ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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.

Coulomb field and geometric charge scale

Example

On the exterior region $r > 0$ of oriented Minkowski four-space, use length coordinates $x^{0} = c t$ and radial coordinate $r$ in metres. For a signed charge parameter $Q_{e}$ in coulombs, consider the smooth Coulomb two-form whose only independent spherical component is

$F_{0 r} = - \frac{Q_{e}}{4 π ε_{0} c r^{2}} .$

The singular source point $r = 0$ is outside the domain; this example records no measurement or distributional source.

Facts & Assumptions

Given: The positive conversion parameters and $F = α F$ of SI and geometrized Einstein–Maxwell parameters, the Lorentzian coordinate convention of Oriented Lorentzian fields, observers and coordinate conventions, and the exact geometric electrovac normalization of Exact geometric electrovac normalization. The inherited geometric setup carries ${A C}_{ω}$ as in [A1].

[F1]

The definitions give $α = \sqrt{κ / (2 μ_{0})}$ , $κ = 8 π G / c^{4}$ , and $ε_{0} μ_{0} c^{2} = 1$ . (SI and geometrized Einstein–Maxwell parameters)

[F2]

The same two-form is rescaled by $F = α F$ in the exact geometric Einstein normalization. (Exact geometric electrovac normalization)

[A1]

The inherited choice assumption is Countable Choice ${A C}_{ω}$ , not full Choice. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

Verification

1.1givenF1A1algebra

From [F1], $α^{2} = 4 π G / (μ_{0} c^{4}) = 4 π G ε_{0} / c^{2}$ . Since $α > 0$ , this gives $α = \sqrt{4 π G ε_{0}} / c$ and hence $α / (4 π ε_{0} c) = \sqrt{G / (4 π ε_{0} c^{4})}$ ; the positivity assumptions justify taking the positive square root.

2.1givenF2step 1.1algebra∎

Multiplying the Coulomb component by $α$ and using the equality in step 1.1 yields $F_{0 r} = α F_{0 r} = - \sqrt{G / (4 π ε_{0} c^{4})} Q_{e} / r^{2} = - q_{g e o} / r^{2}$ . By [F2], this is the geometric charge coefficient in the exact normalized Einstein equation. The equality is valid only on $r > 0$ and introduces no point-source solution at the excluded origin.

Coulomb field and geometric charge scale

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources