Hawkings Theory

Discussion in 'Astronomy, Exobiology, & Cosmology' started by thedemon13666, May 24, 2006.

  1. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    to Blobrama:

    What you called “wrong” was my statement, in post 37 taken out of context by omitting the preceding part I made bold below:

    “…do not confuse the "virtual particles" that modern physics imagines are the solution to the "action at a distance" problem - I.e. the particles that are exchanged between the sources of various forces with the real, but briefly existing particle of the vacuum polarization. Those polarization pair particles normally mutually annihilate within the time allowed by the uncertainty principle for them to exist. The only violation of this is when one of the pair falls inside the EH of a BH and leaves the other as a long lived real electron or positron in our universe. I.e. it is not a "virtual particle" but a real one …”

    Your post 40, responding to my request “What is wrong?” explains my “error” was to state that Hawking Radiation was associated with the capture of only one of the vacuum polarization pair. I.e. your post 40 rely was:

    “…The process is not a electron / positron etc type process.
    Hawkings radiation is essentially from a photon-antiphoton nonannihilate effect
    . …”

    But I have been very clear in several posts that neither the photon-antiphoton nonannihilate (which you seem to prefer) nor the vacuum polarization pair partial annihilation is correct. They are only verbal attempts to give humans, like me and you, who can not follow the GR +QM math a “warm fuzzy felling” as if we did understand what was going on.

    I was only commenting on your confusion about the role and nature of virtual particles (bold above, which you omitted) NOT stating that the vacuum polarization was correct POV, as you assumed I was and then called me “wrong.”

    My post 32 clearly shows that I consider both your preferred “virtual photons” & the “vacuum pair” to be inadequate attempts at VERBAL DESCRIPTIONS, not the correct GR+QM math. In post 32 I state:
    “…Some word descriptions imagine that photons ("thermal radiation”) remove mass for the BH.
    Others imagine that the vacuum polarization (usually electron/position appearing from nothing but for such short time that this is allowed by the QM uncertainty product of Delta E times Delta T). Then if these oppositely charged particles happen to be very close to the EH, one may be captured by the intense gravity gradient near the EH and the other not so it is new energy of long duration in our universe, and that energy must (not well explained in words how) come from the mass of the BH. …

    If I were to try to summarize in words, I would say that the prediction of Hawking's Radiation is "permitted" by the QM uncertainty but not caused by QM. It is caused by the (math as done by Hawking

    SUMMARY: I was not "wrong." - You just incorrectly thought I was supporting the vacuum polarization pair POV. I do not. Neither it nor your photon-antiphoton nonannihilate effect is correct. They are just attempts in words to provide a "warm fuzzy feeling."
    Last edited by a moderator: May 17, 2008
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  3. blobrana Registered Senior Member

    tnx for the clarification,
    i of course was just pointing out the example you gave was wrong. Not that you agreed, or disagreed, with the vacuum polarisation pair POV.
    (i should add that the example was `wrong` because it is probably a simplistic and incomplete solution.)

    As i said, i`m not too clear about the difference.
    I will leave this for others to philosophise about.
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  5. blobrana Registered Senior Member

    But, how is `my` photon-antiphoton nonannihilate effect incorrect? *

    Yes, i can accept that the process can be viewed as just a mathematical trick; but, can you clarify what you mean.
    My view is that the virtual photon-antiphoton model is the standard explanation for Hawking radiation and blackhole evaporation. They were not meant to be "warm fuzzy feeling" words. The physics and calculations behind it are quite simple, (although there have been a few key and quite complex elements that i haven't bothered to mention in this thread for simplicity, but anyone who has studied BHs or looked into the mathematics would already know them).

    *The photons are of course nondestructively interfering.
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  7. 2inquisitive The Devil is in the details Registered Senior Member

    blobrana, QM states all particles in the universe are created in the highly curved spacetime near the event horizon of a black hole. You seem to think photons carry away mass from the black hole. Do you think photons have rest mass energy, or just momentum and kinetic energy? Kinetic energy does not contribute to rest mass energy and does not affect gravity. Rest mass must be lost from the black hole if the singularity is to reduce in mass and thus gravity. In other words, the electrons, positrons, and other massive particles that are theorized to be created are the mechanism by which rest mass is predicted to be lost by Hawking radiation. Photons are mainly a visual method of determining if a black hole is emitting Hawking radiation from a distant location. IF a black hole is created in a collider and evaporates by Hawking radiation, we will measure particles of all descriptions emitted during the evaporation, not just photons.
  8. blobrana Registered Senior Member


    No. Photons can remove mass (*) from a BH. (The devil is in the semantics)

    As i said before photons are their own antiparticle. Why that gem of information is significant i`ll let others discover for themselves.

    Yes. And we could even make predictions on the rate and ratios of those particles. Observation of LHC blackholes would be useful in refining the importance of the other similar quantum mechanisms and constrain a few unknowns.

    (*) Mass energy
  9. 2inquisitive The Devil is in the details Registered Senior Member

    Explain how, please. The 'm' in E=mc2 is rest mass energy. A photon has no rest mass energy. Do you think an object bombarded with photons, the Earth for example, will gain mass from the photons? Do photons transfer mass from the sun to the Earth? Or do they just transfer energy that has no mass energy from the sun to the Earth?
  10. blobrana Registered Senior Member

    One mechanism to remove mass from a black hole is to use a quantum mechanism to generate* a photon/antiphoton pair near a BH event horizon.
    An anti-photon is identical to a photon, but moving backwards in time, and carrying negative mass energy, (You can regard this as just a mathematical trick, `virtual` or `real`, it doesn't matter).
    Normally, the two particles would completely interfere destructively with each other leaving `nothing`; but because one gets swallowed and one escapes from around a BH, we see a photon, (Of course we don't / can't see a virtual photon**).

    And the answer to your question: “Do you think an object bombarded with photons, the Earth for example, will gain mass from the photons?” is No.

    * ( http://www.journaloftheoretics.com/articles/6-6/str.aqa.pdf )
    ** (A virtual photon changes into a `real` photon because the original pair were entangled)
  11. Vkothii Banned Banned

    Hang about.
    I thought photons were their own antiparticle?
    How established is the "antiphoton", except for some mathematical idea or other?
  12. 2inquisitive The Devil is in the details Registered Senior Member

    Notice the link theorizes a photon with negative energy, not negative mass energy. I do not dispute photons have energy and momentum, just mass energy. The theorized negative momentum is sometimes proposed as an mechanism by which rotating Kerr black holes may lose angular momentum over time to eventually transform into non-rotating Schwarzchild black holes. The argument for mass loss by Hawking radiation depends on the conservation of mass-energy. That is why negative mass particles are proposed as a mechanism, to balance the mass-energy scales. The excaping virtual particle must gain mass to become a real particle. The mass is gained from the black hole. If a virtual photon becomes a real photon, no mass is gained by the photon, only energy and momentum. It would be a violation of the conservation of mass-energy if mass were lost by the black hole, but none were gained by the escaping particle.
  13. blobrana Registered Senior Member

    i didn`t see that bit.


    Perhaps. But perhaps it already has positive mass energy. There are two main models for this mechanism - one is that it is a consequence of the spacetime topology around the BH, ie this causes the negative mass energy photons to always fall into the BH; Or that it is a result of a wave function collapse of the virtual entangled pair, ie we will always observe positive mass energy photons on the outside of the BH.

    It is an indirect process that occurs on the outside of the BH. However, a tiny bit of mass is cancelled out in the blackhole.
    We must remember that a the virtual photon/antiphoton pair add up to zero mass.

    In a nutshell.
  14. blobrana Registered Senior Member

    Lets just say `No one has observed Hawking radiation`.
  15. Farsight

    A photon is its own antiparticle because you could combine two out-of-phase photons to be left with nothing. It's just like combining two ordinary waves:


    ..the peak of one coincides with the trough of the other. Note that both these photons have energy, but neither of them have any mass. Also note that negative mass is a nonsense that would involve runaway repulsion. Talking of runaways....

    A photon has no mass, but when a black hole swallows a photon it gains mass because of mass/energy equivalence. If a pair of "virtual" photons are created near a micro black hole, if the black hole swallowed just one of them, the other would comprise what we call Hawking Radiation. But you know what? Most of the time it's going to swallow them both. Then the micro black hole would gain some mass while the universe would gain some gravity, so all's square. This could happen again and again. The black hole never has to repay the energy debt, because there is no debt. This kiddie is his own banker, and he keeps his own time. This means the micro black hole would grow due to QM effects, spitting out Hawking Radiation, but also increasing its mass and gravity, creating both matter/energy stress and the balancing gravitational "tension" out of nothing. That’s a runaway. Sounds familiar doesn't it? It should. Neil Cornish, an astrophysicist at Montana State University, said this:

    Indeed, the fluctuations we see in the CMB are thought to be generated by a process that is closely analogous to Hawking radiation from black holes..

    See Universe Measured: We're 156 Billion Light-years Wide! on http://www.space.com/scienceastronomy/mystery_monday_040524.html.
    Last edited: May 18, 2008
  16. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    BHs radiate as a black body. - If it is so simple, please show me how to derive from the physics (not some result lifted from wiki text, etc.) how the temperature of the BH relates to it mass.
  17. blobrana Registered Senior Member

    The temperature is related to (Planck's constant x c^3) / (Boltzmann constant x G x Mass)
    A quick google search on BH black body radiation or the Stefan-Boltzmann formula will give you the proper formula.

    It basically shows that as the BH loses mass the temperature goes up.
    (ie Brightness = surface area x Temperature^4 x constant)
  18. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    I already knew that too. You said the physics was easy, so I asked to DERIVE these results, specifically telling you not to quote them from wiki.

    As you did not show how these results follow from the physics I assume you cannot and you now admit that your statement about it being simple physic was wrong / in error.

    I cannot derive the temperature of the black hole either. Not sure I can still derive Planck's equation for the form of Black Body radiation anymore - It has been a long time. There are at least two different approaches. The one I once could do is mainly based on statical mechanics. It is much easier, almost classical physics with no GR requirements. Can you at least do this much easier physics, or do you admit that even that is too difficult for you?

    My point is that you should not say the physic is easy if it is far beyond your capabilities. Many great minds struggled with the conflict between the classical law of equal partition of energy and the observed distribution of radiation until Planck resolved this problem. I doubt if you even understand that there was a problem, much less how to solve it. Do not belittle the accomplishments of Planck and Hawking by stating that the "physics is easy" when you do not even understand the problem.
  19. blobrana Registered Senior Member

    The actual formula is T = (h x c^3)/(16pi x Boltzmann constant x G x M)

    Sry i cant be bothered, (it is a lot of work).
    Like i said a quick google search will give you the proper formula (and how it is derived).
    If anyone else is interested on a weblink with an easily understandable step by step formulas and calculation for Hawking radiation just give me a PM.

    Why the hostility?
    I was not belittling their accomplishments.
    By not reinventing the wheel the physics is easily understandable.
  20. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    Interesting that you have time to reply to many via PMs but not time to post derivations or even link to it only once here. Could it be that you are just slinging BS?
  21. blobrana Registered Senior Member

    fair enough, you want to question my understanding of Hawking radiation.
    But why the hostility?
    Try to be a bit more civil.

    All I'm saying is that not too many people here are interested that
    T = {(2 x h x f)/c^2 }x {(c^5/(32pi x G x M x f)}

    and that the formulas are freely available on the net.

    If you do not agree with my point of view then fine. It is no big deal.
    If you think i know nothing then just add me to your ignore list.
    i don't post here to sell a book or gain favour, or to look for a fight, nor aim to do other peoples homework.
  22. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    I am quite civil in my reaction to your post 43 statement below, that belittles the difficulty of what both Planck and Hawkings did.
    Frankly, it is quite evident that you do not even understand the problems they solved and that physic is far from "simple."

    Your "I have not bothered ..." implies you could. That is I am sure is a gross exaggeration. Once I could derive the Planck equation, after a high-level (Graduate School) course in statical mechanics, given at JHU by Franco Rasetti, who work in partnership with Fermi, back in their Italian lab but I could not even follow the GR equations of Hawking, which if I recall correctly were published years ago in Phys. Rev. Section D. (I doubt you have ever even tried to read equations on that level.) It arrogance to suggest you just “don't bother.”

    I do not like unfounded arrogance, and treat it with the respect it deserves.
    Last edited by a moderator: May 19, 2008
  23. blobrana Registered Senior Member

    ok, i`ll change my mind,
    The physic is far from "simple."
    Hawkings photon-antiphoton nonannihilate effect is quite difficult to understand.

    And i`ll state again that it was not my intention to belittle the difficulty of what both Planck and Hawkings did.

    Now can we move on?

    T = m x {(c^5)/(32pi x G x M x f)}

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