From Moduli to Velocities

Part 1, Part 1: Elasticity and the Moduli

Learning objectives

Write the two velocity equations and identify which moduli each one feels
Use the unit convention in which moduli in GPa and density in g/cc give velocity directly in km/s
Explain why a pore-fluid change moves Vp and Vs differently, the germ of fluid discrimination
Trace how density in the denominator lets a stiffening fluid and a heavier rock fight over Vp

The Two Equations

With the two moduli in hand, the velocities are immediate. A P-wave both compresses and shears the rock as it passes, so it feels both moduli: $V_P = \sqrt{\dfrac{K + \tfrac{4}{3}\mu}{\rho}}$ P=sqrtdfracK+tfrac43murho. An S-wave only changes shape, so it feels the shear modulus alone: $V_S = \sqrt{\dfrac{\mu}{\rho}}$ S=sqrtdfracmurho. Both carry the density $\rho$ in the denominator, because inertia resists the motion regardless of which stiffness is driving it. These two equations are the hinge of the whole course: composition and texture set $K$ , $\mu$ , and $\rho$ on the right, and the seismic velocities fall out on the left.

A Convenient Set of Units

The equations look cleanest in a unit system that geophysics adopted for exactly this reason. Put the moduli in gigapascals and the density in grams per cubic centimeter, and the square root comes out directly in kilometers per second, with no conversion factor to remember, because $\sqrt{\text{GPa}/(\text{g/cc})}$ is exactly km/s. Take mineral quartz as the standard check: $K = 36.6$ GPa, $\mu = 45$ GPa, $\rho = 2.65$ g/cc. Then $V_P = \sqrt{(36.6 + \tfrac{4}{3}\cdot 45)/2.65} = 6.04$ P=sqrt(36.6+tfrac43cdot45)/2.65=6.04 km/s and $V_S = \sqrt{45/2.65} = 4.12$ S=sqrt45/2.65=4.12 km/s. Those two numbers reappear in every part of this course; if a frame or fluid model ever predicts a clean quartz sand faster than mineral quartz itself, the model is wrong.

Why the Fluid Shows Up Twice

Now let a fluid enter the pores and watch the two velocities part ways. The fluid raises the bulk modulus $K$ , because it resists the squeeze, and it raises the density $\rho$ , because it has mass, but it leaves $\mu$ exactly where it was. So $V_S = \sqrt{\mu/\rho}$ S=sqrtmu/rho can only fall: its numerator is fixed while its denominator grows, and a saturated rock always has a slightly lower shear velocity than the same rock dry. $V_P$ P is the subtle one, because $K$ and $\rho$ pull it in opposite directions: a stiffer $K$ speeds the P-wave up, a heavier $\rho$ slows it down, and which wins depends on the fluid. That is why the choice of fluid, not merely its presence, changes the seismic answer.

The two headline cases follow directly. Swapping brine for gas softens the saturated $K$ far more than it lightens $\rho$ , so the drop in $K$ wins and $V_P$ P falls sharply, the reason gas often lights up, while $V_S$ S actually edges up a little as the density drops. Swapping brine for oil nudges $K$ and $\rho$ by similar small amounts whose effects on $V_P$ P nearly cancel, so $V_P$ P barely moves and oil is close to invisible on velocity alone. That the same fluid reaches $V_P$ P through two competing channels and $V_S$ S through only one is why no single velocity tells the whole story. The cleanest way to read the pair is their ratio, which cancels the density and exposes the balance of the two moduli. That ratio, and the Poisson's ratio it is equivalent to, is the next section.

References

Mavko, G., Mukerji, T., & Dvorkin, J. (2009). The Rock Physics Handbook (2nd ed.). Cambridge University Press.
Simm, R., & Bacon, M. (2014). Seismic Amplitude: An Interpreter's Handbook. Cambridge University Press.