Contents

490 32 Kerr–Schild metricsTheorem 32.4 A space-time is a special Kerr–Schild space-time (definedby the properties (32.1)–(32.2) and (32.21)–(32.22)) if and only ifits Newman–Penrose coefficients can be transformed to⎛ ⎞ ⎛κτσρ 0∆Y 0 ¯δY ⎞⎝ πνµλ⎠ = ⎝ 0 ν µ 0 ⎠ , S = µεγβαρ = ¯µ¯ρ (32.27)0 γ 0 0with respect to a suitably chosen null tetrad.The details of the proof are given by Debever (1974). To show that (32.27)is sufficient, one first determines the remaining Newman–Penrose coefficientsby means of the Newman–Penrose equations and the commutationrelations, see §§7.1, 7.3. The result readsγ =2 −1 (DS + S(ρ − ¯ρ)), ν = ¯δS − S∆Y, µ = Sρ. (32.28)Then, solving the first Cartan equations (2.76)dω 1 = Y ,a ω a ∧ (ω 4 − Sω 3 )=dω 2 ,dω 3 = Y ,a ω a ∧ ω 1 + Y ,a ω a ∧ ω 2 , (32.29)dω 4 = SY ,a ω a ∧ ω 2 + SY ,a ω a ∧ ω 1 + S ,a ω a ∧ ω 3 ,one obtains the dual basisω 1 =dζ + Y dv, ω 2 =d¯ζ + Y dv,ω 3 = Y dζ + Y d¯ζ + Y Y dv +du, ω 4 = Sω 3 +dv.(32.30)The associated metricds 2 = 2(dζ d¯ζ − du dv) − 2S(Y dζ + Y d¯ζ + Y Y dv +du) 2 (32.31)is clearly of the Kerr–Schild type.If we start with a line element (32.31), then the null congruenceω 3 = −k a dx a = Y dζ + Y dζ + Y Y dv +du (32.32)will be geodesic and shearfree only if Y obeys (32.25). Because ofD = −(Y∂ ζ + Y∂¯ζ − ∂ v − Y Y∂ u ), δ = ∂ ζ − Y∂ u ,∆=S(Y∂ ζ + Y∂¯ζ − ∂ v − Y Y∂ u )+∂ u ,(32.33)these conditions readY ,ζ− YY ,u =0, Y ,v − YY ,ζ =0. (32.34)