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 Incoming links: S-wave

u = A e^{i (K \cdot x - ω t)}

K \times (K \times A) = K (K \cdot A) - K^{2} A

gives

ρ ω^{2} A = (λ + μ) K (K \cdot A) + μ K^{2} A .

A

is parallel to

K

, this becomes

ω^{2} = c_{P}^{2} K^{2}, c_{P} = \frac{λ + 2 μ}{ρ},

and gives a longitudinal P-wave. If

A \cdot K = 0

, it becomes

ω^{2} = c_{S}^{2} K^{2}, c_{S} = \frac{μ}{ρ},

and gives either polarization of an S-wave.

For

K = (k, 0, m)

wave P SV SH A (k, 0, m) (m, 0, - k) (0, 1, 0) .

The SV-wave lies in the

x z

propagation plane, whereas the SH-wave is perpendicular to that plane.

ρ_{0} \frac{\partial ^{2}}{\partial t ^{2}} (\nabla \cdot u) = (λ + 2 μ) \nabla^{2} (\nabla \cdot u) .

Thus the dilatation is a wave of speed

c_{P} = \frac{λ + 2 μ}{ρ _{0}} .

Taking the curl, and using

\nabla \times \nabla f = 0

, gives

ρ_{0} \frac{\partial ^{2}}{\partial t ^{2}} (\nabla \times u) = μ \nabla^{2} (\nabla \times u),

so the rotation propagates at

c_{S} = \frac{μ}{ρ _{0}} .

These are respectively the speeds of P-waves and S-waves.