 Solution

At a welded interface, elastic-interface boundary conditions require continuous SH displacement and shear traction. Incident and reflected vertical slownesses have signs $q_1,-q_1$, givingThus the local displacement coefficient isIt multiplies the leading waveform discontinuity as well as a harmonic displacement. At the interface the incident amplitude from SH ray-tube amplitude transport is $C/\sqrt{r_1}$, where $r_1=h/({\beta }_1{q}_1)$ is the incident path length and $C$ is the source normalization from the previous part.

The transmitted geometrical spreading is not simply its total path length. Differentiate the ray position with respect to $p$ at fixed travel time and project onto the unit direction transverse to the transmitted ray, ${\mathbf{m}}_2=({\beta }_2{q}_2,-{\beta }_{2}p)$. Since $d{x}_I/dp=h/({\beta }_1^2{q}_1^3)$, the transverse width per unit $p$ isThe derivative of $t_1$ contributes only in the ray direction and drops out of this projection. At entry,Within uniform medium 2 the leading amplitude varies as $J_2^{-1/2}$, so the required refraction of a cylindrical SH wavefront amplitude isThe local transmission coefficient and ray-tube spreading together incorporate both impedance contrast and refraction. Indeed the normal energy transmission fraction is ${\mu }_2{q}_2|T_{\mathrm{SH}}{|}^2/({\mu }_1{q}_1)$, while the change of transverse width supplies the refraction factor in the conserved tube flux. If the media coincide, $q_1=q_2$ and $T_{\mathrm{SH}}=1$, reducing this expression to $C/\sqrt{{\beta }_{1}t}$ as required. The formula holds for the stated propagating slownesses, without a critical or head-wave contribution.