Solution for Problem Set 21
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3 Solution for Problem Set 21 (compiled by Nate Bode April 24, 2009 A 21.1 Non-resonant Particle Energy in Wave [by Keith Matthews adapted from Chris Hirata] The rate of change of the non-resonant electron inetic energy is given by du e = d 1 2 m ev 2 F 0 dv Insert eq. (21.20 d F 0 = t F 0 + v z F 0 = v ( D v F 0 We note that D non res is independent of v, integrate by parts twice, and then simply integrate to get du e = m e D F 0 dv = m e nd Insert eq. (21.23 for D = D non res du e = ω i ε d which is the desired eq. ( Energy Conservation [by Alexander Putilin/ 99] The electron inetic energy density and momentum density are given by 1 U e = 2 m ev 2 F 0 dv (1 1 Sz e = 2 m ev 3 F 0 dv (2 1
4 and we have U e t + Se z 1 z = 2 m ev 2 1 = 2 m ev 2 v = ( F0 t + v F 0 z ( D F 0 v dv (3 dv (4 D F 0 v m evdv (5 Now using eq. (21.22 for the resonant electrons D = e2 π ɛ 0 m 2 e de δ(ω r v (6 we find U e t + Se z z = e2 π ɛ 0 m e d E ω r 2 F 0 ( ωr (7 ( which becomes, upon using eq. (21.12 F 0 ωr = 2ɛ 0m e 2 πe 2 ω r ω i, d 2ω i E (8 = t = U w t de Sw z z z de ω r using eq. (21.18 (9 (10 where we ve used the facts that de is the wave s energy density U w, and de ω r is the wave s energy flux Sw z. Thus we finally have U e t which is the energy conservation law. + Se z z = U w Sw z t z ( Cerenov Power in Electrostatic Waves [by Alexander Putilin 99] The emission rate of plasmons is given by (21.43 W = πe2 ω r ɛ 0 2 δ(ω r v (12 Each plasmon has energy ω r, so the radiated power per unit time is P = 1 (2π 3 d 3 W ω r = e2 8π 2 d 3 ω2 r ɛ 0 2 δ(ω r v (13 2
5 The integration is over the region < max (outside this region waves are strongly Landau damped. A good estimate of max is the inverse Debye length: max 1/λ D (see the discussion at the end of Sec Since λ D < 1, we can approximate ω r ( by a constant ω p. Choosing v to point along z-axis, we have P = e2 ω 2 p 8π 2 ɛ 0 = e2 ω 2 p 8π 2 ɛ 0 = e2 ω 2 p 8π 2 ɛ 0 v = e2 ω 2 p 8π 2 ɛ 0 v = e2 ω 2 p 4πɛ 0 v ln d 3 1 < max 2 δ(ω r v (14 d 2 1 d z < max 2 + δ(ω r v (15 2 z d (ω2 p/v 2 (16 2 <2 max (ω2 p /v2 2 max (ω2 p /v2 0 ( max v ω p 2π d (ω2 p/v 2 (17 (18 Note that P depends on max logarithmically. So if v is sufficiently large, it doesn t mae much difference what particular definition we use for max Electron Foer-Planc Equation [by Xinai Wu 02] It s very straightforward to carry out the Taylor expansion in to second order, so we don t bother to write down all the formulae here. Rather we ll just point out a few things worth noticing in the expansion. Fristly you may find it s easier to use component notation rather than tensor notation. Secondly, you may want to group terms coming out of expanding the r.h.s. of eq. (21.47 into f three categories: f terms, v terms, and 2 f j v j v terms. Thirdly, to get the final n answer, notice that in the classical limit both η and W are large, i.e. of order 1/. After all these, you can set 0 and recover eq. ( Three-Wave Mixing [by Xinai Wu 02] (a Diagram (b in Fig gives the rate of creation of B, ( dηb = W C AB η C (1 + η A (1 + η B dv A (2π 3 creation dv C (2π 3 (19 where in the above expression, the unity in (1+η A corresponds to spontaneous emission and η A corresponds to induced emission. Similarly for (1 + η B. Diagram (a in Fig gives the rate of destruction of B, ( dηb destruction = W AB C (η C + 1η A η B dv A (2π 3 3 dv C (2π 3 (20
6 The net rate of change is given by the sum of these two, and by the principle of detailed balance W AB C = W C AB. So we get dη B = W AB C [(1 + η A + η B η C η A η B ] dv A dv C (2π 3 (2π 3 (21 (b Under the approximation stated in the problem, we have η C (1 + η A + η B η A η B η B (η C η A (22 Also let s change notation: B, C, η B ( B η ia (, η A ( A η L ( A, η C ( C η L (, also change ω A and ω C to ω L. Then using the Taylor expansion η L ( η L ( η L ( (23 ω L ( ω L ( ω L ( = V g ( (24 we easily get eq. ( (c [Thorne/Matthews 05] Eq. (21.59 becomes almost identical to eq. (21.42 upon η ia ( η(, and η L ( f(v. Cerenov radiation occurs when the emitting particle travels faster than the group velocity of the emitted wave (provided there exists a process that couples the two. As a result the emitting particle generates a shocwave lie that produced by a supersonic aircraft. This shocwave taes on a form almost independent of the type of source and the waves generated. The formation of this shocwave depends upon the emitted plasmons having much lower energy than the source and following boson statistics. This causes them to have much higher number density than the emitting plasmons which facilitates stimulated emission. In the same manner in which fast electrons generate a shocwave of Langmuir waves, Langmuir wave excitations are plasmons (i.e. particles that travel faster than ion-acoustic modes and generate a shocwave in them. These similarities, exceeding speed, low emitted plasmon energy and stimulated emission are the physical source of the similarity between the two results. (d Keith and Kip hope to provide a solution to this part by Monday of next wee Three-Wave Mixing - Langmuir Evolution [by Xinai Wu 02] (a See Fig. 1 for the four relevant diagrams. The desired rate of change for the Langmuir occupation number is given by the sum/difference of these four diagrams: 4
7 - + (1 (2 - + (3 (4 Figure 1: Diagrams for the change of Langmuir occupation number (we use solid lines to denote Langmuir plasmons, and dashed lines ion acoustic plasmons dη L ( = d 3 [(1 + (2 (3 (4] (25 (2π 6 where (1 = η L ( + [1 + η L ( ][1 + η ia (]R( +,, δ(ω L ( + ω L ( ω ia (26 ( (2 = η L ( η ia ([1 + η L ( ]R(,, δ(ω L ( ω L ( ω ia ( (27 (3 = η L ( [1 + η L ( ][1 + η ia (]R(,, δ(ω L ( ω L ( ω ia (28 ( (4 = η L ( η ia ([1 + η L ( + ]R( +,, δ(ω L ( + ω L ( ω ia ( (29 (b Now under the approximation stated in ex part b, all the δ-functions in the previous part reduce to the same expression δ(ω ia ( V g (. We also have(dropping this δ-function for the moment being (1 (4 = {η L ( + [1 + η L ( + η ia (] η L ( η ia (} R( +,, (30 and [η L ( + η L ( ]R( +,, η ia ( (31 η L ( + R( +,, η ia ( (32 5
8 (2 (3 = {η L ( η ia ( η L ( [1 + η L ( + η ia (]} R(,, (33 [η L ( η L ( ]R(,, η ia ( (34 η L ( R(,, η ia ( (35 where to reach the last expression in the above equation, we ve used the fact that R is symmetric w.r.t. the two Langmuir wave momenta. It s not hard to see that (1 (4 and (2 (3 are the same expression of K evaluated at K = + and respectively So integrating their sum over phase space we immediately get the desired answer dη L ( = [D( η L ( ] (36 with D( d 3 (2π 6 η ia( R(,, δ(ω ia ( V g ( (37 (c See above, ex. 21.5c. For the same reasons that the evolution equation for the emitted particles is characteristic, so is the evolution of the emitting particles. 6
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