Question

Given the Rayleigh wave velocity \(V_r\), shear wave velocity \(V_s\) and the P-wave velocity \(V_p\), which of the following relationships is/are CORRECT?

• A. \(V_r < V_s < V_p\)
• B. \(V_s < V_r < V_p\)
• C. \(V_s < V_p < V_r\)
• D. \(V_s = V_r < V_p\)

Hint:

Rayleigh waves must decay evanescently with depth relative to both P and S potentials, forcing Vr below the smaller of the two body-wave speeds; for a Poisson solid, Vr is about 0.92 Vs.

Answer 1

The Correct Option is A

Rayleigh waves arise from the constructive interference of P and SV waves trapped near a free surface. Their phase velocity \(c = V_r\) is found from the classical Rayleigh secular (free-surface) equation for a homogeneous elastic half-space:

\[ \left(2 - \frac{c^2}{\beta^2}\right)^2 = 4\sqrt{1 - \frac{c^2}{\alpha^2}}\,\sqrt{1 - \frac{c^2}{\beta^2}} \]

where \(\alpha = V_p\) and \(\beta = V_s\) are the P- and S-wave velocities. For a Poisson solid (Poisson's ratio \(\nu = 0.25\), i.e. \(\alpha = \sqrt{3}\,\beta\)), solving this equation numerically for the root \(0 < c < \beta\) gives

\[ c = V_r \approx 0.9194\, V_s \]

So the Rayleigh wave is always slightly slower than the shear wave, \(V_r < V_s\). Also, for any elastic solid with a positive Poisson's ratio, the shear velocity is always less than the compressional velocity, \(V_s = \sqrt{\mu/\rho} < V_p = \sqrt{(K+4\mu/3)/\rho}\), because the P-wave modulus includes the extra bulk-modulus term. Combining both inequalities:

\[ V_r < V_s < V_p \]

This matches option (A) exactly, and it also explains the standard seismogram ordering of arrivals: P first, then S, then the (slowest) Rayleigh surface wave.

\(\boxed{V_r < V_s < V_p}\)