This is essentially the 'no drama' assumption the article is about. It means roughly, that there is no way to detect that you did cross the event horizon. And therefore the brain still works fine.

One of the few "no brainers" in quantum physics.

They don't have to for your brain and nervous system to continue functioning. Suppose you are falling feet first towards the singularity, and your feet send a nerve signal to your head. From your viewpoint, the signal travels up your body normally. But from the "god's-eye" viewpoint, the nerve signal from your feet to your head just falls towards the singularity more slowly than your feet and head do, so your feet fall away from the signal and your head catches up with it.

If I can see the floor, then light from the floor is passing the horizon and entering my eye.

No, it isn't. Consider a ray of light emitted by the floor just below the horizon, and suppose that your eye is just above the horizon when it is emitted. By the time that ray of light reaches your eye, your eye will have fallen below the horizon. So your eye will see the ray of light just fine.

If you are trying to "hover" at a constant altitude above the black hole's horizon, then yes, the acceleration you would feel increases without bound the closer to the horizon you try to hover. But if you're falling into the hole you clearly aren't "hovering"; it is perfectly possible to fall in while feeling only 1 g by adjusting your rocket power properly.

The tidal gravity at the horizon might be very large, or very small, depending on the mass of the hole. If it's large enough, yes, tidal forces would tear you apart by the time you reached the horizon. But that would happen even if your center of mass was in free fall.

I guess my question is how fast you'd be traveling by the time you got near/in the event horizon.

Relative to what? There is no absolute answer to "how fast" you are going; velocity is relative.

In order for that to happen you would have to travel faster than light.

Huh? The light ray and you are moving towards each other. Why would you have to travel faster than light to reach it?

Eg: The light ray came from the floor 1 foot below the event horizon, and your eyes are 3 feet away from it.

No, this is incorrect. You are thinking of the event horizon as a "place in space". It's not. It's an outgoing light ray. And "locations" with a given radius inside the horizon are not places in space either; they are moments of time.

Here's how to unpack what I just said in more detail. What kind of curve is a curve of constant r (and constant theta, phi if we include the angular coordinates)? Our ordinary intuition about how "space" works says that a curve of constant r is timelike; that is, it is a possible worldline for an object. This means that an object can "stay in the same place" by staying at constant r for all time.

If a black hole is present, however, this ordinary intuition is only valid outside the event horizon. Outside the horizon, curves of constant r are timelike. However, the event horizon is a curve of constant r, and it is not timelike: it's null. It's the path of an outgoing light ray. A light ray always moves at the speed of light; it can never "stay in the same place". So you can't think of the event horizon as "a place in space" the way you can think of a curve of constant r outside the horizon.

Inside the horizon, a curve of constant r is spacelike. The physical meaning of a spacelike curve is that it is the set of all events that happen "at the same time" for some observer. So a curve of constant r inside the horizon is "a moment of time" rather than "a place in space".

So a light ray that is emitted "1 foot below the horizon" is really emitted "1 foot to the future of the horizon". Or, since light travels about 1 foot per nanosecond, the light from your feet is emitted 1 nanosecond to the future of the horizon. That light can't travel "upward" to meet your feet; that would mean traveling into the past. Instead, your feet have to fall through the horizon to meet the light.

And don't tell me that it curves away because that would be a violation of conservation of momentum.

It also can't travel directly up then back down, because light can't do that - at the moment of reversal it's not traveling at the speed of light.

And the light particle can't evaporate (i.e. red shift into nothingness) by loosing all its energy to the blackhole because that would be a violation of conservation of angular momentum (every photon caries some angular momentum).

So what happens to it?

I should mention that after studying all the physics that I could understand I do not believe black holes exist. (Super massive objects might, like the observed objects in the center of galaxies, but not as black holes.)

And what does that have to do with conservaton of momentum? If i throw a rock upwards it curves away rather than leaving the solar system — momentum is still conserved and the rock has much more than a photon.

It's the same phenomenon as what causes gravitational lensing.

It's not moving in a curved line, it's moving straight - from the POV of a distant observer as well.

In order to curve it must have some lateral motion relative to "down".

Ordinary masses bend space nowhere nearly enough. The Sun bends space just barely enough for the effect to be observable in a solar eclipse (it was used as a verification of Einstein's theories) by causing stars behind it to appear ever so slightly out of place when its disc is just about to overlap them. The light from these stars is NOT being bent - it keeps going in a straight line, but the Sun's gravity slightly alters the very idea of straight in its immediate vicinity. The effect is a tiny distorsion, about 0.00048 degrees.

And the Sun is not exactly small.... but it's not dense enough to have a strong effect. Nothing in our everyday experience is. If you could stuff all of its mass into a tiny volume then in its immediate vicinity you could experience this.

The "barrier" you speak of (the event horizon) is already traveling outward at the speed of light. Gravity curves spacetime so much at the horizon that a light beam traveling radially outward stays at the same radius forever. Inside the horizon, spacetime is curved even more strongly, so that a light beam traveling radially outward still gets pulled inward towards the singularity.

The Hairy ball theorem requires it.

Are you somehow claiming that traveling in a straight line away from the black hole leads toward the black hole? If that were the case then gravity would be pulling objects away from the black hole, and toward it, at the same time, which makes little sense.

And if it existed anyway, despite making no sense, that would imply there is a point away from the black hole where the gravity of the black hole is exactly balanced (i.e. you could hover there forever), which would a stunningly huge thing to say.

If you mean which way is the event horizon traveling, it's traveling radially outward. That is a well-defined direction all by itself; there is no "left or right" involved.

The Hairy ball theorem requires it.

If you mean the theorem that says there can't be an everywhere non-vanishing vector field on a sphere, that theorem doesn't apply in this case because the manifold of the black hole spacetime is not a sphere (meaning, topologically it's not a sphere).

Are you somehow claiming that traveling in a straight line away from the black hole leads toward the black hole?

No. I'm saying that inside the horizon, even something that is moving as fast as it can away from the black hole still has a radial coordinate that decreases with time. That's because of the way the spacetime is curved; your intuitions about how "space" works don't work in a region of spacetime that is curved as strongly as the region inside the black hole's horizon.

that would imply there is a point away from the black hole where the gravity of the black hole is exactly balanced

How does it imply this? I don't understand your argument here.

> How does it imply this? I don't understand your argument here.

Emit a photon from just inside the event horizon, traveling directly away from "down".

What does the photon do?

Does it go into orbit? (What gave it the angular momentum to do that?) Does it hit the black hole? (How did it turbn around without ever curving?) Does it just travel forever thinking it's moving away from the black hole, but not actually going anywhere? (i.e. redshift into nothingness, since it's constantly fighting gravity)

Something else?

Tell me what it does from the POV of the photon, not the black hole.

That said, from just inside the black hole - what is "down"? Towards the singularity? Does that even make sense when you'll hit the singularity no matter what direction you travel?

From the POV of the photon, nothing special happens. You get emitted, you move towards the singularity, and then we have no idea what happens next. You might happen to take a longer path, but that's it.

The photon is traveling radially outward; that means it is traveling in a direction opposite to a line pointing "down". At least, that's the way it's traveling spatially. In spacetime, a photon at the horizon is traveling along a curve of constant radial coordinate r (and constant angular coordinates theta, phi if you include them). A photon inside the horizon is traveling along a curve of decreasing radial coordinate r; inside the horizon even outgoing null curves (the worldlines of outgoing light beams) have decreasing r.

Emit a photon from just inside the event horizon, traveling directly away from "down". What does the photon do?

According to an observer that is falling inward just inside the event horizon, the photon moves radially outward at the speed of light.

Does it go into orbit?

No. As you say, it has no angular momentum, but that's not the primary reason; the primary reason is that there are no "orbits" inside the horizon. In fact, there are no "orbits" inside a radius of 3/2 the horizon radius; at that radius, a photon can orbit the hole in a circular orbit.

Does it hit the black hole?

Eventually, yes.

(How did it turn around without ever curving?)

It didn't. The spacetime itself is curved inside the horizon to such an extent that even a photon traveling radially outward ends up hitting the singularity.

Once again, your intuitions about how "space" works and how things "travel in space" break down inside the horizon. You are thinking of a point with radius r < 2M, i.e., just inside the horizon, as a "place in space". It isn't. It's more correct to think of it as a "moment of time". The horizon itself is also not a place in space; it's more correct to think of it as an outgoing light beam.

Does it just travel forever thinking it's moving away from the black hole, but not actually going anywhere?

This is one way (but probably not the best way) to visualize what a photon exactly at the horizon does; it is moving radially outward at the speed of light, but because of the curvature of spacetime at the horizon it stays at the horizon.

(i.e. redshift into nothingness, since it's constantly fighting gravity)

"Redshift" is relative; you have to specify what observer is receiving the photon and measuring its frequency.

Tell me what it does from the POV of the photon, not the black hole.

There is no such thing as "the POV of the photon"; photons don't have "rest frames" in the usual sense of that term. I said above what the photon does from the POV of an observer falling into the hole.

No, the light does make it to the horizon.

our math breaks down at the horizon

No, the horizon, and the spacetime inside all the way down to the singularity, can be described perfectly well mathematically.

The larger the black hole, the smaller the tidal forces at the event horizon, so with a large enough black hole, a human body would be fine.

Now, a brain that was only half across would have some trouble as you describe I think. If you crossed it fast enough maybe no part of the brain would have enough time to realize it was isolated? I suspect if you are falling in faster than anything in your body was moving in the opposite direction nothing would be noticed.

First, I'm not sure it's possible for the brain to be half-way across an event horizon. When an in-falling object nears the event horizon, the horizon "dimples" in response to the changing gravitational field caused by the object's mass.

Second, once inside the horizon, it is true that messages cannot be sent opposite the direction of the singularity. However, your entire brain is moving toward the singularity, and therefore neural communication does not require messages to be sent away from the singularity.

>Things past the event horizon should be able to travel in the direction of it, though they will always turn back before actually crossing it.

According to the best of my understanding, if you're inside the event horizon, every direction you look leads to singularity. There are no paths away from the singularity.

Sure it is. But the brain is falling inward, so nothing has to move outward across the horizon for the brain to continue working. Everything falls inward, but signals traveling from one part of the brain to another might fall inward slightly more slowly, so the part of the brain the signal is going to will "catch up" with the signal, from a global perspective.

True, but you are not standing still. You are moving. Accelerating, in fact. No messages need to be sent away from the event horizon for the brain to work.

Maybe that's where we are already :)

The basic mistake in this whole argument is that there is no such thing as an inertial frame or near-inertial frame that crosses an event horizon. An inertial frame outside the event horizon will appear to be fully outside it from within the frame, even if extended.

Event horizons can exist in GR even without black holes, for example, an accelerating observer sees an apparent horizon or Rindler horizon, where signals beyond it will never reach him.

But there is no near-inertial frame that, from within the frame itself, appears to cross the Rindler horizon.

Sure there is. Consider an object falling into the black hole; you can set up a local inertial frame around the event of that object crossing the horizon and it will work just fine even though it straddles the horizon.

What you can't have is a static inertial frame that crosses the horizon; i.e., you can't have one that stays at the same radius, even for an instant. The inertial frame I described above is infalling--i.e., an observer at rest in the frame is falling into the hole.

But there is no near-inertial frame that, from within the frame itself, appears to cross the Rindler horizon.

Yes, there is. This is even easier than the black hole case because you can have a Rindler horizon in flat spacetime, where all inertial frames are global--they cover the entire spacetime, including the region behind the Rindler horizon.

I should note that I do not mean to imply that the "finbot" article linked to is correct; it isn't. As I described above, inertial frames that straddle the horizon can be constructed, and they are perfectly consistent with the equivalence principle.

At this point, my understanding of physics gets hazy so I'm not sure how much this extrapolates to the black hole case.

Actually, the accelerating observer can't "see" the Rindler horizon; light emitted at the Rindler horizon never reaches the accelerating observer (that's the definition of the Rindler horizon). The inertial observer is the one who can actually "see" the horizon, because he passes it.

Physical laws that hold in any inertial frame do not necessarily have to hold when observing from a non-inertial frame (for example ability to signal between any two points).

Any two points within the "accelerating frame" of the accelerating observer can signal to each other. The Rindler horizon is not, strictly speaking, "within" that frame; Rindler coordinates become singular at the Rindler horizon.

That said, I don't see what the properties of the accelerating frame have to do with the statement I made that any inertial frame in flat spacetime will include the Rindler horizon, since all inertial frames in flat spacetime are global.