Re: If a black hole evaporates, where or how does the rotational energy go?
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Originally Posted by
Yora
I belive it's not proven that a real singularity actually exists. Looked at from the outside, the event horizon does not give any indication how the mass and energy inside of it is distributed. All matter that falls into a black hole still needs time to reach the the center. Matter and energy collapsing into a point with zero volume might take an infinite amount of time. So while the ideal mathmatical model has a perfect singularity at the center, perhaps all real black holes are only approaching a true singularity.
Assuming that additional matter falls in from the outside faster than the black hole loses mass to hawking radiation, shouldn't we, in practice, observe infalling matter corssing the event horizon in a finite period of time, due to the fact that even though its position becomes arrested in our reference frame, the infalling matter causes the event horizon to expand (and, crucially, will eventually to expand past the position of the object in question provided that it is given enough infallig mass)
Re: If a black hole evaporates, where or how does the rotational energy go?
Yora's point was about not seeing anything past the event horizon. In the case you describe one would see things falling into the event horizon, but they wouldn't see anything of the distribution of matter within the event horizon.
Re: If a black hole evaporates, where or how does the rotational energy go?
Quote:
Originally Posted by
Quizatzhaderac
Let me stop and make myself 100% clear on a term here. A regular doughnut has a hole. A jelly filled doughnut has a cavity, but has no hole.
I have not been talking about doughnuts. When it comes to eating doughnuts, I prefer by a long way jam doughnuts (though the jam is more like fruit sauce than normal jam), with no hole, and they don't have a cavity because they have jam in them.
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The singularity of a Kerr (spinning) black hole has a hole. The event horizon has a cavity, but no hole (also, unlike the jelly doughnut, the cavity doesn't connect to the outside at all). There is absolutely no way to see the singularity or the cavity.
The event horizon is a surface, how can that have a cavity?
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Adding angular momentum will shrink the event horizon and make the cavity bigger. However the outer and inner event horizon approach being the exact same shape at the exact same place as you add angular momentum. Once they're at the same place, they cancel and there is no event horizon anywhere. THe mass in the singularity would have nothing forcing to be a singularity and would just be mass at that point.
I would imagine the event horizon being a given distance (that distance mostly depending on the mass of the singularity) from the singularity, which would mean that the event horizon moved outward at the equator, and inward at the poles (equator and poles being relative to the rotation of the singularity). So whether or not the singularity could spin up to the point that there was a big enough dip at the poles to go through, the spin of the singularity could be determined from the shape of the event horizon (and ergosphere, and photon horizon, I don't claim to understand that stuff), as well as from frame dragging.
What's interesting me about evaporating black holes is that in the last few seconds a huge amount of mass is being lost very quickly, and if there is any angular momentum there seems to be a strong risk of the angular momentum relative to the mass rising very fast, which by your explanation would give rise to a naked singularity, and by mine would lead to a toroidal event horizon.
Re: If a black hole evaporates, where or how does the rotational energy go?
Quote:
Originally Posted by
halfeye
The event horizon is a surface, how can that have a cavity?
Good question; I was unclear. A spinning black hole has two event horizons and a region between them. Within that region everything must be moving downward, time doesn't work the same way, and kinetic and potential energies have an imaginary component.
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I would imagine the event horizon being a given distance (that distance mostly depending on the mass of the singularity) from the singularity, which would mean that the event horizon moved outward at the equator, and inward at the poles (equator and poles being relative to the rotation of the singularity).
The event horizons of a spinning black hole (and for that matter, every kind of black hole) are perfect spheres.
As the black hole changes from no spin to maximum spin the event horizon shrinks to half it's radius. The ergosphere keeps the same equator as the original event horizon, but it's poles pinch in to meet the poles f the new event horizon.
Re: If a black hole evaporates, where or how does the rotational energy go?
Quote:
Originally Posted by
Quizatzhaderac
The event horizons of a spinning black hole (and for that matter, every kind of black hole) are perfect spheres.
Wikipedia disagrees:
https://en.wikipedia.org/wiki/Black_hole#Event_horizon
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The shape of the event horizon of a black hole is always approximately spherical.[Note 4][86] For non-rotating (static) black holes the geometry of the event horizon is precisely spherical, while for rotating black holes the event horizon is oblate.[
Re: If a black hole evaporates, where or how does the rotational energy go?
Quote:
Originally Posted by
halfeye
Wikipedia disagrees:
This is a case of Wikipedia contracting itself since it also says the event horizon of a spinning black hole is a sphere. I'm personally more inclined to go with the page that has equations, but I'm not a proper authority so feel free to disagree with me.
Re: If a black hole evaporates, where or how does the rotational energy go?
Quote:
Originally Posted by
Quizatzhaderac
I'm not seeing anywhere in that section where it says anything about the event horizon being sphereical in English. It could be Wikipedia is having an edit war about this. I don't know enough about the subject to take sides in an edit war on it.