Use the displacement-potential convention , and harmonic dependence . Choosing the opposite sign for the shear potential changes its column sign but not the P-SV directional impedance matrix. Write , , , and . At a common point the forward potential wave amplitudes giveThe traction formulas follow from and , using and . The velocity matrix has determinant , so elimination gives the P-SV directional impedance matrixIts off-diagonal entries are real and antisymmetric. ThereforeWith real-part phasors, the outward flux is , sofor every nonzero forward displacement field. Mass density and angular frequency are essential to this flux coefficient; both appear correctly in the PDF.
For downward-phase reversal, the P-SV displacement potentials instead produceEliminating the potentials as before givesThe Hermitian part is negative definite, and consequentlyfor a nonzero backward displacement field. The P potential must use vertical wavenumber and the S potential , as in the original PDF; the TeX incorrectly gives the shear wavenumber for both.
Define . Its antisymmetric off-diagonal entries cancel, leavingFrom and , solve explicitly for the P-SV characteristic reconstruction:The prescribed strictly positive make this decomposition unique; a grazing or evanescent branch is outside these assumptions.
In the mean flux, the two mixed directional terms arebecause makes the second scalar minus the complex conjugate of the first. The remaining self terms giveFinally use either directional velocity-potential matrix. Its weighted quadratic form satisfiesIndeed the two P-SV cross terms have coefficients and for the forward field, with the signs reversed for the backward field, so they cancel. Substitution provesThus net normal elastic-wave energy flux is the difference of the independently flux-weighted P and SV directional powers, even for coherent mixtures.
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