3+1 formalism and bases of numerical relativity - LUTh ...

2.3 Hypersurface embedded in spacetime 25
Figure 2.2: Plane Σ as a hypersurface of the Euclidean space R 3 . Notice that the unit normal vector n stays
constant along Σ; this implies that the extrinsic curvature of Σ vanishes identically. Besides, the sum of angles of
any triangle lying in Σ is α + β + γ = π, which shows that the intrinsic curvature of (Σ, γ) vanishes as well.
Example 1 : a plane in R 3
Let us take for Σ the simplest surface one may think of: a plane (cf. Fig. 2.2). Let us
consider Cartesian coordinates (X α ) = (x,y,z) on R 3 , such that Σ is the z = 0 plane.
The scalar function t defining Σ according to Eq. (2.21) is then simply t = z. (x i ) =
(x,y) constitutes a coordinate system on Σ and the metric γ induced by g on Σ has the
components γij = diag(1,1) with respect to these coordinates. It is obvious that this metric
is flat: Riem(γ) = 0. The unit normal n has components n α = (0,0,1) with respect
to the coordinates (X α ). The components of the gradient ∇n being simply given by the
partial derivatives ∇βn α = ∂n α /∂X β [the Christoffel symbols vanishes for the coordinates
(X α )], we get immediately ∇n = 0. Consequently, the Weingarten map and the extrinsic
curvature vanish identically: χ = 0 and K = 0.
Example 2 : a cylinder in R 3
Let us at present consider for Σ the cylinder defined by the equation t := ρ − R = 0, where
ρ := x 2 + y 2 and R is a positive constant — the radius of the cylinder (cf Fig. 2.3). Let
us introduce the cylindrical coordinates (x α ) = (ρ,ϕ,z), such that ϕ ∈ [0,2π), x = r cos ϕ
and y = r sinϕ. Then (x i ) = (ϕ,z) constitutes a coordinate system on Σ. The components
of the induced metric in this coordinate system are given by
γij dx i dx j = R 2 dϕ 2 + dz 2 . (2.45)
It appears that this metric is flat, as for the plane considered above. Indeed, the change of
coordinate η := R ϕ (remember R is a constant !) transforms the metric components into
γi ′ j ′ dxi′ dx j′
= dη 2 + dz 2 , (2.46)