The Black Hole Information Paradox Comes to an End

Are you sure that's right? Thrust implies acceleration, not just velocity.

Suppose you're just inside the event horizon, from the perspective of an external observer. You're traveling outward at a velocity of .95c. That's lower than the escape velocity (which is > c), so you're going to fall back in before you actually get to the external observer. But if you were close to the event horizon you might cross back over it before you fall back in, right? At which point the external observer could see you again.

And once you're outside the event horizon, e.g. to the place where escape velocity is .9c, you might by then have slowed to .8c, but if you still have the capacity to accelerate back above .9c again, you could still get out, couldn't you?

Strictly speaking, no, the usual definition of event horizon is completely observer independent: the event horizon is the boundary of the region of spacetime that cannot send light signals to infinity. That region, and its boundary, are invariant geometric properties of the spacetime, independent of any observer.

> time and space coordinate switch places

This is a common pop science statement, but it can be very misleading. It would be better to say that, inside the horizon, moving "down in space" (meaning, decreasing the surface area of the 2-spheres you pass through, since the radial coordinate that you are using is defined in terms of those areas) is moving forward in time: the "forward in time" direction in spacetime is also the "decreasing areal radius" direction in spacetime.

Even that, however, depends on a particular choice of coordinates. There are other choices of coordinates where it is not true (at least not in the simple form in which I have just stated it).

I mean that there are two separate, disjoint regions of spacetime (one outside the horizon and one inside the horizon), and each one is covered by a separate coordinate chart. Informally people often talk as though those two charts are "the same", but they're not. They're two separate coordinate charts, that just happen to be described using the same symbols.

> we seem to agree that an external observer sees a coordinate singularity at the horizon

No, we don't agree about that. The coordinate singularity is only there for the particular coordinates we are talking about (which cannot cover the horizon). There are plenty of other possible choices of coordinates that do not have a coordinate singularity at the horizon, so a single coordinate patch covers the entire region of spacetime of interest.

There is a certain sense in which the coordinates we have been using (Schwarzschild coordinates) are the "natural" ones for an observer who is at rest relative to the hole. But one has to be very careful not to put too much weight on that; it can easily lead to errors if taken too far.

> I tend to use a hyperbolic coordinate system.

Can you give a reference for this? What you are describing in your next paragraph (about "hyperbolic angle" and so forth) does not seem like any of the coordinate charts I am aware of for Schwarzschild spacetime (Schwarzschild, Painleve, Eddington-Finkelstein, Kruskal-Szekeres, or Penrose).

> In the region on the outside of the horizon, it matches the natural coordinate system for an observer maintaining constant acceleration away from the black hole.

This is just Schwarzschild coordinates; any observer who is at rest relative to the hole (a static observer) has to maintain constant outward proper acceleration to stay that way. Close enough to the horizon, these coordinates can be approximated by Rindler coordinates on flat spacetime, which are a commonly used coordinate chart in special relativity for treating observers with constant proper acceleration.

However, if Schwarzschild coordinates are what you have in mind, first, your next paragraph doesn't seem to be describing them at all, and second, the key property you give here does not hold on the patch inside the horizon, since there are no static observers there.

> Consider just portion of the path that occurs before the horizon. This path crosses infinite coordinate time

No, it doesn't. Every event on the path that is outside the horizon has a finite value of Schwarzschild coordinate time. You even say so later in this paragraph, so you appear to be contradicting yourself.

It is true that, for some cases, you can take limits as Schwarzschild coordinate time increases without bound to compute quantities that are finite on the horizon. In fact, the proper time experienced by an infalling observer from some altitude above the horizon, to the horizon itself, is one of them. But you cannot use this limiting process to justify saying that the Schwarzschild coordinate time itself "is infinite" at the horizon.

> A free falling observer is "external" until they cross the horizon, but they don't see a coordinate singularity there.

Any observer can choose any coordinates, regardless of their state of motion. Whether any observer "sees a coordinate singularity" at the horizon is purely a matter of which coordinates they choose, and has no physical meaning.