A solution in Weyl gravity with planar symmetry

Utah State University From the SelectedWorks of James Thomas Wheeler Spring May 23, 205 A solution in Weyl gravity with planar symmetry James Thomas Wheeler, Utah State University Available at: https://works.bepress.com/james_wheeler/7/

A solution in Weyl gravity with planar symmetry James T. Wheeler May 23, 205 Abstract We solve the Bach equation for Weyl gravity for the case of a static metric with planar symmetry. The solution is not conformal to the solution to the corresponding Einstein equation. Gauge choice We look at a static, planar symmetric line element of the form Define a new z coordinate, ds 2 = e 2fz) dt 2 + e 2gz) dx 2 + dy 2) + dz 2 ds 2 = e 2fz) dt 2 + e 2gz) dx 2 + dy 2) + e 2gz) dz 2 so that dz = e gz) dz. Then a conformal transformation by e 2g leaves us with ds 2 = e 2f g) dt 2 + dx 2 + dy 2 + dz 2 = h 2 Z) dt 2 + dx 2 + dy 2 + dz 2 where h Z) = e fzz)) gzz)). Since Weyl gravity is conformally invariant, we are solving for a conformal equivalence class of metrics, so we may always do this. Let Z z so the metric takes the form ds 2 = h 2 z) dt 2 + dx 2 + dy 2 + dz 2 2 Curvature Define an orthonormal basis, e 0 = hdt e = dx e 2 = dy e 3 = dz Then de 0 = h dz dt de = 0 de 2 = 0 de 3 = 0 Utah State University Dept of Physics email: jim.wheeler@usu.edu

The Cartan structure equation gives us de a = e b ω a b h dz dt = e b ω 0 b = e ω 0 + e 2 ω 0 2 + e 3 ω 0 3 = dx ω 0 + dy ω 0 2 + dz ω 0 3 ω 0 3 = h dt 0 = e b ω b = e 0 ω 0 + e 2 ω 2 + e 3 ω 3 = hdt ω 0 + dy ω 2 + dz ω 3 0 = e b ω b = e 0 ω 2 0 + e ω 2 + e 3 ω 2 3 = hdt ω 2 0 + dx ω 2 + dz ω 2 3 0 = e b ω 3 b = e 0 ω 3 0 = hdt h dt = 0 Thus the only nozero spin connection components are ω 3 0 = ω 0 3 = h dt = h h e0 Then the curvatures are R a b = dωa b ωc b ωa c but all the ω c b ωa c terms will depend on dt dt = 0 so we have only R 0 3 = dω 0 3 = d h dt) = h dz dt = h h e3 e 0 R 0 = R 0 2 = 0 R 2 = R 3 = R 2 3 = 0 and therefore, the Riemann curvature has the only one independent nonvanishing component, R 0 303 = h h2) Notice that the curvature here is equivalent to R 0303 = h h2) The Ricci tensor is R α βµν = Γ α βµ,ν Γ α βν,µ Γ α ρµγ ρ βν + Γα ρνγ ρ βµ R 00 = R 00 + R 2 020 + R 3 030 = h h2) R 33 = h h2) 2

and the Ricci scalar is R = η ab R ab For arbitrary h, find the Weyl curvature tensor, This gives = R 00 + R 33 = 2 h h2) C a bcd = R a bcd 2 δa c R bd δ a dr bc η bc R a d + η bd R a c) + 6 R δa c η bd δ a dη bc ) C 0 303 = R 0 303 R33 + η 33 R 0 2 0) + 6 Rη 33 = h h ) h 2 h h h h 3h = h 3h so any solutions with h nonzero are not conformally flat. The only components which involve R a bcd are this one, and those related by symmetry, e.g., C 3 030 = R 3 030 R00 + η 00 R 3 2 3) + 6 Rη 00 = h h ) h 2 h + h + h h 3h = h 3h All other nonvanishing components are of the form C a bcd = 2 δa c R bd δ a dr bc η bc R a d + η bd R a c) + 6 R δa c η bd δ a dη bc ) for example, = 2 δa c R bd δ a dr bc η bc R a d + η bd R a c) h 3h δa c η bd δ a dη bc ) C 33 = C 2 323 = 2 R 33 + 6 Rη 33 = h 2 h h 3h = 6h h C 00 = C 2 020 = 2 R 00 + 6 Rη 00 = h 2 h + 3h h = 6h h C 22 = h 3h 3

2. The Einstein equation The Einstein equation for this form of metric is trivial. With R 00 = h h2) R 33 = h h2) we have simply h 2) = 0 with linear solutions h = az + b. The curvature vanishes and the spacetime is flat, and therefore is not conformal to any solution with h 2) 0. 3 The Bach equation We may write the Bach equation as The second term on the left is not difficult: D µ D ν C αµβν 2 R µνc αµβν = κt αβ R µν C αµβν = R 00 C α0β0 + R 33 C α3β3 = h h Cα0β0 h = 0 h Cα3β3 Now we need the covariant derivatives. These are much 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 ) where our sign convention for the curvature shows up in the derivative terms. We have the following in an orthonormal basis: R 0 303 = h h2) R 0303 = h h2) R 00 = h h2) R 33 = h h2) In a coordinate basis, R = 2 h h2) R α βµν = e R 0 303 = e α a eβ α 0 eβ b eµ c eν d R a bcd 3 eµ 0 eν 3 R 0 303 = h h2) R 0303 = hh 2) 4

R 00 = hh 2) R 33 = h h2) 3. Connection Return to the metric so that and the only derivative is R = 2 h h2) dτ 2 = h 2 Z) dt 2 dx 2 dy 2 dz 2 g αβ = h 2 g 00,3 = 2hh Therefore from Γ 003 = Γ 030 = Γ 300 = hh we have: The curvatures are: 3.2 Derivatives We need the following derivatives: 3.2. d Alembertian of the Ricci tensor First, expand the derivatives: Γ 0 03 = Γ 0 30 = h h) Γ 3 00 = hh ) 2) R 0303 = hh 2) 3) R 00 = hh 2) 4) R 33 = h h2) 5) R = 2 h h2) 6) D µ D µ R αβ D α D β R D µ D µ R D µ D µ R αβ = g µν D µ D ν R αβ = g µν D µ ν R αβ R ρβ Γ ρ αν R αρ Γ ρ βν ) = g µν µ ν R αβ R ρβ Γ ρ αν R αρ Γ ρ βν ) g µν σ R αβ R ρβ Γ ρ ασ R αρ Γ ρ βσ Γ σ νµ ) g µν ν R σβ R ρβ Γ ρ σν R σρ Γ ρ βν Γ σ αµ g µν ν R ασ R ρσ Γ ρ αν R αρ Γ ρ σν) Γ σ βµ ) 5

For the 00 component this becomes D µ D µ R 00 = g µν µ ν R 00 R ρ0 Γ ρ 0ν R 0ρΓ ρ 0ν ) g µν σ R 00 R ρ0 Γ ρ 0σ R 0ρΓ ρ 0σ ) Γσ νµ g µν ν R σ0 R ρ0 Γ ρ σν R σρ Γ ρ 0ν ) Γσ 0µ g µν ν R 0σ R ρσ Γ ρ 0ν R 0ρΓ ρ σν) Γ σ 0µ = g 33 3 3 R 00 R 00 Γ 0 03 R 00 Γ 0 ) 03 g 00 3 R 00 R 00 Γ 0 03 R 00 Γ 0 03) Γ 3 00 g 00 R 00 Γ 0 30 R 33 Γ 3 00) Γ 3 00 g 33 3 R 00 R 00 Γ 0 03 R 00 Γ 0 03) Γ 0 03 g 00 R 33 Γ 3 00 R 00 Γ 0 30) Γ 3 00 g 33 3 R 00 R 00 Γ 0 03 R 00 Γ 0 03) Γ 0 03 Collecting terms, D µ D µ R 00 = 3 3 R 00 4 3 R 00 Γ 0 03 + h 2 3R 00 Γ 3 00 2R 00 3 Γ 0 03 + 4R 00 Γ 0 03Γ 0 03 4 h 2 R 00Γ 0 03Γ 3 00 2 h 2 R 33Γ 3 00Γ 3 00 Then with the connection and curvatures given by eqs.2-6) this becomes so D µ D µ R 00 = 3 3 hh 2)) 4 3 hh 2)) h h) + h 2 3 hh 2)) hh ) ) 2hh 2) 3 h h) + 4hh 2) h h h) h ) 4 h 2 hh2) h h) hh ) 2 h 2 ) h h2) hhh ) h ) The 33 component: = 3 3 hh 2)) 4 h 3 hh 2)) h ) + h 3 hh 2)) ) h ) 2hh 2) 3 h h) + 4 h h2) h ) h ) 4 h h2) h ) h ) + 2 h h2) h ) h ) = hh 4) + 2h ) h 3) + h 2) h 2) 4h 3) h ) 4 h h2) h ) h ) + h 3) h ) + h h2) h ) h ) 2h 2) h 2) + 2 h h2) h ) h ) + 4 h h2) h ) h ) 4 h h2) h ) h ) + 2 h h2) h ) h ) = hh 4) h 3) h ) h 2) h 2) + h h2) h ) h ) D µ D µ R 00 = hh 4) h 3) h ) h 2) h 2) + h h2) h ) h ) D µ D µ R 33 = g µν µ ν R 33 R ρ3 Γ ρ 3ν R 3ρΓ ρ 3ν ) 6

g µν σ R 33 R ρ3 Γ ρ 3σ R 3ρΓ ρ 3σ ) Γσ νµ g µν ν R σ3 R ρ3 Γ ρ σν R σρ Γ ρ 3ν ) Γσ 3µ g µν ν R 3σ R ρσ Γ ρ 3ν R 3ρΓ ρ σν) Γ σ 3µ = g 33 3 3 R 33 g 00 Γ 3 00 3 R 33 +g 00 R 33 Γ 3 00 + R 00 Γ 0 30) Γ 0 30 +g 00 R 00 Γ 0 30 + R 33 Γ 3 00) Γ 0 30 = 3 3 ) h h2) + h 2 hh) 3 ) h h2) h 2 h h2) hh ) + hh 2) ) h h) h h) h 2 hh 2) h h) ) h h2) hh ) h h) = 3 3 ) h h2) + h h) 3 ) h h2) = h h4) + h 2 h3) h ) + h 2 h2) h 2) h 3 h2) h ) h ) = g + h g h 3.2.2 d Alembertian of the scalar curvature The d Alembertian of R, D µ D µ R, follows immediately: D µ D µ R = g αβ D µ D µ R αβ = g 00 D µ D µ R 00 + g 33 D µ D µ R 33 = h 2 D µd µ R 00 + D µ D µ R 33 = h h4) + h 2 h3) h ) + h 2 h2) h 2) h 3 h2) h ) h ) h h4) + h 2 h3) h ) + h 2 h2) h 2) h 3 h2) h ) h ) = 2 h h4) + 2 h 2 h3) h ) + 2 h 2 h2) h 2) 2 h 3 h2) h ) h ) 3.2.3 Derivatives of the Ricci scalar Finally, we have the second derivative of the Ricci scalar, D α D β R = D α β R) = α β R µ R Γ µ βα D 0 D 0 R = 3 R Γ 3 00 = 3 2 h h2) ) hh ) = 2h 3) h ) 2 h h2) h ) h ) D 3 D 3 R = 3 3 R = 3 3 2 h h2) ) 7

= 3 2 h h3) + 2 h 2 h2) h ) ) Check the d Alembertian, = 2 h h4) + 2 h 2 h3) h ) + 2 h 2 h3) h ) + 2 h 2 h2) h 2) 4 h 3 h2) h ) h ) = 2 h h4) + 4 h 2 h3) h ) + 2 h 2 h2) h 2) 4 h 3 h2) h ) h ) D 3 D 3 R = 3 3 R = 3 3 2 h h2) ) = 2g = 2 h h4) + 4 h 2 h3) h ) + 2 h 2 h2) h 2) 4 h 4 h2) h ) h ) D µ D µ R = g µν D µ D ν R = g 00 D 0 D 0 R + g 33 D 3 D 3 R = h 2 2h 3) h ) 2 h h2) h ) h ) ) This agrees. 2 h h4) + 4 h 2 h3) h ) + 2 h 2 h2) h 2) 4 h 3 h2) h ) h ) = 2 h h4) + 2 h 2 h3) h ) + 2 h 2 h2) h 2) 2 h 3 h2) h ) h ) 3.2.4 Summary of derivatives The derivatives we need are: D µ D µ R = 2 h h4) + 2 h 2 h3) h ) + 2 h 2 h2) h 2) 2 h 3 h2) h ) h ) D 0 D 0 R = 2h 3) h ) 2 h h2) h ) h ) D 3 D 3 R = 2 h h4) + 4 h 2 h3) h ) + 2 h 2 h2) h 2) 4 h 3 h2) h ) h ) D µ D µ R 00 = hh 4) h 3) h ) h 2) h 2) + h h2) h ) h ) D µ D µ R 33 = h h4) + h 2 h3) h ) + h 2 h2) h 2) h 3 h2) h ) h ) 3.3 Algebraic parts We have 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 so we need: R 2 D µ D µ R 3R µν R µν R µν R αµβν RR αβ 8

R 00 = hh 2) R 33 = h h2) Substituting from eqs.2-6), these are R 2 D µ D µ R 3R µν R µν = R = 2 h h2) 2 ) 2 h h2) + 2 h h4) 2 h 2 h3) h ) 2 h 2 h2) h 2) + 2 h 3 h2) h ) h ) 3 h 2 h2) h 2) + ) h 2 h2) h 2) = 2 h h4) 2 h 2 h3) h ) 4 h 2 h2) h 2) + 2 h 3 h2) h ) h ) R µν R 0µ0ν = R 33 R 0303 = h 2) h 2) R µν R 3µ3ν = R 00 R 3030 = h 4 hh2) hh 2) = h 2 h2) h 2) RR 00 = 2h 2) h 2) RR 33 = 2 h 2 h2) h 2) This gives all the elements we need to construct the Bach tensor. 3.4 Combine all terms With W αβ given by 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 we first compute +2R µν R αµβν 2 3 RR αβ W 00 = 3 D 0D 0 R + D µ D µ R 00 + 6 R 2 D µ D µ R 3R µν R µν) g 00 +2R µν R 0µ0ν 2 3 RR 00 = 2h 3) h ) 2h ) 3 h2) h ) h ) + hh 4) h 3) h ) h 2) h 2) + h ) h2) h ) h ) 2 6 h2 h h4) 2 h 2 h3) h ) 4 h 2 h2) h 2) + 2 ) h 3 h2) h ) h ) +2 h 2) h 2)) 2 2h 2) h 2)) 3 = 2 3 hh4) 4 3 h3) h ) h 2) h 2) + 4 3 h h2) h ) h ) 9

Next, and finally, W = W 22 = 6 = 3 R 2 + D µ D µ R 3R µν R µν) h h4) h 2 h3) h ) 2 h 2 h2) h 2) + ) h 3 h2) h ) h ) W 33 = 3 D 3D 3 R + D µ D µ R 33 + 6 R 2 D µ D µ R 3R µν R µν) g 33 3.4. Check the trace +2R µν R 3µ3ν 2 3 RR 33 = 2 3 h h4) 4 3 h 2 h3) h ) 2 3 h 2 h2) h 2) + 4 3 h 3 h2) h ) h ) h h4) + h 2 h3) h ) + h 2 h2) h 2) h 3 h2) h ) h ) 3 h h4) h 2 h3) h ) 2 h 2 h2) h 2) + ) h 3 h2) h ) h ) +2 h 2 h2) h 2) 4 3 h 2 h2) h 2) = 2 3 h 2 h3) h ) + 3 h 2 h2) h 2) + 2 3 h 3 h2) h ) h ) The trace W α α vanishes identically, so we check: g αβ W αβ = g 00 W 00 + g W + g 22 W 22 + g 33 W 33 = 2 h 2 3 hh4) 4 3 h3) h ) h 2) h 2) + 4 ) 3 h h2) h ) h ) + 2 3 h h4) h 2 h3) h ) 2 h 2 h2) h 2) + ) h 3 h2) h ) h ) 2 3 h 2 h3) h ) + 3 h 2 h2) h 2) + 2 3 h 3 h2) h ) h ) = + 2 3 h h4) 2 3 h h4) 2 3 h 2 h3) h ) 2 3 h 2 h3) h ) + 4 3 h 2 h3) h ) 4 3 h 2 h2) h 2) + h 2 h2) h 2) + 3 h 2 h2) h 2) + 2 3 h 3 h2) h ) h ) 4 3 h 3 h2) h ) h ) + 2 3 h 3 h2) h ) h ) = 0 3.4.2 Check the divergence It must be the case that W αβ ;β = 0. This follows from an infinitesimal coordinate transformation. With ˆ δs = W αβ δg αβ gd 4 x and the diffeomorphism symmetry taking the form δg αβ = h α;β + h β;α 0

we have ˆ 0 δs = W αβ h α;β + h β;α ) gd 4 x ˆ = 2 W αβ ;β h α gd 4 x for any h α. We therefore check that where One component at a time: W αβ ;β = 0 W 00 = 2 3 hh4) 4 3 h3) h ) h 2) h 2) + 4 3 h h2) h ) h ) W = W 22 = 3 h h4) h 2 h3) h ) 2 h 2 h2) h 2) + ) h 3 h2) h ) h ) W 33 = 2 3 h 2 h3) h ) + 3 h 2 h2) h 2) + 2 3 h 3 h2) h ) h ) W 0β ;β = β W 0β + W µβ Γ 0 µβ + W 0µ Γ β µβ = 0 W 00 + 2W 0β Γ 0 0β + W 00 Γ β 0β = 0 W β ;β = β W β + W µβ Γ µβ + W µ Γ β µβ = 0 W 2β ;β = 0 W 3β ;β = β W 3β + W µβ Γ 3 µβ + W 3µ Γ β µβ = 3 W 33 + W 00 Γ 3 00 + W 33 Γ 0 30 = 3 2 3 h 2 h3) h ) + 3 h 2 h2) h 2) + 2 ) 3 h 3 h2) h ) h ) + 2 3 h 2 h4) h ) 4 3 h 3 h3) h ) h ) h 3 h2) h 2) h ) + 4 3 h 4 h2) h ) h ) h ) 2 3 h 3 h3) h ) h ) + 3 h 3 h2) h 2) h ) + 2 3 h 4 h2) h ) h ) h ) = 2 3 h 2 h4) h ) 2 3 h 2 h3) h 2) + 4 3 h 3 h3) h ) h ) + 2 3 h 2 h3) h 2) 2 3 h 3 h2) h 2) h ) + 2 3 h 3 h3) h ) h ) + 4 3 h 3 h2) h 2) h ) 2 h 4 h2) h ) h ) h 4) + 2 3 h 2 h4) h ) 4 3 h 3 h3) h ) h ) h 3 h2) h 2) h ) + 4 3 h 4 h2) h ) h ) h ) 2 3 h 3 h3) h ) h ) + 3 h 3 h2) h 2) h ) + 2 3 h 4 h2) h ) h ) h ) = 2 3 h 2 h4) h ) 2 3 h 2 h4) h ) + 2 3 h 2 h3) h 2) 2 3 h 2 h3) h 2) + 2 3 h 3 h3) h ) h ) + 4 3 h 3 h3) h ) h ) 4 3 h 3 h3) h ) h ) 2 3 h 3 h3) h ) h )

2 3 h 3 h2) h 2) h ) + 4 3 h 3 h2) h 2) h ) + 3 h 3 h2) h 2) h ) h 3 h2) h 2) h ) + 4 3 h 4 h2) h ) h ) h ) 2 h 4 h2) h ) h ) h ) + 2 3 h 4 h2) h ) h ) h ) = 0 4 Vacuum solution to the Bach equation We now solve W αβ = 0 where W 00 = 2 3 hh4) 4 3 h3) h ) h 2) h 2) + 4 3 h h2) h ) h ) W = W 22 = 3 h h4) h 2 h3) h ) 2 h 2 h2) h 2) + ) h 3 h2) h ) h ) W 33 = 2 3 h 2 h3) h ) + 3 h 2 h2) h 2) + 2 3 h 3 h2) h ) h ) First, rewrite these in terms of g h h2) using g = 3 3 h h2) ) and The first equation becomes = 3 h h3) h 2 h2) h ) ) = h h4) h 2 h3) h ) h 2 h3) h ) h 2 h2) h 2) + 2 h 3 h2) h ) h ) h h4) = g + 2 h 2 h3) h ) + h 2 h2) h 2) 2 h 3 h2) h ) h ) ) 3 h 2 h2) h ) = h 2 h3) h ) + h 2 h2) h 2) 2 h 3 h2) h ) h ) ) h 2 h3) h ) = 3 h 2 h2) h ) h 2 h2) h 2) + 2 h 3 h2) h ) h ) 0 = 2 3 h h4) 4 3 h 2 h3) h ) h 2 h2) h 2) + 4 3 h 3 h2) h ) h ) = 2 g + 2h 3 2 h3) h ) + h 2 h2) h 2) 2h ) 3 h2) h ) h ) 4 3 h 2 h3) h ) h 2 h2) h 2) + 4 3 h 3 h2) h ) h ) = 2 3 g 3 h 2 h2) h 2) = 2 3 g 3 g2 or The second pair) of equations becomes 0 = 2g g 2 0 = h h4) h 2 h3) h ) 2 h 2 h2) h 2) + h 3 h2) h ) h ) 2

= g + 2 h 2 h3) h ) h 2 h3) h ) + h 2 h2) h 2) 2 h 2 h2) h 2) + h 3 h2) h ) h ) 2 h 3 h2) h ) h ) = g + h 2 h3) h ) h 2 h2) h 2) h 3 h2) h ) h ) ) = g + h h) g 2 h 2 h2) h 2) + h 3 h2) h ) h ) ) = g + h h) g 2g 2 + h 2 gh) h ) = g + h h) g + h h2) g h 2 h) h ) g 2g 2 + h 2 gh) h ) = g + h h g g 2 and the final equation is 0 = 2 h 2 h3) h ) + h 2 h2) h 2) + 2 h 3 h2) h ) h ) = 2 3 h 2 h2) h ) ) + 2 h 2 h2) h 2) 4 h 3 h2) h ) h ) + h 2 h2) h 2) + 2 h 3 h2) h ) h ) = 2 h g h 2 h gh + 2 h 2 gh h + 3g 2 2 h 2 gh h = 2 h g h 2 h gh + 3g 2 0 = 2 h g h g 2 The three equations 0 = 2g g 2 0 = g + h h g g 2 0 = 2 h g h g 2 should be dependent; indeed, substituting the third into the second reproduces the first: 0 = g + h h g g 2 Combining the first and third, = g 2 g2 h h 0 = 2 h g h 2g = g g ln h h 0 = ln g g 0 g = ah Including the original definition of g, this gives us three equations: g = h h 3

g = 2 g2 g = ah The final two imply the first. Differentiate the third to get g = ah and substitute into the second, giving ah = 2 g2 Differentiate this and use the final again ah = gg ah = agh g = h h Therefore, it is sufficient to set h = a g, then solve the first equation, which may be integrated directly: ˆ 0 = 2g g g 2 g = g ) 2 3 g3 + b g = 3 g3 b dg = z 3 g3 b In general this gives an elliptic function. Inverting gives g z), which then gives h = a 3 g3 b. Consider the special case, b = 0. The integral for this special case is simpler, ˆ dg = z g 3/2 3 2 g = 3 z z 0 ) choosing z 0 = 0. Then we have 4 g = 3 z z 0) 2 g = 2 z 2 0 = 2 h g h g 2 = 2 24 h h z 3 44 z 4 0 = h + 3 z h dh h = 3 z dz ln h = ln z 3 h = a z 3 4

We also require 2 z 2 = h h = z3 a = 2 z 2 so we have a consistent solution. The metric for this special case is therefore 4. Singularity 2a z 5 ds 2 = 44 z 4 dt2 + dx 2 + dy 2 + dz 2 There is a coordiate singularity at the plane at z = 0. Check whether this is a curvature singularity. Compute the invariant C αβµν C αβµν Noting that the action, S = C αβµν C αβµν gd 4 x, is equavalent to S = up to a topological invariant, it is sufficient to check ˆ R αβ R αβ 3 R2 ) gd 4 x R αβ R αβ 3 R2 = h 2 h2) h 2) + h 2 h2) h 2) 4 3h 2 h2) h 2) = 2 3 g2 = 96 z 4 The singularity at z = 0 is therefore a curvature singularity. 5