next up previous contents
Next: Radiation Reaction Up: Radiation from Point Charges Previous: Larmor's Formula   Contents

Thomson Scattering of Radiation

Suppose that a plane wave of monochromatic electromagnetic radiation is incident on a free particle of charge $e$ and mass $m$. The particle will experience a force from this field, and will accelerate. As it accelerates, it will emit radiation in different directions, dispersing the incident beam.

For a non-relativistic particle accelerated by a force we can see that:

\begin{displaymath}
\frac{dP}{d\Omega} = \frac{e^2}{16\pi^2\epsilon_0}
\frac{1}...
...psilon$}}^\ast \cdot \dot{\mbox{\boldmath$v$}} \right\vert^2
\end{displaymath} (18.47)

(where $\vert\hat{\mbox{\boldmath$\epsilon$}}^\ast \cdot \dot{\mbox{\boldmath$v$}}\vert^2 = \vert\dot{\mbox{\boldmath$v$}}\vert^2\sin^2\Theta$ for a particular polarization perpendicular to the plane of $\hat{\mbox{\boldmath$n$}}$ and $\dot{\mbox{\boldmath$v$}}$).

The (leading order) acceleration is due to the plane wave electric field with polarization $\hat{\mbox{\boldmath$\epsilon$}}_0$, wave vector $\mbox{\boldmath$k$}_0$, and Newton's Law:

\begin{displaymath}
\dot{\mbox{\boldmath$v$}} = \frac{e}{m} E_0 \hat{\mbox{\bol...
...e^{i\mbox{\boldmath$k$}_0 \cdot \mbox{\boldmath$x$}-\omega t}
\end{displaymath} (18.48)

If the charge moves much less than one wavelength during a cycle (true for all but the lightest particles and strongest fields) then
\begin{displaymath}
\vert\dot{\mbox{\boldmath$v$}}\vert _{\rm av} = \frac{1}{2}...
...t{\mbox{\boldmath$v$}} \cdot
\dot{\mbox{\boldmath$v$}}^\ast)
\end{displaymath} (18.49)

Thus the average power flux distribution is
$\displaystyle \left( \frac{dP}{d\Omega} \right)_{\rm av}$ $\textstyle =$ $\displaystyle \frac{c}{32 \pi^2
\epsilon_0} \left\vert E_0 \right\vert^2 \left(...
...mbox{\boldmath$\epsilon$}}^\ast \cdot \hat{\mbox{\boldmath$\epsilon$}}_0\vert^2$  
  $\textstyle =$ $\displaystyle \left\{\frac{e^2}{4 \pi \epsilon_0 mc^2}\right\}^2
\frac{\epsilon...
...mbox{\boldmath$\epsilon$}}^\ast \cdot \hat{\mbox{\boldmath$\epsilon$}}_0\vert^2$ (18.50)

This is clearly of the same general form as the scattering expressions we described and derived earlier. Since the result contains $E_0^2$ it makes sense to divide out the incident intensity and thus obtain a differential cross section that works for all but the strongest fields. We thus divide out the time-averaged flux of the Poynting vector of the incident plane wave:

\begin{displaymath}
I = \frac{\epsilon_0 c E_0^2}{2}
\end{displaymath} (18.51)

hence
\begin{displaymath}
\frac{d\sigma}{d\Omega} = \left\{\frac{e^2}{4 \pi \epsilon_...
...ilon$}}^\ast \cdot \hat{\mbox{\boldmath$\epsilon$}}_0\vert^2
\end{displaymath} (18.52)

If we let the plane wave be incident along the $z$ axis, let $\hat{\mbox{\boldmath$n$}}$ form an angle $\theta$ with that axis, and pick two polarization directions in and perpendicular to the $(\hat{\mbox{\boldmath$n$}},\hat{\mbox{\boldmath$z$}})$ plane (as before), and average over polarizations then this dot product yields:

\begin{displaymath}
\frac{d\sigma}{d\Omega} = \left\{\frac{e^2}{4 \pi \epsilon_0
mc^2}\right\}^2 \frac{1}{2} (1 + \cos^2 \theta).
\end{displaymath} (18.53)

as it did back in our earlier work on scattering, but now for a point particle.

This is the Thomson formula for scattering of radiation by free charge. It works for X-rays for electrons or $\gamma$-rays for protons. It does not work when the photon momentum and the recoil of the charged particle cannot be neglected. The integral of this,

\begin{displaymath}
\sigma_T = \frac{8 \pi}{3} \left\{\frac{e^2}{4 \pi \epsilon_0
mc^2}\right\}^2
\end{displaymath} (18.54)

is called the Thomson cross-section. It is $0.665 \times
10^{-29}$ m$^2$ for electrons.

The quantity in parentheses has the units of length. If the total ``mass-energy'' of the electron were due to its charge being concentrated in a ball, then this would be the close order of the radius of that ball; it is called the classical electron radius. This number crops up quite frequently, so you should remember it. What it tells us is that even point particles have a finite scattering cross-section that appears in this limit to be independent of the wavelength of the light scattered.

However, this is not really true if you recall the approximations made - this expression will fail if the wavelength is on the same order as the classical radius, which is precisely where pair production becomes a significant process quantum mechanically. In quantum mechanics, if the energy of the incident photon $\hbar \omega \approx mc^2$ for the electron, significant momentum is transferred to the electron by the collision and the energy of the scattered photon cannot be equal to the energy of the incident photon. Whatever a photon is ...

We can actually fix that without too much difficulty, deriving the Compton scattering formula (which takes over from Thomson in this limit). This formula adds a wavelength/angle dependence to Thomson's general result and yields the Klien-Nishina formula, but this is beyond our scope in this course to derive or discuss in further detail.

We are almost finished with our study of electrodynamics. Our final object of study will be to to try to address the following observation:

Accelerated charges radiate. Radiation accelerates charge. Energy must be conserved. These three things have not been consistently maintained in our treatments. We study one, then the other, and require the third to be true in only part of the dynamics.

What is missing is radiation reaction. As charges accelerate, they radiate. This radiation carries energy away from the system. This, then means that a counterforce must be exerted on the charges when we try to accelerate them that damps charge oscillations.

At last the folly of our ways is apparent. Our blind insistence that only retarded fields are meaningful (so that we can imagine the fields to be zero up to some time and then start moving a charge, which subsequently radiates) has left us with only one charge that can produce the field that produces the force that damps applied external forces -- the charge itself that is radiating. No other charge produces a field that can act on this charge ``in time''. We have invented the most sublime of violations of Newton's laws - an object that lift's itself up by its own bootstraps, an Aristotelian object that might even be able to come to rest on its own in the absence of external forces.

Clearly we must investigate radiation reaction as a self-force acting on an electron due to its own radiation field, and see if it is possible to salvage anything like a Newtonian description of even classical dynamics. We already know that Larmor radiation plus stable atoms spells trouble for Newton, but Newton still works classically, doesn't it?

Let's take a look. Uh-oh, you say. Wasn't the, well, wasn't everything singular on a point charge? Won't we get infinities at every turn? How will we realize finite results from infinite fields, potentials, self-energies, and so on?

Yes! I cry with glee. That's the problem. Finally we will learn how to take a singular field, a singular charge, and infinite energy, and make a physically realized (almost) radiation reaction force out of it.


next up previous contents
Next: Radiation Reaction Up: Radiation from Point Charges Previous: Larmor's Formula   Contents
Robert G. Brown 2013-01-04