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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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  • Radius (Score:5, Informative)

    by Moraelin (679338) on Friday November 13, 2009 @09:40AM (#30085900) Journal

    Well, the key isn't just mass, but also radius. Gravity (I'll go newtonian, just because I'm lazy) increased linearly with mass, but decreases with the square of the radius. So for example, if you packed something the mass of Earth in just half the size of Earth, the gravity on the surface would be 4 times that of Earth. Squeeze it into a quarter of the size of Earth and get 16 times the gravity on the surface. Squeeze it small enough and you have a black hole.

    If you do the proper maths, the Schwarzschild radius of a black hole with the mass of Earth is about 9mm.

    Which really means, don't think something that will suck matter and bend light spectacularly all the way to Alpha Centauri. It means that if light happens to go within 9mm of that singularity, it ain't coming out. But farther away, it's still a body with the mass of Earth. The moon's orbit will still have the same radius for example.

  • by Alain Williams (2972) <addw@phcomp.co.uk> on Friday November 13, 2009 @09:48AM (#30085972) Homepage
    To do this they need to hit each other first. Their cross section is tiny (10^-35m, size of an electron is 16^-15m), they will be moving slowly (about 11km/second if they are created with zero velocity at CERN) - so the chance of them hitting each other is small. If they came across an atom - most of that is empty space; the protons & neutrons are mostly empty space (between the quarks) to something as small as the black hole.
  • Re:Evaporate? (Score:5, Informative)

    by gclef (96311) on Friday November 13, 2009 @10:02AM (#30086100)

    Mass = Energy...it evaporates by emitting other forms of energy (light, etc).

  • by necro81 (917438) on Friday November 13, 2009 @10:07AM (#30086142) Journal
    In principle, any mass, if packed densely enough, could become a black hole. For each mass - from a cluster of atoms to an entire galaxy - there is a calculable quantity called the Schwarzschild radius [wikipedia.org]. If you could somehow pack the mass so that it fit inside a volume smaller than that mass's Schwarzschild radius, the force of gravity would invariably overcome all other forces and cause the mass to become a singularity. The Schwarzschild radius also defines the "edge" of the black hole - if anything, including light, gets closer than one Schwarzschild radius from the central mass, it will not be able to escape. In other words, at the Schwarzschild radius, the escape velocity [wikipedia.org] is the speed of light.

    It is easy to see how the core of a really big star could collapse on itself in a supernova - there's just so much mass, coupled with the force of the explosion. However, our own sun could become a black hole - if some as-yet unknown physical process could squeeze its entire mass into a 6-km diameter sphere. The Schwarzschild radius of one solar mass is about 3 km.

    It is important to note that, were this to happen tomorrow, the Earth and the other planets would continue to orbit the black hole sun exactly as they have done for billions of years. The gravity of the sun hasn't changed, because its mass hasn't changed. If you were, however, unfortunate enough to come within 3 km of the center of the black hole sun, that's the last the universe would ever see of you. (As a practical matter, you'd be doomed long before then, simply because no rocket would be powerful enough to bring you away once you got closer than a few thousand kilometers. To escape the black hole sun once you were, say, 3.1 km away, you would need to somehow achieve a speed near to the speed of light, which we simply can't do.)

    It is also important to note that you would not be sucked into a black hole if you came within 3 km of the center of the sun as it exists today, shining hot and bright. This is because 99.999% of the mass of the sun lies outside of that 3 km radius and so "doesn't count" in terms of the force of gravity. Aside from instantly transforming into plasma from the heat, you would actually feel far less gravity than you would on the Moon. (For reasons why, see here [wikipedia.org].) Remember: a black hole would exist only if you could compress the whole mass of the sun into that 3-km radius spherical volume. This can be applied to just about any mass. The Schwarzschild radius of the Earth is about 9 mm - smaller than a grape. This gives you a sense of how densely you'd have to pack things if you wanted to make an Earth-mass black hole. For a pair of protons smashed together at high energies - as in the LHC - I think you need to bring in other areas of physics than just general relativity. Suffice to say the Schwarzschild radius would be much, much, much smaller than the size of a proton, which in turn is much, much, much smaller than the size of an atom, which is much smaller than the distance between atoms in most solids. So in order for a micro-black-hole to accumulate mass, it would need to pass very close, on the order of its Schwarzschild radius, to the nucleus of another atom. At the length scales we are talking about, that's about as likely as me randomly shooting off a bb gun and hitting a passing bird a kilometer away.

    So rest easy, the world isn't about to end.

    I apologize for the long answer, but I hope it has answered your question.
  • by ChowRiit (939581) on Friday November 13, 2009 @10:07AM (#30086144)

    A black hole is any body tightly packed enough that its escape velocity is greater than the speed of light. Because material, as a result of this, can ONLY travel towards the centre of mass (outward travel, sideways travel and staying stationary are all forbidden this therefore HAS to form a singularity, as matter is all forced to head towards and occupy a single point.

    The distance from the object where the escape velocity drops below the speed of light is the event horizon (aka the Schwarzschild radius), within this sphere* no light can escape so we call this sphere the black hole. In the centre of it is the singularity, which is the "true" black-hole.

    All objects have Schwarzschild radii, however this radius is only a "real" radius if it exceeds the radius of the object. Wikipedia claims the Schwarzschild radius of the Earth is 9mm, so Earth would form a black-hole itself if it were compressed to smaller than 9mm in radius.

    The key point is that a "black-hole" is not an object, per se, but a region of space from which light cannot escape. The "object" would be the singularity in the centre. From outside the black-hole, there's no real difference from a star of the same mass in terms of gravity.

    *rotating black holes have a slightly different shape, depends on the speed of rotation.

  • Re:Evaporate? (Score:5, Informative)

    by ChowRiit (939581) on Friday November 13, 2009 @10:12AM (#30086206)

    Particles. A hand-waving description of what happens is as follows:

    Pairs of particles (one matter, one antimatter) form randomly near the event horizon. One quantum-tunnels out of the black-hole and so appears to an observer outside the black-hole to have been emitted. Therefore, to conserve energy, the other particle must have negative energy and thus the black-hole loses a tiny parcel of energy (and thus mass).

    The main point is that, because the particle was formed near the event horizon and didn't come from the black-hole itself, it carries no information out - thus, while the black-hole loses mass, no information can escape.

  • Re:Evaporate? (Score:5, Informative)

    by Maximum Prophet (716608) on Friday November 13, 2009 @10:15AM (#30086242)
    Photons pop out of the vacuum all the time. A photon and an anti-photon (or do they call it a virtual photon) will appear at the same time, and as long as the pair doesn't stick around longer than the mass * Plank's constant, conservation of mass is preserved.

    If the photon and anti-photon appear at the edge of a black hole, sometimes the photon goes off, and the anti-photon gets sucked into the black hole where it cancels some of the mass of the black hole. Thus it looks like the BH is radiating and evaporating, but nothing actual leaves the BH.

    *Note: I've left out some details, and my terminology might be off.
  • by ChowRiit (939581) on Friday November 13, 2009 @10:41AM (#30086606)

    Ah, but that's not the case.

    On small scales, that is true. However, take the moon. Electromagnetically it's neutral, however it exerts a sizeable gravitational pull. As the Schwartzchild radius is proportional to mass (not mass squared or cubed, but mass), if one took instead 8 moons and packed them together in a cuboid arrangement, the mass has increased eight-fold while the radius has only doubled. Therefore if we keep adding mass, there will come a point when the Scwarzschild radius is larger than the radius of the huge moon-array and therefore the whole moon-array has an escape velocity greater than the speed of light and is therefore a black-hole.

    Now imagine if all those moons are positively charged - it still doesn't matter, because no matter the strength of the outward force it cannot give them a velocity greater than that of the speed of light, so they remain a black-hole.

    Gravity is so significant on large scales precisely because of this - with no negative charge, gravity is the most significant force at large distance scales.

  • Let's do the maths (Score:4, Informative)

    by Moraelin (679338) on Friday November 13, 2009 @10:53AM (#30086744) Journal

    Well, yes, any matter you throw at it (and energy converts neatly to matter too) can only cause it to grow. But there's still the problem of how much and how close.

    But, really, let's do some simple maths.

    Let's say we want to produce a black hole the size of a helium atom. You know, big enough to occasionally actually bounce into stuff and gobble it up. (Remember, only matter coming closer than the Schwarzschild radius is actually gobbled up.) It's not a big black hole, but it has the potential to grow. So we apply:

    r = (2G/c^2) * m ... Where the thing in brackets is approx 1.5 * 10^-27 m/kg. We'll want to get a hole measuring 3x10^-11 m. So we'd need a mass of 2x10^15 kg, or two millions of millions of metric tons.

    Yep, that huge a mass will only gobble stuff up if it comes within 3x10^-11m of it. But it's a start, and as an evil genius you may have to start small ;)

    To produce that hole, the protons we throw at it, as a total, will have to have the equivalent of that much mass in energy.

    Let's transform that into MeV though, since we are talking energy. 1MeV is about 1.8x10^-36 Kg. Let's round to 2x10^-36, since we're only doing a back-of-the-napkin calculation, and are only interested in rough ballpark figures. So we're talking about 10^51 MeV

    If we got that energy from uranium, and assuming that we could (A) split every single U235 atom, and (B) capture 100% of the released energy, each atom split releases 180 MeV. (RL reactors don't come even close in both aspects.) Again, let's round it up to 200. (In my fantasy land, reactors are better than 100% efficient;)

    That works out to about 5*10^48 uranium atoms split. Avogadro's number being about 6x10^23, that's about 10^25 moles of uranium. (Again, I'm only interested in the order of magnitude. Plus, we rounded up in the other direction before, so it evens up.) And a mole of U235 weighs 235 grams, or about half a pound or almost a quarter kilo.

    We're talking about 2 to 3 times 10^24 kilos of uranium, or 2 to 3 times 10^21 _tons_ of U235. That's 2-3 thousand billions of billions of tons of U235. Or about a hundred thousands of billions of billions of reactor-grade enriched uranium. Completely used up in a 100% effective reactor.

    So basically yes we _could_ make a bigger black hole by keeping throwing stuff at it, close to the speed of light, but the energy requirements are nuts even to get a hole the size of a helium atom. We don't even _have_ the kind of reactors and capacitors where you could split a hundred thousands of billions of billions of reactor-grade uranium and dump it all into just creating a black hole.

  • Been Done (Score:5, Informative)

    by DynaSoar (714234) on Friday November 13, 2009 @11:01AM (#30086822) Journal

    TFA calculates the likely results based on higher dimensional brane physics. It was done earlier in more classical relativity maths and the results summarized in Alan Boyle's Cosmic Log. The max mass was greater and thus life time longer. Still, mass and accretion never crossed the limit that would allow it to reach whatever they call critical mass for these thing. The example given was that if it were charged and it were trapped within the electron cloud of an atom (both conditions lending it additional life span), it would circulate there on the order of weeks before encountering an electron which it could then consume. Even if it did so it would evaporate before it could hit the run away point, and would likely evaporate before eating even one electron. The specific results were different but the conclusion the same - too small to live long enough to do any damage.

    Another point made in Cosmic Log (I don't recall if it was the same person/calculations) was that quantum black holes (a more correct descriptor than 'mini-') of the mass and life span hypothesized would be likely to occur regularly in the atmosphere due to incoming primary cosmic rays. Those have been impacting the Earth for billions of years, and we're still here. The hypothesized Hawking radiation is not obvious, thus these may not even be occurring. In any case, their creation would be a highly improbable event.

    That last assertion is strictly conjecture based on calculations by my Brambleweeny 57 sub-meson brain. Now if you'll excuse me I'm for a nice hot cup of tea.

  • by Anonymous Coward on Friday November 13, 2009 @11:16AM (#30086998)

    We can't be exactly sure, no.

    However, even basic physics should be enough to determine that microscopic black holes aren't going to be particularly dangerous. The kind of black holes that could be created by the LHC have a very small mass - they're created by smashing a couple of subatomic particles into each other, after all. The total mass of the black hole can not possibly be higher than the total mass of the particles that created it.

    That means that the black hole will have the same gravitational force as the particles that created it. Therefore, the event horizon of the black hole will be very small. Since matter is composed mostly of empty space, the chances of it actually hitting anything are remote, to say the least. In order for it to absorb a particle, it would have to almost collide with it. This is very unlikely, although given enough time probably will happen.

    Worst case scenario - Hawking radiation doesn't exist. The micro black holes will continue to exist indefinitely, and will slowly consume the planet. Before the micro black hole has absorbed even a few kilograms of matter, the Sun will expand, swallowing the planet. The black holes will continue gradually consuming the Sun, and given a few quadrillion years or so (and the entire universe will be long dead by that point) might actually start to do some damage. By this point, I doubt that any humans will still be around to care. If we've managed to survive the destruction of our own planet, the death of our own star, and the death of the universe itself, a puny little black hole shouldn't be a problem.

    More likely scenario - Hawking radiation does exist, and the micro black holes will simply evaporate before they even come close to absorbing anything else. No big deal. If we detect evidence of Hawking radiation, that pretty much confirms the existence of black holes, and Steven Hawking gets a Nobel prize.

  • Re:Evaporate? (Score:5, Informative)

    by crazyeti (1677994) on Friday November 13, 2009 @11:56AM (#30087510)
    There's no such thing as an anti-photon. In the case you are describing - pair production - both of the particles are virtual particles. They can be an electron and a positron (anti-electron), a quark and its anti-quark, etc - any particle/antiparticle pair. However a photon is its own anti-particle. And your explanation of the uncertainly principle is wrong. The time-energy formulation says (uncertainty in time) * (uncertainty in energy) = hbar, so the time limit for the life of the virtual particles is Planck's constant / energy (or Planck's constant divided by mass, since mass and energy are proportional and we measure the mass of these particles in units of electron-volts anyhow). Note that if it's mass * hbar, as you have above, then the higher the mass is, the longer the particles can stick around! That's exactly backwards. It's the tiny little particles that are flickering in and out of existence, not huge massive objects! If it were mass*hbar, you'd have virtual planets, stars and galaxies - the larger the object the more likely it would be to suddenly appear out of nowhere! This is an amusing thought but doesn't accurately describe the reality that we find ourselves living in.
  • Re:Evaporate? (Score:3, Informative)

    by crazyeti (1677994) on Friday November 13, 2009 @12:01PM (#30087586)
    There's no such thing as an anti-photon. In the case you are describing - pair production - both of the particles are virtual particles. They can be an electron and a positron (anti-electron), a quark and its anti-quark, etc - any particle/antiparticle pair. However a photon is its own anti-particle. (See http://en.wikipedia.org/wiki/Antiparticle [wikipedia.org] ) And your explanation of the uncertainly principle is wrong. The time-energy formulation says (uncertainty in time) * (uncertainty in energy) = hbar, so the time limit for the life of the virtual particles is Planck's constant / energy (or Planck's constant divided by mass, since mass and energy are proportional and we measure the mass of these particles in units of electron-volts anyhow). Note that if it's mass * hbar, as you have above, then the higher the mass is, the longer the particles can stick around! That's exactly backwards. It's the tiny little particles that are flickering in and out of existence, not huge massive objects! If it were mass*hbar, you'd have virtual planets, stars and galaxies - the larger the object the more likely it would be to suddenly appear out of nowhere! This is an amusing thought but doesn't accurately describe the reality that we find ourselves living in.
  • by blueg3 (192743) on Friday November 13, 2009 @12:10PM (#30087688)

    This is a typical nonsense argument. You imply that because there are some things we don't know (e.g., questions to be answered by the LHC) that it's reasonably possible that we will encounter aberrant behavior that contradicts previous observation.

    There are few avenues for the MBH to be incorrect. They already assume that we are wrong about Hawking radiation (otherwise an MBH would boil off immediately). The only real options are that energy conservation is violated and the LHC is able to somehow create a heavy black hole, or the gravitational pull of a MBH is somehow enormously higher than its mass-energy would permit. (As the Schwartzchild radius is directly derived from its gravitational pull, there's not really any room for this to be wrong.)

  • by blueg3 (192743) on Friday November 13, 2009 @12:14PM (#30087720)

    You actually picked the weak form of this argument.

    Our planet is small and not particularly dense. There's only one, and something like MBH or strangelets could be fairly rare. We could be lucky.

    Fortunately, there's an enormous field of stars, including large, dense neutron stars. Neutron stars are great at capturing errant particles, producing MBHes, and things like that. Looking at our estimates of the ages of these neutron stars, you can show that micro black holes cannot be responsible for stellar/planetary destruction.

  • by Anonymous Coward on Friday November 13, 2009 @01:24PM (#30088572)

    As soon as matter is packed inside its own Schwarzchild radius it will inevitably collapse to a singularity (at least, if you believe GR is valid all the way to the singularity itself; in reality, some form of quantum corrections will come into play at some point, at which point all bets are off. But the scales that will happen on is *well* inside the event horizon even of holes the LHC would ever create.) Basically, anything that cross the event horizon is doomed to hit the singularity, no matter what, simply because *light* is doomed to hit the singularity, and nothing travels faster than light.

    So you can pack the matter over its event horizon fine, but it's not going to hang around there, it will collapse. From the outside it's a pretty academic poit, though -- the gravitational field is the same whether or not the matter is at the singularity, and what we'd *observe* (if we could see the event horizon, which we can't because no light comes from it) is actually an image of the matter just as it passed over, due to an ever-increasing gravitational time dilation as you approach the horizon.

    I hope that makes sense. If not, basically if you chuck a football at the event horizon of a black hole, you'll see it moving ever more slowly (because of a strong time dilation) and getting ever dimmer (because the light is getting redder and redder), and never quite touching the event horizon. The same applies to the matter that formed the black hole itself.

    Inside, it's less than a microsecond to slam into the singularity, of course...

  • Gravity is not weak! (Score:3, Informative)

    by Chris Burke (6130) on Friday November 13, 2009 @02:03PM (#30089178) Homepage

    Gravity is much, much weaker than the subatomic electrostatic forces that hold subatomic particles apart.

    It really isn't, not in the way that you mean. Yes the Gravitational Constant is much smaller than Coulomb's Constant, and yes the gravitational attraction between two protons is much weaker than the electrostatic repulsion between two protons.

    However as soon as you do anything more complicated than compare two charged particles, things change. The reason is because the two forces bind to different properties of matter, and while the charge property can be both positive and negative, mass is only positive. So while the gravitational force between two hydrogen atoms is very small, it is bigger than the electrostatic force between them because they are carrying no net charge.

    Thus gravity can easily be the stronger force in any given situation, because the forces of opposite charges will cancel, while their masses will only add together. Put enough mass together, and the gravitational force can easily outstrip every other force.

    In essence, what you're claiming in a black hole is a neutron star - a single massive nucleus - packed together as tightly as is physically possible for matter to be packed. This is impossible on the most basic level: the larger an atomic nucleus gets, the more unstable it is. There are no stable atomic nuclei any larger than lead-208.

    Kind of an ironic statement, since the electrostatic force is much, much weaker than the strong nuclear force which holds the protons together, and yet it is exactly because of the electrostatic force overcoming the strong force that these atoms become unstable. Because the strong force is only stronger in the same naive way in which electromagnetism is stronger than gravity.

    Also ironic because gravity overcoming electrostatic forces is also responsible for the existence of all of those large, unstable atoms in the first place. Fusing even two hydrogen atoms requires overcoming the repulsion of their nuclei when very close, and it's the intense heat and pressure in the core of a star -- caused by its immense mass -- which allows this. As the star over time fuses heavier elements the energy released decreases until lead where it crosses over into negative. At this point all the fusion energy that was holding the mass of the star up fails, and all that mass in the outer portions of the star collapses in due to gravity, and that transfer of energy fuses atoms much, much heavier than lead and leads to all the unstable elements we find on earth plus many that don't last long enough to become part of a planet.

    Gravity, the "weakest" force, creates atoms which the strong interaction, the "strongest" force, cannot hold together!

    So, obviously the situation is more complex than just making a blanket statement that one force is stronger than the other.

  • by Anonymous Coward on Friday November 13, 2009 @02:06PM (#30089210)

    Oh, you mean like the possibility of the earths atmosphere igniting and killing all life on earth? This was one possibility of the first atomic bomb test. Still did it anyway. Nothing happened.

  • by ducomputergeek (595742) on Friday November 13, 2009 @02:38PM (#30089784)

    I've seen this one, all you have to do is reverse the polarity......right? Right?

  • by Anonymous Coward on Friday November 13, 2009 @03:27PM (#30090550)

    The sun won't ever go black hole because there's not enough mass to overcome the electron repulsion between iron atoms.

    The sun isn't big enough to become a supernova either, since the remains of the star AFTER burning all the fusion products would have to be heavier than the sun is NOW (and it will lose mass as it reaches red giant stage).

    And since the earth would orbit further away if the sun were lighter but the total energy (gravitational potential + kinetic) were the same for the earth, there's a good chance the earth would spiral out as the sun loses mass into its red giant phase and not get burned inside the sun's larger atmosphere, even though that would extend beyond the current orbit of the earth.

    And there's no such thing as "drag" as you describe it. Photon pressure is about 1.6 pounds per square meter at 1 AU. Total force on the earth would be ~2 million pound force. Since the earth weighs 10^21lbs, the acceleration would naff all.

    So, no, an object put at earth's position with a sun that was dead (solid iron is the only option, but even if it were a black hole, the idea is the same) it wouldn't collapse in because there's no decay in the orbit.

    You've watched Disney's "The Black Hole" and thought it was a documentary, I think.

    There's no such theory as one your message proposes someone else to have thought up.

  • by ChowRiit (939581) on Friday November 13, 2009 @03:37PM (#30090690)

    Short answer: because the escape velocity is greater than the speed of light inside the Schwartzschild radius, matter cannot travel outwards, but also cannot stay stationary (to prove this properly you need some fairly complex general relativity). Because of this it can only head towards the centre of mass, so the matter all converges on a single point. As the escape velocity only increases as one gets closer to the centre, this forces all the matter into a single point of spacetime, the singularity.

    This is, at least, what general relativity tells us. However, it's impossible to know for sure, as one of the properties of black-holes is that it is considered impossible for information to escape the event horizon.

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