# Another layman blackhole question, pulling one end of a string out from behind the event horizon

No long explanation is needed,

What would happen if I were to allow one end of a rope to fall past the event horizon of a black hole while I held the other end?

Would I be able to pull it out? Would the rope feel extremely (infinitely?) heavy?

• You'd have to find a rope that does not snap in half by the tension first. – ACuriousMind Jul 17 '14 at 12:55
• So the rope that is past the horizon would gain so much mass(weight) it would snap the rope? Any rope, a high tension rated rope would just snap a little bit later? – user1596244 Jul 17 '14 at 12:56
• I very much guess so, but I'm sure someone will be along to calculate exactly how and when that happens. ;) – ACuriousMind Jul 17 '14 at 13:00
• Related: physics.stackexchange.com/q/104474/2451 and links therein. – Qmechanic Jul 17 '14 at 14:00

What would happen if I were to allow one end of a rope to fall past the event horizon of a black hole while I held the other end?

As usual, this is in the context of a Schwarzschild black hole.

First, outside the horizon, a object with constant radial coordinate 'feels' a constant proper acceleration, i.e., an accelerometer (think of a weight scale) attached to the object gives a constant, non-zero value.

Second, the proper acceleration increases without bound as the radial coordinate approaches the value of the Schwarzschild radius.

Now, imagine that there is a rope extending from some fixed radius inward to the horizon. But the weight of a section of rope increases without bound as one approaches the horizon. Do you see the essential problem here?

When talking about black holes, you need to take into account time dilation. As you lower a rope into an event horizon, you will see time for the end of the rope slow down. You will not be able to say at some point: "Now the rope has crossed the event horizon", because you would need to wait indefinitely.

The rope, on the other hand (or some observer you placed there), will look back at you and see you age very fast. Then, just before it crosses the event horizon, it will witness the entire (probably infinite) lifetime of the universe.

One should notice, that due to gravitational redshift, the end of the rope will not be visible with the naked eye. Conversely, the observer at the end of the rope will see a blueshift, so near the event horizon, if nothing else kills him, there are still gamma rays from the background radiation.

This isn't exactly an answer to your question, because as it stands your question can't be answered, but I thought I'd post this because the answer really surprised me.

Firstly, the reason your question can't be answered is that you can never get your rope below the event horizon. From the perspective of an observer stationary with respect to the black hole anything dropped into it takes an infinite time to even reach the event horizon, let along cross it. So you could not find yourself holding one end of a rope that had its other end below the black hole - not even if you waited an infinite time.

But provided the bottom end of the rope is above the event horizon then it's perfectly reasonable to ask what force you feel holding the end of the rope, and it's also perfectly reasonable to ask what happens to this force in the limit of reaching the event horizon. So let's do this.

But the force on a rope is hard to calculate because the mass is distributed evenly along its length. To keep things simple replace the rope by some mass $m$ dangling on the end of a weightless rope. With this setup calculating the force is easy.

Suppose the mass $m$ is at a distance $r$ from the centre of a black hole of mass $M$. Twistor59's answer to the question What is the weight equation through general relativity? tells us that relative to a shell observer hovering at a distance $r$ the gravitational acceleration is:

$$a_{shell} = \frac{GM}{r^2} \frac{1}{\sqrt{1 - \frac{r_s}{r}}}$$

where $r_s$ is the radius of the event horizon. But relative to you standing a large distance from the black hole the time of the shell observer is dilated by a factor of:

$$t_r = \frac{1}{\sqrt{1 - \frac{r_s}{r}}}$$

And the acceleration you measure far from the black hole is $a_{shell}$ divided by this factor squared so:

\begin{align} a &= \frac{GM}{r^2} \frac{1}{\sqrt{1 - \frac{r_s}{r}}} \left( 1 - \frac{r_s}{r} \right) \\ &= \frac{GM}{r^2} \sqrt{1 - \frac{r_s}{r}} \end{align}

And the force is simply the acceleration multipled by the mass of your weight $m$:

\begin{align} F &= \frac{GMm}{r^2} \sqrt{1 - \frac{r_s}{r}} \\ &= F_N \sqrt{1 - \frac{r_s}{r}} \end{align}

where $F_N$ is the force predicted by Newtonian gravity i.e. the force you'd measure in the absence of realtivistic effects.

So the force you would feel is actually less than you'd expect from Newtonian gravity and indeed the force goes to zero as the weight approaches the event horizon. To illustrate this I've graphed the force you would feel compared to the force predicted by Newton's equation:

The force is in units of $GMm$. At distances of around four times the event horizon radius and greater the force is similar to the one calculated by Newton's equation, by as the weight approaches the event horizon the force you feel peaks around $1.4r_s$ then falls to zero at the horizon.

• John, I'm concerned that, in this context, this notion of a weightless rope is not remotely physical and thus, I suspect the result isn't physically meaningful. I've been reading this: arxiv.org/abs/1207.3342 "The energy conditions demand that the tension T of a static rope cannot exceed its mass-per-unit-length μ". But we must have tension for there to be a force on the far away observer, thus, there must be a non-zero $\mu$. – Alfred Centauri Jul 18 '14 at 12:55
• Also, from the paper: "The force at infinity required to suspend both box and supporting NEC-obedient rope does diverge as the horizon is approached." – Alfred Centauri Jul 18 '14 at 13:04
• @AlfredCentauri: everyday experience suggests it's possible to dangle weights on ropes in a Schwarzschild metric (well outside $r_s$). If there is a limit beyond which my treatment fails it would be interesting to know more. I'll read the paper you link, thanks :-) – John Rennie Jul 18 '14 at 14:08

In order to not fall straight in, you would have to be orbiting the black hole very quickly, in fact near the speed of light. By definition event horizon is when not even light can escape as it orbits. (Edit: as John Rennie commented, hovering in a rocket is also an option)

So imagine you are whizzing around at nearly the speed of light. You lower Your rope of unlimited strength toward the event horizon... wait, your in orbit that compensated gravity... so you'd have to throw it.

Edit:

• Or imagine you are hovering in a super advanced super high power rocket...
• Or imagine a giant scaffolding around the whole thing (Mike Dunlavey)

As the end of the rope gets closer to the event horizon, it would start to be pulled in by gravity. As it gets closer and closer the force will increase to no limit. That's right, there's no amount of force that could get the rope to just touch the event horizon and hold it there. In short, you would be pulled in.

I'm not even sure how to explain the spacial distortions due to relativity in that orbit. But that's a different question anyway. (Edit: And John Rennie mentioned in the comments that such an orbit would have to be 3x the event horizon to be stable.)

• As long as you stay above the event horizon you can hover above a black hole using a rocket. You don't have to orbit. In fact there is no stable orbit for matter within 3 times the horizon distance. – John Rennie Jul 17 '14 at 15:42
• Very interesting, I didn't know about the 3x ratio. Thanks! – Michael Jul 18 '14 at 13:09
• True! Hovering with a rocket is possible. I'll edit it to make that clear. – Michael Jul 18 '14 at 13:09
• You could just have a stationary scaffold around the entire black hole, and stand on that. – Mike Dunlavey Jul 18 '14 at 14:26
• haha... ok, scaffold is another option – Michael Jul 18 '14 at 15:27

I have an explicit calculation of the tension in the rope in section 8.1, example 5 of my GR book ("A rope dangling in a Schwarzschild spacetime"), which is free online: http://www.lightandmatter.com/genrel/ . I'll just sketch the main results here. Suppose we have a bucket hanging on the end of a rope in the Schwarzshcild spacetime. The tension $$T$$ in the rope obeys the differential equation

$$0=T'+(f'/f)T-(f'/f)\mu,$$

where primes denote differentiation with respect to the Schwarzschild coordinate $$r$$, $$f=\sqrt(1-2m/r)$$, and $$\mu$$ is the mass per unit length. We get a finite result for $$\lim_{r\rightarrow\infty}T$$, even when the bucket is brought arbitrarily close to the horizon. (The solution in this case is just $$T = T_\infty /f$$ , where $$T_\infty$$ is the tension at r = ∞.) However, this is misleading without the caveat that for μ < T , the speed of transverse waves in the rope is greater than c, which is not possible for any known form of matter — it would violate the null energy condition. For realistic forms of matter, the rope will break above the horizon.

This makes sense because the exterior of the black hole is causally disconnected from the interior.

• You may enjoy Greg Egan's analysis of the similar problem involving a Rindler horizon. – PM 2Ring Aug 29 at 5:43