Question 26.5: Muon Decay Examine the process of muon decay μ^- → e^- + νe ...
Muon Decay
Examine the process of muon decay \mu^{-} \rightarrow e^{-}+\bar{\nu}_{e}+\nu_{\mu} from the standpoint of conservation laws.
The particles here are all leptons. With the muon and electron and their corresponding neutrinos involved, we need to consider both electron lepton number L_{e} and muon lepton number L_{\mu} .The appropriate numbers are
\text { Muon and muon neutrino: } L_{\mu}=1.
\text { Electron: } L_{e}=1.
\text { Electron antineutrino: } L_{e}=-1.
Therefore, the muon lepton number is L_{\mu}=1 both before and after the reaction. The electron lepton number is zero before the reaction, and after the reaction the two L_{e}numbers add to zero.
Therefore, both lepton numbers are conserved. Electric charge is also conserved; neutrinos are uncharged, so the charge before and after the reaction is the same: -e.
REFLECT It wasn’t necessary to consider baryon number in this reaction, which only involved leptons. And there’s no problem with energy conservation, because the electron is lighter than the muon and the neutrinos have nearly zero rest energy.