122 The differential elastic cross-section is given by where v$_{1}$ and v$_{2}$ are the velocities of the φ and χ particles before the collision respectively, where v$_{2}$ = c . As we are setting c =1 , the first term in Eq. (B.5) ≈ 1 . Collecting terms together gives Integrating this gives From the 4-momenta of p$_{1}$ and p$_{3}$ , Eqs. (B.1) and (B.3) respectively, we have and E$_{1}$ = √ p 2 + m 2 φ, eff (B.10) E$_{3}$ = √ p ′ 2 + m 2 φ, eff (B.11) σ ≈ − g 4 64 π $^{2}$E$_{1}$ E$_{3}$ ∫∫ 1 pp ′ δ $^{4}$( p$_{1}$ + p$_{2}$ − p$_{3}$ − p$_{4}$ )d 3 $_{p}$′$_{d}$ $^{3}$( − $_{p}$′$_{)}$ (B.7) ≈ − g 4 64 π $^{2}$E$_{1}$ E$_{3}$ ∫ 1 pp ′ δ ( E$_{1}$ + p − E$_{3}$ − p ′$_{)d}$ 3 $_{p}$′ (B.8) ≈ − g 4 64 π $^{2}$E$_{1}$ E$_{3}$ ∫ 1 p δ ( E$_{1}$ + p − E$_{3}$ − p ′$_{)}$ p ′$_{d}$ p ′ ∫ Ω dΩ (B.9) d σ ≈ − g 4 64 π $^{2}$E$_{1}$ E$_{3}$pp ′ δ $^{4}$( p$_{1}$ + p$_{2}$ − p$_{3}$ − p$_{4}$ )d 3 $_{p}$′$_{d}$ $^{3}$( − $_{p}$′$_{)}$ (B.6) d σ = 1 | v$_{1}$ − v$_{2}$ | 1 2 E$_{1}$ 1 2 p ∣ − ig 2 ∣ ∣$^{2}$ d $^{3}$p ′ (2 π ) $^{3}$2 E$_{3}$ d $^{3}$( − p ′ ) (2 π ) $^{3}$2 p ′ (2 π ) $^{4}$δ$^{4}$( p$_{1}$ + p$_{2}$ − p$_{3}$ − p$_{4}$ ) (B.5)