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Compton scattering of photons

  1. May 14, 2010 #1
    A quick question regarding compton scattering: if we consider the situation in which a photon incident on a free electron is scattered through an angle of 180 degrees, its energy essentially does not change. Since energy must be conserved, I assume this means there is no effect whatsoever on the energy, momentum, or velocity of the electron? Seems counter-intuitive somehow...

    EDIT: This is only for the case that [tex]E_{\gamma} = 2m_e[/tex]
     
    Last edited: May 14, 2010
  2. jcsd
  3. May 14, 2010 #2

    phyzguy

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    The photon's energy certainly does change. Compton scattering is essentially elastic scattering, like billiard balls. So if a photon is back-scattered off a stationary electron, it imparts energy and momentum to the electron. So the photon loses energy, and its frequency decreases.
     
  4. May 14, 2010 #3

    tom.stoer

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    But that's wrong; look at the change of the wavelength as a function of the photon scattering angle

    [tex]\Delta\lambda = \lambda_C(1-\cos\theta_\gamma)[/tex]

    So its wavelength and therefore its energy changes!
     
    Last edited: May 14, 2010
  5. May 14, 2010 #4
    Sorry, you're absolutely right. However, I'm talking about the specific case in which the photon energy [tex]E_{gamma} = 2m_e[/tex]. I should have mentioned that earlier.
     
  6. May 14, 2010 #5

    tom.stoer

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    Can you please explain? Wavelength and energy always change.
     
  7. May 14, 2010 #6

    phyzguy

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    Why do you think that special case is "special"? If I look at the Klein-Nishina formula, I find:

    [tex]\frac{Eout}{Ein}=\frac{1}{1+\frac{E_\gamma}{m_ec^2}(1-cos(\theta))}[/tex]

    This says that in the specific case you mentioned the outgoing photon has 1/5 the energy of the incoming, no?

    (Not sure why the tex in you're quote is messed up.)
     
  8. May 14, 2010 #7
    Sure.

    Initial photon energy: [tex]E = 2m_e[/tex]

    So [tex]\lambda = hc/2m_e[/tex].

    Using the compton formula, [tex]\lambda ' = h/m_e c(1-\cos\theta_\gamma) + \lambda [/tex], with \theta = 180:

    lambda' = 2h/m_e c + hc/2m_e = h/m_e (2/c + c/2) = 2h/m_e c

    EDIT: This is ridiculous. Latex refuses to work.
     
    Last edited: May 14, 2010
  9. May 14, 2010 #8

    tom.stoer

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    Fractions (and units)?

    [tex]\frac{2}{1} + \frac{1}{2} = \frac{3}{2}[/tex]

    Please check you calculations carefully.
     
  10. May 14, 2010 #9

    tom.stoer

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    I think we agree that for 180 degrees scattering angle we have

    [tex]\lambda^\prime = \lambda + \lambda_C(1-\cos\theta_\gamma) = \lambda + 2\lambda_c[/tex]

    Now we use

    [tex]\lambda = \frac{c}{\nu} = \frac{hc}{E_\gamma} =\frac{hc}{xm_ec^2} = \frac{1}{x}\lambda_C[/tex]

    where x means the fraction of the electron's rest energy. In your case x=2.

    Then we get

    [tex]\lambda^\prime = \frac{1}{x}\lambda_C + 2\lambda_c = \left(\frac{1}{x} + 2\right)\lambda_C[/tex]

    So again: the energy always changes, even for backward scattering and your special choice of energy
     
    Last edited: May 14, 2010
  11. May 14, 2010 #10
    If you shoot a photon (or laser beam) of energy Elaser head-on at a high-energy electron beam of energy

    Ee = γmec2

    the backscattered Compton photon energy is roughly

    Ebackscatter = 4γ2 Elaser. See Eqn (2) in

    http://physics.princeton.edu/~mcdonald/examples/accel/aoki_nim_a516_228_04.pdf [Broken]

    This is because there are two Lorentz transformations from the lab to center-of-mass coordinates and back. See Sections 37.1 and 37.2 in

    http://pdg.lbl.gov/2002/kinemarpp.pdf

    Bob S
     
    Last edited by a moderator: May 4, 2017
  12. May 15, 2010 #11

    tom.stoer

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    @leonidas24: note that the usual Compton formaula is valid only for scattering in the electron's rest frame. Bob_S' experimental is different.
     
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