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 Incoming links: SV-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.

Let

k_{S} = ω / c_{S}

and

k_{P} = ω / c_{P}

k_{S} sin θ = k_{P} sin ϕ, sin ϕ = \frac{c _{P}}{c _{S}} sin θ .

Choose the reflected SV-wave and converted reflected P-wave as

u_{RS} = Re {R (cos θ, 0, sin θ) e^{i k_{S} (x s i n θ - z c o s θ) - iω t}},

u_{RP} = Re {T (sin ϕ, 0, - cos ϕ) e^{i k_{P} (x s i n ϕ - z c o s ϕ) - iω t}} .

The polarizations are respectively perpendicular and parallel to their wavevectors. Phase matching makes all three exponential factors equal at

z = 0

u_{I} + u_{RS} + u_{RP} = 0

there gives

cos θ (1 + R) + T sin ϕ = 0,

- sin θ + R sin θ - T cos ϕ = 0.

Solving this two-by-two linear system yields

R = - \frac{cos ( θ + ϕ )}{cos ( θ - ϕ )}, T = - \frac{sin 2 θ}{cos ( θ - ϕ )} .

An SV-wave is transverse but polarized in the vertical plane containing the wavevector:

B_{V} = B_{0} (cos θ, 0, - sin θ) .

k

and has

ω = c_{s} κ