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Exact conformal electrovacuum constraint transformation

Statement

Use the smooth conformal seed data defined in this development on an oriented smooth three-manifold, and let $D$ and $\overset{ˉ}{D}$ be the Levi-Civita connections of $h = ψ^{4} \overset{ˉ}{h}$ and $\overset{ˉ}{h}$ . Define $Δ_{\overset{ˉ}{h}} ψ : = {\overset{ˉ}{h}}^{i j} {\overset{ˉ}{D}}_{i} \partial_{j} ψ$ , with the Euclidean convention $Δ = \sum_{i} \partial_{i}^{2}$ in Cartesian coordinates. Use the spatial curvature sign and scalar-curvature contraction fixed for induced data in the preceding Gauss-Codazzi convention. Let $e = α^{- 1} ψ^{- 6} \overset{ˉ}{E}$ and $b = α^{- 1} ψ^{- 6} \overset{ˉ}{B}$ be the SI-convention vector fields associated with the inherited constrained electrovacuum initial-data convention. Then the two Maxwell divergence constraints are equivalent to

${\overset{ˉ}{D}}_{i} {\overset{ˉ}{E}}^{i} = 0, {\overset{ˉ}{D}}_{i} {\overset{ˉ}{B}}^{i} = 0,$

and the SI Hamiltonian constraint in that definition is equivalent to

$- 8 Δ_{\overset{ˉ}{h}} ψ + \overset{ˉ}{R} ψ - ∣ \overset{ˉ}{A} ∣_{\overset{ˉ}{h}}^{2} ψ^{- 7} + (\frac{2}{3} τ^{2} - 2 Λ) ψ^{5} - 2 (∣ \overset{ˉ}{E} ∣_{\overset{ˉ}{h}}^{2} + ∣ \overset{ˉ}{B} ∣_{\overset{ˉ}{h}}^{2}) ψ^{- 3} = 0.$

Here $\overset{ˉ}{R}$ is the scalar curvature of $\overset{ˉ}{h}$ in the stated convention, and the barred norms use $\overset{ˉ}{h}$ to raise and lower indices. The exact vector divergence identities are $D_{i} E^{i} = ψ^{- 6} {\overset{ˉ}{D}}_{i} {\overset{ˉ}{E}}^{i}$ and $D_{i} B^{i} = ψ^{- 6} {\overset{ˉ}{D}}_{i} {\overset{ˉ}{B}}^{i}$ . The separate momentum constraint remains

$D^{j} K_{j i} - D_{i} τ = - 2 ψ^{- 6} (\overset{ˉ}{E} \times_{\overset{ˉ}{h}} \overset{ˉ}{B})_{i},$

where the cross product on the right is lowered with $\overset{ˉ}{h}$ . In particular, if $K = 0$ and $\overset{ˉ}{E} \times_{\overset{ˉ}{h}} \overset{ˉ}{B} = 0$ , the momentum equation holds. The transformed scalar equation alone does not impose the momentum constraint and gives no elliptic existence theorem.

Statement references

Conformal electrovacuum seed data

Constrained electrovacuum initial data

SI and geometrized Einstein–Maxwell parameters

Gauss-Codazzi decomposition in Gaussian normal coordinates and the constraint equations in terms of the induced metric and second fundamental form

Facts & Assumptions

Given: Smooth seed data as in the Statement, with $ψ > 0$ , $α > 0$ , and the SI-convention constraint equations defined on the same three-manifold.

[F1]

Conformal electrovacuum seed data: $h = ψ^{4} \overset{ˉ}{h}$ , $K = (τ / 3) h + ψ^{- 2} \overset{ˉ}{A}$ , $E = ψ^{- 6} \overset{ˉ}{E}$ , and $B = ψ^{- 6} \overset{ˉ}{B}$ , where $\overset{ˉ}{A}$ is $\overset{ˉ}{h}$ -trace-free.

[F2]

Constrained electrovacuum initial data: $D_{i} e^{i} = D_{i} b^{i} = 0$ and $R (h) + ({tr}_{h} K)^{2} - ∣ K ∣_{h}^{2} = 2 Λ + (κ / μ_{0}) (∣ e ∣_{h}^{2} + ∣ b ∣_{h}^{2})$ ; its momentum equation is $D^{j} K_{j i} - D_{i} {tr}_{h} K = - (κ / μ_{0}) (e \times_{h} b)_{i}$ .

[F3]

SI and geometrized Einstein–Maxwell parameters: $α^{2} = κ / (2 μ_{0})$ , with $α > 0$ constant.

[F4]

Gauss-Codazzi decomposition in Gaussian normal coordinates and the constraint equations in terms of the induced metric and second fundamental form: the induced spatial scalar curvature is contracted from the curvature operator with the same sign convention as the ambient operator; the Hamiltonian constraint uses $R (h) + τ^{2} - ∣ K ∣_{h}^{2}$ for $K = + \frac{1}{2} L_{n} h$ .

Proof

1.1givenF1F2algebra

In any positively oriented coordinate chart, $\det h = ψ^{12} \det \overset{ˉ}{h}$ , so $\sqrt{\det h} = ψ^{6} \sqrt{\det \overset{ˉ}{h}}$ . The Levi-Civita formula and the determinant derivative give $Γ^{i}_{i k} = \partial_{k} \log \sqrt{\det h}$ , hence $D_{i} V^{i} = (\sqrt{\det h})^{- 1} \partial_{i} (\sqrt{\det h} V^{i})$ for every smooth vector field $V$ . Substituting $V = E = ψ^{- 6} \overset{ˉ}{E}$ and then $V = B = ψ^{- 6} \overset{ˉ}{B}$ gives $D_{i} E^{i} = ψ^{- 6} {\overset{ˉ}{D}}_{i} {\overset{ˉ}{E}}^{i}$ and $D_{i} B^{i} = ψ^{- 6} {\overset{ˉ}{D}}_{i} {\overset{ˉ}{B}}^{i}$ . Since $e = α^{- 1} E$ , $b = α^{- 1} B$ , and both factors are positive, the two SI divergence constraints are equivalent to the two barred divergence equations.

1.2F1F2F3algebra

The oriented volume forms satisfy ${vol}_{h} = ψ^{6} {vol}_{\overset{ˉ}{h}}$ . In coordinates, $(E \times_{h} B)_{i} = ({vol}_{h})_{i j k} E^{j} B^{k} = ψ^{- 6} (\overset{ˉ}{E} \times_{\overset{ˉ}{h}} \overset{ˉ}{B})_{i}$ . Also $∣ E ∣_{h}^{2} = ψ^{- 8} ∣ \overset{ˉ}{E} ∣_{\overset{ˉ}{h}}^{2}$ and $∣ B ∣_{h}^{2} = ψ^{- 8} ∣ \overset{ˉ}{B} ∣_{\overset{ˉ}{h}}^{2}$ , by inserting $h_{i j} = ψ^{4} {\overset{ˉ}{h}}_{i j}$ and the two vector weights. Therefore $(κ / μ_{0}) (∣ e ∣_{h}^{2} + ∣ b ∣_{h}^{2}) = 2 ψ^{- 8} (∣ \overset{ˉ}{E} ∣_{\overset{ˉ}{h}}^{2} + ∣ \overset{ˉ}{B} ∣_{\overset{ˉ}{h}}^{2})$ and $(κ / μ_{0}) (e \times_{h} b)_{i} = 2 ψ^{- 6} (\overset{ˉ}{E} \times_{\overset{ˉ}{h}} \overset{ˉ}{B})_{i}$ , using [F3].

1.3F1algebra

Taking the $h$ -trace of $K = (τ / 3) h + ψ^{- 2} \overset{ˉ}{A}$ gives ${tr}_{h} K = τ + ψ^{- 6} {\overset{ˉ}{h}}^{i j} {\overset{ˉ}{A}}_{i j} = τ$ , because $\overset{ˉ}{A}$ is trace-free. In the squared norm, the cross term between $(τ / 3) h$ and $ψ^{- 2} \overset{ˉ}{A}$ vanishes by the same trace condition; the pure trace term is $τ^{2} / 3$ , and the trace-free term is $ψ^{- 12} ∣ \overset{ˉ}{A} ∣_{\overset{ˉ}{h}}^{2}$ . Thus $∣ K ∣_{h}^{2} = τ^{2} / 3 + ψ^{- 12} ∣ \overset{ˉ}{A} ∣_{\overset{ˉ}{h}}^{2}$ .

1.4F1F4algebra

Put $u = 2 \log ψ$ , so $h = e^{2 u} \overset{ˉ}{h}$ . In a chart, metric compatibility and torsion-freeness give $Γ [h]^{k}_{i j} = \frac{1}{2} h^{k ℓ} (\partial_{i} h_{j ℓ} + \partial_{j} h_{i ℓ} - \partial_{ℓ} h_{i j})$ , hence $C^{k}_{i j} : = Γ [h]^{k}_{i j} - Γ [\overset{ˉ}{h}]^{k}_{i j} = δ_{i}^{k} u_{j} + δ_{j}^{k} u_{i} - {\overset{ˉ}{h}}_{i j} u^{k}$ . Expanding the curvature definition and contracting gives $Ric (h)_{i j} = Ric (\overset{ˉ}{h})_{i j} + {\overset{ˉ}{D}}_{k} C^{k}_{i j} - {\overset{ˉ}{D}}_{i} C^{k}_{k j} + C^{k}_{i j} C^{ℓ}_{k ℓ} - C^{k}_{i ℓ} C^{ℓ}_{k j}$ . Here $C^{k}_{k j} = 3 u_{j}$ , ${\overset{ˉ}{D}}_{k} C^{k}_{i j} = 2 {\overset{ˉ}{D}}_{i} u_{j} - {\overset{ˉ}{h}}_{i j} \overset{ˉ}{Δ} u$ , and ${\overset{ˉ}{D}}_{i} C^{k}_{k j} = 3 {\overset{ˉ}{D}}_{i} u_{j}$ because $d u$ is closed. Expanding the products gives $C^{k}_{i j} C^{ℓ}_{k ℓ} = 6 u_{i} u_{j} - 3 {\overset{ˉ}{h}}_{i j} ∣ d u ∣_{\overset{ˉ}{h}}^{2}$ and $C^{k}_{i ℓ} C^{ℓ}_{k j} = 5 u_{i} u_{j} - 2 {\overset{ˉ}{h}}_{i j} ∣ d u ∣_{\overset{ˉ}{h}}^{2}$ ; tracing with $h^{i j} = ψ^{- 4} {\overset{ˉ}{h}}^{i j}$ therefore yields $R (h) = ψ^{- 4} (\overset{ˉ}{R} - 4 \overset{ˉ}{Δ} u - 2 ∣ d u ∣_{\overset{ˉ}{h}}^{2})$ . Substitution of $u = 2 \log ψ$ and $\overset{ˉ}{Δ} \log ψ = ψ^{- 1} Δ_{\overset{ˉ}{h}} ψ - ψ^{- 2} ∣ d ψ ∣_{\overset{ˉ}{h}}^{2}$ cancels the gradient-square terms and gives $R (h) = ψ^{- 5} (\overset{ˉ}{R} ψ - 8 Δ_{\overset{ˉ}{h}} ψ)$ .

2.1F2step 1.1step 1.2step 1.3step 1.4algebra

Substituting steps 1.3 and 1.4 into the left side of the SI Hamiltonian constraint [F2], and step 1.2 into its right side, gives $ψ^{- 5} (- 8 Δ_{\overset{ˉ}{h}} ψ + \overset{ˉ}{R} ψ) - ψ^{- 12} ∣ \overset{ˉ}{A} ∣_{\overset{ˉ}{h}}^{2} + \frac{2}{3} τ^{2} = 2 Λ + 2 ψ^{- 8} (∣ \overset{ˉ}{E} ∣_{\overset{ˉ}{h}}^{2} + ∣ \overset{ˉ}{B} ∣_{\overset{ˉ}{h}}^{2})$ . Multiplication by the positive $ψ^{5}$ is reversible and yields exactly the transformed scalar equation in the Statement; step 1.1 gives the two equivalent Maxwell constraints.

3.1F1F2step 1.2algebra∎

By step 1.2, the right side of the SI momentum constraint in [F2] is $- 2 ψ^{- 6} (\overset{ˉ}{E} \times_{\overset{ˉ}{h}} \overset{ˉ}{B})_{i}$ , which proves the displayed retained momentum equation. If $K = 0$ , its left side vanishes; if also $\overset{ˉ}{E} \times_{\overset{ˉ}{h}} \overset{ˉ}{B} = 0$ , its right side vanishes. No step has inferred this vector equation from the scalar constraint.

Exact conformal electrovacuum constraint transformation

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources