﻿ Solution: Pion Decay Energies | Primer on Geometric Algebra | David Hestenes

## Solution: Pion Decay Energies

Problem:  Calculate, if possible, the energies of the decay products when the pion decays from rest.

Solution:

$\Large{\begin{array}{l}{\pi ^ + }\,\, \to \,\,{\mu ^ + }\,\, + \,\,{\nu _\mu }\,\,\,\,\,\,\,\,\,\,\,\,\\\\\\{m_\pi }{c^2}\,\, = \,\,\,139.57\,\,{\rm{MeV}}\,{\rm{,}}\,\,\,~~~~~~~~~~~~\,\,\,\,\,\,\,\,\,{m_\mu }{c^2}\,\, = \,\,\,105.66\,\,{\rm{MeV}}\,{\rm{,}}\,\\\\\\Q\,\,\, \equiv \,\,\,{m_\pi }{c^2}\,\, - \,\,{m_\mu }{c^2}\,\,\, = \,\,\,33.91\,\,{\rm{MeV}}\,{\rm{,}}\,\,~~~~~~~~~~~~\,\,\,\,\,\,\,\,\,\,{m_\nu }\,\, = \,\,0\\\\\\{P_\pi }\,\,\, = \,\,\,{m_\pi }c\,\,\, = \,\,\,{P_\mu }\,\, + \,\,{P_\nu }\,\,\, = \,\,\,\left( {\frac{{{E_\mu }}}{c}\,\, + \,\,{{\bf{p}}_\mu }} \right)\,\, + \,\,\left( {{p_\nu }\,\, + \,\,{{\bf{p}}_\nu }} \right)\\\\\\{{\bf{p}}_\mu }\,\,\, = \,\,\, - \,\,{{\bf{p}}_\nu }\,\,\,\,\,\,~~~~~~~~~~~~\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,~~~~~~~~~~~~\,\,\,\,\,\,{p_\mu }\,\,\, = \,\,\,{p_\nu }\,\,\, \equiv \,\,\,p\\\\\\\\p\,\,\, = \,\,\,{m_\pi }c\,\, - \,\,\frac{{{E_\mu }}}{c}\,\,\, = \,\,\,\frac{{{m_\pi }{c^2}\,\, - \,\,\left( {{m_\mu }{c^2}\,\, + \,\,{K_\mu }} \right)}}{c}\,\,\, = \,\,\,\frac{{Q\,\, - \,\,{K_\mu }}}{c}\\\\\\\\{P_\mu }\widetilde {{P_\mu }}\,\,\, = \,\,\,{m^2}_\mu {c^2}\,\,\, = \,\,\,\frac{{{E^2}_\mu }}{{{c^2}}}\,\, - \,\,{p^2}\,\,\, = \,\,\,\frac{{{{\left( {{m_\mu }{c^2}\,\, + \,\,{K_\mu }} \right)}^2}}}{{{c^2}}}\,\, - \,\,\frac{{{{\left( {Q\,\, - \,\,{K_\mu }} \right)}^2}}}{{{c^2}}}\\\\\\\\\\\\\\\\0\,\,\, = \,\,\,2{m_\mu }{c^2}{K_\mu }\,\, - \,\,{Q^2}\,\, + \,\,2Q{K_\mu }\,\,\, = \,\,\,2{K_\mu }\left( {{m_\mu }{c^2}\,\, + \,\,{m_\pi }{c^2}\,\, - \,\,{m_\mu }{c^2}} \right)\,\, - \,\,{Q^2}\\\\\\\\{K_\mu }\,\,\, = \,\,\,\frac{{{Q^2}}}{{2{m_\pi }{c^2}}}\,\,\, = \,\,\,\frac{{{{\left( {33.91\,\,{\rm{MeV}}} \right)}^2}}}{{2\,\left( {139.57\,\,{\rm{MeV}}} \right)}}\,\,\, = \,\,\,4.12\,\,{\rm{MeV}}\\\\\\\\\\\\{E_\nu }\,\,\, = \,\,\,pc\,\,\, = \,\,\,Q\,\, - \,\,{K_\mu }\,\,\, = \,\,\,33.91\,\,{\rm{MeV}}\,\, - \,\,4.12\,\,{\rm{MeV}}\,\,\, = \,\,\,29.79\,\,{\rm{MeV}}\end{array}}$

Thus, if the pion decays from rest, we know precisely what energies to look for in the decay products.