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Space Science

How To Destroy a Black Hole 364

KentuckyFC writes "The critical concept that makes a black hole black is the event horizon: a theoretical boundary in space through which light and other objects can pass in one direction but not the other. Since light cannot escape the event horizon, it must be black. The event horizon is a nuisance to astrophysicists because it hides the interesting new physics that must go on inside a black hole. What they would like is a way to get rid of the event horizon so that they can see what goes on behind it. It turns out that just such a thing may be possible, say physicists. According to the mathematics of general relativity, the event horizon should disappear if a black hole were fed enough charge and angular momentum relative to its mass. However the calculations are so fiendish (PDF) that nobody knows whether the black hole would shed this extra angular momentum and charge before it could settle into a stable 'naked' state. However, the possibility that the event horizon could be destroyed raises the question of what astrophysicists would see behind this veil. According to some, black holes are regions of spacetime with infinite curvature called singularities. Many believe that 'naked' singularities cannot exist in nature. And yet there are enough question marks to suggest that this mystery is far from settled."
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How To Destroy a Black Hole

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  • by Anonymous Coward on Friday June 11, 2010 @12:24PM (#32536918)
    Christ, you always make the most asinine comments. It's almost impressive.
  • by Red Flayer ( 890720 ) on Friday June 11, 2010 @12:29PM (#32536982) Journal

    On a less serious note, does Rule 34 apply to naked singularities?

    Yes, of course. Rule34 applies to everything (and if not, Rule35 comes into play).

    In this case -- you've seen goatse, right?

  • by NicknamesAreStupid ( 1040118 ) on Friday June 11, 2010 @12:38PM (#32537156)
    Photons cannot escape because they are red-shifted due to time dilation. This means that the horizon will vary depending on the level of energy trying to escape it. For example, an X-ray might escape where an infra-red photon wouldn't. All or part of a huge energy blast may or may not escape, depending on its frequency, level and position. Whether it would affect the hole itself seems problematic.
  • In theory, yes. (Score:5, Informative)

    by maillemaker ( 924053 ) on Friday June 11, 2010 @12:42PM (#32537226)

    One of the problems with approaching a black hole (aside from massive amounts of radiation around ones actively eating matter) is the fact that the force of gravity increases as you approach the mass responsible for the gravity.

    With small black holes, as you approach (feet first) the difference in gravitational pull at your feet would be many times larger than the gravitational pull at your head. You would be literally ripped apart, down to the molecular level. This is known as "Spaghettification". []

    However, with a large enough black hole, you should be able to pass the event horizon before these tidal forces grow large enough to rip you apart. Of course, this does you no good, because once you are inside the event horizon you cannot exert a great enough force to prevent yourself from falling deeper until the forces ARE great enough to rip you apart.

    But for a large black hole, in theory, you could cross the event horizon without being ripped apart.

  • by professionalfurryele ( 877225 ) on Friday June 11, 2010 @12:43PM (#32537246)

    Although no one knows what happens at the end point of black hole evaporation it is unlikely it would leave a naked singularity since the mass of the singularity is what is being 'evaporated'. Besides, even if there was a naked singularity around just before the thing evaporates it would be kicking out so much energy you wouldn't be able to get anywhere near it. A 1kg black hole evaporating would release the equivalent energy to a large thermonuclear weapon in a fraction of a second.

  • by MouseR ( 3264 ) on Friday June 11, 2010 @12:45PM (#32537286) Homepage

    Goatse is not a black hole but a brown dwarf.

  • by alyosha1 ( 581809 ) on Friday June 11, 2010 @01:03PM (#32537626)

    Uh, no. An X-ray photon and an infra-red photon have the same velocity, c. They have different frequencies. Neither will escape a black hole, which is pretty much defined as a body having an escape velocity greater than c.

  • Re:IANAA (Score:5, Informative)

    by Remus Shepherd ( 32833 ) <> on Friday June 11, 2010 @01:06PM (#32537676) Homepage

    (Disclaimer: I Am A Physicist, but this is not my area of expertise, and only the experts understand those equations.)

    It's not feeding it mass that does the trick; it's feeding it charge and angular momentum. The only reason you feed it more mass is because you need mass to carry the charge and momentum into the hole.

    What you get if you feed it charge and angular momentum is a spinning monopole. I think they are postulating that a spinning monopole causes rotational frame dragging [], and if you do it right you can get the charged frame dragging effects to cancel out the gravitational effects -- namely, the event horizon.

    After you do all that, what will be left? Like the article says, nobody knows. That's why it's exciting.

  • by Ktistec Machine ( 159201 ) on Friday June 11, 2010 @01:06PM (#32537680)

    The event horizon isn't wavelength-dependent. It's the place where the escape velocity equals the speed of light.

  • misleading title (Score:4, Informative)

    by bcrowell ( 177657 ) on Friday June 11, 2010 @01:33PM (#32538212) Homepage

    The title of the paper is "Destroying black holes with test bodies," and the language about "destroying" black holes is echoed in both the arxiv blog summary and the /. summary. This may be somewhat misleading. They're actually talking about processes that would strip away the event horizon, leaving behind a naked singularity. The black hole wouldn't have been "destroyed," but just changed into a different form. The authors themselves put the word "destroying" in quotes in the paper.

    The paper doesn't settle this one way or the other. It says shows that if you use a certain set of approximations, the result is that the event horizon can go away. However, there is no particular reason to believe that the approximation is correct.

    The real issue here isn't whether a black hole can actually be transformed this way, it's the question of whether cosmic censorship [] holds. If cosmic censorship fails, then general relativity is fundamentally flawed as a classical field theory, because it fails to make predictions. John Earman's famous way of expressing this is that anything could come out of a naked singularity: lost socks, green slime, even horrible things like Nixon's "Checkers" speech or Japanese monster-movie creatures.

  • by MetricT ( 128876 ) on Friday June 11, 2010 @01:39PM (#32538314)

    I did Ph.D. research on this exact subject a decade ago, and at a quick glance I didn't see anything new in this paper. A spinning and/or charged black hole in theory can be spun or charged to the point where a naked singularity would appear. But, the harder you spin/charge the black hole, the harder it tries to neutralize itself by preferentially emitting particles of a given angular momentum or charge. So the equality probably is a physical limit. I thought someone had proven that years ago, but I've been out of the field for a while.

    This looks kind of like someone wrote a paper so they could go to a conference or something. There doesn't appear to be anything earth-shattering (or black hole-shattering) here.

  • by Attila Dimedici ( 1036002 ) on Friday June 11, 2010 @01:50PM (#32538526)
    There are people who think you are joking, and that is my best guess. The problem with your joke (if it is a joke) is that there are actually people who think there is something racist about using the word "black" in the term "black hole". I believe that there was a story on here about that one to two years ago (no, I'm not going to go search for it). Even if it wasn't on here, there was such a story in that time frame.
  • by $RANDOMLUSER ( 804576 ) on Friday June 11, 2010 @01:55PM (#32538630)

    The event horizon is ... the place where the escape velocity equals the speed of light.

    Ding ding ding - we have a winner. This is the point that so many cosmology shows on Discovery Channel or Science Channel (or whatever) completely fail to mention; they keep describing black holes as "so massive, even light can't escape" without explaining why (Michio Kaku, Alex Fillipenko, (sp) I'm looking at you). See Wikipedia [] for the details, but the important point is that escape velocity is dependent on an object's mass divided by its radius. So if mass goes high enough, or radius low enough, you get an escape velocity greater than the speed of light: AKA an event horizon.

    Say it again, and remember it later:
    The event horizon is ... the place where the escape velocity equals the speed of light.

  • by steelfood ( 895457 ) on Friday June 11, 2010 @03:01PM (#32539824)

    Maybe by the time anything reaches the singularity, the universe would've ended and time would cease to exist. That is, you can never reach a singularity, you (or what's left of you) can only ever continue to spiral around it until literally the end of time.

  • by bcrowell ( 177657 ) on Friday June 11, 2010 @03:44PM (#32540566) Homepage

    I once read a bit about black holes, and one of the things I read was: a black hole doesn't necessarily have to be very dense. It can also be sparse (and the larger, the sparser it can be). For example, if you'd take a lot of stars and planets, and put them together (but not too close together), then at one point if you make this large enough, it'll also be a black hole: there appears an event horizon around all this matter. But inside of it are still stars with gaps between them, maybe some planets orbiting around them, ... So now I wonder, if the above is true: can someone live inside that? Would there be any noticeable difference between being inside of that, and the other side (the outside) of this event horizon?

    No, this is totally incorrect. The Penrose singularity theorem [] forbids this.

  • by Rakishi ( 759894 ) on Friday June 11, 2010 @05:37PM (#32542612)

    You're bringing horribly naive Newtonian physics into a realm they do not belong in. Stop before you make yourself sounds like even more of a fool.

    As you fall into a black hole the event horizon appears to move away from you from your point of view. An external observer on the other hand would never see you cross the horizon due to other effects, just see you falling ever more slowly into it as the light from you takes ever longer to reach them. You'd see some odd effects from crossing the event horizon but you'd continue to observe things. [] []

    The event horizon is precisely the point at which you will be traveling at the speed of light...

    No it's not. Matter cannot go at the speed of light and never does. Nothing can go faster than light. That's the whole bloody point of relativity. Stop watching bad sci-fi movies. There are I believe odd effects for certain observers who might see you but practically speaking you'd just be moving at just something like 0.99c. But that's just due to the acceleration of the black hole and it will continue to increase even after you cross the event horizon.

    As soon as the flashlight passes through the event horizon, it'll disappear: not even the light from it will be able to reach you.

    Nope, you continue seeing the flashlight. Well in some way at least. The event horizon for you has receded. Another way to think of it is that while the light is no longer moving towards you, sort of hovers at the event horizon, you're now moving towards the light. Relative velocity is all that matters. You'll get some odd relativistic distortions I believe, makes me wonder if a human nervous system even survive them, but you'll continue to observe things that went before you through the even horizon.

  • by internic ( 453511 ) on Friday June 11, 2010 @05:41PM (#32542690)

    The Schwarzschild solution to General Relativity applies to a stationary, spherically symmetric (uncharged) distribution of mass with asymptotically Minkowski boundary conditions for space-time. So, it basically assumes that you have a ball of stuff surrounded by empty space, and that's the type of situation where the equation for the Schwatzschild radius applies.

    If we accept the idea that we can approximate the universe by a uniform distribution of matter on a large enough scale (and this is, perhaps, debatable), the Schwarzschild solution doesn't apply. But you can solve GR for this situation and what you get is the Friedman-Robertson-Walker [] metric, which doesn't have any sort of event horizon, no matter how dense the matter within. Of course, it does have the interesting feature of expansion (or contraction) which lead to the beginning of modern cosmology.

    It is interesting, though, if the numbers for the mass and radius of the observable universe come out that way. Perhaps it has some import, but I can't say what it is offhand (but then I don't study GR). My first guess would be that it has to do with the universe being approximately flat, but I don't think that's actually true (because that should depend on the density, Hubble's constant, and the cosmological constant).

Honesty is for the most part less profitable than dishonesty. -- Plato