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Thomson scattering and unpolarized light

  1. Jul 17, 2012 #1

    Wox

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    Elastic scattering from a bound electron is classically described by considering the driven, damped harmonic oscillator model for the motion of a bound electron in a classical em-wave. The (non-relativistic) equation of motion is written as
    [tex]
    m\frac{d^{2}\bar{x}}{dt^{2}}=q\bar{E}(t,\bar{x})-m\omega_{0}^{2}\bar{x}-m\gamma\frac{d\bar{x}}{dt}
    [/tex]
    where the effect of the magnetic field is neglected. For a monochromatic plane wave [itex]\bar{E}(t,\bar{x})=\bar{E}_{0}e^{i(\bar{k\cdot\bar{x}}-\omega t)}[/itex] the solution is given by
    [tex]
    \bar{x}(t)=e^{-\frac{\gamma}{2}t}(C_{1}e^{Ct}+C_{2}e^{-Ct})-\frac{q\bar{E}_{0}}{m(\omega^{2}-\omega_{0}^{2}+ \omega\gamma i)}e^{i(\bar{k\cdot\bar{x}}-\omega t)}
    [/tex]
    where [itex]C=\sqrt{\frac{\gamma^{2}}{4}-\omega_{0}^{2}}[/itex], [itex]C_{1}[/itex] and [itex]C_{2}[/itex] constants depending on the boundary conditions, [itex]\omega_{0}[/itex] is the resonance frequency of the harmonic oscillator and [itex]\gamma[/itex] the damping coefficient. (The first part is known as the transient solution and the second as the steady state solution.)

    Now what would be the solution for unpolarized light?
     
    Last edited: Jul 17, 2012
  2. jcsd
  3. Jul 18, 2012 #2

    I think that these sites seem to cover your questions well :

    http://en.wikipedia.org/wiki/Thomson_scattering
    http://fermi.la.asu.edu/PHY531/thomson/index.html
     
  4. Jul 19, 2012 #3

    Wox

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    The links you gave me follow the usual reasoning. First they derive the result (i.e. the scattered intensity or the cross section) by assuming a monochromatic plane wave. This starts by solving the equation of motion of the electron as given in my original post (for a free electron [itex]\omega_{0}=0[/itex] and [itex]\gamma=0[/itex]). But when extending this to unpolarized radiation, I'm lost. The idea I get when reading it is that I should consider a monochromatic plane wave, but now with a random orientation of the electric field vector in the plane perpendicular to the propagation. That would mean
    [tex]
    \bar{E}(t,\bar{x})=\bar{E}_{0}(t)e^{i(\bar{k}\cdot \bar{x}-\omega t)}
    [/tex]
    with [itex]\bar{E}_{0}(t)[/itex] a vector in the plane perpendicular, with a particular norm but with a random orientation. This then means we can use the solution for scattering of a monochromatic plane wave
    [tex]
    \bar{E}(t,\bar{x})=\bar{E}_{0}e^{i(\bar{k} \cdot\bar{x}-\omega t)}
    [/tex]
    up till the point of time averaging (since [itex]\bar{E}_{0}[/itex] is now time dependent). Two reasons why this is not valid:
    1. I'm not sure the solution of the equation of motion of the electron is analog to the monochromatic plane wave, just replacing [itex]\bar{E}_{0}[/itex] by [itex]\bar{E}_{0}(t)[/itex]
    2. I've been told by several physicists that this is not the correct way of representing unpolarized radiation
    Therefore this reasoning is not valid and my question still stands.
     
  5. Jul 19, 2012 #4

    Ulf

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    Two identical waves with opposite polarization directions result in unpolarized light. So i suggest to try the superposition of a monochromatic polarized plane wave and a second wave with a 180°-phase shift. so that the polarization vector is always in the oppotise direction, resulting in zero net poolarization. but its only a guess, maybe it helps. i will sit down to the problme for myself later. ^^ have fun!
     
  6. Jul 19, 2012 #5
    Unpolarized light is the superposition of many wavelets of different polarization.

    The electric field vectors add linearly, so if in any small volume of space there were many many waves coexisting they would average out to a net result of zero electric field, i.e. no wave.

    Therefore if there is a wave with non-wero amplitude within our small volume, then it must have a polarization.

    Then how can you get unpolarized beams?

    It is more useful to think of "unpolarized" as average over time or over a finite volume much larger than the wave length and thus much larger than the atom. On the atomic scale light is polarized.

    One might make the argument that the electric field vector performs a random walk with non-zero result... in that case there will be a (smaller) resulting electric field vector and thus a resulting polarization.
     
  7. Jul 21, 2012 #6

    Wox

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    I'm reading "Statistical Optics" by Goodman to understand how unpolarized light can be rigorously treated and there is more to it than just the sum of some monochromatic plane waves.
     
  8. Jul 21, 2012 #7
    Born and Wolf treat it using the "coherency matrix" that has nothing to do with coherence; instead it is a density matrix that effectively provides a means of summing up individual plane, perfectly polarized waves.

    One way to rigorously treat scattering of polarization is to use Jones matrices that allow you to calculate how the density matrix (or better the polarization matrix) transform in a scattering process. Jones matrices are derived from the scattering of perfectly polarized waves.

    Transmission (including phase shifts and absoption) is just forward scattering and can be treated using the same formalism.

    http://arxiv.org/abs/1106.4446
     
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