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Space Earth

Micro-Black Holes Make Poor Planet Killers 314

Posted by timothy
from the they-don't-even-make-good-nightclubs dept.
astroengine writes "Physicists are getting excited about the possibility of micro-black holes (MBH) being produced by the LHC and an international group of researchers have done the math to see what kind of impact they could have on the Earth. Unfortunately, if you're a megalomaniac looking for your next globe-eating weapon, you can scrub MBHs off your WMD list. If a speedy MBH is produced, flying through our planet, it will only have a few seconds to accrete the mass of a few atoms. It would then be lost to space where it will evaporate. If a slow MBH is produced, dropping into the Earth where it sits for a few billion years, the results are even more boring."
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Micro-Black Holes Make Poor Planet Killers

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  • by Sockatume (732728) on Friday November 13, 2009 @09:17AM (#30085732)

    Actually it's freshman-level physics. Calculating how quickly a micro-black-hole would accumulate mass strikes me as a great undergrad tutorial question.

  • by jimboindeutchland (1125659) on Friday November 13, 2009 @09:28AM (#30085796) Homepage

    I'm sure there's somebody on /. who can answer this:

    Correct me if I'm wrong, but I thought to be a black hole you had to be 2 things.
    1. a singularity
    2. heavy/massive enough to stop anything from escaping

    If you've got a singularity (worst case in our example) that's the mass of the earth, how's that supposed to stop any light/matter/etc escaping? It's not massive enough!

    or am I missing something.

    Also, please excuse my lack of correct terminology. IANAAP

  • by MacAnkka (1172589) on Friday November 13, 2009 @10:11AM (#30086188)
    Am I the only one who reads MBH as mega black hole, not micro black hole? It's confusing. If the prefix is micro, it would make sense to use a letter that actually means micro, instead of a letter that represents mega.
  • Re:Radius (Score:4, Interesting)

    by Animaether (411575) on Friday November 13, 2009 @10:22AM (#30086330) Journal

    you'd have to figure out a way to suspend it inside your spaceship and travel along with your spaceship.. which would have to be spherical because different gravitational magnitudes being exerted on different parts of your body has got to be uncomfortable.

    Even with it being spherical, your feet would get a noticeably larger gravitational force than your head.. blood circulation might become an issue. Somebody with too much time on their hands could probably work out the 'safe' minimal radius of the sphere surface you'd be walking on.

  • by sjames (1099) on Friday November 13, 2009 @10:36AM (#30086508) Homepage

    The math that suggests that a quantum black hole will evaporate in an instant may be fairly advanced, but the math showing that even if Hawking is completely wrong such a black hole would have no noticeable effect on the earth over a 13 billion year period is not all that advanced.

    Then there's simple logic. While LHC may produce the most powerful collisions ever under our control, nature routinely produces much more powerful collisions including cosmic rays. Clearly, in billions of years none of this has resulted in a planet eating black hole.

  • Re:Evaporate? (Score:3, Interesting)

    by Maximum Prophet (716608) on Friday November 13, 2009 @10:55AM (#30086754)

    The thing I'd add to that is that there are no anti-photons -- photons are their own anti-particle.

    That's correct, but something I've never wrapped my mind around. When the photon and photon-prime are created, then one falls into the black hole, how does the BH know that its photon should cancel mass, rather than increase it?

  • by radtea (464814) on Friday November 13, 2009 @12:29PM (#30087912)

    Another way to put it: if we were so sure that what we know is 100% correct then we wouldn't need to build the LHC to test our theories in the first place.

    There's a nice equivocation in this statement: we can be as sure as we are of anything that LHC black holes won't destroy the Earth. If they did we'd see evidence in the cosmic-ray spectrum due to evaporating black hole signatures and the like, as well as the Earth not actually being here because it would have been destroyed in the past.

    So while we do need to build the LHC to test theories regarding the Higgs boson, we do not need to build it to test theories regarding LHC black holes. That's the thing about science: all sources of experimental knowledge are equally valid, and you don't get to say our knowlege of black holes supposedly created by high energy collisions "doesn't count" because it comes from cosmic rays rather than accelerators.

    Furthermore, I'm not sure why you and others keep bringing up the 100% correct thing. You can't be sure 100% certainty that the act of typing your next post into /. won't invoke some as-yet-to-be-discovered physical law and cause you to grow a second head. But for some reason you won't explain you aren't worried about that, even though you seem to pretend to be worried about LHC black holes destroying the Earth, which has no greater probablity.

    Why is that? Why aren't you posting about all the other things that you can't be 100% sure of not destroying the Earth? Why only the LHC and not hitherto undiscovered physical laws that will cause the DROID phone to result in the death of us all? The "not 100% sure" standard is so silly that you'd have to terrified of damned near everything, if you were remotely intellectually honest.

  • by sznupi (719324) on Friday November 13, 2009 @12:34PM (#30087976) Homepage

    Not disagreeing with you generally of course - but aren't protons & neutrons more accuratelly conceptualised as, indeed, a sphere with each of the quarks not occupying any specific point but being "mixed" together?

  • by clone53421 (1310749) on Friday November 13, 2009 @12:37PM (#30088044) Journal

    The units of the "gravitational field" are not those of veloicty

    No, they are units of acceleration, m/s^2. My point was that it is impossible for a finite gravitational field to accelerate anything to the speed of light. Not that the field itself equaled the speed of light, but that the field integrated over some finite time supposedly equaled the speed of light, which is impossible for any finite field.

  • by CTachyon (412849) <chronos.chronos-tachyon@net> on Friday November 13, 2009 @01:52PM (#30089014) Homepage

    You seem to know your physics. I have a question. If a "stationary" black hole gets hit by an object of comparable mass, and neglecting the effects of gravity between both objects, will the black hole move at all? Will it only get as much kinetic energy as the mass it absorbs had, none at all, or you could actually hit it?

    This is actually quite easy to answer, because momentum is conserved (in both Newtonian physics and General Relativity). In the Newtonian model, which is accurate for the masses and velocities we're dealing with, momentum equals mass times velocity. A stationary black hole has mass m_0 > 0 and velocity v_0 = 0, for a momentum of m_0*v_0 = 0. An incoming object with mass equal to the black hole has mass m_1 = m_0 and velocity v_1 > 0, for a momentum of m_1*v_1 > 0. The joint system, after the black hole has completely absorbed the incoming object, has momentum m_joint*v_joint = (m_0*v_0 + m_1*v_1) = (0 + m_1*v_1) = m_1*v_1 and mass m_joint = (m_0 + m_1) = 2*m_1. Therefore, its velocity is v_joint = (m_joint*v_joint / m_joint) = (m_1*v_1 / 2*m_1) = (v_1 / 2), or exactly half the velocity of the original incoming object, traveling in the same direction that the original incoming object was traveling. This is simply a fully inelastic collision — that is, a collision where the two colliding objects stick together instead of bouncing off each other. The fact that one object is a black hole is immaterial.

We warn the reader in advance that the proof presented here depends on a clever but highly unmotivated trick. -- Howard Anton, "Elementary Linear Algebra"

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