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I read that a pair of entangled quantum particles can spontaneously pop into existence for a brief moment and especially near to the event horizon whereby one of the entangled quantum particle passes through while the other escape, and the quantum particle that falls in is always carry negative energy while the escapee always have positive energy becoming a real particle. Does it means that no matter which one of the entangled pair falls into the black hole it always carry negative energy provided the other pair manage to escape and gets promoted to real particle.

So far not good... a pair of virtual matter-antimatter particles are entangled and cannot be tell apart from each other, when one of the pair happens to fall into the black hole we immediately assume it carries negative energy! Just because the other particle becomes real particle, is that the reason?

user6760
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  • Yeah its conservation of energy – Jaywalker Jan 12 '16 at 07:01
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    Macroscopic black holes do not emit charged particles. They are far too cold for that. Instead they will emit very long wave electromagnetic radiation, so the virtual pair picture is a toy model, or maybe not even that. I don't know where it comes from because it surely does not reflect the physics of semi-classical black holes very well. It is far better to understand Hawking radiation as a special case of Unruh radiation. – CuriousOne Jan 12 '16 at 07:08
  • Already asked and answered: http://physics.stackexchange.com/questions/106882/how-does-negative-energy-from-hawking-radiation-cause-a-black-hole-to-shrink ... Also, I would avoid the use of "negative energy", I don't know who uses that notion, but it is a misnomer for what really occurs (and is understandably creating you confusion). Point is, a particle and anti-particle have energy, and one flies away from the black hole (could be either one), so the total energy remaining at the black hole is less than what it was originally. – Chris Gerig Jan 12 '16 at 07:22

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