Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 40C (Waves)

Taking the divergence of the Cauchy momentum equation and commuting constant-coefficient derivatives givesThus the dilatation is a wave of speedTaking the curl, and using $\nabla \times \nabla f=0$, givesso the rotation propagates atThese are respectively the speeds of P-waves and S-waves.

All wavevectors lie in the $xz$ plane and have the same tangential component. The incident P-wave travels upward toward $z=0$. In the lower solid there are a downward reflected P-wave, polarized parallel to its wavevector, and a downward reflected SV-wave, polarized in the $xz$ plane perpendicular to its wavevector. In the upper solid there are corresponding upward transmitted P- and SV-waves. The SH polarization decouples from this in-plane incident field. Their angles obey the Snell law for elastic and acoustic waves.

For a welded interface, the two in-plane displacement components are continuous:The two in-plane traction components are also continuous:These four independent scalar conditions determine the four reflected and transmitted amplitudes. This is mode conversion at a planar elastic interface.

Tangential phase matching givesA transmitted propagating wave exists when its normal wavenumber is real, equivalentlyFor incidence angles between zero and $\pi /2$, this is automatic if $c_P^{\prime }\le c_P$ and imposes the corresponding critical-angle bound if $c_P^{\prime }>c_P$.

Suppressing the common tangential phase, suitable displacement fields areandFor an inviscid elastic liquid, $\mu =0$, $c_P^2=\lambda /\rho$, and $\mathbf{\sigma }=\lambda (\nabla \cdot \mathbf{u})I$. Continuity of normal displacement and normal traction at $z=0$ givesTogether with the phase-matching relation from part (i), these determine $R$ and $T$.

Eliminating $T$ from the two boundary conditions givesThe Snell law for elastic and acoustic waves impliesTherefore the displacement reflection from an interface between two inviscid elastic liquids is

No reflection requiresUsing $\lambda =\rho c_P^2$, ${\lambda }^{\prime }={\rho }^{\prime }(c_P^{\prime }{)}^2$, and
$\sin {\theta }^{\prime }=(c_P^{\prime }/c_P)\sin \theta$, this reduces to the normal acoustic impedance matching conditionAfter squaring and eliminating ${\theta }^{\prime }$, the required incidence angle satisfiesSuch a no-reflection angle exists only when the right-hand side lies in $[0,1]$ and the transmitted wave is propagating.