# Magnitude

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Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 3 47 - 77 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 3.10

Using equation (1,8) in Table 2.2a, show that the fractional change ${\displaystyle \Delta \sigma /\sigma }$ is not necessarily small when ${\displaystyle \Delta \alpha /\alpha }$, ${\displaystyle \Delta \beta /\beta }$, and ${\displaystyle \Delta p/p}$ are all small.

### Solution

Equation (1,8) in Table 2.2a is

{\displaystyle {\begin{aligned}\beta /\alpha =\left(1-2\sigma \right)/2\left(1-\sigma \right).\end{aligned}}}

Because ${\displaystyle p}$ does not enter into this equation, it has no effect upon ${\displaystyle \Delta \sigma /\sigma }$. The fractions ${\displaystyle \Delta \alpha /\alpha }$, ${\displaystyle \Delta \beta /\beta }$, and ${\displaystyle \Delta \sigma /\sigma }$ are of the form ${\displaystyle \Delta x/x}$ which suggests that we use logs [since ${\displaystyle \Delta \left(Inx\right)=\Delta x/x]}$. Taking logs of both sides of the above equation, we get

{\displaystyle {\begin{aligned}\ln \beta -In\alpha =\ln \left(1-2\sigma \right)-In2-In\left(1-\sigma \right).\end{aligned}}}.

Differentiation gives

{\displaystyle {\begin{aligned}{\frac {\Delta \beta }{\beta }}-{\frac {\Delta \alpha }{\alpha }}={\frac {-2\Delta \sigma }{1-2\sigma }}+{\frac {\Delta \sigma }{1-\sigma }}=\left({\frac {\Delta \sigma }{\sigma }}\right){\frac {-1}{\left(1-2\sigma \right)\left({\frac {1}{\sigma }}-1\right)}}.\end{aligned}}}

Thus,

{\displaystyle {\begin{aligned}\left|{\frac {\Delta \sigma }{\sigma }}\left|=\right|\left({\frac {\Delta \beta }{\beta }}-{\frac {\Delta \alpha }{\alpha }}\right)\left(2\sigma -1\right)\left(1-{\frac {1}{\sigma }}\right)\right|\end{aligned}}}

Since ${\displaystyle 0\leq \sigma \leq +0.5}$, the product of the two ${\displaystyle \sigma }$-factors varies between ${\displaystyle 0}$ (when ${\displaystyle \sigma =0.5}$) and ${\displaystyle +\infty }$ (when ${\displaystyle 2\sigma =0}$). Therefore, even though ${\displaystyle \left({\frac {\Delta \beta }{\beta }}-{\frac {\Delta \alpha }{\alpha }}\right)}$ is small (being the difference between two small quantities), the right-hand side can be large.