Flavour violating muon decay Why is a muon decay into gamma and electron not possible?
Where/how does it violate the lepton flavour as muon and electron both have -1?
Reference: https://en.wikipedia.org/wiki/Lepton_number
 A: Flavour quantum numbers are routinely violated by weak interactions, but we can define a generation quantum number that is almost conserved. For example the number of first generation particles, i.e. number of electrons + number of electron neutrinos, is almost conserved. Likewise the number of muons + muon neutrinos and the number of taus + tau neutrinos.
I say almost conserved because neutrinos oscillate between the three neutrino flavours and this breaks the generation number conservation, but in most scattering events the probability of this happening is so tiny that it can be ignored.
And you can now see why $\mu \to e + \gamma$ is not allowed, because it changes the number of second and first generation particles present. For this to happen you would need something like:


*

*$\mu \to \nu_\mu + W^-$

*$W^- \to e + \bar \nu_e$

*$\nu_\mu \to \nu_e$

*$\nu_e + \bar\nu_e \to \gamma$
This is technically possible but the third step, i.e. the neutrino oscillation, is so improbable as to be effectively impossible.
A: The decay from muon to electron need to happen via the weak nuclear force. The electromagnetic force does nto allow this decay. Hence, during such a decay it needs to radiate a boson of the weak nuclear force. Since the change is not changes (the muon and the electron both carry negative charge) the radiated boson needs to be the $Z_0$ boson, which is neutral. This boson is massive and unstable. So it will then decay into a pair of particles, typically two neutrinos. Since the $Z_0$ boson is neutral. It does not couple to the electromagnetic force. Hence no photon is radiated.
