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Electron-Muon Scattering Cross Section

by boltzman1969
Tags: cross, electronmuon, scattering
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boltzman1969
#1
Mar18-14, 01:35 PM
P: 7
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)?

Thank you in advance for your assistance.
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samalkhaiat
#2
Mar18-14, 04:28 PM
Sci Advisor
P: 910
Quote Quote by boltzman1969 View Post
Hi,

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

Thank you in advance for your assistance.
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: [itex]\delta^{ 4 } ( k - k' - q )[/itex] at the electron vertex and [itex]\delta^{ 4 } ( p - p' + q )[/itex] at the muon vertex. So when you do the integral [itex]\int d^{ 4 }q[/itex] you forced to put [itex]p' - p= q = k' - k[/itex]. After that you can use the Casimir's trick the trace theorems to obtain the [itex]| \langle \mathcal{ M } \rangle |^{ 2 }[/itex].
boltzman1969
#3
Mar19-14, 08:35 AM
P: 7
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


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