Leptonic Decay of Muon and Tau

So the question asks:

(a) Draw the Feynman diagrams showing the dominant leptonic decay mode of the $\mu^-$ and the $\tau^-$.

(b) Assuming $m_\tau \gg m_\mu \gg m_e$, estimate the ratio of the rate of the two decay modes.

Part (a) is simple.

Part (b), I'm not sure what I am meant to do. I know the decay widths for the decay modes are $$\Gamma(\mu^- \to e^-+\bar{\nu}_e+\nu_\mu) = \frac{G_F^eG_F^\mu m_\mu^5}{192 \pi^3},$$ $$\Gamma(\tau^- \to e^-+\bar{\nu}_e+\nu_\tau) = \frac{G_F^eG_F^\tau m_\tau^5}{192 \pi^3},$$ $$\Gamma(\tau^- \to \mu^-+\bar{\nu}_\mu+\nu_\tau) = \frac{G_F^\mu G_F^\tau m_\tau^5}{192 \pi^3}$$

and by lepton universality, $$G_F^e=G_F^\mu = G_F^\tau.$$

I'm not sure how to proceed from here. Is it simply finding the branching ratios and dividing one by the other?

• Note that the question asks for a single ratio. Do you know what ratio is intended and why? Question that might help out: "Is the decay of the tau to a muon distinguishable from that to an electron?" – dmckee --- ex-moderator kitten Dec 10 '15 at 3:19
• I do not know what ratio is intended. As for the question, surely you can distinguish the decay of the tau to a muon or to an electron? The muon is heavier than the electron so the energy loss by Bremsstrahlung for a muon is far smaller than the electron. For example in the CMS experiment muons will travel further than electrons. – John Sweeney Dec 10 '15 at 3:53
• Good. Now what does quantum mechanics tell you about combining distinguishable final states vis a vis combining indistinguishable final states? – dmckee --- ex-moderator kitten Dec 10 '15 at 4:28
• I know about distinguishable and indistinguishable particles. If two particles are distinguishable then the quantum state of the system is simply a tensor product of the two states the particles are in. If they are indistinguishable then there is a linear combination of tensor products, e.g $(|a\rangle |b \rangle + |b\rangle|a \rangle )/ sqrt{2}$. Is this what you are asking for? – John Sweeney Dec 10 '15 at 5:44

1 Answer

You don't need to worry about branching ratios. The two decay modes being referred to are the dominant leptonic decay mode of the muon and the dominant leptonic decay mode for the tau (not "tauon"). You have those decay widths already, so you can just take their ratio. Most of the factors will cancel out, leaving just a power of the muon-tau mass ratio, indicating the heavier tau has faster leptonic decays.

(Of course, what makes real tau decays really interesting is that the tau is heavy enough to have lots of hadronic decay modes, but those are obviously much more complicated.)

• You still have to deal with the fact that there isn't a the dominate leptonic decay of the tau. The two tree level diagrams have essentiality the same branching ratio. – dmckee --- ex-moderator kitten Dec 10 '15 at 4:11
• As a practical matter, yes, there are two similar branching ratios. However, the problem only wants the largest for each case. This is just an issue of semantics (pragmatics, really), not physics. – Buzz Dec 10 '15 at 11:39