v 1 1 8 A ug 1 99 9 Non - Riemannian cosmic walls from gravitational collapse

Two classes of metrics obtained from non-Riemannian gravitational collapse are presented.The first is the Taub planar symmetric exact solutions of Einstein-Cartan field equations of gravity describing torsion walls which are obtained from gravitational collapse of time dependent perturbation of Riemannian Taub symmetric solutions of General Relativity.The second is a modification of the Vilenkin Riemannian planar wall which is obtained from a non-Riemannian planar distribution of spinning matter. Departamento de F́ısica Teórica Instituto de F́ısica UERJ Rua São Fco. Xavier 524, Rio de Janeiro, RJ Maracanã, CEP:20550-003 , Brasil. Gravitational collapse of plane symmetric inhomogeneous distribution of matter leading to cosmic walls have been investigated in detail by Holvorcem and Letelier [1].In this letter we present two examples of metrics which represent cosmic walls obtained from non-Riemannian gravitational collapse.The first is a non-Riemannian wall obtained from a also non-Riemannian planar Taub wall [2] obtained from the General relativistic cosmic wall by a time dependent perturbation.The second is the same time dependent perturbation acting on a Vilenkin Riemannian wall which yields a Riemannian planar thin wall from the gravitational collapse of a non-Riemannian distribution of spin-torsion matter.Let us consider the first metric given by ds = (−dt + dz) (1 + kz) 1 2 + (1 + kz)e β t (dx + dy) (1) Where β is a constant.Note that when time becomes infinity this metric reduces to the well known Taub space-time .Let us now consider the EinsteinCartan field equation in the quasi-Einsteinian form Gij = T EC ij (2) where T ij is the energy-stress tensor similar to the Kalb-Hammond energy momentum tensor given by T ij = SiklS kl j − 1 2 gijSijkS ijk (3) The Einstein Riemannian tensor componenets Gij , where i, j = 0, 1, 2, 3 are given by Gtt = μ 2 t − 2μzz − 3μ 2 z + 2μzνz (4) and Gtz = 2(μtz + μzμt − μtνz) (5) which is responsible for the heat flow and Gzz = 2μtt + 3μ 2 t − 2μz − μ 2 z (6) which yields the orthogonal pressure p| and finally the component Gxx = Gyy which is responsible for the parallel pressure p|| Gxx = Gyy = e (μtt + μ 2 t − μzz − μ 2 z − νzz) (7)