On the other hand, the holographic principle says that the event horizon surface is big enough to contain all the information, so in a way of speaking we seem to be free to model black hole information as though it is "on the event horizon".
That may still be the case, but for that we need a theory of quantum gravity.
While this is a common pop science statement, it is not correct. The correct statement is that the external observer never sees the material cross the event horizon, because, heuristically, it would take an infinite time for light emitted exactly on the event horizon to get out to the external observer. However, this does not equate to the external observer being able to assign an "infinite time" to the material crossing the horizon. The notion of "time" the external observer would have to use to do that is mathematically undefined on the horizon.
> there is matter that starts out inside the horizon
No, there isn't. When an ordinary object like a star collapses to a black hole, there is no horizon at all to start with, and there is no matter starting out inside any horizon.
As you fall into a black hole the event horizon shrinks away below you and from your own POV you never cross it. Thats true of every particle or photon it traps. Of course they get twisted and distorted by the extreme gravitational forces, but in principle there is always structure. I suppose from the internal perspective, Hawking Radiation manifests as a rain of negative energy.
I'm sure I'm wildly off the mark, but it's the best I can do.
You have it backwards. From the POV of an external observer you never cross the event horizon, because that's a function time dilation, and light that you emit being more and more red-shifted as you get closer to the event horizon. From the falling observer POV, nothing really changes - or rather, what happens close to the event horizon or just past the event horizon depends on what the theory of quantum gravity says.
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.
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?
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.
It's right.
> 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.
No, they won't. This is simply wrong. See my other responses upthread.
More precisely, the external observer never sees you cross the horizon. But he can still deduce that you did from other data. So the external observer does not say "you never cross the horizon", period. He just can't see it happen.
In this coordinate system, the time coordinate becomes space-like passed the event horizon and the radial coordinate becomes timelike. There is a coordinate singularity around this switch, which is where the infinite coordinate time comes from.
An observer fammiliar with relativity could compute the proper time of a path traversing the event horizon, which they would find to be finite.
No: the path of the free-falling observer at and beneath the horizon is not covered at all by the coordinate system of the external observer. That coordinate system simply becomes mathematically undefined, which means you can't use it to make any claims about what happens.
> In this coordinate system, the time coordinate becomes space-like passed the event horizon and the radial coordinate becomes timelike.
No: there is a separate coordinate patch that covers the region inside the horizon (but not the horizon itself--the horizon is not covered by either patch), in which the "areal radius" coordinate is timelike, and in which the coordinate corresponding to the extra Killing vector field is spacelike. Calling the former coordinate "radial" is at least justifiable, since it still is the "areal radius" coordinate even though it's timelike. But calling the latter coordinate "the time coordinate" is simply wrong; the fact that the letter t is commonly used for it does not mean it's a "time" coordinate in any meaningful sense.
> There is a coordinate singularity around this switch
Yes.
> which is where the infinite coordinate time comes from.
No, which is what makes statements like "infinite coordinate time" incorrect; the correct statement is that neither coordinate patch covers the horizon at all.
> An observer fammiliar with relativity could compute the proper time of a path traversing the event horizon
Yes. The easiest way to do that is to switch to some other coordinate chart which covers the entire path, above, at, and below the horizon. In such a coordinate chart, the "areal radius" coordinate will not be timelike below the horizon. (In at least one chart commonly used for this, Painleve coordinates, all four coordinates are spacelike below the horizon.)
By "time coordinate" I mean the hyperbolic angle. This corresponds to the proper time of our constantly accelerating observer. This remains a well defined coordinate on the interior of the black hole. As you identify, in the language of relativity, this is a spatial coordinate on the interior of the horizon. I specifically refer to this as "coordinate time" because it is the coordinate that our observer is using for time; not because it is actually a time coordinate in all regions of space-time.
> No, which is what makes statements like "infinite coordinate time" incorrect; the correct statement is that neither coordinate patch covers the horizon at all.
You do not need to consider the horizon itself. Consider just portion of the path that occurs before the horizon. This path crosses infinite coordinate time. Put another way, the singularity in this coordinate system is precisely that coordinate time diverges towards infinity as you approach the horizon. Put another way, if you pick any finite amount of coordinate time, you can find a portion of free-falling path from before the horizon that crosses that amount of coordinate time .
[0] Although I think we need to be a bit more careful about what we mean by external observer for this to be formally true. A free falling observer is "external" until they cross the horizon, but they don't see a coordinate singularity there.
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.
No, this is not correct. To the extent that the horizon is something that can be "seen" at all, you see it when you cross it. It doesn't shrink away below you.
> I suppose from the internal perspective, Hawking Radiation manifests as a rain of negative energy.
This is a common heuristic picture, but it is very limited. No real particles fall into the hole as a result of Hawking radiation. An observer already inside the horizon might be able to deduce that Hawking radiation was being emitted from local observations on, say, the tidal forces in his vicinity, but I'm not sure even that would be true: the models I am familiar with for evaporating black holes do not require any causal connection at all between the emission of Hawking radiation and any particular observer inside the horizon (in more technical language, the two remain spacelike separated for the entire trajectory of the observer inside the horizon).
Tidal forces are low for supermassive black holes, right?
https://en.wikipedia.org/wiki/Supermassive_black_hole#Descri...