You said that from the perspective of a falling observer, they never see themselves cross the event horizon. That's not right. The falling observer certainly will see themselves cross the event horizon. In their frame of reference, time flows normally. For smaller black holes, they may get spaghettified, but for larger ones, crossing event horizon will be an unremarkable occurrence.
There will always be an event horizon below them beyond which light cannot reach them, and they will always be able to observe the external universe. So how will they see themselves crossing the event horizon?
> There will always be an event horizon below them beyond which light cannot reach them
Wrong. See my other response upthread.
> they will always be able to observe the external universe
But the light coming to them from the external universe will be more and more redshifted, and they can calculate from the observed redshift when they have crossed the horizon.
Also, they will only see a limited portion of the future of the universe; anything that happens later than a fairly short time after they cross the horizon, they will never see, because they will hit the singularity inside the hole before light from such events can reach them.
The usual definition of event horizon is from the perspective of the external observer—the falling astronaut can calculate the threshold at which no amount of thrust can ever move them away from the singularity (in fact, this is what is meant that a time and space coordinate switch places—you can now only move 'down' in space, just like you can only move forward in time).
> The usual definition of event horizon is from the perspective of the external observer
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).
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?
> Are you sure that's right? Thrust implies acceleration, not just velocity.
Yes. The simpler answer is that even if you shoot a laser straight up the light can't get out, and the corollary is that you can't get out yourself even with arbitrarily high thrust.
No matter how much thrust you have to fight against gravity, light will always beat you, because light cannot be slowed down by gravity. If light is trapped then you are trapped.
> You're traveling outward at a velocity of .95c.
Let's make it simpler, and calculate what it takes for the outside observer sees you moving away from the black hole at all, even 1mph.
The closer you get to the black hole, the higher your local speed has to be. If you're 10 meters outside the event horizon, you have to get up to 99.999etc. percent of light speed (locally measured) just to slowly inch away from it (as measured by the outside observer), or even just to stay still. Once you fall inside the event horizon, you would have to go faster then light locally for the outside observer to measure you as moving upward.
> Suppose you're just inside the event horizon, from the perspective of an external observer.
Whether or not you are inside the event horizon is not a matter of perspective. All observers will agree on it.
> You're traveling outward at a velocity of .95c.
Relative to what?
Outside the horizon, velocities, such as the escape velocity you mention, are usually meant as velocities relative to static observers--observers who are maintaining the same altitude above the horizon.
But below the horizon, there are no static observers. It is impossible to maintain a constant "altitude" (which here just means the surface area of the 2-sphere at your location, centered on the hole) even for an instant. So "velocity" relative to a static observer has no meaning inside the horizon.
> 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.
No, that's not what happens. Relativity is not Newtonian physics.
The event horizon is not a place in space. It is an outgoing null surface: a surface formed by radially outgoing light rays. Spacetime around the black hole is curved in such a way that this particular outgoing null surface has a constant surface area: in other words, even though there is a full 2-sphere's worth of radially outgoing light rays at the horizon, the surface area of the 2-sphere formed by those light rays is constant--the 2-sphere does not expand, as you would expect a radially outgoing spherical wave front to do, because of the curvature of spacetime.
If you are just a little bit inside the horizon, you are on a slightly smaller 2-sphere, and on that 2-sphere, even radially outgoing light rays cannot stay in the same place, much less expand; they contract--that spherical wave front, even though it is radially outgoing, decreases in area with time, i.e., it is falling. So you, having to move slower than light, will fall even more. You can't move outward at all; you can't even stop moving inward. (The fact that there are no static observers inside the horizon is a consequence of this.) That is true no matter how hard you accelerate, since no amount of acceleration will let you move faster than light.
