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Wave types in a planar geometry We know from the Helmholtz theorem that u = V0 + , (14) and that this relationship among the displacement vector and its two wave poten- tials yields a 3-space result for the displacement. What if we postulate that all wavevectors in our problems will be confined to the rz or 2113 plane? That is, all waves propagate in a single plane only. In that case ke = (kel, 0, kez) and likewise for the other wave type k, = (ksi, 0, k33). With this restriction we find that the plane-wave solutions become, °(, 12, 13,t) = A expli(kami + ke313 – wt) (15) V(21, 12, 13, t) = [0, B, 0 expli(kyl11 + k3313 - wt)], where u = ikçi A expli(katı + k0313 - wt) – iksz B expli(kylmi + k 313 – wt)] (16) U3 = ikiz A expli(kami + k3T3 - wt) +ikyl B expliky + ky323 - wt)]. U2 = 0
IV.2) Verify that for a plane-wave vector potential (11, 12, 13, 1) [B1, B2, B3] exp[i((ks1:21+k1313 -wt)], the uz displacement component will be non-zero, contradicting Eq (16), unless B1 = 0 and B3 = 0.
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