Can Humans Survive Near a Black Hole's Event Horizon?

[mjacobsca]

1) If you could hover over a black hole's event horizon (let's say it was a BIG hole with lower tidal forces and you could hover 1 mile out), and stuck a 1mi long pole over the event horizon such that 1mm of the tip of the 1mi pole were over the horizon, would the entire pole be ripped out of your hands? Could you pull the 1mm tip back out? I assume you couldn't.

2) I'm still slightly confused by Hawking Radiation. Almost all examples I've read give an example of two virtual particles developing OUTSIDE of the event horizon, and one particle being pulled into the event horizon while the other is escaping. I'm unclear as to how the black hole loses mass when it is adding particles from outside. One example I read said that the particle that gets pulled into the horizon gains gravitational energy from the black hole and turns into a real particle, hence the hole loses mass. However, even if the black hole lost gravitational energy to the particle that got sucked in, isn't the energy sucked back into the black hole? How is energy lost. I'm hoping one of you can give a hobbyist physics enthusiast a non-mathematical explanation of the whole phenomena.

Mike

 

[Haelfix]

1) From a free falling observers point of view, he/she does not have any information available to actually know when they pass the event horizon or even where it is. There is nothing unusual about the particular point locally. The observer will still feel whatever tidal forces he felt 1 m before and 1 m after.

The rough analogy is to think of fish swimming down a river where the current is steadily getting faster. The fictitious point where the fish can no longer turn around and go back to his starting point is the 'event horizon', but from its vantage point there is no way off knowing where or when exactly that is, everything is still perfectly uniform and smooth before and after. The fish would need to know the global properties of the river to make that assessment, information that in the case of black holes it does not and cannot have.

2) Technically, the black hole doesn't lose mass like you would think (as in chunks of it coming off). Rather it absorbs 'negative energy'. The explanation you were given is morally correct, the virtual particle 'borrows' energy from the black holes gravitational field (somewhat reminiscent of the Dirac sea analogy). Its a little hard to give the precise details here, b/c the process is still not understood perfectly well and it involves some rather heavy duty mathematical machinery so the word analogy has to stay loose. For instance, exactly what does one mean by 'particle', and what does one mean by 'energy' or the 'gravitational field of a bh' . Its a bit involved to give precise meanings to these rather fuzzy concepts which are very far removed from the classical counterparts.

 

[xantox]

1) If you could hover over a black hole's event horizon (let's say it was a BIG hole with lower tidal forces and you could hover 1 mile out), and stuck a 1mi long pole over the event horizon such that 1mm of the tip of the 1mi pole were over the horizon, would the entire pole be ripped out of your hands? Could you pull the 1mm tip back out? I assume you couldn't

The proper acceleration felt by a hovering observer at radius r tends to infinity when r tends to 2M (the event horizon), for black holes of any size. This means that lower portions of a hovering pole would be subject to increasing and diverging stresses, ie it would break or vaporize as soon as you start to lower it sufficiently near the horizon.

 

[nutgeb]

The proper acceleration felt by a hovering observer at radius r tends to infinity when r tends to 2M (the event horizon), for black holes of any size. This means that lower portions of a hovering pole would be subject to increasing and diverging stresses, ie it would break or vaporize as soon as you start to lower it sufficiently near the horizon.

I agree with this but with a qualification -- I think Haelfix's description is correct, that from the local perspective of the pole, the fact that part of it has crossed the event horizon creates only a (relatively tiny) incremental increase in acceleration, not a stark contrast from one end of the pole to the other. At the radial distance of the event horizon, the tidal discrepency in acceleration as between one end of the pole to the other may be fairly small, particularly if the mass of the BH is large. But if the pole is very, very long, tidal forces could come into play, causing the pole to break.

I think the specific answer to the OP is that the hovering person holding the outer end of the pole would never be able to pull hard enough to bring the inward end of the pole back out over the event horizon -- or even to prevent it from accelerating ever faster toward the BH. If the pole-holder is capable of arbitrarily increasing her pulling force progressively, eventually the material of the pole will stretch to its breaking point, and any broken piece(s) of the pole that extend at all into the event horizon at the time of the break will continue accelerating into the BH. The remaining piece held by the pole-holder would of course stay entirely outside the event horizon.

 

[xantox]

Iand any broken piece(s) of the pole that extend at all into the event horizon at the time of the break

If a pole is hovering, it shall break before any part of it crosses the horizon.

 

[nutgeb]

If a pole is hovering, it shall break before any part of it crosses the horizon.

I don't know where you got the idea that gravitational acceleration becomes infinite at the event horizon. That's not correct. Gravitational acceleration increases smoothly at 1/r² as the horizon is approached and crossed. Nothing unusual happens to the acceleration gradient at the horizon.

Gravitational acceleration approaches infinity at the singularity at the center of the BH, not at the horizon.

 

[JesseM]

I don't know where you got the idea that gravitational acceleration becomes infinite at the event horizon. That's not correct. Gravitational force increases smoothly at 1/r² as the horizon is approached and crossed.

There is no "gravitational force" in GR as there is in Newtonian gravity. A freefalling observer experiences no G-forces anywhere, but non-inertial observers do just like they do in SR, and the G-force experienced by an observer hovering non-inertially at constant Schwarzschild radius does go to infinity as you approach the horizon. This is discussed on this page, in the paragraph that begins with However, this acceleration is expressed in terms of the Schwarzschild radial parameter r, whereas the hovering observer’s radial distance r' must be scaled by the “gravitational boost” factor. The equation they give for the "proper local acceleration" felt by a hovering observer is \frac{-m}{r^2\sqrt{1 - 2m/r}}, which does go to infinity as r approaches the Schwarzschild radius 2m.

Gravitational acceleration approaches infinity at the singularity at the center of the BH, not at the horizon.

It's tidal forces (and curvature) that approach infinity at the singularity, not "gravitational acceleration" (of course it is impossible to calculate the proper acceleration of a hovering observer at any radius less than the Schwarzschild radius, because such an observer would have to be moving faster than light).

 

[nutgeb]

Jesse, I am convinced your statements are wrong, but I will leave it to others to contradict or confirm them.

 

[JesseM]

Jesse, I am convinced your statements are wrong, but I will leave it to others to contradict or confirm them.

Are you convinced that the page I linked to is wrong? It says clearly in that paragraph and multiple subsequent ones that the proper acceleration experienced by a hovering observer (the G-force) goes to infinity as you approach the horizon. You can find the same equation for proper acceleration experienced by a hovering observer in textbooks too--for example, see pages 3-31 to 3-32 of https://www.amazon.com/dp/020138423X/?tag=pfamazon01-20, see pervect's post #4.

I have observed that you tend to make a lot of incorrect statements about GR based on non-quantitative arguments that wrongly blend GR concepts with classical ones (for example, the concept of a gravitational force which drops off according to an inverse-square law, an idea which is not fundamental in GR though it can be recovered in the Newtonian limit). If your response to being corrected is just to say "I disagree" but then not actually explain why you disagree, then you will never learn anything new.

 

[DrGreg]

JesseM is correct. The equation he quotes can also be found in the book

Woodhouse, N M J (2007), General Relativity, Springer, ISBN 978-1-84628-486-1, p.99​

(which applies under the convention that units are chosen where c = 1 and G = 1). You can find the lecture notes on which this book was closely based http://people.maths.ox.ac.uk/~nwoodh/gr/index.html , p.54 (Lecture 12).

You are perhaps being confused by the fact that in Schwartzschild coordinates the radial spatial component of four-acceleration is indeed given m/r². But after applying to local metric, the proper acceleration is the formula given by JesseM (what you actually feel and can be measured by an accelerometer, and which dictates the force required to provide the acceleration).

If the "acceleration due to gravity" (i.e. the proper acceleration of a hovering particle) were finite at the event horizon, it would be possible for massive particles to hover at the event horizon by applying a finite force. Such hovering is impossible; only massless particles can hover at the horizon.

 

[nutgeb]

OK, I stand corrected. Dr Greg, I appreciate the way you give explanations without superfluous strutting of ego.

Question: If the locally measured acceleration becomes infinite as the horizon is reached, then why doesn't the locally measured velocity also immediately become infinite? Presumably the locally measured rate of acceleration (1st derivative of velocity) must return to a finite level once the observer attains a non-zero inward velocity at or inside the horizon.

 

[JesseM]

OK, I stand corrected. Dr Greg, I appreciate the way you give explanations without superfluous strutting of ego.

Strutting of ego? I didn't talk about how great I was, I just criticized you for repeatedly refusing to respond to critiques of your claims (apparently you only do this when the critiques come from me, out of some personal grudge I imagine)

Question: If the locally measured acceleration becomes infinite as the horizon is reached, then why doesn't the locally measured velocity also immediately become infinite?

There's never a physically possible situation where the acceleration measured by an accelerometer is infinite, because it's impossible for any massive observer to hover on the horizon. The locally measured acceleration can become arbitrarily large at smaller and smaller finite distances from the horizon of course, but "locally" means you're talking about an infinitesimally small region of both space and time, so the change in velocity (as measured in the locally inertial coordinate system of a freefalling observer passing the 'hovering' object) should be infinitesimal in this region too regardless of the magnitude of the acceleration.

A non-massive object like a photon can hover on the horizon, though of course it can't carry its own accelerometer. But there is a sense in which the photon's acceleration can be seen as infinite from the perspective of a freefalling observer passing it, as discussed on the first page I linked to:

Interestingly, as the preceding figure suggests, an outward going photon can hover precisely at the event horizon, since at that location the outward edge of the light cone is vertical. This may seem surprising at first, considering that the proper acceleration of gravity at that location is infinite. However, the proper acceleration of a photon is indeed infinite, since the edge of a light cone can be regarded as hyperbolic motion with acceleration “a” in the limit as “a” goes to infinity, as illustrated in the figure below.

http://www.mathpages.com/rr/s7-03/7-03_files/image011.gif

[/URL]

 

[skeptic2]

1) If you could hover over a black hole's event horizon (let's say it was a BIG hole with lower tidal forces and you could hover 1 mile out), and stuck a 1mi long pole over the event horizon such that 1mm of the tip of the 1mi pole were over the horizon, would the entire pole be ripped out of your hands? Could you pull the 1mm tip back out? I assume you couldn't.

If I'm not mistaken, if a person hovering 1 mile above an event horizon (from his perspective) dangles a 1 mile pole towards the event horizon, due to the contraction of spacetime, the pole will not even reach the event horizon.

 

[nutgeb]

Strutting of ego?

The strutting that occurs on forums such as this from time to time can include (a) putting others down, (b) nitpicking casual uses of terminology as a tactic to discredit the other party's knowledge, (c) digressing into off-topic discussions that are of little relevance as a tactic to bolster the credibility of one's on-topic assertions, and (d) needing to have the last word on every subject.

I just criticized you for repeatedly refusing to respond to critiques of your claims ...

Repeatedly? I do try to avoid expanding on my first attempt at an explanation unless and until I'm ready to say something coherent.

The locally measured acceleration can become arbitrarily large at smaller and smaller finite distances from the horizon of course, but "locally" means you're talking about an infinitesimally small region of both space and time, so the change in velocity (as measured in the locally inertial coordinate system of a freefalling observer passing the 'hovering' object) should be infinitesimal in this region too regardless of the magnitude of the acceleration.

Are you specifically saying that the "small region of both space and time" is always exactly the right size such that the "arbitrarily large" acceleration always results in the observer crossing the horizon at a locally measured velocity of exactly c, regardless of the observer's initial velocity?

Also, if an observer hovering infintessimally close outside the horizon gets accelerated across the horizon, they are going to locally measure that their velocity continues to accelerate in a finite way as they approach the center. Since they had locally measured the acceleration to be arbitrarily large and approaching infinite while they were still hovering, then once they attain a non-zero inward velocity they must thereafter locally measure their rate of acceleration (change in velocity) to be finite and no longer arbitrarily large.

 

[Haelfix]

Its probably worth putting numbers in here. 1 mile is not very long, and is pretty localized. Further what's the magnitude of the tidal forces we are talking about (or what's the mass of the black hole in question) which impacts on how much burning an observer on a rocket ship has to accelerate to stay hovering.

 

[JesseM]

The strutting that occurs on forums such as this from time to time can include (a) putting others down, (b) nitpicking casual uses of terminology as a tactic to discredit the other party's knowledge, (c) digressing into off-topic discussions that are of little relevance as a tactic to bolster the credibility of one's on-topic assertions, and (d) needing to have the last word on every subject.

I'm a stickler for correctness and I don't like to see incorrect claims bandied about. This may result in putting down arguments that are incorrect (I haven't put you down as a person because I don't know anything about you), or making an issue of correcting wrong claims others make even if they aren't central to what was being argued (the charge of wanting to have the last word is unfair, since what I really want is for the other person to stick to the topic being discussed and either explain in detail what points of mine they disagree with and why, or else come to an agreement). This isn't ego-strutting, it's just a minor obsession with accuracy in every detail.

I just criticized you for repeatedly refusing to respond to critiques of your claims ...

Repeatedly? I do try to avoid expanding on my first attempt at an explanation unless and until I'm ready to say something coherent.

But if you make some claim that you can't justify in a rigorous way, then you shouldn't be making it on this forum in the first place--the rules state that this forum is not meant as a place for speculative discussion of claims that are not justifiable in mainstream SR and GR (of course the rules allow people to ask questions about whether a particular nonrigorous conceptual argument can be made rigorous in a GR context, but quite often you make definite assertions using ill-defined conceptual arguments). And yes, this has happened repeatedly in discussions I've had with you, as in this thread where you were making various arguments which were ill-defined or outright impossible in a GR context (like talking about the length of an object dragged from one region of Schwarzschild spacetime to another, even though extended rigid bodies are impossible in GR), or this thread where you were claiming the relativistic Doppler effect causes light to become more blueshifted than would be predicted by the classical Doppler formula when as I argued the opposite seems to be true, or this recent thread where you asserted that redshift could be derived from energy considerations and never provided a derivation or acknowledged that you weren't sure that such a derivation would actually be possible. I've also noticed a number of other threads that I didn't participate in where you also seemed to be using ill-defined arguments to try to come to definite conclusions which are not an accepted part of mainstream GR, like this one and this one and this one and this one.

Are you specifically saying that the "small region of both space and time" is always exactly the right size such that the "arbitrarily large" acceleration always results in the observer crossing the horizon at a locally measured velocity of exactly c, regardless of the observer's initial velocity?

Again, the size is understood to be infinitesimal when we are talking about "local" measurements. As measured in the locally inertial frame of a freefalling observer crossing the horizon, the horizon is always measured to be moving outward at exactly c.

Also, if an observer hovering infintessimally close outside the horizon

A massive observer can't hover infinitesimally close the horizon, they can only hover at some finite distance from it (an infinitesimal is smaller than any possible finite real number greater than zero).

they are going to locally measure that their velocity continues to accelerate in a finite way as they approach the center.

You can't locally measure your velocity relative to the center since the center is not in the same infinitesimal region of spacetime as you. Also note that the singularity has a timelike separation from an observer in the horizon, not a spacelike one, so talking about your velocity relative to the center is analogous to talking about your velocity relative to the Big Crunch in a closed universe. The Schwarzschild coordinate system confuses things by having a radial coordinate which is physically timelike inside the horizon (and a time coordinate which is physically spacelike inside the horizon), it's less confusing if you use a coordinate system where the time coordinate is always timelike and the radial coordinate is always spacelike, such as Kruskal-Szekeres coordinates (see the bottom section of this page). Kruskal coordinates have the nice property that light rays always look like straight diagonal lines at 45 degrees in this coordinate system, while timelike worldlines always have an angle that's closer to vertical than 45 degrees. In Kruskal diagrams you can see that the singularity is a timelike future one that everyone inside the horizon will inevitably hit as their worldline takes them forward in time, much like the Big Crunch in a closed universe.

Since they had locally measured the acceleration to be arbitrarily large and approaching infinite while they were still hovering, then once they attain a non-zero inward velocity they must thereafter locally measure their rate of acceleration (change in velocity) to be finite and no longer arbitrarily large.

The thing about acceleration approaching infinity as you approach the horizon is specifically for hovering observers--a freefalling observer's accelerometer will show a reading of zero at every point up to and including the horizon, and accelerating observers who are not hovering will measure different G-forces from hovering observers even when they cross paths at the same point in spacetime. It's simply impossible for a massive observer to hover at any radius inside the horizon, so the question of how the formula for acceleration experienced by hovering observers continues past the horizon isn't a meaningful one.

 

[DrGreg]

OK, I stand corrected. Dr Greg, I appreciate the way you give explanations without superfluous strutting of ego.

That's not my interpretation of JesseM's comments. His style differs from mine, but I think he was just trying to get you to think before you make definitive statements that might actually turn out to be false.

Question: If the locally measured acceleration becomes infinite as the horizon is reached, then why doesn't the locally measured velocity also immediately become infinite? Presumably the locally measured rate of acceleration (1st derivative of velocity) must return to a finite level once the observer attains a non-zero inward velocity at or inside the horizon.

I think JesseM has pretty much answered this, but to clarify:

The acceleration here is the acceleration of a hovering object relative to a local freefalling object, rather than the other way round. To be honest, I haven't done the calculation to see whether these two accelerations are the same or not. I suspect it probably is provided the object is falling from rest as you measure it.

Objects with mass simply cannot hover at or below the event horizon, so there is no answer to what their proper acceleration would be there. Also note that in relativity, the speed of an observer affects the value of any accelerations they measure.

I don't think it's directly relevant, but it's perhaps worth noting that a hovering observer observing a falling object will say the object decelerates and stops at the event horizon, taking an infinite time to get there. This is just making the point that we need to stick with local measurements of acceleration; remote measurements just get distorted by spacetime curvature!

The falling observer measures his own acceleration to be zero at all times, and, once through the event horizon, there cannot be any local hovering landmarks to compare against. Making a remote measurement of the singularity isn't feasible because it's not local.

Although it's pretty meaningless to talk of the proper acceleration of a photon, we can think of a massive particle accelerating towards the speed of light. To reach that limit is a finite time would require a proper acceleration tending to infinity. Therefore, loosely speaking, we could say "the proper acceleration of something at the speed of light is infinite", recognising that this is a bit of an abuse of the terminology.

As a concluding remark, in general relativity, when we mention velocity, acceleration, etc, we need to be clear what is being measured relative to what, and whether it is a "proper" invariant measurement, or a local or remote coordinate measurement.

 

[nutgeb]

Objects with mass simply cannot hover at or below the event horizon. . .

Agreed. I did not mean to suggest otherwise.

I don't think it's directly relevant, but it's perhaps worth noting that a hovering observer observing a falling object will say the object decelerates and stops at the event horizon, taking an infinite time to get there.

I understand that with respect to an observer at infinity. But will a hovering observer who is fairly near the horizon see it that way also? Certainly that hovering observer will see a plunging object pass by him at a high rate of velocity.

The falling observer measures his own acceleration to be zero at all times, and, once through the event horizon, there cannot be any local hovering landmarks to compare against. Making a remote measurement of the singularity isn't feasible because it's not local.

Fair enough. I understand the difficulty of the falling observer measuring his own change in proper velocity as a function of time inside the horizon, given the lack of stationary landmarks and the lack of any feeling of acceleration in the seat of his pants. And any calculation the falling observer does of his proper velocity needs to take into account the "distance boost", as compared to the Schwarzschild r coordinate. Nevertheless, the falling observer measures a finite proper time ("wristwatch time"), to fall from the horizon to the center, which, (disregarding the fact of his own eventual tidal demise), enables a calculation that (a) his proper velocity must have increased progressively after crossing the horizon, and (b) his derived proper acceleration must have been finite and not "arbitrarily large." Such a calculation will show that his proper velocity inside the horizon was "timelike" in the sense that the figure will be faster than c, but I don't see how that precludes the notion that a finite proper distance also has been traversed between the horizon and the center. Or perhaps after accounting for the "distance boost", the falling observer will calculate that the distance he traveled from the horizon to the center was infinite (?)

As a concluding remark, in general relativity, when we mention velocity, acceleration, etc, we need to be clear what is being measured relative to what, and whether it is a "proper" invariant measurement, or a local or remote coordinate measurement.

Agreed. I've tried to do that, and I appreciate you keeping me honest if I get it wrong.

 

[atyy]

The formula JesseM and DrGreg gave is infinite at the event horizon, and I'm not sure it's valid there, but it also doesn't matter, since only hovering above the horizon needs to be considered, and for any finite distance above the horizon, no matter how small, the acceleration is finite.

 

[xantox]

I don't know where you got the idea that gravitational acceleration becomes infinite at the event horizon. That's not correct. Gravitational acceleration increases smoothly at 1/r² as the horizon is approached and crossed.

As already thoroughly answered by the previous posters, this is a general relativistic result – it seems you were applying the Newtonian law. Another non-intuitive fact is that in the hovering frame, radial distances are stretched, so that a hovering pole would need to be very long to actually reach the horizon.

Nothing unusual happens to the acceleration gradient at the horizon.

Static tidal forces diverge too in the limit of the horizon for hovering observers.

Its probably worth putting numbers in here.

I obtain the following for a supermassive black hole of 10000 solar masses. Its horizon radius is 29530 km. Hovering 1 km away from the horizon would imply for an astronaut weighing 70kg on Earth to weigh 10^14kg. Interestingly, the predicted Unruh temperature should still be fairly cold out there at 10^-6K (but this will also diverge in the limit of the horizon), though the radiation flux from the rest of the universe should be quite intense. It would be interesting to calculate it as well as the Hawking luminosity on that frame.

The black hole should probably appear from that hovering frame as a luminous region (per Hawking radiation) encompassing almost all sky excepted for a small tunnel of intense light on the other side.

 

[nutgeb]

The formula JesseM and DrGreg gave is infinite at the event horizon, and I'm not sure it's valid there, but it also doesn't matter, since only hovering above the horizon needs to be considered, and for any finite distance above the horizon, no matter how small, the acceleration is finite.

Thanks, I understand that.

Going back to the OP, it seems that the infinite acceleration does matter for the long pole dipped into the horizon from outside. That indicates that, if the outward end of the pole is held with arbitrarily great strength by the outside hoverer, the pole must break as soon as the inward end approaches the horizon.

One thing that confuses me is that if the pole-holder is infinitely far from the horizon, he will feel the same BH's "surface gravity" pulling on the inward tip of the pole at the horizon at a finite force of only 1/4m (in geometric unit figures). In which case the pole won't necessarily break (?) The Raine & Thomas textbook seems to explain this as a redshifting phenomenon.

 

[DrGreg]

Agreed. I did not mean to suggest otherwise.
I understand that with respect to an observer at infinity. But will a hovering observer who is fairly near the horizon see it that way also? Certainly that hovering observer will see a plunging object pass by him at a high rate of velocity.

Yes, and yes. The point is that the hovering observer is accelerating, and if they are close to the horizon, accelerating at an extremely high rate. You know, even in flat spacetime, with no gravity, if you drop an apple from an accelerating rocket, the apple never "falls" more than a distance of c²/a below the rocket, before it slows down and almost stops, relative to the rocket's accelerating frame. (Look up "Rindler coordinates".) When the acceleration becomes enormous, that "stopping distance" can become very small!

(To digress from this conversation and address the original question of this thread, under such huge proper acceleration, apart from the fact that a human would be crushed to death by the G-force and his rocket will eventually disintegrate under the stress, as you lower the pole, the pole's weight will be enormous, approaching infinity as you lower it, and at some point either you'll let go or it will break under the huge tension force within it.)

Fair enough. I understand the difficulty of the falling observer measuring his own change in proper velocity as a function of time inside the horizon, given the lack of stationary landmarks and the lack of any feeling of acceleration in the seat of his pants.

I'll have to pull you up over the term "proper velocity". There's no such thing in the sense you are using it here. (There is something called "proper velocity" or "celerity" in relativity, but it's not relevant to what's being discussed here.)

In view of this, the rest of your post doesn't really make much sense. I'll have to go back to my last post: when you say "velocity", velocity relative to what?

The proper acceleration of the falling object is zero. That's a coordinate-independent fact. It's the proper acceleration of a succession of floating observers that increases towards infinity as you drop to the horizon, and there aren't any beyond it. The speed of these floaters relative to you increases to c as you drop to the horizon, but there aren't any beyond it.

There are others on this forum who have more expertise in this area than me, but I'm not sure if it really makes any sense to talk of "distance travelled" inside the horizon, again, relative to what exactly?

 

[DrGreg]

The formula JesseM and DrGreg gave is infinite at the event horizon, and I'm not sure it's valid there, but it also doesn't matter, since only hovering above the horizon needs to be considered, and for any finite distance above the horizon, no matter how small, the acceleration is finite.

If any formula gives you an answer of infinity, that's the mathematics' way of saying the formula isn't valid in those circumstances. In this case it's not valid because there are no hovering massive particles to be measured at or below the horizon.

 

[Haelfix]

"The black hole should probably appear from that hovering frame as a luminous region (per Hawking radiation) encompassing almost all sky excepted for a small tunnel of intense light on the other side. "

I don't think I buy that, b/c it breaks the equivalence principle. A local observer should not be able to detect hawking radiation. Instead he/she will detect Unruh radiation (b/c the hovering observer is in an accelerated frame) but he/she cannot know that some of that radiation escapes away to infinity as Hawking radiation.

 

[atyy]

The proper acceleration felt by a hovering observer at radius r tends to infinity when r tends to 2M (the event horizon), for black holes of any size. This means that lower portions of a hovering pole would be subject to increasing and diverging stresses, ie it would break or vaporize as soon as you start to lower it sufficiently near the horizon.

Does the pole break because of the increasing proper acceleration, or because it is not "born rigidly accelerated". Is there any analogue of born rigid acceleration" that would enable the pole to hover stresslessly?

 

[xantox]

I don't think I buy that, b/c it breaks the equivalence principle. A local observer should not be able to detect hawking radiation. Instead he/she will detect Unruh radiation (b/c the hovering observer is in an accelerated frame) but he/she cannot know that some of that radiation escapes away to infinity as Hawking radiation.

Per the equivalence principle these radiations should be equivalent. Ie. they should be two alternate descriptions of the same physical phenomena. The radiation flux being intense, the whole (aberrated) black hole region of the sky would appear luminous to an observer hovering sufficiently near the horizon.

Does the pole break because of the increasing proper acceleration, or because it is not "born rigidly accelerated". Is there any analogue of born rigid acceleration" that would enable the pole to hover stresslessly?

It breaks because of its own weight and because of the tidal forces. Indeed if one had a way to apply the good varying force to each particle of the pole while lowering it then it would be possible to maintain its rigid shape a little closer to the horizon (though still not up to the point of reaching the horizon).

 

[stevebd1]

1) If you could hover over a black hole's event horizon (let's say it was a BIG hole with lower tidal forces and you could hover 1 mile out), and stuck a 1mi long pole over the event horizon such that 1mm of the tip of the 1mi pole were over the horizon, would the entire pole be ripped out of your hands? Could you pull the 1mm tip back out? I assume you couldn't.

It's also worth noting when hovering close to the event horizon of a black hole, the proper distance remains relatively large until you are nearly at the event horizon-

From http://www.mathpages.com/rr/s7-03/7-03.htm -

'..relative to the frame of a particle falling in from infinity, a hovering observer must be moving outward at near light velocity. Consequently his axial distances are tremendously contracted, to the extent that, if the value of Dr is normalized to his frame of reference, he is actually a great distance (perhaps even light-years) from the r=2M boundary, even though he is just 1 inch above r=2M in terms of the Schwarzschild coordinate r. Also, the closer he tries to hover, the more radial boost he needs to hold that value of r, and the more contracted his radial distances become. Thus he is living in a thinner and thinner shell of Dr, but from his own perspective there's a world of room...

Quantitatively, for an observer hovering at a small Schwarzschild distance Dr above the horizon of a black hole, the radial distance Dr' to the event horizon with respect to the observer's local coordinates would be-

\Delta r'=\frac{\Delta r}{\sqrt{1-\frac{2M}{2M+\Delta r}}}

where Δr' is proper distance, Δr is coordinate distance and M=Gm/c²

 

[skeptic2]

Thank you Stevebd1. This is what I was trying to say in post #13 but you said it so much better.

 

[nutgeb]

For what it's worth, the pole-holder hovering close outside the horizon can let a long pole slip axially downward in freefall, and then tighten its grip suddenly. In this way the inward tip of the pole can penetrate any arbitrary distance inside the horizon and it won't break until the holder tightens its grip on the other end. (This assumes the pole doesn't enter regions where the tidal forces alone would break it).

 

[DrGreg]

For what it's worth, the pole-holder hovering close outside the horizon can let a long pole slip axially downward in freefall, and then tighten his grip suddenly. In this way the inward tip of the pole can penetrate any arbitrary distance inside the horizon and it won't break until the holder tightens its grip on the other end. (This assumes the pole doesn't enter regions where the tidal forces alone would break it).

That sounds plausible at first, but actually it's more bizarre than that. Yes, the end of the pole that is "slipping through your hands" will continue to accelerate relative to you, but the bottom end will actually slow down and almost stop, taking an infinite time to reach the horizon in your frame of reference. The explanation is that the extreme acceleration of the pole, relative to you, causes extreme Lorentz contraction, and the two effects cancel each other out so that the bottom end almost stops. The top of the pole will slip through your hands before the bottom reaches the horizon, no matter how close you are hovering from the horizon and no matter how long the pole is!

What seems like a very short distance to the horizon for the hovering observer is a very long distance for a free-falling observer.

I hope I've got this right. This thread shows just how confusing black holes can be!

 

[nutgeb]

That sounds plausible at first, but actually it's more bizarre than that. Yes, the end of the pole that is "slipping through your hands" will continue to accelerate relative to you, but the bottom end will actually slow down and almost stop, taking an infinite time to reach the horizon in your frame of reference. The explanation is that the extreme acceleration of the pole, relative to you, causes extreme Lorentz contraction, and the two effects cancel each other out so that the bottom end almost stops. The top of the pole will slip through your hands before the bottom reaches the horizon, no matter how close you are hovering from the horizon and no matter how long the pole is!

I think I agree with your description.

But what happens from the pole's own perspective? The distance it perceives that it falls to the horizon is far less than its length, and of course its length is not contracted in its own frame. So the pole would predict that most of itself is still in the holder's grip when it crosses the horizon.

This seems like a case of failure of simultaneity in SR, like the ladder/barn paradox.

 

[mjacobsca]

Thanks for the great discussion guys. I've gotten more than my money's worth! My question was more of a thought experiment about what the pole holder would feel when holding onto the pole. I think I expected the answer to be conclusively that the pole would be ripped out of his hands or broken apart, but the concepts of Lorentz contraction and slow-down of time presents intriguing alternatives that I did not think of.

Mike

 

[PeterDonis]

Yes, the end of the pole that is "slipping through your hands" will continue to accelerate relative to you, but the bottom end will actually slow down and almost stop, taking an infinite time to reach the horizon in your frame of reference.

This is not correct. It will take infinite *coordinate time*, if you are using Schwarzschild coordinates as your coordinate time, but that is *not* the same as infinite proper time for you hovering close to the hole's horizon.

If you drop the pole (letting it free fall) while hovering at a given r-coordinate, you can calculate, fairly easily, how much proper time by your clock will elapse before you reach a "point of no return" where no amount of force you exert on the pole by tightening your grip will prevent the pole from falling through the horizon and either breaking (if your grip is stronger than the pole's tensile strength) or slipping through your grip never to return (if your grip is weaker). The details of the calculation are on Greg Egan's science pages (link below); the result is that after a proper time (for the hovering observer) of:

ln(2) \frac{a}{c^2},

where a is the (proper) acceleration felt by the hovering observer. Note that Egan does the calculation in terms of s_0, the "distance to the horizon" in the Rindler coordinates he's using--see my comment below on that--but to properly carry the result over to the black hole scenario, it's easier to rewrite things in terms of the acceleration; the relation between the two is:

a = \frac{c^2}{s_0}.

See http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html" for a good treatment of situations of this sort, with *lots* more details worked through than most treatments I've seen on the web. He does the calculation for an accelerating observer in Minkowski spacetime, but for this particular scenario, the result is the same as for an observer hovering (and therefore accelerating) at a constant r-coordinate above a black hole horizon. (He also does the calculation for a rope being paid out from the accelerating spaceship, which corresponds better to the original post in this thread, "lowering" the pole instead of just letting it free fall.)

 

[DrGreg]

Yes, the end of the pole that is "slipping through your hands" will continue to accelerate relative to you, but the bottom end will actually slow down and almost stop, taking an infinite time to reach the horizon in your frame of reference.

This is not correct. It will take infinite *coordinate time*, if you are using Schwarzschild coordinates as your coordinate time, but that is *not* the same as infinite proper time for you hovering close to the hole's horizon.

But that doesn't make sense. I can use my proper time only to measure myself. To measure the bottom of the pole I have no choice but to use coordinate time, and in this scenario, my coordinates are proportional to Schwarzschild coordinates.

If you drop the pole (letting it free fall) while hovering at a given r-coordinate, you can calculate, fairly easily, how much proper time by your clock will elapse before you reach a "point of no return" where no amount of force you exert on the pole by tightening your grip will prevent the pole from falling through the horizon and either breaking (if your grip is stronger than the pole's tensile strength) or slipping through your grip never to return (if your grip is weaker). The details of the calculation are on Greg Egan's science pages (link below); the result is that after a proper time (for the hovering observer) of:

ln(2) \frac{a}{c^2},

where a is the (proper) acceleration felt by the hovering observer. Note that Egan does the calculation in terms of s_0, the "distance to the horizon" in the Rindler coordinates he's using--see my comment below on that--but to properly carry the result over to the black hole scenario, it's easier to rewrite things in terms of the acceleration; the relation between the two is:

a = \frac{c^2}{s_0}.

See http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html" for a good treatment of situations of this sort, with *lots* more details worked through than most treatments I've seen on the web. He does the calculation for an accelerating observer in Minkowski spacetime, but for this particular scenario, the result is the same as for an observer hovering (and therefore accelerating) at a constant r-coordinate above a black hole horizon. (He also does the calculation for a rope being paid out from the accelerating spaceship, which corresponds better to the original post in this thread, "lowering" the pole instead of just letting it free fall.)

I agree with all of that, but nothing there actually contradicts the statement I made. The (proper) time that it takes until it's too late for me to stop the bottom of the pole fall through the horizon (which is finite) is not the same thing as the (coordinate) time it takes, in my frame, for it to actually fall through (which is infinite).

But thanks for the link. I'd recommend everyone taking part in this thread read it, as it's highly relevant.

 

[PeterDonis]

I can use my proper time only to measure myself. To measure the bottom of the pole I have no choice but to use coordinate time, and in this scenario, my coordinates are proportional to Schwarzschild coordinates.

You can only *measure* your own proper time, but you can *calculate* the proper time elapsed along any world line you like. And only quantities like proper time that are invariant under changes of coordinates have physical meaning; Schwarzschild coordinate time is *not* such a quantity. You're correct that the Schwarzschild coordinate time it takes for an object to free-fall through the horizon is infinite, but that fact by itself has no physical meaning.

(You could give it a kind of physical meaning by noting that Schwarzschild coordinate time is a good approximation of proper time for observers very far away from the hole, so far away that they do not fall appreciably towards the hole for the duration of your scenario. Those observers would indeed have to wait an infinite amount of time to "see" an object cross the horizon. But that does *not* mean that someone hovering close to the horizon would have to wait an infinite amount of time before the pole in free fall would either slip away from them or break when they gripped it. Nor does it mean that an observer free-falling along with the pole would have to wait an infinite amount of time before crossing the horizon. So whatever physical meaning Schwarzschild coordinate time might have isn't relevant to this scenario.)

You can use Schwarzschild coordinates to *calculate* things like the proper time elapsed along a worldline, but only outside the horizon; at the horizon, Schwarzschild coordinates are singular, so you can't calculate with them there. So if you're trying to deal with objects that are crossing the horizon, you need to find another coordinate system that is nonsingular across the horizon. (In Greg Egan's examples that I linked to, when "translated" to the black hole scenario, the relevant coordinates would be Kruskal coordinates, which are the analogue of Rindler coordinates in a black hole scenario. Kruskal coordinates are nonsingular across the horizon, so they work fine for situations like this one.)

 

[nutgeb]

Dr Greg,

The article is very interesting. The section about Eve letting Adam fall toward the horizon with a rope seems pertinent to our discussion.

It suggests to me that when we let our pole slide through the holder's grip (actually, we need to actively propel the pole at a high rate as indicted in the article), the lower end of a pole of finite length can actually cross the horizon while the upper end is still above our grip. This is true even though the pole holder does not see the lower end of the pole ever cross the horizon.

When we eventually tighten our grip, the pole will break when the tension has sufficient time to travel some distance down the length of the pole.

Do you agree?

The article suggests, as a general proposition, that Adam and Eve can (and will) disagree about whether a finite length of rope can cross the horizon. Again, this seems like a failure of simultaneity to me.

 

[PeterDonis]

The article suggests, as a general proposition, that Adam and Eve can (and will) disagree about whether a finite length of rope can cross the horizon. Again, this seems like a failure of simultaneity to me.

I agree in general with your description of what will happen to the pole. However, as a general proposition, whether or not a finite length of rope can cross the horizon is an invariant; it's a proposition about actual physical events (whether or not a given worldline crosses the horizon), and must come out the same for all observers. If Adam and Eve disagree about whether a finite length of rope can cross the horizon, one of them must be wrong.

Egan does say: "There is no finite q coordinate for any point on the Rindler horizon, which means there is no time for Eve when, in her co-moving inertial reference frame, Adam passes through the horizon. In that sense, Eve could claim that Adam never reaches the horizon as far as she's concerned." However, he immediately follows this with: "However, not only is it clear that Adam really does cross the horizon, there is a time \tau_{crit} for Eve after which any signal she sends to Adam will reach him only after he's on the other side." For "Rindler horizon" we can substitute "black hole horizon" and the above still holds good.

In other words, Eve's claim does *not* amount to saying that Adam's worldline does not cross the horizon; she may not be able to *observe* Adam do so, but she can certainly *calculate* that he will do so, and even calculate the time \tau_{crit} after which any signal she sends him will only reach him after he crosses the horizon (so he will be unable to ever get a response back to her).

 

[JesseM]

But that doesn't make sense. I can use my proper time only to measure myself. To measure the bottom of the pole I have no choice but to use coordinate time, and in this scenario, my coordinates are proportional to Schwarzschild coordinates.

Why assume that the outside observer is using Schwarzschild coordinates, rather than some other system like Eddington-Finkelstein coordinates or Kruskal-Szekeres coordinates? Does the problem specify?

 

[atyy]

The Schwarzschild time coordinate is the proper time of the observer at infinity. So there's a coordinate invariant meaning for the observer at infinity if it takes infinite coordinate time for something to happen.

 

[JesseM]

The Schwarzschild time coordinate is the proper time of the observer at infinity. So there's a coordinate invariant meaning for the observer at infinity if it takes infinite coordinate time for something to happen.

I don't think that makes sense--there's a coordinate invariant meaning for anything that happens on the worldline of that observer at infinity, but coordinate times of events at finite radius don't have the same kind of physical meaning (although I think Schwarzschild coordinates are designed to have the property that if a clock hovering at a given radius is running slow by a factor of F relative to Schwarzschild coordinate time, then if it sends light signals to the observer at infinity with every tick, then the observer at infinity will also see the clock running slow by the same factor of F visually). Since all coordinate systems are permissible in general relativity, I'm sure you could design plenty of other coordinate systems which had the property that the time interval between events on the worldline of the observer at infinity would match up with his proper time, yet these coordinate systems would disagree about the time intervals between events at finite radii.

 

[PeterDonis]

Why assume that the outside observer is using Schwarzschild coordinates, rather than some other system like Eddington-Finkelstein coordinates or Kruskal-Szekeres coordinates? Does the problem specify?

A problem can't "specify" what coordinates you have to use to work it out--unless you're taking a class with a particularly evil professor... :-)

You can use any system of coordinates you want; you will still get the same answer when you calculate answers that are invariants, like whether Adam's worldline crosses the horizon. The only caveat, as I said in a previous post, is that if you're dealing with objects that cross the horizon, you have to use coordinates that are nonsingular at the horizon. That rules out Schwarzschild coordinates, but any of the other commonly used systems for black hole spacetimes would work fine.

 

[atyy]

I don't think that makes sense--there's a coordinate invariant meaning for anything that happens on the worldline of that observer at infinity, but coordinate times of events at finite radius don't have the same kind of physical meaning (although I think Schwarzschild coordinates are designed to have the property that if a clock hovering at a given radius is running slow by a factor of F relative to Schwarzschild coordinate time, then if it sends light signals to the observer at infinity with every tick, then the observer at infinity will also see the clock running slow by the same factor of F visually). Since all coordinate systems are permissible in general relativity, I'm sure you could design plenty of other coordinate systems which had the property that the time interval between events on the worldline of the observer at infinity would match up with his proper time, yet these coordinate systems would disagree about the time intervals between events at finite radii.

What I'm suggesting is that DrGreg's use of coordinate-dependent "in his frame" should be just a few steps away from a coordinate-independent "reality" such as what the observer at infinity sees - for example, using the light signal property you mentioned. You can of course calculate in any coordinate system, but the "frame of the observer at infinity" is probably short-hand for something physically meaningful for that observer, such as what he sees.

 

[PeterDonis]

(In Greg Egan's examples that I linked to, when "translated" to the black hole scenario, the relevant coordinates would be Kruskal coordinates, which are the analogue of Rindler coordinates in a black hole scenario. Kruskal coordinates are nonsingular across the horizon, so they work fine for situations like this one.)

Oops, I just realized that I misspoke a bit here. Kruskal coordinates are the analogue of *Minkowski* coordinates in flat spacetime (i.e., the usual "t, x" coordinates in an inertial frame in special relativity, in which Egan's scenarios are set). The analogy is not perfect, of course, since Minkowski coordinates "go to infinity" in all directions, whereas Kruskal coordinates stop at the singularity at r = 0. But for the scenarios we're discussing, that's not relevant.

The analogue in a black hole scenario of Rindler coordinates are actually *Schwarzschild* coordinates, but only for the portion of the spacetime exterior to the hole (just as Rindler coordinates only cover the "wedge" of Minkowski spacetime to the "right"--i.e., positive x-direction--of the null lines t = x and t = -x).

 

[xantox]

It suggests to me that when we let our pole slide through the holder's grip (actually, we need to actively propel the pole at a high rate as indicted in the article), the lower end of a pole of finite length can actually cross the horizon while the upper end is still above our grip. This is true even though the pole holder does not see the lower end of the pole ever cross the horizon.

I don't see a contradiction in respect to what DrGreg said above. If the pole is free-falling, a clock located on the pole will not measure the same thing as a clock located on the hovering ship. Now when you say "the lower end of a pole crosses the horizon while the upper end is above our grip" there is an implicit attempt to determine the simultaneity of two events. However – let's assume the clock on the ship reads noon and the upper end is still above my grip – the question "where is "now" the lower end?" is generally meaningless, as the specific answer will be arbitrarily dependent on the choice of a coordinate system. Even if we don't use (unphysical) coordinate time but clock time (ie. a parameterization on some observable quantity), it will be still an arbitrary choice. Incidentally, the lower end of the pole will cross the horizon at t = ∞ also according to the physical clock in the ship. Without contradiction with the above, it is possible to compute that on the lower end of the pole's own wristwatch a short time will elapse before it crosses the horizon. But still, this computed quantity cannot be compared to ship's clock readings so as to allow saying from the ship "now the lower end has already crossed the horizon".

 

[PeterDonis]

Now when you say "the lower end of a pole crosses the horizon while the upper end is above our grip" there is an implicit attempt to determine the simultaneity of two events. However – let's assume the clock on the ship reads noon and the upper end is still above my grip – the question "where is "now" the lower end?" is generally meaningless, as the specific answer will be arbitrarily dependent on the choice of a coordinate system.

That's true, but the choice isn't completely arbitrary. All coordinate systems are equal, but for some purposes, some are more equal than others. :-)

In the case of a free-falling pole, after a proper time has elapsed since letting go of the pole (by the hovering observer's clock) of

\tau_{crit} = ln(2) \frac{c^2}{a}

where a is the proper acceleration of the hovering observer, no force that the hovering observer can exert on the pole can stop the lower tip of the pole from falling through the horizon. So it seems reasonable to say that the lower end of the pole "has crossed the horizon" at time \tau_{crit} by the hovering observer's clock.

Also, if the proper length of the pole (i.e., its length in its own rest frame) is at least

L_{crit} = \frac{1}{4} \frac{c^2}{a}

then the upper end of the pole will still be at or above the hovering observer's hand when a proper time \tau_{crit} has elapsed by his clock since letting go. So again, it seems reasonable to say that if the pole's proper length is at least L_{crit}, then the pole will still be "passing through the observer's hands when it crosses the horizon."

I agree that you can't *force* someone to accept these definitions, but they tell you useful physical facts about the situation, so the choice to use them isn't completely arbitrary.

 

[PeterDonis]

Just to beat this topic to death a little more :-), we can also calculate the required length of the pole (for a freely falling pole) and the proper time by the hovering observer's clock, if we require that the upper end of the pole just pass through the observer's hands at the same proper time, in the *pole's* rest frame, that the lower end of the pole is just crossing the horizon.

If we write \tau_{pole} for the latter proper time (in the pole's rest frame, and taking \tau_{pole} = 0 at the moment when the pole is released (at that moment the lower end of the pole is just at the observer's hand), L_{pole} for the corresponding proper length of the pole (again in the pole's rest frame), and \tau_{obs} for the proper time in the *observer* frame at the moment the upper end of the pole passes through the observer's hand, then we have (again writing a for the proper acceleration experienced by the hovering observer):

\tau_{pole} = \frac{c^2}{a}

L_{pole} = 0.4143 \frac{c^2}{a}

\tau_{obs} = 0.8814 \frac{c^2}{a}

The above describes another reasonable candidate for the "required length of the pole so that the upper end passes through the observer's hand *at the same time* as the lower end crosses the horizon"--the difference, of course, being that here "at the same time" applies in the pole's rest frame.

Since \tau_{obs} above is greater than \tau_{crit} that we calculated before, it is already too late at this point for the hovering observer to pull back the pole in one piece. In fact, it is fairly easy to see that the maximum length of the pole that can be "saved" at this point is

L_{saved} = \frac{1}{2} L_{pole}

or just half the total length of the pole; the rest of it is doomed at this point to drop below the horizon before any force exerted at the upper end can reach it.

 

[xantox]

So it seems reasonable to say that the lower end of the pole "has crossed the horizon" at time \tau_{crit} by the hovering observer's clock.

This sounds like a metaphor – proper times over different worldlines in curved space-time cannot be compared directly, each worldine has its own private metric structure. It is just enough to say that those signals emitted when e.g. the ship clock reads 12:01 cannot reach the lower end before it will cross the horizon, without the part "the lower end of the pole has crossed the horizon at 12:01". Geometrically, the event of the pole reaching the light-like surface of the black hole horizon cannot happen after any finite time-like interval crossed by a hovering ship at constant radius (or such surface could not be said to be an "event horizon").

 

[PeterDonis]

proper times over different worldlines in curved space-time cannot be compared directly, each worldine has its own private metric structure.

If you're going to take this position, then this

the event of the pole reaching the light-like surface of the black hole horizon cannot happen after any finite time-like interval crossed by a hovering ship at constant radius.

is just a "metaphor" too. Since the event of the pole reaching the horizon is on a different worldline than that of the hovering observer, by your own statement above, you can't compare their proper times directly, so there is no meaning other than "metaphorical" to the statement that the pole crossing the horizon "cannot happen *after* any finite time-like interval crossed by a hovering ship"--because "after" is a comparison of proper times.

All of this is OK with me; I'm willing to admit that statements about comparing proper times along different worldlines are "metaphorical", including the ones I've made in previous posts. (It's different if the worldlines can later converge, as in the twin paradox, so that observers can actually compare their clock readings, but that isn't the case here.)

 

[xantox]

If you're going to take this position, then "the event of the pole reaching the light-like surface of the black hole horizon cannot happen after any finite time-like interval crossed by a hovering ship at constant radius" is just a "metaphor" too. Since the event of the pole reaching the horizon is on a different worldline than that of the hovering observer, by your own statement above, you can't compare their proper times directly,

Here coordinate times are implicitly compared, as events are on different worldlines. The clock on the ship may be used to synchronize a shell coordinate system (locally flat) where each tick of the clock is a tick of shell time – a rescaling of Schwarzschild time. EDIT: The time coordinate in the limit of the pole horizon crossing event tends to infinity. I believe DrGreg was meaning this in comment 34. We clearly agree that this implicit choice of coordinates is arbitrary.

 

[PeterDonis]

Here coordinate times are implicitly compared, as events are on different worldlines. The clock on the ship may be used to synchronize a shell coordinate system (locally flat) where each tick of the clock is a tick of shell time – a rescaling of Schwarzschild time. The time coordinate of the pole horizon crossing event is tshell = ∞. I believe DrGreg was meaning this in comment 34. We clearly agree that this implicit choice of coordinates is arbitrary.

Yes, we do, but remember also that some choices of coordinates will not cover the entire manifold, so you have to be very careful talking about the values of those coordinates for events outside the portion of the manifold they cover. For example, the "ship coordinates" you're using do not cover the entire manifold, in either version (Rindler coordinates in Minkowski spacetime or Schwarzschild coordinates in Schwarzschild spacetime). Those coordinates become singular at the horizon: not just because the time coordinate goes to infinity, but because the metric, written in those coordinates, has a "divide by zero error" (i.e., an infinite value in the denominator of one of the metric coefficients).

So to talk about events either on or inside the horizon, you have to switch to a different coordinate system that remains nonsingular on and inside the horizon (for example, Minkowski coordinates in Minkowski spacetime or Kruskal coordinates in Schwarzschild spacetime). You can still say, if you like, that events like the pole crossing the horizon happen at "t = infinity" in the ship's frame, but the deductions you can draw from that statement, by itself, are *extremely* limited because of the coordinate singularity.