Why is e-e+ -> μ-μ+ such an important reaction?
It's the first example of "real" process worked out in Peskin and Schroeder, and they mention that it's used as a calibration for detectors. Can anyone tell me about this? Specifically, I'd naively expect e^- e^+ -> e^- e^+ to be easier to observe, because the right-hand mass is lower.
It's possible I could answer the question by computing the cross-section myself, or reading more, but I'm a mathematician and don't know anything about experiments so I'd probably get it wrong.
Posted Nov 28, 2016, 6:23 PM UTC
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So one of the reasons it's the first "real" process in Peskin and Schroeder is because it's one of the easiest to calculate. For e- e+ -> e- e+ you have a few diagrams, coming from the t and u channel as well as a s-channel diagram. For e- e+ -> mu- mu+ it's a bit simpler, and therefore reasonable to calculate on pen and paper.
Part of the calibration procedure at detectors (for there are many, not just one) is looking for the Z boson resonance peak; muons are nice experimentally for this because of their properties. They penetrate completely through a typical detector built on a collider and are not colour charged. Therefore they show up as a charged track in the tracking sectors far from the collision and we can be relatively confident in their ID.