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

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Passivity gives a contractive scattering matrix

Physical scope

A conditional classical Maxwell scattering deduction at one real frequency for the finite port model of emx-port-scattering-channels. The material branches are exactly the homogeneous finite Lorentz oscillator or Drude current closures stated here; fields are smooth and periodic, free-current work is zero, wall power is nonnegative, and all unlisted outgoing power is counted. It is not an existence theorem, does not cover arbitrary passive constitutive laws, and makes no empirical claim.

Statement

Assume ${A C}_{ω}$ . Fix the finite port model at a real angular frequency $ω > 0$ and period $T = 2 π / ω$ , with bounded $C^{1}$ device region $Ω$ and scattering matrix $S (ω)$ . For every incoming amplitude $a \in C^{N}$ , let ${\tilde{E}}_{a}, {\tilde{H}}_{a} \in C^{1} (\overline{Ω}; C^{3})$ be that model's unique phasor solution with outgoing amplitude $b = S (ω) a$ , and set $E (x, t) = Re ({\tilde{E}}_{a} (x) e^{- i ω t})$ and $H (x, t) = Re ({\tilde{H}}_{a} (x) e^{- i ω t})$ . Assume for every such input that these real fields, $D, B : \overline{Ω} \times R \to R^{3}$ , and the material states are jointly $C^{\infty}$ and $T$ -periodic, satisfy the smooth source-free macroscopic Maxwell equations with $ρ_{f} = 0$ and $J_{f} = 0$ , and obey the same one of the following two homogeneous material branches with coefficients independent of the input. The units of $E, H, D, B$ are respectively $V m^{- 1}$ , $A m^{- 1}$ , $C m^{- 2}$ , and $T$ .

For real vectors $v, w \in R^{3}$ , the Euclidean product is $v \cdot w : = \sum_{r = 1}^{3} v_{r} w_{r}$ and $∣ v ∣^{2} : = v \cdot v$ .

The port model uses incoming and outgoing vectors $a, b \in C^{N}$ with amplitude units $W^{1 / 2}$ ; for either vector $z$ , $∥ z ∥_{2}^{2} : = \sum_{α = 1}^{N} ∣ z_{α} ∣^{2}$ is its power in W, and the net outward port power is $∥ b ∥_{2}^{2} - ∥ a ∥_{2}^{2}$ .

Finite Lorentz branch. For a finite $M \geq 1$ , each $P_{ℓ} : \overline{Ω} \times R \to R^{3}$ has units $C m^{- 2}$ , $D = ε_{\infty} E + \sum_{ℓ = 1}^{M} P_{ℓ}$ , $B = μ H$ , and $\partial_{t}^{2} P_{ℓ} + γ_{ℓ} \partial_{t} P_{ℓ} + ω_{ℓ}^{2} P_{ℓ} = a_{ℓ} E$ , where $ε_{\infty}, μ, a_{ℓ}, ω_{ℓ} > 0$ and $γ_{ℓ} \geq 0$ .

Drude branch. $P, V : \overline{Ω} \times R \to R^{3}$ have units $C m^{- 2}$ and $A m^{- 2}$ , respectively; $D = ε_{\infty} E + P$ , $B = μ H$ , $\partial_{t} P = V$ , and $\partial_{t} V + γ V = a E$ , where $ε_{\infty}, μ, a > 0$ and $γ \geq 0$ .

The constants have the SI units of the respective adopted closure: $ε_{\infty}$ in $F m^{- 1}$ , $μ$ in $H m^{- 1}$ , each $a_{ℓ}$ or $a$ in $F m^{- 1} s^{- 2}$ , and each $ω_{ℓ}$ or damping rate in $s^{- 1}$ . In the Lorentz branch define the loss-density map $ℓ : \overline{Ω} \times R \to [0, \infty)$ by $ℓ = \sum_{ℓ = 1}^{M} (γ_{ℓ} / a_{ℓ}) ∣ \partial_{t} P_{ℓ} ∣^{2}$ ; in the Drude branch define it by $ℓ = (γ / a) ∣ V ∣^{2}$ . These have units $W m^{- 3}$ . For the solution with the fixed input $a$ , set $P_{m a t} (a) : = T^{- 1} \int_{0}^{T} \int_{Ω} ℓ d V d t$ . Let $P_{\partial} (a) \geq 0$ be its cycle-averaged outward power at passive physical walls, and let $P_{o t h e r} (a) \geq 0$ be all its outgoing radiation or propagating-channel power not represented in $b$ , as defined by the port model. For brevity, in the proof write these three quantities as $P_{m a t}$ , $P_{\partial}$ , and $P_{o t h e r}$ . Then

$∥ a ∥_{2}^{2} - ∥ S (ω) a ∥_{2}^{2} = P_{m a t} (a) + P_{\partial} (a) + P_{o t h e r} (a) \geq 0.$

For complex vectors and matrices, an overline denotes componentwise complex conjugation and a superscript $^{T}$ denotes ordinary transpose; set $z^{†} : = {\overline{z}}^{T}$ for $z \in C^{N}$ and $A^{†} : = {\overline{A}}^{T}$ for $A \in C^{N \times N}$ . A matrix is Hermitian when $A^{†} = A$ , and is unitary when $A^{†} A = A A^{†} = I_{N}$ , where $I_{N}$ is the identity matrix on $C^{N}$ .

Consequently $S (ω)$ is contractive: $∥ S a ∥_{2} \leq ∥ a ∥_{2}$ for every $a$ . If $P_{m a t} (a) = P_{\partial} (a) = P_{o t h e r} (a) = 0$ for every incoming vector $a$ , then $S^{†} S = I_{N}$ ; because the defined map is square, $S$ is unitary. The exact equality and unitarity are conditional deductions of this model, not existence claims about arbitrary scatterers.

Statement references

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

Finite propagating-port scattering map

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$

The divergence and curl of a cross product

Finite Lorentz oscillator closure

Drude current closure and collisionless limit

The macroscopic Maxwell equations in terms of D, H and the free sources

Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$

Divergence on a bounded C1 Euclidean domain

Fubini's theorem for L^1 functions on a sigma-finite product

Facts & Assumptions

Given: ${A C}_{ω}$ ; one fixed $ω > 0$ and $T = 2 π / ω$ ; for every input, the real $E, H$ obtained from that finite scattering model's phasor solution; the same stated homogeneous material branch with its positive parameters and nonnegative damping; smooth periodic real fields and material states for every input; zero free current; and nonnegative passive-wall and unlisted-outgoing powers. Fix an arbitrary input $a$ for the calculation below.

[A1]

The Lorentz postulate has constant $ε_{\infty}, μ, a_{ℓ}, ω_{ℓ}, γ_{ℓ}$ , $D = ε_{\infty} E + \sum P_{ℓ}$ , and the displayed second-order oscillator equations. Finite Lorentz oscillator closure

[A2]

The Drude postulate has constant $ε_{\infty}, μ, a, γ$ , $D = ε_{\infty} E + P$ , $P_{t} = V$ , and $V_{t} + γ V = a E$ . Drude current closure and collisionless limit

[L1]

On the smooth Maxwell branch, $curl E = - \partial_{t} B$ and $curl H = J_{f} + \partial_{t} D$ . The macroscopic Maxwell equations in terms of D, H and the free sources

[F1]

For real $C^{1}$ fields, $div (E \times H) = H \cdot curl E - E \cdot curl H$ . The divergence and curl of a cross product

[F2]

Component products obey the ordinary product rule; the real Euclidean inner product is symmetric and componentwise bilinear. Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$

[F3]

On a bounded $C^{1}$ device, the integral of the divergence of a real $C^{1}$ vector field equals its outward boundary flux under ${A C}_{ω}$ . Divergence on a bounded C1 Euclidean domain

[F4]

Fubini's theorem permits exchanging the order of integration for an integrable function on the finite-volume product $Ω \times [0, T]$ . Fubini's theorem for L^1 functions on a sigma-finite product

[F5]

The port model defines the outward port flux as $∥ b ∥_{2}^{2} - ∥ a ∥_{2}^{2}$ and the total outward boundary power as that port flux plus $P_{\partial} + P_{o t h e r}$ ; it assumes no unlisted incoming channel. Finite propagating-port scattering map

[A3]

Physical wall conditions are either smooth PEC, with zero outward flux, or passive conditions with cycle-averaged outward power $P_{\partial} \geq 0$ . This is an explicit model premise, not a trace theorem. Finite propagating-port scattering map

Proof

technique · derive the local Poynting identity separately for the Lorentz and Drude closures, integrate it over the device and one period, then use the port flux identity and finite-dimensional matrix algebra

1.1givenA1A2L1algebra

In the Lorentz branch set $J_{t o t} = J_{f} + \sum_{ℓ = 1}^{M} \partial_{t} P_{ℓ}$ , and in the Drude branch set $J_{t o t} = J_{f} + V$ . From [L1] and [A1] or [A2], respectively, the field and constitutive equations give $ε_{\infty} \partial_{t} E = curl H - J_{t o t}$ and $μ \partial_{t} H = - curl E$ .

1.2givenA1F2algebra

In the Lorentz branch, dot the $ℓ$ th oscillator equation [A1] with $\partial_{t} P_{ℓ} / a_{ℓ}$ . Since $a_{ℓ} > 0$ , the division is valid; the product rule in [F2] gives $E \cdot \partial_{t} P_{ℓ} = \partial_{t} ⁣ ((∣ \partial_{t} P_{ℓ} ∣^{2} + ω_{ℓ}^{2} ∣ P_{ℓ} ∣^{2}) / (2 a_{ℓ})) + (γ_{ℓ} / a_{ℓ}) ∣ \partial_{t} P_{ℓ} ∣^{2}$ . The last term is nonnegative because $γ_{ℓ} \geq 0$ .

1.3givenA2F2algebra

In the Drude branch, dot $\partial_{t} V + γ V = a E$ from [A2] with $V / a$ . The product rule and $a > 0$ give $E \cdot V = \partial_{t} (∣ V ∣^{2} / (2 a)) + (γ / a) ∣ V ∣^{2}$ , whose last term is nonnegative because $γ \geq 0$ .

2.1step 1.1F1F2algebra

Define the field energy density $u_{f} : \overline{Ω} \times R \to [0, \infty)$ by $u_{f} = (ε_{\infty} ∣ E ∣^{2} + μ ∣ H ∣^{2}) / 2$ . Differentiate it componentwise using [F2] and the field equations of step 1.1. This gives $\partial_{t} u_{f} = E \cdot curl H - E \cdot J_{t o t} - H \cdot curl E$ . Applying [F1] yields $- div (E \times H) - E \cdot J_{f} - \sum E \cdot \partial_{t} P_{ℓ}$ in the Lorentz branch and $- div (E \times H) - E \cdot J_{f} - E \cdot V$ in the Drude branch.

3.1givenstep 1.2step 1.3step 2.1algebra

Define the total field-plus-material energy density $u : \overline{Ω} \times R \to [0, \infty)$ branchwise: in the Lorentz branch, $u : = u_{f} + \sum_{ℓ = 1}^{M} (∣ \partial_{t} P_{ℓ} ∣^{2} + ω_{ℓ}^{2} ∣ P_{ℓ} ∣^{2}) / (2 a_{ℓ})$ ; in the Drude branch, $u : = u_{f} + ∣ V ∣^{2} / (2 a)$ . All terms are nonnegative, have SI units $J m^{- 3}$ , and are jointly smooth and $T$ -periodic under the Statement's hypotheses. Substituting step 1.2 into the Lorentz field balance of step 2.1 cancels every $E \cdot \partial_{t} P_{ℓ}$ term; substituting step 1.3 into its Drude balance cancels $E \cdot V$ . Since the given source-free condition has $J_{f} = 0$ , each branch therefore satisfies $\partial_{t} u + div (E \times H) = - ℓ$ with exactly the loss density $ℓ$ in the Statement.

4.1givenF3F4step 3.1algebra

Integrate the local identity of step 3.1 over $Ω$ at each time and apply [F3] to $E \times H$ . Then integrate in time from $0$ to $T$ . The integrand $\partial_{t} u$ is continuous and integrable on the finite product domain, so [F4] permits integration first in time: $\int_{Ω} \int_{0}^{T} \partial_{t} u d t d V = \int_{Ω} (u (x, T) - u (x, 0)) d V = 0$ by periodicity. Hence the time-integrated outward boundary flux equals $- T P_{m a t}$ .

5.1givenA3F5step 4.1algebra

For the given phasor solution, expand each real part as $(z + \overline{z}) / 2$ in $E \times H$ . The terms proportional to $e^{\pm 2 i ω t}$ have zero average over $[0, T]$ , leaving $T^{- 1} \int_{0}^{T} E \times H d t = \frac{1}{2} Re ({\tilde{E}}_{a} \times \overline{{\tilde{H}}_{a}})$ . Thus the cycle average in step 4.1 is precisely the phasor power used by [F5]. Divide that step by $T$ and use its total boundary-flux decomposition: the port term is $∥ b ∥_{2}^{2} - ∥ a ∥_{2}^{2}$ , the wall term is $P_{\partial} \geq 0$ by [A3], and the remaining outgoing term is $P_{o t h e r} \geq 0$ . Rearrangement gives the power balance for the arbitrary input $a$ . The all-input hypotheses permit this calculation for every incoming vector of the same matrix $S (ω)$ .

6.1givenstep 5.1algebra∎

Since each of $P_{m a t}, P_{\partial}, P_{o t h e r}$ is nonnegative, step 5.1 gives $∥ S a ∥_{2}^{2} \leq ∥ a ∥_{2}^{2}$ for every $a$ ; taking nonnegative square roots proves contractivity. If all three powers vanish for every $a$ , set $K : = I_{N} - S^{†} S$ . Then $K^{†} = K$ because $(S^{†} S)^{†} = S^{†} S$ , and the power balance gives $a^{†} K a = ∥ a ∥_{2}^{2} - ∥ S a ∥_{2}^{2} = 0$ for every $a$ . Let $e_{j}$ denote the $j$ th coordinate vector of $C^{N}$ . Testing $a = e_{j}$ gives $K_{j j} = 0$ ; for $j < k$ , testing $a = e_{j} + e_{k}$ gives $2 Re K_{j k} = 0$ , while testing $a = e_{j} + i e_{k}$ gives $- 2 Im K_{j k} = 0$ . Hence every entry of $K$ is zero, so $S^{†} S = I_{N}$ . The square matrix $S$ is injective, hence surjective by finite-dimensional rank-nullity; therefore $S^{- 1} = S^{†}$ and $S S^{†} = I_{N}$ as well. Thus $S$ is unitary.

Passivity gives a contractive scattering matrix

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources