Electron-Muon Scattering Cross Section

  1. 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.
     
  2. jcsd
  3. samalkhaiat

    samalkhaiat 1,117
    Science Advisor

    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].
     
  4. 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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