# Is this calculation assuming that the neutrino is massless?

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

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

implies

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

and

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

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.

## 1 Answer

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."