What do you mean by that? In a one black hole situation, a particle whose energy comes from the black hole is emitted, right? But both virtual particles fall into different black holes, which one of the pair of virtual particles becomes real? Are 3rd and 4th virtual particles created after absorption of the 1st and 2nd?
Well, let’s first ignore the issue of the black holes merging later (which we’ll get back to).
Suppose that a pair pops out of the vacuum at BH1. One member falls over the horizon, the other goes to BH2. Net effect: BH1 dumps some energy into BH2. So to answer the basic question, if a pair is created at one BH, then it is the particle that crosses the event horizon of BH1 which is virtual, just as before; it doesn’t matter that the other particle enters BH2, Hawking radiation is Hawking radiation.
Now, let’s suppose that we have two BHs right on top of each other. Pretty clearly, they’ll both be radiating, coming into thermal equilibrium with each other. Both holes are going to be radiating, the heavier one should be losing mass to the lighter one. For any particular event, though, it could go either way.
Of course, I should add that double BH systems would presumably have something like a figure 8 event horizon to begin with, would be insanely complicated, and I’m just speculating (I think accurately) without being qualified to deliver an authoritative answer.
Well, actually I wasn’t saying that either.
Energy / time are canonically conjugate variables, and, as I’m sure you know, that means they are inherently uncertain with respect to each other (HUP). You certainly can’t violate energy conservation if you don’t know what the energy is.
As an aside it isn’t absolutely clear that energy is conserved in general relativity.
Shoot. Minor nitpick in my earlier post… when I said x=3, I meant x=-3; obviously, the farther away they start, the longer it takes for them to coalesce. I have no idea where I managed to lose the minus sign.
As far as
I agree that if I don’t know the energy before I create a virtual particle to arbitrary accuracy, I don’t know the energy of the virtual particle to arbitrary accuracy either. But because we enforce conservation of 4-momentum at each interaction, if I could exactly know the energy of each particle that comes in from one side, then I would exactly know the energy of the virtual particles as well. That is, in e[sup]-[/sup] + e[sup]-[/sup] -> [symbol]m[sup]+[/sup] + m[sup]-[/sup][/symbol], if my incoming particles are in plane waves, then the energy of the virtual photon in the case where only one virtual photon is exchanged is known to be precisely E = 2 sqrt(m[sub]e[/sub][sup]2[/sup] + p[sup]2[/sup]) in the center of momentum frame.
As far as energy conservation in GR, I’d never heard it wasn’t clear; it seems like the general coordinate invariance of the Lagrangian automatically implies that 4-momentum is conserved in the theory, but I’d be enchanted to learn otherwise.
Tim Robinson asks:
John Baez (one hell of a physicist) answers:
For more detail see:
http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html
Interesting. Looks like it’s saying that energy conservation in the form of a continuity equation is still a perfectly valid and proper way of thinking about things, but that you can’t integrate it because of curvature. I’d think that the differential form is the more useful definition, but that’s just a personal preference. Thanks! 
Do we know anything about how to determine which way it would go?
You said that if one virtual particle pops into existence, outside the event horizon and the other inside, the one outside becomes real. How does the first particle “know” the second fell into the black hole? I assume that at some point in time, since it didn’t recombine, it becomes a real particle. But then, how did the second (the one that fell into the black hole) “know” to become the energy-destroying particle?
Right. We’re getting rather beyond my level of expertise here. Ring’s explanation is probably a better way of doing things, to be honest, even though I dislike the concept of borrowing energy from the vacuum, as one needn’t try to explain things like this.
Whichever particle comes out has to become real if it’s to go anywhere, as virtual particles don’t go too far. The simplified explanation (which, to be honest, is about the level at which my understanding lies) is essentially that because negative energy particles can’t be real, you can’t have the situation in which the black hole gains mass by Hawking radiation.
Pleonast, I think, was closest to the right answer. I wanted to know a little more about Hawking Radiation when I saw this topic, so I bit the bullet and went straight to the horse’s mouth: Hawking’s 1975 paper “Particle Creation by Black Holes” in Comm. Math. Phys., vol 43. Thought that now, knowing some quantum field theory and some general relativity, I might finally be able to get a hold on it.
I was wrong.
But here’s what I did learn: (1) the positive-negative virtual particle story was stated by Hawking himself, in the paper, as a heuristic aid. (2) It’s not wise to take that explanation too literally.
But in any case, here’s a more complete statement of Hawking’s story of the virtual particles. Two virtual particles can pop into existence just outside the event horizon. One has real, positive energy, and the other has negative energy. The one with negative energy is then in a “classically forbidden zone”–but it can tunnel through the event horizon. On the other side of the event horizon, spacelike and timelike Killing vectors have exchanged places. This may sound like mumbojumbo, but it means that now the negative energy particle, inside the event horizon, can be real. Since, from the outside, we saw a negative energy particle enter the black hole, that means that the hole has less mass-energy than before.
So that’s a more complete picuture of the story, but I’d just like to point out that it’s only a story. Situations like this really highlight the differences between the Model and the Reality. We tell stories about the virtual particles, but they have no bearing on reality–virtual particles, by definition, are not real. A calculation of the magnitude of Hawking radiation would not explicitly involve discussion of any virtual particles–it was just a way of interpreting what’s happening in a way that was already familiar to physicists. Virtual particles have become popular because they’re an easy way to visualize what’s happening in a quantum field theory calculation, and a big help in explaining things at a very elementary level, but they’re just that–tools. In short, virtual particles are just mnemonics for calculations, and sometimes the analogy gets stretched a little far.
For those of you who can (and want to) handle it:
The virtual particle becoming real means that its momentum goes from being spacelike outside the horizon to timelike inside the horizon, due presumably to the inversion of the roles of r and t at the Schwarzschild radius, in the Scharszchild metric, written in Schwarzschild coordinates.
The real (still shortened) story though, as far I’m concerned, is this:
When we quantize a real (say, scalar) field, we decompose the field into positive and negative frequency components, accompanied by creation and annihilation operators. We interpret these positive and negative frequency components as being related to particles and anti-particles. In curved spacetime, however, the positive and negative frequencies do not have invariant meaning, so this interpretation no longer really makes sense. Moreover, if we have an asymptotically flat metric (1), which becomes a curved metric (2), then returns to an asymptotically flat metric (3), the solutions to the field equations in the two flat metrics will be different–this means that the annihilation operators of region (1) will not annihilate the vacuum state |0> of region (3). We can interpret this as a change in the number of particles. Thus, from this view, all the ‘particles’ radiated as Hawking radiation are ‘created’ during the collapse of the black hole–region (2), when the metric is time varying–we see them coming out longer than that because of the divergent time dilation at the hole’s horizon.
I don’t think two black holes can come into thermal equilibrium. Larger black holes emit less radiation. The lighter will be losing mass to the heavier.
Oops! You’re right, of course… I can’t believe I didn’t catch that (memo to self: check both for typos and physics mistakes), but it does make me wonder how far we can take BH thermodynamics if we can’t talk about thermal equilibrium between two BHs. Obviously, thermal equilibrium is going to require the masses to equalize, which would mean energy going from cold to hot, and as you say, that’s not gonna happen. What’s the deal here?