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

Question

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,

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

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

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

Answer 1

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.