Can I stick my arm in past the event horizon?

[Curtph]

So the event horizon is the distance from the singularity where escape velocity exceeds the speed of light, right?
Well let's say I am standing next to the event horizon, RIGHT next to it. What if I stick my arm in past the event horizon, then pull it out.

I can't do that, right? Why not?

Neglecting the "place to stand" thing and the relativistic effects (which, if I understand them, are irrelevant since I am only referring to my time frame, not an outside observer watching my time slow down as I become a holographic splatter on the surface of the event horizon) and the high-energy whizzy photons that are screaming around just inside the event horizon trying to get out, and other amounts of badness, as a thought experiment, what would actually prevent me from just sticking something in, and then pulling it back out again?

I assume no. please educate me :)

 

[PeterDonis]

lets say I am standing next to the event horizon, RIGHT next to it.

You can't. The event horizon is not a place. It's an outgoing lightlike surface: that is, it is moving radially outward at the speed of light. If you are just outside the horizon, the only way to stay outside it is to accelerate radially outward hard enough that the horizon can't catch you. You can't just stand still.

What if I stick my arm in past the event horizon, then pull it out.

I can't do that, right? Why not?

You could stick your arm in, but the rest of you would have to follow. As above, if you're just outside the horizon, and you want to stay outside, you have to be accelerating outward very hard. So if you stick your arm just inside the horizon, you are accelerating very hard away from it--so hard that your arm, once it's inside the horizon, would have to move faster than light to keep up with you. Obviously that's impossible; either you must stop accelerating and fall inside the horizon along with your arm, or you lose your arm.

 

[Curtph]

You can't. The event horizon is not a place. It's an outgoing lightlike surface: that is, it is moving radially outward at the speed of light. If you are just outside the horizon, the only way to stay outside it is to accelerate radially outward hard enough that the horizon can't catch you. You can't just stand still.

I am hung up on "moving radially outward" as that implies a displacement in space, but it is my understanding that the event horizon is a virtual barrier at a fixed (and deterministic) distance from a black hole of given mass/

I think I get it, I was considering a black hole, say, the mass of the Earth which would be diameter of about a marble? At the distance I am now (radius of earth) I would feel the same force but if that mass were concentrated in s space that small and I was sufficiently close to it you are saying the 'a' term in F = ma would be impossibly large such that it could not be exerted even on a massless(?) particle such as a photon?

I realize applying a classical Newtonian formula to a relativity problem has someone spinning in their grave but is that essentially correct?

 

[robphy]

Check out this spacetime diagram of an observer who enters the black hole
from Geroch's General Relativity from A to B [Google book link].

 

[Dale]

I am hung up on "moving radially outward" as that implies a displacement in space, but it is my understanding that the event horizon is a virtual barrier at a fixed (and deterministic) distance from a black hole of given mass

It is located at a fixed coordinate r in a particular popular coordinate system known as the Schwarzschild coordinates. However, coordinates in general relativity have no physical significance, they merely serve as convenient mathematical labels.

Instead, various other coordinate-independent measures are used, and it is this coordinate independent sense in which the event horizon moves at c. It passes any material object at c as measured locally in that objects frame.

If you stick your right arm in, you would have to pull it back faster than c, which is not possible.

 

[Ibix]

Gravity in general relativity is a rather more complex and subtle beast than Newtonian gravity. There is rather a lot of "common sense" stuff that you need to throw out to get an accurate description of what's going on in extreme circumstances like near a black hole. One such issue is that it is entirely possible for the event horizon to be both static and moving at the speed of light - it rather depends what you mean by "moving" and the word can be used in different senses. The event horizon is static in a sense that is convenient a long way from a black hole, but you are talking about a situation where you are right on top of one. In that case, you need to worry about a different sense that is about to kill you. I wouldn't worry about it too much unless you want to learn general relativity properly - in that case, burn it into your brain.

Edit: DaleSpam said the above rather more precisely.

In Newtonian gravity it is possible to have a black hole - an object whose escape velocity exceeds c, as you said. In that case, it is possible (at least in principle) to put your arm across the "event horizon" and then remove it, assuming it's really well stuck on. You don't have to exceed escape velocity to move upwards as far as you like. You only have to exceed escape velocity if you want to move upwards forever with your rockets off.

General relativistic black holes are different. Once you have crossed the event horizon, it does not matter how hard you try you are going into the singularity. It isn't possible to get further away from the singularity, or even stay at the same distance, without exceeding the speed of light locally, which is impossible. That means that it doesn't matter how well stuck on your arm is - you can either follow it into the hole or accept its loss.

 

[rootone]

The gravity of the black hole is trying draw you in, and the only way you could hover just outside of the event horizon would be to apply to yourself an equally massive force in the opposite direction,
This would have be an incomprehensibly large force, highly improbable to achieve in practice and would likely kill a soft organism in several unpleasant ways, but not a physical impossibility.
Once you arm is inside the event horizon, that incomprehensibly large force needed just to stay still becomes infinite, and then it's impossible for your arm to be retrieved.no matter how much force you apply.
Goodbye arm (unless the rest of you follows it into the BH).

 

[PeterDonis]

I am hung up on "moving radially outward" as that implies a displacement in space

It implies a direction of relative motion, but not necessarily a displacement in space; that depends on what coordinates you are using.

In local coordinates--for example, a local inertial frame in which an observer who is freely falling through the horizon is at rest--the horizon does have a displacement in space; it moves outward in this frame at the speed of light, just as an outgoing light ray would in special relativity.

But in the usual global coordinates used to describe a black hole, the horizon has a constant radial coordinate $r$. Or, to put it another way, light rays emitted radially outward just at the horizon--for example, an light ray emitted radially outward by an observer free-falling through the horizon, just as he reaches the horizon--stay at the same $r$ coordinate, because of the curvature of spacetime.

a black hole, say, the mass of the Earth which would be diameter of about a marble?

Yes; the horizon radius would be just short of 1 centimeter.

At the distance I am now (radius of earth) I would feel the same force

If you were maintaining the same altitude (by, for example, being in a spaceship and firing your rockets to keep yourself from falling), yes, you would feel a 1 g acceleration, just as if you were standing on the Earth's surface.

if that mass were concentrated in s space that small and I was sufficiently close to it you are saying the 'a' term in F = ma would be impossibly large

A better way of putting it would be that the acceleration you feel would increase without bound as you got closer and closer to the horizon. F = ma is not a good formula to use because it's not relativistic. But in order to produce that increasing acceleration, yes, something "exerting a force" (like the rocket engine in your spaceship) would have to work harder and harder.

such that it could not be exerted even on a massless(?) particle such as a photon?

That's not a good way to think of it. The reason you can't stay static at the horizon is simply that no object with nonzero rest mass can move on a null (i.e., lightlike) worldline. With respect to a local inertial frame such as the one I described above, this could be stated in the usual special relativistic manner: an object with nonzero rest mass can't move at the speed of light. But the horizon remains a null (lightlike) surface, along which only massless objects can move, regardless of what coordinates you use to describe it. It's a matter of spacetime geometry.

 

[Curtph]

Thank you so much for answering my questions sufficiently! I have recently taken up an interested in physics and my poor engineering degrees are not up to the task. Back to school for me :)

 

[anorlunda]

We poor laymen sometimes feel like ping pong balls regarding EH answers. I have no doubt that Dalespam's and PeterDoni's answers are exactly correct for the question asked about a small BH.

On the other hand we have other answers that Bob would notice nothing special at all when free falling through an EH (even with an outstretched arm). Or that nothing is extreme or even detectable at the EH of supermassive BH. I think I understand it when looking at drawn diagrams, but confidence goes out the window when hearing verbal answers.

 

[Nugatory]

On the other hand we have other answers that Bob would notice nothing special at all when free falling through an EH (even with an outstretched arm).

That is true (if the black hole is large enough that tidal effects won't matter)... but if you're free-falling through the horizon you aren't pulling your arm back out through the horizon, you're following it in.

 

[Dale]

We poor laymen sometimes feel like ping pong balls regarding EH answers. I have no doubt that Dalespam's and PeterDoni's answers are exactly correct for the question asked about a small BH.

On the other hand we have other answers that Bob would notice nothing special at all when free falling through an EH (even with an outstretched arm). Or that nothing is extreme or even detectable at the EH of supermassive BH. I think I understand it when looking at drawn diagrams, but confidence goes out the window when hearing verbal answers.

Yes, a very large black hole and a very small black hole are quite different in terms of tidal forces. The tidal forces near the horizon are much higher for the smaller black hole.

However, the "can't pull your arm back" applies for both. The small black hole will spaghettify you, so your arm will get pulled off which you will most likely notice. The large black hole wouldn't spaghettify you but you simply cannot pull your hand back faster than c. Since you are used to that you probably won't notice.

 

[wabbit]

In a way I think crossing a black hole horizon is more similar to crossing an instant in time than crossing a surface in space : you cannot go back to yesterday, but did you feel something special at midnight ?

 

[PAllen]

In a way I think crossing a black hole horizon is more similar to crossing an instant in time than crossing a surface in space : you cannot go back to yesterday, but did you feel something special at midnight ?

I don't think that is a good description. Locally, a passing a horizon is no different than the light of flash bulb passing you by, except that there is no light.

 

[PAllen]

Anorlunda, here are some more words, that are certainly not intended to confuse.

First, imagine a scenario far out in space, away from everything. You wearing a space suit, in a 'play room' in a rocket that has a hole for sticking your arm out. You stick it out, wiggle it, pull it back, no problem. However, while it is out this (magically powerful, made of magical materials) rocket accelerates to 1 billion gee and continues so accelerating for a while. Poor you - your am separates immediately and the rest of you is reduced to puddle on the floor of the room.

It just happens that near an event horizon of a supermassive BH, as of the moment when your arm is on one side of the (locally undetectable) horizon while the rest of you and the rocket are not, it is certain that:

1) If the rocket does nothing, you feel nothing, and the rest of you and the rocket are across the horizon less than a microsecond later (depends how long the rocket is; for the rest of you, it is a few nanoseconds).

2) If the rocket is to escape to distant space, its only choice is that acceleration to billions of gees within a nanoseconds. This produces the same result as in 'far away' space. You effectively have to outrace light from a small head start.

The why of this is terrible choice is global not local. The closer you get to the supermassive BH, the smaller the fraction of your future light cone stays outside the horizon. Thus, your 'escape clause' gets harsher and harsher.

 

[wabbit]

I don't think that is a good description. Locally, a passing a horizon is no different than the light of flash bulb passing you by, except that there is no light.

Agreed. It is not a description at all actually, only an analogy with another situation where moving in a certain direction is irresistible but doesn't involve feeling a force or any noticeable change, but even then this would apply better to what happens after the crossing, than to the crossing itself.

 

[pervect]

Let's give an example with some numbers of what happens with a large black hole. (Or, equivalently, with the sticking your arm experiment carried out with the Rindler horizon on Einsten's elevator). In either case, the mass of the black hole isn't a factor in the equation.

Lets say you have longish 1-meter arms. Then, to hold station 1 meter away from the event horizon, your body must be accelerating at a proper acceleration of c^2 / distance, which is c^2 / (1 meter) $\approx 9\, 10^{16}$ m/s^2, or about 10^16 gravities.

For those interested in some details of how this equation was derived, http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html has a source, but it might not be the easiest to follow without an already good background.

I will ignore the rest of your arm for now, though one could apply the c^2 / d equation to figure out the proper acceleration required to hold station, an acceleration that becomes infinite as d approaches zero.

Let's pick up the picture with what happens to your hand, stuck through the event horizon. It must move inwards regardless of what forces act on it or how hard it accelerates. So, regardless of how impossibly strong your arm is, it must stretch. Presumably it will eventually break off (though in theory it could continue to stretch indefinitely, I suppose). Of course, in actuality, no human body or substance known to man could withstand 10^16 g's anyway, so the problem is not terribly realistic. Just another example of rigid bodies being a non-physical idealization, an idealization that's known to be incompatible with special relativity, much less general relativity.

The analogy suggested previously of thinking about a light beam hitting your hand, but accelerating your body fast enough that the light that touches your hand never quite catches up with the rest of your body, is also quite valid. I use it myself. On consideration of the audience, though, it may not make sense without a familiarity with hyperbolic motion. Most likely one might not even realize it's possible to stay ahead of a light beam by accelerating hard enough without a significant exposure to the physics beforehand. Greg Egan's web page quoted above and various other sources including wiki will go through a more detailed explanation (search for "hyperbolic motion"). However, it's more than I'd care to attempt in a short post and may require a concerted effort to fully grasp. The advantage of this approach though is t hat it requires only a good knowledge of special relativity, a much easier (though still daunting) task than understanding general relativity.

 

[PAllen]

Let's pick up the picture with what happens to your hand, stuck through the event horizon. It must move inwards regardless of what forces act on it or how hard it accelerates. So, regardless of how impossibly strong your arm is, it must stretch. Presumably it will eventually break off (though in theory it could continue to stretch indefinitely, I suppose). Of course, in actuality, no human body or substance known to man could withstand 10^16 g's anyway, so the problem is not terribly realistic. Just another example of rigid bodies being a non-physical idealization, an idealization that's known to be incompatible with special relativity, much less general relativity.

Yes, I guess if you are elastigirl, you can imagine that you become a pancake with arm disappearing into the singularity, and the rest of you pulled along until all of you is yanked through the hole, following into the singularity.

 

[jartsa]

So the event horizon is the distance from the singularity where escape velocity exceeds the speed of light, right?
Well let's say I am standing next to the event horizon, RIGHT next to it. What if I stick my arm in past the event horizon, then pull it out.

I can't do that, right? Why not?

We have a dynamical situation, not a static one, so I think something like this might happen:

Your arm is moving in the frame of your body, so the arm is length-contracted in the frame of your body.

How much is the arm length-contracted? That depends on how fast the arm moves. As there is such thing as gravitational time dilation, observers near the event horizon are very much time-dilated, and they see a highly length-contracted hand whizz by at high speed.

Length-contraction is a frame-dependent thing. As you and the observers hovering near the event horizon are in the same frame, you agree with the observers: the arm is length-contracted.

No arm is long enough to reach past the event horizon, because when a hand approaches the event horizon, its coordinate-length approaches zero. If the coordinate length does not approach zero, then the proper length of the hand approaches infinity.

 

[PeterDonis]

How much is the arm length-contracted? That depends on how fast the arm moves.

Yes, but the arm does not have to move at nearly the speed of light relative to the hovering observer. It would if it were in free fall downwards, but it doesn't have to be. If the hovering observer reaches his arm towards the horizon, his arm could still be accelerated almost as much as he is, so it could be moving arbitrarily slowly relative to him. (In the local inertial frame of an observer free-falling through the horizon, the hovering observer would be accelerating outward very hard, and the arm would be accelerating outward a little less hard, so that it didn't stay above the horizon, which is moving outward at the speed of light in this frame.)

As there is such thing as gravitational time dilation, observers near the event horizon are very much time-dilated, and they see a highly length-contracted hand whizz by at high speed.

Gravitational time dilation has nothing to do with this. Gravitational time dilation is observed between observers close to the horizon and observers very far away. Here all of the observers in question are close to the horizon.

No arm is long enough to reach past the event horizon, because when part of arm approaches the event horizon, its length approaches zero. If the length does not approach zero, then the proper distance between the molecules of the part of arm approaches infinity.

This is not correct. It's perfectly possible for someone hovering just above the horizon to reach their arm below the horizon; but if they do, as I said in post #2, they will have to, very very quickly, either stop hovering and fall in after their arm, or lose the arm.

 

[PAllen]

No arm is long enough to reach past the event horizon, because when part of arm approaches the event horizon, its length approaches zero. If the length does not approach zero, then the proper distance between the molecules of the part of arm approaches infinity.

This is just wrong. It is not only possible, but true for any object falling into a BH that part of it is inside the horizon and part outside. For a free falling body, you locally have Minkowski spacetime, and the horizon is like a flash of light passing by.

There is only a 'problem' (for a sumpermassive BH), if your require part of the body to remain outside the horizon. There is no need to repeat the discussions above, which cover this scenario.

 

[PeterDonis]

There is only a 'problem' (for a sumpermassive BH), if your require part of the body to remain outside the horizon.

Even then it's not necessarily a problem as regards length contraction, as I said in my previous post.

 

[PAllen]

Even then it's not necessarily a problem as regards length contraction, as I said in my previous post.

I never said it was because of length contraction. I referred to prior discussions of why it was a problem (yours, mine, and pervect's).

 

[pervect]

We have a dynamical situation, not a static one, so I think something like this might happen:

Your arm is moving in the frame of your body, so the arm is length-contracted in the frame of your body.

How much is the arm length-contracted? That depends on how fast the arm moves. As there is such thing as gravitational time dilation, observers near the event horizon are very much time-dilated, and they see a highly length-contracted hand whizz by at high speed.

Length-contraction is a frame-dependent thing. As you and the observers hovering near the event horizon are in the same frame, you agree with the observers: the arm is length-contracted.

No arm is long enough to reach past the event horizon, because when a hand approaches the event horizon, its coordinate-length approaches zero. If the coordinate length does not approach zero, then the proper length of the hand approaches infinity.

I don't think I agree, I would need to see a more detailed calculation to be convinced.

My current thinking is that the Rindler metric is a good approximation to the metric near a large enough black hole, as well as being interesting in its own right. You do have the additional issue that you raise about the dynamics, which I didn't consider. I agree that the time dilation in the Rindler frame becomes infinite (the metric coefficient for time approaches zero) as you try to "reach" for the event horizon.

I did totally ignored the dynamical issue, treating the problem as static. But I believe it would be a fair description of the problem to to say that while it would require you to wait an infinite amount of time to "reach" into the event horizon slowly enough that the you could ignore the Lorentz contraction of your arm, the limit in which you reach slowly enough to reach arbitrarily close to the event horizon still exists, and in that limit the length of your arm is finite. It just takes you forever to do it.

Another way of saying this is that the spatial metric coeffficients in the Rindler metric are unity and well behaved, only the time dilation factor "blows up". So it's not that the distance goes to infinity, it's best to desacribe the mathematics as saying that it takes you an infinite amount of time.

 

[PeterDonis]

I believe it would be a fair description of the problem to to say that while it would require you to wait an infinite amount of time to "reach" into the event horizon slowly enough that the you could ignore the Lorentz contraction of your arm

No, it wouldn't. Look at it in a local inertial frame in which the observer hovering just above the horizon is momentarily at rest at the instant his arm reaches out. One limiting case would be the end of the arm freely falling through the horizon, starting at that instant--i.e., the worldline of the end of the arm is just a straight vertical line in a spacetime diagram, while the observer's body follows a hyperbola. In this limiting case, the observer would be unable to pull his arm back after a fairly short length of his own proper time, but the arm would be moving very fast relative to him by that time, so it would, in his new local inertial frame (at the instant where he can no longer pull his arm back), be greatly length contracted.

The other limiting case would be the arm moving on a trajectory just slightly removed from the hyperbola the observer is following. In this limiting case, it would be a very long time before the observer could not pull his arm back, and its motion relative to him would be extremely slow.

But there are plenty of cases intermediate between these two, where the arm still moves slowly enough that its speed relative to the observer's body does not become relativistic, or at least not extremely so (length contraction might be observable, but not extreme), but it still only takes a reasonable amount of the observer's proper time before he can't pull his arm back.

 

[jartsa]

Yes, but the arm does not have to move at nearly the speed of light relative to the hovering observer. It would if it were in free fall downwards, but it doesn't have to be. If the hovering observer reaches his arm towards the horizon, his arm could still be accelerated almost as much as he is, so it could be moving arbitrarily slowly relative to him. (In the local inertial frame of an observer free-falling through the horizon, the hovering observer would be accelerating outward very hard, and the arm would be accelerating outward a little less hard, so that it didn't stay above the horizon, which is moving outward at the speed of light in this frame.)

Yes, but because there is gravitational time dilation, what is slow motion of a hand for the hand owner, is a fast motion of a hand for those observers that are much more time dilated than the owner of the hand.

Gravitational time dilation has nothing to do with this. Gravitational time dilation is observed between observers close to the horizon and observers very far away. Here all of the observers in question are close to the horizon.

As I understand, gravitational potential difference between event horizon and a person 1 m above the event horizon is exactly c², according to the person. Short distance & very strong gravity -> large potential difference -> large gravitational time dilation difference.

This is not correct. It's perfectly possible for someone hovering just above the horizon to reach their arm below the horizon; but if they do, as I said in post #2, they will have to, very very quickly, either stop hovering and fall in after their arm, or lose the arm.

Yes, but then the arm's experience is one of infinite stretching. Such result I "derived" in the previous post.

 

[PeterDonis]

because there is gravitational time dilation, what is slow motion of a hand for the hand owner, is a fast motion of a hand for those observers that are much more time dilated than the owner of the hand.

What observers would those be? The hand owner is very close to the horizon.

As I understand, gravitational potential difference between event horizon and a person 1 m above the event horizon is exactly c²

I don't know where you're getting this understanding, but it's wrong. "Gravitational potential" has no meaning at the horizon; it only has meaning in a static region of spacetime, and the spacetime at and below the horizon is not static. So there's also no meaning to "gravitational potential difference" between the horizon and any point above it.

the arm's experience is one of infinite stretching. Such result I "derived" in the previous post.

I didn't see you "derive" anything. I saw you make a hand-waving verbal argument based on incorrect premises, leading you to an incorrect conclusion.

 

[jartsa]

This is just wrong. It is not only possible, but true for any object falling into a BH that part of it is inside the horizon and part outside. For a free falling body, you locally have Minkowski spacetime, and the horizon is like a flash of light passing by.

And for a observer hovering near the horizon, the falling body passes by almost at speed of light, and is very length-contracted?

 

[PeterDonis]

for a observer hovering near the horizon, the falling body passes by almost at speed of light, and is very length-contracted?

A freely falling body does, yes. But, as I've said several times in this thread already, there is nothing requiring the arm that the observer hovering just above the horizon is reaching out to be freely falling. It could have a proper acceleration just a little less than that of the observer, which would mean it would be moving slowly relative to that observer.

 

[Stephanus]

Thank you so much for answering my questions sufficiently! I have recently taken up an interested in physics and my poor engineering degrees are not up to the task. Back to school for me :)

You bet, in this forum the more I ask, the more I know that the more that I don't know
Can I add a question here.
I'm doing a simple schwarzshild calculator.
M: Mass, in solar Mass
R: Schwarzshild radius in meter
A: Gravity acceleration in m/s²
V: Escape velocity in m/s

  M            R          A         V
1.00E+00   2.95E+03   3.05E+13   3.00E+08
1.00E+03   2.95E+06   3.05E+10   3.00E+08
1.00E+06   2.95E+09   3.05E+07   3.00E+08
1.00E+09   2.95E+12   3.05E+04   3.00E+08
1.00E+12   2.95E+15   3.05E+01   3.00E+08
[COLOR=#ff0000][B]1.00E+15   2.95E+18   3.05E-02   3.00E+08[/B][/COLOR]

So, for a black hole 1000 trilion times solar mass (still below the mass of the observable universe, right).
Schwarzshild radius is 300 thousands ly. But the gravity acceleration in EH is 3cm/s?
Or am I doing the wrong calculation?
If it's true, sorry, I can't picture this. What is the force of the rocket must have to stay at EH perpendicular to the singularity?
Is it so hard for the astronout to take his arm?
Thanks.

 

[rootone]

By definition, an object inside the EH cannot by any means get outside of it,
To do that in principle would require a greater than infinite amount of force to be applied to the object, and that of course is nonsensical.

 

[Stephanus]

By definition, an object inside the EH cannot by any means get outside of it,
To do that in principle would require a greater than infinite amount of force to be applied to the object, and that of course is nonsensical.

Sorry, I still can picture or understand that.
Force is not identical with speed, altough closely related. The more force you have the faster you move, right?
So, on Earth a 1 ton rocket must have 9800 Newton force just to stay still.
What about the table that I write above.
The Schwarzshild radius for a 1000 trillion solar mass black hole is 300 thousands light year.
And the force needed for a 1 tons rocket just to to stay still at EH is 30 Newton?
Is that right?

 

[rootone]

An object outside of the event horizon experiences the gravity of the black hole.
If there are no other forces being applied to the object this will result in the object accelerating towards the BH.
However we could in principle attach an extremely powerful engine to the object and apply a force opposite to gravity, so the object can escape, (though probably not in great shape).
The closer we get to the EH, the more powerful the engine has to be to provide enough force to counter the gravity.
At the EH itself the amount of force needed goes to infinity.

 

[pervect]

No, it wouldn't. Look at it in a local inertial frame in which the observer hovering just above the horizon is momentarily at rest at the instant his arm reaches out. One limiting case would be the end of the arm freely falling through the horizon, starting at that instant--i.e., the worldline of the end of the arm is just a straight vertical line in a spacetime diagram, while the observer's body follows a hyperbola. In this limiting case, the observer would be unable to pull his arm back after a fairly short length of his own proper time, but the arm would be moving very fast relative to him by that time, so it would, in his new local inertial frame (at the instant where he can no longer pull his arm back), be greatly length contracted.

I was considering the case where the arm was born-rigid, at least in some sort of limit. I can see that I never actually explicitly said that - my bad. I'd agree that if you let the hand free-fall, it will reach the event horizon in a finite amount of (proper) time. But when you let the hand free-fall, while the body is holding station (a static observer), the arm must be stretching.

Part of the solution to the problem is realizing that the arm eventually must stretch, but I wanted to delay the onset of the stretching as much as is physically possible.

The advantage of having a born-rigid arm is that it corresponds naturally to the ruler definition of "distance". We need some operational way of defining what we mean by distance. The radar distance to the event horizon is obviously infinite, but the distance as measured by a ruler - specifically a born-rigid ruler - is finite. At least in the applicable limit, to be precise one would need to say that the limit of the ruler distance as one approaches the event horizon exists and is finite, unlike the radar distance. Of course, one can't actually have the end of the ruler actually reach the event horizon, the only class of objects that can hold station at the event horizon are massless objects such as light.The whole issue of dynamics is a bit of an afterthought, and not all that compatible with the notion of absolute born-rigidity. I don't see any practical problems (other than requiring a proper acceleration of 10^22 gravities, that is) with reaching within 1 micron of the horizon. The time dilation would be a modest million:1 between your body and your idealized pointlike hand, certainly significant but you still wouldn't notice the relativistic effects much if you kept the coordinate velocity of your "hand" below a meter/second.

 

[PeterDonis]

So, for a black hole 1000 trilion times solar mass (still below the mass of the observable universe, right).
Schwarzshild radius is 300 thousands ly. But the gravity acceleration in EH is 3cm/s?
Or am I doing the wrong calculation?

You are doing the wrong calculation. You are evidently using the Newtonian formula, $a = G M / R^2$, to compute this. That formula is not correct; the correct GR formula is

$$
a = \frac{GM}{R^2 \sqrt{1 - \frac{2GM}{c^2 R}}}
$$

According to this formula, $a$ increases without bound as $R \rightarrow 2GM/c^2$, i.e., as the horizon is approached.

What is the force of the rocket must have to stay at EH perpendicular to the singularity?

It is impossible for a rocket, or any object with nonzero rest mass, to remain static at the horizon. Only massless objects, which move at the speed of light, can stay static at the horizon.