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Elastic scattering and momentum conservation

  1. Sep 2, 2009 #1
    Perhaps this question is silly, but I don't entirely understand how elastic scattering of photons is even possible given that the directions of the incident/scattered photon differ. If there is a change in direction of the photons momentum, then there must be some momentum transferred to the scattering atom, hence the scattered photon should have less energy than the incident photon (which is not elastic scattering).

    The only way I could see elastic scattering make sense is if the nucleus mass is large enough so that the nucleus recoil is negligible (which is analogous to a ball bouncing off of a wall of infinite mass), but I find this hard to believe since moving atoms using photons is the basis of laser cooling.

    Another possible explanation is that the energy of the scattered photon *is* decreased, but the decrease is within the linewidth of photon energy distribution (..or something--this is a very iffy argument).

    Anyway, any help would be great. This question came up when talking about Rayleigh scattering, and I actually don't exactly understand how that works either from a quantum standpoint (i.e., not modelling light as a wave which drives a dipole).
  2. jcsd
  3. Sep 2, 2009 #2


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    Homework Helper

    you are correct--"elastic scattering" often means "quasi-elastic scattering". I.e., people neglect the recoil of the nucleus and whatnot. Cheers.
  4. Sep 3, 2009 #3
    Thanks. I'm surprised it was that simple...
  5. Sep 3, 2009 #4
    r tea-
    Not quite that simple, actually. We have to follow the equation for scattering from the plane wave derivation (very long wavelength), all the way through Rayleigh scattering (wavelengths equal roughly to visible light), resonant scattering (usually in the UV), Thomson scattering (long wavelength x-rays), and finally Compton scattering (short wavelength x-rays and high energy photons). I will not be using Jackson's notation, but one in which the classical electron radius is given by

    re = (e2/4 pi ε0 mc2) = 2.828 x 10-13 cm.

    In Slater and Frank "Electromagnetism" and other physics books, it is shown that the total cross section (probability of light scattering off a single atom in 1 cm2 total area) is

    σ = (8 pi/3) re2 (ω/ω0)4 cm2/electron

    where ω = 2 pi times the frequency of the light, and ω0 is the natural resonant electric dipole frequency of the atom. This applies only to frequencies lower than the electric dipole resonance frequency, usually in the UV where the absorpton is high (anomalous dispersion) and the index of refraction drops below n = 1.

    Above this region, the total cross section for long wavelength x-ray Thomson scattering is

    σ = (8 pi/3) re2 cm2/electron

    (about 0.66 x 10-24 cm2 = 0.66 barns where 1 barn = 1 x 10-24 cm2). Finally, when the mass of the electron becomes important, and there is measurable energy loss, the cross section is given by the Klein Nishina formula,


    and the scalttering kinematics is given by the Compton scattering formulas. See


    The fractional energy loss of a photon scattering off a free electron is:

    ω/ω0 = [1 + α(1-cos θ)]-1

    where ω is 2 pi frequency of scattered photon, ω0 is 2 pi frequency of incident photon, θ is angle of scattering, and α = hω0/(2 pi mc2), where mc2 is electron mass (511 KeV/c2), and h = Planck's constant. So at long wavelengths where hω0 << mc2, the frequency shift in ω/ω0 is negligbly small. The Klein Nishina cross section at low energies matches the Thomson cross section above.

    Finally, the discussion above applies to photon scattering on electrons. For scattering on nuclei of charge Z and mass M, the classical electron radius re is replaced by

    rn = Z2(m/M) re, where m and M are mass of electron and nucleus.
    Last edited: Sep 3, 2009
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