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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.

At a welded interface, elastic-interface boundary conditions require continuous SH displacement and shear traction. Incident and reflected vertical slownesses have signs $q_1,-q_1$, givingThus the local displacement coefficient isIt multiplies the leading waveform discontinuity as well as a harmonic displacement. At the interface the incident amplitude from SH ray-tube amplitude transport is $C/\sqrt{r_1}$, where $r_1=h/({\beta }_1{q}_1)$ is the incident path length and $C$ is the source normalization from the previous part.

The transmitted geometrical spreading is not simply its total path length. Differentiate the ray position with respect to $p$ at fixed travel time and project onto the unit direction transverse to the transmitted ray, ${\mathbf{m}}_2=({\beta }_2{q}_2,-{\beta }_{2}p)$. Since $d{x}_I/dp=h/({\beta }_1^2{q}_1^3)$, the transverse width per unit $p$ isThe derivative of $t_1$ contributes only in the ray direction and drops out of this projection. At entry,Within uniform medium 2 the leading amplitude varies as $J_2^{-1/2}$, so the required refraction of a cylindrical SH wavefront amplitude isThe local transmission coefficient and ray-tube spreading together incorporate both impedance contrast and refraction. Indeed the normal energy transmission fraction is ${\mu }_2{q}_2|T_{\mathrm{SH}}{|}^2/({\mu }_1{q}_1)$, while the change of transverse width supplies the refraction factor in the conserved tube flux. If the media coincide, $q_1=q_2$ and $T_{\mathrm{SH}}=1$, reducing this expression to $C/\sqrt{{\beta }_{1}t}$ as required. The formula holds for the stated propagating slownesses, without a critical or head-wave contribution.

The scattering interpretation assumes $0<ϵ<U$, so $k=\sqrt{ϵ}$ is real and $K=\sqrt{U-ϵ}>0$. The incoming amplitudes are $a$ from the left and $b$ from the right; the outgoing amplitudes are $c,d$. The conserved probability current iswhich gives $j=\hslash k(|a{|}^2-|c{|}^2)/m$ on the left and $j=\hslash k(|d{|}^2-|b{|}^2)/m$ on the right. Equality proves $|c{|}^2+|d{|}^2=|a{|}^2+|b{|}^2$. Equal exterior wavenumbers make the flux weights equal.

Continuity of the wavefunction and its derivative at the finite potential steps yields the four relations, with $C=\cosh KL$ and $S=\sinh KL$,Set $\lambda =K/(ik)$. Substituting $e=a+c$, $f=(a-c)/\lambda$ and subtracting the last two relations after dividing the fourth by $ik$ giveswhere $D=C-(\lambda +{\lambda }^{-1})S/2$. HenceReflection about the barrier midpoint interchanges the left and right incidence problems and, with the chosen phase origins, exchanges $a\leftrightarrow b$ and $c\leftrightarrow d$. ThereforeFor incidence from the left alone, the transmission coefficient is the transmitted-to-incident probability flux ratio, $T=|d/a{|}^2=|D{|}^{-2}$. Since $\lambda =-iK/k$, direct calculation givesFor fixed energy inside the barrier range and $KL\gg 1$, ${\sinh }^{2}KL\sim e^{2KL}/4$, soThis is the thick-barrier quantum tunnelling limit, not a uniform approximation as energy approaches the barrier top while $KL$ remains small.

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.

The one-dimensional probability current isAt a finite jump of the potential, both $\psi$ and ${\psi }^{\prime }$ are continuous. A jump in $\psi$ would create a delta derivative in the stationary Schrödinger equation, and integrating that equation across an increasingly small interval then shows continuity of ${\psi }^{\prime }$. Accordingly impose continuity of both quantities at zero and at $a$.

For an incident wave of positive energy, define $k=\sqrt{2mE}/\hslash$. With no wave incident from the right, the left region hasThe interference terms cancel from the imaginary part in the current formula. If $E>V_0$, let $k_0=\sqrt{2m(E-V_0)}/\hslash$. In the right region,For $0<E<V_0$, let ${\kappa }_0=\sqrt{2m(V_0-E)}/\hslash$. Boundedness at positive infinity excludes the growing exponential, soThe decaying solution has zero current even if its amplitude is nonzero. These exterior forms do not require $E$ to lie on a particular side of $V_1$: in the finite middle layer the solution is oscillatory for $E>V_1$, exponential for $E<V_1$, and linear at $E=V_1$; matching determines the amplitudes.

Define the reflection coefficient as reflected current magnitude divided by incident current, and the transmission coefficient as outward right current divided by incident current. ThenThe denominator is $j_{\mathrm{incident}}=\hslash k|A{|}^2/m$. For a real potential, differentiation givesby the stationary Schrödinger equation. Continuity at each interface preserves this conserved current, so $j_{\mathrm{left}}=j_{\mathrm{right}}$. Dividing by incident current yieldsThere can be penetration into the barrier without transmission to infinity, because the entire right half-line remains classically forbidden.

Put $p=\sqrt{2mE}/\hslash$ and $\gamma =mk/{\hslash }^2$. Integrating the Schrödinger equation across the delta potential gives continuity of $\psi$ andFor incidence from the left, write $\psi =e^{ipx}+r{e}^{-ipx}$ on $x<0$ and $\psi =t{e}^{ipx}$ on $x>0$. Matching gives $t=1+r$ and $ip(t-1+r)=2\gamma t$, henceThe incident and transmitted probability currents have the same velocity factor, so the reflection coefficient and transmission coefficient areFor $k<0$, an even bound state has $\psi =A{e}^{-\kappa |x|}$, $\kappa =\sqrt{-2mE}/\hslash$. The derivative jump now gives $-2\kappa A=2\gamma A$, so $\kappa =-\gamma >0$ andThere is exactly one positive value of $\kappa$, hence exactly one even bound state. Its normalized amplitude is $A=\sqrt{\kappa }$. An odd decaying solution would have to vanish at zero and is therefore trivial.

In the shrinking finite square well, its integrated potential is $k=-2Ul$, not $-Ul$: the full width is $2l$. With $Ul$ fixed and $l\to 0$, $0\le (\alpha l{)}^2\le 2mU{l}^2/{\hslash }^2\to 0$. The matching condition then bounds $\beta$ by $(2mUl/{\hslash }^2)(1+o(1))$, so $E$ remains bounded. Consequently $El\to 0$, and the previous matching condition givesUsing $E=-{\hslash }^2{\beta }^2/(2m)$ recovers the same bound-state energy. All other even levels disappear as the width shrinks.

Linearized mass conservation, momentum balance and the isentropic pressure-density relation are ${\rho }_t^{\prime }+{\rho }_0\nabla \cdot \mathbf{u}=0$, ${\rho }_0{\mathbf{u}}_t=-\nabla \tilde{p}$, and $\tilde{p}=c_0^2{\rho }^{\prime }$. Differentiate mass conservation in time and take the divergence of momentum balance to eliminate velocity. This gives the acoustic pressure wave equationFor the convention $\tilde{p}=\mathrm{Re}{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, momentum balance gives the complex velocity amplitude $\mathbf{k}/({\rho }_0\omega )$, with $\omega =c_0|\mathbf{k}|$. Thus the incident velocity field is $\mathrm{Re}[{\mathbf{k}}_I{e}^{i({\mathbf{k}}_I\cdot \mathbf{r}-\omega t)}/({\rho }_1\omega )]$ in the lower gas.

Let $a=k_I\sin \theta$, $l_1=k_I\cos \theta >0$, $\omega =c_1{k}_I$, and $l_2=\sqrt{(\omega /c_2{)}^2-a^2}$ with nonnegative real or imaginary part. This chooses an outgoing propagating wave or an evanescent wave decaying as $y\to +\infty$. Conservation of tangential wavevector is the acoustic Snell law for elastic and acoustic waves. Write the complex pressures asInviscid kinematic matching requires equal normal velocities at the plate, not equal tangential velocities:Define the normal acoustic impedance ratio $\delta ={\rho }_2{l}_1/({\rho }_1{l}_2)$, so $T=\delta (1-R)$. For a propagating transmitted wave this is ${\rho }_2{c}_2\cos \theta /({\rho }_1{c}_1\cos {\theta }_2)$. The dynamic condition giveswhich is exactly the printed dimensionless parameter. Solving yields the pressure reflection coefficient and transmission coefficientAt $l_2=0$ use the preceding velocity equations, or take their limit, rather than substituting an infinite $\delta$. For real $l_2>0$, the reflected energy fraction is $|R{|}^2$, the transmitted fraction is $|T{|}^2/\delta$, and these sum to one. An evanescent transmitted wave carries no mean normal energy flux.

When the gases match, $\delta =1$, so $R=-i\beta /(2-i\beta )$ and $T=2/(2-i\beta )$. For a nongrazing incident wave of nonzero frequency, perfect transmission means $R=0$, equivalent toThe plate's inertial and bending terms cancel: the incident frequency and tangential wavevector coincide with its free flexural wave dispersion relation. Then $T=1$, including its phase. With positive $m,B$, this condition is possible only at angles and frequencies satisfying that relation; for normal incidence at nonzero frequency it cannot occur unless the mass is zero.

Normalize the solution by its incoming amplitude at $+\infty$. The transmission coefficient and reflection coefficient areFor two solutions of the same equation in Sturm-Liouville form, the modified Wronskianis independent of $x$. Take ${\psi }_1={\varphi }_k$ and ${\psi }_2=\overline{{\varphi }_k}$. At $-\infty$ this gives $W=-2ik$, while at $+\infty$ it givesTherefore $|a{|}^2-|b{|}^2=1$, and division by $|a{|}^2$ yields conservation of scattering flux:

Place the junction at $x=0$ and use the complex amplitude convention $e^{-i\omega t}$. With incident displacement wave amplitude $A_i$, writeHere $r$ and $t$ are displacement wave reflection and transmission coefficients, and $k_j=\omega /c_j$. The junction is assumed massless, with no externally applied transverse force, and both strings have the common equilibrium tension $T$. Continuity of displacement and transverse force givesSince $T{k}_j=\omega Z_j$, these become $1+r=t$ and $Z_L(1-r)=Z_{R}t$. Solving,A larger right-hand mechanical impedance produces a reflected displacement with reversed sign, as expected in the fixed-end limit $Z_R\to \infty$. Equal mechanical impedances give $r=0$ and $t=1$.