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Rutherford cross-section from QED

  1. May 20, 2014 #1
    Hi everyone. I have a question about the calculation of Rutherford cross section in the context of QED. I know how to compute it using the usual four potential:
    $$
    A_\mu(q)=(\frac{e}{q^2},0,0,0)
    $$
    and taking the matrix element to be:
    $$
    \mathcal{M}=\bar u_{s'}(p')\gamma_\mu u_s(p)A^{\mu}(q).
    $$

    I was wondering if it is possible to compute it also by considering it as the process e+γ→e, i.e. taking the matrix element to be:
    $$
    \mathcal{M}=e\bar u_{s'}(p')\gamma_\mu u_s(p)\epsilon^\mu,
    $$
    where [itex]\epsilon^\mu[/itex] is the photon polarization.

    In this case I got (if I did everything correctly):
    $$
    \frac{1}{2}\sum_{spin}|\mathcal{M}|^2=2e^2q^2.
    $$

    My question is: how should I now integrate this to obtain the cross section? In order words, what is the phase space for this weird 2→1 process?

    Thanks
     
  2. jcsd
  3. May 20, 2014 #2

    Bill_K

    User Avatar
    Science Advisor

    No, but you can compute it from Møller scattering, that is, e + e → e + e, with exchange of a photon, in which M is something like

    $$
    \mathcal{M}=e^2 \frac{\bar u(p_1')\gamma^\mu u(p_1) \bar u(p_2')\gamma_\mu u(p_2)}{(p_1' - p_1)^2}
    $$

    and then take the nonrelativistic limit.
     
  4. May 20, 2014 #3
    Does it mean just to take the mass of one of the two electrons to be infinite?
     
  5. May 20, 2014 #4

    Bill_K

    User Avatar
    Science Advisor

    I don't think you need to, I think you can just go to the CM system and regard it as one of the electrons scattering off its mirror image.
     
  6. May 20, 2014 #5
    Sounds good, thanks!
     
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