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I Hawking radiation question

  1. Mar 6, 2015 #1
    Hawking radiation talk about particles emitted from black holes.
    Are the particles emitted are protons and electrons?
    (emitted at near light speed and gradually slowdown by gravity pull of black hole as the particles move outward to fill the interstellar space of the galaxy)
    If yes, that means those protons will combine with electrons and form hydrogen atoms.
    And than hydrogen atoms accumulated under gravity and form new stars?
    It look like the black holes absorb energy (all type of waves, including background microwaves) and emit matter(particles) - Isn't these the "reverse process of what is going on in stars "?
     
    Last edited: Mar 7, 2015
  2. jcsd
  3. Mar 7, 2015 #2
    Why would the particles slow down? Friction does not exist, in space unless they run into something then that might slow them down. And to answer your first question the radiation emitted is electromagnetic radiation, "a form of radiant energy released by certain electromagnetic processes".
     
  4. Mar 7, 2015 #3

    Chronos

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    Particles is probably a poor way to view the nature of Hawking radiation.
     
  5. Mar 7, 2015 #4

    Drakkith

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    Some are. I believe they can be any type of particle, with more massive particles appearing in greater numbers as the black hole shrinks and the gravitational gradient becomes steeper.

    I don't believe so, but I can't explain why.

    While its true that black holes do absorb all types of energy, trying to say that a black hole is doing the opposite of a star is an oversimplification in my opinion. Star's don't even convert matter particles to radiation, they convert potential energy into radiation and leave behind a great deal of more compact matter.
     
  6. Mar 7, 2015 #5

    PeterDonis

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    In principle this is possible, but as Drakkith said, this won't happen in any significant quantity until the black hole has shrunk--quite a lot, actually. A black hole of stellar mass or larger has a temperature much, much, much too low to have a significant probability of emitting anything but photons. It would take a time many, many, many orders of magnitude longer than the current age of the universe for such a black hole to shrink to the point where its temperature was high enough to have a significant probability of emitting anything with nonzero rest mass. And by the time it has shrunk to that point, it is close to evaporating away completely, so there isn't much mass left to emit.

    It's worth noting that, even at the point where a black hole has a significant probability of emitting electrons and protons, it won't emit them in equal numbers. The total charge of all emitted particles has to be zero, but there are plenty of other positively charged particles the hole can emit to balance the negative charge of the electrons (positrons would be the most likely).
     
  7. Mar 7, 2015 #6

    Chalnoth

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    Essentially anything can be emitted in principle. In practice, most of the particles emitted are photons. Until the black hole gets close to the end of its life, the temperature is too low for any significant number of massive particles (even electrons) to be emitted.
     
  8. Mar 8, 2015 #7
    Hawking Radiation can only occur with a pair of entangled particles. They are created spontaneously just outside the Event Horizon of a Black Hole. One particle is captured by the Event Horizon and the other, because the momentum of entangled particles is conserved, is ejected into space as a high energy particle. It's more like Cosmic Radiation than Electromagnetic Radiation.
     
    Last edited: Mar 8, 2015
  9. Mar 8, 2015 #8

    PeterDonis

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    No, this is not correct. The pair of virtual particles that gets created is a particle-antiparticle pair, but there is no requirement that they be entangled.

    This isn't really correct either. Virtual particles are off-shell, so their momentum and energy do not have to obey the usual energy-momentum relation; but they also cannot exist for more than a short time (the further off-shell they are, the shorter the time). When one of the particles gets ejected as Hawking radiation, it has to become a real particle, and the only place the energy can come from is the mass of the hole. But its momentum when it is ejected has no connection to the momentum of the other virtual particle (the one that fell into the hole).
     
  10. Mar 8, 2015 #9
    How do you create a virtual pair of particles that are not entangled? If they are not entangled, when what law of nature allows the swallowing of one particle by the Black Hole to effect the momentum of the other?
     
    Last edited: Mar 8, 2015
  11. Mar 8, 2015 #10

    PeterDonis

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    Again, virtual particles do not obey the usual energy-momentum relation. That is equivalent to saying that they do not obey the usual energy and momentum conservation laws. Entanglement only makes sense if the conservation laws are obeyed.

    As I said in my previous post, the momentum of the particle that escapes has no connection to the momentum of the one that is captured by the hole.
     
    Last edited: Mar 8, 2015
  12. Jan 25, 2017 #11
    As I understand it, a particle antiparticle pair form near the event horizon of the black hole. One passes across the event horizon, the other may or may not but assuming it doesn't, it then moves off with considerable kinetic energy and is a 'particle' of Hawking radiation. Now the mass of the black hole is only reduced if the particle that passes across the horizon and is 'swallowed up' is the antiparticle member of the pair. In this way, as I understand it, over eons of time, sufficient antiparticles will be absorbed to cancel the mass of matter inside the hole, hence it 'evaporates' and the (partner) particles are those which formed the Hawking radiation. My question is; surely there is an equal probability of either the antiparticle or the particle passing across the event horizon. If this is the case the hole cannot evaporate. So, surely the whole idea is redundant - unless there is some bias towards more antiparticles crossing the horizon?
     
  13. Jan 25, 2017 #12

    jbriggs444

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    Both particles and anti-particles have positive (or zero) mass. There is no obvious reason to expect an asymmetry in the rate at which particles or anti-particles evaporate off as Hawking radiation.

    As has already been suggested, what comes off is likely to be photons -- which are their own anti-particle.
     
  14. Jan 25, 2017 #13
    Okay, if the particles are photons of zero rest mass, how can they contribute to the 'evaporation' of the B.H. i.e. how is the mass inside the B.H. reduced by this process?
     
  15. Jan 25, 2017 #14

    jbriggs444

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    They carry away energy.

    In [special] relativity, the mass of a system is given by its total energy in the frame of reference where it is at rest (**) according to the equation e=mc2 (*) In general relativity, the situation is more complicated, but similar reasoning applies. If you carry away energy, you carry away mass, even if the entity carrying away the energy is massless.

    In relativity, mass is not an additive property. The mass of a system is not given by the sum of the masses of its components.

    (*) If a system has no rest frame then its mass is zero and its energy is given by E=pc.

    (**) "at rest" should be read as "has zero total momentum".
     
    Last edited: Jan 25, 2017
  16. Jan 25, 2017 #15

    Drakkith

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    You've almost certainly seen the equation ##e=mc^2##. The full form of this equation is actually ##e^2=(mc^2)^2+(pc)^2##, where p is the momentum of the object. For massless objects like photons, ##m## is zero but ##p## is not. So this reduces to ##e^2=(pc)^2## or ##e=pc##.
     
  17. Jan 25, 2017 #16

    Chalnoth

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    And to clarify, this happens any time an object radiates.

    For example, if you heat up a potato, it gains mass due to the added thermal energy. Then, as the potato cools through diffusion, convection, and radiation, that added mass is lost. Of course, the amount of mass gained in this circumstance is so small that you probably couldn't ever measure it.
     
  18. Jan 25, 2017 #17
    Right; to the last three replies, thank you, I am familiar with these concepts / equations. I will go away and think about this. However, my immediate reaction is, I am not sure how adding energy E = pc for the photon that crosses the event horizon and falls into the B.H. reduces the mass of the B.H. given that the other photon (it's pair partner) of Hawking radiation has the same energy.
     
  19. Jan 25, 2017 #18

    jbriggs444

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    Per my understanding (which is worth what you pay for it), the virtual particle picture is heuristic. Using it as a basis for further reasoning is fraught with peril. Heuristically, the particle falling has negative energy and subtracts from the mass of the black hole. This does not entail that it is an anti-particle.
     
  20. Jan 26, 2017 #19
    Thank you, I agree that the particle falling into the B.H. must posses negative energy in order to subtract from its mass. I understand this negative energy arises from the negative root of the Einstein equation quoted above, does it not? Dirac himself interpreted this as the possibility of the existence of antiparticles (or the equivalent idea in the day). Surely the idea of a negative energy particle being equivalent to a particle of opposite charge is the definition of an antiparticle. Although, it is difficult to see what this means for a photon which is neutral. Except that it is a photon with negative energy - is that the accepted view?

    Incidentally how much am I paying you for this advice?:wink:
     
  21. Jan 26, 2017 #20

    Chalnoth

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    I don't think the view of Hawking Radiation as particle/anti-particle pairs with one of the pairs falling into the black hole is necessarily a good one to think about. It does make a sort of sense in the context of quantum field theory. But it may be more confusing than not to non-physicists. And for physicists, it may be just a bad way to describe it because it isn't used in the actual calculations for deriving the temperature of a black hole.

    That said, I'll give a try at describing why this seems reasonable to somebody with some quantum field theory background. The typical picture of virtual particles comes from interactions like the one shown here:
    https://en.wikipedia.org/wiki/Virtual_particle#/media/File:Momentum_exchange.svg

    The lines at the left and right are particles (they could be electrons, protons, photons, etc.), while the dotted line in the center is the virtual particle that mediates the interaction. This could also be any particle. The diagram as a whole must obey the relevant conservation laws. One of those laws is the conservation of energy-momentum. In normal particles, energy and momentum are related to mass via ##m^2 = E^2 - p^2## (in units where ##c = 1##). With virtual particles, the energy ##E## and the momentum ##p## are fixed by the conservation of energy-momentum, and it turns out that in most interactions, the imputed mass of the virtual particle is very different from the mass of the corresponding real particle (e.g., the mass of a virtual photon is generally not zero). These particles are represented as being "off the mass shell", and they very often have negative mass. They can also have negative energy.

    If you move the two interacting particles away from one another, such that the virtual particle has to travel a longer distance, it starts to behave much more like a real particle with a mass consistent with the real particle.

    With this picture, it's really easy to think of the virtual particles that fall into the black hole as having negative mass or energy.
     
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