Poisson's Ratio and Vp/Vs

Part 1, Part 1: Elasticity and the Moduli

Learning objectives

Form the ratio Vp over Vs and show that it cancels density and isolates the K over mu balance
Convert between Vp/Vs and Poisson's ratio, and read the fluid and lithology anchors of the scale
Place dry sand, brine sand, carbonate, shale, and a fluid on the Vp/Vs scale
Explain why gas lowers Vp/Vs, which makes the ratio a fluid indicator

The Ratio That Forgets the Density

Take the ratio of the two velocities and something clean happens: the density cancels. $\dfrac{V_P}{V_S} = \sqrt{\dfrac{K + \tfrac{4}{3}\mu}{\mu}} = \sqrt{\dfrac{K}{\mu} + \tfrac{4}{3}}$ sqrtdfracK+tfrac43mumu=sqrtdfracKmu+tfrac43. The ratio depends only on the balance between the two moduli, $K/\mu$ , and on nothing else. Where an individual velocity confounds stiffness with density, the ratio strips the density away and reports the one quantity a fluid or a lithology change acts on most directly: how bulk-stiff the rock is relative to how shear-stiff it is. That is why $V_P/V_S$ , and the Poisson's ratio equivalent to it, is the workhorse discriminator of quantitative interpretation.

Two Names for One Number

Poisson's ratio is the same information in a different dress: $\nu = \dfrac{V_P^2 - 2V_S^2}{2(V_P^2 - V_S^2)}$ . The two extremes anchor the scale. A fluid, with $\mu = 0$ , has $V_S = 0$ S=0, so $V_P/V_S$ is infinite and $\nu = 0.5$ : that is the ceiling, the most a Poisson's ratio can ever be. At the other end, a dry quartz sand is nearly all shear stiffness with little bulk stiffness once the fluid is gone, so it reads a strikingly low Poisson's ratio, roughly 0.05 to 0.15, with $V_P/V_S$ near 1.5. Real rocks live between these posts, and where they sit is diagnostic.

Reading the Bands

Slide the point along the curve and the rock types sort themselves out. Water-saturated soft sediments sit high, Poisson's ratio around 0.40 to 0.45 and $V_P/V_S$ of 2.5 or more, because a weak frame lets the pore water dominate the bulk stiffness. A consolidated brine sandstone reads far lower, $V_P/V_S$ around 1.6 to 1.8. Carbonates cluster near 1.9, shales spread broadly from about 1.7 to 2.3, and the values overlap enough that the ratio alone rarely names a lithology with certainty. What the ratio does do decisively is respond to gas. Because gas softens $K$ while leaving $\mu$ almost untouched, $V_P$ P drops while $V_S$ S barely moves, so the ratio $V_P/V_S$ falls: a low $V_P/V_S$ against a brine-sand background is the classic gas indicator, and it is why crossplots of the ratio against impedance sit at the center of direct hydrocarbon detection.

One caution travels with this power: the empirical $V_P$ P to $V_S$ S relations that let you predict one velocity from the other, the mudrock line and its lithology-specific cousins, are calibrations, not laws, and Part 7 builds them with their error bars. For now the lesson is that the ratio isolates the $K/\mu$ balance, and that balance is set, before any pore or fluid ever enters, by what the rock is made of. So Part 1 closes where every model begins: the mineral end-members whose moduli anchor the whole subject. That is the next and final section of this part.

References

Mavko, G., Mukerji, T., & Dvorkin, J. (2009). The Rock Physics Handbook (2nd ed.). Cambridge University Press.
Castagna, J. P., Batzle, M. L., & Eastwood, R. L. (1985). Relationships between compressional-wave and shear-wave velocities in clastic silicate rocks. Geophysics, 50(4), 571-581.