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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!

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  • $\begingroup$ Easy enough. Would a solution to Maxwell's equations suffice, provided you treat polarization as an oscillating current of "free" charges? $\endgroup$ – LLlAMnYP Dec 1 '17 at 15:31
  • $\begingroup$ The charges don't "oscillate along the refracted ray" - the only meaningful oscillations that mediate the process are surface currents, caused by oscillating microscopic dipoles on either side that go uncancelled. What's so wrong about starting from the Fresnel formalism to find that current? Keep in mind that you're talking about a monochromatic effect, i.e. one where any initial transients have died out, and any microscopic dipoles are responding to the fully established situation. (cont.) $\endgroup$ – Emilio Pisanty Dec 2 '17 at 17:18
  • $\begingroup$ The question isn't (and cannot be) how, at the onset of refraction, the microscopic dipoles start oscillating, since that would require a broadband formalism instead of a monochromatic one. Instead, the real question is (can only be) how the post-transient situation works and sustains itself, and that is simply (a nontrivial layer of) the interpretation of the Fresnel solutions. $\endgroup$ – Emilio Pisanty Dec 2 '17 at 17:18
  • $\begingroup$ @EmilioPisanty That's really what I had in mind! (i.e. a spatially large, but finite wavepacket) The rest of what you're saying is also intriguing -- there are differences between the surface currents and the currents inside the glass? I'd really appreciate it if you could elaborate. $\endgroup$ – knzhou Dec 2 '17 at 17:25
  • $\begingroup$ @knzhou Imposing a finite wavepacket is a whole other kettle of fish, and it explicitly steps out of the situation you demarcate in the question ("Consider a plane polarized light wave..."), so if you want answers to consider that situation you definitely need to do some substantial surgery on your question. Moreover, it doesn't really matter whether the wavepacket is "large" or not: if you're interested in transients, that just sets the timescale of interest to the length of the wavepacket itself. (Personally, I find that the onset of refraction isn't particularly interesting.) $\endgroup$ – Emilio Pisanty Dec 2 '17 at 17:38
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It seems that what you are really asking for is answered by the Ewald-Oseen extinction theorem.

https://en.wikipedia.org/wiki/Ewald%E2%80%93Oseen_extinction_theorem

The canonical derivation is in Born and Wolf.

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  • $\begingroup$ Unbelievable, that is exactly what I’m looking for! I’ll have a look at the reference as soon as I can, thanks. $\endgroup$ – knzhou Apr 11 '18 at 23:28
  • $\begingroup$ @knzhou I agree that this is what you’re looking for: a classical, microscopic theory of light-dielectric interaction. But to deal with non-normal incidence on a microscopic level, you won’t be able to escape Huygens’s principle. As described above, Huygens’ principle is more subtle than you give it credit, and it is the only way to connect the microscopic absorption/emission of atoms to the macroscopic phenomena of reflection, refraction, and interference. The macroscopic theory of polarizability (leading to Maxwell’s equations) is based on a superposition of individual dipoles. $\endgroup$ – Gilbert Apr 12 '18 at 15:09
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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

enter image description here

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

enter image description here

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.

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  • $\begingroup$ Okay, but from the standpoint of Maxwell's equations, we begin with a plane polarized wave (the incident wave), and after it hits a bunch of charges we end up with a reflected and refracted wave, which means the charges themselves must have generated a wave that canceled the incident wave, in addition to the reflected and refracted waves. Can you explain how the incident wave is canceled? $\endgroup$ – knzhou Dec 2 '17 at 16:01
  • $\begingroup$ By canceling the incident wave, do you mean that there isn't a backward propagating wave in the direction of incident? $\endgroup$ – eranreches Dec 2 '17 at 16:04
  • $\begingroup$ If you shine a light on a mirror, the atoms in the mirror actually emit two waves: one that propagates back out at you (the reflection), and one that propagates forward that exactly cancels the incident wave (so no light goes through the mirror). The latter one is the one I'm talking about; it seems hard to explain microscopically at oblique incidence. $\endgroup$ – knzhou Dec 2 '17 at 16:08
  • $\begingroup$ Oh, I see what you mean. I need to do some calculations. Meanwhile I'll try throwing a guess: maybe the wave we finally observe in the material is just the sum of the incident wave and the wave generated by the lattice atoms? In other words, maybe it is not correct to say that the green wavefronts compose the refracted beam - but the sum of the red and green is? $\endgroup$ – eranreches Dec 2 '17 at 16:17
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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.

We start with a bunch of charges all oscillating in the same direction (presumably), and somehow the charges produce radiation in exactly three directions.

Is the assumption that all the charges in the plane oscillate in the same direction incorrect? Does something different happen right at the interface? Otherwise, how can charges all oscillating the same way produce radiation in three directions, rather than one, two, or infinitely many?

All oscillators oscillate in the same direction, we know this from macroscopic theory where polarization has the same direction everywhere.

The net effect seems to be to cancel the primary wave in the medium and produce another wave of different wavelength and direction. (But this is only apparent effect in macroscopic theory. It does not mean that the oscillators in the medium do not experience force due to primary wave.)

This also happens only under special circumstances: a boundary that is long and smooth; the medium is dense enough, so the oscillators are close to each other so scattering is limited.

Should the boundary be rough or comparable in length to wavelength, the resulting radiation would be probably much more complex than two waves. Also if the medium was low density gas or dust, there would not be a single refracted plane wave, but the radiation would be more diffuse in all directions.

I know you are not interested in macroscopic reasons, but they are the most reliable explanation as they do not use any particular model of the medium. They are the guiding information to use when setting up and analyzing a microscopic model.

The above conditions translate into details of the microscopic model - oscillators are located in half space, they are distributed with uniform density high enough so mutual distances are much smaller than wavelength; their positions are limited by a planar boundary.

I do not know how to analyze mutual interactions of many particles under those conditions and answer the question: why in the macroscopic description the primary wave is not present and why the refracted awave has different wavelength and direction. Part of the answer is finding a reasonable connection between the microscopic field and macroscopic field for this kind of model, which is not easy (they have different wavelengths and the macroscopic field has to mimic the field of force experienced by the oscillators).

But it is plausible that if the oscillators are close enough and have uniform density (uniform index of refraction) and are excited by a single plane wave (primary wave), then the elementary secondary waves add up with random phases and tend to cancel each other except one direction, where they will reinforce each other.

It is similar to how diffraction on special gratings results in radiation being transmitted only in certain directions, or to phase-locked antenna arrays, which are designed to radiate in a few (single) desired directions.

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As is typical for these problems, let's assume a solution and then show that it satisfies Maxwell's equations. We'll have the interface of vacuum and a medium at the $x=0$ plane. The electric field shall be taken as

$$ \vec{E}(\vec{r_{x<0}}, t) = \exp(i(k y \sin\theta - \omega \tau)) \left\{0, 0, \exp(i k x\cos\theta) + r \exp(-i k x\cos\theta) \right\} $$

i.e. the sum of the incident and reflected wave in vacuum. On the other side of the boundary it shall be

$$ \vec{E}(\vec{r_{x>0}}, t) = \exp(-i \omega \tau) \left\{0, 0, \exp(i (x k_x + y k_y)\right\} $$

i.e. the refracted wave. The corresponding $\vec{B}$ fields are easily found from $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$.

$$ \vec{B}(\vec{r_{x<0}}, t) = \frac{k}{\omega}\exp(i(k y \sin\theta - \omega \tau)) \times \\ \left\{(\exp(i k x \cos\theta) + r \exp(-ikx\cos\theta))\sin\theta , (-\exp(i k x \cos\theta) + r \exp(-ikx\cos\theta))\cos\theta, 0 \right\} $$

$$ \vec{B}(\vec{r_{x>0}}, t) = \frac{1}{\omega}\exp(i(k_x x + k_y y - \omega \tau)) \left\{k_y t, k_x t, 0\right\} $$

By matching the $\vec{E}$ fields at $x=0$ we find $t=1+r$ and $k_y = k \sin\theta$. By matching the magnetic fields we also find

$$ t = \frac{2k\cos\theta}{k_x + k\cos\theta} $$

With some inspection we can realize that we have reached a possible valid solution which consists of three waves and satisfies Maxwell's equations. But why must it be so? Why can't we have $r=0, t=1, k_x=k \cos\theta$? The incident wavevector is given (i.e. $\omega/k=c$ and $\theta$ is defined).

We need to relate the wavevector inside a medium to the frequency and material properties. A classical approach says that our medium is a somewhat polarizable mix of negative and positive charge carriers which can be displaced by an electric field, optionally with friction and a restoring force. Putting an equation for a harmonic oscillator together with two of Maxwell's equations we get

$$ k \vec{P} + \gamma \dot{\vec{P}} + m \ddot{\vec{P}} = q^2 \vec{E} $$

$$ -\frac{\partial\vec{B}}{\partial x} = \mu_0 \dot{\vec{P}} + \dot{\vec{E}}/c^2$$

$$ \frac{\partial \vec{E}}{\partial x} = -\dot{\vec{B}} $$

Where $q$ is the effective charge density (depends on frequency, could be shell electrons, could be the ions, etc), $k, \gamma, m$ are the effective spring, damping, and mass constants normalized to unit volume. Finding the steady state solution of this set of equations will give us for a given $\omega$ a wavevector which is generally different from $\omega/c$, so $k_x$ will generally be different from $k \cos\theta$ and thus a solution for a wave incident on a surface will necessarily need both a refracted and a reflected wave.

It should not be too hard to reproduce this logic for a $p$ polarized wave.

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  • $\begingroup$ Thanks for the answer, but I've prefer to avoid the use of boundary conditions because they reflect something different and nontrivial happening right at the interface, i.e. the surface currents and the bound currents really play distinct roles. Is there any chance you could solve for the surface currents? What part of the fields do they produce? $\endgroup$ – knzhou Dec 4 '17 at 13:28
  • $\begingroup$ @knzhou But I didn't use any boundary conditions and didn't assume any surface currents, only volume currents ($\dot{\vec{P}}$). I simply demanded continuity of $\vec{E}$ and $\vec{B}$. I'll admit, though, that p-polarized light might actually admit a new level of complication since an x-component of the electric field would induced surface charge. $\endgroup$ – LLlAMnYP Dec 4 '17 at 13:32
  • $\begingroup$ I guess what I'm getting at is that I want a causal story of 'original field drives charges, so charges make new fields'. Just writing down a solution to Maxwell's equations totally skips over everything the charges are doing. It makes it unclear to me how the charges would know to oscillate along the refracted ray when originally we have nothing but an incident ray. $\endgroup$ – knzhou Dec 4 '17 at 13:37
  • $\begingroup$ @knzhou All I can really do here is show using Maxwell's equation that a self-consistent arrangement of fields and current, given a specific incident wave is what it is. It is also possible to numerically (but not analytically) integrate the fields of each individual dipole to find what the field is, but the problem with this approach is that we need to assume how exactly they oscillate. I'm not sure how feasible a full-blown integration with an arriving wavepacket is (in contrast to a steady-state solution). $\endgroup$ – LLlAMnYP Dec 4 '17 at 14:47

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