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A nondispersive wave has a linear dispersion relation $\omega =vk$ over the relevant range, so every Fourier transform component has the same phase velocity $v_p=\omega /k=v$ and group velocity $v_g=d\omega /dk=v$. A wave packet then translates without spreading in an ideal homogeneous lossless medium.

For wave dispersion, $v_p$ depends on frequency or wavenumber. Different components accumulate different phases, and in general the group velocity differs from the phase velocity; variation of group velocity across the spectrum produces spreading of a wave packet. The membrane material can be nondispersive while confinement in a membrane waveguide introduces wave dispersion along its length.

For a plane wave wavefunction ${\psi }_k(x)\propto e^{ikx}$, impose the cyclic or periodic boundary condition ${\psi }_k(x+L)={\psi }_k(x)$. It requiresAdjacent allowed wavenumbers on the full signed $k$ axis are spaced by $\Delta k=2\pi /L$. Ignoring spin, the density of states per unit $k$ is consequentlyBoth positive and negative $k$ are included in this density. It is not yet the density per unit magnitude $|k|$, which has an extra factor of two away from zero. In $d$ independent dimensions the same periodic boundary conditions give one spatial state per $k$-space cell of measure $(2\pi /L{)}^d$.

Guided electromagnetic waves propagate along a structure whose boundaries select transverse field patterns. At a perfect conductor the tangential electric field vanishes. In a hollow rectangular guide, TE modes have $E_z=0$ and TM modes have $H_z=0$. Their longitudinal wavenumber obeysBelow the waveguide cutoff frequency ${\omega }_c=v{k}_{\perp }$, that mode is evanescent rather than freely propagating. TM modes require both indices positive; TE modes may have one zero but not both. For $a>b$, the lowest cutoff is the TE${}_{10}$ mode, with $f_c=v/(2a)$. A hollow single-conductor guide has no TEM mode, whereas a two-conductor transmission line can support one without this geometrical cutoff.^[ref]

For a propagating TE or TM mode, $v_p=v/\sqrt{1-{\omega }_c^2/{\omega }^2}$ and $v_g=v\sqrt{1-{\omega }_c^2/{\omega }^2}$. The phase velocity can exceed $v$ without superluminal information transfer. Frequency-dependent group delay can broaden pulses; excitation of several modes adds intermodal dispersion.

Use a cold, unmagnetized, collisionless plasma with a sharp interface, and take the incident medium to be vacuum. Its relative permittivity is ${ϵ}_r=1-{\omega }_p^2/{\omega }^2$, so at the stated frequencies $n=\sqrt{1-(10/12{)}^2}=\sqrt{11}/6$. Conservation of tangential wavenumber givesFor oblique total reflection at an unmagnetized plasma, require $k_{z,\mathrm{plasma}}^2<0$, or $\cos {\theta }_i<5/6$. Thus total reflection occurs for ${\theta }_i>33.56^{\circ }$ measured from the normal, up to grazing incidence. At the ideal critical angle ${\theta }_c={\cos }^{-1}(5/6)$ the transmitted normal energy flux also vanishes, so the lossless reflectance reaches unity there. Collisions or an imposed magnetic field change this simple criterion. The constitutive response used here follows the electron-motion derivation in the cited university notes.^[ref]

Assume a hollow square electromagnetic waveguide with perfectly conducting walls, filled with vacuum or air. For side $a$, the waveguide cutoff frequency isA transverse electric mode can have one zero index, but not both. A transverse magnetic mode requires both indices positive. Thus the lowest cutoff belongs to the two degenerate transverse electric modes ${\mathrm{TE}}_{10}$ and ${\mathrm{TE}}_{01}$, withThe corresponding cutoff vacuum wavelength is $2a=2.0\mathrm{cm}$. Strictly, a travelling mode carrying power along the guide needs $f>f_c$: at the cutoff itself, the axial wavenumber and group velocity are zero. The number above is therefore the lower frequency threshold, not a propagating mode with nonzero longitudinal energy flux exactly at that frequency. A dielectric filling would replace $c$ by the wave speed in that medium.

For a cold, unmagnetized, collisionless plasma, take an equilibrium electron density $n_e$, electrons of mass $m_e$ and charge $-e$, and approximately stationary ions. Use fields proportional to $e^{-i\omega t}$. The electron displacement satisfiesIts electric polarization is $\mathbf{P}=-n_{e}e\mathbf{x}$, hence the cold collisionless plasma dielectric responseThe negative susceptibility arises from the inertial electron response, rather than a restoring force binding an electron to a particular atom.

A transverse electromagnetic wave then obeysFor $\omega >{\omega }_p$, the plasma supports propagation, with refractive index $n_r=\sqrt{1-{\omega }_p^2/{\omega }^2}<1$. The phase velocity is greater than $c$, while the group velocity is $c\sqrt{1-{\omega }_p^2/{\omega }^2}<c$; their product is $c^2$, and no superluminal signal is implied. Below the plasma frequency, the bulk wavenumber is imaginary and the field is evanescent. A sufficiently thick plasma largely reflects such radiation in the lossless model.

Including an electron momentum-relaxation rate ${\nu }_c$ replaces the denominator by $\omega (\omega +i{\nu }_c)$:Its imaginary part describes absorption with the stated time convention. Finite electron temperature also produces plasma screening of slowly varying electrostatic fields, with ${\lambda }_D=\sqrt{{ϵ}_0{k}_B{T}_e/(n_e{e}^2)}$. That spatial, longitudinal screening problem is distinct from simply taking $\omega =0$ in the cold transverse-wave formula. Magnetic bias, ion motion and kinetic resonances require more general dielectric responses.

Distributed circuit model. A transmission line must be treated as a distributed system when its propagation time is appreciable compared with the signal timescale. Divide a uniform two-conductor line into elements of length $dx$. Each has series resistance $R^{\prime }dx$ and inductance $L^{\prime }dx$, and shunt electrical conductance $G^{\prime }dx$ and capacitance $C^{\prime }dx$. The primes denote quantities per unit length, with units $\mathrm{\Omega }{\mathrm{m}}^{-1}$, $\mathrm{H}{\mathrm{m}}^{-1}$, $\mathrm{S}{\mathrm{m}}^{-1}$ and $\mathrm{F}{\mathrm{m}}^{-1}$, respectively.

Kirchhoff's voltage and current laws give the Telegrapher's equations:For a lossless line, $R^{\prime }=G^{\prime }=0$, differentiation gives $V_{xx}=L^{\prime }{C}^{\prime }{V}_{tt}$ and the same wave equation for $I$. The propagation speed and characteristic impedance areA general lossless signal has $V=V^{+}(t-x/u)+V^{-}(t+x/u)$ and $I=[V^{+}(t-x/u)-V^{-}(t+x/u)]/Z_0$. Thus characteristic impedance is the voltage/current ratio of one travelling wave, not simply the resistance of the wires. It can be real even when the line dissipates no energy, because energy is carried away by propagation.

For sinusoidal quantities with convention $e^{i\omega t}$, the general lossy-line formulas areChoose the square-root branches for forward attenuation and positive forward power. The real part of ${\gamma }_{\mathrm{prop}}$ is attenuation per unit length; its imaginary part is the phase wavenumber. For small losses, attenuation is approximately $R^{\prime }/(2Z_0)+G^{\prime }{Z}_0/2$, using the lossless real $Z_0$ in this estimate.

Fields and geometry. Two separated conductors support a transverse electromagnetic mode, with electric and magnetic fields transverse to the propagation axis. For a homogeneous dielectric and negligible conductor penetration, both examples below have $L^{\prime }{C}^{\prime }=\mu ϵ$ and hence $u=1/\sqrt{\mu ϵ}$. The distributed circuit model follows from the field energy and charges in each slice.

For a coaxial cable with inner radius $a$ and inner surface of the outer conductor at $b$, Gauss's law and Ampère's law giveThe electric field is radial and the magnetic field circles the axis. Integration of voltage and field energy givesThe equal and opposite return current on the outer conductor confines the ideal external field to the annulus.

For wide parallel plates of width $w$ separated by $d\ll w$, neglect fringing. The electric field crosses the gap, $E\simeq V/d$, while the magnetic field lies in the gap parallel to the plates and perpendicular to propagation, $H\simeq I/w$. ThereforeFor a forward wave, $\mathbf{E}\times \mathbf{H}$ points along the line. Integrating this Poynting vector over either cross-section gives instantaneous power $VI$. For complex peak amplitudes, average power is $\mathrm{Re}(V{I}^{\ast })/2$, or $|V^{+}{|}^2/(2Z_0)$ for a single forward wave on a lossless line. The conductors guide an energy flux predominantly in the dielectric between them.

Terminations and reflection. At a load $Z_L$, require $V_L=Z_L{I}_L$. If $V^{+}$ and $V^{-}$ are the incident and reflected voltage amplitudes there, the total current is $(V^{+}-V^{-})/Z_0$. ThusA matched load, $Z_L=Z_0$, absorbs the incident power with no reflection. An open circuit has ${\Gamma }_V=+1$, a current node and a doubled voltage at the end. A short circuit has ${\Gamma }_V=-1$, a voltage node and a doubled current there. A purely reactive load on a lossless line has $|{\Gamma }_V|=1$ and accepts no average power. For real $Z_0$, the reflected power fraction is $|{\Gamma }_V{|}^2$ and the accepted fraction is $1-|{\Gamma }_V{|}^2$.

Incident and reflected waves produce a standing-wave pattern with voltage standing wave ratio $(1+|{\Gamma }_V|)/(1-|{\Gamma }_V|)$. The input impedance of a lossless length $\ell$ isThis gives $i{Z}_0\tan (k\ell )$ for a short and $-i{Z}_0\cot (k\ell )$ for an open. A quarter-wave section gives $Z_{\mathrm{in}}=Z_0^2/Z_L$, enabling impedance transformation. After a source launches a pulse, a distant termination affects it only after the reflected pulse returns; the return time is $2\ell /u$. Real conductor resistance, dielectric loss and departures from a TEM field profile limit the ideal model at high frequency.

Here $V(x)$ is potential energy, $E$ is the stationary state's total energy, and $\psi (x)$ is its spatial probability amplitude. The wave function has time factor $e^{-iEt/\hslash }$; $|\psi {|}^2$ is a position probability density for a normalized state. Stationary plane-wave scattering solutions are idealizations described by probability-current ratios, or components of a normalizable wave packet.

Above an upward step, a classical particle is fully transmitted with reduced speed. Quantum mechanically the abrupt wavelength change generally produces both reflected and transmitted amplitudes even though both regions are classically allowed. With unit incident amplitude and no wave arriving from the right,The reflected component has signed wavevector $-k_1$ and the same wavenumber magnitude as the incident component. A finite potential jump with constant mass and no delta interaction requires continuity of $\psi$ and ${\psi }^{\prime }$:The probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ gives probability reflection and transmission $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(k_2/k_1)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. The amplitude $t$ may exceed one without violating conservation of total probability, because the transmitted group speed is smaller.

For $0<E<V_0$, let $\kappa =\sqrt{2m(V_0-E)}/\hslash$. The bounded right-hand solution is $t{e}^{-\kappa x}$, an evanescent wave tail with amplitude penetration scale $1/\kappa$. It carries no outgoing probability current in an infinitely long forbidden half-space, so reflection is total. This differs from transmission through a finite potential barrier, where a second boundary can support quantum tunnelling.

For a downward step, replace $k_2$ by $\sqrt{2m(E+V_0)}/\hslash >k_1$. The same boundary conditions give a nonzero negative $r$, corresponding to a reflected amplitude with phase reversal. If reflection were forbidden, $r=0$ would require both $t=1$ and $k_1=k_2$, impossible for a nonzero step. Reflection therefore does not require a classical potential barrier; the discontinuity of wavenumber mismatch is enough.

For the specified electron energies, $k_2/k_1=\sqrt{40/100}=\sqrt{0.4}$, so the requested amplitude coefficients areThe corresponding probability coefficients are approximately $\mathcal{R}=0.05069$ and $\mathcal{T}=0.94931$. electron mass cancels from these ratios; there is no need to invent an additional material effective mass.

Use a stationary wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$ solving the time-independent Schrödinger equation. For energy above the positive step potential, the positive wavenumbers areIncidence only from the left permits a right-moving incident plane wave and a left-moving reflected plane wave for $x\le 0$, but only a right-moving transmitted plane wave for $x>0$:The omitted right-incident component would describe a different scattering experiment. Wavefunction matching at a finite potential step requires continuity of both $\psi$ and $d\psi /dx$ for a finite jump and the same mass on either side:Eliminate $B=C-A$ from the second condition:The first ratio is a transmission amplitude; a transmitted probability current also depends on its different propagation velocity.

For $E\gg V_0$, the two wavenumbers nearly coincide:Consequently $B/A\simeq V_0/(4E)\to 0$ and $C/A\to 1$. To leading order the complete wavefunction isThe small fractional change in kinetic energy scarcely changes the propagation wavenumber, so reflection vanishes and the particle behaves approximately as a free right-moving plane wave. A more accurate phase at large positive $x$ retains $k_{2}x$ on the transmitted side; even a small wavenumber difference accumulates over a sufficiently long distance.

For a downward step potential, $V_0=-|V_0|$, the region on the right is classically allowed at every positive incident energy. The same matching calculation applies withAs $E\to 0^{+}$, $k_1\to 0$ and $k_2\to k_{\ast }=\sqrt{2m|V_0|}/\hslash >0$. Henceand the flux coefficients satisfyThis quantum reflection at a downward step occurs even though a classical particle would simply accelerate into the lower-potential region. The abrupt change in wavenumber causes strong reflected-wave interference.

Both this limit and part (d) give total reflection, but their probability amplitudes and spatial densities differ. At the upward-step threshold, $B/A\to +1$ and $C/A\to 2$: an antinode meets the origin and the right-hand density stays finite while its propagation velocity vanishes. Here $B/A\to -1$ and $C/A\to 0$: the incident and reflected waves nearly cancel at the origin, and the transmitted density vanishes while the right-hand propagation velocity remains finite. In both descriptions the incident probability amplitude is held fixed for comparison.

The coefficients above are limiting flux ratios for $E>0$. At exactly $E=0$ the incident probability current vanishes, so directly dividing zero-energy fluxes would not define a scattering probability.

The time-independent Schrödinger equation is the energy eigenvalue equation $\hat{H}\psi =E\psi$ for a particle of mass $m$. The Hamiltonian operator is the sum of the kinetic operator $-({\hslash }^2/2m)d^2/d{x}^2$ and the prescribed potential energy $V(x)$. Here $\hslash$ is the reduced Planck constant. A corresponding stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. The Born rule identifies $|\Psi {|}^2=|\psi {|}^2$ as probability density for a state with wavefunction normalization $\int |\Psi {|}^{2}dx=1$; integrating it over an interval gives a position probability. A plane wave scattering state is not a normalizable wavefunction on the whole line, but its probability currents give the reflection and transmission fractions of an incident wave packet.

There is a notation discrepancy in the sources: the original PDF prints an unbarred $h^2$ in the kinetic term, whereas the local TeX uses ${\hslash }^2$. The standard physical equation uses the reduced Planck constant $\hslash =h/(2\pi )$, where $h$ is the Planck constant; that convention is used below. Reading the printed $h$ literally as the Planck constant would scale all lengths in this model by $2\pi$ relative to the standard answers, but would leave the amplitude ratios unchanged.

For $E>V_0$, both sides allow oscillatory wavefunctions. A classical particle always crosses the potential step and slows because its kinetic energy falls from $E$ to $E-V_0$. In quantum mechanics there is generally both reflection and transmission even though the particle is above the step. Define the positive wavenumbersWith no wave incident from the right, writeA finite potential step with constant mass makes both the wavefunction and its first derivative continuous. The two boundary conditions at $x=0$ areSolving gives the reflected and transmitted wavefunctionsFor the specified electron energies, $q/k=\sqrt{10/50}=1/\sqrt{5}$, so the relative probability amplitudes areThe transmitted probability amplitude exceeding one is compatible with conservation of quantum probability, since the outgoing speed is smaller. In fact the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ givesThese are the quantum reflection probability and transmission probability, which must be distinguished from the requested relative amplitudes.

For nonzero meridional wavenumber,At least one growing zonally varying mode can exist only ifThe upper value is a supremum approached as $k\to 0$, where the growth itself tends to zero. Thus not every permitted meridional wavenumber has the same fastest-growing zonal scale; nonzero $l$ restricts the available band of $k$ and weakens the growth factor $k/\sqrt{k^2+l^2}$ at fixed $\eta$.

Using $N=\sqrt{N^2}$, not the tabulated squared value itself, givesThe planetary radii put these into angular context: $l_{\sup }R$ is about 280 for Earth and 149 for Jupiter, corresponding to minimum meridional angular wavelengths about $1.28^{\circ }$ and $2.42^{\circ }$, respectively. They are reciprocal-length wave numbers; the dimensionless angular counts should not be substituted for $l$ in the Cartesian dispersion relation. If a channel admits a half-wave across its width, the corresponding minimum width is $\pi /l_{\sup }$ rather than a full wavelength.

The oceanic scale is mesoscale, while the much larger deformation length makes the Jovian modes thousands of kilometres across. Both are small fractions of the planetary radius, consistent with using a local patch as a first approximation. Real boundaries, planetary-vorticity gradients, varying stratification and jets can change these limits; this simple model alone does not predict the complete observed flow pattern.

Finally, the source's phrase $c^2<0$ must be interpreted in the mean-flow frame. The actual criterion is $D<0$, or $(c-\Lambda H/2{)}^2<0$. In the original frame $c$ has nonzero real part, so $c^2$ is generally complex, not an ordered negative real number. For example, at $\eta =1$ and $\Lambda H=1$, $c=0.5+0.2511i$ and $c^2\simeq 0.1870+0.2511i$. This is an unstable wave even though the literal printed inequality is inapplicable.

Potential temperature and a parcel. For a dry ideal gas of fixed composition, the first law of thermodynamics per unit mass isA reversible adiabatic process has $\delta q=0$. Taking $c_p,R_d$ constant gives $d\ln T=\kappa d\ln p$, where $\kappa =R_d/c_p$. Integrating to a fixed reference pressure $p_0$ defines the potential temperatureIt is the temperature a parcel would have after reversible dry adiabatic process taking it to $p_0$. Equivalently $ds=c_{p}d\ln \Theta$ under these assumptions, making potential temperature a thermodynamic entropy label useful for diagnosing stratification and air-mass motion. Latent heating, radiation, mixing and variable composition require modifying the conservation statement.

An air parcel is a small material volume followed with the flow. The parcel model treats it as internally approximately uniform, large compared with molecular scales and small compared with environmental variation scales. Rapid acoustic wave adjustment makes its pressure approximately equal to the surrounding pressure, while slower heat and mass exchange can be neglected during an idealized displacement. It is not a rigid container or an isolated molecule.

At the parcel's height, environmental hydrostatic pressure supplies an upward pressure force ${\rho }_{e}g$ per unit volume. Dividing by parcel density gives the upward accelerationThe last steps use equal pressure, equal gas constants and the same reference pressure. Moisture-dependent density requires virtual temperature or its potential-temperature counterpart. Nonhydrostatic pressure perturbations and entrainment are excluded from this elementary buoyancy expression.

The convective available potential energy is the positive buoyancy work per unit mass between the level of free convection and the equilibrium level:One sums positive-buoyancy layers when there is more than one. Negative work below free convection contributes convective inhibition rather than positive CAPE. In a lossless parcel estimate, conversion of CAPE to vertical kinetic energy gives $w_{\max }\simeq \sqrt{2\mathrm{CAPE}}$; pressure work beyond this model, mixing and precipitation can change that speed.

Linear stability. Let $\xi$ denote displacement from a neutrally buoyant reference altitude. Dry motion through an adiabatic process retains ${\Theta }_p={\Theta }_e(0)$, while ${\Theta }_e(\xi )={\Theta }_e(0)+\xi {\Theta }_e^{\prime }(0)+O({\xi }^2)$. The linear vertical momentum equation is thereforeFor $N_0^2>0$, $\xi =C\cos N_{0}t+D\sin N_{0}t$: stable buoyancy oscillations with period $2\pi /N_0$. For $N_0^2=0$, $\xi =C+Dt$: neutral stratification with no restoring force. For $N_0^2<0$, growing and decaying exponentials occur, indicating convective instability. The growing solution is generic unless the initial data select only the decaying branch. These are local small-displacement results with an approximately constant environmental gradient.

Lapse-rate estimate. Define $\Gamma =-dT/dz$. Hydrostatic balance and the ideal gas law imply $p^{\prime }/p=-g/(R_{d}T)$, soIn the free-parcel-oscillator interpretation, identifying the ridge encounter interval with the buoyancy period gives $P=10^4/20=500\mathrm{s}$ and $N_0=2\pi /P=0.01257{\mathrm{s}}^{-1}$. With $T=293.15\mathrm{K}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $c_p\simeq 1004\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$,The frequency identification is an extra modeling assumption. A stationary mountain gravity wave generally obeys ${\hat{\omega }}^2=N_0^2{k}_h^2/(k_h^2+m_z^2)$ without rotation, with $\hat{\omega }=U{k}_h$. Ridge spacing fixes $\hat{\omega }$, not the vertical wavenumber $m_z$. For example, $m_z=k_h$ allows the same spacing and wind with $N_0=\sqrt{2}\hat{\omega }$, and a lapse rate about $0.3\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$. Thus $5.1$ is the requested parcel-period estimate, not a uniquely identifiable lapse rate for every possible mountain-wave field; a resolved wave field needs its vertical structure as well.

Detector noise and graphical reading. Noise-equivalent power specifies the incident optical power producing signal-to-noise ratio one, with the electrical bandwidth and averaging convention stated. A spectral NEP has units $\mathrm{W}\mathrm{H}{\mathrm{z}}^{-1/2}$; a band-integrated NEP has units W. Specific detectivity normalizes the latter by detector area and noise bandwidth:Its usual unit, the Jones, is $\mathrm{cm}\mathrm{H}{\mathrm{z}}^{1/2}{\mathrm{W}}^{-1}$. Removing the expected square-root area and bandwidth dependence helps compare different detector sizes and readout bandwidths, provided noise conditions and spectral response are comparable. This does not normalize away every background, temperature or implementation difference.

Reading the native embedded voltage plot gives approximate means $V_C\simeq 0.331\mathrm{V}$, $V_S\simeq 0.406\mathrm{V}$ and $V_H\simeq 0.493\mathrm{V}$. The one-second root mean square scatter is about $0.006\mathrm{V}$; the plotted peak-to-peak excursions are larger and must not be substituted for an root mean square noise. Independent graphical digitization of the blue trace gives means near $0.331,0.406,0.493$ V and scatters about $0.006$–$0.007$ V, consistent with this estimate. These are readings from a drawing, not exact original data.

The supplied Planck approximation gives $B_C\simeq 0.08590$ and $B_H\simeq 0.12836$ in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}(\mathrm{c}{\mathrm{m}}^{-1}{)}^{-1}$. Thus the noise-equivalent radiance of one one-second sample isHere the calibrating means are used to determine the response slope; their own uncertainty is included explicitly in the final scene estimate below.

Optical throughput and detectivity. A paraxial $f/F$ cone has half-angle approximately $1/(2F)$ and projected solid angle approximately $\pi /(4F^2)$. With $F=2$ and $A_d=10^{-6}{\mathrm{m}}^2$,Using the exact cone angle $\tan \theta =1/(2F)$ gives projected etendue $A_d\pi {\sin }^2\theta =1.85\times 10^{-7}{\mathrm{m}}^2\mathrm{sr}$, within the intended approximation. The radiance noise corresponds to detector power noiseThe wavenumber bandwidth is $1\mathrm{c}{\mathrm{m}}^{-1}$ and must match the radiance's per-wavenumber unit; it is not an electrical noise bandwidth. For the common order-of-magnitude assignment $\Delta f\sim 1\mathrm{Hz}$ to a one-second sample,Area in the last expression is $1\mathrm{m}{\mathrm{m}}^2=0.01\mathrm{c}{\mathrm{m}}^2$. For a specified one-second boxcar with white noise, the one-sided equivalent noise bandwidth is instead $1/(2t)=0.5\mathrm{Hz}$. That convention gives spectral NEP $\simeq 2.5\times 10^{-10}\mathrm{W}\mathrm{H}{\mathrm{z}}^{-1/2}$ and $D_{\ast }\simeq 3.9\times 10^8$ Jones. The source does not specify the electrical filter, so the two conventions should not be mixed or assigned spurious precision.

Two-point retrieval and its uncertainty. Assume linear, stable response for the three views. The radiometer two-point calibration isWith target radiances treated as known, differentiating givesNotice $b<0$ for an intermediate scene: raising the hot voltage while keeping the scene voltage fixed reduces the inferred scene radiance. Numerically $a\simeq 0.2621$, $b\simeq -0.1214$ in radiance units per volt.

The scene interpolation fraction is about $0.463$, giving $B_S\simeq 0.1056$ andFor independent white noise in each view, each hundred-sample mean has standard error ${\sigma }_{\bar{V}}=0.006/10=0.0006\mathrm{V}$. The radiance error isUsing the given temperature sensitivity,A scatter estimate closer to $0.007$ V changes this to roughly $0.14$ K. These are random standard errors under the averaging assumptions. Target-temperature/emissivity uncertainty, response drift and atmospheric or surface-emissivity corrections are separate: the retrieved quantity is a brightness temperature, not automatically the true sea-surface temperature.

Potential-vorticity dynamics. The quasi-geostrophic potential-vorticity equation expresses material conservation of a balanced potential vorticity by a horizontally nondivergent geostrophic flow. Its terms represent relative vorticity, planetary vorticity and the layer-thickness/free-surface contribution. The Rossby deformation radius $L_d$ measures the scale at which that thickness contribution competes with the horizontal-vorticity term. The quasi-geostrophic approximation filters fast acoustic waves and gravity waves, making the slower balanced evolution accessible to early digital forecasts. The 1950 barotropic-vorticity forecasting study demonstrated practical twenty-four-hour forecasts from observed mid-tropospheric fields, while identifying discretization and omitted baroclinicity as error sources.^[ref]

Near the north pole put $\delta =\pi /2-\theta$ and $r\simeq a\delta$. ThenUsing projected rather than arc distance gives the same leading polar expansion. It is a north-polar expression; the sign structure is changed consistently at the opposite pole.

In cylindrical coordinates, horizontal advection is $u_r{\partial }_{r}Q+(u_{\varphi }/r){\partial }_{\varphi }Q$. Substituting the specified streamfunction givesThe factors of $1/r$ are required by cylindrical geometry.

An exact traveling eigenfunction. This is a polar gamma-plane Rossby wave. The radial equation implies ${\nabla }_h^2\psi -\psi /L_d^2=-{\alpha }^2\psi$, so $Q=f-{\alpha }^2\psi$. Since $J(\psi ,\psi )=0$, its nonlinear self-advection vanishes exactly, regardless of amplitude. Also $f_r=-\gamma r$, giving $J(\psi ,f)=\gamma {\psi }_{\varphi }$. HenceFor $\psi =F(r)\cos (\omega t-n\varphi )$, this yieldsThe wave is retrograde relative to the balanced mean flow, since $\omega /n<0$ in this convention.

Multiplying the radial equation by $r^2$ givesFor $K^2={\alpha }^2-L_d^{-2}>0$, setting $y=Kr$ gives the standard Bessel equation with derivatives with respect to $y$. ThusRegularity at the pole excludes $Y_n$ for the usual nonnegative integers. If $K^2<0$, the corresponding general solutions are modified Bessel functions; at $K=0$ they are $r^{\pm n}$, with a constant and $\ln r$ for $n=0$. The printed Bessel hint mixes its independent-variable labels; using $y=Kr$ and derivatives in $y$ resolves that notation.

What the hexagon fixes. With the north-polar approximation and the supplied rotation period,The intrinsic phase speed at that radius is $c_{\varphi }=r_0\omega /n=-\gamma r_0/{\alpha }^2$. Therefore the measured $-4\mathrm{m}{\mathrm{s}}^{-1}$ fixesThe polar-wave deformation-radius identifiability issue is that a unique $L_d$ additionally requires a radial wavenumber or boundary condition. The quoted peak-to-peak meander is an amplitude, not a radial wavelength: multiplying the exact eigenfunction by a constant changes its amplitude without changing $K,\alpha$ or $\omega$. Thus it cannot be silently converted into $K=2\pi /\Delta r$.

A transparent order-of-magnitude closure is a slowly varying radial envelope near $r_0$, so $F^{\prime \prime }+F^{\prime }/r$ is small relative to $n^{2}F/r_0^2$. Then $K^2\simeq n^2/r_0^2$ andThis is a conditional estimate, not a uniquely determined observation. The general real-$K$ bound is $L_d\ge {\alpha }^{-1}\simeq 1.7\times 10^6$ m. Different radial structures give different values: for example $K=3.0\times 10^{-7}$ and $5.0\times 10^{-7}{\mathrm{m}}^{-1}$ yield about $2.0\times 10^6$ and $3.2\times 10^6$ m, respectively, at the same measured phase speed. Their amplitudes remain free. The given $5^{\circ }$ meander corresponds to a radial half-amplitude about $2.5\times 10^6$ m and helps characterize the pattern's size, but alone removes none of this eigenvalue ambiguity.

Use $u$ as a spectroscopic wavenumber, in the same units throughout. Write the mass absorption coefficient of the absorbing species asHere $S$ is the spectral line strength, the integrated mass absorption coefficient; it is not the line-centre coefficient. If ${\kappa }_{\nu }$ has units ${\mathrm{m}}^2\mathrm{k}{\mathrm{g}}^{-1}$ and $\nu$ is in $\mathrm{c}{\mathrm{m}}^{-1}$, then $S$ has units ${\mathrm{m}}^2\mathrm{k}{\mathrm{g}}^{-1}\mathrm{c}{\mathrm{m}}^{-1}$. For absorber mass density ${\rho }_a$, the volume coefficient in Beer-Lambert law is ${\rho }_a{\kappa }_{\nu }$.

Lifetime broadening. Finite lifetimes of the upper and lower states destroy optical coherence, even for an isolated molecule. Exponential coherence decay gives a Lorentzian line shape. For population decay rates ${\Gamma }_u,{\Gamma }_l$ and no extra dephasing, its wavenumber half-width is $({\Gamma }_u+{\Gamma }_l)/(4\pi c)$, with $c$ expressed in the length unit reciprocal to $\nu$. This sets a natural linewidth and can matter at extremely low collision rates; it is generally much smaller than atmospheric collisional or thermal widths.