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Why is energy/ mass taken away from a black hole? Doesn't the energy coming from the virtual particle come from the vacuum energy? ...if this is so, why does the black hole have to pay the energy debt?

If the black hole creates the particles by its energy, what energy was used from the black hole to create them?

HDE 226868
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Christophe
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  • The trick is that, relative to an observer at infinity, the infalling particle has negative energy. Therefore, when it goes behind the horizon, then the mass of the horizon will be its initial mass, minus the negative infalling energy. This enables the outgoing particle to go on-shell and have positive energy. – Zo the Relativist Sep 11 '14 at 00:12
  • @JerrySchirmer : There is a change of signature in the Schwarzschild coordinates (time, radius) at the horizon, so the "energy" of the ingoing particle, seen by an observer at infinity, seems to be more a (negative) radial momentum. Is is not possible to consider a process with $3$ actors (black hole, ingoing particle, outgoing particle), where, by conservation of energy, the energy has to be lost by the black hole itself? – Trimok Sep 11 '14 at 09:06
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    @Trimok: the energy is a well-defined thing. I like the derivation of Hawking radiation in terms of in and out states at infinity better, anyway, where you don't directly deal with this "virtual particles" at the horizon interpretation. – Zo the Relativist Sep 11 '14 at 12:56
  • Related: http://physics.stackexchange.com/questions/30597/black-holes-and-positive-negative-energy-particles – mpv Jan 19 '15 at 07:55

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The easiest way to see this is the third law of thermodynamics: every object in the universe has to have a temperature >0K, including black holes. Objects that have a temperature radiate. So if the temperature of the cosmic microwave background radiation is lower than the temperature of a black hole, the black hole has to radiate more energy into the universe than it receives from it (assuming that there isn't any mass to fall into it), which means that it loses net energy. That energy comes (by means of the equivalence of mass and energy) from the enormous mass of a black hole. There are many more complicated ways to formulate and refine this, but in essence there are two fundamental things behind the process: thermodynamics and the equivalence of mass and energy.

CuriousOne
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Samir Mathur gives a very accessible (heuristic?) account of Hawking radiation and why it leads to the black hole losing mass. You can access the page here. It's a work in progress and is updated about once every couple of months.

To summarize, the lifetime of particles is dictated by $\Delta E \Delta t < \frac{\hbar}{2}.$ Normally this means that the more massive the virtual particle-antiparticle pair, the shorter lived it is. However close to the horizon of a black hole, the gravitational potential energy is so negative that the total energy of the pair (rest energy plus gravitational potential energy) is essentially zero. So the pair can become long lived. Of the pair, one of them escapes to infinity, and the other falls in past the horizon. The one that falls in will have a total energy that is negative, because the negative gravitational potential energy outweighs any positive rest mass energy or kinetic energy. As a result, negative energy goes into the black hole and positive energy comes out. From a distance, we see it as the black hole losing mass. Note that it doesn't matter whether it is the particle or the antiparticle that escapes - the negative energy happens because of the very negative gravitational potential energy, no matter whether it is a particle or anti-particle that falls in. After all, both particles and anti-particles have positive rest mass energy.

So that's a rough (hand-wavy) explanation of how Hawking radiation leads to a black hole losing mass.

G. Paily
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