(ed.). Gravitational waves (IOP, 2001)(422s).

Solutions to exercises 87The gauge transformation for a perturbation in linearized theory isSubstituting (iii) into (i) and (ii), we obtainh ′ αβ = h αβ − ξ α,β − ξ β,α . (iii) (7.15)(i) = 1 2 ηµσ (h ′ σβ,αν + ξ σ,βαν + h ′ σα,βν + ξ σ,αβν − h ′ αβ,σ ν )(ii) = 1 2 ηµσ (h ′ σν,αβ + ξ σ,ναβ + h ′ σα,βν + ξ σ,αβν − h ′ αν,σβ ).If we find the difference between the two formulae above we getChapter 5Exercise (e)R µ αβν= (ii) − (i) = Ɣ′µαν,β − Ɣ′µ αβ,ν = R′µ αβν .The action principle is:∫ √ ∫δ(R −g)δI =h µν d 4 x =−δg µνG µν√ −gh µν d 4 x = 0. (i) (7.16)If we perform an infinitesimal coordinate transformation x µ → x µ + ξ µ withoutotherwise varying the metric, then the action I must not change:∫0 = δI = G µν (ξ µ;ν + ξ ν;µ ) √ −g d 4 x∫= 2 G µν ξ µ;ν d 4 x.This can be transformed in the following way:∫∫δI = (G µν √ξ µ ) ;ν −g d 4 x − (G µν ;νξ µ ) √ −g d 4 x = 0.The first integral is a divergence and vanishes.arbitrariness of ξ µ , gives the Bianchi’s identities:G µν ;ν = 0.The second, because of theExercise (f)The two polarization components are h + xx yy=−h+ = A +e −ik(t−z) and h xy× =A × e −ik(t−z) . The energy flux is the negative of〈T (GW)0z〉= 132π 〈hij ,0h ij,z 〉=− k216π (A2 + + A2 × )〈sin2 k(t − z)〉=− k232π (A2 + + A2 × ).