Scientists May Get Best View Yet of a Black Hole in Action

That's because the blog post itself is not very well written, and doesn't state correctly what relativity actually says.

Please be way more clear and short and to the point

I'll give it another shot below.

They agree that frame X in the blog post is validly defined and validly used in the thought experiment. Their quote in the blog makes that clear.

It does no such thing. Taylor and Wheeler do not say that the blog post's "Law J" and "Law K" are correct statements of what relativity actually says. They're not; they're not even stated precisely enough to have a definite meaning that can be compared with what relativity says.

Rather than try to go through the blog post in detail, let me try stating two simpler puzzles that bring out what seem to me to be the essential points. Here's the setting for both puzzles: an astronaut is falling through the event horizon of a black hole. Just before reaching the horizon, he launches a probe outward at escape velocity, which will be just a smidgen less than the speed of light.

Now consider the astronaut's local inertial frame (LIF) as he crosses the horizon: i.e., the origin (t = 0, x = 0) of this frame is the event at which the astronaut crosses the horizon, and the time axis of the frame is the astronaut's worldline as he falls in. According to relativity, the following are all true statements:

(1) The horizon is an outward-moving lightlike curve that passes through the frame's origin; i.e., it is the line t = x in that frame.

(2) The probe's worldline starts at t = some value just a little bit less than zero, x = 0, and moves in the positive x direction at just a bit less than the speed of light.

(3) Therefore, in this LIF, the horizon is moving outward faster than the probe.

The first puzzle is simple: if all of the above are true, how can the probe ever escape? Won't the horizon catch it? (Meaning, won't it end up below the horizon, not escaping to infinity?)

The resolution of this puzzle is equally simple: the size of the LIF is much, much smaller than the predicted distance, extrapolated from within the LIF, that it will take for the horizon to catch the probe. Therefore, within the LIF, the probe remains outside the horizon. And once we're outside the LIF, tidal gravity is not negligible, and it will "pull" the horizon down with respect to the probe, so the horizon will never catch the probe.

The calculations underlying what I just said about the size of the LIF vs. the distance required for the horizon to catch the probe are on PhysicsForums here:

http://www.physicsforums.com/showpost.php?p=4285946&postcoun...

The second puzzle is a bit more subtle. Consider the following statements, which are also true according to relativity:

(4) With respect to a global coordinate chart describing the black hole spacetime, the horizon is at a constant radial coordinate r.

(5) With respect to the same global coordinate chart, the probe's radial coordinate r is increasing.

(6) However, the distance from the probe to the horizon, in the astronaut's LIF, is decreasing, at least while the probe and horizon remain within the LIF. (This is obvious from statement #3 given earlier.)

The second puzzle then is, how can #5 and #6 be reconciled? How can the probe have increasing r when it's getting closer to the horizon, which is at constant r?

The best way I know of to see the resolution this second puzzle is to actually draw a spacetime diagram of the LIF, with curves of constant r correctly drawn. If you do that, you will see that the curves of constant r are drawn in such a way that the probe's worldline has increasing r even though its distance from the horizon, in terms of the LIF's x coordinate, decreases. I can't draw such a diagram here, but the key fact that makes this true can be seen by considering statement #4 above, combined with statement #1: the horizon is a curve of constant r, but in the LIF, it is a 45 degree line moving up and to the right! (It's the line t = x in the LIF, per #1.) A curve of constant r lying just outside the horizon will slope up and to the right, a bit more vertical than 45 degrees (which makes it timelike); whereas a curve of constant r just inside the horizon will slope up and to the right a bit more horizontal than 45 degrees (which makes it spacelike). The slope of the lines of constant r that pass through the probe's worldline, which also slopes up and to the right at almost 45 degrees (since the probe is moving outward at almost the speed of light) turn out to be sloped a little closer to 45 degrees than the probe's worldline is; so the probe ends up crossing curves with gradually increasing values of r.

No, it doesn't. The event horizon is an outgoing null geodesic; it has zero proper acceleration. It is certainly not anything like the surface of the Earth.

The horizon forms a hyperbola in any such inertial frame

No, it doesn't. The horizon is an outgoing lightlike surface, so in any local inertial frame that contains it, it will be a straight line moving up and to the right at 45 degrees. (See my post in response to fargolime upthread for a more detailed description.)

in relativity, test particles moving with constant acceleration can be "uncatchable" even though they never reach c. See Rindler coordinates for more.

It's true that the Rindler horizon gives a good flat spacetime analogy for many key features of the event horizon of a black hole. But you appear to have things backwards: the object moving at constant acceleration is not the Rindler horizon; it is "uncatchable" by the Rindler horizon! In other words, the Rindler horizon is a light ray moving in the same direction as the accelerating object, but which never quite catches up with it (because the test object has just enough of a head start).

This illustrates why an outgoing test particle cannot cross the event horizon.

No, it doesn't. The reason an outgoing test particle inside the horizon can't cross it is simple: the horizon is moving outward at the speed of light, and nothing can go faster than light.

There are no inertial frames that cross the horizon

This is not correct; there are, just as Taylor and Wheeler say. There are plenty of wrong statements in the blog post, but this is not one of them.

This is true for a black hole with mass a few times the mass of the Sun, the sort we expect to be formed by the gravitational collapse of stars. However, a much larger black hole would have much less tidal gravity at the horizon. Some of the supermassive black holes that are believed to be at the centers of quasars would have less tidal gravity at their horizon than you feel on the surface of the Earth.

The confounding factor is that if you're falling into a black hole the acceleration can get your velocity to nearly light speed

Velocity is relative. You will be moving at the speed of light relative to observers who are "hovering" just outside the hole's horizon; but other people falling into the hole just ahead of or behind you could have much smaller velocities relative to you, even well after you cross the horizon (depending on how close they were to you to start with).

and at that velocity your perception of time slows, to the point of nearly stopping

Your perception of time would be normal; you would notice nothing unusual in the behavior of clocks you carried with you, even well after you fell inside the horizon (assuming the tidal gravity was bearable--see above).

Also, "time" as you're using it here is relative; there is no absolute notion of "perception of time".

while people watching you fall in might see a burst of xrays as your physical being converted into energy, "you" might perceive nothing at all

They would only see this if it happened outside the horizon, which it might if the hole's tidal gravity was large enough outside the horizon. But in this case, while it would be true that you would perceive nothing at all, that would simply be because you would be turned into x-rays and destroyed; it would have nothing to do with any change in your "perception of time" due to relativity (see above).

Once you reach the horizon, even if you emit x-rays, nobody outside the hole will ever see them, since light emitted at or inside the horizon can never get back out. But your "perception of time" will continue just fine, assuming again that tidal gravity is bearable (see above).

Most theories do not currently postulate a 'far side' of a black hole

By "far side" do you mean a region inside the event horizon? If so, you are wrong; our current theories most certainly do predict (not postulate, it's not an assumption, it's derived as a theorem) that there is spacetime inside the horizons of black holes.

In theory its really just a point where the numbers go out of whack because the equations have a divide by zero error there.

This is only true in a particular system of coordinates; it is not true as a statement about the actual physics. That is, there is no actual problem with spacetime at the horizon; all physical quantities are perfectly finite there. The "divide by zero error" is a purely mathematical problem with one coordinate chart, which can be fixed simply by using different coordinates.