-a2cos2θ, which is always larger than the horizonr+ (except atthe polesθ= 0, π). So,inside the ergosphere, we must havedϕ/dt >0: frame dragging is so strong that anyparticle (massive, massless, geodesic or not), must have a positivedϕ/dt.The Penrose processSuppose we drop a particle from a far distance, with conserved energyEin=-pt.Now suppose we make surethat this particle decays into two particles inside the ergosphere. Conservation of 4-momentum implies, in particular,Ein=E(1)+E(2)≡ -p(1)t-p(2)tSince the ergosphere is outside the horizon, it is possible for at least one of the particles to escape back to infinity,say particle 1. Suppose particle 2 plunges towards the horizon. Since the Killing vector field∂tis spacelike insidethe ergosphere (its norm squared isgtt>0) and 4-momenta are timelike, it is in principle possible to arrange it sothatE(2)=-p(2)μ∂μt<0 – ask yourself why this would not be possible in general, the answer is deeper than you maythink! With this setup, we would then haveEout=E(1)> Ein. So it is, in principle, possible to extract energy froma Kerr black hole!
5FIG. 4. Schematic representation of the ergosphere and outer horizon of a Kerr black hole