Past exam of the sub-atomic physics course of the University of Oxford / 2016 / 4 / e / Solution

The new light active leptons open one additional $W\to \sigma N$ channel and two $Z$ channels. At the stated equal-partial-width approximation,The quoted $501$ MeV is the total invisible width into the three known neutrino flavors, not the width into one flavor. The invisible Z width per active neutrino fixes this interpretation. The notation in the numerical input is ambiguous; the Standard Model normalization fixes the new active-neutrino contribution to $501/3=167$ MeV. Adding the new charged-lepton width therefore givesThis interpretation is consistent with the usual determination of the effective number of active neutrinos from the invisible width divided by the predicted width of a single species.^[ref]

Keep the original light-fermion production and observed-decay partial decay widths, masses and prefactors fixed. New leptonic channels alter their branching fractions through the total widths. From the peak formula in d,Consequently,The relatively small change is a cancellation between the increase in the two total widths; each separately changes substantially more. This assumes that the new quarks do not contribute and that adding leptons leaves the old partial decay widths unchanged.

If the source's $501$ MeV were instead imposed literally as the width of a single neutrino flavor, the algebra would give ${\Gamma }_Z^{\prime }=3080$ MeV and $R^{\prime }/R=1.23401$. That would be a different numerical premise, inconsistent with the Standard Model interpretation of the supplied invisible-width value. It is recorded to make the normalization issue explicit, rather than silently treating the three-flavor sum as a one-flavor partial decay width.