# Why the $\beta^-$ decay releases $e^-+\nu_e$ and not $\tau$ or $\mu$ generations?

The $\beta^-$ decay is

$$n \rightarrow p + e^- + \bar\nu_e \tag{1}$$

It can be broken down into

\begin{align} d \rightarrow & \ u + W^- \tag{2}\\ W^- \rightarrow & \ e^- + \bar \nu_e \tag{3} \end{align}

Why (3) cannot decay into $(\mu^- + \bar \nu_\mu)$ or $(\tau^- + \bar \nu_\tau)$ ?

• It can and does unless forbidden by energy conservation (which it is in this case). – knzhou Jan 30 '18 at 14:51
• @knzhou, Do you mean that the $W^-$ boson, in this case, has not enough energy to produce a heavy lepton? But then, how does it come that the light quark $d$ can decay into a heavy lepton $W^-$? – Stefano Jan 30 '18 at 14:56
• The $W$ is a virtual particle. – knzhou Jan 30 '18 at 14:56
• The conservation of energy must hold between the intial proton state and the final three-body state. Muon and tau are simply too heavy. – Arnaldo Maccarone Jan 30 '18 at 15:21

Mass of neutron

mass of proton

The mass difference 939.6-938.3 is 1.3 MeV and baryon number is conserved anyway, so the neutron can only decay into a proton after and the excess energy is 1.3 MeV .

The mass of the muon is 107 MeV and the mass of the tau 1777MeV. Energetically only the electron with its small mass of 0.5 MeV is available.