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Black Hole Thought Experiment

shadron

Philosopher
Joined
Sep 2, 2005
Messages
5,918
OK, I was inspired by a blurb from a NOVA program by Dr. David Brin: here is the 30-second audio:

http://www.pbs.org/wgbh/nova/blackhole/expl-brin.html

So. You have some kind of marvelous spacecraft that can set you a yard out from the event horizon of a singularity. You reach out and poke your finger through the event horizon (hopefully its not wobbling around too much) and you draw it back out. What do you see?

Elsewhere I've read that here is nothing spectacular at the event horizon; it is possible to cross it without noticing it - I'd expect to be able to draw the finger back, or perhaps it would suck me in in-toto. Except nothing ever comes back from the event horizon excepting quantum evaporation, right? Elsewhere I've read that immediately inside the horizon is a huge amount of radiation trying to get out of the hole and just not being able to make it, eternally dressed up with no place to go, and so it accumulates there. I'd expect that finger to be toast. I know that the tidal forces may not be high enough at the event horizon to spaghettify the finger; that is a function of the BH's mass, and happens in closer proximity to he singularity, I assume.

So what does happen?

Overactive imaginations want to know...
 
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My guess would be that you could not pull your finger out. Wouldn't you need infinite energy to overcome the gravity at the event horizon?
 
For an observer passing through the event horizon there is no event horizon. But this is an ideal observer with no extent. This situation is an observer with extent (you and your finger). I would treat that as 2 observers, i.e. imagine that there is a tiny observer sitting on your finger. As far as your finger is concerned there is no effect of the event horizon. You though will never see your finger pass the event horizon.

There is no "huge amount of radiation trying to get out of the hole" at the event horizon. Whatever you read may be confused with Hawking radiation which is tiny for macroscopic black holes.
 
Elsewhere I've read that here is nothing spectacular at the event horizon; it is possible to cross it without noticing it

IIRC, you would certainly notice the crossing of a "normal" blackhole horizon and you would be spaghettified. But you can cross the EH of a supermassive blackhole and not feel any discomfort. Not that anyone has ever tried.
 
Physics isn't my field (I'm an Engineer) but my thinking is that you wouldn’t notice much of anything at all. It seems to me an objects center of mass would need to cross the event horizon for that object to trapped permanently. Assuming the black hole is large enough that you don’t get ripped apart simply from being near the event horizon is should look like normal flat space and you wouldn’t really see anything different.

My concern about this guess is that you could bring information across the event horizon, is that supposed to be possible? You could also attach to something else inside the event horizon, and as long as your combined center of mass is outside the event horizon you could drag it out, which seems like it shouldn’t be possible.
 
So. You have some kind of marvelous spacecraft that can set you a yard out from the event horizon of a singularity. You reach out and poke your finger through the event horizon (hopefully its not wobbling around too much) and you draw it back out. What do you see?

If you tried that, and you had managed to hold yourself together up to that point, you would lose your finger in a particularly painful way.

Elsewhere I've read that here is nothing spectacular at the event horizon; it is possible to cross it without noticing it - I'd expect to be able to draw the finger back, or perhaps it would suck me in in-toto.

It works like this. It is true that nothing special happens to the spacetime near the event horizon of a large black hole. However the horizon is not really a stationary surface in the usual sense - it's a little better to think of it as a surface accelerating out away from the hole with infinite acceleration. Because the spacetime as a whole is curved, the horizon remains in one place as viewed by distant observers. But locally, it's equivalent to a surface undergoing infinite acceleration.* The surfaces slightly outside it are also accelerating out, but with large but finite acceleration. So in order to stay a fixed distance outside the event horizon you would need to accelerate away from it by an amount that goes to infinity as that distance goes to zero.

In other words, to stay outside your horizon you'd need to have a rocket pack strapped to your back, thrusting you away from the hole. If you turned it off you'd fall peacefully through, noticing nothing much until you arrived near the singularity. But if you want to stay outside, you have to keep your rockets going. When you stretch out your finger, there will be an additional force on it because it's closer to the horizon (so it has to accelerate even more to stay out). For a small distance from you that could be provided by the tensile strength of your skin and bones etc., but the force will go to infinity as the finger crosses the horizon.

Except nothing ever comes back from the event horizon excepting quantum evaporation, right? Elsewhere I've read that immediately inside the horizon is a huge amount of radiation trying to get out of the hole and just not being able to make it, eternally dressed up with no place to go, and so it accumulates there. I'd expect that finger to be toast.

That's also true. But for a real finger and a large hole I think it would break off before the temperature reached any significant level.


*Think of a river as it approaches a waterfall. The velocity of the water gets larger and larger as you approach the falls, and at some point it's equal to the speed a lightfish could swim in still water. A lightfish at that point would have to propel herself (accelerate) upstream as fast as she could just to remain stationary. That's the lightfish horizon - any lightfish that floats past it will go over the falls. The main difference with a light horizon is that because the speed of light is the maximum speed attainable even after an infinite amount of acceleration, a black hole horizon requires infinite acceleration to remain stationary at.
 
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So. You have some kind of marvelous spacecraft that can set you a yard out from the event horizon of a singularity. You reach out and poke your finger through the event horizon (hopefully its not wobbling around too much) and you draw it back out. What do you see?
One thing you're overlooking is that as you get closer to the event horizon, your "marvelous spacecraft" would have to work harder and harder to keep you at a constant "altitude". In the limit where the distance from the event horizon goes to zero, the force from the rocket engine goes to infinity. This implies among other things that you will be crushed like a bug against the floor as you descend slowly towards the event horizon, or in this case (you seem to be holding your arm below the rocket) your body would get torn to pieces.

But OK, let's pretend that you can survive, and that your arm muscles are stronger than the rocket engines so that you can pull your arm back. Your finger would definitely be gone.

Someone at another forum asked if you fall in head first, can you wiggle your toes? I think that's a pretty good question. The answer is yes. If you decide to wiggle your toes when your head is inside the horizon and the rest of your body outside, the nerve signals won't reach your toes until a long time after your feet are inside too. (If something is preventing your feet from falling in, your body gets torn to pieces).

Elsewhere I've read that here is nothing spectacular at the event horizon; it is possible to cross it without noticing it
What Tubbythin said, but note that this applies only when you're falling in. It doesn't apply when you're using a rocket to keep you from falling. A rocket that prevents you from falling in will kill you (if you're close enough) because of the acceleration required to stay at a constant distance. The required acceleration goes to infinity as the distance goes to zero.

Elsewhere I've read that immediately inside the horizon is a huge amount of radiation trying to get out of the hole and just not being able to make it, eternally dressed up with no place to go, and so it accumulates there.
I don't think I've heard that claim before. It isn't true. Anything that passes the event horizon will fall into the singularity in a very short time.
 
Except nothing ever comes back from the event horizon excepting quantum evaporation, right? Elsewhere I've read that immediately inside the horizon is a huge amount of radiation trying to get out of the hole and just not being able to make it, eternally dressed up with no place to go, and so it accumulates there. I'd expect that finger to be toast.
That's also true.
I'm confused. Does your answer mean that you agree with the part I colored red? If you do, can you explain why there's lots of radiation "immediately inside the horizon"?
 
Physics isn't my field (I'm an Engineer) but my thinking is that you wouldn’t notice much of anything at all. It seems to me an objects center of mass would need to cross the event horizon for that object to trapped permanently.
So a 1 kg weight is trapped permanently if it falls in by itself, but isn't trapped permanently if it's attached to a 100 kg weight that's still outside the horizon by a thin piece of string? ;)

No, anything that ends up inside is gone.

Assuming the black hole is large enough that you don’t get ripped apart simply from being near the event horizon is should look like normal flat space and you wouldn’t really see anything different.
This is correct, when you're falling.

My concern about this guess is that you could bring information across the event horizon, is that supposed to be possible? You could also attach to something else inside the event horizon, and as long as your combined center of mass is outside the event horizon you could drag it out, which seems like it shouldn’t be possible.
You can't drag anything out. You can't even send messages. If you fall in holding one member of a pair of telephones with a long cable between them, and the other phone is on the outside, the message you send will never get outside. The only things that can happen are: 1. the other phone falls in too, 2. the cable breaks before any information gets out.
 
I'm confused. Does your answer mean that you agree with the part I colored red? If you do, can you explain why there's lots of radiation "immediately inside the horizon"?

Well, it doesn't mean much to talk about how much radiation there is inside the horizon - because there's no unique answer. The point is, the amount of radiation one detects is a function not just of the spacetime, but also of the trajectory of the observer through that spacetime. If the observer accelerates she measures radiation even in flat, empty space. The amount of radiation is proportional to the amount of acceleration, and since staying outside the horizon of a black hole requires an acceleration that goes to infinity as you get closer, the amount of radiation you measure also goes to infinity right at the horizon (all of those statements apply to observers accelerating at a constant rate so as to remain a fixed distance from the horizon).

So why can't we ask about inside? Well, no matter how much you accelerate you cannot remain close to the horizon from the inside - all observers will fall further in. So there's no special class of observers to concentrate on, and how much radiation you measure will depend on how hard you try (and fail) to stay close to the horizon.
 
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The event horizon is a graduation mark on something that actually increases gradually on the way in. The escape velocity gradually gets higher and higher on the way up to c and then beyond. Time dilation also gradually gets stronger and stronger until it essentially stops and then goes beyond (making time inside the black hole not backward, but something else that isn't forward either, like imaginary numbers). So if your finger tip is inside it, then the part of your finger that's an inch back from there is just an inch away and facing extreme time dilation. It's not that it's impossible to pull that part of your finger back; it's just that it will take so long that it will never happen anyway.
 
So. You have some kind of marvelous spacecraft that can set you a yard out from the event horizon of a singularity. You reach out and poke your finger through the event horizon (hopefully its not wobbling around too much) and you draw it back out. What do you see?

A nub.

Elsewhere I've read that here is nothing spectacular at the event horizon;

There is something special about it, even if you don't notice it when falling. In certain metrics, it is also a coordinate singularity, but not a real singularity.

it is possible to cross it without noticing it

Yes: if you let yourself fall in. If you try to NOT fall in, you will notice.

I'd expect to be able to draw the finger back, or perhaps it would suck me in in-toto.

Depends on how hard you pull. If you can't pull hard enough, your finger will pull you in after it. If you can pull hard enough, you will pull back a nub.
 
Is there any reason the space craft couldn't be orbitting ome meter above the black hole (ignoring the issue of what it might run into)?

It seems like the real issue is not with the event horizon but rather with the tidal forces on the orbiting craft. If those are too great then stretching anything towards or away from the black hole is going to cause the destruction of the stretched out thing.

Could somebody provide a quantitative estimate of the variation in gravity across one meter for something that was very close to the event horizon of a large black hole. Presumably the bigger the black hole the smaller the gravity gradient, but I sit back and wait for edification on that.

Assuming for a second that the black hole is large enough the gravity gradient is small enough across a one meter space craft that things aren't completely shredded then my guess is that from the point of view of the guy in the space craft he can put his arm out and pull it back. But from the point of view of a distant observer pulling the arm back takes an infinite amount of time.
 
OK, so there is an infinite force gradient across the event horizon if you are accelerating away from it (essentially the amount of the acceleration required), and it will procure the proffered pfinger. If you are going with the (gravity) flow, you don't see/feel it until you get much closer to the singularity and he spaghettification starts in. Fair?

BTW, I believe that spaghettification from tidal forces has no direct relationship to the event horizon. For small black holes it happens outside the horizon, for large ones inside.

I don't know where I read the stuff about radiation inside the event horizon piling up as it attempted to exit outwards. Sorry.
 
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Assuming for a second that the black hole is large enough the gravity gradient is small enough across a one meter space craft that things aren't completely shredded then my guess is that from the point of view of the guy in the space craft he can put his arm out and pull it back.
Definitely not. The spaceship needs to have its engines push with a force that's absolutely ridiculous to remain 1 m from the horizon of a big black hole (here "big" means "massive enough to make the tidal forces at the horizon survivable"). That means that anything in the ship is going through an absolutely ridiculous acceleration. If you put your arm outside, it will get ripped from your body. If you keep it inside, your entire body gets crushed against the floor. (Note that this is what always happens if you're in a rocket with engines that are pushing that hard, no matter if you're near a black hole or not).
 
Please note that I am not a physicist and could be wrong, but I do not see any reason why you would automatically lose your finger, provided that the black hole is large enough for the gravitational gradient to be small. Think of a galaxy sized black hole.

I don't think there is any magical property of the event horizon. It is simply the distance at which the escape velocity of the hole exceeds the speed of light. Every massive object has an escape velocity. The EV of Earth is approximately 7 miles/sec.

Using Earth as an example, most people think that you must exceed the escape velocity to get away from the earth, such as a vehicle with 'DESTINATION MARS' painted on it. We envision such a vehicle atop massive rocket boosters that will accelerate the vehicle to a speed beyond the escape velocity to start it on its voyage. However, this is misleading. The EV is actually the velocity in which an object has enough energy because of its velocity alone to escape the gravitational field.

If we had a cannon that could fire cannon balls at a velocity of 4 miles/sec, those cannon balls would fall back to the earth. If we had a cannon that could impart a velocity of 7 miles/sec or greater, a projectile fired from it will not return to earth. We could actually shoot at Mars, and hit it if our aim was exact enough. Of course, if we fire our cannon ball at a velocity of say, 6 miles/sec, the cannon ball might not return to earth for a long time because it exceeds the orbital velocity of Earth, which is around 5 miles/sec.

The point is, we do not have to exceed the escape velocity to leave the gravitational well. If we had plenty of energy and reaction mass available, we could travel to the orbit of Mars at a leisurely velocity, even without exceeding our terrestrial highway speed limits if we wished, by simply keeping our thrusters firing. Another way we could leave Earth without reaching EV is by taking the space elevator. If we had a cable attached to earth and to an orbiting object out in space, we could pull ourselves away from the earth at very low velocity, and low cost too!

So, compare the situation with the space elevator with the question from the OP. And remember that the caveat is that NO SPAGHETTIFICATION ALLOWED near the event horizon as in super duper massive black holes.
 
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The point is, we do not have to exceed the escape velocity to leave the gravitational well.

But you still have to apply force -- even if you are climbing a ladder at a snail's pace. You will feel the pressure of the rungs against your feet. At Earth gravity, that won't be much. Near the event horizon of a black hole it will be huge. (Not that such a ladder could be fixed to the event horizon -- but anyway!)

The same things happens if you're in a jet pack hovering above the surface of the Earth -- it can be comfortable because it only counteracts 1g. Over an event horizon, it must counteract much more than 1g.
 
Someone at another forum asked if you fall in head first, can you wiggle your toes? I think that's a pretty good question. The answer is yes. If you decide to wiggle your toes when your head is inside the horizon and the rest of your body outside, the nerve signals won't reach your toes until a long time after your feet are inside too. (If something is preventing your feet from falling in, your body gets torn to pieces).
.

Can you explain this to me? I understand that if your head is past the EH and you decide to wiggle your toes, you won't be able to because the original nerve signal is also past the EH and therefore can't escape.

However, I don't understand how that would change once your entire body was past the EH. Surely as long as your head stayed closer to the singularity than your feet, the nerve signal would never be able to reach your feet.

(I'm not being sarcastic, btw. I'm not a physicist at all and am genuinely curious how this could work).
 
I don't think there is any magical property of the event horizon. It is simply the distance at which the escape velocity of the hole exceeds the speed of light.

But, in the theory of relativity, the speed of light is somewhat magical, because exceeding it is impossible.

If the speed of your fingertip needs to exceed c in order to leave the black hole, it isn't leaving it.

The rest of your post is basically correct, because Earth's escape velocity is slow enough that the physics is nearly classical.

So, compare the situation with the space elevator with the question from the OP. And remember that the caveat is that NO SPAGHETTIFICATION ALLOWED near the event horizon as in super duper massive black holes.

A very massive black hole won't stretch you, all by itself, while you're outside it. But, here, there's a spaceship helping out. The black hole pulls your fingertip one way, and the spaceship pulls the rest of you the other way.

Tidal forces on the Earth aren't very strong either---you wouldn't feel stretched if you were falling---but you certainly feel stretched when doing a chin-up, because the Earth pulls down on your body and the chin-up bar pulls up on your hands.
 
Can you explain this to me? I understand that if your head is past the EH and you decide to wiggle your toes, you won't be able to because the original nerve signal is also past the EH and therefore can't escape.

However, I don't understand how that would change once your entire body was past the EH. Surely as long as your head stayed closer to the singularity than your feet, the nerve signal would never be able to reach your feet.

As you fall head-first, your feet catch up to where your head was when it sent the nerve signal, and then they even pass that position. So there's no reason why they couldn't get the signal. If you like: the signal doesn't reach them, they reach the signal.
 

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