Extracting rotational energy from a black hole: why does the direction in which we drop stuff in matter?

Imagine an advanced civilization living around a black hole, extracting energy from it for their needs. Two ways to do this:

1. Drop matter in and use the light it emits as it heats up.
2. Extract the rotational energy of the black hole.

For the second mechanism, if we drop in the matter in a way that it increases the angular momentum ("in the direction of rotation"), we'll have more rotational energy to extract. On the other hand, if we drop it in a way it decreases the angular momentum, we'll have less rotational energy to extract. It puzzles me that this decision of how we throw the matter in seems to make a difference in the energy we have available.

If I think of a regular massive object, the paradox is resolved by the fact that throwing matter in a way that it slows down the rotation of the massive object will generate more friction and heat. Is this then true for a black hole as well? Does throwing matter in such that it reduces the angular momentum create more heat than throwing it in such that it increases it?

The process you are referring to is called the Penrose Process, which is essentially a calculation on the Kerr Solution (a rotating black hole in the context of general relativity). There are more subtle technicalities to this problem than you think which I'm going to explain in detail! Actually, the short answer is that if you want to extract energy from such a solution, then the essential condition for the object falling inside the black hole would be to decrease the black hole's angular momentum! I am going to explain the argument now, but if you are not familiar with concepts such as Killing Vector, Metric, Asymptotic Behavior, etc. I suggest that you read useful books such as Spacetime and Geometry: An Introduction to General Relativity by Sean M. Carroll, or any other book on this matter.

In the case of the Kerr metric, the norm of $$\xi_{t}=\partial_{t}$$ which is also a Killing vector for this solution, is equal to zero in a region that is different from where the outer horizon is! We call this surface Ergosphere.

$$|\partial_{t}|^{2}=g_{tt}=-\frac{\Delta-a^{2}\sin^{2}\theta}{\Sigma}=0$$

As you can see in this equation, this is somewhere different from the place of horizon. As a matter of fact, there is a region between the outer horizon and this surface where $$g_{tt}>0 , r>r_{+}.$$ We can find the exact formula for this surface $$r^{2}-2Mr+a^{2}\cos^{2}=0$$ and $$r=M\pm\sqrt{M^{2}-a^{2}\cos^{2}\theta}$$ We call the region between the outer horizon and the Ergosphere, The Ergoregion. In this region, $$\partial_{t}$$ is spacelike as seen by the observer sitting at infinity. The asymptotic observer sees that this Killing vector is timelike outside this region, null on this surface and spacelike inside the Ergoregion. As a matter of fact, Ergosphere is the infinite red-shift surface, which is not on the horizon. This region leads to amazing consequences. As $$|\partial_{t}|^{2}<0$$ inside the ergo-region, the sign of energy is negative from a distant observer's point of view. Now consider an object with energy $$E>0$$ falling into the ergo-sphere. Due to the fact that ergo-region is not a trapped surface, this object can still scape from it and come outside. Now suppose that the object explodes (or decays or any other process that can make it into two pieces) before coming back and turns into two pieces. One falling toward the horizon and the other coming back. From the asymptotic observer's point of view, the first piece has energy $$E_{1}<0$$ because it is still on the other side of the ergo-sphere, and the second piece has energy $$E_{2}>0$$ because it came back from the region and is now outside. Now lets write the conservation law for this system $$E=E_{1}+E_{2}.$$ Since $$E_{1}$$ is negative, then $$E_{2}>E$$. Which means that we have extracted energy from the black hole. If we take a more careful look at this problem, we notice that, we should make sure that the particle with energy $$E_{1}$$, falls inside the black hole and does not come out. These are the definitions of energies $$E=-P.\xi_{t}$$ $$E_{1}=-P_{1}.\xi_{t}$$ $$E_{2}=-P_{2}.\xi_{t}$$ Where $$P=P_{1}+P_{2}$$ and $$E,E_{2}>0$$. Now, we should make sure that object number 1, falls inside the horizon. The condition satisfying this is $$-P_{1}.\xi\geq0 ; \xi=\partial_{t}+\Omega_{+}\partial_{\phi}$$ Where $$\xi$$ is the null Killing vector on the horizon. We also know that $$-P_{1}.\xi_{t}<0$$, which means that $$E_{1}-\Omega_{+}J_{1}\geq0$$ and because $$E_{1}<0$$ then $$J_{1}<0$$. This means that if we start with a black hole with mass $$M$$ and angular momentum $$J$$, we would have a black hole with mass $$M+\delta M$$ and angular momentum $$J+\delta J$$, after the object falls inside it. Therefore $$\delta M-\Omega_{+}\delta J\geq0$$ where $$\Omega_{+}=\frac{a}{r^{2}+a^{2}}=\frac{\frac{J}{M}}{(M+\sqrt{M^{2}-\frac{J^{2}}{M^{2}}})^{2}+\frac{J^{2}}{M^{2}}}\Rightarrow \delta(M^{2}+\sqrt{M^{4}-J^{2}}\geq0)$$ So until this last relation holds, we can extract energy from the black hole. In conclusion, we can extract energy from a Kerr black hole in the cost of decreasing it's angular momentum. So, until it's angular momentum is non-zero, we can extract energy from it through a process like this. Note that just because an observer sitting at infinity says that there is no time-like Killing vector in ergo-region, does not mean that the geometry lacks a time-like Killing vector there.

• Thanks, this helps. Will take a while to understand it. But I believe you've described how to extract rotational energy from a BH. My question is slightly different. Let's say we simply throw a mass into it. For a regular spinning object, you could throw it in a way that it increases the angular momentum or throw it in a way that it reduces the angular momentum. Are there similarly two ways to throw an object into a rotating black hole? And if so, will more heat and light be generated when its thrown in a way that it reduces the angular momentum? Nov 23 '21 at 21:33
• The simple answer is that if you throw something inside a black hole that increases its angular momentum, then you would not be able to extract energy from it. You just increase its angular momentum! If you wish to extract energy from it, the process requires the black hole to somehow "transform" its angular momentum, to the energy you want from it! Nov 23 '21 at 22:25
• Sure, let's say we don't want to extract energy, we're doing this purely as an experiment. Is it true that we can throw matter into it in a way it either reduces or increases its angular momentum? Nov 24 '21 at 0:06
• Sure. But what is the point? Nov 24 '21 at 17:31
• 1) P. K. Townsend. Black holes: Lecture notes. Preprint: gr-qc/9707012. 2) S. M. Carroll. Spacetime and geometry: An introduction to general relativity. San Francisco, USA: Addison-Wesley (2004). 3) Roy P. Kerr. Gravitational field of a spinning mass as an example of algebraically special metrics. Phys.Rev.Lett., 11:237-238, 1963. 4) James M. Bardeen, B. Carter, and S.W. Hawking. The Four laws of black hole mechanics. Commun.Math.Phys., 31:161-170, 1973. Or basically any paper or book regarding black holes! Nov 30 '21 at 10:10