Why don't merging black holes disprove the no-hair theorem?
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The no-hair theorem of black holes says they're completely categorised by their charge and angular momentum and mass.
But imagine two black holes colliding. At some point their event horizons would merge and I imagine the combined event horizon would not be spherical.
You could even imagine 50 black holes merging. Then the combined event horizon would be a very odd shape.
Why does this not disprove the no-hair theorem? Since the information about the shape of the event horizon is surely more than just charge, angular momentum and mass?
general-relativity black-holes collision event-horizon no-hair-theorem
edited 25 mins ago
David Z♦
63.4k23136252
asked 4 hours ago
zoobyzooby
1,389514
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add a comment |
$begingroup$
The no-hair theorem of black holes says they're completely categorised by their charge and angular momentum and mass.
But imagine two black holes colliding. At some point their event horizons would merge and I imagine the combined event horizon would not be spherical.
You could even imagine 50 black holes merging. Then the combined event horizon would be a very odd shape.
Why does this not disprove the no-hair theorem? Since the information about the shape of the event horizon is surely more than just charge, angular momentum and mass?
general-relativity black-holes collision event-horizon no-hair-theorem
edited 25 mins ago
David Z♦
63.4k23136252
asked 4 hours ago
zoobyzooby
1,389514
$endgroup$
add a comment |
$begingroup$
The no-hair theorem of black holes says they're completely categorised by their charge and angular momentum and mass.
But imagine two black holes colliding. At some point their event horizons would merge and I imagine the combined event horizon would not be spherical.
You could even imagine 50 black holes merging. Then the combined event horizon would be a very odd shape.
Why does this not disprove the no-hair theorem? Since the information about the shape of the event horizon is surely more than just charge, angular momentum and mass?
general-relativity black-holes collision event-horizon no-hair-theorem
edited 25 mins ago
David Z♦
63.4k23136252
asked 4 hours ago
zoobyzooby
1,389514
$endgroup$
The no-hair theorem of black holes says they're completely categorised by their charge and angular momentum and mass.
But imagine two black holes colliding. At some point their event horizons would merge and I imagine the combined event horizon would not be spherical.
You could even imagine 50 black holes merging. Then the combined event horizon would be a very odd shape.
Why does this not disprove the no-hair theorem? Since the information about the shape of the event horizon is surely more than just charge, angular momentum and mass?
general-relativity black-holes collision event-horizon no-hair-theorem
general-relativity black-holes collision event-horizon no-hair-theorem
edited 25 mins ago
David Z♦
63.4k23136252
asked 4 hours ago
zoobyzooby
1,389514
edited 25 mins ago
David Z♦
63.4k23136252
asked 4 hours ago
zoobyzooby
1,389514
edited 25 mins ago
David Z♦
63.4k23136252
edited 25 mins ago
David Z♦
63.4k23136252
edited 25 mins ago
David Z♦
63.4k23136252
63.4k23136252
asked 4 hours ago
zoobyzooby
1,389514
asked 4 hours ago
zoobyzooby
1,389514
asked 4 hours ago
zoobyzooby
1,389514
1,389514
add a comment |
add a comment |
1 Answer
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No. The no-hair conjecture applies to stable solutions of the Einstein-Maxwell equations. In the case of merging black holes, it applies to the end state of the merger into a single quiescent black hole, after the “ringdown” has stopped.
answered 4 hours ago
G. SmithG. Smith
6,97611123
$endgroup$
$begingroup$
It's an odd theorem because most black holes aren't stable, their either evaporating or each particles from the background space.
$endgroup$
– zooby
4 hours ago
$begingroup$
those processes are slow enough that a black hole that is evaporating or accreting gas (not planets or other big objects) can be considered stable.
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– niels nielsen
4 hours ago
4
$begingroup$
Black hole evaporation is a quantum effect. The no-hair conjecture is a conjecture in classical General Relativity since we don’t have any consensus about quantum gravity. Quantum black holes probably do have some kind of hair that encodes all of the information that fell into them. Physicists are trying to figure out how that might work.
$endgroup$
– G. Smith
4 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. The no-hair conjecture applies to stable solutions of the Einstein-Maxwell equations. In the case of merging black holes, it applies to the end state of the merger into a single quiescent black hole, after the “ringdown” has stopped.
answered 4 hours ago
G. SmithG. Smith
6,97611123
$endgroup$
$begingroup$
It's an odd theorem because most black holes aren't stable, their either evaporating or each particles from the background space.
$endgroup$
– zooby
4 hours ago
$begingroup$
those processes are slow enough that a black hole that is evaporating or accreting gas (not planets or other big objects) can be considered stable.
$endgroup$
– niels nielsen
4 hours ago
4
$begingroup$
Black hole evaporation is a quantum effect. The no-hair conjecture is a conjecture in classical General Relativity since we don’t have any consensus about quantum gravity. Quantum black holes probably do have some kind of hair that encodes all of the information that fell into them. Physicists are trying to figure out how that might work.
$endgroup$
– G. Smith
4 hours ago
add a comment |
$begingroup$
No. The no-hair conjecture applies to stable solutions of the Einstein-Maxwell equations. In the case of merging black holes, it applies to the end state of the merger into a single quiescent black hole, after the “ringdown” has stopped.
answered 4 hours ago
G. SmithG. Smith
6,97611123
$endgroup$
$begingroup$
It's an odd theorem because most black holes aren't stable, their either evaporating or each particles from the background space.
$endgroup$
– zooby
4 hours ago
$begingroup$
those processes are slow enough that a black hole that is evaporating or accreting gas (not planets or other big objects) can be considered stable.
$endgroup$
– niels nielsen
4 hours ago
4
$begingroup$
Black hole evaporation is a quantum effect. The no-hair conjecture is a conjecture in classical General Relativity since we don’t have any consensus about quantum gravity. Quantum black holes probably do have some kind of hair that encodes all of the information that fell into them. Physicists are trying to figure out how that might work.
$endgroup$
– G. Smith
4 hours ago
add a comment |
$begingroup$
No. The no-hair conjecture applies to stable solutions of the Einstein-Maxwell equations. In the case of merging black holes, it applies to the end state of the merger into a single quiescent black hole, after the “ringdown” has stopped.
answered 4 hours ago
G. SmithG. Smith
6,97611123
$endgroup$
No. The no-hair conjecture applies to stable solutions of the Einstein-Maxwell equations. In the case of merging black holes, it applies to the end state of the merger into a single quiescent black hole, after the “ringdown” has stopped.
answered 4 hours ago
G. SmithG. Smith
6,97611123
answered 4 hours ago
G. SmithG. Smith
6,97611123
answered 4 hours ago
G. SmithG. Smith
6,97611123
answered 4 hours ago
G. SmithG. Smith
6,97611123
6,97611123
$begingroup$
It's an odd theorem because most black holes aren't stable, their either evaporating or each particles from the background space.
$endgroup$
– zooby
4 hours ago
$begingroup$
those processes are slow enough that a black hole that is evaporating or accreting gas (not planets or other big objects) can be considered stable.
$endgroup$
– niels nielsen
4 hours ago
4
$begingroup$
Black hole evaporation is a quantum effect. The no-hair conjecture is a conjecture in classical General Relativity since we don’t have any consensus about quantum gravity. Quantum black holes probably do have some kind of hair that encodes all of the information that fell into them. Physicists are trying to figure out how that might work.
$endgroup$
– G. Smith
4 hours ago
add a comment |
$begingroup$
It's an odd theorem because most black holes aren't stable, their either evaporating or each particles from the background space.
$endgroup$
– zooby
4 hours ago
$begingroup$
those processes are slow enough that a black hole that is evaporating or accreting gas (not planets or other big objects) can be considered stable.
$endgroup$
– niels nielsen
4 hours ago
4
$begingroup$
Black hole evaporation is a quantum effect. The no-hair conjecture is a conjecture in classical General Relativity since we don’t have any consensus about quantum gravity. Quantum black holes probably do have some kind of hair that encodes all of the information that fell into them. Physicists are trying to figure out how that might work.
$endgroup$
– G. Smith
4 hours ago
$begingroup$
It's an odd theorem because most black holes aren't stable, their either evaporating or each particles from the background space.
$endgroup$
– zooby
4 hours ago
$begingroup$
It's an odd theorem because most black holes aren't stable, their either evaporating or each particles from the background space.
$endgroup$
– zooby
4 hours ago
$begingroup$
those processes are slow enough that a black hole that is evaporating or accreting gas (not planets or other big objects) can be considered stable.
$endgroup$
– niels nielsen
4 hours ago
$begingroup$
those processes are slow enough that a black hole that is evaporating or accreting gas (not planets or other big objects) can be considered stable.
$endgroup$
– niels nielsen
4 hours ago
4
4
$begingroup$
Black hole evaporation is a quantum effect. The no-hair conjecture is a conjecture in classical General Relativity since we don’t have any consensus about quantum gravity. Quantum black holes probably do have some kind of hair that encodes all of the information that fell into them. Physicists are trying to figure out how that might work.
$endgroup$
– G. Smith
4 hours ago
$begingroup$
Black hole evaporation is a quantum effect. The no-hair conjecture is a conjecture in classical General Relativity since we don’t have any consensus about quantum gravity. Quantum black holes probably do have some kind of hair that encodes all of the information that fell into them. Physicists are trying to figure out how that might work.
$endgroup$
– G. Smith
4 hours ago
add a comment |
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