How a P wave shares its energy at a rock boundary

At normal incidence a P wave only reflects and transmits as P. At any other angle it also makes S waves, one going up and one going down. Snell’s law gives the four angles. The amplitudes follow from requiring the two rocks to move together and push on each other equally across the boundary. Convert amplitudes to energy flux and the four shares always add up to the incident energy.

Sandbox Physics · 2026-10-30

One P wave in, four waves out

Light at a glass surface keeps its type. Seismic waves do not. An incident P wave or an incident SV wave, whose motion lies in the plane of the ray, produces both P and SV waves on reflection and on transmission whenever it arrives at an angle. Saavedra and Moro make this point in their Section 2. A horizontally polarized SH wave stays on its own and is left out here.

So an incident P wave has four unknown outgoing amplitudes: reflected P, reflected S, transmitted P and transmitted S. Their angles already follow from the shared ray parameter of the last chapter.

Four continuity conditions fix four amplitudes

Two rocks in welded contact cannot slide or separate. The horizontal and vertical displacement must therefore be the same on both sides of the boundary. Newton’s third law says the traction, the force per area across the boundary, must also match: both its normal part and its shear part.

Saavedra and Moro write these four conditions as Eqs. A2.1 to A2.4 of their Appendix 2, then use Snell’s law to turn them into four linear equations for the four amplitudes, Eqs. A2.5 to A2.8. These are the Zoeppritz equations. The interface Lab solves the same four equations as a complex 4 by 4 system at every angle. Appendix A of Weibull and Ahmed gives a closed form for the reflected P and reflected S coefficients (Eqs. 36 to 40). A separate program written from that closed form matches the Lab’s solver to round-off for the converted S wave (Eq. 37). For the reflected P wave (Eq. 36) it matches only after the sign of the numerator term proportional to the squared ray parameter is flipped, which points to a sign slip in the printed equation of version 2.

\big[u_x\big]_{z=0}=\big[u_z\big]_{z=0}=\big[\sigma_{zz}\big]_{z=0}=\big[\sigma_{xz}\big]_{z=0}=0

Head-on, the answer is short

At normal incidence there is no shear across the boundary, so no S waves are made and only two conditions remain. The result depends only on the acoustic impedance of each rock, its density times its P speed. Saavedra and Moro give the reflection coefficient as their Eq. 1.

Impedance alone cannot tell density from speed: two rocks with the same impedance reflect nothing head-on even if their speeds differ. At oblique angles the speeds matter separately, which is why amplitude changes with angle carry extra information. Their Figure 1b shows exactly this.

R=\frac{Z_2-Z_1}{Z_2+Z_1},\qquad T=\frac{2Z_1}{Z_2+Z_1},\qquad Z=\rho\alpha

Count energy flux, not amplitude

An amplitude ratio larger than one does not create energy, because energy flux also depends on density, speed and direction. A plane wave carries energy across the boundary at a rate proportional to its density, its speed, the cosine of its angle and its amplitude squared. Divide each outgoing flux by the incident flux and the four fractions add up to one.

A wave past its critical angle runs along the boundary and its amplitude decays away from it. Its amplitude stays finite, yet its energy fraction is exactly zero, because its energy flows parallel to the boundary.

E_k=\frac{\rho_k\,c_k\cos\theta_k}{\rho_1\,\alpha_1\cos i_1}\,\lvert A_k\rvert^2,\qquad E_{\rm RP}+E_{\rm RS}+E_{\rm TP}+E_{\rm TS}=1

Worked example

Split the energy at two angles

Use the first preset of the interface Lab: the upper rock has a density of 2100 kg per cubic metre and a P speed of 2000 m/s; the lower rock 2500 kg per cubic metre and 4000 m/s. Their impedances are 4.20 and 10.0 million kilograms per square metre per second.

Head-on, the reflection coefficient is 5.80 over 14.2, which is 0.4085, and the transmission coefficient is 8.40 over 14.2, which is 0.5915. The reflected energy is 0.4085 squared, 0.167. The transmitted energy is the impedance ratio 2.381 times 0.5915 squared, 0.833. The two add up to one.

Turn the angle to the Lab’s default of 40 degrees. That is past the 30-degree critical angle, so the transmitted P fraction is zero. The 4 by 4 solve gives 0.143 for reflected P, 0.420 for reflected S and 0.437 for transmitted S. Their sum is 1.000. More than half the energy now leaves as S waves, although only a P wave came in.

R=\frac{10.0-4.20}{10.0+4.20}=0.408,\quad R^2+\frac{Z_2}{Z_1}T^2=0.167+0.833=1;\qquad 40^\circ:\ 0.143+0.420+0+0.437=1.000

An experiment to think through

Predict, change, explain

In the interface Lab, choose P turns into S and set the angle to zero. Predict the two S energy fractions, then predict whether they grow or shrink as you raise the angle to 40 degrees.
Raise the incidence angle to 40 degrees and watch the energy chart.
Explain why the S fractions start at zero and why the reflected P fraction stays small in this preset at every angle.

Split one P wave into four at a rock boundary

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. Show that at normal incidence the reflected and transmitted energy fractions add up to one.

Read an explained answer

Write both over the square of the impedance sum. The reflected part is the squared difference of the impedances; the transmitted part is four times their product. The squared difference plus four times the product is the squared sum, so the total is one.

\frac{(Z_2-Z_1)^2+4Z_1Z_2}{(Z_1+Z_2)^2}=1

2. In Saavedra and Moro’s Figure 1b the two media have equal impedance but different P speeds. Why is there no reflection head-on, yet a reflection at oblique angles?

Read an explained answer

Head-on, only the impedances enter, and they are equal. At an angle, the transmitted wave bends to a different angle and S waves are made, so the boundary conditions involve the speeds and rigidities separately. They no longer balance without a reflected wave.

3. The Water over rock preset at 45 degrees sends 0.054 of the energy back up as P and none into the rock as P. Where does the rest go, and why?

Read an explained answer

The P critical angle for water over this rock is 35.2 degrees, so at 45 degrees the transmitted P wave is evanescent. The remaining 0.946 leaves as a transmitted S wave in the rock. Water has no rigidity, so there is no reflected S wave.

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.
W. W. Weibull and N. Ahmed, Seismic amplitude-variation-with-offset inversion using an algebraically equivalent formulation of the exact Zoeppritz equations and a matrix-free adjoint-state framework, arXiv:2512.13172v2
Appendix A restates the classical Knott–Zoeppritz solution for an incident P wave: the shared ray parameter (Eq. 35) and closed forms for the reflected P and converted S coefficients (Eqs. 36 to 40). The main text fits reflection amplitudes with an adjoint-state gradient and a limited-memory quasi-Newton method.