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You may neglect Ti (but not Te) and assume that all leptons followthe Boltzmann relation.4.12. For electromagnetic waves, show that the index of refraction is equal to thesquare root of the appropriate plasma dielectric constant (cf. Problem 4.4).4.13. In a potassium Q-machine plasma, a fraction κ of the electrons can bereplaced by negative Cl ions. The plasma then has n0 K+ ions, κn0 Cl ions,and (1 κ)n0 electrons per m3. Find the critical value of n0 which will cut offa 3-cm wavelength microwave beam if κ ¼ 0.6.4.14. An 8-mm microwave interferometer is used on an infinite plane-parallelplasma slab 8 cm thick (Fig. P4.13).(a) If the plasma density is uniform, and a phase shift of 1/10 fringe isobserved, what is the density? (Note: One fringe corresponds to a 360phase shift.)(b) Show that if the phase shift is small, it is proportional to the density.4.14Electromagnetic Waves Perpendicular to B0113Fig.
P4.134.14Electromagnetic Waves Perpendicular to B0We now consider the propagation of electromagnetic waves when a magnetic fieldis present. We treat first the case of perpendicular propagation, k ⊥ B0. If we taketransverse waves, with k ⊥ E1, there are still two choices: E1 can be parallel to B0or perpendicular to B0 (Fig. 4.33).4.14.1 Ordinary Wave, E1 ║ B0If E1 is parallel to B0, we may take B0 ¼ B0^z , E1 ¼ E1^z , k ¼ k^x : In a realexperiment, this geometry is approximated by a beam of microwaves incident ona plasma column with the narrow dimension of the waveguide in line with B0(Fig. 4.34).Fig. 4.33 Geometry forelectromagnetic wavespropagating at right anglesto B01144 Waves in PlasmasFig.
4.34 An ordinarywave launched from awaveguide antenna towarda magnetized plasmacolumnThe wave equation for this case is still given by Eq. (4.81):ω2 c2 k2 E1 ¼ iω j1 =ε0 ¼ in0 eωve1 =ε0ð4:92Þm∂vez =∂t ¼ eEzð4:93ÞSince E1 ¼ E1^z ; we need only the component vez. This is given by the equation ofmotionSince this is the same as the equation for B0 ¼ 0, the result is the same as we hadpreviously for B0 ¼ 0:ω2 ¼ ω2p þ c2 k2ð4:94ÞThis wave, with E1 ║ B0, is called the ordinary wave. The terminology “ordinary”and “extraordinary” is taken from crystal optics; however, the terms have beeninterchanged. In plasma physics, it makes more sense to let the “ordinary” wave bethe one that is not affected by the magnetic field.
Strict analogy with crystal opticswould have required calling this the “extraordinary” wave.4.14Electromagnetic Waves Perpendicular to B0115Fig. 4.35 The E-vector ofan extraordinary wave iselliptically polarized. Thecomponents Ex and Eyoscillate 90 out of phase,so that the total electric fieldvector E1 has a tip thatmoves in an ellipse once ineach wave period4.14.2 Extraordinary Wave, E1 ⊥ B0If E1 is perpendicular to B0, the electron motion will be affected by B0, and thedispersion relation will be changed. To treat this case, one would be tempted to takeE1 ¼ E1 ^y and k ¼ k^x (Fig. 4.33).
However, it turns out that waves with E1 ⊥ B0tend to be elliptically polarized instead of plane polarized. That is, as such a wavepropagates into a plasma, it develops a component Ex along k, thus becoming partlylongitudinal and partly transverse. To treat this mode properly, we must allow E1 tohave both x and y components (Fig.
4.35):E1 ¼ Ex ^x þ E y ^yð4:95ÞThe linearized electron equation of motion (with KTe ¼ 0) is nowimωve1 ¼ eðE þ ve1 B0 Þð4:96ÞOnly the x and y components are nontrivial; they areie Ex þ v y B0mωie E y v x B0vy ¼mωð4:97Þ1e ωc ω2iEx E y 1 c2mωωω1e ωc ω2ciE y Ex 1 2vy ¼mωωωð4:98Þvx ¼The subscripts 1 and e have been suppressed. Solving for vx and vy as usual, we findvx ¼1164 Waves in PlasmasThe wave equation is given by Eq. (4.80), where we must now keep the longitudinalterm k · E1 ¼ kEx:ω2 c2 k2 E1 þ c2 kEx k ¼ iω j1 =ε0 ¼ in0 ωeve1 =ε0ð4:99ÞSeparating this into x and y components and using Eq.
(4.98), we have1iωn0 e e ωc ω2ciEx þω Ex ¼ Ey 1 2ε0 mωωω1 2iωn0 e e ωc ω2iE y Ex 1 c2ω c2 k 2 E y ¼ ε0 mωωω2ð4:100ÞIntroducing the definition of ωp, we may write this set asw2 1 −w 2 − c 2k 2 1 −w 2p wcw 2c2w+i−EEy = 0xpww2w 2cw2− w 2p Ey − iw 2p wcwð4:101ÞEx = 0These are two simultaneous equations for Ex and Ey which are compatible only ifthe determinant of the coefficients vanishes:A Bð4:102ÞC D ¼ 0Since the coefficient A is ω2 ω2h ; where ωh is the upper hybrid frequency definedby Eq. (4.60), the condition AD ¼ BC can be written ω2ω2 ωc 2ω2 ω2h c2 k2 1 c2¼ ωpω2 22222ω ωh ω p ωc =ω = ω ωhω2 ω2h2 2c k¼ω2ð4:103Þω2 ω2cThis can be simplified by a few algebraic manipulations.
Replacing the first ω2h onthe right-hand side by ω2c þ ω2p and multiplying through by ω2 ω2h ; we have4.15Cutoffs and Resonances1172 ω2p ω2 ω2h þ ω4p ω2c =ωc2 k 2¼1ω2ω2 ω2c ω2 ω2hω2p ω2 ω2 ω2h þ ω2p ω2c¼1 2 2ω ω ω2c ω2 ω2hω2p ω2 ω2 ω2c ω2p ω2 ω2c¼1 2ωω2 ω2c ω2 ω2hω2p ω2 ω2pc2 k 2 c2¼¼1ω2ω2 ω2 ω2hv2ϕð4:104ÞThis is the dispersion relation for the extraordinary wave. It is an electromagneticwave, partly transverse and partly longitudinal, which propagates perpendicular toB0 with E1 perpendicular to B0.4.15Cutoffs and ResonancesThe dispersion relation for the extraordinary wave is considerably more complicated than any we have met up to now.
To analyze what it means, it is useful todefine the terms cutoff and resonance. A cutoff occurs in a plasma when the index ofrefraction goes to zero; that is, when the wavelength becomes infinite, since n~ ¼ ck/ω. A resonance occurs when the index of refraction becomes infinite; that is, whenthe wavelength becomes zero. As a wave propagates through a region in which ωpand ωc are changing, it may encounter cutoffs and resonances. A wave is generallyreflected at a cutoff and absorbed at a resonance.The resonance of the extraordinary wave is found by setting k equal to infinity inEq.
(4.104). For any finite ω, k ! 1 implies ω ! ωh, so that a resonance occurs at apoint in the plasma whereω2h ¼ ω2p þ ω2c ¼ ω2ð4:105ÞThis is easily recognized as the dispersion relation for electrostatic waves propagating across B0 (Eq. (4.60)). As a wave of given ω approaches the resonance point,both its phase velocity and its group velocity approach zero, and the wave energy isconverted into upper hybrid oscillations.
The extraordinary wave is partly electromagnetic and partly electrostatic; it can easily be shown (Problem 4.14) that atresonance this wave loses its electromagnetic character and becomes an electrostatic oscillation.1184 Waves in PlasmasThe cutoffs of the extraordinary wave are found by setting k equal to zero inEq. (4.104). Dividing by ω2 ω2p ; we can write the resulting equation for ω asfollows:1¼ω2p1hi2ω 1 ω2 = ω2 ω 2cpð4:106ÞA few tricky algebraic steps will yield a simple expression for ω:1ω2pω2c¼ω2 ω2p ω21ω2pω2c =ω2¼ω2 1 ω2 =ω2pω2p1 2ω1!2ω2cω2ω2pωc¼ωω2ω2 ωωc ω2p ¼ 0ð4:107ÞEach of the two signs will give a different cutoff frequency; we shall call these ωRand ωL.
The roots of the two quadratics are1=2 1ωc þ ω2c þ 4ω2p21=2 122ωL ¼ ωc þ ωc þ 4ω p2ωR ¼ð4:108ÞWe have taken the plus sign in front of the square root in each case because we areusing the convention that ω is always positive; waves going in the x direction willbe described by negative k. The cutoff frequencies ωR and ωL are called the righthand and left-hand cutoffs, respectively, for reasons which will become clear in thenext section.The cutoff and resonance frequencies divide the dispersion diagram into regionsof propagation and non propagation. Instead of the usual ω–k diagram, it is moreenlightening to give a plot of phase velocity versus frequency; or, to be precise, aplot of ω2/c2k2 ¼ 1/~n2 vs. ω (Fig.
4.36). To interpret this diagram, imagine that ωc isfixed, and a wave with a fixed frequency ω is sent into a plasma from the outside. Asthe wave encounters regions of increasing density, the frequencies ωL, ωp, ωh, andωR all increase, moving to the right in the diagram. This is the same as if the density4.15Cutoffs and Resonances119Fig. 4.36 The dispersion of the extraordinary wave, as seen from the behavior of the phasevelocity with frequency. The wave does not propagate in the shaded regionsFig.
4.37 A similardispersion diagramfor the ordinary wavewere fixed and the frequency ω were gradually being decreased. Taking the latterpoint of view, we see that at large ω (or low density), the phase velocity approachesthe velocity of light. As the wave travels further, vϕ increases until the right-handcutoff ω ¼ ωR is encountered. There, vϕ becomes infinite.
Between the ω ¼ ωR andω ¼ ωh layers, ω2/k2 is negative, and there is no propagation possible. At ω ¼ ωh,there is a resonance, and vϕ goes to zero. Between ω ¼ ωh and ω ¼ ωL, propagationis again possible. In this region, the wave travels either faster or slower thanc depending on whether ω is smaller or larger than ωp. From Eq. (4.104), it isclear that at ω ¼ ωp, the wave travels at the velocity c.
For ω < ωL, there is anotherregion of nonpropagation. The extraordinary wave, therefore, has two regions ofpropagation separated by a stop band.By way of comparison, we show in Fig. 4.37 the same sort of diagram for theordinary wave. This dispersion relation has only one cutoff and no resonances.1204.164 Waves in PlasmasElectromagnetic Waves Parallel to B0Now we let k lie along the z axis and allow E1 to have both transverse componentsEx and Ey:k ¼ k^zE1 ¼ Ex ^x þE y ^yð4:109ÞThe wave equation (4.99) for the extraordinary wave can still be used if we simplychange k from k^x to k^z : From Eq. (4.100), the components are nowω2piωcEExy1 ω2c =ω2ω 2ω2piωc2 2ExEy þω c k Ey ¼ω1 ω2c =ω2ω2 c2 k2 Ex ¼ð4:110ÞUsing the abbreviationαω2 p1 ω2c =ω2ð4:111Þwe can write the coupled equations for Ex and Ey asωcω2 c2 k2 α Ex þ iα E y ¼ 0ω 2ωcω c2 k2 α E y iα Ex ¼ 0ωð4:112ÞSetting the determinant of the coefficients to zero, we have2ω2 c2 k2 α ¼ ðαωc =ωÞ2ω2 c2 k2 α ¼ αωc =ωð4:113Þð4:114ÞThusω2ωc ω p 1 cω2 c 2 k 2 ¼ α 1 ¼22ωω1 ωc =ωω2p1 ðωc =ωÞ¼¼ ω2p 1 ðωc =ωÞ1 þ ðωc =ωÞ 1 ðωc =ωÞð4:115Þ4.16Electromagnetic Waves Parallel to B0121The sign indicates that there are two possible solutions to Eq.
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