Can I stick my arm in past the event horizon?

1. Jun 4, 2015

Curtph

So the event horizon is the distance from the singularity where escape velocity exceeds the speed of light, right?
Well lets 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 :)

2. Jun 4, 2015

Staff: Mentor

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.

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.

3. Jun 4, 2015

Curtph

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?

4. Jun 4, 2015

5. Jun 4, 2015

Staff: Mentor

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.

6. Jun 4, 2015

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 in to the hole or accept its loss.

7. Jun 4, 2015

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).

Last edited: Jun 4, 2015
8. Jun 4, 2015

Staff: Mentor

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.

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

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.

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.

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.

9. Jun 5, 2015

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 :)

10. Jun 5, 2015

Staff: Mentor

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.

11. Jun 5, 2015

Staff: Mentor

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.

12. Jun 5, 2015

Staff: Mentor

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.

13. Jun 5, 2015

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 ?

14. Jun 5, 2015

PAllen

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.

15. Jun 5, 2015

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.

16. Jun 5, 2015

wabbit

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.

17. Jun 5, 2015

pervect

Staff Emeritus
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.

18. Jun 5, 2015

PAllen

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.

19. Jun 5, 2015

jartsa

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.

20. Jun 5, 2015

Staff: Mentor

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.)
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.

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.

21. Jun 5, 2015

PAllen

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.

22. Jun 5, 2015

Staff: Mentor

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

23. Jun 5, 2015

PAllen

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).

24. Jun 5, 2015

pervect

Staff Emeritus
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.

25. Jun 5, 2015

Staff: Mentor

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.