What is a photon's speed inside a dieletric?

We know that EM waves are slowed down in a dielectric. But at what speed does the photons that make up the wave travel?

Do they always travel at the speed $c$, but colliding/being absorbed and re-emitted by the molecules in the medium causes the photons' average speed to be reduced?

Or does the electromagnetic field created by the polarization in the material somehow affects the photons at a quantum level to give them mass? If they do acquire mass, would the formulae $E = \hbar \omega$ and $E=\gamma m c^2$ work?

I suppose it's not absurd that an electromagnetic field gives a particle a mass – there is, after all, the famous Higgs field – but what is special about the electric field produced by a polarized dielectric? For example, the electric field of a point charge doesn't slow light down.

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Photons slow down because of collisions in the same way that it's hard to run down a busy sidewalk but easy to run down an empty one. Photons always move at $c$. –  santa claus Feb 13 '13 at 19:04
@Qmechanic The top answer to that question simply says that the apparent speed of light is lower due to nebulous "interactions [of the photons] with the atoms of the materials". "Interactions" can be anything. Are we talking about collisions like in the Drude model? Or do the photons get absorbed, then re-irradiated? Or something else altogether? –  Sandra Feb 13 '13 at 19:17
@Sandra: Did you also look at the 6 linked posts inside that question? –  Qmechanic Feb 13 '13 at 21:53

I imagine that one of my students asks me this very interesting question. This is how I would try to explain it to her/him, without the use of complex mathematics and concepts involving ‘screening currents’ and Higgs mechanisms etc.

It is known that in most part matter is empty space (vacuum) between the atoms. Therefore, although photons can be subjected to various 'obstacles' inside any tansparent solid, as they travel through the solid, from atom to atom, they do so at the normal speed of light as we know it $c=3\times 10^8 ms^{-1}$. However there are processes (the obstacles) the photons are subjected to as they travel. They are of quantum mechanical origin and these can be:

(i) absorption by an atom, and subsequent reemission by that atom

(ii) Compton scattering by electrons in the material.

(iii) Photoelectric effect

(iv) Excitation of vibrational modes in the solid which implies phonon excitation, hence energy conversion into heat.

The processes that involve absorption and reemission are the ones that propagate the photon that travels in the solid, and these may not change its frequency. However, these processes are characterised by their excitation and relaxation time. They are not taking place instantly. Therefore, although the photon travels at the normal speed of light when it is reemitted, it is delayed because these transitions take time to occur. This results in an effective speed of light inside the material which is macroscopically represented by the dielectric constant of the material, i.e. we can write $\epsilon=\epsilon_0\epsilon_r$. It is as if these processes change the electric permitivity of space, but in actual fact it is a collective phenomenon of the condensed matter. Therefore, the frequency of light does not change, as one would expect. In extremely dense materials such as some boson condensates, the effective speed of light is so slow that you can ‘watch’ the light diffusing through the material in a similar way you watch the milk diffusing in your tea (or coffee), so to speak.

Here is a practical analogy, which I hope will make my student understand something about processes in materials, and the effects these can have on their optical properties:

Practical Example Imagine two chains of people, chain 1 has 1000 people and chain 2 has also 1000. Now imagine you give one tennis ball (this the 'photon') to the first person of each chain, and instruct them to pass the ball to the next person and so on. Once the ball leaves any person in either chain it goes at the same speed c. However, imagine the people in chain 2 are slower in the passing of the ball than people in chain 1. The result will be that the ball will reach the other end slower in chain 2. Effectively the speed of 'light' is slower in chain 2 compared to chain 1.

Conclusion: The photons travel at the speed of light c between atoms inside the material , but the processes of propagation (absorption reemission etc ) make it appear as if it slows down.

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But from what I understand, an EM wave is a collection of zillions of "coherent" photons, which people generally explain to mean they are all "synched" (I'm not sure this is correct?). When a photon is absorbed by an atom, wouldn't it be later re-emitted at a random direction? And if all photons are being re-emitted at random directions, wouldn't this break the coherence and destroy the EM wave? –  Sandra Feb 16 '13 at 18:25
@Sandra Not all EMW are coherent. You are probably talking about LASER light. Coherent means all components in the field have same frequency and are at same phase-like a group of people walking in pace. When talking about absorption and reemission, or any particle interaction, momentum is conserved, so most likey the photon will be reemitted in the initial over all direction like in the Compton scattering. But yes, it is possible, by other mechanisms, to take other direction or be converted into heat or be reflected. This is how light gets attenuated going through matter, even glass. –  JKL Feb 16 '13 at 19:27

Photon in a dielectric is a quasi-particle, i.e., it is a collective mode. Many atoms (charges) take part in creating such a wave in a coherent way. So this wave may propagate slower than in vacuum.

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