# How much energy does lowering an object into a black hole generate?

An object of mass m is slowly lowered into a black hole of mass 1000 m. Is the amount of braking energy larger than $0.6 mc^2$?

Now what if, after lowering the mass close to the event horizon, we wait the black hole to shrink to mass 100 m, by emitting Hawking radiation, and then we lower the mass some more? Is the braking energy of the second lowering process larger than $0.5 mc^2$?

I wonder can we extract more than $E=mc^2$ energy from mass m?

(In this thought experiment we use a wire and a winch, or anything that works)

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The total mass/energy of the suspended object goes like $$M(r)=\sqrt{g_{00}(r)} M_\infty$$ so near the event horizon, this goes to zero. Consequently, almost 100% of the $E=M_\infty c^2$ latent energy may be extracted by lowering it near the horizon.

If you're willing to wait for the black hole to Hawking evaporate, you may obtain the whole $E_{\rm BH} = M_{\rm BH}c^2$ in the form of radiation. However, this energy isn't coming from a particular object that you lowered; it comes from all of them, from converting the black hole itself to energy.

Even if you wanted to attribute the energy from the "extra lowering" to the particular suspended object, you won't be able to get more than $E=M_\infty c^2$ out of the object. The first formula of this answer is still true. So if you keep the object at the same "depth" but you allow the black hole to shrink a bit so that the suspended object is suddenly further from the event horizon, it means that $g_{00}$ (the red shift factor) will be increased again, which will reduce the energy of the suspended object.

In other words, if you want to keep the (highly reduced) energy of the suspended object constant (near zero), you should allow it to keep a small distance from the event horizon. It means that if the black hole shrinks by Hawking evaporation, the suspended object will have constant energy if it is lowered as the black hole shrinks.

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Isn't it like this: if you keep the object at the same "depth" but you allow the black hole to shrink a bit so that the suspended object is suddenly further from the event horizon, it means that g00 (the red shift factor) will be DECREASED again, which will INCREASE the energy of the suspended object – kartsa Jan 17 '12 at 2:32
Is it so that an object that has been lowered close to an event horizon has lost most of its weight? Then electro-magnetic forces have to be reduced too, so that things look normal: Thin thread must break, when an object that was heavy at upper location, is hanging on it – kartsa Jan 17 '12 at 4:21
I see redshift factor is a SMALL number, when there is a LARGE redshift. Redshift becomes SMALLER when event horizon moves further away, doesn't it, so energy of an object INCREASES when black hole shrinks under the object. – kartsa Jan 17 '12 at 4:30

With this method you can gather an unlimited amount of energy, you just need a black hole big enough. This is how it works:

1: You lower a mass m close to event horizon, you get energy less than $mc^2$, because the event horizon moved a little bit upwards.

2: You wait while the black hole shrinks. During this time the black hole produces redshifted Hawking radiation. The gravity field of the mass being lowered causes the redshift. Because the mass of the object being lowered has become almost zero, the redshift is almost zero. But energy that is "missing" from the Hawking radiation goes into our object, which gains mass, which means that the redshift increases.

3: Repeat many times

The energy you get comes from the object and from the Hawking radiation.

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