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I Intuition for Rayleigh Scattering

  1. Sep 12, 2016 #1
    Is there some way to - from an intuition standpoint - justify the fact that there should be a factor of ##a^{6}##, (where ##a ## is the particle diameter) in the Rayleigh Scattering formula? I've seen a few sources hint that there should be. I can follow the derivation from e.g a Lorentz atom, but I don't see why I should immediately be thinking of the factor of ##a^{6}##? [Is it somehow related to a dipole moment?]

    Rayleigh Scattering Formula:

    $$ I \propto I_{0} \lambda^{-4} a^{6} $$
     
  2. jcsd
  3. Sep 13, 2016 #2

    Bystander

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    Square of volume?
     
  4. Sep 13, 2016 #3
    Your question got me interested to look around a bit. I found the following paper, Particle Optics in the Rayleigh Regime: http://patarnott.com/pdf/Moosmuller2009JAWMA.pdf

    It's more than I wanted to read right now, and I'm not sure it meets your requirement of being intuitive, but on page 1029 they state:

    Thus the r6 (for spherical particles) or more general V2 size dependence of Rayleigh particle scattering has been obtained from two simple facts: (1) the scattering cross section is proportional to the number of identical scatterers squared (i.e., n2); and (2) the number of scatterers (or molecules) in a particle is proportional to its volume, or to its radius cubed for a spherical particle.
     
  5. Sep 21, 2016 #4
    Thanks! Sorry for the slow reply - I've been away from a connection for a while. I guess as Bystander says it does make intuitive sense that you should have a factor of volume squared for scattering of two particles. But in that case why should the photon be seen as having the same volume as the scatterer? (as much as it makes sense for a photon to have a volume...) I think the second argument is convincing. Though I've also seen another derivation now in terms of the polarization of a sphere - where the scattering intensity is proportional to the square of the amplitude reflected field, the reflected field depends on the instantaneous dipole moment, and so we get ## a^{6}## again - I can no longer find it, which is a pain, and that's all I remember - but perhaps you can get something out of that?
     
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