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Scattering of light ?

  1. Jul 5, 2010 #1
    Why does a light scatter when it interacts with a nitrogen molecule.
    I have read about Rayleigh and mie scattering , I was told that the photons are not getting absorbed and re-emitted by the nitrogen, So when i shoot photos into a jar of nitrogen why does it change the photons path , is it interacting with the nucleus or the electron cloud
    and how is it interacting with them.
  2. jcsd
  3. Jul 5, 2010 #2

    Good question; Crager.

    The answer is that (according to current modeling) optical light is primarily 'scattered' by the electrons in gases, (I assume you mean nitrogen gas and not liguid nitrogen).
    The optical frequencies are lower than the natural absorption / emission frequency of Nitrogen atoms (which is in UV), so no absorption; but the electric field of the incident light wave induces an oscillation in the electron, and this induced dipole moment radiates (since it is acceleration of charge), and thus more precisely it scatters the incident wave. = Rayleigh scattering
    Last edited: Jul 5, 2010
  4. Jul 5, 2010 #3
    The picture of the electron accelerating is how I've heard it explained as well. Classically, it makes perfect sense. But quantum mechanically, is it so simple? It seems to me that if we say an electron oscillates, it means that the wavefunction is slightly distorted, such that the electron is no longer in an eigenstate. Then the electron must be in a superposition of eigenstates, ie some of the other states of the system are mixed in. This starts to sound like there is some sort of quasi-absorption process going on, with finite occupational probabilities in other states.

    My QM is not exactly advanced. Am I way off the mark here, or is there any validity to this line of thinking?
  5. Jul 5, 2010 #4


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    Can this same mechanism in Rayleigh scattering occur in a solid, and be in part responsible for the refractive index ?
  6. Jul 6, 2010 #5
    yes you can have Rayleigh scattering in a solid , so we are saying that the electric field component of the photon is affecting the electron in the nitrogen molecule and altering its path .
  7. Jul 6, 2010 #6


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    Well, it's about half right I'd say. Think about the Zeeman and Stark effects - the levels will shift in response to an electrical or magnetic field (i.e. a photon). And as cragar said - it's mostly the electrical field. So even if you don't absorb the photon, you still have a response in the form of a perturbation of the electronic state. But absorption does not occur, and when the photon leaves, the system returns to where it was. (if it doesn't then it's Raman, not Rayleigh scattering) So you can look at it as if the photon 'bounces' elastically off the electron cloud.

    Raman (inelastic) scattering, on the other hand,you have excitation to a 'virtual' level followed by immediate re-emission. But as with virtual particles, this is just a way of interpreting the math - the virtual level does not actually exist.

    Finally, fluorescence is absorption and re-emission from a real level, and this occurs on a much longer timescale than scattering.
  8. Jul 6, 2010 #7
    Of course, moorobay; in fact, I think Rayleigh's original formulation was for a solid, namely, the aether (which was assumed to be solid).

    Rayleigh Scattering is typically elastic so the wavelegth of the incoming is unchanged upon scattering. In solids the situation is generally somewhat different and can result in inelastic (Raman) scattering so that energy is absorbed by the medium and the scattered radiation is of a different frequency,(as alxm mentioned).

    Yes, the two are intimately related. There are many ways to formulate Rayleigh Scattering....My comments were from a semi-classical description.

    The induced electric dipole from the light wave can also be thought of as a polarization of the medium which is in effect determines the refractive index of the medium.
    In fact the INTENSITY of the scattering light can be written in terms of the "polarizability" of the medium.... see 3rd page here for the formula ..
    http://physicsx.pr.erau.edu/Courses/CoursesF2008/PS495C/Rayleigh.ppt [Broken]

    Notice also that the intensity is dependent upon the frequency of the incident light, varying as inverse of the 4th power of the wavelength, meaning... in general, shorter wavelengths are scattered much more intensely.

    Last edited by a moderator: May 4, 2017
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