Paper 2, Section II, B

General Relativity | Part II, 2010

A vector field kak^{a} which satisfies

ka;b+kb;a=0k_{a ; b}+k_{b ; a}=0

is called a Killing vector field. Prove that kak^{a} is a Killing vector field if and only if

kcgab,c+k,bcgac+k,acgbc=0k^{c} g_{a b, c}+k_{, b}^{c} g_{a c}+k_{, a}^{c} g_{b c}=0

Prove also that if VaV^{a} satisfies

V;baVb=0V_{; b}^{a} V^{b}=0

then

(Vaka),bVb=0\left(V^{a} k_{a}\right)_{, b} V^{b}=0

for any Killing vector field kak^{a}.

In the two-dimensional space-time with coordinates xa=(u,v)x^{a}=(u, v) and line element

ds2=du2+u2dv2d s^{2}=-d u^{2}+u^{2} d v^{2}

verify that (0,1),ev(1,u1)(0,1), e^{-v}\left(1, u^{-1}\right) and ev(1,u1)e^{v}\left(-1, u^{-1}\right) are Killing vector fields. Show, by using ()(*) with VaV^{a} the tangent vector to a geodesic, that geodesics in this space-time are given by

αev+βev=2γu1\alpha e^{v}+\beta e^{-v}=2 \gamma u^{-1}

where α,β\alpha, \beta and γ\gamma are arbitrary real constants.

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