# 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 ${\displaystyle {\mathcal {P}}}$ of 5 atmospheres (${\displaystyle 5.0\times 10^{5}}$ Pa). If the compressibility of water is ${\displaystyle 4.5\times 10^{-10}}$/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) ${\displaystyle E}$ is equal to [see Sheriff and Geldart, 1995, equation (2.22)]

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

{\displaystyle {\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

{\displaystyle {\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 ${\displaystyle E}$ are the same as those of stress, since strains are dimensionless. Thus, stress units are ${\displaystyle {\rm {N/m}}^{2}={\rm {Nm/m}}^{3}={\rm {J/m}}^{3}}$.]

## Problem 2.3b

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

### Solution

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

{\displaystyle {\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 ${\displaystyle k={\mathrm {\lambda } }+{\frac {2}{3}}{\mathrm {\mu } }={\frac {5}{3}}{\mathrm {\lambda } }}$ so that

{\displaystyle {\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}}}