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434 28 Robinson–Trautman solutionshave been found by Kowalczyński(1978, 1985). Some are covered by Theorem28.5. The remaining solutions areQ = Q 0 ζ √ ]u, A = Q 0 Q 0[ln(1 + εζζ/2) − (ln u)/4 ,[h = h = εQ 0 / (2 + εζζ) √ ](28.60)u , ε = ±1,and2A = Q 0 Q 0 {ln[1 − ζζ/2] − ln[(1 + ζ/ √ 2)(1 + ζ/ √ 2)] − (ln u)/4},Q = Q 0√ u, h =Q 0√2u1+ζ/ √ 2(2 − ζζ)(1 + ζ/ √ 2) , ε = −1.(28.61)Like most of the solutions with a ζ–ζ -space of constant negative curvature,the solution (28.61) can be interpreted in terms of a tachyon’s worldline.If P ,u = Q ,u = h ,u =0, then the field equations (28.37a)–(28.37e) canpartially be integrated and give (with real constants a 0 ,b 0 and m 0 andthe complex constant ψ 0 )Q = b 0 β(ζ)+ψ 0 ,h = −β ′ (ζ),m = a 0 u − 2κ 0 b 0 ββ − 4κ 0 Re (ψ 0 β)+m 0 ,(28.62)the only surviving equation beingP 2 ∂ ζ ∂ ζP 2 ∂ ζ ∂ ζln P = a 0 + κ 0 P 2 β ′ (ζ)β ′ (ζ) (28.63)(Herlt and Stephani1984). For a 0 ̸= 0, the only known solution isP =1, β = β 0 ζ, a 0 = −κ 0 β 0 β 0 . (28.64)For a 0 =0, (28.63) can be integrated twice and yieldsP 2 ∂ ζ ∂ ζln P = κ 0 β(ζ)β(ζ)+ϕ(ζ)+ϕ(ζ). (28.65)Some Einstein–Maxwell fields belonging to this class areP =(ζ + ζ − 2 3 κ 0ββ) 3/2 , β = β 0 ζ, (28.66)P =(2/3κ 0 c 0 ) 1/2 (κ 0 ζζ + c 0 ) 3/2 , β = ζ, (28.67)P =(2κ 0 ζζ) 1/2 [δ ′ (ζ)δ ′ (ζ)] −1/2 [1 + δ(ζ)δ(ζ)/2], β = ζ. (28.68)The solutions (28.66) and (28.68) generalize (28.16) and (28.53), respectively.