 Solution

Use the displacement-potential convention $u_x={\partial }_x\varphi -{\partial }_z\psi$, $u_z={\partial }_z\varphi +{\partial }_x\psi$ and harmonic dependence $e^{-i\omega t}$. Choosing the opposite sign for the shear potential changes its column sign but not the P-SV directional impedance matrix. Write $p=k_{\alpha }>0$, $s=k_{\beta }>0$, $D=k^2+ps$, and $H=\rho {\omega }^2-2\mu k^2$. At a common point the forward potential wave amplitudes giveThe traction formulas follow from ${\sigma }_{xz}=\mu (u_{x,z}+u_{z,x})$ and ${\sigma }_{zz}=\lambda \nabla \cdot \mathbf{u}+2\mu u_{z,z}$, using $(\lambda +2\mu ){\omega }^2/{\alpha }^2=\rho {\omega }^2$ and $\mu (s^2-k^2)=H$. The velocity matrix has determinant ${\omega }^{2}D>0$, so elimination gives the P-SV directional impedance matrixIts off-diagonal entries are real and antisymmetric. ThereforeWith real-part phasors, the outward $z$ flux is $-\frac{1}{2}\mathrm{Re}({\mathbf{v}}^{†}\mathbf{t})$, sofor every nonzero forward displacement field. Mass density and angular frequency are essential to this flux coefficient; both appear correctly in the PDF.