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Refraction of Light

  1. Aug 3, 2004 #1
    Is the refraction of light a photon absorption-emission process?
    If so, why doesn't the beam scatter in all directions?
     
  2. jcsd
  3. Aug 4, 2004 #2
    Classically, it is not considered as absorption. It is n and related to the speed of light in the media, while absoption is kappa. the relation is N = n + ikappa. This is a from the 100% wave point point of view.

    Quantum mechanically, I would say that the fact that there is a plane boundary between the two media is important, and that the specular direction is closely related to the shape of the boundary. (This would also be true using the classical/plane wave point of view.).
     
  4. Aug 4, 2004 #3

    Claude Bile

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    The process is most definately not absorption, the incoming photon stimulates (or drives) the electron at the frequency of the photon, which in turn generates another photon (the reflected photon). Note that this process is not absorption as it does not promote the electron to a higher energy state.

    The direction of the reflected photon is due to the parallel component of the momentum of the photon being conserved.

    Claude.
     
  5. Aug 4, 2004 #4

    Claude Bile

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    Sorry, I wrote reflection instead of refraction.

    The same argument can be applied to refraction. The presence of a higher refractive index causes the parallel component of the photon's k vector to reduce by a factor of n (Where n is the refractive index of the glass). This is embodied in Snell's law.
     
  6. Aug 5, 2004 #5
    Can someone explain this to me in English?
     
  7. Aug 5, 2004 #6
    I agree with the first sentence, this is classical optics. But when the electron is "stimulated/driven", doesn't it gain energy? From a quantum point of view, doesn't it have to change energy level? Why isn't this absorption?
     
  8. Aug 5, 2004 #7

    Claude Bile

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    Why should the electron have to change an energy level from a quantum point of view?

    The energy from the incoming photon goes into emitting a new photon, not the electron.
     
  9. Aug 5, 2004 #8
    Because it seems to have done so from a classical point of view. It was relaxing, a photon came, it oscillated, a photon went, it relaxed again. During the seemingly brief time it oscillated out the next photon, it seems to have been affected somehow or done something, since the before and after photons are not the same.
     
  10. Aug 6, 2004 #9

    Claude Bile

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    I understand your confusion, but this is very difficult to explain without the aid of a diagram.

    The picture that of 'relaxed' and 'excited' electrons and that they spontaneously transform from one into the other is an oversimplification. If you are interested, look up pertubation theory, as it gives a better insight into what happens when a photon is incident on an electron.

    Ask yourself the following questions -

    How does light propagate through glass?
    What happens if an incoming photon has an energy that does not correspond to an allowed energy transition (and thus cannot be absorbed), does the photon simply ignore the electron?

    Claude.
     
  11. Aug 6, 2004 #10
    Excellent, thanks.
     
  12. Aug 14, 2004 #11
    What happens to the incoming photon when it "stimulates" the electron? If it is not absorbed, then you will end up with 2 photons for every 1 that goes in. The only way this can happen is if those 2 have half the frequency of the first. We know this cant be true because when you shine a certain colored light into glass, the same color comes out.
     
  13. Aug 14, 2004 #12

    Gokul43201

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    Just keep in mind that you don't really have distinct energy levels in solids like you do in individual atoms or molecules. You have bands. In a conductor or semi-conductor, the photon can excite the electron into the conduction band - though this is not required - but remember that most transmitters of light are insulators or dielectrics. But if the photon can not excite an electron, it can still excite a phonon or plasmon mode.

    I'm sure you'll find a decent explanation in Johnson or even Griffiths (perhaps) - or whatever is the standard E&M text where you are.

    "Why doesn't the beam scatter ?" - short answer : to conserve momentum.
     
    Last edited: Aug 14, 2004
  14. Aug 15, 2004 #13
    hmmm....thats over my head. Is that supposed to answer my question?
     
  15. Aug 15, 2004 #14
    A mode is a way to vibrate (a note on a guitar string, or a harmonic of that note, 2f, 3f, etc.). In a crystal, the atoms can vibrate in an organized manner and also have modes. When they vibrate in such a mode, we say a phonon is propagating. Gokul43201 is saying that if the energy of an incoming photon is not sufficient to change the energy (frequency) of an electron, it may still change the energy (frequency) of vibration of the atoms (associated "particle" : phonon, in analogy to photon, but for vibrating matter = sound = phono, phonograph, phonetics etc.). (Plasmon : not very familiar with this but probably the same for a plasma, or free electron gas in a a solid lattice).

    I don't see why you're saying that "If it is not absorbed, then you will end up with 2 photons for every 1 that goes in." This is not necessary, nor true IMO, except in non-linear frequency doubling.
     
  16. Aug 15, 2004 #15

    Claude Bile

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    Gonzolo, you are describing a photon/phonon interaction. This type of interaction usually involves some momentum transfer and thus a direction and/or wavelength shift of the incident photon.

    It is entirely possible (probable in some cases) that the photon will simply pass through the electron cloud. The E-field of the photon interacts with the atom by polarising it. This polarisation oscillates in phase with the E-field of the photon and propagates through the medium as a polarisation wave. The photon then emerges from the medium, unchanged in direction, wavelength or polarisation (Assuming that exotic effects such as the Faraday effect, or nonlinear effects are assumed not to be present).

    Claude.
     
  17. Aug 16, 2004 #16

    ZapperZ

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    Humm... However, when this occur, doesn't the polarization itself take in energy from the field? If you have a bunch of randomly-oriented dipoles, and you pass a field through it that will cause a net torque on each dipole to cause a specific orientation, this requires that work being down by the field. So each time the field oscillates, the act of polarization will use up some energy out of the field. So if what you said is the mechanism, there will be energy loss after passing through the cloud.

    I'm guessing that this scenario is similar to the beam-loading effect in particle accelerators. If you have a large amount of charge particles (electrons) in a part of the beamline, the RF field that you put in isn't the same as the RF field that is transmitted, because the charge particles literally suck up some of the RF field that is causing it to oscillate in the RF cavity. It is why dark currents in accelerating structures and photoinjectors are not desirable.

    Zz.
     
  18. Aug 16, 2004 #17
    Could this change in momentum, direction and wavelength be linked to what a change in n does at each interface? Varying index of refraction do change a beam's direction and wavelength.

    When polarisation occurs, doesn't anything change quantum mechanically? It also seems to me that the electron/atom/dipole/primitive cell did gain energy, because when the photon is through, the dipole will spring back to its original state if the photon polarisation stretched it. Also, photon beams are not always polarized.

    Nonlinear effects are related to saturation effects when very high fields (many many photons) or "sensitive" materials are involved. The question here as I understand it is only about one photon or the common linear regime in general.
     
    Last edited by a moderator: Aug 16, 2004
  19. Aug 16, 2004 #18

    Claude Bile

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    The direction change at an interface is due to continuity conditions imposed by Maxwell's equations at a boundary.

    Quantum mechanically, the photon does perturb the electron slightly, but the perturbation is over a very short time scale (femtoseconds or less). It is entirely possible that the electron experiences a change in energy over this time frame, but this energy is quickly removed as it propagates through the medium.

    And no, photon beams are not always polarised (in fact, obtaining perfect polarisation is quite impossible), a single photon will posess a polarisation.

    ZapperZ, I am not sure the exact details, as this is beginning to move outside my field of knowledge. Your question is a good one, I have a thought;

    the situation you describe would seem to indicate that a photon passing through a medium would leave a trail of aligned dipoles in its wake. This (to my knowledge) is not observed in practice. Perhaps there is a quantum mechanical explanation?

    Claude.
     
  20. Aug 17, 2004 #19
    What I understand from perturbation theory is that it is a change in the ground energy. Like when the classic mass-on-a-spring is in a G field. If this is the case, I'm almost convinced that polarizing an electron cloud doesn't necessarily change the electron's energy level.

    Just like the mass-on-a-spring doesn't change energy when it is switched from one G field to another.
     
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