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Suppose we have a Schwarzschild black hole of mass $M$ at a distance $R$ to an observer. Due to the curvature of space, the apparent angle (or half-angle) of occlusion $\theta_a$ of the silhouette of the blackhole will be larger than the angle described by its Schwarzschild radius, $\theta_s$. But how much larger?

blackhole observer

Taking no relativistic effects into account, the angle subtended by the center of a sphere to the edge of a sphere with a radius the same size as the Schwarzschild radius $r_s=\frac{2GM}{c^2}$ can be deduced to be $\theta_s=\arcsin\frac{r_s}{R}=\arcsin\frac{2GM}{Rc^2}$. However, a light ray cast from the observer at an even larger angle (maximally $\theta_a$) may find it's way falling into orbit or directly into the blackhole, causing the apparent size of the blackhole to be larger than its Schwarzschild radius.

How is $\theta_a$ described by $R$ and $M$? Is it the angle (or half-angle) subtended by the photon sphere?

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Any light ray that crosses the photon sphere will hit the horizon, and any light ray that doesn't, doesn't. So, your guess that it would be the angle subtended by the photonsphere is pretty good. However, in reality the angle is slightly bigger, since two light rays that start parallel at infinity will have bend further inwards by the time they reach the photon sphere.

The quantity you want to look at as the critical impact parameter $b_{crit}$. This is the distance between two light rays that start parallel at infinity at are tangent to the photon sphere.

For a Schwarzschild black hole with mass $M$:

$$b_{crit} = 6\sqrt{3}\frac{G M}{c^2} $$

So for a sufficiently distant observer at distance $R$ (ignoring any cosmological effects due to expansion) the angle $\theta$ subtended by the black hole is given by:

$$ \theta \sim 6\sqrt{3}\frac{G M}{ R c^2}.$$

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  • $\begingroup$ Fascinating. Cheers. $\endgroup$
    – Graviton
    Jun 1, 2021 at 6:04

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