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I’ve been reading questions and answers to this topic and have conducted a few experiments as well to try and understand what happens to light in a medium. Somethings just don’t seem to make sense to me. I’m not requesting that all of the questions below be answered, even just one would be appreciated.

  1. Why does the reemission photon travel in the same direction as the photon that caused the reemission?

  2. How could the medium, glass, water etc. vibrate at visible light frequencies. Seems too fast.

  3. What info do the photons pass along the way that allows for the photons to exit the material in the same direction as those entering the material?

  4. Is there a physical cause for the reduced wavelength. I know it is necessary for conservation of energy, but how is the wavelength actually decreased?

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  • $\begingroup$ You seem to already know that light travels through glass with wavelength shorter than in vacuum. Could you reformulate the title of your question accordingly? You also seem to have a model in mind involving "readmission". Do you mean reemission ? Why do you adopt this model? Most physicists use the concept of polarisation. $\endgroup$ – my2cts Jul 2 '18 at 17:38
  • $\begingroup$ @my2cts Thanks for the spelling info. Not sure what another title would be. $\endgroup$ – Lambda Jul 2 '18 at 17:46
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    $\begingroup$ See physics.stackexchange.com/questions/247084/…, where I've discussed absorption/reemission, and why it is not the process found in transparent materials; then I give a version of Feynman's explanation for transmission through a transparent material. It should answer all of the points from your question. $\endgroup$ – Peter Diehr Jul 2 '18 at 18:09
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    $\begingroup$ @my2cts I’m looking for the interactions that create the events that we observe when light transits through a medium. What happens inside the water that makes the wavelength shorten? Why does the reemission photo follow the same path, etc $\endgroup$ – Lambda Jul 2 '18 at 23:00
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    $\begingroup$ Possible duplicate of Why is not everything transparent? $\endgroup$ – Stéphane Rollandin Jul 5 '18 at 20:08
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I will answer your questions but first we need to clarify:

there are three things that can happen when a photon interacts with an atom:

  1. elastic scattering, the photon keeps its energy and phase and changes direction

  2. inelastic scattering, the photon gives part of its energy to the atom and changes direction

  3. absorption, the photon gives all its energy to the atom, and the valence electron moves to a higher energy level as per QM

In the case of glass, this is elastic scattering, this is the only way (just like with a mirror) to keep the photons' energy and phase an to create a mirror image.

Now in the case of glass, the direction is the same as the original (in the case of mirror it is opposite), but the photons inside the glass are traveling perpendicular (or at a different angle then the original) to the glass's surface, and they are regaining the original angle of travel when exiting the glass.

enter image description here

Now your questions:

  1. the direction does change when the photons enter and exit the media. What is the same is the angle of refraction when entering and exiting.

  2. it is not absorption.

  3. it is not only the photons that make the angle of refraction, but the molecular and atomic structure of glass (the difference between the two medias)

  4. what gets reduced is the speed of the wavefront in the denser media. each photon still travels at speed c inside the media (since photons always travel in vacuum).

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  • $\begingroup$ A photon scattering is a process of absorption and reemission. For example, in the Compton scattering of a photon on an electron, the process is mediated by a virtual electron that carries the energy of the absorbed photon from the point of absorption to the point of emission of a new photon. You can clearly see this in the Feinmann's diagrams here: redberry.cc/documentation:tutorials:compton_scattering_in_qed $\endgroup$ – safesphere Jul 2 '18 at 19:01
  • $\begingroup$ @safesphere: but coherent transmission thru transparent material is modeled on Thomson scattering, not Compton scattering! $\endgroup$ – Peter Diehr Jul 2 '18 at 20:52
  • $\begingroup$ @PeterDiehr "Thomson scattering is the elastic scattering of electromagnetic radiation by a free charged particle, as described by classical electromagnetism. It is just the low-energy limit of Compton scattering" - en.m.wikipedia.org/wiki/Thomson_scattering $\endgroup$ – safesphere Jul 2 '18 at 21:52
  • $\begingroup$ Sorry, but both Compton and Thomson is wrong for this case. Compton scattering and Thomson scattering is when the wavelength is small compared to the atoms. Rayleigh scattering is when the wavelength is big compared to the atoms. For visible light (which the question is about), 400-700nm is bigger then the atoms. $\endgroup$ – Árpád Szendrei Jul 3 '18 at 0:27
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By mentioning photons in your question, you implicitly require a quantum description of the process. It the must fundamental level the quantum decription of the interaction of a photon with anything else is given by quantum electrodynamics (QED). In QED the photon is always absorbed and re-emitted.

However, it would be extremely cumbersome to analyse the propagation of a photon propagating through a piece of glass, using QED in its most fundamental way. It would require coherently summing up all the different possible paths that a photon can take scattering from atom to atom (i.e. by being absorbed and re-emitted by each of these atoms). For that reason we use some approximations, such as effective medium theory.

What we can also do is to say that a single photon is just a single excitation of the wave function for the electromagnetic field. If there are no nonlinearities, then the behaviour of the wave function and that of the classical optical field is described by exactly the same equations. For that reason, we can know exactly how a photon propagates through a piece of glass by studying how a classical field propagates through a piece of glass. Assuming that you already know that, you also already know what a photon does.

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