= Electric-displacement angle from birefringent walk-off
{title2=$|\angle(\mathbf E,\mathbf D)|=|\angle(\mathbf S,\mathbf k)|$}
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.
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