 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-82/4/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

For downward-phase reversal, the P-SV displacement potentials instead produce

v = ω (k - p s k) (ϕ_{-} ψ_{-}), t = (2 μ k p - H H 2 μ k s) (ϕ_{-} ψ_{-}) .

Eliminating the potentials as before gives

Z_{-} = \frac{1}{ω D} (- ρ ω^{2} p - k (2 μ D - ρ ω^{2}) k (2 μ D - ρ ω^{2}) - ρ ω^{2} s) = - Z_{+}^{†} .

The Hermitian part is negative definite, and consequently

⟨ S_{z} ⟩_{-} = - \frac{ρ ω}{2 D} (p ∣ v_{x} ∣^{2} + s ∣ v_{z} ∣^{2}) < 0

for a nonzero backward displacement field. The P potential must use vertical wavenumber

k_{α}

and the S potential

k_{β}

, as in the original PDF; the TeX incorrectly gives the shear wavenumber for both.

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