Generating Einstein-scalar solutions

Abstract: In this comment on a recent paper by Dunajski a method of generating solutions of the Einstein-scalar field equations from Einstein metrics is presented. Two spherically symmetric examples are presented. 1 In a recent paper,[1], Maciej Dunajski used the standard Kaluza-Klein reduction ansatz from five to four dimensions to construct new four-metrics which satisfy the Einstein-Maxwell-Dilaton field equations from Einstein four-metrics. The aim of this comment is to point out a way in which Lorentzian solutions of the Einstein-scalar equations can be similarly con-structed from any known Lorentzian Einstein four-metric which admits a non-null hypersurface orthogonal Killing vector field. The construction fol-lows the same main lines as Dunajski’s but differs in some details. It leads to Einstein-scalar four metrics which do not necessarily possess continuous isometries. The zero rest-mass scalar fields are minimally coupled in the standard (non-phantom) way and the cosmological constant is zero. First let ds2(4) be the line element of a Lorentzian four-metric on a four dimensional manifold M, with signature (3,1), which admits a hyper-surface orthogonal, non-null killing vector field K. It is assumed to be Einstein so that it satisfies the equations Gαβ = −Λgαβ. (1) In coordinates adapted to K the line element takes the form ds2(4) = gijdx idxj + V (dx4)2, (2) where K = ∂/∂x4and ∂/∂x4(gij) = ∂/∂x 4V = 0. Two cases need to be considered. (i) The function V <0, the signature of gij is Euclidean and K is time-like. (ii) The function V> 0,the signature of gij is (2,1) and K is space-like. In the first case when V is replaced by −V the resulting Riemannian metric also satisfies Equation (1) and it this metric that is employed in the following construction. To emphasize this let |V | = U2. Now it is known that any Einstein four-metric on M can be used to construct a Ricci flat five-metric on M ×R, [2]. Here the Ricci- flat five-metric can be written as ds2(5) = (dx 0)2 + (ax0 + b)2gijdx idxj + (ax0 + b)2U2(dx4)2. (3) Here | | = 1 and a and b are constants with a2 = Λ/3. (4) Consequently and a non-zero Λ must have the same sign. The five-metric is Lorentzian, as is chosen to be the case here, when = −1 in case (i) and = 1 in case (ii)