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Strict charged barotropic tensor system — Examples

1 · Prerequisites

2 · Summary

The linear equation of state gives explicit density and enthalpy coordinates. On a compact boundaryless slice, the sourced Gauss constraint is incompatible with a charge-per-number-density field of one strict sign; the counterexample includes a spatially uniform candidate and checks its pointwise Gauss failure.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck pendingOpen item page →

Linear barotropic equation of state and its enthalpy coordinates

Example

Fix a dimensionless constant $w$ with $0 < w < 1$ , take $J = (0, \infty)$ , choose a reference energy density $ϵ_{*} > 0$ , and set

$p (ϵ) = w ϵ .$

Choose $N_{*} > 0$ with units $m^{- 3}$ as in Strict charged barotropic tensor system. Then $ϵ + p = (1 + w) ϵ > 0$ and $p^{'} = w \in (0, 1)$ on $J$ , so this is an instance of the strict EOS. The EOS-defined number-density and enthalpy coordinates are

$N (ϵ) = N_{*} {(\frac{ϵ}{ϵ_{*}})}^{1 / (1 + w)}, r (ϵ) = \frac{\sqrt{w}}{1 + w} \log (\frac{ϵ}{ϵ_{*}}) .$

Here the ratios and $r$ are dimensionless. The map $r : J \to R$ is strictly increasing and onto; in these variables $N = N_{*} e^{r / \sqrt{w}}$ . These are mathematical EOS variables, not a microscopic derivation of particle number.

Facts & Assumptions

Given: $0 < w < 1$ , $ϵ > 0$ , $ϵ_{*} > 0$ , $N_{*} > 0$ , and the EOS-defined functions of Strict charged barotropic tensor system and Strict charged-fluid symmetric reduction.

[F1]

Strict charged barotropic tensor system: $N (ϵ) = N_{*} \exp (\int_{ϵ_{*}}^{ϵ} (z + p (z))^{- 1} d z)$ , with $N_{*} > 0$ .

[F2]

Strict charged-fluid symmetric reduction: $r (ϵ) = \int_{ϵ_{*}}^{ϵ} \sqrt{p^{'} (z)} / (z + p (z)) d z$ and $r^{'} = \sqrt{p^{'}} / (ϵ + p) > 0$ for the strict EOS.

Proof

technique · direct integral evaluation and differentiation

1.1givenalgebra

The derivative of $p (ϵ) = w ϵ$ is $p^{'} = w$ . Since $w > 0$ and $ϵ > 0$ , $p^{'} = w \in (0, 1)$ and $ϵ + p = (1 + w) ϵ > 0$ on $J$ . Thus the strict EOS hypotheses hold.

1.2F1givenalgebra

For every $z > 0$ , $(z + p (z))^{- 1} = ((1 + w) z)^{- 1}$ . Substituting this in [F1] and integrating gives $N = N_{*} \exp ((1 + w)^{- 1} \log (ϵ / ϵ_{*})) = N_{*} (ϵ / ϵ_{*})^{1 / (1 + w)}$ . The logarithm is dimensionless because both arguments are energy densities.

2.1F2givenstep 1.2algebra∎

Since $\sqrt{p^{'} (z)} = \sqrt{w}$ , [F2] gives $r = (\sqrt{w} / (1 + w)) \log (ϵ / ϵ_{*})$ . Differentiating yields $r^{'} = \sqrt{w} / ((1 + w) ϵ) > 0$ , so $r$ is strictly increasing. As $ϵ ↓ 0$ its logarithm tends to $- \infty$ , and as $ϵ \to \infty$ it tends to $+ \infty$ ; the positive prefactor preserves these limits, so its range is all of $R$ . Solving the displayed formula gives $ϵ / ϵ_{*} = \exp ((1 + w) r / \sqrt{w})$ ; inserting this in step 1.2 yields $N = N_{*} e^{r / \sqrt{w}}$ .

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck pendingjudge pass (gpt-6-luna)Open item page →

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.

Strict charged barotropic tensor system — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

Linear barotropic equation of state and its enthalpy coordinates

Example

Facts & Assumptions

Proof

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

Statement refuted

Example

Facts & Assumptions

Counterexample

Sources