Feynman Diagrams For Pedestrians - Herbstschule Maria Laach

with the “hadronic tensor”H µν (q 1 , q 2 , k) = ∑spins,ɛ∵ with the same trick as in problem 21, we can show thatJ µ (q 1 , q 2 , k, ɛ)J ν,∗ (q 1 , q 2 , k, ɛ) (157)[q µ 1 + qµ 2 + kµ ] J µ (q 1 , q 2 , k, ɛ) = 0 (158)∴ using the center of mass momentum p = p 1 + p 2 = q 1 + q 2 + kp µ H µν (q 1 , q 2 , k) = p ν H µν (q 1 , q 2 , k) = 0 (159)• angular dependence of H µν (q 1 , q 2 , k) contains a lot of information . . .∵ . . . but the energy dependence is much simpler∴ integrate over the anglesx 1 = 2q 1 p/p 2 , x 2 = 2q 2 p/p 2 , x 3 = 2kp/p 2 (160)∫˜dq 1˜dq2 ˜dk (2π) 4 δ 4 (q 1 + q 2 + k − p) f(x 1 , x 2 , x 3 )afterwards, thre result will depend only on p and the x i .∵ from (159) we find∫( )pd ˜Ω H µν µ p ν(q 1 , q 2 , k) = − g µνp 2∴˜H(p, x 1 , x 2 ) = − 1 3= s128π 3 ∫dx 1 dx 2 f(x 1 , x 2 , 2 − x 1 − x 2 ) (161)˜H(p, x 1 , x 2 ) (162)∫d ˜Ω H µ µ(q 1 , q 2 , k) (163)• energy conservation• momentum conservation for massless particleswith equality for parallel q 1 und q 2x 1 + x 2 + x 3 = 2(q 1 + q 2 + k)pp 2 = 2 (164)x 1 + x 2 x 3 = 2 − x 1 − x 2 (165)∴x 1 + x 2 1 (165 ′ )33