View source for Interrelationships among elastic constants

{{publication
 | name    = Problems in Exploration Seismology and their Solutions
 | image   = [[File:897.Def.Lz-cropped.png|frameless|180px|link=Problems in Exploration Seismology and their Solutions]]
 | width   = 120px
 | series  = Geophysical References Series
 | title   = [[Problems in Exploration Seismology and their Solutions]]
 | part    = 
 | chapter = 2
 | pages  = 7 - 46
 | author  = [[Lloyd P. Geldart]] and [[Robert E. Sheriff]]
 | doi     = 10.1190/1.9781560801733
 | pdf     = 
 | store   = https://shop.seg.org/shop/products/detail/897
 | isbn    = ISBN 9781560801153
}}
__TOC__
== Problem 2.2 ==
The entries in Table 2.2a express, for isotropic media, the quantities at the heads of the columns in terms of the pairs of elastic constants or velocities at the left ends of the rows. The first three entries in the ninth row are equations (2.1j), (2.1a), and (2.1k) and the next two entries in the same row are formulas for the P- and S- wave velocities <math>{\alpha} </math> and <math>{\beta} </math> (see [[General solutions of the wave equation|problem 2.5]]). Starting from these five relations, derive the other relations in the table.

=== Background ===
For isotropic media, any two of the elastic constants can be considered as independent and the others can be expressed in terms of these two. The P- and S-wave velocities, <math>{\mathrm{\alpha}} </math> and <math>{\mathrm{\beta}} </math>, given by the equations

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\rho}}{\mathrm{\alpha}} ^{2} =\left({\mathrm{\lambda}} +2{\mathrm{\mu}} \right),{\quad} {\mathrm{\rho}}\beta ^{2} =\mu
\end{align}
</math>|{{EquationRef|2.2a}}}}

[see Sheriff and Geldart, 1995, equations (2.28) and (2.29)] can also be expressed in terms of any two elastic constants (plus the density <math>{\mathrm{\rho}}</math>).

=== Solution ===

Denoting the equations by row and column (as for matrix elements) and using a comma instead of a period, we use equations (9,1) to (9,3) to derive the equations that do not involve <math>{\mathrm{\alpha}} </math> and <math>{\mathrm{\beta}} </math>, then we use equations (9,6) and (9,7) (see equation (2.2a)) to derive the rest. From equation (9,1),

{{NumBlk|:|<math>
\begin{align}
{{\mathrm{\lambda}} \left(E-3{\mathrm{\mu}} \right)=2{\mathrm{\mu}} ^{2} -{\mathrm{\mu}} E,}\\
{{\mathrm{\lambda}} ={\mathrm{\mu}} \left(E-2{\mathrm{\mu}} \right)/\left(3{\mathrm{\mu}} -E\right).}
\end{align}
</math>|{{EquationRef|3,5}}}}

From equation (9,2),

{{NumBlk|:|<math>
\begin{align}
{{\mathrm{\lambda}} \left(2{\mathrm{\sigma}} -1\right)+2{\mathrm{\mu}} {\mathrm{\sigma}} =0,}\\
{{\mathrm{\lambda}} =2{\mathrm{\mu}} {\mathrm{\sigma}} /\left(1-2{\mathrm{\sigma}} \right).}
\end{align}
</math>|{{EquationRef|5,5}}}}

Solving equation (9,2) for <math>{\mathrm{\mu}} </math>, we have <math>2{\mathrm{\mu}} {\mathrm{\sigma}} ={\mathrm{\lambda}} \left(1-2{\mathrm{\sigma}} \right)</math>,

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\mu}} ={\mathrm{\lambda}} \left(1-2{\mathrm{\sigma}} \right)/2{\mathrm{\sigma}}.
\end{align}
</math>|{{EquationRef|6,4}}}}

From equation (9,3),

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\lambda}} =k-\frac{2}{3} {\mathrm{\mu}}.
\end{align}
</math>|{{EquationRef|7,5}}}}

Equating <math>{\mathrm{\mu}} </math> from equations (9,2) and (9,3) [or from equations (6,4) and (7,5)] gives

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\mu}} ={\mathrm{\lambda}} \left(1-2{\mathrm{\sigma}} \right)/2{\mathrm{\sigma}} =\frac{3}{2} \left(k-{\mathrm{\lambda}} \right),
\end{align}
</math>|{{EquationRef|8,4}}}}
thus
{{NumBlk|:|<math>
\begin{align}
k={\mathrm{\lambda}} \left(1-2{\mathrm{\sigma}} \right)/3{\mathrm{\sigma}} +{\mathrm{\lambda}} ={\mathrm{\lambda}} \left(1+{\mathrm{\sigma}} \right)/3\sigma,
\end{align}
</math>|{{EquationRef|6,3}}}}
and
{{NumBlk|:|<math>
\begin{align}
{\mathrm{\lambda}} =\frac{3{\mathrm{\sigma}} k}{\left(1+{\mathrm{\sigma}} \right)}.
\end{align}
</math>|{{EquationRef|4,5}}}}

Solving equation (4,5) for <math>{\mathrm{\sigma}}</math> gives

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\sigma}} =\frac{{\mathrm{\lambda}} }{3k-{\mathrm{\lambda}} }.
\end{align}
</math>|{{EquationRef|8,2}}}}

Substituting equation (4,5) into equation (8,4), we get,

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\mu}} =\frac{3}{2} \left(k-\frac{3{\mathrm{\sigma}} k}{\left(1+{\mathrm{\sigma}} \right)} \right)=\frac{3k\left(1-2{\mathrm{\sigma}} \right)}{2\left(1+{\mathrm{\sigma}} \right)}.
\end{align}
</math>|{{EquationRef|4,4}}}}

We use equation (7,5) to eliminate <math>{\mathrm{\lambda}} </math> from equation (6,3):

<math>
\begin{align}
k=\left(k-\frac{2}{3} {\mathrm{\mu}} \right)\left(\frac{1+{\mathrm{\sigma}} }{3{\mathrm{\sigma}} } \right)=\frac{k\left(1+{\mathrm{\sigma}} \right)}{3{\mathrm{\sigma}} } -\frac{2{\mathrm{\mu}} \left(1+{\mathrm{\sigma}} \right)}{9{\mathrm{\sigma}} },
\end{align}
</math>

that is, 

<math>
\begin{align}
k\left(\frac{1+{\mathrm{\sigma}} }{3{\mathrm{\sigma}} } -1\right)=\frac{2{\mathrm{\mu}} \left(1+{\mathrm{\sigma}} \right)}{9{\mathrm{\sigma}} },
\end{align}
</math>

so

{{NumBlk|:|<math>
\begin{align}
k=\frac{2{\mathrm{\mu}} \left(1+{\mathrm{\sigma}} \right)}{3\left(1-2{\mathrm{\sigma}} \right)}.
\end{align}
</math>|{{EquationRef|5,3}}}}

Solving for <math>{\mathrm{\sigma}}</math>, we get

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\sigma}} =\frac{\left(3k-2{\mathrm{\mu}} \right)}{2\left(3k+{\mathrm{\mu}} \right)}.
\end{align}
</math>|{{EquationRef|7,2}}}}

[[file:ch02_tab2-2a.png|thumb|{{figure number|2.2a)}} Relations between elastic constants and velocities for isotropic media.]]

We use equation (8,4) to express <math>{\mathrm{\mu}} </math> in equation (9,1) in terms of <math>\left(k,\; {\mathrm{\lambda}} \right)</math>. Thus

{{NumBlk|:|<math>
\begin{align}
E=\frac{\frac{3}{2}\left(k-{\mathrm{\lambda}} \right)\left[3{\mathrm{\lambda}} +3\left(k-{\mathrm{\lambda}} \right)\right]}{{\mathrm{\lambda}} +\frac{3}{2} \left(k-{\mathrm{\lambda}} \right)} =\frac{9k\left(k-{\mathrm{\lambda}} \right)}{\left(3k-{\mathrm{\lambda}} \right)}.
\end{align}
</math>|{{EquationRef|8,1}}}}

Solving equation (8,1) for <math>{\mathrm{\lambda}} </math> gives

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\lambda}} =3k\left(3k-E\right)/\left(9k-E\right).
\end{align}
</math>|{{EquationRef|2,5}}}}

We now eliminate <math>{\mathrm{\lambda}} </math> from equation (9,1) using equation (7,5)

{{NumBlk|:|<math>
\begin{align}
{E = \frac{{\mu \left( {3k - 2\mu  + 2\mu } \right)}}{{\left( {k - 2\mu /3 + \mu } \right)}}}\\
{ = 9k\mu /\left( {3k + \mu } \right).}
\end{align}
</math>|{{EquationRef|7,1}}}}

Solving equation (7,1) for <math>k</math> and <math>{\mathrm{\mu}} </math> gives

{{NumBlk|:|<math>
\begin{align}
k ={\mathrm{\mu}} E/3\left(3{\mathrm{\mu}} -E\right),
\end{align}
</math>|{{EquationRef|3,3}}}}

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\mu}} =3kE/\left(9k-E\right).
\end{align}
</math>|{{EquationRef|2,4}}}}

Next we use equation (6,4) to eliminate <math>{\mathrm{\mu}} </math> from equation (9,1):

{{NumBlk|:|<math>
\begin{align}
E={\mathrm{\mu}} \left(\frac{6{\mathrm{\mu}} {\mathrm{\sigma}} +2{\mathrm{\mu}} -4{\mathrm{\mu}} {\mathrm{\sigma}} }{2{\mathrm{\mu}} {\mathrm{\sigma}} +{\mathrm{\mu}} -2{\mathrm{\mu}} {\mathrm{\sigma}} } \right)={\mathrm{\lambda}} \left(1-2{\mathrm{\sigma}} \right)\left(1+{\mathrm{\sigma}} \right)/{\mathrm{\sigma}}.
\end{align}
</math>|{{EquationRef|6,1}}}}

Using equations (9,1) and (5,5) we get

{{NumBlk|:|<math>
\begin{align}
{E ={\mathrm{\mu}} \left(6{\mathrm{\mu}} {\mathrm{\sigma}} \right)/\left(2{\mathrm{\mu}} +{\mathrm{\mu}} -2{\mathrm{\mu}} {\mathrm{\sigma}} \right)} \\
{ {\quad} =2{\mathrm{\mu}} ^{2} \left(1+{\mathrm{\sigma}} \right)/{\mathrm{\mu}} =2{\mathrm{\mu}} \left(1+{\mathrm{\sigma}} \right).}
\end{align}
</math>|{{EquationRef|5,1}}}}

Solving equation (5,1) for <math>{\mathrm{\sigma}}</math> and <math>{\mathrm{\mu}} </math> gives

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\sigma}} =E/2{\mathrm{\mu}} -1=\left(E-2{\mathrm{\mu}} \right)/2\mu,
\end{align}
</math>|{{EquationRef|3,2}}}}

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\mu}} =E/2\left(1+{\mathrm{\sigma}} \right).
\end{align}
</math>|{{EquationRef|1,4}}}}

Using equations (4,4) and (4,5) to replace <math>\left({\mathrm{\lambda}}, \; {\mathrm{\mu}} \right)</math> in equation (9,1) by <math>\left({\mathrm{\sigma}}, \; k\right)</math> gives

{{NumBlk|:|<math>
\begin{align}
E=\frac{3k\left(1-2{\mathrm{\sigma}} \right)}{2\left(1+{\mathrm{\sigma}} \right)} \times \frac{\left(9{\mathrm{\sigma}} k+3k-6{\mathrm{\sigma}} k\right)}{\left[3{\mathrm{\sigma}} k+\left(3k-6{\mathrm{\sigma}} k\right)/2\right]} =3k\left(1-2{\mathrm{\sigma}} \right).
\end{align}
</math>|{{EquationRef|4,1}}}}

Solving equation (4,1) for <math>k</math> and <math>{\mathrm{\sigma}}</math>, we get

{{NumBlk|:|<math>
\begin{align}
k=E/3\left(1-2{\mathrm{\sigma}} \right),
\end{align}
</math>|{{EquationRef|1,3}}}}

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\sigma}} =\left(3k-E\right)/6k.
\end{align}
</math>|{{EquationRef|2,2}}}}

The last equation (of this group) can be obtained by substituting equation (1,3) into equation (4,5):

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\lambda}} =\frac{3{\mathrm{\sigma}} k}{\left(1+{\mathrm{\sigma}} \right)} =\frac{3{\mathrm{\sigma}} E}{3\left(1-2{\mathrm{\sigma}} \right)\left(1+{\mathrm{\sigma}} \right)} =\frac{{\mathrm{\sigma}} E}{\left(1+{\mathrm{\sigma}} \right)\left(1-2{\mathrm{\sigma}} \right)}.
\end{align}
</math>|{{EquationRef|1,5}}}}

Equations (10,1) to (10,3) express <math>E</math>, <math>{\mathrm{\sigma}}</math>, and <math>k</math> in terms of the P- and S-wave velocities, <math>{\mathrm{\alpha}} </math> and <math>{\mathrm{\beta}} </math> (see equation 2.2a). To introduce <math>{\mathrm{\alpha}} </math> and <math>{\mathrm{\beta}} </math>, we write equations (10,1) to (10,3) in terms of <math>\left({\mathrm{\lambda}} +2{\mathrm{\mu}} \right)</math> and <math>{\mathrm{\mu}} </math>. Thus,

{{NumBlk|:|<math>
\begin{align}
{E ={\mathrm{\mu}} \left(\frac{3{\mathrm{\lambda}} +2{\mathrm{\mu}} }{{\mathrm{\lambda}} +{\mathrm{\mu}} } \right)={\mathrm{\mu}} \left[\frac{3\left({\mathrm{\lambda}} +2{\mathrm{\mu}} \right)-4{\mathrm{\mu}} }{\left({\mathrm{\lambda}} +2{\mathrm{\mu}} \right)-{\mathrm{\mu}} } \right] }\\
{ {\qquad}{\quad}=\frac{{\mathrm{\rho}}\beta ^{2} \left(3{\mathrm{\rho}}{\mathrm{\alpha}} ^{2} -4{\mathrm{\rho}}\beta ^{2} \right)}{{\mathrm{\rho}}\left({\mathrm{\alpha}} ^{2} -\beta ^{2} \right)} =\frac{{\mathrm{\rho}}\beta ^{2} \left(3{\mathrm{\alpha}} ^{2} -4\beta ^{2} \right)}{{\mathrm{\alpha}} ^{2} -\beta ^{2} }},
\end{align}
</math>|{{EquationRef|10,1}}}}

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\sigma}} =\frac{{\mathrm{\lambda}} }{2\left({\mathrm{\lambda}} +{\mathrm{\mu}} \right)} =\frac{\left({\mathrm{\lambda}} +2{\mathrm{\mu}} \right)-2{\mathrm{\mu}} }{2\left({\mathrm{\lambda}} +2{\mathrm{\mu}} \right)-2{\mathrm{\mu}} } =\frac{{\mathrm{\alpha}} ^{2} -2\beta ^{2} }{2\left({\mathrm{\alpha}} ^{2} -\beta ^{2} \right)},
\end{align}
</math>|{{EquationRef|10,2}}}}

{{NumBlk|:|<math>
\begin{align}
k={\mathrm{\lambda}} +\frac{2{\mathrm{\mu}} }{3} =\left({\mathrm{\lambda}} +2{\mathrm{\mu}} \right)-\frac{4}{3} {\mathrm{\mu}} ={\mathrm{\rho}}\left({\mathrm{\alpha}} ^{2} -\frac{4}{3} \beta ^{2} \right).
\end{align}
</math>|{{EquationRef|10,3}}}}

Equation (10,4) is the second equation in equation (2.2a). To get equation (10,5), we write

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\lambda}} =\left({\mathrm{\lambda}} +2{\mathrm{\mu}} \right)-2{\mathrm{\mu}} ={\mathrm{\rho}}\left({\mathrm{\alpha}} ^{2} -2\beta ^{2} \right).
\end{align}
</math>|{{EquationRef|10,5}}}}

To verify column 6, we start with equation (9,6) and express <math>{\mathrm{\lambda}} </math> and <math>{\mathrm{\mu}} </math> in terms of the required pair of constants. Thus, using equations (1,4) and (1,5) we get

{{NumBlk|:|<math>
\begin{align}
{{\mathrm{\rho}}{\mathrm{\alpha}} ^{2} =\left({\mathrm{\lambda}} +2{\mathrm{\mu}} \right)=\frac{E{\mathrm{\sigma}} }{\left(1+{\mathrm{\sigma}} \right)\left(1-2{\mathrm{\sigma}} \right)} +\frac{2E}{2\left(1+{\mathrm{\sigma}} \right)}}\\
{{\qquad}=\left(\frac{E}{1+{\mathrm{\sigma}} } \right)\left(\frac{{\mathrm{\sigma}} }{1-2{\mathrm{\sigma}} } +1\right)=\frac{E\left(1-{\mathrm{\sigma}} \right)}{\left(1+{\mathrm{\sigma}} \right)\left(1-2{\mathrm{\sigma}} \right)}}.
\end{align}
</math>|{{EquationRef|1,6}}}}

Following the same procedure, using equations (2,4) and (2,5), we get

{{NumBlk|:|<math>
\begin{align}
{{\mathrm{\rho}}{\mathrm{\alpha}} ^{2} =3k\left(\frac{3k-E}{9k-E} \right)+\left(\frac{6kE}{9k-E} \right) }\\
{{\qquad}=\left(\frac{3k}{9k-E} \right)\left(3k-E+2E\right)=\frac{3k\left(3k+E\right)}{9k-E}.}
\end{align}
</math>|{{EquationRef|2,6}}}}

In the same way, we get the following results:


{{NumBlk|:|<math>
\begin{align}
{{\mathrm{\rho}}{\mathrm{\alpha}} ^{2} ={\mathrm{\mu}} \left(\frac{E-2{\mathrm{\mu}} }{3{\mathrm{\mu}} -E} \right)+2{\mathrm{\mu}} =\left(\frac{{\mathrm{\mu}} }{3{\mathrm{\mu}} -E} \right)\left(E-2{\mathrm{\mu}} +6{\mathrm{\mu}} -2E\right) }\\
{{\qquad} =\left(\frac{{\mathrm{\mu}} }{3{\mathrm{\mu}} -E} \right)\left(4{\mathrm{\mu}} -E\right),}
\end{align}
</math>|{{EquationRef|3,6}}}}

{{NumBlk|:|<math>
\begin{align}
{{\mathrm{\rho}}{\mathrm{\alpha}} ^{2} =\left(\frac{3k{\mathrm{\sigma}} }{1+{\mathrm{\sigma}} } \right)+3k\left(\frac{1-2{\mathrm{\sigma}} }{1+{\mathrm{\sigma}} } \right)=\left(\frac{3k}{1+{\mathrm{\sigma}} } \right)\left({\mathrm{\sigma}} +1-2{\mathrm{\sigma}} \right)} \\
{{\qquad} =\frac{3k\left(1-{\mathrm{\sigma}} \right)}{\left(1+{\mathrm{\sigma}} \right)},}
\end{align}
</math>|{{EquationRef|4,6}}}}

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\rho}}{\mathrm{\alpha}} ^{2} =\left(\frac{2{\mathrm{\mu}} {\mathrm{\sigma}} }{1-2{\mathrm{\sigma}} } \right)+2{\mathrm{\mu}} =2{\mathrm{\mu}} \left(\frac{{\mathrm{\sigma}} }{1-2{\mathrm{\sigma}} } +1\right)=2{\mathrm{\mu}} \left(\frac{1-{\mathrm{\sigma}} }{1-2{\mathrm{\sigma}} } \right),
\end{align}
</math>|{{EquationRef|5,6}}}}

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\rho}}{\mathrm{\alpha}} ^{2} ={\mathrm{\lambda}} +\frac{{\mathrm{\lambda}} \left(1-2{\mathrm{\sigma}} \right)}{{\mathrm{\sigma}} } =\frac{{\mathrm{\lambda}} }{{\mathrm{\sigma}} } \left(1-{\mathrm{\sigma}} \right),
\end{align}
</math>|{{EquationRef|6,6}}}}

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\rho}}{\mathrm{\alpha}} ^{2} =\left(k-\frac{2}{3} {\mathrm{\mu}} \right)+2{\mathrm{\mu}} =k+\frac{4}{3} {\mathrm{\mu}},
\end{align}
</math>|{{EquationRef|7,6}}}}

{{NumBlk|:|<math>
\begin{align}
{\mathrm{\rho}}{\mathrm{\alpha}} ^{2} ={\mathrm{\lambda}} +3\left(k-{\mathrm{\lambda}} \right)=3k-2\lambda.
\end{align}
</math>|{{EquationRef|8,6}}}}

Column 7 is merely the square root of column 4 after dividing by <math>{\mathrm{\rho}}</math>. Column 8 is obtained by dividing column 7 by column 6.

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[[Problems_in_Exploration_Seismology_and_their_Solutions|Table of Contents (book)]]
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== Also in this chapter ==
* [[The basic elastic constants]]
* [[Magnitude of disturbance from a seismic source]]
* [[Magnitudes of elastic constants]]
* [[General solutions of the wave equation]]
* [[Wave equation in cylindrical and spherical coordinates]]
* [[Sum of waves of different frequencies and group velocity]]
* [[Magnitudes of seismic wave parameters]]
* [[Potential functions used to solve wave equations]]
* [[Boundary conditions at different types of interfaces]]
* [[Boundary conditions in terms of potential functions]]
* [[Disturbance produced by a point source]]
* [[Far- and near-field effects for a point source]]
* [[Rayleigh-wave relationships]]
* [[Directional geophone responses to different waves]]
* [[Tube-wave relationships]]
* [[Relation between nepers and decibels]]
* [[Attenuation calculations]]
* [[Diffraction from a half-plane]]

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