Geodesics of the Kerr metric

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then The equation ) 2 (r − Ma2 ω 2 + 4Ma ω − 2M =0. (4.85) r 2 r 2 r 2 (r 3 − Ma 2 )ω 2 +2Maω − M =0 (4.86) has discriminant M 2 a 2 + M(r 3 − Ma 2 )=Mr 3 (4.87) and solutions ω ± = −Ma ± √ Mr 3 = ± √ M r3/2 ∓ a √ M r 3 − Ma 2 r 3 − Ma 2 = ± √ r 3/2 ∓ a √ M M (r 3/2 + a √ M)(r 3/2 − a √ M) √ M = ± r 3/2 ± a √ M . (4.88) This is the relation between angular velocity and radius of circular orbits, and reduces, in Schwarzschild limit a =0,to which is Kepler’s 3 rd law. ω ± = ±√ M r 3 , (4.89) 4.2 General geodesic motion: the Carter constant To study generic geodesics in Kerr spacetime, we need to apply the Hamilton-Jacobi approach, whichallowstoidentifyafurther constant of motion (not related to anyisometryofthemetric). To apply the Hamilton-Jacobi approach, we first have to express our system with an Hamiltonian description. Given the Lagrangian of the system L(x µ , ẋ µ )= 1 2 g µνẋ µ ẋ ν (4.90) 84
we have defined the conjugate momenta 2 p µ = ∂L ∂ẋ µ = g µνẋ ν . (4.91) By inverting (4.91), we can express ẋ µ in terms of the conjugate momenta: ẋ µ = g µν p ν . (4.92) The Hamiltonian is a functional of the coordinate functions x µ (λ) and of their conjugate momenta p µ (λ), defined as In our case, we have H(x µ ,p ν )=p µ ẋ µ (p ν ) −L(x µ , ẋ µ (p ν )) . (4.93) H = 1 2 gµν p µ p ν . (4.94) The geodesics equation is equivalent to the Euler-Lagrange equations for the Lagrangian functional, which are equivalent to the Hamilton equations for the Hamiltonian functional: ẋ µ = ∂H ∂p µ ṗ µ = − ∂H ∂x µ . (4.95) To solve equations (4.95) is as difficult as to solve the Euler-Lagrange equations. Anyway, it is possible to apply the Hamilton-Jacobi approach, which here we briefly recall. In the Hamilton-Jacobi approach, we look for a function of the coordinates and of the curve parameter λ, S = S(x µ ,λ) (4.96) which is solution of the Hamilton-Jacobi equation ( H x µ , ∂S ) + ∂S =0. (4.97) ∂x µ ∂λ In general such a solution will depend on four integration constants. It can be shown that, if S is a solution of the Hamilton-Jacobi equation, then ∂S ∂x µ = p µ . (4.98) 2 Not to be confused with the four-momentum of the particle, which we denote with P µ . 85
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 and 10: of the orbits. To this purpose it i
Page 11 and 12: ackwards in time. Furthermore, from
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: Figure 4.2: Geodesics of Massive pa
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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