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

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Ordinary and extraordinary sheets

Physical scope

Conditional classification of nonzero real-wavevector source-free bulk plane-wave modes in a homogeneous lossless uniaxial dielectric with positive real principal permittivities and positive scalar permeability, in the material-rest frame and SI units. The general uniaxial classification uses the xz-plane section; the isotropic specialization covers all real-wavevector directions. Physical use is restricted to a declared finite positive-frequency band of effective phase response. Interfaces, absorption, spatial dispersion and empirical material claims are outside scope.

Statement

In the homogeneous lossless plane-wave model, let $ε = diag (ε_{⊥}, ε_{⊥}, ε_{∥})$ with $ε_{⊥}, ε_{∥} > 0$ in $F m^{- 1}$ , let $μ > 0$ be a scalar in $H m^{- 1}$ , and let $ω > 0$ be in $s^{- 1}$ . For a nonzero source-free mode with real wavevector $k = (k_{x}, 0, k_{z}) \neq 0$ in $m^{- 1}$ , the nonzero-mode wavevectors lie on exactly the following two sheets:

$k_{x}^{2} + k_{z}^{2} = μ ε_{⊥} ω^{2}, \frac{k_{x}^{2}}{ε_{∥}} + \frac{k_{z}^{2}}{ε_{⊥}} = μ ω^{2} .$

Away from an intersection the first sheet is the ordinary sector, whose electric amplitude is parallel to $\hat{y}$ , and the second is the extraordinary sector, whose electric amplitude lies in the $x z$ plane; each then has a one-dimensional polarization space. At an intersection the two sectors span a two-dimensional polarization space. If $ε_{⊥} = ε_{∥} = ε$ , the isotropic specialization also allows arbitrary real $k \neq 0$ , without the $x z$ -plane restriction: its nonzero-mode wavevectors lie in every direction on the common sphere $∣ k ∣^{2} = μ ε ω^{2}$ , and at every such wavevector the polarization space is the two-dimensional complex subspace ${e \in C^{3} : k \cdot e = 0}$ . In the $x z$ section the ordinary/extraordinary labels then denote the two coordinate sectors, not distinct dispersion branches. If $ε_{⊥} \neq ε_{∥}$ , the sheets intersect only for $k_{x} = 0$ and $k_{z}^{2} = μ ε_{⊥} ω^{2}$ ; at those optic-axis directions the polarization space is again two-dimensional. The zero amplitude is not a mode in this statement.

The symbol classification holds algebraically for every $ω > 0$ . For physical interpretation, declare a finite band $I_{ω} = [ω_{-}, ω_{+}]$ with $0 < ω_{-} < ω_{+} < \infty$ , in $s^{- 1}$ , and use the constant-coefficient model only for $ω \in I_{ω}$ , as in its defining material closure.

Statement references

Anisotropic Maxwell symbol

Facts & Assumptions

Given: The fixed material-rest frame, right-handed Cartesian axes, SI units, positive coefficients, real wavevector, positive angular frequency and nonzero plane-wave amplitude stated above.

[L1]

Anisotropic Maxwell symbol defines the source-free anisotropic plane-mode equation as $μ^{- 1} k \times (k \times e) + ω^{2} ε e = 0$ , with $d = ε e$ , $k \cdot d = 0$ , and $e \neq 0$ .

Proof

1.1givenL1algebra

Write $e = (e_{x}, e_{y}, e_{z})$ . Substituting the stated diagonal tensor and $k = (k_{x}, 0, k_{z})$ into [L1], and expanding the cross products componentwise, gives $(\begin{array}{ccc} ω^{2} ε_{⊥} - k_{z}^{2} / μ & 0 & k_{x} k_{z} / μ \\ 0 & ω^{2} ε_{⊥} - (k_{x}^{2} + k_{z}^{2}) / μ & 0 \\ k_{x} k_{z} / μ & 0 & ω^{2} ε_{∥} - k_{x}^{2} / μ \end{array}) (\begin{array}{c} e_{x} \\ e_{y} \\ e_{z} \end{array}) = 0$ .

1.2givenL1algebra

If $e_{y} \neq 0$ , the middle row gives $k_{x}^{2} + k_{z}^{2} = μ ε_{⊥} ω^{2}$ . If $(e_{x}, e_{z}) \neq (0, 0)$ , the $x z$ block must be singular, and its determinant is $\frac{ω^{2} ε_{⊥} ε_{∥}}{μ} (μ ω^{2} - \frac{k_{x}^{2}}{ε_{∥}} - \frac{k_{z}^{2}}{ε_{⊥}})$ , so it vanishes exactly on the second sheet. Conversely, on the first sheet $(0, 1, 0)$ is a nonzero kernel vector, and on the second sheet the singular $x z$ block has a nonzero kernel vector. Since the matrix separates the $y$ and $x z$ sectors, these are all nonzero modes. Away from an intersection, the $x z$ block is invertible on the first sheet, so its kernel is exactly the $y$ line. On the second sheet, away from an intersection, the middle coefficient is nonzero and the singular $x z$ block has rank one: its off-diagonal entry is nonzero when $k_{x} k_{z} \neq 0$ , while on either coordinate axis its other diagonal entry is respectively $ω^{2} ε_{∥}$ or $ω^{2} ε_{⊥}$ , both positive. Thus both polarization spaces are one-dimensional away from intersections.

2.1givenL1step 1.1step 1.2algebra∎

At an intersection with $ε_{⊥} \neq ε_{∥}$ , dividing the first sheet by $ε_{⊥}$ and subtracting the second gives $k_{x}^{2} (1 / ε_{⊥} - 1 / ε_{∥}) = 0$ , hence $k_{x} = 0$ ; either sheet then gives $k_{z}^{2} = μ ε_{⊥} ω^{2}$ . Conversely, substituting these values into both sheets verifies an intersection. The matrix in step 1.1 is then $diag (0, 0, ω^{2} ε_{∥})$ , so its kernel is the complex span of $\hat{x}, \hat{y}$ . For $ε_{⊥} = ε_{∥} = ε$ , take any real $k \in R^{3} ∖ {0}$ . Expanding the cross products in [L1] gives $M (k, ω) = (ε ω^{2} - ∣ k ∣^{2} / μ) I_{3} + k k^{T} / μ$ . Every complex amplitude has the direct decomposition $e = e_{∥} + e_{⊥}$ , where $e_{∥} = k (k \cdot e) / ∣ k ∣^{2}$ and $k \cdot e_{⊥} = 0$ ; the denominator is positive since $k$ is real and nonzero. The symbol acts by $M e_{∥} = ε ω^{2} e_{∥}$ and $M e_{⊥} = (ε ω^{2} - ∣ k ∣^{2} / μ) e_{⊥}$ . Because $ε ω^{2} > 0$ , a nonzero kernel occurs exactly on $∣ k ∣^{2} = μ ε ω^{2}$ , and there it is precisely $k^{⊥} = {e : k \cdot e = 0}$ . This complex space has dimension two: choose an index with $k_{i} \neq 0$ and solve its one linear equation for $e_{i}$ in terms of the other two complex coordinates. For every real unit direction $n$ , taking $k = ω \sqrt{μ ε} n$ gives a point on that sphere. In the $x z$ section this kernel is the sum of the $y$ and transverse $x z$ coordinate sectors, verifying the common-circle section as well as the full-sphere claim. This proves the stated degeneracies and completes the conditional model classification.

Diagram

For the representative positive-uniaxial ratio $ε_{∥} = 2 ε_{⊥}$ , normalize both wavevector coordinates by $ω \sqrt{μ ε_{⊥}}$ . The ordinary cross-section is the unit circle; the extraordinary cross-section is the ellipse $u_{x}^{2} / 2 + u_{z}^{2} = 1$ . They intersect at the two optic-axis points $(0, \pm 1)$ . The drawing is a normalized cross-section of the two exact equations in the Statement; the proof does not rely on it.

\begin{tikzpicture}[scale=1.25,>=stealth]
  \draw[->] (-1.8,0) -- (1.8,0) node[right] {$u_x$};
  \draw[->] (0,-1.45) -- (0,1.45) node[above] {$u_z$};
  \draw[thick,blue] (0,0) circle (1);
  \draw[thick,red,dashed] (0,0) ellipse [x radius=1.414,y radius=1];
  \fill (0,1) circle (1.4pt);
  \fill (0,-1) circle (1.4pt);
  \node[blue] at (-1.05,0.75) {ordinary};
  \node[red] at (1.15,0.78) {extraordinary};
  \node[below left] at (0,0) {$0$};
\end{tikzpicture}

Depends on

Anisotropic Maxwell symbol