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448 29 Twistingvacuum solutionsThe coordinate transformations that preserve (29.51) areζ ′ =(p 1 ζ + q 1 )/(p 2 ζ + q 2 ), u ′ = uk(ζ,ζ)+h(ζ,ζ). (29.52)Because of the form (29.22) of the Weyl tensor components, m +iM =0,together with I = 0, would give flat space-time. So we have to assumem +iM ̸= 0; that excludes Petrov typesIII,N and O.The integration procedure of the field equations runs as follows. Asthe function G = ∂ ζ ln P − L ,u has an inhomogeneous transformationlaw because of (29.26), G can always be made non-zero. The differentialequations (29.51) are then integrated byG(ζ,ζ) =−[ζ + g(ζ)] −1 ,L(ζ,ζ,u)=(G + ∂ ζ ln P )u + P −1 Gl(ζ,ζ).(29.53)Having thus evaluated the conditions (29.51), we can now attack the threefield equations.The first field equation (29.16) shows that m +iM does not depend onu, and can therefore be written as m +iM =2P 3 G 3 A(ζ,ζ). The secondfield equation (29.20b) then yields A = A(ζ), i.e.m +iM =2P 3 G 3 A(ζ). (29.54)With these results, the last field equation (29.20c) can be written in theform[]Im ∂ ζ ∂ ζ (AG − ∂ ∂V ) =0=(AG − ∂ ∂V ) ,ζζ − (A G − ∂∂V ) ,ζ ζ. (29.55)This is exactly the condition for the existence of a real function B(ζ,ζ)such thatAG − ∂ ∂V = B ,ζ ζ. (29.56)Because of V ,u = P and P ,u =0,Vhas the structure V = Pu+ v(ζ,ζ),with an arbitrary real function v. Inserting this result into (29.56), andusing (29.54), we obtainAG + G l ,ζ= B ,ζ ζ+ v ,ζ ζ. (29.57)If we choose v = −B, then (29.57) is solved by∫l(ζ,ζ) =− A(ζ)G(ζ,ζ)G −1 (ζ,ζ)dζ + l 1 (ζ). (29.58)To get the explicit form of the metric, one has to prescribe the functionsP (ζ,ζ), A(ζ), g(ζ) and l 1 (ζ). The formulae (29.53), (29.54) and (29.58)will give the functions L and m +iM, which together with P give all the