Up: PHY531-PHY532 Classical Electrodynamics
Department of Physics and Astronomy
Arizona State University
Tempe, AZ 85287
Thomson scattering is the scattering of electromagnetic radiation by
a point particle of charge q and mass m . The scattering takes
place at a frequency and field strength such that the particle is
displaced very little during the period of the incoming wave.
The calculation can be broken down into these steps.
Write down the electric field of a plane wave.
Calculate the motion of a free charge in this electric field ignoring
radiation and the magnetic field. After calculating this motion, it is
a good idea to check
that the magnetic field's effect on the motion of the charge really
is negligible and that the motion of the charge really is small.
Write down the vector potential in terms of the current and the retarded
Evaluate the vector potential for a harmonic time dependence then evaluate
at a large distance from the source, keeping all terms that contribute
to the radiation.
For the case where the motion of the particle is much smaller than a
wave length, expand the exponential keeping only the leading order 1 term.
Use the divergence theorem to write the integral of the current in
terms of the dipole moment.
Evaluate the oscillating dipole moment of the oscillating charge, and
therefore derive the vector potential for the radiated fields.
Calculate the cross section from the definition that it is the
Energy per unit time radiated in a solid angle d divided
by the energy per unit time per unit area incident from the plane wave.
A plane wave traveling in the
direction with polarization
i and peak electric field E0 has an electric field
given by the real part of
where E0 is a complex amplitude: the magnitude is the peak amplitude of
the physical electric field, and the phase gives the phase of the wave
relative to zero at t = 0 . To satisfy Maxwell's equations,
i must be orthogonal. The magnitude
|| = /c .
Maxwell's equations also show that the magnetic induction is
A free charge at
p(t) will have the equation of motion
Let's assume that the particle is initially at rest, and the plane wave
is either turned on slowly, or there is sufficient damping that after
a long enough time the particle oscillates around the origin. Further,
let's assume that the applied field is weak enough that the amplitude
of the charge's oscillations is small. In that case,
are small and the charge's velocity is small; the magnetic field term
is the product of two small quantities and will therefore be much smaller
than the electric field term in this limit. We start by dropping the
magnetic field term. Since
p(t) is small, we can Taylor
series expand the electric field about the origin, and only the
(0,t) term is first order in small quantities. Therefore in this
limit, the equation of motion is
m = q.
Since this in now a linear equation, a particular solution with the
harmonic driving force is
m = q(0,t) .
where to get the physical position, we need to multiply by
e - it and take the real part. This solution satisfies the boundary
conditions that the particle oscillates around the origin. A general
solution would add A + Bt to this in each coordinate direction
where A and B are arbitrary constants.
Notice the general solution describes a particle moving at a velocity B
and starting position A . If this were the physical situation, we could
transform to a new frame where the particle was at rest at the origin.
Therefore, our assumption of our particle oscillating around the origin
is just a particular choic of coordinate system for our solution.
p = -
Before continuing let's make a quick order of magnitude check on our
|rp| = .
Air breaks down at around
Emax = 104 Volts/cm
or about 104/300 StatVolts/cm. mc 2 for an electron is 0.511 MeV,
With these values
|rp| (cm ) . Therefore rp will
be much smaller than a wave length as long as the wave length is much
smaller than a centimeter even for field strengths that would ionize air
at DC. Optical wavelengths are about a factor of 105 smaller, so
there, this approximation should be excellent. The magnetic field
force term is down by a factor of order
rp/c , which is
order rp/ , and again for optical frequencies this is down
by a factor of 105 or so even at these high fields. Therefore, our
approximations should be excellent.
Here we calculate the radiation from a known current distribution
which has a harmonic time depdence. That is the current is the
real part of
is a complex vector amplitude.
In Lorentz gauge, the vector potential is
which for a sinusoidal current becomes
() is now the complex amplitude for the vector
potential which needs to be multiplied by
e - it and the real
part taken to get the physical vector potential.
Since we primarily interested in the radiation field, we can expand
the vector potential for large r . In that limit
Radiation fields fall off like r - 1, so only the r term needs to
be kept in the denominator of the integrand. For r'/c arbitrary,
the next term cannot be neglected in the exponent since it is independent
of r as r becomes large. The higher order terms go to zero for
large r and can be dropped. The result for large r is
| - '|
r[1 - 2r - 1 ' + O(r'2/r 2)]1/2
r - ' + O(r'2/r 2)] .
( ) = d 3r'(')e - i '
Before going on, it is interesting to compare this expression to the
Fourier transform of the current amplitude
We see that the vector potential for radiation is the Fourier transform
of the current evaluated at
= /c multiplied
by an outgoing spherical wave. Even though the current has many other
Fourier components, the radiation field is only sensitive to the part
that is ``on the energy shell.'' That is which satisfies
k = /c ,
as required for plane wave solutions to Maxwell's source free equations.
You can also get to this point using Coulomb gauge. Since the scalar
potential satisfies Laplace's equation, the electric field from it
drops off like 1/r 2 or faster. Therefore in the radiation zone,
the scalar potential can be ignored. The vector potential is
calculated exactly as for the Lorentz gauge except that only the
transverse current is used.
The longitudinal current in Fourier space
is the component in the
direction. The fourier transform
of the transverse current is therefore just the component perpendicular
The coulomb gauge vector potential at large distances is therefore
If the region where the current is nonzero is small compared to the
wavelength, that is if
|r'|/c 1 , for all r' where
is nonnegligible, the exponential can be expanded in
powers of its argument and all but the 1 term can be dropped.
Since we assumed that the charged particle's oscillations were small,
this should be an excellent approximation. In that case, we can use
the standard trick which is essentially integration by parts, writing the
i th component of the current as
The integral of the current is then
(xi) - xi
(xi) - ixi()
where the divergence theorem and the vanishing of
distances lets us drop the first term in the expression for Ji ,
is the amplitude of the oscillating dipole moment.
d 3r() = - id 3r() = - i
The vector potential at large distances in this limit is
The magnetic induction is given by the curl
To give terms that go like 1/r , the gradient can only act on the exponent,
so the radiation field is
Maxwell's equations then give the electric field
( ) = - [ x ( x )] .
Similarly in Coulomb gauge
and the electric field is
c( ) = i[ x ( x )] .
which agrees with the Lorentz gauge result as expected.
The time averaged flux, which is the
energy incident per unit time per unit area, or equivalently the
power incident per unit area, is given by the time averaged Poynting
vector. Using complex notation,
For a plane wave, the magnitude of
are the same
and they are perpendicular, so that
Far from the an outgoing spherical wave, the same relationship can be used.
The total power radiated per solid angle d is therefore
s is the scattered field.
f is the polarization detected, only the portion
of the power corresponding to this polarization will be detected. That power
f must be perpendicular to
The differential cross section is the power radiated per solid angle divided
by the power incident per unit area, which for our case is
Evaluating the dipole moment,
The Thomson cross section becomes
Es() = - [ x ( x i] .
r0 = e 2/mc 2 is called the classical radius of the electron.
If two independent polarizations are measured, or, since the electric
field has a definite polarization, a single measurement with
along the polarization direction gives the polarization averaged cross
, and two perpendicular polarization directions
form a complete set, summing the cross section over the polarizations is
equivalent to summing over a complete set and subtracting the
For the incoming polarization along
, the cross section is
r02i* ff* i
For unpolarized incident light, we average over the incoming polarizations.
This can be done by averaging over all possible polarizations, or equivalently,
averaging over two independent polarizations. Let's use the second method.
If the incoming light is polarized along
, the cross section is
and averaging with the result along
The total cross section for unpolarized light is then
[1 + ]
Problem: What is the differential
cross section as a function of and , for a positive
helicity (i.e. circularly polarized) incoming plane wave to scatter
with outgoing positive helicity? What is the cross section for
a negative outgoing helicity?
Up: PHY531-PHY532 Classical Electrodynamics