Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)
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Periodic smooth Gauss excludes nonzero net charge

Statement refuted

Assume Countable Choice, L,ϵ0>0, and the periodic Coulomb-gauge Gauss equation of Gauss multiplier and smooth periodic Coulomb solver. For every smooth periodic charge density ρ on TL3, there exists a smooth periodic potential ϕ satisfying −Δϕ=ρ/ϵ0.

Construction

Assume Countable Choice and the periodic Gauss constraint of Gauss multiplier and smooth periodic Coulomb solver. On TL3, take the nonzero constant smooth charge density ρ(x)=ρ0 with ρ0≠0. There is no smooth periodic potential satisfying −Δϕ=ρ0/ϵ0. More generally, every smooth periodic solution of Gauss's equation in Coulomb gauge requires zero total charge. A boundary, puncture, or compensating background changes the hypotheses and is outside this counterexample.

Facts & Assumptions

Given: Countable Choice; L,ϵ0>0; a smooth periodic potential; and the Coulomb-gauge Gauss equation.

[A1]
[F1]

In Coulomb gauge the periodic Gauss constraint is −Δϕ=ρ/ϵ0. Gauss multiplier and smooth periodic Coulomb solver

Counterexample

Given: [A1], [F1], ρ0≠0, and periodic boundary conditions.

1.1F1givenalgebra

Integrating [F1] over [0,L]3 gives ∫[0,L]3−Δϕ d3x=−∑j=13∫0L ⁣∫[0,L]2∂j2ϕ d2xj^ dxj=0, because for each j the integral in xj is the difference of the periodic values of ∂jϕ at L and 0.

2.1A1F1step 1.1given∎

The right side of the same integrated equation is ϵ0−1∫[0,L]3ρ0 d3x=ϵ0−1ρ0L3≠0, contradicting step 1.1. Thus no smooth periodic solution exists for this nonzero total charge; the same integration proves neutrality is necessary for any periodic solution.

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