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giant_bog
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Homework Statement
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 traveling 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.
Homework Equations
The amplitude is given as -\frac{4\pi\alpha}{(p_1-p_3)^2} \bar{u}_3\gamma^\mu u_1 \bar{u}_4\gamma_\mu u_2.
The Attempt at a Solution
As far as I can figure, p_1 = (E_1/c,0,0,p), p_3 = (E_1/c,0,0,-p), p_2 = (E_2/c,0,0,-p), and p_4 = (E_2/c,0,0,p), for the incoming and outgoing electron, and incoming and outgoing muon, respectively. That would make the spinors equal to:
u_1 = n (1,0,p/n^2,0)^T
u_3 = n (1,0,-p/n^2,0)^T
u_2 = N (1,0,-p/N^2,0)^T
u_4 = N (1,0,p/N^2,0)^T
where n=\frac{E_1+m_e c^2}{c} and N=\frac{E_2+m_\mu c^2}{c}
That makes the adjoints
\bar{u}_3 = u_3^\dagger\gamma^0 = n (1,0,p/n^2,0) and
\bar{u}_4 = u_4^\dagger\gamma^0 = N (1,0,-p/N^2,0)
These are all in the form of (A, 0, B, 0)\gamma^\mu(A, 0, B, 0)^T. It appears to work out that this is always 0 unless \mu=0. But \bar{u}_3 looks exactly like u_1^\dagger, so \bar{u}_3\gamma^0 u_1 = u_1^\dagger\gamma^0 u_1 = \bar{u}_1 u_1 = 2m_e c. The muon term works out the same way to 2m_\mu c. That makes the spinor contribution 4m_e m_\mu c^2
The (p_1-p_3)^2 term in the denominator should be ((E_1/c,0,0,p)-(E_1/c,0,0,-p))^2 = -4p^2 assuming the incoming and outgoing energies are the same, giving the final result M = \frac{4\pi\alpha m_e m_\mu c^2}{p^2}.
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.