# Would someone falling into or orbiting a black hole see themselves?

Spacetime is extremely curved in the vicinity of a black hole. When a light ray is targeted close to a black hole, it is curved to a significant angle. This angle increases as the target is pointed closer and closer. Is it possible that this angle reaches $$\sim 180^\circ$$, allowing an observer at orbit or falling into the black hole see themselves in front of them? How far away from the event horizon would this happen? Would the image appear to become closer or further to the observer?

Please ignore all the complications of interference from accretion disc, performance of a camera, the observer receiving fatal radiation dose, et cetera.

• Some years ago, Scientific American did an excellent article on exactly this topic: search for it, and all your questions will be answered- and none of your answers will be questioned! May 30, 2023 at 18:04

Is it possible that this angle reaches $$\sim 180^\circ$$, allowing an observer at orbit or falling into the black hole [to] see themselves in front of them?

Certainly! The light deflection angle near a BH (black hole) can be arbitrarily high, and you don't need to be close to the BH to have a light ray come back to you. (In this answer, I'll only discuss Schwarzschild BHs, rotating BHs are a little more complicated).

As I said in a previous answer on this topic,

The Schwarzschild radius $$r_s = \frac{2GM}{c^2}$$ is the natural distance unit to use when discussing black holes. It's convenient to work in units where $$r_s=1$$.

We can describe a photon trajectory in terms of the impact parameter, $$b$$, which is the perpendicular distance from the centre of the black hole to the asymptote of the trajectory. In other words, $$b$$ is the minimum distance (in Schwarzschild coordinates) from the photon trajectory to the centre of the black hole if the trajectory were not deflected by gravity.

Here's a graph showing the photon deflection angle for a range of impact parameters which result in deflections ranging from $$90°$$ to $$360°$$.

Here's the Sage / Python script I used to create that plot, running on the SageMathCell server. It can be used to create high-precision plots for arbitrary impact parameter ranges. It can also show the deflection angle given by the simple formula which is often used when $$b$$ is large.

Here's a typical example BH photon trajectory, from an answer I posted on Astronomy.SE

In that diagram, $$b=4$$ and the deflection angle is slightly more than $$49.2°$$. The blue curve shows the in-going light trajectory and the red curve shows the out-going trajectory, but the motion is symmetric. The green circle at $$r=1.5$$ is the photon sphere, the dotted grey circle at $$\sqrt{27}/2$$shows the critical impact parameter: a ray with this $$b$$ feeds into the photon sphere.

Photon orbits in the photon sphere are not stable. Such orbits are like a knife balanced on its edge, and the tiniest perturbation will kick the photon out of the photon sphere. However, a photon can loop around the BH multiple times if it's slightly outside (or slightly inside) the photon sphere.

Bodies with non-zero mass cannot orbit in the photon sphere, since the orbit speed there is $$c$$. The closest stable circular orbit to a BH for massive bodies is the Innermost stable circular orbit (ISCO).

As I mentioned in my previous answer, Albert Sneppen introduces a convenient parameter $$\delta$$, where $$b=b_0+\delta$$ and $$b_0$$ is the critical impact parameter.

Here's a diagram for $$b=2.678479, \delta=0.080403$$, which gives a deflection of $$180°$$ (and a closest approach of $$r=1.760309$$), so the outgoing ray is parallel to the incoming ray. However, that trajectory won't let you see yourself. Here's the trajectory for $$b=2.625165, \delta=0.027089$$, which gives a more useful deflection of $$240°$$ (with $$r=1.639672$$).

With this deflection, we're still fairly close to the BH. For a practical experiment we'd need to be a bit further away, unless the BH is huge, to avoid getting spaghettified by the tidal effect. ;) So we'd probably prefer a deflection just a little over $$180°$$.

The trajectories shown in this answer are more accurate than the trajectories in my earlier answer. The plotted points were computed via Carlson's elliptic integral AGM algorithm using 128 bit precision arithmetic. (The curves connecting the plotted points are just cubic Bézier curves, but they are accurate to within a half a pixel).

• Here are more precise values of the parameters for 180° deflection: b = 2.678479315370894344, delta = 0.080403104017578404, r = 1.760308981575963230 Aug 13, 2023 at 0:25
• And for 540°: b = 2.598216750552862354, delta = 0.000140539199546414, r = 1.509080184418068980 Aug 13, 2023 at 22:27

This is the effective potential for a photon close to a black hole. (Source) It has a maximum at $$r = \frac{3GM}{c^2}$$ which is called the photon sphere. At this distance, photons can actually orbit the black hole multiple times. Since it's a maximum in the potential, the orbits are unstable, such that sooner or later, the photon will either fly out or fall into the black hole. So at this distance, you would be able to see the back of your own head.

• The observer is falling, so I don't think that is (obviously) true. May 30, 2023 at 21:52

The term you're looking for is the photon sphere, and it's the last "stable" orbit a photon can take around a black hole. "Stable" is used here very inaccurately and more in a mathematical than physical sense, as the border between escaping the black hole's gravity and falling into it is infinitesimally thin. But, theoretically, this is the orbit where you could "see" the back of your head from light that orbits the black hole and then enters your eye.

For a Schwarzschild black hole, the photon sphere has radius $$r = \frac{3r_S}{2},$$ where $$r_S$$ is the black hole's Schwarzschild radius.