This excellent answer mentions several ways to try to observationally measure the spin of a rotating black hole. The third one is intriguing, but I don't understand how this works:

  • The spin of a black hole also affects how it deflects light. Consequently, the pictures of a black hole's shadow such as taken by the event horizon telescope can be used to determine the spin of the black hole (if we happen to view it under the right angle).

Question: What is a black hole's "shadow" and the best angle to view it to measure the BH's spin?


1 Answer 1


The black hole shadow is basically the image of the event horizon. As you know the event horizon is the geometric locus of the points from which a light ray pointing in the opposite side of the singularity (the center of the black hole) can't no longer escape from it. Any other light ray emmited in any other direction from those points would never reach an external observer since the one pointing directly outside the black hole was the one with the best chances to do so.

So we might think that because of the absence of light coming from these region inside the event horizon an external observer might see a black ball of 1 Schwarzschild radius (which is the physical size of the even horizon), but this is not correct. Even if the event horizon marks the physical boundary between both regions the fact is that due to the extreme light bending the actual image of the event horizon (its "shadow") is a distorted view of this surface. For a classical non-rotating black hole this "shadow" (we call a shadow the image created by the absence of light rays but we can trace it with hypothetical "dark rays" that behave in the same way) looks like a ball with 2.6 Schwarzschild radii in size. Much larger than the actual event horizon! To visualize this I can only point to this fantastic explanation by Derek Muller from whom I took this animation.

As you can see we are throwing light rays from infinity into the black hole (that's why they are all parallel in the beginning). Our light rays fall reach the event horizon even if they were not directly pointed there, because they curve. Since light paths can be reversed and the physics still hold we can talk about light rays coming from different parts of the event horizon and reaching the outside observer in the reversed paths. So as you can see not only light rays emmited just barely outside the event horizon pointing towards us will reach us but also rays coming from "the back" of the event horizon can reach us. And as you can see those rays would create an enlarged image of the event horizon since they seem to come from a region located farther than the event horizon itself. So when you look to a real black hole from outside you will see this "black shadow", which is a map reprojection of the surface of the true event horizon where you can see even the 100% of the surface of it from a single vantage point.

This phenomenon (which is called relativistic light deflection) is also noticiable in other compact objects like Neutron stars. The intense gravitational distortion around them allows for light rays coming from its bright surface to deflect when going outwards and reach your eye even if they where emmited in regions close to "the back of the neutron star". Even if that region of the star shouldn't be accesible to an observer if light rays moved in straight lines (since those parts of the surface lie behind the curve of the star) you still can see them (which is something that can mess with the calculations of their true brightness).

You can make a square grid on top of the surface of the neutron star and see how much of it you can see from far away in this representation:

enter image description here

As you can see we are able to see more than an hemisphere (more than 50% the surface of the star). In fact you can see both polar regions and their surroundings. Well this is the same thing happening in a black hole but in that case the reprojected map is all black (since the event horizon is a uniform featureless surface all around) and you can see the 100% of that surface not only a small extra percent.

Now, this all changes if the black hole is rotating. Why? Because of relativistic frame-dragging. According to Einstein field equations mass-energy does not only curve spacetime but is also able to "twist it" if the object is rotating. We have measured this "twisting" of the surrounding spacetime in our own planet using exquisite instruments onboard the GRACE satellites.

In our case the important thing is that a non-rotating black hole (a Schwarzschild black hole) has a region outside the event horizon where orbiting the black hole in a stable manner is possible, we call it the innermost stable circular orbit (or ISCO). Getting closer to ISCO makes your orbit unstable and you end up falling into the event horizon. But if the black hole is rotating (a Kerr black hole), then ISCO is different if your orbit is prograde (orbit in the same direction as the black hole rotation) or retrograde (goes in the opposite direction around the black hole) because frame-dragging alters the solution. If you orbit prograde the fact that spacetime is been dragged in the same direction allows you to have some push by the black hole and your orbit can be mantained even much closer to the event horizon in a stable manner. On the contrary if you orbit the black hole in the opposite direction you are fighting against the drag of spacetime and thus you will decay more easily, making the ISCO for retrograde orbits a lot higher than the ISCO for prograde orbits.

If you apply this reasoning to photons you can start to notice something interesting. Light coming from far away stars behind the black hole as viewed from an outside observed gets bent in different ways if it is coming from one side or the other due to this frame-dragging effect. If the light ray is coming parallel over the rotating surface of the black hole it is going to be helped by the black hole itself, and some of the angular momentum is going to be transfered to that light ray from the Kerr black hole. If instead the light ray comes anti-parallel to the rotation (which is going to happen in the other side of the black hole), then it might never reach the observer. This reasoning can be applied to the "dark rays" (which do not exist but is a way to trace the shadow which is the absence of light rays) coming from the event horizon and thus the shadow of the black hole is no longer a perfect black disk but an asymetric D-shaped black region, that tells you if the black hole is rotating clockwise or counter-clockwise.

In this animation you can see the appearance of the shadow of the black hole when we increase its rotational speed, as you can see it goes of-center and asymetric as we increase it.

Since frame-dragging goes as the mass rotates you can't notice it so strongly if you watch the black hole from another inclination. In fact the projected rotational speed if you watch a black hole from the poles ($i = 0^\circ$) is zero, and thus the black hole would look just like a non-rotating one. Here you can see the dependance of the shape of the shadow (in red) of a Kerr black hole with a fixed rotational speed when you see it from different inclinations (from the equator to the poles). The event horizon is represented in blue (but remember, you don't see that, you only see the shadow)

As you can see the effect is the same in both cases (changing view angle for a fixed rotational speed vs. changing rotational speed for a fixed inclination), which means that you can't tell the real rotational speed of a black hole just by measuring the shape of the shadow (a disk like shadow could mean a non-rotating black hole or a Kerr black hole as viewed from above for example), but at least it gives you a minumim estimate for the rotation. To study the exact rotational speed we need some independent measurements like for example the inclination of a disk of material around it. In those cases you would have complete information about the angular momentum of the black hole.

Finally here you have a beautifull simulation on what you would see from orbit around a Kerr black hole (I don't know how to embed YouTube videos so...).

  • $\begingroup$ Wow, thank you for such a thorough answer! It will take me some time to read through it carefully, but +1 for the moment! fyi answers to Could a trajectory around a large mass ever deflect by more than 180 degrees due to general relativistic effects? are only slightly related as these are orbit trajectories of projectiles rather than light rays. $\endgroup$
    – uhoh
    Sep 15, 2019 at 12:35
  • $\begingroup$ I'm sorry, English is not my mother tongue and I've never lived in an English-speaking country. I fixed those. If you spot any other error please tell me. $\endgroup$
    – Swike
    Sep 16, 2019 at 9:26
  • 3
    $\begingroup$ The shadow is interior to the photon ring, at 1.5 $r_s$, not the event horizon. It is the photon ring which is magnified by a factor $(1-r_s/r)^{-1/2}$ to give $2.6r_s$. The spin of a black hole can also be estimated from the asymmetry of brightness in any photon ring, not just from asymmetry in the shape of the ring itself. $\endgroup$
    – ProfRob
    Sep 16, 2019 at 9:51
  • $\begingroup$ Even if the photon sphere is higher up than the event horizon, the image of the photon ring can be surpassed by the distorted image we call the shadow of the black hole, right? At least that what it looks like in Fig. 3 of this paper arxiv.org/pdf/1906.00873.pdf Maybe I'm missunderstanding some things since this is not my field. $\endgroup$
    – Swike
    Sep 16, 2019 at 14:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .