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ElectronMuon Scattering Cross Section 
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#1
Mar1814, 01:35 PM

P: 7

Hi,
I am selfteaching 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 238239: In calculating the uchannel electronmuon scattering amplitude at the onephoton exchange, one can simplify the calculation by introducing the electron and muon tensors: L^{μγ}M_{μγ}, where L^{μγ} = 2[k'_{μ}k_{γ}+k'_{γ}k_{μ}+(q^{2}/2)g^{μγ}] (electron tensor) and M^{μγ} = 2[p'_{μ}p_{γ}+p'_{γ}p_{μ}+(q^{2}/2)g^{μγ}] (muon tensor) Now q^{μ} = (kk')^{μ} = (pp')^{μ} is the 4momentum of the exchanged photon; p, p' are the intial and final momenta of the muon; k, k' the initial and final 4momenta of the electron. It is claimed that q^{μ}L_{μγ} = q^{γ}L_{μγ} = 0, which is fine because L is the product of 2 4currents and q^{μ}j_{μ}^{e} = 0. However, according to the text that I am reading, the q^{μ}L_{μγ} = q^{γ}L_{μγ} condition implies that we can replace p' in the muon tensor with (p+q); ie, M_{effective} = 2[2p_{μ}p_{γ} + (q^{2}/2)g^{μγ}. Does anyone know how to go from the condition q^{μ}L_{μγ} = 0 to the constraint condition p' = (p+q)? Thank you in advance for your assistance. 


#2
Mar1814, 04:28 PM

Sci Advisor
P: 892

There is only one diagram in the process, so the amplitude has two momentumconserving 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]. 


#3
Mar1914, 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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