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Up: PHY531-PHY532 Classical Electrodynamics

Thomson Scattering

K.E. Schmidt
Department of Physics and Astronomy
Arizona State University
Tempe, AZ 85287

1 Introduction

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.

2 The motion of a charge in a plane wave field

A plane wave traveling in the $\mbox{\boldmath$\space {\hat {k}} $}$ direction with polarization $\mbox{\boldmath$\space {\hat {\epsilon}} $}$i and peak electric field E0 has an electric field given by the real part of

E0$\displaystyle\mbox{\boldmath$ {\hat {\epsilon}} $}$ie i$\scriptstyle\mbox{\boldmath$ {k} $}$ $\scriptstyle\cdot$ $\scriptstyle\mbox{\boldmath$ {r} $}$ - i$\scriptstyle\omega$t . (1)

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, $\mbox{\boldmath$\space {\hat {k}} $}$ and $\mbox{\boldmath$\space {\hat {\epsilon}} $}$i must be orthogonal. The magnitude |$\mbox{\boldmath$\space {k} $}$| = $\omega$/c . Maxwell's equations also show that the magnetic induction is

$\displaystyle\mbox{\boldmath$ {B} $}$ = $\displaystyle\mbox{\boldmath$ {\hat {k}} $}$ x $\displaystyle\mbox{\boldmath$ {E} $}$ . (2)

A free charge at $\mbox{\boldmath$\space {r} $}$p(t) will have the equation of motion

m$\displaystyle{\frac{d^2 \mbox{\boldmath$ {r} $}_p (t)}{dt^2}}$ = q$\displaystyle\left [ \mbox{\boldmath$ {E} $}(\mbox{\boldmath$ {r} $}_p(t),t) + ...
 ...)}{dt} \times
\mbox{\boldmath$ {B} $}(\mbox{\boldmath$ {r} $}_p(t),t) \right]\,$. (3)

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, $\mbox{\boldmath$\space {E} $}$ and $\mbox{\boldmath$\space {B} $}$ 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 $\mbox{\boldmath$\space {r} $}$p(t) is small, we can Taylor series expand the electric field about the origin, and only the $\mbox{\boldmath$\space {E} $}$(0,t) term is first order in small quantities. Therefore in this limit, the equation of motion is

m$\displaystyle{\frac{d^2 \mbox{\boldmath$ {r} $}_p (t)}{dt^2}}$ = q$\displaystyle\mbox{\boldmath$ {E} $}$(0,t) . (4)

Since this in now a linear equation, a particular solution with the harmonic driving force is

$\displaystyle\mbox{\boldmath$ {r} $}$p = - $\displaystyle{\frac{q \mbox{\boldmath$ {\epsilon} $}_i E_0}{m \omega^2}}$ (5)

where to get the physical position, we need to multiply by e - i$\scriptstyle\omega$t 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.

Before continuing let's make a quick order of magnitude check on our approximations. Write |rp| = ${\frac{4 \pi^2 q E \lambda^2}{m c^2}}$ . 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| $\sim$ $\lambda^{2}_{}$(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$\omega$/c , which is order rp/$\lambda$ , 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.

3 The radiated field

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

$\displaystyle\mbox{\boldmath$ {J} $}$($\displaystyle\mbox{\boldmath$ {r} $}$)e - i$\scriptstyle\omega$t (6)

where $\mbox{\boldmath$\space {J} $}$ is a complex vector amplitude.

In Lorentz gauge, the vector potential is

$\displaystyle\mbox{\boldmath$ {A} $}$($\displaystyle\mbox{\boldmath$ {r} $}$,t) = $\displaystyle{\textstyle\frac{1}{c}}$$\displaystyle\int$d 3r'$\displaystyle{\frac{ \mbox{\boldmath$ {J} $}(\mbox{\boldmath$ {r} $},t-c^{-1}\v...
 ...h$ {r} $}'\vert}{\vert\mbox{\boldmath$ {r} $} - \mbox{\boldmath$ {r} $}'\vert}}$ (7)

which for a sinusoidal current becomes

$\displaystyle\mbox{\boldmath$ {A} $}$($\displaystyle\mbox{\boldmath$ {r} $}$) = $\displaystyle{\textstyle\frac{1}{c}}$$\displaystyle\int$d 3r'$\displaystyle{\frac{ \mbox{\boldmath$ {J} $}(\mbox{\boldmath$ {r} $},t-c^{-1}\v...
 ...h$ {r} $}'\vert}{\vert\mbox{\boldmath$ {r} $} - \mbox{\boldmath$ {r} $}'\vert}}$ (8)

where $\mbox{\boldmath$\space {A} $}$($\mbox{\boldmath$\space {r} $}$) is now the complex amplitude for the vector potential which needs to be multiplied by e - i$\scriptstyle\omega$t 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

|$\displaystyle\mbox{\boldmath$ {r} $}$ - $\displaystyle\mbox{\boldmath$ {r} $}$'| = $\displaystyle\sqrt{r^2 -2 \mbox{\boldmath$ {r} $} \cdot \mbox{\boldmath$ {r} $}' + r'^2}$   
  = r[1 - 2r - 1$\displaystyle\mbox{\boldmath$ {\hat {r}} $}$ $\displaystyle\cdot$ $\displaystyle\mbox{\boldmath$ {r} $}$' + O(r'2/r 2)]1/2   
  = r - $\displaystyle\mbox{\boldmath$ {\hat {r}} $}$ $\displaystyle\cdot$ $\displaystyle\mbox{\boldmath$ {r} $}$' + O(r'2/r 2)] . (9)
Radiation fields fall off like r - 1, so only the r term needs to be kept in the denominator of the integrand. For $\omega$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

$\displaystyle\mbox{\boldmath$ {A} $}$($\displaystyle\mbox{\boldmath$ {r} $}$ $\displaystyle\rightarrow$ $\displaystyle\infty$) = $\displaystyle{\frac{e^{i \frac{\omega}{c} r}}{cr}}$$\displaystyle\int$d 3r'$\displaystyle\mbox{\boldmath$ {J} $}$($\displaystyle\mbox{\boldmath$ {r} $}$')e - i$\scriptstyle{\frac{\omega}{c}}$$\scriptstyle\mbox{\boldmath$ {\hat {r}} $}$ $\scriptstyle\cdot$ $\scriptstyle\mbox{\boldmath$ {r} $}$' (10)

Before going on, it is interesting to compare this expression to the Fourier transform of the current amplitude

$\displaystyle\tilde{\mbox{\boldmath$ {J} $}}$($\displaystyle\mbox{\boldmath$ {k} $}$) = $\displaystyle\int$d 3r$\displaystyle\mbox{\boldmath$ {J} $}$($\displaystyle\mbox{\boldmath$ {r} $}$)e - i$\scriptstyle\mbox{\boldmath$ {k} $}$ $\scriptstyle\cdot$ $\scriptstyle\mbox{\boldmath$ {r} $}$ . (11)

We see that the vector potential for radiation is the Fourier transform of the current evaluated at $\mbox{\boldmath$\space {k} $}$ = $\mbox{\boldmath$\space {\hat {r}} $}$$\omega$/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 = $\omega$/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 $\mbox{\boldmath$\space {\hat {k}} $}$ direction. The fourier transform of the transverse current is therefore just the component perpendicular to $\mbox{\boldmath$\space {\hat {k}} $}$. The coulomb gauge vector potential at large distances is therefore

$\displaystyle\mbox{\boldmath$ {A} $}$c($\displaystyle\mbox{\boldmath$ {r} $}$ $\displaystyle\rightarrow$ $\displaystyle\infty$) = - $\displaystyle{\frac{e^{i \frac{\omega}{c} r}}{cr}}$$\displaystyle\left [ {\mbox{\boldmath$ {\hat {r}} $}} \times \left ({\mbox{\bol...
 ...ox{\boldmath$ {\hat {r}} $}} \cdot \mbox{\boldmath$ {r} $}'} \right ) \right]\,$. (12)

4 The Electric Dipole Approximation

If the region where the current is nonzero is small compared to the wavelength, that is if $\omega$|r'|/c $\ll$ 1 , for all r' where $\mbox{\boldmath$\space {J} $}$ 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
Ji = $\displaystyle\mbox{\boldmath$ {\nabla} $}$ $\displaystyle\cdot$ (xi$\displaystyle\mbox{\boldmath$ {J} $}$) - xi$\displaystyle\mbox{\boldmath$ {\nabla} $}$ $\displaystyle\cdot$ $\displaystyle\mbox{\boldmath$ {J} $}$   
  = $\displaystyle\mbox{\boldmath$ {\nabla} $}$ $\displaystyle\cdot$ (xi$\displaystyle\mbox{\boldmath$ {J} $}$) - i$\displaystyle\omega$xi$\displaystyle\rho$($\displaystyle\mbox{\boldmath$ {r} $}$) (13)
The integral of the current is then

$\displaystyle\int$d 3r$\displaystyle\mbox{\boldmath$ {J} $}$($\displaystyle\mbox{\boldmath$ {r} $}$) = - i$\displaystyle\omega$$\displaystyle\int$d 3r$\displaystyle\mbox{\boldmath$ {r} $}$$\displaystyle\rho$($\displaystyle\mbox{\boldmath$ {r} $}$) = - i$\displaystyle\omega$$\displaystyle\mbox{\boldmath$ {d} $}$ (14)

where the divergence theorem and the vanishing of $\mbox{\boldmath$\space {J} $}$ at large distances lets us drop the first term in the expression for Ji , and $\mbox{\boldmath$\space {d} $}$ is the amplitude of the oscillating dipole moment.

The vector potential at large distances in this limit is

$\displaystyle\mbox{\boldmath$ {A} $}$($\displaystyle\mbox{\boldmath$ {r} $}$ $\displaystyle\rightarrow$ $\displaystyle\infty$) = - i$\displaystyle\omega$$\displaystyle\mbox{\boldmath$ {d} $}$$\displaystyle{\frac{e^{i \frac{\omega}{c} r}}{rc}}$ . (15)

The magnetic induction is given by the curl

$\displaystyle\mbox{\boldmath$ {B} $}$($\displaystyle\mbox{\boldmath$ {r} $}$ $\displaystyle\rightarrow$ $\displaystyle\infty$) = i$\displaystyle\omega$$\displaystyle\mbox{\boldmath$ {d} $}$ x $\displaystyle\mbox{\boldmath$ {\nabla} $}$$\displaystyle{\frac{e^{i \frac{\omega}{c} r}}{rc}}$ . (16)

To give terms that go like 1/r , the gradient can only act on the exponent, so the radiation field is

$\displaystyle\mbox{\boldmath$ {B} $}$($\displaystyle\mbox{\boldmath$ {r} $}$ $\displaystyle\rightarrow$ $\displaystyle\infty$) = - $\displaystyle\omega^{2}_{}$$\displaystyle\mbox{\boldmath$ {d} $}$ x $\displaystyle\mbox{\boldmath$ {\hat {r}} $}$$\displaystyle{\frac{e^{i \frac{\omega}{c} r}}{rc^2}}$ . (17)

Maxwell's equations then give the electric field

$\displaystyle\mbox{\boldmath$ {E} $}$($\displaystyle\mbox{\boldmath$ {r} $}$ $\displaystyle\rightarrow$ $\displaystyle\infty$) = - $\displaystyle\omega^{2}_{}$[$\displaystyle\mbox{\boldmath$ {\hat {r}} $}$ x ($\displaystyle\mbox{\boldmath$ {\hat {r}} $}$ x $\displaystyle\mbox{\boldmath$ {d} $}$)]$\displaystyle{\frac{e^{i \frac{\omega}{c} r}}{rc^2}}$ . (18)

Similarly in Coulomb gauge

$\displaystyle\mbox{\boldmath$ {A} $}$c($\displaystyle\mbox{\boldmath$ {r} $}$ $\displaystyle\rightarrow$ $\displaystyle\infty$) = i$\displaystyle\omega$[$\displaystyle\mbox{\boldmath$ {\hat {r}} $}$ x ($\displaystyle\mbox{\boldmath$ {\hat {r}} $}$ x $\displaystyle\mbox{\boldmath$ {d} $}$)]$\displaystyle{\frac{e^{i \frac{\omega}{c} r}}{rc}}$ . (19)

and the electric field is
$\displaystyle\mbox{\boldmath$ {E} $}$($\displaystyle\mbox{\boldmath$ {r} $}$ $\displaystyle\rightarrow$ $\displaystyle\infty$) = i$\displaystyle{\frac{\omega}{c}}$$\displaystyle\mbox{\boldmath$ {A} $}$c($\displaystyle\mbox{\boldmath$ {r} $}$ $\displaystyle\rightarrow$ $\displaystyle\infty$)      (20)
which agrees with the Lorentz gauge result as expected.

5 Cross Section

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,

$\displaystyle\langle$$\displaystyle\mbox{\boldmath$ {S} $}$$\displaystyle\rangle$ = $\displaystyle{\frac{c}{8\pi}}$Re $\displaystyle\mbox{\boldmath$ {E} $}$ x $\displaystyle\mbox{\boldmath$ {B} $}$* . (21)

For a plane wave, the magnitude of $\mbox{\boldmath$\space {E} $}$ and $\mbox{\boldmath$\space {B} $}$ are the same and they are perpendicular, so that

$\displaystyle\langle$$\displaystyle\mbox{\boldmath$ {S} $}$$\displaystyle\rangle$ = $\displaystyle{\frac{c}{8\pi}}$|$\displaystyle\mbox{\boldmath$ {E} $}$|2 . (22)

Far from the an outgoing spherical wave, the same relationship can be used. The total power radiated per solid angle d$\Omega$ is therefore

$\displaystyle\lim_{r \rightarrow \infty}^{}$$\displaystyle{\frac{c}{8\pi}}$|$\displaystyle\mbox{\boldmath$ {E} $}$s(r,$\displaystyle\theta$,$\displaystyle\phi$)|2r 2d$\displaystyle\Omega$ , (23)

where $\mbox{\boldmath$\space {E} $}$s is the scattered field. If $\mbox{\boldmath$\space {\hat {\epsilon}} $}$f is the polarization detected, only the portion of the power corresponding to this polarization will be detected. That power is

$\displaystyle\lim_{r \rightarrow \infty}^{}$$\displaystyle{\frac{c}{8\pi}}$|$\displaystyle\mbox{\boldmath$ {E} $}$s(r,$\displaystyle\theta$,$\displaystyle\phi$) $\displaystyle\cdot$ $\displaystyle\mbox{\boldmath$ {\hat {\epsilon}} $}$f|2r 2d$\displaystyle\Omega$ . (24)

Note that $\mbox{\boldmath$\space {\hat {\epsilon}} $}$f must be perpendicular to $\mbox{\boldmath$\space {\hat {r}} $}$.

The differential cross section is the power radiated per solid angle divided by the power incident per unit area, which for our case is

$\displaystyle{\frac{d\sigma}{d\Omega}}$ = $\displaystyle{\frac{
\lim_{r \rightarrow \infty}
\vert\mbox{\boldmath$ {E} $}_s...
 ...heta,\phi) \cdot {\mbox{\boldmath$ {\hat {\epsilon}} $}}_f \vert^2 r^2}{E_0^2}}$ (25)

Evaluating the dipole moment,

$\displaystyle\lim_{r \rightarrow \infty}^{}$Es($\displaystyle\mbox{\boldmath$ {r} $}$) = - $\displaystyle{\frac{q^2 E_0}{mc^2}}$[$\displaystyle\mbox{\boldmath$ {\hat {r}} $}$ x ($\displaystyle\mbox{\boldmath$ {\hat {r}} $}$ x $\displaystyle\mbox{\boldmath$ {\hat {\epsilon}} $}$i]$\displaystyle{\frac{e^{i \frac{\omega}{c}r}}{r}}$ . (26)

The Thomson cross section becomes
$\displaystyle{\frac{d\sigma}{d\Omega}}$ = $\displaystyle{\frac{q^4}{m^2 c^4}}$|$\displaystyle\mbox{\boldmath$ {\hat {\epsilon}} $}$i $\displaystyle\cdot$ $\displaystyle\mbox{\boldmath$ {\hat {\epsilon}} $}$f|2   
  = r02|$\displaystyle\mbox{\boldmath$ {\hat {\epsilon}} $}$i $\displaystyle\cdot$ $\displaystyle\mbox{\boldmath$ {\hat {\epsilon}} $}$f|2 (27)
where r0 = e 2/mc 2 is called the classical radius of the electron.

6 Some Special Cases

If two independent polarizations are measured, or, since the electric field has a definite polarization, a single measurement with $\mbox{\boldmath$\space {\hat {\epsilon}} $}$f along the polarization direction gives the polarization averaged cross section. Since $\mbox{\boldmath$\space {\hat {r}} $}$, 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 $\mbox{\boldmath$\space {\hat {r}} $}$ component contribution,
$\displaystyle\sum_{\rm polarizations}^{}$$\displaystyle{\frac{d\sigma}{d\Omega}}$ = r02$\displaystyle\sum_{\rm polarizations}^{}$$\displaystyle\mbox{\boldmath$ {\hat {\epsilon}} $}$i* $\displaystyle\cdot$ $\displaystyle\mbox{\boldmath$ {\hat {\epsilon}} $}$f$\displaystyle\mbox{\boldmath$ {\hat {\epsilon}} $}$f* $\displaystyle\cdot$ $\displaystyle\mbox{\boldmath$ {\hat {\epsilon}} $}$i   
  = r02$\displaystyle\left [ 1 - \vert{\mbox{\boldmath$ {\hat {\epsilon}} $}}_i \cdot {\mbox{\boldmath$ {\hat {r}} $}}\vert^2 \right]\,$. (28)
For the incoming polarization along $\mbox{\boldmath$\space {\hat {x}} $}$, the cross section is

$\displaystyle\sum_{\rm polarizations}^{}$$\displaystyle{\frac{d\sigma}{d\Omega}}$ = r02$\displaystyle\left [ 1- \sin^2 \theta \cos^2 \phi \right]$ (29)

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 $\mbox{\boldmath$\space {\hat {y}} $}$, the cross section is

$\displaystyle\sum_{\rm polarizations}^{}$$\displaystyle{\frac{d\sigma}{d\Omega}}$ = r02$\displaystyle\left [ 1- \sin^2 \theta \sin^2 \phi \right]$ (30)

and averaging with the result along $\mbox{\boldmath$\space {\hat {x}} $}$ gives
$\displaystyle\sum_{\rm polarizations}^{}$$\displaystyle\left . \frac{d\sigma}{d\Omega}
\right\vert _$unpolarized = r02$\displaystyle\left [ 1- \frac{1}{2}\sin^2 \theta \right]$   
  = $\displaystyle{\frac{r_0^2}{2}}$[1 + $\displaystyle\cos^{2}_{}$$\displaystyle\theta$] (31)
The total cross section for unpolarized light is then

$\displaystyle\sigma$ = $\displaystyle{\frac{8\pi r_0^2}{3}}$ . (32)

Problem: What is the differential cross section as a function of $\theta$ and $\phi$ , 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?

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