Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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FALSE: a field with vanishing divergence has zero outward flux through the boundary of every solid it surrounds

Statement

False claim: if a vector field has vanishing divergence wherever it is defined, then its outward flux through the boundary of every solid it surrounds is zero.

The claim looks like the divergence-free corollary on the A page, but it quietly weakens the hypothesis. The proved corollary requires the field to be C1 on an open set containing the whole solid, not merely away from a singularity inside it.

Facts & Assumptions

Given: The inverse-square field F(x,y,z)=(x,y,z)/(x2+y2+z2)3/2 on R3∖{0}.

[L1]

The outward flux of this field through the sphere of radius R centred at the origin is 4π (The outward flux of the inverse-square field through a sphere centred at the origin is 4π).

[L2]

The divergence of this field is zero at every point where it is defined (The inverse-square field is divergence free, and its flux through the sphere bounding the translated unit ball vanishes).

[L3]

If a finite gluing of elementary solid regions is given and a C1 field on an open set containing its union has vanishing divergence, then its outward boundary flux is zero (A field with vanishing divergence has zero outward flux through the boundary of a glued elementary solid).

[F1]

The divergence of a C1 field on an open subset of R3 is the sum of its coordinate partial derivatives (Divergence and curl of a C1 vector field).

[F2]

Flux is computed against the oriented area vector of a patch (Unit normal fields, orientations, and flux through a regular surface patch).

[F3]

For a finite patch presentation, total flux is the sum of the patch fluxes (Finitely patched regular surfaces, their area, scalar integrals, and flux).

Refutation

technique · direct
1.1

By [L2] and [F1], the witness field has vanishing divergence at every point where it is defined.

givenL2F1
1.2

By [L1], its outward flux through any sphere centred at the origin is 4π, so in particular it is not zero.

L1F2F3
2.1

Steps 1.1 and 1.2 contradict the claim, so the claim is false.

step 1.1step 1.2
3.1

What fails is not the divergence theorem or the corollary [L3], but the weakened hypothesis: the field is not C1 on any open set containing the solid bounded by a sphere centred at the origin.

step 2.1L3F1
4.1

The same field on a sphere whose enclosed ball misses the origin does satisfy the corollary and has zero flux there, exactly as The inverse-square field is divergence free, and its flux through the sphere bounding the translated unit ball vanishes records.

step 3.1L2L3∎

Remarks

  • The example separates two different statements that are often conflated: vanishing divergence on the punctured domain, and the existence of an open neighbourhood of the solid on which the field is C1.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources