Why isn't the black hole white? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) What stellar content do we want to share with Twitter?Does matter accumulate just outside the event horizon of a black hole?Ramifications of black hole stellar systemHow does an accreting black hole acquire magnetic fields?Could we verify the structure of a black hole by observing an orbiting object?Black hole without singularity?Black Hole growthShouldn't we not be able to see some black holes?M87 Black hole. Why can we see the blackness?What part of the EM spectrum was used in the black hole image?Why don't we see the gas behind the black hole?

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Why isn't the black hole white?



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
What stellar content do we want to share with Twitter?Does matter accumulate just outside the event horizon of a black hole?Ramifications of black hole stellar systemHow does an accreting black hole acquire magnetic fields?Could we verify the structure of a black hole by observing an orbiting object?Black hole without singularity?Black Hole growthShouldn't we not be able to see some black holes?M87 Black hole. Why can we see the blackness?What part of the EM spectrum was used in the black hole image?Why don't we see the gas behind the black hole?










5












$begingroup$


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?










share|improve this question







New contributor




Cristian M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 2




    $begingroup$
    See Veritasium’s explanation at 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 youtube.com/watch?v=Aj6Kc1mvsdo.
    $endgroup$
    – Roman Odaisky
    2 days ago










  • $begingroup$
    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.
    $endgroup$
    – Fattie
    yesterday










  • $begingroup$
    @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.
    $endgroup$
    – Rob Jeffries
    yesterday










  • $begingroup$
    thanks @RobJeffries I'll research the existing QA !
    $endgroup$
    – Fattie
    yesterday















5












$begingroup$


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?










share|improve this question







New contributor




Cristian M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 2




    $begingroup$
    See Veritasium’s explanation at 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 youtube.com/watch?v=Aj6Kc1mvsdo.
    $endgroup$
    – Roman Odaisky
    2 days ago










  • $begingroup$
    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.
    $endgroup$
    – Fattie
    yesterday










  • $begingroup$
    @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.
    $endgroup$
    – Rob Jeffries
    yesterday










  • $begingroup$
    thanks @RobJeffries I'll research the existing QA !
    $endgroup$
    – Fattie
    yesterday













5












5








5





$begingroup$


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?










share|improve this question







New contributor




Cristian M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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?







black-hole matter disk






share|improve this question







New contributor




Cristian M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




Cristian M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






New contributor




Cristian M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









Cristian MCristian M

282




282




New contributor




Cristian M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Cristian M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Cristian M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 2




    $begingroup$
    See Veritasium’s explanation at 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 youtube.com/watch?v=Aj6Kc1mvsdo.
    $endgroup$
    – Roman Odaisky
    2 days ago










  • $begingroup$
    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.
    $endgroup$
    – Fattie
    yesterday










  • $begingroup$
    @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.
    $endgroup$
    – Rob Jeffries
    yesterday










  • $begingroup$
    thanks @RobJeffries I'll research the existing QA !
    $endgroup$
    – Fattie
    yesterday












  • 2




    $begingroup$
    See Veritasium’s explanation at 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 youtube.com/watch?v=Aj6Kc1mvsdo.
    $endgroup$
    – Roman Odaisky
    2 days ago










  • $begingroup$
    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.
    $endgroup$
    – Fattie
    yesterday










  • $begingroup$
    @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.
    $endgroup$
    – Rob Jeffries
    yesterday










  • $begingroup$
    thanks @RobJeffries I'll research the existing QA !
    $endgroup$
    – Fattie
    yesterday







2




2




$begingroup$
See Veritasium’s explanation at 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 youtube.com/watch?v=Aj6Kc1mvsdo.
$endgroup$
– Roman Odaisky
2 days ago




$begingroup$
See Veritasium’s explanation at 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 youtube.com/watch?v=Aj6Kc1mvsdo.
$endgroup$
– Roman Odaisky
2 days ago












$begingroup$
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.
$endgroup$
– Fattie
yesterday




$begingroup$
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.
$endgroup$
– Fattie
yesterday












$begingroup$
@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.
$endgroup$
– Rob Jeffries
yesterday




$begingroup$
@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.
$endgroup$
– Rob Jeffries
yesterday












$begingroup$
thanks @RobJeffries I'll research the existing QA !
$endgroup$
– Fattie
yesterday




$begingroup$
thanks @RobJeffries I'll research the existing QA !
$endgroup$
– Fattie
yesterday










3 Answers
3






active

oldest

votes


















5












$begingroup$

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 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.






share|improve this answer











$endgroup$












  • $begingroup$
    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....
    $endgroup$
    – Alchimista
    yesterday


















2












$begingroup$

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, 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). 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 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.



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.






share|improve this answer











$endgroup$












  • $begingroup$
    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?
    $endgroup$
    – Cristian M
    2 days ago










  • $begingroup$
    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.
    $endgroup$
    – S. Imp
    2 days ago






  • 1




    $begingroup$
    There is no answer here to the question asked.
    $endgroup$
    – Rob Jeffries
    2 days ago







  • 1




    $begingroup$
    @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.
    $endgroup$
    – David Richerby
    yesterday






  • 1




    $begingroup$
    @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.
    $endgroup$
    – Fattie
    yesterday


















2












$begingroup$

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.






share|improve this answer









$endgroup$












  • $begingroup$
    The question was why don't we see the accretion flow in front of the black hole?
    $endgroup$
    – Rob Jeffries
    2 days ago











  • $begingroup$
    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?)
    $endgroup$
    – Fattie
    yesterday










  • $begingroup$
    @Fattie All that would change is the orientation of the brightness asymmetry.
    $endgroup$
    – Rob Jeffries
    yesterday










  • $begingroup$
    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?
    $endgroup$
    – Fattie
    yesterday











  • $begingroup$
    BTW I was going to and will ask a QA about this!
    $endgroup$
    – Fattie
    yesterday











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

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 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.






share|improve this answer











$endgroup$












  • $begingroup$
    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....
    $endgroup$
    – Alchimista
    yesterday















5












$begingroup$

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 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.






share|improve this answer











$endgroup$












  • $begingroup$
    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....
    $endgroup$
    – Alchimista
    yesterday













5












5








5





$begingroup$

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 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.






share|improve this answer











$endgroup$



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 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.







share|improve this answer














share|improve this answer



share|improve this answer








edited 2 days ago

























answered 2 days ago









Rob JeffriesRob Jeffries

54.6k4113175




54.6k4113175











  • $begingroup$
    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....
    $endgroup$
    – Alchimista
    yesterday
















  • $begingroup$
    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....
    $endgroup$
    – Alchimista
    yesterday















$begingroup$
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....
$endgroup$
– Alchimista
yesterday




$begingroup$
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....
$endgroup$
– Alchimista
yesterday











2












$begingroup$

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, 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). 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 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.



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.






share|improve this answer











$endgroup$












  • $begingroup$
    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?
    $endgroup$
    – Cristian M
    2 days ago










  • $begingroup$
    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.
    $endgroup$
    – S. Imp
    2 days ago






  • 1




    $begingroup$
    There is no answer here to the question asked.
    $endgroup$
    – Rob Jeffries
    2 days ago







  • 1




    $begingroup$
    @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.
    $endgroup$
    – David Richerby
    yesterday






  • 1




    $begingroup$
    @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.
    $endgroup$
    – Fattie
    yesterday















2












$begingroup$

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, 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). 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 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.



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.






share|improve this answer











$endgroup$












  • $begingroup$
    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?
    $endgroup$
    – Cristian M
    2 days ago










  • $begingroup$
    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.
    $endgroup$
    – S. Imp
    2 days ago






  • 1




    $begingroup$
    There is no answer here to the question asked.
    $endgroup$
    – Rob Jeffries
    2 days ago







  • 1




    $begingroup$
    @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.
    $endgroup$
    – David Richerby
    yesterday






  • 1




    $begingroup$
    @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.
    $endgroup$
    – Fattie
    yesterday













2












2








2





$begingroup$

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, 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). 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 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.



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.






share|improve this answer











$endgroup$



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, 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). 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 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.



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.







share|improve this answer














share|improve this answer



share|improve this answer








edited 2 days ago

























answered 2 days ago









S. ImpS. Imp

20618




20618











  • $begingroup$
    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?
    $endgroup$
    – Cristian M
    2 days ago










  • $begingroup$
    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.
    $endgroup$
    – S. Imp
    2 days ago






  • 1




    $begingroup$
    There is no answer here to the question asked.
    $endgroup$
    – Rob Jeffries
    2 days ago







  • 1




    $begingroup$
    @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.
    $endgroup$
    – David Richerby
    yesterday






  • 1




    $begingroup$
    @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.
    $endgroup$
    – Fattie
    yesterday
















  • $begingroup$
    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?
    $endgroup$
    – Cristian M
    2 days ago










  • $begingroup$
    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.
    $endgroup$
    – S. Imp
    2 days ago






  • 1




    $begingroup$
    There is no answer here to the question asked.
    $endgroup$
    – Rob Jeffries
    2 days ago







  • 1




    $begingroup$
    @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.
    $endgroup$
    – David Richerby
    yesterday






  • 1




    $begingroup$
    @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.
    $endgroup$
    – Fattie
    yesterday















$begingroup$
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?
$endgroup$
– Cristian M
2 days ago




$begingroup$
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?
$endgroup$
– Cristian M
2 days ago












$begingroup$
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.
$endgroup$
– S. Imp
2 days ago




$begingroup$
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.
$endgroup$
– S. Imp
2 days ago




1




1




$begingroup$
There is no answer here to the question asked.
$endgroup$
– Rob Jeffries
2 days ago





$begingroup$
There is no answer here to the question asked.
$endgroup$
– Rob Jeffries
2 days ago





1




1




$begingroup$
@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.
$endgroup$
– David Richerby
yesterday




$begingroup$
@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.
$endgroup$
– David Richerby
yesterday




1




1




$begingroup$
@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.
$endgroup$
– Fattie
yesterday




$begingroup$
@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.
$endgroup$
– Fattie
yesterday











2












$begingroup$

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.






share|improve this answer









$endgroup$












  • $begingroup$
    The question was why don't we see the accretion flow in front of the black hole?
    $endgroup$
    – Rob Jeffries
    2 days ago











  • $begingroup$
    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?)
    $endgroup$
    – Fattie
    yesterday










  • $begingroup$
    @Fattie All that would change is the orientation of the brightness asymmetry.
    $endgroup$
    – Rob Jeffries
    yesterday










  • $begingroup$
    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?
    $endgroup$
    – Fattie
    yesterday











  • $begingroup$
    BTW I was going to and will ask a QA about this!
    $endgroup$
    – Fattie
    yesterday















2












$begingroup$

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.






share|improve this answer









$endgroup$












  • $begingroup$
    The question was why don't we see the accretion flow in front of the black hole?
    $endgroup$
    – Rob Jeffries
    2 days ago











  • $begingroup$
    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?)
    $endgroup$
    – Fattie
    yesterday










  • $begingroup$
    @Fattie All that would change is the orientation of the brightness asymmetry.
    $endgroup$
    – Rob Jeffries
    yesterday










  • $begingroup$
    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?
    $endgroup$
    – Fattie
    yesterday











  • $begingroup$
    BTW I was going to and will ask a QA about this!
    $endgroup$
    – Fattie
    yesterday













2












2








2





$begingroup$

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.






share|improve this answer









$endgroup$



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.







share|improve this answer












share|improve this answer



share|improve this answer










answered 2 days ago









Steve LintonSteve Linton

2,6831320




2,6831320











  • $begingroup$
    The question was why don't we see the accretion flow in front of the black hole?
    $endgroup$
    – Rob Jeffries
    2 days ago











  • $begingroup$
    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?)
    $endgroup$
    – Fattie
    yesterday










  • $begingroup$
    @Fattie All that would change is the orientation of the brightness asymmetry.
    $endgroup$
    – Rob Jeffries
    yesterday










  • $begingroup$
    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?
    $endgroup$
    – Fattie
    yesterday











  • $begingroup$
    BTW I was going to and will ask a QA about this!
    $endgroup$
    – Fattie
    yesterday
















  • $begingroup$
    The question was why don't we see the accretion flow in front of the black hole?
    $endgroup$
    – Rob Jeffries
    2 days ago











  • $begingroup$
    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?)
    $endgroup$
    – Fattie
    yesterday










  • $begingroup$
    @Fattie All that would change is the orientation of the brightness asymmetry.
    $endgroup$
    – Rob Jeffries
    yesterday










  • $begingroup$
    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?
    $endgroup$
    – Fattie
    yesterday











  • $begingroup$
    BTW I was going to and will ask a QA about this!
    $endgroup$
    – Fattie
    yesterday















$begingroup$
The question was why don't we see the accretion flow in front of the black hole?
$endgroup$
– Rob Jeffries
2 days ago





$begingroup$
The question was why don't we see the accretion flow in front of the black hole?
$endgroup$
– Rob Jeffries
2 days ago













$begingroup$
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?)
$endgroup$
– Fattie
yesterday




$begingroup$
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?)
$endgroup$
– Fattie
yesterday












$begingroup$
@Fattie All that would change is the orientation of the brightness asymmetry.
$endgroup$
– Rob Jeffries
yesterday




$begingroup$
@Fattie All that would change is the orientation of the brightness asymmetry.
$endgroup$
– Rob Jeffries
yesterday












$begingroup$
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?
$endgroup$
– Fattie
yesterday





$begingroup$
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?
$endgroup$
– Fattie
yesterday













$begingroup$
BTW I was going to and will ask a QA about this!
$endgroup$
– Fattie
yesterday




$begingroup$
BTW I was going to and will ask a QA about this!
$endgroup$
– Fattie
yesterday










Cristian M is a new contributor. Be nice, and check out our Code of Conduct.









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Cristian M is a new contributor. Be nice, and check out our Code of Conduct.












Cristian M is a new contributor. Be nice, and check out our Code of Conduct.











Cristian M is a new contributor. Be nice, and check out our Code of Conduct.














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