 Solution

The observer is static at $r_0$, with four-velocity $U=f_0^{-1/2}{\partial }_t$, where $f_0=1-2m/r_0$. Angles are measured in the observer's rest frame, using the positive spatial metric tensoron the equatorial plane. One cannot directly normalize a photon tangent with the full Lorentzian metric to define an angle, because that tangent is null.

At the observer the spatial direction of an outward-moving grazing ray is proportional to $w=A{\partial }_r+{\partial }_{\varphi }$, with $A=(dr/d\varphi {)}_{r_0}$. Its angle $\psi =\alpha /2$ to the outward radial direction equals the corresponding sky angle to the inward center of the image. Taking the spatial inner product with $e_r={\partial }_r$,ConsequentlyThe limiting value $A=0$ is understood by continuity.

For a grazing ray the areal radius has a turning point at the stellar surface $R$, so $\dot{r}=0$ there. The radial equation givesSince $R=4m>3m$, the surface lies outside the photon sphere. The function $f/r^2$ decreases for $r>3m$, so this tangential ray can escape to every $r_0>R$ and correctly defines the image's rim. For the specified surface,Also $dr/d\varphi =-r^{2}du/d\varphi$, henceIts sign depends on the orientation of the ray and of $\varphi$, and does not affect the angular diameter. Substitution givesThis apparent angular radius of a spherical star in Schwarzschild spacetime has useful limiting checks. As $r_0\downarrow R$, the star occupies a hemisphere, $\alpha \to \pi$. Far away, $\alpha \simeq 2b/r_0=8\sqrt{2}m/r_0$: Schwarzschild light deflection increases the apparent radius by a factor $\sqrt{2}$ relative to the flat-space value $R/r_0$.