= Solution
Measure $\alpha$ in the <orthonormal tetrad> of a static observer at $r>2\mu$. This specifies the local frame needed to define an angle. If $E_\infty$ is the conserved photon energy, <gravitational redshift> gives $E_{\rm loc}=E_\infty/\sqrt{1-2\mu/r}$. The tangential local <momentum> is $E_{\rm loc}\sin\alpha/c$, so the <impact parameter> is
$$
b=\frac{r\sin\alpha}{\sqrt{1-2\mu/r}}.
$$
Inside the <photon sphere>, an outward ray must have $b<3\sqrt3\mu$ to cross the potential barrier. The <Schwarzschild photon escape cone> therefore has critical half-angle
$$
\boxed{\alpha_{\rm crit}(r)=\arcsin\left[\frac{3\sqrt3\mu}{r}\sqrt{1-\frac{2\mu}{r}}\right],\qquad 0\le\alpha<\alpha_{\rm crit}.}
$$
The upper bound is strict: a critical ray only approaches the circular orbit. At $r=3\mu$, the critical half-angle is $\pi/2$, and exactly tangential light is circular. As $r\downarrow2\mu$, the cone shrinks to zero opening. No static observer exists exactly on the <Schwarzschild event horizon>, and no future-directed photon emitted there escapes to exterior infinity; the zero-angle result is an exterior limit.
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