<p>The picture is of the central region of M87, taken at a wavelength where the gas is "optically thin".</p>

<p>The ring of bright light is pretty much exactly where it is expected to be for the synchrotron radiation emitted by the hot gas to have been gravitationally lensed by a black hole with the same mass as deduced previously from the motion of stars close to the centre of M87.</p>

<p>It <em>is</em> a "picture of a black hole", though all you see is the way it warps the light around it.</p>

<p>Your analogies are poor. It isn't like taking a picture of someone in a dark room. The room here is illuminated and the effects of the black hole on that light have apparently been seen quite clearly. That, by the way is the only method by which you can see someone in an illuminated room, unless you have eyes that detect the mid-infrared light they emit. Of course that option is not available for a black hole, since they emit no light at all.</p>

<p>Your other analogy is also curious. A cursory glance at the sky reveals it is full of air molecules. That is why it is blue. Your view though lacks the angular resolution to see individual air molecules, so you would not be able to see them.</p>

<p>Here, the resolution of the "telescope" is just good enough to reveal the essential structure and size of the photon ring.</p>

<p>Finally, why isn't it a picture of an accretion disk? As I said, it is expected that the disk is optically thin and actually much bigger than the imaged area. Emission from the disk would not be expected to show anything but a haze with a vague central concentration. Given the spatial resolution of the image, what has actually been seen is a sharp, bright ring.</p>

<p>The accretion disk around a black hole should be terminated at or slightly beyond the innermost stable circular orbit. This is at 3 times the Schwarzschild radius, but gravitational lensing would make it <em>appear</em> to have an inner radius of 3.7 times the Schwarzschild radius. The bright ring has a radius of 2.6 Schwarzschild radii; correct for a lensed photon sphere explanation, incorrect for the inner wall of the accretion disk (unless the black hole is 70% of the mass previously determined from stellar motions).</p>

<p>The orientation might be just about ok for an accretion disk explanation. If the rotation axis is aligned with the jet of M87, the disk should be about 17 degrees from face on. However, in order to see the inner parts having such a circular shape and well defined structure would require a geometrically thin and optically thick disk, the opposite of what is deduced from the physical conditions in the plasma.</p>

<p>To see more context, here is a wider view of M87 taken by Chandra, and from the <a href="http://chandra.si.edu/blog/node/719" rel="nofollow noreferrer">Chandra team's blog post</a>:</p>

<blockquote>
  <p>Chandra was used to observe M87 and other targets during the EHT campaign. While Chandra can’t see the shadow itself, its field of view is much larger than the EHT’s, so Chandra can view the full length of the jet of high-energy particles launched by the intense gravitational and magnetic fields around the black hole. This jet extends more than 1,000 light years from the center of the galaxy.</p>
</blockquote>

<p><a href="https://i.sstatic.net/mgGVB.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/mgGVB.jpg" alt="enter image description here"></a></p>

CC BY-SA 4.0
2019-04-13T08:47:59.493Z
Nice and fill my curiosity about the "would be" accretion disk there.
Alchimista
CC BY-SA 4.0
2019-04-14T02:47:38.653Z
“The room here is illuminated and the effects of the black hole on that light have apparently been seen quite clearly.” Except it’s not. The authors of the image state that there really is no orange or red light; these were just assigned color values to represent the radio frequencies capture from the source. So the black hole being “illuminated” seems very false.
8protons
CC BY-SA 4.0
2019-04-14T08:03:47.387Z
@8protons this reduce to the channel use to "see". Analogue to observe something in medical x ray, with an infrared camera, or taking a beautiful picture in electron microscopy. Point is the black hole would have a striking resemblance to the radio picture when seen in the visible. At least one very recent Q here or Physics SE refers to that.
Alchimista
CC BY-SA 4.0
2019-04-14T08:55:09.780Z
@8protons Radio waves (actually these are microwaves) are light. Illumination refers to light. If you are complaining that we haven't seen the black hole because it isn't observable at visible wavelengths then that isn't stated in your question (and would attract downvotes).
ProfRob
CC BY-SA 4.0
2019-04-14T13:48:38.690Z
Anyone know what the scale of the Chandra images is?
Steve Linton
CC BY-SA 4.0
2019-04-13T08:04:17.287Z
2019-04-13T09:18:39.080Z
2019-04-13T09:18:39.080Z
ProfRob
<p>Consider an extremely transparent lens. If you photograph the lens, what the camera really picks up where the lens covers is the distorted image of what is behind the lens. Would you say that is still a photograph of the lens? I would say yes.</p>
<p>If you take a picture of an object coated with Vantablack, the amount of light entering the camera from that object is negligible. With rounding, all the pixels covering the object might actually be zero. Is it still a picture of that object? I would say yes.</p>
<p>As I understand it, that black spot in the middle is bigger than the event horizon, but that doesn't mean it's not what the black hole looks like. Seeing anything is really seeing how that thing has influenced what light is coming to you from its direction. Most things reflect light, but diffuse reflection is a very different thing from mirror reflection. Some things allow light to pass through but refract it in a way you can tell. Some things soak it up so you don't see any light from that direction. As long as it can create a contrast from the light around it, we say that we are seeing that thing.</p>

CC BY-SA 4.0
2021-02-18T20:15:51.540Z
I like this way of looking at it (pun intended)
uhoh
CC BY-SA 4.0
2021-02-18T15:31:03.603Z
2021-02-18T15:31:03.603Z
Mark Foskey
<p>There's some discussion that the image composed by the Event Horizons Telescope is <em>really</em> 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 gases within the accretion disc. Since black holes do not allow light to escape, is this actually an image of a black hole?</p>

<p>The analogy I've heard is that it would be like taking a picture of an object in a completely dark room, showing someone the photograph taken, and saying, "Hey look at this picture I took of this object." The other person could ask, "Where is the object in this entirely black photo?" to which the other person might respond, "It's somewhere in the picture." </p>

<p>Was a picture of the object really taken if you can't even see it? If you photograph the sky, have you "captured an image of air molecules?"</p>

<p><a href="https://i.sstatic.net/inpbk.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/inpbk.png" alt="enter image description here"></a></p>

CC BY-SA 4.0
2019-04-13T01:08:30.533Z
"How did the victim in the horror movie know the murderer was in front of her, when all she could see was his blackened silhouette?"
Ingolifs
CC BY-SA 4.0
2019-04-13T01:09:38.387Z
@Ingolifs Knowing the murderer is there and taking a picture of the murderer you can’t see are not the same things.
8protons
CC BY-SA 4.0
2019-04-13T01:12:46.777Z
"What murderer?"
Ingolifs
<black-hole><light><photography><accretion-discs><event-horizon-telescope>
If a black hole does not emit light, how can one take a picture of the black hole itself?
8protons
# If a black hole does not emit light, how can one take a picture of the black hole itself?

(Asked by: username_0 on 2019-04-13)

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 gases within the accretion disc. Since black holes do not allow light to escape, is this actually an image of a black hole?

The analogy I've heard is that it would be like taking a picture of an object in a completely dark room, showing someone the photograph taken, and saying, "Hey look at this picture I took of this object." The other person could ask, "Where is the object in this entirely black photo?" to which the other person might respond, "It's somewhere in the picture."

Was a picture of the object really taken if you can't even see it? If you photograph the sky, have you "captured an image of air molecules?"

[![enter image description here](https://i.sstatic.net/inpbk.png)](https://i.sstatic.net/inpbk.png)

**Tags:** black-hole, light, photography, accretion-discs, event-horizon-telescope

> **Comments:**
> **username_1**: "How did the victim in the horror movie know the murderer was in front of her, when all she could see was his blackened silhouette?"  
> **username_0**: @username_1 Knowing the murderer is there and taking a picture of the murderer you can’t see are not the same things.  
> **username_1**: "What murderer?"  

---

## 2 answers

### Answer by username_2 on 2019-04-13. Score: 6

The picture is of the central region of M87, taken at a wavelength where the gas is "optically thin".

The ring of bright light is pretty much exactly where it is expected to be for the synchrotron radiation emitted by the hot gas to have been gravitationally lensed by a black hole with the same mass as deduced previously from the motion of stars close to the centre of M87.

It *is* a "picture of a black hole", though all you see is the way it warps the light around it.

Your analogies are poor. It isn't like taking a picture of someone in a dark room. The room here is illuminated and the effects of the black hole on that light have apparently been seen quite clearly. That, by the way is the only method by which you can see someone in an illuminated room, unless you have eyes that detect the mid-infrared light they emit. Of course that option is not available for a black hole, since they emit no light at all.

Your other analogy is also curious. A cursory glance at the sky reveals it is full of air molecules. That is why it is blue. Your view though lacks the angular resolution to see individual air molecules, so you would not be able to see them.

Here, the resolution of the "telescope" is just good enough to reveal the essential structure and size of the photon ring.

Finally, why isn't it a picture of an accretion disk? As I said, it is expected that the disk is optically thin and actually much bigger than the imaged area. Emission from the disk would not be expected to show anything but a haze with a vague central concentration. Given the spatial resolution of the image, what has actually been seen is a sharp, bright ring.

The accretion disk around a black hole should be terminated at or slightly beyond the innermost stable circular orbit. This is at 3 times the Schwarzschild radius, but gravitational lensing would make it *appear* to have an inner radius of 3.7 times the Schwarzschild radius. The bright ring has a radius of 2.6 Schwarzschild radii; correct for a lensed photon sphere explanation, incorrect for the inner wall of the accretion disk (unless the black hole is 70% of the mass previously determined from stellar motions).

The orientation might be just about ok for an accretion disk explanation. If the rotation axis is aligned with the jet of M87, the disk should be about 17 degrees from face on. However, in order to see the inner parts having such a circular shape and well defined structure would require a geometrically thin and optically thick disk, the opposite of what is deduced from the physical conditions in the plasma.

To see more context, here is a wider view of M87 taken by Chandra, and from the [Chandra team's blog post](http://chandra.si.edu/blog/node/719):

> Chandra was used to observe M87 and other targets during the EHT campaign. While Chandra can’t see the shadow itself, its field of view is much larger than the EHT’s, so Chandra can view the full length of the jet of high-energy particles launched by the intense gravitational and magnetic fields around the black hole. This jet extends more than 1,000 light years from the center of the galaxy.

[![enter image description here](https://i.sstatic.net/mgGVB.jpg)](https://i.sstatic.net/mgGVB.jpg)

> **Comments:**
> **username_3**: Nice and fill my curiosity about the "would be" accretion disk there.  
> **username_0**: “The room here is illuminated and the effects of the black hole on that light have apparently been seen quite clearly.” Except it’s not. The authors of the image state that there really is no orange or red light; these were just assigned color values to represent the radio frequencies capture from the source. So the black hole being “illuminated” seems very false.  
> **username_3**: @username_0 this reduce to the channel use to "see". Analogue to observe something in medical x ray, with an infrared camera, or taking a beautiful picture in electron microscopy. Point is the black hole would have a striking resemblance to the radio picture when seen in the visible. At least one very recent Q here or Physics SE refers to that.  
> **username_2**: @username_0 Radio waves (actually these are microwaves) are light. Illumination refers to light. If you are complaining that we haven't seen the black hole because it isn't observable at visible wavelengths then that isn't stated in your question (and would attract downvotes).  
> **username_4**: Anyone know what the scale of the Chandra images is?  

---

### Answer by username_5 on 2021-02-18. Score: 3

Consider an extremely transparent lens. If you photograph the lens, what the camera really picks up where the lens covers is the distorted image of what is behind the lens. Would you say that is still a photograph of the lens? I would say yes.

If you take a picture of an object coated with Vantablack, the amount of light entering the camera from that object is negligible. With rounding, all the pixels covering the object might actually be zero. Is it still a picture of that object? I would say yes.

As I understand it, that black spot in the middle is bigger than the event horizon, but that doesn't mean it's not what the black hole looks like. Seeing anything is really seeing how that thing has influenced what light is coming to you from its direction. Most things reflect light, but diffuse reflection is a very different thing from mirror reflection. Some things allow light to pass through but refract it in a way you can tell. Some things soak it up so you don't see any light from that direction. As long as it can create a contrast from the light around it, we say that we are seeing that thing.

> **Comments:**
> **username_6**: I like this way of looking at it (pun intended)  

---

