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If two particles are created at the event horizon of a black hole (in popular language), the particle moving towards the hole will increase the mass of the hole. The event horizon though decreases in area. But an increasing mass of the hole implies that the horizon gets bigger. So is the curvature of space reduced when a particle anti-particle pair is produced, by extracting energy from the spacetime surrounding the hole and in such a way that the mass of the hole decreases (though a particle is added to the hole)?

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    $\begingroup$ The mass of the BH decreases, it does not increase. The naïve way of thinking of this is that the particle which crosses the horizon has negative energy, but the real story is more complicated. $\endgroup$ – tfb Feb 1 '17 at 22:49
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When Virtual particles are created, they usually vanish right after. During this period, Due to energy conservation the particles have to borrow energy from somewhere. But when 2 particles form at the event horizon of a black hole, one gets sucked in while the other escapes.Hence, the energy must be derived from the black hole. This is like saying you have a certain number of biscuits. Your friend takes 2 biscuits ( borrows energy) from you and then takes one and gives you one back. So, end up losing one biscuit(losing energy). So, the more this happens, the more energy or biscuits you keep losing and eventually, you will run out of biscuits. That's exactly what happens. Since the energy and mass go down, the radius of the event horizon gets smaller. If you mean curvature at the point that used to be the event horizon, the curvature decreases. But, if you mean curvature at a point on the new and smaller event horizon, it remains the same as the event horizon is a boundary where all possible paths lead to the singularity. To be more clear,

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Far away from the black hole a particle can move in any direction. It is only restricted by the speed of light.

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Closer to the black hole spacetime starts to deform. In some convenient coordinate systems, there are more paths going towards the black hole than paths moving away

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Inside the event horizon all paths bring the particle closer to the center of the black hole. It is no longer possible for the particle to escape.

Light or any other particle for that matter cannot escape if the curvature of space goes above a beyond point. So, the curvature of all the event horizons is the same. The things that vary are the radius of the horizon( distance from the singularity to reach a point where you have the required curvature, to make light not be able to escape), and the rate of change of curvature as we move radially outward from the singularity.

Hence, to sum it all up:

  • If you mean curvature at the point that previously used to be the event horizon, the curvature decreases. But, if you mean curvature at a point on the new and smaller event horizon, it remains the same as the event horizon is a boundary where all possible paths lead to the singularity.
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  • $\begingroup$ The question is actually about the curvature of spacetime. Perhaps you can add what happens to it. $\endgroup$ – flippiefanus Feb 2 '17 at 4:29
  • $\begingroup$ I've edited the answer as I was ambiguous. Hope this clears stuff up $\endgroup$ – Chandrahas Feb 2 '17 at 6:30
  • $\begingroup$ @Chandrahas-Got it! Thanks! Very good explanation. So at the moment the particles are created they take energy(to become real) from the curvature of spacetime? $\endgroup$ – descheleschilder Feb 2 '17 at 7:59
  • $\begingroup$ Yeah. They use the black holes energy (or as you say, the spacetime curvature surrounding them) $\endgroup$ – Chandrahas Feb 2 '17 at 10:45

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