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Set $E=0$ in part (a), substitute the two conserved quantities, and multiply the normalization by $f$:Consequently the null effective potential in Schwarzschild spacetime isOnly radii where the total radial energy is at least this effective potential are accessible.

At $2\mu <r\le 3\mu$, measure the emission angle $\alpha$ from the outward radial direction in a static observer's orthonormal tetrad. Local photon energy is $E_{\mathrm{loc}}=E_{\infty }/\sqrt{1-2\mu /r}$, and its tangential momentum is $E_{\mathrm{loc}}\sin \alpha /c$. Thus its impact parameter is $b=r\sin \alpha /\sqrt{1-2\mu /r}$. Combining this with the null effective potential in Schwarzschild spacetime gives escape precisely for outward directions $0\le \alpha <{\alpha }_{\mathrm{crit}}$; the limiting ray approaches the photon sphere and does not escape.

As $r\downarrow 2\mu$ the cone shrinks to zero opening. A static observer does not exist on the horizon itself, and no future-directed ray starting on the Schwarzschild event horizon reaches exterior infinity. At $r=3\mu$, the critical half-angle is $\pi /2$; the exactly tangential critical ray stays on the unstable circular orbit. Escape cones for a moving observer are changed by relativistic aberration.