Muon Decay Lagrangian

Question

I am working on figuring out the following problem

The muon decays to an electron and two neutrinos through an intermediate massive particle called the W$^-$ boson. The muon, electron and W$^-$ all have charge -l.

(a)Write down a Lagrangian that would allow for $\mu^-\rightarrow e^-\bar{\nu_{e}}\nu_\mu$ Assume the W and other particles are all scalars, and the e$^-$, $\nu_e$ and $\nu_\mu$ are massless. Call the coupling g.

So I know that I will have some free L for all particles

$$L_{free}=\frac{1}{2}(\partial \phi([\mu])^2+\frac{1}{2}(\partial \phi[W])^2+\frac{1}{2}(\partial \phi[e])^2+\frac{1}{2}(\partial \phi[\nu_e])^2+\frac{1}{2}(\partial \phi[\nu_\mu])^2+\frac 12 m_\mu^2\phi[\mu]+\frac 12 m_W^2\phi[W]$$

But for the interaction, is it enough to say that $$L_{int}=-g\phi[\mu]\phi[W]\phi[e]\phi[\nu_e]\phi[\nu_\mu]$$ Or do I have to have different interactions for the decay shown in the image below?

Answer 1

Note that the Feynman diagram includes two different types of interaction vertex, so each one will need a term in the Lagrangian:$$-\mathcal{L}_{int}=g \bar{\nu}_\mu W^+ \mu + g \bar{\nu}_e W^+ e + \mathrm{h.c.}$$ Both terms have the same coupling $g$ because of the universality of weak interactions.

Usually, the energy of the initial and final particles is very small compared to the mass of the $W$ boson (around 80 GeV). Therefore, we can integrate out the $W$ boson in order to obtain an effective Lagrangian that describes only the low energy degrees of freedom of the theory: $$-\mathcal{L}_{eff} = G [\bar{\nu}_\mu \mu][\bar{e}\nu_e]$$ where $G$ is a dimensionful coupling constant (while $g$ is dimensionless) which depends on the mass of the $W$ boson as $G\sim M_W^{-2}$.