wikipedia (Muon - Muon decay) says that:

The dominant muon decay mode (sometimes called the Michel decay after Louis Michel) is the simplest possible: the muon decays to an electron, an electron antineutrino, and a muon neutrino.

but a muon has 207 times more energy than electron, so how do you account for the remaining 206 masses-energy just in exchange of two neutrinos?


1 Answer 1


The energy that is not in the masses of the decay products becomes their kinetic energies.

  • $\begingroup$ but how can a tiny neutrino absorb and carry such a huge amount of KE the energy of one tenth of a proton?a neutrino reaches 1/2 c with less than one eV, right? $\endgroup$
    – user337596
    Sep 14, 2022 at 9:06
  • $\begingroup$ @charlie I'm not sure what you mean by "how" - due to how relativity works, anything can have essentially infinite kinetic energy and still not quite reach the speed of light. $\endgroup$
    – ACuriousMind
    Sep 14, 2022 at 9:09
  • $\begingroup$ Thanks, but has a neutrino with 0.1 GeV ever been recorded? My question on the issue got no answers: physics.stackexchange.com/questions/726673/…, which implies the question: how do you figure out the mass of the neutrino separating it from its KE? $\endgroup$
    – user337596
    Sep 14, 2022 at 9:45
  • $\begingroup$ Until recently, neutrinos were believed to be completely massless, always travelling at the speed of light; and for most practical purposes, in the most common processes, they should still be treated as such. Nowadays we know, from evidence of neutrino oscillations (flavours of neutrinos changing into each other), that they possess (very small) mass; however, for most processes such as the muon decay, this mass is negligible, and neutrinos should be treated as massless. $\endgroup$
    – printf
    Sep 14, 2022 at 12:44
  • $\begingroup$ @charlie I'm not sure why you're focussing on the neutrino here - the kinetic energy can just as well be kinetic energy of the electron; note that I said "decay products", not specifically "neutrino". Which proportion of the energy ends up where depends on the detailed kinematics of this reaction. $\endgroup$
    – ACuriousMind
    Sep 14, 2022 at 12:49

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