As it is commonly defined {https://en.wikipedia.org/wiki/R_(cross_section_ratio), the cross section ratio is given by: $$R = \frac{\sigma(e^+e^-\to \mathrm{hadrons})}{\sigma(e^+ e^- \to \mu^+ \mu^-)} $$ Out of curiosity, why is the denominator chosen to be the cross section of the interaction $e^+ e^- \to \mu^+ \mu^-$ and not $e^+ e^- \to \ell^+ \ell^-$ for $\ell$ as another lepton flavour? Thanks guys.
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$\begingroup$ Can you write down the tree level diagrams involved? Do the electron and positron have to annihilate to a photon in your hypothetical if the product is $e^+e^-$? As for $\tau^+\tau^-$, that lepton was discovered after the popularity of R was established, and has a messier decay and detection. $\endgroup$– Cosmas ZachosApr 3, 2018 at 0:07
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Two things:
First of all, $e^+ + e^- \to e^+ + e^-$ has a very fundamental problem in that you don't know if a reaction we are interested in happened, or if a simple scattering event occurred instead.
Secondly muons are experimentally easy. Very, very easy. They are long-lived, highly penetrating, and have distinctive end-states; a combination that makes them about the most easily IDed particle there is.
Basically, if you can do it with muons it's going to be harder with any other particle.
answered Apr 3, 2018 at 2:39
dmckee --- ex-moderator kittendmckee --- ex-moderator kitten
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1$\begingroup$ Shouldn't you know that something happened by the change in momentum? Unfortunately, that something could involve t-channel exchange of a photon instead of s-channel annihilation followed by pair (re)creation. $\endgroup$ Apr 3, 2018 at 21:24
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$\begingroup$ @BertBarrois Uhm. Yeah. We don't know if the right reaction occured. $\endgroup$ Apr 4, 2018 at 18:57