<p>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 <em>can</em> 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 <em>then</em> head in your direction.</p>
<p>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 &quot;ring&quot; 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 &quot;frame dragging&quot; caused by the blackhole rotation (which is why the shadow is &quot;off-centre&quot;).</p>
<p>A rather academic description of the phenomenon is given by <a href="https://iopscience.iop.org/article/10.1086/312423/meta" rel="nofollow noreferrer">Falcke et al. (2000)</a> and <a href="https://academic.oup.com/mnras/article/379/3/833/1036973" rel="nofollow noreferrer">Huang et al. (2007)</a>.</p>
<p>You can observe the effects of &quot;shadowing&quot; for both Kerr and Schwarzschild blackholes at this <a href="https://www.esa.int/gsp/ACT/projects/blackhole_shadows/" rel="nofollow noreferrer">website</a>.</p>

CC BY-SA 4.0
2019-04-10T14:52:22.707Z
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
CC BY-SA 4.0
2019-04-10T14:53:33.470Z
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
CC BY-SA 4.0
2019-04-10T15:06:13.060Z
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
ProfRob
CC BY-SA 4.0
2019-04-10T15:11:31.433Z
That is brilliant. Thanks for the edit. Your link to the shadowing effect is brilliant!
MCG
CC BY-SA 4.0
2019-04-10T16:12:56.237Z
Rob I think I can ask via comment: are the faint tilted & lateral features the accretion disk? We shouldn't see it right from pole because intuitively there should be less or no asymmetry, so I was expecting a kind of bar like ø
Alchimista
CC BY-SA 4.0
2019-04-11T15:38:13.840Z
@Alchimista I do not believe we are "seeing" the accretion disk, which would be much bigger than this picture.
ProfRob
CC BY-SA 4.0
2019-04-11T15:59:34.613Z
That is a point that I would like to understand. Also the conference was a bit cryptic in reference to rotational verse.  But I am going through papers IV and V and I am confident I would extract quite enough from them.
Alchimista
CC BY-SA 4.0
2019-04-15T08:19:52.263Z
@RobJeffries I see you put up a bounty and mentioned the answer by pela. I will admit, it was very tempting to replace the accepted answer to his. When I last viewed it, his answer seemed like a complement to yours. Coming back a few days later, the additional information that has been added has swayed me to change. I do however want to say that yours is indeed still an excellent answer with great information. I appreciate all the answers here. Also, how does the bounty work? Do you give it seeing as you put it up?
MCG
CC BY-SA 4.0
2019-04-17T08:33:31.367Z
Oh _you_ put up a bounty, Rob? Thanks a lot! Also thanks to both you and @MCG, I learned a lot in this process of answering. Cheers!
pela
CC BY-SA 4.0
2019-04-17T08:49:40.017Z
@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
CC BY-SA 4.0
2021-10-15T23:35:48.010Z
The ESA link at the end of your answer is dead.
PM 2Ring
CC BY-SA 4.0
2021-10-16T07:00:55.083Z
@PM2Ring found it again, but doesn't work on my phone.
ProfRob
CC BY-SA 4.0
2021-10-16T11:53:45.640Z
It doesn't work on my phone (Galaxy S9, using the Samsung browser), either. It says the image is probably too large. Oh well.
PM 2Ring
CC BY-SA 4.0
2021-10-16T13:07:30.970Z
@PM2Ring it's fine on a proper computer.
ProfRob
CC BY-SA 4.0
2019-04-10T14:48:22.387Z
2021-10-16T06:59:50.117Z
2021-10-16T06:59:50.117Z
ProfRob
<p>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.</p>

CC BY-SA 4.0
2019-04-10T14:49:24.757Z
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
CC BY-SA 4.0
2019-04-10T15:05:29.083Z
@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
Steve Linton
CC BY-SA 4.0
2019-04-10T14:45:11.657Z
2019-04-10T14:45:11.657Z
Steve Linton
<p>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.</p>

CC BY-SA 4.0
2019-04-10T14:47:04.743Z
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
CC BY-SA 4.0
2019-04-10T16:21:30.870Z
It is not clear whether the bright region is an accretion disc or the jet.
Vladimir F Героям слава
CC BY-SA 4.0
2019-04-11T07:10:00.920Z
@MCG I don't understand the question. The accretion disk is a disk, not a hollow sphere around the horizon.
Parrotmaster
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2019-04-11T07:29:05.353Z
Could someone point out what about my answer is wrong?
Parrotmaster
CC BY-SA 4.0
2019-04-11T10:17:28.387Z
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
CC BY-SA 4.0
2019-04-11T13:15:58.717Z
@pela Isn't that what the event horizon is?
Parrotmaster
CC BY-SA 4.0
2019-04-11T14:22:05.420Z
Well, yes. You're right that we're seeing the event horizon. But when you say "_the EH stops the light_", it sounds (to me at least, and maybe to the downvoters) like the black region is just the EH blocking light from behind, like any object in front of something shiny. In fact, it is the EH viewed from all sides at once (and multiple times), surrounded by shiny stuff. Even if you were viewing the accretion disk edge-on, you'd see light all around the "shadow", because your sightlines bend around the hole. Sorry, this is a bit confusing, the drawing in my post may explain it better.
pela
CC BY-SA 4.0
2019-04-11T15:04:18.897Z
@pela I think you've misunderstood me then. I don't see a difference between what I was saying and your drawing.
Parrotmaster
CC BY-SA 4.0
2019-04-11T15:43:58.677Z
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)!
ProfRob
CC BY-SA 4.0
2019-04-11T16:54:18.817Z
@RobJeffries Yes you're right, that drawing doesn't correspond to M87. I'll make a new drawing at some point.
pela
CC BY-SA 4.0
2019-04-11T16:56:40.740Z
@pela Just label S at the top and it is ok.
ProfRob
CC BY-SA 4.0
2019-04-10T14:42:42.983Z
2019-04-10T14:42:42.983Z
Parrotmaster
<p>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.</p>

<h3>The photon sphere</h3>

<p>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 <span class="math-container">$r = 2GM/c^2 \equiv r_\mathrm{S}$</span> (the <a href="https://en.wikipedia.org/wiki/Schwarzschild_radius" rel="noreferrer">Schwarzschild radius</a>) from the BH marks the region from which no photons may escape if emitted <em>radially</em>, photons on partially tangential orbit will fall back out to a distance of <span class="math-container">$r = 1.5r_\mathrm{S}$</span>, where photons traveling fully tangentially will stay on the <a href="https://en.wikipedia.org/wiki/Photon_sphere" rel="noreferrer">photon sphere</a> (although this is an unstable orbit).</p>

<h3>The innermost stable orbit and the accretion disk</h3>

<p>Ordinary matter will spiral inward out to twice this distance; hence, inside the <a href="https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit" rel="noreferrer">innermost stable circular orbit</a> (ISCO) at <span class="math-container">$r=3r_\mathrm{S}$</span>, matter is pretty much bound to be absorbed. Outside this region matter may orbit, forming the <a href="https://en.wikipedia.org/wiki/Accretion_disk" rel="noreferrer">accretion disk</a>, 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 <a href="https://www.google.com/search?q=interstellar+black+hole" rel="noreferrer">depicted in the movie Interstellar</a>; rather a thick "cloud" surrounding most of the BH.</p>

<p>Photons emitted tangentially <em>just</em> outside the photon sphere will spiral around the BH many times, slowly increasing their distance, until eventually they escape at a projected distance of <span class="math-container">$\sqrt{27/4}r_\mathrm{S} \simeq 2.6r_\mathrm{S}$</span> from the BH (e.g. <a href="http://adsabs.harvard.edu/abs/1998bhp..book.....F" rel="noreferrer">Frolov &amp; Novikov 1998</a>).</p>

<h3>The shadow</h3>

<p>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 <em>closer</em> than (a projected distance of) <span class="math-container">$2.6r_\mathrm{S}$</span> to the BH will, eventually, end up on the EH, even if taking several orbits around the BH. These sightlines comprise the so-called <em>shadow</em> (<a href="http://adsabs.harvard.edu/abs/2000ApJ...528L..13F" rel="noreferrer">Falcke et al.(2000)</a>; <a href="http://adsabs.harvard.edu/abs/2019ApJ...875L...1E" rel="noreferrer">Event Horizon Telescope Collaboration et al.(2019a)</a>, ). 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 <em>don't</em> 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. <a href="http://adsabs.harvard.edu/abs/2019ApJ...875L...5E" rel="noreferrer">Event Horizon Telescope Collaboration et al.(2019b)</a>). This bright ring just outside the shadow is called the <em>photon ring</em>, or the <em>emission ring</em>.</p>

<h3>The drawing</h3>

<p>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 <em>again</em> see the front of the EH, and so on ad infinitum until you reach the photon ring.</p>

<p><a href="https://i.sstatic.net/2Kio5.jpg" rel="noreferrer"><img src="https://i.sstatic.net/2Kio5.jpg" alt="BHshadow"></a></p>

<h3>The observation</h3>

<p>Despite the observational resolution being an astonishing <span class="math-container">$\sim25$</span> 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".</p>

<p>Unless the accretion disk is viewed exactly face-on, half of the accretion disk<span class="math-container">$^\dagger$</span> has a velocity component toward you, making it brighter than the other half through a special relativistic effect called <a href="https://en.wikipedia.org/wiki/Relativistic_beaming" rel="noreferrer">beaming</a>. This is seen in the southern part of the M87 BH.</p>

<p>The figure below (from <a href="http://adsabs.harvard.edu/abs/2019ApJ...875L...5E" rel="noreferrer">Event Horizon Telescope Collaboration et al.(2019b)</a>) 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.</p>

<p><a href="https://i.sstatic.net/KtLpj.png" rel="noreferrer"><img src="https://i.sstatic.net/KtLpj.png" alt="BHobs"></a></p>

<hr>

<p><span class="math-container">$^\dagger$</span><sub>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.</sub></p>

CC BY-SA 4.0
2019-04-11T21:59:07.417Z
The disk is geometrically thick and optically thin. Nothing like the simulation in interstellar, which is the opposite.
ProfRob
CC BY-SA 4.0
2019-04-11T22:03:48.407Z
@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
CC BY-SA 4.0
2019-04-11T22:05:04.790Z
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.
ProfRob
CC BY-SA 4.0
2019-04-11T22:11:52.410Z
The orientation of the accretion disk does not affect the observability of a shadow.
ProfRob
CC BY-SA 4.0
2019-04-11T22:18:07.907Z
@RobJeffries Thanks, I didn't realize that. I'll remove the part about Interstellar.
pela
CC BY-SA 4.0
2019-04-12T07:59:17.683Z
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
CC BY-SA 4.0
2019-04-27T03:30:46.393Z
Beautiful answer, and great drawing!
Max0815
CC BY-SA 4.0
2019-04-27T05:49:58.933Z
You are invited to post and answer and your original work as another answer to [Could a trajectory around a large mass ever deflect by more than 180 degrees due to general relativistic effects?](https://physics.stackexchange.com/q/467144/83380) ;-)
uhoh
CC BY-SA 4.0
2020-09-27T06:18:57.683Z
The answer incorrectly says that sight lines in the shadow terminate on the event horizon. No sight lines from outside the horizon intersect the horizon. They "terminate" on matter that fell into the hole. The light from the matter is absurdly redshifted so it appears perfectly black. I think it's an important distinction because the idea that event horizons are like "glass ceilings" at a particular spatial location leads to a lot of confusion and prompts a lot of unnecessary questions on Physics.SE.
benrg
CC BY-SA 4.0
2020-09-27T10:10:25.953Z
Sightlines are not physical. You’re of course right that *matter* doesn’t actually reach  the horizon but it comes arbitrarily close so I think saying that the sightline terminates at the EH is okay in this context. But I don’t mind if you want to edit the phrasing :)
pela
CC BY-SA 4.0
2021-10-16T19:16:37.387Z
Brillant answer! One question: in the (non-blurred) model, there is a streaky spiral structure inside and outside the bright ring. Do I understand correctly that these are the distorted images of steaks of matter in the accretion disk?
uUnwY
CC BY-SA 4.0
2021-10-18T05:32:15.690Z
@StephanMatthiesen Yes, I think it's just spiral patterns of accreting matter, but they don't seem distorted. I'm a bit uncertain of how this model incorporates the bent sightlines that would distort the image.
pela
CC BY-SA 4.0
2019-04-11T12:06:31.313Z
2019-04-12T21:02:12.563Z
2019-04-12T21:02:12.563Z
pela
<p>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 <em>seeing</em> 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?</p>

<p>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. </p>

<p>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?</p>

<black-hole>
M87 Black hole. Why can we see the blackness?
MCG
# M87 Black hole. Why can we see the blackness?

(Asked by: username_0 on 2019-04-10)

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?

**Tags:** black-hole

---

## 4 answers

### Answer by username_3 on 2019-04-11. Score: 16 | Accepted answer

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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/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](https://www.google.com/search?q=interstellar+black+hole); 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](http://adsabs.harvard.edu/abs/1998bhp..book.....F)).

### The shadow

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)](http://adsabs.harvard.edu/abs/2000ApJ...528L..13F); [Event Horizon Telescope Collaboration et al.(2019a)](http://adsabs.harvard.edu/abs/2019ApJ...875L...1E), ). 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)](http://adsabs.harvard.edu/abs/2019ApJ...875L...5E)). 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.

[![BHshadow](https://i.sstatic.net/2Kio5.jpg)](https://i.sstatic.net/2Kio5.jpg)

### 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](https://en.wikipedia.org/wiki/Relativistic_beaming). This is seen in the southern part of the M87 BH.

The figure below (from [Event Horizon Telescope Collaboration et al.(2019b)](http://adsabs.harvard.edu/abs/2019ApJ...875L...5E)) 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.

[![BHobs](https://i.sstatic.net/KtLpj.png)](https://i.sstatic.net/KtLpj.png)

---

$^\dagger$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.

> **Comments:**
> **username_1**: The disk is geometrically thick and optically thin. Nothing like the simulation in interstellar, which is the opposite.  
> **username_3**: @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 $\tau$?  
> **username_1**: 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.  
> **username_1**: The orientation of the accretion disk does not affect the observability of a shadow.  
> **username_3**: @RobJeffries Thanks, I didn't realize that. I'll remove the part about Interstellar.  
> **username_0**: 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  
> **username_8**: Beautiful answer, and great drawing!  
> **username_9**: You are invited to post and answer and your original work as another answer to [Could a trajectory around a large mass ever deflect by more than 180 degrees due to general relativistic effects?](https://physics.stackexchange.com/q/467144/83380) ;-)  
> **username_10**: The answer incorrectly says that sight lines in the shadow terminate on the event horizon. No sight lines from outside the horizon intersect the horizon. They "terminate" on matter that fell into the hole. The light from the matter is absurdly redshifted so it appears perfectly black. I think it's an important distinction because the idea that event horizons are like "glass ceilings" at a particular spatial location leads to a lot of confusion and prompts a lot of unnecessary questions on Physics.SE.  
> **username_3**: Sightlines are not physical. You’re of course right that *matter* doesn’t actually reach  the horizon but it comes arbitrarily close so I think saying that the sightline terminates at the EH is okay in this context. But I don’t mind if you want to edit the phrasing :)  
> **username_11**: Brillant answer! One question: in the (non-blurred) model, there is a streaky spiral structure inside and outside the bright ring. Do I understand correctly that these are the distorted images of steaks of matter in the accretion disk?  
> **username_3**: @StephanMatthiesen Yes, I think it's just spiral patterns of accreting matter, but they don't seem distorted. I'm a bit uncertain of how this model incorporates the bent sightlines that would distort the image.  

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### Answer by username_1 on 2019-04-10. Score: 10

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)](https://iopscience.iop.org/article/10.1086/312423/meta) and [Huang et al. (2007)](https://academic.oup.com/mnras/article/379/3/833/1036973).

You can observe the effects of "shadowing" for both Kerr and Schwarzschild blackholes at this [website](https://www.esa.int/gsp/ACT/projects/blackhole_shadows/).

> **Comments:**
> **username_0**: 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  
> **username_0**: 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"?  
> **username_1**: 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  
> **username_0**: That is brilliant. Thanks for the edit. Your link to the shadowing effect is brilliant!  
> **username_2**: Rob I think I can ask via comment: are the faint tilted & lateral features the accretion disk? We shouldn't see it right from pole because intuitively there should be less or no asymmetry, so I was expecting a kind of bar like ø  
> **username_1**: @Alchimista I do not believe we are "seeing" the accretion disk, which would be much bigger than this picture.  
> **username_2**: That is a point that I would like to understand. Also the conference was a bit cryptic in reference to rotational verse.  But I am going through papers IV and V and I am confident I would extract quite enough from them.  
> **username_0**: @RobJeffries I see you put up a bounty and mentioned the answer by pela. I will admit, it was very tempting to replace the accepted answer to his. When I last viewed it, his answer seemed like a complement to yours. Coming back a few days later, the additional information that has been added has swayed me to change. I do however want to say that yours is indeed still an excellent answer with great information. I appreciate all the answers here. Also, how does the bounty work? Do you give it seeing as you put it up?  
> **username_3**: Oh _you_ put up a bounty, Rob? Thanks a lot! Also thanks to both you and @MCG, I learned a lot in this process of answering. Cheers!  
> **username_0**: @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  
> **username_4**: The ESA link at the end of your answer is dead.  
> **username_1**: @PM2Ring found it again, but doesn't work on my phone.  
> **username_4**: It doesn't work on my phone (Galaxy S9, using the Samsung browser), either. It says the image is probably too large. Oh well.  
> **username_1**: @PM2Ring it's fine on a proper computer.  

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### Answer by username_5 on 2019-04-10. Score: 2

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.

> **Comments:**
> **username_0**: 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?  
> **username_5**: @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  

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### Answer by username_6 on 2019-04-10. Score: -2

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.

> **Comments:**
> **username_0**: 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?  
> **username_7**: It is not clear whether the bright region is an accretion disc or the jet.  
> **username_6**: @MCG I don't understand the question. The accretion disk is a disk, not a hollow sphere around the horizon.  
> **username_6**: Could someone point out what about my answer is wrong?  
> **username_3**: 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.  
> **username_6**: @pela Isn't that what the event horizon is?  
> **username_3**: Well, yes. You're right that we're seeing the event horizon. But when you say "_the EH stops the light_", it sounds (to me at least, and maybe to the downvoters) like the black region is just the EH blocking light from behind, like any object in front of something shiny. In fact, it is the EH viewed from all sides at once (and multiple times), surrounded by shiny stuff. Even if you were viewing the accretion disk edge-on, you'd see light all around the "shadow", because your sightlines bend around the hole. Sorry, this is a bit confusing, the drawing in my post may explain it better.  
> **username_6**: @pela I think you've misunderstood me then. I don't see a difference between what I was saying and your drawing.  
> **username_1**: 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)!  
> **username_3**: @RobJeffries Yes you're right, that drawing doesn't correspond to M87. I'll make a new drawing at some point.  
> **username_1**: @pela Just label S at the top and it is ok.  

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