# Movement of lone electron in a plane-wave electromagnetic field

1. Jul 24, 2011

### Robin-Whittle

I would appreciate some help regarding the movement of an electron in a plane-wave linearly polarized electromagnetic field. Its simple enough question, but I am not sure what approach to take.

Let's say the electron is at rest in our frame of reference and the plane wave is continuous 300GHz, so it has a wavelength of 1mm. The electron is alone in space - there's no other particles for at least a few mm around. The field strength is not so high as to accelerate the electron to relativistic velocities.

The electrostatic field will move the electron a little side-to-side, coupling some of its energy to the electron, but the electron gives some or all of it back to the field, with a phase lag, just as a weight on a spring follows a sinusoidal motion, with a phase lag. Or does the electron's movement scatter some or all of this energy into different directions than that of the plane wave which excites it?

How much does the electron move? Where does the momentum come from to move it, since there is only the oscillating field in the vicinity? There's no other particle with momentum to push against.

If there were a thousand electrons all quite close to each other, say within 0.1mm of each other, I doubt that they would move as far as if there was just one, so the movement of even one electron will be affecting the field in its vicinity, weakening or at least somewhat phase-shifting it.

In high-power laser research, the electron's movement may be referred to with terms such as "quiver velocity", "quiver energy", "excursion amplitude" and "excursion length". But there, the electron has just been ejected from an atom, so there are other particles within a wavelength of the electron, so the papers I am turning up with these search terms are not very helpful.

Quantum electrodynamics, I think, would say that the electromagnetic wave doesn't interact with lone electrons, because that is not what photons do. But I am interested in classical theory which is applicable to this situation.

There's plenty of material on Thompson scattering, but that seems to be concerned with the behaviour of the radiation which is scattered, rather than the movement of the electron or whether it is within a few wavelengths of other charged particles.

- Robin

2. Jul 26, 2011

### chrisbaird

If the photon has enough energy, is will inelastically scatter with the electron. This means that the photon becomes a photon with lower energy (and frequency) and the electron gains kinetic energy. This is similar to Compton scattering. If the photon is low energy, the scattering will be elastic: the electron is momentarily excited and wiggles (the electric field of the photon exerts a force on the charged particle and accelerates it), but then relaxes again so that the scattered photon has the same energy and frequency as the incident one.

3. Jul 26, 2011

### chrisbaird

A lone electron (not bound to an atom or smashing into other electrons) is similar to the free ions in a perfect plasma: there is no damping and no local restoring force, so the electron effectively has a resonant frequency of zero.

4. Jul 26, 2011

### Robin-Whittle

Hi Chris,

Thanks for your replies. I am seeking a classical electrodynamic theoretical approach to this situation, in part because I don't believe in photons.

I agree that there is no restoring force or resonance for the lone electron. In the classical theoretical framework, the electron is moved first to the left and then to the right by the electrical field of the electromagnetic radiation. This repeats for every cycle of the radiation which passes by. The question is how to quantify that movement, and how that movement depends on the presence of other particles in the vicinity. This movement is surely continuous in a continuous field, not something "momentary" as you suggest in your first reply. That movement will definitely have an impact on the field, since the field is giving some of its energy to the movement, and also since some of that energy is lost as "scattered" radiation in a donut-shaped lobe of emission.

- Robin