# Magnitude of disturbance from a seismic source

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 2.3a

Firing an air gun in water creates a pressure transient a small distance away from the air gun with peak pressure ${\mathcal {P}}$ of 5 atmospheres ($5.0\times 10^{5}$ Pa). If the compressibility of water is $4.5\times 10^{-10}$ /Pa, what is the peak energy density?

### Background

Air guns (see problem 7.7) suddenly inject a bubble of high‐pressure air into the water to generate a seismic wave.

Stresses acting upon a medium cause energy to be stored as strain energy, because the stresses are present while the medium is being displaced. Strain energy density (energy/unit volume) $E$ is equal to [see Sheriff and Geldart, 1995, equation (2.22)]

 {\begin{aligned}{E={\frac {1}{2}}\sum \limits _{i}\sum \limits _{j}{\mathrm {\sigma } }_{ij}{\mathrm {\varepsilon } }_{ij}={\frac {1}{2}}{\mathrm {\lambda } }{\mathrm {\Delta } }^{2}+{\mathrm {\mu } }\left({\mathrm {\varepsilon } }_{xx}^{2}+{\mathrm {\varepsilon } }_{yy}^{2}+{\mathrm {\varepsilon } }_{zz}^{2}\right)}\\{{\qquad }{\qquad }{\qquad }{\qquad }+{\frac {1}{2}}{\mathrm {\mu } }\left({\mathrm {\varepsilon } }_{xy}^{2}+{\mathrm {\varepsilon } }_{yz}^{2}+{\mathrm {\varepsilon } }_{zx}^{2}\right).}\end{aligned}} (2.3a)

### Solution

From problem 2.1c, we see that $k=1/\left(4.5\times 10^{-10}\right)=2.2\times 10^{9}$ Pa. Also, ${\mathrm {\mu } }=0$ for water, so $E={\frac {1}{2}}{\mathrm {\lambda } }{\mathrm {\Delta } }^{2}$ . From equation (7,5) of Table 2.2a we find that ${\mathrm {\lambda } }=k$ when ${\mathrm {\mu } }=0$ . Also, $k=-{\mathcal {P}}/{\mathrm {\Delta } }$ (see equation (2.1f)), so

{\begin{aligned}\Delta =\left|{\cal {P}}/k\right|=5.0\times 10^{5}/2.2\times 10^{9}=2.3\times 10^{-4}.\end{aligned}} Using equation (2.3a) we find that

{\begin{aligned}E={\frac {1}{2}}\left(2.2\times 10^{9}\times 2.3^{2}\times 10^{-8}\right)=58\,\,\,\mathrm {J/m} ^{3}.\end{aligned}} [The dimensions of $E$ are the same as those of stress, since strains are dimensionless. Thus, stress units are ${\rm {N/m}}^{2}={\rm {Nm/m}}^{3}={\rm {J/m}}^{3}$ .]

## Problem 2.3b

If the same wave is generated in rock with ${\mathrm {\lambda } }={\mathrm {\mu } }=3.0\times 10^{10}$ Pa, what is the peak energy density? Assume a symmetrical $P$ -wave with ${\mathrm {\varepsilon } }_{xx}={\mathrm {\varepsilon } }_{yy}={\mathrm {\varepsilon } }_{zz},{\mathrm {\varepsilon } }_{ij}=0$ for $i\neq j$ .

### Solution

We have ${\mathrm {\Delta } }={\mathrm {\varepsilon } }_{xx}+{\mathrm {\varepsilon } }_{yy}+{\mathrm {\varepsilon } }_{zz}=3{\mathrm {\varepsilon } }_{xx}$ , ${\mathrm {\varepsilon } }_{ij}=0,i\neq j$ , so equation (2.3a) becomes

{\begin{aligned}E={\frac {1}{2}}{\mathrm {\lambda } }{\mathrm {\Delta } }^{2}+3{\mathrm {\mu } }{\mathrm {\varepsilon } }_{xx}^{2}={\mathrm {\lambda } }{\mathrm {\Delta } }^{2}\left({\frac {1}{2}}+{\frac {1}{3}}\right)={\frac {5}{6}}{\mathrm {\lambda } }({\mathcal {P}}/k)^{2}.\end{aligned}} Equation (9,3) in Table 2.2a gives $k={\mathrm {\lambda } }+{\frac {2}{3}}{\mathrm {\mu } }={\frac {5}{3}}{\mathrm {\lambda } }$ so that

{\begin{aligned}E={\frac {5}{6}}{\mathrm {\lambda } }{\mathcal {P}}^{2}\left/\right.\left({\frac {5}{3}}{\mathrm {\lambda } }\right)^{2}=0.3\times (5.0\times 10^{5})^{2}/3\times 10^{10}=2.5\ \mathrm {J/m} ^{3}.\end{aligned}} 