Anisotropic Dielectrics and the Permittivity Tensor

Set a clear calcite crystal on a printed page and you see two copies of every letter. Rotate the crystal and one copy circles the other. Nothing in the last eighteen articles can explain this: we have treated every dielectric as if it responded to a field the same way in every direction, one number χe\chi_e per material. Calcite refuses. Its carbonate groups all lie in parallel planes, so bound charge slides easily within those planes and reluctantly across them. Push with a field at an angle and the medium pushes back in a different direction — the polarization, and with it D\mathbf{D}, tilts away from E\mathbf{E}. This article closes the dielectrics chapter by making that direction-dependence precise. The single number χe\chi_e grows into a 3×33\times3 matrix — the permittivity tensor — and everything we know follows it: why energy conservation forces the tensor to be symmetric, how diagonalization hands every crystal a natural frame of principal axes with just three numbers, and how lattice symmetry prunes those numbers down (cubic crystals, remarkably, back to one). The microscopic ingredients are already on our shelf: the molecular polarizability tensor promised in Induced Dipoles and Atomic Polarizability, and the fragile lattice sum of Local Field and the Lorentz Cavity, whose failure under a 5% tetragonal stretch was anisotropy announcing itself. Master the tensor and you hold the static foundation of crystal optics — birefringence, wave plates, the liquid-crystal pixel you are probably reading this on — and the vocabulary (principal axes, eigenvalues, quadratic forms) the rest of physics reuses for inertia, stress, and conductivity.

Key takeaways
  • In a crystal the dielectric response depends on direction: Di=εijEjD_i=\varepsilon_{ij}E_j with a 3×33\times3 permittivity tensor, and D\mathbf{D} is generally not parallel to E\mathbf{E}.
  • Energy bookkeeping forces the tensor to be symmetric, εij=εji\varepsilon_{ij}=\varepsilon_{ji}, for any lossless, non-gyrotropic medium — otherwise you could harvest energy by cycling the field.
  • A real symmetric tensor can always be rotated to diagonal form: the eigenvectors are the crystal's principal axes, the eigenvalues its three principal permittivities εx,εy,εz\varepsilon_x,\varepsilon_y,\varepsilon_z.
  • Crystal symmetry prunes the count: triclinic keeps 6 independent components, orthorhombic 3 (biaxial), tetragonal/hexagonal/trigonal 2 (uniaxial), cubic exactly 1 — a cubic crystal is optically isotropic, by theorem not by accident.
  • In calcite (εo=2.75 ε0\varepsilon_o=2.75\,\varepsilon_0, εe=2.21 ε0\varepsilon_e=2.21\,\varepsilon_0 at optical frequency) a field at 45∘45^\circ to the optic axis produces a D\mathbf{D} tilted 6.2∘6.2^\circ away from E\mathbf{E} — the static seed of double refraction.
  • The energy density is still u=12E⋅Du=\tfrac12\mathbf{E}\cdot\mathbf{D}; its constant-energy surfaces are ellipsoids whose semi-axes encode the principal indices ni=εi/ε0n_i=\sqrt{\varepsilon_i/\varepsilon_0}.
  • Key idea: in an anisotropic dielectric the linear response is a tensor relation Di=εijEjD_i=\varepsilon_{ij}E_j; symmetry εij=εji\varepsilon_{ij}=\varepsilon_{ji} lets you rotate to principal axes where three numbers — and the geometry of an ellipsoid — carry all the physics.
The core insight

In an anisotropic dielectric the linear response is a tensor relation Di=εijEjD_i=\varepsilon_{ij}E_j; the symmetry εij=εji\varepsilon_{ij}=\varepsilon_{ji} lets you rotate to principal axes where the tensor is diagonal, reducing the physics to three principal permittivities and their refractive indices.

Where this lives today

The permittivity tensor is the most commercially deployed object in this chapter. Every liquid-crystal display pixel is a vat of rod-like molecules forming a tunable uniaxial dielectric: the nematic director (the local optic axis) reorients in an applied field because the molecules' Δε=ε∥−ε⊥\Delta\varepsilon=\varepsilon_\parallel-\varepsilon_\perp makes alignment energetically favorable, and the reoriented tensor then rotates the polarization of transmitted light between crossed polarizers. A billion screens steer their tensors, pixel by pixel, sixty times a second. At the engineering frontier the game is designing εij\varepsilon_{ij} rather than accepting what nature grows. Metamaterials assemble sub-wavelength cells whose effective tensors take values no crystal offers, including principal values of opposite sign (hyperbolic media); transformation optics runs our chapter backwards — choose the field pattern you want and solve for the spatially varying tensor that produces it. Even the humble circuit board participates: woven fiberglass laminate is measurably biaxial, and high-speed designers route differential pairs at an angle to the weave to dodge the resulting skew.

The physical picture

Picture a bound electron in a crystal as a mass caged by springs — stiff springs along one crystallographic direction, soft springs along another. A field applied exactly along a spring direction stretches only those springs, and the induced dipole points along the field. But push at 45∘45^\circ and the charge cheats: it slides farther along the soft direction than the stiff one, so the dipole moment — and, summed over the lattice, the polarization P\mathbf{P} — leans away from E\mathbf{E} toward the compliant axis. The permittivity tensor is nothing more than the bookkeeping matrix of these direction-dependent stiffnesses.

Figure
An anisotropic spring cage tilts the response — calcite numbers, drawn to scale
the cage: stiff along the optic axis few coils = stiff (axis, low eps_e) many coils = soft (plane, high eps_o) z = optic axis is vertical optic axis z E at 45.0 deg D at 51.2 deg P at 55.4 deg rho = 6.2 deg calcite, optical-frequency tensor eps_o = 2.75 eps0 (easy plane) eps_e = 2.21 eps0 (stiff axis) bound charge slides farther in the easy plane, so P and D lean away from the optic axis relative to E tan(theta_D) = (eps_o/eps_e) tan(theta_E) walk-off peaks at 6.26 deg near 41.9 deg

Both microscopic ingredients of this cage are already on our shelf. Induced Dipoles and Atomic Polarizability warned that for anything less symmetric than a sphere, α\alpha is itself a tensor — a CO2_2 molecule polarizes about twice as easily along its O=C=O axis as across it, a debt we pay in full below. And Local Field and the Lorentz Cavity showed that even perfectly isotropic molecules acquire an anisotropic response from an anisotropic lattice: the near-field sum that vanishes identically on a cubic lattice revives at the 14% level under a mere 5% tetragonal stretch, turning the local-field correction — and hence the susceptibility — into a direction-dependent object. Molecular shape and lattice geometry are the two roads into the tensor; most real crystals travel both. The theme of this article is that the tensor is far tamer than nine arbitrary numbers. Energy conservation makes it symmetric; symmetry makes it diagonalizable; and the crystal's own point symmetry then squeezes the three surviving eigenvalues, sometimes all the way back to one.

From molecule to crystal: the tensor law

For an isotropic linear dielectric we wrote P=ε0χeE\mathbf{P}=\varepsilon_0\chi_e\mathbf{E}, one scalar multiplying the total macroscopic field. To let the medium respond differently in different directions, we keep linearity — still valid for the small-field Taylor reasons of Linear Dielectrics, Susceptibility and Permittivity — but promote the single coefficient to a matrix,

Pi=ε0 χij Ej(1)P_i=\varepsilon_0\,\chi_{ij}\,E_j \qquad (1)

where: PiP_i — components of the polarization density (C/m²); EjE_j — components of the total macroscopic field, the repeated index jj summed over x,y,zx,y,z; χij\chi_{ij} — the electric susceptibility tensor, nine dimensionless numbers. Read it as a machine: feed in a field along y^\hat{\mathbf{y}} and the column (χxy,χyy,χzy)(\chi_{xy},\chi_{yy},\chi_{zy}) tells you the polarization it produces, off-axis components included. Diagonal χij=χeδij\chi_{ij}=\chi_e\delta_{ij} recovers the isotropic case exactly. The displacement keeps its universal definition D=ε0E+P\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P} — that equation, from The Electric Displacement D, is bookkeeping and holds in any medium whatsoever. Only the constitutive link closing the equations changes:

Di=ε0Ei+Pi=εijEj,εij=ε0(δij+χij)(2)D_i=\varepsilon_0 E_i+P_i=\varepsilon_{ij}E_j,\qquad \varepsilon_{ij}=\varepsilon_0\left(\delta_{ij}+\chi_{ij}\right) \qquad (2)

where: εij\varepsilon_{ij} — the permittivity tensor (F/m); δij\delta_{ij} — the Kronecker delta carrying the vacuum part. Because χij\chi_{ij} need not be proportional to δij\delta_{ij}, D\mathbf{D} and E\mathbf{E} point in different directions unless E\mathbf{E} lies along one of the special directions we construct below. Gauss's law in dielectrics is untouched: ∇⋅D=ρf\nabla\cdot\mathbf{D}=\rho_f cares only about D\mathbf{D}'s divergence, not about which way it leans relative to E\mathbf{E}. Where do the nine χij\chi_{ij} come from? Sum the microscopic responses, as in the isotropic chapter but keeping directions. If each molecule carries a polarizability tensor αij\alpha_{ij} (so pi=αijEjlocp_i=\alpha_{ij}E^{\mathrm{loc}}_j) and sits in a lattice whose local-field correction is itself directional, χij\chi_{ij} inherits anisotropy from both factors. For CO2_2: α∥≈4.8×10−40 C m2/V\alpha_\parallel\approx4.8\times10^{-40}\ \mathrm{C\,m^2/V} along the molecular axis versus α⊥≈2.5×10−40\alpha_\perp\approx2.5\times10^{-40} across it (polarizability volumes 4.34.3 and 2.2 A˚32.2\ \text{Å}^3) — two-to-one anisotropy in a single molecule. In the gas this washes out: tumbling molecules present the average αˉ=(α∥+2α⊥)/3\bar{\alpha}=(\alpha_\parallel+2\alpha_\perp)/3 and the bulk stays isotropic, as we verify numerically in the worked examples. Freeze the same molecules into an ordered lattice and the anisotropy survives to the macroscale — that is the crystal.

Why the tensor must be symmetric

Nine components suggest nine material constants. Thermodynamics immediately strikes three: for any medium that stores field energy without loss, εij=εji\varepsilon_{ij}=\varepsilon_{ji}. This is not a convenience assumption — its violation would let you build a perpetual-motion machine out of a capacitor, and the derivation shows exactly how.

Derivation ASymmetry of the permittivity tensor from energy bookkeeping

We want to show that εij=εji\varepsilon_{ij}=\varepsilon_{ji} for any lossless dielectric, using nothing but the energy ledger of Energy Stored in Dielectric Systems. The strategy is the same exact-differential argument that proves conservative forces have curl zero: charge the crystal to the same final field along two different paths in "field space," demand that the work invested be the same, and see what that forces on the tensor. Watch for where losslessness enters — a medium that dissipates is under no such obligation, which is why absorbing and gyrotropic media escape the theorem.

Step 1. The work delivered per unit volume by the sources while the displacement grows by dD\mathrm{d}\mathbf{D} is du=E⋅dD\mathrm{d}u=\mathbf{E}\cdot\mathrm{d}\mathbf{D} — this is the same increment we integrated to get u=12D⋅Eu=\tfrac12\mathbf{D}\cdot\mathbf{E} in the isotropic chapter, and its derivation (trickling free charge onto conductors) nowhere assumed D∥E\mathbf{D}\parallel\mathbf{E}. With Di=εijEjD_i=\varepsilon_{ij}E_j linear, du=εijEi dEj\mathrm{d}u=\varepsilon_{ij}E_i\,\mathrm{d}E_j. Why field space? The state of the dielectric is specified by the three numbers (Ex,Ey,Ez)(E_x,E_y,E_z). Charging the crystal is a path from the origin to the final field in this abstract space, and uu is the line integral of εijEi dEj\varepsilon_{ij}E_i\,\mathrm{d}E_j along that path.

Step 2. Charge to the final state (E1,E2,0)(E_1,E_2,0) two ways. Path A: raise ExE_x from 00 to E1E_1 first (cost 12εxxE12\tfrac12\varepsilon_{xx}E_1^2), then raise EyE_y from 00 to E2E_2 while Ex=E1E_x=E_1 sits fixed (cost 12εyyE22+εxyE1E2\tfrac12\varepsilon_{yy}E_2^2+\varepsilon_{xy}E_1E_2, the cross term because the xx-field rides along while DxD_x shifts). Path B: same moves in the opposite order, giving the cross term εyxE1E2\varepsilon_{yx}E_1E_2 instead.

Step 3. Subtract: uB−uA=(εyx−εxy)E1E2u_B-u_A=(\varepsilon_{yx}-\varepsilon_{xy})E_1E_2. If this is nonzero, run path A forward and path B backward: the field returns to zero, the dielectric is exactly as it started, yet the cycle has handed the external circuit the energy (εyx−εxy)E1E2(\varepsilon_{yx}-\varepsilon_{xy})E_1E_2 — a machine that mints energy from nothing, every cycle. Why losslessness is the premise. In an absorbing medium the state at the end of a cycle is not the initial state (heat has been generated), so path-independence is not required and the argument stops. Formally εij\varepsilon_{ij} becomes complex; a static magnetic field (Faraday-active media) adds an antisymmetric part while remaining lossless because B\mathbf{B} breaks time-reversal — both escape the theorem honestly.

Step 4. Forbidding the free lunch forces εxy=εyx\varepsilon_{xy}=\varepsilon_{yx}, and the identical argument in the other two coordinate planes gives εij=εji\varepsilon_{ij}=\varepsilon_{ji}: six independent components, not nine. Equivalently, du\mathrm{d}u is an exact differential and u=12εijEiEju=\tfrac12\varepsilon_{ij}E_iE_j exists as a state function, with εij=∂2u/∂Ei∂Ej\varepsilon_{ij}=\partial^2u/\partial E_i\partial E_j symmetric because mixed partials commute. Verified numerically: charging to (Ex,Ey)=(1.0,2.0)×105(E_x,E_y)=(1.0,2.0)\times10^5 V/m through a symmetric test tensor (εxy=εyx=0.5 ε0\varepsilon_{xy}=\varepsilon_{yx}=0.5\,\varepsilon_0) costs 0.7083350.708335 J/m³ along both paths (equal to 6 digits); repeating with an asymmetric tensor of the same symmetric part (εxy=0.8 ε0\varepsilon_{xy}=0.8\,\varepsilon_0, εyx=0.2 ε0\varepsilon_{yx}=0.2\,\varepsilon_0) gives 0.7610.761 vs 0.6550.655 J/m³ — a cycle would harvest 0.1060.106 J/m³ per lap, the advertised perpetual-motion machine.

With symmetry in hand, the energy density itself comes along for free. Since du=εijEi dEj\mathrm{d}u=\varepsilon_{ij}E_i\,\mathrm{d}E_j is exact precisely when the tensor is symmetric, it integrates to

u=12 εijEiEj=12 E⋅D(3)u=\tfrac12\,\varepsilon_{ij}E_iE_j=\tfrac12\,\mathbf{E}\cdot\mathbf{D} \qquad (3)

where: uu — electrostatic energy density (J/m³), a positive-definite quadratic form for any passive dielectric. The familiar 12E⋅D\tfrac12\mathbf{E}\cdot\mathbf{D} of the isotropic chapter survives verbatim — with the scalar product now doing real work, because the two vectors are no longer parallel. The which-energy honesty of Energy Stored in Dielectric Systems also carries over unchanged: this uu is the free-charge ledger, springs included.

Principal axes: the crystal's own frame

A real symmetric matrix is the friendliest object in linear algebra: it can always be diagonalized by a pure rotation. Physically, every crystal — however low its symmetry — hides three mutually perpendicular directions along which the response is parallel to the push.

Go deeper · Derivation B · 2 minPrincipal axes and the energy ellipsoid
Derivation BPrincipal axes and the energy ellipsoid

We want to show that there exists an orthogonal frame in which εij\varepsilon_{ij} is diagonal, and to extract the geometry that frame carries: constant-energy surfaces that are ellipsoids whose axes encode the three principal permittivities. The route is the spectral theorem, but we walk it concretely — via the quadratic form u(E)u(\mathbf{E}) — because that is the version that generalizes to the inertia tensor and the stress tensor later in the curriculum. Watch for why symmetry is indispensable: an asymmetric matrix need not be diagonalizable by any rotation.

Step 1. Consider the energy u(E)=12εijEiEju(\mathbf{E})=\tfrac12\varepsilon_{ij}E_iE_j on the sphere ∣E∣=E0|\mathbf{E}|=E_0. A continuous function on a sphere attains a maximum; call the maximizing direction e^1\hat{\mathbf{e}}_1. At a constrained maximum the gradient of uu must be parallel to the constraint normal: εijEj=λEi\varepsilon_{ij}E_j=\lambda E_i. The maximizing direction is therefore an eigenvector — a direction where D=λE\mathbf{D}=\lambda\mathbf{E} exactly, response parallel to push.

Step 2. Here symmetry earns its keep: for symmetric εij\varepsilon_{ij}, the subspace perpendicular to e^1\hat{\mathbf{e}}_1 is mapped into itself. Take any v⊥e^1\mathbf{v}\perp\hat{\mathbf{e}}_1; then e^1⋅(εv)=v⋅(εe^1)=λ1 v⋅e^1=0\hat{\mathbf{e}}_1\cdot(\boldsymbol{\varepsilon}\mathbf{v})=\mathbf{v}\cdot(\boldsymbol{\varepsilon}\hat{\mathbf{e}}_1)=\lambda_1\,\mathbf{v}\cdot\hat{\mathbf{e}}_1=0 — the first equality is exactly εij=εji\varepsilon_{ij}=\varepsilon_{ji}. So we can repeat Step 1 inside the perpendicular plane to find e^2\hat{\mathbf{e}}_2, and e^3\hat{\mathbf{e}}_3 follows as the remaining normal. Three mutually orthogonal eigenvectors exist. Why an asymmetric tensor fails. Without εij=εji\varepsilon_{ij}=\varepsilon_{ji} the perpendicular subspace leaks — εv\boldsymbol{\varepsilon}\mathbf{v} acquires a component along e^1\hat{\mathbf{e}}_1 — and the construction stalls after one step. Orthogonal principal axes are a privilege of symmetric tensors.

Step 3. In the frame (e^1,e^2,e^3)(\hat{\mathbf{e}}_1,\hat{\mathbf{e}}_2,\hat{\mathbf{e}}_3) the tensor is diag(εx,εy,εz)\mathrm{diag}(\varepsilon_x,\varepsilon_y,\varepsilon_z): the three eigenvalues are the principal permittivities, all positive for a passive dielectric (each is 2u/E02>02u/E_0^2>0 along its axis). The whole 3×33\times3 machine reduces to component-by-component multiplication: D1=εxE1D_1=\varepsilon_xE_1, D2=εyE2D_2=\varepsilon_yE_2, D3=εzE3D_3=\varepsilon_zE_3. Every anisotropic problem strategy in this subject is now visible: rotate in, scale, rotate out.

Step 4. The geometry: in the principal frame the constant-energy condition u=constu=\mathrm{const} reads εxEx2+εyEy2+εzEz2=2u\varepsilon_xE_x^2+\varepsilon_yE_y^2+\varepsilon_zE_z^2=2u, an ellipsoid in E\mathbf{E}-space with semi-axes ∝1/εi\propto1/\sqrt{\varepsilon_i}. Re-expressed in D\mathbf{D} (using Ei=Di/εiE_i=D_i/\varepsilon_i), ∑iDi2/εi=2u\sum_iD_i^2/\varepsilon_i=2u — an ellipsoid whose semi-axes are ∝εi\propto\sqrt{\varepsilon_i}, i.e. proportional to the principal refractive indices ni=εi/ε0n_i=\sqrt{\varepsilon_i/\varepsilon_0}. Normalized, this D\mathbf{D}-space ellipsoid ∑ixi2/ni2=1\sum_ix_i^2/n_i^2=1 is the index ellipsoid of crystal optics; here it is simply the level surface of our energy ledger. Verified: for ε=diag(2.75,2.75,2.21) ε0\boldsymbol{\varepsilon}=\mathrm{diag}(2.75,2.75,2.21)\,\varepsilon_0 (calcite) and E\mathbf{E} at 45 deg with E=2.0×105E=2.0\times10^5 V/m, the three routes to the energy — the quadratic form 12εijEiEj\tfrac12\varepsilon_{ij}E_iE_j, the dot product 12E⋅D\tfrac12\mathbf{E}\cdot\mathbf{D}, and 12∣E∣∣D∣cos⁡6.23°\tfrac12|\mathbf{E}||\mathbf{D}|\cos6.23° — all give 0.4390.439 J/m³.

One geometric consequence is worth carrying in your head: since Di=∂u/∂EiD_i=\partial u/\partial E_i, D\mathbf{D} points along the normal to the constant-uu ellipsoid at the tip of E\mathbf{E} — and an ellipse's normal is parallel to its radius only on the axes. That is the picture behind every walk-off angle we compute:

Figure
Diagonalization is a rotation — the energy ellipse in a generic frame and in the principal frame
arbitrary lab frame: 6 numbers E D (normal) eps has off-diagonal entries; D leaves the E direction rotate principal frame: 3 numbers prop. to n_x prop. to n_z eps = diag(eps_x, eps_y, eps_z); on an axis, D is parallel to E

What crystal symmetry does to the tensor

Six independent components is the worst case — triclinic, no symmetry beyond the lattice itself. Every symmetry operation the crystal possesses is a constraint, by Neumann's principle: a macroscopic property tensor must be invariant under every point-symmetry operation of its crystal, R ε RT=εR\,\boldsymbol{\varepsilon}\,R^{\mathsf{T}}=\boldsymbol{\varepsilon} for each symmetry rotation or reflection RR — the property cannot be choosier than the crystal that hosts it. The payoff is a complete classification, and one genuinely surprising theorem.

Go deeper · Derivation C · 2 minHow symmetry prunes the tensor and why cubic crystals are exactly isotropic
Derivation CHow symmetry prunes the tensor and why cubic crystals are exactly isotropic

We want the count of independent permittivity components for each crystal family, and in particular the cubic surprise: a crystal with manifestly special directions (a cube's edges are nothing like its diagonals) whose dielectric response is nonetheless perfectly isotropic. The strategy is to apply Neumann's principle operation by operation, starting from the general symmetric tensor with entries ε11,ε22,ε33,ε12,ε13,ε23\varepsilon_{11},\varepsilon_{22},\varepsilon_{33},\varepsilon_{12},\varepsilon_{13},\varepsilon_{23}. The thing to watch is what each symmetry kills: reflections and 2-fold axes kill off-diagonals; higher-fold axes go further and equate diagonals.

Step 1. Orthorhombic (three perpendicular 2-fold axes): a 180° rotation about zz sends (x,y,z)→(−x,−y,z)(x,y,z)\to(-x,-y,z), so ε13→−ε13\varepsilon_{13}\to-\varepsilon_{13} and ε23→−ε23\varepsilon_{23}\to-\varepsilon_{23}; invariance forces both to vanish. The xx-axis rotation kills ε12\varepsilon_{12}. The tensor is diagonal in the crystallographic frame with three distinct entries — the crystal axes are the principal axes, and the crystal is biaxial: 3 independent components.

Step 2. Tetragonal, hexagonal, trigonal (one axis of order n≥3n\ge3, along zz): invariance under rotation by 2π/n2\pi/n about zz requires the 2×22\times2 block in the xyxy-plane to commute with a non-trivial rotation. A symmetric 2×22\times2 matrix invariant under rotation by anything other than 180°180° must be proportional to the identity — its two eigenvalues would otherwise pick out two special perpendicular directions in the plane, which the nn-fold axis forbids. Hence ε11=ε22≡εo\varepsilon_{11}=\varepsilon_{22}\equiv\varepsilon_o, off-diagonals zero, ε33≡εe\varepsilon_{33}\equiv\varepsilon_e free: 2 independent components, the uniaxial family, with the nn-fold axis as the optic axis. Why order 3 already suffices. A 3-fold axis rotates by 120°; the only symmetric tensors unmoved by 120° in a plane are isotropic in that plane. This is why trigonal calcite is uniaxial even though its symmetry looks "lower" than tetragonal.

Step 3. Cubic: the zz-axis 4-fold makes the xyxy block isotropic as in Step 2, so ε11=ε22\varepsilon_{11}=\varepsilon_{22}; but a cube has equivalent 4-fold axes along xx and yy too, forcing ε22=ε33\varepsilon_{22}=\varepsilon_{33} by the same argument. All three eigenvalues coincide: ε=ε 1\boldsymbol{\varepsilon}=\varepsilon\,\mathbb{1}, one independent component. Rock salt, diamond, and silicon are dielectrically indistinguishable from glass — every direction, not just the symmetry directions, sees the same ε\varepsilon, because a symmetric tensor with three equal eigenvalues is a multiple of the identity, full stop. Why the cube's "special directions" leave no trace. The permittivity tensor is a quadratic-form probe — it can only distinguish directions by assigning them different eigenvalues, and the cube's symmetry leaves no room for unequal eigenvalues. Cubic anisotropy is real but invisible at rank 2; it resurfaces in rank-4 properties like elasticity, where single-crystal silicon is measurably direction-dependent.

Step 4. The census: triclinic 6, monoclinic 4 (its single 2-fold kills two off-diagonals), orthorhombic 3 (biaxial: mica, topaz), tetragonal/hexagonal/trigonal 2 (uniaxial: quartz, calcite, rutile), cubic 1 (isotropic: NaCl, diamond). Melt or powder any of them and orientational averaging returns the scalar — which is why this entire chapter could ignore the tensor until its final article. Verified: the counts 6/4/3/2/1 match Nye's standard crystallographic tables (Physical Properties of Crystals, Table 3), and the uniaxial assignments match the handbook data quoted below.

Figure
The dielectric census — one ellipsoid per crystal family
cubic: sphere, 1 number NaCl, diamond, Si — isotropic optic axis uniaxial: spheroid, 2 numbers calcite 2.75/2.21 - quartz 2.38/2.41 biaxial: 3 distinct axes mica, topaz — two optic axes constant-energy surfaces of u = D-squared components over eps_i (drawn in cross-section) more symmetry = rounder ellipsoid: the crystal family is readable from the shape alone

Anchor numbers, at 589 nm where the anisotropic tensor is best tabulated: calcite is negative uniaxial with εo=no2 ε0=2.75 ε0\varepsilon_o=n_o^2\,\varepsilon_0=2.75\,\varepsilon_0, εe=2.21 ε0\varepsilon_e=2.21\,\varepsilon_0 (no=1.658n_o=1.658, ne=1.486n_e=1.486); quartz is weakly positive (no=1.544n_o=1.544, ne=1.553n_e=1.553); rutile dramatically so (no=2.616n_o=2.616, ne=2.903n_e=2.903). One honesty note from Linear Dielectrics: ε=ε0n2\varepsilon=\varepsilon_0n^2 is the optical-frequency tensor. At DC the ionic lattice adds its own stretch response — calcite's static values sit near 8.5 ε08.5\,\varepsilon_0 and 8.0 ε08.0\,\varepsilon_0 — and the full frequency story is dispersion, deferred to the waves chapter.

Limiting cases

CaseResultMeaning
Isotropic / cubicεij=ε δij\varepsilon_{ij}=\varepsilon\,\delta_{ij}, D∥E\mathbf{D}\parallel\mathbf{E} alwaysThe whole chapter until now; also any melt, glass, gas, or powder by orientational averaging.
E\mathbf{E} along a principal axisD=εiE\mathbf{D}=\varepsilon_i\mathbf{E}, walk-off ρ=0\rho=0Anisotropy hides: each axis behaves as an ordinary scalar dielectric with its own εi\varepsilon_i.
Uniaxial (εx=εy=εo≠εe\varepsilon_x=\varepsilon_y=\varepsilon_o\neq\varepsilon_e)one optic axis; tan⁡θD=(εo/εe)tan⁡θE\tan\theta_D=(\varepsilon_o/\varepsilon_e)\tan\theta_ECalcite at 45°45°: D\mathbf{D} leans 6.2°6.2° from E\mathbf{E}; maximum 6.26°6.26° at θE=41.9°\theta_E=41.9°.
Weak anisotropy (εe→εo\varepsilon_e\to\varepsilon_o)ρmax⁡≈(εo−εe)/2εo\rho_{\max}\approx(\varepsilon_o-\varepsilon_e)/2\varepsilon_o radQuartz's 0.6%0.6\% index split gives walk-off of only ∼0.3°\sim0.3° — why calcite, not quartz, shows double images to the naked eye.
Lossy or magneto-active mediumεij\varepsilon_{ij} complex / non-symmetricThe energy theorem's premise fails; Hermitian and gyrotropic parts appear (Faraday effect) — beyond statics.

Worked examples

Example 1 — calcite: how far does D lean? A slab of calcite carries a uniform field E=2.0×105E=2.0\times10^5 V/m at 45.0°45.0° to the optic axis (optical-frequency tensor: εo=2.749 ε0\varepsilon_o=2.749\,\varepsilon_0, εe=2.208 ε0\varepsilon_e=2.208\,\varepsilon_0). Find D\mathbf{D}, the walk-off angle, and the stored energy density.

Step 1. Set up in the principal frame, optic axis along zz, field in the xzxz-plane: Ex=Esin⁡45°=1.414×105E_x=E\sin45°=1.414\times10^5 V/m, Ez=Ecos⁡45°=1.414×105E_z=E\cos45°=1.414\times10^5 V/m.

Step 2. Apply the diagonal tensor component by component: Dx=εoEx=2.749×(8.854×10−12)×1.414×105=3.44×10−6D_x=\varepsilon_oE_x=2.749\times(8.854\times10^{-12})\times1.414\times10^5=3.44\times10^{-6} C/m², and Dz=εeEz=2.77×10−6D_z=\varepsilon_eE_z=2.77\times10^{-6} C/m². The magnitude is ∣D∣=4.42×10−6|\mathbf{D}|=4.42\times10^{-6} C/m².

Step 3. Direction of D\mathbf{D}: tan⁡θD=Dx/Dz=(εo/εe)tan⁡θE=1.245\tan\theta_D=D_x/D_z=(\varepsilon_o/\varepsilon_e)\tan\theta_E=1.245, so θD=51.2°\theta_D=51.2° — the displacement leans toward the easy plane. The walk-off angle is ρ=θD−θE=\rho=\theta_D-\theta_E= 6.2°. (Sweeping θE\theta_E shows the lean maximizes at 6.26°6.26° for θE=41.9°\theta_E=41.9°, the analytic maximum arctan⁡[(r−1)/2r ]\arctan[(r-1)/2\sqrt{r}\,] with r=εo/εer=\varepsilon_o/\varepsilon_e.)

Step 4. Energy density: u=12E⋅D=12(ExDx+EzDz)=u=\tfrac12\mathbf{E}\cdot\mathbf{D}=\tfrac12(E_xD_x+E_zD_z)= 0.439 J/m³. Cross-checks: the quadratic form 12(εoEx2+εeEz2)\tfrac12(\varepsilon_oE_x^2+\varepsilon_eE_z^2) and the projection 12∣E∣∣D∣cos⁡6.2°\tfrac12|\mathbf{E}||\mathbf{D}|\cos6.2° give the same 0.4390.439 J/m³ — the cos⁡ρ\cos\rho is the anisotropy tax on the naive 12∣E∣∣D∣\tfrac12|E||D|.

Example 2 — CO2, from one molecule to a gas. The CO2_2 polarizability tensor has α∥=4.80×10−40\alpha_\parallel=4.80\times10^{-40} C·m²/V along the molecular axis and α⊥=2.46×10−40\alpha_\perp=2.46\times10^{-40} C·m²/V across it. (a) A field E=3.0×106E=3.0\times10^6 V/m is applied at 45°45° to a molecule's axis — where does p\mathbf{p} point? (b) Predict the dielectric constant of CO2_2 gas at 20 °C, 1 atm, and compare with the measured κ=1.000922\kappa=1.000922.

Step 1. Single molecule, principal frame along the O=C=O axis: p∥=α∥Ecos⁡45°=1.02×10−33p_\parallel=\alpha_\parallel E\cos45°=1.02\times10^{-33} C·m, p⊥=α⊥Esin⁡45°=5.2×10−34p_\perp=\alpha_\perp E\sin45°=5.2\times10^{-34} C·m. Direction: tan⁡θp=(α⊥/α∥)tan⁡45°=0.513\tan\theta_p=(\alpha_\perp/\alpha_\parallel)\tan45°=0.513, so p\mathbf{p} sits 27.2°27.2° from the axis — tilted 17.8°17.8° away from E\mathbf{E}, toward the easy axis. A single molecule is a miniature calcite.

Step 2. In the gas the molecules tumble, and the orientation average of the tensor is the scalar αˉ=(α∥+2α⊥)/3=3.24×10−40\bar{\alpha}=(\alpha_\parallel+2\alpha_\perp)/3=3.24\times10^{-40} C·m²/V — one soft axis, two stiff ones, weighted equally. The macroscopic response is isotropic even though no single molecule is.

Step 3. Number density at 20 °C, 1 atm: N=p/kBT=101325/(1.381×10−23×293.15)=2.50×1025N=p/k_BT=101325/(1.381\times10^{-23}\times293.15)=2.50\times10^{25} m⁻³. The gas is dilute enough (Nαˉ/3ε0∼10−4N\bar{\alpha}/3\varepsilon_0\sim10^{-4}) that the Clausius–Mossotti denominator is negligible and χe≈Nαˉ/ε0=9.16×10−4\chi_e\approx N\bar{\alpha}/\varepsilon_0=9.16\times10^{-4}.

Step 4. Prediction: κ=1+χe=\kappa=1+\chi_e= 1.000916, against the measured 1.0009221.000922 — agreement to better than 1% of the susceptibility. The anisotropy has not vanished, only hidden: light scattered from the gas is measurably depolarized because individual molecules still respond off-axis, which is exactly how α∥−α⊥\alpha_\parallel-\alpha_\perp is measured. Freeze the gas into an oriented crystal and the 17.8°17.8° tilt of Step 1 would survive to the macroscale.

A crystal's history

Origins · 1 minAugustin-Jean Fresnel and how this was discovered
AF
Augustin-Jean Fresnel
1788–1827 · École des Ponts et Chaussées; the wave theory of light

Fresnel built the physics of anisotropic media a full generation before anyone could write a tensor. Working from Bartholin's double images and Huygens's two wavefronts in calcite, he made the decisive leap in 1821: light is a transverse vibration, so a crystal can assign different speeds to different vibration directions. From that single idea he reduced every known crystal-optical effect to three principal velocities along three perpendicular axes — principal-axis decomposition of a symmetric tensor, discovered as geometry seventy-seven years before Voigt coined the word "tensor" for exactly this kind of object. A road engineer who did optics between postings and died at 39, Fresnel is why the index ellipsoid, the walk-off construction, and the uniaxial index formula all carry his fingerprints.

History
1669
— Erasmus Bartholin publishes the double refraction of "Iceland spar" (calcite): one ray obeys Snell, the other refuses
1690
— Huygens's Traité de la Lumière derives the extraordinary ray from a spheroidal secondary wavefront — the uniaxial ellipsoid, avant la lettre
1808
— Malus discovers polarization by reflection; the two calcite images are recognized as two polarizations
1821
— Fresnel argues light is transverse and founds crystal optics on three principal axes with three principal speeds
1898
— Woldemar Voigt coins the word "tensor" in his crystal-physics lectures — the permittivity tensor finally gets its name
1971
— Schadt and Helfrich demonstrate the twisted-nematic liquid-crystal cell: a voltage-steered permittivity tensor in every pixel

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §4.4 (linear dielectrics; anisotropic remark) · Jackson, Classical Electrodynamics, 3rd ed., §4.5 and §7.10 (crystal optics) · Zangwill, Modern Electrodynamics, §6.5 (dielectric tensor) · Born & Wolf, Principles of Optics, Ch. 15 (crystal optics, index ellipsoid) · Landau & Lifshitz, Electrodynamics of Continuous Media, §13, §97-98 · Nye, Physical Properties of Crystals, Ch. 1-3 (Neumann's principle, tensor census)

Provenance. Tensor constitutive relation, symmetry from energy, diagonalization, index ellipsoid, and the uniaxial Fresnel equation follow Jackson §4.5 & §7.10, Born & Wolf Ch. 15, and Landau-Lifshitz ECM §13,97. Verifier should recompute: (1) symmetry of εij\varepsilon_{ij} from exactness of du=Ei dDi\mathrm{d}u=E_i\,\mathrm{d}D_i; (2) the uniaxial index formula 1/ne2(θ)=cos⁡2θ/no2+sin⁡2θ/ne21/n_e^2(\theta)=\cos^2\theta/n_o^2+\sin^2\theta/n_e^2 from the plane-wave wave equation; (3) calcite values no≈1.658n_o\approx1.658, ne≈1.486n_e\approx1.486 at 589 nm (standard handbook values); (4) the component counts (cubic 1, uniaxial 2, biaxial 3) against Nye's crystal-symmetry tables.

Builds on: Linear Dielectrics, Susceptibility and Permittivity · The Electric Displacement D · Polarization Density P · Induced Dipoles and Atomic Polarizability · Gauss's Law in Dielectrics · Local Field and the Lorentz Cavity Leads to: Boundary Conditions for D and E · Magnetostatics (the chapter that opens next) · Birefringence and Crystal Optics · Electro-optic Effect (Pockels and Kerr)

Is D still defined as eps0 E + P in a crystal, or does the tensor replace that definition?

The definition never changes: D=ε0E+P\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P} is pure bookkeeping and holds in every medium. What the tensor replaces is only the constitutive link — the material's answer to "what P does this E produce?", which was ε0χeE\varepsilon_0\chi_e\mathbf{E} and is now ε0χijEj\varepsilon_0\chi_{ij}E_j. Everything built on the definition alone — ∇⋅D=ρf\nabla\cdot\mathbf{D}=\rho_f, the pillbox condition on D⊥D_\perp, bound-charge formulas — survives untouched.

If D and E point in different directions, which one do I trust for Gauss's law and field lines?

Both, for their own jobs. ∇⋅D=ρf\nabla\cdot\mathbf{D}=\rho_f still holds exactly, so D\mathbf{D}-lines begin and end only on free charge; ∇×E=0\nabla\times\mathbf{E}=0 still holds in statics, so E\mathbf{E} remains the gradient of a potential. The novelty is that these are now genuinely different vector fields whose line maps do not coincide. The Gauss's Law in Dielectrics ritual still works for D\mathbf{D}, but its last step E=D/ε\mathbf{E}=\mathbf{D}/\varepsilon becomes the matrix solve Ei=(ε−1)ijDjE_i=(\varepsilon^{-1})_{ij}D_j.

How can a cubic crystal be isotropic when a cube obviously has special directions?

The cube's specialness is real but invisible to a rank-2 tensor, which can only express anisotropy by giving different eigenvalues to different directions — and the cube's rotations force all three equal. After that, every direction (edge, face diagonal, body diagonal) gives D=εE\mathbf{D}=\varepsilon\mathbf{E} identically. Higher-rank properties have more room: the rank-4 elastic tensor of cubic silicon keeps three independent constants and is measurably anisotropic, which is why wafer flats mark crystal orientation even though silicon's permittivity is a scalar.

My handbook lists refractive indices, not permittivities. How do I get the tensor from n_o and n_e?

At optical frequencies εi=ε0ni2\varepsilon_i=\varepsilon_0n_i^2 along each principal axis, so calcite's no=1.658n_o=1.658, ne=1.486n_e=1.486 give εo=2.75 ε0\varepsilon_o=2.75\,\varepsilon_0, εe=2.21 ε0\varepsilon_e=2.21\,\varepsilon_0 — the values used throughout this article. The honesty clause: that is the electronic response only. At low frequency the ionic lattice also stretches, and calcite's static tensor is near 8.5 ε0/8.0 ε08.5\,\varepsilon_0/8.0\,\varepsilon_0 — three times the optical values. Which tensor you need depends on which field you apply; the frequency dependence (and why n2=εrn^2=\varepsilon_r only at matching frequencies) is the dispersion story of the waves chapter.

Does an anisotropic crystal have a permanent polarization?

Not by virtue of anisotropy alone. The susceptibility tensor describes response: switch the field off and P\mathbf{P} returns to zero, however anisotropic χij\chi_{ij} may be. Spontaneous polarization is a stronger condition — it needs a polar crystal class, the subject of Ferroelectrics and Spontaneous Polarization. The properties do interact: all ferroelectrics are strongly anisotropic, but most anisotropic crystals (calcite, quartz) are not ferroelectric.

Where does the number six come from for the general tensor?

Count the entries of a symmetric 3×33\times3 matrix: three diagonal plus three independent off-diagonal (each equal to its mirror partner) — six. The energy argument cut nine to six; crystal symmetry then cuts to 3 (orthorhombic), 2 (uniaxial), or 1 (cubic). Six is also why a triclinic crystal needs its principal axes found by diagonalization: nothing in the lattice hands them to you.

Is the 6-degree walk-off in the worked example the same thing as birefringence?

Same root, different phenomena. We computed the static misalignment of D\mathbf{D} and E\mathbf{E}. Birefringence is the wave consequence: a wave in a uniaxial crystal splits into two polarizations with different indices (non_o and ne(θ)n_e(\theta)), and the extraordinary ray's energy walks off from its wavefront normal by an angle governed by exactly our ρ\rho. The calcite double image is the wave phenomenon; its size is set by the static tensor computed here. The full machinery — index-ellipsoid slicing, the Fresnel equation, wave plates — is crystal optics in the electromagnetic-waves chapter, where we hand this off.

Before this

Where this leads