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A time-independent ray Hamiltonian conserves the stationary-observer absolute frequency. A horizontally uniform background also conserves horizontal wavenumber. The intrinsic frequency $\hat{\omega }=\omega -kU(z)$ nevertheless changes when the ray moves through shear. These follow from the Hamiltonian ray-tracing equations and are essential to identifying a critical level of an internal gravity wave.

For $0<k<1$, a neutral mode of the equal-width Hazel model behaves as $w\sim z^{1-k}$ near $z=0$. Its derivative diverges as $z^{-k}$, so it is not a classical continuously differentiable mode across the critical level of an internal gravity wave. The horizontal velocity is proportional to $w^{\prime }/k$; its local kinetic energy is finite only for $k<1/2$, because ${\int }_0^{ϵ}{z}^{-2k}dz$ converges precisely then. One possible neutral-mode convention is the boundary value from $U-c$ with $c_i\downarrow 0$, which fixes the phase of the power on the negative-$z$ side. The Miles–Howard theorem concerns growing modes with nonreal $c$ and does not rule out such singular neutral limits.

The zero-phase velocity ansatz $w=(\mathrm{sech}z{)}^k(\tanh z{)}^{1-k}$ solves the Taylor–Goldstein equation away from its critical level of an internal gravity wave exactly when $J=k(1-k)$. Let $y=\tanh z$ and $a=1-k$. Then $w^{\prime }/w=a/y-y$ and $w^{\prime \prime }/w=a(a-1)/y^2-a-1+2y^2$, so the differential-equation residual divided by $w$ is $[J-k(1-k)](y^{-2}-1)$. The curve has maximum $1/4$ at $k=1/2$. The endpoint $k=0$ does not decay at infinity; fractional powers at the critical level require a branch and regularity qualification.

Let $p$ be kinematic pressure, with the resting hydrostatic part subtracted, and let $\sigma$ be the buoyancy perturbation about background buoyancy $N^{2}z$. The nonrotating inviscid Boussinesq equations, with no buoyancy diffusion, areHere the third equation is mass conservation for incompressible flow, and the last expresses buoyancy conservation following a parcel. Linearization about rest removes the products of disturbances. For a plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$, the resulting Linearized Boussinesq equations becomeWith $K=(k^2+m^2{)}^{1/2}$ and $k\ne 0$, the first and third equations give $\hat{p}=-m\omega \hat{w}/k^2$. Substitution in vertical momentum and buoyancy gives the dispersion relationSince the velocity is the time derivative of the displacement field, $\hat{\mathbf{u}}=-i\omega \hat{\mathbf{\xi }}$. Thus the complete buoyancy polarization of a plane internal gravity wave isBoth velocity and displacement are perpendicular to the wavevector; pressure and velocity are in quadrature with buoyancy and displacement.

For upward/rightward phase velocity, choose $k,m>0$ and $\omega =Nk/K$. Differentiating this dispersion relation gives the group velocity, while the vector phase velocity points along the wavevector:Their dot product is zero and ${\mathbf{c}}_p+{\mathbf{c}}_g=(N/K,0)$. Draw ${\mathbf{c}}_g$ from the tip of ${\mathbf{c}}_p$: the right-triangle geometry of internal-wave velocities has horizontal hypotenuse $N/K$. This hypotenuse is distinct from the horizontal phase velocity $\omega /k$.

At an instant with real complex amplitude $\hat{\sigma }=A$, write $\varphi =kx+mz-\omega t$. The spatial fields areThe constant-phase lines of an internal gravity wave slope down to the right, with $dz/dx=-k/m$. Buoyancy maxima lie a quarter wavelength from pressure maxima; the oscillating displacement and velocity run along the phase lines. The group velocity points down and right along them, despite upward/rightward motion of the wave phase.

Let $P=\overline{{\xi }_{x}u+{\zeta }_{x}w}$, using horizontal averaging over a wavelength. This is the displacement pseudomomentum of an internal gravity wave, whose sign convention is opposite to $k\overline{E}/\omega$. For complex coefficients, $\overline{ab}=\frac{1}{2}\mathrm{Re}(\hat{a}{\hat{b}}^{\ast })$. The above wave polarization consequently givesSince $c_{gz}=-\omega m/K^2$, these obey $\overline{{\zeta }_{x}p}=c_{gz}P$. With weak kinematic viscosity, each leading monochromatic velocity component satisfies ${\nabla }^{2}u=-K^{2}u$, ${\nabla }^{2}w=-K^{2}w$. The dissipative term is therefore $-\nu K^{2}P$, so the supplied averaged balance reduces toFor constant $k,m,N$ and a steady wave envelope, $c_{gz}{P}_z=-\nu K^{2}P$. As $P$ is proportional to squared wave amplitude, viscous attenuation of an internal gravity wave givesHere $c_{gz}<0$, so amplitude decreases below a source at $z_s$: the source is above the receiving level. The same conclusion follows directly because downward group velocity carries energy from the source downward. Upward phase velocity does not locate the source.

For a slowly varying mean shear flow $U(z)$, the local intrinsic frequency is $\hat{\omega }=\omega -kU(z)$. A local WKB approximation uses ${\hat{\omega }}^2=N^2{k}^2/(k^2+m(z{)}^2)$, with varying vertical wavenumber and group velocity. The appropriate transported quantity is wave action, or horizontal wave pseudomomentum $kE/\hat{\omega }$; its vertical flux has viscous loss, and the changing polarization factors must be retained. In particular, $P/|\hat{\sigma }{|}^2$ is no longer constant.

Near a critical level of an internal gravity wave, $\hat{\omega }\to 0$, so $|m|\sim N|k|/|\hat{\omega }|$ and $|c_{gz}|\sim {\hat{\omega }}^2/(N|k|)$. The damping per vertical distance, of order $\nu K^2/|c_{gz}|$, grows rapidly. The waves are strongly attenuated over a narrow altitude range and transfer momentum to the mean flow there or before reaching the critical level. The inviscid WKB approximation eventually fails; viscosity and possibly internal-wave breaking regulate the small scales.

Start from $\rho ({\partial }_{t}u+u\cdot \nabla u)=-\nabla p-\rho g{e}_z$, ${\partial }_t\rho +u\cdot \nabla \rho =0$, and $\nabla \cdot u=0$. The stationary background obeys $p_0^{\prime }=-g{\rho }_0$. Linearize in small perturbations and use the Boussinesq approximation: replace the density by a reference value ${\rho }_{\ast }$ in inertia but retain the density perturbation in the buoyancy force. With $\pi =p^{\prime }/{\rho }_{\ast }$ and $b=-g{\rho }^{\prime }/{\rho }_{\ast }$,This is the Brunt–Väisälä frequency. Its assumed constancy requires a locally uniform stable density gradient in this approximation; rotation and viscosity are neglected.

For a plane wave with wavevector $K=(k,\ell ,m)$, project the momentum equation onto the space perpendicular to $K$ to eliminate pressure. Its vertical component is $-i\omega w=b[1-m^2/|K{|}^2]$, while $-i\omega b+N^{2}w=0$. Eliminating $b$ yieldsFor $0<\omega <N$, a small oscillating body radiates upward and downward conical internal gravity wave beams. In a vertical section the beams make a St Andrew's cross, with angle $\alpha$ to the horizontal satisfying $\sin \alpha =\omega /N$. Indeed the group velocity is perpendicular to the wavevector and has ratio of vertical to horizontal magnitudes $\sqrt{k^2+{\ell }^2}/|m|$. Thus wave energy follows the cones, rather than the direction normal to a fixed-radius spherical wavefront. There are no such propagating beams for $\omega >N$.

For the shear-flow calculation choose the positive intrinsic-frequency branch and $k_0>0$, conditions needed for the printed critical-level formula. The slowly varying local dispersion relation isStationarity and horizontal homogeneity conserve $\Omega$ and $k=k_0$. The Hamiltonian ray equations for a local dispersion relation giveThus $m_0<0$ stays negative and the ray propagates upward. Conservation of frequency givesIf the shear reaches the required velocity difference, its first zero is the critical level of an internal gravity wave,As $z\uparrow z_c$, the intrinsic frequency tends to zero, $m\to -\infty$, the vertical wavelength tends to zero and the vertical group velocity tends to zero. The ideal ray cannot cross this level. For finite $U_c^{\prime }>0$, $\dot{z}\sim k_0(U_c^{\prime }{)}^2(z_c-z{)}^2/N$, so it approaches the level only asymptotically in time. The disturbance develops short scales; even weak viscosity can then absorb it, and nonlinear effects can invalidate the small-amplitude approximation.

The printed sign and existence hypotheses are incomplete. On the positive-frequency branch, $k_0<0$ instead gives $\dot{m}>0$ and eventually a turning point when the intrinsic frequency reaches $N$. As a concrete counterexample to the stated universal height bound, take $N=1$, $k_0=-1$, $m_0=-3$, and $U(z)=\epsilon z$ with any small $\epsilon >0$. At the printed height $z_c=1/(\epsilon \sqrt{10})$, the intrinsic frequency is $2/\sqrt{10}<1$ and $m=-\sqrt{3/2}<0$, so the vertical group velocity is still positive and the ray crosses that height. Also, $U^{\prime }>0$ alone does not ensure that the displayed velocity difference is ever reached; an increasing bounded $U$ may have no finite critical level. The qualified derivation above gives the intended positive-$k_0$ result whenever that level exists.

Let $b=-g{\rho }^{\prime }/{\rho }_0$ be the buoyancy perturbation and $p$ the pressure perturbation divided by ${\rho }_0$. For stable stratification, $N>0$. The nonrotating Linearized Boussinesq equations areEliminating $u,p,b$ gives $({\partial }_x^2+{\partial }_z^2)w_{tt}+N^2{w}_{xx}=0$. Since $w={\zeta }_t$, a nonzero-frequency plane wave obeys the same equation for its displacement. Substituting its phase yields the dispersion relation for a plane internal gravity wave:Thus the frequency depends on the wavevector direction rather than its magnitude.

Advection of the background density gives ${\rho }^{\prime }=-\zeta {\bar{\rho }}_z$ to first order. With constant ${\bar{\rho }}_z<0$, the instantaneous density gradient is ${\rho }_z={\bar{\rho }}_z(1-{\zeta }_z)$. A region has unstable density stratification when this becomes positive, namely when ${\zeta }_z>1$. The maximum of ${\zeta }_z$ is $|m{A}_{\zeta }|$, so the monochromatic internal-wave overturning criterion isEquality gives a locally vanishing gradient. This is the prediction of the displacement field extrapolated to overturning; the small-amplitude approximation itself ceases to be reliable there.

For the rising packet, distinguish its conserved absolute frequency $\omega$ from its actual intrinsic frequency $\sigma =\omega -kU(z)$. The printed terminology calls $\omega$ intrinsic while also assigning it to a stationary observer; the stationary-observer interpretation is the one consistent with the displayed Doppler shift. On the positive-frequency branch, the ray Hamiltonian isThe Hamiltonian ray-tracing equations giveThe last identity follows also by differentiating the Hamiltonian along its canonical trajectory: the spatial and wavevector terms cancel in pairs. Thus absolute-frequency conservation in steady shear gives constant $k$, constant $\omega$, and constant stationary-observer horizontal phase speed $c_x=\omega /k$. In contrast, $\sigma =\omega -ksz$ decreases as the packet rises. At its initial height,This is the critical level of an internal gravity wave. In fact $m(t)=m_0-kst$ and $z(t)=[\omega -\sigma (k,m(t))]/(ks)$, so the inviscid ray approaches $z_c$ as $t\to \infty$, rather than reaching it at a finite time.

Write $\theta =|\Theta |=\arctan (|m|/k)$, so $\sigma =N\cos \theta$ with $0<\theta <\pi /2$. The intrinsic internal-wave phase and group velocity calculation givesThe observer-frame horizontal ray velocity is $U+c_{gx}$. Dividing it by $c_{gz}$ proves the internal-wave ray in uniform vertical shear:The angle increases toward $\pi /2$ and the vertical group speed tends to zero near the critical level.

The wave-action conservation law fixes the prescribed upward flux. For a nonzero packet, $B>0$, and the given flux relation impliesApply the monochromatic internal-wave overturning criterion, using $|m|=k\tan \theta$. After multiplying by the positive trigonometric factors, the exact instability condition isAt marginal overturning near a critical level, $\sin \theta \simeq 1$, so the wave-action criterion for critical-level overturning givesWith fixed $B,k,N$, this is the requested quarter-power order estimate; the prefactor supplies the dimensions suppressed in that notation. It is an onset balance, not a replacement for $\cos \theta =\sigma /N$. Combining the two relations instead gives ${\cos }^3\theta =(B{k}^3/N^2)\sin \theta$ at onset. Since $m^2{A}_{\zeta }^2$ diverges as ${\sigma }^{-3}$ toward $z_c$, any nonzero packet flux eventually violates the linear overturning criterion before reaching that level, within this nondissipative ray model.

The WKB approximation treats the vertical wavenumber as locally constant over one vertical wavelength. Write $w=\Re \{A(z)e^{ikx+i{\int }^{z}m(s)ds}\}$, with $|m^{\prime }|\ll m^2$ and $|A^{\prime }|\ll |mA|$. For the stationary internal-wave WKB solution, $A\propto m^{-1/2}$ away from singular or turning levels. A small terrain slope and small perturbations are also needed for Linearized Boussinesq equations.

Let $p$ be the perturbation fluid pressure divided by reference mass density, and $b$ the buoyancy perturbation. The steady Linearized Boussinesq equations and incompressibility condition areFor an $e^{ikx}$ component, elimination gives $u=-w^{\prime }/(ik)$, $b=-N^{2}w/(ikU)$ and $p=(U{w}^{\prime }-U^{\prime }w)/(ik)$. Substitution into vertical momentum gives the exact stationary Taylor–Goldstein equationThe leading WKB approximation therefore identifiesThe kinematic boundary condition fixes the horizontal wavenumber to $k=k_T$ for the sinusoidal forcing.

For explicit heights assume a constant positive buoyancy frequency $N$, $0<k{U}_0<N$, and $|S|/N\ll 1$. The latter is a large gradient Richardson number. The linear shear flow has $U^{\prime \prime }=0$, so the propagation condition is $|U|<N/k$.

For $S<0$, the first obstruction is the critical level of an internal gravity wave atAs $z\uparrow z_c$, $m\sim N/U\to \infty$. The critical-height approach of a stationary wave in linear shear has upward group velocity $c_{gz}\sim k{U}^2/N$, so the travel time to $z_c$ diverges: the height is a supremum, approached in the ideal ray model. Here $|m^{\prime }|/m^2\to |S|/N$, which can remain small; it would be incorrect to say that the short-wavelength WKB approximation necessarily fails solely because $m$ diverges. Nevertheless $w\propto U^{1/2}$ and displacement $w/(ikU)\propto U^{-1/2}$, so a finite disturbance eventually violates linearity. Dissipation and overturning also matter near this singular level.

For $S>0$, the turning height of a stationary wave in linear shear isAt $z_t$, $m=0$ and the WKB approximation fails. An Airy function matching region describes reflection and an upper evanescent wave. Thus $z_t$ is the highest height of vertical propagation, rather than a sharp cut-off of every disturbance.

The problem initially allows $N(z)$ to vary. Without the extra constant-$N$ assumption, replace the explicit turning height by the first root of $N(z)=k|U_0+Sz|$ encountered on a propagating branch. A declining $N(z)$ can cause a turning level before the negative-shear critical level. The two closed-form heights require the stated constant-stratification assumption.

Set $y=\tanh z$ and $a=1-k$, so $dy/dz=1-y^2$ and $\hat{w}=(1-y^2{)}^{k/2}{y}^a$. Away from $y=0$, the logarithmic derivative givesFor $c=0$, ${\overline{U}}^{\prime \prime }/\overline{U}=-2(1-y^2)$ and $N^2/{\overline{U}}^2=J(1-y^2)/y^2$. The Taylor–Goldstein equation residual divided by $\hat{w}$ consequently simplifies toThus the neutral mode of the equal-width Hazel model requiresThe qualifier “solution” needs care at the critical level of an internal gravity wave $z=0$. For $0<k<1$, $\hat{w}\sim z^{1-k}$ and ${\hat{w}}^{\prime }\sim (1-k)z^{-k}$: it is a solution separately on either side, not a classical smooth eigenfunction across the critical level of an internal gravity wave. A branch for negative $\tanh z$ is also needed. For example, a limit from $c_i>0$ assigns $(\tanh z{)}^{1-k}=|\tanh z{|}^{1-k}{e}^{-i\pi (1-k)}$ on $z<0$.

The critical-layer regularity of a neutral Hazel mode further gives local finite horizontal kinetic energy only for $0<k<1/2$, since incompressible flow gives $\hat{u}=i{\hat{w}}^{\prime }/k$ and ${\int }_0^{ϵ}{z}^{-2k}dz$ then converges. At $k=1$, $J=0$ and $\hat{w}=\mathrm{sech}z$ is smooth after the removable ${\overline{U}}^{\prime \prime }/\overline{U}$ quotient is continued. At $k=0$, $J=0$ and the formal $\tanh z$ profile does not decay at infinity, so it fails the remote endpoint condition. These qualifications prevent interpreting the whole closed parameter interval as a family of classical decaying modes.

For the rapidly varying wave phase, define $k_i={\partial }_i\theta$ and $\omega =-{\partial }_t\theta$, with the common scaling by $\epsilon$ absorbed into these local variables. A packet moves with group velocity $c_{g,i}=\partial \Omega /\partial k_i$. Along that trajectory, $d/dt={\partial }_t+c_g\cdot {\nabla }_x$. Compatibility gives ${\partial }_t{k}_i=-{\partial }_i\omega$ and ${\partial }_i{k}_j={\partial }_j{k}_i$. Differentiating the local dispersion relation then cancels its implicit $k$ derivatives, yieldingThe last equality follows also by differentiating $\omega =\Omega (k,x,t)$ along the ray and cancelling the two Hamiltonian cross terms. The partial derivatives of $\Omega$ hold its other arguments fixed.

A mean flow adds the Doppler shift: the intrinsic frequency is $\hat{\omega }=\omega -kU(z)$, soon the positive-intrinsic-frequency branch launched here. The medium is stationary and independent of $x$, so $\omega$ and $k$ are constant. Initially $m=0,U=0$, hence $\omega =N$. Moreover $\dot{m}=-k{U}^{\prime }<0$, and then $\dot{z}=-Nkm/(k^2+m^2{)}^{3/2}>0$. The initial vertical group velocity is zero, but $m$ immediately becomes negative and the ray accelerates upwards.

If $U$ reaches $N/k$ at a finite height $z_c$, this is the critical level of an internal gravity wave, with $U(z_c)=\omega /k$. Below it,Assuming $0<U^{\prime }(z_c)<\infty$, put $\delta =z_c-z$. Then $N-kU\sim k{U}^{\prime }(z_c)\delta$, givingThus the ray cannot cross this level in finite time within the geometrical-optics approximation. The hypothesis $U^{\prime }>0,U(0)=0$ alone does not guarantee a finite $z_c$: for example $U(z)=\frac{N}{2k}(1-e^{-z})$ on $z\ge 0$ never reaches $N/k$. Existence of a finite critical level is an additional assumption. Very close to one, diverging wavenumber can invalidate the inviscid, linear, slowly varying approximation; the result is the formal ray prediction.

Put $D={\partial }_t+U{\partial }_x$, write $p$ for perturbation pressure divided by ${\rho }_0$, and set $b=-g{\rho }^{\prime }/{\rho }_0$. The linearized Boussinesq approximation, incompressibility and material derivative giveTaking the curl eliminates pressure; differentiating the resulting vorticity equation in $x$ and using incompressibility gives $D{\nabla }^{2}w-U^{\prime \prime }{w}_x=b_{xx}$. A further application of $D$ therefore yieldsFor a stationary nonzero horizontal Fourier mode, $D=U{\partial }_x$, so cancellation of ${\partial }_x^2$ gives the stationary Taylor–Goldstein equationThis division requires $U\ne 0$ on the interval considered: a zero of $U$ is a critical level of an internal gravity wave. The mean profiles must be sufficiently smooth for the displayed derivatives, with background hydrostatic pressure and stable density stratification, $N^2>0$, for propagating internal gravity waves. The horizontally uniform component is excluded from the cancellation. For a horizontal wavenumber $k$, local vertical propagation additionally requires $N^2/U^2-U^{\prime \prime }/U>k^2$; this is distinct from the smoothness restrictions. If stability of the background against other disturbances is needed, the Miles–Howard theorem supplies the sufficient condition $N^2\ge (U^{\prime }{)}^2/4$, rather than a necessary condition for deriving the equation.

Take $U_0,N_0,k>0$ and assume an initially propagating mode, $k<N_0/U_0$. Put $s=N_0/U_0$ and $\delta (z)=1+\gamma z$. Neglecting $U^{\prime \prime }/U$ as stipulated, the Scorer parameter and stationary internal-wave WKB solution areThe positive root $m>0$ selects upward group velocity for the negative intrinsic frequency branch $\hat{\omega }=-kU$. The WKB approximation needs a slowly varying background relative to the local vertical wavelength, in particular $|m^{\prime }|\ll m^2$ and $|m^{\prime \prime }|\ll |m{|}^3$, away from a turning level or singular mean profile.

The local dispersion relation is $\omega =kU-Nk/(k^2+m^2{)}^{1/2}=0$. Its intrinsic and laboratory group velocities are, with $K=(k^2+m^2{)}^{1/2}$,Thus intrinsic energy propagation is upstream and upward, while laboratory energy propagation is downstream and upward, parallel to $(k,m)$. Constant-phase lines of an internal gravity wave have slope $dz/dx=-k/m$ and normal spacing $2\pi /K$; the horizontal spacing remains $2\pi /k$.

If $\gamma <0$, $m$ grows with height. The wavevector and laboratory energy ray become more vertical; phase lines become flatter and closer together. For finite nonzero $N$, the heightis a critical level of an internal gravity wave: $U\to 0$ and the vertical wavenumber diverges. The laboratory group velocity tends to zero. Importantly, $|m^{\prime }|/m^2\to |\gamma |/s$, so the formal WKB condition need not deteriorate before this level if that ratio is small. Nevertheless the stationary equation is singular at the level, the phase has no finite limit, and the horizontal perturbation velocity grows as $m^{1/2}$; the inviscid linear wave cannot be continued uniformly through it without additional physics.

If $\gamma >0$, $m$ decreases to zero atThe wavevector and laboratory ray become horizontal, the phase lines become vertical, and the normal spacing increases. Above $z_t$, $m^2<0$ and the disturbance is an evanescent wave. Here $|m^{\prime }|/m^2$ diverges, so the WKB approximation fails in a turning region and must be replaced by a local connection solution.

Use a common factor $e^{i(kx-\omega t)}$, with $k=k_0>0$ and $\omega ={\omega }_0$. For the upward propagating branches with $0<{\hat{\omega }}_2=\omega -k{U}_2<N_2$, defineThese are the inclinations of upward-sloping constant-phase lines of an internal gravity wave to the horizontal. Write the incident, reflected and transmitted complex velocity amplitudes as $A_0,A_r,A_t$. Incompressibility gives $(u_i,w_i)=A_0(\cos {\theta }_1,\sin {\theta }_1)$, $(u_r,w_r)=A_r(\cos {\theta }_1,-\sin {\theta }_1)$, and $(u_t,w_t)=A_t(\cos {\theta }_2,\sin {\theta }_2)$.

The jump conditions for stratified inviscid shear flow require a common interface displacement, not a common vertical velocity: $w=-i\hat{\omega }\eta$ on each side. Since the background mass density is continuous, pressure is continuous without a hydrostatic jump. The horizontal momentum equation gives $p/{\rho }_0=(\hat{\omega }/k)u$. HenceThese two matching equations determine internal-wave transmission across a velocity jump:The amplitude formula printed in the PDF is inconsistent with these material-interface matching conditions. In particular, identical layers have $A_t/A_0=1$ and $A_r/A_0=0$, whereas its expression labelled “reflected” equals one. The displayed results above distinguish reflection from transmission and retain the intrinsic-frequency factors required by the kinematic boundary condition and pressure balance.

For equal buoyancy frequencies and the specified opposing current, ${\theta }_1=\pi /6$, ${\theta }_2=\arcsin (3/4)$, andThe incident phase lines have slope $1/\sqrt{3}$, the reflected lines slope $-1/\sqrt{3}$, and the transmitted lines slope $3/\sqrt{7}$: they steepen above the interface. For the other specified current, ${\hat{\omega }}_2=0$. The transmitted vertical wavenumber then diverges and there is no regular propagating upper-layer wave of that frequency: this is the critical level of an internal gravity wave limit. Approaching it from ${\hat{\omega }}_2>0$ gives $A_r/A_0\to -1$ and vanishing transmitted vertical energy flux; setting the intrinsic frequency to zero directly is outside the regular matching calculation.