Electric-displacement angle from birefringent walk-off ($|\angle (\mathbf{E},\mathbf{D})|=|\angle (\mathbf{S},\mathbf{k})|$)

For a lossless nonmagnetic uniaxial optical medium, the electric displacement field of a plane electromagnetic wave is perpendicular to its wavevector. The magnetic field is perpendicular to the plane containing the electric field and wavevector, while the Poynting vector is perpendicular to the electric field in that same plane. Thus the angle between $\mathbf{E}$ and $\mathbf{D}$ equals the walk-off angle between energy flow and phase propagation. At normal incidence on a parallel slab of thickness $L$, a lateral separation $\Delta$ of the ordinary and extraordinary exiting rays gives ${\alpha }_e=\arctan (|\Delta |/L)$, while ${\alpha }_o=0$. The output rays can be parallel even though their internal energy paths differ.