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Compact no-boundary normal charge compatibility

Statement

Let $Σ$ be a compact oriented smooth three-manifold without boundary with smooth Riemannian metric $h$ . Let $n$ be a future unit normal, let $e$ be the electric vector determined by the two-form convention of Strict charged barotropic tensor system, and suppose smooth initial data for that system satisfy the sourced electric Gauss constraint

${div}_{h} e = - μ_{0} g (j, n) .$

Define the SI normal-charge integral

$Q_{n} : = - \frac{1}{c} \int_{Σ} g (j, n) {vol}_{h} = \int_{Σ} N Z (- g (u, n)) {vol}_{h} .$

Then $Q_{n} = 0$ . The result is conditional on the stated Gauss constraint; it does not assert existence of initial data satisfying the remaining Einstein or magnetic constraints.

Facts & Assumptions

Given: The compact oriented boundaryless Riemannian three-manifold and smooth charged-fluid initial data of Strict charged barotropic tensor system, with its sourced Gauss constraint.

[F1]

Strict charged barotropic tensor system: $j^{a} = c N Z u^{a}$ , $N > 0$ , and $g (u, u) = - 1$ .

[A1]

The specified initial electric Gauss equation is ${div}_{h} e = - μ_{0} g (j, n)$ ; it is an explicit hypothesis of this proposition.

[F2]

Riemannian divergence: The Riemannian divergence is the scalar determined in each oriented chart by $L_{e} {vol}_{h} = ({div}_{h} e) {vol}_{h}$ .

[F3]

Coordinate formula for riemannian divergence: If $h_{i j}$ is the coordinate metric and $e = e^{i} \partial_{i}$ , then ${div}_{h} e = (\det h)^{- 1 / 2} \partial_{i} ((\det h)^{1 / 2} e^{i})$ .

[F4]

Finite chart localization gives choice-free integration and compact Stokes: For a compactly supported smooth $2$ -form $η$ on an oriented boundaryless smooth three-manifold, its finite chart integral satisfies $\int_{Σ} d η = 0$ .

Proof

technique · direct

1.1F2F3algebra

On an oriented coordinate chart write ${vol}_{h} = \sqrt{\det h} d x^{1} \land d x^{2} \land d x^{3}$ . Direct contraction gives $ι_{e} {vol}_{h} = e^{1} \sqrt{\det h} d x^{2} \land d x^{3} - e^{2} \sqrt{\det h} d x^{1} \land d x^{3} + e^{3} \sqrt{\det h} d x^{1} \land d x^{2}$ . Differentiating and reordering the wedge factors gives $d (ι_{e} {vol}_{h}) = \partial_{i} (\sqrt{\det h} e^{i}) d x^{1} \land d x^{2} \land d x^{3} = ({div}_{h} e) {vol}_{h}$ by [F3].

2.1step 1.1A1F4algebra

Since $Σ$ is compact, $ι_{e} {vol}_{h}$ is compactly supported. Apply [F4] to step 1.1 to obtain $\int_{Σ} {div}_{h} e {vol}_{h} = 0$ . Integrating [A1] therefore gives $0 = - μ_{0} \int_{Σ} g (j, n) {vol}_{h}$ , and $μ_{0} > 0$ implies $\int_{Σ} g (j, n) {vol}_{h} = 0$ .

3.1step 2.1F1algebra∎

By [F1], $j = c N Z u$ and hence $- c^{- 1} g (j, n) = N Z (- g (u, n))$ pointwise. The definition of $Q_{n}$ and step 2.1 give $Q_{n} = 0$ . No sign assumption on $Z$ is needed.

Compact no-boundary normal charge compatibility

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources