  The lowest-order expression for the partial $W$ width to $e \nu ,~\Gamma (W
\to e \nu) = G_\mu M_W^3 /(6 \pi \sqrt{2})$, has no oblique radiative
corrections from new physics if the measured $W$ mass is used. Here $G_\mu =
(1.16639 \pm 0.00002) \times 10^{-5}$ GeV/$c^2$ is the muon decay constant. For
the present value of $M_W = (80.14 \pm 0.27)$ GeV/$c^2$, and with $m_t = 140$
GeV$/c^2$, one expects $\Gamma (W \to e \nu) = (224.4 \pm 2.3)$ MeV. The total
width $\Gamma_{\rm tot}(W)$ is also expected to lack oblique corrections from
new physics, so that $\Gamma_{\rm tot} (W)/ \Gamma (W \to e \nu) = 3 + 6 [1 +
\{\alpha_s (M_W)/\pi \}]$. Present data are consistent with this prediction.
