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15.3 Spherical and plane symmetry 229Theorem 15.3 Perfect fluid (with T ik ̸= Λg ik ) and dust solutions cannotadmit a group G 3 on timelike orbits T 2 .In the coordinate system (15.3) (upper signs) the 4-velocity u of a perfectfluid has the form u = u 3 ∂ 3 +u 4 ∂ 4 and from the invariance of u underisometries one infers that the components u 3 , u 4 cannot depend on x 1 andx 2 . Therefore u is hypersurface-orthogonal and there is a transformationof x 3 and x 4 which takes u intou =e −ν ∂ 4 (15.8)(comoving coordinates) and, simultaneously, preserves the form of themetric (15.3).15.3 Spherical and plane symmetryThe group G 3 on V 2 contains two special cases of particular physicalinterest: spherical and plane symmetry.In the first four decades of research on general relativity the majorityof exact solutions were obtained by solving the field equations under theassumption of spherical symmetry. The exterior and interior Schwarzschildsolutions and the Friedmann model of relativistic cosmology arewell-known examples.Originally, problems with spherical and plane symmetry were treatedmore or less intuitively. In the modern literature the group theoreticalapproach is preferred and spherical and plane symmetry are invariantlydefined by theDefinition: A space-time V 4 is said to be spherically- (plane-) symmetricif it admits a group G 3 IX (G 3 VII 0 ) of motions actingon spacelike2-spaces S 2 and if the non-metric fields inherit the same symmetry.Each orbit S 2 has constant positive (zero) Gaussian curvature, k =1(k = 0). The isotropy group I 1 represents a spatial rotation in the tangentspace of S 2 . According to Theorem 15.1, spherically- and plane-symmetricspace-times are of Petrov type D or O.In the expressions (15.6) for the components of the Einstein tensorwe have to choose the upper signs, and k =1ork =0.Forsphericalsymmetry (k = 1), we can specialize the metric (15.3) tods 2 = Y 2 (r, t)(dϑ 2 +sin 2 ϑ dϕ 2 )+e 2λ(r,t) dr 2 − e 2ν(r,t) dt 2 . (15.9)For plane symmetry (k = 0), we often use Cartesian coordinates in theorbits:ds 2 = Y 2 (z,t)(dx 2 +dy 2 )+e 2λ(z,t) dz 2 − e 2ν(z,t) dt 2 ,ξ 1 = ∂ x , ξ 2 = ∂ y , ξ 3 = x∂ y − y∂ x .(15.10)