The Moduli That Matter

Part 1, Part 1: Elasticity and the Moduli

Learning objectives

Define the bulk modulus K as resistance to volume change under hydrostatic pressure, and the shear modulus mu as resistance to change of shape
Show that Young's modulus, the Lame parameter, and Poisson's ratio are combinations of K and mu, carrying no new information
Explain why a fluid has a shear modulus of exactly zero, and what that forces on S-waves
Squeeze and shear a solid and a fluid, and watch which modulus each deformation tests

Two Ways to Deform a Rock

There are only two independent ways to deform a small piece of an isotropic rock, and each has its own stiffness. Squeeze it from all sides with a hydrostatic pressure and it changes volume; the resistance is the bulk modulus $K$ , defined by $\dfrac{\Delta V}{V} = -\dfrac{P}{K}$ , so a stiffer rock loses less volume under the same pressure. Deform it so that its shape changes while its volume does not, a pure shear, and the resistance is the shear modulus $\mu$ . These two numbers are the entire elastic description of an isotropic material. Everything a P-wave or an S-wave does in that rock, it does through $K$ , $\mu$ , and the density.

Everything Else Is a Combination

The literature is full of other elastic constants, and they can be intimidating until you see that they carry no new information. Young's modulus $E = \dfrac{9K\mu}{3K + \mu}$ is the stiffness you would measure by stretching a rod. Poisson's ratio $\nu = \dfrac{3K - 2\mu}{2(3K + \mu)}$ is how much that rod bulges sideways as you stretch it. The Lame parameter $\lambda = K - \tfrac{2}{3}\mu$ is the combination that appears in the wave equation. Every one of them is a function of $K$ and $\mu$ alone: give me those two and the rest follow, so an isotropic rock has exactly two elastic degrees of freedom, not five. This course keeps $K$ and $\mu$ as the currency because they split cleanly along the one distinction that matters most for fluids.

The Fact That Runs the Whole Subject

Set the demonstration to a fluid and watch the shear panel go slack. A fluid has a shear modulus of exactly zero: it cannot store any energy in a change of shape, because its molecules simply flow to relieve the stress. This is not an approximation, it is what makes a fluid a fluid. Two consequences run through everything ahead. First, a shear wave deforms the rock purely in shape, so it has nothing to propagate through in a fluid, which is why S-waves die the instant they reach one and why $V_S$ S probes only the solid frame. Second, when you change the fluid in a rock's pores you change how hard the rock is to squeeze, its bulk modulus, but you cannot change its resistance to shear: the pore fluid contributes nothing to $\mu$ .

That second consequence is the quiet engine of fluid detection, and it is the heart of Gassmann's theory in Part 4: the fluid moves $K$ and leaves $\mu$ untouched. Before we can use it, we have to turn these two moduli into the velocities a seismic wave actually carries. That is the next section: the two equations that convert $K$ , $\mu$ , and density into $V_P$ P and $V_S$ S.

References

Mavko, G., Mukerji, T., & Dvorkin, J. (2009). The Rock Physics Handbook (2nd ed.). Cambridge University Press.
Gassmann, F. (1951). Über die Elastizität poröser Medien. Vierteljahrsschrift der Naturforschenden Gesellschaft in Zürich, 96, 1-23.