# How does a $\mu_R^-/\mu_L^+$ decay?

While studying SM, I was taught that weak force bosons $$V=\{W^\pm,Z^0\}$$ do not interact with right/left-chiral fermions/antifermions. For this reason, we cannot observe right-handed neutrinos (if they do exist). However, we do observe right-chiral charged leptons as $$e^-_R, \mu^-_R\dots$$ through EM interactions (thanks to their charge).

But the decay of heavily charged leptons as $$\mu$$ and $$\tau$$ must involve weak force otherwise lepton family number would be violated,

$$$$\label{eq1} \mu^-_L\rightarrow W^-+\nu_{\mu L} \rightarrow e^-_L+\bar{\nu}_{eR}+\nu_{\mu L} \quad \text{Involving weak bosons}$$$$

$$\mu\rightarrow e+\gamma \quad \text{Cannot happen since LFN is violated}$$

If right-chiral heavy fermions as $$\mu_R^-$$ and $$\tau_R^-$$ cannot decay through weak force? then how do they decay?

A physical muon is a mass eigenstate. This is a superposition of the left and right chirality components. Because of this, the physical muon can always interact via the weak currents through its left-chiral component, and as such it (and a physical $$\tau$$) will always decay.
Directly addressing your question - the right-chiral muon $$\mu_R$$ is stable (because it cannot decay via weak currents or EM, as you point out), but this is not not very physically relevant. You cannot stabilize a $$\mu$$ by isolating a right-chiral $$\mu_R$$, because the muon's mass causes mixing between the components. That is, the coefficients of the above-mentioned superposition are in general not constant in time - instead they have some phase factor with a frequency related to the muon's mass.
• @A.M.MElsayed马克 I did not say anything about how the components are affected by the interaction - only that the left-chiral component must not be zero in order for the interaction to occur. The $W$ interaction occurring basically constitutes a chirality measurement, taking the right-chiral component to zero, if you want to think about it that way. In the decay frame immediately after the interaction you must have a purely left-chiral electron (although it will be subject to the same chiral oscillations in its future) Commented Jun 13 at 15:40