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A Penrose Process

  1. Jun 5, 2016 #1
    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?
     
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  3. Jun 5, 2016 #2

    PeterDonis

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    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.
     
  4. Jun 6, 2016 #3
    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?
     
  5. Jun 6, 2016 #4

    PeterDonis

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    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.

    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.
     
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