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Loss and reciprocity coexist

Physical scope

A conditional single-oscillator Lorentz phasor response with strictly positive damping and a nonzero local electric amplitude, together with a real positive scalar permeability. It shows positive cycle-averaged work on the oscillator contribution to the polarization-current density under the stated response model while the coefficient tensors satisfy the local reciprocity condition. It asserts no measured material response, global field existence, or time-reversal statement.

Example

Assume ${A C}_{ω}$ . Choose a one-oscillator Lorentz response with $ε_{\infty} > 0$ in $F m^{- 1}$ , $a > 0$ in $F m^{- 1} s^{- 2}$ , $ω_{0} > 0$ and $γ > 0$ in $s^{- 1}$ , and fix $ω > 0$ in $s^{- 1}$ . In the $e^{- i ω t}$ convention its scalar absolute permittivity is $ε (ω) = ε_{\infty} + a / (ω_{0}^{2} - ω^{2} - i γ ω)$ . Let $I_{3}$ denote the $3 \times 3$ identity matrix, and let the tensor coefficients be $ε (ω) = ε (ω) I_{3}$ and $μ = μ I_{3}$ with real $μ > 0$ in $H m^{- 1}$ . These coefficients satisfy the transpose-symmetry condition of the reciprocal phasor model, while $Im ε (ω) > 0$ . For any nonzero local electric amplitude $e \in C^{3}$ in $V m^{- 1}$ , define the real electric field at a fixed material point by $E (t) = Re (e e^{- i ω t})$ . Let $p_{1} = (ε (ω) - ε_{\infty}) e$ be the oscillator polarization amplitude from the one-oscillator response, in $C m^{- 2}$ ; put $P_{1} (t) = Re (p_{1} e^{- i ω t})$ , and define its polarization-current contribution by $J_{o s c} (t) : = \partial_{t} P_{1} (t)$ . Its phasor amplitude is $j_{o s c} = - i ω (ε (ω) - ε_{\infty}) e$ in $A m^{- 2}$ . The cycle-averaged power density transferred from the electric field to the oscillator through this current contribution, in $W m^{- 3}$ , is

${\overline{q}}_{o s c} : = \frac{1}{T} \int_{0}^{T} J_{o s c} (t) \cdot E (t) d t = \frac{ω}{2} Im ε (ω) ∣ e ∣^{2} > 0, T = \frac{2 π}{ω}, ∣ e ∣^{2} : = e^{*} \cdot e = \sum_{r = 1}^{3} ∣ e_{r} ∣^{2} .$

Here $e^{*}$ is componentwise complex conjugation. The conjugate used in this cycle average is distinct from the bilinear, unconjugated product in the reciprocity identity.

Thus the coefficient condition that supplies the Lorentz reciprocity identity is compatible with positive cycle-averaged work on the oscillator-current contribution in this adopted model.

Statement references

Reciprocity in reciprocal linear media

Analytic susceptibility and dispersion restrictions

Fixed-frequency reciprocal tensor medium

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

Facts & Assumptions

Given: ${A C}_{ω}$ , a single Lorentz oscillator with the strictly positive coefficients and damping specified above, a fixed real $ω > 0$ , real scalar $μ > 0$ , and a nonzero local harmonic electric amplitude $e$ .

[A1]

${A C}_{ω}$ is explicitly assumed by the supplier statement used for the oscillator response (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For the $e^{- i ω t}$ convention and one Lorentz oscillator driven by electric-field amplitude $e$ , the oscillator polarization amplitude is $p_{1} = a e / (ω_{0}^{2} - ω^{2} - i γ ω) = (ε (ω) - ε_{\infty}) e$ . Also $ε (ω) = ε_{\infty} + a / (ω_{0}^{2} - ω^{2} - i γ ω)$ and $Im ε (ω) = a γ ω / ((ω_{0}^{2} - ω^{2})^{2} + γ^{2} ω^{2}) \geq 0$ (Analytic susceptibility and dispersion restrictions, harmonic-response calculation and pointwise-loss clause).

[F2]

The phasor medium is reciprocal exactly when $ε^{T} = ε$ and $μ^{T} = μ$ ; its products in the Lorentz identity are bilinear without conjugation (Fixed-frequency reciprocal tensor medium).

[L1]

Two C1 phasor solutions of the reciprocal model obey the bounded-domain Lorentz bilinear identity; the theorem does not require the coefficients to be Hermitian or lossless (Reciprocity in reciprocal linear media).

Verification

1.1givenA1F1F2L1algebra

The denominator $(ω_{0}^{2} - ω^{2})^{2} + γ^{2} ω^{2}$ is strictly positive because $γ ω > 0$ . Therefore [F1] gives $Im ε (ω) = a γ ω / ((ω_{0}^{2} - ω^{2})^{2} + γ^{2} ω^{2}) > 0$ . A scalar multiple of $I_{3}$ equals its ordinary transpose even when the scalar is complex, and the real scalar $μ$ also equals its transpose; hence [F2] places these coefficients in the reciprocal class. The reciprocity identity [L1] is consequently applicable whenever the corresponding C1 phasor solutions exist.

2.1givenA1F1step 1.1algebra∎

By [F1], $P_{1} (t) = Re (p_{1} e^{- i ω t})$ has amplitude $p_{1} = (ε - ε_{\infty}) e$ , so $J_{o s c} = \partial_{t} P_{1}$ has phasor amplitude $j_{o s c} = - i ω p_{1}$ . For any complex scalar amplitudes $u, v$ , write the product as $\frac{1}{4} (u v e^{- 2 i ω t} + u v^{*} + u^{*} v + u^{*} v^{*} e^{2 i ω t})$ . The period integrals of $e^{\pm 2 i ω t}$ are zero because $ω > 0$ , so the average is $\frac{1}{2} Re (u v^{*})$ . Applying this componentwise gives ${\overline{q}}_{o s c} = \frac{1}{2} Re (j_{o s c} \cdot e^{*}) = \frac{ω}{2} Im (ε - ε_{\infty}) ∣ e ∣^{2} = \frac{ω}{2} Im ε ∣ e ∣^{2} > 0$ , since $ε_{\infty}$ is real and $e \neq 0$ . Positive ${\overline{q}}_{o s c}$ is work transferred from the field to the oscillator-current contribution in the adopted Maxwell sign convention. This is a model deduction, not an observation.

Remarks

This calculation establishes coexistence of transpose-symmetric reciprocal coefficients and positive cycle-averaged work on the oscillator-current contribution in the stated model. It makes no claim about a time-reversal transformation.

Loss and reciprocity coexist

Example

Statement references

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources