$$\tau \rightarrow \nu_{\tau}\pi^+\pi^-\pi^-$$

This is a possible decay mode of the $\tau$ lepton (see Wikipedia). Given that a neutrino is present this must be the weak interaction and therefore we can imagine the Feynman diagram will look something like this:

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What would we draw in place of "pions"? Is it possible for a W boson to produce 6 quarks? I don't see why not, but this is not something I have ever seen drawn before. If not, through what interaction would all 6 quarks be produced at this vertex?


The $W^-$ gives a $d \overline u$ quark-antiquark pair. That's the only relevant diagram: extra $W$, $Z$ and $\gamma$ lines push the rate right down.

For this particular $\tau$ decay the quark and antiquark have their spins aligned, $S=1$, they have orbital angular momentum $L=1$, and the two combine to give $J=1$. This gives the $a_1$ meson (at least, that's the dominant process). The $a_1$ has a mass of 1260 MeV and a width of about 450 MeV (so it's very broad). It decays to 3 pions through $\rho \pi$ intermediate states. This is a low energy strong process and can't really be described in terms of gluons (it would need an infinite number of diagrams in a non-converging series).


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