Linearized Conformal gravity

Transcript

1 Utah State University From the SelectedWors of James Thomas Wheeler Winter January 28, 206 Linearized Conformal gravity James Thomas Wheeler Available at:

2 Linearized conformal gravity James T Wheeler January 28, 206 Abstract We examine the linearization of Weyl conformal gravity, showing that the only solutions are also solutions to linearized general relativity. The Bach equation The Weyl gravity action is S C αµβν C αµβν gd 4 x It leads, in vacuum, to the Bach equation, D µ D ν C αµβν 2 R µνc αµβν 0 It is generally easier to compute from the alternate form of the Bach equation, W αβ 3 D αd β R + D µ D µ R αβ + 6 R 2 D µ D µ R 3R µν R µν) g αβ +2R µν R αµβν 2 3 RR αβ ) which is found by expanding the Weyl curvatures in the action in terms of the Riemann curvature, then using the Gauss-Bonnet expression for the Euler character to eliminate all but the Ricci and Ricci scalar terms. 2 Linearized gravity The linearization of general relativity is well established, either by perturbing around flat space or about a fixed bacground. We consider first a flat bacground, writing g αβ η αβ + h αβ g αβ η αβ h αβ where the components of h αβ are regarded as perturbatively small and h αβ η αµ η βν h µν. Then at first order, h αβ acts as a field on flat spacetime. The connection is Γ α βµ 2 ηαν h νβ,µ + h νµ,β h βµ,ν ) Utah State University Dept of Physics jim.wheeler@usu.edu

3 The curvature becomes R α βµν Γ α βµ,ν Γ α βν,µ Γ α ρµγ ρ βν + Γα ρνγ ρ βµ 2 ηασ h σβ,µν + h σµ,βν h βµ,σν ) 2 ηασ h σβ,µν + h σν,βµ h βν,σµ ) 2 ηασ h σµ,βν h βµ,σν h σν,βµ + h βν,σµ ) It is useful to define the trace-reversed variable, The Ricci tensor becomes h αβ h αβ 2 η αβh h η µν h µν R α βαν 2 ηασ h σα,βν h βα,σν h σν,βα + h βν,σα ) 2 2 h,βν 2h α ν,βα + h βν ) The infinitesimal coordinate freedom of the metric, h,βν + 2 η βν h h α β,αν h α ν,αβ + h βν )) 2 η βνh allows us to change h αβ, to mae it divergence free, g αβ g αβ + α;β + β;α h αβ h αβ + α,β + β,α 0 h αβ,β h αβ,β + α + β,α,β An additional transformation may mae it traceless. The fully transverse-traceless mode for h αβ allows us to write it as 0 h αβ h + h h h + 0 Assuming the transverse traceless conditions hold for h αβ, the Riemann tensor, Ricci tensor and Ricci scalar reduce to R α βµν 2 ηασ h σµ,βν h βµ,σν h σν,βµ + h βν,σµ ) R αβ h αβ R 0 3 Substituting into the Bach tensor Now substitute into W αβ 3 D αd β R + D µ D µ R αβ + 6 R 2 D µ D µ R 3R µν R µν) g αβ +2R µν R αµβν 2 3 RR αβ D µ D µ R αβ h αβ 2

4 We see vacuum solutions for W αβ 0. Setting or equivalently, the pair h αβ 0 h + 0 h 0 we may solve for each component of h ± in the usual manner, by Fourier expansion, h ± d 4 A ± µ ) e i βx ν δ β ) ) β d 4 A ± µ ) e iνxν δ ω 2 )) d3 2ω A ± i ) e i x ωt) + B ± i ) e i x+ωt)) where ω ω ) This solution to the wave equation now acts as source for the same wave equation. To solve, we need the Green s function. Choosing vanishing boundary conditions at infinity this is straightforward to derive. We need G x α, x α 0 ) 4πδ 4 x α x α 0 ) Taing the Fourier transform, 4π 2 d 4 G ν ) e iαxα x α 0 ) 4π 6π 4 d 4 e iαxα x α 0 ) 4π 2 d 4 β β G ν ) + ) e iαxα x α 0 ) 0 π G ν ) π β β Therefore, G x α, x α 0 ) 4π 2 4π 3 4π 3 d 4 G ν ) e iαxα x α 0 ) d 4 α β β eiαx x α 0 ) d 3 e ix x0) dω e iωt t0) ω 2 Infinitesimally offsetting the poles by replacing ω ω + iε, we complete contours by half circles at infinity. e iωt e dω ω 2 dω iωt ω + ) ω ) e iωt+εt dω ω + ) + iε ω ) + iε 3

5 For t < 0, we complete the contour in the upper half plane and there are no poles enclosed. For t > 0, we complete the contour with a half circle at infinity in the lower half plane and there are two poles. Letting ε 0, the integral becomes dω e iωt ω 2 where ω ) +. We are left with G x α, x α 0 ) 4π 2 i Θ t t 0) 4π 2 i Θ t t 0) For the remaining integrals, d 3 ei x x0) ωt t0)) ω ) 2π 2π ) e iω)t 2πiΘ t) 2ω ) + e iω)t 2ω ) 2π e iω)t ω Θ t) e iω)t ) 2i 2π Θ t) sin ω ) t ω d3 ei x x0) e iω)t t0) e iω)t t0)) ω ) e i x x0) ωt t0)) e i x x0)+ωt t0))) d3 ω ) 2 ddϕd cos θ) ei x x0 cos θ t t0)) 2 dd cos θ) e i x x 0 e i x x0 ) e it t0) 2 d 2π i x x 0 2π i x x 0 2π ei x x0 cos θ t t0)) i 2 x x 0 d e i x x0 t t0)) e i x x0 +t t0))) δ t t 0 ) δ t t 0 + )) giving the advanced and retarded solutions. For clarity we have put in the speed of light explicitly. G x α, x α 0 ) 4π 2 i Θ t t 0) d3 ω ) eix x0) ωt t0)) 4π 2 i Θ t) d3 ω ) eix x0)+ωt t0)) 4π 2 i Θ t t 2π 0) i x x 0 2π δ t t 0 ) δ t t 0 + )) 4π 2 i Θ t t 2π 0) i x x 0 2π δ t t 0 ) δ t t 0 + )) Θ t t 0 ) x x 0 δ t t 0 ) +Θ t t 0 ) x x 0 δ t t 0 ) Something s wrong with signs and/or BC here, but let s just use the retarded part: With this, the solution to G x α, x α 0 ) Θ t t 0) x x 0 δ t t 0 x x 0 ) h ± x) J ± x) J ± x) d3 A ± i ) e i x ωt) + B ± i ) e i x+ωt)) 2ω 4

6 is h ± x) d 4 x 0 G x, x 0, t, t 0 ) J ± x 0 ) 4π 2 d 4 x 0 d 4 G ) e iµxµ x µ 0 ) J± x 0 ) d 4 G ) e iµxµ 4π 2 d 4 x 0 J ± x 0 ) e iµxµ 0 d 4 G ) J ± )) e iµxµ This shows that h ± x) 0 d 4 G ) J ± )) e iµxµ whenever h ± x) 0 But clearly, h ± x) 0 if h ± x) 0 so we conclude that: h ± x) 0 iff h ± x) 0 This means we have a solution to linearized Weyl gravity in vacuum if and only if it is also a solution to linearized general relativity in vacuum. 5