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The change in Wk due to the wave is found bysubtracting the first term, which is the original energy. The average energy changeis thenFrom Eq. (7.77), we have1 hΔW k i ¼ m nu v21 þ 2un1 v12ð7:81Þ 1e2 E21,nu v21 ¼ nu2 m2 ðω kuÞ2ð7:82Þthe factor 12 representing cos 2 ðkx ωtÞ .
Similarly, from Eq. (7.79), we have2uhn1 v1 i ¼ nue2 E21 kum2 ðω kuÞ3ð7:83ÞConsequently,1e2 E212ku1þhΔW k i ¼ mnu4ðω kuÞm2 ðω kuÞ2nu e2 E21 ω þ ku¼4 m ðω kuÞ3ð7:84ÞThis result shows that hΔWki depends on the frame of the observer and that itdoes not change secularly with time. Consider the picture of a frictionless block2367 Kinetic TheoryFig. 7.22 Mechanical analogy for an electron moving in a moving potentialFig. 7.23 The quadraticrelation between kineticenergy and velocity causesa symmetric velocityperturbation to give riseto an increased averageenergy.sliding over a washboard-like surface (Fig. 7.22). In the frame of the washboard,ΔWk is proportional to (ku)2, as seen by taking ω ¼ 0 in Eq.
(7.84). It is intuitivelyclear that (1) hΔWki is negative, since the block spends more time at the peaks than atthe valleys, and (2) the block does not gain or lose energy on the average, once theoscillation is started. Now if one goes into a frame in which the washboard is movingwith a steady velocity ω/k (a velocity unaffected by the motion of the block, since wehave assumed that nu is negligibly small compared with the density of the wholeplasma), it is still true that the block does not gain or lose energy on the average, oncethe oscillation is started.
But Eq. (7.84) tells us that hΔWki depends on the velocityω/k, and hence on the frame of the observer. In particular, it shows that a beam hasless energy in the presence of the wave than in its absence if ω ku < 0 or u > vϕ;and it has more energy if ω ku > 0 or u < vϕ. The reason for this can be traced backto the phase relation between n1 and v1. As Fig. 7.23 shows, Wk is a parabolicfunction of v.
As v oscillates between u |v1| and u + |v1|, Wk will attain an averagevalue larger than the equilibrium value Wk0, provided that the particle spends anequal amount of time in each half of the oscillation. This effect is the meaning of thefirst term in Eq. (7.81), which is positive definite. The second term in that equation isa correction due to the fact that the particle does not distribute its time equally. InFig. 7.21, one sees that both electron a and electron b spend more time at the top of7.5 The Meaning of Landau Damping237the potential hill than at the bottom, but electron a reaches that point after a period ofdeceleration, so that v1 is negative there, while electron b reaches that point after aperiod of acceleration (to the right), so that v1 is positive there. This effect causeshΔWki to change sign at u ¼ vϕ.7.5.2The Effect of Initial ConditionsThe result we have just derived, however, still has nothing to do with linear Landaudamping.
Damping requires a continuous increase of Wk at the expense of waveenergy, but we have found that hΔWki for untrapped particles is constant in time. Ifneither the untrapped particles nor particle trapping is responsible for linear Landaudamping, what is? The answer can be gleaned from the following observation: IfhΔWki is positive, say, there must have been a time when it was increasing. Indeed,there are particles in the original distribution which have velocities so close to vϕthat at time t they have not yet gone a half-wavelength relative to the wave. Forthese particles, one cannot take the average hΔWki. These particles can absorbenergy from the wave and are properly called the “resonant” particles. As time goeson, the number of resonant electrons decreases, since an increasing number willhave shifted more than 12λ from their original positions.
The damping rate, however,can stay constant, since the amplitude is now smaller, and it takes fewer electrons tomaintain a constant damping rate.The effect of the initial conditions is most easily seen from a phase-spacediagram (Fig. 7.24). Here, we have drawn the phase-space trajectories of electrons,and also the electrostatic potential eϕ1 which they see. We have assumed that thiselectrostatic wave exists at t ¼ 0, and that the distribution f0(v), shown plotted in aplane perpendicular to the paper, is uniform in space and monotonically decreasingwith |v| at that time.
For clarity, the size of the wave has been greatly exaggerated.Of course, the existence of a wave implies the existence of an f1(v) at t ¼ 0.However, the damping caused by this is a higher-order effect neglected in the lineartheory. Now let us go to the wave frame, so that the pattern of Fig. 7.24 does notmove, and consider the motion of the electrons.
Electrons initially at A start out atthe top of the potential hill and move to the right, since they have v > vϕ. Electronsinitially at B move to the left, since they have v < vϕ. Those at C and D start at thepotential trough and move to the right and left, respectively. Electrons starting onthe closed contours E have insufficient energy to go over the potential hill and aretrapped. In the limit of small initial wave amplitude, the population of the trappedelectrons can be made arbitrarily small. After some time t, short enough that none ofthe electrons at A, B, C or D has gone more than half a wavelength, the electronswill have moved to the positions marked by open circles. It is seen that the electronsat A and D have gained energy, while those at B and C have lost energy.
Now, iff0(v) was initially uniform in space, there were originally more electrons at A than atC, and more at D than at B. Therefore, there is a net gain of energy by the electrons,and hence a net loss of wave energy. This is linear Landau damping, and it is2387 Kinetic TheoryFig. 7.24 Phase-space trajectories (top) for electrons moving in a potential wave (bottom). Theentire pattern moves to the right. The arrows refer to the direction of electron motion relative to thewave pattern.
The equilibrium distribution f0(v) is plotted in a plane perpendicular to the paper.critically dependent on the assumed initial conditions. After a long time, theelectrons are so smeared out in phase that the initial distribution is forgotten, andthere is no further average energy gain, as we found in the previous section. In thispicture, both the electrons with v > vϕ and those with v < vϕ when averaged over awavelength, gain energy at the expense of the wave. This apparent contradictionwith the idea developed in the picture of the surfer will be resolved shortly.7.6A Physical Derivation of Landau DampingWe are now in a position to derive the Landau damping rate without recourse tocontour integration.
As before, we divide the plasma up into beams of velocityu and density nu, and examine their motion in a waveE ¼ E1 sin ðkx ωtÞð7:85Þ7.6 A Physical Derivation of Landau Damping239From Eq. (7.77), the velocity of each beam isv1 ¼ eE1 cos ðkx ωtÞω kumð7:86ÞThis solution satisfies the equation of motion (7.76), but it does not satisfy the initialcondition v1 ¼ 0 at t ¼ 0. It is clear that this initial condition must be imposed;otherwise, v1 would be very large in the vicinity of u ¼ ω/k, and the plasma wouldbe in a specially prepared state initially. We can fix up Eq.
(7.86) to satisfy theinitial condition by adding an arbitrary function of kx kut. The composite solutionwould still satisfy Eq. (7.76) because the operator on the left-hand side ofEq. (7.76), when applied to f(kx kut), gives zero. Obviously, to get v1 ¼ 0 att ¼ 0, the function f(kx kut) must be taken to be cos(kx kut). Thus we have,instead of Eq. (7.86),v1 ¼eE1 cos ðkx ωtÞ cos ðkx kutÞω kumð7:87ÞNext, we must solve the equation of continuity (7.78) for n1, again subject to theinitial condition n1 ¼ 0 at t ¼ 0. Since we are now much cleverer than before, wemay try a Solution of the formn1 ¼ n1 ½ cos ðkx ωtÞ cos ðkx kutÞð7:88ÞInserting this into Eq.
(7.78) and using Eq. (7.87) for v1, we findn1 sin ðkx ωtÞ ¼ nueE1 k sin ðkx ωtÞ sin ðkx kutÞmðω kuÞ2ð7:89ÞApparently, we were not clever enough, since the sin(kx ωt) factor does notcancel. To get a term of the form sin(kx kut), which came from the added termin v1, we can add a term of the form At sin(kx kut) to n1. This term obviouslyvanishes at t ¼ 0, and it will give the sin(kx kut) term when the operator on theleft-hand side of Eq. (7.78) operates on the t factor. When the operator operates onthe sin(kx kut) factor, it yields zero. The coefficient A must be proportional to(ω ku)1 in order to match the same factor in ∂v1/∂x.
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