Electromagnetic Waves in Anisotropic

Dielectric Media

Introduction

Maxwell's equations are fundamental to understanding the behavior of

electromagnetic waves. In isotropic media, the permittivity (ε) and permeability (μ) are
scalar quantities, meaning that the electric displacement field (D) is always parallel to
the electric field (E), and the magnetic field (B) is always parallel to the magnetic field
intensity (H). However, in anisotropic media, such as crystals, the material properties
depend on the direction of propagation. This directional dependence is captured by
representing permittivity and permeability as tensors. This leads to fascinating
phenomena like birefringence, where an electromagnetic wave splits into two distinct
waves with different polarizations and propagation speeds.

This document will demonstrate, using Maxwell's equations, how two sets of solutions
for different polarizations of an electromagnetic wave exist when travelling at an angle
from one of the principal axes of a linear anisotropic dielectric medium. We will derive
the wave equation in such a medium and then express the solutions for the electric
and magnetic fields for both the ordinary and extraordinary waves.

Maxwell's Equations in Anisotropic Media

Maxwell's equations in a source-free, linear, and non-magnetic medium are given by:

1. Faraday's Law: ∇ × E = − ∂B
∂t
​

∂D
2. Ampere's Law (with Maxwell's addition): ∇ × H = ∂t
​

3. Gauss's Law for Electric Fields: ∇ ⋅ D =0

4. Gauss's Law for Magnetic Fields: ∇ ⋅ B =0

In an anisotropic dielectric medium, the relationship between the electric

displacement field D and the electric field E is tensorial, meaning D is not necessarily
parallel to E . This relationship is expressed as:

D = ϵE

where ϵ is the dielectric permittivity tensor. For a non-magnetic medium, the magnetic
permeability is a scalar, B = μ0 H . ​

For a time-harmonic plane wave propagating in the medium, we can assume the fields
have the form ej(k⋅r−ωt) , where k is the wave vector and ω is the angular frequency.
∂
Under this assumption, the time derivatives become ∂t = −jω and the spatial
​

derivatives (curl and divergence) become ∇ = jk .

Substituting these into Maxwell's equations:

1. j k × E = −jωμ0 H ⟹ k × E = ωμ0 H (Faraday's Law)

​ ​

2. j k × H = −jω D ⟹ k × H = −ωD (Ampere's Law)

3. j k ⋅ D = 0 ⟹ k ⋅ D = 0 (Gauss's Law for Electric Fields)
4. j k ⋅ B = 0 ⟹ k ⋅ B = 0 (Gauss's Law for Magnetic Fields)

From Gauss's Law for Electric Fields, we see that the electric displacement vector D is
always perpendicular to the wave vector k . Similarly, from Gauss's Law for Magnetic
Fields, the magnetic field B (and thus H ) is also perpendicular to k . However, due to
the anisotropic nature of the medium, E is generally not perpendicular to k . This is a
key difference from isotropic media.

Derivation of the Wave Equation

To derive the wave equation, we can take the curl of Faraday's Law:

∇ × (∇ × E ) = ∇ × (−jωμ0 H ) ​

Using the vector identity ∇ × (∇ × A) = ∇(∇ ⋅ A) − ∇2 A, and substituting ∇ =

jk :

j k × (j k × E ) = −jωμ0 (jω D)
​

− k × (k × E ) = ω 2 μ 0 D​

Now, substitute D = ϵE :
− k × (k × E ) = ω 2 μ 0 ϵ E ​

Using the vector identity A × (B × C ) = B (A ⋅ C ) − C (A ⋅ B ) with A = k , B = k

, and C = E:

−[k (k ⋅ E ) − E (k ⋅ k )] = ω 2 μ0 ϵE ​

k (k ⋅ E ) − k 2 E + ω 2 μ 0 ϵ E = 0 ​

This is the general wave equation for E in an anisotropic medium. This equation is a
system of three coupled linear equations for the components of E . The existence of
non-trivial solutions for E requires the determinant of the coefficient matrix to be zero,
which leads to the dispersion relation and the possible polarizations. This is often
referred to as the Christoffel equation or Fresnel's equation of wave normals.

Two Sets of Solutions: Ordinary and Extraordinary

Waves

To simplify the analysis and demonstrate the two solutions, we consider a uniaxial
anisotropic medium. In such a medium, two of the principal dielectric constants are
equal, say ϵx = ϵy = ϵo (ordinary permittivity) and ϵz = ϵe (extraordinary
​ ​ ​ ​ ​

permittivity). The dielectric tensor in its principal axes system is:

$\epsilon = \begin{pmatrix} \epsilon_o & 0 & 0 \ 0 & \epsilon_o & 0 \ 0 & 0 & \epsilon_e
\end{pmatrix}$

Let's assume the wave vector k lies in the x-z plane, so k = (kx , 0, kz ). The wave ​ ​

equation can be written in component form. Recall that k ⋅ D = 0, which means

kx Dx + kz Dz = 0. Also, Dx = ϵo Ex , Dy = ϵo Ey , and Dz = ϵe Ez .
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​

Substituting these into the wave equation k (k ⋅ E ) − k 2 E + ω 2 μ 0 ϵ E = 0: ​

For the x-component: kx (kx Ex ​ ​ ​ + kz Ez ) − k 2 Ex + ω 2 μ 0 ϵ o Ex = 0

​ ​ ​ ​ ​ ​

For the y-component: ky (kx Ex + kz Ez ) − k 2 Ey + ω 2 μ0 ϵo Ey = 0 Since ky = 0, this

​ ​ ​ ​ ​ ​ ​ ​ ​

simplifies to: −k 2 Ey + ω 2 μ0 ϵo Ey = 0 Ey (ω 2 μ0 ϵo − k 2 ) = 0
​ ​ ​ ​ ​ ​ ​

For the z-component: kz (kx Ex ​ ​ ​ + kz Ez ) − k 2 Ez + ω 2 μ 0 ϵ e Ez = 0

​ ​ ​ ​ ​ ​

From the y-component equation, we have two possibilities:

Case 1: Ordinary Wave (o-wave)

If Ey ​ = 0, then (ω 2 μ0 ϵo − k 2 ) = 0. This gives the dispersion relation for the ordinary

 ​ ​

wave:

k 2 = ω 2 μ0 ϵo ​ ​

1
Since k , we have k 2 and c0 = ω 2 μ0 ϵ0 n2 = ω 2 μ0 ϵo . Thus, the refractive
ωn
μ0 ϵ 0
= c0 ​
​ ​ = ​
​

​ ​
​ ​ ​ ​

index for the ordinary wave is no = ϵ0 .

ϵo
​
​

​
​ ​

For the ordinary wave, the electric field E is polarized along the y-axis, i.e., E =
(0, Ey , 0). This means E is perpendicular to the plane containing k (x-z plane) and the
​

optical axis (z-axis). The ordinary wave behaves as if it were propagating in an isotropic
medium with permittivity ϵo , regardless of the direction of propagation. The magnetic ​

field H can be found from k × E = ωμ0 H . ​

Case 2: Extraordinary Wave (e-wave)

If Ey ​
= 0, then the electric field lies in the x-z plane, i.e., E = (Ex , 0, Ez ). The x and z ​ ​

components of the wave equation become coupled:

k x (k x E x + k z E z ) − k 2 E x + ω 2 μ 0 ϵ o E x = 0
​ ​ ​ ​ ​ ​ ​ ​ ​ k z (k x E x + k z E z ) − k 2 E z +
​ ​ ​ ​ ​ ​

ω 2 μ 0 ϵ e Ez = 0 ​ ​ ​

These can be rewritten as:

(kx2 − k 2 + ω 2 μ0 ϵo )Ex + (kx kz )Ez = 0 (kz kx )Ex + (kz2 − k 2 + ω 2 μ0 ϵe )Ez = 0

​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​

For a non-trivial solution for Ex and Ez , the determinant of the coefficient matrix must ​ ​

be zero:

$\begin{vmatrix} k_x^2 - k^2 + \omega^2 \mu_0 \epsilon_o & k_x k_z \ k_z k_x & k_z^2
- k^2 + \omega^2 \mu_0 \epsilon_e \end{vmatrix} = 0$

Expanding the determinant and simplifying (using k 2 = kx2 + kz2 and k02 = ω 2 μ0 ϵ0 ):
​ ​ ​ ​ ​

(kx2 − kx2 − kz2 + k02 n2o )(kz2 − kx2 − kz2 + k02 n2e ) − (kx kz )2 = 0
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​

(−kz2 + k02 n2o )(−kx2 + k02 n2e ) − kx2 kz2 = 0

​ ​ ​ ​ ​ ​ ​ ​

kz2 kx2 − kz2 k02 n2e − kx2 k02 n2o + k04 n2o n2e − kx2 kz2 = 0
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​
−kz2 k02 n2e − kx2 k02 n2o + k04 n2o n2e = 0
​ ​ ​ ​ ​ ​ ​ ​

Dividing by k02 n2o n2e : ​ ​ ​

2
kx2
− nkz2 − + k02 = 0
​ ​

n2e
​ ​ ​

o ​ ​

kx2 kz2
+ = k02
​ ​

n2e n2o
​ ​ ​

​ ​

This is the dispersion relation for the extraordinary wave. The refractive index n(θ) for
the extraordinary wave depends on the angle θ between the wave vector k and the
optical axis (z-axis). If kx = k sin θ and kz = k cos θ , and k = k0 n(θ), then: ​ ​ ​

k 2 sin2 θ k 2 cos2 θ
n2e ​
​ + n2o ​
​ = k02 ​

k02 n(θ)2 sin2 θ k02 n(θ)2 cos2 θ

+ = k02
​ ​

n2e n2o
​ ​ ​

​ ​

Dividing by k02 n(θ)2 : ​

sin2 θ cos2 θ 1
n2e ​
​ + n2o ​
​ = n(θ)2 ​

This shows that the phase velocity of the extraordinary wave depends on its direction
of propagation. For the extraordinary wave, the electric field E is generally not
perpendicular to k , but the electric displacement field D is. The magnetic field H can
be found from k × E = ωμ0 H . ​

In summary, the two solutions arise from the anisotropic nature of the permittivity,
leading to two distinct refractive indices and polarization states for waves propagating
in a given direction. The ordinary wave has its electric field perpendicular to the plane
containing the wave vector and the optical axis, while the extraordinary wave has its
electric field in this plane. This phenomenon is known as birefringence.

Diagrams and Illustrations

Index Ellipsoid

The index ellipsoid is a geometric representation that helps visualize the refractive
indices in an anisotropic crystal. For a uniaxial crystal, it is an ellipsoid of revolution, as
shown below. The semi-axes of the ellipsoid correspond to the principal refractive
indices. For a uniaxial crystal, two of the axes are equal (no ), and the third is different (
​

ne ).
​

Wave Propagation and Polarization

When an unpolarized electromagnetic wave enters a uniaxial crystal, it generally splits

into two distinct waves: the ordinary (o-wave) and the extraordinary (e-wave). These
waves travel with different phase velocities and are orthogonally polarized.

Ordinary Wave: The electric field of the ordinary wave is always polarized
perpendicular to the optical axis (and perpendicular to the plane containing the
wave vector and the optical axis). It experiences the ordinary refractive index (no
), which is independent of the direction of propagation.

Extraordinary Wave: The electric field of the extraordinary wave is polarized in

the plane containing the optical axis and the wave vector. It experiences the
extraordinary refractive index (ne ), which depends on the angle between the
​

wave vector and the optical axis.