Reflection and transmission coefficients

Problems in Exploration Seismology and their Solutions

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

Problem 3.7a

Calculate the reflection and transmission coefficients, $R$ and $T$ of equations (3.6a,b), for a sandstone/shale interface (for a wave incident from the sandstone) for the following:

1. $V_{ss}=2.43$ km/s, $V_{sh}=2.02$ km/s, $\rho _{ss}=2.08$ g/cm$^{3}$, and $\rho _{sh}=2.23$ g/cm$^{3}$;
2. $V_{ss}=3.35$ km/s, $V_{sh}=3.14$ km/s,

    $\rho _{ss}=2.21$ g/cm$^{3}$, and $\rho _{sh}=2.52$ g/cm$^{3}$.
   

3. What are the corresponding values in nepers and decibels?

Background

A wave has kinetic energy due to the velocity of the medium and potential energy due to the strains in the medium. For a harmonic wave $\psi =A\sin \omega t$, the particle velocity is $\partial \psi /\partial t=\omega A\cos \omega t$ and the kinetic energy/unit volume is ${\frac {1}{2}}\rho \omega ^{2}A^{2}\cos ^{2}\omega t$, $\rho$ being the density. As the wave progresses, the energy changes back and forth between kinetic and potential. When the potential energy is zero, the kinetic energy is a maximum and therefore equals the total energy. The maximum value comes when $\cos ^{2}\omega t=+1$, so the total energy/unit volume${}=E={\frac {1}{2}}\rho \omega ^{2}A^{2}$, where $E$ is the energy density (problem 2.3).

The coefficients $R$ and $T$ in equations (3.6a,b) give ratios of the relative amplitudes of the reflected and transmitted waves. We denote the fractions of the incident energy that are reflected and transmitted by $E_{R}$ and $E_{T}$; “energy” as used here denotes the amount of energy flowing through a unit area normal to the wave direction per unit time (the intensity). The energy density (i.e., energy/unit volume) is given by equation (3.3k). The energy flowing through a unit area normal to the wave direction per unit time is equal to the energy density times the velocity, that is, $\left({\frac {1}{2}}\rho \omega ^{2}A^{2}\right)\alpha$ for a P-wave. Therefore

${\begin{aligned}E_{R}=R^{2},\qquad E_{T}={\frac {\rho _{2}\alpha _{2}\omega ^{2}A_{2}^{2}}{\rho _{1}\alpha _{1}\omega ^{2}A_{0}^{2}}}=\left({\frac {Z_{2}}{Z_{1}}}\right)T^{2}={\frac {4Z_{1}Z_{2}}{(Z_{1}+Z_{2})^{2}}}.\end{aligned}}$

(3.7a)

The coefficients $E_{R}$ and $E_{T}$ are sometimes referred to as reflection and transmission energy coefficients to distinguish them from $R$ and $T$. Note that both $E_{R}$ and $E_{T}$ are independent of the direction of travel through the interface.

Solution

i) Using equations (3.6a,b), we have

${\begin{aligned}Z_{ss}=2.08\times 2.43=5.05,\qquad Z_{sh}=2.23\times 2.02=4.50\end{aligned}}$

(where the units are g. km/cm³. s). Then,

${\begin{aligned}R=\left(4.50-5.05\right)/\left(4.50+5.05\right)=-0.55/9.55=-0.058.\end{aligned}}$

(The minus sign denotes a phase reversal; see problem 3.6.) Because the incident wave is in the sandstone, we have

${\begin{aligned}T=2\times 5.05/9.55=1.06.\end{aligned}}$

Note that the amplitude of the transmitted wave is larger than that of the incident wave when $R$ is negative (see problem 3.6c).

ii) $Z_{ss}=2.21\times 3.35=7.40,\ Z_{sh}=2.52\times 3.14=7.91$.

Then,

${\begin{aligned}R&=\left(7.91-7.40\right)/\left(7.91+7.40\right)=0.51/15.3=0.033,\\T&=2\times 7.40/15.3=0.967.\end{aligned}}$

iii) From problem 2.17, we have nepers ln(amplitude ratio) and 1 neper ${}=8.686$ dB. Therefore,

for (i),

${\begin{aligned}R&=\ln \left(0.058\right)=-2.8\ \mathrm {nepers} =-24\ \mathrm {dB} ,\\T&=\ln \left(1.06\right)=0.058\ \mathrm {nepers} \ =0.51\ \mathrm {dB} .\end{aligned}}$

For (ii),

${\begin{aligned}R&=\ln \left(0.033\right)=-3.4\ \mathrm {nepers} \ =-30\ \mathrm {dB} ,\\T&=\ln \left(0.967\right)=-0.034\ \mathrm {nepers} \ =-0.29\ \mathrm {dB} .\end{aligned}}$

Negative values of nepers and $dB$ merely mean that the values are less than unity.

Problem 3.7b

Calculate the energy coefficients $E_{R}$ and $E_{T}$ for cases (i) and (ii) in part (a).

Solution

We use equation (3.7a) to get,

for (i),

${\begin{aligned}E_{R}&=R^{2}=(0.058)^{2}=0.0034,\\E_{T}&=\left(Z_{2}/Z_{1}\right)T^{2}=\left(4.50/5.05\right)\times 1.06^{2}=1.00.\end{aligned}}$

For (ii),

${\begin{aligned}E_{R}&=0.033^{2}=0.001,\\E_{T}&=(\left(7.91/7.40\right)\times 0.967^{2}=1.00.\end{aligned}}$

Note that $E_{R}+E_{T}=1$ (within the accuracy of the calculations).

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Table of Contents (book)

Also in this chapter

• General form of Snell’s law
• Reflection/refraction at a solid/solid interface and displacement of a free surface
• Reflection/refraction at a liquid/solid interface
• Zoeppritz’s equations for incident SV- and SH-waves
• Reinforcement depth in marine recording
• Complex coefficient of reflection
• Amplitude/energy of reflections and multiples
• Reflection/transmission coefficients at small angles and magnitude
• Magnitude
• AVO versus AVA and effect of velocity gradient
• Variation of reflectivity with angle (AVA)

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