Feynman Diagrams For Pedestrians - Herbstschule Maria Laach

• momentum conservation:s + t + u = p 2 1 + p 2 2 + q 2 1 + q 2 2 =4∑m 2 i (115)i=1• center of mass system (CMS) at high energiesp 1/2 = (E; 0, 0, ±E)q 1/2 = (E; ±E sin θ cos φ, ±E sin θ sin φ, ±E cos θ)(116a)(116b)s = 4E 2 , t = −2E 2 (1 − cos θ) , u = −2E 2 (1 + cos θ) (117)3.5 Phase Space• two particles in the final state:∫d 3 ⃗q 1(2π) 3 2E 1d 3 ⃗q 2(2π) 3 2E 2(2π) 4 δ 4 (q 1 + q 2 − P)= 116π 2 ∫ |⃗q1 | 2 d|q 1 |dΩ 1E 1 E 2δ(E 1 (|⃗q 1 |) + E 2 (|⃗q 1 |) − E)= 1 ∫ |⃗q1 |E 1 dE 1 dΩ 1δ(E16π 2 1 + E 2 (E 1 ) − E)= 1 ∫|⃗q 1 |d cos θE 1 E 2 16π 2 1 dφ 1E(118)• the second equality in (118) follows from E 2 = |⃗q| 2 +m 2 , which yields |⃗q|d|⃗q| = EdEindependent of the masses• in the third equality we have used that E 2 depends on E 1 through momentumconservation and dispersion relationsd(E 1 + E 2 (E 1 ) − E)dE 1= 1 + E 1 /E 2 = E/E 2 (119)• special case: high energy limit in the center of mass frame: |⃗q 1 | = |⃗q 2 | = E/2 +O(m/|⃗q 2 | 2 ).d cos θ 1 dφ 1 / (32π 2 ) (118 ′ )23