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Death of a black hole

  1. Dec 6, 2008 #1
    I've heard that black holes dissapear when they are done. Doesn't it make sense to say it would end in a form like a neutron star?
     
  2. jcsd
  3. Dec 6, 2008 #2

    mgb_phys

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    Small blackholes can evaporate to nothing by a process called hawking radiation.

    Large black holes are pretty much permanent.
     
  4. Dec 8, 2008 #3
    I don't think they would end up like a neutron star. Their enormous mass collapsed beyond that point and they haven't reached a singularity yet because of time dilation. They are frozen in time relative to us and time is the key here, they're done - gone, never to come back.
     
  5. Dec 8, 2008 #4
    So it is accepted now that large black holes will last forever?
     
  6. Dec 8, 2008 #5

    mgb_phys

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    Forever is a bit tricky - it depends on your end of universe model - but basically yes, there is no obvious way to get rid of a large blackhole.
     
  7. Dec 22, 2008 #6
    Hello!

    What do you mean when you say the black hole 'evaporates'. Does that mean...it breaks up...or does that mean it turns into 'nothing'. Is there something that can go from something to nothing?
     
  8. Dec 22, 2008 #7

    George Jones

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    Forever is a long time. Current observation and theoretical evidence indicate that the universe will expand forever. If this is true, and if the ideas behind Hawking radiation are true, then all black holes, no matter how large, eventually will evaporate. New observational or theoretical developments could change our expectations, though.

    John Baez has written an interesting essay on the development of the universe far into the future, and in it he talks about black holes.

    http://math.ucr.edu/home/baez/end.html
    Through a quantum process, black holes radiate particles (and anti-particles), and thus lose mass. As they lose mass, that rate at which they radiate increases.

    Hawking stunned the physics community by showing that the tidal force near a black hole can tear the quantum vacuum to shreds, producing pairs of particles in which one member of each pair has positive energy while the other has negative energy. If the negative energy particle falls into the black hole, the positive energy particle can sometimes escape. The negative energy particle lowers the energy of the black hole, thus decreasing its mass, while the positive energy particle appears as radiation, called Hawking radiation, from the black hole. As a result, the black hole steadily radiates its mass away.

    For more details, see

    http://www.physics.ucdavis.edu/Text/Carlip.html#Hawkrad.

    I wrote a treatment at about the second-year physics level,

    https://www.physicsforums.com/showthread.php?t=205711.
     
  9. Dec 22, 2008 #8
    Thank you! :smile:

    If the black hole looses mass by radiation, it must loose its mass at a slower rate than it is gaining mass right? At least in order for the black hole to grow.

    Anything that goes past event horizon would contribute to the mass, while the radiation takes some of the mass away. Correct?

    What kind of factors would determine whether the rate of evaporation is higher than the rate of mass growth?

    And would the evaporation be what they call high energy cosmic rays? The kind that does not come from the sun of course.

    I did my decree 14 years ago...physics was not my major..and astronomy wasnt in my course. Sorry if i seem a bit 'out of it'.:shy:

    If matter goes into the hole and radiation comes out in the form of particles, then do those particles have a speed higher than that of light? I figured such a thing cant be found.

    Heneni
     
  10. Dec 22, 2008 #9

    mgb_phys

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    Much slower, the rate of mass loss depends on the mass. Larger holes lose at a slower rate.
    The rate of mass increase of a black hole depends on what material is available around it.
    There is also the 'temperature' of the black hole, solar mass black holes have a 'temperature' lower than the microwave background and so can continually absorb CMB photons.

    The time to totally evaporate a black hole is (according to wiki)
    t = 5100 pi G^2 M^3 / h c^4 (seconds)

    I don't know what energy they would have, or even if they have a particular energy.
    George Jones is an expert on this - I'm not.

    The particles don't come out of the black hole. They are created from the energy near the event horizon. One of the pair goes into the blackhole - one escapes. The escaping one effectively removes some energy.
     
  11. Dec 22, 2008 #10

    George Jones

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    The expansion of the universe plays a large role. If the universe continues to expand, then stuff (including the cosmic microwave background form shortly after the big bang) "thins out" so much that evaporation probably wins. From the John Baez essay to which I gave a line in my previous post:
     
  12. Dec 22, 2008 #11

    Janus

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    The More massive the black hole, The lower its rate of evaporation. The radiation it loses gives the Black hole what is called a Hawking temperature. You can determine the HAwking temperature from the black hole mass. Even a Black hole far removed from other sources of radiation would subject to the cosmic Background radiation at 3°k. If the Hawking temp of the black hole is less than that of the CMB, then the black hole can not shrink, As it is taking in more radiation then it loses it.

    Now, as pointed out by George Jones, if the Universe continues to expand, it will continue to cool and the temp of the CMB will drop. So every black hole would eventually become "warmer" than the CMB and begin to shrink through evaporation.
     
  13. Dec 22, 2008 #12
    Thank you all kindly! I will look more intelligent at parties after this! HE HE....

    So energy at the (edge of the???) horison creates this pair of particles, some of the energy is lost as one pair of the particles moves away from the horizon. Ok...so according to E = mc2 if the black hole looses energy by the escaping particle then it will loose mass.

    Its just...im trying to wrap my head around a particle that is created from the energy of the blackhole horizon, yet being able to escape it, carrying with it some energy, and therefore some mass.

    Did i miss something?
     
  14. Dec 22, 2008 #13

    mgb_phys

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    Pretty much yes.

    The event horizon is the distance from a black hole where you cannot escape if you get closer than this. In classical terms it's the distance at which the escape velocity is > the speed of light.
    In reality it's a little more complicated, if the particle pair were simply created from energy falling into the black hole then all it would mean is that the black hole only grew by half as much.
    But in quantum mechanics the idea of exact position and possible/impossible is all a bit fuzzy so some energy that is already 'just' inside the black hole can form a pair of particles and one of them can escape - even though classically this would be impossible.
     
  15. Dec 22, 2008 #14
    Not exactly, I believe that the 'Big Rip' scenario is the most realistic one.
     
  16. Dec 22, 2008 #15

    George Jones

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    I don't think that many people believe that [itex]p/\rho = w < -1[/itex], as required for a Big Rip.
     
  17. Dec 22, 2008 #16
    Hi, I tried to find a reference to a very interesting article I saw about 2 days ago about our brane changing the signature from Lorentz to Euclidean geometry. To the internal observers this process manifests as dark energy, leading to the singularity in a finite time, then :) time dissapears (becomes space like)

    So DE could be not about the gravity at all, but about the topology of our brane.
     
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