Uniaxial optical medium ($\mathbf{\epsilon }={\epsilon }_0\mathrm{diag}(n_o^2,n_o^2,n_e^2)$)

A transparent, nonmagnetic linear anisotropic dielectric is uniaxial when two principal permittivities agree. With the optic axis along $z$, its permittivity tensor has the displayed diagonal matrix form. Here $n_o$ and $n_e$ are principal refractive indices at a specified frequency; $n_e$ is not the extraordinary wave's index in every direction.^[ref]

The symmetry direction of a uniaxial optical medium. For a wavevector along it, the two refractive indices coincide: there is no birefringence for that propagation direction, even though $n_e\ne n_o$.

The ordinary wave's electric field is perpendicular to the plane containing the wavevector and optic axis. Its refractive index $n_o$ is direction independent at fixed frequency.

The extraordinary wave's electric displacement field lies in that plane and is perpendicular to the wavevector; its electric field need not be. For angle $\theta$ between the wavevector and optic axis,Thus $n_{\mathrm{ex}}(0)=n_o$ and $n_{\mathrm{ex}}(\pi /2)=n_e$.^[ref]

A short derivation starts from the existing plane wave in an anisotropic dielectric equation. Put $\mathbf{k}=(k_x,0,k_z)$ and $k_0=\omega /c$, with angular frequency $\omega$ and vacuum speed of light $c$. The $y$ component gives the ordinary wave relation $k_x^2+k_z^2=n_o^2{k}_0^2$. The determinant of the $x,z$ system givesSubstituting $k_x=n_{\mathrm{ex}}{k}_0\sin \theta$ and $k_z=n_{\mathrm{ex}}{k}_0\cos \theta$ yields the displayed directional refractive index.

The extraordinary wave's Poynting vector can differ in direction from its wavevector. The latter specifies wave phase propagation, while the former specifies energy flux. This angular separation is walk-off.^[ref]

In the lossless, nondispersive model above, take the gradient of $\omega =c\sqrt{k_x^2/n_e^2+k_z^2/n_o^2}$ to obtain the group velocity, parallel to the Poynting vector. If its angle to the optic axis is $\varphi$, this gives $\tan \varphi =(n_o^2/n_e^2)\tan \theta$. Consequently the extraordinary wave generally carries energy away from its wavevector direction. An internal wavevector normal to a slab face therefore does not guarantee a normal energy path. Birefringent walk-off vanishes at $\theta =0$ and $\theta =\pi /2$; the latter direction still has unequal refractive indices. This is a useful waveplate geometry: overlapping components can acquire phase retardation without spatial splitting. A thin specimen, insufficient imaging resolution, or excitation of only one linear polarization can also hide a double image, for different reasons.