(ed.). Gravitational waves (IOP, 2001)(422s).

Einstein theory 251where δ ˜g µν = 2δẽ αµẽ αν has been used. The metric is diagonal in this gauge, sothe equations for the conformal factors become(traceless part of){− 1 2 λ−1 ∂ µ λ∂ ν ρ + 1 4 ∂ µ∂ ν ρ}=matter. (14.45)We have not explicitly written the matter sector yet. This will be done in the nextsection.The pure gravity action written in two dimensions in the conformal gaugereads− 1 4 e(3) R (3) =− 1 2 λ−1 ∂ µ λ∂ µ ρ (14.46)where the equation £ρ = 0 has been used to make the second term of (14.36)vanish. The whole Lagrangian is thenÄ E =− 1 2 λ−1 ∂ µ λ∂ µ ρ + 1 8 ρ−2 (∂ µ ∂ µ + ∂ µ B∂ µ B) (14.47)where the second term with the Kaluza–Klein scalar and the dual of the Kaluza–Klein vector has been obtained considering only the two-dimensional part of theaction. The subscript E stands for Ehlers, who did this analysis for the first timein the 1950s.Here we have treated one possible way of performing dimensional reduction:we have seen it consists of many steps. One first reduces from four to three; then,dualizes the vector field and performs the reduction to two dimensions. However,this is not the whole story: actually, it is possible also to get the two-dimensionalLagrangian directly from the three-dimensional one, without dualization.The procedure for doing the calculation is as follows: first we express theKaluza–Klein vector in the formB m = (B µ , B 2 ≡ ˜B) (14.48)and then we perform directly the dimensional reduction in conformal gauge byusing the previous choice of the three-bein in triangular form. Proceeding inthis way, we meet two electromagnetic fields in two dimensions, A µ and B µ ,which can be set to zero because they do not propagate (we have already used thisargument before) and there is no cosmological constant.Let us note now the various steps of the calculation. The newwriting for the Kaluza–Klein vector and the considerations on two-dimensionalelectromagnetism implyB µ = 0 → B µν = 0. (14.49)So the only non-vanishing terms of the three-dimensional Lagrangian before thereduction areÄ =− 1 4 e(3) R (3) (e) − 1 8 e(3) 2 B µ2 B µ2 + 1 8 e(3) g mn −2 ∂ m ∂ n . (14.50)Then we reduce the dimensions as before (we set A µ = 0) and choose theconformal gauge. Keeping in mind thatg µν = λ 2 η µν , g 22 =−ρ 2 (14.51)