# Homework Help: Pion decay in W rest frame

1. Jan 12, 2014

### gildomar

1. The problem statement, all variables and given/known data

"(a) Draw a quark-flow diagram for the weak decay $\pi^{-}\rightarrow\mu^{-}+\overline{\nu_{\mu}}$ . Explicitly include the appropriate intermediate vector boson. (b) By considering the production of $\mu^{-}$ and $\overline{\nu_{\mu}}$ in the rest frame of the vector boson, show from the necessary parity nonconservation that the boson is indeed a vector type, that is, that it has spin one."

2. Relevant equations

3. The attempt at a solution

I know that the intermediary step for the decay is the pion turning into a charged W boson, which in turn decays into the muon and antineutrino. I also don't have a problem drawing the reaction in terms of the quark content. The part that I'm unsure about is what, if any, role the pion has in the interaction in the W boson's rest frame. That is, I'm not whether to include it or not, since I know that W bosons prefer to decay to just leptons, but it seems wrong to leave it out. Whether or not to include obviously will also affect the parity nonconservation test. Though the wording for that part is a little strange, since it would seem to make more sense to just look at the spins to show that the spin of the W is 1 instead of checking parity.

2. Jan 12, 2014

### Staff: Mentor

The pion is at rest in the same frame as the W boson. It "decays" to an off-shell W boson, and that "decays" to muon+neutrino. Only the second process should be relevant for parity non-conservation.

3. Jan 12, 2014

### gildomar

I was beginning to suspect that. Now I just need to figure out how the parity non-conservation leads to the W boson being spin 1 when it emits the muon and antineutrino. The closest that I got is that it has something to do with the muon and antineutrino being emitted directly opposite each other compared to the W boson so as to conserve momentum. Possibly something also about them having opposite angular momentum as well to conserve that too.