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A uniform one-sign charged fluid cannot satisfy Gauss on a compact torus

Statement refuted

The claim that smooth sourced-Gauss initial data on a compact boundaryless slice may have a charge-per-number-density scalar $Z$ that is nonzero and has one strict sign everywhere.

Example

Take $Σ = T_{2 π}^{3}$ with its standard orientation, any smooth Riemannian metric $h$ , a future unit normal $n$ , smooth strict-EOS density $ϵ_{0}$ , and smooth future unit velocity $u_{0}$ . If $Z_{0}$ is everywhere strictly positive or everywhere strictly negative, the sourced electric Gauss constraint cannot hold for any electric field $e$ : compact charge compatibility forces the total normal charge to vanish, whereas the strict charged-fluid current gives it the same strict sign as $Z_{0}$ . In particular, the spatially uniform candidate $h = δ$ , $u_{0} = n$ , $ϵ_{0} = ϵ_{*} > 0$ , $Z_{0} = Z_{*} > 0$ and $e = b = 0$ fails Gauss pointwise.

Facts & Assumptions

Given: A nonempty compact oriented boundaryless three-manifold $T_{2 π}^{3}$ , the strict charged-fluid model, and its sourced electric Gauss constraint.

[F1]

Strict charged barotropic tensor system: $j_{0} = c N (ϵ_{0}) Z_{0} u_{0}$ , $N (ϵ_{0}) > 0$ , $c > 0$ , and $u_{0}$ is future unit timelike.

[F2]

Compact no-boundary normal charge compatibility: If the sourced electric Gauss constraint holds on a compact oriented boundaryless slice, then $Q_{n} = - c^{- 1} \int_{Σ} g (j_{0}, n) {vol}_{h} = \int_{Σ} N Z_{0} (- g (u_{0}, n)) {vol}_{h} = 0$ .

[F3]

The observer rest-space decomposition and the spatial projection: For a unit timelike observer $n$ , every vector decomposes as $u_{0} = a n + X$ with $a = - g (u_{0}, n)$ , $X ⊥ n$ , and the induced spatial metric satisfies $g (X, X) = h (X, X) \geq 0$ .

[F4]

Classification of future and past causal vectors by a timelike field: The future timelike component consists exactly of timelike vectors whose inner product with the fixed time-orientation field is negative.

[F5]

The timelike set has exactly two convex components, described in orthonormal coordinates: In an orthonormal basis whose time vector is future, the future timelike component is $v^{0} > ∣ v_{v e c} ∣$ .

[F6]

Every Lorentz form admits an ordered basis with matrix diag(-1,1,1,1): A Lorentzian vector space admits an ordered orthonormal basis of signature $(-, +, +, +)$ .

[F7]

Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors: For Euclidean spatial coordinate vectors $x, y$ , $∣ x \cdot y ∣ \leq ∣ x ∣ ∣ y ∣$ .

Counterexample

technique · direct contradiction with the global charge compatibility identity and pointwise Gauss check for the displayed candidate

1.1F1F2F3F4F5F6F7givenalgebra

Suppose such data have $Z_{0}$ of one strict sign. At a point choose an orthonormal basis by [F6], and choose its time vector in the future component; [F4] fixes that component by the time-orientation field and [F5] gives its coordinate form. Therefore the future unit vectors $u_{0}$ and $n$ have coordinates $u_{0}^{0} > ∣ u_{0, v e c} ∣$ and $n^{0} > ∣ n_{v e c} ∣$ . By [F7], $g (u_{0}, n) = - u_{0}^{0} n^{0} + u_{0, v e c} \cdot n_{v e c} \leq - u_{0}^{0} n^{0} + ∣ u_{0, v e c} ∣ ∣ n_{v e c} ∣ < 0$ . Put $a = - g (u_{0}, n) > 0$ and $X = u_{0} - a n$ ; then [F3] gives $X ⊥ n$ and $g (X, X) = h (X, X) \geq 0$ . The normalization $g (u_{0}, u_{0}) = - 1$ now gives $- 1 = - a^{2} + g (X, X)$ , so $a^{2} = 1 + g (X, X) \geq 1$ and $a \geq 1$ . Since $N (ϵ_{0}) > 0$ by [F1], the integrand $N Z_{0} a$ has the same strict sign as $Z_{0}$ at every point. Its integral is nonzero: at any point its absolute value is positive, and continuity makes it bounded away from zero on a neighborhood of positive Riemannian volume. This contradicts [F2], which forces the integral to be zero. Therefore no such sourced-Gauss data exist.

2.1F1givenalgebra∎

For the displayed uniform candidate, $N (ϵ_{*}) = N_{*}$ and $j_{0} = c N_{*} Z_{*} n$ . Hence $- μ_{0} g (j_{0}, n) = μ_{0} c N_{*} Z_{*} > 0$ , whereas ${div}_{h} e = 0$ for $e = 0$ . The sourced Gauss equation is therefore false for that candidate, independently of whether its Einstein constraints hold.

A uniform one-sign charged fluid cannot satisfy Gauss on a compact torus

Statement refuted

Example

Facts & Assumptions

Counterexample

Depends on

Dependency tree · 2 levels

Sources