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

Table of contents
- Solution b

Solution

Put

μ = r_{+} (r_{+}^{2} + a^{2})

. At large

r

, the one-form dual to

K

has

K^{♭} = g_{tt} d t + g_{tϕ} d ϕ, g_{tt} = - 1 + \frac{μ}{r Σ} = - 1 + \frac{μ}{r ^{3}} + O (r^{- 5}) .

Only the

d r \land d t

term contributes to the constant-

t, r

Komar integral, and

d K^{♭} = - \frac{3 μ}{r ^{4}}, d r \land d t + O (r^{- 5}) .

With the stated orientation, the asymptotic Hodge star operator gives

⋆ (d r \land d t) = r^{4} sin θ cos^{2} θ sin λ d θ \land d ϕ \land d λ \land d ψ .

The angular integral is the area of the unit four-sphere,

\int sin θ cos^{2} θ sin λ d θ, d ϕ, d λ, d ψ = \frac{8 π ^{2}}{3} .

Consequently

\int ⋆ d K^{♭} = - 8 π^{2} μ, M = - \frac{1}{12 π} \int ⋆ d K^{♭} = \frac{2 π}{3} r_{+} (r_{+}^{2} + a^{2}) .

This is the Komar mass of the Singly rotating six-dimensional Myers-Perry black hole in the units of the question.

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