Question

A K⁺ meson at rest decays into a π⁺ meson and π⁰ meson. The π⁺ meson decays into a μ meson and a neutrino. What is the maximum energy of the final μ meson? What is its minimum energy? 

(m_K = 493.5 MeV/c², m_π+ = 139.5 MeV/c², m_π⁰ = 135 MeV/c², m_μ = 106 MeV/^c2, m_ν = 0)

Answer 1

The total energy carried by π⁺ in the rest frame of K⁺ can be calculated from 

E^π+ = (mK² + m_π² − m_ν²)/2m_k = 265MeV  ....(1) 

γ_c = γ^∗_π+ = 265/139.5 = 1.9 

and β_c = β_π^∗ = (γ^∗2_π+ − 1)^1/2/γ^∗_π+ 

In the rest frame of pion, the total energy of muon is obtained again by (1) 

E_μ+² = (m_μ+² + m_π+² − 0)/2m_π+ = 110MeV 

γ_μ^∗ = 110/106 = 1.0377 

β_μ^∗ = (γ^∗2_μ − 1)^1/2/γ_μ^∗ = 0.267 

The Lorentz factor γ_μ for the muon in the LS is obtained from γ_μ = γ_cγ^∗_μ(1+ β^∗_μβ_ccos θ^∗_μ) where θ^∗_μ is the emission angle of the motion in the rest frame of pion. Put θ^∗_μ = 0 to obtain γ_μ(max) and θ^∗_μ = 180^◦ to obtain γ_μ(min). The maximum and minimum kinetic energies are 150 and 55.5 MeV.