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Electric amplitude has a longitudinal component

Physical scope

A conditional bulk plane-wave example in a homogeneous lossless uniaxial dielectric with positive principal permittivities, positive scalar permeability, a real nonzero wavevector on the extraordinary sheet, and nonzero transverse and axial wavevector components. Physical interpretation is restricted to a declared finite positive-frequency band of effective phase response; the algebraic example holds for every positive frequency. It makes no interface, finite-beam or empirical claim.

Example

Let $ε = diag (ε_{⊥}, ε_{⊥}, ε_{∥})$ with $ε_{⊥}, ε_{∥} > 0$ and $ε_{⊥} \neq ε_{∥}$ , and let $μ > 0$ . Fix $ω > 0$ and real $k = (k_{x}, 0, k_{z})$ with $k_{x} k_{z} \neq 0$ on the extraordinary sheet $k_{x}^{2} / ε_{∥} + k_{z}^{2} / ε_{⊥} = μ ω^{2}$ . Choose $d_{0} \in C ∖ {0}$ in $C m^{- 1}$ and set the displacement amplitude $d = d_{0} (k_{z}, 0, - k_{x})$ and electric amplitude $e = ε^{- 1} d$ . Then $k \cdot d = 0$ but $k \cdot e = d_{0} k_{x} k_{z} (1 / ε_{⊥} - 1 / ε_{∥}) \neq 0$ . Thus the extraordinary electric amplitude is not orthogonal to the wavevector even though the displacement amplitude is.

The algebraic example holds for every $ω > 0$ . For physical interpretation, declare a finite angular-frequency 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 the supplying uniaxial classification.

Statement references

Ordinary and extraordinary sheets

Facts & Assumptions

Given: The positive uniaxial coefficients, $ω > 0$ , $k_{x} k_{z} \neq 0$ , and the extraordinary dispersion relation stated above, all in SI units.

[L1]

The extraordinary sheet and its nonzero source-free polarization modes are those of Ordinary and extraordinary sheets; the displacement amplitude is $d = ε e$ in the same model.

Verification

1.1givenL1algebra

Since $ε_{⊥}, ε_{∥} > 0$ , the diagonal matrix $ε$ is invertible, so $e = ε^{- 1} d = d_{0} (k_{z} / ε_{⊥}, 0, - k_{x} / ε_{∥})$ is defined and nonzero. Direct contraction gives $k \cdot d = d_{0} (k_{x} k_{z} - k_{z} k_{x}) = 0$ and $k \cdot e = d_{0} k_{x} k_{z} (1 / ε_{⊥} - 1 / ε_{∥}) \neq 0$ by the nonzero-component and unequal-permittivity hypotheses.

2.1givenL1step 1.1algebra∎

The extraordinary dispersion relation gives $k_{x}^{2} / ε_{∥} + k_{z}^{2} / ε_{⊥} = μ ω^{2}$ . Hence the $x$ and $z$ components of $k \times (k \times e)$ are $- d_{0} k_{z} (k_{x}^{2} / ε_{∥} + k_{z}^{2} / ε_{⊥}) = - μ ω^{2} d_{0} k_{z}$ and $d_{0} k_{x} (k_{x}^{2} / ε_{∥} + k_{z}^{2} / ε_{⊥}) = μ ω^{2} d_{0} k_{x}$ , while the $y$ component is zero. Thus $μ^{- 1} k \times (k \times e) + ω^{2} d = 0$ , so $e$ is an extraordinary symbol mode and the longitudinal-component claim follows.

Depends on

Ordinary and extraordinary sheets

Dependency tree · 2 levels

Sources

Jin Au Kong, Electromagnetic Wave Theory, last edition 2008, PIERS free release 2021