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Author Topic: In hawking radiation, does the black hole always absorb the antiparticle?  (Read 2570 times)


yor_on

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There are several things to it. One is the definition of pair production at the event horizon. In it we read 'anti particles' are what falls in, annihilating as they meet 'particles'. Assuming the possibility of there being 'particles' falling in you get a addition as you so rightly point out. I seem to remember a discussion we had about this here, with a guy explaining the concept behind why all 'particles' passing the event horizon by necessity becomes 'anti particles', but I can't seem to find it?

I've seen a definition though in where it's the 'energy' it cost to create and separate those particles in the Black Holes gravitational field that will make the Black Hole lose mass. Looking at it that way you may be able to account for the addition of energy inside the EH (Event Horizon) as the annihilation takes place too, maybe? Because, even if annihilating it still will leave a excess of 'energy' inside that EH, as I think of it?

Maybe the last one can be ignored if we assume the 'energy' to be something that pass the EH, much like gravity does? But the radiation released can't leave though, it all goes back to what 'energy' means to me. (the question there is what it cost to 'transform' and if the 'heat' released is equivalent to the cost? If it is so then 'energy' is transformations and the conservation laws is upheld. In the other case, if 'something more' is lost?)
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Thinking of it, assuming that this is a correct definition, the amount of particles or anti particles infalling shouldn't matter for it. You can always use statistics and define a proportionality to the amount, and doing so that proportionality must still be less, than the 'energy' stolen from the Black Holes Event Horizon (4 big letters in a row :) as the pair production happen. Anyway, it imply that the cost for creating is bigger, for the BH gravitational field, than the energy/heat/radiation added through what is annihilated inside. Which should be correct considering that proximately half of the particles gets added to our side of the universe, the other half either adding, or getting destroyed releasing 'energy', inside that Even Horizon.
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Rereading I better point out that 'gravity' is a geometry in relativity, not any 'energy' per se, although you have gravitational waves taking with it 'energy', as (possibly) proven by binary stars losing energy due to gravitational waves propagating from them as they rotate around each other. We know that all energy adapt to gravitation though so there must be a close relation between them anyway.
« Last Edit: 01/10/2012 12:19:48 by yor_on »

imatfaal

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Hello Dave - Two answers an you can choose your favourite

1.  The particles are born through quantum fluctuation - one is pulled into blackhole and leaving it's pair still free and leaving the blackhole; a distant observer will see the blackhole lose mass and a particle leave the black hole.  For the blackhole to lose mass or energy then the particle falling in (from the viewpoint of a distant observer) must have negative energy. I think you can get around the problem of negative energy through weird maths beyond the event horizon; space becomes timelike, time becomes spacelike - and negative energy particles tunnelling into the blackhole can be considered as postive energy particles on a reserved time.
Quote from: Stephen Hawking p 202
Instead of thinking of negative energy particles tunnelling through the horizon in the positive sense of time one could regard them as positive energy particles crossing the horizon on past directed world-lines
The above is how some sources will explain Hawking radiation ....

2.  The problem is that the above explanation, as you realised, is a bit dodgy!  The actual physics and maths behind hawking radiation is fiendish - and to make it clear that I am not making this up - another quote from the original article concerning the particle pair explanation
Quote from: Stephen Hawking p 202
It should be emphasized that these pictures of the mechanism responsible for the thermal emission and area decrease are heuristic only and should not be taken too literally.

I have tried to get my head round many simplified explanations of Hawking radiation (my physics and maths is not good enough to follow the explanation in the quoted paper) - so far without success.  The fact that I have found about half a dozen explanations - all of which have common factors but which have distinct and crucial differences makes me think that we are still awaiting a nice explanation! Remember this is quantum mechanics thrust up against Gravity by the extreme conditions at the event horizon of a black hole - and we do not have a functioning theory or idea that connects the super-small world of quantum mechanics with the (normally) large scale world of Gravity. 


Quotes from:
S. W. Hawking Particle Creation by Black Holes Commun. math. Phys. 43, 199220 (1975)

evan_au

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For any black hole of macroscopic size, the most likely particles to escape are photons - and photons with quite low energy, in the microwave region or below. Since the photon is its own anti-particle, it doesn't matter which one escapes.

Quote
A black hole of one solar mass has a temperature of only 60 nanokelvins; in fact, such a black hole would absorb far more cosmic microwave background radiation than it emits. A black hole of 4.5 1022 kg (about the mass of the Moon) would be in equilibrium at 2.7 kelvin, absorbing as much radiation as it emits. Yet smaller primordial black holes would emit more than they absorb, and thereby lose mass

http://en.wikipedia.org/wiki/Hawking_radiation#Overview

imatfaal

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For any black hole of macroscopic size, the most likely particles to escape are photons - and photons with quite low energy, in the microwave region or below. Since the photon is its own anti-particle, it doesn't matter which one escapes.

Quote
A black hole of one solar mass has a temperature of only 60 nanokelvins; in fact, such a black hole would absorb far more cosmic microwave background radiation than it emits. A black hole of 4.5 1022 kg (about the mass of the Moon) would be in equilibrium at 2.7 kelvin, absorbing as much radiation as it emits. Yet smaller primordial black holes would emit more than they absorb, and thereby lose mass

http://en.wikipedia.org/wiki/Hawking_radiation#Overview

Ah but Evan - that would lead to a violation of conservation of energy - photon is emitted, and photon goes to black hole; that's two extra photons with no balance (and that was the basis of Dave's question).  The balance seems to be that a photon is emitted and the black hole gets smaller - but if a normal particle (photon or whatever) is added to a blackhole it will get larger, why does the blackhole get smaller?  The particle going to the blackhole needs to remove energy/mass from the blackhole - or the blackhole needs to provide the energy for the pair production in the first place.  That is the quandary - I have tried (and I think failed) to outline some ideas above

yor_on

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I actually think that the explanation I gave above is the one closest to the 'truth' here. It is simple, and it doesn't care about whether it is a particle or a anti particle falling pass the EH. But it is also very strange in that it assumes some sort of symmetry to gravity-energy. Think of a black hole and the space around such a one, could you call a empty space 'compressed' as in having a higher energy per cubic cm? Seems so wrong to me, but? In what manner would you do that? As we now assume that a space around a black hole must be sizzling with 'energy' making that spontaneous pair production more or less inevitable.

If it is correct that you will have more energy around that Black hole, then it follows logically that a gravity and 'energy' must be connected. From that assumption you then can define a 'cost' for the energy disappearing out in space (from the balancing of 'energy' relative how 'compressed' a space might be, if you like :) and doing so you will find that the gravitational field generated by the black holes mass must lose 'energy', and as 'energy' is mass? QED :)

 

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