$W^+/W^-$ bosons ratio in proton-proton collision

Question

I'm studying my first-year physics in college, and I'm having to write a report of some proton-proton collisions that were registered in the LHC of CERN years ago. The main goal is to identify different bosons (W and Z) that are decaying into other elemental particles. I've been asked to get the W+/W- ratio in these events and discuss the results. I've done some research and I've found out that "it should be 1.4, according to the proton quarks model ", which is quite close to the results I've got after analyzing many collisions. I have tried so hard to understand what that quote means, but I can't. Could someone explain to me the meaning of it and the explanation of the result?

Answer 1

Basically you are looking at:

$$ u + \bar d \rightarrow W^+ $$

and

$$ d + \bar u \rightarrow W^- $$

where the quark (sea or valence) from one proton interacts with a sea antiquark in the other proton.

The cross sections should be different, since a proton has twice as many valence up quarks than valence down quarks. Since the magnitude of the weak isospin is the same for $u$ and $d$ quarks, maybe you'd expect $R=2$ for valence quarks, and $R=1$ for the sea quark contribution, with an average of 1.4?

I have seen $R=1.6$ in the literature, where the theoretical predictions requires Monte Carlo analysis with next-to-next-to leading order Parton distribution function (NNLO PDFs) and leading order electroweak calculations all within specific renormalization schemes for the lepton signals, meaning: this is so far beyond the level of undergraduate physics, it advanced graduate seminar level for partitioners in the field.