[millstone]

The simplest way to show the essential physical error is via this picture, from the article: http://finbot.files.wordpress.com/2008/03/t9.png

This picture shows the event horizon as having constant position in a free-falling frame. But the event horizon accelerates in all nearby inertial frames, exactly like how the surface of the earth accelerates (towards you) in every inertial frame near the earth. So the line representing the event horizon cannot be straight: it must curve. It must be hyperbolic. That is the key feature that the article gets wrong.

> Then law J shows that a locally inertial frame relative to which the particle is at rest can’t even in principle extend below the horizon

The horizon forms a hyperbola in any such inertial frame. Because the horizon accelerates, whether a spatial point is below the horizon depends on time.

If you have a test particle moving with constant acceleration, can you reach it with a signal at some point after it leaves? In Newtonian physics you can: just send the signal fast enough. But in relativity, test particles moving with constant acceleration can be "uncatchable" even though they never reach c. See Rindler coordinates for more.

This illustrates why an outgoing test particle cannot cross the event horizon. Picture the event horizon as a hyperbola, and the test particle as an asymptote: it can never intersect it.

> a test of the laws of physics can distinguish X from an inertial frame in an idealized, gravity-free universe

Well duh. There are no inertial frames that cross the horizon. Technically any two separated points in our frame will be accelerating with respect to each other. We approximate this acceleration as zero, but this approximation can break down near the speed of light, as things tend to do.

[pdonis]

the event horizon accelerates in all nearby inertial frame

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.

[fargolime]

> This picture shows the event horizon as having constant position in a free-falling frame. But the event horizon accelerates in all nearby inertial frames

The line represents the horizon at a single moment in the life of frame X, during which it has a specific position relative to X and is not accelerating relative to X.

> If you have a test particle moving with constant acceleration, ...

The test particles are defined as freely falling. The frame X is defined as inertial. No free particle can accelerate in an inertial frame (at least not measurably), including in relativity. Check out Taylor and Wheeler's quote at the bottom of the blog post: "Keys, coins, and coffee cups continue to move in straight lines with constant speed in such a local free-float frame." According to you, those things would not move at constant speed; rather they'd be "moving with constant acceleration".

> Well duh. There are no inertial frames that cross the horizon.

This again disagrees with Taylor and Wheeler's quote. They are clear that not only can an inertial frame cross a horizon, but also that frame is "a free-float frame like any other".

I didn't respond to everything you said because it hinges on the same disagreement. Before we could get further in the discussion you would need to accept Taylor and Wheeler there, or show how I'm misunderstanding the conflict between what you said and their quote that seems to explicitly disagree with you.