# I Thought experiment- Dropping a rope into a Black hole

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1. May 23, 2017

### YZEDY1one

Hi, Physics Forum!

I've been wondering what would happen in the following thought experiment. I am not physicist and may have made some incorrect assumptions. Please explain if any of these assumptions are wrong.

What would happen if you threw a piece of (very long and strong) rope into a black hole? If I stood outside the black hole, how fast would I observe the rope to go after time 't'? Surely if nothing can exit the event horizon once it has entered, the force exerted on the rope is near infinite, leading to an infinite acceleration? Is the issue with this, the assumption that a rope that is strong enough to withstand this force is even possible? What would be the limiting factor, at what point would the rope snap?

Also, couldn't you connect one end of the rope to a generator (assuming long long rope) and drop the other end into the black hole (spinning the generator). How much power could you potentially generate, what would be the limit? If the force required to pull the rope out of the event horizon is infinite, surely the energy requirement is infinite? I don't believe it would create energy but could it at least be in theory an efficient way to generate energy from a given mass of rope?

Thanks!

2. May 23, 2017

### Drakkith

Staff Emeritus
The rope would indeed break. You would need an infinite amount of force to pull the rope back up from beyond the event horizon, a feat which is obviously impossible.

Offhand I would say that you're only limited by how much rope you have. However, you're going to have to expend energy to keep yourself from being pulled towards the black hole as well since much of the force on the rope will be transferred to your ship, gradually pulling it closer. I'm not sure if there would be a net gain of energy or not.

3. May 23, 2017

### BvU

Hello YZ,

Original ideas ! Keep them coming and don't worry about objections (anyone can do the latter)

Such as:
What would you mount the generator on ? It's being pulled towards the hole, complete with stand and whatever more (ship, planet?) that it's on

4. May 23, 2017

### Staff: Mentor

5. May 23, 2017

### PeroK

@YZEDY1one it's not that different an idea from hydroelectric power, which uses water falling under gravity. The extension to a black hole environment would, of course, create additional engineering challenges!

6. May 23, 2017

### Staff: Mentor

Not from a rope and mechanical generator, but the idea of generating energy by tossing stuff into black holes has been thought of before and comes up sometimes in sci fi. I believe the Romulins power their ships that way...

7. May 23, 2017

### Staff: Mentor

Greg Egan has done a detailed treatment of a similar scenario, "dropping" an object from a rocket accelerating in flat spacetime:

http://www.gregegan.net/SCIENCE/Rindler/RindlerHorizon.html
If you are close enough to the horizon of a black hole, the two scenarios are basically identical, and the flat spacetime one is a lot easier to analyze and to build intuition about.

How fast at your location, where it is being dropped? That depends on how fast you pay it out as you drop it.

How fast down closer to the horizon? You will see (as in, the actual light rays coming back to you from the rope as it falls) the rope falling slower and slower as it gets closer to the horizon, without ever reaching it. But that is an optical illusion; it doesn't mean the rope doesn't actually reach the horizon and fall on through. It does.

If you are just letting the rope free-fall, there is no force on it. If you try to exert force on it to stop it from falling, and a portion of the rope is below the horizon, then the rope will break at some point above the horizon--the exact point will depend on details like the rope's tensile strength and mass per unit length.
First, bear in mind that, in order to generate power, you need to be "hovering" at some finite altitude above the hole's horizon, and doing that requires rocket power. (In the flat spacetime case, this is obvious, since if the rocket's engine is not firing, the rocket won't accelerate.) So you are only "generating" power at the expense of using even more power to hover using the rocket.

Leaving that aside, if we let the rope unroll and spin a generator, the total energy generated will depend on the length of the rope (how much fits on the roll). The power (rate of energy generation) will depend on how fast you let the rope unroll.

Once again, if you're letting the rope free-fall, there is no force on it. If you do exert force on the rope, then as Drakkith pointed out, you will need to increase the rocket power you exert to hover, so there's no net gain to doing so.

8. May 23, 2017

### Staff: Mentor

There couldn't be. Consider the flat spacetime case: an accelerating rocket and a rope being unrolled from it in free fall. If force is exerted on the rope, that has to increase the rocket power expended by at least enough to maintain the same proper acceleration with increased mass (since the rope is now being pulled along). If there were a net gain of energy, that would amount to saying that exerting a force on the rope somehow provides rocket power of its own.

9. May 23, 2017

### jbriggs444

Is there an effect which prevents you from building a self-supporting structure around a black hole at a distance and paying a rope out from this structure?

10. May 23, 2017

### puzzled fish

No, I would put my station in a geodesic orbit around the black hole and let my rope unfold towards the center of black hole from there. No power needed for hovering thus.

11. May 23, 2017

### Staff: Mentor

You could, as long as the structure's minimum radial coordinate was at least $9M / 4$, i.e., 9/8 of the horizon radius. Below that, the structure must collapse and fall into the hole. That far away, the approximation used in Greg Egan's page, using Rindler coordinates, would not be valid.

Practically speaking, any actual structure would have to be higher, since the limiting factor is the stress in the material of the structure, and as the minimum radius approaches $9M / 4$, the stress in the structure increases without bound.

Note that the above assumes that the structure itself has negligible mass. If it doesn't, the "M" in the above has to be the mass of the hole plus the mass of the structure, i.e., for purposes of determining the minimum altitude of the structure, we calculate a "horizon radius" based on the total mass present, not just the mass of the hole, even though the actual horizon is determined by only the mass of the hole.

First, there are no geodesic orbits lower than $r = 3M$ (3/2 of the horizon radius), and the ones from $r = 3M$ to $r = 6M$ are unstable. That is much too far away from the hole for the approximation using Rindler coordinates, as in Greg Egan's page, to be valid.

Second, a geodesic orbit will have a nonzero angular velocity, which means the rope can't just pay out radially, and the whole problem gets much more complicated.

12. May 23, 2017

### Ibix

Presumably the rope must disintegrate at 9M/4, too. So you can't extract energy after that point.

Am I right in thinking that one can define a potential for the Schwarzschild spacetime, since it's stationary? If so can we get a maximum energy by considering the difference in potential at jbriggs444's ringworld radius and 9M/4? Or is that hopelessly naive?

13. May 23, 2017

### SlowThinker

I'm puzzled by you saying this and others discussing it as a real problem. This is an engineering problem, not a physics one. I immediately imagined a large construction around the BH, paying 2 ropes from opposite sides.

If there is a point where the rope must disintegrate, you can probably extract unbounded energy at that point, whether it's event horizon or 9/8 of it.

I hope we can settle for "things can be arranged such that the rope is pulled just below its tensile strength, at any speed you desire", is that correct?

14. May 23, 2017

### Drakkith

Staff Emeritus
That would seem to solve the problem then.

15. May 23, 2017

### jbriggs444

Extracting unbounded energy output from finite energy input strikes me as unlikely.

16. May 23, 2017

### Staff: Mentor

No. The $9M / 4$ limit applies to objects supporting their own weight statically. If the rope is falling into the hole, it can stay together just fine.

If it is being suspended from some structure that is static at higher than $9M / 4$, I think the rope can stay together lower than $9M / 4$ in that case too, because the rope's weight is being supported by tension from above, not by compression from below, which is the assumption under which the $9M / 4$ limit is derived. But I haven't actually tried to work through the math for the tension case.

Yes.

I don't think so, because the potential difference is the difference in energy per unit mass for stationary (static in this case) objects. In other words, if we start with an object at rest at some radius $r_1$ and we lower it by some process that ends with it at rest at radius $r_0 < r_1$, and we make sure to extract all the energy possible during the process, then the total energy extracted will be $m \left( \phi(r_1) - \phi(r_0) \right)$, where $m$ is the rest mass of the object and $\phi$ is the potential (in units of energy per unit mass).

Here we're talking about something different: a rope on a reel that unrolls as the rope free-falls off of our structure. The energy we can extract from this process, as far as I can tell, doesn't depend at all on what happens to the rope once it's left the reel, provided that the lower end doesn't hit something before the upper end has left the reel. It only depends on the weight of the rope per unit length, as measured at the reel, and the length of the rope. The weight per unit length is just the proper acceleration at the reel times the rope's rest mass per unit length; so the total work done is just the mass of the rope times the proper acceleration at the reel (actually I think there's another factor of the rope length in there somewhere). So the only "gravitational" factor involved is the proper acceleration, which depends on how close the reel is to the horizon and how massive the hole is. If we assume that the reel is at $r = 9M / 4$, then the proper acceleration is

$$a = \frac{M}{r^2 \sqrt{1 - 2M / r}} = \frac{M}{(9M / 4)^2 \sqrt{1 - 8/9}} = \frac{48}{81M}$$

17. May 23, 2017

### SlowThinker

Maybe this is where the "weak energy condition" comes into play? I learned about the concept just yesterday so I'm not sure if it's correct but it seems reasonable to say that the tensile strength of the rope cannot be higher than some multiple of its mass (perhaps $\sigma=\rho c^2$?) so that we cannot extract more than $mc^2$ when dropping mass $m$ of the rope.

18. May 23, 2017

### Staff: Mentor

Basically, yes. The more precise statement of the weak energy condition is that the stress in the rope can't be greater than its energy density.