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Electron Muon Scattering

  1. Apr 11, 2009 #1
    1. The problem statement, all variables and given/known data

    I'm having trouble with e-muon scattering. Tree level, no loops. (This is problem 7.26 in Griffiths Intro to Elem Particles). I get that the amplitude is as stated in the text, but I am having problems coming up with a number when the momenta and spins are added in.

    This is in the COM frame with the electron travelling up the z axis with momentum p and the muon going down the z axis with the same momentum. They then repel and go back the way they came. All helicities are assumed to be +1.

    2. Relevant equations

    The amplitude is given as [tex]-\frac{4\pi\alpha}{(p_1-p_3)^2} \bar{u}_3\gamma^\mu u_1 \bar{u}_4\gamma_\mu u_2[/tex].

    3. The attempt at a solution

    As far as I can figure, [tex]p_1 = (E_1/c,0,0,p)[/tex], [tex]p_3 = (E_1/c,0,0,-p)[/tex], [tex]p_2 = (E_2/c,0,0,-p)[/tex], and [tex]p_4 = (E_2/c,0,0,p)[/tex], for the incoming and outgoing electron, and incoming and outgoing muon, respectively. That would make the spinors equal to:

    [tex]u_1 = n (1,0,p/n^2,0)^T[/tex]
    [tex]u_3 = n (1,0,-p/n^2,0)^T[/tex]
    [tex]u_2 = N (1,0,-p/N^2,0)^T[/tex]
    [tex]u_4 = N (1,0,p/N^2,0)^T[/tex]

    where [tex]n=\frac{E_1+m_e c^2}{c}[/tex] and [tex]N=\frac{E_2+m_\mu c^2}{c}[/tex]
    That makes the adjoints
    [tex]\bar{u}_3 = u_3^\dagger\gamma^0 = n (1,0,p/n^2,0)[/tex] and
    [tex]\bar{u}_4 = u_4^\dagger\gamma^0 = N (1,0,-p/N^2,0)[/tex]

    These are all in the form of [tex](A, 0, B, 0)\gamma^\mu(A, 0, B, 0)^T[/tex]. It appears to work out that this is always 0 unless [tex]\mu=0[/tex]. But [tex]\bar{u}_3[/tex] looks exactly like [tex]u_1^\dagger[/tex], so [tex]\bar{u}_3\gamma^0 u_1 = u_1^\dagger\gamma^0 u_1 = \bar{u}_1 u_1 = 2m_e c[/tex]. The muon term works out the same way to [tex]2m_\mu c[/tex]. That makes the spinor contribution [tex]4m_e m_\mu c^2[/tex]

    The [tex](p_1-p_3)^2[/tex] term in the denominator should be [tex]((E_1/c,0,0,p)-(E_1/c,0,0,-p))^2 = -4p^2[/tex] assuming the incoming and outgoing energies are the same, giving the final result [tex]M = \frac{4\pi\alpha m_e m_\mu c^2}{p^2}[/tex].

    I know this is wrong, because the probability shouldn't be dependent on the initial momentum: I could make that momentum as low as I want and make the amplitude as high as I want. Besides, it differs from the answer given in the text and also the answer given in the Physics Bowl episode of The Big Bang Theory! Not sure where I went wrong, though.
     
  2. jcsd
  3. Apr 11, 2009 #2
    post deleted
     
    Last edited: Apr 11, 2009
  4. Apr 17, 2009 #3
    Now I got it. Never mind, y'all.
     
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