Rayleigh wave ($(1+{\beta }^2{)}^2=4\alpha \beta$)

A Rayleigh wave propagates along a stress-free boundary of an isotropic elastic half-space and decays into its interior. It combines an evanescent longitudinal wave with an evanescent vertically polarized shear wave. If its speed is $0<c<c_S<c_P$, the decay factors are $\alpha =\sqrt{1-c^2/c_P^2}$ and $\beta =\sqrt{1-c^2/c_S^2}$. Vanishing shear and normal traction gives the displayed determinant condition.

For a Rayleigh wave at a free surface of a homogeneous isotropic elastic half-space, define $s=c^2/c_T^2$ and $\eta =c_T^2/c_L^2$. Cancelling the two surface tractions between the constituents with Rayleigh wave decay exponents gives the displayed equation with $0<s<1$. Equivalently $(b_T^2+k^2{)}^2=4k^2{b}_L{b}_T$. Squaring gives a factor $s$ times $P(s)=s^3-8s^2+(24-16\eta )s-16(1-\eta )$. For stable Lamé parameters, $0<\eta <3/4$, so $P(0)<0$, $P(1)=1$, and $P^{\prime \prime }(s)=6s-16<0$ on $(0,1)$. Its concavity ensures exactly one crossing in this interval. Both sides of the unsquared equation are positive, so this root gives the physical speed. The zero root is a degenerate static cancellation, not a propagating Rayleigh wave.

In a homogeneous isotropic elastic half-space $Y>0$, a Rayleigh wave with factor $e^{ik(X-ct)}$ has two depth factors $e^{-b_{L}Y}$ and $e^{-b_{T}Y}$. Substitution into the Navier-Cauchy equation factors the amplitude determinant into longitudinal and transverse factors, giving the displayed exponents. Strict exponential decay requires $0<c<\min (c_L,c_T)$. The longitudinal constituent has $(A,B)\propto (ik,-b_L)$; the transverse constituent has $ikA-b_{T}B=0$. The free-surface traction conditions select a combination of both constituents.