Physics theoremStatement: AI-adaptedProof: AI-generatedverified 2026-10-22 (gpt-6.1-sol)

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Vacuum duality and its domain

Physical scope

Smooth source-free vacuum Maxwell fields in the fixed flat oriented SI laboratory chart, with the Lorentzian Hodge star of signature (-+++). The duality does not assert invariance of a material constitutive law or a boundary condition, and does not introduce magnetic charge.

Statement

On a region of the fixed oriented Minkowski laboratory chart with $P = M = 0$ and free-current form $J = 0$ , let $F$ be a $C^{1}$ electromagnetic field-strength 2-form and let $θ$ be a constant real number. Then

$F_{θ} = \cos (θ) F + \sin (θ) ⋆_{η} F$

again satisfies $d F_{θ} = 0$ and $d ⋆_{η} F_{θ} = 0$ . This is a symmetry of the source-free vacuum equations only. With nonzero free-current form $J$ , a rotation with $\sin (θ) \neq 0$ generally fails the same no-magnetic-source equation; no duality invariance is asserted for a material closure or boundary condition without a separate compatibility proof. The companion conductor example gives a concrete boundary failure. No magnetic charge sector is adopted.

Statement references

Metric-free Maxwell equations and metric-dependent vacuum closure

Lorentzian Hodge star and periods

Lorentzian duality sign

The microscopic vacuum Maxwell equations: smooth and weak branches

Facts & Assumptions

Given: A C^1 vacuum field-strength form F, a constant theta, and the fixed orientation, signature, and Hodge star of the Statement.

[A1]

The microscopic vacuum Maxwell equations: smooth and weak branches adopts the smooth Maxwell equations without magnetic-source terms.

[F1]

Metric-free Maxwell equations and metric-dependent vacuum closure gives $d F = 0$ and $d G = J$ in vacuum, with $G = (μ_{0} c)^{- 1} ⋆_{η} F$ ; the source-free case has $J = 0$ .

[F2]

Lorentzian Hodge star and periods defines $⋆_{η}$ as a linear map on 2-forms for orientation $d x^{0} \land d x^{1} \land d x^{2} \land d x^{3}$ and signature $(- + + +)$ .

[F3]

Lorentzian duality sign gives $⋆_{η}^{2} = - id$ on 2-forms.

Proof

Given: The field and assumptions in [A1] and [F1]–[F3].

1.1A1F1F2F3givenalgebra

By linearity of $d$ and [F1], $d F_{θ} = \cos (θ) d F + \sin (θ) d ⋆_{η} F = 0$ . By [F2] and [F3], $⋆_{η} F_{θ} = \cos (θ) ⋆_{η} F - \sin (θ) F$ , so $d ⋆_{η} F_{θ} = \cos (θ) d ⋆_{η} F - \sin (θ) d F = 0$ . Thus the rotated form solves both source-free vacuum equations.

2.1A1F1givenalgebra∎

In a vacuum region with nonzero free-current form $J$ , [F1] gives $d ⋆_{η} F = μ_{0} c J$ . Therefore $d F_{θ} = \sin (θ) μ_{0} c J$ , which is nonzero whenever $\sin (θ) J$ is nonzero. It then fails the adopted equation $d F_{θ} = 0$ ; restoring a duality symmetry would require changing the source sector, which this framework does not adopt. The theorem makes no claim that any material closure or boundary condition is preserved.

Vacuum duality and its domain

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources