Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 311 / 2 / d / Solution

For this horizon generator, direct evaluation of

ξ^{b} \nabla_{b} ξ^{a} = κ ξ^{a}

, or the radial derivative of

Δ

, gives

κ = \frac{Δ ^{'} ( r _{+} )}{2 ( r _{+}^{2} + a ^{2} )} .

Because

Δ = r^{2} + a^{2} - μ / r

and

μ = r_{+} (r_{+}^{2} + a^{2})

,

Δ^{'} (r_{+}) = 2 r_{+} + \frac{μ}{r _{+}^{2}} = \frac{3 r _{+}^{2} + a ^{2}}{r _{+}},

and therefore

κ = \frac{3 r _{+}^{2} + a ^{2}}{2 r _{+} ( r _{+}^{2} + a ^{2} )} .

The induced horizon cross-section has volume element

d A = r_{+}^{2} (r_{+}^{2} + a^{2}) sin θ cos^{2} θ sin λ d θ, d ϕ, d λ, d ψ .

Using the unit-four-sphere integral from part b gives

A_{H} = \frac{8 π ^{2}}{3} r_{+}^{2} (r_{+}^{2} + a^{2}) .

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