Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 309 / 2 / c / ii

Table of contents
- Solution ii

Solution

For

V = r \partial_{r}

, Cartan's magic formula gives

L_{V} d r = d (i_{V} d r) + i_{V} d (d r) = d (r) = d r,

whereas contraction and exterior differentiation both vanish for

d u, d θ, d ϕ

. Thus

L_{V} d u = L_{V} d θ = L_{V} d ϕ = 0.

Applying the Lie derivative to

g = - d u^{2} - 2, d u, d r + r^{2} d Ω^{2}

yields

L_{V} g = - 2, d u, d r + 2 r^{2} d Ω^{2} .

Here

n_{a} = - (d u + d r)_{a}

and

V_{a} = - r (d u)_{a}

, so comparison with

2 g

gives

(L_{V} g)_{ab} = 2 g_{ab} + \frac{2}{r} n_{(a} V_{b)}, α = 2 .

Using tracelessness from part (b)(i),

\nabla_{a} (T^{ab} V_{b}) = \frac{1}{r} T^{ab} n_{a} V_{b} \geq 0

by the dominant energy condition, because

n

is future timelike and

V

is future null.

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