Cross Section Difference Bhabha Scattering and Muon Pair Production

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SUMMARY

The discussion clarifies the differences in angular distributions between Bhabha scattering and muon pair production. Bhabha scattering, represented as e+ e- → e+ e-, involves an additional Feynman diagram allowing for both scattering and pair annihilation/production. In contrast, muon pair production, e+ e- → μ+ μ-, occurs solely through pair annihilation. The scattering amplitudes are defined as Mμ = (u2 + t2)/s2 and Mel = (s2 + u2)/t2 + 2u2/(ts) + Mμ, with cross-sections proportional to the squares of these amplitudes.

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Phileas.Fogg
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Hello,
how do the angular distributions of Bhabha Scattering

e^+ e^- \Rightarrow e^+ e^-

and Muon pair production

e+ e- \Rightarrow \mu^+ \mu^-

differ?

Regards,
Phileas Fogg
 
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The difference is caused by the fact that there is an additional Feynman diagram in Bhabha scattering. (The electron process can occur by scattering OR pair annihilation/production, whereas the muon process can only occur by the latter.)

This gives a scattering amplitude (in terms of the Mandelstam variables s,t,u) of

M_mu = (u^2 + t^2)/s^2

and

M_el = (s^2 + u^2)/t^2 + 2u^2/(ts) + M_mu

The cross-section is proportional to the squares of these quantities. In the center-of-mass frame (same as the lab frame for a two-beam collider), s has no angular dependence, and

t ~ 1-cos theta

and

u ~ 1+cos theta

So the relative angular dependence isn't hard to work out (though I don't know it offhand).
(These values are all from Quarks and Leptons by Halzen & Martin)

The rest is algebra. Note that this is a high-energy approximation, which assumes that the particle masses are negligible compared to the beam energy.

(Hope this is what you were looking for!) :smile:
 

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