Feynman Diagrams For Pedestrians - Herbstschule Maria Laach

Solution 14.L µν = tr [(/p + m)γ µ (/q − m)γ ν ] = tr [/pγ µ /qγ ν ] − m 2 tr [γ µ γ ν ]= 4 · (p µ q ν − g µν pq + p ν q µ ) − 4m 2 · g µν = 4 · (pµ q ν + p ν q µ − (pq + m 2 )g µν)(210)Solution 15.L ρ2 ρ 1L ρ 1ρ 2= 16 · (p 1,ρ2 p 2,ρ1 + p 1,ρ1 p 2,ρ2 − p 1 p 2 g ρ2 ρ 1)(q ρ 11 qρ 22+ q ρ 21 qρ 12− q 1 q 2 g ρ 1ρ 2)())= 8 · 2(p 1 q 2 )2(p 2 q 1 ) + 2(p 1 q 1 )2(p 2 q 2 ) = 8 ·(u 2 + t 2 (211)NB: cross terms cancel:(p 1,ρ2 p 2,ρ1 + p 1,ρ1 p 2,ρ2 )q 1 q 2 g ρ 1ρ 2+ p 1 p 2 g ρ2 ρ 1(q ρ 11 qρ 22+ q ρ 21 qρ 12 ) = p 1p 2 g ρ2 ρ 1q 1 q 2 g ρ 1ρ 2Solution 16.thereforet = − s 2 (1 − cos θ) , u = −s (1 + cos θ) (212)2t 2 + u 2s 2= 1 + cos2 θ2(213)dσ =therefore11 1√∏4 (p 1 p 2 ) 2 − m 2 1 m2 i n i! 4 |T|2˜dq1˜dq2 (2π) 4 δ 4 (p 1 + p 2 − q 1 − q 2 )21=4 √ 1 11 1(s/2) 2 4 |T|2 d cos θdφ =32π2 64π 2 s 4 |T|2 dΩ (214)dσdΩ = 1 ( 64π 2 s e4 1 + cos 2 θ ) = α 2 1 (1 + cos 2 θ ) (215)4sSolution 17.∫σ = dΩ dσ ∫ 1dΩ = α24s 2π d cos θ ( 1 + cos 2 θ ) =(2 α2−14s 2π + 2 )3comparewithσ =σ == 4πα23s(216)4πα 23( √ s/TeV) 2 0.39 nb (19′′ )α = e24π = 1137.0359895(61)(217)87 fb( √ s/TeV) = 8.7 pb2 ( √ (19 ′′ )s/100 GeV) 243