First things first, from how I understand it if we assume a black hole is non-rotating and uncharged, and does not emit any Hawking radiation, and an object fell into the black hole, then from the point of view of an outside observer they will never, after any period of time, visibly see the object pass the event horizon. This is because the light emitted by the object takes progressively longer to travel from the object to the observer the closer the object gets to the horizon, and at the exact moment it passes the horizon the time it takes for the light to travel to the observer will be infinite. There’s also the fact that the light gets redshifted in the process, and gets dimmer.
Now, I’m under the impression that the time it takes for the object to actually, literally, pass the event horizon is finite regardless of reference frame. It’s just that you could never see it from the outside because the light would never reach you. If you looked at the black hole millions of years after the object started falling, you would still see some residual light from it showing it apparently hovering above the surface, but it actually fell in a while ago even in your reference frame, and what you’re seeing is essentially an echo or ghost.
I arrived at this conclusion through a simple thought experiment. Let's say someone builds a truly massive space station, millions of kilometers across, relatively near a black hole, though far enough away that its gravitational influence on the station is negligible. Let’s then say the entire side of this station is a massive display, that shows the current time aboard the station.
Let's say an astronaut then falls into the black hole. From the astronaut’s point of view, while she might see the clock on the side of the station speed up to some degree, she does not see it speed up infinitely. In fact based on what I can find I’m pretty sure it wouldn’t even seem to speed up that much. She falls through the horizon, and eventually impacts the singularity, and from how I understand it she’s able to see and read the time shown on the side of the station the entire way down. Now let's look from the station’s point of view. They see her fall towards the event horizon, slow down, and eventually stop, and her light gets dimmer and dimmer. Their clock runs to infinity before they observe her fall past the horizon.
Therefore, the astronaut experiences crossing over the horizon, and experiences time continuing to move forwards after doing so, while from an outside observer they never see her cross the horizon even after infinite time has passed.
As far as I can tell, there are two possible explanations for this. The first is that from the station’s reference frame, she does actually fall past the horizon and impact the singularity, they just can’t visibly see it with their own eyes because the photons will never reach them.
The second is that I was wrong about some of what I said earlier, and from the astronaut’s reference frame she doesn't in fact ever cross the horizon or hit the singularity, and the entire heat death of the universe does in fact play out above her, she just can’t see it because the light from it never manages to catch up with her because she essentially has a head start on it. If the black hole emits Hawking radiation, then it evaporates out from under her before she reaches the horizon. If it doesn't emit said radiation then from an outside reference frame she hangs there for eternity, and from her reference frame, well, I don’t know.
The first solution seems to make more sense and is what I concluded, but of course I might be missing something or there might be a third or fourth solution I haven’t thought of. In any case, the big question I have is firstly, am I on the right track with all that - do objects actually pass through the horizon from an external reference frame and we simply can't visibly see it -, and secondly, if I am, how, exactly, would one go about calculating how much time has passed from an external reference frame between an object starting falling towards a black hole and it passing through the event horizon?
EDIT:
Okay, so, I’ve read the comments that have appeared so far and tried my best to look up the terms they used and figure out what they mean.
Here's basically my understanding at this point;
Let’s imagine a world with two spatial dimensions (x and y) and one temporal dimension (z). Let’s visualize this as a 3d volume where horizontal movement is across x and y and vertical movement as across z. In this world, the event horizon of a black hole would be a vertical cylinder, and the singularity would be a vertical line at the center of that cylinder. Outside the cylinder, objects are free to move as they wish horizontally in x and y but must move upwards across z at a fixed speed. Inside the cylinder however this is reversed, they can move as they wish up and down z but must move at a fixed speed in the x and y plane towards the center.
It is worth noting that if you were to take a “slice” of time, outside the cylinder this would appear as a horizontal slice of the volume (with a hole in it). But inside the cylinder, because space is the vertical axis, a slice of time would actually be the surface of another cylinder of smaller diameter. If you were to take a horizontal slice inside the cylinder it would instead show everything that occupied the same position at various different times.
So while you could take a slice of time of the entire universe, inside and outside the horizon, and nothing would technically be in two positions at once, the position of the stuff outside the horizon would be mapped to the surface of a horizontal plane with a hole in it and the stuff inside would be mapped to a vertical cylinder. And as such nothing can move from one to another, even light, except in the edge case where the cylinder’s edge touches the horizontal plane, which represents the event horizon itself, and that can only be crossed in one direction.
This is really weird behavior, but I think I can visualize and understand it now. It has really wild implications if you try to imagine tracing the path an object would take as it falls into the cylinder. It also means that in a way both options were right, when you fall into a black hole you do cross the horizon and eventually hit the singularity, and the entire universe does undergo heat death above you, but you’ll never see it happen because the photons from that are lagging behind where you are in your descent? I think? This is really hard to figure out.
It’s also worth noting that if I’m understanding this right, an FTL drive which simply lets you change your position as fast as you like would not actually help you leave a black hole, because your position is mapped (in the 3d example from earlier) to the vertical axis, and to escape you would have to travel across the horizontal axis, which requires going back in time. An alcubierre drive might work though because those operate (theoretically) by warping spacetime, so in theory you could just un-curve the spacetime inside the horizon until it’s all flat again, essentially deleting the black hole from existence.
All this, however, doesn't really answer my original question. For the reference frame of an outside observer, which is stationary relative to the black hole and a very large distance away from it, is the time taken for an object to fall from an arbitrary altitude down to the event horizon finite or infinite - note that I am NOT asking for the time taken for the photons to get from the object right as it passes the horizon back to the observer, I know that should be infinite. And if it is finite, how does one go about calculating it?
