymbol{k})$ yet. Hence, $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{k},t)=a(\boldsymbol{k})\unicode[STIX]{x1D700}^{-1}(\boldsymbol{k},t)$ . Consider a specific representation for $a(\boldsymbol{k})$ so that the external forcing function is monochromatic: $a(\boldsymbol{k})=a\unicode[STIX]{x1D6FF}(\boldsymbol{k}-\boldsymbol{k}_{0})(2\unicode[STIX]{x03C0})^{3}\text{ and }\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{r},t)=a\text{e}^{\text{i}\boldsymbol{k}_{0}\boldsymbol{\cdot }\boldsymbol{r}}\unicode[STIX]{x1D6FF}(t)$ .

Theorem. The pulse-response solution for the perturbed electric potential using (2.55) is then

(2.56) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r},t)=\frac{1}{(2\unicode[STIX]{x03C0})^{3}}\int \text{d}^{3}\boldsymbol{k}\text{e}^{\text{i}\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{r}}a(\boldsymbol{k})\unicode[STIX]{x1D700}^{-1}(\boldsymbol{k},t)=a\text{e}^{\text{i}\boldsymbol{k}_{0}\boldsymbol{\cdot }\boldsymbol{ r}}\unicode[STIX]{x1D700}^{-1}(\boldsymbol{k}_{0},t).\end{eqnarray}$$

The stability or instability of the pulse response is determined by $\unicode[STIX]{x1D700}^{-1}(\boldsymbol{k},t)$ . The contour integration in (2.55) is illustrated in figure 3. We make use of analytic continuation and depress the integration contour: $\unicode[STIX]{x1D714}=\unicode[STIX]{x1D714}^{\prime }+\text{i}b$ to remove the line integration and only leave the poles. In so doing, we hope that there are no vertical branch cuts. The integration of the deformed contour integration becomes

(2.57) $$\begin{eqnarray}\unicode[STIX]{x1D700}^{-1}(\boldsymbol{k},t)=\int _{-\infty }^{\infty }\frac{\text{d}\unicode[STIX]{x1D714}^{\prime }}{2\unicode[STIX]{x03C0}}\frac{\exp (-\text{i}\unicode[STIX]{x1D714}^{\prime }t-bt)}{\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714}^{\prime }-\text{i}b)}-\frac{2\unicode[STIX]{x03C0}\text{i}}{2\unicode[STIX]{x03C0}}\mathop{\sum }_{\ell }\frac{\exp (-\text{i}\unicode[STIX]{x1D714}_{\ell }(\boldsymbol{k})t)}{\left.\displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\right|_{\unicode[STIX]{x1D714}_{\ell }}}.\end{eqnarray}$$

Consider a perturbative solution of the zeros of the linear dielectric function, $\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})\approx \unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714}_{\ell })+(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{\ell })\left.\unicode[STIX]{x2202}\unicode[STIX]{x1D700}/\unicode[STIX]{x2202}\unicode[STIX]{x1D714}\right|_{\unicode[STIX]{x1D714}_{\ell }}$ where $\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714}_{\ell })=0$ defines the pole at $\unicode[STIX]{x1D714}_{\ell }$ . Define the complex frequency at the pole as $\unicode[STIX]{x1D714}_{\ell }=\unicode[STIX]{x1D6FA}_{\ell }+\text{i}\unicode[STIX]{x1D6FE}_{\ell }$ so that $\exp (-\text{i}\unicode[STIX]{x1D714}_{\ell }(\boldsymbol{k})t)=\exp (-\text{i}\unicode[STIX]{x1D6FA}_{\ell }(\boldsymbol{k})t+\unicode[STIX]{x1D6FE}_{\ell }t)$ .

Figure 3. Contour integration for pulse response showing the depressed contour and poles of $\unicode[STIX]{x1D700}^{-1}$ in the region of analytic continuation.

Theorem. A pole in the upper half-plane corresponds to instability $\unicode[STIX]{x1D6FE}_{\ell }>0$ , and a pole in the lower half-plane is a stable root $\unicode[STIX]{x1D6FE}_{\ell }<0$ .

For $b>\unicode[STIX]{x1D6FE}$ the first term on the right-hand side of (2.57) damps out after a suitable length of time, which leads to

(2.58) $$\begin{eqnarray}\unicode[STIX]{x1D700}^{-1}(\boldsymbol{k},t)\Rightarrow -\text{i}\mathop{\sum }_{\ell }\frac{\exp \left(-\text{i}\unicode[STIX]{x1D714}_{\ell }(\boldsymbol{k})t\right)}{\left.\displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\right|_{\unicode[STIX]{x1D714}_{\ell }}}.\end{eqnarray}$$

Definition. $g(\boldsymbol{v})\equiv \sum _{s}\unicode[STIX]{x1D714}_{s}^{2}g^{s}(\boldsymbol{v})/\sum _{s}\unicode[STIX]{x1D714}_{s}^{2}$ and $\unicode[STIX]{x1D714}_{p}^{2}\equiv \sum _{s}\unicode[STIX]{x1D714}_{s}^{2}$ .

The linear dielectric function becomes

(2.59) $$\begin{eqnarray}\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})=1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k^{2}}\int \text{d}^{3}\boldsymbol{v}\frac{\boldsymbol{k}\boldsymbol{\cdot }\displaystyle \frac{\unicode[STIX]{x2202}g}{\unicode[STIX]{x2202}\boldsymbol{v}}}{\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{v}-\unicode[STIX]{x1D714}}=1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k^{2}}\int \text{d}^{3}\boldsymbol{v}\frac{\hat{\boldsymbol{k}}\boldsymbol{\cdot }\displaystyle \frac{\unicode[STIX]{x2202}g}{\unicode[STIX]{x2202}\boldsymbol{v}}}{\hat{\boldsymbol{k}}\boldsymbol{\cdot }\boldsymbol{v}-\unicode[STIX]{x1D714}/k}.\end{eqnarray}$$

We note the integral on the right-hand side of (2.59) can be simplified using the definition $u\equiv \hat{\boldsymbol{k}}\boldsymbol{\cdot }\boldsymbol{v}$ so that

$$\begin{eqnarray}\hat{\boldsymbol{k}}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}g}{\unicode[STIX]{x2202}\boldsymbol{v}}=\frac{\unicode[STIX]{x2202}g(u,\boldsymbol{v}_{2},\boldsymbol{v}_{3})}{\unicode[STIX]{x2202}u}.\end{eqnarray}$$

This allows the velocity-space integral over two of the three dimensions in (2.59) to be done immediately,

$$\begin{eqnarray}g(u)=\int \text{d}v_{2}\text{d}v_{3}g(u,v_{2},v_{3})\quad \text{where }g(u)\equiv \int \text{d}^{3}\boldsymbol{v}g(\boldsymbol{v})\unicode[STIX]{x1D6FF}(\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{v}-\unicode[STIX]{x1D714})\text{ and }\int _{-\infty }^{\infty }\text{d}u\,g(u)=1.\end{eqnarray}$$

Equation (2.54) then becomes

(2.60) $$\begin{eqnarray}\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})=1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k^{2}}\int _{-\infty }^{\infty }\text{d}u\frac{g^{\prime }(u)}{u-v_{p}},\quad v_{p}=\frac{\unicode[STIX]{x1D714}}{k},\end{eqnarray}$$

where $v_{p}=\unicode[STIX]{x1D714}/k$ is the phase velocity; and in the following few expressions we drop the ‘ $p$ ’ subscript.

Theorem. The eigenvalues of the linearized Vlasov–Poisson system are determined by the roots of the linear dielectric function $\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714}_{\ell })=0\rightarrow \unicode[STIX]{x1D714}_{\ell }(\boldsymbol{k})$ , which are the eigenfrequencies. This follows from $\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})\unicode[STIX]{x1D719}(\boldsymbol{k},\unicode[STIX]{x1D714})=\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{k},\unicode[STIX]{x1D714})=0$ and corresponds to free, undriven oscillations.

Definition. The Hilbert transform of $g$ is

(2.61) $$\begin{eqnarray}\int _{-\infty }^{\infty }\text{d}u\frac{g_{\hat{\boldsymbol{k}}}(u)}{u-v}\equiv \text{Hilbert transform of }g\equiv Z_{\hat{\boldsymbol{k}}}(v)\quad (\operatorname{Im}v>0,k\equiv |\boldsymbol{k}|>0).\end{eqnarray}$$

One can differentiate the Hilbert transform with respect to $v$ to obtain,

(2.62) $$\begin{eqnarray}\frac{\text{d}}{\text{d}v}Z_{\hat{\boldsymbol{k}}}=\int _{-\infty }^{\infty }\text{d}ug_{\hat{\boldsymbol{ k}}}(u)\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}v}\frac{1}{u-v}\equiv \int _{-\infty }^{\infty }\text{d}u\,\frac{g_{\hat{\boldsymbol{k}}}^{\prime }(u)}{u-v},\quad g_{\hat{\boldsymbol{k}}}^{\prime }\equiv \frac{\text{d}}{\text{d}u}g_{\hat{\boldsymbol{k}}}(u)\end{eqnarray}$$

using $(\unicode[STIX]{x2202}/\unicode[STIX]{x2202}v)(u-v)^{-1}=-(\unicode[STIX]{x2202}/\unicode[STIX]{x2202}u)(u-v)^{-1}$ and integrating by parts. Equation (2.60) then yields the following.

Theorem. The dielectric function is

(2.63) $$\begin{eqnarray}\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})=1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k^{2}}Z_{\hat{\boldsymbol{k}}}^{\prime }(v).\end{eqnarray}$$

2.5.4 Examples of linear dielectrics for a few simple velocity distributions

Table 1 presents examples of five model velocity distributions and the corresponding dielectric responses. We note that the Cauchy velocity distribution does not exist in nature because its energy moment diverges.

Table 1. Examples of velocity distributions and resulting dielectric responses.

2.5.5 Inverse Fourier transform to obtain spatial and temporal response – Green’s function for pulse response and cold-plasma example

From (2.55) we have an integral relation for $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{k},t)$ , i.e.

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{k},t)=\int _{-\infty }^{\infty }\text{d}\unicode[STIX]{x1D70F}\unicode[STIX]{x1D700}^{-1}(\boldsymbol{k},\unicode[STIX]{x1D70F})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{k},t-\unicode[STIX]{x1D70F}).\end{eqnarray}$$

In configuration space, equation (2.55) becomes

(2.64) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r},t) & = & \displaystyle \int \text{d}^{3}s\int _{-\infty }^{\infty }\text{d}\unicode[STIX]{x1D70F}\unicode[STIX]{x1D700}^{-1}(\boldsymbol{s},\unicode[STIX]{x1D70F})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{r}-\boldsymbol{s},t-\unicode[STIX]{x1D70F}),\nonumber\\ \displaystyle & & \displaystyle \text{where }\unicode[STIX]{x1D700}^{-1}(\boldsymbol{s},\unicode[STIX]{x1D70F})=\int \frac{\text{d}^{3}\boldsymbol{k}}{(2\unicode[STIX]{x03C0})^{3}}\int \frac{\text{d}\unicode[STIX]{x1D714}}{2\unicode[STIX]{x03C0}}\unicode[STIX]{x1D700}^{-1}(\boldsymbol{k},\unicode[STIX]{x1D714})\exp (\text{i}\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{s}-\text{i}\unicode[STIX]{x1D714}\unicode[STIX]{x1D70F}).\qquad\end{eqnarray}$$

Exercise. Prove $\unicode[STIX]{x1D700}^{-1}(\boldsymbol{k},\unicode[STIX]{x1D70F})=0$ for $\unicode[STIX]{x1D70F}<0$ . See § 2.5.3.

Exercise. Simplify $\unicode[STIX]{x1D700}(k,\unicode[STIX]{x1D714})$ for hot electrons and cold ions in the limit that $\unicode[STIX]{x1D714}/k=V_{p}\ll v_{\text{th}}$ where $v_{\text{th}}^{2}=T_{e}/m_{e}$ is the square of the electron thermal speed, $T_{e}$ is the electron temperature and $m_{e}$ is the electron mass. Introduce the definition of the Debye length $\unicode[STIX]{x1D706}_{e}\equiv v_{\text{th}}/\unicode[STIX]{x1D714}_{e}$ . Derive the dispersion relation for the ion-acoustic wave,

(2.65) $$\begin{eqnarray}\unicode[STIX]{x1D714}^{2}=\frac{\unicode[STIX]{x1D714}_{i}^{2}}{1+{\displaystyle \frac{1}{k^{2}\unicode[STIX]{x1D706}_{e}^{2}}}}={\displaystyle \frac{k^{2}c_{s}^{2}}{1+k^{2}\unicode[STIX]{x1D706}_{e}^{2}}},\quad c_{s}^{2}\equiv \frac{m_{e}}{m_{i}}v_{e}^{2}=\frac{k_{B}T_{e}}{m_{i}},\end{eqnarray}$$

where $c_{s}$ is the ion sound speed for cold ions. The ions provide the inertia while the electrons provide pressure, and the electric field binds the motion of the two species.

We return to (2.64) to calculate the Green’s function for the pulse response of the electric potential in a cold plasma with dielectric function $\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})=1-\unicode[STIX]{x1D714}_{p}^{2}/\unicode[STIX]{x1D714}^{2}$ . If there is no $\boldsymbol{k}$ dependence in $\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})$ then

(2.66) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D700}^{-1}(\boldsymbol{s},\unicode[STIX]{x1D70F}) & = & \displaystyle \unicode[STIX]{x1D6FF}(\boldsymbol{s})\int \frac{\text{d}\unicode[STIX]{x1D714}}{2\unicode[STIX]{x03C0}}\exp (-\text{i}\unicode[STIX]{x1D714}\unicode[STIX]{x1D70F})\left(1+\frac{\unicode[STIX]{x1D714}_{p}^{2}}{(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{p})(\unicode[STIX]{x1D714}+\unicode[STIX]{x1D714}_{p})}\right)\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6FF}(\boldsymbol{s})\left[\unicode[STIX]{x1D6FF}(\unicode[STIX]{x1D70F})+\unicode[STIX]{x1D714}_{p}^{2}\times \left\{0\,\unicode[STIX]{x1D70F}<0,-\frac{1}{\unicode[STIX]{x1D714}_{p}}\sin \unicode[STIX]{x1D714}_{p}\unicode[STIX]{x1D70F}\,\unicode[STIX]{x1D70F}>0\right\}\right].\end{eqnarray}$$

The integral over $\unicode[STIX]{x1D714}$ has poles at $\pm \unicode[STIX]{x1D714}_{p}$ . The term involving $\unicode[STIX]{x1D6FF}(\unicode[STIX]{x1D70F})$ is the vacuum response, and the rest is associated with the plasma shielded response.

Theorem. Using (2.66), (2.64) becomes

(2.67) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r},t)=\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{r},t)-\int _{0}^{\infty }\text{d}\unicode[STIX]{x1D70F}\unicode[STIX]{x1D714}_{p}\sin (\unicode[STIX]{x1D714}_{p}\unicode[STIX]{x1D70F})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{r},t-\unicode[STIX]{x1D70F}).\end{eqnarray}$$

The excitation in $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r},t)$ depends on all the other excitations and the external potential at $\boldsymbol{r}$ from earlier times. This is partly due to the absence of damping in the cold plasma.

Example. The impulse response for $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\unicode[STIX]{x1D700}\text{ext}}=\unicode[STIX]{x1D6FF}(\boldsymbol{r})\unicode[STIX]{x1D6FF}(t)$ is $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r},t)=-\unicode[STIX]{x1D714}_{p}\unicode[STIX]{x1D6FF}(\boldsymbol{r})\sin (\unicode[STIX]{x1D714}_{p}t)$ for $t>0$ . For this response $\unicode[STIX]{x1D714}(\boldsymbol{k})=\unicode[STIX]{x1D714}_{p}$ and $\text{d}\unicode[STIX]{x1D714}/\text{d}\boldsymbol{k}=0$ . The plasma response to the $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\unicode[STIX]{x1D700}\text{ext}}$ impulse is negative over the first half-cycle of the plasma oscillation as the plasma tries to neutralize $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\unicode[STIX]{x1D700}\text{ext}}$ (figure 4).

Figure 4. Impulse response $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(t)$ for $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\unicode[STIX]{x1D700}\text{ext}}=\unicode[STIX]{x1D6FF}(\boldsymbol{r})\unicode[STIX]{x1D6FF}(t)$ .

Example. $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r},t)$ due to a moving test particle. Consider the external potential for a moving charge: $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{r},t)=e_{0}/(\boldsymbol{r}-\boldsymbol{r}_{0}(t))$ with $\boldsymbol{r}_{0}(t)=\boldsymbol{v}_{0}t$ and charge $e_{0}>0$ . From Poisson’s equation one obtains the total charge density,

(2.68) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}(\boldsymbol{r},t) & = & \displaystyle -\left(\frac{1}{4\unicode[STIX]{x03C0}}\right)\unicode[STIX]{x1D6FB}^{2}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r},t)\nonumber\\ \displaystyle & = & \displaystyle -\left(\frac{1}{4\unicode[STIX]{x03C0}}\right)\unicode[STIX]{x1D6FB}^{2}\left[-\unicode[STIX]{x1D714}_{p}\int _{0}^{\infty }\text{d}\unicode[STIX]{x1D70F}\sin (\unicode[STIX]{x1D714}_{p}\unicode[STIX]{x1D70F})\frac{e_{0}}{\boldsymbol{r}-\boldsymbol{r}_{0}(t-\unicode[STIX]{x1D70F})}+\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{r},t)\right]\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6FF}(x)\unicode[STIX]{x1D6FF}(y)\left(-\frac{e_{0}\unicode[STIX]{x1D714}_{p}}{v_{0}}\right)\sin \!\left(-\frac{\unicode[STIX]{x1D714}_{p}z}{v_{0}}\right)+e_{0}\unicode[STIX]{x1D6FF}(x)\unicode[STIX]{x1D6FF}(y)\unicode[STIX]{x1D6FF}(z-v_{0}t).\end{eqnarray}$$

A schematic of the one-dimensional charge density and the local charge under the curve in the wake of the moving test particle is presented in figure 5. The total charge exclusive of the test particle is $-e_{0}$ as can be shown from the integral of $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D70C}$ .

(2.69) $$\begin{eqnarray}\lim _{\unicode[STIX]{x1D706}\rightarrow 0}\int _{-\infty }^{0}\text{d}z^{\prime }\left(-\frac{e_{0}\unicode[STIX]{x1D714}_{p}}{v_{0}}\right)\sin \left(\frac{\unicode[STIX]{x1D714}_{p}z^{\prime }}{v_{0}}\right)\exp (-\unicode[STIX]{x1D706}z^{\prime })=-e_{0}.\end{eqnarray}$$

Thus, the total charge is zero. We can also calculate the dipole moment $\unicode[STIX]{x1D64B}$ of the plasma response similarly by weighting the integrand in (2.69) with $z$ to show that $\unicode[STIX]{x1D64B}=0$ .

Figure 5. One-dimensional charge density in the wake of a moving test particle with charge $e_{0}$ .

Example. For the more general case where $\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})$ has $\boldsymbol{k}$ dependence, we use Cauchy’s theorem and (2.57), (2.58) and (2.64) to obtain

(2.70) $$\begin{eqnarray}\unicode[STIX]{x1D700}^{-1}(\boldsymbol{s},\unicode[STIX]{x1D70F})=-\text{i}\int \frac{\text{d}^{3}\boldsymbol{k}}{(2\unicode[STIX]{x03C0})^{3}}\mathop{\sum }_{\ell }\frac{\exp (\text{i}\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{s}-\text{i}\unicode[STIX]{x1D714}_{\ell }(\boldsymbol{k})\unicode[STIX]{x1D70F})}{\displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}|_{\unicode[STIX]{x1D714}_{\ell }(\boldsymbol{k})}},\end{eqnarray}$$

where the summation in (2.70) is over the poles in the complex $\unicode[STIX]{x1D714}$ plane. However, for warm plasmas it is difficult to obtain explicit formulae because we cannot do the integrals in most cases of physical interest.

2.6 Streaming instabilities

Consider an infinite uniform plasma with two or more beams:

$$\begin{eqnarray}f_{s}^{0}(\boldsymbol{v})=n_{s}\unicode[STIX]{x1D6FF}(\boldsymbol{v}-\boldsymbol{u}_{s})\quad \text{and}\quad \unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})=1-\mathop{\sum }_{s}\frac{\unicode[STIX]{x1D714}_{s}^{2}}{(\unicode[STIX]{x1D714}-\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{u}_{s})^{2}},\quad \text{with }\unicode[STIX]{x1D714}_{s}^{2}\equiv \frac{4\unicode[STIX]{x03C0}n_{s}e_{s}^{2}}{m_{s}}.\end{eqnarray}$$

Restrict this linear analysis to one spatial dimension: $k\hat{\boldsymbol{k}}\boldsymbol{\cdot }\boldsymbol{u}_{s}=ku_{s}$ . Then the linear dielectric can be rewritten as

(2.71) $$\begin{eqnarray}\unicode[STIX]{x1D700}(k,\unicode[STIX]{x1D714})=1-\mathop{\sum }_{s}\frac{\unicode[STIX]{x1D714}_{s}^{2}}{(\unicode[STIX]{x1D714}-ku_{s})^{2}}=1-\frac{1}{k^{2}}\mathop{\sum }_{s}\frac{\unicode[STIX]{x1D714}_{s}^{2}}{(V-u_{s})^{2}},\quad V\equiv \unicode[STIX]{x1D714}/k.\end{eqnarray}$$

The solution of $\unicode[STIX]{x1D700}(k,\unicode[STIX]{x1D714})=0$ for the normal modes $\unicode[STIX]{x1D714}_{k}$ is then determined by solving the following order $2N$ polynomial equation for $V=\unicode[STIX]{x1D714}/k$ :

(2.72) $$\begin{eqnarray}k^{2}=F(V)=\mathop{\sum }_{s=1}^{N}\frac{\unicode[STIX]{x1D714}_{s}^{2}}{(V-u_{s})^{2}}.\end{eqnarray}$$

Example. Consider the graphical solution of (2.72) for $N=3$ .

Figure 6. Schematic of solution of (2.72 $k^{2}=F(V)$ for $N=3$ .

There are resonances at $V=u_{i},i=1$ , 2 and 3 and three branches of the dispersion relation with $Re\,\unicode[STIX]{x1D714}\sim \mathit{ku}_{i}\pm \unicode[STIX]{x1D714}_{s}$ . For very large values of $k^{2}$ , e.g. $k_{1}^{2}$ , there are $2N=6$ intersections of $k^{2}$ with $F(V)$ , i.e. 6 stable roots. If for smaller values of $k^{2}$ , e.g. $k_{2}^{2}$ , the $k_{2}^{2}$ horizontal line falls below the locus of $F(V)$ , such that there are fewer than $2N$ intersections; and then there are one or more pairs of complex-conjugate roots. The complex root(s) with $\operatorname{Im}\unicode[STIX]{x1D714}>0$ is the unstable root, and the conjugate root with $\operatorname{Im}\unicode[STIX]{x1D714}<0$ is a damped normal mode. In a finite plasma, boundary conditions must be defined; and there is a lower limit on $k$ corresponding to $2\unicode[STIX]{x03C0}/L$ , where $L$ is the length of the plasma.

2.6.1 Examples – two-stream and weak-beam instabilities

Example. $N=2$ , two-stream instability. Suppose $\unicode[STIX]{x1D714}_{1}=\unicode[STIX]{x1D714}_{2}$ and select a reference frame with $u_{1}=-u_{2}$ . The infinite-medium normal modes are determined by the solutions of the dispersion relation,

(2.73) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D700}(k,\unicode[STIX]{x1D714}_{\ell }) & = & \displaystyle 1-\frac{\unicode[STIX]{x1D714}_{1}^{2}}{(\unicode[STIX]{x1D714}-ku_{1})^{2}}-\frac{\unicode[STIX]{x1D714}_{1}^{2}}{(\unicode[STIX]{x1D714}+ku_{1})^{2}}=0\nonumber\\ \displaystyle & \rightarrow & \displaystyle \frac{\unicode[STIX]{x1D714}_{\ell }^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}=\frac{1}{2}\left[1+2\frac{k^{2}u_{1}^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}\pm \sqrt{1+\frac{8k^{2}u_{1}^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}}\right]\quad \text{with }\unicode[STIX]{x1D714}_{p}^{2}=\unicode[STIX]{x1D714}_{1}^{2}+\unicode[STIX]{x1D714}_{2}^{2}=2\unicode[STIX]{x1D714}_{1}^{2}.\qquad\end{eqnarray}$$

Exercise. Sketch $\unicode[STIX]{x1D714}^{2}$ versus $k^{2}$ , $Re\unicode[STIX]{x1D714}$ versus $k$ and $\operatorname{Im}\unicode[STIX]{x1D714}$ versus $k$ . Show that for $k<k_{c}$ there is instability; what is $k_{c}$ ? Show that $\max (\operatorname{Im}\unicode[STIX]{x1D714})=\unicode[STIX]{x1D714}_{p}/\surd 8$ for $k_{\max }=(3/8)^{\text{1/2}}\unicode[STIX]{x1D714}_{p}/u_{1}$ . Evaluate $\unicode[STIX]{x1D700}^{\text{-1}}(x,t)$ approximately and sketch. Show that $\unicode[STIX]{x1D700}^{-1}(x,t)=0$ for $|x/t|>u_{1}$ using analyticity.

Figure 7. (a) Schematic of the solution of $k^{2}=F(V)$ for $N=2$ and $\unicode[STIX]{x1D714}_{2}=\unicode[STIX]{x1D714}_{b}\ll \unicode[STIX]{x1D714}_{1}=\unicode[STIX]{x1D714}_{p}$ , weak-beam instability. (b) Schematic showing the crossing of the plasma frequency and beam branches.

Example. $N=2$ , weak-beam instability. Assume $\unicode[STIX]{x1D714}_{b}\ll \unicode[STIX]{x1D714}_{1}=\unicode[STIX]{x1D714}_{p}$ where the ‘first’ component is the ‘plasma.’ Select the frame in which the plasma component is at rest. The weak-beam instability is diagrammed in figure 7. The dispersion relation for the normal modes is given by the solution of

(2.74) $$\begin{eqnarray}\unicode[STIX]{x1D700}(k,\unicode[STIX]{x1D714})=1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{\unicode[STIX]{x1D714}^{2}}-\frac{\unicode[STIX]{x1D714}_{b}^{2}}{(\unicode[STIX]{x1D714}-ku_{b})^{2}}=\unicode[STIX]{x1D700}_{p}(k,\unicode[STIX]{x1D714})-\frac{\unicode[STIX]{x1D714}_{b}^{2}}{(\unicode[STIX]{x1D714}-ku_{b})^{2}}=0,\end{eqnarray}$$

where the beam and plasma wave branches cross, one of the real roots can disappear, giving rise to a pair of complex-conjugate roots and instability. If the plasma is warm, the plasma branches acquire curvature. In general, one uses the warm-plasma dielectric for $\unicode[STIX]{x1D700}_{p}$ . For $\unicode[STIX]{x1D714}_{b}^{2}/\unicode[STIX]{x1D714}_{p}^{2}\ll 1$ we solve (2.74) perturbatively,

(2.75) $$\begin{eqnarray}\displaystyle 0 & = & \displaystyle \unicode[STIX]{x1D700}_{p}(k,\unicode[STIX]{x1D714})-\frac{\unicode[STIX]{x1D714}_{b}^{2}}{(\unicode[STIX]{x1D714}-ku_{b})^{2}}\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D700}_{p}(k,\unicode[STIX]{x1D714}_{0})+\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714}\left.\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D700}_{p}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\right|_{\unicode[STIX]{x1D714}_{0},k_{0}}+\unicode[STIX]{x1D6FF}k\left.\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D700}_{p}}{\unicode[STIX]{x2202}k}\right|_{\unicode[STIX]{x1D714}_{0},k_{0}}+\frac{1}{2}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714}^{2}\left.\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D700}_{p}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}^{2}}\right|_{\unicode[STIX]{x1D714}_{0},k_{0}}\nonumber\\ \displaystyle & & \displaystyle +\,\cdots -\frac{\unicode[STIX]{x1D714}_{b}^{2}}{(\unicode[STIX]{x1D714}_{0}+\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714}-k_{0}u_{b}-\unicode[STIX]{x1D6FF}ku_{b})^{2}},\nonumber\\ \displaystyle & & \displaystyle \unicode[STIX]{x1D700}_{p}(k,\unicode[STIX]{x1D714}_{0})=0,\quad \unicode[STIX]{x1D714}=\unicode[STIX]{x1D714}_{0}+\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714},\quad k=k_{0}+\unicode[STIX]{x1D6FF}k.\end{eqnarray}$$

At resonance $\unicode[STIX]{x1D714}_{0}-k_{0}u_{b}=0$ , and (2.75) becomes

(2.76) $$\begin{eqnarray}\displaystyle & & \displaystyle (\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714}-u\unicode[STIX]{x1D6FF}k)^{2}(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714}\unicode[STIX]{x1D700}_{\unicode[STIX]{x1D714}}+\unicode[STIX]{x1D6FF}k\unicode[STIX]{x1D700}_{k}+O(\unicode[STIX]{x1D6FF}^{2}))=\unicode[STIX]{x1D714}_{b}^{2},\quad \unicode[STIX]{x1D700}_{\unicode[STIX]{x1D714}}\equiv \left.\unicode[STIX]{x2202}\unicode[STIX]{x1D700}/\unicode[STIX]{x2202}\unicode[STIX]{x1D714}\right|_{\unicode[STIX]{x1D714}_{0},k_{0}}\nonumber\\ \displaystyle & & \displaystyle \qquad (\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714}-u\unicode[STIX]{x1D6FF}k)^{2}(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714}\unicode[STIX]{x1D700}_{\unicode[STIX]{x1D714}}+\unicode[STIX]{x1D6FF}k\unicode[STIX]{x1D700}_{k}+O(\unicode[STIX]{x1D6FF}^{2}))=\unicode[STIX]{x1D714}_{b}^{2},\quad \unicode[STIX]{x1D700}_{\unicode[STIX]{x1D714}}\equiv \left.\unicode[STIX]{x2202}\unicode[STIX]{x1D700}_{p}/\unicode[STIX]{x2202}\unicode[STIX]{x1D714}\right|_{\unicode[STIX]{x1D714}_{0},k_{0}}\!.\end{eqnarray}$$

(i) For $\unicode[STIX]{x1D6FF}k=0:\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714}^{3}+O(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714}^{4})=\unicode[STIX]{x1D700}_{\unicode[STIX]{x1D714}}^{-1}\unicode[STIX]{x1D714}_{b}^{2}$ . For small $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714}$ ,

(2.77) $$\begin{eqnarray}\displaystyle & \displaystyle \left(\frac{\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714}}{\unicode[STIX]{x1D714}_{0}}\right)^{3}=\frac{\unicode[STIX]{x1D714}_{b}^{2}}{\unicode[STIX]{x1D714}_{0}^{2}}\frac{1}{\unicode[STIX]{x1D714}_{0}\unicode[STIX]{x1D700}_{\unicode[STIX]{x1D714}}}\equiv \unicode[STIX]{x1D702}\approx \frac{1}{2}\frac{\unicode[STIX]{x1D714}_{b}^{2}}{\unicode[STIX]{x1D714}_{0}^{2}}\ll 1 & \displaystyle \nonumber\\ \displaystyle & \displaystyle \rightarrow \frac{\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714}}{\unicode[STIX]{x1D714}_{0}}=\unicode[STIX]{x1D702}^{1/3}=\left|\unicode[STIX]{x1D702}^{1/3}\right|\left\{1,-\frac{1}{2}\pm \text{i}\sqrt{\frac{3}{2}}\right\}. & \displaystyle\end{eqnarray}$$

We note for a cold plasma, $\unicode[STIX]{x1D700}_{p}=1-\unicode[STIX]{x1D714}_{p}^{2}/\unicode[STIX]{x1D714}^{2}$ and $\unicode[STIX]{x1D700}_{\unicode[STIX]{x1D714}}=2/\unicode[STIX]{x1D714}_{p}$ . There are three solutions for the frequency shift $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714}$ in (2.77): the coupling of the plasma wave to the weak beam produces a small frequency shift, a damped mode and a weak instability.

(ii) For $\unicode[STIX]{x1D6FF}k\neq 0$ , equation (2.76) is solved in the beam frame $\unicode[STIX]{x1D714}^{\prime }\equiv \unicode[STIX]{x1D714}-ku_{b}=\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D714}-\unicode[STIX]{x1D6FF}ku_{b}$ . If the plasma is cold, then $\unicode[STIX]{x1D714}^{\prime }$ is given by

(2.78) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D714}^{\prime } & = & \displaystyle \unicode[STIX]{x1D714}_{p}\left[\left|\unicode[STIX]{x1D702}^{1/3}\right|\exp \!\left(\text{i}\frac{2\unicode[STIX]{x03C0}}{3}\right)-\frac{1}{3}\frac{\unicode[STIX]{x1D6FF}k}{k_{0}}+\frac{1}{9}\left(\frac{\unicode[STIX]{x1D6FF}k}{k_{0}}\right)^{2}\left|\unicode[STIX]{x1D702}^{1/3}\right|\exp \left(-\text{i}\frac{2\unicode[STIX]{x03C0}}{3}\right)+O(\unicode[STIX]{x1D6FF}k^{3})\right]\nonumber\\ \displaystyle \operatorname{Re}\unicode[STIX]{x1D714}^{\prime } & = & \displaystyle \unicode[STIX]{x1D714}_{p}\left[-\frac{1}{2}\left|\unicode[STIX]{x1D702}^{1/3}\right|-\frac{1}{3}\frac{\unicode[STIX]{x1D6FF}k}{k_{0}}-\frac{1}{18}\left(\frac{\unicode[STIX]{x1D6FF}k}{k_{0}}\right)^{2}\left|\unicode[STIX]{x1D702}^{1/3}\right|+\cdots \right]\nonumber\\ \displaystyle \operatorname{Im}\unicode[STIX]{x1D714}^{\prime } & = & \displaystyle \unicode[STIX]{x1D714}_{p}\left[\frac{\sqrt{3}}{2}\left|\unicode[STIX]{x1D702}^{1/3}\right|+\frac{1}{9}\left(-\frac{\sqrt{3}}{2}\right)\left(\frac{\unicode[STIX]{x1D6FF}k}{k_{0}}\right)^{2}\left|\unicode[STIX]{x1D702}^{1/3}\right|+\cdots \right].\qquad\end{eqnarray}$$

In the beam frame there is a small negative shift of the phase velocity given by $\operatorname{Re}\unicode[STIX]{x1D714}^{\prime }/k_{0}=-(1/2)|\unicode[STIX]{x1D702}|^{1/3}u_{b}$ . There is also a shift in the group velocity of the plasma wave due to the coupling with the beam, which is given by $\text{d}\operatorname{Re}\unicode[STIX]{x1D714}^{\prime }/\text{d}k=\text{d}\operatorname{Re}\unicode[STIX]{x1D714}^{\prime }/\text{d}\unicode[STIX]{x1D6FF}k=-(1/3)u_{b}$ , which is also negative but is not small. We note that the dispersion in the real frequency of the instability is given by $\text{d}^{2}\operatorname{Re}\unicode[STIX]{x1D714}^{\prime }/\text{d}\unicode[STIX]{x1D6FF}k^{2}=-(1/9)|\unicode[STIX]{x1D702}|^{1/3}(\unicode[STIX]{x1D714}_{p}/k_{0}^{2})$ , which is small in $|\unicode[STIX]{x1D702}|^{1/3}$ . From the solution for $\operatorname{Im}\unicode[STIX]{x1D714}^{\prime }$ we see that the growth rate is peaked at a value of $(\surd 3/2)\unicode[STIX]{x1D702}^{1/3}\unicode[STIX]{x1D714}_{p}$ for $\unicode[STIX]{x1D6FF}k=0$ ; and

$$\begin{eqnarray}\unicode[STIX]{x1D6FE}_{kk}\equiv \left|\frac{\text{d}^{2}\unicode[STIX]{x1D6FE}}{\text{d}k^{2}}\right|=\frac{\unicode[STIX]{x1D6FE}_{0}}{k_{0}^{2}}\frac{2}{9}\left|\unicode[STIX]{x1D702}^{-2/3}\right|,\quad \unicode[STIX]{x1D6FE}_{0}=\frac{\sqrt{3}}{2}\unicode[STIX]{x1D714}_{p}\left|\unicode[STIX]{x1D702}^{1/3}\right|.\end{eqnarray}$$

From the square root of the ratio of the peak growth rate to $-\text{d}^{2}\operatorname{Im}\unicode[STIX]{x1D714}^{\prime }/\text{d}\unicode[STIX]{x1D6FF}k^{2}$ at $k_{0}$ we can estimate the half-width in $\unicode[STIX]{x1D6FF}k_{1/2}$ of the peak in the growth rate, which has a scaling $\unicode[STIX]{x1D6FF}k_{1/2}/k_{0}\sim |\unicode[STIX]{x1D702}^{1/3}|\ll 1$ . Thus, a very small range in $k$ -space is involved in the weak-beam instability; and a long wave packet can be formed that remains coherent over a long distance.

All of the results so far in § 2.6 were derived for a cold plasma. For a ‘hot’ plasma only numerical coefficients will change for the examples considered. The generalization of the results will involve formulae like the following:

(2.79) $$\begin{eqnarray}\operatorname{Re}\unicode[STIX]{x1D714}^{\prime }=k_{0}V_{p}^{\prime }+\unicode[STIX]{x1D6FF}kV_{g}^{\prime }+\frac{1}{2}\unicode[STIX]{x1D6FF}k^{2}\frac{\text{d}^{2}}{\text{d}\unicode[STIX]{x1D6FF}k^{2}}\unicode[STIX]{x1D714}^{\prime }\quad \unicode[STIX]{x1D6FE}\equiv \operatorname{Im}\unicode[STIX]{x1D714}^{\prime }=\unicode[STIX]{x1D6FE}_{0}+\frac{1}{2}\unicode[STIX]{x1D6FF}k^{2}\frac{\text{d}^{2}}{\text{d}\unicode[STIX]{x1D6FF}k^{2}}\unicode[STIX]{x1D6FE},\end{eqnarray}$$

where $V_{p}^{\prime }=\unicode[STIX]{x1D714}^{\prime }(k_{0})/k_{0}$ , $V_{g}^{\prime }=\text{d}\unicode[STIX]{x1D714}^{\prime }/\text{d}k_{}$ at $k_{0}$ and $\unicode[STIX]{x1D6FE}_{0}$ is the peak growth rate occurring at $k_{0}$ . Finite-temperature effects naturally lead to dispersion, i.e. the phase velocity $V_{p}$ is a function of wavenumber $k$ . Thus, in a hot plasma we expect that the $x-t$ response of a growing disturbance will exhibit a growing and spreading wave packet travelling at the group velocity $V_{g}$ . The $x-t$ response depends on the content of (2.79) and makes explicit how fast the growing wave packet spreads compared to its advection. The next sub-section elaborates the $x-t$ response of an unstable disturbance.

2.6.2 Definition of convective and absolute instability

How fast a growing wave packet spreads compared to how fast it advects past a fixed observation point is an important distinction.

Definition. Absolute versus convective instability. If an unstable wave packet advects faster than it spreads, an observer at a fixed observation point will see a growing signal as the front of the wave packet passes followed by peaking and then decay of the signal. As the packet advects the peak signal continues to grow exponentially in time. The foregoing corresponds to a convective instability. An absolute instability corresponds to when the spreading of the growing response exceeds the advection at the group velocity $V_{g}$ so that the signal at a fixed observation point continues to grow without cessation (until the linear assumption fails and nonlinear effects may come into the problem).

Exercise. Sketch a growing and advecting pulse at two distinct times in one spatial dimension for a convective instability, and make the corresponding sketch for an absolute instability.

The dielectric pulse response in one spatial dimension for an unstable root of the dispersion relation follows from (2.70). To obtain the results in (2.80) we make use of the Taylor-series expansion of the dispersion relation as given in (2.79) for small $\unicode[STIX]{x1D6FF}k$ . Because we perform the integral with respect to $\unicode[STIX]{x1D6FF}k$ by the method of steepest descents and take advantage of the rapid convergence of that integral with respect to large $\unicode[STIX]{x1D6FF}k$ due to the term $\exp (-{\textstyle \frac{1}{2}}\unicode[STIX]{x1D6FF}k^{2}|\unicode[STIX]{x1D6FE}_{kk}|t)$ , there is no conflict between the limits of the $\unicode[STIX]{x1D6FF}k$ integration being $(-\infty ,\infty )$ and Taylor-series expansion in small $\unicode[STIX]{x1D6FF}k$ . The pulse response is then

(2.80) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D700}^{-1}(x,t) & = & \displaystyle \int _{-\infty }^{\infty }\frac{\text{d}k}{2\unicode[STIX]{x03C0}}\int _{-\infty }^{\infty }\frac{\text{d}\unicode[STIX]{x1D714}}{2\unicode[STIX]{x03C0}}\frac{\exp (\text{i}kx-\text{i}\unicode[STIX]{x1D714}t)}{\unicode[STIX]{x1D700}}+\text{c.c.}=-\text{i}\int _{-\infty }^{\infty }\frac{\text{d}k}{2\unicode[STIX]{x03C0}}\frac{\exp (\text{i}kx-\text{i}\unicode[STIX]{x1D714}t)}{\displaystyle \left.\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D700}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\right|_{\unicode[STIX]{x1D714}_{k}(k)}}+\text{c.c.}\nonumber\\ \displaystyle & {\approx} & \displaystyle -\text{i}\frac{1}{\left.\displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D700}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\right|_{\unicode[STIX]{x1D714}_{k}(k_{0})}}\int _{-\infty }^{\infty }\frac{\text{d}(\unicode[STIX]{x1D6FF}k)}{2\unicode[STIX]{x03C0}}\nonumber\\ \displaystyle & & \displaystyle \times \,\exp \!\left[\text{i}k_{0}x+\text{i}\unicode[STIX]{x1D6FF}kx-\text{i}\!\left(k_{0}V_{p}+\unicode[STIX]{x1D6FF}kV_{g}+\textstyle \frac{1}{2}\unicode[STIX]{x1D714}_{kk}\unicode[STIX]{x1D6FF}k^{2}\right)t+\left(\unicode[STIX]{x1D6FE}_{0}-\textstyle \frac{1}{2}\left|\unicode[STIX]{x1D6FE}_{kk}\right|\unicode[STIX]{x1D6FF}k^{2}\right)t\right]+\text{c.c.}\nonumber\\ \displaystyle & {\approx} & \displaystyle -\text{i}\frac{\exp (\text{i}k_{0}(x-V_{p}t))\exp \unicode[STIX]{x1D6FE}_{0}t}{\left.\unicode[STIX]{x1D700}_{\unicode[STIX]{x1D714}}\right|_{\unicode[STIX]{x1D714}_{k}(k_{0})}}\int _{-\infty }^{\infty }\frac{\text{d}(\unicode[STIX]{x1D6FF}k)}{2\unicode[STIX]{x03C0}}\nonumber\\ \displaystyle & & \displaystyle \times \,\exp \!\left[\text{i}\unicode[STIX]{x1D6FF}k(x-V_{g}t)\right]\exp \!\left[-\textstyle \frac{1}{2}\unicode[STIX]{x1D6FF}k^{2}(\left|\unicode[STIX]{x1D6FE}_{kk}\right|+\text{i}\unicode[STIX]{x1D714}_{kk})t\right]+\text{c.c.}\nonumber\\ \displaystyle & {\approx} & \displaystyle -\text{i}\frac{\exp (\text{i}k_{0}(x-V_{p}t))\exp \unicode[STIX]{x1D6FE}_{0}t}{\left.\unicode[STIX]{x1D700}_{\unicode[STIX]{x1D714}}\right|_{\unicode[STIX]{x1D714}_{k}(k_{0})}}\frac{1}{\sqrt{2\unicode[STIX]{x03C0}}}\frac{1}{\sqrt{(\left|\unicode[STIX]{x1D6FE}_{kk}\right|+\text{i}\unicode[STIX]{x1D714}_{kk})t}}\nonumber\\ \displaystyle & & \displaystyle \times \,\exp \!\left[\frac{-\frac{1}{2}(x-V_{g}t)^{2}}{(\left|\unicode[STIX]{x1D6FE}_{kk}\right|+\text{i}\unicode[STIX]{x1D714}_{kk})t}\right]+\text{c.c.}\nonumber\\ \displaystyle & {\approx} & \displaystyle -\text{i}\frac{\exp (\text{i}k_{0}(x-V_{p}t))\exp \unicode[STIX]{x1D6FE}_{0}t}{\left.\unicode[STIX]{x1D700}_{\unicode[STIX]{x1D714}}\right|_{\unicode[STIX]{x1D714}_{k}(k_{0})}}\frac{1}{\sqrt{2\unicode[STIX]{x03C0}}}\frac{1}{\sqrt{(\left|\unicode[STIX]{x1D6FE}_{kk}\right|+\text{i}\unicode[STIX]{x1D714}_{kk})t}}\nonumber\\ \displaystyle & & \displaystyle \times \,\exp \!\left[\frac{-\textstyle \frac{1}{2}(x-V_{g}t)^{2}\left|\unicode[STIX]{x1D6FE}_{kk}\right|}{(\left|\unicode[STIX]{x1D6FE}_{kk}\right|^{2}+\unicode[STIX]{x1D714}_{kk}^{2})t}+\text{i}\frac{\frac{1}{2}(x-V_{g}t)^{2}\left|\unicode[STIX]{x1D714}_{kk}\right|}{(\left|\unicode[STIX]{x1D6FE}_{kk}\right|^{2}+\unicode[STIX]{x1D714}_{kk}^{2})t}\right]+\text{c.c.}\end{eqnarray}$$

We note that the spatial width of the pulse scales as $\unicode[STIX]{x0394}x\sim t^{1/2}(|\unicode[STIX]{x1D6FE}_{kk}|^{2}+\unicode[STIX]{x1D714}_{kk}^{2})^{1/2}/|\unicode[STIX]{x1D6FE}_{kk}|_{}$ . Hence, the larger (and steeper) $|\unicode[STIX]{x1D6FE}_{kk}|$ , the faster the spreading in configuration space. Because the pulse width in the space–time domain is increasing as $t^{1/2}$ , the pulse in the frequency–wavenumber dual domain is decreasing as $t^{-1/2}$ ; hence, the wave packet is lengthening in space–time and becoming purer in its spectral content as it advects and grows.

To determine whether an instability is absolute or convective, we must compare the spreading $\sqrt{\text{D}t}=\sqrt{(|\unicode[STIX]{x1D6FE}_{kk}|^{2}+\unicode[STIX]{x1D714}_{kk}^{2})t/\unicode[STIX]{x1D6FE}_{kk}}$ and the exponential growth $\exp (\unicode[STIX]{x1D6FE}_{0}t)$ with the advection of the pulse at the group velocity $V_{g}$ . In the frame advecting with the wave packet, $|\unicode[STIX]{x1D700}^{-1}(x,t)|\propto \exp (\unicode[STIX]{x1D6FE}_{0}t-x^{2}/2\,\text{D}t)$ ; and the pulse grows exponentially and spreads. In the plasma frame one obtains

(2.81) $$\begin{eqnarray}\displaystyle & & \displaystyle \left|\unicode[STIX]{x1D700}^{-1}(x,t)\right|\propto \exp \left(\unicode[STIX]{x1D6FE}_{0}t-\frac{(x-V_{g}t)^{2}}{2\text{D}t}\right)\nonumber\\ \displaystyle & & \displaystyle \quad \equiv \exp \left(\unicode[STIX]{x1D70F}-\frac{(\unicode[STIX]{x1D709}-\unicode[STIX]{x1D70E}\unicode[STIX]{x1D70F})^{2}}{2\unicode[STIX]{x1D70F}}\right),\quad \unicode[STIX]{x1D709}\equiv \sqrt{\frac{\unicode[STIX]{x1D6FE}_{0}}{D}}x,\unicode[STIX]{x1D70F}\equiv \unicode[STIX]{x1D6FE}_{0}t,\unicode[STIX]{x1D70E}\equiv V_{g}\sqrt{\frac{1}{\unicode[STIX]{x1D6FE}_{0}D}}\nonumber\\ \displaystyle & & \displaystyle \quad \rightarrow \ln \left|\unicode[STIX]{x1D700}^{-1}(x,t)\right|\sim \unicode[STIX]{x1D6FE}_{0}t-\frac{(x-V_{g}t)^{2}}{2\,\text{D}t}=\unicode[STIX]{x1D70F}-\frac{(\unicode[STIX]{x1D709}-\unicode[STIX]{x1D70E}\unicode[STIX]{x1D70F})^{2}}{2\unicode[STIX]{x1D70F}}.\end{eqnarray}$$

Theorem. For fixed $\unicode[STIX]{x1D709}$ , the asymptotics of $|\unicode[STIX]{x1D700}^{-1}(x,t)|$ in (2.81) as $\unicode[STIX]{x1D70F}\rightarrow \infty$ determines whether the pulse is growing absolutely or convectively, i.e.

(2.82) $$\begin{eqnarray}\lim _{\unicode[STIX]{x1D70F}\rightarrow \infty }\left[\ln |\unicode[STIX]{x1D700}^{-1}(x,t)|\sim \unicode[STIX]{x1D70F}-\frac{(\unicode[STIX]{x1D709}-\unicode[STIX]{x1D70E}\unicode[STIX]{x1D70F})^{2}}{2\unicode[STIX]{x1D70F}}\right]\rightarrow \unicode[STIX]{x1D70F}\left(1-\frac{\unicode[STIX]{x1D70E}^{2}}{2}\right).\end{eqnarray}$$

For $\unicode[STIX]{x1D70E}>\surd 2,\ln \!|\unicode[STIX]{x1D700}^{-1}|\rightarrow -\infty$ as $\unicode[STIX]{x1D70F}\rightarrow \infty$ , i.e. the instability is convective; and the disturbance grows at first and then dies away. For $\unicode[STIX]{x1D70E}<\surd 2,\ln |\unicode[STIX]{x1D700}^{-1}|\rightarrow \infty$ as $\unicode[STIX]{x1D70F}\rightarrow \infty$ , i.e. the instability is absolute and continues to grow exponentially at any fixed position. The condition for absolute (or convective) instability is then

(2.83) $$\begin{eqnarray}V_{g}<({>})\sqrt{2\unicode[STIX]{x1D6FE}_{0}D}=\sqrt{2\unicode[STIX]{x1D6FE}_{0}\left(\frac{|\unicode[STIX]{x1D6FE}_{kk}|^{2}+\unicode[STIX]{x1D714}_{kk}^{2}}{|\unicode[STIX]{x1D6FE}_{kk}|}\right)}.\end{eqnarray}$$

Exercise. Show that in the plasma frame the group velocity for the weak beam in a cold-plasma instability is $(2/3)u_{b}$ and the instability is convective using (2.83) and the analysis in § 2.6.1.

2.6.3 Peter Sturrock’s method for analysing absolute instability (reference)

Stanford Professor P. A. Sturrock introduced a method for analysing whether a growing instability is convective or absolute (Sturrock 1958). Sturrock’s method is reviewed in (Clemmow & Dougherty 1989) and was assigned as reading but not covered in class lectures.

2.6.4 Bers and Briggs’ method for analysing absolute instability

Bers (1973) and Briggs (1964) derived a method for calculating the impulse response using contour integration and analytic continuation, from which absolute and convective instability can be distinguished. The calculation begins with consideration of (2.80) in one spatial dimension transformed to the moving reference frame $x=\mathit{wt}$ ,

(2.84) $$\begin{eqnarray}\unicode[STIX]{x1D700}^{-1}(x,t)=\int \frac{\text{d}k}{2\unicode[STIX]{x03C0}}\int \frac{\text{d}\unicode[STIX]{x1D714}}{2\unicode[STIX]{x03C0}}\frac{\exp (\text{i}kx-\text{i}\unicode[STIX]{x1D714}t)}{\unicode[STIX]{x1D700}(k,\unicode[STIX]{x1D714})}=\int \frac{\text{d}k}{2\unicode[STIX]{x03C0}}\int \frac{\text{d}\unicode[STIX]{x1D714}}{2\unicode[STIX]{x03C0}}\frac{\exp (\text{i}\unicode[STIX]{x1D714}_{w}t)}{\unicode[STIX]{x1D700}(k,\unicode[STIX]{x1D714}_{w}+kw)},\end{eqnarray}$$

where $\unicode[STIX]{x1D714}_{w}=\unicode[STIX]{x1D714}-kw$ , which is the Doppler-shifted frequency. There are two contour integrals to perform in (2.84). These are diagrammed in figure 8 for a convective instability and in figure 9 for an absolute instability. In the complex $k$ plane the locus of poles in $k$ of the integrand in (2.84) for fixed $\unicode[STIX]{x1D714}$ is shown by $C_{\unicode[STIX]{x1D714}}$ , and the integration contour is $C_{k}$ . In the complex $\unicode[STIX]{x1D714}$ plane the locus of poles in $\unicode[STIX]{x1D714}$ of the integrand in (2.84) for fixed $k$ is shown by $C_{k}$ , and the integration contour is $C_{\unicode[STIX]{x1D714}}$ . Making use of analytic continuation, the integration contour in the complex $\unicode[STIX]{x1D714}$ plane is depressed to lower values on the imaginary axis as shown in figure 6 until poles on $C_{k}$ from below the $C_{\unicode[STIX]{x1D714}}$ contour are encountered. The poles $C_{k}$ in the complex $\unicode[STIX]{x1D714}$ plane are parametrized by the complex value of $k$ swept out by the $C_{k}$ contour in the complex $k$ plane. In order to depress the $C_{\unicode[STIX]{x1D714}}$ contour to lower values of $\operatorname{Im}\unicode[STIX]{x1D714}$ , we must move the $C_{k}$ contour down in the complex $k$ plane. In the complex $k$ plane there are a loci of poles $C_{\unicode[STIX]{x1D714}}$ above and below the $C_{k}$ integration contour. For the convectively unstable case, the $C_{\unicode[STIX]{x1D714}}$ contour in the complex $\unicode[STIX]{x1D714}$ plane can be depressed to values of $\operatorname{Im}\unicode[STIX]{x1D714}$ ${<}$ 0; and $\unicode[STIX]{x1D700}^{-1}(x,t)$ decays (figure 8). In the absolutely unstable case, the $C_{\unicode[STIX]{x1D714}}$ contour in the complex $\unicode[STIX]{x1D714}$ plane can be depressed and deformed to lower values of $\operatorname{Im}\unicode[STIX]{x1D714}$ everywhere except as one approaches the pinch point $P$ in figure 9(c) because the $C_{k}$ integration contour in the $k$ plane becomes trapped (‘pinched’) between the $C_{\unicode[STIX]{x1D714}}$ contours above and below it; and part of the $C_{k}$ loci of poles has some values $\operatorname{Im}\unicode[STIX]{x1D714}_{k}$ ${>}$ 0 in figure 9(d). The asymptotic response for $\unicode[STIX]{x1D700}^{-1}(x,t)$ exhibits exponential growth at the value of $\operatorname{Im}\unicode[STIX]{x1D714}_{k}$ ${>}0$ at the pinch point.

Figure 8. Convective instability: diagrams of contour integral paths $C_{k}$ in complex $k$ plane (a,c), and contour integral paths $C_{\unicode[STIX]{x1D714}}$ in complex $\unicode[STIX]{x1D714}$ plane (b) and depressed in (d) showing the loci of poles of the integrand in (2.84).

Figure 9. Absolute instability: diagrams of contour integral paths $C_{k}$ in complex $k$ plane (a,c), and contour integral paths $C_{\unicode[STIX]{x1D714}}$ in complex $\unicode[STIX]{x1D714}$ plane (b) and depressed in (d) showing the loci of poles of the integrand in (2.84). A pinching of roots occurs at point $P$ in (c).

Exercise. Examine the conditions for absolute versus convective instability for the cold beam – cold-plasma instability investigated in § 2.6.1 in the (a) plasma frame in which $\unicode[STIX]{x1D700}(k,\unicode[STIX]{x1D714})=1-\unicode[STIX]{x1D714}_{p}^{2}/\unicode[STIX]{x1D714}^{2}-\unicode[STIX]{x1D714}_{b}^{2}/(\unicode[STIX]{x1D714}-ku_{b})^{2}$ and (b) the beam frame in which $\unicode[STIX]{x1D700}(k,\unicode[STIX]{x1D714})=1-\unicode[STIX]{x1D714}_{p}^{2}/(\unicode[STIX]{x1D714}+ku_{b})^{2}-\unicode[STIX]{x1D714}_{b}^{2}/\unicode[STIX]{x1D714}^{2}$ , and there is absolute instability.

Theorem. The condition for absolute instability is equivalent to obtaining two pinching roots for $(\unicode[STIX]{x1D714},k)$ from the simultaneous solution of $\unicode[STIX]{x1D700}(\unicode[STIX]{x1D714},k)=0$ and $\unicode[STIX]{x2202}\unicode[STIX]{x1D700}(\unicode[STIX]{x1D714},k)/\unicode[STIX]{x2202}k=0$ with $\operatorname{Im}\unicode[STIX]{x1D714}>0$ , provided that $\unicode[STIX]{x2202}\unicode[STIX]{x1D700}(\unicode[STIX]{x1D714},k)/\unicode[STIX]{x2202}\unicode[STIX]{x1D714}\neq 0$ .

Exercise. Prove this theorem.

2.7 Linear steady-state response to a fixed-frequency disturbance

We consider here the response of a plasma to a steady-state force at a fixed frequency. We assume that the plasma is quiescent and non-turbulent. There may be transient convective instabilities but no absolute instability. We implant a localized fixed-frequency disturbance and derive the plasma response. Let the implanted disturbance be a planar disturbance, for example a biased grid connected to an oscillator,

(2.85) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{r},t) & = & \displaystyle \left\{0~t<0,\unicode[STIX]{x1D6FF}(z)\sin (\unicode[STIX]{x1D714}_{0}t)\quad t>0\right\}\nonumber\\ \displaystyle & \rightarrow & \displaystyle \unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{k},\unicode[STIX]{x1D714})=-(2\unicode[STIX]{x03C0})^{2}\unicode[STIX]{x1D6FF}(k_{x})\unicode[STIX]{x1D6FF}(k_{y})\frac{\unicode[STIX]{x1D714}_{0}}{\unicode[STIX]{x1D714}^{2}-\unicode[STIX]{x1D714}_{0}^{2}}.\end{eqnarray}$$

The plasma potential is then given by (2.50) and (2.85)

(2.86) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{k},\unicode[STIX]{x1D714}) & = & \displaystyle \frac{\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{k},\unicode[STIX]{x1D714})}{\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})}\nonumber\\ \displaystyle & \rightarrow & \displaystyle \unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r},t)=-\int \frac{\text{d}^{3}k}{(2\unicode[STIX]{x03C0})^{3}}\int \frac{\text{d}\unicode[STIX]{x1D714}}{2\unicode[STIX]{x03C0}}\nonumber\\ \displaystyle & & \displaystyle \times \,\exp \text{i}(k_{x}x+k_{y}y+k_{z}z-\unicode[STIX]{x1D714}t)\frac{(2\unicode[STIX]{x03C0})^{2}\unicode[STIX]{x1D6FF}(k_{x})\unicode[STIX]{x1D6FF}(k_{y})\displaystyle \frac{\unicode[STIX]{x1D714}_{0}}{\unicode[STIX]{x1D714}^{2}-\unicode[STIX]{x1D714}_{0}^{2}}}{\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})}\nonumber\\ \displaystyle & \rightarrow & \displaystyle -\int \frac{\text{d}k_{z}}{2\unicode[STIX]{x03C0}}\int \frac{\text{d}\unicode[STIX]{x1D714}}{2\unicode[STIX]{x03C0}}\exp (\text{i}k_{z}z-\text{i}\unicode[STIX]{x1D714}t)\frac{1}{\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})}\frac{\unicode[STIX]{x1D714}_{0}}{\unicode[STIX]{x1D714}^{2}-\unicode[STIX]{x1D714}_{0}^{2}}.\end{eqnarray}$$

We let any transients associated with initial conditions and any instabilities convect or decay away.

Theorem. Given that the dielectric is an even function of $z$ , then $\unicode[STIX]{x1D700}(-k_{z},-\unicode[STIX]{x1D714}_{r}+\text{i}\unicode[STIX]{x1D714}_{i})=\unicode[STIX]{x1D700}(k_{z},\unicode[STIX]{x1D714}_{r}+\text{i}\unicode[STIX]{x1D714}_{i})$ .

2.7.1 Response for a sinusoidally driven stable system

The contour integrations in (2.86) are diagrammed in figure 10. For every value of $k$ the $C_{k}$ contour in figure 10(b) is below the $\operatorname{Re}\unicode[STIX]{x1D714}$ axis, which dictates that $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r},t)$ is stable. The $\unicode[STIX]{x1D714}$ integration in (2.86) is performed using Cauchy’s theorem to obtain,

(2.87) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r},t)=\frac{\text{i}}{2}\int \frac{\text{d}k_{z}}{2\unicode[STIX]{x03C0}}\frac{\exp (\text{i}k_{z}z-\text{i}\unicode[STIX]{x1D714}_{0}t)}{\unicode[STIX]{x1D700}(k_{z},\unicode[STIX]{x1D714}_{0})}+\text{c.c.}\end{eqnarray}$$

and transients that are evanescent. The $k_{z}$ integration is now performed.

Figure 10. Steady-state response to a fixed-frequency disturbance: diagrams of the contour integral path $C_{k}$ in complex $k$ plane (a) and the contour integral paths $C_{\unicode[STIX]{x1D714}}$ in complex $\unicode[STIX]{x1D714}$ plane (b) showing the loci of poles of the integrand in (2.86).

Consider the poles of the integrand in (2.87), i.e. the solutions of $\unicode[STIX]{x1D700}(k_{z},\unicode[STIX]{x1D714}_{0})=0$ . For the example of a velocity distribution that is a square in velocity space (see table 1) then $\unicode[STIX]{x1D714}_{0}^{2}=\unicode[STIX]{x1D714}_{p}^{2}+k_{z}^{2}c^{2}\rightarrow k_{z}=\pm \sqrt{\unicode[STIX]{x1D714}_{0}^{2}-\unicode[STIX]{x1D714}_{p}^{2}}/c$ , which roots lie on the real $k_{z}$ axis. A real physical system will have some finite dissipation. Hence, $\unicode[STIX]{x1D714}(k)=\pm \sqrt{\unicode[STIX]{x1D714}_{p}^{2}+k^{2}c^{2}}-\text{i}\unicode[STIX]{x1D708}$ and $(\unicode[STIX]{x1D714}+\text{i}\unicode[STIX]{x1D708})^{2}=\unicode[STIX]{x1D714}_{p}^{2}+k^{2}c^{2}$ , where $\unicode[STIX]{x1D708}$ is a damping rate, e.g. due to weak collisions, so that free oscillations are damped and transients will indeed die away. In this circumstance the solutions for $k_{z}$ become $k_{z}=\pm \sqrt{(\unicode[STIX]{x1D714}_{0}+\text{i}\unicode[STIX]{x1D708})^{2}-\unicode[STIX]{x1D714}_{p}^{2}}/c$ . The sign of $k_{z}$ is selected based on whether values of $z$ are negative or positive so that the response of the system dies away for $|z|\rightarrow \infty$ . For $\unicode[STIX]{x1D714}_{0}<\unicode[STIX]{x1D714}_{p}~k_{z}$ is purely imaginary, and the plasma response is evanescent; and modes do not propagate.

For $z>$ 0 the contour integral in (2.87) is closed counter-clockwise in the upper half-plane in figure 10(a), and we sum over pole contributions,

(2.88) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}(\boldsymbol{r},t)=-\frac{1}{2}\mathop{\sum }_{\ell }\exp (\text{i}k_{zl}^{\prime }z-k_{zl}^{\prime \prime }z-\text{i}\unicode[STIX]{x1D714}_{0}t)\frac{1}{\left.\displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D700}(k_{z},\unicode[STIX]{x1D714}_{0})}{\unicode[STIX]{x2202}k_{z}}\right|_{\ell }}+\text{c.c.},\end{eqnarray}$$

where $k_{zl}=k_{zl}^{\prime }+\text{i}k_{zl}^{\prime \prime }$ is a solution of $\unicode[STIX]{x1D700}(k_{zl},\unicode[STIX]{x1D714}_{0})=0$ with $k_{zl}^{\prime \prime }>0$ . For $z<0$ the contour integral is closed clockwise in the lower half-plane with $k_{zl}^{\prime \prime }<0$ ; and the overall sign of the right-hand side of (2.88) changes. For $z$ positive or negative the plasma response is sinusoidal in time with frequency $\unicode[STIX]{x1D714}_{0}$ and damps away from the origin in $z$ .

2.7.2 Response for a sinusoidally driven convectively unstable system

One can calculate the sinusoidally driven response for a convectively unstable system in the same manner as in § 2.7.1 Because the system is convectively unstable, the integration contour $C_{\unicode[STIX]{x1D714}}$ can be depressed as in figure 10; and the results in (2.87) and (2.88) pertain. However, the solutions of $\unicode[STIX]{x1D700}(k_{z},\unicode[STIX]{x1D714}_{0})=0$ for $k_{z}$ can have $\operatorname{Im}k_{z}<0$ for $z>0$ (and $\operatorname{Im}k_{z}>0$ for $z<0$ ) leading to spatially growing solutions away from the origin in $z$ .

Exercise. Consider the beam-plasma system in the plasma reference frame (§ 2.6.1) which is convectively unstable. Solve $\unicode[STIX]{x1D700}(k_{z},\unicode[STIX]{x1D714}_{0})=0$ based on (2.74) for $k_{z}$ and evaluate (2.88). Contrast this to the stable case of a beaming plasma,

$$\begin{eqnarray}\unicode[STIX]{x1D700}(k,\unicode[STIX]{x1D714})=1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{(\unicode[STIX]{x1D714}+\text{i}\unicode[STIX]{x1D708}-ku)^{2}}=0.\end{eqnarray}$$

Exercise. Consider an external driving potential that is a spherical waveform,

$$\begin{eqnarray}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D719}^{\text{ext}}(\boldsymbol{r},t)=\left\{0~t<0,\unicode[STIX]{x1D6FF}(r)\sin (\unicode[STIX]{x1D714}_{0}t)~t>0\right\}\end{eqnarray}$$

in the rest frame of a cold plasma. Find the response for $\unicode[STIX]{x1D714}_{0}>\unicode[STIX]{x1D714}_{p}$ and $\unicode[STIX]{x1D714}_{0}<\unicode[STIX]{x1D714}_{p}$ . Contrast this to the case of a mono-energetic velocity distribution: $f_{0}=\unicode[STIX]{x1D6FF}(|\boldsymbol{v}|-v_{0})$ /norm.

2.8 Linear stability or instability for a few simple velocity distributions

We return to the consideration of Vlasov stability for more general velocity distributions based on (2.63):

(2.89) $$\begin{eqnarray}\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})=1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k^{2}}Z^{\prime }(v)=1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k^{2}}\int \text{d}u\frac{g_{\hat{\boldsymbol{k}}}^{\prime }(u)}{u-v},\end{eqnarray}$$

where $v=\unicode[STIX]{x1D714}/k$ and $\int \text{d}u\,g(u)=1$ . In figure 11 are shown four examples of velocity distributions. Figure 11(a) depicts two warm beams, which is always linearly unstable. A single-humped velocity distribution is shown in figure 11(b), which is stable. The velocity distribution function shown in figure 11(c) can be stable or unstable depending on the relative heights $b$ and $a$ . In the limit $b\rightarrow 0$ , the velocity distribution is single humped and stable. In the limit $b\rightarrow a$ , the velocity distribution corresponds to two counter-propagating warm beams and is unstable. Figure 11(d) depicts a double Cauchy beam. When the thermal spread $c$ is larger than the beam centroid velocity $u_{0}$ , the plasma is stable; and when $c<u_{0}$ , the plasma is unstable. The results of these examples give one a hint toward a more general stability condition. A general condition is that it is necessary for $g(u)$ to have a minimum for instability.

Figure 11. Four examples of finite-temperature velocity distribution functions.

Exercise. Show that $\unicode[STIX]{x1D714}(k,c)=\unicode[STIX]{x1D714}(k,0)-\text{i}kc$ for the double Cauchy beam, which illustrates the stabilizing influence of the thermal spread.

2.9 General analysis of the dielectric response

2.9.1 Perturbative expansion for a fast wave

Equation (2.89) can be evaluated easily when thermal effects are weak. For a ‘fast’ wave, $V\gg v_{\text{th}}$ , where $v_{\text{th}}$ is a characteristic thermal velocity, $v_{\text{th}}^{2}=T/m$ . one can expand

(2.90) $$\begin{eqnarray}\frac{1}{u-v}=-\frac{1}{v}\left(\frac{1}{1-\displaystyle \frac{u}{v}}\right)=-\frac{1}{v}\left(1+\frac{u}{v}+\displaystyle \frac{u^{2}}{v^{2}}+\frac{u^{3}}{v^{3}}\ldots \right).\end{eqnarray}$$

Theorem. From (2.61) and (2.90) the first thermal correction to $Z(v)$ is then

(2.91) $$\begin{eqnarray}Z(v)=-\frac{1}{v}\left(1+\frac{v_{\text{th}}^{2}}{v^{2}}+O\left(\frac{v_{\text{th}}^{3}}{v^{3}}\right)\right),\end{eqnarray}$$

where $v=\unicode[STIX]{x1D714}/k$ and $[1,0,v_{\text{th}}^{2}]=\int \!\,\text{d}u[1,u,u^{2}]g(u)$ , and

(2.92) $$\begin{eqnarray}Z^{\prime }(v)=\frac{1}{v^{2}}+\frac{3v_{\text{th}}^{2}}{v^{4}}+\cdots \,.\end{eqnarray}$$

Hence,

(2.93) $$\begin{eqnarray}\unicode[STIX]{x1D700}(k,\unicode[STIX]{x1D714})=1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{\unicode[STIX]{x1D714}^{2}}-\frac{\unicode[STIX]{x1D714}_{p}^{2}3k^{2}v_{\text{th}}^{2}}{\unicode[STIX]{x1D714}^{4}}+O(\unicode[STIX]{x1D714}^{-5})\end{eqnarray}$$

and the Bohm–Gross dispersion relation for the electron plasma wave results from $\unicode[STIX]{x1D700}(k,\unicode[STIX]{x1D714})=0$ ,

(2.94) $$\begin{eqnarray}\unicode[STIX]{x1D714}^{2}=\unicode[STIX]{x1D714}_{p}^{2}+\frac{3k^{2}v_{\text{th}}^{2}}{\unicode[STIX]{x1D714}^{2}}\unicode[STIX]{x1D714}_{p}^{2}+\cdots \approx \unicode[STIX]{x1D714}_{p}^{2}+3k^{2}v_{\text{th}}^{2},\end{eqnarray}$$

with the inclusion of thermal effects the electron plasma wave has acquired dispersion.

Theorem. The product of the phase and group velocities for the electron plasma wave is given by $v_{p}v_{g}=3v_{\text{th}}^{2}$ for $kv_{\text{th}}\ll \unicode[STIX]{x1D714}_{p}$ , i.e. $k\unicode[STIX]{x1D706}_{D}\ll 1$ , where $\unicode[STIX]{x1D706}_{D}=v_{\text{th}}/\unicode[STIX]{x1D714}_{p}$ is the Debye length. With $v_{p}\gg v_{\text{th}}$ the group velocity satisfies $v_{g}\ll v_{\text{th}}$ .

2.9.2 Use of the Hilbert transform in deriving the dielectric function

We return to the consideration of $Z(v)$ , equation (2.61),

(2.95) $$\begin{eqnarray}\displaystyle Z(v) & = & \displaystyle \int _{-\infty }^{\infty }\text{d}u\frac{g_{\hat{\boldsymbol{k}}}(u)}{u-v_{R}-\text{i}v_{I}}\nonumber\\ \displaystyle & = & \displaystyle \int _{-\infty }^{\infty }\text{d}u\frac{g_{\hat{\boldsymbol{k}}}(u)}{(u-v_{R})^{2}+v_{I}^{2}}\left[u-v_{R}+\text{i}v_{I}\right],\quad v=v_{R}+\text{i}v_{I},\quad v_{I}>0\nonumber\\ \displaystyle & \rightarrow & \displaystyle \operatorname{Re}Z(v)=\int _{-\infty }^{\infty }\text{d}u\frac{g_{\hat{\boldsymbol{k}}}(u)}{(u-v_{R})^{2}+v_{I}^{2}}\left(u-v_{R}\right),\nonumber\\ \displaystyle & & \displaystyle \operatorname{Im}Z(v)=v_{I}\int _{-\infty }^{\infty }\text{d}u\frac{g_{\hat{\boldsymbol{k}}}(u)}{(u-v_{R})^{2}+v_{I}^{2}}.\end{eqnarray}$$

One notes that $\operatorname{Re}Z(v)$ has odd symmetry with respect to $u-v_{R}$ and goes to zero for $v_{I}=0$ and for $v_{R}\rightarrow \pm \infty$ .

Theorem. As $v_{I}\rightarrow 0$ there is a singularity in $\operatorname{Re}Z(v)$ at $v_{R}=0$ while $\operatorname{Im}Z(v)\propto v_{I}\rightarrow 0$ . In the limit $v_{I}\rightarrow 0$ , $Z(v)$ takes on the following special forms:

(2.96) $$\begin{eqnarray}Z\left(v_{R}\right)=P\int \text{d}u\frac{g(u)}{u-v_{R}}+\text{i}\int \text{d}u\,g(u)\unicode[STIX]{x03C0}\unicode[STIX]{x1D6FF}(u-v_{R}),\end{eqnarray}$$

where $P$ denotes the Cauchy principal value of the integral that immediately follows it. The development of (2.96) has a basis in distribution theory (Lighthill 1958),

(2.97) $$\begin{eqnarray}\lim _{y\rightarrow 0\pm }\frac{1}{x+\text{i}y}=P\left(\frac{1}{x}\right)\mp \text{i}\unicode[STIX]{x03C0}\unicode[STIX]{x1D6FF}(x).\end{eqnarray}$$

Corollary. Hilbert transform

(2.98) $$\begin{eqnarray}Z\left(v\right)=\text{i}\int _{0}^{\infty }\text{d}s\,\exp (\text{i}vs)\int _{-\infty }^{\infty }\text{d}u\,g(u)\exp (-\text{i}us)=\int _{-\infty }^{\infty }\text{d}u\frac{g(u)}{u-v},\end{eqnarray}$$

where the first integral is a Laplace transform and the second integral is a Fourier transform.

Exercise. Invert the Hilbert transform to obtain $g(u)$ given $Z(v)$ .

Theorem. Relation of the dielectric function to $Z^{\prime }(v)$ from (2.89)

(2.99) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714}) & = & \displaystyle 1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k^{2}}Z^{\prime }(v)=1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k^{2}}\int \text{d}u\frac{g_{\hat{\boldsymbol{k}}}^{\prime }(u)}{u-v}\nonumber\\ \displaystyle & = & \displaystyle 1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k^{2}}P\int \text{d}u\,\frac{g_{\hat{\boldsymbol{k}}}^{\prime }(u)}{u-v}-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k^{2}}\int \text{d}u\,g_{\hat{\boldsymbol{k}}}^{\prime }(u)\text{i}\unicode[STIX]{x03C0}\unicode[STIX]{x1D6FF}(u-v),\quad v\equiv \frac{\unicode[STIX]{x1D714}}{k}.\end{eqnarray}$$

Exercise. Show that $\unicode[STIX]{x1D700}_{R}=\operatorname{Re}\unicode[STIX]{x1D700}$ is an even function of $\unicode[STIX]{x1D714}$ and $\unicode[STIX]{x1D700}_{I}=\operatorname{Im}\unicode[STIX]{x1D700}$ is an odd function of $\unicode[STIX]{x1D714}$ .

2.9.3 Dispersion relation for weak damping or growth rate compared to the real part of the frequency

Consider $\unicode[STIX]{x1D714}=\unicode[STIX]{x1D714}_{R}+\text{i}\unicode[STIX]{x1D6FE}$ and conditions such that $|\unicode[STIX]{x1D6FE}|=|\!\operatorname{Im}\unicode[STIX]{x1D714}|\ll |\!\operatorname{Re}\unicode[STIX]{x1D714}|$ and Taylor-series expand $\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})=0$ for small $\unicode[STIX]{x1D6FE}$ ,

(2.100) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714}_{k}) & = & \displaystyle \unicode[STIX]{x1D700}_{R}(\boldsymbol{k},\unicode[STIX]{x1D714}_{k})+\text{i}\unicode[STIX]{x1D700}_{I}(\boldsymbol{k},\unicode[STIX]{x1D714}_{k})+\text{i}\unicode[STIX]{x1D6FE}\left.\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})\right|_{\unicode[STIX]{x1D714}_{R}}+O(\unicode[STIX]{x1D6FE}^{2})=0\nonumber\\ \displaystyle & \rightarrow & \displaystyle \unicode[STIX]{x1D700}_{R}(\boldsymbol{k},\unicode[STIX]{x1D714}_{k})-\unicode[STIX]{x1D6FE}\left.\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\unicode[STIX]{x1D700}_{I}(\boldsymbol{k},\unicode[STIX]{x1D714})\right|_{\unicode[STIX]{x1D714}_{R}}+O(\unicode[STIX]{x1D6FE}^{2})=0\quad \unicode[STIX]{x1D700}_{I}(\boldsymbol{k},\unicode[STIX]{x1D714}_{k})+\unicode[STIX]{x1D6FE}\left.\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\unicode[STIX]{x1D700}_{R}(\boldsymbol{k},\unicode[STIX]{x1D714})\right|_{\unicode[STIX]{x1D714}_{R}}\nonumber\\ \displaystyle & & \displaystyle +\,O(\unicode[STIX]{x1D6FE}^{2})=0.\end{eqnarray}$$

Theorem. For weak damping or weak growth rates the solution of $\unicode[STIX]{x1D700}(\boldsymbol{k},\unicode[STIX]{x1D714})=0$ yields solutions for $\unicode[STIX]{x1D714}_{R}(\boldsymbol{k})$ and $\unicode[STIX]{x1D6FE}(\boldsymbol{k})$ ,

(2.101) $$\begin{eqnarray}\unicode[STIX]{x1D700}_{R}\left(\boldsymbol{k},\unicode[STIX]{x1D714}_{R}\right)=0\rightarrow \unicode[STIX]{x1D714}_{R}=\cdots \unicode[STIX]{x1D6FE}(\boldsymbol{k})=-\frac{\unicode[STIX]{x1D700}_{I}\left(\boldsymbol{k},\unicode[STIX]{x1D714}_{R}\right)}{\left.\displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\unicode[STIX]{x1D700}_{R}\left(\boldsymbol{k},\unicode[STIX]{x1D714}\right)\right|_{\unicode[STIX]{x1D714}_{R}}},\end{eqnarray}$$

$\unicode[STIX]{x1D714}_{\boldsymbol{k}}$ is defined equal to $\unicode[STIX]{x1D714}_{R}(\boldsymbol{k})$ . The ratio of the damping rate to the frequency $\unicode[STIX]{x1D6FE}(\boldsymbol{k})/\unicode[STIX]{x1D714}_{R}(\boldsymbol{k})$ is a measure of the ratio of the dissipation (negative or positive) to the wave energy,

(2.102) $$\begin{eqnarray}\frac{\unicode[STIX]{x1D6FE}(\boldsymbol{k})}{\unicode[STIX]{x1D714}_{R}}=-\frac{\unicode[STIX]{x03C0}g^{\prime }(v)}{\displaystyle \frac{\unicode[STIX]{x1D714}_{R}}{k}P\int \text{d}u\,\frac{g^{\prime \prime }(u)}{u-v}}=-\frac{\unicode[STIX]{x03C0}g^{\prime }(v)}{vZ_{R}^{\prime \prime }(v)}.\end{eqnarray}$$

Exercise. Derive the derivative of the principal part of the Hilbert transform.

2.9.4 Maxwellian velocity distribution function – electron Landau damping in fast and slow waves

We next construct the dispersion relation for a Maxwellian electron velocity distribution function,

(2.103) $$\begin{eqnarray}g(u)=\frac{1}{\sqrt{2\unicode[STIX]{x03C0}v_{\text{th}}^{2}}}\exp \left(-\frac{u^{2}}{2v_{\text{th}}^{2}}\right),\quad g^{\prime }(u)=-\frac{u}{v_{\text{th}}^{3}\sqrt{2\unicode[STIX]{x03C0}}}\exp \left(-\frac{u^{2}}{2v_{\text{th}}^{2}}\right),\quad v_{\text{th}}^{2}=\frac{T_{e}}{m_{e}}.\end{eqnarray}$$

Waves in a warm plasma can be classified into three categories: fast, intermediate and slow depending on the ratio of the phase velocity to the thermal velocity. For a fast wave $v=\unicode[STIX]{x1D714}/k\gg v_{\text{th}},\unicode[STIX]{x1D6FE}\propto \exp (-v^{2}/2v_{\text{th}}^{2})$ is exponentially small. For intermediate waves $v\sim v_{\text{th}},\unicode[STIX]{x1D6FE}\sim \unicode[STIX]{x1D714}_{k}$ ; and $\unicode[STIX]{x1D6FE}$ is relatively large, which invalidates the Taylor-series expansion in (2.100) leading to (2.101) and (2.102). For slow waves $v\ll v_{\text{th}},\unicode[STIX]{x1D6FE}\propto v/v_{\text{th}}$ and is linearly small.

Example. Electron plasma wave. For a fast wave in a Maxwellian plasma, the phase velocity falls far out on the high energy tail of the velocity distribution function. From (2.101)–(2.103) one obtains

(2.104) $$\begin{eqnarray}\unicode[STIX]{x1D700}_{R}\rightarrow 1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{\unicode[STIX]{x1D714}^{2}},\quad \unicode[STIX]{x1D714}\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D700}_{R}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\approx 2\frac{\unicode[STIX]{x1D714}_{p}^{2}}{\unicode[STIX]{x1D714}^{2}}\approx 2,\quad \frac{\unicode[STIX]{x1D6FE}(\boldsymbol{k})}{\unicode[STIX]{x1D714}_{R}}=-\sqrt{\frac{\unicode[STIX]{x03C0}}{8}}\left(\frac{v}{v_{\text{th}}}\right)^{3}\exp \left(-\frac{v^{2}}{2v_{\text{th}}^{2}}\right).\end{eqnarray}$$

The damping rate given in (2.104) is the Landau damping rate for the electron plasma wave in a Vlasov plasma (collisionless). As elaborated in the next section, particles with velocity nearly equal to the phase velocity (the ‘resonant’ particles) contribute positive dissipation, because there are fewer particles faster than the wave (which particles are decelerated) than particles slower than the wave (which particles are accelerated). Equation (2.94) gives the thermal correction to the real part of the frequency for the electron plasma wave. For small $k\unicode[STIX]{x1D706}_{D}\ll 1$ , the phase velocity $v=\unicode[STIX]{x1D714}/k\rightarrow \infty$ as $k\rightarrow 0$ , while $v_{g}\rightarrow 0$ ; and $\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D714}\rightarrow 0$ because the number of resonant particles in the tail is exponentially small. For increasing $k$ , the wave frequency increases while the phase velocity decreases more rapidly, which increases the damping rate and ultimately invalidates the fast-wave assumption.

Example. Ion-acoustic wave. For a slow wave in a Maxwellian plasma, $v\ll v_{\text{the}}$ , the phase velocity falls near the peak of the electron velocity distribution at low velocities. We assume that the ions are singly charged and relatively cold, $T_{i}\ll T_{e}$ , and $v_{i}\ll v\ll v_{\text{the}}$ . In these limits $v=c_{s}=(T_{e}/m_{i})^{1/2}$ , ion Landau damping is exponentially small and the electron Landau damping is linearly small. In this two-species plasma $g(u)=\sum _{s}\!\unicode[STIX]{x1D714}_{s}^{2}g_{s}(u)/\unicode[STIX]{x1D714}_{p}^{2}$ . For cold ions the ion-acoustic wave dispersion relation is derived from

(2.105) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D700}_{R}\left(\boldsymbol{k},\unicode[STIX]{x1D714}\right) & = & \displaystyle 1-\frac{\unicode[STIX]{x1D714}_{\text{pi}}^{2}}{\unicode[STIX]{x1D714}^{2}}+\frac{1}{k^{2}\unicode[STIX]{x1D706}_{e}^{2}}=0\rightarrow \unicode[STIX]{x1D714}_{k}^{2}=\frac{k^{2}\unicode[STIX]{x1D706}_{e}^{2}\unicode[STIX]{x1D714}_{\text{pi}}^{2}}{1+k^{2}\unicode[STIX]{x1D706}_{e}^{2}}=\frac{k^{2}c_{s}^{2}}{1+k^{2}\unicode[STIX]{x1D706}_{e}^{2}}\nonumber\\ \displaystyle \left.\unicode[STIX]{x1D714}\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D700}_{R}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\right|_{\unicode[STIX]{x1D714}_{k}} & = & \displaystyle 2\left(1+\frac{1}{k^{2}\unicode[STIX]{x1D706}_{e}^{2}}\right)\frac{\unicode[STIX]{x1D6FE}(k)}{\unicode[STIX]{x1D714}(k)}=-\sqrt{\frac{\unicode[STIX]{x03C0}}{8}}\left(\frac{m_{e}}{m_{i}}\right)^{1/2}\left(\frac{1}{1+k^{2}\unicode[STIX]{x1D706}_{e}^{2}}\right)^{3/2}.\end{eqnarray}$$

The electron Landau damping for the ion-acoustic wave is small for all values of $k\unicode[STIX]{x1D706}_{e}$ .

Figure 12. Velocity distribution function for the bump-on-tail instability.

2.9.5 Bump-on-tail velocity distribution and resonant instability

The bump-on-tail velocity tail velocity distribution is diagrammed in figure 12. This velocity distribution is obviously related to the weak-beam case studied in § 2.6. However, here both the plasma and the beam have acquired finite thermal spreads. If the phase velocity of an electron plasma wave falls on an interval of the velocity distribution of the beam with positive slope there can be instability. Equations (2.101) and (2.102) can be used to compute the growth or damping rate for the bump-on-tail distribution,

(2.106) $$\begin{eqnarray}\frac{\unicode[STIX]{x1D6FE}}{\unicode[STIX]{x1D714}_{k}}\approx \frac{\unicode[STIX]{x1D714}_{p}^{2}}{k^{2}}\frac{\unicode[STIX]{x03C0}g^{\prime }(v)}{2},\end{eqnarray}$$

where we have used $\unicode[STIX]{x1D714}_{k}\unicode[STIX]{x2202}\unicode[STIX]{x1D700}_{R}/\unicode[STIX]{x2202}\unicode[STIX]{x1D714}=2\unicode[STIX]{x1D714}_{p}^{2}/\unicode[STIX]{x1D714}_{k}^{2}\sim 2$ , which is good for a relatively small bump on the tail. If $g^{\prime }(v)>0$ then $\unicode[STIX]{x1D6FE}>0$ , i.e. there is instability.

Definition. Resonant instability or damping. When there are particles whose unperturbed velocity is coincident or nearly coincident with the phase velocity of a wave (velocity resonance), the particles can resonantly interact with the wave and exchange energy. In the wave frame, $\unicode[STIX]{x1D714}-\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{v}=0$ , the resonant particle sees a constant phase and can be steadily accelerated or decelerated by the waves electric field. When there are more resonant particles giving energy to the wave due to deceleration than particles extracting energy from the wave due to acceleration, the wave can grow; and there is resonant instability. The slope of the velocity distribution dictates whether there are more or fewer particles faster or slower than the wave phase velocity. If the slope of the velocity distribution is negative, then there are more particles with velocities slower than the wave phase velocity that are accelerated by the wave; and the wave experiences resonant (Landau) damping.

We can consider the resonant particle interaction with a wave from the quantum mechanical perspective. Suppose there are exchanges of momentum and energy with the wave given by $\unicode[STIX]{x0394}\boldsymbol{p}=\pm \hbar \boldsymbol{k}$ and $\unicode[STIX]{x0394}\unicode[STIX]{x03B5}=\pm \hbar \unicode[STIX]{x1D714}$ that are small compared to the particle’s momentum and energy ( $h$ is Planck’s constant, and the over bar indicates $h/2\unicode[STIX]{x03C0}$ ). Then $\unicode[STIX]{x0394}\unicode[STIX]{x03B5}=m\boldsymbol{v}\boldsymbol{\cdot }\unicode[STIX]{x0394}\boldsymbol{v}=\boldsymbol{v}\boldsymbol{\cdot }\unicode[STIX]{x0394}\boldsymbol{p}$ , and substituting the expressions for $\unicode[STIX]{x0394}\boldsymbol{p}$ and $\unicode[STIX]{x0394}\unicode[STIX]{x03B5}$ one immediately obtains $\unicode[STIX]{x1D714}=\boldsymbol{k}\boldsymbol{\cdot }\boldsymbol{v}$ , which is the resonance condition.

[Editor’s note: there is a rich literature regarding physical considerations associated with Landau damping and interpretations of Landau damping in terms of the exchange of energy between resonant particles and waves. No attempt here is made to review the literature on Landau damping, but we give two illustrative examples: D.G. Swanson addresses the applicability of energy-related arguments to linear Landau damping in §§ 4.2.3 and 4.2.4 D.G. Swanson, Plasma Waves, Second Edition, Series in Plasma Physics, Institute of Physics Publishing, 2003; and D.D. Ryutov gives an historical perspective on Landau damping in D.D. Ryutov, Landau damping: half a century with the great discovery, Plasma Phys. Control. Fusion 41, A1–A12 (1999).]

3 Vlasov–Maxwell plasma formulation

In this section the totality of Maxwell’s equations are introduced in the context of a collisionless Vlasov plasma theory.

3.1 Wave energy and Poynting theorem

Maxwell’s equations in a plasma take the form

(3.1) $$\begin{eqnarray}\unicode[STIX]{x1D735}\times \boldsymbol{B}-\frac{1}{c}\frac{\unicode[STIX]{x2202}\boldsymbol{E}}{\unicode[STIX]{x2202}t}=\frac{4\unicode[STIX]{x03C0}}{c}\boldsymbol{j}\quad \unicode[STIX]{x1D735}\times \boldsymbol{E}+\frac{1}{c}\frac{\unicode[STIX]{x2202}\boldsymbol{B}}{\unicode[STIX]{x2202}t}=0,\end{eqnarray}$$

where the current $\boldsymbol{j}$ is the sum of externally applied currents and the currents due to free charges in the plasma. We will ignore gravity here but otherwise will be as general as possible. We introduce a notation to emphasize that the wave phenomena are spatially and temporally varying perturbations $\unicode[STIX]{x1D6FF}\boldsymbol{B},\unicode[STIX]{x1D6FF}\boldsymbol{E}$ and $\unicode[STIX]{x1D6FF}\boldsymbol{j}$ . If we compute the dot product of $c\unicode[STIX]{x1D6FF}\boldsymbol{E}$ with the perturbed Ampere’s law and combine with the dot product of $c\unicode[STIX]{x1D6FF}\boldsymbol{B}$ with Faraday’s law, we obtain

(3.2) $$\begin{eqnarray}-\!\unicode[STIX]{x1D735}\boldsymbol{\cdot }\left(\frac{c}{4\unicode[STIX]{x03C0}}\unicode[STIX]{x1D6FF}\boldsymbol{E}\times \unicode[STIX]{x1D6FF}\boldsymbol{B}\right)=\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\left(\frac{\left|\unicode[STIX]{x1D6FF}\boldsymbol{E}\right|^{2}}{8\unicode[STIX]{x03C0}}+\frac{\left|\unicode[STIX]{x1D6FF}\boldsymbol{B}\right|^{2}}{8\unicode[STIX]{x03C0}}\right)+\unicode[STIX]{x1D6FF}\boldsymbol{j}\boldsymbol{\cdot }\unicode[STIX]{x1D6FF}\boldsymbol{E}.\end{eqnarray}$$

The left-hand side of (3.2) is the divergence of the Poynting flux. The last term on the right is the rate of work done by the electromagnetic fields on the particles. In obtaining (3.2) only terms that are bilinear in the perturbed fields and currents are retained, and linear terms like $\unicode[STIX]{x1D6FF}\boldsymbol{B}\boldsymbol{\cdot }\boldsymbol{B}_{0}$ are dropped because they have no finite average value when time averaged over the cycle of the perturbed fields.

Compute the volume integral of (3.2) with vanishing surface terms at $\infty$ to obtain

(3.3) $$\begin{eqnarray}\frac{\text{d}}{\text{d}t}\int \text{d}^{3}\boldsymbol{r}\frac{\left|\unicode[STIX]{x1D6FF}\boldsymbol{E}\right|^{2}+\left|\unicode[STIX]{x1D6FF}\boldsymbol{B}\right|^{2}}{8\unicode[STIX]{x03C0}}=-\int \text{d}^{3}\boldsymbol{r}\unicode[STIX]{x1D6FF}\boldsymbol{j}\boldsymbol{\cdot }\unicode[STIX]{x1D6FF}\boldsymbol{E}.\end{eqnarray}$$

Equation (3.3) describes a balance between the change of stored electromagnetic field energy and work done by the electric field on the plasma.

3.2 Conductivity tensor

The conductivity tensor relates the perturbed current $\unicode[STIX]{x1D6FF}\boldsymbol{j}$ to the perturbed electric field $\unicode[STIX]{x1D6FF}\boldsymbol{E}$ . This is a linear relation. The analytic construction of the conductivity tensor typically begins with the assumption that the unperturbed system is stationary, but there are notable exceptions, e.g. when there are slowly varying background fields as in the solar wind and the ionosphere, or when there are rapidly oscillatory finite-amplitude fields as in laser–plasma interactions. As in the development of the dielectric response in § 2.5, we begin by employing causality and writing down the general expression

(3.4) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\boldsymbol{j}(r,t)=\int _{0}^{\infty }\text{d}\unicode[STIX]{x1D70F}\int \text{d}^{3}\boldsymbol{r}^{\prime }\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}(\boldsymbol{r};\boldsymbol{r}^{\prime };\unicode[STIX]{x1D70F})\boldsymbol{\cdot }\unicode[STIX]{x1D6FF}\boldsymbol{E}(\boldsymbol{r}^{\prime },t-\unicode[STIX]{x1D70F}).\end{eqnarray}$$

Next we assume that perturbed electric field can be represented as a superposition of normal modes,

(3.5) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\boldsymbol{E}(\boldsymbol{r},t)=\mathop{\sum }_{\ell }\exp (-\text{i}\unicode[STIX]{x1D714}_{\ell }t)\boldsymbol{E}_{\ell }(\boldsymbol{r},t),\end{eqnarray}$$

where $\unicode[STIX]{x1D714}_{\ell }$ is real, $\boldsymbol{E}_{\ell }$ is slowly varying in time, $\unicode[STIX]{x1D714}_{-\ell }=-\unicode[STIX]{x1D714}_{\ell }$ and $\boldsymbol{E}_{-\ell }=-\boldsymbol{E}_{\ell }$ . Use of (3.5) in (3.4) yields

(3.6) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FF}\boldsymbol{j}(r,t) & = & \displaystyle \mathop{\sum }_{\ell }\exp (-\text{i}\unicode[STIX]{x1D714}_{\ell }t)\int _{0}^{\infty }\text{d}\unicode[STIX]{x1D70F}\int \text{d}^{3}\boldsymbol{r}^{\prime }\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}(\boldsymbol{r};\boldsymbol{r}^{\prime };\unicode[STIX]{x1D70F})\boldsymbol{\cdot }\boldsymbol{E}_{\ell }(\boldsymbol{r}^{\prime },t-\unicode[STIX]{x1D70F})\exp (\text{i}\unicode[STIX]{x1D714}_{\ell }\unicode[STIX]{x1D70F})\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{\ell }\exp (-\text{i}\unicode[STIX]{x1D714}_{\ell }t)\int _{0}^{\infty }\text{d}\unicode[STIX]{x1D70F}\int \text{d}^{3}\boldsymbol{r}^{\prime }\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}(\boldsymbol{r};\boldsymbol{r}^{\prime };\unicode[STIX]{x1D70F})\nonumber\\ \displaystyle & & \displaystyle \boldsymbol{\cdot }\left[\boldsymbol{E}_{\ell }(\boldsymbol{r}^{\prime },t)-\unicode[STIX]{x1D70F}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\boldsymbol{E}_{\ell }(\boldsymbol{r}^{\prime },t)+O(\unicode[STIX]{x1D70F}^{2}\ddot{\boldsymbol{E}}_{\ell })\right]\exp (\text{i}\unicode[STIX]{x1D714}_{\ell }\unicode[STIX]{x1D70F})\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{\ell }\exp (-\text{i}\unicode[STIX]{x1D714}_{\ell }t)\int \text{d}^{3}\boldsymbol{r}^{\prime }\nonumber\\ \displaystyle & & \displaystyle \times \left[\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}(\boldsymbol{r};\boldsymbol{r}^{\prime };\unicode[STIX]{x1D714}_{\ell })\boldsymbol{\cdot }\boldsymbol{E}_{\ell }(\boldsymbol{r}^{\prime },t)+\text{i}\left.\frac{\unicode[STIX]{x2202}\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}(\boldsymbol{r};\boldsymbol{r}^{\prime };\unicode[STIX]{x1D714})}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\right|_{\unicode[STIX]{x1D714}_{\ell }}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\boldsymbol{E}_{\ell }(\boldsymbol{r}^{\prime },t)+\cdots \,\right]\!,\end{eqnarray}$$

where we have made use of the following definition and identity.

Definition. The Fourier transformed conductivity tensor is given by

$$\begin{eqnarray}\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}(\boldsymbol{r},\boldsymbol{r}^{\prime };\unicode[STIX]{x1D714})=\int _{0}^{\infty }\text{d}\unicode[STIX]{x1D70F}\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}(\boldsymbol{r},\boldsymbol{r}^{\prime };\unicode[STIX]{x1D70F})\exp (\text{i}\unicode[STIX]{x1D714}\unicode[STIX]{x1D70F})\quad \text{with }\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}(\boldsymbol{r},\boldsymbol{r}^{\prime };\unicode[STIX]{x1D70F}<0)=0.\end{eqnarray}$$

Theorem.

$$\begin{eqnarray}(\unicode[STIX]{x2202}/\unicode[STIX]{x2202}\unicode[STIX]{x1D714})F_{\unicode[STIX]{x1D70F}\rightarrow \unicode[STIX]{x1D714}}(g(\unicode[STIX]{x1D70F}))=F_{\unicode[STIX]{x1D70F}\rightarrow \unicode[STIX]{x1D714}}(\text{i}\unicode[STIX]{x1D70F}g(\unicode[STIX]{x1D70F}))\end{eqnarray}$$

where $F_{\unicode[STIX]{x1D70F}\rightarrow \unicode[STIX]{x1D714}}(g(\unicode[STIX]{x1D70F}))$ is the Fourier transform from $\unicode[STIX]{x1D70F}$ to $\unicode[STIX]{x1D714}$ .

The conductivity is a three-dimensional tensor and a kernel, but it is not Hermitian in general. One can decompose any tensor into Hermitian and anti-Hermitian parts,

Theorem.

$$\begin{eqnarray}\boldsymbol{A}=\boldsymbol{A}^{\unicode[STIX]{x1D704}}+\text{i}\boldsymbol{A}^{\unicode[STIX]{x1D704}\unicode[STIX]{x1D704}}\quad A_{\unicode[STIX]{x1D707}\unicode[STIX]{x1D708}}^{\unicode[STIX]{x1D704}}=\frac{1}{2}(A_{\unicode[STIX]{x1D707}\unicode[STIX]{x1D708}}+A_{\unicode[STIX]{x1D707}\unicode[STIX]{x1D708}}^{\ast })~A_{\unicode[STIX]{x1D707}\unicode[STIX]{x1D708}}^{\unicode[STIX]{x1D704}}=(A_{\unicode[STIX]{x1D708}\unicode[STIX]{x1D707}}^{\unicode[STIX]{x1D704}})^{\ast }\quad A_{\unicode[STIX]{x1D707}\unicode[STIX]{x1D708}}^{\unicode[STIX]{x1D704}\unicode[STIX]{x1D704}}=(A_{\unicode[STIX]{x1D708}\unicode[STIX]{x1D707}}^{\unicode[STIX]{x1D704}\unicode[STIX]{x1D704}})^{\ast }.\end{eqnarray}$$

For real $\unicode[STIX]{x1D714}$ , $\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D707}\unicode[STIX]{x1D708}}^{\unicode[STIX]{x1D704}}(\boldsymbol{r},\boldsymbol{r}^{\prime };\unicode[STIX]{x1D714})\equiv \frac{1}{2}(\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D707}\unicode[STIX]{x1D708}}(\boldsymbol{r},\boldsymbol{r}^{\prime };\unicode[STIX]{x1D714})+\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D708}\unicode[STIX]{x1D707}}(\boldsymbol{r}^{\prime },\boldsymbol{r};\unicode[STIX]{x1D714}))$ .

Use of (3.6) in (3.3) yields

(3.7) $$\begin{eqnarray}\displaystyle \frac{\text{d}}{\text{d}t}\int \text{d}^{3}\boldsymbol{r}\frac{|\unicode[STIX]{x1D6FF}\boldsymbol{E}|^{2}+|\unicode[STIX]{x1D6FF}\boldsymbol{B}|^{2}}{8\unicode[STIX]{x03C0}} & = & \displaystyle -\int \text{d}^{3}\boldsymbol{r}\unicode[STIX]{x1D6FF}\boldsymbol{j}\boldsymbol{\cdot }\unicode[STIX]{x1D6FF}\boldsymbol{E}\nonumber\\ \displaystyle & = & \displaystyle -\int \text{d}^{3}\boldsymbol{r}\int \text{d}^{3}\boldsymbol{r}^{\prime }\mathop{\sum }_{\ell }\exp (\text{i}\unicode[STIX]{x1D714}_{\ell }t)\boldsymbol{E}_{\ell }^{\ast }(\boldsymbol{r},t)\boldsymbol{\cdot }\mathop{\sum }_{\ell ^{\prime }}\exp (-\text{i}\unicode[STIX]{x1D714}_{\ell ^{\prime }}t)\nonumber\\ \displaystyle & & \displaystyle \times \,\left[\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}(\boldsymbol{r};\boldsymbol{r}^{\prime };\unicode[STIX]{x1D714}_{\ell ^{\prime }})\boldsymbol{\cdot }\boldsymbol{E}_{\ell ^{\prime }}(\boldsymbol{r}^{\prime },t)+\text{i}\left.\frac{\unicode[STIX]{x2202}\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}(\boldsymbol{r};\boldsymbol{r}^{\prime };\unicode[STIX]{x1D714})}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\right|_{\unicode[STIX]{x1D714}_{\ell ^{\prime }}}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\boldsymbol{E}_{\ell ^{\prime }}(\boldsymbol{r}^{\prime },t)\right]\!.\hspace{-10.00002pt}\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

If we perform a coarse-grain time average of (3.7), i.e. we average over the characteristic time scales ${\sim}1/\unicode[STIX]{x1D714}_{\ell }$ , but retain the slow time scale variations, and make use of

$$\begin{eqnarray}\left\langle \exp \left(-\text{i}(\unicode[STIX]{x1D714}_{\ell }-\unicode[STIX]{x1D714}_{\ell ^{\prime }})t\right)\right\rangle =\unicode[STIX]{x1D6FF}_{\ell \ell ^{\prime }}\end{eqnarray}$$

we can delete the sum over $\ell ^{\prime }$ and set $\ell =\ell ^{\prime }$ . Thus, (3.7) simplifies to

(3.8) $$\begin{eqnarray}\displaystyle \frac{\text{d}}{\text{d}t}\int \text{d}^{3}\boldsymbol{r}\frac{\left|\unicode[STIX]{x1D6FF}\boldsymbol{E}\right|^{2}+\left|\unicode[STIX]{x1D6FF}\boldsymbol{B}\right|^{2}}{8\unicode[STIX]{x03C0}} & = & \displaystyle -\int \text{d}^{3}\boldsymbol{r}\unicode[STIX]{x1D6FF}\boldsymbol{j}\boldsymbol{\cdot }\unicode[STIX]{x1D6FF}\boldsymbol{E}\nonumber\\ \displaystyle & = & \displaystyle -\int \text{d}^{3}\boldsymbol{r}\int \text{d}^{3}\boldsymbol{r}^{\prime }\mathop{\sum }_{\ell }\boldsymbol{E}_{\ell }^{\ast }(\boldsymbol{r},t)\nonumber\\ \displaystyle & & \displaystyle \boldsymbol{\cdot }\left[\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}(r,r^{\prime };\unicode[STIX]{x1D714}_{\ell })\boldsymbol{\cdot }\boldsymbol{E}_{\ell }(\boldsymbol{r}^{\prime };t)+\text{i}\left.\frac{\unicode[STIX]{x2202}\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}(\boldsymbol{r};\boldsymbol{r}^{\prime };\unicode[STIX]{x1D714})}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\right|_{\unicode[STIX]{x1D714}_{\ell }}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\boldsymbol{E}_{\ell }(\boldsymbol{r}^{\prime },t)\right]\nonumber\\ \displaystyle & = & \displaystyle -\int \text{d}^{3}\boldsymbol{r}\int \text{d}^{3}\boldsymbol{r}^{\prime }\mathop{\sum }_{\ell }\boldsymbol{E}_{\ell }^{\ast }(\boldsymbol{r},t)\nonumber\\ \displaystyle & & \displaystyle \boldsymbol{\cdot }\left[\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}_{R}(r,r^{\prime };\unicode[STIX]{x1D714}_{\ell })\boldsymbol{\cdot }\boldsymbol{E}_{\ell }(\boldsymbol{r}^{\prime };t)-\left.\frac{\unicode[STIX]{x2202}\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}_{I}(\boldsymbol{r};\boldsymbol{r}^{\prime };\unicode[STIX]{x1D714})}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\right|_{\unicode[STIX]{x1D714}_{\ell }}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\boldsymbol{E}_{\ell }(\boldsymbol{r}^{\prime },t)\right]\nonumber\\ \displaystyle & = & \displaystyle -\int \text{d}^{3}\boldsymbol{r}\int \text{d}^{3}\boldsymbol{r}^{\prime }\mathop{\sum }_{\ell }\boldsymbol{E}_{\ell }^{\ast }(\boldsymbol{r},t)\boldsymbol{\cdot }\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}_{R}(r,r^{\prime };\unicode[STIX]{x1D714}_{\ell })\boldsymbol{\cdot }\boldsymbol{E}_{\ell }(\boldsymbol{r}^{\prime };t)\nonumber\\ \displaystyle & & \displaystyle +\,\frac{1}{2}\frac{\text{d}}{\text{d}t}\int \text{d}^{3}\boldsymbol{r}\int \text{d}^{3}\boldsymbol{r}^{\prime }\mathop{\sum }_{\ell }\boldsymbol{E}_{\ell }^{\ast }(\boldsymbol{r},t)\boldsymbol{\cdot }\left.\frac{\unicode[STIX]{x2202}\overset{\leftrightarrow }{\unicode[STIX]{x1D70E}}_{I}(\boldsymbol{r};\boldsymbol{r}^{\prime };\unicode[STIX]{x1D714})}{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}\right|_{\unicode[STIX]{x1D714}_{\ell }}\boldsymbol{\cdot }\boldsymbol{E}_{\ell }(\boldsymbol{r}^{\prime },t),\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where $\unicode[STIX]{x1D748}=\unicode[STIX]{x1D748}_{R}+\text{i}\unicode[STIX]{x1D748}_{I}$ .

Exercise. Prove (3.8) using the hermiticity of $\unicode[STIX]{x1D748}_{R}$ and $\unicode[STIX]{x1D748}_{I}$ , and the right-hand side of (3.8) is necessarily real.

3.3 Energy conservation

Theorem. Equation (3.8) can be rewritten in the form that expresses energy conservation between the electromagnetic field energy and the plasma

(3.9)