How Does the Penrose Process Allow for Energy Extraction from a Black Hole?

[astrolollo]

Hi everyone,

I am a bit confused about the Penrose Process. Let's say a particle with energy E at infinity arrives in the ergoregion of a Kerr black hole and then it decays into two photons. One of them has L_1<0 and E_1<0 and hence it falls towards the singularity, while the other has L_2>0 and E_2>0. The latter photon can reach infinity (if E_2 is sufficiently high) wher an observer would detect it. By energy conservation, E_2=E-E_1 >E since E_1<0. Thus we have received a photon with an energy greater than the energy of the particle we had sent into the black hole. I mathematically understand this, but I am still puzzled. The ergosphere (the boundary of the ergoregion) is a infinite-redshift surface. Hence a photon departing from there arrives at infinity with frequency equal to zero. So, since $E=h \nu$ , I wouldn't detect anything. I can't figure out what happens to the redshift as soon as a photon comes out from INSIDE the ergoregion. Moreover, I wouldn't have more energy than the one of the particle I had sent towards the black hole. Another inconsistency is that E must be a conserved quantity (related to the Killing vector $\partial_{t}$ ), and in fact it is not if the photon is created ON the ergosphere (it would need to have E>0 to escape the black hole, but since the light is redshifted it has E=0 at infinity). Can anyone help me?

 

[PeterDonis]

The ergosphere (the boundary of the ergoregion) is a infinite-redshift surface.

Not for all photons. It's a surface at which a photon that is moving purely radially outward will have infinite redshift. (Note that this is a heuristic description and has significant limitations, but it will do for this discussion.) But a photon coming out from the ergosphere will not be moving purely radially. It will have a positive angular velocity, i.e., it will have a component of its motion which is angular, in the same direction as the rotation of the hole. So it can escape to infinity just fine.

 

[astrolollo]

Ok thanks. I guess though that the process is efficient only for very high energetic photons, since they will undergo a gravitational redshift anyway. Is this correct?

 

[PeterDonis]

I guess though that the process is efficient only for very high energetic photons

The process works for photons of any energy. More precisely, any photon that has a positive energy at infinity can escape. So, in the terminology of your OP, any E2 that is positive will work.

since they will undergo a gravitational redshift anyway

The energy at infinity of the photon already takes into account all of the gravitational redshift it will undergo as it escapes. So if E2 > 0, that means the photon will still have some energy left after it has escaped to infinity.