5
$\begingroup$

Whenever there is a picture of a black hole, (even though they are artist depictions, this is how they are described) most are shown kind of in a flat disk like this. enter image description here

My question is, why do we see the black hole itself? If it is sucking in material from all directions, then shouldn't the material be kind of covering the black hole, not just spinning around it to be sucked in from only the sides. Especially in the case that a star would be sucked in, then wouldn't all of its light & matter in the accretion disk be encasing the black hole? In other words, why is the accretion disk flat, and not a sphere surrounding the black hole? So if this were true, we shouldn't be able to see some black holes, only their accretion disks around them.

(unrevised/old post: If a black hole is spherical and has infinite density, then it would be sucking in objects from all directions because the gravity would be the same all around. In other words, the event horizon should be all around it!

I understand everything would orbit the black hole a little before 'falling in', but since the amount of matter and light is so large, shouldn't it encase the black hole, thus making it covered in a ball of light? The event horizon would be a shell so we could not see some black holes, which would mean there could be billions more out there.

As for the other black holes which we have seen/proven, if the information above is true, then maybe they are the result of black holes with poles with low force/gravity/magnetivity/energy. Possibly they could be a disk of some sort or have an unequal distribution of density, either of which could lead to the flat event horizon, maybe even both.

Poles on a black hole would also help account for the quasars shooting out of some. Since the gravitational pull would not be as great from the poles of a black hole, then this would allow the quasars to beam out (otherwise they would have to be going faster than the escape velocity, which is the speed of light).

Maybe the quasars are even the black holes that I was talking about earlier, and this is why you would see no event horizon, only a shell of bright light.

I am no expert in astronomy but if any of these theories sound plausible, please let me know! Just looking for some explanation, I thought of this earlier and couldn't figure it out, I could be completely wrong though. Thanks!)

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ Can you rephrase this to ask a single specific and clear question? The question posed in the title doesn't seem to be in the body of the post. $\endgroup$
    – James K
    Commented Feb 1, 2019 at 20:59
  • 1
    $\begingroup$ Hi Kimberly, welcome to Astronomy. Your question "why do we see the black hole itself?" is impossible to answer, as it's based on an incorrect assumption. Can you give an example of where we "see" a black hole? I'm not aware of any. We detect black holes by inference: stellar orbits, intense radiation from an accretion disk, gravitational waves, etc. Artists' impressions are simply that: impressions, not real observation. Wikipedia is a good start for a better understanding. :-) $\endgroup$
    – Chappo Hasn't Forgotten
    Commented Feb 2, 2019 at 4:33
  • 1
    $\begingroup$ Perhaps your real question is about the shape of the accretion disk. In fact there's already exactly that question on our site - see Accretion disks - why are they disk-shaped, rather than spherical?. $\endgroup$
    – Chappo Hasn't Forgotten
    Commented Feb 2, 2019 at 4:40
  • 1
    $\begingroup$ Thanks! No, I was aware we cannot see any black holes, but they are almost always described in the way as in the picture, which is why I put quotes around 'see'. I was not trying to reference that we actually could see them! And yes, that is the question I was trying to figure out thank you very much!! $\endgroup$
    – Kimberly Bailey
    Commented Feb 3, 2019 at 3:03
  • 1
    $\begingroup$ @Chappo we do have an image of at least one black hole - that at the centre of M87. It is also a fair question to ask, why, when simulations are produced or artists impressions produced are black holes shown in this way? $\endgroup$
    – ProfRob
    Commented Sep 17, 2019 at 16:21

2 Answers 2

Reset to default
7
$\begingroup$

Well it doesn't exactly work the way you describe it. Matter doesn't just fall into a black hole.

A black hole still has a finite mass, which means other matter orbits its just like it would a comparable object of the same mass, like a star. In fact the gravitational disturbance it causes in this way is usually how we determine that there are, for example, supermassive blackholes in the center of galaxies.

Now the light you speak of is an accretion disk. When a star orbits a black hole, the star can get torn apart due to gravitational variation in the star as a consequence of the nearby black hole. Matter from the star starts orbiting the black hole, but it cant just do that in any direction. Due to conservation of angular momentum (the star itself was rotating) it is forced to rotate in one specific circular path, thats what we call the accretion disk and is the reason it doesn't become a huge sphere of light. As for light itself, it can't orbit, it either gets deflected or it gets trapped inside.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ Okay! Yes, that I can understand thank you! The thing I don't get is if the angular momentum of each star/object is different (and not all are, but most), wouldn't there be a whole bunch of varying orbits around the black hole, not just the flat accretion disk with pretty much the same orbits? Also, yes you are correct light does not orbit sorry I did not mean to say that. I was kinda trying to decribe gravitational lensing but couldn't remember the word. thanks! $\endgroup$
    – Kimberly Bailey
    Commented Feb 2, 2019 at 2:14
  • 1
    $\begingroup$ @Kimberly But light can orbit a black hole, in the photon sphere, although as that article mentions, those orbits are rarely stable. And of course the photons in the photon sphere aren't visible to observers outside the photon sphere. $\endgroup$
    – PM 2Ring
    Commented Feb 2, 2019 at 8:11
  • 1
    $\begingroup$ @Kimberly Bailey Ah I see what you mean. The answer is that space is actually really empty. For an accretion disk to form a star had to orbit and thus be in its neighbourhood. To give you a bit of a perspective: when our galaxy collides with andromeda, chances are pretty high that no stars will collide at all, just because the distances between them are so immensly vast. The accretion disk dissapears before there is even a chance for a second one to appear. $\endgroup$
    – Jeroen
    Commented Feb 2, 2019 at 8:41
  • 1
    $\begingroup$ ^^ Oh wow, that is very interesting I did not know that!! $\endgroup$
    – Kimberly Bailey
    Commented Feb 3, 2019 at 3:06
  • 2
    $\begingroup$ Okay, that makes sense! So really, the chances of even one star oribiting a black hole are pretty low, and multiple objects/stars is close to none? That would explain a lot, thanks for all the information and help! $\endgroup$
    – Kimberly Bailey
    Commented Feb 3, 2019 at 3:10
2
$\begingroup$

To be entirely covered by material, it'd have to be surrounded by material in the first place.

It's my understanding that most matter in the universe tends to organize in a plane or some description (maybe because it's a vastly simpler orientation; or most material came from stars/supernovae, and in anything rotating in multiple directions at once those rotations sum together to be a single one).

Heck, there may be black holes out there that're currently (minus light ToF) surrounded by accretion disks. But it's far less likely to happen (how many disks do you need to cover a ball?), and you wouldn't know because you can't see it.

Only way to know for sure is to wait for such a black hole to get discovered due to its mass, and then for a bright set of disks to be seen obscuring it. (Though, with distance and a sufficiently bright disk, this could happen with just one.)

Improve this answer
$\endgroup$
7
  • 1
    $\begingroup$ What do you mean by "surrounded by accretion disks"? $\endgroup$
    – PM 2Ring
    Commented Sep 16, 2019 at 11:09
  • $\begingroup$ That enough be around the black hole that the number direct paths (not intersecting a disk) between the event horizon and the observer is low or zero. That is assuming that intersecting disks can exist for long enough to be noticed before interfering with each other. (and enough are being formed that more than one even happens at the same time.) No idea how much of my understanding is just geometry that's completely wrong at this scale. $\endgroup$
    – bobsburner
    Commented Sep 16, 2019 at 11:14
  • 1
    $\begingroup$ Accretion disks are a bit like the rings of Saturn. They rotate in the black hole's equatorial plane, so you can't really have 2 intersecting accretion disks like that. Also, unlike Newtonian gravity, where you can have orbits in pretty much any plane, orbits around a rotating BH are strongly constrained. Even stuff that approaches the BH from a weird angle gets forced into an equatorial orbit because of frame dragging. $\endgroup$
    – PM 2Ring
    Commented Sep 16, 2019 at 11:39
  • $\begingroup$ You make a good point. How long should it take for everything to get pulled into line with the equatorial plane? If I think of a decent way to phrase it, I'll edit my answer. Feel free, if anyone thinks they can improve it. $\endgroup$
    – bobsburner
    Commented Sep 16, 2019 at 11:55
  • 1
    $\begingroup$ With black holes, "How long?" questions are tricky, due to time dilation, so processes look slower to distant observers. But in the time frame of bodies falling towards the BH, stuff happens pretty quickly, since everything's moving at relativistic speeds. Not everything that falls towards the BH gets pulled inside the event horizon. It can get deflected into a hyperbolic trajectory. Or it can collide with matter in the accretion disk. Such collisions are extremely violent, and tend to eject a fair bit of "shrapnel", some of that may be thrown out in jets by the disk's powerful magnetic field. $\endgroup$
    – PM 2Ring
    Commented Sep 16, 2019 at 12:11

You must log in to answer this question.

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