A Simple Way to Estimate the Value of $\bar{\alpha }\equiv \alpha (m_Z^2)$

Abstract

To obtain the value of electromagnetic coupling constant at $q^2=m_Z^2$, $\bar{\alpha }$, which plays a key role in electroweak physics one has to integrate the cross-section of $e^{+}{e}^{-}$-annihilation into hadrons divided by $(s-m_Z^2)$ over $s$ from threshold to infinity. By combining, for each flavor channel, the contribution of lowest resonance with the perturbative QCD continuum, we obtain $1/\bar{\alpha }=128.90\pm 0.06$ a result which is close to known result obtained with purely experimental inputs, i.e. $1/\bar{\alpha }=128.87\pm 0.12$.

Comments: LaTex, 6 pages, preprint # FTUV/94-27, IFIC/94-25