Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 311 / 1 / d

Solution

The normal to a surface of constant

r

n_{a} = (d r)_{a}

, and the inverse ingoing metric gives

n^{a} n_{a} = g^{rr} = f (r) .

Hence

r = r_{+}

ξ = k + Ω_{+} m = \partial_{v} + Ω_{+} \partial_{χ}, Ω_{+} = Ω (r_{+}) = \frac{r _{-}}{L r _{+}} .

This constant linear combination of Killing vector fields is Killing. Its squared norm is

ξ^{2} = - f + r^{2} (Ω - Ω_{+})^{2},

which vanishes at

r_{+}

. More strongly, lowering its index in the ingoing metric gives

ξ_{a} = (d r)_{a}

on the horizon. It is therefore the null normal and generator, so the surface is a Killing horizon.

Using

\nabla_{a} (ξ^{2}) = - 2 κ ξ_{a}

on a Killing horizon, the squared term contributes no first derivative at

r_{+}

and

κ = \frac{1}{2} f^{'} (r_{+}) .

Differentiating

f

at its simple outer root gives

f^{'} (r_{+}) = \frac{2 ( r _{+}^{2} - r _{-}^{2} )}{L ^{2} r _{+}},

κ = \frac{r _{+}^{2} - r _{-}^{2}}{L ^{2} r _{+}} .

It vanishes in the extremal case

r_{+} = r_{-}

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