Does light actually travel through glass?

Question

I am currently reading about the interactions between light and matter, but I keep coming across conflicting explanations.

My initial understanding (using classical electrodynamics) was that light (EM) waves do not actually travel through glass. Instead, they are absorbed by the atoms and/or electrons within the glass, which then emit new EM waves. The new EM waves emitted by the energised particles make up the forward-radiation, which is the light observed from the opposite side of the glass. So it appears as if light is travelling through the glass, but it's actually being absorbed and then re-emitted with (mostly) the same properties.

The article How Does Light Travel Through Glass? by Chad Orzel, based on a related SE question, confirms the above:

To understand the propagation of a wave through a medium, you can think of each component of the medium– atoms, in the case of a glass block– as being set into motion by the incoming wave, and then acting as a point source of its own waves.

But an answer to a different but related question, If light is made of particles, how does it pass through glass? conflicts with the above:

the electrons in glass are tightly bound to atoms so they are not free to move like the electrons in a metal and therefore they do not absorb the photons.

I know the above is referring to the quantum electrodynamics understanding, but I think it still conflicts as it claims that EM waves are not absorbed by the electrons in glass because they are too tightly bound to their atoms.

Also, the Wikipedia page on reflection provides a small explanation of the mechanics behind refraction within glass, which kind of conflicts with both of the above:

In the case of dielectrics such as glass, the electric field of the light acts on the electrons in the material, and the moving electrons generate fields and become new radiators. The refracted light in the glass is the combination of the forward radiation of the electrons and the incident light. The reflected light is the combination of the backward radiation of all of the electrons.

From my understanding, the above claims that some of the light is absorbed and re-emitted by the electrons (which conflicts with the linked answer), and that the rest of the incident light travels through the glass (makes up the rest of the refracted light) (which I think conflicts with the linked article, claiming the light is absorbed (although the article doesn't technically state whether all of the light is absorbed and re-emitted, so maybe that's where I'm going wrong(?))

I'm sure this is my misunderstanding, so I would be very grateful if you could clear it up for me:

When a light wave is incident on a sheet of glass, do the particles that make up the glass absorb the waves and re-emit them in the forward direction? Or do the light waves manage to travel through the glass, without being absorbed and then exit the glass?

Answer 1

The short answer to the core of your question,

do the particles that make up the glass absorb the waves and re-emit them in the forward direction? Or do the light waves manage to travel through the glass, without being absorbed and then exit the glass?

is that both processes are at play, though there are important subtleties involved in the term 'absorb'. The transmitted light is produced by the interference of the original beam and the radiation produced by the excitations it produces inside the glass, and it is this interference that produces the additional phase encoded in the glass's refractive index, which then trickles out to the phase velocity and the change in the wavelength.

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To see why this is the case, then the best place to start is with the macroscopic-electrodynamics description of the glass slab, in which the essential concept is that of the glass's polarization density $\mathbf P$. Electrodynamics is a linear theory, and it sees the problem of light propagating through glass as a superposition of two different sources:

• the initial light beam, which is present at the glass slab as well as beyond it, and
• the charge oscillations that initial light beam excites in the glass, which primarily emit (because they are essentially a phased array) in the initial beam's direction.
• (There are also surface terms produced by those charge oscillations, which produce reflections at the boundaries, but I'll ignore those here.)

The field after the glass slab is produced by the interference of the initial beam and the radiation produced by the oscillating charges inside the glass.

However, here is the important thing: the glass can do this induced-oscillations-then-emission game without absorbing any energy at all. In the steady state, the charge oscillations inside the glass are exactly 90° out of phase with the electric field that drives them, which means that the net power that they emit is being put in, in equal amounts, by the driver.

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If you want to go further than that and talk about photons, though, then you need to be a good bit more careful. If you're thinking about the light as a quantized object, then it is a quantized object that is coupled to the matter it is travelling through: in other words, it is a joint excitation of the EM field and the charge oscillations it produces in the glass.

In essence, then, the photon briefly becomes a polariton while it is in the glass ─ though the term is generally reserved for situations with much stronger coupling to resonant energy levels where there is a substantial population in those excited energy levels (or, more precisely, a substantial probability amplitude of excitation of each individual emitter to those energy levels) than what happens in an everyday dispersive dielectric. In glass, the probability of excitation (i.e. the "probability that the atom has absorbed a photon") is negligible as far as its contributions to the internal energy go (so it is often called a 'virtual transition'), but that probability amplitude does contribute to charge oscillations which re-shape the phase of the beam without altering its amplitude.

That said, of course, you still do need to provide the initial bit small of energy to take that small population to the excited states, and this is normally taken from the leading edge of the pulse while the monochromatic situation is being set up ─ and generally returned back to the field when the pulse leaves, as long as the medium is transparent. (Up to a point, of course - there's some strict constraints on how dispersive a medium can be without absorbing energy. However, there's plenty of transparent media where the absorption is extremely small and this can be disregarded.)

Answer 2

When a light wave is incident on a sheet of glass, do the particles that make up the glass absorb the waves and re-emit them in the forward direction? Or do the light waves manage to travel through the glass, without being absorbed and then exit the glass?

This answer is from the standpoint of non-quantum EM theory (no photons).

If this question is to make sense, we must assume that EM fields can be distinguished by their source. This may seem impossible to do in practice, but is actually mathematically possible assumption due to linearity of Maxwell's equations.

Then, your question can be reformulated in this way: does the primary field due to the light source stop at the surface of the glass and from there on the field inside the glass is that of the glass particles?

The answer to that question is definitely no. The principal reason is that in EM theory with the above assumption of identification of field with its source, the primary field cannot be blocked by other particles put in its way. The primary field in the form of outgoing wave just moves on as if no glass was there at all, determined by Maxwell's equations from the motion of the source particles and the usual assumption that fields are retarded (or some other assumption that singles out solution of the Maxwell equations consistent with the motion of the source particles).

Other particles like those in the glass may interact with this field and thus change their motion. They may emit their own EM field as a result of this interaction (secondary field) and thus influence the total EM field in the glass which is sum of the primary and secondary field.

EDIT

If the other particles interact with the primary field, cannot they not absorb the energy from it? And if so, can they not absorb the entire wave before it 'exits' the glass?

The particles can absorb energy from the EM field energy, but this does not mean the primary wave has to be altered (it cannot be, since it obeys Maxwell's equations with source terms that do not contain the glass particles). In macroscopic theory, the energy question is explained as follows. EM energy in any region is a function of the total EM field, it cannot be thought of as sum of independent contributions due to independent fields defined above (by connection to the source). Mathematically, this is because EM energy contains "interaction terms" each of which depends on two fields, in our case there will be cross term for primary and secondary field. We know the EM energy will change as the glass particles will absorb some energy from it. But as they do, they produce their secondary field that changes in time and makes EM energy change. This is how EM energy changes near the glass particles, even if the primary field is as if the glass wasn't there. The change in the secondary field is enough.

Also, how can the primary wave just move on as if no glass was there at all? Surely the particles at the surface of the glass also interact with and absorb some of the wave?

The particles interact in the sense they experience force. But they do not change the primary wave directly. This is because the primary wave obeys, by definition, Maxwell's equations where the source terms (charge and current density) are those of the light source only, there is no contribution due to the glass particles.

The glass particles do change the total field via their own field (secondary field). It is this total field that we observe to decay in amplitude (and describe by word 'absorption'), as the wave propagates into the glass medium.

Answer 3

A bit of both. Theres no way every photon could be absorbed by every glass molecule, so some do just pass through the empty space between molecules. But the light that actually strikes the glass molecules absorbs the photon and releases it again in the forward direction. Think of the atomic structure's rigidity ony allowing it to release the exact energy it receives, rather than portions of it with more kinetically energetic electrons.

Answer 4

I believe your confusion stems from the fact that the first reference uses a picture from quantum field theory (QFT) while the second uses a picture from quantum mechanics (QM).

In QFT the electron is allowed to "borrow" energy for a "short" amount of time. Hence, it can absorb the photon, even if the glass does not posses an energetically adequate electronic state -- an adequate electronic state would be a state, which fulfills energy conservation. However, in my opinion, it is not really helpful to think of this QFT process as a real absorption of the photon, but merely as an interaction between the electron and the photon. If you do so the two pictures become coherent: Both explain why we obtain an index of refraction $n$.