Muon decays had been studied for many years, and it was well established that the charged secondary is an electron. How do we know there are two of them (neutrino)?
The phase space for a two or three body decay is very different. From the energy distribution of the electron it can be infered that it has to be a three-body decay.
Theoretically, the Standard Model predicts that the muon decays to a nuon-neutrino and a W gauge boson, which consequently decays to an electron and a electron-neutrino.
The lepton number is conserved in the most common decay of the muon into an electron toghether with a neutrino and anti neutrino. Thre neutrinos would violate lepton number conservation.
First of all, there will be 2 neutrinos due to Lepton Flavor conservation. And since its a multi body decay, you will get energy distribution for electron, which is the one you gonna measure due to its sharing of energy with other neutrinos. Once you gets energy distribution for electron, make sure that no other particle that you "observed", and from the conservation law, you can clearly says 2 neutrinos (1 electron anti neutrino and 1 muon neutrino) will also be there...{ For two body decay, you wont get energy distribution.)
you can easily exclude the possibility of one 1 neutrino + electron, since in the decay with 2-body final state, you will have constant electron energy sepectrum due to energy and momentum conservation. And Avelino is right, you need even netruino number due to angular momentum conservation.
In principle , you can have more than 2 neutrinos. For example, mu->Z*mu*->nuetrino pairs+(normal products), or mu->W*W*mu*->2*(neutrino+charged lepton pair)+normal product. But the rate is incredibly low. The decay with emission of a gamma plus electron and 2 neutrino is the order of 1%. The process with emission a virtual Z should be some 15-20 orders of magnitude less that that, making it quite impossible to detect in experiment.
I think there is some way to determine how many missing objects in the final states in experiment but I haven't fount it.
This is a very good question. The key is that gauge interactions do not change flavor. In the first vertex, therefore a muon neutrino is produced along with a W boson. In the second vertex, the W boson goes to electron and electron anti neutrino. It could have gone to tau lepton and tau anti neutrino in principle. But, because tau lepton is heavier than muon this channel is forbidden. The following diagram is taken from wikipedia
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