# Rayleigh-wave relationships

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 2 7 - 46 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 2.14a

Show that, when ${\displaystyle {\mathbf {\sigma =0.25} }}$, the Rayleigh‐wave potentials, equation (2.14c), become

{\displaystyle {\begin{aligned}{\mathbf {\phi =Ae^{-0.847\kappa z}e^{\mathrm {j} \kappa \left(x-V_{R}t\right)}} },\end{aligned}}}

and

{\displaystyle {\begin{aligned}{\mathbf {\chi =1.466jAe^{-0.394{\kappa z}}e^{j\kappa \left(x-V_{R}t\right)}} },\end{aligned}}}

and that the displacements ${\displaystyle {u}}$ and ${\displaystyle {w}}$ at depth ${\displaystyle {z}}$ are

 {\displaystyle {\begin{aligned}{u}&={\mathbf {\kappa A\left(-e^{-0.847\kappa z}+0.578e^{-0.394{\kappa z}}\right)\sin \kappa \left(x-V_{R}t\right)} },\end{aligned}}} (2.14a)

 {\displaystyle {\begin{aligned}{w}&={\mathbf {\kappa A\left(-0.847e^{-0.847\kappa z}+1.466e^{-0.394{\kappa z}}\right)\cos \kappa \left(x-V_{R}t\right)} }.\end{aligned}}} (2.14b)

### Background

When a medium is divided by the ${\displaystyle xy}$-plane into two semi-infinite media having different properties, surface waves are propagated parallel to the ${\displaystyle xy}$-plane, the amplitude decreasing with increasing distance from the plane. When one medium is a solid and the other a vacuum, the surface wave is known as a Rayleigh wave. The near-equivalent at the surface of the real earth, a pseudo-Rayleigh wave, is called ground roll.

We take the potential functions of equation (2.9b) in the form

 {\displaystyle {\begin{aligned}\mathrm {\phi } =Ae^{-m\mathrm {\kappa } z}e^{j\mathrm {\kappa } \left(x-V_{R}t\right)}\;,\;\chi =Be^{-n\mathrm {\kappa } z}e^{j\mathrm {\kappa } \left(x-V_{R}t\right)},\end{aligned}}} (2.14c)

where the ${\displaystyle z}$-axis is positive downward, ${\displaystyle m}$ and ${\displaystyle n}$ are real positive constants (so that the amplitudes decrease as ${\displaystyle z}$ increases) and ${\displaystyle V_{R}}$ is the Rayleigh-wave velocity. We can take either the real or the imaginary parts of the functions as a solution, the only difference being the phase. When we substitute these functions in the P- and S-wave equations [equation (2.5a) with ${\displaystyle \mathrm {\psi } }$ replaced with ${\displaystyle \Delta }$ and ${\displaystyle \mathrm {\theta } _{y}}$, respectively), we find that ${\displaystyle m}$ and ${\displaystyle n}$ must satisfy the equations

 {\displaystyle {\begin{aligned}m^{2}=\left[1-\left(V_{R}/\mathrm {\alpha } \right)^{2}\right]\;,\;n^{2}=\left[1-\left(V_{R}/\mathrm {\beta } \right)^{2}\right].\end{aligned}}} (2.14d)

Because both ${\displaystyle m}$ and ${\displaystyle n}$ must be real, ${\displaystyle V_{R}<\mathrm {\beta } <\mathrm {\alpha } }$.

When we apply the boundary conditions of problem 2.10b, we find that

{\displaystyle {\begin{aligned}B=\left({\frac {2\mathrm {j} m}{1+n^{2}}}\right)A,\end{aligned}}}

and that the velocity ${\displaystyle V_{R}}$ is a root of the equation

 {\displaystyle {\begin{aligned}x^{3}-8x^{2}+\left[24-16(\mathrm {\beta } /\mathrm {\alpha } )^{2}\right]x+16\left[\left(\mathrm {\beta } /\mathrm {\alpha } \right)^{2}-1\right]=0,\end{aligned}}} (2.14e)

where ${\displaystyle x=(V_{R}/\mathrm {\beta } )^{2},(\mathrm {\beta } /\mathrm {\alpha } )^{2}=\left(1-2\mathrm {\sigma } \right)/2\left(1-\mathrm {\sigma } \right)}$ from equation (1,8) in Table 2.2a.

The left-hand side of equation (2.14e) is negative when ${\displaystyle x=0}$, positive for ${\displaystyle x=+1}$, so a root must exist in the range ${\displaystyle 0. When ${\displaystyle \mathrm {\sigma } =0.25}$, the root is ${\displaystyle V_{R}=0.919\mathrm {\beta } }$.

The angle ${\displaystyle \mathrm {\theta } }$ in Figure 2.14a is given by the equation

 {\displaystyle {\begin{aligned}\mathrm {\theta } =\tan ^{-1}\left(-w/u\right)\end{aligned}}} (2.14f)

(the minus sign is necessary because we have taken ${\displaystyle w}$ positive downward).

### Solution

When ${\displaystyle \mathrm {\sigma } =0.25,V_{R}/\mathrm {\beta } =0.919}$, ${\displaystyle (\mathrm {\beta } /\mathrm {\alpha } )^{2}=\left(1-2\mathrm {\sigma } \right)/2\left(1-\mathrm {\sigma } \right)=1/3}$, and ${\displaystyle \left(V_{R}/\mathrm {\alpha } \right)=\left(V_{R}/\mathrm {\beta } \right)\times \left(\mathrm {\beta } /\mathrm {\alpha } \right)=0.919/{\sqrt {3}}=0.531}$. Using these values, equation (2.14d) gives ${\displaystyle m=0.848}$, ${\displaystyle n=0.393}$; also, ${\displaystyle B=2\mathrm {j} m/\left(1+n^{2}\right)A=1.468\mathrm {j} A}$; the j indicates that ${\displaystyle \mathrm {\phi } }$ and ${\displaystyle \mathrm {\chi } }$ are ${\displaystyle 90^{\circ }}$ out of phase. The potential functions are now

{\displaystyle {\begin{aligned}\mathrm {\phi } =Ae^{-0.848\mathrm {\kappa } z}e^{\mathrm {j} \mathrm {\kappa } \left(x-V_{R}t\right)},\end{aligned}}}

and

{\displaystyle {\begin{aligned}\mathrm {\chi } =1.468\mathrm {j} Ae^{-0.393\mathrm {\kappa } z}e^{\mathrm {j} \mathrm {\kappa } \left(x-V_{R}t\right)}.\end{aligned}}}

From equations (2.9d) and (2.9e) we have

{\displaystyle {\begin{aligned}u=\mathrm {\phi } _{x}+\mathrm {\chi } _{z}=\mathrm {j} \mathrm {\kappa } \mathrm {\phi } -0.393\mathrm {\kappa } \mathrm {\chi } \;,\;w=\mathrm {\phi } _{z}-\mathrm {\chi } _{x}=-0.848\mathrm {\kappa } \mathrm {\phi } -\mathrm {j} \mathrm {\kappa } \mathrm {\chi } .\end{aligned}}}

Taking the real part of the solution (see Sheriff and Geldart, 1995, Section 15.1.5) we obtain

 {\displaystyle {\begin{aligned}u&=\mathrm {\kappa } A\{e^{-0.848\mathrm {\kappa } z}\left[-\sin \mathrm {\kappa } \left(x-V_{R}t\right)\right]\\&\quad +0.393\times 1.468e^{-0.393\mathrm {\kappa } z}\sin \mathrm {\kappa } \left(x-V_{R}t\right)\}\\&=\mathrm {\kappa } A\left(-e^{-0.848\mathrm {\kappa } z}+0.577e^{-0.393\mathrm {\kappa } z}\right)\sin \mathrm {\kappa } \left(x-V_{R}t\right),\end{aligned}}} (2.14g)

 {\displaystyle {\begin{aligned}w&=\mathrm {\kappa } A\left(-0.848e^{-0.848\mathrm {\kappa } z}+1.468e^{-0.393\mathrm {\kappa } z}\right)\cos \mathrm {\kappa } \left(x-V_{R}t\right)\end{aligned}}} (2.14h)

## Problem 2.14b

What are the values of ${\displaystyle {u}}$, ${\displaystyle {w}}$, and ${\displaystyle {\theta }}$ (i) when ${\displaystyle {z=0}}$ ; (ii) when ${\displaystyle {z=1/2\mathrm {\kappa } }}$; (iii) when ${\displaystyle {z=1/\mathrm {\kappa } }}$?

### Solution

i) At the surface ${\displaystyle \mathrm {z} =0}$, and equations (2.14g,h) give

{\displaystyle {\begin{aligned}u&=-0.423\mathrm {\kappa } A\sin \mathrm {\kappa } \left(x-V_{R}t\right),\\w&=0.620\mathrm {\kappa } A\cos \mathrm {\kappa } \left(x-V_{R}t\right).\end{aligned}}}

From equation (2.14f), ${\displaystyle \tan \mathrm {\theta } =1.465\cot \mathrm {\kappa } \left(x-V_{R}t\right)}$

ii) When ${\displaystyle z=1/2\mathrm {\kappa } }$, equations (2.14f,g,h) give for ${\displaystyle u}$, ${\displaystyle w}$, and ${\displaystyle \theta }$ {\displaystyle {\begin{aligned}u&=\mathrm {\kappa } A\left(-e^{-0.424}+0.578e^{-0.197}\right)\sin \mathrm {\kappa } \left(x-V_{R}t\right)\\&=-0.180\mathrm {\kappa } A\sin \mathrm {\kappa } \left(x-V_{R}t\right),\\w&=\mathrm {\kappa } A\left(-0.847e^{-0.424}+1.466e^{-0.197}\right)\cos \mathrm {\kappa } \left(x-V_{R}t\right)\\&=0.650\mathrm {\kappa } A\cos \mathrm {\kappa } \left(x-V_{R}t\right),\\\tan \mathrm {\theta } &=\left(0.650/0.180\right)\cot \mathrm {\kappa } \left(x-V_{R}t\right)=3.61\cot \mathrm {\kappa } \left(x-V_{R}t\right).\end{aligned}}}

iii) For ${\displaystyle z=1/\mathrm {\kappa } }$, we get

{\displaystyle {\begin{aligned}u&=\mathrm {\kappa } A\left(-e^{-0.847}+0.578e^{-0.394}\right)\sin \mathrm {\kappa } \left(x-V_{R}t\right)\\&=-0.039\mathrm {\kappa } A\sin \mathrm {\kappa } \left(x-V_{R}t\right),\\w&=\mathrm {\kappa } A\left(-0.847e^{-0.847}+1.466e^{-0.394}\right)\cos \mathrm {\kappa } \left(x-V_{R}t\right)\\&=0.625\mathrm {\kappa } A\cos \mathrm {\kappa } \left(x-V_{R}t\right),\\\tan \mathrm {\theta } &=\left(0.625/0.039\right)\cot \mathrm {\kappa } \left(x-V_{R}t\right)=16.0\cot \mathrm {\kappa } \left(x-V_{R}t\right).\end{aligned}}}

## Problem 2.14c

Is the motion retrograde for all values of ${\displaystyle {z}}$?

### Solution

When ${\displaystyle x}$ and ${\displaystyle z}$ are fixed, the argument of ${\displaystyle \cot \mathrm {\kappa } \left(x-V_{R}t\right)}$ decreases as ${\displaystyle t}$ increases, and so the cotangent increases, that is, the angle ${\displaystyle \mathrm {\theta } }$ in Figure 2.14a also increases. Because the wave is progressing in the positive direction of the ${\displaystyle x}$-axis, this counter-clockwise rotation is said to be retrograde. For the motion not to be retrograde, ${\displaystyle \mathrm {\theta } }$ must change sign, and thus ${\displaystyle \left(-w/u\right)}$ must also change sign, that is, either ${\displaystyle u}$ or ${\displaystyle w}$ but not both must change signs. For ${\displaystyle \mathrm {\kappa } z=0}$, ${\displaystyle u}$ is negative while ${\displaystyle w}$ is positive. For ${\displaystyle u}$ to change sign, it must pass through zero; in this case

{\displaystyle {\begin{aligned}u=0=-e^{-0.847\mathrm {\kappa } z}+0.578e^{-0.394\mathrm {\kappa } z}\end{aligned}}}

or ${\displaystyle e^{0.453\mathrm {\kappa } z}=1.73,\mathrm {\kappa } z=1.21.}$ For ${\displaystyle \mathrm {\kappa } z>1.21}$, ${\displaystyle u}$ is positive. For ${\displaystyle w}$ to change sign,

{\displaystyle {\begin{aligned}w=0=-0.847e^{-0.847\mathrm {\kappa } z}+1.466e^{-0.394\mathrm {\kappa } z},\end{aligned}}}

or ${\displaystyle e^{0.453\mathrm {\kappa } z}=0.578,\mathrm {\kappa } z=-1.21}$. Since ${\displaystyle \mathrm {\kappa } z}$ is always positive, ${\displaystyle w}$ can never be zero. Consequently, the motion is retrograde in the interval ${\displaystyle 0<\mathrm {\kappa } z<1.21}$ and prograde when ${\displaystyle \mathrm {\kappa } z>1.21}$.

## Problem 2.14d

What are the values of ${\displaystyle {\mathbf {V_{R}} }}$, the Rayleigh-wave velocity, when ${\displaystyle {\sigma =0.4}}$ and when ${\displaystyle {\sigma =0.2}}$? What are the corresponding values of the constants in the expressions for ${\displaystyle {u}}$ and ${\displaystyle {w}}$ in equation (2.14g,h)?

### Solution

From Figure 2.14b we find that ${\displaystyle V_{R}/\mathrm {\beta } \approx 0.95}$ when ${\displaystyle \mathrm {\sigma } =0.4}$ and ${\displaystyle V_{R}/\mathrm {\beta } \approx 0.92}$ when ${\displaystyle \mathrm {\sigma } =0.2}$. To get more accurate values, we solve equation (2.14e) using one of the standard methods of solving cubic equations.

Figure 2.14b  ${\displaystyle V_{R}}$ as a function of ${\displaystyle \mathrm {\sigma } }$.

i) For ${\displaystyle \mathrm {\sigma } =0.4,(\mathrm {\beta } /\mathrm {\alpha } )^{2}=(1-2\mathrm {\sigma } )/2\left(1-\mathrm {\sigma } \right)=1/6}$. Equation (2.14e) now becomes

{\displaystyle {\begin{aligned}x^{3}-8x^{2}+\left(64/3\right)x-\left(40/3\right)=0=x^{3}+px^{2}+qx+r=0,\end{aligned}}}

where ${\displaystyle p=-8}$, ${\displaystyle q=64/3}$, ${\displaystyle r=-40/3}$. Next we eliminate the ${\displaystyle x^{2}}$-term by substituting ${\displaystyle x=y-p/3=y+8/3}$. This gives ${\displaystyle y^{3}+ay+b=0}$, where ${\displaystyle a=\left(q-p^{2}/3\right)=0,b={\frac {1}{27}}(2p^{3}-9pq+27r)=152/27}$ To check on the nature of the roots, we calculate ${\displaystyle ({\frac {b^{2}}{4}}+{\frac {a^{3}}{27}})}$; the value is ${\displaystyle {\frac {1}{4}}({\frac {152}{27}})^{2}}$, that is, it is positive, so we have one real root and two complex ones.

We now calculate the quantities

{\displaystyle {\begin{aligned}C=\left[-{\frac {b}{2}}+\left({\frac {b^{2}}{4}}+{\frac {a^{3}}{27}}\right)^{1/2}\right]^{1/3},\quad D=\left[-{\frac {b}{2}}-\left({\frac {b^{2}}{4}}+{\frac {a^{3}}{27}}\right)^{1/2}\right]^{1/3}.\end{aligned}}}

The three roots of the equation are ${\displaystyle \left(C+D\right),-{\frac {1}{2}}\left(C+D\right)\pm 3j/{\sqrt {2}}}$. The last two roots are complex, so we are left with only the first root, that is, ${\displaystyle y=\left(C+D\right)}$. Substituting the values of ${\displaystyle a}$ and ${\displaystyle b}$, we get ${\displaystyle C=0}$, ${\displaystyle D=-1.78}$. Thus, ${\displaystyle y=-1.78=x-8/3}$, ${\displaystyle x=0.887=(V_{R}/\mathrm {\beta } )^{2}}$, so ${\displaystyle V_{R}/\mathrm {\beta } =0.942}$.

To get the values of the constants in part (a) for ${\displaystyle \mathrm {\sigma } =0.4}$, we have

{\displaystyle {\begin{aligned}V_{R}/\beta &=0.942,\;\quad n=[1-(V_{R}/\mathrm {\beta } )^{2}]^{1/2}=0.336,\\V_{R}/\mathrm {\alpha } &=\left(V_{R}/\mathrm {\beta } \right)\left(\mathrm {\beta } /\mathrm {\alpha } \right)=0.942(1/6)^{1/2}=0.385,\\m&=[1-(V_{R}/\mathrm {\alpha } )^{2}]^{1/2}=0.923,\\B/A&=2\mathrm {j} m/\left(1+n^{2}\right)=1.659\mathrm {j} ,\\\mathrm {\phi } &=Ae^{-0.923\mathrm {\kappa } z}e^{\mathrm {j} \mathrm {\kappa } \left(x-V_{R}t\right)},\\\mathrm {\chi } &=1.659\mathrm {j} Ae^{-0.336\mathrm {\kappa } z}e^{j\mathrm {\kappa } \left(x-V_{R}t\right)},\\u&=\mathrm {\phi } _{x}+\mathrm {\chi } _{z}=j\mathrm {\kappa } A\left(e^{-0.923\mathrm {\kappa } z}-0.557e^{-0.336\mathrm {\kappa } z}\right)e^{\mathrm {j} \mathrm {\kappa } \left(x-V_{R}t\right)},\\w&=\mathrm {\phi } _{z}-\mathrm {\chi } _{x}\\&=\mathrm {\kappa } A\left(-0.923e^{-0.923\mathrm {\kappa } z}+1.659e^{-0.336\mathrm {\kappa } z}\right)e^{\mathrm {j} \mathrm {\kappa } \left(x-V_{R}t\right)}.\end{aligned}}}

ii) For ${\displaystyle \mathrm {\sigma } =0.2}$, ${\displaystyle (\mathrm {\beta } /\mathrm {\alpha } )^{2}=\left(1-2\mathrm {\sigma } \right)/2\left(1-\mathrm {\sigma } \right)=3/8}$. This gives

{\displaystyle {\begin{aligned}x^{3}-8x^{2}+18x-10=0,\quad \mathrm {so} \ p=-8,\;q=18,\;r=-10.\end{aligned}}}

Let ${\displaystyle x=y+8/3}$, so ${\displaystyle y^{3}+ay+b=0}$, where ${\displaystyle a=q-p^{2}/3=-10/3}$, ${\displaystyle b=r-pq/3+\left(2/27\right)p^{3}=2/27}$.

The discriminant ${\displaystyle \left({\frac {b^{2}}{4}}+{\frac {a^{3}}{27}}\right)<0}$, so there are three real unequal roots; in this case a trigonometric solution is convenient. We find the value of

{\displaystyle {\begin{aligned}\cos \mathrm {\gamma } =-{\frac {b}{2}}\left(-{\frac {27}{a^{3}}}\right)^{1/2}=-0.0316,\quad \gamma =91.8^{\circ }\;,\quad \gamma /3=30.6^{\circ }.\end{aligned}}}

Next we calculate ${\displaystyle 2{\sqrt {-a/3}}=2.11}$. The roots are now

{\displaystyle {\begin{aligned}y&=2.11\cos 30.6^{\circ },\;2.11\cos \left(30.6^{\circ }+2\pi /3\right),\;2.11\cos \left(30.6^{\circ }+4\pi /3\right)\\&=1.82,\;-1.84,\;0.022;\\x&=y+8/3=4.49,\;0.830,\;2.69.\end{aligned}}}

But ${\displaystyle x<1}$, so the only valid root is

{\displaystyle {\begin{aligned}x=0.830=(V_{R}/\mathrm {\beta } )^{2}\;,\;\left(V_{R}/\mathrm {\beta } \right)=0.911.\end{aligned}}}

Hence, for ${\displaystyle \mathrm {\sigma } =0.2,}$

{\displaystyle {\begin{aligned}V_{R}/\mathrm {\beta } &=0.911\;,\quad \;n=[1-(V_{R}/\mathrm {\beta } )^{2}]^{1/2}=0.412,\\V_{R}/\mathrm {\alpha } &=0.911\left(\mathrm {\beta } /\mathrm {\alpha } \right)=0.911(3/8)^{1/2}=0.558,\\m&=(1-0.558^{2})^{1/2}=0.830,\\B/A&=2\mathrm {j} m/\left(1+n^{2}\right)=1.419\mathrm {j} ,\\\phi &=Ae^{-0.830\mathrm {\kappa } z}e^{\mathrm {j} \mathrm {\kappa } \left(x-V_{R}t\right)},\\\chi &=1.419\mathrm {j} Ae^{-0.412\mathrm {\kappa } z}e^{\mathrm {j} \mathrm {\kappa } \left(x-V_{R}t\right)},\\u&=\phi _{x}+\chi _{z}=\mathrm {j} \mathrm {\kappa } A\left(e^{-0.830\mathrm {\kappa } z}-0.585e^{-0.412\mathrm {\kappa } z}\right)e^{\mathrm {j} \mathrm {\kappa } \left(x-V_{R}t\right)},\\w&=\phi _{z}-\chi _{x}=\mathrm {\kappa } A\left(-0.830e^{-0.830\mathrm {\kappa } z}-1.419e^{-0.412\mathrm {\kappa } z}\right)e^{\mathrm {j} \mathrm {\kappa } \left(x-V_{R}t\right)}.\end{aligned}}}