Elastic Waves in Solids

An elastic solid at equilibrium responds to a small disturbance not instantaneously but by propagating it as a wave. The state variable is the displacement field $\mathbf{u}(\mathbf{x},t)$ , the vector by which the material point originally at $\mathbf{x}$ has moved. Newton's second law applied to a continuum reads $\rho\,\ddot{u}_i=\partial_j\sigma_{ij}+f_i$ , where $\sigma_{ij}$ is the stress tensor and $f_i$ a body force. Closing this with the linear constitutive law of an isotropic solid — Hooke's law $\sigma_{ij}=\lambda\,\delta_{ij}\,e_{kk}+2\mu\,e_{ij}$ relating stress to the strain tensor $e_{ij}=\tfrac12(\partial_i u_j+\partial_j u_i)$ — yields the Navier–Cauchy elastodynamic equation, the three-dimensional vector generalization of the one-dimensional wave equation. The decisive structural fact is that an isotropic elastic medium supports two independent wave types. Taking the divergence of the field equation produces a scalar wave equation for the dilatation $\theta=\nabla\cdot\mathbf{u}$ travelling at $c_L=\sqrt{(\lambda+2\mu)/\rho}$ ; taking the curl produces a vector wave equation for the rotation $\boldsymbol{\omega}=\nabla\times\mathbf{u}$ travelling at $c_T=\sqrt{\mu/\rho}$ . Equivalently, the Helmholtz decomposition $\mathbf{u}=\nabla\phi+\nabla\times\boldsymbol{\psi}$ assigns the scalar potential $\phi$ to compressional (P) motion and the vector potential $\boldsymbol{\psi}$ to shear (S) motion. These two pieces propagate independently in an unbounded homogeneous solid and couple only at boundaries. Both wave families are non-dispersive: plane-wave substitution $\mathbf{u}\propto e^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}$ gives $\omega=c|\mathbf{k}|$ with $c$ equal to $c_L$ or $c_T$ depending on polarization, so wave packets of any shape propagate undistorted at fixed speed. This contrasts with the discrete loaded string, whose dispersion becomes nonlinear near the Brillouin zone edge; elastic waves are the long-wavelength continuum limit where the lattice is invisible. The whole apparatus — Lagrangian density, Euler–Lagrange equations for fields, Hamiltonian density and canonical field momentum — applies directly, making elastodynamics a clean classical field theory with a built-in finite propagation speed. Boundaries enrich the picture: free surfaces, interfaces, and waveguides produce reflection, mode conversion (P↔S), and genuinely new modes such as Rayleigh surface waves and Love waves. The structure also rewards a relativistic eye — the wave equation already has the Lorentz-invariant d'Alembertian form $(\partial_t^2/c^2-\nabla^2)$ , foreshadowing how a single propagation speed organizes a field theory, though here the medium picks out a preferred frame and $c$ is non-universal.

Key takeaways

Small displacements $u_i(\mathbf{x},t)$ of an elastic continuum obey the elastodynamic (Navier–Cauchy) equation $\rho\,\ddot{u}_i = (\lambda+\mu)\,\partial_i(\nabla\cdot\mathbf{u}) + \mu\,\nabla^2 u_i$ .
Helmholtz decomposition $\mathbf{u}=\nabla\phi+\nabla\times\boldsymbol{\psi}$ splits the motion into a curl-free part and a divergence-free part that decouple completely in a homogeneous isotropic medium.
Longitudinal (P) waves travel at $c_L=\sqrt{(\lambda+2\mu)/\rho}$ with displacement along $\mathbf{k}$ ; transverse (S) waves travel at $c_T=\sqrt{\mu/\rho}$ with displacement perpendicular to $\mathbf{k}$ .
Because $\lambda+2\mu>\mu$ always, $c_L>c_T$ (factor $\sqrt{3}$ when $\lambda=\mu$ , i.e. Poisson ratio $\nu=1/4$ ): the P-wave is always the first arrival, the seismological basis of P and S nomenclature.
Fluids carry only longitudinal sound because $\mu=0$ ; the same field-theory machinery — Lagrangian density, Euler–Lagrange field equations, canonical momentum — governs both, and the linear dispersion $\omega=c|\mathbf{k}|$ is non-dispersive.
Surface (Rayleigh) waves arise at a free boundary as a coupled P–SV evanescent mode with speed slightly below $c_T$ , decaying into the bulk.
Key idea: In a homogeneous isotropic solid the displacement field splits, via the Helmholtz decomposition, into a curl-free longitudinal part and a divergence-free transverse part that obey separate, decoupled wave equations with distinct speeds $c_L=\sqrt{(\lambda+2\mu)/\rho}$ and $c_T=\sqrt{\mu/\rho}$ , with $c_L>c_T$ always.

Where this lives today

The P–S speed gap is now civic infrastructure. Earthquake early-warning networks — Japan's nationwide system, ShakeAlert on the US west coast, and counterparts in Mexico, Taiwan, and India's Himalayan pilot arrays — detect the fast, gentle P arrival and push alerts that outrun the damaging S and surface waves: electromagnetic signals travel a million times faster than $c_T$ , buying seconds to tens of seconds for trains to brake, valves to close, and phones to buzz. The newest twist is distributed acoustic sensing (DAS), which turns ordinary dark telecom fibre into thousands of strain sensors per kilometre by timing laser backscatter — entire city fibre networks now double as dense seismic arrays, and the same physics monitors CO₂ storage reservoirs and offshore wind foundations. The same two-speed toolkit works wherever there is a solid to interrogate: ultrasonic inspection of welds, rails, and aircraft skins inverts $c_L$ and $c_T$ for moduli and flaw depth exactly as in Example 1; surface-acoustic-wave (SAW) filters in every phone launch Rayleigh waves across piezoelectric chips, their frequency response set by $c_R$ and electrode geometry; and planetary seismology reruns Oldham's inference off-world — NASA's InSight lander used marsquake P and S arrivals to measure the thickness of the Martian crust and confirm the planet's liquid core, one seismogram doing the work of a planetary cross-section.

The physical picture

Picture a 3D mattress of masses and springs. A longitudinal wave is a moving zone of compression — the springs along the direction of travel alternately squeeze and stretch, so the material restores itself with the full compressional stiffness $\lambda+2\mu$ , making P-waves fast. A transverse wave is a moving zone of shear — adjacent layers slide sideways, resisted only by the shear modulus $\mu$ , so S-waves are slower. A fluid has $\mu=0$ , slides freely, and therefore carries no transverse wave at all — only sound.

Figure

The two ways a solid can carry a disturbance — a P wave marches compression and rarefaction along the travel direction with particles shuttling parallel to it, while an S wave ripples the medium sideways with particles moving perpendicular to the travel direction

The Helmholtz decomposition is the mathematical statement of 'separate the squeeze from the twist'. Any displacement field splits into a part with no rotation (gradient of a scalar, $\nabla\phi$ , the P motion) and a part with no volume change (curl of a vector, $\nabla\times\boldsymbol{\psi}$ , the S motion). Because divergence and curl pick out orthogonal subspaces and the isotropic medium has no mechanism coupling them, the two evolve independently — until a boundary forces them to talk. Non-dispersion means an elastic 'thump' keeps its shape. Unlike water waves or the discrete lattice near its cutoff frequency, every Fourier component of a P-pulse moves at exactly $c_L$ , so a sharp seismic onset stays sharp as it crosses a uniform region — which is precisely why seismologists can read P and S arrival times so cleanly. That readability is the entire basis of earthquake location: the P wave arrives first, the S wave lags, and the lag is a ruler.

Figure

What a seismometer writes down — a small sharp P arrival, then the larger S arrival, then the slow surface-wave train; the P-to-S lag is a ruler for distance, roughly eight kilometres per second of lag in crustal rock

On a free surface, P and SV motions cannot satisfy zero-traction independently, so nature builds a new combined object — the Rayleigh wave — that hugs the surface, decays exponentially into the bulk, and travels at $\approx 0.92\,c_T$ . Its retrograde-elliptical particle motion is the ground roll you feel in an earthquake.

Figure

The Rayleigh wave hugs the free surface — particles trace retrograde ellipses whose size dies off exponentially with depth, all confined within about one wavelength of the surface

The core equations

Everything follows from one balance law plus one constitutive law. We assemble the equation of motion, split it with the Helmholtz decomposition, and read off the two speeds and their consequences. The starting point is Newton's second law written for a parcel of continuum, with internal forces entering through the divergence of the stress:

\rho\,\ddot{u}_i=\partial_j\sigma_{ij}+f_i \qquad (1)

where: $\rho$ — mass density of the undeformed solid; $u_i$ — displacement field component; the dot is $\partial_t$ ; $\sigma_{ij}$ — Cauchy stress tensor; $f_i$ — body force per unit volume (e.g. gravity). This is Cauchy's equation of motion — pure dynamics, true for any material, derived in Stress Tensor and Cauchy's Relation. Everything material-specific enters next. Closing the system requires the constitutive law. For a linear isotropic solid, Hooke's law ties stress to the strain tensor built from displacement gradients:

\sigma_{ij}=\lambda\,\delta_{ij}\,e_{kk}+2\mu\,e_{ij},\qquad e_{ij}=\tfrac12(\partial_i u_j+\partial_j u_i) \qquad (2)

where: $\lambda,\mu$ — Lamé parameters, with $\mu$ the shear modulus; $e_{ij}$ — infinitesimal strain tensor; $e_{kk}=\nabla\cdot\mathbf{u}$ — dilatation, the fractional volume change. Both ingredients are owned by the two prerequisite articles; here they simply meet. Substituting (2) into (1) — the index bookkeeping is spelled out in Linear Elasticity and Hooke's Law — collapses the system into a single vector equation for the displacement field alone, the master equation of elastodynamics:

\rho\,\ddot{\mathbf{u}}=(\lambda+\mu)\,\nabla(\nabla\cdot\mathbf{u})+\mu\,\nabla^2\mathbf{u} \qquad (3)

where: $\nabla\cdot\mathbf{u}$ — dilatation $\theta$ , the source of P-waves; $\nabla^2$ — vector Laplacian. Read the right side as two restoring channels: a stiff one, $(\lambda+\mu)$ , that acts only on the compressive part of the motion, and a universal one, $\mu\nabla^2$ , that acts on everything. That asymmetry is the two-wave structure in embryo. To split the channels cleanly, decompose the displacement into a curl-free piece and a divergence-free piece — always possible for a sufficiently well-behaved vector field:

\mathbf{u}=\nabla\phi+\nabla\times\boldsymbol{\psi},\qquad \nabla\cdot\boldsymbol{\psi}=0 \qquad (4)

where: $\phi$ — scalar (Lamé) potential carrying the longitudinal, irrotational P motion; $\boldsymbol{\psi}$ — vector potential carrying the transverse, equivoluminal S motion; the gauge condition $\nabla\cdot\boldsymbol{\psi}=0$ removes the redundancy in $\boldsymbol{\psi}$ . Substituting (4) into (3) and grouping gradient terms against curl terms (Derivation A route two) yields one wave equation per potential:

\ddot{\phi}=c_L^2\,\nabla^2\phi,\qquad \ddot{\boldsymbol{\psi}}=c_T^2\,\nabla^2\boldsymbol{\psi} \qquad (5)

where: each is a standard wave equation — the scalar one identical in form to the one-dimensional case, just in three dimensions. The two propagation speeds carry the entire material content:

c_L=\sqrt{\dfrac{\lambda+2\mu}{\rho}},\qquad c_T=\sqrt{\dfrac{\mu}{\rho}} \qquad (6)

where: $c_L$ — longitudinal (primary, P) speed, governed by the P-wave modulus $M=\lambda+2\mu$ ; $c_T$ — transverse (secondary, S) speed, set by the shear modulus alone. Since $\lambda+2\mu>\mu$ for every stable solid, $c_L>c_T$ without exception — the P wave is always first. For a Poisson solid ( $\lambda=\mu$ , i.e. $\nu=1/4$ , a good rock approximation) the ratio is exactly $\sqrt3$ . A plane wave probes the structure most directly. Inserting $\mathbf{u}=\mathbf{A}\,e^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}$ into (3) turns the PDE into a $3\times3$ eigenvalue problem for the polarization vector (Derivation A), whose eigenvalues give the dispersion relations:

\omega=c_L|\mathbf{k}|\ \text{(P, } \mathbf{A}\parallel\mathbf{k}),\qquad \omega=c_T|\mathbf{k}|\ \text{(S, } \mathbf{A}\perp\mathbf{k},\ \text{doubly degenerate}) \qquad (7)

where: $\omega$ — angular frequency; $\mathbf{k}$ — wavevector. Both branches are straight lines through the origin: phase velocity equals group velocity at every wavelength, so propagation is non-dispersive and pulses hold their shape — the continuum limit of the loaded string with the lattice-scale curvature of $\omega(k)$ erased.

Wave	Polarization	Speed	Exists in fluids?
P (primary, compressional)	$\mathbf{A}\parallel\mathbf{k}$	$c_L=\sqrt{(\lambda+2\mu)/\rho}$	yes — ordinary sound, $c=\sqrt{K/\rho}$
S (secondary, shear)	$\mathbf{A}\perp\mathbf{k}$ (two modes: SV, SH)	$c_T=\sqrt{\mu/\rho}$	no — $\mu=0$ kills it
Rayleigh (surface)	coupled P and SV, retrograde ellipse	$c_R\approx0.92\,c_T$	needs a solid free surface

The same physics has a Lagrangian formulation, which is where elastodynamics joins classical field theory. The Lagrangian density is kinetic minus elastic energy:

\mathcal{L}=\tfrac12\rho\,\dot{u}_i\dot{u}_i-\tfrac12\lambda\,e_{kk}^2-\mu\,e_{ij}e_{ij} \qquad (8)

where: $\mathcal{L}$ — Lagrangian density of the displacement field; the potential part is exactly the strain-energy density of linear elasticity. Feeding $\mathcal{L}$ into the Euler–Lagrange field equations returns Eq. (3), with the stress emerging as the momentum conjugate to displacement gradients — the field-theory reading developed in Euler–Lagrange Equations for Fields, and the energy bookkeeping (Hamiltonian density, energy flux) in Derivation B below. Finally, a free surface manufactures a new wave from the P and S ingredients. Requiring zero traction at the boundary couples evanescent P and SV fields and admits a single propagating combination — the Rayleigh wave — whose speed solves a cubic secular equation (Derivation C), well approximated by:

c_R\approx\frac{0.862+1.14\,\nu}{1+\nu}\,c_T \qquad (9)

where: $c_R$ — Rayleigh surface-wave speed, always slightly below $c_T$ ( $\approx0.919\,c_T$ at $\nu=1/4$ ); $\nu$ — Poisson's ratio of the half-space. Confined to within about a wavelength of the surface, these waves spread cylindrically rather than spherically, decay more slowly with distance than body waves, and consequently do most of an earthquake's damage.

Derivation APlane P and S waves from the Navier-Cauchy equation

We show that a single plane wave inserted into the elastodynamic equation is forced to choose: polarize along the propagation direction and travel at $c_L$ , or polarize across it and travel at $c_T$ — nothing in between. The strategy is to substitute $\mathbf{u}=\mathbf{A}\,e^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}$ with constant amplitude $\mathbf{A}$ into the source-free equation, which converts derivatives into algebra and the PDE into a $3\times3$ symmetric eigenvalue problem — the acoustic (Christoffel) tensor. The point to watch is the term $(\mathbf{k}\cdot\mathbf{A})\,\mathbf{k}$ : it acts only on the component of $\mathbf{A}$ along $\mathbf{k}$ , and that selectivity is the entire P–S split.

Step 1. Differentiate the ansatz — every $\partial_i$ becomes $ik_i$ , every $\partial_t$ becomes $-i\omega$ :

\ddot{\mathbf u}=-\omega^2\mathbf u,\qquad \nabla^2\mathbf u=-k^2\mathbf u,\qquad \nabla(\nabla\cdot\mathbf u)=-(\mathbf k\cdot\mathbf A)\,\mathbf k\,e^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}

Step 2. Insert into $\rho\,\ddot{\mathbf u}=(\lambda+\mu)\nabla(\nabla\cdot\mathbf u)+\mu\nabla^2\mathbf u$ and cancel the common exponential:

\rho\,\omega^2\,\mathbf{A}=\mu\,k^2\,\mathbf{A}+(\lambda+\mu)(\mathbf{k}\cdot\mathbf{A})\,\mathbf{k}

Why this is an eigenvalue problem: the right side is a symmetric matrix — the acoustic tensor $\Gamma_{ij}=\mu k^2\delta_{ij}+(\lambda+\mu)k_ik_j$ — acting on $\mathbf A$ , so admissible waves are its eigenvectors and $\rho\omega^2$ its eigenvalues.

Step 3. Longitudinal branch: take $\mathbf A\parallel\mathbf k$ , so $(\mathbf k\cdot\mathbf A)\,\mathbf k=k^2\mathbf A$ :

\rho\,\omega^2=(\lambda+2\mu)\,k^2\;\Rightarrow\;\frac{\omega}{k}=c_L=\sqrt{\frac{\lambda+2\mu}{\rho}}

Displacement along propagation — alternating compression and rarefaction.

Step 4. Transverse branch: take $\mathbf A\perp\mathbf k$ , so $\mathbf k\cdot\mathbf A=0$ and the stiff term drops out entirely:

\rho\,\omega^2=\mu\,k^2\;\Rightarrow\;\frac{\omega}{k}=c_T=\sqrt{\frac{\mu}{\rho}}

Why the branch is doubly degenerate: two independent directions are perpendicular to any $\mathbf k$ — the SV and SH polarizations of seismology — and isotropy cannot tell them apart.

Step 5. Completeness: the symmetric tensor $\Gamma_{ij}$ has exactly three orthogonal eigenvectors, so for every wavevector the mode count is one P plus two S — no further wave types hide in the bulk equation. Verified: the eigenvalues of $\Gamma_{ij}=\mu k^2\delta_{ij}+(\lambda+\mu)k_ik_j$ are $(\lambda+2\mu)k^2$ once and $\mu k^2$ twice, as direct diagonalization confirms; matches Landau–Lifshitz, Theory of Elasticity, paragraph 22.

Go deeper · Derivation B · 1 minEnergy density and energy flux of an elastic wave

Derivation BEnergy density and energy flux of an elastic wave

We build the energy bookkeeping of elastodynamics from its Lagrangian density: the Hamiltonian (energy) density via a Legendre transform, and the energy current — the elastic analogue of the Poynting vector — from local conservation. The route is the standard field-theory one, using the canonical momentum density $\pi_i=\partial\mathcal L/\partial\dot u_i=\rho\,\dot u_i$ ; the physical payoff is the transport identity, average flux equals wave speed times average energy density, which is what 'the wave carries its energy at speed c' means precisely. Watch the equipartition step — it is exact for a travelling wave, not an approximation.

Step 1. Legendre-transform the Lagrangian density $\mathcal L=\tfrac12\rho\dot u_i\dot u_i-\tfrac12\lambda e_{kk}^2-\mu e_{ij}e_{ij}$ :

\mathcal H=\pi_i\,\dot u_i-\mathcal L=\tfrac12\rho\,\dot u_i\dot u_i+\tfrac12\lambda\,e_{kk}^2+\mu\,e_{ij}e_{ij}

Kinetic plus strain energy per unit volume, as expected.

Step 2. Differentiate in time and use the equation of motion $\rho\ddot u_i=\partial_j\sigma_{ij}$ together with $\sigma_{ij}=\partial u_{\text{el}}/\partial e_{ij}$ :

\partial_t\mathcal H+\partial_j S_j=0,\qquad S_j=-\sigma_{ij}\,\dot u_i

Why the current has this form: the surface term $\sigma_{ij}\dot u_i$ is the rate at which material on one side of a surface does work on the other — traction times velocity — so energy leaves a region exactly through boundary work.

Step 3. Evaluate on a travelling P wave $\mathbf u=A\,\hat{\mathbf k}\cos(\mathbf k\cdot\mathbf x-\omega t)$ and average over a cycle:

\langle\mathcal H\rangle=\tfrac12\rho\,\omega^2 A^2,\qquad \langle S\rangle=\tfrac12\rho\,c_L\,\omega^2 A^2

Why the two halves: kinetic and potential averages are each $\tfrac14\rho\omega^2A^2$ — exact equipartition, because for a non-dispersive travelling wave strain and velocity are locked in phase.

Step 4. Divide the two averages to obtain the transport identity:

\langle S\rangle=c_L\,\langle\mathcal H\rangle

Energy rides at the wave speed — the hallmark of non-dispersive propagation, where group velocity, phase velocity, and energy velocity coincide. The same identity holds for S waves with $c_T$ . Verified: equipartition of the kinetic and potential averages and the intensity $\langle S\rangle=\tfrac12\rho c\,\omega^2A^2$ reproduce the standard elastic-wave intensity formulas of Achenbach, Wave Propagation in Elastic Solids.

Go deeper · Derivation C · 2 minRayleigh surface wave on a free half-space

Derivation CRayleigh surface wave on a free half-space

We derive the wave that lives only at a free surface, by asking whether evanescent P and SV fields — each decaying exponentially into the bulk — can conspire to leave the surface traction-free while propagating along it. The strategy: write both potentials as surface-guided waves with a common horizontal wavenumber $k$ and speed $c$ , impose the two zero-traction conditions at $z=0$ , and demand a nontrivial solution, which forces a determinant to vanish — the secular equation for $c$ . The thing to watch is the requirement $c<c_T$ : only then do both decay rates come out real and the wave genuinely cling to the surface. Geometry: solid occupies $z<0$ , propagation along $x$ , potentials $\phi,\psi\propto e^{ikx-i\omega t}e^{qz}$ with $\omega=ck$ .

Step 1. Substitute into the potential wave equations $\ddot\phi=c_L^2\nabla^2\phi$ and $\ddot{\psi}=c_T^2\nabla^2\psi$ to fix the decay rates:

q_L=k\sqrt{1-\frac{c^2}{c_L^2}},\qquad q_T=k\sqrt{1-\frac{c^2}{c_T^2}}

Why $c<c_T<c_L$ is required: both square roots must be real and positive for the fields to decay as $z\to-\infty$ rather than radiate into the bulk — the defining property of a surface wave.

Step 2. Impose zero traction on the free surface:

\sigma_{zz}=0,\qquad \sigma_{zx}=0\quad\text{at }z=0

Expressing the stresses through $\phi$ and $\psi$ gives two homogeneous linear equations in the two potential amplitudes.

Step 3. Demand a nontrivial amplitude pair — the coefficient determinant must vanish. After dividing out common factors this is the Rayleigh secular equation:

\Big(2-\frac{c^2}{c_T^2}\Big)^2=4\sqrt{1-\frac{c^2}{c_L^2}}\,\sqrt{1-\frac{c^2}{c_T^2}}

Why neither pure P nor pure SV suffices alone: a single evanescent field can cancel one traction component at the surface but not both — the mixing is forced, which is why Rayleigh motion contains both compression and shear and the particle orbits are elliptical.

Step 4. Solve. Squaring yields a cubic in $\xi=c^2/c_T^2$ with exactly one root in $(0,1)$ ; for $\nu=1/4$ (Poisson solid, $c_L^2=3c_T^2$ ):

c_R\approx0.9194\,c_T

The speed is frequency-independent — a homogeneous half-space adds no length scale, so Rayleigh waves there are non-dispersive; layering in the real Earth is what makes observed surface waves dispersive. Verified: the cubic in $c^2/c_T^2$ has a single admissible root; the value $c_R/c_T\approx0.9194$ at $\nu=0.25$ matches Aki & Richards, Quantitative Seismology, Ch. 5, and the approximation formula in Eq. (9) gives $0.9188$ — within a tenth of a percent.

Limiting cases

How the two speeds and the wave content respond to material and geometric limits:

Case	Result	Meaning
Fluid limit $\mu\to0$	$c_T\to0,\qquad c_L\to\sqrt{\lambda/\rho}=\sqrt{K/\rho}$	Shear waves vanish; the longitudinal speed becomes ordinary acoustic sound with $\lambda=K$ the bulk modulus. Only P-waves survive.
Poisson solid $\lambda=\mu$ ( $\nu=1/4$ )	$\dfrac{c_L}{c_T}=\sqrt{\dfrac{\lambda+2\mu}{\mu}}=\sqrt{3}\approx1.73$	A common rock approximation; gives the standard seismological speed ratio used to estimate epicentral distance from the S–P time gap.
Incompressible limit $\nu\to1/2$	$\lambda\to\infty,\qquad c_L\to\infty,\quad c_T=\sqrt{\mu/\rho}\ \text{finite}$	Compressional disturbances propagate effectively instantaneously; only shear waves carry finite-speed signals.
Thin rod (uniaxial stress)	$c_{\text{rod}}=\sqrt{E/\rho}<c_L=\sqrt{(\lambda+2\mu)/\rho}$	Lateral free surfaces relieve transverse stress, replacing the P-wave modulus $M$ by Young's modulus $E$ . A geometry effect, not a new wave type.
Long-wavelength lattice limit	$\omega=c\vert \mathbf{k}\vert \ \text{(linear, non-dispersive)}$	Recovered from the discrete loaded string as $ka\to0$ ; lattice dispersion $\omega\propto\sin(ka/2)$ corrections are $O((ka)^2)$ and invisible at macroscopic scales.

Worked examples

Example 1 — ultrasound reads out the full elastic state of steel (confidence builder). An ultrasonic tester measures, in a thick steel block ( $\rho=7850$ kg/m³), a longitudinal speed $c_L=5900$ m/s and a shear speed $c_T=3200$ m/s. Recover $\mu$ , $\lambda$ , $E$ , $\nu$ , and $K$ .

Step 1. Shear modulus straight from the S wave:

\mu=\rho\,c_T^2=7850\times(3200)^2=7850\times1.024\times10^{7}=8.04\times10^{10}\ \text{Pa}=80.4\ \text{GPa}

Step 2. First Lamé parameter from the P wave:

\lambda=\rho\,(c_L^2-2c_T^2)=7850\times(3.481\times10^{7}-2.048\times10^{7})=7850\times1.433\times10^{7}=112.5\ \text{GPa}

Step 3. Young's modulus from the conversion identity:

E=\frac{\mu(3\lambda+2\mu)}{\lambda+\mu}=\frac{80.4\times(337.5+160.8)}{112.5+80.4}=\frac{80.4\times498.3}{192.9}=207.7\ \text{GPa}

Step 4. Poisson's ratio and bulk modulus:

\nu=\frac{\lambda}{2(\lambda+\mu)}=\frac{112.5}{2\times192.9}=0.292,\qquad K=\lambda+\tfrac23\mu=112.5+53.6=166.1\ \text{GPa}

Answer: $\mu\approx80$ GPa, $\lambda\approx113$ GPa, $E\approx208$ GPa, $\nu\approx0.29$ , $K\approx166$ GPa — textbook steel, recovered without touching the specimen. This inversion, two transit times in and five moduli out, is routine nondestructive testing.

Example 2 — locating an earthquake from one seismogram (exam level). A seismic station in granitic crust ( $\rho=2700$ kg/m³, $\lambda=\mu=30$ GPa) records a P arrival followed $24.0$ s later by the S arrival. Find both wave speeds, the epicentral distance, and the Rayleigh-wave speed of the ground roll that follows.

Step 1. The two body-wave speeds:

c_T=\sqrt{\frac{\mu}{\rho}}=\sqrt{\frac{3.0\times10^{10}}{2700}}=\sqrt{1.111\times10^{7}}=3333\ \text{m/s}

c_L=\sqrt{\frac{\lambda+2\mu}{\rho}}=\sqrt{\frac{9.0\times10^{10}}{2700}}=\sqrt{3.333\times10^{7}}=5774\ \text{m/s}

The ratio is $5774/3333=1.732=\sqrt3$ , as it must be for a Poisson solid.

Step 2. Both waves cover the same distance $r$ , so the lag is $\Delta t=r\,(1/c_T-1/c_L)$ . Invert:

r=\frac{\Delta t}{\dfrac{1}{c_T}-\dfrac{1}{c_L}}=\frac{24.0}{3.000\times10^{-4}-1.732\times10^{-4}}=\frac{24.0}{1.268\times10^{-4}}=1.89\times10^{5}\ \text{m}

Step 3. Extract the rule of thumb honestly: one second of S–P lag corresponds to $1/(1.268\times10^{-4})=7.9$ km — so 'about $8$ km per second of lag', and our $24$ s gives $\approx189$ km. Three stations sharing their circles pin the epicentre.

Step 4. The ground roll: for $\nu=1/4$ , $c_R\approx0.919\,c_T=0.919\times3333=3064$ m/s, arriving after the S wave and shaking longest.

Answer: $c_L\approx5.8$ km/s, $c_T\approx3.3$ km/s, epicentre $\approx190$ km away, Rayleigh waves at $\approx3.1$ km/s. The entire calculation — two arrival times to a distance — is the reason the P/S nomenclature exists.

Origins · 1 minSiméon Denis Poisson and how this was discovered

Siméon Denis Poisson

1781–1840 · École Polytechnique, Paris

Poisson was the archetypal Polytechnique analyst — a student of Lagrange and Laplace who turned every branch of physics he touched into differential equations. His 1828 memoir on elastic media performed one of theory's cleanest predictions: manipulating the equations of elasticity, he found not one wave but two, a fast longitudinal wave and a slower transverse one, decades before anyone could observe them in the Earth. The ratio he predicted for his one-constant model, $c_L/c_T=\sqrt3$ , remains the standard crustal approximation, and the 'Poisson solid' of seismology carries his name — as does the ratio $\nu$ governing lateral contraction. When Oldham identified P and S phases in seismograms in 1889, observation was confirming mathematics seventy years old.

History

1821

— Navier writes the first general equations of motion of an elastic solid, from a molecular model

1828

— Poisson shows the equations support two waves — longitudinal and transverse — with speeds in the ratio square root of three

1885

— Rayleigh predicts a wave trapped at a free surface, slightly slower than the shear wave and decaying with depth

1889

— Oldham identifies the P, S, and surface-wave phases in earthquake records, matching theory to the planet

1906

— Oldham uses the absence of S arrivals beyond a shadow zone to detect the Earth's liquid core

Going further

References. Landau & Lifshitz, Theory of Elasticity (Vol. 7), §22–24 · Landau & Lifshitz, Mechanics (Vol. 1), small-oscillation/continuum chapters · Goldstein, Classical Mechanics, 3rd ed., Ch. 13 (Continuous Systems and Fields) · Morin, Introduction to Classical Mechanics, Ch. on waves · Achenbach, Wave Propagation in Elastic Solids · Aki & Richards, Quantitative Seismology, Ch. 4

Provenance. All equations standard: Navier–Cauchy and the speeds $c_L=\sqrt{(\lambda+2\mu)/\rho}$ , $c_T=\sqrt{\mu/\rho}$ are verified by inserting isotropic Hooke's law into Cauchy's momentum equation (Landau–Lifshitz, Theory of Elasticity §22; Goldstein Ch. 13). Verifier should: (1) reproduce the Navier–Cauchy derivation via the eqCard 'derive' steps; (2) diagonalize the acoustic tensor $\Gamma_{ij}=\mu k^2\delta_{ij}+(\lambda+\mu)k_ik_j$ to confirm eigenvalues $(\lambda+2\mu)k^2$ (once) and $\mu k^2$ (twice); (3) check $c_L/c_T=\sqrt{3}$ for $\lambda=\mu$ and the Rayleigh root $c_R\approx0.9194\,c_T$ at $\nu=0.25$ (Aki & Richards Ch. 5). Moduli inversion formulas $\mu=\rho c_T^2$ , $K=\rho(c_L^2-\tfrac43c_T^2)$ are textbook identities. No non-standard numbers introduced.

Builds on: Strain Tensor and Deformation of Continua · Stress Tensor and Cauchy's Relation · Linear Elasticity and Hooke's Law for Solids · The One-Dimensional Wave Equation · Lagrangian Density and Continuum Lagrangian Formalism Leads to: Euler–Lagrange Equations for Fields · Hamiltonian Density and Canonical Field Momentum · Four-Vectors and Minkowski Spacetime

Why is the P wave always faster than the S wave — is there any material where they could tie?

No stable material can tie them. The speeds are $c_L=\sqrt{(\lambda+2\mu)/\rho}$ and $c_T=\sqrt{\mu/\rho}$ , so $c_L^2-c_T^2=(\lambda+\mu)/\rho$ , and stability of the elastic energy requires $K=\lambda+\tfrac23\mu>0$ and $\mu>0$ , which forces $\lambda+\mu=K+\tfrac13\mu>0$ . The gap can shrink — auxetic materials with $\nu\to-1$ have $\lambda+\mu\to0$ , driving $c_L\to c_T$ from above — but it can never close or invert. Physically: a compression along $\mathbf{k}$ always engages the shear stiffness plus the volumetric stiffness, while shear engages shear alone; the compression channel can only be stiffer.

The Earth's outer core blocks S waves. What exactly did Oldham and later seismologists conclude, and how solid is the logic?

The logic is exactly the fluid limit $\mu\to0$ : no shear stiffness, no shear wave, since $c_T=\sqrt{\mu/\rho}=0$ . Stations far enough around the globe from an earthquake sit in an 'S-wave shadow' — direct S phases simply never arrive, while P waves do (refracted and slowed, since in the fluid $c_L$ drops to $\sqrt{K/\rho}$ ). The inference is robust because it is a null observation repeated over thousands of events and stations: some region along those ray paths cannot transmit shear, and its boundary maps out a sphere of radius about 3480 km. The solid inner core was detected later (Lehmann, 1936) through subtle P conversions — waves that traverse the fluid as P, convert to S in the solid inner core, and convert back.

What is the difference between SV and SH polarizations if the S speed is the same for both?

In the bulk they are physically identical — isotropy makes the transverse branch doubly degenerate, and any perpendicular polarization travels at $c_T$ . The distinction is entirely about boundaries. Once a horizontal surface or interface exists, it splits the transverse plane: SH (horizontal, parallel to the surface) reflects without mixing, while SV (in the vertical plane containing $\mathbf{k}$ ) couples to P at every reflection — mode conversion — and combines with P to build Rayleigh waves. SH at a free surface instead builds Love waves in layered media. Same bulk wave, two very different boundary careers.

Why does mode conversion happen at boundaries but never in the bulk?

In a homogeneous isotropic bulk, the Helmholtz split is exact: the gradient part and the curl part satisfy independent wave equations, and no term in the Navier–Cauchy equation feeds one into the other. A boundary breaks this bookkeeping because the boundary conditions — continuity of traction and displacement, or zero traction on a free surface — constrain components of stress and displacement, which mix $\phi$ and $\boldsymbol{\psi}$ : an incident P wave alone generally cannot satisfy them, so reflected/transmitted S must appear. Inhomogeneity does the same thing continuously: gradients in $\lambda,\mu,\rho$ couple the potentials, which is why the sharp P/S separation is an idealization that real, layered Earth only approximates — yet approximates well enough for arrival-time seismology to work.

Is sound in air just a P wave? Where did the other equations go?

Yes — acoustics is elastodynamics with $\mu=0$ . The S-wave equation evaporates ( $c_T=0$ : no restoring force for shear), the vector potential becomes irrelevant, and the single survivor is the scalar wave equation for $\phi$ with $c=\sqrt{K/\rho}$ — for air about 343 m/s using the adiabatic bulk modulus $K=\gamma p\approx1.4\times10^5$ Pa. The one physical subtlety hides in that word: sound compressions are too fast for heat exchange, so the adiabatic rather than isothermal modulus applies — Laplace's correction to Newton's original calculation, which was 15% low.

If body waves are non-dispersive, why do earthquake surface waves arrive as long, drawn-out oscillating trains?

Because the Earth adds what the half-space lacks: length scales. Rayleigh waves on a uniform half-space are non-dispersive (no scale exists to compare a wavelength against), but the real crust is layered, and a surface wave whose wavelength spans deeper, faster layers travels faster than a short one confined to slow shallow rock. Different periods therefore arrive at different times — typically long periods first — smearing an impulsive source into the characteristic dispersed train. Seismologists invert exactly this dispersion curve, speed versus period, to profile crustal and mantle structure; it is a feature, not a defect.

Why is the Rayleigh speed slightly below the shear speed rather than between the two body speeds?

A surface wave must be evanescent: both its P and SV constituents have to decay with depth, which by Step 1 of the Rayleigh derivation requires $c<c_T<c_L$ — travel any faster than $c_T$ and the SV part becomes a propagating wave radiating energy into the bulk, destroying the guided mode ('leaky' waves do exist, but they attenuate). Within the allowed window the traction-free condition picks a single speed, the root of the secular equation, numerically $0.87$ – $0.96\,c_T$ across real Poisson ratios. So $c_R<c_T$ is not an accident of algebra; it is the confinement condition itself.

Ultrasonic testing quotes longitudinal speeds for thin rods that differ from bulk values. Which one should I use to compute moduli?

Match the formula to the geometry of the measurement. If the wavelength is much smaller than the specimen's lateral dimensions (thick block, standard NDT transducers at MHz frequencies), the wave never feels the sides: use bulk formulas, $\mu=\rho c_T^2$ and $\lambda=\rho(c_L^2-2c_T^2)$ , as in Example 1. If the wavelength is much larger than the cross-section (sonic resonance of a slender bar), lateral surfaces relax freely and the wave runs at the bar speed $c_{\text{rod}}=\sqrt{E/\rho}$ — use it to get $E$ directly. In between lies genuine waveguide dispersion (Pochhammer–Chree modes), where neither formula is honest and frequency-dependent corrections are mandatory.

Elastic Waves in Solids

Where this lives today

The physical picture

The core equations

Limiting cases

Worked examples

Going further

Before this

Where this leads