Why a solid rock carries two kinds of seismic wave

Rock pushes back when you squeeze it and also when you shear it sideways. Each kind of stiffness carries its own wave. The squeezing wave, P, always travels faster than the shearing wave, S. Measure both speeds and you can work out how stiff the rock is in each way. Water has no shear stiffness at all, so it carries only P.

Sandbox Physics · 2026-10-30

Two ways to deform a block of rock

Take a small cube of rock and push on it. It can change volume, like a sponge being squeezed, or it can change shape at fixed volume, like a deck of cards sliding sideways. Hooke’s law for an isotropic solid says the stress is linear in the strain and needs only two constants for this, called the Lamé constants.

The second constant is the shear modulus or rigidity. It measures resistance to shear and is zero in water and in the liquid iron of the outer core. The first constant has no simple meaning on its own, but together with the rigidity it gives the bulk modulus, the resistance to a change of volume: the first constant plus two thirds of the rigidity. The MIT notes for 12.201 set this out in Section 4.5.

\sigma_{ij}=\lambda\,(\nabla\!\cdot\mathbf u)\,\delta_{ij}+\mu\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right),\qquad \kappa=\lambda+\tfrac23\mu

Split the motion into squeezing and shearing

Put Hooke’s law into Newton’s second law for each small cube. For a uniform rock the result is one vector equation for the displacement. It mixes two effects, but it can be split in two.

Take the divergence of both sides. The divergence of the displacement is the fractional change of volume, and it obeys an ordinary wave equation whose speed squared is the first Lamé constant plus twice the rigidity, divided by the density. This is the P wave, and the rock moves back and forth along the direction of travel.

Take the curl instead. The curl measures local rotation, and it obeys a wave equation whose speed squared is the rigidity divided by the density. This is the S wave, and the rock moves across the direction of travel. The MIT notes derive both results in Section 4.6, Eqs. 4.32 to 4.37. The only assumptions are small strains and a rock with the same properties everywhere and in every direction.

\rho\,\ddot{\mathbf u}=(\lambda+\mu)\nabla(\nabla\!\cdot\mathbf u)+\mu\nabla^2\mathbf u\ \Rightarrow\ \alpha=\sqrt{\frac{\lambda+2\mu}{\rho}},\qquad \beta=\sqrt{\frac{\mu}{\rho}}

Worked example

Read the stiffness of one rock from its two speeds

The lower rock in the interface Lab’s first preset has a P speed of 4000 m/s, an S speed of 2200 m/s and a density of 2500 kg per cubic metre. Run the speed formulas backwards to find its elastic constants.

The rigidity is the density times the S speed squared: 2500 times 2200 squared gives 12.1 GPa. The density times the P speed squared gives the first Lamé constant plus twice the rigidity, 40.0 GPa. Subtract twice the rigidity, 24.2 GPa, and the first Lamé constant is 15.8 GPa. The bulk modulus is then 15.8 plus two thirds of 12.1, which is 23.9 GPa.

Poisson’s ratio, how much the rock narrows when stretched, depends only on the ratio of the two speeds. Here the P speed is 1.818 times the S speed and Poisson’s ratio is 0.283. The upper rock of the same preset, 2000 m/s and 1000 m/s, has a speed ratio of exactly two and a Poisson’s ratio of one third.

Seawater in the last preset has a P speed of 1500 m/s and a density of 1000 kg per cubic metre. Its bulk modulus is 2.25 GPa and its rigidity is zero, so its S speed is zero: no S wave can travel in it.

\nu=\frac{\lambda}{2(\lambda+\mu)}=\frac{\alpha^2-2\beta^2}{2(\alpha^2-\beta^2)},\qquad \frac{\alpha}{\beta}=\frac{4000}{2200}=1.818\ \Rightarrow\ \nu=0.283

Why the P wave always arrives first

Divide the two speed formulas. The square of the speed ratio is two plus the first Lamé constant divided by the rigidity. A stable solid has a positive bulk modulus, so the first Lamé constant is larger than minus two thirds of the rigidity. The square of the ratio is then larger than four thirds, and P is always faster than S. Real crustal rock has a ratio near the square root of three, the value for equal Lamé constants. The location Lab uses exactly that ratio.

Because the two waves leave the source together and travel at different speeds, the gap between their arrivals grows with distance. That gap is the first clue to where an earthquake happened, and the location chapter uses it.

\frac{\alpha^2}{\beta^2}=2+\frac{\lambda}{\mu}>\frac43\quad\text{when}\quad \kappa=\lambda+\tfrac23\mu>0

Predict, change, explain

In the interface Lab, choose Past the critical angle and show the P wavefield only, then the S wavefield only. Predict which front is ahead in the lower rock and by roughly what factor.

Choose Water over rock and show the S wavefield only.

Explain why no S front moves through the water, yet an S front still runs through the rock below.

The Labs use flat, isotropic, elastic or acoustic rock with made-up but realistic speeds; none of them reproduces a real survey. Each Lab states its model and limits in its notes and can export the data behind its readouts.

Try explaining it yourself

1. A granite has a P speed of 6.0 km/s, an S speed of 3.5 km/s and a density of 2700 kg per cubic metre. Find its rigidity and Poisson’s ratio.

Read an explained answer

The rigidity is 2700 times 3500 squared, 33.1 GPa. For Poisson’s ratio use the speeds in km/s: the P speed squared minus twice the S speed squared is 36 minus 24.5, or 11.5. Twice the difference of the squares is twice 23.75, or 47.5. The ratio is 0.242.

2. Show that a solid with equal Lamé constants has a speed ratio of the square root of three and a Poisson’s ratio of one quarter.

Read an explained answer

With equal constants the P speed squared is three times the rigidity over the density, and the S speed squared is the rigidity over the density, so their ratio is three. Poisson’s ratio is the first constant divided by twice the sum of the two constants, which is one divided by four.

\lambda=\mu\ \Rightarrow\ \frac{\alpha}{\beta}=\sqrt3,\quad \nu=\frac{\mu}{2(2\mu)}=\frac14

3. Seismic stations on the far side of the Earth record P waves that crossed the core but no direct S waves through it. What does that say about the outer core?

Read an explained answer

An S wave needs rigidity. If the outer core has none, it is a fluid, and S waves cannot pass through it. The MIT notes give the rigidity of liquid iron in the outer core as zero.

Sources to read alongside

MIT OpenCourseWare, 12.201 Essentials of Geophysics (Fall 2004), lecture notes, Chapter 4: Seismology

Section 4.6 separates the elastic wave equation into P and S waves and gives their speeds (Eqs. 4.34 and 4.37). Sections 4.14 to 4.16 derive Snell’s law from Fermat’s principle (Eqs. 4.91 to 4.96) and show that the ray parameter is the slope of the travel-time curve (Eq. 4.98). Section 4.19 notes that in travel-time tomography the matrix entries are the ray lengths in each block.

H. G. Saavedra and R. Moro, Zoeppritz equations: from seismology to medical exploration, Revista ECIPerú 22(2) (2025), arXiv:2510.26668v1

Equation 1 gives the normal-incidence reflection coefficient from the two acoustic impedances. Appendix 2 writes the four continuity conditions for an incident P wave (A2.1 to A2.8) and for an incident SV wave (A2.9 to A2.12). Section 3.2 and Figure 3 show the phase change of reflected wavelets past the critical angle, about 40 degrees for skin over bone.