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For homogeneous stable linear elasticity, the three positive eigenvalues of the acoustic tensor give three phase velocities for each wavefront normal. Plotting the corresponding three vectors $\mathbf{p}=\mathbf{n}/v$ gives the three sheets of the elastic slowness surface. Sheets can meet at degeneracies. The elastic-wave energy flux is normal to each regular sheet: differentiating its eigenvalue equation along a sheet tangent $d\mathbf{p}$ gives $c_{ijkl}{a}_i{a}_k{p}_{l}d{p}_j=0$.

These characteristic wave amplitudes split a one-dimensional elastic wave into forward and backward components. Their instantaneous net elastic-wave energy flux is ${\varphi }_{+}^2-{\varphi }_{-}^2$. For real-part harmonic phasors the mean flux is $(|{\varphi }_{+}{|}^2-|{\varphi }_{-}{|}^2)/2$. In a static graded medium with real positive seismic impedance, put $g=(\log Z{)}^{\prime }/2$; then ${\partial }_x{\varphi }_{\pm }\pm {\beta }^{-1}{\partial }_t{\varphi }_{\pm }=g{\varphi }_{\mp }$. Seismic impedance gradients couple the two directions while preserving net flux.

For a homogeneous lossless elastic solid, insert the plane wave $u_i=\mathrm{Re}\{U{a}_i{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}\}$ into the equations of linear elasticity. Put $\mathbf{k}=\kappa \mathbf{n}$, $|\mathbf{n}|=1$, and $v=\omega /\kappa$. The resulting eigenvalue problem isThe minor and major symmetries make this acoustic tensor a real symmetric matrix. For a stable material it is positive-definite, so its three positive eigenvalues give three real positive phase velocities and therefore three slownesses $s_a=1/v_a$. Coincident eigenvalues are possible, so there need not be three distinct speeds. For distinct eigenvalues the displacement polarization are orthogonal; in an anisotropic solid they are generally quasi-longitudinal and quasi-transverse rather than exactly longitudinal and transverse. Stability, or at least strong ellipticity in elasticity, is required for the assertion about positive real speeds.

The elastic slowness surface is the locus of the three vectors $\mathbf{p}=\mathbf{n}/v_a(\mathbf{n})=\mathbf{k}/\omega$. Its equation isFor each wavefront normal there are three radial sheets, which may meet at degeneracies. To obtain the energy flux directly, multiply the elastic equation by ${\dot{u}}_i$. Forthe symmetry of stress gives ${\partial }_{t}W={\partial }_j({\sigma }_{ij}{\dot{u}}_i)$. Thus the instantaneous elastic-wave energy flux is $F_j=-{\sigma }_{ij}{\dot{u}}_i$. With a real branch polarization $\mathbf{a}$, harmonic averaging givesNow differentiate $c_{ijkl}{p}_j{p}_l{a}_k=\rho a_i$ along a regular slowness surface sheet and contract with $a_i$. The terms containing $d\mathbf{a}$ cancel by symmetry and the original eigenvalue equation, leavingEvery tangent $d\mathbf{p}$ is consequently perpendicular to $⟨\mathbf{F}⟩$: the energy flux is normal to the slowness surface. At a degenerate point this statement is made on the limiting branches, rather than assuming a unique normal.

For completeness, equality of the mean kinetic and strain energies gives $⟨W⟩=\rho {\omega }^2|U{|}^2|\mathbf{a}{|}^2/2$. Differentiating the acoustic tensor equation in the wavevector variables similarly givesThus the group velocity is the elastic energy velocity. The homogeneous elastic dispersion relation is degree one in $\mathbf{k}$, so differentiating $\omega (\lambda \mathbf{k})=\lambda \omega (\mathbf{k})$ at $\lambda =1$ gives $\mathbf{p}\cdot \mathbf{g}=1$.

The elastic wave surface is the unit-time locus of these group velocities. From an impulsive point source, the propagating wavefronts at time $t$ are its sheets scaled to $\mathbf{x}=t\mathbf{g}$. A particular source may fail to excite a mode in some directions, and its amplitudes require the source polarization and geometrical spreading; the wave surface determines the arrival geometry. It does not specify all near-field motion between the radiative fronts. Geometrically it is the envelope of the planes $\mathbf{p}\cdot \mathbf{x}=1$: tangent variation gives $d\mathbf{p}\cdot \mathbf{x}=0$, so $\mathbf{x}$ is the appropriately normalized normal to the slowness surface.

In a symmetry-plane cross-section write $\mathbf{p}=r(\theta )\mathbf{n}$, $\mathbf{n}=(\cos \theta ,\sin \theta )$ and $\mathbf{t}=(-\sin \theta ,\cos \theta )$. The two envelope conditions giveThe numerator is the signed curvature numerator of the slowness surface cross-section. At a simple inflection point, $h=v+v^{\prime \prime }=0$ and $h^{\prime }\ne 0$. Then ${\mathbf{g}}^{\prime \prime }=h^{\prime }\mathbf{t}$ and ${\mathbf{g}}^{\prime \prime \prime }=h^{\prime \prime }\mathbf{t}+2h^{\prime }{\mathbf{t}}^{\prime }$ are linearly independent. These are precisely the local conditions for an ordinary cusp. The slowness curvature and wave-surface cusps relation explains why a smooth nonconvex slowness surface can generate a cusped wavefront and multiple ray arrivals.

The example below uses a positive orthotropic stiffness in normalized units: $\rho =1$, $C_{11}=C_{22}=C_{33}=4$, $C_{44}=C_{55}=1.2$, $C_{66}=1$, and all normal off-diagonal stiffnesses zero. In the symmetry plane $n_3=0$, its two in-plane acoustic tensor eigenvalues are $v_{\pm }^2=(5\pm \sqrt{5+4\cos 4\theta })/2$, and its out-of-plane shear speed is $\sqrt{1.2}$. The in-plane shear sheet is nonconvex and produces the marked cusps; the construction uses an actual stable elastic tensor.

At normal incidence the P wave and the two S wave polarizations decouple. In each uniform layer the relevant displacement satisfies $\rho u_{tt}=C{u}_{zz}$, with $C=\lambda +2\mu$ for a longitudinal wave and $C=\mu$ for a shear wave. Its speed is $v=\sqrt{C/\rho }$ and its seismic impedance is $Z=\rho v$. A welded interface preserves particle velocity and traction. For displacement or velocity amplitudes, incidence from medium 0 onto medium 1 therefore givesReflection reverses the displacement sign when the receiving seismic impedance is larger. The energy-flux fractions areThe impedance factor matters: a displacement transmission coefficient larger than one does not imply energy amplification. Equal impedance eliminates reflection even when the densities and speeds differ.

For a stack, every interface creates reflected waves which can return to all earlier interfaces. A seismic layer transfer matrix incorporates those multiple reflections exactly. Use harmonic time dependence $e^{-i\omega t}$, particle velocity $V$ and compressive traction $P=-{\sigma }_{zz}$ for a P-wave, or the negative relevant shear traction for an S-wave. Through layer $j$ of thickness $h_j$,This follows by solving $V_z=i\omega P/C_j$ and $P_z=i\omega {\rho }_{j}V$. Continuity means that the state passes unchanged across each welded interface. Put $M=M_N\cdots M_1=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, with $ad-bc=1$, and let the exterior impedances be $Z_L,Z_R$. A unit incident velocity gives top state $(1+r,Z_L(1-r))$ and bottom state $(t,Z_{R}t)$. Eliminating these states yieldsFor real moduli and no attenuation, conservation of elastic-wave energy flux gives $|r{|}^2+(Z_R/Z_L)|t{|}^2=1$. Reciprocal incidence gives equal energy transmission in the reverse direction; flux-normalized amplitudes express that reciprocity without unequal exterior impedances obscuring it.

The wave phase accumulated in each layer determines constructive or destructive interference. For one layer between identical exterior media, an integral number of half-wavelengths makes $M_j=\pm I$ and gives perfect transmission, regardless of the layer impedance. At quarter-wave phase thickness, a single matching layer with $Z_j=\sqrt{Z_L{Z}_R}$ also gives zero reflection: the layer transforms the terminal impedance to $Z_j^2/Z_R=Z_L$. Away from these matching conditions there are transmission peaks, reflection maxima and rapid frequency-dependent phase changes. In the time domain these are a direct transmitted pulse followed by delayed internal reverberations and reflected echoes. Long residence in a resonant stack produces narrow peaks and long ringing, without creating energy.

For a periodic two-layer stack, the half-trace of one cell isThis is the periodic seismic multilayer stop band criterion. When the absolute right side exceeds one, the cell eigenvalues are a reciprocal growing/decaying pair and a long finite stack has exponentially small transmission. The missing transmitted energy is reflected, not absorbed. Large impedance contrast and repeated quarter-wave layers strengthen such stop bands. In pass bands the cell modes propagate, and a finite stack superposes transmission resonances on that propagation. An irregular stack need not have equally spaced peaks: the individual travel times and impedance contrasts set its interference pattern.

At sufficiently low frequency a finite stack has $M\to I$, so its leading reflection is that of the two exterior media alone. For many thin repeated layers with wavelength much longer than a cell, their effective density is the volume-weighted density and their normal modulus is the harmonic mean, $C_{\mathrm{eff}}^{-1}=⟨C^{-1}⟩$. This follows by expanding each propagation matrix to first order and averaging its two coefficients. At high frequency, sharp interfaces still reflect: there is no general disappearance of reflection, only increasingly rapid interference fringes. Smooth impedance grading instead suppresses short-wavelength reflection when its variation scale is long compared with the wavelength.

These calculations provide a reproducible investigation: vary the layer travel times, impedances and number of cells; compute the matrix product; plot reflected/transmitted fluxes and check their sum. Adding attenuation makes the flux sum smaller than one and reduces reverberation peaks, while oblique incidence introduces Snell law for elastic waves and P-SV mode conversion absent from this one-dimensional setting. Impedance contrasts generate reflections, propagation phases organize their interference, and energy conservation separates reflection from genuine absorption.

Choose depth $z$ positive downwards and harmonic time dependence $e^{-i\omega t}$. At normal incidence, P waves and the two polarizations of S waves are independent scalar problems: a P wave uses the longitudinal modulus $\rho {\alpha }^2$, and an S wave uses the shear modulus $\rho {\beta }^2$. In a uniform layer with wave speed $c$ and seismic impedance $q=\rho c$, write velocity and compressive traction asAt a perfectly bonded interface the velocity and traction are continuous. For incident velocity wave amplitude one from medium 1 into medium 2, these two conditions are $1+r_v=t_v$ and $q_1(1-r_v)=q_2{t}_v$. ThusReflection is determined by seismic impedance, not by speed alone. A step to larger seismic impedance reverses the reflected velocity/displacement field waveform; the pressure reflection coefficient has the opposite sign. Its transmitted pressure coefficient is $t_p=2q_2/(q_1+q_2)$. Equal seismic impedance eliminates the reflection even when speed and wavelength change. At a free surface traction vanishes and velocity reflects with coefficient $+1$; at a rigid surface velocity reflects with coefficient $-1$.

The mean elastic-wave energy flux is $\frac{1}{2}\mathrm{Re}(P{V}^{\ast })=q(|A{|}^2-|B{|}^2)/2$ for positive real $q$. Therefore the reflected and transmitted energy fractions areA displacement field transmission coefficient greater than one is compatible with conservation of energy; its flux must be weighted by the receiving seismic impedance.

For many layers, use the seismic layer transfer matrix. A layer of thickness $h_j$ has wave phase thickness ${\delta }_j=\omega h_j/c_j$ andThis follows either by eliminating $A,B$ between the two faces or by integrating $V^{\prime }=i\omega P/(\rho c^2)$ and $P^{\prime }=i\omega \rho V$. For the complete stack put $M=M_N\cdots M_1=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. With upper and lower half-space impedances $q_a,q_b$, an outgoing lower wave satisfies $P_b=q_b{V}_b$. Solving this condition gives the input seismic impedanceHere $T$ is lower-face velocity divided by upper-face incident velocity, including the propagation wave phase. For lossless layers and real frequency, $|R{|}^2+(q_b/q_a)|T{|}^2=1$. This supplies a sensitive check on any spreadsheet or numerical implementation; it is not correct simply to multiply the individual interface transmissions and discard the elastic waves.

The physical reason is repeated wave interference. In one layer between media 0 and 2, let $r_{ij},t_{ij}$ denote the velocity coefficients above and $\delta =\omega h/c_1$. The first transmitted wave has wave amplitude $t_{01}{t}_{12}{e}^{i\delta }$. Every extra round trip supplies $r_{10}{r}_{12}{e}^{2i\delta }$. Summing the geometric series givesIn the time domain the successive arrivals are delayed by $2h/c_1$, producing reverberation. In the frequency domain the same delayed terms produce resonant maxima and minima. For a layer between identical half-spaces of impedance $q_0$,At $\delta =n\pi$ the layer transmits all the incident energy: the internally elastic waves cancel the direct reflection. Halfway between these resonances, a large impedance contrast gives strong reflection. For unequal exterior impedances, a quarter-wave layer with $q_1^2=q_a{q}_b$ transforms its lower load to $Z_{\mathrm{in}}=q_1^2/q_b=q_a$, again giving zero reflection at its design frequency. This matching is narrow-band.

Periodic alternation produces a periodic seismic multilayer stop band. For a cell of two layers with total thickness $d$, the cell matrix has reciprocal eigenvalues $e^{\pm iKd}$, soWhen the absolute value of this expression exceeds one, the infinite periodic medium has no propagating cell mode; a long finite stack has exponentially small transmission. The energy is reflected, not absorbed. Pass bands instead contain oscillatory transmission peaks, becoming more numerous as the stack length increases. Disorder removes the exact cell periodicity but not multiple scattering or frequency-selective interference.

For a concrete reproducible investigation, take $q_H/q_L=4$, equal one-way travel times $\tau$ in both layer types, exterior impedance $q_L$, and frequency coordinate $\nu =2\omega \tau /\pi$. At $\nu =1$ each layer is a quarter wavelength; one cell has diagonal matrix entries $-4,-1/4$. For $N$ cells,It is about $0.22145$ for one cell, $6.10\times 10^{-5}$ for four cells and $9.31\times 10^{-10}$ for eight cells. The infinite-cell stop band is $2\arcsin (0.8)/\pi <\nu <2-2\arcsin (0.8)/\pi$. The accompanying calculation also shows the first transmitted arrivals for a single high-impedance slab: their wave amplitudes are $0.64(0.36{)}^n$ at times $(2n+1)\tau$. Their sum is one, consistent with unit zero-frequency transmission. These calculations illustrate both the spectrum and the pulse train without assuming access to the course's spreadsheet.

Finally, as $\omega \to 0$ for a fixed finite stack, every $M_j\to I$, so the response approaches that of a direct interface between the exterior half-spaces. Equal exterior impedances give $R\to 0,T\to 1$. For wavelengths much longer than a repeating cell, expansion of $M_j$ to first order instead gives an effective medium with averaged mass density ${\rho }_{\mathrm{eff}}=\sum h_j{\rho }_j/d$ and harmonic-mean modulus $M_{\mathrm{eff}}^{-1}=\sum h_j/M_j^{\mathrm{mod}}/d$, where $M_j^{\mathrm{mod}}={\rho }_j{c}_j^2$. Its speed is $\sqrt{M_{\mathrm{eff}}/{\rho }_{\mathrm{eff}}}$, not generally the arithmetic mean of the layer speeds. Attenuation makes the energy balance an inequality, while for a fixed oblique horizontal slowness the scalar SH calculation uses $q_j={\mu }_j{p}_{\beta ,j}$ and ${\delta }_j=\omega p_{\beta ,j}{h}_j$. Oblique P-SV waves require coupled displacement field and traction matrices because of mode conversion at a planar elastic interface. Thick evanescent layers require scattering or impedance recursion rather than exponentially ill-conditioned transfer-matrix products.

Put $Z=\rho \beta =\sqrt{\rho \mu }>0$, the seismic impedance. In a uniform medium, define $w_{+}=v-\sigma /Z$ and $w_{-}=v+\sigma /Z$. The field equations giveThus $w_{+}=2F(x-\beta t)$ and $w_{-}=2G(x+\beta t)$, with arbitrary independent profiles $F,G$. Reconstruction givesThe two characteristic components therefore have $v_{+}=F$, ${\sigma }_{+}=-Z{v}_{+}$ and $v_{-}=G$, ${\sigma }_{-}=Z{v}_{-}$.

The kinetic plus elastic energy mass density is $e=\rho v^2/2+{\sigma }^2/(2\mu )$. Multiplying the equations by $v$ and $\sigma /\mu$ yields $e_t=(v\sigma {)}_x$. Hence the elastic-wave energy flux is $S=-v\sigma$. For a pure forward wave $S_{+}=Z{v}_{+}^2\ge 0$; for a pure backward wave $S_{-}=-Z{v}_{-}^2\le 0$. In the mixture the cross terms cancel, giving $S=Z(v_{+}^2-v_{-}^2)$. Strict sign holds for nonzero components.

At a sharp interface, continuity of velocity and traction gives, for unit incident velocity from the left,Thus $r_v=(Z_a-Z_b)/(Z_a+Z_b)$ and $t_v=2Z_a/(Z_a+Z_b)$. Normalizing the transmitted velocity by its elastic-wave energy flux gives $T_{ba}=\sqrt{Z_b/Z_a}{t}_v$, exactly the result above. Right incidence gives the reversed reflection sign and the same flux-normalized transmission. A stress reflection coefficient has the opposite sign to the velocity reflection coefficient; the convention matters.

Put $r=R_{aa}$ and $t=T_{ab}$. The incoming-to-outgoing matrix is $S_0=\left(\begin{array}{cc}r& t\\ t& -r\end{array}\right)$, with $r^2+t^2=1$. Therefore $S_0^{†}{S}_0=I$, proving equality of incoming and outgoing time-averaged flux for arbitrary coherent incident wave amplitudes, not just incidence from one side.

For all real frequencies the harmonic equations instead giveTheir right sides are equal; the diagonal terms $\pm i\omega /\beta$ have zero real contribution. ConsequentlyThe mean flux for real-part phasors is $(|{\varphi }_{+}{|}^2-|{\varphi }_{-}{|}^2)/2$. Evaluating it at the two endpoints proves exactly that total outgoing flux equals total incoming flux at every real frequency. This relies on real lossless moduli and mass density; absorption would change the identity.

In the bilinear identity choose field $a$ to be the complex conjugate of the field at frequency $\omega$, equivalently its $-\omega$ partner, and field $b$ to be the field itself. The squared frequencies are equal, so the mass density terms cancel:Multiply by $i\omega$ and use ${\hat{v}}_i=-i\omega {\hat{u}}_i$, ${\hat{v}}_i^{\ast }=i\omega {\hat{u}}_i^{\ast }$ to obtainFor the real fields defined in this question as the sum of positive and negative frequency components, time averaging cancels the $e^{\pm 2i\omega t}$ terms. Their mean body-force work mass density is exactly ${\hat{f}}_i^{\ast }{\hat{v}}_i+{\hat{v}}_i^{\ast }{\hat{f}}_i$, and their outward mean elastic-wave energy flux is $-t_i^{\ast }{\hat{v}}_i-{\hat{v}}_i^{\ast }{t}_i$. Thus mean body-force power equals mean outward boundary power. Mean stored energy does not accumulate in a periodic steady field.

There is no extra factor $1/2$ here: the specified real field is $\hat{f}{e}^{-i\omega t}+{\hat{f}}^{\ast }{e}^{i\omega t}$, not $\mathrm{Re}(\hat{f}{e}^{-i\omega t})$. For the latter phasor convention, both averages instead have the usual factor $1/2$. At $\omega =0$ velocity and both powers vanish.

Define $K=Z_{+}-Z_{-}$. Its antisymmetric off-diagonal entries cancel, leavingFrom $\mathbf{v}={\mathbf{v}}^{+}+{\mathbf{v}}^{-}$ and $\mathbf{t}=-Z_{+}{\mathbf{v}}^{+}-Z_{-}{\mathbf{v}}^{-}$, solve explicitly for the P-SV characteristic reconstruction:The prescribed strictly positive $p,s$ make this decomposition unique; a grazing or evanescent branch is outside these assumptions.

In the mean flux, the two mixed directional terms arebecause $Z_{-}=-Z_{+}^{†}$ makes the second scalar minus the complex conjugate of the first. The remaining self terms giveFinally use either directional velocity-potential matrix. Its weighted quadratic form satisfiesIndeed the two P-SV cross terms have coefficients $-kps$ and $+kps$ for the forward field, with the signs reversed for the backward field, so they cancel. Substitution provesThus net normal elastic-wave energy flux is the difference of the independently flux-weighted P and SV directional powers, even for coherent mixtures.

Average the instantaneous elastic-wave energy flux over one temporal period at fixed position. For complex amplitudes,Write the evanescent P displacement amplitude asFor an isotropic linear elastic material with Lamé parameters $\lambda ,\mu$, its stress and velocity amplitudes areApart from a real common factor,which is purely imaginary. Its real part vanishes, and henceThe converted P-wave is a reactive near-boundary field and transports no time-averaged acoustic energy in the normal direction.

For a one-dimensional shear elastic wave, the seismic impedance is the magnitude of the ratio of stress to particle velocity in a progressive wave. With the positive coordinate aligned with elastic wave propagation and tension-positive stress, $\sigma =-Zv$ for the forward wave and $\sigma =Zv$ for the backward wave. Its instantaneous elastic-wave energy flux is $-v\sigma$.