 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 $E_{\mathrm{loc}}=E_{\infty }/\sqrt{1-2\mu /r}$. The tangential local momentum is $E_{\mathrm{loc}}\sin \alpha /c$, so the impact parameter isInside the photon sphere, an outward ray must have $b<3\sqrt{3}\mu$ to cross the potential barrier. The Schwarzschild photon escape cone therefore has critical half-angleThe 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 $r\downarrow 2\mu$, 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.