We know that the pion decays into a lepton/neutrino pair while the muon decays into a muon neutrino, electron and electron neutrino. How do we distinguish them experimentally? Would the muon decay have two 'missing widths' compared to the pion decay?

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  • $\begingroup$ What do you want to distinguish here ? Decaying muon from decaying pions ? $\endgroup$
    – agemO
    Commented Jun 10, 2015 at 13:23

1 Answer 1


It is the classic two body versus three body decay.

For a two body decay the energy in the center of mass of the two bodies is fixed for the individual decay products, a delta function for the muon, in the above example, within uncertainties.

For a three body decay there is a variable spectrum for the energy of the electron. (Analogous in kinematics to the beta decay of a neutron) . Thus the missing energy has to be carried by something, in this case the electron neutrino was postulated and contributed to the lepton number conservation rule.

The distinction comes from measuring the momenta and fitting the whole interaction to the hypothesis of the outgoing particles, for three bodies a 1 constraint fit, although it is obvious in the picture below:


The green is the pion loosing energy with ionization and it stops and then decays to a muon, a tiny line because the differences in the masses of pion and muon are only about 30 meV and there is a three body decay at the end. These momenta can be measured and can be fitted with a one constraint fit , in principle.

  • $\begingroup$ This and the fact that we observe both flavor components in neutrino beams. $\endgroup$ Commented Jun 10, 2015 at 14:10

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