W and Z at SPS:
Since W's have spin 1 and select only the left-handed fermions and right-handed anti-fermion, in a dbar-u collision the spin of the W+ is the direction of the dbar, and so of the anti-proton beam. As a consequence, the angular distribution of the l+ goes like (1+cosθ) i.e. it's peaked at θ=0, i.e. along the proton beam. Instead, the l- (from W-) is peaked along the anti-proton beam.
In the cross-section calculation for W and Z, one has to divide by 3 because the probability of finding a color matching between the initial state quarks is 1/Nc.
σ(W)~10*σ(Z) because:
- for the Z there is a factor sin2θw~0.23 (roughly a factor of 4-5 suppression) (*)
- taking into account only valence quarks, there are 4 possible initial states that give a W (u1+dbar, u2+dbar, ubar1+d, ubar2+d), and 5 that give a Z (u1+ubar1, u1+ubar2, u2+ubar1, u2+ubar2, d+dbar)
- since BR(W->u+dbar)=Nc/9=33%, and BR(Z->u/d+ubar/dbar)~70%/5~14%, there is a further factor of 2 suppression for Z wrt W
all these, combined, give a rough factor of 10 suppression.
Note: at LHC (pp collisions) the only different thing is the second one: assuming that the quark is valence and the anti-quark comes from the sea (and so all flavours have equal possibility), the equivalent initial states are 3 for W (u1+dbar, u2+dbar, d+ubar) and 3 also for Z (u1+ubar, u2+ubar, d+dbar). The factor of 10 remains. Of course for sea-sea collisions there is equality, too.
(*) θw is defined by MW=cosθw*MZ.