Does the horizon shrink down to nothing? or does it evaporate at a certain radius?
As the BH evaporates, the horizon shrinks; eventually, it shrinks to zero radius, and at that instant, the BH has completely evaporated.
However, all this happens long *after* the infalling particle has crossed the horizon, reached the singularity, and been destroyed.
So, the partial reaches the horizon at the same instant that the BH evaporates?
No. The distant observer *sees* the particle reaching the horizon at the same time that he *sees* the BH finally evaporate. But that's because the light that was emitted outward by the particle as it crossed the horizon ends up on the same outgoing trajectory as the light from the final evaporation of the BH. It is *not* because the BH's final evaporation happens at the same event as the particle crossing the horizon.
It is *really* important to keep clear the distinction between what the distant observer sees, as in what light signals he receives, and what happens down at the horizon, because the horizon is an outgoing null surface, so light that is emitted outward at the horizon stays at the horizon, for as long as the horizon exists. That means that, when the BH finally evaporates, all the light that was emitted at the horizon, for the entire time of the horizon's existence, is now mingled with the light from the BH's final evaporation, and all of that light goes outward together.
What is observed happening to the particle?
The distant observer never sees anything of the particle once it has crossed the horizon, as I said before. The actual light images he receives will show the BH evaporating away along with the particle falling through the horizon; but as noted above, that does *not* mean that is what actually happened. The light from events that happened in distinct portions of spacetime gets mingled together because of the way the spacetime is curved.
I am having trouble seeing how from the particles frame, there is a discontinuity as it hits the singularity
Because the singularity has infinite curvature. But as noted, the distant observer can never see that, or indeed anything inside the horizon. See below.
but from the distant observers frame, there is no discontinuity, but it is never observed falling into the hole?
Because the distant observer can never see anything inside the horizon. That's what the horizon *is*--the boundary of a region of spacetime that can't send signals out to distant observers.
Surly either, there is a discontinuity some where, from the distant observers point of view, or the particle doesn't hit the singularity. Am I wrong?
Yes. You've left out a third possibility: that the region of spacetime inside the horizon, even though it exists, can't send light signals out to the distant observer. The reason it can't is that the curvature of spacetime in that region is great enough to force even outgoing light to fall inward towards the singularity.
The below is from http://en.wikipedia.org/wiki/Apparent_horizon, not that I necessarily trust Wikipedia 100%
A wise policy. In this case, the Wiki page is correct in what it says, but what it says does not contradict anything I've said.
I have not been able to read the actual papers that are referenced in that article, since they're behind paywalls, but AFAIK, whatever spacetime slicings would not show an apparent horizon, the ones that are relevant to our discussion--the spacetime slicing of the distant observer, and the spacetime slicing of the infalling observer--both *do* show apparent horizons. (Others who have commented in this thread may have read the actual papers and be able to correct me if I've misstated something, or add more clarification.)
Also, as I've noted before, the boundary of the region that can't send light signals to the distant observer is the *absolute* horizon, not the apparent horizon; and strictly speaking, what I said above about light emitted outward at the horizon mingling with light from the BH's final evaporation applies to the absolute horizon. (As I've noted before, for the case we're discussing the difference between the two kinds of horizons, for the obsevers we're talking about, is not significant.)
Finally, just to clear up a possible misunderstanding of the term "apparent horizon": the word "apparent" in this term does not refer directly to whether, or for how long, the observer actually sees light from the horizon. It is just a way of distinguishing it from the absolute horizon. What has been said about the horizon "appearing" to be in front of the infalling observer, even after he has actually fallen past it, applies to both kinds of horizons, apparent and absolute.
Anyway, as I have been tought, if you fall into a bh, you will see a bh in front of you, all the way to the singularity.
You will *see* the horizon in front of you, yes, because of the bending of light emitted *inward* from the horizon by the curvature of spacetime inside the hole. That does not mean you are actually outside the horizon.
There will always be a horizon in front of you from which light cannot reach you. Or is that wrong?
Yes. Light from the horizon appears to be coming from in front of you, but the curvature of spacetime inside the horizon is extreme enough that you can't trust your intuitions about what that light is telling you.
If not then you could see the singularity couldn't you?
No. You can never see the singularity, because it's in your future; you can't see it for the same reason you can't see tomorrow.