Pipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Arbitrary-positive-H conformastatic Ricci and Einstein components

1 · Prerequisites

2 · Summary

Scope

This page fixes a global length-coordinate chart for the metric $g_{H} = - H^{- 2} (d x^{0})^{2} + H^{2} δ_{i j} d x^{i} d x^{j}$ , its coordinate orientation, and the time direction selected by $H \partial_{0}$ . It proves the Lorentzian frame and derives the Ricci, scalar, and Einstein components for every positive $C^{3}$ function $H$ , without imposing harmonicity. For smooth $H > 0$ with $Δ H \leq 0$ , it then verifies an exact formal Einstein–Maxwell–charged-dust family and derives the regular spherical profile, proper mass and charge, and dilute bounds. These are mathematical deductions under the stated formal equations, signature, charge sign, and SI bookkeeping; they do not assert physical realization, stability, or general initial-value existence.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Positive-scalar conformastatic chart data

Definition

Let $D \subseteq R^{3}$ be open and let $H : D \to (0, \infty)$ be of class $C^{3}$ , meaning that every partial derivative of total order at most three exists and is continuous. Set $M : = R_{x^{0}} \times D$ , an open subset of $R^{4}$ , with global coordinates $(x^{0}, x^{1}, x^{2}, x^{3})$ ; each coordinate is assigned length units, with $x^{0} = c t$ when a time coordinate is wanted. The function $H$ depends only on $(x^{1}, x^{2}, x^{3})$ and is dimensionless.

Give $M$ the coordinate orientation determined by $d x^{0} \land d x^{1} \land d x^{2} \land d x^{3}$ .

Define the symmetric coordinate tensor $g_{H}$ by $g_{H} = - H^{- 2} (d x^{0})^{2} + H^{2} \sum_{i = 1}^{3} (d x^{i})^{2},$ where $H^{- 1} : = 1 / H$ and $H^{- 2} : = (1 / H)^{2}$ . This exact coordinate form is called the positive-scalar conformastatic chart tensor in this item; no field equation or physical solution property is part of the definition.

For this $C^{3}$ chart tensor, a time orientation means a continuous selection at each $p \in M$ of one of the two connected components of ${v \in T_{p} M : g_{H} (v, v) < 0}$ , with the topology from the coordinate identification $T M = M \times R^{4}$ . Specify the selection globally by $C_{p}^{+} : = {v = v^{0} \partial_{0} + \sum_{i = 1}^{3} v^{i} \partial_{i} : g_{H} (v, v) < 0, v^{0} > 0} .$ Here the selection is continuous in the explicit sense that the fiberwise linear coordinate map $(p, v) \mapsto (p, v^{0} / H (p), H (p) v^{1}, H (p) v^{2}, H (p) v^{3})$ is a continuous bundle isomorphism carrying $C^{+}$ to the constant cone ${(u^{0}, u) : u^{0} > ∥ u ∥_{2}}$ ; this map and its inverse are $C^{3}$ . That cone and its negative are convex and nonempty, and a timelike vector has nonzero $u^{0}$ , so they are exactly the two timelike components. Define a nonzero causal vector by $g_{H} (v, v) \leq 0$ and call it future-directed when $v^{0} > 0$ . Set $N_{H} : = H \partial_{0}$ , a $C^{3}$ vector field (smooth if $H$ is smooth). The Lorentzian lemma below verifies $g_{H} (N_{H}, N_{H}) = - 1$ , so $N_{H} \in C_{p}^{+}$ everywhere. No smoothness beyond $C^{3}$ is required for this definition.

Because $H > 0$ , the functions $H^{- 1}$ and $H^{- 2}$ are compositions and products of $H$ with positive-base powers of exponents $- 1$ and $- 2$ . The derivative formula for these powers, repeated through any finite order, and closure of $C^{k}$ functions under composition and finite products show that the displayed coefficients are $C^{3}$ when $H$ is $C^{3}$ , and smooth when $H$ is smooth.

For later component formulas define $H_{i} : = \partial_{i} H$ , $H_{i j} : = \partial_{i} \partial_{j} H$ , $q : = \sum_{i = 1}^{3} H_{i}^{2}$ , and $Δ H : = \sum_{i = 1}^{3} H_{i i}$ .

Remarks

The tensor coefficients are $C^{3}$ by the stated regularity and positivity of $H$ . Its nondegeneracy, Lorentzian signature, and the timelike property needed for the chosen time orientation are proved by The positive-scalar conformastatic chart tensor is Lorentzian ↗; this definition alone does not assert them.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

The positive-scalar conformastatic chart tensor is Lorentzian

Statement

For every $D, H, g_{H}$ in Positive-scalar conformastatic chart data, the coordinate tensor $g_{H}$ is a C3 nondegenerate Lorentzian metric of signature $(-, +, +, +)$ . Its inverse components are $g^{00} = - H^{2}$ and $g^{i j} = H^{- 2} δ_{i j}$ , its determinant is $- H^{4}$ , and $\sqrt{∣ \det g ∣} = H^{2}$ . The vector fields $n : = H \partial_{0}$ and $e_{i} : = H^{- 1} \partial_{i}$ form a global orthonormal frame with $g (n, n) = - 1$ , $g (n, e_{i}) = 0$ , and $g (e_{i}, e_{j}) = δ_{i j}$ . The chosen time orientation makes $n$ future-directed, and the frame $(n, e_{1}, e_{2}, e_{3})$ is positively oriented.

Facts & Assumptions

Given: An open set $D \subseteq R^{3}$ , a positive C3 function $H$ independent of $x^{0}$ , and the coordinate tensor $g_{H}$ .

[F1]

Positive-scalar conformastatic chart data defines the coordinate matrix $diag (- H^{- 2}, H^{2}, H^{2}, H^{2})$ , with $H > 0$ . It fixes the coordinate orientation $d x^{0} \land d x^{1} \land d x^{2} \land d x^{3}$ and the future timelike cone $C_{p}^{+} = {v : g_{H} (v, v) < 0, v^{0} > 0}$ . Thus a timelike vector is future-directed exactly when its zeroth coordinate is positive; for $n = H \partial_{0}$ , that coordinate is $H > 0$ .

Proof

1.1givenF1algebra

The coordinate matrix is diagonal and symmetric. Since $H > 0$ , all entries are nonzero, its inverse is $diag (- H^{2}, H^{- 2}, H^{- 2}, H^{- 2})$ , and multiplying the two diagonal matrices gives the identity. The determinant is $(- H^{- 2}) (H^{2})^{3} = - H^{4} < 0$ , so the matrix has exactly one negative and three positive diagonal entries and no zero eigenvalue. Thus it is nondegenerate of signature $(-, +, +, +)$ at every point. Its coefficients and inverse are C3 because $H$ is C3 and the reciprocal map $u \mapsto 1 / u$ is smooth on $(0, \infty)$ .

2.1givenF1step 1.1algebra∎

With the frame fields defined in the Statement, direct contraction against the diagonal matrix gives $g (n, n) = H^{2} (- H^{- 2}) = - 1$ , $g (n, e_{i}) = 0$ , and $g (e_{i}, e_{j}) = H^{- 2} H^{2} δ_{i j} = δ_{i j}$ . Each frame vector is a nonzero multiple of a distinct coordinate vector, so the four vectors form a basis. Their coordinate change matrix is $diag (H, H^{- 1}, H^{- 1}, H^{- 1})$ , with positive determinant $H^{- 2}$ ; hence the frame is positively oriented. The time orientation is the one specified in [F1], so $n$ is future-directed. The already computed metric determinant gives $\sqrt{∣ \det g ∣} = H^{2}$ .

Remarks

The result is pointwise and coordinate-local. It does not assert geodesic completeness, asymptotic flatness, or any Einstein–Maxwell equation.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Arbitrary-positive-H conformastatic Ricci and Einstein components

Statement

Let $D \subseteq R^{3}$ be open, let $H \in C^{3} (D, (0, \infty))$ , and take the C3 coordinate metric $g_{H}$ from Positive-scalar conformastatic chart data, which is Lorentzian by The positive-scalar conformastatic chart tensor is Lorentzian. All indices range over $0, 1, 2, 3$ , spatial indices $i, j, k$ range over $1, 2, 3$ , and repeated indices are summed. Use the coordinate connection $Γ^{α}_{β γ} : = \frac{1}{2} g^{α δ} (\partial_{β} g_{δ γ} + \partial_{γ} g_{β δ} - \partial_{δ} g_{β γ})$ and the explicit curvature convention $R^{α}_{β μ ν} : = \partial_{μ} Γ^{α}_{ν β} - \partial_{ν} Γ^{α}_{μ β} + Γ^{α}_{μ λ} Γ^{λ}_{ν β} - Γ^{α}_{ν λ} Γ^{λ}_{μ β}, {Ric}_{β ν} : = R^{α}_{β α ν}, S : = g^{β ν} {Ric}_{β ν}, G_{μ ν} : = {Ric}_{μ ν} - \frac{1}{2} S g_{μ ν} .$ The displayed Christoffel coefficients define the unique torsion-free, $g_{H}$ -compatible connection on the global coordinate chart, and the displayed curvature is formed from that connection. These are the standard coordinate formulas with the Riemann sign $R (X, Y) Z = \nabla_{X} \nabla_{Y} Z - \nabla_{Y} \nabla_{X} Z - \nabla_{[X, Y]} Z$ ; the C3 regularity suffices for every displayed derivative. With $H_{i}, H_{i j}, q, Δ H$ as in Positive-scalar conformastatic chart data, the only nonzero Christoffel types are $Γ^{0}_{0 i} = Γ^{0}_{i 0} = - \frac{H_{i}}{H}, Γ^{i}_{00} = - \frac{H_{i}}{H^{5}}, Γ^{i}_{j k} = \frac{δ_{i j} H_{k} + δ_{i k} H_{j} - δ_{j k} H_{i}}{H} .$ The Ricci components, scalar curvature, and orthonormal-frame Einstein components are ${Ric}_{00} = \frac{q}{H^{6}} - \frac{Δ H}{H^{5}}, {Ric}_{0 i} = 0, {Ric}_{i j} = - \frac{δ_{i j} Δ H}{H} + \frac{δ_{i j} q - 2 H_{i} H_{j}}{H^{2}},$ $S = - \frac{2 Δ H}{H^{3}}, G (n, n) = \frac{q - 2 H Δ H}{H^{4}}, G (n, e_{i}) = 0, G (e_{i}, e_{j}) = \frac{δ_{i j} q - 2 H_{i} H_{j}}{H^{4}},$ where $n = H \partial_{0}$ and $e_{i} = H^{- 1} \partial_{i}$ are the frame fields of The positive-scalar conformastatic chart tensor is Lorentzian. No harmonicity condition on $H$ is imposed.

Facts & Assumptions

Given: The open set $D$ , positive C3 function $H$ , the coordinate tensor $g_{H}$ , and the definitions of its connection, curvature, and contractions in the Statement.

[F1]

Positive-scalar conformastatic chart data supplies $g_{00} = - H^{- 2}$ , $g_{i j} = H^{2} δ_{i j}$ , $g_{0 i} = 0$ , and $H_{i}, H_{i j}, q, Δ H$ .

[F2]

The positive-scalar conformastatic chart tensor is Lorentzian supplies $g^{00} = - H^{2}$ , $g^{i j} = H^{- 2} δ_{i j}$ , $\sqrt{∣ \det g ∣} = H^{2}$ , and the frame $n = H \partial_{0}$ , $e_{i} = H^{- 1} \partial_{i}$ .

[F3]

Positive-base powers have derivative $(u^{α})^{'} = α u^{α - 1}$ for $u > 0$ , and the ordinary one-variable chain and product rules apply. (Continuity and derivatives of positive-base real powers, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ )

[F4]

Since $H$ is C3, its second coordinate partials commute: $\partial_{i} H_{j} = \partial_{j} H_{i}$ . (Clairaut--Schwarz theorem for continuous second partial derivatives)

[A1]

$H$ is C3, positive everywhere on $D$ , and independent of $x^{0}$ . No condition $Δ H = 0$ is assumed.

Proof

1.1givenF1F2F3algebra

Because $M$ has one global coordinate chart, the displayed $Γ^{α}_{β γ}$ define a connection by $\nabla_{\partial_{β}} \partial_{γ} = Γ^{α}_{β γ} \partial_{α}$ and the connection Leibniz rule. The formula is symmetric in $β, γ$ , so the connection is torsion-free. Lowering its upper index gives $g_{ρ α} Γ^{α}_{β γ} = \frac{1}{2} (\partial_{β} g_{ρ γ} + \partial_{γ} g_{ρ β} - \partial_{ρ} g_{β γ})$ ; adding the expressions with $(ρ, γ)$ interchanged cancels opposite derivative terms and yields $\partial_{β} g_{ρ γ}$ , which is metric compatibility. Applying the formula to $g_{00} = - H^{- 2}$ and $g_{i j} = H^{2} δ_{i j}$ gives $Γ^{0}_{0 i} = \frac{1}{2} (- H^{2}) (2 H^{- 3} H_{i}) = - H_{i} / H$ , $Γ^{i}_{00} = - \frac{1}{2} H^{- 2} (2 H^{- 3} H_{i}) = - H_{i} / H^{5}$ , and $Γ^{i}_{j k} = (δ_{i j} H_{k} + δ_{i k} H_{j} - δ_{j k} H_{i}) / H$ . Every other type vanishes because $g$ is diagonal and time-independent. The derivatives use [F3] and the positivity of $H$ .

2.1F2step 1.1algebra

Any other torsion-free, $g_{H}$ -compatible connection $\hat{\nabla}$ differs from $\nabla$ by the tensor $C (X, Y) : = {\hat{\nabla}}_{X} Y - \nabla_{X} Y$ . Torsion-freeness gives $C (X, Y) = C (Y, X)$ , while subtracting the two compatibility identities gives $C_{X Y Z} : = g_{H} (C (X, Y), Z) = - C_{X Z Y}$ . Thus $C_{X Y Z} = C_{Y X Z} = - C_{Y Z X} = - C_{Z Y X} = C_{Z X Y} = C_{X Z Y} = - C_{X Y Z}$ , so $C = 0$ by nondegeneracy of $g_{H}$ . Therefore the displayed connection is unique. Its trace is $Γ^{α}_{α i} = Γ^{0}_{0 i} + \sum_{k = 1}^{3} Γ^{k}_{k i} = - H_{i} / H + 3 H_{i} / H = 2 H_{i} / H$ , while $Γ^{α}_{α 0} = 0$ .

3.1givenF1F2step 1.1step 2.1algebra

In the Ricci formula, $\partial_{i} Γ^{i}_{00} = - Δ H / H^{5} + 5 q / H^{6}$ by differentiating $- H_{i} H^{- 5}$ . The trace-product term is $Γ^{i}_{00} Γ^{α}_{α i} = - 2 q / H^{6}$ . The final product contraction $Γ^{α}_{0 β} Γ^{β}_{0 α}$ has two nonzero orders for each spatial $i$ , namely $Γ^{0}_{0 i} Γ^{i}_{00}$ and $Γ^{i}_{00} Γ^{0}_{i 0}$ ; each is $H_{i}^{2} / H^{6}$ , so its subtraction contributes $- 2 q / H^{6}$ . Thus ${Ric}_{00} = q / H^{6} - Δ H / H^{5}$ .

3.2givenF1step 1.1step 2.1algebra

For ${Ric}_{0 i}$ , the derivative terms vanish because all coefficients are time-independent and $Γ^{α}_{α 0} = 0$ . In the trace-product term, a nonzero $Γ^{α}_{0 i}$ requires $α = 0$ , but then $Γ^{β}_{β 0} = 0$ . In the last contraction, the only potentially nonzero first factors are $Γ^{0}_{0 j}$ or $Γ^{j}_{00}$ ; their paired factors have a lower time-space connection slot that vanishes. Hence ${Ric}_{0 i} = 0$ .

3.3givenF1F2F3F4step 1.1step 2.1algebra

For spatial $i, j$ , $Γ^{k}_{i j} = (δ_{k i} H_{j} + δ_{k j} H_{i} - δ_{i j} H_{k}) / H$ . Differentiating and summing $k$ gives $\partial_{k} Γ^{k}_{i j} = 2 H_{i j} / H - 2 H_{i} H_{j} / H^{2} - δ_{i j} Δ H / H + δ_{i j} q / H^{2}$ . The trace term is $- \partial_{j} Γ^{α}_{α i} = - 2 H_{i j} / H + 2 H_{i} H_{j} / H^{2}$ , using the commuting mixed partials of [F4]. The other trace product is $Γ^{k}_{i j} Γ^{α}_{α k} = (4 H_{i} H_{j} - 2 δ_{i j} q) / H^{2}$ .

4.1step 1.1step 3.3algebra

In the last Ricci contraction, the time-index product is $Γ^{0}_{i 0} Γ^{0}_{j 0} = H_{i} H_{j} / H^{2}$ . The spatial numerator is $(δ_{a i} H_{b} + δ_{a b} H_{i} - δ_{i b} H_{a}) (δ_{b j} H_{a} + δ_{b a} H_{j} - δ_{j a} H_{b})$ . Summing over $a, b$ gives the nine terms $H_{i} H_{j}, H_{i} H_{j}, - δ_{i j} q, H_{i} H_{j}, 3 H_{i} H_{j}, - H_{i} H_{j}, - δ_{i j} q, - H_{i} H_{j}, H_{i} H_{j}$ , hence the spatial product is $(5 H_{i} H_{j} - 2 δ_{i j} q) / H^{2}$ . Including the time product, the full product before the Ricci minus sign is $(6 H_{i} H_{j} - 2 δ_{i j} q) / H^{2}$ . Combining this with step 3.3 gives ${Ric}_{i j} = - δ_{i j} Δ H / H + (δ_{i j} q - 2 H_{i} H_{j}) / H^{2}$ .

4.2step 3.1step 3.3F2algebra

Contracting the Ricci table with $g^{00} = - H^{2}$ and $g^{i j} = H^{- 2} δ_{i j}$ gives $g^{00} {Ric}_{00} = - q / H^{4} + Δ H / H^{3}$ and $\sum_{i j} g^{i j} {Ric}_{i j} = q / H^{4} - 3 Δ H / H^{3}$ . Their sum is $S = - 2 Δ H / H^{3}$ .

5.1step 3.2step 3.1step 3.3step 4.1step 4.2F2algebra∎

The coordinate Einstein component is $G_{00} = {Ric}_{00} - \frac{1}{2} S g_{00} = q / H^{6} - 2 Δ H / H^{5}$ , so $G (n, n) = H^{2} G_{00} = (q - 2 H Δ H) / H^{4}$ . For spatial indices, $G_{i j} = {Ric}_{i j} - \frac{1}{2} S g_{i j} = (δ_{i j} q - 2 H_{i} H_{j}) / H^{2}$ because the $δ_{i j} Δ H / H$ terms cancel; therefore $G (e_{i}, e_{j}) = H^{- 2} G_{i j} = (δ_{i j} q - 2 H_{i} H_{j}) / H^{4}$ . The mixed components vanish by step 3.2.

Remarks

For constant $H$ , all displayed curvature components vanish. For nonconstant harmonic $H$ , the scalar curvature is zero but the Einstein tensor need not vanish. The table is not restricted to electrovacuum, and it is not a physical field equation.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Exact smooth superharmonic counterpoised dust family

Statement

Assume ${A C}_{ω}$ as in The Axiom of Countable Choice ( ${A C}_{ω}$ ), as required by the currently inspected general geometric convention in Smooth formal matter, EM tensors and observer quantities. Let $D \subseteq R^{3}$ be open and let $H \in C^{\infty} (D, (0, \infty))$ satisfy $Δ H \leq 0$ . Use $M = R \times D$ , its coordinate orientation, and the time orientation for which $n : = H \partial_{0}$ is future, all as fixed in Positive-scalar conformastatic chart data and verified in The positive-scalar conformastatic chart tensor is Lorentzian. Use the metric $g_{H}$ and coordinates $x^{0} = c t, x^{i}$ of Positive-scalar conformastatic chart data. Fix formal constants $c, μ_{0}, G_{N} > 0$ , put $ϵ_{0} = (μ_{0} c^{2})^{- 1}$ , $κ = 8 π G_{N} / c^{4}$ , and define the positive charge-to-mass parameter $Q_{s} : = \sqrt{4 π ϵ_{0} G_{N}}$ using the positive root of Explicit rational and real power regimes. For a sign $s \in {- 1, + 1}$ set $A : = s \frac{c}{Q_{s}} H^{- 1} d x^{0}, F : = d A, U : = c H \partial_{0}, ρ : = - \frac{c^{2} Δ H}{4 π G_{N} H^{3}}, σ : = s Q_{s} ρ, j : = σ U, T_{m}^{a b} : = ρ U^{a} U^{b} .$ Let $a^{b} : = U^{a} \nabla_{a} U^{b}$ be the covariant four-acceleration of the timelike field $U$ . Then $U = c n$ is future timelike with $g_{H} (U, U) = - c^{2}$ , all displayed fields are smooth, and $ρ \geq 0$ . On ${ρ > 0}$ the charge-to-mass ratio is $σ / ρ = s Q_{s}$ ; no ratio is defined by division at $ρ = 0$ . The tuple satisfies the formal sourced Maxwell equations $d F = 0$ , $\nabla_{a} F^{a b} = - μ_{0} j^{b}$ , the dust matter exchange equation $\nabla_{a} T_{m}^{a b} = F^{b}_{c} j^{c}$ , and the formal Einstein equation with $Λ = 0$ , $G_{a b} = κ (T_{m, a b} + T_{a b}^{E M}),$ where $T^{E M}$ is exactly the tensor of Smooth formal matter, EM tensors and observer quantities. Its total stress satisfies WEC, NEC, SEC, and DEC. The coordinate four-acceleration of the static dust field is $a^{0} = 0$ , $a^{i} = - c^{2} H_{i} / H^{3}$ , and its Lorentz force per unit mass balances it wherever $ρ > 0$ ; the undivided exchange identity also holds where $ρ = 0$ .

This is a conditional mathematical solution family for the stated formal equations. It makes no claim of physical realization, equilibrium stability, or coupled-field existence for arbitrary initial data.

Facts & Assumptions

Given: The open set $D$ , smooth positive $H$ with $Δ H \leq 0$ , constants $c, μ_{0}, G_{N} > 0$ , sign $s \in {- 1, + 1}$ , and the coordinate data of the Statement.

[F1]

Smooth formal matter, EM tensors and observer quantities defines the pointwise formal sourced Maxwell equations $d F = 0$ , $\nabla_{a} F^{a b} = - μ_{0} j^{b}$ , the matter-exchange equation $\nabla_{a} T_{m}^{a b} = F^{b}_{c} j^{c}$ , and the formal Einstein equation $G_{a b} + Λ g_{a b} = κ (T_{m, a b} + T_{a b}^{E M})$ . It also defines a local potential by $d A = F$ . This current draft definition is used only at these exact equation interfaces.

[F2]

Explicit rational and real power regimes defines the unique positive square root and rational powers for positive real bases. Here $Q_{s}$ is the unique positive root of $4 π ϵ_{0} G_{N}$ , and negative integer powers of $H$ mean reciprocals.

[F3]

Charged dust balance equivalence and advected charge ratio defines $a^{b} : = U^{a} \nabla_{a} U^{b}$ ; on a region where $ρ > 0$ , the dust exchange equation is equivalent to continuity $\nabla_{a} (ρ U^{a}) = 0$ and force balance $ρ a^{b} = σ F^{b}_{c} U^{c}$ . Its statement makes no division-based conclusion at $ρ = 0$ . This is the current draft theorem and is used only on its stated positive-density domain.

[F4]

EM, dust and perfect-fluid energy inequalities: the Maxwell tensor and a charged-dust tensor with $ρ \geq 0$ satisfy WEC, NEC, SEC, and DEC, and their sum retains each condition. This is the current draft theorem; no general matter positivity is assumed.

[F5]

Positive-scalar conformastatic chart data defines $M$ , $g_{H}$ , $H_{i}$ , $q$ , and $Δ H$ , with coordinate orientation and time orientation selected by $n = H \partial_{0}$ .

[F6]

The positive-scalar conformastatic chart tensor is Lorentzian gives the inverse metric, nondegeneracy, the future unit vector $n = H \partial_{0}$ , the orthonormal spatial frame $e_{i} = H^{- 1} \partial_{i}$ , and their orientation.

[F7]

Arbitrary-positive-H conformastatic Ricci and Einstein components gives the coordinate connection and the arbitrary-positive- $H$ Ricci, scalar, and Einstein table in the curvature convention stated there.

[F8]

Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields: for a smooth nondegenerate metric, there is a unique torsion-free metric-compatible connection with the Christoffel formula in the statement; its induced tensor derivative has the displayed component rule, including $\nabla_{c} F^{a b} = \partial_{c} F^{a b} + Γ^{a}_{d c} F^{d b} + Γ^{b}_{d c} F^{a d}$ .

[F9]

Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ : differentiable real functions obey the product rule, so differentiating $H \cdot H^{- 1} = 1$ gives the reciprocal derivative used below.

[F10]

The local coordinate formula for the exterior derivative: for a coordinate one-form $A = A_{a} d x^{a}$ , $(d A)_{a b} = \partial_{a} A_{b} - \partial_{b} A_{a}$ .

[F11]

The exterior derivative squares to zero: for every smooth differential form $A$ , $d (d A) = 0$ .

[F12]

Smooth formal matter, EM tensors and observer quantities defines $T_{E M}^{a b} = μ_{0}^{- 1} (F^{a c} F^{b}_{c} - \frac{1}{4} g^{a b} F_{c d} F^{c d})$ and in a positively oriented orthonormal tetrad uses $F_{0 i} = - E_{i} / c$ .

[F13]

Smooth formal matter, EM tensors and observer quantities defines charged dust by $T_{m}^{a b} = ρ U^{a} U^{b}$ and $j^{a} = σ U^{a}$ , with $ρ, σ \in C^{1}$ , $ρ \geq 0$ , and future-timelike $U$ satisfying $g (U, U) = - c^{2}$ .

[A1]

The adopted background includes only ${A C}_{ω}$ as stated in The Axiom of Countable Choice ( ${A C}_{ω}$ ). The explicit global coordinates on $R \times D$ and every finite component calculation below use no stronger choice principle.

Proof

1.1givenF2F5F6F13A1algebra

Since $H$ is smooth and positive, its reciprocal is smooth by Positive-scalar conformastatic chart data; $ϵ_{0} > 0$ and $G_{N} > 0$ make the positive root $Q_{s}$ well-defined by [F2]. The chart tensor is a smooth Lorentzian metric with the stated orientation by [F5] and [F6]. Thus $U = c n$ is future timelike, and the formulas define smooth $A, F, U, ρ, σ, j, T_{m}$ ; they have the charged-dust form [F13], and $Δ H \leq 0$ gives $ρ \geq 0$ .

1.2givenF1F6F9F10F11algebra

Write $f = H^{- 1}$ . Differentiating $H f = 1$ in each spatial coordinate and using [F9] gives $\partial_{i} f = - H_{i} / H^{2}$ . The coordinate formula [F10] therefore gives the only nonzero covariant components $F_{i 0} = - s (c / Q_{s}) H_{i} / H^{2}$ and $F_{0 i} = s (c / Q_{s}) H_{i} / H^{2}$ . Raising indices with [F6] gives $F^{i 0} = s (c / Q_{s}) H_{i} / H^{2}$ and $F^{0 i} = - F^{i 0}$ ; all $F^{i j}$ vanish. Finally [F11] gives $d F = d (d A) = 0$ .

1.3F7F8

The connection used in the formal equations is the unique torsion-free metric-compatible connection by [F8]. The coefficients and compatibility of the local connection in [F7] were established in that item, so the two connections coincide; hence its component table may be used in the sourced equations below.

2.1givenF7F8step 1.2algebra

By the tensor derivative formula [F8], $\nabla_{a} F^{a b} = \partial_{a} F^{a b} + Γ^{a}_{a c} F^{c b} + Γ^{b}_{a c} F^{a c}$ , and the final term is zero because $Γ^{b}_{a c}$ is symmetric in $a, c$ while $F^{a c}$ is antisymmetric. From [F7], $Γ^{a}_{a i} = 2 H_{i} / H$ and $Γ^{a}_{a 0} = 0$ . Hence $\nabla_{a} F^{a 0} = \partial_{i} (s \frac{c}{Q_{s}} \frac{H_{i}}{H^{2}}) + \frac{2 H_{i}}{H} s \frac{c}{Q_{s}} \frac{H_{i}}{H^{2}} = s \frac{c}{Q_{s}} \frac{Δ H}{H^{2}}$ , because the two terms proportional to $q / H^{3}$ cancel.

2.2givenF6F7F8F13step 1.2algebra

The field $U = c H \partial_{0}$ has $g (U, U) = - c^{2}$ by [F13]. Its divergence is $\nabla_{a} (ρ U^{a}) = \partial_{0} (ρ c H) + Γ^{a}_{a 0} ρ c H = 0$ , since all fields are static. Its acceleration is $a^{i} = (c H)^{2} Γ^{i}_{00} = - c^{2} H_{i} / H^{3}$ and $a^{0} = 0$ . Moreover $F^{i}_{0} = g^{i j} F_{j 0} = - s (c / Q_{s}) H_{i} / H^{4}$ , hence $σ F^{i}_{c} U^{c} = s Q_{s} ρ F^{i}_{0} c H = - ρ c^{2} H_{i} / H^{3} = ρ a^{i}$ ; the temporal components of both sides are zero. This undivided relation holds at every point, including $ρ = 0$ .

2.3givenF2F6F12step 1.2algebra

In the positively oriented orthonormal frame $n = H \partial_{0}$ , $e_{i} = H^{- 1} \partial_{i}$ from [F6], the electric components are $E_{i} = c F (e_{i}, n) = - s c^{2} H_{i} / (Q_{s} H^{2})$ and the magnetic components vanish because $F_{i j} = 0$ . In that frame $F_{\hat{0} \hat{i}} = - E_{i} / c$ , $F_{c d} F^{c d} = - 2 ∣ E ∣^{2} / c^{2}$ , $F^{\hat{0} c} F^{\hat{0}}_{c} = ∣ E ∣^{2} / c^{2}$ , and $F^{\hat{i} c} F^{\hat{j}}_{c} = - E_{i} E_{j} / c^{2}$ . Substitution into the stress-tensor formula [F12] yields $T_{E M}^{\hat{0} \hat{0}} = \frac{ϵ_{0} ∣ E ∣^{2}}{2} = \frac{ϵ_{0} c^{4} q}{2 Q_{s}^{2} H^{4}}$ , $T_{E M}^{\hat{0} \hat{i}} = 0$ , and $T_{E M}^{\hat{i} \hat{j}} = \frac{ϵ_{0} c^{4}}{2 Q_{s}^{2} H^{4}} (δ_{i j} q - 2 H_{i} H_{j})$ . Using $κ = 8 π G_{N} / c^{4}$ and $Q_{s}^{2} = 4 π ϵ_{0} G_{N}$ gives $κ T_{E M}^{\hat{0} \hat{0}} = q / H^{4}$ and $κ T_{E M}^{\hat{i} \hat{j}} = (δ_{i j} q - 2 H_{i} H_{j}) / H^{4}$ .

3.1givenF1F2F13step 2.1algebra

The right side of the time component of the sourced equation [F1] is $- μ_{0} j^{0} = - μ_{0} s Q_{s} ρ c H = s μ_{0} Q_{s} c^{3} Δ H / (4 π G_{N} H^{2})$ by the definitions in the Statement. The definitions give $Q_{s}^{2} = 4 π G_{N} / (μ_{0} c^{2})$ by [F2], so $μ_{0} Q_{s} c^{3} / (4 π G_{N}) = c / Q_{s}$ ; this agrees with step 2.1. For each spatial $i$ , $F^{0 i}$ is static, $F^{j i} = 0$ , $Γ^{a}_{a 0} = 0$ , and the symmetric-connection contraction is zero, so $\nabla_{a} F^{a i} = 0 = - μ_{0} j^{i}$ . Thus all sourced Maxwell components hold.

3.2givenF1F3F8F13step 2.2algebra

On the open set ${ρ > 0}$ , step 2.2 verifies the two equivalent conditions in [F3], so the matter exchange equation [F1] follows there. At a point with $ρ = 0$ , the already verified divergence in step 2.2 makes $U^{b} \nabla_{a} (ρ U^{a}) = 0$ , while $ρ a^{b} = 0$ and $σ F^{b}_{c} U^{c} = 0$ because $σ = s Q_{s} ρ = 0$ by [F13]. The product rule $\nabla_{a} (ρ U^{a} U^{b}) = U^{b} \nabla_{a} (ρ U^{a}) + ρ a^{b}$ therefore proves the exchange equation on the zero-density set too.

3.3givenF1F7F13step 2.3algebra

In the same frame, $T_{m}^{\hat{0} \hat{0}} = ρ c^{2}$ and $T_{m}^{\hat{0} \hat{i}} = T_{m}^{\hat{i} \hat{j}} = 0$ by [F13], so $κ T_{m}^{\hat{0} \hat{0}} = 8 π G_{N} ρ / c^{2} = - 2 Δ H / H^{3}$ by the density in the Statement. Consequently the total components are $κ (T_{m} + T_{E M})^{\hat{0} \hat{0}} = (q - 2 H Δ H) / H^{4}$ and $κ (T_{m} + T_{E M})^{\hat{i} \hat{j}} = (δ_{i j} q - 2 H_{i} H_{j}) / H^{4}$ , with zero mixed components. In an orthonormal frame these numerical values equal the corresponding covariant components (the two time-index signs cancel), and they are exactly $G (n, n)$ , $G (e_{i}, e_{j})$ and $G (n, e_{i})$ from [F7]. Thus the formal Einstein equation [F1] holds componentwise.

4.1givenF4F12F13step 1.1step 3.3∎

Since $ρ \geq 0$ , [F4] applies to the Maxwell and dust tensors [F12, F13] separately and states that their sum satisfies WEC, NEC, SEC, and DEC. This is a conditional mathematical consequence of the formal model equations and the pointwise tensor definitions; it is not an empirical claim.

Remarks

The calculation checks every component of the formal Maxwell, dust-exchange, and $Λ = 0$ Einstein equations. At $Δ H = 0$ , $ρ = σ = 0$ and the formulas reduce to the source-free Maxwell branch without division by the density. The result does not supply PDE existence, stability, or a physical postulate.

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Globally regular sphere, proper integrals and dilute bounds

Statement

Assume ${A C}_{ω}$ as stated in The Axiom of Countable Choice ( ${A C}_{ω}$ ). Let $a, b > 0$ have length units, let $s \in {- 1, + 1}$ , and on all of $R^{3}$ set $r^{2} = (x^{1})^{2} + (x^{2})^{2} + (x^{3})^{2}, Ψ (x) : = \sqrt{r^{2} + b^{2}} > 0, H (x) : = 1 + \frac{a}{Ψ (x)} .$ Use the constants and exact formal coupled family of Exact smooth superharmonic counterpoised dust family. The function $H$ is smooth, positive, and satisfies $Δ H = - \frac{3 a b^{2}}{(r^{2} + b^{2})^{5 / 2}} < 0, ρ (x) = \frac{3 c^{2} a b^{2}}{4 π G_{N} H (x)^{3} (r^{2} + b^{2})^{5 / 2}} .$ On the slice $x^{0} = constant$ the induced spatial metric is $h = H^{2} δ_{i j} d x^{i} d x^{j}$ . Define the candidate proper mass and signed charge by $M_{p r o p} : = \int_{R^{3}} ρ H^{3} d λ_{3}, Q_{p r o p} : = \int_{R^{3}} σ H^{3} d λ_{3} .$ The proof establishes that the first is a finite nonnegative Lebesgue integral and the second is a finite signed Lebesgue integral. They equal $M_{p r o p} = \frac{c^{2} a}{G_{N}}, Q_{p r o p} = s Q_{s} \frac{c^{2} a}{G_{N}} .$ The scalar curvature is $S_{g} = - 2 Δ H / H^{3} > 0$ , the total stress satisfies WEC, NEC, SEC, and DEC, and the smooth matter density is positive at every point and has infinite spatial extent. For $a = δ b$ with $0 < δ \leq 1$ , put $h_{0} (x) : = b / \sqrt{r^{2} + b^{2}}$ and $ρ_{l i n} (x) : = \frac{3 c^{2} δ b^{3}}{4 π G_{N} (r^{2} + b^{2})^{5 / 2}} .$ The same exact coupled solutions obey the pointwise Cartesian-coordinate bounds $∣ g_{00} + (1 - 2 δ h_{0}) ∣ \leq 3 δ^{2}, ∣ g_{i j} - (1 + 2 δ h_{0}) δ_{i j} ∣ \leq δ^{2}, 0 \leq 1 - \frac{ρ}{ρ_{l i n}} \leq 3 δ .$ These are ideal mathematical estimates, not measurements or stability statements.

Facts & Assumptions

Given: $a, b > 0$ , $s \in {- 1, + 1}$ , ${A C}_{ω}$ , the positive-root convention, the smooth family, and the Lebesgue measure and integral definitions.

[F1]

Explicit rational and real power regimes defines the unique positive square root and reciprocal integer powers for positive bases.

[F2]

The positive square-root branch is smooth: for $q > 0$ , the unique positive root $s (q)$ is smooth and $s^{'} (q) = 1 / (2 s (q))$ .

[F3]

Exact smooth superharmonic counterpoised dust family supplies the exact smooth formal solution family and the density $ρ = - c^{2} Δ H / (4 π G_{N} H^{3})$ for smooth positive $H$ with $Δ H \leq 0$ .

[F4]

EM, dust and perfect-fluid energy inequalities: the Maxwell tensor plus any nonnegative charged-dust tensor satisfies WEC, NEC, SEC, and DEC.

[F5]

Arbitrary-positive-H conformastatic Ricci and Einstein components gives $S_{g} = - 2 Δ H / H^{3}$ and the arbitrary-positive- $H$ Einstein table.

[F6]

Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ defines $λ_{3}$ , and The nonnegative Lebesgue integral defines the integral of a nonnegative measurable function as the supremum of its simple-function integrals.

[F7]

Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma: under ${A C}_{ω}$ , for every nonnegative Borel $f$ on $R^{3}$ , $\int_{R^{3}} f d λ_{3} = \int_{0}^{\infty} r^{2} \int_{S^{2}} f (r ω) d ς_{2} (ω) d r,$ where $ς_{2}$ is the polar surface measure of The polar surface set function on the unit sphere. The symbol $ς_{2}$ is used here to distinguish surface measure from the charge density $σ$ .

[F8]

Sphere integrals in angle coordinates: under ${A C}_{ω}$ , $ς_{2} (S^{2}) = 4 π$ .

[F9]

Continuous functions on Euclidean spaces are Borel measurable and Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable: under ${A C}_{ω}$ , continuous Euclidean functions are Borel and Borel sets are Lebesgue measurable; in particular the nonnegative integrand below is measurable for both suppliers.

[F10]

A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions applies under ${A C}_{ω}$ to a $C^{1}$ diffeomorphism of open Euclidean sets and transforms a nonnegative Lebesgue integral by the absolute Jacobian. Every singleton in $R$ is Lebesgue null by Every affine hyperplane of $R^{n}$ , and hence every proper linear subspace, is Lebesgue null.

[F11]

Integrable real and complex functions, and their integrals defines a real signed integral by positive and negative parts and says integrability is equivalent to finiteness of $\int ∣ f ∣$ ; The Lebesgue integral is linear on $L^{1} (μ)$ gives linearity on integrable functions.

[F12]

Positive-base powers have the stated derivative; the one-variable product, quotient and chain rules apply; sine and cosine have the derivatives, signs, ranges, Pythagorean identity and quarter-turn values used below. A continuous function on an interval assumes intermediate values, and a one-dimensional C1 map with nonzero derivative has local C1 inverses. (Continuity and derivatives of positive-base real powers, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , The derivatives of sine and cosine are cosine and minus sine, Quarter-turn values and shifts by pi/2 and pi, Signs, monotonicity intervals, and ranges of sine and cosine, Parity and the Pythagorean identity for sine and cosine, The Euclidean inverse function theorem, Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ )

[F13]

Continuous functions are Riemann integrable on a compact interval; their bounded Riemann and Lebesgue integrals agree under ${A C}_{ω}$ ; and Newton--Leibniz evaluates an integral from a continuous antiderivative with an integrable interior derivative. (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral, Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative)

[F14]

For Riemann-integrable real functions on $[u, v]$ with $u < v$ , pointwise nonnegativity implies a nonnegative integral, pointwise order is preserved by integration, and a bound $m \leq h \leq M$ implies $m (v - u) \leq \int_{u}^{v} h \leq M (v - u)$ (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[A1]

Countable choice is limited to the indexed Axiom ${A C}_{ω}$ of The Axiom of Countable Choice ( ${A C}_{ω}$ ); no full Axiom of Choice is assumed.

Proof

1.1givenA1F1F2F3F9F12algebra

Since $r^{2} + b^{2} \geq b^{2} > 0$ , [F1] gives the unique positive root $Ψ$ , and [F2] makes it smooth with $\partial_{i} Ψ = x^{i} / Ψ$ . The chain, product and quotient rules [F12] then give $H_{i} = - a x^{i} / Ψ^{3}$ and $\partial_{i} H_{i} = - a (Ψ^{- 3} - 3 (x^{i})^{2} Ψ^{- 5})$ , so $Δ H = - a \sum_{i = 1}^{3} (Ψ^{- 3} - 3 (x^{i})^{2} Ψ^{- 5}) = - a (3 Ψ^{- 3} - 3 r^{2} Ψ^{- 5}) = - 3 a b^{2} / Ψ^{5} < 0$ . The integrand $ρ H^{3} = 3 c^{2} a b^{2} / [4 π G_{N} (r^{2} + b^{2})^{5 / 2}]$ is nonnegative and continuous on $R^{3}$ , hence Borel and Lebesgue measurable by [F9]. The coupled-dust theorem [F3] applies and supplies the smooth fields and exact formal solution.

2.1givenF6step 1.1algebra

On $x^{0} = constant$ the metric pullback is $h_{i j} = H^{2} δ_{i j}$ , so $\det h = H^{6}$ and its positive volume density is $\sqrt{\det h} = H^{3}$ . Substituting the density from step 1.1 gives $ρ H^{3} = 3 c^{2} a b^{2} / [4 π G_{N} (r^{2} + b^{2})^{5 / 2}]$ , a nonnegative Borel integrand for [F6].

2.2step 1.1F4F5algebra

The curvature formula [F5] and strict negativity in step 1.1 give $S_{g} = 6 a b^{2} / (H^{3} Ψ^{5}) > 0$ . The energy-condition theorem [F4] applies because $ρ \geq 0$ . Since the density formula in step 1.1 is strictly positive for every $x \in R^{3}$ , the set of points where it is nonzero is all of $R^{3}$ ; no compact subset of $R^{3}$ contains all those points.

2.3givenF1F12F13F14step 1.1algebra

If $a = δ b$ , then $H = 1 + z$ with $z = δ h_{0}$ and $0 < h_{0} \leq 1$ , since $Ψ \geq b$ . For $f (u) = (1 + u)^{- 2}$ , $f (0) = 1$ , $f^{'} (0) = - 2$ , and $0 < f^{''} (u) = 6 (1 + u)^{- 4} \leq 6$ for $u \geq 0$ by [F12]. Fix $z > 0$ and define $G_{z} (u) : = (z - u) f^{'} (u) + f (u)$ on $[0, z]$ . The product rule [F12] gives $G_{z}^{'} (u) = (z - u) f^{''} (u)$ on its interior. This derivative is continuous on $[0, z]$ , hence Riemann integrable by [F13]; $G_{z}$ is continuous there. Newton--Leibniz [F13] therefore gives $\int_{0}^{z} (z - u) f^{''} (u) d u = G_{z} (z) - G_{z} (0) = f (z) - 1 + 2 z .$ The integrand lies between $0$ and $6 (z - u)$ . These continuous comparison functions are integrable by [F13], so [F14] and Newton--Leibniz applied to $6 z u - 3 u^{2}$ give $0 \leq f (z) - (1 - 2 z) \leq \int_{0}^{z} 6 (z - u) d u = 3 z^{2} \leq 3 δ^{2} .$ At $z = 0$ the same remainder and bound are zero directly; no theorem requiring a nondegenerate interval is applied there. Since $g_{00} = - H^{- 2}$ , this proves the stated time-component bound.

3.1givenA1F6F7F8step 2.1algebra

Apply the polar formula [F7] to the integrand in step 2.1; it is constant on each sphere of radius $r$ . The area factor [F8] then gives $M_{p r o p} = (3 c^{2} a b^{2} / G_{N}) \int_{0}^{\infty} r^{2} (r^{2} + b^{2})^{- 5 / 2} d r$ . All uses of the polar formula are under exactly [A1]'s ${A C}_{ω}$ hypothesis.

4.1givenA1F10F12F13step 3.1algebra

The radial integrand has finite value at $r = 0$ , and that singleton has measure zero by [F10], so its integral over $[0, \infty)$ equals its integral over $(0, \infty)$ . On $(0, π / 2)$ , $\cos θ > 0$ , and quotient differentiation with the sine/cosine derivatives and Pythagorean identity in [F12] gives $(b \tan θ)^{'} = b \sec^{2} θ > 0$ . Also $\sin$ increases and $\cos$ decreases while remaining positive on this interval [F12], so $\tan = \sin / \cos$ is strictly increasing. Their continuity and quarter-turn values give limits $0$ and $+ \infty$ at the interval's endpoints. For each $y > 0$ , choose interior points $θ_{-} < θ_{+}$ with $T (θ_{-}) < y < T (θ_{+})$ ; continuity on $[θ_{-}, θ_{+}]$ and the intermediate value theorem [F12] give a preimage, and strict monotonicity gives uniqueness. Hence $T (θ) = b \tan θ$ is a bijection $(0, π / 2) \to (0, \infty)$ . Since $T^{'}$ never vanishes, the inverse function theorem [F12] gives C1 local inverses, which agree by bijectivity to form a global C1 inverse. Thus [F10] applies, and using $\cos θ > 0$ to simplify gives $\int_{0}^{\infty} r^{2} (r^{2} + b^{2})^{- 5 / 2} d r = b^{- 2} \int_{0}^{π / 2} \sin^{2} θ \cos θ d θ$ . The integrand is continuous on $[0, π / 2]$ , so it is Riemann integrable and its Lebesgue integral agrees by [F13]; the endpoints have measure zero by [F10]. The function $G (θ) = \sin^{3} θ / 3$ has derivative $\sin^{2} θ \cos θ$ on the interior by [F12], and Newton--Leibniz [F13] with $\sin 0 = 0$ , $\sin (π / 2) = 1$ yields $1 / 3$ . Thus the radial integral is $1 / (3 b^{2})$ , and step 3.1 gives $M_{p r o p} = c^{2} a / G_{N} < \infty$ .

5.1step 4.1F3F11algebra

Put $f = ρ H^{3}$ . By step 4.1, $f \geq 0$ and $\int f = M_{p r o p} < \infty$ , so $f$ is real integrable under [F11]. Since $σ H^{3} = s Q_{s} f$ and $s Q_{s}$ is a finite real constant, [F11] gives integrability of the signed charge density and $Q_{p r o p} = \int_{R^{3}} σ H^{3} d λ_{3} = s Q_{s} \int_{R^{3}} f d λ_{3} = s Q_{s} M_{p r o p}$ .

6.1givenF1F12F13F14step 2.3algebra∎

The spatial metric is exactly $g_{i j} = H^{2} δ_{i j} = (1 + 2 δ h_{0} + δ^{2} h_{0}^{2}) δ_{i j}$ , so the component error is $δ^{2} h_{0}^{2} ∣ δ_{i j} ∣ \leq δ^{2}$ . Also $ρ / ρ_{l i n} = H^{- 3} = (1 + z)^{- 3}$ . By [F12], the function $P (u) : = 1 - (1 + u)^{- 3}$ is continuously differentiable for $u \geq 0$ with $P^{'} (u) = 3 (1 + u)^{- 4}$ . For $z > 0$ , [F13] makes this derivative integrable on $[0, z]$ and gives $1 - (1 + z)^{- 3} = P (z) - P (0) = \int_{0}^{z} 3 (1 + u)^{- 4} d u .$ Its integrand lies between $0$ and $3$ , so [F14] yields $0 \leq 1 - ρ / ρ_{l i n} \leq 3 z \leq 3 δ$ . At $z = 0$ both sides vanish by substitution. This proves the density bound.

Remarks

The masses here are the integrals defined from the induced spatial volume density; no ADM-mass identification is used. The density's nonzero set is all of $R^{3}$ , and the family has a fixed signed charge-to-mass ratio. The dilute estimates compare exact coupled solutions with explicitly stated linear profiles; they do not assert stability or a general perturbation theorem.

5 · Examples, experiments and counterexamples

None yet.

Arbitrary-positive-H conformastatic Ricci and Einstein components

1 · Prerequisites

2 · Summary

Scope

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

Positive-scalar conformastatic chart data

Definition

Remarks

The positive-scalar conformastatic chart tensor is Lorentzian

Statement

Facts & Assumptions

Proof

Remarks

Arbitrary-positive-H conformastatic Ricci and Einstein components

Statement

Facts & Assumptions

Proof

Remarks

Exact smooth superharmonic counterpoised dust family

Statement

Facts & Assumptions

Proof

Remarks

Globally regular sphere, proper integrals and dilute bounds

Statement

Facts & Assumptions

Proof

Remarks

5 · Examples, experiments and counterexamples

Sources