= Distinct-flavor quark scattering color factor
{title2=$\overline{|\mathcal M|^2}=4g_s^4(s^2+u^2)/(9t^2)$}
For massless distinguishable <quarks> exchanging one <gluon>, summing the squared generator contraction gives two with $\operatorname{Tr}(T^aT^b)=\delta^{ab}/2$. Averaging nine initial colors and four initial spins leaves the coefficient $1/18$ multiplying the two Dirac <traces>. Their contraction is $8(s^2+u^2)$, yielding the displayed tree-level result. Identical <quarks> require an additional exchange amplitude and its interference.
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