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







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    4












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 25 mins ago









      David Z♦

      63.4k23136252




      63.4k23136252










      asked 4 hours ago









      zoobyzooby

      1,389514




      1,389514






















          1 Answer
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          $begingroup$

          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.






          share|cite|improve this answer









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          • $begingroup$
            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











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          1 Answer
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          1 Answer
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          active

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          active

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          6












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          6












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          6












          6








          6





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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