<p>I think, that's a bit semantics, and a question of what do you call 'I see something': If you see the shadow cast by an object on a bright background - do you see the object? Do you see Mercury or Venus when it passes in front of the disk of our sun? Similarly, if you look an object through an atomic force microscope - do you see the object, or do you just see some force indicator giving a value?</p>

<p>So yes, in this general sense, you see something when you have an instrument to detect it which allows to deduce properties of the object. So you also can see an object when you just see its shape in front of a bright background like a black hole within its accretion disk - even when all light you see comes from the latter.</p>

CC BY-SA 4.0
2020-04-14T17:23:56.010Z
2020-04-15T13:27:54.117Z
2020-04-15T13:27:54.117Z
planetmaker
<p>No, you cannot "see" a black hole, only the way that it interacts with objects or light in its vicinity.</p>

<p>In terms of interacting with other objects, the classic observations that betray the presence of a black hole are the very rapid motions of stars and gas close to an object with a large mass, but which cannot be directly seen. Examples include low- and high-mass X-ray binary systems like <a href="https://en.wikipedia.org/wiki/A0620-00" rel="noreferrer">A0620-00</a> and <a href="https://en.wikipedia.org/wiki/Cygnus_X-1" rel="noreferrer">Cygnus X-1</a> respectively. Supermassive black holes have been detected by measuring the rapid motion of gas at the centres of many galaxies, and the 4 million solar-mass example at the centre of our Galaxy is revealed by watching the motion of stars that <a href="https://www.eso.org/public/videos/eso1825e/" rel="noreferrer">orbit it</a>.</p>

<p>In terms of interacting with light, the strong gravity of black holes will distort objects seen in the background of a black hole, but this effect is yet to be seen in detail. Instead, using a clever interferometric imaging technique, the central region of the galaxy M87 has been observed at microwave wavelengths. The famous image you have doubtless seen, is a projection of the <a href="https://en.wikipedia.org/wiki/Photon_sphere" rel="noreferrer">photon sphere</a> of the supermassive black hole, which is illuminated by hot gas surrounding it. Much more detail on this result can be found in several other Astronomy SE questions - notably <a href="https://astronomy.stackexchange.com/questions/30317/m87-black-hole-why-can-we-see-the-blackness?noredirect=1&amp;lq=1">M87 Black hole. Why can we see the blackness?</a> Note that the photon ring is <em>not</em> the accretion disk.</p>

CC BY-SA 4.0
2020-04-15T19:09:13.513Z
"No, you cannot "see" a black hole, only the way that it interacts with objects or light in its vicinity.". @user253751 no mention of distance.
ProfRob
CC BY-SA 4.0
2020-04-16T20:17:11.333Z
@user253751 what about my answer contradicts that? You appear not to have read it
ProfRob
CC BY-SA 4.0
2020-04-14T20:05:33.250Z
2020-04-14T20:05:33.250Z
ProfRob
<p><strong>Yes,</strong> we can, indeed, "see" a black hole and we just did not too long ago.</p>

<p>You see, being able to "see" an object is not solely about "reflection" alone: rather, you "see" things by them <em>interacting</em> in <em>any</em> way with the light that reaches your eye (or other "seeing" instrument), and then by your eye interacting with that object-interacted radiation. Reflection is just <em>one</em> way. Absorption (the <em>removal</em> of light rays that would otherwise reach your eye) is another, and you don't need a black hole for that: any object that is <em>not</em> totally transparent, shiny, or white, under white light, must have absorbed some radiation for it to look that way.</p>

<p>If you took a big ball and painted it black and hung it in front of a light background, would you say you were "seeing" that black ball? If so, that is <em>not</em> very much different. If you agree you can "see" that, then you can "see" a black hole, just same. It does the same thing: it <em>absorbs</em> the light from any luminous background, distinguishing itself because those rays that were sent from the points on the background you'd see were it not there, <em>fail</em> to reach your eye. Likewise, if you want to argue you "cannot 'see'" the black hole, then you cannot "see" that black ball either, by the same reasoning.</p>

<p>Now, of course, the <em>mechanism</em> of the absorption is different in each case, but the result is the same in terms of what happens to light rays. Rays that would have been otherwise able to reach your eyes (or here, cameras) were the object not there, now don't.</p>

<p>You also mention that the picture is "colorized". This is true, but it's not because black holes "can't be seen", but rather because that the telescope is "seeing" in a different <em>wavelength range</em> than your eyes do and that's because those wavelengths have advantages for detecting them at this level of resolution and at this distance compared to visible light, <em>not</em> because that we could not see the black hole with our own eyes were we there (and suitably protected from the intense energy). Colors are <em>always</em> arbitrary, being assigned by our brains. Light itself doesn't have "colors". Since our eyes, and so our brains, aren't set up with (kind of tautologically) receptors and hence color channels for wavelengths we can't see, we've got to reuse the ones we've got somehow. This doesn't make it any less a "real photograph" though, any more than taking a greyscale photograph of a scene with an ordinary visible-light camera does (and if anything, I'd have put it in grey, but I suspect the red is because they want to make it look "fiery", so it's a little artistic license.).</p>

CC BY-SA 4.0
2020-04-15T19:20:07.197Z
2020-04-15T19:32:06.727Z
2020-04-15T19:32:06.727Z
The_Sympathizer
<p><a href="https://skyandtelescope.org/astronomy-news/the-appearance-of-a-black-holes-shadow/" rel="noreferrer">This article</a> explains what the image shows.</p>

<blockquote>
  <p>This distinct structure is a result of the warped spacetime around massive objects like black holes. The ring of light is comprised of photons from the hot, radiating gas that surrounds the black hole, whose paths have been bent around the black hole before arriving at our telescopes. The dark region in the center is termed the black hole’s “shadow”; this is the collection of paths of photons that did not escape, but were instead captured by the black hole.</p>
</blockquote>

<p>It's also worth saying that the  telescopes were not measuring visible light, but rather very short wave radio energy (or, if you prefer, very long wave infrared light). Furthermore the image was not made as it is in an optical telescope by bringing all the light to a focus on a detector, but rather by a more complex process which involves recording the signals at multiple telescopes and then combining them later with a supercomputer.  However, the pixels of the image do correspond to the intensity of the radio energy coming from the (slightly) different directions to the different parts of the black hole and its close surroundings, so in a sense the image is (apart from the colour, which is just to loop pretty) what you would see if you had ridiculously sharp eyesight and could see radio waves.</p>

CC BY-SA 4.0
2020-04-14T18:13:45.017Z
2020-04-14T18:18:50.870Z
2020-04-14T18:18:50.870Z
Steve Linton
<p>If you were really close, you could perhaps see <a href="https://en.wikipedia.org/wiki/Hawking_radiation" rel="nofollow noreferrer">Hawking Radiation</a>. Two caveats:</p>

<ul>
<li>It is predicted to exist, but was never experimentally observed.</li>
<li>The bigger the black hole, the "colder" it appears, so you would have to find a very small black hole in order to see it emitting visible light.</li>
</ul>

CC BY-SA 4.0
2020-04-16T07:27:48.527Z
A black hole needs to be tiny to emit enough Hawking radiation in the visible light spectrum to be actually visible to a human. See https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator There *might* be primordial black holes that small, but black holes formed in supernovae are far too large (and much colder than the CMB, so they absorb far more radiation than they emit).
PM 2Ring
CC BY-SA 4.0
2020-04-16T11:38:03.223Z
That's what I mean by "very small", sorry for being imprecise. I just wanted to add to the resounding "no" from the other answers; that under some *very* peculiar conditions OP could have they wanted. Not for Messier 87* or any other black hole of stellar mass or bigger, though.
mjm
CC BY-SA 4.0
2020-04-16T06:10:14.267Z
2020-04-16T06:10:14.267Z
mjm
<p>Can one see a black hole on earth?</p>

<p>If a black hole does not emit light, how can one take a picture of the black hole itself?</p>

<p>There's some discussion that the image composed by the Event Horizons Telescope is really just an accretion disk. To "take an image" of something, you need the light reflecting off of the surface of the object within frame. The "light" in the image is really just color-mapped ...
Can one see a black hole on earth?</p>

CC BY-SA 4.0
2020-04-14T19:52:36.873Z
See https://astronomy.stackexchange.com/questions/30317/m87-black-hole-why-can-we-see-the-blackness?noredirect=1&lq=1
ProfRob
CC BY-SA 4.0
2020-04-15T16:28:59.647Z
I guess you can't see mirrors, the text on this page (if you have the light theme), or ravens either...
Jason Goemaat
CC BY-SA 4.0
2020-04-16T12:57:15.250Z
The light coming from the sun is clearly not reflected, so the statement "you need the light reflecting off of the surface" is false. But one big issue with your question is: What do you mean by "black hole"?? Strictly spoken, it is the point-shaped singularity which indeed we cannot "see"  in any meaningful sense of the word. If you think of the space inside the event horizon: Hawking radiation *does* make that sphere (or whatever, for rotating holes) "visible", however faintly in whatever low frequencies.
Peter - Reinstate Monica
CC BY-SA 4.0
2020-04-16T21:54:18.777Z
Can you see the wind? Can you see a magnetic field? Can you see the gravitational field of the earth or the sun? They’re all variations of the same philosophical questions.
Fogmeister
CC BY-SA 4.0
2020-04-16T22:38:34.273Z
@Fogmeister : Wind?  Yes, via [Schlieren photography](https://en.wikipedia.org/wiki/Schlieren_photography).
Eric Towers
<black-hole><corona>
Can one see a black hole?
Ebai Emmanuel Ewangi
# Can one see a black hole?

(Asked by: username_0 on 2020-04-14)

Can one see a black hole on earth?

If a black hole does not emit light, how can one take a picture of the black hole itself?

There's some discussion that the image composed by the Event Horizons Telescope is really just an accretion disk. To "take an image" of something, you need the light reflecting off of the surface of the object within frame. The "light" in the image is really just color-mapped ...
Can one see a black hole on earth?

**Tags:** black-hole, corona

> **Comments:**
> **username_1**: See https://astronomy.stackexchange.com/questions/30317/m87-black-hole-why-can-we-see-the-blackness?noredirect=1&lq=1  
> **username_2**: I guess you can't see mirrors, the text on this page (if you have the light theme), or ravens either...  
> **username_3**: The light coming from the sun is clearly not reflected, so the statement "you need the light reflecting off of the surface" is false. But one big issue with your question is: What do you mean by "black hole"?? Strictly spoken, it is the point-shaped singularity which indeed we cannot "see"  in any meaningful sense of the word. If you think of the space inside the event horizon: Hawking radiation *does* make that sphere (or whatever, for rotating holes) "visible", however faintly in whatever low frequencies.  
> **username_4**: Can you see the wind? Can you see a magnetic field? Can you see the gravitational field of the earth or the sun? They’re all variations of the same philosophical questions.  
> **username_5**: @username_4 : Wind?  Yes, via [Schlieren photography](https://en.wikipedia.org/wiki/Schlieren_photography).  

---

## 5 answers

### Answer by username_6 on 2020-04-14. Score: 22

I think, that's a bit semantics, and a question of what do you call 'I see something': If you see the shadow cast by an object on a bright background - do you see the object? Do you see Mercury or Venus when it passes in front of the disk of our sun? Similarly, if you look an object through an atomic force microscope - do you see the object, or do you just see some force indicator giving a value?

So yes, in this general sense, you see something when you have an instrument to detect it which allows to deduce properties of the object. So you also can see an object when you just see its shape in front of a bright background like a black hole within its accretion disk - even when all light you see comes from the latter.


---

### Answer by username_1 on 2020-04-14. Score: 12

No, you cannot "see" a black hole, only the way that it interacts with objects or light in its vicinity.

In terms of interacting with other objects, the classic observations that betray the presence of a black hole are the very rapid motions of stars and gas close to an object with a large mass, but which cannot be directly seen. Examples include low- and high-mass X-ray binary systems like [A0620-00](https://en.wikipedia.org/wiki/A0620-00) and [Cygnus X-1](https://en.wikipedia.org/wiki/Cygnus_X-1) respectively. Supermassive black holes have been detected by measuring the rapid motion of gas at the centres of many galaxies, and the 4 million solar-mass example at the centre of our Galaxy is revealed by watching the motion of stars that [orbit it](https://www.eso.org/public/videos/eso1825e/).

In terms of interacting with light, the strong gravity of black holes will distort objects seen in the background of a black hole, but this effect is yet to be seen in detail. Instead, using a clever interferometric imaging technique, the central region of the galaxy M87 has been observed at microwave wavelengths. The famous image you have doubtless seen, is a projection of the [photon sphere](https://en.wikipedia.org/wiki/Photon_sphere) of the supermassive black hole, which is illuminated by hot gas surrounding it. Much more detail on this result can be found in several other Astronomy SE questions - notably [M87 Black hole. Why can we see the blackness?](https://astronomy.stackexchange.com/questions/30317/m87-black-hole-why-can-we-see-the-blackness?noredirect=1&lq=1) Note that the photon ring is *not* the accretion disk.

> **Comments:**
> **username_1**: "No, you cannot "see" a black hole, only the way that it interacts with objects or light in its vicinity.". @user253751 no mention of distance.  
> **username_1**: @user253751 what about my answer contradicts that? You appear not to have read it  

---

### Answer by username_7 on 2020-04-15. Score: 6

**Yes,** we can, indeed, "see" a black hole and we just did not too long ago.

You see, being able to "see" an object is not solely about "reflection" alone: rather, you "see" things by them *interacting* in *any* way with the light that reaches your eye (or other "seeing" instrument), and then by your eye interacting with that object-interacted radiation. Reflection is just *one* way. Absorption (the *removal* of light rays that would otherwise reach your eye) is another, and you don't need a black hole for that: any object that is *not* totally transparent, shiny, or white, under white light, must have absorbed some radiation for it to look that way.

If you took a big ball and painted it black and hung it in front of a light background, would you say you were "seeing" that black ball? If so, that is *not* very much different. If you agree you can "see" that, then you can "see" a black hole, just same. It does the same thing: it *absorbs* the light from any luminous background, distinguishing itself because those rays that were sent from the points on the background you'd see were it not there, *fail* to reach your eye. Likewise, if you want to argue you "cannot 'see'" the black hole, then you cannot "see" that black ball either, by the same reasoning.

Now, of course, the *mechanism* of the absorption is different in each case, but the result is the same in terms of what happens to light rays. Rays that would have been otherwise able to reach your eyes (or here, cameras) were the object not there, now don't.

You also mention that the picture is "colorized". This is true, but it's not because black holes "can't be seen", but rather because that the telescope is "seeing" in a different *wavelength range* than your eyes do and that's because those wavelengths have advantages for detecting them at this level of resolution and at this distance compared to visible light, *not* because that we could not see the black hole with our own eyes were we there (and suitably protected from the intense energy). Colors are *always* arbitrary, being assigned by our brains. Light itself doesn't have "colors". Since our eyes, and so our brains, aren't set up with (kind of tautologically) receptors and hence color channels for wavelengths we can't see, we've got to reuse the ones we've got somehow. This doesn't make it any less a "real photograph" though, any more than taking a greyscale photograph of a scene with an ordinary visible-light camera does (and if anything, I'd have put it in grey, but I suspect the red is because they want to make it look "fiery", so it's a little artistic license.).


---

### Answer by username_8 on 2020-04-14. Score: 5

[This article](https://skyandtelescope.org/astronomy-news/the-appearance-of-a-black-holes-shadow/) explains what the image shows.

> This distinct structure is a result of the warped spacetime around massive objects like black holes. The ring of light is comprised of photons from the hot, radiating gas that surrounds the black hole, whose paths have been bent around the black hole before arriving at our telescopes. The dark region in the center is termed the black hole’s “shadow”; this is the collection of paths of photons that did not escape, but were instead captured by the black hole.

It's also worth saying that the telescopes were not measuring visible light, but rather very short wave radio energy (or, if you prefer, very long wave infrared light). Furthermore the image was not made as it is in an optical telescope by bringing all the light to a focus on a detector, but rather by a more complex process which involves recording the signals at multiple telescopes and then combining them later with a supercomputer. However, the pixels of the image do correspond to the intensity of the radio energy coming from the (slightly) different directions to the different parts of the black hole and its close surroundings, so in a sense the image is (apart from the colour, which is just to loop pretty) what you would see if you had ridiculously sharp eyesight and could see radio waves.


---

### Answer by username_9 on 2020-04-16. Score: 0

If you were really close, you could perhaps see [Hawking Radiation](https://en.wikipedia.org/wiki/Hawking_radiation). Two caveats:

* It is predicted to exist, but was never experimentally observed.
* The bigger the black hole, the "colder" it appears, so you would have to find a very small black hole in order to see it emitting visible light.

> **Comments:**
> **username_10**: A black hole needs to be tiny to emit enough Hawking radiation in the visible light spectrum to be actually visible to a human. See https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator There *might* be primordial black holes that small, but black holes formed in supernovae are far too large (and much colder than the CMB, so they absorb far more radiation than they emit).  
> **username_9**: That's what I mean by "very small", sorry for being imprecise. I just wanted to add to the resounding "no" from the other answers; that under some *very* peculiar conditions OP could have they wanted. Not for Messier 87* or any other black hole of stellar mass or bigger, though.  

---

