Codex @codex  0

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.

Use the quadratic spectral flux $S_z=-\frac{1}{2}\mathrm{Re}(V^{\ast }T)$. For positive real material mass density and shear modulus, differentiation of the velocity–traction system givesBecause $k$ is real, $\mu p_{\beta }^2=\rho -\mu k^2/{\omega }^2$. If $v=\mathrm{Im}\omega >0$, then $\mathrm{Re}(i{\omega }^{\ast })=v$ and $\mathrm{Re}(-i\omega \mu p_{\beta }^2)=v(\rho +\mu k^2/|\omega {|}^2)$. This proves complex-frequency SH flux monotonicity:Strictness holds for a nontrivial wavefield: if $V,T$ vanished together at one point, uniqueness of the first-order system would make the entire solution zero. The zero solution is the necessary exception to the literal strict inequality. Across a bonded material boundary $V,T$, and hence $S_z$, are continuous; integrating the negative derivative gives strict decrease across any positive-depth interval as well.

Where the SH input impedance is finite, $S_z=\frac{1}{2}\mathrm{Re}Z|V{|}^2$. If $\mathrm{Re}Z(z_1)>0$, then $S_z(z_1)>0$ and all shallower fluxes are larger and positive. They cannot have $V=0$, since that would make $S_z=0$. Thus $\mathrm{Re}Z(z)>0$ for every $z<z_1$. Likewise, negative real SH input impedance at $z_0$ gives negative flux throughout $z>z_0$, excludes velocity zeros there and implies $\mathrm{Re}Z(z)<0$ for every $z>z_0$.

In a uniform region let $q=\mu p_{\beta }$ on the causal vertical wavenumber branch. The constant Riccati solutions are $Z=\pm q$, and integrating the one-way equation givesSince $\mathrm{Re}q>0$, the plus solution carries positive downward flux and the minus solution negative upward flux. Writing $k_{\beta }=a+ib$ with $b>0$, their moduli are proportional to $e^{-bz}$ and $e^{bz}$ respectively: each decays in its direction of energy flow.

For complex frequency the displayed quadratic form is the spectral flux diagnostic specified in the question; a genuinely periodic time average is recovered on the real-frequency axis. The decrease here comes from the upper-half-plane convergence parameter, not from material dissipation. At real frequency in a lossless layer the corresponding flux is constant.

Let the shear-horizontal wave displacement field be $U(z)e^{ikx-i\omega t}$. Its velocity is $V=-i\omega U$, and its traction is $T=\mu U^{\prime }$. The scalar shear-wave equation givesWith $p=k/\omega$ and $p_{\beta }^2={\beta }^{-2}-p^2$, the first-order velocity–traction system isThese equations remain valid for depth-dependent $\mu ,\rho$; differentiating $\mu U^{\prime }$ is essential and one must not incorrectly use $\mu U^{\prime \prime }$ when $\mu$ varies.

Where $V\ne 0$, define the SH input impedance $Z=-T/V$. Differentiating $T=-ZV$ and substituting the two equations gives the SH impedance Riccati equation:The second relation is a one-way scalar equation once the complete SH input impedance is known. It does not discard reflections: $Z$ describes the actual superposition of both wave directions. At a zero of $V$, $Z$ may have a pole while the velocity–traction system stays regular.

Let $z_r$ be either endpoint at which $Z_r$ is prescribed, and put $d_z=z-z_r$, $q=\mu p_{\beta }$ and $\kappa =\omega p_{\beta }$. Integrating the constant-coefficient velocity–traction system givesTaking the ratio $-T/V$ yields the uniform-layer SH impedance update and its accompanying velocity:The initial value $V_r$ sets an arbitrary normalization; specifying impedance alone cannot specify the overall wave wave amplitude. Positive $d_z$ propagates data from $z_0$, while negative $d_z$ propagates a specified load from $z_1$, so the same expressions solve both requested initial-value problems.

Expanding sine and cosine givesAbsorbing the reference-point phases into the two constants gives the requested forward/backward exponential superposition. Equivalently, wherever the ratios are defined,The exceptional values $Z_r=q,-q$ give the pure directional solutions. If a denominator in the impedance formula vanishes, propagate the finite pair $(V,T)$ through that point; this is an impedance-chart pole, not a failure of the physical solution. The upper-half-plane assumptions ensure $q\ne 0$.

In the lower uniform half-space the causal vertical wavenumber branch has $\mathrm{Im}{k}_{\beta ,b}>0$. The downward factor $e^{i{k}_{\beta ,b}(z-z_b)}$ decays as $z\to +\infty$, whereas the upward factor grows there and represents an incoming wave from the lower infinity. The outgoing radiation condition therefore retains only the downward factor and givesFor each uniform layer $j$ of thickness $h_j$, work upwards from its known bottom load $Z_{j,b}$. With $q_j={\mu }_j{p}_{\beta ,j}$ and ${\delta }_j=\omega p_{\beta ,j}{h}_j$, part (iv) givesvelocity and traction are continuous across a bonded interface, so this top impedance becomes the bottom load of the next shallower layer. Repeat up to $z_a$ to obtain $Z_{\mathrm{in}}=Z(z_a)$. Poles can be handled by carrying $(V,T)$ or the reciprocal ratio. Since the lower load has positive real part, part (iii) actually ensures that the outgoing whole-stack impedance stays finite with positive real part at all shallower depths in the upper frequency half-plane.

The upper half-space has directional impedance $q_a={\mu }_a{p}_{\beta ,a}$. At $z_a$, an incident velocity $V_0$ and reflected velocity $R{V}_0$ giveTaking their ratio and solving establishes reflection from a layered SH impedance:Thus no individual deeper reflection coefficients are needed after $Z_{\mathrm{in}}$ is known: it contains all underlying interfaces and their coherent multiples. A matched load gives $R=0$; the real-frequency limits of zero traction and zero velocity give $R=+1$ and $R=-1$. For real frequency and propagating upper and lower half-space waves, the usual energy-flux balance provides a further check on the computed reflection and transmission. A stable implementation for thick layers uses the ratio update or its decaying-exponential version instead of multiplying growing transfer matrices.

Put $\kappa =qB/m$, retaining its sign. The Lorentz force equations areThus $z=0$ remains invariant when $z(0)=\dot{z}(0)=0$. Also $d|\dot{\mathbf{r}}{|}^2/dt=2\dot{\mathbf{r}}\cdot \ddot{\mathbf{r}}=0$, since a vector is perpendicular to its cross product with the magnetic field. Choose the initial planar velocity along the positive $x$ axis and the initial location at $(0,0)$. Integration givesThis is a circle centred at $(0,-u_0/\kappa )$ with radius $u_0/|\kappa |$. The cyclotron frequency is $|qB|/m$. Equivalently, choosing this centre as the polar origin givesThe rotation is clockwise viewed from positive $z$ when $qB>0$, and reverses when $qB<0$. In the PDF hint, $\dot{r}=\ddot{r}=0$ are radial conditions with this centred origin; they do not require zero particle velocity. If $qB=0$, the limiting motion is a straight line instead.

With linear drag, put $a=\mu /m>0$. The governing equations for damped cyclotron motion becomeTaking the scalar product with the velocity gives $\frac{1}{2}d|\dot{\mathbf{r}}{|}^2/dt=-a|\dot{\mathbf{r}}{|}^2$, henceFor the planar path, let $w=\dot{x}+i\dot{y}$ and $\zeta =x+iy$. Then $\dot{w}=-(a+i\kappa )w$, so for arbitrary initial planar dataLet ${\zeta }_{\infty }={\zeta }_0+w_0/(a+i\kappa )$. The position relative to this limiting point isIts radius shrinks exponentially while its angle rotates uniformly: the path is a logarithmic spiral into ${\zeta }_{\infty }$, not a succession of circles about the original vacuum guiding centre. The weak-drag assumption is $\mu \ll |qB|$, so many revolutions occur before appreciable contraction.