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RE: Super Cool Science S#!t #24 - The Destroyer of Worlds
If the negatively charged antiparticle is the one pulled in, the black hole's total energy will decrease, and in turn, so too will the total mass of the black hole, because Einstein said so... (E=MC2)
I don't get this part. To me, it would seem that it doesn't matter which particle is sucked in, since both the particle and the anti-particle have a certain amount of mass. The difference in charge shouldn't change the amount of energy, according to my understanding about this.
It's very well possible that I misunderstood things, but I simply don't get this part :P
That's a great question, and I suppose I could have explained it a bit better than "because Einstein said so" :D
When an antiparticle collides with a "regular" particle at the event horizon, they eliminate each other, giving off energy. Since the antiparticle came from a pair outside the black hole, it decreases the total number of particles within.
That help?
Alright, so it meets a particle at the event horizon, giving off energy... But wouldn't that mean the energy is still there in the black hole? Or does it happen at the edge, meaning a part of it leaves and a part of it stays?