# M87 Black hole. Why can we see the blackness?

So, as the title states, why are we able to actually see the 'blackness' of the black hole? I get that what we are actually seeing is the event horizon, or accretion disk. But should this not extend all the way round? Surely the black hole is not a 2D thing so we can "look into it from above" (I use that term loosely as obviously there are no directions in space!), so why are we able to actually see the blackness?

The only thing I can think of is that a black hole, much like our solar system has some sort of ecliptic, that the vast majority of matter is orbiting, and everywhere else just doesn't have enough matter for the light to be visible, kind of like why we are unable to see the Oort cloud.

I hope that made sense, and I could be way off, but it is the only thing I could think of to explain it. If it is the case, then would we be able to get a similar image of Sagittarius A, seeing as we could be in this "ecliptic" of it, so surely we would only be able to see the heated matter around the event horizon and not the blackness?

Rob Jeffries' answer is excellent, I just wanted to add this picture trying to explain the geometry. Here, I assume a non-rotating black hole (BH); for a rotating BH the exact numbers are slightly different.

### The photon sphere

Photons move on straight lines, but in the heavily curved space around a BH these straight lines appear curved. Although the event horizon (EH) at a distance of $$r = 2GM/c^2 \equiv r_\mathrm{S}$$ (the Schwarzschild radius) from the BH marks the region from which no photons may escape if emitted radially, photons on partially tangential orbit will fall back out to a distance of $$r = 1.5r_\mathrm{S}$$, where photons traveling fully tangentially will stay on the photon sphere (although this is an unstable orbit).

### The innermost stable orbit and the accretion disk

Ordinary matter will spiral inward out to twice this distance; hence, inside the innermost stable circular orbit (ISCO) at $$r=3r_\mathrm{S}$$, matter is pretty much bound to be absorbed. Outside this region matter may orbit, forming the accretion disk, but since friction between the particles will cause them to lose energy, they will slowly approach the ISCO, after which they will rapidly fall into the BH. Note that the M87 BH doesn't have a thin accretion disk like the one depicted in the movie Interstellar; rather a thick "cloud" surrounding most of the BH.

Photons emitted tangentially just outside the photon sphere will spiral around the BH many times, slowly increasing their distance, until eventually they escape at a projected distance of $$\sqrt{27/4}r_\mathrm{S} \simeq 2.6r_\mathrm{S}$$ from the BH (e.g. Frolov & Novikov 1998).

Just as the path of light rays are curved around the BH, so are the sightlines from you toward the BH (you can think of sightlines as reversed photons). That means that all sightlines that are closer than (a projected distance of) $$2.6r_\mathrm{S}$$ to the BH will, eventually, end up on the EH, even if taking several orbits around the BH. These sightlines comprise the so-called shadow (Falcke et al.(2000); Event Horizon Telescope Collaboration et al.(2019a), ). On the other hand, along sightlines that a farther away, you see the radiation emitted from the matter falling into the BH, both in front of and behind the BH. And since the first sightlines that don't terminate at the EH circle the photon sphere many times, those sightlines are actually very long paths through matter shining its last light before being engulfed, and hence they look exceptionally bright (e.g. Event Horizon Telescope Collaboration et al.(2019b)). This bright ring just outside the shadow is called the photon ring, or the emission ring.

### The drawing

The drawing below may help understand. All the red lines are sightlines toward the BH. Only the uppermost one just grazes the photon sphere (and the luminous matter behind). The rest terminate at the EH, and hence look black (except for luminous matter in front). Close to the center, you see the front of the EH; farther out you actually see the back of the EH; even farther out you again see the front of the EH, and so on ad infinitum until you reach the photon ring.

### The observation

Despite the observational resolution being an astonishing $$\sim25$$ micro-arcseconds, the photon ring is smeared out over a larger region, resulting in the doughnut shape you've seen. That is, what you see in that image is not "the EH in front of an accretion disk", but rather "the EH seen from all sides at the same time and enlarged, with light emitted from the photons ring".

Unless the accretion disk is viewed exactly face-on, half of the accretion disk$$^\dagger$$ has a velocity component toward you, making it brighter than the other half through a special relativistic effect called beaming. This is seen in the southern part of the M87 BH.

The figure below (from Event Horizon Telescope Collaboration et al.(2019b)) shows, from left to right, the actual observation, a model where you see the rather sharp photon ring, and this model blurred to match the resolution of the observation.

$$^\dagger$$At least the material just before it plunges into the BH, which follows the rotation of the BH. Farther out, the rest of the accretion disk may in principle rotate the other way.

• The disk is geometrically thick and optically thin. Nothing like the simulation in interstellar, which is the opposite. Apr 11, 2019 at 21:59
• @RobJeffries Really? But if it's optically thin, why is the M87 BH so bright then? Because the sightlines take many turns around the BH, increasing the total $\tau$?
– pela
Apr 11, 2019 at 22:03
• Indeed the only reason a photon ring is visible is because the plasma is optically thin. Sight lines passing close to the photon ring have larger optical depths. Apr 11, 2019 at 22:05
• This is also a brilliant answer. It complements the answer by Rob brilliantly. If I could accept both, I would! Thank you for including a drawing too, that makes things easier to imagine. I will have a read through those links you posted, the only ones I am familiar with are the Schwarzschild radius and accretion disk. Seems I have a lot to read up on! Thank you again
– MCG
Apr 12, 2019 at 7:59
• Beautiful answer, and great drawing! Apr 27, 2019 at 3:30

You have to think about how the light is going to get to you from where it is produced close to the black hole event horizon. Light produced between you and the black hole can get to you. Light produced immediately behind the black hole cannot get to you (or at least it does not come to you from that direction). Light produced at other positions can get to you via various routes, one of which is to orbit the black hole and then head in your direction.

As a result of this there is a concentration of the observed light into an apparent ring around the black hole and a dark(er) circle inside it which marks the region from which light cannot travel directly to you, but instead either falls into the black hole or loops around it. Asymmetries in the photon "ring" are going to be caused by the relativistic orbital motion of material that has the effect of boosting emission in the forward direction and also by the "frame dragging" caused by the blackhole rotation (which is why the shadow is "off-centre").

A rather academic description of the phenomenon is given by Falcke et al. (2000) and Huang et al. (2007).

You can observe the effects of "shadowing" for both Kerr and Schwarzschild blackholes at this website.

• Thanks for that, this is a good explanation! Do you happen to have any sources or links to back this up? Especially about light taking alternative routes to get to us. Just to clarify, I am not asking for links because of skepticism, I would just like something to read into a bit more as I can imagine it would be difficult to put it all into a single answer here! And I would like to understand it a bit better
– MCG
Apr 10, 2019 at 14:52
• And also, does this mean that even if this particular black hole was oriented differently, we would still see the same phenomena? As in we would still see an accretion disk circling it, and we would still see a circular "blackness"?
– MCG
Apr 10, 2019 at 14:53
• Small edit applied. The orientation would only matter in so far that the brightness asymmetry would change. The "shadows" are all nearly circular if GR is correct. @MCG Apr 10, 2019 at 15:06
• @pela as did I! I ended up reading through the EHT papers too. A lot of it went over my head (I'm an engineer, not an astrophysicist lol) but I was able to learn a lot from it. Following up on the links both you and Rob put up was really helpful
– MCG
Apr 17, 2019 at 8:49
• @PM2Ring it's fine on a proper computer. Oct 16, 2021 at 13:07

The paths light takes near a black hole are not anything like the ones it takes in empty space. Basically we're seeing the "shadow" of the black hole. Much of the light that we would expect to be coming towards us from that particular direction has been diverted elsewhere by the hole's gravity.

• So (if I understand correctly), what you are saying is that there may well be matter on the "side" facing towards us, but because the light of it is distorted by the black hole, we are unable to see it, thus allowing the illusion of us seeing "in" to it? Hence why we see the accretion disk appear to be almost '2D' from our perspective?
– MCG
Apr 10, 2019 at 14:49
• @MCG more or less. We may see light from that matter, but it doesn't come straight to us (or at least not much of it does) so we don't see light in that place on the image Apr 10, 2019 at 15:05

This black hole has an accretion disk, which is a disk of matter that rotates around the black hole at extreme speeds, causing it to heat up. The orange colour you see in the picture is that matter. The matter seems "thicker" on one side because the bottom of the disk is tilted slightly towards us. The "blackness" you see is simply the event-horizon stopping the light from that region from escaping.

• I understand what we are seeing is the accretion disk (as stated in the question), but it is why we get to see the actual blackness, as surely there should be matter around the part tilted towards us? Why can we not see that?
– MCG
Apr 10, 2019 at 14:47
• It is not clear whether the bright region is an accretion disc or the jet. Apr 10, 2019 at 16:21
• Hi @Parrotmaster, the bright region is not because the disk is tilted in that direction, but because matter is rotating toward us at that point, increasing the brightness through relativistic beaming. Also, the "blackness" is not really the event horizon "stopping light", but a somewhat (2.6×, to be specific) larger region, consisting of all sightlines ending at the horizon.
– pela
Apr 11, 2019 at 10:17
• You seem to be suggesting that the emission maps the accretion disk. It doesn't. The orientation of the plane of the accretion disk, which is much bigger than the picture is almost N-S (as Pela has sketched it in fact). Unfortunately, Pela has it rotating the wrong way (if N is to the top)! Apr 11, 2019 at 15:43
• @pela Just label S at the top and it is ok. Apr 11, 2019 at 16:56