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Question about relativistic quantum hawking radiation.

  1. Dec 16, 2013 #1
    So as I understand it Stephen Hawking and one other person I can't remember the name of worked on an equation to described the "temperature" of a black hole, that you can actually get information out of it. I know that there is one way it could happen with those vacuum particles in vacuum energy that for whatever reason appear out the the nothingness of space and then annihilate itself and near a black hole one get's sucked in while the other doesn't. But, I was wondering I guess more about the other type of particle pair it describes, photons.
    As I understand it, a lower frequency photon will be more delocalized, so it would make sense that photons of extremely low frequency would have an uncertainty in position that can exceed the boundaries of a black hole in purely 3 dimensional space. But, when this happens, apparently one photon somehow suddenly becomes two photons. A photon fluctuates against the event horizon from the interior of the black hole and somehow get's separated into two photons, similar to a quantum tunneling effect. But, how does a photon actually "reach" the event horizon from the interior of a black hole when there's length contraction past a zero metric and how do photons of that low of energy exist in a black hole when it has huge concentrations of energy compacted into a small dense space and should be very very very hot?
     
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  3. Dec 16, 2013 #2

    phinds

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    Although I can't give you a specific citation (not because there isn't one but just because I don't have a link handy), Hawking himself said that the "virtual particle" explanation was just a sort of layman's description to what really happens and it isn't REALLY what happens, it's just that English does not lend itself to a simple description of what the math really says.

    SO ... don't get wrapped up in the virtual particle explanation of Hawking Radiation.

    Towards the end of your post you talk about a particle reaching the EH from inside the BH. that can't happen. The virtual particle explanation is based on particles OUTSIDE the EH, not inside.
     
  4. Dec 16, 2013 #3
    Hmm so how is it that black holes lose energy and evaporate then? When I talked to a different physicist about it they seemed to suggest a tunneling effect from the interior that allows a black hole to lose energy, like when you shoot a particle at a wall and very small probability passes through the other side.
    I guess the problem comes from the fact that photons aren't actually "traveling" backwards to get out of black hole, they are merely following the correlations that describe their probability. I don't get how it happens when space is that contracted though, it seems like a photon would have to have a wavelength of hundreds of miles to do it, which I guess would explain why it would take millions of years for a black hole to evaporate.
     
  5. Dec 16, 2013 #4

    phinds

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    No, the virtual particle explanation says that a pair forms just outside the BH and one leaves the vicinity and the other falls into the BH and somehow (and this is where I really don't get it) the one that falls in always has negative energy and thus reduces the mass of the BH.
     
  6. Dec 16, 2013 #5
    But negative energy as far as I know only hypothetically occurs in rotating black holes within the Ergo-sphere, otherwise there's really no reason to assume "negative" energy exists. Also, where does the energy go? If I combine matter and anti matter, the net matter is 0, but I still have "stuff" left over in the form of photons. Are you suggesting that somehow that negative energy just somehow causes non-existence?
    But I also know that Hawking didn't like the notion that you can't extract information from the interior of a black hole and so it would make sense that he came up with a way to do it.
     
  7. Dec 16, 2013 #6
    None of those asserations represent current theory.

    Kip Thorne, BLACK HOLES AND TIME WARPS, PGS 435-440
    Wikipedia describes it this way:

    Emission process
    Quoting from "Quantum Fields in Curved Space" by Birrell and Davies,

     
  8. Dec 16, 2013 #7
    But that seems to suggest that the uncertainty of the thermal energy of a black hole exceeds the boundary of the event horizon...which still suggests particles aren't actually "traveling", just "appearing" as it says.
    I'm really asking more about the transition from finite length contraction to whatever is on the other side.
     
  9. Dec 16, 2013 #8
    Here is another view...I was looking for Leonard Susskind's decription but this is virtually identical.

    from a prior discussion in these forums, Mitchell Porter posts:

    edit: Keep in mind these various descriptions are in part coordinate based descriptions based on idealized spacetimes....static spacetimes, views from infinity, etc, etc. There is certainly some ambigutiy about what the math means 'physically'....I have seen Penrose statements say 'there is no mass inside a BH' which makes sense because the simple solution is a vacuum solution of the Einstein equations.....elsewhere, others say nobody knows for sure....Is there pressure, vacuum energy.....it seems the event horizon is 'coordinate independent' and therefore 'physical' but just how idealized is that??

    I think everybody agrees so far BH seem like the ultimate roach motel: you can get in but you can't get out.

    You can read more under "Holographic Principle' of t'hooft and Susskind......say, in wikipedia...
     
    Last edited: Dec 16, 2013
  10. Dec 16, 2013 #9
    Yes I've come cross that so called "halogram" theory, but from my knowledge it doesn't make sense. One is that it seems based on an old view of falling towards a black-hole from Shwarzild's equations, which I've actually read in some books that says "something falling towards a blck hole appears completely stopped at the event horizon". If this was physically true, a black hole would never gain mass. What actually happens is someone later came up with better equations and it was shown that an observer would keep traveling normally from their frame of reference no problem right past the event horizon supposedly without really noticing it, assuming they aren't torn apart, meanwhile the outside world sees you traveling velocity but becoming indefinitely red-shifted, eventually turning black or invisible as you cross the event horizon while your clock comes to a halt. Although one thing I don't known is how to balance tidal forces and metric contraction. Near the event horizon, you should appear squished, but as you approach the black hole, tidal forces rip you apart. So do you actually stretch out or appear indefinitely squished as you cross the event horizon?
    And then, the event horizon of a black hole grows. I a black hole gains mass, it's even horizon won't grow uniformally, it's event horizon will sort of "stretch" and form a bulge and gradually form a perfect hairy-sphere again, and so the spontaneous-uniform growth of the event horizon would have to mess that up.
     
  11. Dec 16, 2013 #10
    A virtual particle energy and momentum is not required to follow the usual energy-momentum relationship: [itex]E^2=(cp)^2+(mc^2)^2.[/itex] That's means that a virtual particle can have negative energy.
     
  12. Dec 16, 2013 #11
    The one that falls must have negative energy because the one that leaves has positive energy and energy is conserved. The one that leaves must have positive energy because it is a real particle and only virtual particles are allowed to have negative energy.
     
  13. Dec 16, 2013 #12
    And you can also have imaginary mass too. I also don't understand where they come from in the first place if not from the uncertainty in thermal energy.

    But didn't both particles start out as "virtual"? And why is it only the one with negative energy that leaves? And doesn't "virtual" only describe a particle's limitation in space caused by it's own limitations in uncertainty from relative mass like with strong and weak force bosons?
     
  14. Dec 16, 2013 #13

    Nugatory

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    http://www.itp.uni-hannover.de/~giulini/papers/BlackHoleSeminar/Hawking_CMP_1975.pdf is fairly heavy going, but it's also a hint as to just how oversimplified the popular virtual particle pairs explanation of what's going at the event horizon is.
     
  15. Dec 16, 2013 #14
    The spaghettification (That's the technical term for the stretching you're describing) happens near the singularity. If the blackhole is large enough, it is safe to cross the horizon which lies away from the singularity itself.
     
  16. Dec 16, 2013 #15

    phinds

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    OK, but if the one that leaves becomes real, why doesn't the one that falls in?
     
  17. Dec 16, 2013 #16
    A virtual particle is one that doesn't obey the equation I quoted. In empty space away from the B-hole it is impossible to to produce real particles out of the vacuum because there is no way to produce particles that will both obey conservation of energy (even virtual particles do) and the dispersion relation for energy and momentum. The black hole makes it possible.
     
  18. Dec 16, 2013 #17
    Because it has negative energy. a real particle must have positive energy that obeys the dispersion relation equation. virtual particle are free from those requirements but are shortlived.
     
  19. Dec 16, 2013 #18
    Ok so I think I better understand the situation with virtual particles, but why do those virtual particles appear in the first place? And why "can't" the situation I described take place anyway? As long as a photon isn't actually "traveling", it should be fine, just as an electron isn't actually "traveling" when it changes orbitals at technically infinite speed.
     
  20. Dec 16, 2013 #19
    They appear all the time everywhere continuously.
     
  21. Dec 16, 2013 #20
    Yeah and they annihilate themselves upholding the conservation outside of a black hole right? So...why do they appear in the first place? And how exactly do we measure the difference in vacuum "pressure" in the casimir effect if the virtual pairs annihilate themselves?
     
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