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An ideal PEC boundary condition is not preserved by a vacuum duality quarter-turn

Physical scope

A smooth source-free vacuum field in a half-space with an ideal perfect-electric-conductor boundary condition imposed as model data. The calculation concerns that boundary condition only; it makes no claim about finite conductivity or measurements.

Example

On $Ω = {(x_{1}, x_{2}, x_{3}) : x_{3} > 0}$ , consider the constant vacuum field $E = 0$ and $B = B_{0} (1, 0, 0)$ , where $B_{0} \neq 0$ has units T. Let $Π = {x_{3} = 0}$ , with unit normal $n = (0, 0, 1)$ . The constant fields are understood with their continuous traces on $Π$ . In this example, a perfect-electric-conductor (PEC) boundary means the ideal boundary model imposing $E_{∥} = E - (E \cdot n) n = 0$ on $Π$ ; this condition is imposed as model data and no finite-conductivity claim is made. For $θ = π / 2$ , the vacuum duality rotation $F_{θ} = ⋆_{η} F$ has electric field $E_{θ} = - c B_{0} (1, 0, 0)$ , which is nonzero and parallel to $Π$ . Thus the source-free vacuum duality rotation does not preserve this same PEC boundary condition.

Example references

Vacuum duality and its domain

Excitation and free-current forms

Lorentzian Hodge star and periods

Framework-local SI units, the vacuum constants and the exact/measured status of their values

The microscopic vacuum Maxwell equations: smooth and weak branches

Facts & Assumptions

Given: The half-space, plane, normal, constant field, ideal PEC boundary condition, and quarter-turn angle in the Example.

[A1]

The PEC condition used here is exactly E_parallel = E - (E dot n)n = 0 on Pi. It is an ideal boundary assumption, not a material constitutive law.

[F1]

Excitation and free-current forms defines $F = B - d t \land E$ and fixes $F_{0 i} = - E_{i} / c$ .

[F2]

Vacuum duality and its domain states that the quarter-turn $F \mapsto ⋆_{η} F$ maps a source-free vacuum solution to another source-free vacuum solution.

[F3]

Lorentzian Hodge star and periods gives $⋆_{η} (d x^{2} \land d x^{3}) = d x^{0} \land d x^{1}$ in the fixed $(- + + +)$ orientation.

[F4]

Framework-local SI units, the vacuum constants and the exact/measured status of their values gives $x^{0} = c t$ and $c > 0$ in SI units.

[F5]

The microscopic vacuum Maxwell equations: smooth and weak branches adopts the four smooth vacuum Maxwell equations and excludes magnetic-source terms.

Verification

1.1A1F5givenalgebra

The fields are constant, so all their spatial and time derivatives vanish. With $P = M = 0$ and $ρ = J = 0$ , they satisfy the four vacuum Maxwell equations on $Ω$ by [F5]. The original tangential electric field is $E_{∥} = 0$ , so it satisfies the imposed PEC condition [A1].

2.1A1F1F2F3F4givenstep 1.1algebra∎

The field form is $F = B_{0} d x^{2} \land d x^{3}$ . By [F3], $⋆_{η} F = B_{0} d x^{0} \land d x^{1} = c B_{0} d t \land d x^{1}$ . Comparing with the defining form $B_{θ} - d t \land E_{θ}$ in [F1] gives $E_{θ} = - c B_{0} (1, 0, 0)$ . Since $E_{θ} \cdot n = 0$ , its tangential component equals $E_{θ}$ and is nonzero. Hence the rotated vacuum solution from [F2] fails the same PEC boundary condition. The unit $c B_{0}$ is V m $^{- 1}$ , as required for an electric field by [F4].

An ideal PEC boundary condition is not preserved by a vacuum duality quarter-turn

Example

Example references

Facts & Assumptions

Verification

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