**Gnargenox:**You cannot build a perpetual motion machine. No matter how clever the methods you use. But we can use your questions to understand more about black holes and how energy, and energy conversion, works near them.

You

*can*extract useful energy from matter falling into a black hole, or from the black hole itself (especially rotating ones). The amount of energy you can extract cannot be infinite, however. It cannot be greater than the total mass-energy the system (black hole + whatever you have around it) had to begin with.

Gnargenox wrote:Source of the post Could I use the force generated from the gravitational pull of a Black Hole that is pulling on a long string, which goes into and feeds the hole, replacing whatever mass is lost from Hawking Radiation, without the need for it to be infinity strong?

Let's simplify this scenario (don't worry, we won't lose any of the physics, or energy converted!)

Imagine dropping an object of mass

**m**, from very far away, into a black hole of mass

**M**. If you do, then the black hole becomes more massive by exactly

**m**. No more, no less, because the total energy of the object is conserved all along the way. And you get no useful energy from this because it all went into the black hole.

Now to extract energy. Let's say you have built a shell of some (incredibly strong) material, and locate it near the event horizon at a radius

**r**. On the shell is some device which catches the falling mass and brings it to rest, converting all of its energy of motion into useful energy. That could then be used to power a laser, beaming the energy back up to a safe distance to be captured and used as you see fit.

_{o}**We'll assume this whole process of capturing the kinetic energy and beaming it back by laser is 100% efficient.**How much energy could you extract by this process?

According to general relativity, when that mass is brought to rest on that shell, an observer on that shell will measure its mass to be

This is less than the original mass m. The difference, multiplied by c

^{2}, was the energy released by bringing it to rest. We capture it and then beam it back to us by laser.

Now we must consider the gravitation yet again. The same gravitational field that accelerated the mass (converting potential energy into kinetic if we think in Newtonian terms) is redshifting the laser light as it climbs away from the black hole. The photons decrease in energy by the same factor that the energy of the object increased (according to observers on shells measuring it locally as it passes them).

**Gravity is a conservative force!**

*Furthermore, the pulse of laser light will take a very long time to reach us, due to the gravitational time dilation.*

**To maximize the amount of energy converted and beamed back to us, we must build the shell and energy converter very close to the event horizon. In that limit, the photons are redshifted to infinity, the time it takes for the pulse of laser light to reach us is infinite, and the total energy we receive (over that infinite time!) is exactly equal to the total energy (mc**

^{2}) of the object we dropped in. In fact the time is not really infinite, but equal to the evaporation time of the black hole. Basically we've done no better than turning the object to Hawking radiation, which is exactly what would happen if we had not bothered to capture and convert the energy at all and instead just let the object fall freely into the event horizon!By the way,

**this is the**

**best we can do**

*--*the maximum energy conversion we can

*possibly*have from a non-rotating black hole. Which should make sense from physical laws.

Here's the fraction of the rest mass we can get, as a function of the radius from the black hole that we build the shell/converter. This curve is completely independent of the mass of the black hole.

Gnargenox wrote:Source of the post Can we make a piezoelectric string large enough it wont snap?

No. Any real string or rope lowered slowly into a black hole will quickly cease to be one, due to the tension which will trend to infinity near the horizon. Why? Because of the (locally measured) gravitational acceleration near the horizon. This acceleration is similar to the Newtonian formula, but divided by a factor of (1-r/r

_{s})

^{1/2}due to general relativity, where r

_{s}is the event horizon radius. The acceleration trends to infinity at the event horizon, but how quickly it trends to infinity depends on the mass of the black hole. For more massive black holes it increases more gently. Here it is for the black hole SgrA* at the center of our galaxy:

If the idea of the locally measured acceleration trending to infinity feels weird, then it is at least consistent with another property of the motion at the event horizon: all observers located on the event horizon will observe all objects falling into the horizon to have a velocity of exactly c, regardless of how far away from the black hole they were dropped! (Of course no observer can actually hover on the horizon, but we can imagine this as a sort of limit that they are exceedingly close to the horizon. It also means the observer must be moving at the speed of light to hover on the horizon, which again requires infinite acceleration for anything with mass.)

Gnargenox wrote:Source of the post I guess from our point of view, it never even actually passes over the Event Horizon, so our limit to tensile strength doesn't need to be infinite as well.

We'll say it never crosses the horizon, true, but it does approach the horizon so quickly from our point of view that it makes no practical difference. If you lower the string in, the tension in it will increase very dramatically, and then more and more of it will become irrecoverable. Pull the string back up, and the end of it will be missing.

Gnargenox wrote:Source of the post This is kinetic energy, the acceleration of the string as it reaches the event horizon. This should not be converted to mass in the Black Hole

Incorrect. If we take an object and drop it into a black hole from far away, then we add more mass to the black hole than if we slowly lowered it down to just above the event horizon and

*then*dropped it in.

In fact there is no generalized sense of distinguishing the potential energy, kinetic energy, and rest-mass energy of an object falling into a black hole. Different observers will have different notions about what these energies are and how they are converted. What is true for all observers is that the total energy remains constant for an object freely falling in.

Gnargenox wrote:Source of the post Is this mass that leaves the end of the string actually happening in real time? Is the energy actually lost to the dilation of time?

What do you mean by happening in real time? What do you mean by mass leaving the end of the string?

What you are doing is converting mass to energy. In the Newtonian terms, it is gravitational potential energy converted into... whatever you want to convert it to.

For an analogy, water spilling over a cliff also converts gravitational potential energy into kinetic energy as it descends. Then that kinetic energy is converted into sound and some heat when it hits the bottom. The mass of the water

*does*decrease by this. Where did it go? Converted to other forms of energy. Is time dilation involved? You can think of it that way if you like, since time passes more slowly at the bottom of the falls than at the top. But the falling water, or a person going over the falls themself, would have no notion of time slowing down.