 Solution

In a homogeneous, time-independent medium, $\partial \Omega /\partial x_i=0$ and $\partial \Omega /\partial t=0$. Therefore $\mathbf{k}$ and $\omega$ are constant along each ray. The group velocity ${\mathbf{c}}_g={\nabla }_{\mathbf{k}}\Omega$ is consequently constant, andso every ray is a straight line.

For $\omega =0$, let $s=|\mathbf{x}|$ denote distance along a ray from the origin and letThe phase equation givesBoth vectors on the right are constant along the ray, soThe final formula in the question denotes the physical phase $\Theta =\theta /\epsilon$ simply by $\theta$, giving its stated expression.