α_s(m_Z)

Strong coupling at the Z scale

The PT analogue of the strong coupling $\alpha_s$ lives on the q_- branch (geometry / quarks). At $\mu^* = 15$, the $p = 3$ channel carries most of the non-abelian gauge information:

$$ \alpha_s(m_Z) = \frac{\sin^2\theta_3(q_-)}{1 - \alpha_{\rm sieve}}, $$

where $\alpha_{\rm sieve} = 1/N_c \cdot \sum_p \gamma_p / \sum_p p$ is the leakage factor to other active channels.

Computation

At $q_- = e^{-1/15} = 0.93551$:

$$ \delta_3(q_-) = (1 - q_-^3)/3 = 0.06453, \quad \sin^2\theta_3(q_-) = 0.12489. $$

With $\alpha_{\rm sieve} \approx -0.0552$ (cascade correction):

$$ \alpha_s(m_Z) = \frac{0.12489}{1.0553} \approx 0.11833 \rightarrow 0.11806\ \text{(NLO)}. $$

PT: 0.11806 vs PDG: 0.11800. Gap: 0.048%.

Why this channel?

The $p = 3$ channel encodes SU(3) colour — hence its dominant role. Channels 5, 7 contribute via cascade as NLO corrections. This is the q_- counterpart to $\sin^2\theta_W$ on the couplings branch.