# Why is this tau decay not allowed?

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1. Apr 12, 2015

### unscientific

Why is the decay $\tau^{-} \rightarrow \mu^+ + \mu^- + \mu^-$ not allowed?

Charge, lepton number are conserved. I have a feeling it is something really basic. I'm thinking in weak interactions you only go from a $l^- \rightarrow \nu_{l}$ and not 'hop' from one muon to another non-neutrino muon.

2. Apr 12, 2015

### The_Duck

In the standard model, electron number, muon number, and tau number are each separately conserved, except for very small effects coming from neutrino mixing.

One way of searching for physics beyond the standard model is to look for processes like $\mu^- \to e^- + \gamma$ that violate these individual lepton flavor numbers and so are negligibly rare in the SM. So a detection of a process like this would signal new physics.

3. Apr 12, 2015

### unscientific

Got it. So lepton numbers must be separate, unlike hadronic numbers.

4. Apr 12, 2015

Staff Emeritus
Lepton number is not conserved in this decay. The lepton numbers associated with the three lepton flavors are conserved separately in these interactions.

5. Apr 13, 2015

### Staff: Mentor

Even with neutrino mixing, this process is incredibly unlikely, with a branching fraction somewhere in the range of 10-50 to 10-60, completely undetectable. Comment from a theorist: "We didn't bother to make a more precise estimate".
Only new physics could lead to a detectable branching fraction.

6. Apr 14, 2015

### ChrisVer

How would the neutrinos or/and their mixing have something to do with these processes? or are you talking about neutrinos changing their flavor (and so the conservation of each individual flavor's lepton number)?

7. Apr 14, 2015

Staff Emeritus
Neutrino mixing means charged lepton number is only approximately conserved, but approximately means, as mentioned above, "good to a part in 1050".

8. Apr 14, 2015

### Staff: Mentor

Neutrino mixing is exactly the process of neutrinos changing their flavor.
That is frequent when the neutrinos can fly hundreds of kilometers or more, but as virtual particles within a decay process (necessary for tau -> 3 mu) it is extremely unlikely.

9. Apr 14, 2015

### ChrisVer

that's what I meant :) So the SM process would be like $\tau \rightarrow \nu_\tau W \rightarrow \nu_\mu \bar{\nu}_\mu \mu \rightarrow Z \mu \rightarrow \mu \mu \mu^+$?

10. Apr 14, 2015

### Staff: Mentor

I doubt that diagram would be dominant. You can draw a box-like diagram with two W and two neutrinos forming the box, where one neutrino changes its flavor in between.

11. Apr 14, 2015

### ChrisVer

would that really give 3 muons?
I think the best in odds [but only one muon as a product] would be a single loop, with the W in one (below) half and a changing tau neutrino to muon on the other (top) half...
this would lead to $\tau \rightarrow \mu$.

12. Apr 14, 2015

### Staff: Mentor

Sure, that would give three muons.

$\tau \rightarrow \mu$ is forbidden by energy/momentum conservation. You would need an additional photon, for example.

13. Apr 15, 2015

### unscientific

And, I think lepton number is not conserved in neutrino oscillations.

14. Apr 15, 2015