1. Nov 24, 2003

Silverious

I think I have the general idea of how it works.

But how does one particle get negative energy when it falls into the event horizon?

2. Nov 24, 2003

LURCH

I've often wondered about that myself, as it doesn't seem to be covered in any of the literature written for laypeople. Here's what I've come up with from my own musings;

For virtual partical pairs (VPP's) to form, one must have negative energy and the other positive in order for energy to be conserved, right? From there I make the leap hat, because these paerticals are virtual, the distinction between which is regular (positive) energy and which is negative (unleaded?) remains undetermined. If this is the case, an observer could say "A is positive and B is negative" or "A is negative and B is positive", either way is equally correct. If this is an accurate discription of the state of VPP's, then the question becomes easy to resolve. Once one of the particals escapes to infinity, and becomes "real" (positive enrgy travelling through the universe), the partical that does not escape is relegated to be the negative partical sort-of ex post facto.

I must re-state, however, that's just what I came up with in my twisted and feavered imagination. Hopefully one of the other members can tell you if it's anywhere near correct.

3. Nov 24, 2003

Nommos Prime (Dogon)

Black Hole Thermodynamics (Quantum Microstates)

Old-school black hole physics considers the Event Horizon to be a prison from which even light (mass-less particle) cannot escape. This is wrong. Why? Because the Einstein equations used to describe the Classical Properties of black holes failed to take into account quantum mechanics (woops!).
Quantum uncertainty and its implications (as aptly described in an example by Lurch) creates quantum fluctuations in a vacuum state (ie. virtual particle/antiparticle creation/destruction cycles/fluxes). Now, Hawking Radiation brings with it the “phantom-like” Quantum Microstates (you really have to be into String Theory to understand Black Hole Thermodynamics). Now if we assume that energy is conserved in quantum gravity states, then when a black hole decays and hurls or “evaporates” a particle, strangely, the particle will have a Thermal Distribution.

Now to flow on from Lurch. He is correct in stating that once one particle escapes the Event Horizon, it becomes a “real” particle within our observable Universe, and the antiparticle “disappears” below the Event Horizon. The particle which disappears below the Event Horizon remains in a “virtual” state and MUST restore its Conservation of Energy by endowing itself with a negative mass-energy. When the virtual particle does this, the Black Hole will then absorb the virtual particle’s negative mass-energy, and therefore, SHRINK! (because of the resultant loss of mass). In fact, you can actually do the equations for this process by simply working out the mass of the Black Hole. Once you have its mass, then apply this formula:

“The rate of power emission (radiation) from a Black Hole’s Event Horizon is proportional to the inverse square of the Black Hole’s real mass.”
Easy.

However, again expanding on Lurch, there is another possibility. BOTH particles (particle/antiparticle) are capable of escaping the Black Hole’s Event Horizon and becoming REAL!
I can expand on it, if anybody’s interested.

4. Nov 25, 2003

LURCH

Re: Black Hole Thermodynamics (Quantum Microstates)

I am! How can the virtual particals avoid annihilating with one another, without the event Horizon to separate them?

5. Nov 25, 2003

Labguy

Re: Re: Black Hole Thermodynamics (Quantum Microstates)

Because the distance between them, with high velocity during their 80 attosecond "virtual" existance, doesn't leave them close enough to each other to interact. So then, you have two "real" particles. There is no loss of energy, or "negative" energy since the mass of the BH decreases as per e = Mc2

Labguy

Last edited: Nov 25, 2003
6. Nov 25, 2003

Ambitwistor

Re: Re: Re: Black Hole Thermodynamics (Quantum Microstates)

You made that up. There isn't any equation for that process within the framework of quantum field theory in curved spacetime. LURCH is correct: the net energy of a vacuum-produced pair is required to be zero in QFT; that's why a positive-energy escaping particle has to be balanced by a negative-energy infalling particle. (If you disagree, present the calculation: handwaving not accepted.)

For a somewhat more detailed answer to the original question, see:

http://groups.google.com/groups?selm=aic1qh$sdn$1@woodrow.ucdavis.edu

7. Nov 26, 2003

ranyart

Back to back virtual pair production can also lead one to ask the inevitable question : Where has all the Anti-Matter gone?

Starting with this statement;Antimatter poses another riddle the LHC will help us to solve. At the Universe's birth, the Big Bang, matter and antimatter are believed to have been created in equal amounts, yet today we live in a Universe apparently made entirely of matter. So where has all the antimatter gone? It was once thought that antimatter was a perfect "reflection" of matter - that if you replaced matter with antimatter and looked at the result in a mirror, you would not be able to tell the difference. We now know that the reflection is imperfect, and this could have led to the matter-antimatter imbalance. Link here:http://www.clab.edc.uoc.gr/materials/pc/dive/questions.html

One can progress to here:http://www.superstringtheory.com/blackh/blackh3.html Take note to the animation of Particle and its Anti Particle, Where have all the Anti-particles gone?

It is widely accepted that Blackholes are fundemental to every Galaxy in the observed Universe, the are what hold Galaxies together, they are at Galactic centre's. There is evidence that if one rewinds the Einstein Field Equations, each Galaxy returns into the Blackholes at thier core, and very early times all the Galactic Matter 'hovers' around the Blackhole Horizon, now according to my own theory, the particles emerging from the very Early Universe of Blackhole Horizons are infact the Size of Large "STAR-LIKE-ENTITIES", infact they are Stars.

We have to understand that our perception of the Early Universe is being made from within 'full-blown' Galaxies with full-blown Stars. Some models (the main big-bang model) predicts that Stars Create Blackholes, according to the Equations when a Star explodes to a certain limit, A remnant Blackhole is created?

Blackholes create Stars!, its not that Stars creates Blackoles. We have to understand now that at a reversed model in Einsteins field Equations, the Universe is the space between Galaxies, and at a precise moment, for example for that of our Milkyway, it contracts not to a singularity of the Whole Universe, but just to the Coupled Blackhole singularity which is central to 'our' milkyway, this Takes the Anti-Matter into another whole different context?

As our Galaxy wanes away, the Vacuum Field is left jsut as it is now, between other Galaxies that are still around, the Particle-Antiparticles produce the Energy background in just the same way.

So in a simplistic way, all the Anti-matter went into the Blackhole, which created the Stars, which formed our Galaxy and every other Galaxy, which exist in a Vacuum Field, the density of which is fundemental to the production of Blackholes, or the tearing of Space, this is what a near perfect expanding Vacuum is, like the 'zipper'on a garment, it can sometime releave the pressure after a wholesome festive meal.

This is my simplistic view of course, leaving out a lot of corraborating data.

8. Nov 26, 2003

Labguy

Re: Re: Re: Re: Black Hole Thermodynamics (Quantum Microstates)

No, I didn't make that up! Sometimes a single virtual particle "escapes" and sometimes both particles escape, as someone mentioned above. Each and every release of Hawking radiation is not identical to every other particular release. Sometimes it is Photon pairs, sometimes Positron-Electron pairs, and even more massive pairs if enough energy is available at the EH. A few quotes from a few HR sites:

"If one virtual particle falls into the black hole and the other one escapes, it escapes as Hawking radiation from the black hole. This radiation shows the same allocation as black body radiation. This assumes that black holes do have a temperature too. In fact this was proven by Bekenstein in 1972, before Hawking discovered that black holes also emit temperature radiation.".... and:

"If one particle of a pair of virtual particles falls into a black hole and the other one doesn't they can't react back to energy and the escaping one becomes a real particle leaving behind a lack of energy in the vacuum. Somehow this "hole" in the vacuum energy has to be filled again - even in quantum physics the law of energy preservation can't be disobeyed for a longer time. So this "hole" draws energy from that black hole. But what kind of energy does a black hole without any spin or charge have? It's mass! Thus a black hole loses some of its mass after Einstein's famous formula E = mc².".... and:

"Since virtual photons have such a short lifetime they won't get very far during that (24 nanometers for a virtual photon of orange light). Photons always move at the speed of light, so the distance a virtual photon travels in its lifetime is given by: (a formula showing a lifetime of 80 attoseconds for orange light,(500 THz))."

It (the formula) is on a referenced website. It is (lifetime for a virtual particle): 1/(8Pi)(frequency). For orange light, that becomes: 8.0x(10-17s) = 80as (attoseconds). I'm not going to wade through them all for you, but a few to browse are here:

http://online.itp.ucsb.edu/online/gravity99/jacobson/
http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html
http://superstringtheory.com/blackh/blackh3.html
http://relativity.livingreviews.org/Articles/lrr-2001-6/ [Broken]

In addition, I happen to own a few, small books; you know, those paper things where you don't read off of a CRT.

So, if two virtual particles just happen to travel at V=c, and if lifetime is long enough (based on Hz) for seperation to be enough for no "mutual annihilation", then (1) one particle escapes and one re-enters the EH, or, (2) both escape. If you are saying that there is not enough time for (2) to take place, then you would also be saying that (1) cannot happen. That would be incorrect.

And, some of the citations could be incorrect, Hawking could be incorrect, Lurch could be incorrect, Einstein could be incorrect and I could be incorrect, but I certainly don't just make this crap up without at least looking about to see what other, more qualified scientists write in their books or post to various websites!

I think that the first, short sentence of your post is at least insulting and more likely arrogant. Slam other posters as idiots if you like, but don't do it again to me!

Labguy

Last edited by a moderator: May 1, 2017
9. Nov 26, 2003

Ambitwistor

Re: Re: Re: Re: Re: Black Hole Thermodynamics (Quantum Microstates)

Like I said: prove it. Write down the equation for the latter process.

So? That has nothing to do with the previous statement.

Those quotes don't support your point. In fact, they support mine: they all speak of the black hole losing energy when one particle falls in.

Maybe you should have waded through them all. I did, and none of them support your contention either.

Do you own a few of these paper things that might happen to perform the actual calculation you're claiming exists? You know, like Birrell & Davies's Quantum Fields in Curved Space, or Wald's Quantum Field Theory in Curved Spacetime? I do. If so, can you quote the relevant calculation that shows Hawking radiation producing a pair of particles from vacuum that both escape to infinity, because I didn't see it in my copy.

It's not a matter of time; it's a matter of the fact that vacuum pair production produces real particles whose total energy is zero. If one escapes and has positive energy, the other falls in and has negative energy, relative to the same external observer. No observer ever sees a pair of real particles produced from vacuum.

(If that could happen, then it would be able to happen away from a horizon too, but you don't get real radiation from vacuum in the absence of some kind of horizon -- black hole event horizon, cosmological horizon, Rindler horizon, etc. It's the existence of a horizon which can divide the pair that allows for real radiation to be produced.)

I hate to break it to you, but none of your sources actually said what you claimed. So can you tell me where this claim actually came from, if not from you? Maybe all the books and papers I've read are wrong, but I'd like to see an actual calculation, by you or anybody else, that demonstrates what you claim.

10. Nov 26, 2003

Labguy

Re: Black Hole Thermodynamics (Quantum Microstates)

I'm not good enough to transfer them from "there" to here accurately. The latter can happen around an EH with high energy. Either way, the "lost" energy comes from the BH mass, as in the website quote I used above. The formula is e = Mc2, isn't that familiar?

Sure it does, it only is meant to show that all HR is not just virtual photons. Other, more massive particles can be created, depending again on the energy at the EH.

Yes they do. The second quote I listed specifically, again, mentions the BH losing energy by the loss of mass.

Same story, the second quote I listed.

No, my books are more on the order of "Dick and Jane See Black Holes".

Yes, it is a matter of time. Allow no time and you would always have instant annilihation and no HR at all. Why do you think the third quote I posted was from a site explaining HR?

No it wouldn't, because there is no strong gravitational "tug" away from an EH to seperate the particles and prevent mutual wipeout. It is the strong gravity, the defined "EH" radius, and the time allowed that can create two particles and eat one of them. Isn't that what you said?

Start at:

then click on "The core" and then "virtual particles" and keep on clicking around long enough to find something useful. Back up and go to "Hawking Radiation" and click around again. These are called links. They even have some really neato formulae, considerably more than on the sub-Dick & Jane "Google" link you last provided.

My own "favorites" on IE 6.0 has 669 url's in 95 folders, all on cosmology and stellar evolution, neatly catagorized. But, trust me, I didn't look all 669 of them up and read them for this or my last post. Also, believe it or not, I don't have all their content memorized; I actually have to look that stuff up every so often.

You have several times demanded proof in the form of formulae, but I can't seem to find the formulae you posted here.(?)

And finally, whether I have a concept wrong or not, I still have not seen any other "PF Advisor" answer as rudely as you did before. I have seen Phobos, and especially Janus, correct someone with a reasonable explanation and a polite "nudge" in the right direction, but I have never seen either one of them flat-out call the poster an (indirect) idiot or liar. I noticed that your last reply with my quotes did not quote my last paragraph. Why not? You have to realize that most of the PF members are students and sometimes come here to actually learn something and ask an occassional question. A response like your earlier one to me could easily cause posters to avoid PF for fear of being insulted or called ignorant. Don't reply to any more of my black hole comments, I can go look enough up on the internet and eventually stumble across an explanation. Respond instead to this paragraph.

Labguy.

Last edited by a moderator: Apr 20, 2017
11. Nov 26, 2003

Ambitwistor

Re: Re: Black Hole Thermodynamics (Quantum Microstates)

Sorry, waving your hands and quoting e = Mc2 does not constitute a calculation. You cannot get nonzero total energy from the zero energy of the vacuum. In quantum field theory, the vacuum is, by definition, the ground state: its energy does not change. (Ignoring metastable false vacuums that play no role here...) If real particles are produced, you have the vacuum, plus two new particles. They must conserve energy, which means that the vacuum+particle system has to still have zero energy. The total energy of the vacuum (interior plus exterior of the black hole) remains zero, which means that the real particles also have to have zero energy, which means that you can't have both real particles escaping to infinity.

Whether black holes can produce particles other than photons has nothing to do with whether two real particles can escape to infinity.

Read it again. The second quote you listed specifically, again, mentions the second particle falling into the black hole.

That the black hole loses mass-energy is not in question. Of course the black hole loses mass-enegy. In fact, a negative-energy particle falling into the hole is the reason why the hole loses mass. In the virtual pair picture of Hawking radiation we're talking about, if a negative-mass particle doesn't fall into the hole, then the hole cannot lose energy.

(It's also possible to view Hawking radiation from a field perspective, as Wald's Living Review mentions, in which case the explanation of Hawking radiation is different, but also does not involve pairs of real particles being emitted to infinity.)

I'm not saying that time is not involved in Hawking radiation. I'm saying that the matter of whether two real particles can escape to infinity has nothing to do with time: it's a conservation of energy issue. Once you're talking about real particle production, you can ignore the energy-time Heisenberg uncertainty relations that permit violation of conservation of energy for virtual particles, because for real particles the conservations laws have to be satisfied for all time.

It's the actual presence of an event horizon, and not just a strong gravitational field, that is needed. I can make the gravitational field outside of a compact body arbitrarily similar to the field outside a black hole of the same mass, but I won't get any Hawking radiation. It's only when the horizon appears that I get Hawking radiation, and that's because one particle crosses it and the other does not.

But anyway, tidal forces are irrelevant. It doesn't matter how strong they are (as long as they're finite): a virtual pair can only exist for a limited time before recombining -- regardless of whether there is any field, gravitational or otherwise, "trying to separate them" -- since it violates the conservation laws; only the Heisenberg uncertainty principle allows it to exist for any time at all, and that only briefly. The only way that the particles can be permanently separated and become real is if one of them acquires a negative energy, and for that you need the horizon.

Your original "the particles don't recombine during their virtual existence because they're not left close enough to interact" argument is also invalid, by the way. You can write down Feynman diagrams for virtual particles to separate to arbitrarily large distances before recombining. (They can do that by travelling at arbitrarily large speeds; virtual particles are not limited to be on shell and travel at speeds <= c.)

You will note the conspicuous absence of any formulae describing the spontaneous vacuum emission of two real particles to infinity, because that process exists only in your own mind. The Carlip article to which I referred you is correct by the way, formulas or no. Moreover, it gives a more careful description of the process than does the ThinkQuest site, mentioning, for instance, the oft-neglected but important point of why the real particle that falls into the hole acquires a negative energy. (It has positive energy relative to the timelike coordinate inside the horizon, but negative energh with respect to an external timelike coordinate.) In fact, that was the question that kicked off this thread.

In order to find that quote, start at: http://library.thinkquest.org/C007571/english/advance/english.htm?tqskip1=1&tqtime=0602
then click on "The core" and then "virtual particles" and keep on clicking around long enough to find something useful. Back up and go to "Hawking Radiation" and click around again. These are called links.

I can't figure out why you persist in referring me to sites that contradict your own argument. Do you even read these links to which you refer me?

And while you're at it, try reading some real books, like the aforementioned Birrell & Davies or Wald, which are far above your Dick and Jane page produced by high school students, filled with crude quasi-nonrelativistic approximations. (Not that it's a bad site; it's quite good for that level ... and you can't go wrong quoting a Niven story.) Or heck, try reading your own links, like the one to Wald's Living Review. None of them support your claims.

I didn't post any formulae here. Did I need to? You already agree with me that the one-particle-falls-in process exists; it's just that you are claiming that there is some extra process, so you need to back that up. If you want formulas for the ordinary Hawking process we both agree on, the web pages you cited already have formulae, and none of them predict the emission of two real particles to infinity. But if you'd like a formula, here's one: E = constant. If you want the works, which (unlike the handwavy calculation you see in lay descriptions like the ThinkQuest site, or in some undergrad textbooks like Schutz) show how the real calculation goes in quantum field theory, try the texts I mentioned; there aren't too many web sites which show real calculations in curved-spacetime QFT. Wald's Living Review summarizes some of them.

Last edited by a moderator: Apr 20, 2017
12. Nov 26, 2003

Labguy

Interesting, and long. You didn't post one, single thing that hasn't already been agreed upon, except my total ignorance. Not much of a question about one virtual particle falling back into the EH, I guess the problem is only my claim that both can escape in some instances. I will look around awhile and find a citation or two that you need where it shows that both can escape. I can state at least two examples of this happening, but I know you will come back with a "show me the math" response. It might be easier if you "showed the math" that it can't. That way, everyone reading this now ridiculous thread can benifit from your infinite knowledge

Your response on the "PF Mentor(advisor?)" question speaks very poorly on the state of this forum. My ignorance will not argue with arrogance that has reached perfection.

EDIT: CORRECTION:
Gee, I see that you did finally have to agree that time does have to be considered...

13. Nov 26, 2003

Labguy

Re: Black Hole Thermodynamics (Quantum Microstates)

If you know the citations or sources offhand, it would save me a lot of time from searching around what I already know is out there somewhere, but don't remember where.. (Duh!) However, I wouldn't blame you at all if you wern't exactly wild about jumping back in on this one, now.

Labguy

14. Nov 26, 2003

Nommos Prime (Dogon)

Its All Theory (Not Dogma)

For the First Scenario, use;
E=(hc^3)/16piGM. Where the assumption is that the virtual particle pair is created near the Black Hole (and NOT from the ¡°further out¡± Event Horizon). The virtual particles then ¡°lend¡± energy from the vacuum medium, whilst remaining ¡°bound¡±, and are accelerated by General Relativity toward the Event Horizon. When the virtual pair ¡°recombines¡± they will now have a higher energy (and therefore must separate). Now the ¡°borrowed¡± energy will be greater than the energy required for the virtual pairs creation. The excess energy acquired by the vacuum medium is irradiated by the Black Hole.
OR
Because the spectrum of outgoing virtual (and real) particles is thermal, radiation and particles will be emitted EXACTLY the same as if the Black Hole were a Hot Body at a temperature of TH=©¤c3/8¥ðkbGM (40.2). In this expression kb is Boltzmann's constant, M is the mass and TH is called the Hawking temperature.

For the Second Scenario (which I started, and seems to have caused a bit of agro between a couple of members). I would put forward that the creation of virtual pairs of particles/antiparticles will not violate any conservation of energy laws. Why? Because they exist for periods way below Planck Time and tidal forces will be inconsequential. When a violation of law occurs within Heisenberg scales there are no implications for laws on atomic scales. short length scale for tidal forces. Just because virtual particles do not become ¡°observable¡±, does not mean that their existence cannot be detected. A theoretical effect, similar to the Casimir Effect can reveal their existence by observing miniscule changes in pressure. Hawking Radiation and Black Holes violate numerous physical laws (eg. Baryon Number Conservation, ie. ¡°they consume information¡±). A new Special Theory of Relativity or completed String Theory like Susskind¡¯s) is needed to resolve these paradoxes.

As for Labguy¡¯s calculations, I agree.

15. Nov 26, 2003

Nommos Prime (Dogon)

Doh!

Damn! All my calculations and maths symbols won't seem to post. Sorry guys, I lost the lot...

16. Nov 27, 2003

Ambitwistor

Except for the fact that the vacuum cannot produce two real particles, which was what more than half of my post was about.

Are you talking about examples in curved or flat spacetime?

I did: E = constant. I'm not being facetious. I already explained this. The vacuum state is the ground state, by definition: the the state in which the system has the lowest possible energy. By conservation of energy, the system still has to be in the ground state after the real particles are produced (conservation can only be violated if the particles never become real). The vacuum energy can't drop below the ground state energy, by definition; if the Hamiltonian is bounded below, then all states with particle excitations have energy greater than the vacuum state. Thus, the only way real particles can be produced from the vacuum is if they contribute no net energy to the system.

As I also said before: I never claimed that time doesn't have to be considered. I said that "whether there is enough time for both particles to escape" has nothing to do with whether both particles escape. You can give them all the time in the world, but two real particles with positive energy as measured by the same observer still can't be produced by the vacuum.

You know, this discussion would go a lot smoother if you read things carefully, such as my posts and the links you cite, before responding.

17. Nov 27, 2003

Ambitwistor

Re: Its All Theory (Not Dogma)

Yes, it's fine for virtual particles to temporarily pop out of vacuum, violating conservation of energy, because of the Heisenberg uncertainty principle. But that's not what this debate is about. This debate is about the virtual particles becoming real and both of them reaching an observer at infinity.

Yes, vacuum polarization can have indirectly observable effects, as in the Casimir effect. But that's not the same thing as actually measuring Hawking radiation in a particle detector. For that, the radiation must be real.

Yes, black holes present some strange paradoxes. However, Hawking radiation was derived within the context of quantum field theory in curved spacetime (not full quantum gravity) --- which, at least for free fields, is a mathematically rigorous and consistent theory --- and as such, must obey the laws of that theory. Those laws preclude the possibility of vacuum-producing a pair of real particles that escape to infinity, for reasons already given.

18. Nov 27, 2003

Labguy

I really don't think that this should become a name-calling thread so it should end soon; maybe there is no "firm" answer at this point.(?)

Perhaps there is a confusion about the semantics of the term "both particles can escape". Maybe it should be changed into "Two particles can escape", but several sites do mention that "both particles can escape", such as seen in:

http://www.alcyone.com/max/physics/laws/h.html
where it states:
The theory that black holes emit radiation like any other hot body. Virtual particle-antiparticle pairs are constantly being created in supposedly empty space. Occasionally, a pair will be created just outside the event horizon of a black hole. There are three possibilities:
both particles are captured by the hole;
both particles escape the hole;
one particle escapes while the other is captured."

In that simple quote, the "both particles can escape" stands out like a sore thumb. The apparent problem is that most sites, like the one above, always mention that case # 2 is resolved by all that has been described so far, with one of the particles re-entering the EH and returning the "borrowed" energy, just like Ambitwistor has been saying all along. On that most common description I will agree 100%.

But, if changed into "two particles can escape", I know that I have read somewhere (that I can't find now) that the following (verbal) scenario can, and sometimes does, happen. This is from old memory and I have no math to "back it up":

(1) Two virtual particles are created just outside the EH, very short-lived based on frequency and seperation; they exist for a very short time.

(2) Both particles re-combine and annihilate. No energy has yet been "returned" to the BH and is still unaccounted for at this point.

(3) On annihilation, the energy gained from the BH is converted into an electron and a positron, each travelling away in different directions. The BH is still waiting for equilibrium. The electron and positron are now real particles.

(4) Any "lost" energy caused by the escape of the real particles is then returned to equilibrium by the loss of mass of the BH equivalent to the energy of the escaping real particles. With both particles escaping, it just happens to be a little more mass than if just one had escaped; same process. At this point the black hole is back to equilibrium, and happy.

(5) Real particles which can be observed can only have positive energies. QM only allows the creation and destruction of virtual particles with arbitrary energies. Virtual particles are not in the final state; They must finish their interactions before energies can be measaured, and the time scale is given by Heisenberg's uncertainty principle, delta E * delta t >= h/2pi. Negative-energy particles cannot "propagate". (This part I did find!)

(6) Since both have positive energies, there is no requirement for a replacement of any negative energy anywhere. That has already been done by the BH's mass loss.

That scenario is as much as I can remember on the subject. I might have left out an important point or three, or it could be that this has since been dumped or disproven. Either way, that is as far as I can go on this subject without more references.

Labguy

19. Nov 28, 2003

Ambitwistor

Actually, what he goes on to say is: "The first two cases are straightforward; the virtual particle-antiparticle pair recombine and return their energy back to the void via the uncertainty principle." For case #2 (two particles escape), there is no necessity for one to enter the event horizon. IOW, he's talking about a virtual vacuum bubble, in which the virtual pair is created and annihilated entirely outside of the horizon. This doesn't result in real particles at all.

In QFT, as soon as the vacuum bubble closes, everything is balanced: there are no particles (real or virtual) whose presence is unbalancing anything.

Also, there is no energy "lost" from the black hole, even virtually, because the pair was produced outside of the EH.

To get everything to balance, you either need two real particles with positive net energy being balanced by a loss of energy by the vacuum, or you need two real particles with zero net energy. But the former process can't happen, because the vacuum can't lose energy: it's already the lowest-energy state. The black hole loses mass by absorbing a particle of negative energy, not by the vacuum losing energy.

The above considerations aside, nothing "escaped" in your scenario, in the sense that something left the black hole: you said the pair was produced outside of the EH. So how does the hole lose any energy, if everything is going on outside of it and nothing passes through its horizon in either direction?

Well, if you ever find any references, I'll be happy to go to the library and look them up to see what's going on, because the process you describe bears no resemblance to the calculation of Hawking radiation as I have seen it, nor to the laws of QFT as I know them.

20. Nov 28, 2003

Ok.