Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 38A (Waves)

The physical rapid phase isDefine the local wavevector, angular frequency, and group velocity byThe equality of mixed partial derivatives gives the phase-compatibility equationsA ray is defined to move with the local group velocity:

Differentiate the local dispersion relationwith respect to $x_i$. Along a ray,The last two terms cancel because $\mathbf{k}$ is a gradient. Similarly,Finally,Thus the Hamiltonian ray-tracing equations are

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.

The obstacle and wave pattern are stationary, so $\omega =\Omega =0$. Withthe dispersion relation becomesHence $\cos \varphi >0$, so $-\pi /2<\varphi <\pi /2$, and

Differentiating the dispersion relation with respect to the wavevector giveswhere the zero-frequency relation
$\alpha /\sqrt{\kappa }=V\cos \varphi$ was used. Its first component is always negative, so the ray angle $\psi$ lies inMoreover,

Put $q=|\tan \varphi |$. The magnitude of the slope relative to the negative $x_1$-axis isIts derivative isso its global maximum occurs at $q=1/\sqrt{2}$ and equalsTherefore all group-velocity rays lie in the stationary square-root-dispersion wakea wedge of semi-angle ${\tan }^{-1}(2^{-3/2})$ extending in the negative $x_1$-direction.