It is a given that black holes radiate their mass away in the form of hawking radiation.
Now, virtual particle pairs are constantly created spontaneously and then self anihilate before the rest of the universe notices their existance. Only when one of the particles crosses over the hole’s event horizon and is lost does the other particle become real, thus imparting a net loss of mass to the overall mass of the singularity.
How?
I mean, I figure countless trillions of particles each instant fall into the black hole, and since we’re talking particle/anti-particle pairs, there should be an equal chance for both kinds of particles being drawn into the hole, therefore shouldn’t there be a net loss of zero?
Basically the idea is that you need a pretty tiny black hole because the separation that’s acheived between part/anitpart pairs is downright miniscule. The deal is you need to have a creation of the pair so close to the event horizon that one falls in and the other has enough energy to escape universal death.
Since this is a particle that hasn’t existed before we have to get its mass from somewhere, and that somewhere happens to be the blackhole.
The quantum mechanical details of Hawking Radiation are much more complicated, and frankly take a bit too long to parse over in a thread. Just realize that when we talk about Hawking radiation we’re talking about tiny, tiny, tiny amounts of energy emitted from tiny, tiny, tiny blackholes. The only place we find tiny, tiny, tiny blackholes is in the period immediately following the big bang, and so that’s where we expect this phenonmenon to occur. The point of the matter is that eventually the Hawking radiation can syphon off enough of the matter so that the black hole tain’t quite there anymore, thus the only known way to return the matter in the black hole to the observable universe.
Not something that we need worry about unless you’re into that sort of far-out there concerns (like why on earth are string theorists so danged impossible to understand?)
My simplistic understanding of this phenomenon is that virtual particle pair production goes on every where all the time via HUP vacuum fluctuations. The particles can only separate by one wavelength, but if this wavelength happens to be about the same as
the holes circumference then the extreme tidal gravitation can pull them apart with enough force that at a quarter circumference they have gained enough energy to become real long lived particles, and at the same time pay back the energy debt to the nearby negative energy regions of space.
At this point one particle can be captured by the hole and the other escape. In effect the hole has supplied enough energy to create two particles but only gets the energy of one back, in effect radiating a particle.
Keep in mind that this process is not well understood by many very top-flight physicists. For instance check out what John Baez has to say about it.
My understanding of the process is that the entirely random event of pair creation is ongoing for all black holes, but becomes relevant only because the Schwartzchild radius defines a surface, which curves. For any point at the horizon, vectors away from that surface are slightly more likely, so that one particle can escape. For large radii holes the difference is very small, so the sublimation is very slow. For very small radii black holes (Such as the quantum black holes so beloved of science fiction writers) the difference is much higher, so the rate of sublimation becomes quite high.
Tris
“It should be possible to explain the laws of physics to a barmaid.” ~ Albert Einstein ~
“You should see the bar where Einstein used to go drink!” ~ Triskadecamus ~
Hmm, from what I’ve read the effect is noticed by all black holes, regardless of size. I thought what was crucial was that the fluctuaion occur at the horizon itself, regardless of radius or circumference or anything.
As far as why the antimatter gets pulled in instead, that’s an interesting question. The question is moot with virtual photons (since they are their own antiparticle) but I don’t know about the others.
It surely can’t be any attraction from the matter inside the hole since none can be there, but there are electrical alines of force surrounding some black holes, I thought, especially the spinning ones, so maybe this accounts for it?
It doesn’t matter which particle falls through the EH, the fact still remains that the Hole has created two particles but only gets one of them back. Hence it loses energy.
Don’t confuse “negative energy density” with antiparticle. The energy density must be referred to the local gravitational potential energy, so a “negative energy” is “bound” and a “positive energy” is “unbound”
The following is what physicists refer to as a heuristic argument. Simplify “countless trillions” to two pairs. Both consist of a particle and antiparticle briefly popping out of vacuum (technically, they’re a virtual pair). In one pair, the particle falls into the black hole, in the other the antiparticle does. This satisfies the equal chance point. Either the remaining particle and antiparticle escape, in which case the black hole appears to radiate two bodies, or they annihilate each other, in which case the hole appears to radiate the resulting photon. Neither case is a net result of zero.
[The sharp may object that the annihilation here is no different from the usual fate of a particle-antiparticle virtual pair. But the timescale for the annihilation in the double pair + black hole example is unrelated to that for the single pair case. There’s an intrinsic difference and this leads to the effect.]
More rigorous arguments involve recognising that the presence of the black hole is changing the boundary conditions of the vacuum. For those who know some quantum physics, an analogy would be the Casimir effect: parallel plates eliminate otherwise possible modes from the vacuum and the result is an observable force pressing the plates together.
Anything more rigorous than this involves scary maths.
Quite true. If Hawking is right (and correct me if I’m wrong, but I haven’t heard of any experimental verification of the Hawking process yet), all black holes give off what’s known as thermal radiation with a temperature inversely proportional to the mass. That means that the larger and heavier the black hole is, the slower it radiates.
However, it’s also true that larger and heavier black holes won’t radiate massive particles, and all you’ll get are photons and the like. The super tiny black holes that JS Princeton was talking about are the only kinds that’ll give off things like electrons.
This topic is fairly complicated. I tried to read Hawking’s original paper, and let’s just say that some of it was a bit opaque to me. He says in the paper that the model of thermal emission is not to be taken literally, incidentally. I have some more info about this on my Mad Scientist page: http://www.badastronomy.com/mad/2000/hawking.html.
Agreed. The pair starts off virtual, but becomes real once one falls into the black hole. The remaining one is then free to annihilate another real particle/antiparticle to produce a real photon.