I was trying to explain something about Brewster's angle and realized I don't completely understand how reflection and refraction work on the microscopic, classical level.
Consider a plane polarized light wave impinging on glass. The charges inside the glass oscillate in some way so that the original wave is canceled, and both a refracted and reflected wave are produced. Thinking just in terms of how charges make radiation, this is quite confusing. We start with a bunch of charges all oscillating in the same direction (presumably), and somehow the charges produce radiation in exactly three directions.
Moreover, the charges in the bulk don't even oscillate in the direction of the electric field of the incident ray. They oscillate along the field of the refracted ray.
This complicated pattern occurs because there are two distinct currents: the surface currents and the bulk currents. I'd like to know how these two currents collectively cancel the incident ray and produce the reflected and refracted ray. Which way do the surface currents move? Do they produce the reflected ray and cancel the incident ray alone, or does the bulk also contribute? How does this whole process start up dynamically for a finite wavepacket? Do the bulk charges always oscillate along the refracted ray or do some of them 'feel' the incident ray? All of this is hidden in the typical treatment which starts from Maxwell's equations in media and boundary conditions, which circumvent everything about what the charges are actually doing.
This isn't a duplicate of any of the many questions about reflection and refraction, because:
- I'm not interested in a quantum explanation, because we should be able to understand it classically.
- I'm not interested in an explanation from Huygens' principle, as it's too general — it never uses the fact that electromagnetic waves are polarized and transverse. I think the polarization structure here is important and the answer may differ for $s$-polarized and $p$-polarized waves. It also doesn't explain the mechanism by which the incident wave is canceled.
- I'm not interested in anything using the Fresnel equations, or really anything that starts from the electromagnetic boundary conditions. These are just consequences of how the charges in the glass are moving, so we shouldn't need them.
- I'm not interested in an explanation that only works at normal incidence; it's the three separate directions at oblique incidence that confuse me here.
I really hope there's a nice, fully classical explanation here, at the level of the charges!
Best Answer
Although you stated you are not interested in Huygens' principle, I want to add a note on this explanation. I will need in my answer the expression for the electric field of a radiating dipole
$$\boldsymbol{\rm E}\left(\boldsymbol{r},t\right)=-\frac{\omega^{2}\mu_{0}p_{0}}{4\pi}\sin\theta\frac{e^{i\omega\left(\frac{r}{c}-t\right)}}{r}\hat{\theta}$$
This expression assumes the dipole oscillates in the $\hat{z}$ direction. Now look for example at this picture
taken from here. This illustration seems to discard the polarization of the incoming wave (as you said), but if you think about it more - it turns out it doesn't. The radiation in the plane of incident is circular only for $s$-polarized light, since then each dipole is oscillating in the $z$ direction (in and out the page) and radiating a field given by
$$\boldsymbol{\rm E}\left(r,\theta=\frac{\pi}{2},\varphi,t\right)=-\frac{\omega^{2}\mu_{0}p_{0}}{4\pi}\frac{e^{i\omega\left(\frac{r}{c}-t\right)}}{r}\hat{z}$$
independent of $\varphi$ and $s$-polarized too. If, on the other hand, you want to treat $p$-polarized light, then each lattice point should radiate like in this image
taken from here, and it will definitely have other consequences from the $s$-polarized dipoles. A popular example is the existence of the Brewster's angle, which is the result of the dipole not radiating on its oscillations axis. Also, as before, you can see that the polarization of the far field radiation is parallel to the direction of the oscillations of the dipole. This means that the $p$-polarization is maintained.