A Poisson-like crust has and

PHYSICS · Multi-part · Fermi University Physics · Oscillations, Waves & Acoustics · Earth, Climate & Energy Physics · Seismology, tectonics and geodynamics

The Solid Earth, Ch Surface waves, free oscillations and seismic tomography

Problem

A Poisson-like crust has vS=3.4 km s−1v_S = 3.4\,\mathrm{km\,s^{-1}} and vP=6.0 km s−1v_P = 6.0\,\mathrm{km\,s^{-1}}. A Rayleigh wave of period T=15 sT=15\,\mathrm{s} propagates on its surface with speed cR=0.9194 vSc_R = 0.9194\,v_S. ((a)) The P and S displacement potentials beneath a Rayleigh wave decay with depth zz as e−kz1−cR2/vP2e^{-kz\sqrt{1-c_R^2/v_P^2}} and e−kz1−cR2/vS2e^{-kz\sqrt{1-c_R^2/v_S^2}} respectively, where k=2π/λk = 2\pi/\lambda and λ=cRT\lambda = c_R T. Find the depth at which each potential has fallen to 1/e1/e of its surface value, and state which of the two decays more slowly with depth. ((b)) Elsewhere, in a region where a body-wave speed model increases linearly with depth, v(z)=v0+gzv(z) = v_0 + gz with v0=6.0 km s−1v_0=6.0\,\mathrm{km\,s^{-1}} and g=0.005 s−1g=0.005\,\mathrm{s^{-1}} (zz in km\mathrm{km}, vv in km s−1\mathrm{km\,s^{-1}}). A ray with flat-Earth ray parameter p=0.15 s km−1p=0.15\,\mathrm{s\,km^{-1}} (defined by Snell's law sin⁡θ(z)/v(z)=p\sin\theta(z)/v(z)=p for the angle θ\theta from the vertical) descends through this medium. Using ds=dz/cos⁡θ(z)\mathrm{d}s = \mathrm{d}z/\cos\theta(z) for the arc length along the ray, find the length of ray path lying between z=40 kmz=40\,\mathrm{km} and z=80 kmz=80\,\mathrm{km}: this length is the entry GijG_{ij} that this ray contributes to the row of the tomographic sensitivity matrix for the cell occupying that depth range.

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