Tau Decay: More Pions, Less Ratio?

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SUMMARY

Tau decay exhibits a significantly higher branching ratio to a charged pion, neutral pion, and tau neutrino compared to a charged pion and tau neutrino. This phenomenon occurs due to the stronger coupling of the virtual W boson to the rho meson than to the pion. The decay process can be described as τ → ρν, followed by ρ± → π±π0. Understanding this requires consideration of multiple factors, including the vector nature of the rho meson and the complexities of semi-leptonic decays involving hadrons and strong interactions.

PREREQUISITES
  • Understanding of tau decay processes in particle physics
  • Familiarity with the electroweak standard model
  • Knowledge of quantum field theory (QFT) and quantum chromodynamics (QCD)
  • Concept of vector-meson dominance in hadronic interactions
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  • Study the role of vector and axial-vector currents in particle decays
  • Explore effective hadronic models derived from chiral symmetry in QCD
  • Research the implications of the vector-meson-dominance model in tau physics
  • Examine perturbation theory applications in electroweak interactions
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Particle physicists, researchers in quantum field theory, and students studying tau decay and hadronic interactions will benefit from this discussion.

BillKet
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Hello! Tau decay has a branching ratio to a charged pion + neutral pion + tau neutrino much bigger than to a charged pion and a tau neutrino. Based on consideration of available phase space, I would imagine that adding an extra pion would decrease the branching ratio. Why is this happening? Thank you!
 
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The decay is actually \tau \rightarrow \rho \nu followed by \rho^\pm \rightarrow \pi^\pm \pi^0. The reason is that the virtual W couples more strongly to the rho than the pion, but I don't think there is an I-level explanation for why. There's not even a single simple A-level explanation: it's multiple factors.
 
Vanadium 50 said:
The decay is actually \tau \rightarrow \rho \nu followed by \rho^\pm \rightarrow \pi^\pm \pi^0. The reason is that the virtual W couples more strongly to the rho than the pion, but I don't think there is an I-level explanation for why. There's not even a single simple A-level explanation: it's multiple factors.
Thanks a lot! So the W boson decays into an up/down pair, but for some reason that pair has a higher probability to form a rho meson than a pion? Iso ne of the reason related to the fact that the rho is a vector while the pion is a scalar, while W is a vector?
 
BillKet said:
Iso ne of the reason related to the fact that the rho is a vector while the pion is a scalar, while W is a vector?

It's not that simple. I don't think there is an I-level explanation for why. There's not even a single simple A-level explanation: it's multiple factors.
 
While the purely leptonic decays are quite easy to understand by just using perturbation theory to the electroweak standard model, QFD, the semi-leptonic decays involving hadrons and the strong interaction, is not so easy to understand from simple analytic principles. An important theoretical tool is to use the approximate chiral symmetry of QCD in the light-quark sector to derive effective hadronic models. Finally there are also some phenomenological models. In connection with the electromagnetic (and maybe also for the weak) interaction of hadrons, particularly pions, the socalled vector-meson-dominance model is quite successful. If I remember right, for the decay via vector and axial-vector currents you need both, direct decays as well as decays vial ##\rho## and ##a_1## mesons. Above, I've quoted a review paper which seems to me pretty up to date concerning ##\tau## physics.
 

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