Geodesics of the Kerr metric

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Note that this equation does not depend on L, thusthevalueofr max is independent of L. The solution of (4.48) is a decreasing function of a, and,inparticular, r max = 3M for a =0 r max = M for a = M r max = 4M for a = −M . (4.49) Therefore, while for a Schwarzschild black hole the unstable circular orbit for a photon is located at r = 3M, for a Kerr black hole it can be located much closer to the black hole, in particular for large values of a; in the case of an extremal (a = M) blackhole, r max = M, whichisthepositionofthehorizonforsuchablackhole (see also Figure 4.2). A photon coming from infinity with constant of motion E > V + (r max ), falls inside the horizon, whereas if 0 0. It remains to consider the case when E
ackwards in time. Furthermore, from a quantum-mechanical point of view, the existence of negative energy particle states would lead to an unstable vacuum, since it would be energetically favourable the creation of more and more of such particles. The energy measured by a different observer is different, but its sign is the same: indeed, if u µ =(γ,γV) withγ =(1− V 2 ) −1/2 , then E (u) = −γ(P 0 + P i V i ) (4.51) which has the same sign as −P 0 since |P i V i | < |P 0 |.Wecanconclude that, in special relativity, particles must have positive energy with respect to all observers; and that if the energy is positive as measured by any observer (for instance the static observer), it is positive as measured by all observers. In general relativity, we still define the energy of a particle measured by an observer u µ as in eq. (4.50). Indeed, eq. (4.50) is a tensor equation which holds in a locally inertial frame, and consequently in any other frame by the principle of general covariance. For the same reason, it must be E (u) > 0 , (4.52) and it is sufficient to prove this for any observer. Let us now consider a particle in the Kerr spacetime, located at radial infinity (where a static observer with u µ st =(1, 0, 0, 0) can always be defined). According to the definition E (u) = −u µ P µ ,the energy measured by the static observer will be E (ust) = −P 0 and, according to eq. (3.21) 1 E (ust) = −P 0 = E (4.53) We conclude that for a particle coming from radial infinity, the constant of motion E can be identified with the energy as measured by a static observer at infinity, and consequently orbits with negative values of E are not allowed. Referring to figure 4.1, let us now consider a massless particle, say a photon, that moves in the ergoregion, i.e. between r + and r S+ . A static observer cannot exist in this region, therefore we need to consider a different observer, for instance the ZAMO, u µ ZAMO = C(1, 0, 0, Ω) (4.54) 1 Note that eq. (4.53) holds for a massless particle; in the case of a massive particle, where E is the energy at infinity per mass unit, we have to multiply E with the particle mass. 77
Page 1 and 2: Chapter 4 Geodesics of the Kerr met
Page 3 and 4: κ =1 forspacelikegeodesics κ =0 f
Page 5 and 6: B ≡ 2Ma r C ≡ r 2 + a 2 + 2Ma2
Page 7 and 8: 4.1.1 Null geodesics In the case of
Page 9: of the orbits. To this purpose it i
Page 13 and 14: 4.1.3 The Penrose process A short c
Page 15 and 16: If we compute the integral at a tim
Page 17 and 18: Figure 4.2: Geodesics of Massive pa
Page 19 and 20: we have defined the conjugate momen
Page 21 and 22: [ ] (r 2 + a 2 ) 2 − − a 2 sin
Page 23: Then we have dθ dλ ′ = ± √

Geodesics of the Kerr metric

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