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

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.

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

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.

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.

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. 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.

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.

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

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.

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

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 Reference Sturrock1958). Sturrock’s method is reviewed in (Clemmow & Dougherty Reference Clemmow and Dougherty1989) and was assigned as reading but not covered in class lectures.

2.6.4 Bers and Briggs’ method for analysing absolute instability

Bers (Reference Bers, Auer and Cap1973) and Briggs (Reference Briggs1964) 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).

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.

2.9.1 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.

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 Reference Lighthill1958),

(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.

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

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

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

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.

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).]

4.2.1 Right-hand circularly polarized waves: whistler, magnetosonic and extraordinary waves

Consider a right-circularly polarized wave, $l=+1$ , with $\unicode[STIX]{x1D714}>0$ and no collisions $\unicode[STIX]{x1D708}=0$ : the dispersion relation is

(4.13) $$\begin{eqnarray}\displaystyle n_{R}^{2}(\unicode[STIX]{x1D714})=\unicode[STIX]{x1D700}_{+}(\unicode[STIX]{x1D714})=1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D6FA}_{e})(\unicode[STIX]{x1D714}+|\unicode[STIX]{x1D6FA}_{i}|)}. & & \displaystyle\end{eqnarray}$$

Figures 14 and 15 sketch solutions for the dispersion relation for right-circularly polarized waves propagating parallel to a uniform applied magnetic field in a cold plasma. The cutoff frequency $\unicode[STIX]{x1D714}_{1}=\unicode[STIX]{x1D714}_{RC}$ in figure 15.

Figure 14. Schematic of the solutions of (4.13) for waves right-circularly polarized wave propagating parallel to $\boldsymbol{B}_{0}$ .

Figure 15. Parallel modes with positive helicity. Electron cyclotron resonance $(k_{\Vert }\rightarrow \infty )$ occurs at $\unicode[STIX]{x1D714}\lesssim |\unicode[STIX]{x1D6FA}_{e}|$ . The non-dispersive low-frequency mode $\unicode[STIX]{x1D714}\ll \unicode[STIX]{x1D6FA}_{i}(\ll |\unicode[STIX]{x1D6FA}_{e}|)$ is the Alfvén mode described by $\unicode[STIX]{x1D714}=k_{\Vert }V_{A}$ (due to A. Hirose). A. Hirose, Physics 862 lecture notes, chap. 6, University of Saskatchewan; http://physics.usask.ca/∼hirose/P862/notes.htm.

4.2.2 Left-hand circularly polarized waves: Alfvén, ion cyclotron and ordinary

Consider left-circularly polarized waves propagating parallel to the magnetic field. Equation (4.8) yields the following dispersion relation.

Theorem. Left-circularly polarized waves, $\ell =-1$ , with $\unicode[STIX]{x1D714}>0$ and no collisions $\unicode[STIX]{x1D708}=0$ , resonate with ions and satisfy the dispersion relation

(4.14) $$\begin{eqnarray}\displaystyle n_{R}^{2}(\unicode[STIX]{x1D714})=\unicode[STIX]{x1D700}_{-}(\unicode[STIX]{x1D714})=1-\frac{\unicode[STIX]{x1D714}_{p}^{2}}{(\unicode[STIX]{x1D714}+\unicode[STIX]{x1D6FA}_{e})(\unicode[STIX]{x1D714}-|\unicode[STIX]{x1D6FA}_{i}|)}. & & \displaystyle\end{eqnarray}$$

Figures 16 and 17 sketch the solutions of the dispersion relation in (4.14). $\unicode[STIX]{x1D714}_{2}$ is the left-circularly polarized cutoff frequency $\unicode[STIX]{x1D714}_{2}=\unicode[STIX]{x1D714}_{LC}=\sqrt{(\unicode[STIX]{x1D6FA}_{e}/2)^{2}+\unicode[STIX]{x1D714}_{p}^{2}}-|\unicode[STIX]{x1D6FA}_{e}|/2$ .

Figure 16. Schematic of the solutions of (4.14) for waves left-circularly polarized wave propagating parallel to $\boldsymbol{B}_{0}\boldsymbol{\cdot }\unicode[STIX]{x1D714}_{LC}=\unicode[STIX]{x1D714}_{2}$ .

Figure 17. Parallel modes with negative helicity. The Alfvén mode $\unicode[STIX]{x1D714}=V_{A}k_{\Vert }$ with negative helicity exists in the low-frequency region $\unicode[STIX]{x1D714}\ll \unicode[STIX]{x1D6FA}_{i}$ . The ion-cyclotron resonance occurs at $\unicode[STIX]{x1D714}\lesssim \unicode[STIX]{x1D6FA}_{i}$ . The cutoff frequency $\unicode[STIX]{x1D714}_{2}$ is given by $\unicode[STIX]{x1D714}_{2}=(\sqrt{4\unicode[STIX]{x1D714}_{\text{pe}}^{2}}-|\unicode[STIX]{x1D6FA}_{e}|)/2$ (due to A. Hirose). A. Hirose, Physics 862 lecture notes, chap. 6, University of Saskatchewan; http://physics.usask.ca/∼hirose/P862/notes.htm.

4.2.3 Electron plasma waves

With $\ell =0$ the solution of (4.8) yields longitudinal waves with $\unicode[STIX]{x1D6FF}\boldsymbol{E}||\boldsymbol{B}_{0},\boldsymbol{k}$ and $\unicode[STIX]{x1D700}_{0}(\unicode[STIX]{x1D714})=0=1-\unicode[STIX]{x1D714}_{p}^{2}/\unicode[STIX]{x1D714}^{2}\rightarrow \unicode[STIX]{x1D714}=\pm \unicode[STIX]{x1D714}_{p}$ , i.e. electron plasma waves.

4.5.1 WKB eikonal method

In the WKB (Wentzel–Kramers–Brillouin or Wentzel–Kramers–Brillouin–Jeffreys) method, all of the field quantities are decomposed in terms of a slowly varying amplitude and a rapidly varying phase factor,

Definitions. $\unicode[STIX]{x1D6FF}\boldsymbol{E}(\boldsymbol{x})=A(\boldsymbol{x})\hat{\boldsymbol{e}}(\boldsymbol{x})\exp (\text{i}\unicode[STIX]{x1D6F7}(\boldsymbol{x}))$

(4.22) $$\begin{eqnarray}\displaystyle \begin{array}{@{}l@{}}\displaystyle \text{(i) }\hat{\boldsymbol{e}}(\boldsymbol{x})\text{ is the complex-valued polarization vector and is slowly varying.}\\ \displaystyle \text{(ii) }A(\boldsymbol{x})\text{ is a slowly varying amplitude.}\\ \displaystyle \text{(iii) }\exp (\text{i}\unicode[STIX]{x1D6F7}(\boldsymbol{x}))\text{ is a phase factor that is a rapidly varying function of }\boldsymbol{x}.\\ \displaystyle \text{(iv) The slowly varying spatial scale is }L\gg \unicode[STIX]{x1D706}\text{the rapidly varying spatial scale}.\qquad \end{array} & & \displaystyle\end{eqnarray}$$

Using the definitions above we obtain

(4.23) $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D735}\times \unicode[STIX]{x1D6FF}\boldsymbol{E}\approx \text{i}\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}\times \unicode[STIX]{x1D6FF}\boldsymbol{E}=\text{i}\boldsymbol{k}(\boldsymbol{x})\times \unicode[STIX]{x1D6FF}\boldsymbol{E}(\boldsymbol{x})\nonumber\\ \displaystyle & & \displaystyle \unicode[STIX]{x1D735}\times \unicode[STIX]{x1D735}\times \unicode[STIX]{x1D6FF}\boldsymbol{E}=\unicode[STIX]{x1D735}(\unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D6FF}\boldsymbol{E})-\unicode[STIX]{x1D6FB}^{2}\unicode[STIX]{x1D6FF}\boldsymbol{E}\approx \text{i}\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}\times (\text{i}\boldsymbol{k}\times \unicode[STIX]{x1D6FF}\boldsymbol{E})\nonumber\\ \displaystyle & & \displaystyle \quad =-\boldsymbol{k}\times \boldsymbol{k}\times \unicode[STIX]{x1D6FF}\boldsymbol{E}=-\boldsymbol{k}\boldsymbol{k}\boldsymbol{\cdot }\unicode[STIX]{x1D6FF}\boldsymbol{E}+k^{2}\unicode[STIX]{x1D6FF}\boldsymbol{E}.\end{eqnarray}$$

Theorem. Using (4.5.2), (4.5.3) and (4.5.4), it follows that

(4.24) $$\begin{eqnarray}\displaystyle [\stackrel{\leftrightarrow }{\unicode[STIX]{x1D700}}(\unicode[STIX]{x1D714},\boldsymbol{x})-n^{2}(\boldsymbol{x})(\stackrel{\leftrightarrow }{\unicode[STIX]{x1D644}}-\hat{\boldsymbol{k}}(\boldsymbol{x})\hat{\boldsymbol{k}}(\boldsymbol{x}))]\boldsymbol{\cdot }\unicode[STIX]{x1D6FF}\boldsymbol{E}(\boldsymbol{x})\equiv \,\stackrel{\leftrightarrow }{\boldsymbol{K}}(\unicode[STIX]{x1D714},\boldsymbol{k}(\boldsymbol{x}),\boldsymbol{x})\boldsymbol{\cdot }\unicode[STIX]{x1D6FF}\boldsymbol{E}(\boldsymbol{x})=0, & & \displaystyle\end{eqnarray}$$

where $n^{2}\equiv k^{2}c^{2}/\unicode[STIX]{x1D714}^{2}$ and we require that $D(\unicode[STIX]{x1D714},\boldsymbol{k}(\boldsymbol{x}),\boldsymbol{x})\equiv \text{det}|\stackrel{\leftrightarrow }{\boldsymbol{K}}|=0$ .

Introduce a variation in the transverse direction $x$ in the example of the plane-stratified medium considered above. Fourier analyse in $x$ and introduce $k_{x}$ . From the eikonal dispersion relation one obtains

(4.25) $$\begin{eqnarray}\displaystyle \frac{\text{d}z}{\text{d}t} & = & \displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}{\unicode[STIX]{x2202}k_{z}}=\frac{k_{z}(z)c^{2}}{\unicode[STIX]{x1D714}}=c\sqrt{1-\frac{\unicode[STIX]{x1D714}_{p}^{2}(z)}{\unicode[STIX]{x1D714}^{2}}-\frac{k_{x}^{2}c^{2}}{\unicode[STIX]{x1D714}^{2}}}\nonumber\\ \displaystyle \frac{\text{d}x}{\text{d}t} & = & \displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}{\unicode[STIX]{x2202}k_{x}}=\frac{k_{x}c^{2}}{\unicode[STIX]{x1D714}}=c\sqrt{1-\frac{\unicode[STIX]{x1D714}_{p}^{2}(z)}{\unicode[STIX]{x1D714}^{2}}-\frac{k_{z}^{2}c^{2}}{\unicode[STIX]{x1D714}^{2}}}\nonumber\\ \displaystyle \frac{\text{d}z}{\text{d}x} & = & \displaystyle \frac{k_{z}(z)}{k_{x}}.\end{eqnarray}$$

Theorem. The eikonal dispersion relation $D(\unicode[STIX]{x1D714},\boldsymbol{k}(\boldsymbol{x}),\boldsymbol{x})=0$ is equivalent to $\unicode[STIX]{x1D714}(\boldsymbol{k},\boldsymbol{x})=\unicode[STIX]{x1D714}_{0}$ and one can insist the total time derivative of $\unicode[STIX]{x1D714}$ vanishes,

(4.26) $$\begin{eqnarray}\displaystyle & & \displaystyle 0=\frac{\text{d}\unicode[STIX]{x1D714}(\boldsymbol{k},\boldsymbol{x})}{\text{d}t}=\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}{\unicode[STIX]{x2202}\boldsymbol{k}}\boldsymbol{\cdot }\frac{\text{d}\boldsymbol{k}}{\text{d}t}+\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}{\unicode[STIX]{x2202}\boldsymbol{x}}\boldsymbol{\cdot }\frac{\text{d}\boldsymbol{x}}{\text{d}t}\rightarrow \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}{\unicode[STIX]{x2202}\boldsymbol{k}}\boldsymbol{\cdot }\left(-\frac{\text{d}\unicode[STIX]{x1D714}}{\text{d}\boldsymbol{x}}\right)+\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}{\unicode[STIX]{x2202}\boldsymbol{x}}\boldsymbol{\cdot }\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}{\unicode[STIX]{x2202}\boldsymbol{k}}\equiv 0\nonumber\\ \displaystyle & & \displaystyle \text{if }\frac{\text{d}\boldsymbol{x}}{\text{d}t}=\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}{\unicode[STIX]{x2202}\boldsymbol{k}}\text{ and }\frac{\text{d}\boldsymbol{k}}{\text{d}t}=-\frac{\text{d}\unicode[STIX]{x1D714}}{\text{d}\boldsymbol{ x}},\text{ the WKBJ wave packet equations}.\end{eqnarray}$$

For only $z$ dependence, $(\text{d}\boldsymbol{k}/\text{d}t)=(\text{d}k_{z}/\text{d}t)\hat{\boldsymbol{k}}=-(\text{d}\unicode[STIX]{x1D714}/\text{d}z)\hat{\boldsymbol{k}}$ . Because $(\text{d}\unicode[STIX]{x1D714}/\text{d}x)=0$ then $(\text{d}k_{x}/\text{d}t)=0$ , i.e. $k_{x}$ is a constant. The equations in (4.26) are Hamilton’s equations. We note that by multiplying the wave packet equations of motion through by $\hbar$ we have the equations of motion for quantum mechanical wave packets. If we identify the Hamiltonian $H=\hbar \unicode[STIX]{x1D714}$ , we note that $(\text{d}H/\text{d}t)=(\unicode[STIX]{x2202}H/\unicode[STIX]{x2202}t)=\hbar (\unicode[STIX]{x2202}\unicode[STIX]{x1D714}(\boldsymbol{k},\boldsymbol{x},t)/\unicode[STIX]{x2202}t)$ . There is a nice discussion of the WKBJ wave packet equations in T.H. Stix, The Theory of Plasma Waves, 1962.

Consider an infinite-length wave train in the absence of sources and sinks. Then energy flux conservation dictates

(4.27) $$\begin{eqnarray}\displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}U(z)=-\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}z}S_{z}=0,\quad S_{z}=\text{const.}=U(z)\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D714}}{\unicode[STIX]{x2202}k}(z), & & \displaystyle\end{eqnarray}$$

where $U(z)$ is the wave energy density, and $S_{z}$ is the $z$ -component of the wave energy flux. In consequence of (4.27), the square of the wave packet amplitude scales as $A^{2}\sim U(z)\sim S/|\unicode[STIX]{x2202}\unicode[STIX]{x1D714}/\unicode[STIX]{x2202}k_{z}|\sim \text{const.}/|\unicode[STIX]{x2202}\unicode[STIX]{x1D714}/\unicode[STIX]{x2202}k_{z}|$ ; thus, $A(z)\sim 1/\sqrt{|}\unicode[STIX]{x2202}\unicode[STIX]{x1D714}/\unicode[STIX]{x2202}k_{z}|$ , which is useful up to a reflection point where $S_{z}$ and $\unicode[STIX]{x2202}\unicode[STIX]{x1D714}/\unicode[STIX]{x2202}k_{z}$ vanish.

4.5.2 Reflection, refraction, turning points, Bohr–Sommerfeld quantization

In WKB either reflection or refraction only can occur. Partial reflection, refraction, and transmission at a sharp boundary like a glass plate cannot be described by WKB. How good is WKB? WKB can be applied to a medium with spatial variation that is slow on the scale of the characteristic wavelength of the wave. Consider a one-dimensional example (ref. Landau and Lifshitz, Quantum Mechanics, Sec. 23, Problem 3) wherein there is a transition in the square of the index of refraction from a higher value $n_{0}$ to a lower value $n_{1}$ over a length scale $L$ . Assume the dispersion relation for an electromagnetic wave in an unmagnetized spatially varying plasma dielectric varies as

(4.28) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D700}(x)=\frac{n_{0}^{2}+n_{1}^{2}e^{x/L}}{1+e^{x/L}}. & & \displaystyle\end{eqnarray}$$

Theorem. For a wave packet incident from $+\infty$ in the profile in (4.28), the reflection coefficient, defined as the reflected power divided by the incident power, is given by (Eckart Reference Eckart1930)

(4.29) $$\begin{eqnarray}\displaystyle R=\frac{\sinh ^{2}\left[\unicode[STIX]{x03C0}{\displaystyle \frac{L}{\bar{\unicode[STIX]{x1D706}}}}(n_{0}-n_{1})\right]}{\sinh ^{2}\left[\unicode[STIX]{x03C0}{\displaystyle \frac{L}{\bar{\unicode[STIX]{x1D706}}}}(n_{0}+n_{1})\right]},\quad \bar{\unicode[STIX]{x1D706}}\equiv \frac{c}{\unicode[STIX]{x1D714}}. & & \displaystyle\end{eqnarray}$$

The WKB limit corresponds to $L\gg \bar{\unicode[STIX]{x1D706}}$ , and $R\rightarrow \text{e}^{-4\unicode[STIX]{x03C0}L/\bar{\unicode[STIX]{x1D706}}}$ ; so the reflectivity is exponentially small in this limit. In the sharp-boundary limit, $L\ll \bar{\unicode[STIX]{x1D706}}$ , $R\rightarrow (n_{0}-n_{1})^{2}/(n_{0}+n_{1})^{2}$ ; and the WKB estimate is a very poor approximation.

There are many treatments of reflection of waves in spatially varying media. There are elegant discussions in J. Heading, Introduction to Phase-Integral Methods (Methuen, 1962); K. G. Budden, Radio Waves in the Ionosphere (Dover, 1961); and E. R. Tracy, A. J. Brizard, A. S. Richardson, and A. N. Kaufman, Ray Tracing and Beyond: Phase-space Methods in Plasma Wave Theory (Cambridge University Press, 2014). A simpler approach is presented in J. Mathews and R. Walker, Mathematical Methods of Physics, 2nd ed. (Benjamin, 1970). Here we follow the treatment in Mathews and Walker. Consider the one-dimensional limit of (4.23) and (4.24),

(4.30) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D735}\times (\unicode[STIX]{x1D735}\times \unicode[STIX]{x1D6FF}\boldsymbol{E}(\boldsymbol{x}))=\frac{\unicode[STIX]{x1D714}^{2}}{c^{2}}\stackrel{\leftrightarrow }{\unicode[STIX]{x1D700}}(\boldsymbol{x})\boldsymbol{\cdot }\unicode[STIX]{x1D6FF}\boldsymbol{E}(\boldsymbol{x})=\unicode[STIX]{x1D735}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D6FF}\boldsymbol{E}(\boldsymbol{x})-\unicode[STIX]{x1D6FB}^{2}\unicode[STIX]{x1D6FF}\boldsymbol{E}. & & \displaystyle\end{eqnarray}$$

Consider a purely transverse wave $\unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D6FF}\boldsymbol{E}=0$ in one dimension and introduce a specific polarization to obtain a scalar wave equation,

(4.31) $$\begin{eqnarray}\displaystyle -\frac{\text{d}^{2}}{\text{d}z^{2}}\unicode[STIX]{x1D6FF}E(z)=\frac{\unicode[STIX]{x1D714}^{2}}{c^{2}}\unicode[STIX]{x1D700}(z)\unicode[STIX]{x1D6FF}E(z)\rightarrow k(z)=\frac{2\unicode[STIX]{x03C0}}{\unicode[STIX]{x1D706}(z)}=\frac{\unicode[STIX]{x1D714}}{c}\sqrt{\unicode[STIX]{x1D700}(z)}. & & \displaystyle\end{eqnarray}$$

Figure 22 illustrates an electromagnetic wave incident from the right in a spatially varying medium with plasma cutoff at $z=0$ . Regions away from $z=0$ are accurately described by WKB. Near $z=0$ if the dielectric $\unicode[STIX]{x1D700}$ goes through zero linearly in $z$ , solving the differential equation in (4.31) yields an Airy function solution, and the Airy function is related to Bessel functions and modified Bessel functions of fractional order $1/3$ .

Figure 22. (a) Schematic of the dielectric function as function of position $z$ for a wave incident from the right with cutoff at $z=0$ . (b) Schematic of electromagnetic wave amplitude for wave incident from the right with cutoff at $z=0$ .

Example. Consider $\unicode[STIX]{x1D716}(z)=1-\unicode[STIX]{x1D714}_{p}^{2}(z)/\unicode[STIX]{x1D714}^{2}$ and assume for a model that the WKB dispersion relation is $\unicode[STIX]{x1D714}_{p}^{2}=-n_{1}^{2}\text{tanh}(z/L)\unicode[STIX]{x1D714}^{2}+\unicode[STIX]{x1D714}^{2}$ . For the moment select units such that $(\unicode[STIX]{x1D714}/c)=1$ and recall that the WKB solution for the phase function satisfies $\text{d}\unicode[STIX]{x1D6F7}/\text{d}z\equiv k(z)$ . The WKB solutions of (4.31) are then given by

(4.32) $$\begin{eqnarray}\displaystyle W_{\pm }(z)=\frac{1}{\unicode[STIX]{x1D700}^{1/4}(z)}\exp \left(\pm \text{i}\int _{0}^{z}\text{d}z^{\prime }\sqrt{\unicode[STIX]{x1D700}(z^{\prime })}\right). & & \displaystyle\end{eqnarray}$$

The asymptotic WKB solutions for the wave amplitude and phase far from $z=0$ are given by

(4.33) $$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}l@{}}\displaystyle z\geqslant \unicode[STIX]{x1D706}:\unicode[STIX]{x1D6FF}E(z)=\frac{\unicode[STIX]{x1D6FD}}{\unicode[STIX]{x1D700}^{1/4}(z)}\cos \left(\frac{\unicode[STIX]{x1D714}}{c}\int _{0}^{z}\text{d}z^{\prime }\sqrt{\unicode[STIX]{x1D700}(z^{\prime })}-\unicode[STIX]{x1D6FC}\right)\\[10.0pt] \displaystyle z\leqslant -\unicode[STIX]{x1D706}:\unicode[STIX]{x1D6FF}E(z)=\frac{1}{\unicode[STIX]{x1D700}^{1/4}(z)}\exp \left(-\frac{\unicode[STIX]{x1D714}}{c}\int _{z}^{0}\text{d}z^{\prime }\sqrt{|\unicode[STIX]{x1D700}(z^{\prime })|}\right).\end{array}\right\} & & \displaystyle\end{eqnarray}$$

The constants $\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D6FD}$ are to be determined. The normalization of the wave amplitude is arbitrary. We rewrite the differential equation (4.31) as

(4.34) $$\begin{eqnarray}\displaystyle \left[-\frac{\text{d}^{2}}{\text{d}z^{2}}+\frac{\unicode[STIX]{x1D714}^{2}}{c^{2}}(-\unicode[STIX]{x1D700}(z))\right]\unicode[STIX]{x1D6FF}E(z)=0, & & \displaystyle\end{eqnarray}$$

which is analogous to the quantum mechanical problem

(4.35) $$\begin{eqnarray}\displaystyle [H]\unicode[STIX]{x1D713}=[K+V]\unicode[STIX]{x1D713}=0. & & \displaystyle\end{eqnarray}$$

With $V\propto -\unicode[STIX]{x1D700}$ , figure 22(a) can be used as a diagram for $V(z)$ (with the sign flipped from that of $\unicode[STIX]{x1D700}(z)$ ). With total energy $E=0$ , the kinetic energy $K$ is positive for $z>0$ ; and there is a turning point at $z=0$ for $V(0)=0$ . We return to the WKB solution of (4.31) with units such that $\unicode[STIX]{x1D714}/c=1$ . Assume that

(4.36) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FF}E(z)=A_{+}(z)W_{+}(z)+A_{-}(z)W_{-}(z)\quad W_{\pm }\propto \exp \left(\mp \int _{0}^{z}\text{d}z^{\prime }|\unicode[STIX]{x1D716}|^{1/2}\right). & & \displaystyle\end{eqnarray}$$

Here $W_{+}$ is exponentially large in the evanescent region $z<0$ , and $W_{+}$ is exponentially small in the evanescent region; $A_{\pm }$ are rapidly varying in the neighbourhood of the turning point. We constrain $A_{+}^{\prime }W_{+}+A_{-}^{\prime }W_{-}=0$ so that $(\text{d}/\text{d}z)\unicode[STIX]{x1D6FF}E(z)=A_{+}W_{+}^{\prime }+A_{-}W_{-}^{\prime }$ . Now substitute (4.36) into (4.31) to obtain

(4.37) $$\begin{eqnarray}\displaystyle \frac{\text{d}A_{+}}{\text{d}z} & = & \displaystyle -\frac{\text{i}}{2}\frac{\unicode[STIX]{x0394}(z)}{\unicode[STIX]{x1D700}^{1/2}(z)}\left[A_{+}+A_{-}\frac{W_{-}}{W_{+}}\right]\quad \text{and}\quad \frac{\text{d}A_{-}}{\text{d}z}=\frac{\text{i}}{2}\frac{\unicode[STIX]{x0394}(z)}{\unicode[STIX]{x1D700}^{1/2}(z)}\left[A_{-}+A_{+}\frac{W_{+}}{W_{-}}\right]\nonumber\\ \displaystyle & & \displaystyle \text{where }\unicode[STIX]{x0394}(z)^{2}\equiv \frac{1}{4}\frac{\text{d}^{2}\unicode[STIX]{x1D700}/\text{d}z^{2}}{\unicode[STIX]{x1D700}}-\frac{5}{16}\left(\frac{\text{d}\unicode[STIX]{x1D700}/\text{d}z}{\unicode[STIX]{x1D700}}\right)^{2}.\end{eqnarray}$$

Consider integration of the first-order differential equations in (4.37) in the complex $z$ plane making use of analytic continuation. Far from the turning point at $z=0$ in the evanescent region select a point no. 1 on the real $z$ axis where $A_{+}=0$ and $A_{-}=1$ , while in the propagation region $z>0$ far from $z=0$ , select a point no. 2 where $A_{+}=1$ and $A_{-}=0$ . We can integrate $\text{d}A_{+}/\text{d}z$ from no. 1 on an arc approaching the imaginary axis, and it remains exponentially small. However, $\text{d}A_{-}/\text{d}z\approx \text{i}\unicode[STIX]{x0394}/(2\unicode[STIX]{x1D700}^{1/2}(z))$ is small, but not exponentially so. Near the imaginary $z$ axis, $z=\text{i}|z|$ , $\unicode[STIX]{x1D700}\sim z$ and $W_{\pm }\sim \exp (\pm z^{3/2})\sim \exp (\pm \text{i}^{3/2}|z|^{3/2})$ . Thus, $W_{+}$ is exponentially small near the imaginary axis and $W_{-}$ is exponentially large near the imaginary axis. The ratio $W_{-}/W_{+}$ remains exponentially large until the real axis. We can examine the integration of (4.37),

(4.38a ) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}A_{+}}{\text{d}z}=-\frac{\text{i}}{2}\frac{\unicode[STIX]{x0394}(z)}{\unicode[STIX]{x1D700}^{1/2}(z)}\left[A_{+}(\text{starts small})+A_{-}({\sim}1)\frac{W_{-}}{W_{+}}(\text{exponentially large})\right]\quad & \displaystyle\end{eqnarray}$$

(4.38b ) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}A_{-}}{\text{d}z}=\frac{\text{i}}{2}\frac{\unicode[STIX]{x0394}(z)}{\unicode[STIX]{x1D700}^{1/2}(z)}\left[A_{-}({\sim}1)+A_{+}(\text{exponentially small})\frac{W_{+}}{W_{-}}({\sim}\text{exponentially small})\right]. & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

$A_{+}$

$A_{+}$

$A_{-}$

$z$

$\unicode[STIX]{x1D6FF}E\sim W_{-}(z)$

$\unicode[STIX]{x1D6FF}E\sim W_{-}(z)+A_{+}W_{+}(z)$

$|A_{+}|=1$

$\unicode[STIX]{x1D6FF}E$

$z$

(4.39) $$\begin{eqnarray}\displaystyle \frac{\exp \left(-\displaystyle \int _{0}^{|z|}\text{d}z^{\prime }|\unicode[STIX]{x1D700}|^{1/2}(z^{\prime })\right)}{\text{e}^{\text{i}\unicode[STIX]{x03C0}/4}|\unicode[STIX]{x1D700}|^{1/4}(z)}\rightarrow \frac{\exp \left(-\text{i}\displaystyle \int _{0}^{z}\text{d}z^{\prime }\unicode[STIX]{x1D700}^{1/2}(z^{\prime })\right)+A_{+}\exp \left(+\text{i}\displaystyle \int _{0}^{z}\text{d}z^{\prime }\unicode[STIX]{x1D700}^{1/2}(z^{\prime })\right)}{\unicode[STIX]{x1D700}^{1/4}(z)}. & & \displaystyle\end{eqnarray}$$

We multiply both sides of (4.39) by $\text{e}^{\text{i}\unicode[STIX]{x03C0}/4}$

(4.40) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{\exp \left(-(\unicode[STIX]{x1D714}/c)\displaystyle \int _{0}^{|z|}\text{d}z^{\prime }|\unicode[STIX]{x1D700}|^{1/2}(z^{\prime })\right)}{|\unicode[STIX]{x1D700}|^{1/4}(z)}\nonumber\\ \displaystyle & & \displaystyle \quad \rightarrow \frac{\text{e}^{\text{i}\unicode[STIX]{x03C0}/4}\exp \left(-\text{i}(\unicode[STIX]{x1D714}/c)\displaystyle \int _{0}^{z}\text{d}z^{\prime }\unicode[STIX]{x1D700}^{1/2}(z^{\prime })\right)+A_{+}\text{e}^{\text{i}\unicode[STIX]{x03C0}/4}\exp \left(+\text{i}(\unicode[STIX]{x1D714}/c)\displaystyle \int _{0}^{z}\text{d}z^{\prime }\unicode[STIX]{x1D700}^{1/2}(z^{\prime })\right)}{\unicode[STIX]{x1D700}^{1/4}(z)}\nonumber\\ \displaystyle & & \displaystyle \quad \rightarrow \frac{2\cos \left[{\displaystyle \frac{\unicode[STIX]{x1D714}}{c}}\displaystyle \int _{0}^{z}\text{d}z^{\prime }\unicode[STIX]{x1D700}^{1/2}(z^{\prime })-\unicode[STIX]{x03C0}/4\right]}{\unicode[STIX]{x1D700}^{1/4}(z)}.\end{eqnarray}$$

Because the left-hand side of (4.40) is real, the two terms in the intermediate expression must be complex conjugates, which leads to the final expression on the right-hand side. We see that the phase factor $\unicode[STIX]{x03C0}/4$ has emerged. Figure 22(b) captures the qualitative behaviour of the solution in (4.40) characteristic of single turning-point behaviour. The peak of the wave at relative value of 2 occurs at $z\approx (\unicode[STIX]{x03C0}/2)c/\unicode[STIX]{x1D714}$ .

If the variation of the dielectric function is such that propagating waves can be trapped between two turning points, the analysis corresponding to (4.40) leads to an eigenvalue problem for the frequency $\unicode[STIX]{x1D714}$ . Given a length $L$ between the turning points, only particular values of $\unicode[STIX]{x1D714}$ will allow the wave forms from either side to link up. Imagine picking a value of $z$ and reflecting the solution for $\unicode[STIX]{x1D6FF}E(z)$ symmetrically in figure 22(b) so that $\unicode[STIX]{x1D6FF}E(z)$ and its spatial derivative are continuous. This can only be satisfied if the spatial derivative of $\unicode[STIX]{x1D6FF}E(z)$ vanishes at the mirror point, which quantizes the relationship between the values of $\unicode[STIX]{x1D714}$ and $L$ . For the two-turning-point problem we take the solution for the wave amplitude on the right-hand side of (4.40) associated with the turning point at $z_{0}$ and match it to the corresponding mirror image waves associated with the turning point at $z_{1}$ and introduce constant $C$ :

(4.41) $$\begin{eqnarray}\displaystyle C=\frac{2\cos \left[{\displaystyle \frac{\unicode[STIX]{x1D714}}{c}}\displaystyle \int _{z}^{z_{1}}\text{d}z^{\prime }\unicode[STIX]{x1D700}^{1/2}(z^{\prime })-\unicode[STIX]{x03C0}/4\right]}{\unicode[STIX]{x1D700}^{1/4}(z)}=\frac{2\cos \left[{\displaystyle \frac{\unicode[STIX]{x1D714}}{c}}\displaystyle \int _{0}^{z}\text{d}z^{\prime }\unicode[STIX]{x1D700}^{1/2}(z^{\prime })-\unicode[STIX]{x03C0}/4\right]}{\unicode[STIX]{x1D700}^{1/4}(z)}. & & \displaystyle\end{eqnarray}$$

For the left- and right-hand sides of (4.41) to be equal, the arguments of the cosines must be equal or the negative of one another so that $C=1$ or $C=-1$ . Hence, we arrive at the relations,

(4.42) $$\begin{eqnarray}\displaystyle & & \displaystyle \hspace{-10.00002pt}\frac{\unicode[STIX]{x1D714}}{c}\int _{z}^{z_{1}}\text{d}z^{\prime }\unicode[STIX]{x1D700}^{1/2}(z^{\prime })-\unicode[STIX]{x03C0}/4=-\left[\frac{\unicode[STIX]{x1D714}}{c}\int _{0}^{z}\text{d}z^{\prime }\unicode[STIX]{x1D700}^{1/2}(z^{\prime })-\unicode[STIX]{x03C0}/4\right]+2\unicode[STIX]{x03C0}\ell ,\quad C=1,\ell =0,1,2,\ldots \nonumber\\ \displaystyle & & \displaystyle \hspace{-10.00002pt}\quad \rightarrow \frac{\unicode[STIX]{x1D714}}{c}\int _{0}^{z_{1}}\text{d}z^{\prime }\unicode[STIX]{x1D700}^{1/2}(z^{\prime })=\unicode[STIX]{x03C0}/2+2\unicode[STIX]{x03C0}\ell \nonumber\\ \displaystyle & & \displaystyle \hspace{-10.00002pt}\quad \text{or }\frac{\unicode[STIX]{x1D714}}{c}\int _{0}^{z_{1}}\text{d}z^{\prime }\unicode[STIX]{x1D700}^{1/2}(z^{\prime })=\unicode[STIX]{x03C0}/2+(2\ell +1)\unicode[STIX]{x03C0},\quad C=-1\nonumber\\ \displaystyle & & \displaystyle \hspace{-10.00002pt}\quad \Rightarrow \frac{\unicode[STIX]{x1D714}}{c}\int _{0}^{z_{1}}\text{d}z^{\prime }\unicode[STIX]{x1D700}^{1/2}(\unicode[STIX]{x1D714},z^{\prime })=\left(\ell +\frac{1}{2}\right)\unicode[STIX]{x03C0}\quad \text{or}\quad \frac{\unicode[STIX]{x1D714}}{c}\oint \text{d}z^{\prime }\unicode[STIX]{x1D700}^{1/2}(\unicode[STIX]{x1D714},z^{\prime })=(2\ell +1)\unicode[STIX]{x03C0}.\end{eqnarray}$$

The final integral in (4.42) is from the turning point $\unicode[STIX]{x1D700}=0$ at $z=0$ to the turning point at $z=z_{1}$ and return. We recognize this expression as the Bohr–Sommerfeld quantization rule. Equation (4.42) is an eigenequation that determines the eigenvalue $\unicode[STIX]{x1D714}$ for each value of $l$ .