Thomson scattering in a magnetic field
An electron moves under the action of an uniform and constant magnetic field B0 = B0 ẑ and of a
monochromatic plane wave which electric field is given by
E(z, t) = Ei ŷeikz−iωt ,
(1)
with Ei ≪ B0 c.
a) Find the solution of the electron equation of motion in steady conditions, negleting friction terms
and the effect of the magnetic field of the wave.
b) Calculate the power radiated by the electron at the frequency ω, and discuss the dependence on
ω of the spectrum and the angular distribution of the emitted radiation in the limits ω ≪ ωc and
ω ≫ ωc , where ωc = eB0 /me .
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Solution
a) Within the assumptions made, the equation of motion is
me
dv
= (−e)(E + v × B0 ),
dt
(2)
i.e..
e
dvx
= − vy B 0 ,
dt
me
dvy
e
= − (Ei e−iωt − vx B0 ),
dt
me
dvz
= 0.
dt
The third equation tells us that the electrons does not moves in the ẑ direction, thus we pose z = 0.
The first and second equation can be derived with respect to t and rearranged as
d 2 vx
e
+ ωc2 vx = ωc Ei e−iωt ,
2
dt
me
d 2 vy
e
+ ωc2 vy = iω Ei e−iωt ,
2
dt
me
(3)
with solution
vx =
ωc2
e
ωc
Ei e−iωt ,
2
− ω me
vy =
ωc2
e
iω
Ei e−iωt .
2
− ω me
(4)
b) The cycle-averaged radiated power is
hP i =
k0 e4
ω2
2
|E
|
(ω 2 + ω 2 ),
i
3m2e c3
(ωc2 − ω 2 )2 c
(5)
which has obviously a maximum at the cyclotron resonance, ω = ωc . In the limit ω/ωc ≪ 1,
hP i ∝ ω 2 /ωc2 , while in the opposite limit ω/ωc ≫ 1 the power is independent of frequency (“white”
spectrum).
Since the components of the electric dipole p = −er are out of phase by π/2, the direction of the
dipole (along which there is no emitted radiation) oscillates in time. However in the limit ω ≪ ωc ,
since vy ∼ (ω/ωc )vx ≪ vx the electron mostly oscillates along ŷ, so that the radiation emission is
strongest in the xz plane. In the opposite limit ω ≫ ωc , since vx ∼ (ωc /ω)vy ≪ vy the radiation
emission is strongest in the yz plane.
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