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Finite spacelike slab and normal flux charge — Examples

1 · Prerequisites

2 · Summary

The example is a smooth, positive scalar density carried by a constant future unit timelike vector in flat Minkowski coordinates. Its divergence is exactly zero, while its slice charge decreases across a finite open box. The complete outward side flux is positive and equals the lost charge. It includes the slab duration $T$ ; the instantaneous flux rate does not. This is a mathematical current example, not a solution of the Einstein–fluid equations or a physical entropy postulate.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

CounterexampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Zero divergence with decreasing slice charge in an open region

Statement refuted

The following universal assertion is false: for every slab and current of Finite spacelike slab and normal flux charge, pointwise nonnegative production $\nabla_{a} J^{a} \geq 0$ implies nondecreasing slice charge $Q_{J} (t_{1}) \geq Q_{J} (t_{0})$ even when outward side flux is unrestricted.

Facts & Assumptions

Given: Positive real parameters $L, L_{2}, L_{3}, T, a, s_{0}$ with $a T < 1$ , and a speed ratio $0 < β < 1$ . Put $γ = (1 - β^{2})^{- 1 / 2}$ . The coordinates below are mathematical real coordinates with no assigned physical units.

[A1]

Use the ${A C}_{ω}$ tangent-bundle convention recorded in The Axiom of Countable Choice ( ${A C}_{ω}$ ) and Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model. No additional choice is used in the explicit construction.

[F1]

In signature $(-, +, +, +)$ , a vector is future timelike when it is timelike and has negative inner product with the supplied future timelike orientation field Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model.

[F2]

The C¹ divergence satisfies $\nabla_{a} J^{a} = w^{- 1} \partial_{a} (w J^{a})$ in a chart (Metric divergence density identity for a C¹ current).

[F3]

The slice charge and net outward side flux are the coordinate integrals specified in Finite spacelike slab and normal flux charge.

[F4]

On the same finite slab, the exact balance is $Q_{J} (t_{1}) - Q_{J} (t_{0}) + Φ_{J} (B) = \int_{K} \nabla_{a} J^{a} {v o l}_{g}$ (Complete finite-slab production and outward-flux balance).

[F5]

The open and mixed-endpoint coordinate products in this example are Jordan sets, and the continuous densities on them are Riemann integrable (Finite coordinate boxes and their faces are compact Jordan sets). Continuous product integrands may be integrated by sections (Fubini over a bounded Jordan set when all but a content-zero family of sections are integrable), and the one-dimensional FTC evaluates the affine-coordinate integrals (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ). The same box lemma states that the boundary of each closed coordinate box has content zero.

[F6]

The chosen four-form $d x^{0} \land d x^{1} \land d x^{2} \land d x^{3}$ supplies the orientation of the flat chart (Oriented smooth manifolds and oriented charts).

Counterexample

1.1givenA1F1F6algebra

Let $M = R^{4}$ with coordinates $(x^{0}, x^{1}, x^{2}, x^{3})$ , metric $η = diag (- 1, 1, 1, 1)$ , orientation $d x^{0} \land d x^{1} \land d x^{2} \land d x^{3}$ , and future direction $\partial_{0}$ . Let $Ω = (0, L) \times (0, L_{2}) \times (0, L_{3})$ and $K = [0, T] \times Ω$ , with compact closure $\overline{K} = [0, T] \times [0, L] \times [0, L_{2}] \times [0, L_{3}]$ . Define the constant vector $u = (γ, γ β, 0, 0)$ and scalar $s = s_{0} [1 + a (x^{1} - β x^{0})]$ . Since $0 < β < 1$ and $a T < 1$ , $1 + a (x^{1} - β x^{0}) \geq 1 - a β T > 0$ on $\overline{K}$ , so $s$ is positive there and extends smoothly to a neighborhood of $\overline{K}$ . Also $η (u, u) = - γ^{2} (1 - β^{2}) = - 1$ and $u$ is future-directed.

1.2F3F5algebra

Let $Q$ be the closed rectangular hull of either $Ω$ or a side-face parameter rectangle $Ω_{i}$ , and let $R$ be the corresponding endpoint-variant product. For a continuous density $f : Q \to R$ , the zero-extension convention in [F3] represents $\int_{R} f$ by a Riemann integral on $Q$ . The extension differs from $f$ only on $\partial Q$ , which has content zero by [F5]. If $∣ f ∣ \leq M$ , finite box covers of $\partial Q$ with arbitrarily small total volume bound the change in the upper and lower Darboux sums by $2 M$ times that volume; hence the two integrals agree as the volume tends to zero.

2.1step 1.1F2algebra

Put $J = s u$ , so $J^{0} = γ s$ and $J^{1} = γ β s$ . In this chart $w = 1$ , and [F2] gives $\nabla_{a} J^{a} = \partial_{0} J^{0} + \partial_{1} J^{1} = γ (- s_{0} a β) + γ β (s_{0} a) = 0$ throughout the slab; the remaining components vanish. Thus local production is zero everywhere.

2.2step 1.1step 1.2F3F5algebra

By step 1.2, [F3] and [F5], the integral of the slice density over the open rectangle equals its ordinary integral over the closed rectangular hull. Integrating the affine density gives $Q_{J} (t) = γ s_{0} L L_{2} L_{3} (1 + \frac{a L}{2} - a β t),$ so $Q_{J} (T) - Q_{J} (0) = - γ s_{0} a β T L L_{2} L_{3} < 0.$ The slice charge strictly decreases.

3.1step 1.2step 2.1step 2.2F3F4F5algebra∎

Only the $x^{1}$ side faces carry flux. Their upper-minus-lower density is $J^{1} (t, L, x^{2}, x^{3}) - J^{1} (t, 0, x^{2}, x^{3}) = γ β s_{0} a L$ , independent of $t, x^{2}, x^{3}$ . By step 1.2, the integral over the mixed-endpoint face parameter rectangle equals the integral over its closed rectangular hull, so the integrated net outward side flux is $Φ_{J} (B) = γ s_{0} a β T L L_{2} L_{3} > 0.$ The factor $T$ is the duration of the slab. By [F4], $Q_{J} (T) - Q_{J} (0) + Φ_{J} (B) = 0$ , in agreement with the zero production calculation. The example refutes the unrestricted monotonicity assertion and shows that the side-flux hypothesis is essential.

Remarks

This is an explicit mathematical current example. The scalar $s$ is entropy-like only if a separate physical model assigns that meaning. The construction does not solve an Einstein–fluid system, adopt a second-law postulate, or report an observation.