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Hencedωωωc 2cdkω2p4.21!1=2/ ω1=21ω2 c2 k2 E1 ¼ iω j1 ðEq:½4-81Þ20 v p is the positron velocityj1 ¼ n0 e v p veFrom the equation of motion,456Appendix D: Answers to Some Problems1ieiωcω2Ex Ey1 c2mωωω1ieiωcω2Ey vy ¼Ex1 c2mωωω 1 21ieω2c2 2∴ ω c k Ex ¼ iω ðn0 eÞð1 þ 1ÞEx 1 220mωω22ω p¼Ex1 ω2c =ω2vx ¼the Ey term canceling out. Similarly, 2ω c2 k 2 E y ¼2ω2pEy1 ω2c =ω2the Ex term cancelling out.

Both equations give2ω2pc2 k 2¼1ω2 ω2cω2The R and L waves are degenerate and have the same phase velocities—hence, no Faraday rotation.4.22 Since the phase difference between the R and L waves is twice the angle ofrotation,ðLðkL kR Þ dz ¼ π0k R, L ¼ k 0ω2p =ω211 ωc =ω!1=2To get a simple expression for kLkR, we wish to expand the square root.Let us assume we can, and then check later for consistency:k R, L k 01 ω2p =ω212 1 ωc =ω!Appendix D: Answers to Some Problems4571 ω2p11kL kR ¼ k0 22 ω 1 ωc =ω 1 þ ωc =ω1 ω2p 2ωc =ω¼ k0 22 ω 1 ω2c =ω2π ¼ LðkL kR Þ ¼ k0 Lω2p ¼πc 2ω ω2cLωcω2p ωc12ω ω ω2cf 2p ¼k0 ¼ωcc f 2 f 2c2L f cf c ¼ 2:8 1010 ð0:1ÞHzc3 108¼ 3:75 1010 Hz¼λ0 8 1033 108 1:41 1021 7:8 1018f 2p ¼ð 2Þ ð 1Þ2:8 109¼ 7:5 1019 ¼ 92 nn ¼ 9:3 1017 m3f ¼To justify expansion, note that fc f, so thatω2p =ω2fp27:5 109 2¼2 ¼ 0:05 11 ωc =ω f3:75 10104.24 12.7 .4.25 (a) The X-wave cutoff frequencies are given by Eq. (4.107).

Thus,ω2p ¼ ωðω ωc Þ ¼ncx ¼4πne2mmωð ω þ ωc Þ4πe2We choose the (+) sign, corresponding to the L cutoff, because that givesthe higher density.458Appendix D: Answers to Some Problems(b)The left branch is the one that has a cutoff at ω ¼ ωL. One might worrythat this branch is inaccessible if the wave is sent in from outside theplasma. However, if ω is kept less than ωc, the stopband between ωh andωR is avoided completely.4.28 (a)1=2pffiffiffif p ¼ 9 n ¼ ð9Þ 1015¼ 2:85 108 Hz f c ¼ 28GHz=T ¼ 2:8 1010 102 ¼ 2:8 108 Hzf ¼ 1:6 108 Hz∴ω p =ω > 1 ωc =ω > 11=2 1pffiffiffi 122ωL ¼ ωc ωc þ 4ω p ωc þ 5ωc22¼ 0:62ωc for ωc ω pf L ¼ ð0:62Þ 2:8 108 ¼ 1:73 108 > fAlso, f > all ion frequencies.(b) The R-wave (whistler mode) is the only wave that propagates here.Appendix D: Answers to Some Problems4.29 (a)vA ¼459B1¼ 1=2ðμ0 nMÞ1:26 106 1019 1:67 1027¼ 6:9 106 m=s1:6 1019 ð1ÞeB ¼ 9:58 107 rad=s¼ Ωc ¼M1:67 1027ω ¼ 0:1Ωc ¼ 9:58 106 rad=sω ¼ kvA ¼ 2πvA =λIf λ ¼ 2 L,L¼(b)πvA π 6:9 106¼¼ 2:26 mω9:58 106L / vA =ω / vA =Ωc / BðnMÞ1=2 B1 M / ðM=nÞ1=21=2 133 1=2 1019∴L ¼ ð2:26Þ¼ 82 m11018This is why Alfvén waves cannot be studied in Q-machines, regardless of B.4.30 (a)ω2 ¼ ω2p þ c2 k22ω dω ¼ c2 2k dkvg ¼ dω=dk ¼ c2 k=ω!1=2ω2pck¼ 1ωω!1=2!ω2p1 ω2pc 1∴vg ¼ c 1 for ω2 ω2p2 ωωvg t ¼ x ∴ t ¼ x=vg!2!ω2pdtx1 ω2px ω2p¼1dω c2 ω2c ω3ω2∴(b)dfc f3 2dtxfp33 108 8 107cf3d f 1¼x¼ 2 ¼ 1:9 1018 mdtfpð9Þ2 2 105 5 1061¼ 1:9 1018 3 1016¼ 63 parsec460Appendix D: Answers to Some Problemsð 2Þð1Þ4.31 (a) Let n0 ¼ ð1 2Þn0 , n0 ¼ 2n0 , ne ¼ n0 eϕ=kT eð1Þð2ÞPoisson: ikE1 ¼ k2 ϕ ¼ 210 e ni þ ni ne(Assume zl,2 ¼ 1, since the ion charge is not explicitly specified.)k ð1Þ ð2Þk ð2Þð1ÞContinuity: n1 ¼ ð1 2Þn0 v1 , n1 ¼ 2n0 v1ωωEquation of motion:!1Ω2c je kϕ 1 2¼ð4:68ÞMjωω"1ek2 eΩ2c12∴k ϕ ¼ð1 2Þn0 21 220ωω M1#1k2 eΩ2c2eþ2n0 2ϕ 0 ðplasma approximationÞ1 2 n0kT eω M2ωð jÞv11 ¼ ð1 2 Þk2 v2s1k2 v2s2þ2(2ω2 Ωc1ω2 Ω2c2(b) There are two roots, one near ω ¼ Ωc1 and one near ω ¼ Ωc2.

If E ! 0,the root near Ωc2 approaches Ωc2 to keep the last term finite. The usualroot, near Ωc1, is shifted by the presence of the M2 species:2ω Ω2c1¼k2 v2s12 22 22 2 ω Ωc1 2 k vs1 k vs2 2ω Ω2c2In the last term, we may approximate ω2 by Ω2c1 þ k2 v2s1 : Thus,2ω Ω2c1þk2 v2s1k2 v2s2 1 k2 v2s1þ2 2Ωc1 Ω2c2Appendix D: Answers to Some Problems(c)4611 k2 v2sD1 k2 vsTþ2 ω2 ω2cD 2 ω2 Ω2cT ¼ KT e =MD ¼ 104 1:6 1019 =ð2Þ 1:67 1027 ¼ 4:79 10111¼v2sDΩcD2v2sT ¼ v2sD ¼ 3:19 10113¼ eB=MD ¼ 1:6 1019 ð5Þ=ð2Þ 1:67 1027 ¼ 2:40 1082ΩcT ¼ ΩcD ¼ 1:60 108 k ¼ 100 m13 2 1 ω Ω2cD ω2 Ω2cT ¼ k2 v2sD ω2 Ω2cT þ v2sT ω2 Ω2cD212 222242ω ω ΩcD þ ΩcT þ k vsD þ vsT21þ Ω2cD Ω2cT þ k2 v2sD Ω2cT þ v2sT Ω2cD ¼ 021642ω ω 8:32 10 þ 3:99 1015 þ 1:47 1033 þ 1:53 1032 ¼ 0ω4 8:72 1016 ω2 þ 1:63 1033 ¼ 01=2 i1hω2 ¼ 8:72 1016 7:60 1033 6:52 10332¼ 6:0 1016 , 2:72 1016ω ¼ 2:45, 1:65 108 sec 14.321 2emveEve ¼2imω e2 ∴ v2e ¼ 2 2 E2m ω 1e2 2 20 ω2p E2E ¼ n0 m 2 2 E ¼2 m ω2ω2 But ω2 ¼ ω2p ∴E ¼ 12 20 E2 :DE1vi E1 =B0E ¼ n0 Mv2i2 1∴E ¼ Mn0 E21 =B0 : But∇ E1 ¼ B_ 1 ∴ E21 ¼ ω2 =k2 B212Mn0 ω2 2 E¼B :2B20 k2 1E ¼ n04.33f ¼ 39 and 26:3 MHzFor Alfvén wave, 2Bω2B20¼ 1¼22μμnMk00 0462Appendix D: Answers to Some Problems4.34 (a) With the L-wave, the cutoff occurs at ω ¼ ωL, so that one requiresω2L < 2ω2 .

Since ωL < ωp if n0 is fixed (Problem 4.15), one can go tohigher values of n0 (for constant 2ω2) with the L-wave than with theO-wave.(b) For the L-cutoff,ω2pωc20 mω2 ωc ∴n1þ¼¼1þce2ωωω2Thus, to double the usual cutoff density of 20 mω2 =e2 , one must havefc ¼ fc3 108¼¼ 8:9 1011 Hzλ 337 1068:9 1011f c ¼ 28 109 Hz=T∴B0 ¼¼ 31:8 T28 109f ¼This would be unreasonably expensive.(c) The plasma has a density maximum at the center, so it behaves like aconvex lens. Such a lens focuses if en > 1 and defocuses if en < 1.

Thewhistler wave always travels with vϕ < c (Problem 4.19), soen ¼ c=vϕ > 1, and the plasma focuses this wave.(d) The question is one of accessibility. If ω < ωc everywhere, the whistlerwave will propagate regardless of n0. However, if ω > ωc, the wave willbe cut off in regions of low density. From (b) above, we see that a field of31.8 T is required; this seems too large for the scheme to be practical.4.35 The answer should come out the same as for cold plasma.4.36 The linearized equation of motion for either species isiωmn0 v1 ¼ qn0 ðE þ v1 B0 Þ γkT i kn1Thusiωmn0 k:v1 ¼ qn0 ðk E þ k v1 B0 Þ γkT i k2 n1 :But k · E ¼ 0 for transverse wave, and k · (v1 B0) ¼ v1 · (k B0) ¼ 0 byassumption. The linearized equation of continuity isiωn1 þ n0 ik v1 ¼ 0Substituting for k · v1, we haveiω2 mn1 ¼ iγkTk2 n1Thus n1 is arbitrary, and we may take it to be 0.

Then the ∇p term vanishes forboth ions and electrons.Appendix D: Answers to Some Problems4634.44 For a given density, the highest cutoff frequency is ωR. Thus the lowest boundfor n is given by ω ¼ ωR.ω2pf 2p1:6 1019 36 104ωc ¼ 0:16¼1¼ 2 ¼1ω2ω0:91 1030 ð2π Þ 1:2 108f2n ¼ f 2p =q2 ¼ ð0:16Þ 1:2 108 q2 ¼ 2:8 1013 m34.46 Let ω ¼ ωR at r1 where n ¼ n1, ωp ¼ ωp1; and ω ¼ ωh at r2, where n ¼ n2,ωp ¼ ωp2 Thenω2p2 ¼ ω2 ω2cð4:105Þω2p1 ¼ ω2 ωωcð4:107ÞThusω2p2 ω2p1 ¼ ωc ðω ωc Þ ¼ ðn2 n1 Þe2 =20 mButSon2 n1 dj∂n=∂r j n1 d=r 0 ¼ 20 m=e2 ðωÞðω ωc Þðd=r 0 Þd ðωc =ωÞr 04.47 (a) The accessible resonance is on the far side, past the density maximum.(b) Let ωc0 be ωc at the left boundary, and ωc be the value at the resonancelayer, where ω ¼ ωp. Then we require464Appendix D: Answers to Some Problemsωc0 > ω, where ω2 ¼ ω2c þ ω2pThusω2c0 > ω2c þ ω2p ω2c0 ω2c > ω2pðωc0 þ ωc Þðωc0 ωc Þ 2ωc Δωc > ω2pω2pΔωc ΔB0¼> 22ωcωcB04.48 These are the upper and lower hybrid frequencies and right- and left-handcutoff frequencies with ion motions included.

Note that ω2p =ωc ¼ Ω2p =Ωc :Resonance:ω4 ω2p þ ω2c þ Ω2p þ Ω2c þ ω2p Ω2c þ ω2c Ω2p þ ω2c Ω2c ¼ 0ω2þ ω2h þ Ω2p 1 ω2c =ω2h ðupper hybridÞω2 ω2c Ω2p =ω2h or111ðlower hybridÞ¼þω2 ωc Ωc Ω2pCutoff:ω2p ωc Ωc¼11ω2ωωRcutoffLThis is more easily obtained, without approximation, from the form given inProblem 4.50.5.1 (a) De ¼ KTe/mv2σ ¼ ð6π Þ 0:53 10"10 ¼ 5:29 1020#m2 1=2ð2Þð2Þ 1:6 10192E¼v¼m9:11 1031¼ 8:39 105 m=sFrom Problem 1.1b, n0 ¼ 3:3 1019 103 ¼ 3:3 1022 m3v ¼ n0 σv ¼ n0 σv ¼ 3:3 1022 5:29 1020 8:39 105¼ 1:46 109 s1Appendix D: Answers to Some Problems465ð2Þ 1:6 1019 ¼ 2:4 102 m2 =sDe ¼ 9:11 1031 1:46 109(b) j ¼ μneEE¼5.21:6 1019 2:4 102μe ¼ eDe =KT e ¼ð2Þ 1:6 1019¼ 1:2 102 m2 =V sj2 103 ¼ 1:04 104 V=m¼μne1:2 102 1016 1:6 1019∂n¼ D∇ 2 n αn2∂t2 π 2 π 2∂ nπx¼ DD∇ 2 n ¼ D 2 ¼ Dn0cosn ¼ αn2∂x2L2L2LD π 20:4 π 2¼ 15¼ 1:1 1018 m3∴n ¼α 2L0:06105.4 (a) From Problem 5.1a, ven ¼ 1.46 109 s1.

We need to find whether μe⊥/μi⊥is large or small:μe Mvin¼μi mven1=2v jn ¼ nn σv j / vth j / m jsince σ is approximately the same for ion–neutral and electron–neutralcollisions. Thus 1=2M¼ ð4 1, 836Þ1=2 ¼ 85:7m1:6 1019 ð0:2ÞeB¼¼ 3:52 1010ωc ¼m9:11 10313:52 1010 24 1 þ ω2c τ2en ¼ 580ωc τen ¼1:46 109 m M1=2¼ ð24Þð85:7Þ1 ¼ 0:28Ωc τin ¼ ωc τenM mμeμiμe⊥ μe 1 þ Ω2c τ2in1:08¼ 0:16 1¼¼ ð85:7Þ580μi⊥ μi 1 þ ω2c τ2enμ De⊥ þ μe⊥ Di⊥μ∴Da⊥ ¼ i⊥ De⊥ þ e⊥ Di⊥mi⊥μi⊥ þ μe⊥¼ De⊥ þ 0:16Di⊥466Appendix D: Answers to Some ProblemsButD¼Di⊥∴De⊥KTμeμ Ti1 0:1¼ 0:3¼ i⊥ ¼μe⊥ T e 0:16 2∴Da⊥ ¼ De⊥ ½1 þ ð0:16Þð0:3Þ ¼ 1:05De⊥ De⊥(b)aðDτÞ1=2¼ 2:4∴τ ¼11τ¼2 2 De⊥2:4 10De⊥ ¼5.5 a 2 12:4 Da⊥2:4 102¼ 0:4140 ðfrom Problem 5:1Þ580∴τ ¼ 42 μsΓ ¼ D dn=dx n ¼ n0 ð1 x=LÞΓ ¼ Dn0 =L ðx > 0ÞQ ¼ 2Γ ¼ 2Dn0 =L∴n0 ¼ QL=2D5.7λei vthe τei ¼ vthe =νeiBut vthe / T e1=2 and νei / T e3=2∴λei / T e1=2 =T e3=2 / T 2e5.8lnΛΩ-m ðassume Z ¼ 1Þ3=2T ev5:2 105 ð10Þ¼ 4:65 108 Ω-m¼3=2ð500Þ j ¼ I=A ¼ 2 105 = 7:5 103 ¼ 2:67 107 A=m2E ¼ ηk j ¼ 4:65 108 2:67 107 ¼ 1:2V=mηk ¼ 5:2 105Appendix D: Answers to Some Problems5.9 (a)(b)(c)467KT i ¼ 20 keV KT e ¼ 10 keV n ¼ 1012 m3ηnðKT i þ KT e ÞB ¼ 5T D⊥ ¼B2 3 ln Λ10 ð10Þη⊥ ¼ ð2:0Þ 5:2 105 3=2 ¼ 3=2T ev1049¼ 1:0 10 Ω-m1:0 109 1021 3 104 1:6 1019D⊥ ¼524 2¼ 3:0 10 m =sdN∂n¼ 2πrLΓr Γr ¼ D⊥dt∂r∂nn¼r ¼ 0:50m L ¼ 100 m∂r 0:1 dN¼ ð2π Þð0:50Þ 102 2:0 104 1021 =0:10 ¼ 6 1020 s1dtNnπr 2 L¼r effective ¼ 0:55 mdN=dt dN=dt 21 10 ðπÞð0:55Þ2 102¼ 150sτ¼6 1020τ¼5.13ηk ¼ 5:2 105η j2 ¼ 1:6 10 10ln ΛΩ-m ¼ 5:2 1053=2T ev103=25105 2¼ 1:6 105 Ω-m¼ 1:6 105 W=m3¼ 1:6 105 J=ðm3 -sÞ ¼ 1:6 10 = 1:6 1019 ¼ 1024 eV=m3 -s¼5_dEevdt3dEev 3 dT ev¼ nE ¼ nKT e ∴22 dtdt468Appendix D: Answers to Some ProblemsdT ev 2 1¼1024 ¼ 0:67 105 eV=s ¼ 0:067eV=μs3 1019dt5.15 (a)en(E≠q0 −virB) − ∇≠q0 pi − e 2n 2h(viq − veq) = 00−en(E≠q −verB) − ∇≠q0 pe + e 2n 2h(viq − veq) = 0add:vir B þ ver B ¼ 0∴vir ¼ ver(This shows ambipolar diffusion.)(b)0pi− e 2n 2h(vir − ver) = 0rp−en(Er + veq B) − e + e 2n 2h(vir − ver) = 0rEr1 piviq = −+= vE + vDiB enB rE1 peveq = − r −= vE + vDeB enB ren(Er +iq B)−(c) From the first equation in (a),e2 n2 ηðviθ veθ ÞenBenη 1 ∂ pi ∂ peη ∂p¼ verþ¼¼ 2B enB ∂r∂rB ∂rvir ¼ (This shows the absence of cross-field mobility.)5.17 (a)∂v1¼ j1 B0∂tE1 þ v1 B0 ¼ η j1ρ0∇ E1 ¼ B_ 1 ∇ B1 ¼ μ0 j1∇ ∇ E1 ¼ ∇ B_ 1 ¼ μ0 j10− k(k • E) + k 2E1 = iwm 0 j1k E ¼ 0 ðtransverse waveÞSolve for v1 in Eq.

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