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536 34 Application of generation techniques to general relativityAs α is nilpotent, we have, for any matrix X, αXα = α Tr(Xα) =αTr(αX). To exploit this identity in solving (34.69), we defineΘ(λ, σ) =λ [][]λ − σ Tr F −1 (σ)F (λ)α ⇒ Θ(σ, σ) =σTr F −1 (σ)F ′ (σ)α .The equation for Θ(σ, σ) and its solution become(34.70)∂ ɛ Θ(σ, σ) =Θ(σ, σ) 2 ⇒ Θ(σ, σ) =[1− ɛΘ 0 (σ, σ)] −1 Θ 0 (σ, σ). (34.71)The subscript ‘0’ refers to the corresponding quantity for the seed solution.From (34.68) we find)∂ ɛ(F (σ)αF −1 (σ) = F (σ)αF −1 (σ)Θ(σ, σ),⇒ F (σ)αF −1 (σ) =[1− ɛΘ 0 (σ, σ)] −2 F 0 (σ)αF0 −1 (σ). (34.72)This can now be inserted into (34.66) which can be solved for F (λ). Thesolution is[F (λ) =1 + ɛλλ − σF 0 (σ)αF0 −1 ](σ)1 − ɛΘ 0 (σ, σ)(F 0 (λ) 1 − ɛλα ). (34.73)λ − σRecall that the real part of H contains the metric functions and thatthe lower left element of H is the E potential. H can be calculated from(34.56). Note that the quantities σ, ɛ and the components of α appear asparameters in the final solution.The transformation can be iterated. The result can conveniently bewritten in matrix notation by introducing the vectors and matrices()Θ=(Θ 0 (σ i ,σ k )ɛ k ) , K =(F 0 (σ i )αɛ i ) , L = F0 −1 (σ i )σi−1 (34.74)(i, k =1,...,N). One findsH = H 0 +iK T (1 − Θ) −1 L. (34.75)N.b. the components of the vectors K and L are themselves matrices; butthe transposition does not affect the components of K.The Ernst potential E is the lower left element of H. If the originalmetric is static, i.e. F 0 (λ) is given by (34.61), and α is chosen as α =(0100z i = σ i /2,), one finds (Dietz 1983b), dropping the subscript ‘0’ and denotingE =e 2U D − D −1+ , D ± = det(δ ik + γ ± (z i ,z k )ɛ k ),γ ± (η, ζ) =ie 2β(ζ) S(ζ) −1 ([S(η) − S(ζ)]/[η − ζ] ± 1), (34.76)S(ζ) =(W 2 − (ζ − V ) 2 ) 1/2 ,S(ζ)∇β(ζ) =(ζ − V )∇U − W ˜∇U.