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Refraction index (or speed of light in a transparent medium) is often explaned as light interacting with atoms in a way such that the atoms create a secondary wave that once added up to the primary ends up delaying it.

This supposes that light is able to excite atoms one way or another. Considering that transparency is not quantized in the sense that a large bandwidth of colors can pass through a window, how is this atom excitation described ?

Manu de Hanoi
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    Does this answer your question? What is the mechanism behind the slowdown of light/photons in a transparent medium? – Thomas Fritsch Mar 28 '22 at 11:24
  • @ThomasFritsch I'm interested in this mysterious "non resonant vibration" mentionned in that page and poorly explained, that has to be both an energy level of the atom, and yet is not quantized. It seems contradictory...as explained in my question – Manu de Hanoi Mar 28 '22 at 11:37
  • @ThomasFritsch did you just vote to close my question without even reading my feedback ? – Manu de Hanoi Mar 28 '22 at 11:42
  • Your feedback came after my close-vote. I have retracted my vote now, since the other question seems to be not an exact duplicate. – Thomas Fritsch Mar 28 '22 at 13:33
  • The pointer by @ThomasFritsch is good. You are concerned that the light is interacting with the solid in an energy region away from an energy level? It's a necessary condition for transparency that the energy of the light be away from energy level. If it matched an energy level there would be absorption, so no transparency. No absorption does not be no interaction. The atoms respond by oscillating at the same frequency as the incident light. The oscillating atom is an oscillating dipole. The cited Q and A fills in the rest of the story. – garyp Mar 28 '22 at 21:34
  • @garyp the link doesnt answer my question, i'll rephrase it plainly for you :how can you excite an atom if it doesnt match any energy level ? It doesnt make sense. Energy levels are here ********precisely********** to tell us how an atom can be excited. ok ? – Manu de Hanoi Mar 28 '22 at 21:43
  • Two things. Energy levels are not precise. They are precise only in the idealized world where there is nothing in the universe except electrons and protons. But in fact there are other things in the universe. Atoms couple to these other things and what was an discrete energy level spreads out a little. Or a lottle, depending on circumstances. There is a classical analog that's mathematically just about identical: two coupled pendula. Uncoupled, both have the same frequency. When coupled, the system has two frequencies, one higher the other lower than the uncoupled frequency. (...) – garyp Mar 28 '22 at 23:06
  • @garyp There you go, you can post that as an answer if you think it's good enough – Manu de Hanoi Mar 29 '22 at 07:57
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    Answered and expanded, as suggested. – garyp Mar 29 '22 at 11:34

2 Answers2

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Two things. Energy levels are not precise. They are precise only in the idealized world where there is nothing in the universe except electrons and protons. But in fact there are other things in the universe. Atoms couple to these other things and what was an discrete energy level spreads out a little. Or a lottle, depending on circumstances. There is a classical analog that's mathematically just about identical: two coupled pendula. Uncoupled, both have the same frequency. When coupled, the system has two frequencies, one higher the other lower than the uncoupled frequency. In a solid, there are a very large number of atom-oscillators, and what was a single natural frequency for a single atom, becomes a very large number of oscillators that spread to become a band.

Even in a single atom far from other atoms in a vacuum the energy levels are not perfectly narrow. They have a small amount of spread. This is also due to coupling, in this case coupling between the atom and the electromagnetic field.

(Aside #1: recall two coupled pendula. If I start one oscillating while the other is at rest, the other pendulum will start to oscillate. Energy is transferred from one to the other under those conditions. Similarly, an excited atom can transfer energy to the objects that it is coupled to, that is, the EM field. This is a semi-classical [atom quantum, field classical] way of describing spontaneous emission)

(Aside #2: in a solid the electronic states also couple to states of motion of the nuclei, so-called phonons. Another mechanism for broadening.)

The second point is somewhat semantic. What do we mean by excite. Usually we mean drive an atom into a higher energy level. This is clearly not happening in a transparent solid. Actually, there is a tiny absorption/excitation on account of the coupling and spreading. The lack of (this meaning of) excitation does not mean that there is no interaction. If you admit a different, non-standard, definition of excite to mean interact you will have "excitation". The oscillating EM field can drive the electron cloud in the atom to slosh back and forth while the much heavier, lattice-locked nucleus remains relatively quiet. An oscillating dipole. And from there we go to the cited Q&A above.

garyp
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  • ok so in short, the matter forms a band (which one ? how is it called for transparent glass ?) of energy levels, that band, despite what you say, is "excited" ("sloshed"=excited), and emits an EM wave (you must be excited to emit EM radiation), therefore energy is tranfered from the incident light to the electrons, then the electrons create new light that adds up such that light appears slowed. Is that correct ? – Manu de Hanoi Mar 29 '22 at 13:05
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    The lowest electronic band in transparent media has an energy above the visible range of frequencies. Two things can happen upon exposure to radiation. The radiation can be taken in by the atom, which will eventually (but in short order) release the energy. It can be released to phonons, to lower-energy photons (fluorescence), or back to the radiation field at the same energy but with a phase that bears no relationship to the phase of the incoming radiation. The loss of "phase memory" occurs by an interaction that does not change the energy, e.g. a mild collision. These are absorption. – garyp Mar 29 '22 at 16:10
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    The second thing that can happen is the energy taken in is released back to the radiation field but maintaining a fixed phase relationship with the incident light. This is scattering. The channels that release lower-energy photons have no bearing on the question of transparency beyond the fact that some light of the original frequency is removed from the system. The part of the energy released with a phase relationship fixed with respect to the original radiation does so with a phase delay, i.e. a time delay. Light from all atoms interferes constructively. Transmission occurs. – garyp Mar 29 '22 at 16:19
  • The energy release with no phase relationship with the incident light cannot cause interference and does not contribute to transmission. For good transparent materials, this channel is very weak. Note that I'm being careful with my words. I am avoiding the word "excitation" and "emission". Those words are causing problems for you. My wording is still not perfect, as I'm not clear here about distinctions among absorption, coherent scattering, and incoherent scattering. You'll have to take more physics courses! – garyp Mar 29 '22 at 16:23
  • Thanks for the clarifications, so in short, transparent transmission is due to scattering of light by electrons, can you specify which flavor of scattering you mean (ex: rayleigh, compton etc...) ? – Manu de Hanoi Mar 29 '22 at 16:50
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    I think the closest would be Rayleigh, although one does not normally use the term in the context of condensed matter. The real key is that it's coherent (fixed phase relationship). Rayleigh scattering is of that type. What takes some math to show is that despite the scattering occurring in all directions, in a transparent solid all of the coherent scattering interference "conspires" to result in a forward traveling wave. – garyp Mar 29 '22 at 18:39
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As you are asking about atoms, a transparent medium is a lattice of atoms, and quantum mechanically the photons interact with the whole lattice. If the medium is truly transparent and keeps the frequencies , i.e. images are transmitted and color is not changed it means that the "photon -lattice" scattering is elastic , in the center of mass of the system "photon-lattice ". Because of the mass of the lattice, the laboratory system practically coincides with the center of mass system, very very small energy is needed to go from the center of mass to the laboratory system. Thus colors (energy) and phases are practically unchanged.

transparency is not quantized in the sense that a large bandwidth of colors can pass through a window,

As the interaction is with the lattice and the process practically elastic a large band width, depending on the material, is allowed. The limits would come from the wavelength of the light, i.e. the energy of the photon. High energy photons (xray) would interact with individual atoms and very low energy ones would be absorbed in heat, vibrational levels of the lattice.

In opaque solids, the photons interact with the surface atoms, there is no organized lattice to have an elastic scattering with.

anna v
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  • are you saying that the change of light speed is caused by elastic scattering ? Elastic scattering works on particles, and demand an exchange of energy between the particle scattered and the medium. However a photon, being quantized, can only exchange energy due to relativistic effects (change of frequency) or due to absobtion, how do you see this energy exchange happening ? – Manu de Hanoi Mar 28 '22 at 12:00
  • Elastic scattering Photon with a field in the center of mass has no energy exchange. Light speed and photon speed are two different variables. The photon always travels by c, for transparent media it is 1 in this list , coherent scattering http://ozradonc.wikidot.com/photon-interactions . Light is a confluence of zillion of photons, the speed of a light ray (classical) comes from the coherent addition of photons scattering elastically in all angles. Look at this single photon double slit experiment to see how the classical interference builds up from the quantum /photon additions. – anna v Mar 28 '22 at 12:06
  • https://www.sps.ch/artikel/progresses/wave-particle-duality-of-light-for-the-classroom-13/ – anna v Mar 28 '22 at 12:07
  • that aside, there are problems with scattering, the light would be dispersed, and it's not, as explained here : https://www.youtube.com/watch?v=CUjt36SD3h8 – Manu de Hanoi Mar 28 '22 at 12:08
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    It is not dispersed because it is in a quantum mechanical coherent state, each ray is a zilion photons in a quantum mechanical coherent state, Light is not the same as a photon, – anna v Mar 28 '22 at 12:16
  • You talk about scattering that doesnt scatter and light matter interaction that doesnt exchange energy. Well, thanks for your explanation, but I can't understand it. – Manu de Hanoi Mar 28 '22 at 12:25
  • In elastic scattering no energy is exchanged, only direction changes, which makes the individual photon path length longer than the geometric path of classical em. Look at the double slit single photon the color does not change. – anna v Mar 28 '22 at 13:24
  • perhaps you meant no NET energy is exchanged, that doesnt mean energy isnt exchanged temporarily like with anything elastic.... – Manu de Hanoi Mar 28 '22 at 13:48