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Compton Scattering

  1. Dec 27, 2006 #1
    1. The problem statement, all variables and given/known data
    The energy E and momentum P of a relativistic electron and energy ε and momentum p of a photon are related by:
    [tex]E^2 - P^2c^2 = m_{e}^2c^4 and \epsilon = pc[/tex]

    let [tex]E_{i}, P_{i} and E_{f}, P_{f}[/tex] denote the inital and final energies and momenta of the electron, and let [tex]\epsilon_{i}, p_{i} and \epsilon_{f}, p_{f}[/tex] denote the initial and final energies and momenta of the photon. Assume the electron is initially at rest, so that [tex]E_{i} = m_{e}c^2[/tex] and [tex]P_{i} = 0[/tex], and assume that the photon is scattered through the angle [tex]\theta[/tex].

    By considering the conservation of momentum show that:
    [tex]\epsilon_{i}^2 - 2\epsilon_{i}\epsilon_{f}cos\theta + \epsilon_{f}^2 = E_{f}^2 - m_{e}^2c^4[/tex]


    2. Relevant equations
    momentum before = momentum after


    3. The attempt at a solution

    initial momentum = [tex]P_{i} + p_{i} = \frac{\epsilon_{i}}{c}[/tex]
    final momentum = [tex]P_{f} + p_{f} = \sqrt{\frac{E_{f}^2-m_{e}^2c^4}{c^2}} + \frac{\epsilon_{f}}{ccos\theta}[/tex]

    therefore: [tex]\frac{\epsilon_{i}}{c} = \sqrt{\frac{E_{f}^2-m_{e}^2c^4}{c^2}} + \frac{\epsilon_{f}}{ccos\theta}[/tex]

    After a bit of rearranging I can get to:

    [tex]\epsilon_{i}^2 - \frac{\epsilon_{f}^2}{cos^2\theta} = E_{f}^2 - m_{e}^2c^4[/tex], where the RHS is as required, but I can't rearrange the LHS to get what they want. I get the feeling I may have left a cosθ out somewhere, or that I've forgotten quite an important (yet simple) technique in here somewhere, to rearrange the LHS. I can see that it could be written as the difference of 2 squares, but I cannot see how that would help me.

    Any hints appreciated.
     
    Last edited: Dec 27, 2006
  2. jcsd
  3. Dec 27, 2006 #2
    I can directly see a mistake in the way you got rid off the square root in :

    [tex]\frac{\epsilon_{i}}{c} = \sqrt{\frac{E_{f}^2-m_{e}^2c^4}{c^2}} + \frac{\epsilon_{f}}{ccos\theta}[/tex]

    you need to do SQUARE THIS :

    [tex]\frac{\epsilon_{i}}{c} - \frac{\epsilon_{f}}{ccos\theta}= \sqrt{\frac{E_{f}^2-m_{e}^2c^4}{c^2}}[/tex]

    But still, there is something wrong with the position of the cosine.

    marlon

    edit : beware, i did not check the entire algebra or the application of the conservation laws. I just denoted an obvious mistake :)
     
    Last edited: Dec 27, 2006
  4. Dec 27, 2006 #3
    Oh, ok so I can't just square all the terms then? I'll give it a go and see what happens.
     
  5. Dec 27, 2006 #4
    Yes you can, only you need to do it correctly :)

    you did (A+B)²=C² --> A² + B² = C²

    This is, ofcourse, wrong


    marlon
     
  6. Dec 27, 2006 #5
    So the cos is in the wrong place? Well I'll work through and see where it should go. On the top of the fraction maybe?
     
  7. Dec 27, 2006 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Remember that the initial and final momenta are vectors.
     
  8. Dec 27, 2006 #7
    Let's start all over.

    1) apply conservation of energy

    2) apply conservation of momentum

    From 1) write that equation as [tex](P_f c)^2[/tex] =

    From 2) we get : [tex]\vec {P_f} = \vec {p_i} - \vec {p_f} [/tex]

    The trick is to write 2) as [tex](P_f c)^2[/tex] = ; so that we can get rid off the electron parameters [tex](P_f c)^2[/tex] . To do that (AND THIS IS WHERE YOU MADE YOUR MISTAKE WITH THE COSINE) you need to be aware of the fact that the momenta are VECTORS.

    So [tex] (P_f c)^2 = c^2 \vec{P_f} \cdot \vec{P_f} = c^2 ( \vec {p_i} - \vec {p_f}) \cdot (\vec {p_i} - \vec {p_f}) [/tex]

    THIS IS A SCALAR PRODUCT !!!

    marlon
     
    Last edited: Dec 27, 2006
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