This is a calculation that I don't understand in Griffiths Intro to Elementary Particles page 105 second edition.

This the decay of a pion into a moun and a neutrino.

From Conservation of energy and momentum,

the momentum four vector

$p_{neutrino} = p_{pion} - p_{muon}$


$p_{neutrino}^2 = p_{pion}^2 + p_{muon}^2 - 2 p_{pion} \cdot p_{muon}$


$p_{neutrino}^2=0 :p_{pion}^2= m_{pion}^2 c^2 , p_{muon}^2 = m_{muon}^2 c^2 $

Ok. Why is $p_{neutrino}^2=0$?

Is it just because he's taking the mass of the neutrino to be 0? But it isn't 0. It's very small.


In his "Formulas and Constants" table on XIII, Griffiths says, under the Lepton table, "Neutrino masses are extremely small, and for most purposes can be taken to be zero; for details see Chapter 11."


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