1 Lecture

1.1 First Example Problem

A pion, initially moving with speed v = βc, decays into a muon and a neutrino, π^{+} → μ^{+}ν_{μ}. If the neutrino emerges at 90^{∘} to the pion direction, what angle does the muon make to the pion?

1.2 Second Example Problem: Perkins problem 1-3

The values of mc^{2} for the pion π^{+} and muon μ^{+} are 139.57 MeV and 105.66 MeV respectively. Find the kinetic energy of the muon in the decay π^{+} → μ^{+}ν_{μ} assuming that the neutrino is massless. For a neutrino of finite but very small mass m_{ν} show that, compared with the case of a massless neutrino, the muon momentum would be reduced by the fraction

\frac{Δp}{p} = −\frac{m_{ν}^{2} (m_{π}^{2} + m_{μ}^{2})}{(m_{π}^{2} − m_{μ}^{2})^{2}} ≈ −\frac{4m_{ν}^{2}}{10^{4}}

where m_{ν} is in MeV.

When m_{ν} = 0 we have shown that

|p| = p = \frac{m_{π}^{2} − m_{μ}^{2}}{2m_{π}} .

The kinetic energy is given by the total energy minus the mass:

T_{μ} = E_{μ}−m_{μ} = \sqrt{m_{μ}^{2} + p^{2}}−m_{μ} = \sqrt{m_{μ}^{2} + \frac{m_{π}^{4} − 2m_{μ}^{2}m_{π}^{2} + m_{μ}^{4}}{4m_{π}^{2}}} −m_{μ} = \frac{m_{π}^{2} + m_{μ}^{2}}{2m_{π}} −m_{μ} = \frac{(m_{π} − m_{μ})^{2}}{2m_{π}} .

With a neutrino of finite mass it is still true that p′_{μ} = −p′_{ν} = p′, but now

E_{μ} + E_{ν} = \sqrt{m_{μ}^{2} + p′^{2}} + \sqrt{m_{ν}^{2} + p′^{2}} = m_{π} .

Rearranging terms and squaring yields:

m_{μ}^{2} + p′^{2} = m_{π}^{2} + m_{ν}^{2} + p′^{2} − 2m_{π}\sqrt{m_{ν}^{2} + p^{2}} .

Rearranging terms yields

\sqrt{m_{ν}^{2} + p′^{2}} = \frac{m_{π}^{2} + m_{ν}^{2} − m_{μ}^{2}}{2m_{π}} .

Squaring both sides of this expression yields

p′^{2} = (\frac{m_{π}^{2} + m_{ν}^{2} − m_{μ}^{2}}{2m_{π}} )^{2} − m_{ν}^{2} = \frac{(m_{π}^{2} − m_{μ}^{2})^{2} − 2m_{π}^{2}m_{ν}^{2} − 2m_{μ}^{2}m_{ν}^{2} + m_{ν}^{4}}{4m_{π}^{2}}

Since m_{ν} ≪ m_{μ} or m_{π}, I keep only the lowest order corrections in terms of m_{ν}:

|p′| = p′ ≈ \frac{m_{π}^{2} − m_{μ}^{2}}{2m_{π}} \sqrt{1 − 2m_{ν}^{2} \frac{m_{π}^{2} + m_{μ}^{2}}{(m_{π}^{2} − m_{μ}^{2})^{2}}} ≈ \frac{m_{π}^{2} − m_{μ}^{2}}{2m_{π}} (1 − m_{ν}^{2} \frac{m_{π}^{2} + m_{μ}^{2}}{(m_{π}^{2} − m_{μ}^{2})^{2}} )

This is a fractional momentum change of

\frac{Δp}{p} = \frac{p′ − p}{p} = −\frac{m_{ν}^{2}(m_{π}^{2} + m_{μ}^{2})}{(m_{π}^{2} − m_{μ}^{2})^{2}}

Inserting values for m_{π} and m_{μ} yields a fractional change of the momentum of the muon of:

\frac{Δp}{p} = −\frac{m_{ν}^{2}}{2 × 10^{4}}

somewhat in disagreement with what was given in the problem.