Particle at event horizon as black hole evaporates

[Curtnos]

If you are observing a particle enter a black hole, you watch its proper time go to zero at the event horizon as it is 'frozen' there from your frame of reference. What happens in your reference frame as the black hole evaporates? While you can't illuminate where the particle is from your frame, if we pretend that there is some way of knowing its location, does its frozen position converge on the singularity as the event horizon shrinks, and what happens when the black hole is completely evaporated?

 

[PeterDonis]

If you are observing a particle enter a black hole, you watch its proper time go to zero at the event horizon as it is 'frozen' there from your frame of reference.

This description, while it is common, is not really correct; and that is important here because its incorrectness is leading you to misunderstand what happens as a black hole evaporates.

Also, the description you give technically only applies to a black hole that does not evaporate--i.e., a purely classical black hole, where we assume there are no quantum effects happening such as the ones that cause black holes to evaporate.

Here is a better description of what happens in the scenario you describe:

(1) You watch a particle fall into a black hole. As it gets closer and closer to the horizon, light emitted from it takes longer and longer to get to you. The light also gets more and more redshifted; in practical terms, it will get so redshifted that you can't even detect it, while it is still some distance above the horizon. However, we'll ignore this here and assume that you can detect light from the object no matter how much it is redshifted.

(2) As the particle reaches the horizon, its light continues to take longer and longer to get to you. If the hole were not going to evaporate, light emitted by the particle at the horizon would never reach you at all. However, because the hole is going to evaporate, that is not the case. Instead, what happens is that, when the hole evaporates, all of the light that was "trapped" at the horizon starts outward again--in effect the final evaporation of the hole releases a flash that contains all of the light that was trapped at the horizon.

(3) So what you see from far away is that the object becomes very faint and appears to approach the horizon very slowly--then, very far in the future (we are assuming you are immortal so you are still around then), you see a flash of light that includes the image of the particle actually crossing the horizon. But that's the last you see of it--any light emitted by the particle after it crossed the horizon still ended up in the singularity at the center of the black hole, and you will never see it. (Note that that also means there is no way of even assigning a "location" to the particle, in your reference frame, once it's below the horizon.)

Note that the above is based on Hawking's original model of an evaporating black hole, but we don't know if that model is actually correct. There are other proposed models which differ in important respects concerning what happens at and below the horizon. However, all of them would look more or less the same as the description above from far away.

 

[Curtnos]

Thank you! That makes much more sense than the model that the lecturer has been using so far, hopefully in the coming weeks we get to strip off some of the simplifications