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 Feb 11 at 4:03
David Z♦
63.7k23136252
asked Feb 10 at 23:43
zoobyzooby
1,537716
$endgroup$
add a comment |
32
$begingroup$
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 Feb 11 at 4:03
David Z♦
63.7k23136252
asked Feb 10 at 23:43
zoobyzooby
1,537716
$endgroup$
add a comment |
32
32
32
5
$begingroup$
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 Feb 11 at 4:03
David Z♦
63.7k23136252
asked Feb 10 at 23:43
zoobyzooby
1,537716
$endgroup$
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 Feb 11 at 4:03
David Z♦
63.7k23136252
asked Feb 10 at 23:43
zoobyzooby
1,537716
edited Feb 11 at 4:03
David Z♦
63.7k23136252
asked Feb 10 at 23:43
zoobyzooby
1,537716
edited Feb 11 at 4:03
David Z♦
63.7k23136252
edited Feb 11 at 4:03
David Z♦
63.7k23136252
edited Feb 11 at 4:03
David Z♦
63.7k23136252
63.7k23136252
asked Feb 10 at 23:43
zoobyzooby
1,537716
asked Feb 10 at 23:43
zoobyzooby
1,537716
asked Feb 10 at 23:43
zoobyzooby
1,537716
1,537716
add a comment |
add a comment |
1 Answer
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active
oldest
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51
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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 Feb 10 at 23:55
G. SmithG. Smith
8,67511427
$endgroup$
1
$begingroup$
It's an odd theorem because most black holes aren't stable, their either evaporating or each particles from the background space.
$endgroup$
– zooby
Feb 11 at 0:01
4
$begingroup$
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
Feb 11 at 0:09
42
$begingroup$
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
Feb 11 at 0:09
$begingroup$
@zooby or to put the same point another way, it can be seen as a timescale separation. If you assume that the black hole is big enough then it makes sense to ask what the state will look like after a long enough time that all classical transient effects have stopped, but short enough that evaporation isn't significant. (Of course, the no hair theorem is still only a claim about the macroscopic spatial scale in that case - there could still be 'hair' on a microscopic quantum scale that can't be modelled by general relativity alone.) (+1 to both question and answer.)
$endgroup$
– Nathaniel
Feb 11 at 7:21
1
$begingroup$
@G.Smith: Could you tell me if this layman intuition is correct (I'm not a physicist)? My intuition behind the no-hair theorem is that, if everyone agrees on where the event horizon is, then consequently nobody can observe what is behind it (even via gravitational effects), i.e. what is behind it becomes unobservable and hence not something that can characterize the black hole. But for everybody to agree on what's inside a black hole, the system needs to have settled into a static state, which is not the case here. Hence the no-hair theorem doesn't apply. Is this correct?
$endgroup$
– Mehrdad
Feb 11 at 8:38
|
show 5 more comments
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1 Answer
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1 Answer
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51
$begingroup$
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 Feb 10 at 23:55
G. SmithG. Smith
8,67511427
$endgroup$
1
$begingroup$
It's an odd theorem because most black holes aren't stable, their either evaporating or each particles from the background space.
$endgroup$
– zooby
Feb 11 at 0:01
4
$begingroup$
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
Feb 11 at 0:09
42
$begingroup$
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
Feb 11 at 0:09
$begingroup$
@zooby or to put the same point another way, it can be seen as a timescale separation. If you assume that the black hole is big enough then it makes sense to ask what the state will look like after a long enough time that all classical transient effects have stopped, but short enough that evaporation isn't significant. (Of course, the no hair theorem is still only a claim about the macroscopic spatial scale in that case - there could still be 'hair' on a microscopic quantum scale that can't be modelled by general relativity alone.) (+1 to both question and answer.)
$endgroup$
– Nathaniel
Feb 11 at 7:21
1
$begingroup$
@G.Smith: Could you tell me if this layman intuition is correct (I'm not a physicist)? My intuition behind the no-hair theorem is that, if everyone agrees on where the event horizon is, then consequently nobody can observe what is behind it (even via gravitational effects), i.e. what is behind it becomes unobservable and hence not something that can characterize the black hole. But for everybody to agree on what's inside a black hole, the system needs to have settled into a static state, which is not the case here. Hence the no-hair theorem doesn't apply. Is this correct?
$endgroup$
– Mehrdad
Feb 11 at 8:38
|
show 5 more comments
51
$begingroup$
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 Feb 10 at 23:55
G. SmithG. Smith
8,67511427
$endgroup$
1
$begingroup$
It's an odd theorem because most black holes aren't stable, their either evaporating or each particles from the background space.
$endgroup$
– zooby
Feb 11 at 0:01
4
$begingroup$
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
Feb 11 at 0:09
42
$begingroup$
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
Feb 11 at 0:09
$begingroup$
@zooby or to put the same point another way, it can be seen as a timescale separation. If you assume that the black hole is big enough then it makes sense to ask what the state will look like after a long enough time that all classical transient effects have stopped, but short enough that evaporation isn't significant. (Of course, the no hair theorem is still only a claim about the macroscopic spatial scale in that case - there could still be 'hair' on a microscopic quantum scale that can't be modelled by general relativity alone.) (+1 to both question and answer.)
$endgroup$
– Nathaniel
Feb 11 at 7:21
1
$begingroup$
@G.Smith: Could you tell me if this layman intuition is correct (I'm not a physicist)? My intuition behind the no-hair theorem is that, if everyone agrees on where the event horizon is, then consequently nobody can observe what is behind it (even via gravitational effects), i.e. what is behind it becomes unobservable and hence not something that can characterize the black hole. But for everybody to agree on what's inside a black hole, the system needs to have settled into a static state, which is not the case here. Hence the no-hair theorem doesn't apply. Is this correct?
$endgroup$
– Mehrdad
Feb 11 at 8:38
|
show 5 more comments
51
51
51
$begingroup$
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 Feb 10 at 23:55
G. SmithG. Smith
8,67511427
$endgroup$
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 Feb 10 at 23:55
G. SmithG. Smith
8,67511427
answered Feb 10 at 23:55
G. SmithG. Smith
8,67511427
answered Feb 10 at 23:55
G. SmithG. Smith
8,67511427
answered Feb 10 at 23:55
G. SmithG. Smith
8,67511427
8,67511427
1
$begingroup$
It's an odd theorem because most black holes aren't stable, their either evaporating or each particles from the background space.
$endgroup$
– zooby
Feb 11 at 0:01
4
$begingroup$
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
Feb 11 at 0:09
42
$begingroup$
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
Feb 11 at 0:09
$begingroup$
@zooby or to put the same point another way, it can be seen as a timescale separation. If you assume that the black hole is big enough then it makes sense to ask what the state will look like after a long enough time that all classical transient effects have stopped, but short enough that evaporation isn't significant. (Of course, the no hair theorem is still only a claim about the macroscopic spatial scale in that case - there could still be 'hair' on a microscopic quantum scale that can't be modelled by general relativity alone.) (+1 to both question and answer.)
$endgroup$
– Nathaniel
Feb 11 at 7:21
1
$begingroup$
@G.Smith: Could you tell me if this layman intuition is correct (I'm not a physicist)? My intuition behind the no-hair theorem is that, if everyone agrees on where the event horizon is, then consequently nobody can observe what is behind it (even via gravitational effects), i.e. what is behind it becomes unobservable and hence not something that can characterize the black hole. But for everybody to agree on what's inside a black hole, the system needs to have settled into a static state, which is not the case here. Hence the no-hair theorem doesn't apply. Is this correct?
$endgroup$
– Mehrdad
Feb 11 at 8:38
|
show 5 more comments
1
$begingroup$
It's an odd theorem because most black holes aren't stable, their either evaporating or each particles from the background space.
$endgroup$
– zooby
Feb 11 at 0:01
4
$begingroup$
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
Feb 11 at 0:09
42
$begingroup$
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
Feb 11 at 0:09
$begingroup$
@zooby or to put the same point another way, it can be seen as a timescale separation. If you assume that the black hole is big enough then it makes sense to ask what the state will look like after a long enough time that all classical transient effects have stopped, but short enough that evaporation isn't significant. (Of course, the no hair theorem is still only a claim about the macroscopic spatial scale in that case - there could still be 'hair' on a microscopic quantum scale that can't be modelled by general relativity alone.) (+1 to both question and answer.)
$endgroup$
– Nathaniel
Feb 11 at 7:21
1
$begingroup$
@G.Smith: Could you tell me if this layman intuition is correct (I'm not a physicist)? My intuition behind the no-hair theorem is that, if everyone agrees on where the event horizon is, then consequently nobody can observe what is behind it (even via gravitational effects), i.e. what is behind it becomes unobservable and hence not something that can characterize the black hole. But for everybody to agree on what's inside a black hole, the system needs to have settled into a static state, which is not the case here. Hence the no-hair theorem doesn't apply. Is this correct?
$endgroup$
– Mehrdad
Feb 11 at 8:38
1
1
$begingroup$
It's an odd theorem because most black holes aren't stable, their either evaporating or each particles from the background space.
$endgroup$
– zooby
Feb 11 at 0:01
$begingroup$
It's an odd theorem because most black holes aren't stable, their either evaporating or each particles from the background space.
$endgroup$
– zooby
Feb 11 at 0:01
4
4
$begingroup$
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
Feb 11 at 0:09
$begingroup$
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
Feb 11 at 0:09
42
42
$begingroup$
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
Feb 11 at 0:09
$begingroup$
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
Feb 11 at 0:09
$begingroup$
@zooby or to put the same point another way, it can be seen as a timescale separation. If you assume that the black hole is big enough then it makes sense to ask what the state will look like after a long enough time that all classical transient effects have stopped, but short enough that evaporation isn't significant. (Of course, the no hair theorem is still only a claim about the macroscopic spatial scale in that case - there could still be 'hair' on a microscopic quantum scale that can't be modelled by general relativity alone.) (+1 to both question and answer.)
$endgroup$
– Nathaniel
Feb 11 at 7:21
$begingroup$
@zooby or to put the same point another way, it can be seen as a timescale separation. If you assume that the black hole is big enough then it makes sense to ask what the state will look like after a long enough time that all classical transient effects have stopped, but short enough that evaporation isn't significant. (Of course, the no hair theorem is still only a claim about the macroscopic spatial scale in that case - there could still be 'hair' on a microscopic quantum scale that can't be modelled by general relativity alone.) (+1 to both question and answer.)
$endgroup$
– Nathaniel
Feb 11 at 7:21
1
1
$begingroup$
@G.Smith: Could you tell me if this layman intuition is correct (I'm not a physicist)? My intuition behind the no-hair theorem is that, if everyone agrees on where the event horizon is, then consequently nobody can observe what is behind it (even via gravitational effects), i.e. what is behind it becomes unobservable and hence not something that can characterize the black hole. But for everybody to agree on what's inside a black hole, the system needs to have settled into a static state, which is not the case here. Hence the no-hair theorem doesn't apply. Is this correct?
$endgroup$
– Mehrdad
Feb 11 at 8:38
$begingroup$
@G.Smith: Could you tell me if this layman intuition is correct (I'm not a physicist)? My intuition behind the no-hair theorem is that, if everyone agrees on where the event horizon is, then consequently nobody can observe what is behind it (even via gravitational effects), i.e. what is behind it becomes unobservable and hence not something that can characterize the black hole. But for everybody to agree on what's inside a black hole, the system needs to have settled into a static state, which is not the case here. Hence the no-hair theorem doesn't apply. Is this correct?
$endgroup$
– Mehrdad
Feb 11 at 8:38
|
show 5 more comments
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