# Far- and near-field effects for a point 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 2.13a

Show that for harmonic waves of the form

 {\displaystyle {\begin{aligned}\phi =\left(A/r\right)\cos[\omega (r/V-t)],\end{aligned}}} (2.13a)

the displacement ${\displaystyle u\left(r,\;t\right)}$ is

 {\displaystyle {\begin{aligned}u\left(r,\;t\right)=-\left({\frac {A}{r^{2}}}\right)\cos \left[\omega \left(r/V-t\right)\right]-\left({\frac {A}{r}}\right)\left({\frac {\omega }{V}}\right)\sin[\omega (r/V-t)].\end{aligned}}} (2.13b)

### Background

If we set ${\displaystyle \mathrm {\chi } =0}$ in equation (2.9a), we obtain the result ${\displaystyle \mathrm {\zeta } =\nabla \mathrm {\phi } }$, where ${\displaystyle \mathrm {\phi } }$ is a solution of the P-wave equation. Furthermore if ${\displaystyle \mathrm {\phi } }$ is independent of latitude and longitude, the wave equation reduces to equation (2.5c), and the solution of problem 2.5c shows that equation (2.13a) is a P-wave solution of equation (2.5c).

### Solution

Since ${\displaystyle \mathrm {\phi } }$ in equation (2.13a) is a solution of equation (2.5c), it represents a spherically symmetrical P-wave and therefore the only displacement is along ${\displaystyle {r}}$. If we take the x-axis along ${\displaystyle {r}}$, equation (2.9d) shows that

{\displaystyle {\begin{aligned}u\left(r,\;t\right)={\frac {\partial \phi }{\partial r}}=-\left({\frac {A}{r^{2}}}\right)\cos \left[\omega \left(r/V-t\right)\right]-\left({\frac {A}{r}}\right)\left({\frac {\omega }{V}}\right)\sin[\omega (r/V-t)].\end{aligned}}}

## Problem 2.13b

Show that the two terms in equation (2.13b), which decay at different rates, are of equal importance at distance ${\displaystyle {\mathbf {r=\lambda /2\pi } }}$.

### Solution

The two terms are of equal importance when the two amplitudes are equal, that is, when ${\displaystyle A/r^{2}=\left(A\omega /rV\right)}$ or ${\displaystyle r=V/\mathrm {\omega } =\mathrm {\lambda } /2\pi }$.