# Electron-Muon Scattering Cross Section

1. Mar 18, 2014

### boltzman1969

Hi,

I am self-teaching Quantum Elctrodynamics, and have come across something which I do not understand. I would appreciate feedback from anyone on this specific issue from Atchison & Hey, "Guage Theories in Particle Physics" pg 238-239:

In calculating the u-channel electron-muon scattering amplitude at the one-photon exchange, one can simplify the calculation by introducing the electron and muon tensors:

LμγMμγ, where

Lμγ = 2[k'μkγ+k'γkμ+(q2/2)gμγ] (electron tensor) and

Mμγ = 2[p'μpγ+p'γpμ+(q2/2)gμγ] (muon tensor)

Now qμ = (k-k')μ = (p-p')μ is the 4-momentum of the exchanged photon; p, p' are the intial and final momenta of the muon; k, k' the initial and final 4-momenta of the electron.

It is claimed that qμLμγ = qγLμγ = 0, which is fine because L is the product of 2 4-currents and qμjμe- = 0. However, according to the text that I am reading, the qμLμγ = qγLμγ condition implies that we can replace p' in the muon tensor with (p+q); ie, Meffective = 2[2pμpγ + (q2/2)gμγ.

Does anyone know how to go from the condition qμLμγ = 0 to the constraint condition p' = (p+q)?

2. Mar 18, 2014

### samalkhaiat

The condition qμLμγ = 0 does not imply p' = (p + q). It is the opposite which is true.
There is only one diagram in the process, so the amplitude has two momentum-conserving delta functions: $\delta^{ 4 } ( k - k' - q )$ at the electron vertex and $\delta^{ 4 } ( p - p' + q )$ at the muon vertex. So when you do the integral $\int d^{ 4 }q$ you forced to put $p' - p= q = k' - k$. After that you can use the Casimir's trick the trace theorems to obtain the $| \langle \mathcal{ M } \rangle |^{ 2 }$.

3. Mar 19, 2014

### boltzman1969

Samalkhaiat,

Thank you for that reply. It is much clearer to me know how to obtain the scattering amplitude.

I was scratching my head trying to figure out the implication in the other direction.

The text was somewhat unclear in this regard.

Boltzman1969