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What do you mean by at the same time? Those two photons reach Bob at exactly the same instant? We had already agreed that General Relativity should imply Bob n

What do you mean by at the same time? Those two photons reach Bob at exactly the same instant? We had already agreed that General Relativity should imply Bob never sees Alice reach EH, regardless of quantum effects (at least not in finite time). We also agreed that Bob does see the BH evaporate (in finite time to any desired fraction), as seemingly dictated by QM. Thus it cannot be that the two are observed simultaneously! Moreover, there cannot be a chain of events whereby Alice evaporates into photons (emanating from EH), and those photons (originating from Alice's mass-energy) hit Bob before the photons of Alice when she was near, but did not reach the EH yet (again because Alice cannot be observed to reach EH). It would violate causality. It can't be that both Alice falls into the EH and we receive Alice's energy from evaporation: either the BH does not evaporate or Alice doesn't cross the EH (even in her own observation). Assuming Hawking radiation really does exist, then some kind of evaporation would need to occur as objects approach the EH, it cannot come directly from the EH boundary. Could it be that Alice herself is seen to evaporate -- the evaporation might stem directly from Alice, before reaching the EH? My claim that the EH/singularity never forms would be valid. I could be terribly mistaken of course, but your arguments did not convince me unfortunately (quite the contrary). I wonder if what I'm missing could be related to the expansion of the EH as a body approximates.

pdonis 8y ago
> What do you mean by at the same time? Those two photons reach Bob at exactly the same instant? In the original idealized model of Hawking, yes. That is what the spacetime geometry says. One way of thinking about it is that, with this spacetime geometry, the path of any light ray emitted outward at the event horizon travels along that horizon into the future, but the area of the horizon decreases as the hole evaporates. At the instant of the hole's final evaporation, all of the light rays emitted anywhere along the horizon are at the same point as the final evaporation of the hole, and they all go outward from there together. This is a highly idealized model; but that's what the model says. > We had already agreed that General Relativity should imply Bob never sees Alice reach EH, regardless of quantum effects No, we haven't. You claimed that, and I pointed out that you were wrong. Bob never sees Alice reach the EH in the classical GR model, with no quantum effects. Quantum effects change the model. In the model including quantum effects, Bob does see Alice reach the EH. See above and my previous posts. > there cannot be a chain of events whereby Alice evaporates into photons (emanating from EH), and those photons (originating from Alice's mass-energy) hit Bob before the photons of Alice when she was near This is correct. And it is also what the model of a hole evaporating by Hawking radiation says. All of the light signals emitted by Alice before she reaches the EH will reach Bob before he sees the hole's final evaporation (and the light signal emitted by Alice when she crossed the EH). > I could be terribly mistaken of course, but your arguments did not convince me You are terribly mistaken, and you didn't read my arguments carefully enough. See above. > the expansion of the EH as a body approximates I don't know what you're talking about here. As a black hole evaporates by Hawking radiation, the area of its horizon decreases. It doesn't increase.

pdonis 8y ago

> What do you mean by at the same time? Those two photons reach Bob at exactly the same instant? In the original idealized model of Hawking, yes. That is what the spacetime geometry says. One way of thinking about it is that, with this spacetime geometry, the path of any light ray emitted outward at the event horizon travels along that horizon into the future, but the area of the horizon decreases as the hole evaporates. At the instant of the hole's final evaporation, all of the light rays emitted anywhere along the horizon are at the same point as the final evaporation of the hole, and they all go outward from there together. This is a highly idealized model; but that's what the model says. > We had already agreed that General Relativity should imply Bob never sees Alice reach EH, regardless of quantum effects No, we haven't. You claimed that, and I pointed out that you were wrong. Bob never sees Alice reach the EH in the classical GR model, with no quantum effects. Quantum effects change the model. In the model including quantum effects, Bob does see Alice reach the EH. See above and my previous posts. > there cannot be a chain of events whereby Alice evaporates into photons (emanating from EH), and those photons (originating from Alice's mass-energy) hit Bob before the photons of Alice when she was near This is correct. And it is also what the model of a hole evaporating by Hawking radiation says. All of the light signals emitted by Alice before she reaches the EH will reach Bob before he sees the hole's final evaporation (and the light signal emitted by Alice when she crossed the EH). > I could be terribly mistaken of course, but your arguments did not convince me You are terribly mistaken, and you didn't read my arguments carefully enough. See above. > the expansion of the EH as a body approximates I don't know what you're talking about here. As a black hole evaporates by Hawking radiation, the area of its horizon decreases. It doesn't increase.