We ended the previous part by making a change of variable:
x = k vo t
I = ò ¥-¥ dvo /vo2 {2 cos (kvo t') - 1] + k vo t sin (k vo t)}
= kt ò ¥-¥ dx /x2 [2 (cos x -1) + x sin x]
è
I = kt -2 ò ¥-¥ dx /x2 + 2 ò (cos x dx/x2 + ò sin x dx/ x
I= -kt ò ¥-¥ sin x dx/ x = - kt p
The other term yields: - 2 p
So:
I= - 2p - kt p
dW(t)R = -p / 2 (mv f / k) f ' (v f) (eE1 /m) 2 t
Using: v f = w e / k
dW(t)R = -p / 2 (m w e / k) 2 f ' (v f) t E1 f ' (v f) E1 2 (e/m) 2
Now introduce substitutions:
f ' (v f) = no g'(v f)
and: vo = v - v f
(between wave and lab frame where v is lab velocity, vo is wave velocity)
Make variable order change: v = vo + v f
In wave frame we have:
< dv(t)>xo =
(eE1 /m) 2 k/2 (1/ k 3 vo3) {2 cos (-kvo t') - 1] + k vo t sin (k vo t)}
For energy change in lab frame:
d e = m/2 (v + dv)2 - m/2 (v)2
energy change in wave frame:
(d e)o = m/2 ( vo+ dv)2 - m vo(dv) + O (dv2 )
N.B. dv = v(t) - v(0) = {vo(t) + v f} - {vo(0) + v f}
= vo(t) - vo(0)
Then: dW R = ò ¥-¥ dv fo(v) < d e >xo
= ò ¥-¥ dv fo(v) mv < d v>xo
(dW R in wave frame)
Then after the wave-lab velocity change:
dW R = m ò ¥-¥ dv fo(vo + v f) (vo + v f)< d v>xo
Note the interaction is strongest near the phase velocity (v f):
Here a beam of electrons, all with speed v with respect to the laboratory frame and speed vo = v - v f in respect to the moving (wave) frame.
We then do a Taylor Expansion around v f to obtain:
fo(vo + v f) = fo (v f) + vo f' (v f) + ......
->
(vo + v f)fo(vo + v f) = vo f' (v f) + vo2 f' (v f) + v f fo (v f) + vo v f f'o v f)
Then: dW R = m ò ¥-¥ m dvo [vo f (v f) + vo v f f'o v f)] < d v>xo
{ }
The dominant term of the weak field (bracketed above) is Landau -damped.
Recall: w e2 = 4p no e2 /m
Thence:
dW R = - w e3/8k2 g'(v f) E1 2 t
No comments:
Post a Comment