# Interrelationships among elastic constants

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 2.2

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 ${\displaystyle {\alpha }}$ and ${\displaystyle {\beta }}$ (see 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, ${\displaystyle {\mathrm {\alpha } }}$ and ${\displaystyle {\mathrm {\beta } }}$, given by the equations

 {\displaystyle {\begin{aligned}{\mathrm {\rho } }{\mathrm {\alpha } }^{2}=\left({\mathrm {\lambda } }+2{\mathrm {\mu } }\right),{\quad }{\mathrm {\rho } }\beta ^{2}=\mu \end{aligned}}} (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 ${\displaystyle {\mathrm {\rho } }}$).

### 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 ${\displaystyle {\mathrm {\alpha } }}$ and ${\displaystyle {\mathrm {\beta } }}$, then we use equations (9,6) and (9,7) (see equation (2.2a)) to derive the rest. From equation (9,1),

 {\displaystyle {\begin{aligned}{{\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{aligned}}} (3,5)

From equation (9,2),

 {\displaystyle {\begin{aligned}{{\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{aligned}}} (5,5)

Solving equation (9,2) for ${\displaystyle {\mathrm {\mu } }}$, we have ${\displaystyle 2{\mathrm {\mu } }{\mathrm {\sigma } }={\mathrm {\lambda } }\left(1-2{\mathrm {\sigma } }\right)}$,

 {\displaystyle {\begin{aligned}{\mathrm {\mu } }={\mathrm {\lambda } }\left(1-2{\mathrm {\sigma } }\right)/2{\mathrm {\sigma } }.\end{aligned}}} (6,4)

From equation (9,3),

 {\displaystyle {\begin{aligned}{\mathrm {\lambda } }=k-{\frac {2}{3}}{\mathrm {\mu } }.\end{aligned}}} (7,5)

Equating ${\displaystyle {\mathrm {\mu } }}$ from equations (9,2) and (9,3) [or from equations (6,4) and (7,5)] gives

 {\displaystyle {\begin{aligned}{\mathrm {\mu } }={\mathrm {\lambda } }\left(1-2{\mathrm {\sigma } }\right)/2{\mathrm {\sigma } }={\frac {3}{2}}\left(k-{\mathrm {\lambda } }\right),\end{aligned}}} (8,4)

thus

 {\displaystyle {\begin{aligned}k={\mathrm {\lambda } }\left(1-2{\mathrm {\sigma } }\right)/3{\mathrm {\sigma } }+{\mathrm {\lambda } }={\mathrm {\lambda } }\left(1+{\mathrm {\sigma } }\right)/3\sigma ,\end{aligned}}} (6,3)

and

 {\displaystyle {\begin{aligned}{\mathrm {\lambda } }={\frac {3{\mathrm {\sigma } }k}{\left(1+{\mathrm {\sigma } }\right)}}.\end{aligned}}} (4,5)

Solving equation (4,5) for ${\displaystyle {\mathrm {\sigma } }}$ gives

 {\displaystyle {\begin{aligned}{\mathrm {\sigma } }={\frac {\mathrm {\lambda } }{3k-{\mathrm {\lambda } }}}.\end{aligned}}} (8,2)

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

 {\displaystyle {\begin{aligned}{\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{aligned}}} (4,4)

We use equation (7,5) to eliminate ${\displaystyle {\mathrm {\lambda } }}$ from equation (6,3):

{\displaystyle {\begin{aligned}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{aligned}}}

that is,

{\displaystyle {\begin{aligned}k\left({\frac {1+{\mathrm {\sigma } }}{3{\mathrm {\sigma } }}}-1\right)={\frac {2{\mathrm {\mu } }\left(1+{\mathrm {\sigma } }\right)}{9{\mathrm {\sigma } }}},\end{aligned}}}

so

 {\displaystyle {\begin{aligned}k={\frac {2{\mathrm {\mu } }\left(1+{\mathrm {\sigma } }\right)}{3\left(1-2{\mathrm {\sigma } }\right)}}.\end{aligned}}} (5,3)

Solving for ${\displaystyle {\mathrm {\sigma } }}$, we get

 {\displaystyle {\begin{aligned}{\mathrm {\sigma } }={\frac {\left(3k-2{\mathrm {\mu } }\right)}{2\left(3k+{\mathrm {\mu } }\right)}}.\end{aligned}}} (7,2)
Figure 2.2a)  Relations between elastic constants and velocities for isotropic media.

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

 {\displaystyle {\begin{aligned}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{aligned}}} (8,1)

Solving equation (8,1) for ${\displaystyle {\mathrm {\lambda } }}$ gives

 {\displaystyle {\begin{aligned}{\mathrm {\lambda } }=3k\left(3k-E\right)/\left(9k-E\right).\end{aligned}}} (2,5)

We now eliminate ${\displaystyle {\mathrm {\lambda } }}$ from equation (9,1) using equation (7,5)

 {\displaystyle {\begin{aligned}{E={\frac {\mu \left({3k-2\mu +2\mu }\right)}{\left({k-2\mu /3+\mu }\right)}}}\\{=9k\mu /\left({3k+\mu }\right).}\end{aligned}}} (7,1)

Solving equation (7,1) for ${\displaystyle k}$ and ${\displaystyle {\mathrm {\mu } }}$ gives

 {\displaystyle {\begin{aligned}k={\mathrm {\mu } }E/3\left(3{\mathrm {\mu } }-E\right),\end{aligned}}} (3,3)

 {\displaystyle {\begin{aligned}{\mathrm {\mu } }=3kE/\left(9k-E\right).\end{aligned}}} (2,4)

Next we use equation (6,4) to eliminate ${\displaystyle {\mathrm {\mu } }}$ from equation (9,1):

 {\displaystyle {\begin{aligned}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{aligned}}} (6,1)

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

 {\displaystyle {\begin{aligned}{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{aligned}}} (5,1)

Solving equation (5,1) for ${\displaystyle {\mathrm {\sigma } }}$ and ${\displaystyle {\mathrm {\mu } }}$ gives

 {\displaystyle {\begin{aligned}{\mathrm {\sigma } }=E/2{\mathrm {\mu } }-1=\left(E-2{\mathrm {\mu } }\right)/2\mu ,\end{aligned}}} (3,2)

 {\displaystyle {\begin{aligned}{\mathrm {\mu } }=E/2\left(1+{\mathrm {\sigma } }\right).\end{aligned}}} (1,4)

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

 {\displaystyle {\begin{aligned}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{aligned}}} (4,1)

Solving equation (4,1) for ${\displaystyle k}$ and ${\displaystyle {\mathrm {\sigma } }}$, we get

 {\displaystyle {\begin{aligned}k=E/3\left(1-2{\mathrm {\sigma } }\right),\end{aligned}}} (1,3)

 {\displaystyle {\begin{aligned}{\mathrm {\sigma } }=\left(3k-E\right)/6k.\end{aligned}}} (2,2)

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

 {\displaystyle {\begin{aligned}{\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{aligned}}} (1,5)

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

 {\displaystyle {\begin{aligned}{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{aligned}}} (10,1)

 {\displaystyle {\begin{aligned}{\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{aligned}}} (10,2)

 {\displaystyle {\begin{aligned}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{aligned}}} (10,3)

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

 {\displaystyle {\begin{aligned}{\mathrm {\lambda } }=\left({\mathrm {\lambda } }+2{\mathrm {\mu } }\right)-2{\mathrm {\mu } }={\mathrm {\rho } }\left({\mathrm {\alpha } }^{2}-2\beta ^{2}\right).\end{aligned}}} (10,5)

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

 {\displaystyle {\begin{aligned}{{\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{aligned}}} (1,6)

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

 {\displaystyle {\begin{aligned}{{\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{aligned}}} (2,6)

In the same way, we get the following results:

 {\displaystyle {\begin{aligned}{{\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{aligned}}} (3,6)

 {\displaystyle {\begin{aligned}{{\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{aligned}}} (4,6)

 {\displaystyle {\begin{aligned}{\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{aligned}}} (5,6)

 {\displaystyle {\begin{aligned}{\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{aligned}}} (6,6)

 {\displaystyle {\begin{aligned}{\mathrm {\rho } }{\mathrm {\alpha } }^{2}=\left(k-{\frac {2}{3}}{\mathrm {\mu } }\right)+2{\mathrm {\mu } }=k+{\frac {4}{3}}{\mathrm {\mu } },\end{aligned}}} (7,6)

 {\displaystyle {\begin{aligned}{\mathrm {\rho } }{\mathrm {\alpha } }^{2}={\mathrm {\lambda } }+3\left(k-{\mathrm {\lambda } }\right)=3k-2\lambda .\end{aligned}}} (8,6)

Column 7 is merely the square root of column 4 after dividing by ${\displaystyle {\mathrm {\rho } }}$. Column 8 is obtained by dividing column 7 by column 6.