 Solution

The displacement equation is $\rho {\mathbf{u}}_{tt}=(\lambda +\mu )\nabla (\nabla \cdot \mathbf{u})+\mu \Delta \mathbf{u}$. With $\mathbf{u}=\nabla \varphi +\nabla \times \mathbf{\psi }$ and divergence-free vector-potential gauge, divergence and curl give wave equations withThus ${\varphi }_{tt}=c_P^2\Delta \varphi$ and ${\mathbf{\psi }}_{tt}=c_S^2\Delta \mathbf{\psi }$, up to gauge constants. The P wave displacement is parallel to its propagation direction; S wave displacement is perpendicular, with two transverse polarizations.

Take the plane of incidence to be $xy$. Let $k$ be the shared tangential wave number, $h=({\omega }^2/c_P^2-k^2{)}^{1/2}$ and $s=({\omega }^2/c_S^2-k^2{)}^{1/2}$. Incident P, reflected P and reflected SV waves have respectively wave vectors $(k,h)$, $(k,-h)$ and $(k,-s)$. Thus the reflected P angle equals $\theta$ and $\sin {\theta }_S=(c_S/c_P)\sin \theta$. No SH wave is excited by this planar P incidence.

Use scalar potentials $\mathbf{u}=({\varphi }_x+{\psi }_y,{\varphi }_y-{\psi }_x)$, with incident potential amplitude one, reflected P amplitude $R$, and reflected SV amplitude $Q$. At the surface their phases are common. Vanishing $xy$ and $yy$ tractions gives, with $D=s^2-k^2$,Eliminating $Q$ givesThere is no reflected P wave precisely when $4k^{2}hs=D^2$. If $\beta =c_S^2/c_P^2$ and $\sigma =k^2{c}_S^2/{\omega }^2=\beta {\sin }^2\theta$, then $h=(\omega /c_S)\sqrt{\beta -\sigma }$, $s=(\omega /c_S)\sqrt{1-\sigma }$ and $D=({\omega }^2/c_S^2)(1-2\sigma )$. Substitution provesThe condition applies only to incidence angles in the physical range $0\le \sigma \le \beta$; it need not have a solution for every elastic-modulus ratio.