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

It is a solid point though that the tidal forces on the event horizon of a really big black hole would be minimal. You'd be just as dead though. But I stand by my assertion that "Something pedantically accurate like 'nothing alive'" would be pretty boring and not really convey the interesting aspects of gravitational theory that result in singularities on what most folks consider "normal" space-time.

Pizazz is fine, but not at the expense of truth.

You'd be just as dead though

Eventually, yes. Not at the horizon.

singularities on what most folks consider "normal" space-time.

There is a singularity at the "center" of the black hole, yes. (I put "center" in scare-quotes because a black hole doesn't have a "center" in the usual sense; but we don't have a better word for it.) But not at the horizon.

> 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.

I'm discussing the simple picture in the blog and its caption. Not a more complex puzzle, or Laws J & K. The cloud of particles is demanded by GR to be splitting in two along the horizon, when all the particles above the horizon are let to be escaping. This "splitting in two" contradicts the equivalence principle and occurs in every arbitrarily short period of time in the life of frame X. There is always an arbitrarily short duration of time available in any LIF, so frame X's size in spacetime is not an issue.

When you talk about "global coordinate chart" and a "lightlike curve" you're being unnecessarily complex, I say. We're talking a simple inertial frame of special relativity here, plus one basic prediction of GR as it relates to the horizon, namely that everything below the horizon moves inexorably inward toward the singularity. This discussion can be much simpler than you're making it. After reading your explanation I shouldn't be left wondering which sentence in the blog is incorrect or doesn't follow from its premises.

The cloud splits in two and that's a "bug". To prove the blog wrong you need to show that the cloud doesn't split in two, that all the cloud's particles can in fact move outward in formation. Can you do that in a way that clearly shows which sentence in the picture's caption is incorrect?

Agreed.

> It does not mean that everything below the horizon must decrease its x coordinate in the LIF.

Agreed.

> ...it is perfectly possible for the first particle (inside the horizon) to be moving in the positive x direction in the LIF faster than the second particle.

Agreed.

> And this can be true even if the second particle (the one outside the horizon) is moving at "escape velocity".

Agreed.

> And it can also be true even though the first particle's r coordinate is decreasing and the second particle's r coordinate is increasing.

Disagree. There is no way you could show this for an inertial frame falling in the Earth's atmosphere, like for a skydiver (ignoring air friction). The EP demands that the laws of physics in the skydiver's frame and frame X are the same, so what you say here should be the same for the skydiver. Of course in the skydiver's frame you wouldn't use terms like "global coordinate chart" and a "lightlike curve", as that would be unnecessarily complex.

In the skydiver's LIF under the conditions above, the first particle (the lower particle) would always be moving in the positive x direction in the LIF slower than the second (upper) particle, regardless of the skydiver's speed relative to the Earth, and regardless of the second particle's speed relative to the Earth (i.e. it doesn't need to be escaping).

I'm responding to this separately because it was easier than trying to cram this plus my other responses into one post. If you are going to talk about what is or is not the same in the skydiver LIF and the LIF falling through the black hole's horizon, you have to first make sure the initial conditions are set up the same. Here's how you would do that:

(1) The LIF is in free fall, i.e., the astronaut/skydiver who is at rest in the LIF is freely falling in the gravitational field of some central body.

(2) At time t = 0 in the LIF, the astronaut/skydiver meets an outgoing light ray. (In the LIF falling through the black hole's horizon, this outgoing light ray is the horizon; in the skydiver LIF, it's just whatever outgoing light ray happens to be passing him at t = 0. Within the LIFs, there is no way to distinguish the two.)

(3) At some time t = minus epsilon in the LIF, the astronaut/skydiver releases a probe that flies outward at nearly the speed of light. (This is a key point that I don't think you understand: the initial condition in the LIF is that the relative velocity of the probe and the astronaut/skydiver must be the same. It is not that the probe's initial velocity is escape velocity. "Escape velocity" is a global concept, not a local concept; it has no meaning within the LIF. It so happens that, in the LIF falling through the black hole horizon, the first probe gets launched at a velocity that, globally, is just sufficient for it to escape to infinity, whereas in the skydiver LIF, the probe's initial velocity is way, way more than needed for it to escape; but there's no way to tell that from within the LIF.)

(4) At some time t = plus epsilon in the LIF, the astronaut/skydiver releases a second probe that flies outward at a speed even closer to the speed of light than the first probe.

These conditions are perfectly possible to set up in both LIF's (the skydiver LIF and the LIF falling through the black hole's horizon), and within the LIF's, there is no way to tell which LIF you are in; every observation within the LIF will be the same for both. The second probe will move closer to the first probe (while they are both within the LIF); but the second probe will be falling behind the light ray that passes the astronaut/skydiver at t = 0.

It's true that, once all these objects exit the LIF, things will be very different in the two cases. In the skydiver case, the outgoing light ray will catch up with and pass the first probe. In the black hole horizon case, it won't. But there's no way to tell that from within the LIF.

It's also true that the r coordinates of these objects behave very differently, even within the LIF. In the skydiver case, all three of the objects that are moving outward (the first probe, the light ray, and the second probe) are increasing their r coordinates (and rather rapidly at that). In the black hole case, the first probe has (very slowly) increasing r, the light ray (the horizon) has constant r, and the second probe has (very slowly) decreasing r. But as I said in the other post I made in response to your latest, the r coordinate is a global coordinate, not a coordinate in the LIF; and there is no law of physics that says the relationship between local coordinates within an LIF and global coordinates must be the same for every LIF. In fact, it obviously can't be, because the whole point of the equivalence principle is that LIFs that look the same locally can occur in parts of spacetime that look very different on a global scale.