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

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:

enter image description here

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?


1 Answer 1


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.