Question

A particle of mass m collides elastically with another identical particle at rest. Show that for a relativistic collision 

tan θtanϕ = 2/(γ + 1) 

where θ,ϕ are the angles of the out-going particles with respect to the direction of the incident particle and γ is the Lorentz factor before the collision. Also, show that θ + ϕ ≤ π/2 where the equal sign is valid in the classical limit

Answer 1

tanθ = sinθ^∗/γ_c(cosθ^∗ + β_c/β₁^∗) = (1/γ_c) tanθ^∗/2  ..(1) 

(Because β_c = β₁^∗) 

Also tanϕ = (1/γ_c) tanθ^∗/2  ...(2)

 Where θ^∗ and ϕ^∗ are the corresponding angles in the CM system 

Multiply (1) and (2) 

tan θ tan ϕ = (1/γ_c²). tan θ^∗/2. tan ϕ^∗/2 

But ϕ^∗ = π − θ^∗ and for m₁ = m₂, γ_c = ((γ + 1)/2)^1/2 

tanθtanϕ = 2/(γ + 1) 

In the classical limit, γ → 1 and tan θ tan ϕ = 1 

But since tan(θ + ϕ) = (tan θ + tan ϕ)/(1 − tan θ tan ϕ) 

tan(ϕ + θ) → ∞ 

i.e. θ + ϕ = π/2 

For γ > 1, tan θ tan ϕ < 1. For tan(θ + ϕ) to be finite (θ + ϕ) < π/2 

Hence, for γ > 1; θ + ϕ < π/2