Why do we refer the cross section ratios to muons?

Question

For electron-positron interactions, we have different cross sections, depending on the available reaction energy. To get an overview how many particles of a certain type have been created, we can calculate cross section ratios, f.ex. \begin{equation} R_{had}=\frac{\sigma \left ( e^-e^+ \rightarrow \text{hadrons} \right ) }{\sigma \left ( e^-e^+ \rightarrow \mu^-\mu^+ \right ) } \end{equation}

Is there a reason why this ratio is mostly referred to (normalized to) the muon pair production and not f.ex. to the electron pair production?

Answer 1

Muons are easy to measure. They're extremely long-lived relative to most other particles, and they're pretty massive compared to electrons, which means they have substantial penetrating power that is essentially unmatched by other charged particles. There's a reason that muon detection systems are put on the very outer edges of detectors - muons can punch through every calorimeter and keep going. Electrons are light enough that they typically get stopped in the electromagnetic calorimeter, if they even make it to that point. As a result, they're typically harder to distinguish from other particles with similar signatures.

Answer 2

To first order, muon pair production:

$$ e^+ + e^- \rightarrow \mu^+ + \mu^- $$

only proceeds via the $s$-channel. That is, the electron position pair annihilate into a virtual photon or Z-boson, which then decays to the final state.

Meanwhile:

$$ e^+ + e^- \rightarrow e^+ + e^- $$

has both an $s$ and $t$ channel amplitude, making comparison with $q\bar q$ processes less clear. In the $t$-channel, the particles scatter by exchanging a photon. Colloquially, one would say the detected particles are the same ones from the colliding beam, but that's a bit deceptive, since all electrons (positrons) are identical. Identical particles means the $s$ and $t$ amplitudes interfere, further complicating comparison with:

$$ e^+ + e^- \rightarrow q + \bar q $$

Neglecting QCD effect, $R$ should be independent of $\sqrt s$ (kinematically), and only depend on the number of quarks available in the final state giving the total collision (threshold effects):

$$ R_{QED} = \frac{\sum_q{e^2_q}}{e^2_{\mu}}$$

where $q$ runs over quark flavors and $e_q$ ($e_{\mu}$) are the quark (muon) charges.