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The difficulty comesfrom our use of the fluid equations. Any real plasma has a finite temperature, and6.6 Two-Stream Instability201Fig. 6.9 The functionF(x, y) in the two-streaminstability, when the plasmais unstablethermal effects should be taken into account by a kinetic-theory treatment. Aphenomenon known as Landau damping (Chap. 7) will then occur for v0 ≲ vth,and no instability is predicted if v0 is too small.This “Buneman” instability, as it is sometimes called, has the following physicalexplanation. The natural frequency of oscillations in the electron fluid is ωp, and thenatural frequency of oscillations in the ion fluid is Ωp ¼ (m/M )1/2ωp.
Because of theDoppler shift of the ωp oscillations in the moving electron fluid, these two frequencies can coincide in the laboratory frame if kv0 has the proper value. The densityfluctuations of ions and electrons can then satisfy Poisson’s equation. Moreover, theelectron oscillations can be shown to have negative energy. That is to say, the totalkinetic energy of the electrons is less when the oscillation is present than when it is12absent. In the undisturbed beam, the kinetic energy per m3 is mn0 v20 . When there is12an oscillation, the kinetic energy is mðn0 þ n1 Þðv0 þ v1 Þ2 .
When this is averaged12over space, it turns out to be less than mn0 v20 because of the phase relation betweennl and v1 required by the equation of continuity. Consequently, the electron oscillations have negative energy, and the ion oscillations have positive energy. Bothwaves can grow together while keeping the total energy of the system constant. Aninstability of this type is used in klystrons to generate microwaves. Velocitymodulation due to E1 causes the electrons to form bunches. As these bunchespass through a microwave resonator, they can be made to excite the naturalmodes of the resonator and produce microwave power.Problems6.6.
(a) Derive the dispersion relation for a two-stream instability occurring whenthere are two cold electron streams with equal and opposite v0 in a1background of fixed ions. Each stream has a density n0 .2(b) Calculate the maximum growth rate.6.7. A plasma consists of two uniform streams of protons with velocities þv0 x^ and2313v0 x^ ; and respective densities n0 and n0 .
There is a neutralizing electronfluid with density n0 and with v0e. ¼ 0. All species are cold, and there is no2026 Equilibrium and Stabilitymagnetic field. Derive a dispersion relation for streaming instabilities in thissystem.6.8. A cold electron beam of density δn0 and velocity u is shot into a cold plasma ofdensity n0 at rest.(a) Derive a dispersion relation for the high-frequency beam-plasma instability that ensues.(b) The maximum growth rate γm is difficult to calculate, but one can make areasonable guess if δ 1 by analogy with the electron–ion Bunemaninstability. Using the result given without proof in Eq. (6.35), give anexpression for γm in terms of δ.16.9. Let two cold, counter-streaming ion fluids have densities n0 and velocities2v0y^ in a magnetic field B0z^ and a cold neutralizing electron fluid.
The fieldB0 is strong enough to confine electrons but not strong enough to affect ionorbits.(a) Obtain the following dispersion relation for electrostatic waves propagating in the ^y direction in the frequency range Ω2c << ω2 << ω2c :Ω2p2ðω kv0 Þþ2Ω2p2ðω þ kv0 Þ2¼ω2pþ1ω2c(b) Calculate the dispersion ω(k), growth rate γ(k), and the range of wavenumbers of the unstable waves.6.7The “Gravitational” InstabilityIn a plasma, a Rayleigh–Taylor instability can occur because the magnetic field actsas a light fluid supporting a heavy fluid (the plasma).
In curved magnetic fields, thecentrifugal force on the plasma due to particle motion along the curved lines offorce acts as an equivalent “gravitational” force. To treat the simplest case, considera plasma boundary lying in the y–z plane (Fig. 6.10). Let there be a density gradientFig. 6.10 A plasma surface subject to a gravitational instability6.7 The “Gravitational” Instability203Fig. 6.11 Physical mechanism of the gravitational instability∇n0 in the x direction and a gravitational field g in the x direction. We may letKTi ¼ KTe. ¼ 0 for simplicity and treat the low-β case, in which B0 is uniform.
In theequilibrium state, the ions obey the equationMn0 ðv0 ∇ Þv0 ¼ env0 B0 þ Mn0 gð6:36ÞIf g is a constant, v0 will be also; and (v0 · ∇)v0 vanishes. Taking the cross product ofEq. (6.36) with B0, we find, as in Sect. 2.2,v0 ¼M g B0g¼ y^e B20Ωcð6:37ÞThe electrons have an opposite drift which can be neglected in the limit m/M ! 0.There is no diamagnetic drift because KT ¼ 0, and no E0 B0 drift becauseE0 ¼ 0.If a ripple should develop in the interface as the result of random thermalfluctuations, the drift v0 will cause the ripple to grow (Fig. 6.11). The drift of ionscauses a charge to build up on the sides of the ripple, and an electric field developswhich changes sign as one goes from crest to trough in the perturbation.
As can beseen from Fig. 6.11, the E1 B0 drift is always upward in those regions wherethe surface has moved upward, and downward where it has moved downward. Theripple grows as a result of these properly phased E1 B0 drifts.To find the growth rate, we can perform the usual linearized wave analysis for wavespropagating in the y direction: k ¼ k^y. The perturbed ion equation of motion is∂M ð n0 þ n1 Þ ð v0 þ v1 Þ þ ð v0 þ v1 Þ ∇ ð v0 þ v1 Þ∂t¼ eðn0 þ n1 Þ½E1 þ ðv0 þ v1 Þ B0 þ Mðn0 þ n1 Þgð6:38ÞWe now multiply Eq.
(6.36) by 1 + (n1/n0) to obtainMðn0 þ n1 Þðv0 ∇ Þv0 ¼ eðn0 þ n1 Þv0 B0 þ Mðn0 þ n1 Þgð6:39Þ2046 Equilibrium and StabilitySubtracting this from Eq. (6.38) and neglecting second-order terms, we have thelinearized equationMn0∂v1þ ðv0 ∇ Þv1 ¼ en0 ðE1 þ v1 B0 Þ∂tð6:40ÞNote that g has cancelled out. Information regarding g, however, is still contained inv0. For perturbations of the form exp [i(ky ωt)], we haveMðω kv0 Þv1 ¼ ieðE1 þ v1 B0 Þð6:41ÞThis is the same as Eq. (4.96) except that ω is replaced by ω kv0, and electronquantities are replaced by ion quantities. The solution, therefore, is given byEq. (4.98) with the appropriate changes. For Ex ¼ 0 andΩ2c >> ðω kv0 Þ2ð6:42Þthe solution isvix ¼EyB0ω kv0 E yΩ c B0viy ¼ ið6:43ÞThe latter quantity is the polarization drift in the ion frame.
The correspondingquantity for electrons vanishes in the limit m/M ! 0. For the electrons, we thereforehavevex ¼ E y =B0vey ¼ 0ð6:44ÞThe perturbed equation of continuity for ions is∂n1þ ∇ ðn0 v0 Þ þ ðv0 ∇ Þn1 þ n1 ∇ v0∂tþ ðv1 ∇ Þn0 þ n0 ∇ v1 þ ∇ ðn1 v1 Þ ¼ 0ð6:45ÞThe zeroth-order term vanishes since v0 is perpendicular to ∇n0, and the n1 ∇ · v0term vanishes if v0 is constant. The first-order equation is, therefore,0iωn1 þ ikv0 n1 þ vix n0 þ ikn0 viy ¼ 0ð6:46Þ0where n0 ¼ ∂n0 =∂x. The electrons follow a simpler equation, since ve0 ¼ 0 andvey ¼ 0:0iωn1 þ vex n0 ¼ 0ð6:47Þ6.7 The “Gravitational” Instability205Note that we have used the plasma approximation and have assumed ni1 ¼ ne1.
Thisis possible because the unstable waves are of low frequencies (this can be justified aposteriori). Equations (6.43) and (6.46) yieldðω kv0 Þn1 þ iEy 0ω kv0 E yn þ ikn0¼0B0 0Ω c B0ð6:48ÞEquations (6.44) and (6.47) yieldωn1 þ iEy 0n ¼0B0 0E y iωn1¼ 0B0n0ð6:49ÞSubstituting this into Eq. (6.48), we haveω kv0 ωn1¼0ðω kv0 Þn1 n0 þ kn00Ωcn0kn0 ω kv0ω kv0 1 þω¼00Ω c n000ωðω kv0 Þ ¼ v0 Ωc n0 =n0Substituting for v0 from Eq.
(6.37), we obtain a quadratic equation for ω: 0ω2 kv0 ω g n0 =n0 ¼ 0ð6:50Þð6:51Þð6:52ÞThe solutions are 01=211ω ¼ kv0 k2 v20 þ g n0 =n024ð6:53ÞThere is instability if ω is complex; that is, if10gn0 =n0 > k2 v204ð6:54Þ0From this, we see that instability requires g and n0 /n0 to have opposite sign.This is just the statement that the light fluid is supporting the heavy fluid; otherwise,ω is real and the plasma is stable. Since g can be used to model the effects ofmagnetic field curvature, we see from this that stability depends on the sign of thecurvature. Configurations with field lines bending in toward the plasma tend to bestabilizing, and vice versa. For sufficiently small k (long wavelength), the growthrate is given byh 0i1=2γ ¼ Im ðωÞ g n0 =n0ð6:55Þ2066 Equilibrium and StabilityFig.
6.12 A “flute”instabilityNote that the real part of ω is1kv .2 0Since v0 is an ion velocity, this is a1is merely alow-frequency oscillation, as previously assumed. The factor of2consequence of neglecting v0e. The wave is stationary in the frame in which thedensity-weighted average of all the v0’s is zero, which in this case is the framemoving at 12v0 . The laboratory frame has no particular significance in this case.This instability, which has k ⊥ B0, is sometimes called a “flute” instability forthe following reason.
In a cylinder, the waves travel in the θ direction if the forcesare in the r direction. The surfaces of constant density then resemble fluted Greekcolumns (Fig. 6.12).6.8Resistive Drift WavesA simple example of a universal instability is the resistive drift wave. In contrast togravitational flute modes, drift waves have a small but finite component of k alongB0.
The constant density surfaces. Therefore, resemble flutes with a slight helicaltwist (Fig. 6.13). If we enlarge the cross section enclosed by the box in Fig. 6.13 andstraighten it out into Cartesian geometry it would appear as in Fig. 6.14. The onlydriving force for the instability is the pressure gradient KT ∇n0 (we assumeKT ¼ constant for simplicity).
In this case, the zeroth-order drifts (for E0 ¼ 0) are0vi0 ¼ vDi ¼KT i n0y^eB0 n0ð6:56Þ0ve0 ¼ vDe ¼ KT e n0y^eB0 n0ð6:57Þ6.8 Resistive Drift Waves207Fig. 6.13 Geometry of adrift instability in acylinder. The region in therectangle is shown in detailin Fig. 6.15Fig. 6.14 Physicalmechanism of a drift waveFrom our experience with the flute instability, we might expect drift waves tohave a phase velocity of the order of vDi or vDe. We shall show that ω/ky isapproximately equal to vDe.2086 Equilibrium and StabilitySince drift waves have finite kz, electrons can flow along B0 to establish athermodynamic equilibrium among themselves (cf. discussion of Sect. 4.10).They will then obey the Boltzmann relation (Sect. 3.5):n1 =n0 ¼ eϕ1 =KT eð6:58ÞAt point A in Fig. 6.14 the density is larger than in equilibrium, nl is positive, andtherefore ϕ1 is positive. Similarly, at point B, n1 and ϕ1 are negative.
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