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For the reflection of a P-wave from a rigid plane, use the displacement convention $u_x={\varphi }_x+{\psi }_y$, $u_y={\varphi }_y-{\psi }_x$, where $\varphi$ is the P potential and $\psi$ the SV potential. Put $\omega =k{c}_p$, $\kappa =k\sin \theta$, $a=k\cos \theta$, and $b=\sqrt{{\omega }^2/c_s^2-{\kappa }^2}$. The reflected waves must share the tangential wavenumber and frequency and propagate towards negative $y$, so writeThe rigid barrier requires both displacement components to vanish. On $y=0$ this gives $\kappa (1+A)-bB=0$ and $a(1-A)-\kappa B=0$. Solving,The total P potential is the given incident wave plus ${\varphi }_R$, and the S potential is ${\psi }_R$. Reversing the sign convention for $\psi$ reverses $B$ but leaves the displacement unchanged. At normal incidence $A=1,B=0$.

An evanescent reflected S wave would require $b^2<0$, or $\sin \theta >c_p/c_s$. In a stable isotropic elastic solid $c_p>c_s$, so this is impossible for real incidence angles: $b^2=k^2[(c_p/c_s{)}^2-{\sin }^2\theta ]>0$. The reflected S wave is never evanescent in the physical isotropic case. The inequality identifies the hypothetical critical-angle case if the speed ordering were reversed; it is S-to-P conversion, not this P-to-S conversion, that can become evanescent in an ordinary solid.