Download Polarization effects on Thomson scattering

Polarization effects on Thomson scattering
An electron is in the field of a plane wave of frequency ω, elliptically polarized. The electric field of
the wave can be written as
E0
E= √
[x̂ cos(kz − ωt) + ǫŷ sin(kz − ωt)] ,
1 + ǫ2
(1)
where 0 < ǫ < 1.
Neglecting for the moment the effect of the magnetic force term −ev × B,
a) characterize the radiation scattered by the electron by determining the frequency and the polarization observed along each axis (x, y, z), and find a direction along which the radiation is circularly
polarized;
b) calculate the total (cycle-averaged) scattered power and discuss its dependence on ǫ;
Now consider the effect of the force term −ev × B on the scattering.
c) Evaluate the −ev × B magnetic force term by calculating the B field from Eq.(1) and using the
result of point a) for v. Discuss the direction and frequency of the magnetic force and its dependence
on ǫ as well.
d) Discuss how the scattering of the incident wave is modified by the magnetic force by specifying which new frequencies are observed, in which direction and with which polarization, and the
modification of the scattered power.
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Solution
a) In steady conditions the velocity of the electron has components
vx =
eE0
√
sin(kz − ωt),
me ω 1 + ǫ
vy = −
eE0 ǫ
√
cos(kz − ωt).
me ω 1 + ǫ
(2)
Since the electric dipole p̈ = −ev̇ = (e2 /m)E, the electron radiates at frequency ω. The polarization
in the generic n̂ direction is along the projection of the dipole momentum in the plane perpendicular
to n̂, i.e. p⊥ = (p̈ × n̂) × n̂. Thus the polarization is linear if n̂ = x̂ or n̂ = ŷ, and elliptical n̂ = ẑ.
If n = (sin θ, 0, cos θ) with cos θ = ǫ, the projections of px and py have equal modulus and are out
of phase of π/2. Thus, the polarization is ciricular.
b) The scattered power is
P =
2k0 2 2k0 e4 2 |p̈|
=
|E| ,
3c3
3m2e c3
(3)
where
1 E02
1
(1 + ǫ2 ) = E02 .
|E|2 = (Ex2 + Ey2 ) =
2
21+ǫ
2
(4)
Thus, the total power does not depend on ǫ and can be written as P = (4π/3)rc2 I where I = cε0 E02
is the intensity of the incident wave.
c) The magnetic field of the wave is
E0
[ŷ cos(kz − ωt) − ǫx̂ sin(kz − ωt)] .
B= √
c 1 + ǫ2
(5)
The only non-vanishing component of v × B is in the ẑ direction, and the force can be written as
Fz = −e(v × B)z = −e(vx By − vy Bx ) =
e2 E02 1 − ǫ2
sin(2kz − 2ωt) .
me cω 1 + ǫ2
(6)
The force vanishes for ǫ = 1, i.e. for circular polarization.
d) The −e(v × B) force drives dipole oscillations along z at 2ω frequency. Thus, in addition to
the above characterized scattering of radiation at ω frequency, we observe scattered radiation at 2ω
frequency, angularly distributed as sin2 φ around the z axis.
Since the dipole oscillating at 2ω is perpendicular to the dipole at ω, we can sum up the
corresponding powers. That due to the 2ω dipole is
2
2 2
2e2 k0 e6 k0
2k0 4π 2 eE0
(1 − ǫ2 )2
2
2
4 (1 − ǫ )
|F
|
=
P2ω = 3 |p̈2ω | =
E
=
r
I.
(7)
z
3c
3m2e c3
3m4e c5 ω 2 0 (1 + ǫ2 )2
3 c me ωc
(1 + ǫ2 )2
Notice that (eE0 /me ωc) is the ratio between the oscillation velocity and c.
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