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## Summary:

- Does Hawking radiation at the event horizon of a black hole result in the reduction of the mass of the black hole?

Does Hawking radiation at the event horizon of a black hole result in the reduction of the mass of the black hole?

Quantum mechanics requires the creation and destruction of particle-antiparticle pairs throughout space-time. When this effect occurs at the event horizon of a black hole, sometimes one of the newly created particles ends up falling into the black hole while the other particle moves away from the event horizon.* The accumulated particles that move away from the event horizon are called “Hawking radiation.”

When the pair-particle that falls into the black hole is an antiparticle, it is annihilated by combining in the black hole with what is (or was?) a particle. At the same time, its partner the particle moves away from the event horizon. The combined effects of this particle-antiparticle pair’s end-state is to reduce the mass of the black hole.

When the pair-particle that falls into the black hole is a particle, it adds to the mass of the black hole. Its partner the antiparticle moves away from the event horizon and is annihilated when it meets a particle. The combined effects of this particle-antiparticle pair’s end-state is to add to the mass of the black hole.

On the assumption that each of these scenarios is equally likely, would not the net effect of quantum mechanics pair-particle creation at the event horizon be neutral despite the existence of Hawking radiation? If this is true, black holes would not evaporate away due to Hawking radiation.

*On the assumption that the pair-particle that moves away from the event horizon requires an escape velocity to do so, this requires the event horizon itself to have a “thickness”. Consider the location of a quantum mechanics event that occurs at x-y-z-t, and that one pair-particle created at that location has sufficient velocity v to move away from the event horizon. Then consider a location x’-y’-z’-t that is slightly different from x-y-z-t and that one pair-particle created at that different location also has sufficient velocity v to move away from the event horizon. That means that there is a non-zero distance between x-y-z-t and z’-y’-z’-t that defines a non-zero thickness for the event horizon itself.

Quantum mechanics requires the creation and destruction of particle-antiparticle pairs throughout space-time. When this effect occurs at the event horizon of a black hole, sometimes one of the newly created particles ends up falling into the black hole while the other particle moves away from the event horizon.* The accumulated particles that move away from the event horizon are called “Hawking radiation.”

When the pair-particle that falls into the black hole is an antiparticle, it is annihilated by combining in the black hole with what is (or was?) a particle. At the same time, its partner the particle moves away from the event horizon. The combined effects of this particle-antiparticle pair’s end-state is to reduce the mass of the black hole.

When the pair-particle that falls into the black hole is a particle, it adds to the mass of the black hole. Its partner the antiparticle moves away from the event horizon and is annihilated when it meets a particle. The combined effects of this particle-antiparticle pair’s end-state is to add to the mass of the black hole.

On the assumption that each of these scenarios is equally likely, would not the net effect of quantum mechanics pair-particle creation at the event horizon be neutral despite the existence of Hawking radiation? If this is true, black holes would not evaporate away due to Hawking radiation.

*On the assumption that the pair-particle that moves away from the event horizon requires an escape velocity to do so, this requires the event horizon itself to have a “thickness”. Consider the location of a quantum mechanics event that occurs at x-y-z-t, and that one pair-particle created at that location has sufficient velocity v to move away from the event horizon. Then consider a location x’-y’-z’-t that is slightly different from x-y-z-t and that one pair-particle created at that different location also has sufficient velocity v to move away from the event horizon. That means that there is a non-zero distance between x-y-z-t and z’-y’-z’-t that defines a non-zero thickness for the event horizon itself.