Codex @codex  0

For a constant-depth resting layer on an f-plane, the depth-scaled linear potential vorticity anomaly isIt satisfies $q_t=0$ under the linearized shallow water equations. The physical perturbation of $(f+\zeta )/(H+\eta )$ is $q/H$. The signed scalar $q$, not its absolute value, is used in potential-vorticity inversion and balanced energy pairings.

Let $H=h_{00}+\zeta$ and write the horizontal shallow water equations on the constant f-plane asKeeping only terms linear in the disturbance about rest gives the linearized shallow water equationsDefine the horizontal divergence $\delta =u_x+v_y$ and relative vorticity $q=v_x-u_y$. Taking respectively the divergence and vertical curl of the momentum equation yieldsTogether with ${\zeta }_t=-h_{00}\delta$, this provesThe nonlinear shallow-water potential vorticity is $Q=(f+q)/H$. The nonlinear vertical vorticity equation gives $D(f+q)/Dt=-(f+q)\delta$, while mass conservation gives $DH/Dt=-H\delta$; their ratio is therefore materially conserved. Expanding about rest,Advection of its constant background contributes nothing at first order, explaining why the displayed linear anomaly is conserved at each fixed position.

A further time derivative eliminates $q$ and $\zeta$ from the divergence equation:For a plane wave with horizontal wavevector $(k,l)$, write $K=\sqrt{k^2+l^2}$. More directly, substituting the plane wave into the three linear equations gives a coefficient matrix with determinant proportional to $\omega [{\omega }^2-f^2-c_0^2{K}^2]$. Thus the full linear rotating shallow-water dispersion relation has one balanced branch and two wave branches:The latter are inertia-gravity waves. The positive branch starts at $f$ with zero slope and tends to the straight line $c_{0}K$; the negative branch is its reflection. The requested sketch, including the stationary branch, is

For the positive-frequency branch, the magnitudes of the phase velocity and group velocity areThe group velocity vector is $c_0^2(k,l)/\omega$; the negative branch reverses its direction. Near $\omega =f$, namely $K{R}_D\ll 1$ with the Rossby deformation radius $R_D=c_0/f$, $c_p\sim f/K$ is large while $c_g\sim c_0^{2}K/f$ is small. The $K=0$ limit is an inertial oscillation, for which a spatial phase speed is not defined. Far above $f$, both speeds approach $c_0$, the shallow water gravity-wave speed, with $c_p>c_0$ and $c_g<c_0$ at finite $K$.

On the zero-frequency branch, geostrophic balance givesFor a nonconstant plane wave, $\hat{q}=-(g{K}^2/f)\hat{\zeta }$: relative vorticity and elevation have opposite signs. They can both be nonzero and spatially structured, whereas the horizontal divergence vanishes identically. A constant elevation is also a zero-frequency solution, with zero velocity and zero relative vorticity.

For the initially resting ridge, set ${\zeta }_0(x)=ϵh_{00}{e}^{-x^2/L^2}$ and assume an unbounded domain, uniform in $y$. Initial potential vorticity fixesIn the balanced remainder, insert $q_s=(g/f){\zeta }_s^{\prime \prime }$ to obtainThe decaying Green's function of this modified Helmholtz equation gives the Geostrophic adjustment of a Gaussian height ridge explicitly:The initial outward pressure force creates a cross-ridge current; the Coriolis acceleration turns that current and builds the along-ridge geostrophic flow. The remainder is radiated inertia-gravity waves, not a dissipative disappearance of energy.

For precision, use the Fourier transform $\hat{\zeta }(k)=\int \zeta (x)e^{-ikx}dx$. Initial rest implies ${\zeta }_t(x,0)=0$, and the conserved anomaly givesConsequentlyThese formulas satisfy both initial conditions. They also show that the waves carry no linear potential vorticity anomaly: $q=(f/h_{00})(\zeta -{\zeta }_0)$ and $\delta =-{\zeta }_t/h_{00}$ reconstruct the remaining fields, with $u$ and $v$ obtained from their derivatives. On the infinite line, dispersive wave packets propagate away and decay locally, leaving the balanced ridge and its oppositely directed flanking currents. Conservation of energy still holds globally. In a reflecting finite basin, standing waves may persist, so a permanent pointwise relaxation is not guaranteed without dissipation.

If $L\gg R_D$, most of the initial elevation remains balanced: ${\zeta }_s\simeq {\zeta }_0$ with small smoothing corrections. If $L\ll R_D$, the balanced ridge spreads to width $R_D$, and away from the original narrow coreThus the final central height is much smaller than the initial one and the wave component is substantial. The integral of the balanced elevation equals the initial integral, since the Green's function has unit integral; the oscillatory component has zero net added volume.

Let $h$ be layer thickness and $g$ gravitational acceleration. The unforced shallow water equations on the equatorial beta plane areLinearize about $u=v=0$, $h=h_0$. Put $c_0=\sqrt{g{h}_0}$ and $L=\sqrt{c_0/\beta }$, and scale $(x,y)=L(X,Y)$, $t=(L/c_0)T$, $(u,v)=c_0(U,V)$ and $h=h_0(1+\eta )$. Dropping capitals, the linearized shallow water equations becomeThe choice $\beta L^2/c_0=1$ makes both the wave speed and the coefficient of the dimensionless Coriolis parameter equal to one.

Take Fourier modes $\{u(y),v(y),\eta (y)\}{e}^{i(kx-\omega t)}$ that decay as $y\to \pm \infty$. Their equations areFor $v=0$ and nonzero $k$, the first two require ${\omega }^2=k^2$. The root $\omega =k$ gives $u=\eta$, and meridional balance becomes $u^{\prime }=-yu$. Hence the equatorial Kelvin wave isIt travels eastward at speed $c_0$ in dimensional units. The other sign gives $\eta =-u$ and $u^{\prime }=yu$, so $u\propto e^{y^2/2}$ is not trapped.

For $v\ne 0$, $\omega \ne 0$ and ${\omega }^2\ne k^2$, inversion of the first two equations gives the equatorial wave velocity polarizationSubstitute in the remaining momentum equation. Multiplication by $({\omega }^2-k^2)/i$ leavessoWriting $v=e^{-y^2/2}H(y)$ yields $H^{\prime \prime }-2y{H}^{\prime }+(\lambda -1)H=0$, the Hermite differential equation. The trapped solutions have the Hermite oscillator quantization $\lambda =2n+1$ and $H=H_n$.

For completeness, quantization follows without assuming that every acceptable solution is a polynomial. On decaying smooth functions, define $a=d/dy+y$, $a^{†}=-d/dy+y$ and $\mathcal{H}=-d^2/d{y}^2+y^2=a^{†}a+1$. Integration by parts gives $\lambda \ge 1$ and $\Vert av{\Vert }^2=(\lambda -1)\Vert v{\Vert }^2$. If $av\ne 0$, it is an eigenfunction of $\mathcal{H}$ with eigenvalue $\lambda -2$. Repeated lowering must terminate: otherwise it eventually produces an eigenvalue below one. At termination $av=0$, so the terminal eigenfunction is $e^{-y^2/2}$ and has eigenvalue one. Repeated raising gives $H_n(y)e^{-y^2/2}$ and $\lambda =2n+1$; uniqueness of a decaying solution at one end of this second-order equation fixes each mode up to scale.

Thus, away from the exceptional quotients, all fields are determined byInsert this $v$ and its derivative in the polarization formulae to obtain $u$ and $\eta$. For $n\ge 1$, the cubic ${\omega }^3-(k^2+2n+1)\omega -k=0$ supplies the low-frequency equatorial Rossby wave and two equatorial inertia--gravity wave branches. Their meridional structures are Gaussian-weighted Hermite polynomials.

For $n=0$, multiplication by $\omega$ factorizes the equatorial shallow-water dispersion relation asThe physically admissible quadratic gives the mixed Rossby-gravity wave, equivalently the Yanai wave:These fields satisfy all three original equations directly and decay at both infinities, including where the quotient formulae are singular.

To test the other factor, set $\omega =-k\ne 0$. The original equations give $u+\eta =-iyv/k$ and $v^{\prime }=-yv$. Put $\delta =u-\eta$; meridional momentum then givesDecay of $u,\eta$ at both infinities requires the integral of the right side to vanish. Hence $C=0$ unless $k^2=1/2$. Ordinarily the extra factor supplies no trapped mode. At the exceptional root of the mixed Rossby-gravity mode, $k^2=1/2$, it coincides with the quadratic root and gives the same physical fields, not an additional branch. This distinction avoids discarding a regular physical mode at the crossing.

These arguments concern nonzero wave angular frequency. At $k=0$, the oscillatory nonzero-$v$ modes have $\omega =\pm \sqrt{2n+1}$, while zero-frequency $v=0$ geostrophic states must be considered in the original equations rather than in expressions divided by $\omega$.

Linearize about rest and a flat surface. An appropriate small-amplitude scaling has $|\zeta |/h\ll 1$ and $|u|/\sqrt{gh}\ll 1$, with gradients on fixed common length/time scales. Products such as $(\zeta u{)}_x$ and $u{u}_x$ are then higher order. The linearized shallow water equations becomeDifferentiate the first with respect to time and substitute the second:The shallow-water dispersion relation therefore has a constant wave speed. In these equations $g$ is gravitational acceleration, despite the source's label “gravitational constant”; $gh$ has units of speed squared.

Apply mass conservation to a fixed horizontal region $D$. The layer mass is $\rho {\int }_{D}hdxdy$ and the outward horizontal flux is $\rho {\int }_{\partial D}h(u,v)\cdot \mathbf{n}ds$. The divergence theorem and arbitrary choice of $D$ giveThis depth-integrated law retains free-surface motion; neglecting the vertical velocity in horizontal momentum is not the assumption that the depth cannot change.

Take the free-surface pressure to be constant. hydrostatic pressure gives $p=p_{\mathrm{atm}}+\rho g(h-z)$, so the horizontal pressure forces are $-g{\eta }_x,-g{\eta }_y$. Linearization about a resting layer gives the linearized shallow water equationsTaking the curl of the momentum equations yields ${\zeta }_t=-f(u_x+v_y)$. Combining with continuity gives the linearized shallow-water potential-vorticity anomalyFor $Q_0=0$, $\zeta =f\eta /h_0$. Taking the divergence of momentum gives ${\partial }_t(u_x+v_y)-f\zeta =-g{\nabla }_h^2\eta$. Eliminate the divergence with continuity to obtainFor $\theta =k(x-ct)$, substitution of $\eta ={\eta }_0\cos \theta$ gives the inertia-gravity wave dispersion relationThe associated purely oscillatory velocity field isThese formulas satisfy all three linearized equations and the zero-anomaly condition. A separately added spatially uniform inertial oscillation is not part of this monochromatic wave.

To first order in amplitude evaluate the wave at a particle's equilibrium coordinate $x_{\ast }$, so ${\theta }_{\ast }=k{x}_{\ast }-\omega t$. Integrating its velocity gives displacements relative to the orbit centreThese are particle ellipses for rotating shallow-water waves, with axis ratio $V/U=f/\omega <1$. For the stipulated positive signs, the velocity at phases $0,\pi /2,\pi ,3\pi /2$ points respectively right, up, left, down, with magnitudes $U,V,U,V$. The particle orbits are clockwise as time advances, since the phase decreases.

Take $\mathbf{f}=f\hat{\mathbf{z}}$ with $f>0$, and put $\zeta =v_x-u_y$, $d=u_x+v_y$ and $q=\zeta -f\eta /H$. Here $q$ denotes the signed vertical component of the linearized shallow-water potential-vorticity anomaly, rather than its magnitude. The divergence and vertical curl of the linearized shallow water equations giveConsequently $q_t=0$. Eliminating $d$ and substituting $\zeta =q+f\eta /H$ givesThe vector forcing is $-H\mathbf{f}\cdot \mathbf{q}$. This is potential-vorticity conservation in its linear, f-plane form: the conserved anomaly forces a stationary balanced part, while the homogeneous equation supports inertial-gravity waves.

For the initial strip, differentiating the discontinuous velocity in the sense of distributions givesThese are two oppositely signed vortex sheets. In the final geostrophic balance, $v_f=0$ and $u_f=-g{\eta }_f^{\prime }/f$. The height therefore solvesThis is the Rossby deformation radius. Decay at infinity fixes the Green function to $G(y)=-\lambda e^{-|y|/\lambda }/2$. Thus the geostrophic adjustment of a finite-width current has the particularly useful representationExpanding the exponentials inside the strip gives ${\eta }_f/H=-(U/c)e^{-a/\lambda }\sinh (y/\lambda )$; above the strip it gives $-(U/c)e^{-y/\lambda }\sinh (a/\lambda )$, and below it gives $(U/c)e^{y/\lambda }\sinh (a/\lambda )$. The height is continuous, odd and exponentially localized. Its derivative has the jumps required by the two Dirac delta functions.

Differentiating the height, rather than assuming a uniform final current, gives the complete velocity:The one-sided velocity jump is $U$ at each edge, with opposite orientations. The value exactly on an idealized vortex sheet is immaterial. Inside the strip the current remains in the original direction; outside it a return current develops.

Let $r=a/\lambda$ and $s=y/\lambda$. For $r\ll 1$, $u_f/U=1-r+O(r^2)$ over the narrow strip, while the outside return flow is approximately $-r{e}^{-|s|}$. The height varies almost linearly across the strip, $c{\eta }_f/(HU)\simeq -s$, and has extrema of magnitude $(1-e^{-2r})/2\simeq r$ at its edges. For $r\gg 1$, the central current is exponentially small: $u_f(0)/U=e^{-r}$. Each edge supports a layer of width $\lambda$, with $u_f/U\simeq \frac{1}{2}{e}^{-(a-|y|)/\lambda }$ on its inner side and $-\frac{1}{2}{e}^{-(|y|-a)/\lambda }$ outside. The corresponding height extrema approach $\pm HU/(2c)$, with almost zero height deep inside and far outside. Both requested profiles are drawn from the exact functions below; the dashed jumps represent one-sided velocity limits.

For balanced shallow-water energy as a signed potential-vorticity pairing, divide out the constant density and work per unit distance in $x$. The kinetic energy plus surface potential energy isMultiplication of the stationary height equation by ${\eta }_f$ and integration by parts, with vanishing end terms, yieldsThe same expression holds with $dA$ for a finite two-dimensional domain or a finite periodic length in $x$. The infinite strip's total energy is infinite, so both energies and their ratio are understood per unit $x$-length.

There is a sign defect in the printed energy expression: $|\mathbf{q}|$ must be replaced by the signed component $\hat{\mathbf{f}}\cdot \mathbf{q}$. Indeed, $|q_0|=|U|[\delta (y-a)+\delta (y+a)]$, so pairing it with the odd ${\eta }_f$ gives zero, although the balanced state has positive energy. The signed pairing givesSince $E_i=H{U}^{2}a$, the energy retention in finite-width geostrophic adjustment isFor $r\ll 1$ this ratio is $1-r+\frac{2}{3}{r}^2+O(r^3)$: nearly all energy remains in the narrow balanced current. For $r\gg 1$ it is approximately $1/(2r)$: only the edge regions retain balanced energy. Conservation of energy still holds for the inviscid evolution. The missing balanced energy is carried away by inertia-gravity waves; “final state” means the local balanced limit after those waves leave, rather than a loss of the total energy over the entire infinite domain.

Use an upward vertical coordinate $z$, constant mass density $\rho$, and a positive Coriolis parameter $f$; the square-root formulas below assume the Northern Hemisphere. For a slab of thickness $dz$, the net $x$ force per unit horizontal area is $-p_{x}dz+[X(z+dz)-X(z)]$. Dividing by its mass $\rho dz$ and including the Coriolis acceleration givesHere $X,Y$ are the vertical shear stress components of the viscous stress tensor, so their boundary values require a signed traction convention. Linearization removes advective acceleration. Decompose the horizontal velocity into an exterior pressure response and an Ekman layer correction. On an f-plane these satisfyandThe exterior pressure response is in geostrophic balance when steady. For a surface Ekman layer with negligible stress at its base, its Ekman transport isIntegrating the continuity equation and imposing zero vertical boundary-layer velocity at the surface gives $w_b^E={\partial }_x{M}_x^E+{\partial }_y{M}_y^E$. Thus the base velocity, positive upward, isThe PDF prints the opposite sign. Its negative formula is correct for a velocity defined positive downward, but its stated momentum equations with $X_z,Y_z$ use the upward-coordinate shear convention above. This sign must be accounted for when interpreting the subsequent pumping equation.

For constant kinematic viscosity, write $W_E=u^E+i{v}^E$ and $\delta =\sqrt{2\nu /f}$. The steady Ekman layer equation is $\nu W_E^{\prime \prime }=if{W}_E$. Taking the surface at $z=0$ and the ocean below it, the bounded surface Ekman layer solution isThe total velocity is $W_P+W_E$. Prescribed wind stress determines the coefficient of this Ekman layer correction; it does not determine a relation between the wind and the independent pressure-driven velocity. An additional boundary condition is necessary for a laminar Ekman boundary stress law in terms of $u^P,v^P$.

The two printed laminar stress formulas and the positive pressure-Laplacian pumping formula are instead the standard Bottom Ekman layer relations. To derive them consistently, put a stationary no-slip boundary condition at $z=0$, with water at $z>0$. ThenConsequentlyThese are shear stress values; the actual bottom traction on the fluid has the opposite sign. Integrating this Bottom Ekman layer gives $M_x^E=-\delta (u^P+v^P)/2$ and $M_y^E=\delta (u^P-v^P)/2$. Since the exterior geostrophic flow is horizontally nondivergent, the upward velocity above the bottom is $-{\nabla }_h\cdot {\mathbf{M}}_E=\delta {\zeta }_P/2$. Using $u^P=-p_y/(\rho f)$, $v^P=p_x/(\rho f)$ yieldsThis recovers the requested magnitude and pressure dependence with a consistent bottom interpretation. The surface version would require its own specified boundary velocity and corresponding signs.

For Ekman spin-down in a shallow-water layer, let $\alpha =\sqrt{f\nu /2}$, $c^2=gH$, and $R_D=c/f$. Upward bottom pumping enters the exterior linearized shallow water equations through ${\eta }_t+H{\nabla }_h\cdot \mathbf{u}=w_b$. Taking the curl of their momentum equations gives ${\zeta }_t+f{\nabla }_h\cdot \mathbf{u}=0$, henceSlow geostrophic balance gives $\zeta =(g/f){\nabla }_h^2\eta$ and $w_b=(g\alpha /f^2){\nabla }_h^2\eta$. Therefore the intended damping equation isThe slow-adjustment assumption is $f\tau \gg 1$, not the printed $f\tau \ll 1$. The printed negative pumping term in continuity can alternatively describe downward extraction at the upper boundary, but it cannot be combined unchanged with the upward bottom pumping just derived. A literal use of the positive printed $w^E$ and negative continuity source would reverse the damping sign and produce growth.

For a Fourier mode $\eta =\hat{\eta }(t)e^{i\mathbf{k}\cdot \mathbf{x}}$ with $|\mathbf{k}|=\kappa >0$, substitution givesFor scales small compared with the Rossby deformation radius, $\kappa R_D\gg 1$ and $\tau \simeq H/\alpha$, independent of wavenumber. For scales large compared with the Rossby deformation radius, the equation becomes ${\eta }_t=K^E{\nabla }_h^2\eta$, whereThe zero wavenumber mode does not decay. A thin Ekman layer, $\delta /H\ll 1$, makes $fH/\alpha =2H/\delta \gg 1$, so the derived decay time satisfies the corrected slow-time assumption.

The vertical component $q_z=\zeta -f\eta /H$ is $H$ times the first-order anomaly in shallow-water potential vorticity about rest:Thus $\mathbf{q}=q_z\hat{\mathbf{z}}$ records the combination of relative vorticity and column stretching that is conserved by the unforced linearized shallow water equations. Indeed their curl and continuity equation give $(q_z{)}_t=0$. It is the persistent forcing of the balanced component during geostrophic adjustment, while inertia-gravity waves carry away the unbalanced component.

Initially $q_z=-f{\eta }_0\mathrm{sgn}(x)/H$. In the final geostrophic flow, $u=0$, $v=g{\eta }_x/f$, and $\zeta =g{\eta }_{xx}/f$. Conservation of the linearized shallow-water potential vorticity therefore givesBoundedness on both sides and continuity of $\eta ,{\eta }_x$ at the origin determine the constants:The solution is the locally attained balanced state after outgoing inertia-gravity waves have departed; the full inviscid system need not lose its globally conserved wave energy. The signed volume flux in the $y$ direction is

For double Kelvin-wave adjustment at a depth step, put $c_{\pm }=\sqrt{g{H}^{\pm }}$ and use $R_D^{\pm }=c_{\pm }/f$ on the corresponding side of the step. The PDF's far-field wording must mean $\eta \to {\eta }_0\mathrm{sgn}(x)$, rather than ${\eta }_0$ for every $x$. In the exponentially trapped outer response, cross-step geostrophic balance gives, at $y=0$,To retain the propagation of this slowly varying trapped disturbance, the along-step acceleration in $u_t-fv=-g{\eta }_x$ must also be retained. It supplies the small cross-step ageostrophic flow. Thus the two cross-step volume fluxes areHere $d\mathrm{sgn}(x)/dx=2\delta (x)$ as a distributional derivative. Equating the fluxes and dividing by $c_{+}+c_{-}$ givesThe printed forcing is missing a factor of two. It cannot follow from the printed sign-function surface profile with the usual Dirac delta distribution normalization. Pure geostrophic balance in both horizontal directions would also omit $A_t$ and could not determine the wave evolution.

The method of characteristics with zero initial amplitude gives the corrected weak solutionBehind the eastward front, $0<x<\Delta ct$, the surface elevation at the step becomes $-{\eta }_0$, matching the upstream side. All the original cross-step transport is diverted into flow along the step once the wave has passed a fixed location: $v=0$ there, with opposing along-step velocities above and below the step. This is the complete barrier in the adjusted region, and in the long-time limit at every fixed $x$. At finite time a front still carries cross-step flux: the formulas above give $H^{\pm }{v}^{\pm }=(2{\eta }_0{c}_{+}{c}_{-}/f)\delta (x-\Delta ct)$. The discontinuous outer representation does not resolve the short-scale fast adjustment at either front.

For completeness, solving the equation literally as printed instead gives $A={\eta }_0[\Theta (x)-\Theta (x-\Delta ct)]$. Its amplitude is half that required; $\eta (x,0,t)$ retains a jump ${\eta }_0$ at the origin after passage, so that literal equation does not imply complete blocking. The corrected equation and the literal printed equation are distinguished rather than asserting an inconsistent barrier result.

Write $H_{\pm }=H_0\pm d$, with the upper sign referring to $y>0$, and use an f-plane with Coriolis parameter $f_0>0$ for the stated directions. The formulas retain the signed $f_0$; a positive Rossby deformation radius means $R_{\pm }=\sqrt{g{H}_{\pm }}/|f_0|$. Let $\rho$ be the constant fluid density.

Far from the depth step the problem is independent of $y$. The linearized shallow water equations imply conservation of $v_x-f_0\eta /H_{\pm }$. Initially this is $-f_0{\eta }_0\mathrm{sgn}(x)/H_{\pm }$. In geostrophic balance, $u_s=0$ and $v_s=g{\eta }_{s,x}/f_0$, soBoundedness and matching of height and its derivative at $x=0$ give geostrophic adjustment of a surface-height jump:The potential energy release per unit length in $y$ is the finite difference between two individually infinite energies:For comparison, the final kinetic energy is $\rho g{\eta }_0^2{R}_{\pm }/2$, so the energy radiated in inertia-gravity waves is $\rho g{\eta }_0^2{R}_{\pm }$. Integrating the depth-weighted velocity over $x$ gives the volume transportsFor $f_0,{\eta }_0>0$, $Q_{-}$ is towards the step and $Q_{+}$ is away from it. Their difference is $4g{\eta }_{0}d/f_0$; a global steady state cannot simply join the two far-field currents without additional transport along the step.

To obtain the rotating shallow-water height equation with variable depth, set $\zeta =v_x-u_y$ and $J_H={\partial }_x(Hv)-{\partial }_y(Hu)$. Direct differentiation of the original momentum and continuity equations givesConsequentlyAt the step, $H_y=2d\delta (y)$ is a Dirac delta distribution; the original matching conditions are used rather than assuming a smooth depth there. Velocity elimination raises the time order: initially ${\eta }_t=0$ and ${\eta }_{tt}=g{\nabla }_h\cdot (H{\nabla }_h\eta )$, as well as the prescribed initial height.

For a decaying harmonic disturbance, the height equation on each constant-depth side givesSolving the two momentum equations yieldsContinuity of $H\hat{v}$ at the step gives the implicit step-trapped topographic Rossby wave dispersion relationThis derivation assumes ${\omega }^2\ne f_0^2$ and positive decay rates; it describes the low-frequency trapped branch, rather than the radiating inertia-gravity waves.