# Homework Help: Electron Muon Scattering

1. Apr 11, 2009

### giant_bog

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 $$-\frac{4\pi\alpha}{(p_1-p_3)^2} \bar{u}_3\gamma^\mu u_1 \bar{u}_4\gamma_\mu u_2$$.

3. 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 $$N=\frac{E_2+m_\mu c^2}{c}$$
$$\bar{u}_3 = u_3^\dagger\gamma^0 = n (1,0,p/n^2,0)$$ 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.

2. Apr 11, 2009

### giant_bog

post deleted

Last edited: Apr 11, 2009
3. Apr 17, 2009

### giant_bog

Now I got it. Never mind, y'all.