Question: The centre-of-mass differential cross-section for the reaction e + e μ + μ due to the one-photon exchange diagram in Figure 10.2(a) is given in

The centre-of-mass differential cross-section for the reaction e+eˆ’†’ μ+μˆ’due to the one-photon exchange diagram in Figure 10.2(a) is given in Equation (7.15),

do (s, cos 0) (1+cos² 0), TQ? σ( 6,θ) - d cos 0 2s

where s = E2CM, θ is the scattering angle and we have, as usual, neglected the lepton masses compared to ECM. When Z0 exchange is taken into account, this expression is modified to become

σ (s,θ ) - πα? [F (s) (1 + cos θ ) + G(s) cos θ , 2s

With

do (s, cos 0) (1+cos 0), TQ? ( 6,) - d cos

and

0 2s (s, ) - ? [F (s) (1 + cos )

where we have assumed M2ˆ’ s >> ΓZ and sin2 θW = 1/4. The terms in σ(s, θ) proportional to α2, G2F and αGF arise from photon exchange, Z0 exchange and the interference between the two diagrams of Figure 10.2, respectively. Use these formulas to obtain expressions for the total cross-section σtot(s) and the forward€“ backward asymmetry parameter

+ G(s) cos , 2s

in terms of F(s) and G(s), where

Hence evaluate the magnitude of A(s) and the percentage correction to the QED total cross-section σγ = 4πα2/3s at ECM = Mz /3 ‰ˆ 30 GeV generated by the Z0 exchange contribution.


Figure 10.2

do (s, cos 0) (1+cos 0), TQ? ( 6,) - d cos 0 2s (s, ) - ? [F (s) (1 + cos ) + G(s) cos , 2s

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