Can black holes grow on their own?

Discussion in 'Astronomy, Exobiology, & Cosmology' started by kaneda, Sep 3, 2009.

  1. prometheus viva voce! Registered Senior Member

    Firstly, let me just say that the virtual particle interpretation of black hole radiation is really the pop science method of understanding what is happening. The QFT calculation that you can do is pretty enigmatic about where the particles actually come from.

    Secondly, Walter is nitpicking again with points that are of very minor relevance to the overall concept. If I were you I'd simply ignore him - you won't miss much in the way of particle physics.

    The Dirac sea is a (not very good) way of thinking about antimatter, not virtual particles, if it were true the vacuum would have a charge. Not only would there be a charge but it would be infinite in magnitude. The Dirac sea really arises from treating the Dirac equation as an equation for a quantum particle which is not really what it is - it's the equation for a quantum field.

    In quantum field theory you can think of virtual particles like this. Suppose you have a loop diagram, for example

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    You can measure the energy of each of the external legs so that is constrained. However, you can't measure the energy around the loop (the middle electron and the photon). To calculate the diagram you integrate over all possible energies going around the loop so the diagram doesn't represent a single event but a class of events that have that loop structure (that maybe obvious to you, but it took me a little while to get it). By all possible energies I really mean all possible energies, so virtual particles don't have to obey the equation of motion - the lingo is that they are "off shell."

    To conclude, we don't really think about virtual particles as single particles - they come out of QFT in sets that have average properties.
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  3. kurros Registered Senior Member

    Hmm yes, ok so perhaps I should be avoiding thinking about them as single particles. What about this negative energy stuff though? For instance, the Parikh and Wilczek paper says:
    "The idea is that the energy of a particle changes sign as it crosses the horizon, so that a pair
    created just inside or just outside the horizon can materialize with zero total energy, after one member of the pair has tunneled to the opposite side."
    Perhaps they explain it more thoroughly somewhere, but my quick read of it didn't reveal an explanation. They certainly treat anti-particles as negative frequency states travelling backwards in time but I couldn't see exactly how this related to the particles "materialising with zero net energy".
    I'll try go through the math properly, I suppose it is in there somewhere.
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  5. AlphaNumeric Fully ionized Registered Senior Member

    Kurros, you might find the section on black hole thermodynamics here of interest. The Hawking temperature is derived in two ways, by doing transformations on Fourier modes of quantum fields or taking a statistical mechanics approach involving Greens functions. They are lecture notes for 4th years so perhaps a little more accessible than working from published papers.
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  7. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    I agree completely with this and think Dirac would have too.
    There are two classes of solutions to the equation, but only the positive energy one is observed. Dirac was just constructing a rationalization telling why the negative energy solution was not. He did not really think the negative sea of electrons existed. Unfortunately I no longer have my copy of his book (Called: quantum physics or theory) but my comments are based on my memory of it. Unless my memory is confused, in some ways Dirac was the epitome of a theoretical physicist. - More concerned with the math than reality. - I think it was Dirac who once wrote a requisition (or in some way asked for) a gram of neutrons.

    Dirac also I am sure offered an explanation as to why all electrons are identical: They are all the same one. I.e. due to the invariance of solutions if both time and charge are reversed, a positron can be considered to be an electron traveling from the future towards the past. Perhaps that electron was scattered in space time in the future and we see it now. Likewise in the past an electron could be scattered forward in time and be in our present to become that electron which in the future will be the one that scattered back to our time to be the positron we now see in the present. I.e. the reason Dirac suggested why all electrons and positrons are identical is that there really is only one which is many times bounced back and forth from past to future, to past, to future, zillions of times and seen zillions of times in our present in different locations.

    Dirac was happy to suggest things like this and the negative electron sea as they answered questions his math put to him. That does not mean Dirac was betting his life on them being real. - As I said, Dirac was the epitome of a theoretical physicist - not very concern about any things that happened in the lab. It was the math than was important for Dirac.

    PS to Walter - I had not notice that you were also correct the absence of a decimal point (I am sure that was just a typo - I am slightly dyslexic so read it correctly even though point was not there.)

    If the negative sea of electrons were real then one can make an electron positron pair with much less energy than the rest mass (1.022MeV) of the pair - just excite one* of the ALREADY EXISTING electrons from a negative energy state in the sea up to a unoccupied positive energy state - now it is an obsservabel electron (real) and the hole in the sea is an observable positron. - that was why I took exception to your statement that 1.022 MeV was required to remove an electron from the negative energy sea of them.

    *One less than 1.022MeV deep in the negative sea.
    Last edited by a moderator: Sep 7, 2009
  8. Walter L. Wagner Cosmic Truth Seeker Valued Senior Member

    Now I understand where you're coming from. Particles are animate. You refer to them as such (..."a thermal spectrum who's temperature ..."). And, as usual, your grammar is off. ( who is = who's, rather than the correct grammar of 'whose').

    It would be nice also if you were correct in your assertions about what an observer sees. Not a single observer has ever seen black hole 'Hawking radiation'. Not one. This is a theoretical postulation which you believe to be true - not something that an observer sees, or has seen.

    Conversely, we do know that black holes are real. We observe their effect at the center of our galaxy, for example, where the stars are in orbit (based on their red-shift/blue-shift) about a large mass we can't see - i.e. a black hole.

    I like Ben's comment about what happens if the OTHER virtual particle falls in - then the mass would increase. Again this would be theoretical. - this is my take on Ben's comment about a particle tunneling out of the black hole.
    Last edited: Sep 7, 2009
  9. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    Let's stick to physics, not grammar.

    I have many times and years ago (even before Ben was here) pointed out that Hawkin radiation is ONLY a math prediction that few can even follow, and noted that there are two, mutually contradictory, "dumbed down" attempts to make it plausible to us physicists who lack the necessary math ability to really understand the prediction.

    Namely in one "dumbed down" version, one and only one*, of the vacuum polarization pair (usually electron / positron) falls inside, leaving a stable mass energy in our observable universe and this mass energy is removed from the BH to conserve energy. (How it gets out is never explained.)

    In the other "dumbed down" version the fact that a temperature of the EH surface exists in the theory is used. This temperature is a rapidly increasing inverse function of the BH mass. Since the BH's EH is a perfect absorber, it radiates as a black body. IR mainly for massive BHs but even gamma rays when the HB is very small.

    The "mutual condradiction" in the"dumbed down" versions is that one has EM wave energy added to our universe as the BH shrinks and the other has new mass energy added to our observable universe. - This is a clear indication at neither is correct. I think a good example that the following rule has been violated:

    Everything should be explained as simply as possible, but NOT MORE SO.
    *If both fall in, which surely is the more common case than only one, the mass of the BH would NOT increase. While it gains 1.022MeV of mass energy it must pay the -1.022Mev debt that exists in the brief period allowed by the uncertainity principle (delta E times delta T). The total energy of the space volume that made the pair was zero (except for the "zero point energy" if that volume is considered to be a box with rigid walls) and it still is zero when the pair briefly exists. When far from the EH a vacuum pair "pops into existence" and soon mutually annhilates, there are not two 0.511MeV gamma rays produced (thank God or we would not live). Their energy is used to pay the debt of 1.022 MeV.
    Last edited by a moderator: Sep 7, 2009
  10. AlphaNumeric Fully ionized Registered Senior Member

    I'd rather my grammar be off than my physics being off.

    Managed to win the lottery yet Walter, there's a 50/50 change you will win remember

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    That isn't true, the effect of the gravitational field on the particles is such that the infalling particle takes with it an energy 'debt' which is to pay for the energy (and charge and momentum etc) which the emitted particle is carrying. While many of the processes involved are symmetric between charges, momenta, matter/antimatter etc the crucial difference is that the gravitational potential is something which breaks this. In flat space there's no preferred or special direction so vacuum bubble processes (ie those of Feynman diagrams with no external legs) are always treating the two particles in a symmetric way. With the gravitational potential of the black hole the radial direction is a special direction and so those climbing up the potential will not be treated in the same manner as those falling in it. Those falling always carry the energy debt of the escaping particle so Hawking radiation can only lower the mass of the black hole. Otherwise you'd have something gaining mass while emitting energy and matter, which is inconsistent with thermodynamics/conservation laws.

    You, long ago, claimed that the Hawking process could feed a black hole and I asked you for your justification. It was never forthcoming....
    Last edited: Sep 7, 2009
  11. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    The same point I made in foot note of prior post 26, but more accurately done.

    I also noted this "debt" prevents the emission of two 0.511MeV gamma rays even if they "Pop out of the vacuum" far from the EH of the BH. Here your arguement about a gravity gradient bias does not apply. Can you replace it for the lack of gamma rays with annhilation in "flat space" ? Or are you, like me, reduced to just knowning that energy is conserved?
    Last edited by a moderator: Sep 7, 2009
  12. AlphaNumeric Fully ionized Registered Senior Member

    I'm not entirely sure (and neither are the other guys in the office) but I heavily suspect there's no contribution from photon-photon bubble diagrams in any cross sections. Electrons and positrons can form in pairs and then vanish, you can construct them via usual 2-point Feynman diagrams and let the incoming and outgoing photons go to zero 4-momentum (so there's nothing before and nothing after). You can't do this with the photon-photon loop as the photon doesn't couple to itself and so there's no way to construct, in QED, a photon bubble diagram and so they are expressly excluded from happening. If I'm wrong in that logic then I would expect the diagram to still not contribute by virtue of it being evaluated to zero (ie putting in a mass regulator and sending it to zero). You can have photons involved in bubble diagrams, such as a positron-electron pairing exchanging one (or many) photons before they annihilate one another again.

    Does that help?
  13. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    Some, but Feynman diagrams have never been much help to me. - Never as a basis for calculation, which I understand they greatly help with, at least in making sure little of significance is forgotten. As far as QED, is concerned, simple continium Maxwellian EM was all I could handle, and even then the math of some boundary value problems was tough for me if not a simple case.
  14. prometheus viva voce! Registered Senior Member

    I've thought for quite a long time about how to respond to this post. I'm pretty sure that if I just knocked a post out when I read it it would be pretty rude. I'm going to do my best not to be too rude.

    You're ascribing something to my statement that I'm not saying. The spectrum is a property of the ensemble of particles. I can either say that every time I want to talk about the temperature or I can just say "whose" or "who's."

    I agree with what AN has already said about this, and I would add that if I was good at English and not Physics I would have done English instead of what I did. It must feel fairly bad to know that the only recourse you have is to criticise my grammar when you can't touch my physics. I learned about Hawking radiation when I was an undergrad. It's pretty basic stuff by today's standards.

    You're misusing the terminology. No one is claiming that Hawking radiation is experimentally verified (although I believe there are experiments in preparation involving the Unruh effect that will verify the Hawking effect). In observer in this context, as you should know is a a theoretical observer, just like the twins in Einsteins twins paradox and many many other examples.

    Also, I would point out that the reason that Hawking radiation has not been seen is not because people have looked and not found it, but because the temperature of the radiation we could see from the black holes we have available is lower than the temperature of the CMB, so it's experimentally impossible to see Hawking radiation from black holes that come from stellar collapse.

    This, in a similar way to your views on plasma / liquid nature of QGP is not the end of the plot. If we observe a dark area that seems to have mass then it may be a black hole, but it also may not be. There are lots of objects that pack a high mass into a reasonably small space and do not emit light and yet are not black holes - neutron stars and white dwarfs for example. That's what you get when you glean all of your modern physics knowledge from pop science books - you don't get the full picture.

    That's not what Ben was saying. The criticisms of the tunnelling approach have no effect on the first order (the temperature) result, but only to the higher order effects that provide deviations from black body behaviour.

    Finally, I should ask that if you are so convinced that the Hawking effect is wrong why can't you find the error in the theoretical reasoning? If what we know about gravity and quantum field theory is right then we expect black holes to have a temperature. If you can find a reason why this would not be the case then that would certainly be worthy for your first publication in physics.

    PS any luck producing a baby with two male adults yet? It could either happen or not right? I also fully expect you to ignore this post like you do when you can't keep up. Being represented in court by you must be a truly interesting experience.
  15. AlphaNumeric Fully ionized Registered Senior Member

    In QED, and any other QFT, its often convenient to work with Lagrangians because they tell you how the different fields couple to one another or themselves (ie interact). In QED the Lagrangian is this and that tells you that photons can only interact with charged particles and not themselves and you can only construct interactions and processes using electron, positron and photon propogators, which are the lines in Feynman diagrams, and a vertex where a photon, electron and positron all meet. If, for a given process, you can't make a diagram out of those which gives that process then the probability of that process is zero.

    The electron-positron pair production in a vacuum can be made from these. You start with the diagram which involves a photon splitting into the pair of particles, which then recombine into the photon. It's process. If you then stipulate that the energy and momentum of the photon coming in (which must also be the energy and momentum of the outgoing photon) goes to zero you'll basically end up with just the circle. That's the electron-positron vacuum fluctuation to one loop.

    Now the issue you raise is what about the equivalent but with a photon making the circle. Since photons don't interact the incoming and out going particle must not be a photon. But also it cannot be an electron because you're only allowed to have the vertex which has 1 photon and 2 charged particles on it, but the photon loop would need two photon 'legs' on each vertex, which isn't allowed. So, providing that logic is sound, you can't have a pair of photons appearing and then disappearing, you have to have charged pairs appearing who then exchange some photons, before annihilating one another, back to nothing.
  16. DRZion Theoretical Experimentalist Valued Senior Member

    From what i read on these forums I come to this conclusion-

    It is my opinion that the density of quantum fluctuations is associated with the curvature of space time. This would explain why the mass of quantum fluctuations does not overtake all other mass.
    It is quantum fluctuations that dictate the speed of light. [edit: I suppose this cannot be true since nothing moves faster than light and the speed of light is constant; however, if you compare this to Kurros' post on Einstein's hole argument, it would make sense that space without anything (not even quantum fluctuations) would not be space at all??]

    Here is proof that the speed of light is not a constant in various reference frames:
    "The speed of light in meters per given interval of "proper time" is a constant, however the travel time of any electromagnetic wave, or signal, moving at 299,792,458 meters per "second" is affected by the gravitational time dilation in regions of spacetime through which it travels. This is because the coordinate time and proper time diverge as the gravitational field strength increase"

    Near bodies of mass you have curvature of space time and you get more quantum fluctuations and so light travels more slowly.
    As the universe expands the voids (which have very low curvature) allow light to travel at perhaps faster than 300,000 km/s and so we see objects that are 40 billion light years away where light only traveled 14 billion years.
    Near a black hole the fluctuations are more intense due to the very steep curvature of space-time; light begins moving more slowly the closer you get to the core of the black hole. Also, it would suggest that a part of the mass of a black hole may be produced through virtual particles inside of the event horizon. So perhaps black holes ARE more massive than all the mass that falls into them.

    When a particle travels through a curvature in space time it will have more pressure on one side than another. It goes back to very simple mechanics that such a particle will change it's direction.

    [It is interesting to explore how one can explain all of gravity by using only the time-dilation principle. Any particle with a finite volume will experience a difference in energy across it's volume. Its kind of like heating a bimetallic plate- one side expands more slowly and so the plate bends in a particular direction.]
    Last edited: Sep 10, 2009
  17. DRZion Theoretical Experimentalist Valued Senior Member

    Is this statement incorrect?
    Wouldn't this mean it is possible to build a machine that moves through space faster than 300,000,000 m/s ? All one has to do is dilate time and then the relativistic speed cap is raised, right??? :xctd:
  18. D H Some other guy Valued Senior Member

    No, it just means you aren't using the right clock / meter. The same effect (gravitational time dilation) means that the clocks on the GPS satellites run a bit faster than those on the surface of the Earth.
  19. DRZion Theoretical Experimentalist Valued Senior Member

    Right, I completely confused the two things. When time dilates light moves more slowly than the other way. Which is why the clocks on GPS satellites move more quickly; because time is more dilated on earth than in orbit.

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