# Ghost amplitude and energy

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 6 181 - 220 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 6.7a

If the source depth is $c\lambda$ (where $\lambda$ is the wavelength) and $0 in equation (6.7a), discuss the conditions under which the amplitude of $\psi _{P}$ is zero.

### Background

The low-velocity layer (LVL) is discussed in problem 4.16, ghosts in problem 3.8. When a ghost is superimposed on a downgoing wave, it affects not only the waveshape but also the directivity. In Figure 6.7a, $S$ is a point source at a depth $c\lambda$ and $I$ is the image point (see problem 4.1) for energy reflected at the surface. For a ghost arriving at point $P$ , the virtual path is $IP$ . If the source emits the wave $\psi _{P}=A{\rm {\;cos\;}}(\kappa r-\omega t)$ and the reflection coefficient at the surface is $-1$ , the combined primary wave plus ghost at point $P$ is

{\begin{aligned}\psi _{P}=A{\rm {\;cos\;}}(\kappa r_{1}-\omega t)-A{\rm {\;cos\;}}(\kappa r_{2}-\omega t).\end{aligned}} Using the identity ${\rm {\;cos\;}}x-{\rm {\;cos\;}}y=-2{\rm {\;sin\;}}[(x+y)/2]{\rm {\;sin\;}}[(x-y)/2]$ , we get

{\begin{aligned}\psi _{P}=-2A{\rm {\;sin\;}}[\kappa (r_{1}+r_{2})/2-\omega t]{\rm {\;sin\;}}[\kappa (r_{1}-r_{2})/2].\end{aligned}} When $r\gg c\lambda$ , $r_{1}\approx r-c\lambda {\rm {\;cos\;}}\theta$ , $r_{2}\approx r+c\lambda {\rm {\;cos\;}}\theta$ ; also $c\lambda =2\pi c/\kappa$ , so we get

 {\begin{aligned}\psi _{P}=2A{\rm {\;sin\;}}(\kappa r-\omega t){\rm {\;sin\;}}(2\pi c{\rm {\;cos\;}}\theta )\\\qquad =[2A{\rm {\;sin\;}}(2\pi c\ {\rm {cos}}\ \theta )]{\rm {\;cos\;}}(\kappa r-\omega t-\pi /2).\end{aligned}} (6.7a)

Transmissivities $T\uparrow$ and $T\downarrow$ are defined in problem 3.6 where equation (3.6c) shows that

 {\begin{aligned}T\uparrow +T\downarrow =2,\;T\uparrow T\downarrow =E_{T}.\end{aligned}} (6.7b)

Absorption is discussed in problem 2.18.

### Solution

Equation (6.7a) gives for the amplitude of the primary wave plus ghost,

 {\begin{aligned}A^{*}=2A{\rm {\;sin\;}}(2\pi c\ {\rm {cos}}\ \theta ).\end{aligned}} (6.7c)

For $A^{*}=0$ , $2\pi c{\rm {\;cos\;}}\theta =n\pi$ , that is, ${\rm {\;cos\;}}\theta =n/2c,n=0,\pm 1,\pm 2$ , .... For $n=0$ , ${\rm {\;cos\;}}\theta =0$ , i. e., $\theta =\pm 90^{\circ }$ , and the waves are traveling horizontally. When $|n|\geq 2$ and $c<1$ , ${\rm {\;cos\;}}\theta >1$ so there are no appropriate values of $\theta$ .

## Problem 6.7b

For a source below the base of the LVL, compare the amplitude and energy of ghosts generated at the base of the LVL and at the surface of the ground, given that the velocities and densities just below and within the LVL are $V_{H}=1.9\ {\rm {km/s}}$ , $\rho _{H}=2.0\ {\rm {g/cm}}^{3}$ , $V_{W}=0.40\ {\rm {km/s}}$ and $\rho _{W}=1.6\ {\rm {g/cm}}^{3}$ , respectively.

### Solution

We assume small incidence angles so that equations (3.6a,b) are valid. Then

{\begin{aligned}Z_{1}=2.0\times 1.9=3.8;Z_{2}=0.40\times 1.6=0.64\ ({\rm {g.km/cm^{3}s}}).\end{aligned}} At the base of the LVL,

{\begin{aligned}R=(0.64-3.8)/(0.64+3.8)=-3.2/4.4=-0.71,\end{aligned}} {\begin{aligned}E_{R}=R^{2}=0.50,\end{aligned}} {\begin{aligned}T\uparrow =2\times 3.8/(3.8+0.64)=1.71\;,\;T\downarrow =0.29,\end{aligned}} {\begin{aligned}E_{T}=T\uparrow \times T\downarrow =1.71\times 0.29=0.50.\end{aligned}} Assuming equal (unit) amplitudes for waves leaving the source in different directions, the ghost produced at the base of the LVL has amplitude $-0.71$ and energy 0.50. The ghost produced at the surface has amplitude $T\uparrow T\downarrow \times (-1)=-0.50$ and energy $E_{T}=0.50^{2}=0.25$ . The amplitude of the ghost from the base of the LVL is $0.71/0.50=1.4$ times that of the ghost from the surface while the ratio of the energies is $0.50/0.25=2.4.$ ## Problem 6.7c

Assume that the LVL is ${\frac {1}{2}}\lambda$ in thickness and that $\eta \lambda =0.6\ {\rm {dB}}$ for the LVL; now what are the ratios of the ghost amplitudes and energies?

### Solution

The surface ghost has to travel a distance $\lambda$ farther than the ghost from the base of the LVL during which its amplitude is reduced by the factor $e^{-\eta x}=e^{-\eta \lambda }=e^{-0.60}=0.55$ . The previous amplitude was $-0.47$ , so with absorption this becomes $-0.26$ , and energy becomes 0.22 $\times 0.55^{2}=0.067$ . The ratios of the amplitudes and energies in part (b) now become $0.73/0.26=2.8$ and $0.53/0.067=7.9$ , the dB values being 8.9 and 18.0 dB, respectively.