<p>The black part in the centre of the image genuinely represents some directions from which less energy is arriving at the telescopes. I believe the intensity in the middle of it is about 10 times lower than the intensity in the bright ring around it.  So in that sense it really is a dark spot.</p>

<p>It is described in the papers as the "shadow" of the black hole, although even that is stretching the word shadow a little. Basically the light that "would have" been coming to us from that direction has been bent away by gravity.</p>

CC BY-SA 4.0
2019-04-11T21:43:32.400Z
The question was why don't we see the accretion flow in front of the black hole?
ProfRob
CC BY-SA 4.0
2019-04-12T12:38:24.173Z
Steve, thanks for much for this, WOULD THE IMAGE LOOK THE SAME from all directions?  (Or by an amazing coincidence are we normal to the plane of a donut-shape?)
Fattie
CC BY-SA 4.0
2019-04-12T12:44:02.083Z
@Fattie All that would change is the orientation of the brightness asymmetry.
ProfRob
CC BY-SA 4.0
2019-04-12T12:59:39.957Z
hi @RobJeffries - gotchya; TBC what I had in mind was the major feature - the "donut hole" in the middle.  Just TBC, you're saying that: from all directions, it would look like what we see: "a donut" shaped pseudoimage.  **Less/no radio emission in the middle** and a ring of radio emission from around the sides.  Again - no matter what angle viewed from you mysteriously see none coming from the "middle".  Is that about right?
Fattie
CC BY-SA 4.0
2019-04-12T13:00:04.173Z
BTW I was going to and will ask a QA about this!
Fattie
CC BY-SA 4.0
2019-04-11T19:48:55.520Z
2019-04-11T19:48:55.520Z
Steve Linton
<p>The black hole image that you saw is not a photograph in the traditional sense. A traditional photograph is created when visible light strikes a digital sensor or film in a camera.</p>

<p>The image of M87 that you saw was created by an elaborate, complex, globe-spanning and labor-intensive operation.</p>

<p>To start with, the telescopes used were radio telescopes. These radio telescopes "see" radio waves with a 1.3mm wavelength like we see visible light, which has a wavelength from violet (380 nanometers) to red (700 nanometers). Think of it like infrared vision or night vision.</p>

<p>Eight radio telescopes all over the globe were pointed toward the same black hole at the same time over the course of four days. Each radio telescope collected an enormous amount of data by observing the black hole in the <strong>radio</strong> spectrum, not the visible light spectrum. According to the <a href="https://eventhorizontelescope.org/" rel="nofollow noreferrer">Event Horizons Telescope website</a>, each of the eight radio telescopes produced about 350 terabytes of data per day. This radio spectrum observation data was timestamped using an extremely accurate atomic clock at each location.</p>

<p>The data was then shipped to supercomputing centers to be processed by an algorithm. The computer algorithms work by lining up the data from the different telescopes using the accurate timestamps and by aligning the observed data spatially. This process is called <a href="https://en.wikipedia.org/wiki/Very-long-baseline_interferometry" rel="nofollow noreferrer">very-long-baseline interferometry (VLBI)</a>. The basic idea is that if you take two distinct observations of the same object from different locations and synchronize them. This helps to eliminate atmospheric noise by seeing what both different images have in common. While it may seem like a highly artificial process, it is very effective at eliminating noise.</p>

<p>The image you see is therefore the result of a computer algorithm processing radio observations of the black hole from 8 different locations on earth. The choice of color was likely a choice made by the computer scientists who wrote the algorithm, but it appears to bear some relation to the intensity of the radio energies observed: more energy shows as brighter, while darker spots with less energy are deeper red or black. This choice of color might make sense because the strong gravity of a black hole will tend to <a href="https://en.wikipedia.org/wiki/Redshift" rel="nofollow noreferrer">redshift</a> any light emanating from nearby or passing close to the black hole. On the other hand, black holes are said to be enormously energetic in a broad spectrum of electromagnetic energy. First-hand accounts of nuclear explosions, which unleash X Rays and Gamma Rays and other far-ultraviolet energy have been described as having <a href="https://www.newyorker.com/tech/annals-of-technology/the-first-light-of-the-trinity-atomic-test" rel="nofollow noreferrer">all kinds of exotic color</a>. </p>

<p>While this may seem a bit disappointing, it's worth noting that visible light does not propagate through space as well as radio. Radio penetrates dust and gas and all the intervening detritus in space much better. Visible light tends to be blocked and re-absorbed. Think of Wifi signals (radio) versus visible light. The Wifi penetrates walls, while the visible light gets blocked by walls, curtains, etc. The EHT website has some helpful <a href="https://eventhorizontelescope.org/infographics" rel="nofollow noreferrer">infographics on VLBI</a>.</p>

CC BY-SA 4.0
2019-04-11T17:16:37.737Z
So, if I understood correctly, we can see a black disk because the matter in front was not "visible" in the wavelength used by the radio telescopes. If so, why?
Cristian M
CC BY-SA 4.0
2019-04-11T17:31:56.640Z
The ring-like appearance of the image is so exciting because it matches theoretical predictions so closely. These theoretical predictions are complex, but say that the visible appearance of the black hole is about 2.5 times the size of the black hole's event horizon. I won't pretend to know all the specifics, but suspect the black hole's gravity traps a lot of light, and at the sides we see more because we're looking at a deeper/thicker cross section of the black hole's surrounding mass. Kinda similar idea to the sun look red at sunrise/sunset because it's traveling through more atmosphere.
S. Imp
CC BY-SA 4.0
2019-04-11T17:35:23.850Z
Also keep in mind that the image has a certain dynamic range. If you've ever adjusted brightness/contrast on an image you might get what I mean. Or like when you take a picture with your phone. The phone adjusts the exposure to the subject of the photo. If your subject is bright, it adjusts to let less light in. If your subject is dark, it adjusts to let more light in. My guess is that the computer algorithm was configured to adapt its dynamic range to the immediate vicinity of the black hole, making the background black and the brightest spot white, so that other colors fall in between.
S. Imp
CC BY-SA 4.0
2019-04-11T17:42:14.867Z
Actually, that sunset/sunrise example is probably not helpful. Think instead of how in pictures from the Int'l Space Station, you can only see the cloudy atmosphere when you look at the edge of the earth. If you look straight down, you're looking thru just a thin layer of atmosphere. If you look toward the edge of earth, you're looking through more atmosphere along the edge of the earth.
S. Imp
CC BY-SA 4.0
2019-04-11T21:39:55.437Z
There is no answer here to the question asked.
ProfRob
CC BY-SA 4.0
2019-04-11T21:53:40.343Z
@RobJeffries I respectfully disagree. The question posted refers specifically to the EHT image posted yesterday. My explanation is long, but does include the short answer "The choice of color was likely a choice made by the computer scientists who wrote the algorithm".
S. Imp
CC BY-SA 4.0
2019-04-12T07:44:07.063Z
Would be the same with a photograph? Eventually hot matter in front of the disk might reduce the contrast but the same applies.
Alchimista
CC BY-SA 4.0
2019-04-12T09:39:21.217Z
@S.Imp OK but that's lost in all the text. I suggest you add a TLDR along the lines of "It's a false-colour image made from radio observations" or at least boldface the sentence that contains the actual answer.
David Richerby
CC BY-SA 4.0
2019-04-12T13:04:00.417Z
@S.Imp dude the question is *"why does the middle part have no emission and there's a ring of emission around the outside"*  Of course, obviously, the "white" is just a chosen representational color in the pseudoimage.
Fattie
CC BY-SA 4.0
2019-04-11T16:56:31.027Z
2019-04-11T17:08:41.393Z
2019-04-11T17:08:41.393Z
S. Imp
<p>What's going on here is that you have been misled into thinking the ring-like structure has anything to do with the accretion disk. It doesn't, or at least only indirectly.</p>

<p>The disk is referred to as geometrically thick, but optically thin (see sections 1 and 2 of <a href="https://iopscience.iop.org/article/10.3847/2041-8213/ab0f43" rel="noreferrer">paper V</a> issued by the Event Horizon Telescope collaboration on 10 April 2019). This is actually the <em>opposite</em> of the disk visualised in the film "Interstellar".</p>

<p>Because the disk is geometrically thick, it covers the whole picture. Because it is optically thin, we can see through it to the black hole. That is the basic answer to your question.</p>

<p>In an optically thin plasma, the brightness you will see is proportional to the optical path length (physical length multiplied by an absorption coefficient) along that sightline. The photon ring marks radiation travelling towards us that has been bent around the black hole, or has even orbited it several times. Hence those sight lines have larger optical path lengths and that is why we see it as a bright ring.</p>

<p>Radiation travelling towards us from plasma in front of the black hole, has a small optical path length and is not very bright. In addition, the sight lines to plasma behind the black hole cannot travel through the black hole or even close to the event horizon. Hence the circular "shadow" inside the photon ring.</p>

CC BY-SA 4.0
2019-04-12T07:52:01.130Z
It is similar to what I asked you before. I realise that the looking of the accretion disk might vary upon the wl used to observe it. Shouldn't looking in Vis reveal more matter in the surrounding? A bar as in interstellar must be light directed to us and not synchrotron rad. The lstter should inevitably gives a glory ring....
Alchimista
CC BY-SA 4.0
2019-04-11T22:28:38.367Z
2019-04-11T22:40:32.633Z
2019-04-11T22:40:32.633Z
ProfRob
<p>I recently saw the first image of a black hole. As I understood, it is covered with bright, hot matter. In this case, how can we see the black disk (event horizon), instead of a bright disk due to the matter surrounding the black hole? Or is the matter around the black hole disposed in a disk shape and the astronomers got lucky that the disk was not oriented towards Earth?</p>

CC BY-SA 4.0
2019-04-11T23:59:38.197Z
See Veritasium’s explanation at https://www.youtube.com/watch?v=zUyH3XhpLTo. As for why it’s an accretion _disk_ and not an “accretion sphere”, see the video by PBS Space Time explaining why gravity turns some objects into spheres and others into disks at https://www.youtube.com/watch?v=Aj6Kc1mvsdo.
Roman Odaisky
CC BY-SA 4.0
2019-04-12T13:00:55.747Z
Something is drastically, drastically wrong.  If it's an accretion **disk**, it's just not feasible that, magically, Earth is directly, perfectly, normal to the plane of the disk.  Not a chance.
Fattie
CC BY-SA 4.0
2019-04-12T13:50:50.163Z
@Fattie you are not reading the answers to this and at least a couple of other questions on the topic. The disk is geometrically thick and optically thin. You cannot see it in the picture. The ring is caused by gravitational lensing.
ProfRob
CC BY-SA 4.0
2019-04-12T13:53:07.463Z
thanks @RobJeffries I'll research the existing QA !
Fattie
<black-hole><matter><disk>
Why isn't the black hole white?
Cristian M
# Why isn't the black hole white?

(Asked by: username_0 on 2019-04-11)

I recently saw the first image of a black hole. As I understood, it is covered with bright, hot matter. In this case, how can we see the black disk (event horizon), instead of a bright disk due to the matter surrounding the black hole? Or is the matter around the black hole disposed in a disk shape and the astronomers got lucky that the disk was not oriented towards Earth?

**Tags:** black-hole, matter, disk

> **Comments:**
> **username_1**: See Veritasium’s explanation at https://www.youtube.com/watch?v=zUyH3XhpLTo. As for why it’s an accretion _disk_ and not an “accretion sphere”, see the video by PBS Space Time explaining why gravity turns some objects into spheres and others into disks at https://www.youtube.com/watch?v=Aj6Kc1mvsdo.  
> **username_2**: Something is drastically, drastically wrong.  If it's an accretion **disk**, it's just not feasible that, magically, Earth is directly, perfectly, normal to the plane of the disk.  Not a chance.  
> **username_3**: @username_2 you are not reading the answers to this and at least a couple of other questions on the topic. The disk is geometrically thick and optically thin. You cannot see it in the picture. The ring is caused by gravitational lensing.  
> **username_2**: thanks @RobJeffries I'll research the existing QA !  

---

## 3 answers

### Answer by username_3 on 2019-04-11. Score: 6 | Accepted answer

What's going on here is that you have been misled into thinking the ring-like structure has anything to do with the accretion disk. It doesn't, or at least only indirectly.

The disk is referred to as geometrically thick, but optically thin (see sections 1 and 2 of [paper V](https://iopscience.iop.org/article/10.3847/2041-8213/ab0f43) issued by the Event Horizon Telescope collaboration on 10 April 2019). This is actually the *opposite* of the disk visualised in the film "Interstellar".

Because the disk is geometrically thick, it covers the whole picture. Because it is optically thin, we can see through it to the black hole. That is the basic answer to your question.

In an optically thin plasma, the brightness you will see is proportional to the optical path length (physical length multiplied by an absorption coefficient) along that sightline. The photon ring marks radiation travelling towards us that has been bent around the black hole, or has even orbited it several times. Hence those sight lines have larger optical path lengths and that is why we see it as a bright ring.

Radiation travelling towards us from plasma in front of the black hole, has a small optical path length and is not very bright. In addition, the sight lines to plasma behind the black hole cannot travel through the black hole or even close to the event horizon. Hence the circular "shadow" inside the photon ring.

> **Comments:**
> **username_6**: It is similar to what I asked you before. I realise that the looking of the accretion disk might vary upon the wl used to observe it. Shouldn't looking in Vis reveal more matter in the surrounding? A bar as in interstellar must be light directed to us and not synchrotron rad. The lstter should inevitably gives a glory ring....  

---

### Answer by username_4 on 2019-04-11. Score: 3

The black part in the centre of the image genuinely represents some directions from which less energy is arriving at the telescopes. I believe the intensity in the middle of it is about 10 times lower than the intensity in the bright ring around it. So in that sense it really is a dark spot.

It is described in the papers as the "shadow" of the black hole, although even that is stretching the word shadow a little. Basically the light that "would have" been coming to us from that direction has been bent away by gravity.

> **Comments:**
> **username_3**: The question was why don't we see the accretion flow in front of the black hole?  
> **username_2**: username_4, thanks for much for this, WOULD THE IMAGE LOOK THE SAME from all directions?  (Or by an amazing coincidence are we normal to the plane of a donut-shape?)  
> **username_3**: @username_2 All that would change is the orientation of the brightness asymmetry.  
> **username_2**: hi @RobJeffries - gotchya; TBC what I had in mind was the major feature - the "donut hole" in the middle.  Just TBC, you're saying that: from all directions, it would look like what we see: "a donut" shaped pseudoimage.  **Less/no radio emission in the middle** and a ring of radio emission from around the sides.  Again - no matter what angle viewed from you mysteriously see none coming from the "middle".  Is that about right?  
> **username_2**: BTW I was going to and will ask a QA about this!  

---

### Answer by username_5 on 2019-04-11. Score: 2

The black hole image that you saw is not a photograph in the traditional sense. A traditional photograph is created when visible light strikes a digital sensor or film in a camera.

The image of M87 that you saw was created by an elaborate, complex, globe-spanning and labor-intensive operation.

To start with, the telescopes used were radio telescopes. These radio telescopes "see" radio waves with a 1.3mm wavelength like we see visible light, which has a wavelength from violet (380 nanometers) to red (700 nanometers). Think of it like infrared vision or night vision.

Eight radio telescopes all over the globe were pointed toward the same black hole at the same time over the course of four days. Each radio telescope collected an enormous amount of data by observing the black hole in the **radio** spectrum, not the visible light spectrum. According to the [Event Horizons Telescope website](https://eventhorizontelescope.org/), each of the eight radio telescopes produced about 350 terabytes of data per day. This radio spectrum observation data was timestamped using an extremely accurate atomic clock at each location.

The data was then shipped to supercomputing centers to be processed by an algorithm. The computer algorithms work by lining up the data from the different telescopes using the accurate timestamps and by aligning the observed data spatially. This process is called [very-long-baseline interferometry (VLBI)](https://en.wikipedia.org/wiki/Very-long-baseline_interferometry). The basic idea is that if you take two distinct observations of the same object from different locations and synchronize them. This helps to eliminate atmospheric noise by seeing what both different images have in common. While it may seem like a highly artificial process, it is very effective at eliminating noise.

The image you see is therefore the result of a computer algorithm processing radio observations of the black hole from 8 different locations on earth. The choice of color was likely a choice made by the computer scientists who wrote the algorithm, but it appears to bear some relation to the intensity of the radio energies observed: more energy shows as brighter, while darker spots with less energy are deeper red or black. This choice of color might make sense because the strong gravity of a black hole will tend to [redshift](https://en.wikipedia.org/wiki/Redshift) any light emanating from nearby or passing close to the black hole. On the other hand, black holes are said to be enormously energetic in a broad spectrum of electromagnetic energy. First-hand accounts of nuclear explosions, which unleash X Rays and Gamma Rays and other far-ultraviolet energy have been described as having [all kinds of exotic color](https://www.newyorker.com/tech/annals-of-technology/the-first-light-of-the-trinity-atomic-test).

While this may seem a bit disappointing, it's worth noting that visible light does not propagate through space as well as radio. Radio penetrates dust and gas and all the intervening detritus in space much better. Visible light tends to be blocked and re-absorbed. Think of Wifi signals (radio) versus visible light. The Wifi penetrates walls, while the visible light gets blocked by walls, curtains, etc. The EHT website has some helpful [infographics on VLBI](https://eventhorizontelescope.org/infographics).

> **Comments:**
> **username_0**: So, if I understood correctly, we can see a black disk because the matter in front was not "visible" in the wavelength used by the radio telescopes. If so, why?  
> **username_5**: The ring-like appearance of the image is so exciting because it matches theoretical predictions so closely. These theoretical predictions are complex, but say that the visible appearance of the black hole is about 2.5 times the size of the black hole's event horizon. I won't pretend to know all the specifics, but suspect the black hole's gravity traps a lot of light, and at the sides we see more because we're looking at a deeper/thicker cross section of the black hole's surrounding mass. Kinda similar idea to the sun look red at sunrise/sunset because it's traveling through more atmosphere.  
> **username_5**: Also keep in mind that the image has a certain dynamic range. If you've ever adjusted brightness/contrast on an image you might get what I mean. Or like when you take a picture with your phone. The phone adjusts the exposure to the subject of the photo. If your subject is bright, it adjusts to let less light in. If your subject is dark, it adjusts to let more light in. My guess is that the computer algorithm was configured to adapt its dynamic range to the immediate vicinity of the black hole, making the background black and the brightest spot white, so that other colors fall in between.  
> **username_5**: Actually, that sunset/sunrise example is probably not helpful. Think instead of how in pictures from the Int'l Space Station, you can only see the cloudy atmosphere when you look at the edge of the earth. If you look straight down, you're looking thru just a thin layer of atmosphere. If you look toward the edge of earth, you're looking through more atmosphere along the edge of the earth.  
> **username_3**: There is no answer here to the question asked.  
> **username_5**: @RobJeffries I respectfully disagree. The question posted refers specifically to the EHT image posted yesterday. My explanation is long, but does include the short answer "The choice of color was likely a choice made by the computer scientists who wrote the algorithm".  
> **username_6**: Would be the same with a photograph? Eventually hot matter in front of the disk might reduce the contrast but the same applies.  
> **username_7**: @username_5 OK but that's lost in all the text. I suggest you add a TLDR along the lines of "It's a false-colour image made from radio observations" or at least boldface the sentence that contains the actual answer.  
> **username_2**: @username_5 dude the question is *"why does the middle part have no emission and there's a ring of emission around the outside"*  Of course, obviously, the "white" is just a chosen representational color in the pseudoimage.  

---

