Why don't merging black holes disprove the no-hair theorem?
$begingroup$
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
$endgroup$
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
$endgroup$
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
$endgroup$
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 18 mins ago
David Z♦
63.4k23136252
63.4k23136252
asked 4 hours ago
zoobyzooby
1,389514
1,389514
add a comment |
add a comment |
1 Answer
1
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.
$endgroup$
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f460041%2fwhy-dont-merging-black-holes-disprove-the-no-hair-theorem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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.
$endgroup$
$begingroup$
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.
$endgroup$
$begingroup$
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.
$endgroup$
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
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 |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f460041%2fwhy-dont-merging-black-holes-disprove-the-no-hair-theorem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown