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A compressional wave is a longitudinal wave whose propagation produces changes of volume or density. In an isotropic elastic solid its displacement is parallel to the wavevector; substituting that polarization into the dynamic Navier-Cauchy equation gives the displayed speed. In a fluid without static shear stiffness the analogous speed is $\sqrt{K/\rho }$, with the appropriate compressional bulk modulus. Reflection at a rigid elastic boundary can convert part of its energy into a shear wave.

In an isotropic half-space $y<0$ with zero displacement at $y=0$, an incident longitudinal wave directed along $(\cos \alpha ,\sin \alpha )$ reflects into both longitudinal and shear waves. Let $c_p^2=(\lambda +2\mu )/\rho$ and $c_s^2=\mu /\rho$. Horizontal phase matching gives $\sin \beta =(c_s/c_p)\cos \alpha$ for a reflected shear wavevector proportional to $(\sin \beta ,-\cos \beta )$ and polarization $(\cos \beta ,\sin \beta )$. The two zero-displacement equations at the boundary give the displayed coefficients. Since $c_s<c_p$ in a stable isotropic solid, both reflected branches are propagating for every genuine incident angle $0<\alpha <\pi$, with no critical-angle obstruction for this incident-wave type. Changing to a stress-free boundary changes the amplitude equations and coefficients.

Decomposing a plane displacement into components parallel and perpendicular to its wavevector gives compressional and shear speedsThe incident polarization is longitudinal. Its reflected longitudinal wave reverses the vertical wavevector component while retaining its horizontal component. The converted shear wave has wavevector $k_s(\sin \beta ,-\cos \beta )$ perpendicular to polarization $(\cos \beta ,\sin \beta )$.

Every term must have the same phase along the rigid boundary. ThusChoose $\cos \beta >0$ so the reflected shear wave propagates downward into $y<0$. At $y=0$, both displacement components must vanish for every boundary phase:Solving this two-by-two system yieldsFor a stable isotropic material $0<c_s/c_p<1$. An incident wave directed toward the boundary has $0<\alpha <\pi$; choosing $\beta \in (-\pi /2,\pi /2)$ by the sine relation gives real reflected waves throughout this range and no zero denominator. In the usual convention restricting incident horizontal propagation to positive $x$, this is $0<\alpha \le \pi /2$. Grazing endpoints have no incoming normal energy flux and are limiting cases. There is no critical angle for incident compressional-to-shear conversion here, because the converted wave is slower; the opposite incident-wave type has different critical-angle behaviour. At normal incidence $\alpha =\pi /2$, $\beta =0$, $R_1=1$ and $R_2=0$, cancelling the incident displacement with the reversed longitudinal polarization. This is compressional-wave reflection from a rigid elastic boundary.

A shear wave in an isotropic elastic material has displacement perpendicular to its wavevector, with zero divergence for a plane wave. The Navier-Cauchy equation with inertia then reduces to $\rho {\mathbf{u}}_{tt}=\mu \Delta \mathbf{u}$, giving the displayed wave speed. At a boundary it can be generated by mode conversion from a compressional wave, even when the incident displacement is purely longitudinal.