According to (for example) this document in calculating $R$ as the sum of charges squared, you get 11/9 for u, d, s, c and b quarks, which you multiply by 3 to factor in the 3 colors. Wouldn't you expect an additional factor of 2 to account for quark/antiquark?
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No, because the annihilation process involves through the creation of a fermion-antifermion pair. This can be $e^{-}+e^{+}\rightarrow\mu^{-}+\mu^{+}$ (used as the benchmark) or $e^{-}+e^{+}\rightarrow q+\bar{q}$, but it always involves the creation of a fermion-antifermion pair (possibly followed by soft hadronization events, when the fermion species involved are quarks). So there when calculating the cross section $e^{-}+e^{+}\rightarrow\,{\rm hadrons}$, there are not separate hard processes involving quarks or antiquarks to sum.
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