As I understand it, Hawking Radiation occurs when a virtual pair is created just on the event horizon and the anti-particle member of the virtual pair is drawn into the black hole and the ‘normal’ particle member of the virtual pair escapes.*
My question is this: What makes anti-particles more appealing to a black hole? Because if they aren’t more appealing, why doesn’t the influx of ‘normal’ particles from virtual pairs cancel out that of the anti-particles?
While posting in this thread, I came across the phenomenon Hawking Radiation again. Originally I’d seen it in a book written by Timothy Zahn. Initially, his explanation made sense, but upon re-examination I’m confused. Naturally, I searched for ‘Hawking Radiation’ in GQ and came across this thread, but it didn’t answer my question. Rather than dredge that oldie up, I figured it would be best to make a new one.
*Interestingly enough, according to this article (Provided by MikeS in this thread), that is not necessarily how Hawking Radiation works…but that article is several leagues over my head in terms of mathematical familiarity.
The virtual particles being formed are specifically photons, IIRC. Since photons are their own antiparticle, the result is the same, no matter which one of the pair escapes.
Not just photons escape from black holes. Massive particles also escape.
For massive particle/anti-particle pairs, it doesn’t matter whether it’s the particle or anti-particle that escapes. Either way, energy is leaving the black hole, reducing its mass.
I think you are using the unstated misunderstanding that anti-matter is also negative energy. It is not. Both matter and anti-matter are positive energy, and either one can remove energy from an event horizon vicinity.
Not precisely…at least, that’s now how I’d word it.
The understanding, incorrect it may be, that I was working off of was that the anti-particle would be drawn into the black hole where it would annhilate a corresponding amount of ‘normal’ matter that was part of the black hole… the process would repeat until the mass of the hole was insufficient to maintain its status as a black hole.
The “standard” theory is that black holes have only three properties: Mass, angular momentum and charge. A black hole made from antimatter would be indistinguishable from one made from regular matter, or from one made from just photons (not that that would happen naturally). Black hole evaporation doesn’t depend on matter and anti-matter annihilating inside the black hole where you can’t see it anyway, although I wouldn’t be surprised if you’ve read it described like that this somewhere. It just depends on particles leaving the vicinity of the black hole. If energy is leaving a black hole, its mass has to decrease.
The one that falls in can be considered to have negative energy, since it’s not escaping (gravitational potential energy is negative). The negative energy particle falling in balances the positive energy particle escaping, and the black hole loses mass by swallowing that negative energy particle. But the one that falls in is equally likely to be the particle or the antiparticle, and in fact, since photons are (probably) the particle most often emitted, it doesn’t matter which (since the photon is its own antiparticle).
Incidentally, there is no known lower bound for the mass of a black hole. There is a minimum mass required to form a black hole in the current Universe (a few times the mass of the Sun), but if a hole were to radiate away enough mass, it could drop below that mass and still be a black hole. It’s possible that once a black hole drops below about the Planck mass (about a microgram) it would stop being a black hole, but that’s just a back-of-the-envelope guess, and we would need a working theory of quantum gravity (which we don’t yet have) to say for sure. It’s also possible that there could be lightweight black holes in the Universe dating back to the Big Bang, when holes of any size could have formed.
Also incidentally, it’s impossible in the current Universe for a “normal” black hole (that is to say, one formed from a star or stars) to lose mass on net by Hawking radiation. The temperature of a black hole’s Hawking radiation is inversely proportional to its mass, and a stellar black hole has a temperature of at most a millionth of a degree above absolute zero. Since the background of the Universe has a temperature of about three degrees above absolute zero, heat will flow from the rest of the Universe into a black hole. In other words, even if you put a black hole in the middle of one of the great intergalactic voids, the least dense places in the Universe, it would still gain more energy from eating stray microwave photons than it would lose via Hawking radiation.
Ahhh, I think I’m getting it a bit better now. It’s still unclear, but I think I’m just going to have to do some reading before I really understand it… I think my initial mistake was over-simplifying black holes (IE, assuming that black hole ‘evaporation’ was just matter being stolen from it).
My thanks to everyone who replied, it’s been quite enlightening.
If I understand your problem (and what’s going on) correctly, the difficulty is that you think that when an antiparticle and particle annihilate, that the mass is completely lost. In fact, (as I understand it) it is simply converted to energy, and an observer of a black hole outside the event horizon would never be able to tell that it happened inside the event horizon. However, if either a photon, particle, or anti particle escapes, that reduces the total energy/mass of the black hole. We are not concerned with what falls back in, but the fact that something trickles out is reducing the black hole.
More fundamental (again if I have this right) is that when a particle/antiparticle pair forms, energy is consumed. Sort of the reverse of two particles annihilating each other and releasing energy in the form of gamma radiation. If one goes in the hole and one out, then half the energy that went into creating the particles is lost from the black hole, whichever particle/antiparticle it was.
But now I have a question about the second paragraph - isn’t the scenario you describe there dependant upon the virtual pair being formed by the black hole?
As little as I know about Hawking Radiation, I know even less about virtual pair formation…
Sort of. The energy of the particles comes from the gravitational energy of the hole. When one of them gets away, the gravitational energy( and hence its mass) gets decreased by a little. The hole stores an enormous amount of energy ( and energy equals mass) in the warping of spacetime around it.