Penrose process is a mechanism by which energy and angular momentum can be harvested from a black hole. It relies on the fact that particles in a Kerr metric can have negative energy. This means that if such particles are absorbed by the black hole, they will lower its mass. This is different from the Newtonian notion of negative energy, which just implies that the particle is bound. Since such trajectories happen to be counter rotating, then they will also leech angular momentum.

If negative energy trajectories exist, the following scheme can be used to extract energy from them:

  1. Two particles start out on some positive energy trajectory
  2. Some process splits them
  3. One particle ends up in a negative energy trajectory
  4. The other particle escapes with more energy than both particles had initially (at the expense of the black hole).

The Kerr metric in the equatorial plane, in Boyer - Lindquist coordinates

$ d \tau^2 = \left( 1- \frac{2 M}{r} \right) dt^2 + \frac{4 M a}{r} dt d\phi - \frac{dr^2}{1 - \frac{2 M}{r} + \frac{a^2}{r^2}} - \left(1+ \frac{a^2}{r^2}+\frac{2 M a^2}{r^3} \right) r^2 d\phi^2 $

where $ \tau $ is the proper time, $ t $ is the time measured by a distant observer, $ M $ is the Schwartzschild radius of the black hole, $ a $ is the spin parameter, $ \phi $ is the azimuthal angle. The parameter $ r $ is monotonous in the real radius, and approaches it at very large distances. To further simplify the problem, we assume a maximally rotating black hole $ a = 1 $. In this case the metric reduces to

$ d \tau^2 = \left( 1- \frac{2 M}{r} \right) dt^2 + \frac{4 M}{r} dt d\phi - \frac{dr^2}{1 - \frac{2 M}{r} + \frac{1}{r^2}} - \left(1+ \frac{1}{r^2}+\frac{2 M}{r^3} \right) r^2 d\phi^2 $

We will simplify things even further by considering trajectories at equal radius ( $ dr = 0 $ ). The energy of a particle is given by

$ \frac{E}{m} = \left( 1 - \frac{2 M}{r} \right) \frac{dt}{d \tau} + \frac{2 M^2}{r}\frac{d \phi}{d \tau} $

It is negative if

$ \frac{d \phi}{d t} < - \frac{r - 2 M}{2 M^2} $

The particle has to be above the horizon, so $ r > 2 M $. This means that it has to rotate in the opposite direction to the black hole. Since the tangential velocity is bounded between zero and the speed of light, negative energy trajectories occur in a bounded range of $ r $.

Further reading Edit