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A photon of wavelength $\lambda$ has energy $E=h\nu =hc/\lambda$ and momentum $\mathbf{p}=(h/\lambda )\hat{\mathbf{n}}$ along its propagation direction.

The specified Compton scattering formula requires the electron to be initially at rest in the frame where the wavelengths and angle are measured. This initial-rest assumption is implicit in the printed request; it is not valid for arbitrary initial electron momentum. Use four-momentum with Minkowski metric signature $(+---)$: initially $P=(mc,0)$, while $K=(E/c,(E/c)\hat{\mathbf{n}})$ and $K^{\prime }=(E^{\prime }/c,(E^{\prime }/c){\hat{\mathbf{n}}}^{\prime })$ are null. Conservation gives $P^{\prime }=P+K-K^{\prime }$, and the outgoing electron satisfies the same mass-shell condition $P^{\prime 2}=P^2=m^2{c}^2$. HenceAfter dividing by $E{E}^{\prime }$ and substituting the photon energy relation,The scale $h/(mc)$ is the electron Compton wavelength. For a moving initial electron, the earlier invariant identity remains valid, but $P\cdot K$ includes its initial velocity and the wavelength shift is different.

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.

Linearizing the ideal magnetohydrodynamic induction equation about a static magnetic field and using $u^{\prime }={\partial }_t\xi$ gives the fluid Lagrangian displacement-generated magnetic field perturbation $b=\nabla \times (\xi \times B)$. The two solenoidal vector field conditions reduce the curl identity toIts divergence vanishes because it is a curl. Directly, the two products of first derivatives in ${\partial }_i{b}_i$ cancel after exchanging their indices, while the remaining terms differentiate $\nabla \cdot \xi$ or $\nabla \cdot B$, both zero. Similarly the integrated linear continuity equation gives ${\rho }^{\prime }=-\nabla \cdot (\rho \xi )$, so $\boxed{{\rho }^{\prime }=-\xi \cdot \nabla \rho }$.

Let $\Pi =p+B^2/(2{\mu }_0)$ be the magnetohydrodynamic total pressure. Magnetostatic equilibrium and the momentum equation in the linearized ideal magnetohydrodynamic equations areThere is no perturbed gravitational potential here because the field is fixed. Multiply by ${\xi }_i^{\ast }$ and integrate. The pressure term vanishes by the assumed boundary cancellation and $\nabla \cdot {\xi }^{\ast }=0$. Thuswhere $\mathcal{M}=-\int {\xi }_i^{\ast }(B_j{\partial }_j{b}_i+b_j{\partial }_j{B}_i)dV$.

Integrate the first magnetic term along $B$, using $\nabla \cdot B=0$, then insert the expression for $b$. This gives the positive term $\int |(B\cdot \nabla )\xi {|}^{2}dV$ and two mixed derivative terms. For clarity, the troublesome term transforms asThe first term on the right cancels the other mixed term. Combining the remaining derivative with the term containing ${\partial }_j{B}_k{\partial }_k{B}_i$ givesDifferentiating the equilibrium relation now combines the gradient of mass density and magnetostatic equilibrium terms:We obtain the incompressible magnetic displacement energy identityThe two Hessian matrices are real symmetric, making their Hermitian form evaluations real. The last term is a squared norm, and $I>0$ for a nonzero displacement in fluid with positive mass density. Thus $\boxed{{\sigma }^2\text{ is real}}$. It may be negative, in which case one temporal branch has exponential growth; real squared frequency is not itself a proof of stability.

For the atmosphere with a horizontal magnetic field, $\Phi =gz$ has zero Hessian matrix and the equilibrium equation isConsequently the interchange stability of an incompressible magnetized atmosphere is governed by the numeratorFor ${\xi }_z=\sin ky$, the magnetic tension term is zero: these fluid Lagrangian displacements interchange neighboring magnetic field lines without varying along them. A mass density increasing with height gives negative energy and instability; decreasing mass density gives positive energy for these vertical interchanges. If ${\rho }_z\le 0$ everywhere, the displayed form is nonnegative for every admitted incompressible displacement. If the gradient changes sign, suitably localized divergence-free interchange trials detect an unstable inverted layer; For example, take ${\xi }_z=f(z)e^{iky}$ and ${\xi }_y=i{f}^{\prime }(z)e^{iky}/k$, with $f$ supported in that layer. Their divergence is zero, and neither component varies along the magnetic field.

For ${\xi }_z=\sin kx$, magnetic field lines are bent. Averaging over a full horizontal wavelength gives the trial Rayleigh quotientunder the stipulated admissible boundary/integrability conditions. It is a trial quotient, not a claim that this displacement is an exact eigenfunction for every profile. mass density decreasing upward is stable here too. For inverted mass density, magnetic tension can stabilize sufficiently short waves along the magnetic field, but does not remove the transverse interchange instability. Locally the two contributions have the familiar scale $-g{\rho }_z/\rho +k^2{v}_A^2$, with the Alfvén speed $v_A=B/\sqrt{{\mu }_0\rho }$.

Measure $z$ from the rigid tube wall towards the air core, so $0<z<h$. To leading order in $h_0/a$ and small interface slope, the curvature entering the Young–Laplace equation isThe air pressure is spatially uniform at the order retained. Thus the linear liquid pressure perturbation associated with thickness amplitude $H$ is $p_1=-{\sigma }_0(a^{-2}-k^2)H$. The circumferential curvature favors growth at long wavelength; the curvature opposes it at short wavelength.

The lubrication theory momentum equation is $\mu u_{zz}=p_x$. With the wall no-slip boundary condition $u(0)=0$ and zero surface shear stress, $u_z(h)=0$, its solution and flux areLeading thin-film mass conservation is $h_t+q_x=0$. For a mode $e^{\beta t+ikx}$, this gives the capillary instability of an annular liquid liningTherefore the unstable band is $\boxed{0<k^2<a^{-2}}$. Maximization with respect to $k^2$ gives $k_{\ast }^2=1/(2a^2)$ andThese curvature coefficients use $a$ instead of $a-h_0$ consistently at leading order in the thin-lining approximation.

At the threshold $P_{\ast }=3+2\sqrt{2}$, the determinant quadratic has a double zero atConsequently the scaled disturbance length in $\cos (s/L)$ isThe physical length is ${\ell }_u{L}_{\mathrm{crit}}=\sqrt{{\ell }_u{\ell }_v}$ at threshold; its full spatial wavelength is $2\pi$ times this length, not this length itself. For much shorter lengths, the large negative diffusion terms smooth both components. For much longer lengths, diffusion contributes little and the stable homogeneous reaction feedback, especially the inhibitor response, suppresses disturbances. The selected length lies between those two stabilization mechanisms.

Use the observed wavelength in $d\lambda /\lambda =dv/c$. Since ${\lambda }_{\mathrm{obs}}=(1+z){\lambda }_{\mathrm{rest}}$,For an Einstein-de Sitter universe at $z=3$, $H(z)=H_0(1+z{)}^{3/2}=8H_0$, and the redshift drift is $\dot{z}=4H_0-8H_0=-4H_0$. With the rounded numerical values supplied in the paper,Therefore the spectroscopic velocity drift accumulated over ten years isThe negative sign means a shift toward shorter observed wavelength. Omitting the factor $1+z$ would overestimate the apparent velocity shift by four.

The Lyman-alpha forest supplies many relatively narrow absorption features along each bright distant quasar sightline. These arise in intervening intergalactic hydrogen rather than in the quasar's rapidly varying emission region. A repeat high-resolution spectrum can be fitted against the same absorption template, and the information from many lines and many sightlines can be combined to estimate a common spectroscopic velocity drift. Diffuse intergalactic absorbers also generally have smaller peculiar accelerations than gas close to galactic nuclei, although their peculiar accelerations and changing line profiles still require control.

The preceding calculation requires a relative wavelength displacement $|\Delta \lambda |/\lambda \simeq 6\times 10^{-10}$ over a decade. This is much smaller than one instrumental resolution element. Accurate line centroids can nevertheless be measured to a fraction of a resolution element when the signal-to-noise ratio and template information are sufficient. The real obstacles are long-term wavelength calibration, instrumental drift, sufficient photon counts and corrections for observer motion, along with astrophysical variability. Statistical averaging reduces independent errors; a common calibration error remains common to all lines. The Sandage-Loeb test therefore needs large collecting area, stable high-resolution spectroscopy and a long time baseline. The original absorption-forest strategy is described in Loeb's 1998 proposal.

The measurement is feasible in principle but demands exceptional long-term stability. It directly probes the change of redshift with observer time, giving $H_0(1+z)-H(z)$ rather than a distance inferred from a standard candle. Measuring its sign and redshift dependence would distinguish the expansion histories in part (ii).

Use geometrized units, and denote the beacon's proper time by $s$ to distinguish it from the short emitted duration $\tau$. Take $s=0$ at release. The period $M$, the duration $\tau$, and the wavelength $\lambda$ are measured by the beacon's own clock. Its radial timelike geodesic has conserved Killing energy per unit rest mass $E=fdt/ds$, where $f=1-2M/r$. Initially $dr/ds=0$ at $r=8M$, so $E=\sqrt{f(8M)}=\sqrt{3}/2$. Normalization of the four-velocity givesFor the Schwarzschild radial fall from rest at a finite radius, introduceIndeed $ds/d\eta =8M(1+\cos \eta )$ and $dr/d\eta =-4M\sin \eta$ reproduce the inward radial equation. At the Schwarzschild event horizon, $r=2M$ and $\cos {\eta }_H=-1/2$. HenceThis is the finite falling-clock time to the event horizon. By contrast, Schwarzschild time obeys $dt/ds=E/f$ and diverges at crossing. Near the horizon $r-2M\simeq E(s_H-s)$, so $dt/ds\simeq 2M/(s_H-s)$ and $t\simeq -2M\log ((s_H-s)/M)+\text{constant}$.

Every outward optical pulse emitted while $r>2M$ can reach a distant observer, and no pulse emitted after entry into the future black-hole region can do so. The infinite coordinate crossing time does not produce infinite emissions, because the pulse clock runs in proper time. If there is a pulse at release, the escape times are $s_n=nM$ for $n=0,\dots ,23$. Thus 24 pulses are received, counting the pulse at release; 23 are emitted subsequently during the fall. More generally the PDF does not specify the phase of the pulse clock. For a first emission at $s=\delta$ with $0\le \delta <M$, the finite pulse count for a falling Schwarzschild beacon isThe ceiling handles a pulse scheduled exactly at crossing correctly: it cannot escape. This count concerns pulses associated with the fall, not any pulses emitted before release, and assumes an ideal observer able to detect arbitrarily redshifted signals.

To find the stretching of an optical pulse, use the Schwarzschild tortoise coordinate $r_{\ast }=r+2M\log (r/(2M)-1)$. An outward radial null geodesic has constant Schwarzschild retarded null coordinate $u=t-r_{\ast }$. At an observer's fixed large radius $R$, its arrival time is $t_R=u+r_{\ast }(R)$; therefore only the varying emission value of $u$ contributes to the interval between arriving wavefronts. Differentiating along the falling source givesThe outgoing pulse redshift from a radial Schwarzschild infaller includes both gravitational redshift and the emitter's inward Doppler shift. A short pulse emitted at radius $r_e$ has, to leading order in its duration,For a finite receiver radius, replace $\mathcal{D}$ by $\sqrt{f(R)}\mathcal{D}$ to measure the receiver's proper time and frequency; this correction is negligible for $R\gg M$. The inverse scaling of frequency follows because the number of carrier cycles is unchanged when the pulse is stretched. Here $\nu =1/\lambda$ is ordinary frequency, not angular frequency, in units with light speed one.

Under the release-phase convention, the last emission is at $s_e=23M$. Solve $8({\eta }_e+\sin {\eta }_e)=23$ to obtain ${\eta }_e\simeq 1.94374184$, $r_e\simeq 2.54255983M$, and $\mathcal{D}(r_e)\simeq 7.49122492$. Because $\tau \ll M$ and this pulse begins $0.683364M$ before crossing, the local short-pulse approximation applies. Thus its approximate observed duration and frequency areWithout fixing that phase, these numerical coefficients are not unique: put $s_e=\delta +(N(\delta )-1)M$, find $r_e$ from the parameterization, and use the boxed general formulas. For a last pulse close to crossing, with $\Delta =s_H-s_e\ll M$ but $\tau \ll \Delta$, the horizon redshift of a Schwarzschild infaller gives the useful estimatesIf a phase instead places the pulse within one pulse-duration of crossing, $\Delta \lesssim \tau$, a single constant stretching factor is inappropriate. The received extent of a complete escaping pulse is exactly $u(s_e+\tau )-u(s_e)$; a pulse straddling crossing has an escaping tail extending to arbitrarily late arrival times, whose instantaneous frequency tends to zero. The finite count remains valid, while its last-pulse duration then has no finite constant multiple of $\tau$. This identifies precisely what the unspecified pulse phase can change.

Use geometrized units and assume $M>0$, $|Q|\le M$, so the Reissner-Nordstrom spacetime contains a black hole. Its outer event horizon is $r_{+}=M+\sqrt{M^2-Q^2}$. The classical absorption cross-section is obtained in geometrical optics, by counting incident null geodesics captured from infinity. Spherical symmetry lets us use the equatorial plane. Write $f(r)=1-2M/r+Q^2/r^2$, and let a dot denote differentiation with respect to an affine parameter. The conserved Killing energy and angular momentum are $E=f\dot{t}$ and $L=r^2\dot{\varphi }$. Null normalization giveswhere the impact parameter is $b=L/E$ and $W=f/r^2$. An incoming ray turns around if the angular-momentum barrier reaches $E^2$ before the ray reaches the event horizon. The limiting captured ray approaches the exterior unstable circular null geodesic, at the maximum of $W$.

The stationary-point equation isThe outer root is the photon sphere radiusIt lies outside $r_{+}$: for $|Q|\le M$ it ranges from $3M$ down to $2M$, whereas $r_{+}$ ranges from $2M$ down to $M$. More precisely $r_{\mathrm{ph}}>2M$ except in the extremal case, and $r_{+}\le 2M$; even at extremality $r_{\mathrm{ph}}=2M>r_{+}=M$. At this root,so the circular null geodesic is unstable. The smaller stationary root lies at or inside the outer horizon and does not set capture from infinity. For $b<b_c$, the radial equation has no exterior turning point and the ray enters the event horizon; for $b>b_c$, it turns around. The critical impact parameter is thereforeThe stationary-point equation gives $Q^2=(3M{r}_{\mathrm{ph}}-r_{\mathrm{ph}}^2)/2$, hence $f(r_{\mathrm{ph}})=(r_{\mathrm{ph}}-M)/(2r_{\mathrm{ph}})$. Since a parallel incident beam is captured over a disk of radius $b_c$, the photon capture cross-section of a Reissner-Nordstrom black hole isThe equality case approaches the unstable orbit indefinitely, but it is a measure-zero boundary of the captured disk and does not change the cross-section.

For the suggested parameterization, choose $\psi$ with $Q=3M\sin \psi /\sqrt{8}$ and $\cos \psi \ge 0$. Then $\sqrt{9M^2-8Q^2}=3M\cos \psi$, so the same cross-section isThe black-hole range has $|\sin \psi |\le \sqrt{8}/3$, hence $\cos \psi \ge 1/3$. For $Q=0$, $r_{\mathrm{ph}}=3M$, $f(r_{\mathrm{ph}})=1/3$, and $b_c=3\sqrt{3}M$, givingthe classical Schwarzschild black hole light-capture result. As an additional limit check, $|Q|=M$ gives $r_{\mathrm{ph}}=2M$, $b_c=4M$, and ${\sigma }_{\mathrm{abs}}=16\pi M^2$. These are geometric capture areas, larger than the corresponding horizon areas; the relevant barrier is the photon sphere, not the horizon's apparent disk. The result applies to light in the short-wavelength limit, while a finite-frequency wave absorption cross-section depends on the greybody factors.

In the Boussinesq approximation, density departures from a constant reference density ${\rho }_0$ are small enough to neglect in inertia and mass continuity, but their gravitational effect is retained. Write the total buoyancy acceleration as $b=-g(\rho -{\rho }_0)/{\rho }_0$, choose a hydrostatic background $\bar{b}(z)$, and set $\sigma =b-\bar{b}$. Thus $\sigma$ is the buoyancy perturbation, positive for a parcel lighter than the background at its present height. The squared buoyancy frequency isStable stratification has $N^2>0$. A small vertical displacement $\xi$ of a parcel conserving its buoyancy gives $\sigma =-N^2\xi$, explaining the restoring acceleration and the name of $N$.

Let $P=p^{\prime }/{\rho }_0$ be pressure with the background hydrostatic pressure removed, $\mathbf{u}=(u,v,w)$, and $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ the material derivative. The ideal, nonrotating Boussinesq equations areThe last equation is equivalent to $Db/Dt=0$. There is neither viscosity nor buoyancy diffusion. The reference density occurs in the pressure normalization, while the small density anomaly occurs only through buoyancy.

For the two-dimensional disturbance, linearization about rest givesA plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$ therefore satisfiesThese internal-wave polarization relations and the dispersion relation determine the individual responses in parts i and ii. A radiating response has no incoming internal gravity wave from $x=+\infty$; an evanescent response is bounded there. The harmonic solutions describe the response after excluding arbitrary initial transients.

For the localized oscillatory boundary, define $\alpha =\omega /\sqrt{N^2-{\omega }^2}$ and $I(z)=ϵ{\int }_0^{\infty }{\hat{u}}_0(m)e^{imz}dm=f(z)+ig(z)$. Both signs of vertical wavenumber must radiate into $x>0$. Their horizontal wavenumber is $k=\alpha |m|$, so the positive-$m$ contribution involves $Z_1=z+\alpha x$ and the negative-$m$ contribution involves $Z_2=z-\alpha x$. Consequently the complex horizontal velocity isTaking its real part provesThe positive-$m$ internal-wave polarization has $w=-\alpha u$ and the negative-$m$ one has $w=+\alpha u$. HenceIn particular, $2f(z)=ϵu_0(z)$, so the boundary condition is recovered exactly. This is two-beam radiation from a localized oscillating boundary.

Take a smooth localized profile, for example $u_0(z)=e^{-z^2/H^2}$. Its effective support is a neighbourhood of width a few $H$; increasing that neighbourhood makes the tails arbitrarily small. The boundary displacement is $ϵu_0(z)\sin \omega t/\omega$ to first order, so at $t=0$ it is flat and moving at maximum speed. At that instant,There are two narrow oblique beams centered on $z=-\alpha x$ and $z=+\alpha x$. Away from their overlap, the velocity is parallel to the relevant beam, in the directions $(1,-\alpha )$ and $(1,+\alpha )$. The figure uses $\alpha =3/4$ and measures velocities in units of $ϵ$.

At other phases the $g$ terms contribute. For the positive-frequency convention above, $g$ is the Hilbert transform of $f$; it generally has longer tails than $f$. Thus even when the boundary is nearly stationary, the interior need not be stationary or confined to the bright cores shown here. Half a period later all velocities reverse. The beam directions remain the same, but the instantaneous velocity profile and its quadrature component change.

A short pulse has a broad frequency spectrum. Frequencies below $N$ produce outgoing internal gravity wave packets with frequency-dependent ray slopes $\pm \omega /\sqrt{N^2-{\omega }^2}$; frequencies above $N$ contribute an evanescent near-boundary response. Moreover the group velocity depends on wavelength as well as direction. The radiated packet therefore spreads into a dispersive fan or wake, with different frequencies traveling at different angles and different scales separating along those directions. After the forcing ends the propagating packet travels away, rather than maintaining the two permanently driven monochromatic beams.

Define the quasi-geostrophic streamfunction by $p^{\prime }/{\rho }_0=f_0\psi$. Leading horizontal geostrophic balance, $f_0\hat{\mathbf{z}}\times {\mathbf{u}}_g=-{\nabla }_{H}P$, gives $(u_g,v_g)=(-{\psi }_y,{\psi }_x)$. The hydrostatic approximation gives $P_z=\sigma$, henceThus $D_g$ is the material derivative along the leading geostrophic flow, and the vertical derivative of pressure determines the buoyancy perturbation. This also gives thermal wind: $u_{gz}=-{\sigma }_y/f_0$ and $v_{gz}={\sigma }_x/f_0$.

The balanced regime is favored by small Rossby number, a shallow aspect ratio, and a Burger number $N^2{H}^2/(f_0^2{L}^2)$ of order one. Equivalently the stratified Froude number $U/(NH)=\mathrm{Ro}/\sqrt{\mathrm{Bu}}$ is small. On a beta plane, require $\beta L/|f_0|=O(\mathrm{Ro})$, and take slow, initially balanced motion away from the equator. Under the traditional approximation, the horizontal Coriolis force generated by vertical velocity is small because vertical velocity is small, while the vertical rotational force is small compared with the leading hydrostatic pressure and gravity terms. Strong stratification and small aspect ratio justify retaining only the vertical component of Earth's rotation in this regime; this argument is not uniform near the equator.

Put $A(z)=f_0^2/N^2(z)$. The buoyancy equation gives $w=-f_0{D}_g{\psi }_z/N^2$. Substitution into the vertical vorticity equation yieldsThere is an important commutator cancellation. With $J_{xy}(a,b)=a_x{b}_y-a_y{b}_x$,Also $D_{g}f=\beta {\psi }_x$ for $f=f_0+\beta y$. Adding terms therefore proves potential-vorticity conservation:This is the three-dimensional quasi-geostrophic potential vorticity.

At an impermeable stationary lower boundary $z=b(x,y)$, the kinematic condition is $w={\mathbf{u}}_H\cdot {\nabla }_{H}b=D_{g}b$. For small elevation and slope, evaluate this at the reference level $z=0$ to leading order. Combining it with the buoyancy equation gives the rigid-boundary buoyancy condition for quasi-geostrophic wavesThe quantity in parentheses is the boundary displacement plus its buoyancy-equivalent displacement.

Now take constant $N,f$ and the mean quasi-geostrophic streamfunction $\bar{\psi }=-(U_0+\Lambda z)y$. Its mean buoyancy perturbation is $-f\Lambda y$, consistent with thermal wind, and its interior potential vorticity is constant. In the boundary condition, the perturbation velocity $v^{\prime }={\psi }_x^{\prime }$ advects the mean boundary buoyancy gradient. Keeping this contribution givesIn particular the $-\Lambda {\psi }_x^{\prime }$ term must not be omitted.

The linear interior potential-vorticity equation is $({\partial }_t+U(z){\partial }_x)Q^{\prime }=0$. Starting with zero interior perturbation potential vorticity, and with forcing only through the boundary, it remains zero. For a steady solution this is also the regular zero-interior-PV choice, continued through any isolated zero of $U$. The inversion equation for a horizontal mode isThe printed convention $K>0$ takes $f>0$; for either hemisphere the decay rate is defined using $|f|$. Boundedness aloft selects the decaying mode. Writewith real part understood. The linear boundary condition becomesFor $K{U}_0+\Lambda \ne 0$, a steady quasi-geostrophic mountain wave is thereforeOn $y=0$, its real meridional velocity isFor the positive-parameter convention $k,ϵ,f,U_0>0$ and $K{U}_0+\Lambda >0$, the velocity minimum lies one quarter wavelength downstream of each topographic maximum; its maximum lies one quarter wavelength downstream of each topographic minimum. If the velocity prefactor changes sign, maxima and minima interchange. The condition $K{U}_0+\Lambda >0$ alone does not fix that prefactor if $U_0$ is allowed to be negative. The upper panel shows the positive-current convention.