From acoustic impedance to the elastic rock

Recap: acoustic impedance and the reflection coefficient

The acoustic impedance of a rock is the product of its bulk density and its P-wave velocity:

$Z_P = \rho\, V_P$

With $V_P$ in km/s and $\rho$ in g/cm³, $Z_P$ comes out in $10^6\,\mathrm{kg\,m^{-2}\,s^{-1}}$: a shale with $V_P$ = 2.92 km/s and $\rho$ = 2.325 g/cm³ has $Z_P$ = 6.79.

At a boundary between two rocks, the share of the incident amplitude that reflects at normal incidence (zero offset) is the reflection coefficient:

$R_0 = \dfrac{Z_2 - Z_1}{Z_2 + Z_1}$

where $Z_1$ is the impedance of the layer above and $Z_2$ that of the layer below. $R_0$ lies between −1 and +1, and between sedimentary rocks it rarely exceeds 0.3 in size. A positive $R_0$ is a peak on SEG normal polarity, a negative one a trough.

The formula depends only on the ratio of the two impedances: a 10% step from 5 to 5.5 reflects exactly as much as a 10% step from 10 to 11. It also depends on nothing else. Two rocks with the same $Z_P$ give no reflection at normal incidence however different they are, and that is the problem this section is about.

The missing piece: shear waves

Acoustic impedance describes how a rock resists compression. A rock also resists shearing, a change of shape at constant volume, with a separate stiffness, the shear modulus $\mu$, which sets the speed of S-waves:

$V_S = \sqrt{\mu / \rho}$

Two properties of S-waves matter here:

• S-waves are slower than P-waves in every rock; in sediments $V_S$ is typically $V_P/1.5$ to $V_P/3$.
• S-waves do not travel through fluids: water, oil and gas have $\mu = 0$. Replacing one pore fluid with another therefore leaves the rock's shear modulus unchanged, while its bulk modulus can change a great deal. Section 5.3 (Gassmann substitution) builds on this.

The shear impedance is the analogue of acoustic impedance:

$Z_S = \rho\, V_S$

The ratio $V_P/V_S$ is one of the most useful single numbers in quantitative interpretation. Along the trends in Figure 5.1, shale runs from about 3.0 when young and soft to 1.8 when hard, brine sand from 2.2 to 1.6, and gas sand stays near 1.55 to 1.6, because gas softens a rock in compression but not in shear. Section 5.2 shows why.

The full elastic description

An isotropic elastic rock is fully described by three independent numbers:

• P-wave velocity $V_P$ (km/s)
• S-wave velocity $V_S$ (km/s)
• bulk density $\rho$ (g/cm³)

Everything else, the acoustic and shear impedances, $V_P/V_S$, the bulk and shear moduli, the Lamé parameters, Young's modulus and Poisson's ratio, follows from those three. A rock-physics model that gives $(V_P, V_S, \rho)$ for a porosity, mineralogy, pore fluid and saturation therefore predicts every seismic quantity, and Part 5 builds that workflow.

Figure 5.1 puts one rock over another. Eight kinds of rock follow published compaction trends, and a slider sets how far along its trend each rock is, from soft and shallow to hard and deep. The slider is a position along each kind's range of $V_P$, not a depth, so the same setting on two kinds of rock need not mean the same burial. Plate (a) shows acoustic impedance alone and plate (b) adds $V_P/V_S$; plate (c) gives the boundary's reflection against incidence angle from the exact Zoeppritz equations, and plate (d) shows the same boundary as synthetic seismic traces. Each time the figure also finds the lower rock's impedance twin: another kind of rock, solved to exactly the same $Z_P$, whose $V_P/V_S$ differs most.

Exercise: see the rock physics

• The figure starts with shale ($V_P$ 2.92 km/s, $Z_P$ 6.79) over a gas sand ($V_P$ 2.63 km/s, $Z_P$ 5.16). The impedance falls, so $R_0$ = −0.136, a trough, and at 30° it strengthens to −0.181. This is the recipe of the classic gas bright spot.
• Switch the lower rock to brine sand at the same compaction. The frame is the same; only the pore fluid has changed. $Z_P$ climbs to 6.63, close to the shale's, and $R_0$ shrinks to −0.012: against this shale a brine sand is nearly invisible at normal incidence.
• Switch to oil sand. Oil is stiffer and denser than gas, so the oil sand falls between the two: $Z_P$ 5.83 and $R_0$ = −0.076.
• Run the figure's third exercise, which compacts the brine sand until its impedance equals the shale's ($V_P$ 3.06 km/s, $\rho$ 2.22 g/cm³). $R_0$ is zero, yet at 30° the boundary reflects −0.028, because $V_P/V_S$ drops from 2.17 to 1.91. Compare the two gathers in (d).
• Put a limestone under the shale. Halfway along its trend its impedance is 11.79, far above the shale's, so its top is a strong peak, $R_0$ = +0.269; and because the limestone is so much faster, a critical angle falls inside the gather, at 37.9°. Past it the transmitted P wave no longer leaves the boundary and the coefficient turns complex: its size jumps to about 0.9, but the reflection is not total, because S waves still carry energy away.
• Put rock salt under a soft shale, then raise the shale's compaction. Salt hardly compacts, so its impedance stays near 9.7: under a young shale ($Z_P$ 5.49) its top is a strong peak, $R_0$ = +0.278, but under the hardest shale in the figure ($Z_P$ 10.11) it is a weak trough, −0.020. The sign of a reflection belongs to the contrast, not to either rock.

Why one number is not enough

If the reflection coefficient is just $(Z_2 - Z_1)/(Z_2 + Z_1)$, why keep $V_S$ and $\rho$ separately?

At normal incidence only $Z_P$ matters, and plate (a) of Figure 5.1 shows why that is a problem. The impedance spans of shale (4.74 to 10.11), brine sand (5.27 to 11.06), oil sand (4.26 to 10.73) and gas sand (3.47 to 10.39) lie on top of one another; limestone joins them above 7.7 and rock salt near 9.7. Every impedance from 5.27 to 10.11 belongs to at least four kinds of rock, so $Z_P$ alone cannot say whether a layer is shale or sand, or whether a sand holds brine or gas.

Plate (b) separates them. At the gas sand's impedance of 5.16 its twin is a soft shale with $V_P/V_S$ = 2.75, against the gas sand's 1.59. Under the same shale the two boundaries reflect exactly the same $R_0$ = −0.136, because their impedances are equal, but at 30° the gas sand reflects −0.181 and the soft shale −0.084: one brightens with angle and the other dims. Surveys record many angles, and the way a reflection changes with angle (the AVO response, Section 5.4) depends on $V_S$ and $\rho$ separately. That difference is often what tells a hydrocarbon bright spot from a change of lithology.

Where the values come from

Each kind of rock in Figure 5.1 is a published trend, not a single value:

• Shale: $V_P$ from 2.2 to 4.0 km/s as it compacts, $V_S$ from the mudrock line of Castagna, Batzle and Eastwood (1985), $V_P = 1.16\,V_S + 1.36$ km/s, and density from $\rho = 1.75\,V_P^{0.265}$ (Castagna, Batzle and Kan 1993), so $\rho$ rises from 2.16 to 2.53 g/cm³.
• Sandstone: brine-saturated $V_P$ from 2.5 to 4.5 km/s, $V_S$ from Greenberg and Castagna (1992) and density from Castagna, Batzle and Kan (1993); the porosity follows from the density and falls from 33% to 12%.
• Pore fluid: the oil and gas sands are the brine sand's own frame with a different fluid, computed with Gassmann's equation (Section 5.3). In the starting sand, gas lowers $V_P$ from 3.00 to 2.63 km/s and the density from 2.21 to 1.96 g/cm³, while $V_S$ rises slightly, from 1.56 to 1.65 km/s, because the shear modulus is unchanged and the rock is lighter. The stiffer the frame, the smaller the change: at the hard end of the trend gas lowers $V_P$ by only 2%.
• Carbonates and evaporites: limestone and dolomite follow the same two sets of relations; rock salt and anhydrite take the moduli of their minerals (Mavko, Mukerji and Dvorkin 2009).
• Pressure: rising effective stress closes cracks and stiffens a rock, so $V_P$ and $V_S$ both rise; the compaction trends fold this in with burial.

These trends are fits to many rocks, and a real rock may sit well off them. Quantitative interpretation calibrates every relation against well logs from the basin and reservoir under study before it trusts a prediction, which is why it begins with a well tie.