1st Lamé parameter

Gabriel Lamé

The 1st Lamé parameter, sometimes called Lamé's first parameter, but is more usually referred to simply as lambda, λ. It is an elastic modulus and used extensively in quantitative seismic interpretation and rock physics. It was first described by the French mathematician, Gabriel Lamé (right). Lamé's second parameter is equivalent to shear modulus, μ.

It is often said that λ has no physical interpretation, and most people find it hard to visualize.

In terms of V_P and V_S

$\lambda =\rho (V_{\mathrm {P} }^{2}-2V_{\mathrm {S} }^{2})$

Other expressions

λ can also be expressed in terms of Young's modulus, E, and Poisson's ratio, ν. This could be thought of as 'the engineer's perspective':

$\lambda ={\frac {E\nu }{(1+\nu )(1-2\nu )}}$

The 'fluid substitution perspective' casts λ in terms of bulk modulus and shear modulus μ:

$\lambda =K-{\frac {2}{3}}\mu$

Typical values

Rock	λ, GPa
Quartz	8
Feldspar	28
Calcite	56
Dolomite	65
Anhydrite	26
Siderite	90
Pyrite	59
Sandstone, 10 pu	1–3
Limestone, 10 pu	18–53
Shale, 5 pu	3–24
Brine	2.3
Oil, 40°API	1.6

Analysis and interpretation

Goodway and others have described an alternative to (or augmentation of) classic impedance inversion and interpretation. The parameters are closely related:

$\lambda \rho =I_{\mathrm {P} }^{2}-2I_{\mathrm {S} }^{2}$

$\mu \rho =I_{\mathrm {S} }^{2}$

This approach—estimating Lamé's parameters indirectly via impedance—is problematic so Gray recommended estimating λ and μ contrasts directly from seismic data. See discussion in Avseth et al.

How can we understand lambda?

Click SHOW to open > > >

Extend a rod or a linear spring. Its extension (strain) is linearly proportional to its tensile stress σ, by a constant factor, the inverse of its Young's modulus E, hence,

$\varepsilon ={\frac {E}{\sigma }}$.

We can extend this to three dimensions, but then we need Poisson's ratio ν, which accounts for the change in shape in the cross-sectional plane.

$\varepsilon _{1}'={\frac {1}{E}}\sigma _{1},\ \ \ \ \varepsilon _{2}'=-{\frac {\nu }{E}}\sigma _{1},\ \ \ \ \varepsilon _{3}'=-{\frac {\nu }{E}}\sigma _{1}$,

We get similar equations to the loads in directions 2 and 3,

$\varepsilon _{1}''=-{\frac {\nu }{E}}\sigma _{2},\ \ \ \ \varepsilon _{2}''={\frac {1}{E}}\sigma _{2},\ \ \ \ \varepsilon _{3}''=-{\frac {\nu }{E}}\sigma _{2}$,

and

$\varepsilon _{1}'''=-{\frac {\nu }{E}}\sigma _{3},\ \ \ \ \varepsilon _{2}'''=-{\frac {\nu }{E}}\sigma _{3},\ \ \ \ \varepsilon _{3}'''={\frac {1}{E}}\sigma _{3}$.

Summing the three cases together ($\varepsilon _{i}=\varepsilon _{i}'+\varepsilon _{i}''+\varepsilon _{i}'''$) we get

$\varepsilon _{1}={\frac {1}{E}}(\sigma _{1}-\nu (\sigma _{2}+\sigma _{3}))$

$\varepsilon _{2}={\frac {1}{E}}(\sigma _{2}-\nu (\sigma _{1}+\sigma _{3}))$

$\varepsilon _{3}={\frac {1}{E}}(\sigma _{3}-\nu (\sigma _{1}+\sigma _{2}))$

or by adding and subtracting one $\nu \sigma$

$\varepsilon _{n}={\frac {1}{E}}((1+\nu )\sigma _{n}-\nu (\sigma _{1}+\sigma _{2}+\sigma _{3}))$

and further we get by solving $\sigma _{1}$

$\sigma _{1}={\frac {E}{1+\nu }}\varepsilon _{1}+{\frac {\nu }{1+\nu }}(\sigma _{1}+\sigma _{2}+\sigma _{3})$.

Calculating the sum

$\sum _{i=1,2,3}\varepsilon _{i}={\frac {1}{E}}((1+\nu )\sum _{i=1,2,3}\sigma _{i}-3\nu (\sum _{i=1,2,3}\sigma _{i}))={\frac {1-2\nu }{E}}\sum _{i=1,2,3}\sigma _{i}$

$\sigma _{1}+\sigma _{2}+\sigma _{3}={\frac {E}{1-2\nu }}(\varepsilon _{1}+\varepsilon _{2}+\varepsilon _{3})$

and substituting it to the equation solved for $\sigma _{1}$ gives

$\sigma _{1}={\frac {E}{1+\nu }}\varepsilon _{1}+{\frac {E\nu }{(1+\nu )(1-2\nu )}}(\varepsilon _{1}+\varepsilon _{2}+\varepsilon _{3})$,

which simplifies if we substitute $\mu$ and $\lambda$, the Lamé parameters.

$\sigma _{1}=2\mu \varepsilon _{1}+\lambda (\varepsilon _{1}+\varepsilon _{2}+\varepsilon _{3})\$.

Similar treatment of directions 2 and 3 gives the Hooke's law in three dimensions.

${\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{31}\\\sigma _{12}\end{bmatrix}}={\cfrac {E}{(1+\nu )(1-2\nu )}}{\begin{bmatrix}1-\nu &\nu &\nu &0&0&0\\\nu &1-\nu &\nu &0&0&0\\\nu &\nu &1-\nu &0&0&0\\0&0&0&(1-2\nu )/2&0&0\\0&0&0&0&(1-2\nu )/2&0\\0&0&0&0&0&(1-2\nu )/2\end{bmatrix}}{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\2\varepsilon _{23}\\2\varepsilon _{31}\\2\varepsilon _{12}\end{bmatrix}}$

which expression can be simplified thanks to the Lamé constants :

${\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\\\sigma _{23}\\\sigma _{31}\\\sigma _{12}\end{bmatrix}}={\begin{bmatrix}2\mu +\lambda &\lambda &\lambda &0&0&0\\\lambda &2\mu +\lambda &\lambda &0&0&0\\\lambda &\lambda &2\mu +\lambda &0&0&0\\0&0&0&\mu &0&0\\0&0&0&0&\mu &0\\0&0&0&0&0&\mu \end{bmatrix}}{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\\2\varepsilon _{23}\\2\varepsilon _{31}\\2\varepsilon _{12}\end{bmatrix}}$

Can we simplify further by just considering the principal axes?

${\begin{bmatrix}\sigma _{11}\\\sigma _{22}\\\sigma _{33}\end{bmatrix}}={\begin{bmatrix}2\mu +\lambda &\lambda &\lambda \\\lambda &2\mu +\lambda &\lambda \\\lambda &\lambda &2\mu +\lambda \\\end{bmatrix}}{\begin{bmatrix}\varepsilon _{11}\\\varepsilon _{22}\\\varepsilon _{33}\end{bmatrix}}$

External links

• First Lamé parameter — Wikipedia entry

Conversion formulas — edit
The elastic properties of homogeneous isotropic linear elastic materials are uniquely determined by any two moduli. Given any two, the others can thus be calculated. Key reference: Mavko, G, T Mukerji and J Dvorkin (2003), The Rock Physics Handbook, Cambridge University Press. $\dagger \ X={\sqrt {9\lambda ^{2}+2E\lambda +E^{2}}}$
	$(V_{\mathrm {P} },\,V_{\mathrm {S} })$	$(\mu ,\,\lambda )$	$(E,\,\lambda )\,\dagger$	$(E,\,\mu )$	$(K,\,\lambda )$	$(K,\,\mu )$	$(K,\,E)$	$(\nu ,\,\lambda )$	$(\nu ,\,\mu )$	$(\nu ,\,E)$	$(\nu ,\,K)$
P-wave velocity $V_{\mathrm {P} }=\,$	$V_{\mathrm {P} }$	${\sqrt {\tfrac {\lambda +2\mu }{\rho }}}$	${\sqrt {\tfrac {E-\lambda +X}{2\rho }}}$	${\sqrt {\tfrac {\mu \,(E-4\mu )}{\rho \,(E-3\mu )}}}$	${\sqrt {\tfrac {3K-2\lambda }{\rho }}}$	${\sqrt {\tfrac {K+{\frac {4}{3}}\mu }{\rho }}}$	${\sqrt {\tfrac {3K\left(3K+E\right)}{\rho \left(9K-E\right)}}}$	${\sqrt {\tfrac {\lambda (1-\nu )}{\rho \nu }}}$	${\sqrt {\tfrac {2\mu (1-\nu )}{\rho (1-2\nu )}}}$	${\sqrt {\tfrac {E(1-\nu )}{\rho (1+\nu )(1-2\nu )}}}$	${\sqrt {\tfrac {3K(1-\nu )}{\rho (1+\nu )}}}$
S-wave velocity $V_{\mathrm {S} }=\,$	$V_{\mathrm {S} }$	${\sqrt {\tfrac {\mu }{\rho }}}$	${\sqrt {\tfrac {E-3\lambda +X}{4\rho }}}$	${\sqrt {\tfrac {\mu }{\rho }}}$	${\sqrt {\tfrac {3(K-\lambda )}{2\rho }}}$	${\sqrt {\tfrac {\mu }{\rho }}}$	${\sqrt {-{\tfrac {3EK}{\rho \left(E-9K\right)}}}}$	${\sqrt {{\tfrac {\lambda }{2\nu \rho }}-{\tfrac {\lambda }{\rho }}}}$	${\sqrt {\tfrac {\mu }{\rho }}}$	${\sqrt {\tfrac {E}{2\rho (1+\nu )}}}$	${\sqrt {-{\tfrac {3K(2\nu -1)}{2\rho (\nu +1)}}}}$
Velocity ratio $\Gamma =\,$	${\frac {V_{\mathrm {P} }}{V_{\mathrm {S} }}}$	${\sqrt {\tfrac {\lambda +2\mu }{\mu }}}$	${\sqrt {\tfrac {3E+3\lambda +X}{2E}}}$	${\sqrt {\tfrac {E-4\mu }{E-3\mu }}}$	${\sqrt {\tfrac {{\tfrac {4}{3}}\lambda -2K}{\lambda -K}}}$	${\sqrt {\tfrac {K+{\frac {4}{3}}\mu }{\mu }}}$	${\sqrt {\tfrac {E+3K}{E}}}$	${\sqrt {\tfrac {2\nu -2}{2\nu -1}}}$	${\sqrt {\tfrac {2\nu -2}{2\nu -1}}}$	${\sqrt {\tfrac {2\nu -2}{2\nu -1}}}$	${\sqrt {\tfrac {2\nu -2}{2\nu -1}}}$
1st Lamé parameter $\lambda =\,$	$\rho (V_{\mathrm {P} }^{2}-2V_{\mathrm {S} }^{2})$	$\lambda$	$\lambda$	${\tfrac {\mu (E-2\mu )}{3\mu -E}}$	$\lambda$	$K-{\tfrac {2\mu }{3}}$	${\tfrac {3K(3K-E)}{9K-E}}$	$\lambda$	${\tfrac {2\mu \nu }{1-2\nu }}$	${\tfrac {E\nu }{(1+\nu )(1-2\nu )}}$	${\tfrac {3K\nu }{1+\nu }}$
Shear modulus $\mu =\,$	$\rho V_{\mathrm {S} }^{2}$	$\mu$	${\tfrac {E-3\lambda +X}{4}}$	$\mu$	${\tfrac {3(K-\lambda )}{2}}$	$\mu$	${\tfrac {3KE}{9K-E}}$	${\tfrac {\lambda (1-2\nu )}{2\nu }}$	$\mu$	${\tfrac {E}{2(1+\nu )}}$	${\tfrac {3K(1-2\nu )}{2(1+\nu )}}$
Young's modulus $E=\,$	${\tfrac {\rho V_{\mathrm {S} }^{2}(3V_{\mathrm {P} }^{2}-4V_{\mathrm {S} }^{2})}{V_{\mathrm {P} }^{2}-V_{\mathrm {S} }^{2}}}$	${\tfrac {\mu (3\lambda +2\mu )}{\lambda +\mu }}$	$E\$	$E$	${\tfrac {9K(K-\lambda )}{3K-\lambda }}$	${\tfrac {9K\mu }{3K+\mu }}$	$E\$	${\tfrac {\lambda (1+\nu )(1-2\nu )}{\nu }}$	$2\mu (1+\nu )\,$	$E\$	$3K(1-2\nu )\,$
Bulk modulus $K=\,$	$\rho (V_{\mathrm {P} }^{2}-{\tfrac {4}{3}}V_{\mathrm {S} }^{2})$	$\lambda +{\tfrac {2\mu }{3}}$	${\tfrac {E+3\lambda +X}{6}}$	${\tfrac {E\mu }{3(3\mu -E)}}$	$K$	$K$	$K$	${\tfrac {\lambda (1+\nu )}{3\nu }}$	${\tfrac {2\mu (1+\nu )}{3(1-2\nu )}}$	${\tfrac {E}{3(1-2\nu )}}$	$K$
Poisson's ratio $\nu =\,$	${\tfrac {V_{\mathrm {P} }^{2}-2V_{\mathrm {S} }^{2}}{2(V_{\mathrm {P} }^{2}-V_{\mathrm {S} }^{2})}}$	${\tfrac {\lambda }{2(\lambda +\mu )}}$	${\tfrac {-E-\lambda +X}{4\lambda }}$	${\tfrac {E}{2\mu }}-1$	${\tfrac {\lambda }{3K-\lambda }}$	${\tfrac {3K-2\mu }{2(3K+\mu )}}$	${\tfrac {3K-E}{6K}}$	$\nu$	$\nu$	$\nu$	$\nu$
P-wave modulus $M=\,$	$\rho V_{\mathrm {P} }^{2}$	$\lambda +2\mu \,$	${\tfrac {E-\lambda +X}{2}}$	${\tfrac {\mu (4\mu -E)}{3\mu -E}}$	$3K-2\lambda \,$	$K+{\tfrac {4\mu }{3}}$	${\tfrac {3K(3K+E)}{9K-E}}$	${\tfrac {\lambda (1-\nu )}{\nu }}$	${\tfrac {2\mu (1-\nu )}{1-2\nu }}$	${\tfrac {E(1-\nu )}{(1+\nu )(1-2\nu )}}$	${\tfrac {3K(1-\nu )}{1+\nu }}$