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A linear anisotropic dielectric has a direction-dependent linear relation between the electric displacement field and electric field:The permittivity tensor need not be proportional to the identity matrix, so the two fields need not be parallel. A corresponding linear magnetic response is $B_i={\mu }_{ij}{H}_j$, where $\mathbf{\mu }$ is the magnetic permeability tensor.

For an instantaneous, nondispersive, lossless model with real time-independent symmetric matrices, the field energy density isDifferentiation gives ${\partial }_{t}u=\mathbf{E}\cdot {\partial }_t\mathbf{D}+\mathbf{H}\cdot {\partial }_t\mathbf{B}$, the storage term in Poynting theorem in a linear anisotropic medium. Positivity of this quadratic energy density requires positive-definite response tensors in this model. Dispersive or absorptive material response requires a more general energy accounting.

Birefringence is polarization-dependent propagation in an optically anisotropic medium. Its permittivity tensor relates $\mathbf{D}$ to $\mathbf{E}$. In a lossless uniaxial dielectric, take the optic axis along $z$ and the principal relative permittivities as $(n_o^2,n_o^2,n_e^2)$. For a wave vector making angle $\vartheta$ with that axis, the ordinary mode has index $n_o$, while the extraordinary phase-index angular dependence isThe ordinary electric field is perpendicular to the plane containing the wave vector and optic axis. The extraordinary electric displacement field lies in that plane and is perpendicular to the wave vector, but its electric field need not be. Its energy-flow direction can therefore differ from the wave-normal direction: birefringent walk-off. Along the optic axis the two indices coincide. Refraction generally splits an incident beam into two differently polarized rays, producing double images, as in calcite.

A waveplate instead uses the relative phase of two eigenpolarizations. With the axes chosen for the propagation direction, the retardance is $\Delta =2\pi L(n_2-n_1)/{\lambda }_0$. A quarter-wave plate gives $|\Delta |=\pi /2$ modulo $2\pi$ and converts equal-amplitude orthogonal components between linear and circular polarization. A half-wave plate gives $|\Delta |=\pi$ and rotates a linear polarization. This is a phase effect; unequal absorption is a separate property.^[ref]

At normal incidence, continuity of tangential wave phase makes both internal wavevectors normal to the parallel faces. For the ordinary wave in a uniaxial optical medium, the electric field is an eigenpolarization of the permittivity tensor, so $\mathbf{D}\parallel \mathbf{E}$. ThusIts energy path is also normal to the faces. The extraordinary wave differs because $\mathbf{D}\perp \mathbf{k}$ but $\mathbf{E}$ need not be perpendicular to $\mathbf{k}$. In a nonmagnetic crystal, $\mathbf{H}\parallel \mathbf{B}$ and $\mathbf{S}=\mathbf{E}\times \mathbf{H}$ is perpendicular to $\mathbf{E}$. The vectors $\mathbf{D},\mathbf{E},\mathbf{k},\mathbf{S}$ lie in the same meridional plane, with $\mathbf{D}\perp \mathbf{k}$ and $\mathbf{E}\perp \mathbf{S}$. Therefore the electric-displacement angle from birefringent walk-off equals the angular separation between $\mathbf{S}$ and $\mathbf{k}$.

The observed lateral separation is accumulated over thickness $L$, so $\Delta =L\tan {\alpha }_e$. With $L=10\mathrm{mm}$ and $\Delta =0.1\mathrm{mm}$,The two emergent rays are parallel because their outgoing wavevectors are both normal to the faces in air; their internal energy paths can nevertheless be displaced by birefringent walk-off. The angle quoted is the positive angle between the field vectors, so the unspecified lateral direction does not affect its value. The unpolarized light excites both eigenpolarizations; it does not require their energy paths to coincide.

Flint glass is approximately optically isotropic without an applied magnetic field. A longitudinal magnetic field changes the motion of its bound electrons: the magnetic part of the Lorentz force acts with opposite sense on the two circularly rotating components of a driven electronic oscillation. Their optical response, and hence their refractive indices, split. Equivalently, the transverse permittivity tensor acquires a field-odd antisymmetric contribution whose eigenvectors are opposite circular polarizations.

The resulting circular birefringence gives Faraday rotation. For a weak longitudinal bias it is linear in the field:Here ${\rm Y}$ is the Verdet constant and $L$ the path length. The magnetic field breaks time-reversal symmetry, which is why this rotation is nonreciprocal. Reversing the bias reverses the rotation; reversing propagation while keeping the bias fixed does not undo it when physical polarization directions are compared in fixed laboratory axes. A transverse magnetic bias can have different, higher-order effects; the setup here is the longitudinal Faraday effect.

From molecular dipoles to macroscopic fields. A dielectric contains charges which remain bound to its atoms or molecules. An applied electric field displaces positive and negative charge centres and can orient permanent electric dipoles. The electric polarization $\mathbf{P}$ is electric dipole moment per unit volume, in $\mathrm{C}{\mathrm{m}}^{-2}$. Its associated bound charge isat the boundary of the polarized material, where $\hat{\mathbf{n}}$ points outward. A uniform electric polarization has no volume bound charge, but generally has surface bound charge. The electric field remains the force per unit test charge; it includes the fields of both free and bound charges.

Introducing the electric displacement field $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$ packages the free-charge source separately:For a homogeneous linear isotropic dielectric,The electric susceptibility ${\chi }_e$ is dimensionless; the permittivity $ϵ$ has units $\mathrm{F}{\mathrm{m}}^{-1}$. This is a constitutive approximation. An anisotropic dielectric requires a permittivity tensor, while a nonlinear dielectric need not have electric polarization proportional to electric field. The field in these equations is the macroscopic field, not necessarily the local field at an individual molecule.

Interfaces. Let $\hat{\mathbf{n}}$ point from medium 1 into medium 2. Integrating Gauss's law over a thin pillbox, and the electrostatic curl equation around a thin loop, gives the electrostatic interface boundary conditionsThus normal electric displacement is continuous when there is no free surface charge, whereas tangential electric field is continuous even if the permittivity changes. Normal electric field generally jumps because the interface carries bound charge. The electric potential is continuous for an ordinary interface without an ideal dipole sheet. These conditions, together with Laplace's equation in each homogeneous charge-free region, determine the field.

Energy and electrical sources. Reversibly assembling free charge requires incremental work $\delta W=\int \mathbf{E}\cdot \delta \mathbf{D}dV$. For a linear dielectric whose response can be reversibly integrated from zero field, the electrostatic energy isThis includes the material polarization response, rather than simply using ${ϵ}_0{E}^2/2$ everywhere. In a capacitor, $W=Q^2/(2C)=C{V}^2/2$. Inserting a dielectric increases capacitance. At fixed charge the stored energy decreases. At fixed voltage the stored field energy increases, but the voltage source supplies $VdQ$; the appropriate mechanical energy including that source changes by $-V^{2}dC/2$. Both cases predict attraction of dielectric into the capacitor. A force calculation must state which electrical quantity is fixed. A lossy or hysteretic dielectric also dissipates energy, so the reversible quadratic expression does not describe its full history.

Shape and depolarization. Suppose the dielectric is surrounded by vacuum and the distant imposed electric field is ${\mathbf{E}}_0$. Bound surface charge produces a depolarization field. For a uniformly polarized ellipsoid, along a principal axis,Here $N_d$ is the depolarization factor of a dielectric ellipsoid. This self-consistent field explains why the same material responds differently in different shapes.

An infinite slab with electric field normal to its faces has $N_d=1$, giving $E_{\mathrm{in}}=E_0/{ϵ}_r$; its opposite surface bound charges reduce the internal field. For electric field parallel to the faces, $N_d=0$, giving $E_{\mathrm{in}}=E_0$. End effects are neglected.

A very long circular rod in an axial electric field has negligible depolarization away from its ends, so $E_{\mathrm{in}}\simeq E_0$. In a transverse electric field the infinite-cylinder solution has $N_d=1/2$, giving $E_{\mathrm{in}}=2E_0/({ϵ}_r+1)$. To see the latter directly, take the applied field along $x$. The interior and exterior electric potentials can be written ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +A\cos \theta /r$. Continuity of potential and normal electric displacement at $r=a$ gives $B=2E_0/({ϵ}_r+1)$.

For a sphere of radius $a$, $N_d=1/3$ andFor example, write ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +C\cos \theta /r^2$. Continuity of potential gives $B=E_0-C/a^3$, and continuity of normal electric displacement gives $ϵB={ϵ}_0(E_0+2C/a^3)$. Solving yields $B=3E_0/({ϵ}_r+2)$ and $C=a^3{E}_0({ϵ}_r-1)/({ϵ}_r+2)$. The $1/r^2$ potential is that of the electric dipole moment above. For ${ϵ}_r=1$ all these induced effects vanish; as ${ϵ}_r$ becomes very large the normal-field sphere and slab approach the corresponding conductor field exclusion. The ideal infinitely long rod in an exactly axial imposed field has a different, singular end-free limit.

The permittivity tensor is the linear material-response matrix relating electric displacement field to electric field at a specified frequency. An isotropic local medium has ${ϵ}_{ij}=ϵ{\delta }_{ij}$. A magnetically biased transparent medium can have an imaginary antisymmetric transverse contribution; its eigenmodes can have opposite circular polarizations and different refractive indices.

In a homogeneous lossless dielectric, let $\mathbf{D}=\mathbf{\epsilon }\mathbf{E}$ with a real symmetric permittivity tensor, and $\mathbf{B}=\mu \mathbf{H}$ with scalar positive magnetic permeability. A source-free plane electromagnetic wave with angular frequency $\omega \ne 0$, real wavevector $\mathbf{k}$ and electric field amplitude $\mathbf{e}$ satisfiesThe second equation follows from Gauss's law. It makes the electric displacement field transverse; the electric field itself need not be transverse in an anisotropic medium.

If $\mathbf{e}$ is an eigenvector of the permittivity tensor with nonzero eigenvalue ${\epsilon }_e$, then ${\epsilon }_e\mathbf{k}\cdot \mathbf{e}=0$, so $\mathbf{e}\perp \mathbf{k}$. The first equation reduces to the dispersion relation $|\mathbf{k}{|}^2={\omega }^2\mu {\epsilon }_e$. For a propagating solution the time-averaged Poynting vector is $⟨\mathbf{S}⟩=|\mathbf{e}{|}^2\mathbf{k}/(2\omega \mu )$ and is parallel to the wavevector.

A general anisotropic eigenmode need not have its electric field along a principal dielectric direction. Its energy flux can then depart from the wavevector, producing spatial walk-off. The nonzero-${\epsilon }_e$ assumption matters: at a zero permittivity eigenvalue, Gauss's law alone no longer forces $\mathbf{k}\cdot \mathbf{e}=0$.^[ref]