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    Default If a black hole evaporates, where or how does the rotational energy go?

    I'm not sure there's an answer, maybe they don't evaporate is the answer? It is interesting, because if there are quanta (photon equivalents) of gravity, they'd have to be very very small, much smaller than photons of EMR.
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    A black hole evaporates through the process of Hawking radiation, and said radiation will carry away all the energy the black hole has, including rotational.

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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    I think he's asking about rotational momentum.

    My hunch is that it's somehow encoded on the event horizon and ultimately imparted to outgoing hawking radiation, just like any other infalling information. Implying that black hole charge and rotational momentum do go towards zero as it evaporates.

    If I could prove this as more than a hunch, I'd be preparing my nobel acceptance speech instead of posting here, and it'd be one of those weird space news facts that everybody would've heard the story by now.

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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by factotum View Post
    A black hole evaporates through the process of Hawking radiation, and said radiation will carry away all the energy the black hole has, including rotational.
    I understand that's the theory, but there seems to be a limit that stops the singularity spinning slow enough to reach further toward the original position of the event horizon of the non-spinning black hole than something like 1/4, so how does the energy get out from the position of the singularity to the position of the event horizon?

    Quote Originally Posted by Anymage View Post
    I think he's asking about rotational momentum.
    Momentum is surely the same as energy?
    Last edited by halfeye; 2019-12-07 at 06:27 AM.
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    I do not think that it needs to be too complicated and most likely was already solved by someone. As far as I would say, most likely the emited radiation is nicely random and isotropic but only in the rotating frame of reference of the black hole. If you go back to any inertial frame of reference it will naturally show that the radiation carries a non-zero angular momentum. This makes additional sense since a rotating black hole drags the surrounding spacetime in its spin.
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by halfeye View Post
    I understand that's the theory, but there seems to be a limit that stops the singularity spinning slow enough to reach further toward the original position of the event horizon of the non-spinning black hole than something like 1/4, so how does the energy get out from the position of the singularity to the position of the event horizon?
    That's the entire question about Hawking Radiation in a nutshell, rotational aspect doesn't matter. We don't know by what mechanism energy can be removed from the black hole, just that (as far as we know) it happens.

    Mind you, thinking about it--by definition the singularity has zero size, so can it actually have rotational momentum?
    Last edited by factotum; 2019-12-07 at 07:56 AM.

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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    If a black hole has angular momentum, then its Hawking radiation will also, on net, have angular momentum.

    ...so how does the energy get out from the position of the singularity to the position of the event horizon?
    Localizing gravitational energy (which of course is all of the energy of a black hole) is a very difficult and largely unsolved problem in general.
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    I'm going to go with "it doesn't evaporate, it stays inside the hole, spinning up the singularity (how does that work, a point mass spinning?) faster and faster until the hole is so light and the spin so fast that the last remaining bits of matter just fly outwards".

    Not because that's how it works, but because it's kind of funny to imagine.
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by factotum View Post
    Mind you, thinking about it--by definition the singularity has zero size, so can it actually have rotational momentum?
    Rotating black hole singularities are rings, of infinitesimal cross sectional area.
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by Lvl 2 Expert View Post
    Not because that's how it works, but because it's kind of funny to imagine.
    This strikes me as a good policy with regards to science on the fringes of understanding. Especially for people, like me, who are laymen in regards to such topics.
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by Radar View Post
    I do not think that it needs to be too complicated and most likely was already solved by someone. As far as I would say, most likely the emited radiation is nicely random and isotropic but only in the rotating frame of reference of the black hole. If you go back to any inertial frame of reference it will naturally show that the radiation carries a non-zero angular momentum. This makes additional sense since a rotating black hole drags the surrounding spacetime in its spin.
    This is a pretty good way to look at it.

    Also, Hawking Radiation values over the entire lifespan of a black hole can only be approximations given our current knowledge of physics. Once the hole gets small enough, we'd need a full theory of quantum gravity at least, and probably a full theory of everything to get precise values.

    However, it's a pretty good bet that Hawking radiation still happens through a complete decay process, even if we don't know the exact rates. Also, it's a pretty good bet that frame dragging or something approximately close to it happens through the whole process. So, the evaporation process will probably carry away angular momentum through the entire decay timeline(which probably proceeds through to 0 mass with one last quanta or pair of quanta emitted,) even if we don't know precisely how fast that process will occur.
    Quote Originally Posted by Harnel View Post
    where is the atropal? and does it have a listed LA?

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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by factotum View Post
    Mind you, thinking about it--by definition the singularity has zero size, so can it actually have rotational momentum?
    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.
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by Yora View Post
    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.
    Right. The singularity as described in this thread is a feature of solutions of the Einstein field equation for general relativity. This description doesn't take quantum effects into account. At those length and energy scales quantum effects would be important. So, since we don't have a proven theory of quantum gravity yet, we don't really yet know what the central parts of a black hole should be like, or if questions about them have any physical relevance.
    Quote Originally Posted by Harnel View Post
    where is the atropal? and does it have a listed LA?

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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Ha, I had just been thinking "did I just discover the quantum gravity problem?"
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    There seems to be a case that there might be more emissions, or more energetic emissions, in, or approximately in, the plane of rotation.

    There also seems to be some risk that as the black hole shrank, the rate of rotation might vary, depending on whether there were more or less extra emissions in the plane of rotation.
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    As an aside, it's refreshing to see a thread online where nobody is trying to say "quantum gravity says...". As folks have correctly stated in this thread, we don't have any idea what quantum gravity would say.
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by halfeye View Post
    There seems to be a case that there might be more emissions, or more energetic emissions, in, or approximately in, the plane of rotation.

    There also seems to be some risk that as the black hole shrank, the rate of rotation might vary, depending on whether there were more or less extra emissions in the plane of rotation.
    Outbound trajectories in the far-field inertial observer's frame should be frame-dragged away from the rotational poles. So, effectively, yes. As measured some distance away from the black hole, anyway.

    I don't see any particular evidence that initial radiation in the rotating frame would be biased against the poles, though. But any observer far enough away to not be damaged should see more radiation on the "equator" of a large imaginary sphere around the hole than at the "poles."
    Quote Originally Posted by Harnel View Post
    where is the atropal? and does it have a listed LA?

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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Individual photons (and gravitons, and neutrinos, and all fundamental particles) have angular momentum. If the black hole isn't spinning, the hawking radiation, on average, will have spins that cancel out. If the balck hoe is spinning, the hawking radiation will have an average spin to reflect that and slowly take away the angular momentum.

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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    There seems to me to be a paradox here.

    The less angular momentum a black hole has, the faster it rotates, which increases the energy lost through frame dragging, which obviously tends to increase the rate of energy loss, and thus the speed of rotation.
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by halfeye View Post
    There seems to me to be a paradox here.

    The less angular momentum a black hole has, the faster it rotates, which increases the energy lost through frame dragging, which obviously tends to increase the rate of energy loss, and thus the speed of rotation.
    Wait, what? That's the opposite of how angular momentum generally works (ie I have not had a course on angular momentum in black holes). Where is this coming from?
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by Lord Torath View Post
    Wait, what? That's the opposite of how angular momentum generally works (ie I have not had a course on angular momentum in black holes). Where is this coming from?
    It's coming from orbital mechanics. The lower an orbit is, the faster it is, until it hits something. There is no other something inside the singularity of a black hole.
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by halfeye View Post
    It's coming from orbital mechanics. The lower an orbit is, the faster it is, until it hits something. There is no other something inside the singularity of a black hole.
    Things in smaller orbits speed up because of conservation of angular momentum. They keep it, they don't lose it.

    At 0 radius you do indeed wind up with math breaking. Singularities are full of divide by zero errors, this is just one of them. Luckily, those of us in the rest of the universe don't have to deal with the consequences of math barfing like that because singularities are causally disconnected from the rest of the universe. That's what an event horizon is by definition.

    Just outside the event horizon you don't get any similar errors, and even exactly at the horizon the errors you do get are milder and can be fixed by shifting reference frames. So from our perspective on the outside, angular momentum can be generally treated as if the whole thing were one big solid ball instead of just the singularity at its center.

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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by halfeye View Post
    There seems to me to be a paradox here.

    The less angular momentum a black hole has, the faster it rotates, which increases the energy lost through frame dragging, which obviously tends to increase the rate of energy loss, and thus the speed of rotation.
    Frame dragging is proportional to angular momentum, not angular velocity.

    Also, for purposes of determining what goes on outside the black hole, it's not really helpful to think of the singularity as in an orbit.

    For example: consider a neutron star that is close to being black hole. We can (and would) use the same equations as a Kerr black hole to determine what gravity and frame dragging effects are near it. However no part of the neutron star is in an orbit, the star's spin can be decreased or increased (within a limit) without effecting it's size or shape dramatically.

    If we decrease the radius (while keeping the angular momentum and mass constant) frame dragging just outside the surface will increase (just like you implied). However, this is moving the surface. If we check the frame dragging of where the surface was, it wouldn't have changed.

    It doesn't matter what the frame dragging would be inside an event horizon, since it can't affect anything we see.

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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by Quizatzhaderac View Post
    Also, for purposes of determining what goes on outside the black hole, it's not really helpful to think of the singularity as in an orbit.
    How not? A Kerr black hole allegedly has a ring singularity, if the radius of that singularity doesn't vary with rotational energy, where is that energy expressed?
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by halfeye View Post
    How not? A Kerr black hole allegedly has a ring singularity, if the radius of that singularity doesn't vary with rotational energy, where is that energy expressed?
    Supposed you have a sealed black box filled with lego. You want to know what will happen if you drop the box, throw it around, et cetra.

    Does it matter what color the lego are? If they form a polka dot pattern? All that matters for you questions is that you have a block of a certain size, weight, and surface material.

    Likewise we can never see inside the event horizon so it's generally best to just regarding the blackhole as an object the size and shape of the event horizon, with a given mass and angular momentum.

    We could work out the velocity, rapidity, rest mass, and potential energy of the singularity, but 1) that's not gong to help us predict what will happen outside the black hole 2) A lot of these numbers of very screwy and possibly wrong.

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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by Quizatzhaderac View Post
    Supposed you have a sealed black box filled with lego. You want to know what will happen if you drop the box, throw it around, et cetra.

    Does it matter what color the lego are? If they form a polka dot pattern? All that matters for you questions is that you have a block of a certain size, weight, and surface material.

    Likewise we can never see inside the event horizon so it's generally best to just regarding the blackhole as an object the size and shape of the event horizon, with a given mass and angular momentum.

    We could work out the velocity, rapidity, rest mass, and potential energy of the singularity, but 1) that's not gong to help us predict what will happen outside the black hole 2) A lot of these numbers of very screwy and possibly wrong.
    If the radius of the singularity was large enough, there could be a hole through the middle of the event horizon (there seems to be a theory that that can't happen), and that would certainly have implications for space outside the event horizon.
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    The singularity is, by definition, a point. Unless we figure out something different as we piece together quantum gravity, in which case those figurings could easily invalidate things we think we know now.

    A combination of rotation and/or charge could in theory create a situation where the event horizon would shrink down to nothing. This is called a naked singularity, and it would be very bad. A singularity is a math error in the universe. The event horizon usually contains the glitch (because nothing on the inside can affect the outside), but an unshielded glitch would do nasty things.

    The theory that naked singularities can't happen is like the theory that the universe prevents any time travel that isn't self-consistent. Both could make a mess that we could probably see if they happened (even if, like vacuum decay, we'd only "see" it when we suffered spontaneous existence failure), so if they haven't happened within the lifespan of the universe some rule might explicitly forbid them.

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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by Anymage View Post
    The singularity is, by definition, a point.
    Only for non-rotating black holes. For rotating holes, the Kerr metric, the singuarity under general relativity is a ring, with no cross sectional width.

    https://en.m.wikipedia.org/wiki/Rotating_black_hole

    https://en.m.wikipedia.org/wiki/Kerr_metric

    Kerr black holes as wormholesEdit
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    Although the Kerr solution appears to be singular at the roots of Δ = 0, these are actually coordinate singularities, and, with an appropriate choice of new coordinates, the Kerr solution can be smoothly extended through the values of r {\displaystyle r} r corresponding to these roots. The larger of these roots determines the location of the event horizon, and the smaller determines the location of a Cauchy horizon. A (future-directed, time-like) curve can start in the exterior and pass through the event horizon. Once having passed through the event horizon, the r {\displaystyle r} r coordinate now behaves like a time coordinate, so it must decrease until the curve passes through the Cauchy horizon.[28]

    The region beyond the Cauchy horizon has several surprising features. The r {\displaystyle r} r coordinate again behaves like a spatial coordinate and can vary freely. The interior region has a reflection symmetry, so that a (future-directed time-like) curve may continue along a symmetric path, which continues through a second Cauchy horizon, through a second event horizon, and out into a new exterior region which is isometric to the original exterior region of the Kerr solution. The curve could then escape to infinity in the new region or enter the future event horizon of the new exterior region and repeat the process. This second exterior is sometimes thought of as another universe. On the other hand, in the Kerr solution, the singularity is a ring, and the curve may pass through the center of this ring. The region beyond permits closed time-like curves. Since the trajectory of observers and particles in general relativity are described by time-like curves, it is possible for observers in this region to return to their past.[20][21] This interior solution is not likely to be physical and considered as a purely mathematical artefact.[29]
    My bold. I don't understand the maths, but I see a phrase like that about the singularity being a ring in every piece on Kerr black holes I read.
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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    You're right. Most of what I'd looked at on naked singularities glossed over the ring part, and I brainfarted.

    Still, you're unlikely to have a big ring with a donut shaped horizon. You're also not going to get an event horizon with a singularity spinning around outside it, or even just peeking out. With that much spinning you're more likely to have ways to get from the singularity to outside. Again that's a naked singularity, and it's not good.

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    Default Re: If a black hole evaporates, where or how does the rotational energy go?

    Quote Originally Posted by halfeye View Post
    If the radius of the singularity was large enough, there could be a hole through the middle of the event horizon (there seems to be a theory that that can't happen), and that would certainly have implications for space outside the event horizon.
    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.

    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.

    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.

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