Physics theoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Reciprocity in reciprocal linear media

Physical scope

Conditional fixed-frequency reciprocity identity in the local linear macroscopic phasor model on a bounded C1 Euclidean domain, for C1 phasor solutions, continuous sources and pointwise transpose-symmetric possibly complex constitutive tensors. The result requires no passivity or losslessness and asserts no field existence, uniqueness, causal material response or empirical validity.

Statement

Assume ${A C}_{ω}$ . Let $Ω \subset R^{3}$ be a bounded domain with $C^{1}$ boundary and outward unit normal $n$ . Fix a real angular frequency $ω$ in $s^{- 1}$ . Let $E_{j}, H_{j} \in C^{1} (\overline{Ω}; C^{3})$ and $J_{j} \in C^{0} (\overline{Ω}; C^{3})$ , for $j = 1, 2$ , be phasor fields and electric-current densities in $V m^{- 1}$ , $A m^{- 1}$ , and $A m^{- 2}$ , respectively. Suppose both pairs solve the fixed-frequency Maxwell equations there with the same continuous coefficient maps $ε, μ : \overline{Ω} \to C^{3 \times 3}$ and $ε (x)^{T} = ε (x)$ , $μ (x)^{T} = μ (x)$ at every $x$ . Use Euclidean volume measure $d V$ and boundary area measure $d S$ . Then

$\int_{\partial Ω} n \cdot (E_{1} \times H_{2} - E_{2} \times H_{1}) d S = \int_{Ω} (E_{2} \cdot J_{1} - E_{1} \cdot J_{2}) d V .$

The dot and cross products in this identity are the complex-bilinear products defined in the reciprocal phasor model; no complex conjugation occurs. Both integrals have units W. In particular, the real symmetric positive permittivity tensor and positive scalar permeability of the lossless anisotropic plane-wave model satisfy these reciprocal-coefficient hypotheses, so the algebraic identity applies to any two such phasor solutions restricted to a bounded $C^{1}$ subdomain. Physical use of that anisotropic branch is restricted to its declared finite positive-frequency band $I_{ω} = [ω_{-}, ω_{+}]$ , $0 < ω_{-} < ω_{+} < \infty$ ; the reciprocity identity itself asserts no broadband material response.

Statement references

Fixed-frequency reciprocal tensor medium

Anisotropic Maxwell symbol

The Axiom of Countable Choice ( ${A C}_{ω}$ )

The divergence and curl of a cross product

Divergence on a bounded C1 Euclidean domain

Facts & Assumptions

Given: ${A C}_{ω}$ , the bounded $C^{1}$ domain, the real frequency, C1 phasor fields, continuous sources, common complex constitutive tensors and their pointwise transpose symmetry stated above.

[A1]

${A C}_{ω}$ is the axiom that every countable family of nonempty sets has a choice function; it is assumed by the divergence-theorem supplier used here (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For real C1 vector fields $F, G$ , $div (F \times G) = G \cdot curl F - F \cdot curl G$ (The divergence and curl of a cross product).

[F2]

For a bounded C1 domain and $F \in C^{1} (\overline{Ω}; R^{3})$ , $\int_{Ω} div F d V = \int_{\partial Ω} F \cdot n d S$ , under ${A C}_{ω}$ (Divergence on a bounded C1 Euclidean domain).

[L1]

In the reciprocal phasor model, $D_{j} = ε E_{j}$ , $B_{j} = μ H_{j}$ , $curl E_{j} = i ω B_{j}$ , and $curl H_{j} = J_{j} - i ω D_{j}$ ; reciprocal coefficients mean the ordinary pointwise transpose equalities $ε^{T} = ε$ and $μ^{T} = μ$ , even when the entries are complex (Fixed-frequency reciprocal tensor medium).

[L2]

In the lossless anisotropic plane-wave branch, $ε$ is real symmetric positive definite and $μ$ is a positive real scalar; its physical phase-response interpretation is restricted to the declared finite band $I_{ω}$ (Anisotropic Maxwell symbol).

Proof

1.1givenF1L1algebra

Put $X = E_{1} \times H_{2} - E_{2} \times H_{1}$ . For the first product write $E_{1} = U + i V$ and $H_{2} = W + i Z$ with real C1 fields. The complex-bilinear product is $E_{1} \times H_{2} = (U \times W - V \times Z) + i (U \times Z + V \times W)$ ; applying [F1] to each real pair and using the componentwise real-linearity of curl and divergence gives $div (E_{1} \times H_{2}) = H_{2} \cdot curl E_{1} - E_{1} \cdot curl H_{2}$ . The same expansion for $E_{2} \times H_{1}$ yields $div X = H_{2} \cdot curl E_{1} - E_{1} \cdot curl H_{2} - H_{1} \cdot curl E_{2} + E_{2} \cdot curl H_{1}$ . Substituting [L1] yields $E_{2} \cdot J_{1} - E_{1} \cdot J_{2} + i ω (H_{2} \cdot μ H_{1} - H_{1} \cdot μ H_{2} + E_{1} \cdot ε E_{2} - E_{2} \cdot ε E_{1})$ . For any complex vectors $u, v$ , $u \cdot ε v = \sum_{r, s} u_{r} ε_{r s} v_{s} = v \cdot ε u$ by swapping $r, s$ and using $ε_{r s} = ε_{s r}$ ; the same calculation applies to $μ$ . Thus the tensor terms cancel without conjugation and $div X = E_{2} \cdot J_{1} - E_{1} \cdot J_{2}$ .

2.1givenA1F2step 1.1algebra

The field $X$ is C1 on $\overline{Ω}$ . Apply [F2] separately to its real and imaginary parts, which satisfy the theorem's real-valued hypotheses, and combine the two identities complex-linearly. Substituting step 1.1 into the resulting equality gives the displayed boundary identity.

3.1givenL1L2step 2.1∎

By [L2], the lossless anisotropic plane-wave branch has a real symmetric tensor $ε$ and scalar $μ$ ; therefore its ordinary transposes equal themselves, so it is a subcase of the coefficient class in [L1]. Applying step 2.1 to any two C1 phasor solutions in that subcase proves the anisotropic specialization, with its physical use restricted to the finite band in [L2].

Remarks

The symmetric conditions use transpose, not conjugate transpose. Positive absorption can therefore coexist with this reciprocity identity; the identity alone asserts neither losslessness nor time-reversal invariance.

Reciprocity in reciprocal linear media

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources