Vanadium 50 said:
Once the dust crosses a certain radius you go immediately from the old horizon to the new horizon.
That's not correct. The horizon (meaning the
event horizon) smoothly increases in size from the smaller to the larger radius. And it does this
before the new shell of dust falls in. At the instant when the shell of dust is at the new horizon radius, the horizon has already expanded smoothly to that new radius.
Vanadium 50 said:
The observer is orbiting at r, r1 < r < 2, where r1 is the initial Schwarzschild radius and r2 is the final one. To escape, he needs to travel from r to r2 before the outer horizon has formed. If he waits too long, he has to travel faster than c to do it, and that means he is now inside the horizon.
Yes. And by keeping track of at what value of ##r## the observer would have to travel faster than ##c## to escape, as a function of time (although you have to be careful of how "time" is defined here), you can track the smooth expansion of the horizon from ##r_1## to ##r_2##.
Vanadium 50 said:
I was bugged by the shell theorem for a while. How can a spherical shell cause what is effectively an inward force? But the conditions do not apply when spacetime is this curved.
The shell theorem does apply in this case, but all it says is that the shell itself doesn't
affect the spacetime curvature inside it. It doesn't say the spacetime inside has to be flat. And in this case, of course, it isn't.
Also, there is no "inward force". The spacetime curvature at a given radius ##r##
does not change until the shell reaches that radius ##r##. The reason an observer between ##r_1## and ##r_2## can no longer escape at some point, even though the shell has not yet reached him, is that he would have to travel faster than light, according to the spacetime geometry
inside the shell region--i.e., according to the spacetime geometry with the
smaller mass. If the shell weren't falling in, he
would be able to escape from any ##r## greater than ##2M_1## (the smaller horizon radius). But because the shell
is falling in, his global future worldline is different from what it would have been with no shell falling in, even though
locally he sees no difference in the spacetime geometry when he starts his escape attempt.
The difficulty here is that the event horizon is teleological--its location depends on the entire future of the spacetime. There is no local counterpart to it in general. For a stationary black hole (i.e., one that nothing ever falls into), there is an apparent horizon co-located with the event horizon, and the apparent horizon can be locally detected (by radially outgoing light rays not moving outward). But in the case of the infalling shell, the apparent horizon
stays at the "old" horizon radius ##2M_1## until the instant that the shell reaches the "new" horizon radius ##2M_2 > 2M_1##, and then the apparent horizon discontinuously jumps to the new horizon radius. So the observer with ##2M_1 < r < 2M_2## can't conclude that he is safe because he is outside the apparent horizon. He can't watch for
any local change that will tell him the shell is falling in. There is simply no local way for him to know he is safe.
(Strictly speaking, if the shell is made of ordinary matter, the observer inside could
see it, for example by ingoing light rays emitted from it, and act on that information. But in the limiting case of an infalling shell of radiation, he would have no possible warning at all, since no information about the shell's coming can get to him faster than light, i.e., faster than the shell itself.)