# P-wave modulus

In linear elasticity, the P-wave modulus $M$, also known as the longitudinal modulus or the constrained modulus, is one of the elastic moduli available to describe isotropic homogeneous materials.

It is defined as the ratio of axial stress to axial strain in a uniaxial strain state

$\sigma_{zz} = M \epsilon_{zz}$

where all the other strains $\epsilon_{**}$ are zero.

This is equivalent to stating that

$M = \rho V_\mathrm{P}^2$

where VP is the velocity of a P-wave.

## References

• G. Mavko, T. Mukerji, J. Dvorkin. The Rock Physics Handbook. Cambridge University Press 2003 (paperback). ISBN 0-521-54344-4
Conversion formulas
Homogeneous isotropic linear elastic materials have their elastic properties uniquely determined by any two moduli among these; thus, given any two, any other of the elastic moduli can be calculated according to these formulas.
$K=\,$ $E=\,$ $\lambda=\,$ $G=\,$ $\nu=\,$ $M=\,$ Notes
$(K,\,E)$ $K$ $E$ $\tfrac{3K(3K-E)}{9K-E}$ $\tfrac{3KE}{9K-E}$ $\tfrac{3K-E}{6K}$ $\tfrac{3K(3K+E)}{9K-E}$
$(K,\,\lambda)$ $K$ $\tfrac{9K(K-\lambda)}{3K-\lambda}$ $\lambda$ $\tfrac{3(K-\lambda)}{2}$ $\tfrac{\lambda}{3K-\lambda}$ $3K-2\lambda\,$
$(K,\,G)$ $K$ $\tfrac{9KG}{3K+G}$ $K-\tfrac{2G}{3}$ $G$ $\tfrac{3K-2G}{2(3K+G)}$ $K+\tfrac{4G}{3}$
$(K,\,\nu)$ $K$ $3K(1-2\nu)\,$ $\tfrac{3K\nu}{1+\nu}$ $\tfrac{3K(1-2\nu)}{2(1+\nu)}$ $\nu$ $\tfrac{3K(1-\nu)}{1+\nu}$
$(K,\,M)$ $K$ $\tfrac{9K(M-K)}{3K+M}$ $\tfrac{3K-M}{2}$ $\tfrac{3(M-K)}{4}$ $\tfrac{3K-M}{3K+M}$ $M$
$(E,\,\lambda)$ $\tfrac{E + 3\lambda + R}{6}$ $E$ $\lambda$ $\tfrac{E-3\lambda+R}{4}$ $\tfrac{2\lambda}{E+\lambda+R}$ $\tfrac{E-\lambda+R}{2}$ $R=\sqrt{E^2+9\lambda^2 + 2E\lambda}$
$(E,\,G)$ $\tfrac{EG}{3(3G-E)}$ $E$ $\tfrac{G(E-2G)}{3G-E}$ $G$ $\tfrac{E}{2G}-1$ $\tfrac{G(4G-E)}{3G-E}$
$(E,\,\nu)$ $\tfrac{E}{3(1-2\nu)}$ $E$ $\tfrac{E\nu}{(1+\nu)(1-2\nu)}$ $\tfrac{E}{2(1+\nu)}$ $\nu$ $\tfrac{E(1-\nu)}{(1+\nu)(1-2\nu)}$
$(E,\,M)$ $\tfrac{3M-E+S}{6}$ $E$ $\tfrac{M-E+S}{4}$ $\tfrac{3M+E-S}{8}$ $\tfrac{E-M+S}{4M}$ $M$

$S=\pm\sqrt{E^2+9M^2-10EM}$

There are two valid solutions.
The plus sign leads to $\nu\geq 0$.
The minus sign leads to $\nu\leq 0$.

$(\lambda,\,G)$ $\lambda+ \tfrac{2G}{3}$ $\tfrac{G(3\lambda + 2G)}{\lambda + G}$ $\lambda$ $G$ $\tfrac{\lambda}{2(\lambda + G)}$ $\lambda+2G\,$
$(\lambda,\,\nu)$ $\tfrac{\lambda(1+\nu)}{3\nu}$ $\tfrac{\lambda(1+\nu)(1-2\nu)}{\nu}$ $\lambda$ $\tfrac{\lambda(1-2\nu)}{2\nu}$ $\nu$ $\tfrac{\lambda(1-\nu)}{\nu}$ Cannot be used when $\nu=0 \Leftrightarrow \lambda=0$
$(\lambda,\,M)$ $\tfrac{M + 2\lambda}{3}$ $\tfrac{(M-\lambda)(M+2\lambda)}{M+\lambda}$ $\lambda$ $\tfrac{M-\lambda}{2}$ $\tfrac{\lambda}{M+\lambda}$ $M$
$(G,\,\nu)$ $\tfrac{2G(1+\nu)}{3(1-2\nu)}$ $2G(1+\nu)\,$ $\tfrac{2 G \nu}{1-2\nu}$ $G$ $\nu$ $\tfrac{2G(1-\nu)}{1-2\nu}$
$(G,\,M)$ $M - \tfrac{4G}{3}$ $\tfrac{G(3M-4G)}{M-G}$ $M - 2G\,$ $G$ $\tfrac{M - 2G}{2M - 2G}$ $M$
$(\nu,\,M)$ $\tfrac{M(1+\nu)}{3(1-\nu)}$ $\tfrac{M(1+\nu)(1-2\nu)}{1-\nu}$ $\tfrac{M \nu}{1-\nu}$ $\tfrac{M(1-2\nu)}{2(1-\nu)}$ $\nu$ $M$