Where is the intervening light in the M87 black hole?
$begingroup$
The M87 black hole is at the centre, more or less, of an elliptical galaxy of 1 trillion or more stars. So certainly there are stars and other light emitting objects (emission nebula e.g.) in the intervening 53 million light years between earth and the black hole.
How is it that the centre of the black hole image is perfectly dark, as if it is being observed directly, with not one point of light between the black hole and earth? I appreciate that space is mostly empty, but do we really have an unobscured line of sight to a black hole in the centre of an enormous elliptical galaxy?
Does it have something to do with the type of light being observed?
black-hole m87
edited 9 hours ago
Peter Mortensen
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add a comment |
$begingroup$
The M87 black hole is at the centre, more or less, of an elliptical galaxy of 1 trillion or more stars. So certainly there are stars and other light emitting objects (emission nebula e.g.) in the intervening 53 million light years between earth and the black hole.
How is it that the centre of the black hole image is perfectly dark, as if it is being observed directly, with not one point of light between the black hole and earth? I appreciate that space is mostly empty, but do we really have an unobscured line of sight to a black hole in the centre of an enormous elliptical galaxy?
Does it have something to do with the type of light being observed?
black-hole m87
edited 9 hours ago
Peter Mortensen
1556
asked 19 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
1715
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$endgroup$
$begingroup$
I think it's mostly that the field of view is incredibly tiny. The popular published image seems to be about four times the diameter of the central dark region, so representing about 800 billion km at the distance of M87 or about 0.1 light year. At the distance of the edge of our galaxy, it would represent about 1/1000 of that. There simply isn't anything in that very very narrow cone bright enough to show up.
$endgroup$
– Steve Linton
18 hours ago
1
$begingroup$
Also, the observations are made in radio, where intervening gas and dust hardly absorbs anything.
$endgroup$
– pela
18 hours ago
$begingroup$
It isn't perfectly dark.
$endgroup$
– Rob Jeffries
8 hours ago
add a comment |
$begingroup$
The M87 black hole is at the centre, more or less, of an elliptical galaxy of 1 trillion or more stars. So certainly there are stars and other light emitting objects (emission nebula e.g.) in the intervening 53 million light years between earth and the black hole.
How is it that the centre of the black hole image is perfectly dark, as if it is being observed directly, with not one point of light between the black hole and earth? I appreciate that space is mostly empty, but do we really have an unobscured line of sight to a black hole in the centre of an enormous elliptical galaxy?
Does it have something to do with the type of light being observed?
black-hole m87
edited 9 hours ago
Peter Mortensen
1556
asked 19 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
1715
New contributor
Bumptious Q Bangwhistle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The M87 black hole is at the centre, more or less, of an elliptical galaxy of 1 trillion or more stars. So certainly there are stars and other light emitting objects (emission nebula e.g.) in the intervening 53 million light years between earth and the black hole.
How is it that the centre of the black hole image is perfectly dark, as if it is being observed directly, with not one point of light between the black hole and earth? I appreciate that space is mostly empty, but do we really have an unobscured line of sight to a black hole in the centre of an enormous elliptical galaxy?
Does it have something to do with the type of light being observed?
black-hole m87
black-hole m87
edited 9 hours ago
Peter Mortensen
1556
asked 19 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
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edited 9 hours ago
Peter Mortensen
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asked 19 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
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edited 9 hours ago
Peter Mortensen
1556
edited 9 hours ago
Peter Mortensen
1556
edited 9 hours ago
Peter Mortensen
1556
1556
asked 19 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
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asked 19 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
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asked 19 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
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1715
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$begingroup$
I think it's mostly that the field of view is incredibly tiny. The popular published image seems to be about four times the diameter of the central dark region, so representing about 800 billion km at the distance of M87 or about 0.1 light year. At the distance of the edge of our galaxy, it would represent about 1/1000 of that. There simply isn't anything in that very very narrow cone bright enough to show up.
$endgroup$
– Steve Linton
18 hours ago
1
$begingroup$
Also, the observations are made in radio, where intervening gas and dust hardly absorbs anything.
$endgroup$
– pela
18 hours ago
$begingroup$
It isn't perfectly dark.
$endgroup$
– Rob Jeffries
8 hours ago
add a comment |
$begingroup$
I think it's mostly that the field of view is incredibly tiny. The popular published image seems to be about four times the diameter of the central dark region, so representing about 800 billion km at the distance of M87 or about 0.1 light year. At the distance of the edge of our galaxy, it would represent about 1/1000 of that. There simply isn't anything in that very very narrow cone bright enough to show up.
$endgroup$
– Steve Linton
18 hours ago
1
$begingroup$
Also, the observations are made in radio, where intervening gas and dust hardly absorbs anything.
$endgroup$
– pela
18 hours ago
$begingroup$
It isn't perfectly dark.
$endgroup$
– Rob Jeffries
8 hours ago
$begingroup$
I think it's mostly that the field of view is incredibly tiny. The popular published image seems to be about four times the diameter of the central dark region, so representing about 800 billion km at the distance of M87 or about 0.1 light year. At the distance of the edge of our galaxy, it would represent about 1/1000 of that. There simply isn't anything in that very very narrow cone bright enough to show up.
$endgroup$
– Steve Linton
18 hours ago
$begingroup$
I think it's mostly that the field of view is incredibly tiny. The popular published image seems to be about four times the diameter of the central dark region, so representing about 800 billion km at the distance of M87 or about 0.1 light year. At the distance of the edge of our galaxy, it would represent about 1/1000 of that. There simply isn't anything in that very very narrow cone bright enough to show up.
$endgroup$
– Steve Linton
18 hours ago
1
1
$begingroup$
Also, the observations are made in radio, where intervening gas and dust hardly absorbs anything.
$endgroup$
– pela
18 hours ago
$begingroup$
Also, the observations are made in radio, where intervening gas and dust hardly absorbs anything.
$endgroup$
– pela
18 hours ago
$begingroup$
It isn't perfectly dark.
$endgroup$
– Rob Jeffries
8 hours ago
$begingroup$
It isn't perfectly dark.
$endgroup$
– Rob Jeffries
8 hours ago
add a comment |
1 Answer
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$begingroup$
Based on the comment from @SteveLinton
The radius of M87 is given to be 60,000 light years, which is a volume of 900 trillion cubic light years. Given that it contains roughly one trillion stars, there is an average of 900 cubic light years for each star.
Several resources referred to the diameter of the black hole in the range of 1.5 light days (38 billion km). That's 0.004 light years.
If you consider the cylinder1 extending from the black hole to the edge of M87, the cylinder is 0.004 light years in diameter with a height of 60,000 light years. The volume of that cylinder is just over 3 cubic light years. Since there is one star for every 900 cubic light years, it follows that the cylinder encompassing our line of sight between the black hole and the edge of M87 is empty of any stars.
Whether or not any star would even be bright enough to be visible is another matter.
1Technically a conic frustum but I don't think the difference in volume is significant.
answered 18 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
1715
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$endgroup$
$begingroup$
"The radius of M87 is given to be 60,000 light years.". No, more like 32 - 50 kiloparsecs radius = 100,000 - 160,000 light years radius. E.g. "According to Stoyan et al. (2010), the distance of M87 is 54.9 million light years and its diameter is 132,000 light years." and "Diameter: 120,000 ly".
$endgroup$
– Peter Mortensen
10 hours ago
$begingroup$
Doesn't this answer assume that the galaxy is of uniform density?
$endgroup$
– JBentley
10 hours ago
1
$begingroup$
@PeterMortensen I'm not sure I understand. Your sources cite a diameter of 120,000 LY and 132,000 LY. I'm using a radius of 60,000 LY. Radius = 1/2 Diameter.
$endgroup$
– Bumptious Q Bangwhistle
8 hours ago
$begingroup$
@JBentley Yes and I appreciate this is a fallacy (Olbers' Paradox).
$endgroup$
– Bumptious Q Bangwhistle
8 hours ago
add a comment |
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$begingroup$
Based on the comment from @SteveLinton
The radius of M87 is given to be 60,000 light years, which is a volume of 900 trillion cubic light years. Given that it contains roughly one trillion stars, there is an average of 900 cubic light years for each star.
Several resources referred to the diameter of the black hole in the range of 1.5 light days (38 billion km). That's 0.004 light years.
If you consider the cylinder1 extending from the black hole to the edge of M87, the cylinder is 0.004 light years in diameter with a height of 60,000 light years. The volume of that cylinder is just over 3 cubic light years. Since there is one star for every 900 cubic light years, it follows that the cylinder encompassing our line of sight between the black hole and the edge of M87 is empty of any stars.
Whether or not any star would even be bright enough to be visible is another matter.
1Technically a conic frustum but I don't think the difference in volume is significant.
answered 18 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
1715
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Bumptious Q Bangwhistle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
$begingroup$
"The radius of M87 is given to be 60,000 light years.". No, more like 32 - 50 kiloparsecs radius = 100,000 - 160,000 light years radius. E.g. "According to Stoyan et al. (2010), the distance of M87 is 54.9 million light years and its diameter is 132,000 light years." and "Diameter: 120,000 ly".
$endgroup$
– Peter Mortensen
10 hours ago
$begingroup$
Doesn't this answer assume that the galaxy is of uniform density?
$endgroup$
– JBentley
10 hours ago
1
$begingroup$
@PeterMortensen I'm not sure I understand. Your sources cite a diameter of 120,000 LY and 132,000 LY. I'm using a radius of 60,000 LY. Radius = 1/2 Diameter.
$endgroup$
– Bumptious Q Bangwhistle
8 hours ago
$begingroup$
@JBentley Yes and I appreciate this is a fallacy (Olbers' Paradox).
$endgroup$
– Bumptious Q Bangwhistle
8 hours ago
add a comment |
$begingroup$
Based on the comment from @SteveLinton
The radius of M87 is given to be 60,000 light years, which is a volume of 900 trillion cubic light years. Given that it contains roughly one trillion stars, there is an average of 900 cubic light years for each star.
Several resources referred to the diameter of the black hole in the range of 1.5 light days (38 billion km). That's 0.004 light years.
If you consider the cylinder1 extending from the black hole to the edge of M87, the cylinder is 0.004 light years in diameter with a height of 60,000 light years. The volume of that cylinder is just over 3 cubic light years. Since there is one star for every 900 cubic light years, it follows that the cylinder encompassing our line of sight between the black hole and the edge of M87 is empty of any stars.
Whether or not any star would even be bright enough to be visible is another matter.
1Technically a conic frustum but I don't think the difference in volume is significant.
answered 18 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
1715
New contributor
Bumptious Q Bangwhistle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
"The radius of M87 is given to be 60,000 light years.". No, more like 32 - 50 kiloparsecs radius = 100,000 - 160,000 light years radius. E.g. "According to Stoyan et al. (2010), the distance of M87 is 54.9 million light years and its diameter is 132,000 light years." and "Diameter: 120,000 ly".
$endgroup$
– Peter Mortensen
10 hours ago
$begingroup$
Doesn't this answer assume that the galaxy is of uniform density?
$endgroup$
– JBentley
10 hours ago
1
$begingroup$
@PeterMortensen I'm not sure I understand. Your sources cite a diameter of 120,000 LY and 132,000 LY. I'm using a radius of 60,000 LY. Radius = 1/2 Diameter.
$endgroup$
– Bumptious Q Bangwhistle
8 hours ago
$begingroup$
@JBentley Yes and I appreciate this is a fallacy (Olbers' Paradox).
$endgroup$
– Bumptious Q Bangwhistle
8 hours ago
add a comment |
$begingroup$
Based on the comment from @SteveLinton
The radius of M87 is given to be 60,000 light years, which is a volume of 900 trillion cubic light years. Given that it contains roughly one trillion stars, there is an average of 900 cubic light years for each star.
Several resources referred to the diameter of the black hole in the range of 1.5 light days (38 billion km). That's 0.004 light years.
If you consider the cylinder1 extending from the black hole to the edge of M87, the cylinder is 0.004 light years in diameter with a height of 60,000 light years. The volume of that cylinder is just over 3 cubic light years. Since there is one star for every 900 cubic light years, it follows that the cylinder encompassing our line of sight between the black hole and the edge of M87 is empty of any stars.
Whether or not any star would even be bright enough to be visible is another matter.
1Technically a conic frustum but I don't think the difference in volume is significant.
answered 18 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
1715
New contributor
Bumptious Q Bangwhistle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Based on the comment from @SteveLinton
The radius of M87 is given to be 60,000 light years, which is a volume of 900 trillion cubic light years. Given that it contains roughly one trillion stars, there is an average of 900 cubic light years for each star.
Several resources referred to the diameter of the black hole in the range of 1.5 light days (38 billion km). That's 0.004 light years.
If you consider the cylinder1 extending from the black hole to the edge of M87, the cylinder is 0.004 light years in diameter with a height of 60,000 light years. The volume of that cylinder is just over 3 cubic light years. Since there is one star for every 900 cubic light years, it follows that the cylinder encompassing our line of sight between the black hole and the edge of M87 is empty of any stars.
Whether or not any star would even be bright enough to be visible is another matter.
1Technically a conic frustum but I don't think the difference in volume is significant.
answered 18 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
1715
New contributor
Bumptious Q Bangwhistle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 17 hours ago
answered 18 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
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answered 18 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
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answered 18 hours ago
Bumptious Q BangwhistleBumptious Q Bangwhistle
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1715
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New contributor
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$begingroup$
"The radius of M87 is given to be 60,000 light years.". No, more like 32 - 50 kiloparsecs radius = 100,000 - 160,000 light years radius. E.g. "According to Stoyan et al. (2010), the distance of M87 is 54.9 million light years and its diameter is 132,000 light years." and "Diameter: 120,000 ly".
$endgroup$
– Peter Mortensen
10 hours ago
$begingroup$
Doesn't this answer assume that the galaxy is of uniform density?
$endgroup$
– JBentley
10 hours ago
1
$begingroup$
@PeterMortensen I'm not sure I understand. Your sources cite a diameter of 120,000 LY and 132,000 LY. I'm using a radius of 60,000 LY. Radius = 1/2 Diameter.
$endgroup$
– Bumptious Q Bangwhistle
8 hours ago
$begingroup$
@JBentley Yes and I appreciate this is a fallacy (Olbers' Paradox).
$endgroup$
– Bumptious Q Bangwhistle
8 hours ago
add a comment |
$begingroup$
"The radius of M87 is given to be 60,000 light years.". No, more like 32 - 50 kiloparsecs radius = 100,000 - 160,000 light years radius. E.g. "According to Stoyan et al. (2010), the distance of M87 is 54.9 million light years and its diameter is 132,000 light years." and "Diameter: 120,000 ly".
$endgroup$
– Peter Mortensen
10 hours ago
$begingroup$
Doesn't this answer assume that the galaxy is of uniform density?
$endgroup$
– JBentley
10 hours ago
1
$begingroup$
@PeterMortensen I'm not sure I understand. Your sources cite a diameter of 120,000 LY and 132,000 LY. I'm using a radius of 60,000 LY. Radius = 1/2 Diameter.
$endgroup$
– Bumptious Q Bangwhistle
8 hours ago
$begingroup$
@JBentley Yes and I appreciate this is a fallacy (Olbers' Paradox).
$endgroup$
– Bumptious Q Bangwhistle
8 hours ago
$begingroup$
"The radius of M87 is given to be 60,000 light years.". No, more like 32 - 50 kiloparsecs radius = 100,000 - 160,000 light years radius. E.g. "According to Stoyan et al. (2010), the distance of M87 is 54.9 million light years and its diameter is 132,000 light years." and "Diameter: 120,000 ly".
$endgroup$
– Peter Mortensen
10 hours ago
$begingroup$
"The radius of M87 is given to be 60,000 light years.". No, more like 32 - 50 kiloparsecs radius = 100,000 - 160,000 light years radius. E.g. "According to Stoyan et al. (2010), the distance of M87 is 54.9 million light years and its diameter is 132,000 light years." and "Diameter: 120,000 ly".
$endgroup$
– Peter Mortensen
10 hours ago
$begingroup$
Doesn't this answer assume that the galaxy is of uniform density?
$endgroup$
– JBentley
10 hours ago
$begingroup$
Doesn't this answer assume that the galaxy is of uniform density?
$endgroup$
– JBentley
10 hours ago
1
1
$begingroup$
@PeterMortensen I'm not sure I understand. Your sources cite a diameter of 120,000 LY and 132,000 LY. I'm using a radius of 60,000 LY. Radius = 1/2 Diameter.
$endgroup$
– Bumptious Q Bangwhistle
8 hours ago
$begingroup$
@PeterMortensen I'm not sure I understand. Your sources cite a diameter of 120,000 LY and 132,000 LY. I'm using a radius of 60,000 LY. Radius = 1/2 Diameter.
$endgroup$
– Bumptious Q Bangwhistle
8 hours ago
$begingroup$
@JBentley Yes and I appreciate this is a fallacy (Olbers' Paradox).
$endgroup$
– Bumptious Q Bangwhistle
8 hours ago
$begingroup$
@JBentley Yes and I appreciate this is a fallacy (Olbers' Paradox).
$endgroup$
– Bumptious Q Bangwhistle
8 hours ago
add a comment |
Bumptious Q Bangwhistle is a new contributor. Be nice, and check out our Code of Conduct.
Bumptious Q Bangwhistle is a new contributor. Be nice, and check out our Code of Conduct.
Bumptious Q Bangwhistle is a new contributor. Be nice, and check out our Code of Conduct.
Bumptious Q Bangwhistle is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I think it's mostly that the field of view is incredibly tiny. The popular published image seems to be about four times the diameter of the central dark region, so representing about 800 billion km at the distance of M87 or about 0.1 light year. At the distance of the edge of our galaxy, it would represent about 1/1000 of that. There simply isn't anything in that very very narrow cone bright enough to show up.
$endgroup$
– Steve Linton
18 hours ago
1
$begingroup$
Also, the observations are made in radio, where intervening gas and dust hardly absorbs anything.
$endgroup$
– pela
18 hours ago
$begingroup$
It isn't perfectly dark.
$endgroup$
– Rob Jeffries
8 hours ago