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Deriving the Compton effect

  1. Oct 10, 2006 #1
    Hi everyone,

    Right now, I am working on a hw problem asking me to derive the Compton effect, which is given by [tex]\lambda\prime-\lambda=\frac{h}{m_ec}(1-cos\theta)[/tex]

    A diagram of the situation can be found here: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/compton.html#c1 (However, in my diagram, the recoil electron and scattered photon are switched with the electron and [tex]\phi[/tex] below the x axis and the photon and [tex]\theta[/tex] above the x axis.)

    To derive the equation, I use the momentum conservation equations and energy conservation equation.

    There are two equations for the momentum, one for the x component and one for the y component:

    x-component: [tex]\frac{h}{\lambda}=\frac{h}{\lambda\prime}cos\theta+\gamma m u cos\phi[/tex]

    y-component: [tex]0=\frac{h}{\lambda\prime}sin\theta-\gamma mu sin\phi[/tex]

    where [tex]\gamma=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}[/tex]
    m=electron mass
    u=electron velocity after the collision

    My equation for energy conservation is:

    [tex]h\frac{c}{\lambda}+mc^2=h\frac{c}{\lambda\prime}+\gamma m c^2[/tex]


    To start off, I solved the x component of the momentum for cos(phi) and the y component for sin(phi). I then added the equations together and got (since sin^2 + cos^2 = 1):

    [tex]\frac{h^2}{\lambda\prime^2\gamma^2m^2u^2}+\frac{h^2}{\lambda^2\gamma^2m^2u^2}-\frac{2h^2cos\theta}{\lambda\prime\lambda\gamma^2m^2u^2}=1[/tex]

    At this point, I am to now eliminate u using the resulting equation and the equation for the energy conservation as given above. However, I am stuck on this next step because it seems as though any algabraic manipulation I try makes the equation extremely complicated. Could anyone please give me some hints as to what step should be next?
     
    Last edited: Oct 10, 2006
  2. jcsd
  3. Oct 11, 2006 #2

    OlderDan

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  4. Oct 11, 2006 #3

    Meir Achuz

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    The easy way to do it is (in natural units):
    1. Conservation of 4-momentum is k-k'=p'-p, where p=(m;0).
    2. Square each 4-vector, getting 2kk'(1-cos)=2mE'-2m^2.
    3. Conservation of energy is E'-m=k-k', so kk'(1-cos)=m(k-k').
    4. Divide by kk' and use lambda=2pi/k. Put in hbar/c if you want.
     
  5. Oct 11, 2006 #4
    Thanks for your replies. However, in the problem wording, it specifically tells us to start by eliminating "phi" first and then eliminate "u" from the 3equations. I have a hunch that this will be accomplished by substituting

    in [tex]\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}[/tex] for [tex]\gamma[/tex]. This eliminates gamma and also results in more "u"s, which may allow me to cancel them out. It's just getting to that point that is the problem.
     
  6. Oct 12, 2006 #5

    OlderDan

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    The link I posted outlines the approach that will get you there. The key is to express the electron energy in terms of its momentum and rest energy instead of the representation you used. Solve your last equation and the energy equation for momentum squared and set them equal. With a bit of Algebra many terms will cancel and you will have the result.
     
    Last edited: Oct 13, 2006
  7. Oct 13, 2006 #6

    Meir Achuz

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    Who would tell you to do it that way?
     
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