Given gravitational lensing, why would black holes visually appear black?

Question

This image from Wikipedia, also referenced in this question, is an artist's vision of how a black hole should look like:

However, given that a black hole is a gravitational lens strong enough to swallow the light rays that come too close to it, shouldn't it also bend light that comes from behind, slightly off the event horizon, so that it visually appears to come from the black disc area? If so, there should be no black disc but rather an area of distorted light from background stars and galaxies, right?

Answer 1

There is still a black disk because no photon can reach the observer directly from the event horizon. Even though light may circle the black hole many times, resulting in multiple images of the same distant stars, it still has to stay outside the horizon (in fact, the photon sphere). Since, as ProfRob's answer rightfully mentions, light trajectories are always curved inward, this means that there is still a "cone" of angles from which no photons reach the observer and this is what is seen as the black disk. This illustration shows the trajectories of the photons.

Since the trajectory of light is reversible (in the case of Schwarzschild spacetime), it may be helpful to think of the observer as emitting light rays in all possible directions rather than thinking of the possible directions from which light can be received. From this perspective, it should be clear that there is a "cone" of angles within which the rays will enter the horizon, which means that light cannot be received from these directions.

Answer 2

In order for a light ray to reach you that looks like it came from the direction of the "black disc", a light ray would have to curve away from the black hole. That doesn't happen; the curvature is always towards the black hole.

What this means, is that even light that grazes just above the radial coordinate $r = 3r_s/2$ (below which it must fall into the black hole, regardless of the direction from which it came), must appear to come to the observer from a radial coordinate $r> 3\sqrt{3}r_s/2$ - the apparent radius of the black disc.

The only light that can appear to come from the direction of the black disc would be from light that is emitted in the foreground of the black hole. i.e. From something that is in front of the black disc, with light travelling directly to the observer.

Answer 3

I have several existing answers on photon trajectories around Schwarzschild black holes, eg https://physics.stackexchange.com/a/805213 but I suppose a few more photon trajectory diagrams may help to show why there's a BH (black hole) "shadow".

I use units where the Schwarzschild radius $r_s=1$, so these diagrams apply to any Schwarzschild BH.

We can describe the trajectory in terms of the impact parameter $b$, which is the closest distance from the trajectory to the centre of the BH if there were no gravity. A light ray approaching the BH stays close to the impact parameter line until it gets near the BH.

As I explained here, the critical impact parameter $b_0$ is $\sqrt{27}/2$. A ray with $b=b_0$ feeds into the photon sphere at radius $1.5$. If $b>b_0$, the ray is deflected by the BH. And if $b<b_0$ then the ray is doomed to cross the photon sphere and hit the event horizon.

It's convenient to introduce a parameter $\delta$ with $b=b_0+\delta$. So only rays with positive $\delta$ can escape the BH.

Here's the trajectory for $\delta=7.5$, $b\approx 10.09808$, which gets deflected by $\theta\approx 13.3737°$.

The grey filled circle is the black hole, with its event horizon at radius $1$ The greenish circle at $1.5$ is the photon sphere. The radius of the dashed circle is $b_0$, so it's the black hole "shadow". The dashed black lines are the impact parameter lines. If the impact parameter line of a light ray crosses the $b_0$ circle then that ray will hit the event horizon.

The photon trajectory is in red and blue, with a purple dot at the point where it makes its closest approach to the BH. A dashed purple line connects that point to the centre of the BH. That line also passes through the point where the impact parameter lines cross. The trajectory is perfectly symmetrical about the purple line. So we can think of it as the path of a ray coming from either the red side or the blue side.

Here's the trajectory for $\delta=0.025, \theta\approx 244.499°$.

If you were at the point where the trajectory crosses itself, you could shoot a laser beam at the BH and it would return to you.

Here's the trajectory for $\delta=0.005, \theta\approx 335.668°$.

If our distant observer is looking at the BH from the right side of the diagram, at the blue end of the trajectory, they can receive light from all sorts of locations, even behind them, as long as $\delta>0$. Any trajectory with $\delta<0$ will hit the event horizon. And by symmetry, if they shoot a ray towards the BH it will get deflected if $\delta>0$, but it will get captured if $\delta<0$.

It's easier to see the "shadow cone" effect if we plot multiple trajectories on one diagram, and rotate them so they pass through one point. In the following diagrams, all trajectories pass through the point on the $X$ axis at radius $10$.

Here are trajectories for $\delta=0.5$ to $\delta=7.5$, with a step size of $0.5$.

$\delta=0.05$ to $\delta=0.5$, with a step size of $0.05$.

$\delta=0.005$ to $\delta=0.05$, with a step size of $0.005$.

Finally, here's a symmetrical path with $\delta = 1.55192567033486682549e-7, \theta = 930.090449414992°$. The path loops around the photon sphere twice before returning to the observer. The angle where the path crosses itself is $30.090449414992°$, so that's virtually the angular diameter of the black hole "shadow" at that distance.

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All of these trajectories were plotted by evaluating the elliptic integrals using Carlson's algorithms, with 80 bit arithmetic, except the final plot, which used 128 bit arithmetic.