Question

A particle of mass m₁ travelling with a velocity v = βc collides elastically with the particle m₂ at rest. The scattering angles of m₁ in the LS and CMS are θ and θ^∗. Show that

(a) γ_c = (γ + ν)/(1 + 2γν + ν²)^1/2 

(b) γ^∗ = (γ + 1/γ )/√(1 + 2γ/ν + 1/ν²) 

(c) tanθ = sinθ^∗/γ_c(cos θ^∗ + β_c/β^∗) 

(d) tan θ^∗ = sinθ/γ_c(cos θ − βc/β^∗) 

where βc is the CMS velocity, β^∗c is the velocity of m₁ in CMS, 

γ_c = (1 − β_c²)^−1/2, γ^∗ = (1 − β^∗2) −1/2, ν = m₂/m₁

Answer 1

(a), (b) In the CMS, m₂ will move with the velocity β_c in a direction opposite to that of m₁. By definition, the total momentum in the CMS before and after the collision is zero. In natural units c = 1.