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I saw an answer to "how to identify a muon neutrino and electron neutrino" as being

"Weak interaction conserve not only the global lepton number, but the lepton flavor numbers as well. And that is how we identify their flavors: electron neutrinos participate in reactions that involve electrons and muon neutrinos participate in reaction that involve muons."

My question is regarding the energy of muon and electron neutrinos.

Do electron neutrinos always have less energy than muon neutrinos?

If not, if you can't identify the source of the neutrino, how can you tell if it is a muon or electron neutrino? Is the answer "by seeing if it reacts with an electron or muon"? Or can you tell by their energies?

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As with other particles, a neutrino's total energy depends on its momentum, and we can construct pairs that have (in a particular frame of reference) whatever relationship between their energies that you want.

Possibly you are trying to ask about the mass of the flavor states (mass being equivalent to energy that is intrinsic to the particle). In that case, you run smack into quantum oddness. The flavor states are not coincident with the mass states, and so they don't have a single and well-defined mass. They do have a reliable probability of being one of the three (as yet unknown) masses, and there is a well defined average mass which you could use to deduce the mass of a large group of neutrinos all in the same flavor state.

Note that because of the mass hierarchy problem we still can't even say if the electron or muon neutrino have a higher average mass, (though theoretical prejudice is for a normal hierarchy, with the muon neutrino being more massive).

But, of course, a neutrino in a composite state of well-defined energy should have three different momenta and therefore three different velocities, right? So how do you deal with that as it propagates over long distances (say through the Earth for "up-going" neutrinos in Super-K or from the sun to the Earth)? Last I heard, that was still a partially open question for which answers are only available at leading order.

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  • $\begingroup$ Hi many thanks, for the excellent answer. I see I need to be a little bit more careful how I frame my questions. $\endgroup$ – andy Dec 21 '15 at 14:33

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