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Next: Physics of Landau Damping Up: Waves in Warm Plasmas Previous: Introduction

Landau Damping

Let us begin our study of the Vlasov equation by examining what appears, at first sight, to be a fairly simple and straight-forward problem. Namely, the propagation of small amplitude plasma waves through a uniform plasma with no equilibrium magnetic field. For the sake of simplicity, we shall only consider electron motion, assuming that the ions form an immobile, neutralizing background. The ions are also assumed to be singly-charged. We shall look for electrostatic plasma waves of the type discussed in Sect. 4.7. Such waves are longitudinal in nature, and possess a perturbed electric field, but no perturbed magnetic field.

Our starting point is the Vlasov equation for an unmagnetized, collisionless plasma:

\frac{\partial f_e}{\partial t} + {\bf v}\!\cdot\!\nabla f_e - \frac{e}{m_e}\,
{\bf E}\!\cdot\!\nabla_v f_e = 0,
\end{displaymath} (985)

where $f_e({\bf r},{\bf v}, t)$ is the ensemble averaged electron distribution function. The electric field satisfies
{\bf E} = -\nabla\phi.
\end{displaymath} (986)

\nabla^2\phi = -\frac{e}{\epsilon_0}\left(n-\int \!f_e\,d^3{\bf v}\right).
\end{displaymath} (987)

Here, $n$ is the number density of ions (which is the same as the number density of electrons).

Since we are dealing with small amplitude waves, it is appropriate to linearize the Vlasov equation. Suppose that the electron distribution function is written

f_e({\bf r}, {\bf v}, t) = f_0({\bf v}) + f_1({\bf r}, {\bf v}, t).
\end{displaymath} (988)

Here, $f_0$ represents the equilibrium electron distribution, whereas $f_1$ represents the small perturbation due to the wave. Note that $\int f_0\,d^3{\bf v}
=n$, otherwise the equilibrium state is not quasi-neutral. The electric field is assumed to be zero in the unperturbed state, so that ${\bf E}$ can be regarded as a small quantity. Thus, linearization of Eqs. (985) and (987) yields
\frac{\partial f_1}{\partial t} + {\bf v}\!\cdot\!\nabla f_1
- \frac{e}{m_e}\,{\bf E}\!\cdot\!\nabla_v f_0 =0,
\end{displaymath} (989)

\nabla^2\phi = \frac{e}{\epsilon_0}\int\! f_1\,d^3{\bf v},
\end{displaymath} (990)


Let us now follow the standard procedure for analyzing small amplitude waves, by assuming that all perturbed quantities vary with ${\bf r}$ and $t$ like $\exp[\,{\rm i}\,({\bf k}\!\cdot\!{\bf r} - \omega\,t)]$. Equations (989) and (990) reduce to

-{\rm i}\,(\omega - {\bf k}\!\cdot\!{\bf v}) f_1
+ {\rm i}\,\frac{e}{m_e}\,\phi \,{\bf k}\!\cdot\!\nabla_v f_0 = 0,
\end{displaymath} (991)

-k^2\,\phi = \frac{e}{\epsilon_0} \int\! f_1\,d^3{\bf v},
\end{displaymath} (992)

respectively. Solving the first of these equations for $f_1$, and substituting into the integral in the second, we conclude that if $\phi$ is non-zero then we must have
1 + \frac{e^2}{\epsilon_0\,m_e\,k^2} \int\frac{ {\bf k}\!\cd...
...bla_v f_0}
{\omega - {\bf k}\!\cdot\!{\bf v}}\,d^3{\bf v} = 0.
\end{displaymath} (993)

We can interpret Eq. (993) as the dispersion relation for electrostatic plasma waves, relating the wave-vector, ${\bf k}$, to the frequency, $\omega$. However, in doing so, we run up against a serious problem, since the integral has a singularity in velocity space, where $\omega={\bf k}\!\cdot\!{\bf v}$, and is, therefore, not properly defined.

The way around this problem was first pointed out by Landau[*] in a very influential paper which laid the basis of much subsequent research on plasma oscillations and instabilities. Landau showed that, instead of simply assuming that $f_1$ varies in time as $\exp(-{\rm i}\,\omega\,t)$, the problem must be regarded as an initial value problem in which $f_1$ is given at $t=0$ and found at later times. We may still Fourier analyze with respect to ${\bf r}$, so we write

f_1({\bf r}, {\bf v}, t) = f_1({\bf v},t)\,{\rm e}^{\,{\rm i}\,{\bf k}\cdot
{\bf r}}.
\end{displaymath} (994)

It is helpful to define $u$ as the velocity component along ${\bf k}$ (i.e., $u= {\bf k}\!\cdot\!{\bf v}/k$) and to define $F_0(u)$ and $F_1(u,t)$ to be the integrals of $f_0({\bf v})$ and $f_1({\bf v},t)$ over the velocity components perpendicular to ${\bf k}$. Thus, we obtain
\frac{\partial F_1}{\partial t} + {\rm i}\,k\,u\,F_1 - \frac{e}{m_e}\,E\,
\frac{\partial F_0}{\partial u} = 0,
\end{displaymath} (995)

{\rm i}\,k\,E =-\frac{e}{\epsilon_0} \int_{-\infty}^{\infty} F_1(u)\,du.
\end{displaymath} (996)

In order to solve Eqs. (995) and (996) as an initial value problem, we introduce the Laplace transform of $F_1$ with respect to $t$:

\bar{F}_1(u,p) = \int_0^\infty F_1(u,t)\,{\rm e}^{-p\,t}\,dt.
\end{displaymath} (997)

If the growth of $F_1$ with $t$ is no faster than exponential then the above integral converges and defines $\bar{F}_1$ as an analytic function of $p$, provided that the real part of $p$ is sufficiently large.

Noting that the Laplace transform of $\partial F_1/\partial t$ is $p\,\bar{F}_1-F_1(u,t=0)$ (as is easily shown by integration by parts), we can Laplace transform Eqs. (995) and (996) to obtain

p\,\bar{F}_1 + {\rm i}\,k\,u\,\bar{F}_1 = \frac{e}{m_e}\,\bar{E}\,\frac{\partial
F_0}{\partial u} + F_1(u,t=0),
\end{displaymath} (998)

{\rm i}\,k\,\bar{E} =-\frac{e}{\epsilon_0}\int_{-\infty}^{\infty}\!\bar{F}_1(u)\,
\end{displaymath} (999)

respectively. The above two equations can be combined to give
{\rm i}\,k\,\bar{E} = -\frac{e}{\epsilon_0}\int_{-\infty}^{\...
...{\rm i}\,k\,u} + \frac{F_1(u,t=0)}{p+{\rm i}\,k\,u}\right] du,
\end{displaymath} (1000)

\bar{E} = -\frac{(e/\epsilon_0)}{{\rm i}\,k\,\epsilon(k,p)}
\int_{-\infty}^\infty \frac{F_1(u,t=0)}{p+{\rm i}\,k\,u}\,du,
\end{displaymath} (1001)

\epsilon(k,p) = 1 + \frac{e^2}{\epsilon_0\,m_e\,k}
...{\infty} \frac{\partial F_0/\partial u}{{\rm i}\,p-
\end{displaymath} (1002)

The function $\epsilon(k,p)$ is known as the plasma dielectric function. Note that if $p$ is replaced by $-{\rm i}\,\omega$ then the dielectric function becomes equivalent to the left-hand side of Eq. (993). However, since $p$ possesses a positive real part, the above integral is well defined.

The Laplace transform of the distribution function is written

\bar{F}_1 = \frac{e}{m_e}\,\bar{E}\, \frac{\partial F_0/\partial u}
{p+ {\rm i}\,k\,u} + \frac{F_1(u,t=0)}{p+ {\rm i}\,k\,u},
\end{displaymath} (1003)

\bar{F}_1 = - \frac{e^2}{\epsilon_0\,m_e\,{\rm i}\,k} \frac{...
...p+ {\rm i}\,k\,u'}\,du' + \frac{F_1(u,t=0)}{p+ {\rm i}\,k\,u}.
\end{displaymath} (1004)

Having found the Laplace transforms of the electric field and the perturbed distribution function, we must now invert them to obtain $E$ and $F_1$ as functions of time. The inverse Laplace transform of the distribution function is given by

F_1(u,t) = \frac{1}{2\pi\,{\rm i}}\int_C \bar{F}_1(u,p)\,{\rm e}^{\,p\,t}\,dp,
\end{displaymath} (1005)

where $C$, the so-called Bromwich contour, is a contour running parallel to the imaginary axis, and lying to the right of all singularities of $\bar{F}_1$ in the complex-$p$ plane (see Fig. 31). There is an analogous expression for the parallel electric field, $E(t)$.

Figure 31: The Bromwich contour.
\epsfysize =3in

Rather than trying to obtain a general expression for $F_1(u,t)$, from Eqs. (1004) and (1005), we shall concentrate on the behaviour of the perturbed distribution function at large times. Looking at Fig. 31, we note that if $\bar{F}_1(u,p)$ has only a finite number of simple poles in the region ${\rm Re}(p)>-\sigma$, then we may deform the contour as shown in Fig. 32, with a loop around each of the singularities. A pole at $p_0$ gives a contribution going as ${\rm e}^{\,p_0\,t}$, whilst the vertical part of the contour goes as ${\rm e}^{-\sigma\,t}$. For sufficiently long times this latter contribution is negligible, and the behaviour is dominated by contributions from the poles furthest to the right.

Figure 32: The distorted Bromwich contour.
\epsfysize =3in

Equations (1001)-(1004) all involve integrals of the form

\int_{-\infty}^\infty \frac{G(u)}{u-{\rm i}\,p/k}\,du,
\end{displaymath} (1006)

which become singular as $p$ approaches the imaginary axis. In order to distort the contour $C$, in the manner shown in Fig. 31, we need to continue these integrals smoothly across the imaginary $p$-axis. By virtue of the way in which the Laplace transform was originally defined, for ${\rm Re}(p)$ sufficiently large, the appropriate way to do this is to take the values of these integrals when $p$ is in the right-hand half-plane, and find the analytic continuation into the left-hand half-plane.

If $G(u)$ is sufficiently well-behaved that it can be continued off the real axis as an analytic function of a complex variable $u$ then the continuation of (1006) as the singularity crosses the real axis in the complex $u$-plane, from the upper to the lower half-plane, is obtained by letting the singularity take the contour with it, as shown in Fig. 33.

Figure 33: The Bromwich contour for Landau damping.
\epsfysize =2.5in

Note that the ability to deform the contour $C$ into that of Fig. 32, and find a dominant contribution to $E(t)$ and $F_1(u,t)$ from a few poles, depends on $F_0(u)$ and $F_1(u,t=0)$ having smooth enough velocity dependences that the integrals appearing in Eqs. (1001)-(1004) can be continued sufficiently far into the left-hand half of the complex $p$-plane.

If we consider the electric field given by the inversion of Eq. (1001), we see that its behaviour at large times is dominated by the zero of $\epsilon(k,p)$ which lies furthest to the right in the complex $p$-plane. According to Eqs. (1004) and (1005), $F_1$ has a similar contribution, as well as a contribution going as ${\rm e}^{-{\rm i}\,k\,u\,t}$. Thus, for sufficiently long times after the initiation of the wave, the electric field depends only on the positions of the roots of $\epsilon(k,p)=0$ in the complex $p$-plane. The distribution function has a corresponding contribution from the poles, as well as a component going as ${\rm e}^{-{\rm i}\,k\,u\,t}$. For large times, the latter component of the distribution function is a rapidly oscillating function of velocity, and its contribution to the charge density, obtained by integrating over $u$, is negligible.

As we have already noted, the function $\epsilon(k,p)$ is equivalent to the left-hand side of Eq. (993), provided that $p$ is replaced by $-{\rm i}\,\omega$. Thus, the dispersion relation, (993), obtained via Fourier transformation of the Vlasov equation, gives the correct behaviour at large times as long as the singular integral is treated correctly. Adapting the procedure which we found using the variable $p$, we see that the integral is defined as it is written for ${\rm Im}(\omega)>0$, and analytically continued, by deforming the contour of integration in the $u$-plane (as shown in Fig. 33), into the region ${\rm Im}(\omega)<0$. The simplest way to remember how to do the analytic continuation is to note that the integral is continued from the part of the $\omega$-plane corresponding to growing perturbations, to that corresponding to damped perturbations. Once we know this rule, we can obtain kinetic dispersion relations in a fairly direct manner via Fourier transformation of the Vlasov equation, and there is no need to attempt the more complicated Laplace transform solution.

In Sect. 4, where we investigated the cold-plasma dispersion relation, we found that for any given $k$ there were a finite number of values of $\omega$, say $\omega_1$, $\omega_2$, $\cdots$, and a general solution was a linear superposition of functions varying in time as ${\rm e}^{-{\rm i}\,\omega_1\,t}$, ${\rm e}^{-{\rm i}\,\omega_2\,t}$, etc. This set of values of $\omega$ is called the spectrum, and the cold-plasma equations yield a discrete spectrum. On the other hand, in the kinetic problem we obtain contributions to the distribution function going as ${\rm e}^{-{\rm i}\,k\,u\,t}$, with $u$ taking any real value. All of the mathematical difficulties of the kinetic problem arise from the existence of this continuous spectrum. At short times, the behaviour is very complicated, and depends on the details of the initial perturbation. It is only asymptotically that a mode varying as ${\rm e}^{-{\rm i}\,\omega\,t}$ is obtained, with $\omega$ determined by a dispersion relation which is solely a function of the unperturbed state. As we have seen, the emergence of such a mode depends on the initial velocity disturbance being sufficiently smooth.

Suppose, for the sake of simplicity, that the background plasma state is a Maxwellian distribution. Working in terms of $\omega$, rather than $p$, the kinetic dispersion relation for electrostatic waves takes the form

\epsilon(k,\omega) = 1 + \frac{e^2}{\epsilon_0\,m_e\,k}
...}^{\infty} \frac{\partial F_0/\partial u}{\omega-
\end{displaymath} (1007)

F_0(u) = \frac{n}{(2\pi\,T_e/m_e)^{1/2}}\,\exp(-m_e\,u^2/2\,T_e).
\end{displaymath} (1008)

Suppose that, to a first approximation, $\omega$ is real. Letting $\omega$ tend to the real axis from the domain ${\rm Im}(\omega)>0$, we obtain
\int_{-\infty}^{\infty} \frac{\partial F_0/\partial u}{\omeg...
...k} \left(\frac{\partial F_0}{\partial u}
\end{displaymath} (1009)

where $P$ denotes the principal part of the integral. The origin of the two terms on the right-hand side of the above equation is illustrated in Fig. 34. The first term--the principal part--is obtained by removing an interval of length $2\,\epsilon$, symmetrical about the pole, $u=\omega/k$, from the range of integration, and then letting $\epsilon\rightarrow 0$. The second term comes from the small semi-circle linking the two halves of the principal part integral. Note that the semi-circle deviates below the real $u$-axis, rather than above, because the integral is calculated by letting the pole approach the axis from the upper half-plane in $u$-space.

Figure 34: Integration path about a pole.
\epsfysize =2in

Suppose that $k$ is sufficiently small that $\omega\gg k\,u$ over the range of $u$ where $\partial F_0/\partial u$ is non-negligible. It follows that we can expand the denominator of the principal part integral in a Taylor series:

\frac{1}{\omega-k\,u} \simeq \frac{1}{\omega}\left(1+ \frac{...
...^2\,u^2}{\omega^2} + \frac{k^3 u^3}{\omega^3} + \cdots\right).
\end{displaymath} (1010)

Integrating the result term by term, and remembering that $\partial F_0/\partial u$ is an odd function, Eq. (1007) reduces to
1-\frac{\omega_p^{~2}}{\omega^2} - 3\,k^2\,\frac{T_e\,\omega...
F_0}{\partial u}\right)_{u=\omega/k} \simeq 0,
\end{displaymath} (1011)

where $\omega_p=\sqrt{n\,e^2/\epsilon_0\,m_e}$ is the electron plasma frequency. Equating the real part of the above expression to zero yields
\omega^2 \simeq \omega_p^{~2}(1+ 3\,k^2\,\lambda_D^2),
\end{displaymath} (1012)

where $\lambda_D =\sqrt{T_e/m_e\,\omega_p^{~2}}$ is the Debye length, and it is assumed that $k\,\lambda_D\ll 1$. We can regard the imaginary part of $\omega$ as a small perturbation, and write $\omega=\omega_0+\delta\omega$, where $\omega_0$ is the root of Eq. (1012). It follows that
2\,\omega_0\,\delta\omega \simeq \omega_0^{~2}\, \frac{e^2}{...
...^2} \left(\frac{\partial
F_0}{\partial u}\right)_{u=\omega/k},
\end{displaymath} (1013)

and so
\delta\omega \simeq \frac{{\rm i}\,\pi}{2} \frac{e^2\,\omega...
F_0}{\partial u}\right)_{u=\omega/k},
\end{displaymath} (1014)

\delta\omega \simeq - \frac{{\rm i}}{2}\sqrt{\frac{\pi}{2}} ...
...\lambda_D)^3} \exp\left[-\frac{1}{2\,(k\,\lambda_D)^2}\right].
\end{displaymath} (1015)

If we compare the above results with those for a cold-plasma, where the dispersion relation for an electrostatic plasma wave was found to be simply $\omega^2=\omega_p^{~2}$, we see, firstly, that $\omega$ now depends on $k$, according to Eq. (1012), so that in a warm plasma the electrostatic plasma wave is a propagating mode, with a non-zero group velocity. Secondly, we now have an imaginary part to $\omega$, given by Eq. (1015), corresponding, since it is negative, to the damping of the wave in time. This damping is generally known as Landau damping. If $k\,\lambda_D\ll 1$ (i.e., if the wave-length is much larger than the Debye length) then the imaginary part of $\omega$ is small compared to the real part, and the wave is only lightly damped. However, as the wave-length becomes comparable to the Debye length, the imaginary part of $\omega$ becomes comparable to the real part, and the damping becomes strong. Admittedly, the approximate solution given above is not very accurate in the short wave-length case, but it is sufficient to indicate the existence of very strong damping.

There are no dissipative effects included in the collisionless Vlasov equation. Thus, it can easily be verified that if the particle velocities are reversed at any time then the solution up to that point is simply reversed in time. At first sight, this reversible behaviour does not seem to be consistent with the fact that an initial perturbation dies out. However, we should note that it is only the electric field which decays. The distribution function contains an undamped term going as ${\rm e}^{-{\rm i}\,k\,u\,t}$. Furthermore, the decay of the electric field depends on there being a sufficiently smooth initial perturbation in velocity space. The presence of the ${\rm e}^{-{\rm i}\,k\,u\,t}$ term means that as time advances the velocity space dependence of the perturbation becomes more and more convoluted. It follows that if we reverse the velocities after some time then we are not starting with a smooth distribution. Under these circumstances, there is no contradiction in the fact that under time reversal the electric field will grow initially, until the smooth initial state is recreated, and subsequently decay away.

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Next: Physics of Landau Damping Up: Waves in Warm Plasmas Previous: Introduction
Richard Fitzpatrick 2011-03-31