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ps6 - Physics 523 Introduction to Relativity Homework 6 Due...

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Physics 523, Introduction to Relativity Homework 6 Due Tuesday, 6 th December 2011 Hans Bantilan Massless Particle near a Kerr Black Hole Consider the Kerr spacetime with mass M and angular momentum per unit mass a such that 0 a M , so that the metric can be expressed in Boyer-Lindquist coordinates ( t, r, θ, φ ) as g ij dx i dx j = - 1 - 2 Mr Σ dt 2 - 4 Mar sin 2 θ Σ dtdφ + Σ Δ dr 2 + Σ 2 + r 2 + a 2 + 2 Ma 2 rsin 2 θ Σ 2 with its inverse given by g ij ∂x i ∂x j = - A ΣΔ ∂t 2 - 4 Mar ΣΔ ∂t ∂φ + Δ Σ ∂r 2 + 1 Σ ∂θ 2 + Δ - a 2 sin 2 θ ΣΔ sin 2 θ ∂φ 2 where Δ( r ) = r 2 - 2 Mr + a 2 Σ( r, θ ) = r 2 + a 2 cos 2 θ A ( r, θ ) = ( r 2 + a 2 ) 2 - a 2 Δ( r ) sin 2 θ. a. We are to find the trajectories of massless particles in Kerr that lie on the equator θ = π/ 2. These particles follow geodesics γ ( λ ) whose tangent vector field we denote by p = ˙ γ ( λ ). Since ∂/∂t and ∂/∂φ are Killing fields of the Kerr spacetime, we can immediately write down two conserved quantities of any geodesic in Kerr - e = g ( ∂t , p ) = p t l = g ( ∂φ , p ) = p φ . For a massless particle, the tangent vectors p are null, so we then have at each point on the geodesic 0 = g ( p, p ) = g tt p t 2 + 2 g p t p φ + g rr p r 2 + g θθ p θ 2 + g φφ p φ 2 = g tt p t 2 + 2 g p t p φ + g rr ( g rr p r ) 2 + g θθ ( g θθ p θ ) 2 + g φφ p φ 2 = - A ΣΔ e 2 + 4 Mar ΣΔ el + Σ Δ dr 2 + Δ - a 2 ΣΔ l 2 . In the last equality we have used the fact that the geodesic lies on the equator θ = π/ 2 with p θ = 0. So Σ 2 dr 2 = Ae 2 - 4 Marel - - a 2 ) l 2 = A e 2 - 4 Mar A el + A (2 Mr - r 2 ) A 2 l 2 and we can simplify by completing the square, which yields 1 Σ 2 dr 2 = A ( r ) ( e - lW + ( r )) ( e - lW - ( r )) where W ± ( r ) = 2 Mar ± r 2 Δ( r ) /A ( r ).

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ps6 - Physics 523 Introduction to Relativity Homework 6 Due...

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