 Solution

ID: past-exam-of-the-general-relativity-ii-course-of-the-university-of-oxford/2017/2/b/iii/solution

K^{a} \nabla_{a} K^{b} = - \frac{1}{2} \nabla^{b} (K^{a} K_{a}) .

Here

K^{2} = - λ^{2}

, so in the regular coordinates

K^{a} \nabla_{a} K^{b} = λ g^{bλ} = - \frac{λ}{M} (\partial_{v})^{b} + \frac{λ ^{3}}{M ^{2}} (\partial_{λ})^{b} .

This vector vanishes at

λ = 0

K = \partial_{v}

is nonzero there, so comparison with the definition of surface gravity yields

κ = 0.

Hence this is a degenerate Killing horizon. The argument uses the vanishing gradient of

K^{2}

on the horizon; the condition

K^{2} = 0

by itself would not force zero surface gravity.

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