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In a spherical Newtonian gravitational potential, use binding energy $E=\psi -v^2/2$ and specific angular momentum $L$. The displayed stationary galactic distribution function has locally Gaussian radial and transverse velocities: ${\sigma }_r^2={\sigma }^2$, ${\sigma }_{\theta }^2={\sigma }_{\varphi }^2={\sigma }^2/(1+r^2/r_a^2)$. Its mass density is $\rho ={\rho }_c{e}^{[\psi (r)-\psi (0)]/{\sigma }^2}/(1+r^2/r_a^2)$. The velocity-anisotropy parameter is ${\beta }_{\mathrm{an}}=r^2/(r^2+r_a^2)$, so the core is nearly isotropic and the halo nearly radial. These exact Gaussian integrals use the untruncated distribution over all velocities; imposing an escape-energy cutoff changes them.

Choose an initially transform-limited pulse with Gaussian power and complex envelopeHere ${\sigma }_0$ is the standard deviation of the power-weighted time distribution; ${\tau }_0$ is its full width at half maximum. Neither is a field-amplitude width. The spectral amplitude is proportional to $e^{-{\sigma }_0^2{\Omega }^2}$, so the spectral power has standard deviation ${\sigma }_{\Omega }=1/(2{\sigma }_0)$. The Gaussian time-bandwidth product in ordinary frequency is ${\tau }_0\Delta \nu =2\ln 2/\pi \simeq 0.4413$, where both widths are power FWHMs.

Multiplication by $e^{i{\varphi }_2{\Omega }^2/2}$ preserves spectral power and energy. Substituting ${\varphi }^{\prime }={\varphi }_2\Omega$ into the moment identity givesThis is an exact result for a Gaussian spectrum with quadratic spectral phase. It also follows by completing the square in its inverse Fourier transform.

For these chosen inputs, ${\sigma }_0=12.7398\mathrm{fs}$. The stretched pulse has $\tau =278.88\mathrm{fs}$ and root mean square width $118.4282\mathrm{fs}$. Adding $-3000\mathrm{f}{\mathrm{s}}^2$ recovers $30\mathrm{fs}$; leaving $300\mathrm{f}{\mathrm{s}}^2$ gives $40.85\mathrm{fs}$. All phase-only cases retain $1\mathrm{nJ}$.

If GDD is cancelled but ${\varphi }_3=20000\mathrm{f}{\mathrm{s}}^3$ remains, the pulse is asymmetric. Gaussian spectral moments giveThus the centroid is $15.4033\mathrm{fs}$ and the root mean square duration is $25.2354\mathrm{fs}$. Converting this last value to a FWHM with the Gaussian factor would be wrong: the power envelope is no longer Gaussian. The calculation's integrated energy agrees in time and frequency through Parseval's identity, while its centroid and variance agree with these independent moment formulas.

A real amplifier also changes spectral amplitude. For a simple small-signal example, stipulate an intensity gain profileThis specifies finite gain bandwidth, with no saturation or extra amplifier phase. Multiplying two Gaussian spectra givesFor chosen ${\sigma }_g={\sigma }_{\Omega }$, the amplified spectrum is narrower by $\sqrt{2}$, the shortest Gaussian power FWHM becomes $30\sqrt{2}=42.43\mathrm{fs}$, and the total energy is $G_{0}U/\sqrt{2}$. Perfect spectral phase correction cannot recreate the suppressed spectral amplitude. Real gain narrowing and saturation require a material-dependent amplifier model; a constant multiplier is only a limited idealization. Gérard Mourou's 2018 lecture discusses gain narrowing as a limit on the shortest amplified pulses.^[ref]

For the Hamiltonian $H=(p^2-x^2)/2$, a Gaussian wave packet $A{e}^{-B{x}^2}$ satisfies $\dot{A}=-i\hslash AB$ and $\dot{B}=-i/(2\hslash )-2i\hslash B^2$. The displayed solution is normalizable for real $\varphi$ exactly when $\sin (2\varphi )>0$; its second moments of a complex Gaussian wave packet areBoth grow as $\hslash e^{2t}/[4\sin (2\varphi )]$, while their difference and hence the mean energy remain constant.

For perturbation $V_0{e}^{-{\beta }^2{x}^2}$ and oscillator inverse length $\alpha =\sqrt{m\omega /\hslash }$, direct Gaussian expectation values give $\Delta E_0=V_0\alpha /({\alpha }^2+{\beta }^2{)}^{1/2}$ and $\Delta E_1=V_0{\alpha }^3/({\alpha }^2+{\beta }^2{)}^{3/2}$. A narrow perturbation samples the odd state's node and suppresses its shift by two extra powers of $\alpha /\beta$. A broad perturbation approaches a constant energy offset for both states.

For an isolated Gaussian pulse, the relation between laser pulse energy $U$ and peak power $P_{\mathrm{pk}}$ depends on the pulse-width convention. Here $P(t)$ is the total incident radiant power, the background is zero, and $\tau >0$ denotes the power full width at half maximum. Integrating $P$ over the entire pulse defines its positive energy $U>0$.

Pulse energy, peak power and average power resolve different information. Consider a Gaussian pulse withThe chosen pulse duration $\tau$ is the full width at half maximum of radiant power: $P(\pm \tau /2)=P_{\mathrm{pk}}/2$. It refers to radiant power, rather than the wave amplitude of the electric field. The Gaussian integral gives the relation between Gaussian pulse energy and peak power:The pulse energy shape factor is $\kappa =U/(P_{\mathrm{pk}}\tau )$. It equals $1.064467$ for this power profile; a rectangular pulse of duration $\tau$ has $\kappa =1$. Knowing only a width and an integrated energy therefore requires a pulse-shape assumption before inferring peak power.

An equivalent form makes the normalization transparent. Write the pulse centre as $t_0$, and let $\sigma$ denote the standard deviation of its energy-normalized pulse profile:The ratio $p(t)=P(t)/U$ is a probability density function in the mathematical sense: it is nonnegative, has units ${\mathrm{s}}^{-1}$, and integrates to one. Here it weights the pulse's energy over time. Its normal distribution profile has mean $t_0$ and variance ${\sigma }^2$. This gives $P_{\mathrm{pk}}=U/(\sqrt{2\pi }\sigma )$ without a separate area estimate.

For a stipulated $\sigma =2\mathrm{ns}$ and $U=1\mathrm{nJ}$, the power full width at half maximum is $4.70964\mathrm{ns}$ and the peak power is $0.19947\mathrm{W}$. Confusing the standard deviation with the full width at half maximum would insert the wrong duration into the conversion.

The Gaussian pulse field-envelope width is a second distinct convention. If a fixed spatial field profile gives $P(t)=K|A(t){|}^2$ for a constant $K>0$, the wave amplitude envelope obeysConsequently its amplitude full width at half maximum is ${\tau }_E=\sqrt{2}\tau$. Inserting ${\tau }_E$ into the formula for a power-width input underestimates the true peak power by a factor $1/\sqrt{2}$. State which measured waveform defines the width.

An electrical photodetector trace can also be broadened by its impulse response. The optical pulse-width inference therefore needs a characterized detector response.^[ref]

The Gaussian pulse time-gate energy fraction is the share of an isolated pulse's laser pulse energy admitted by a finite time window. The result below acts on incident optical radiant power. Let $\tau$ be the power full width at half maximum, and first place the pulse centre and the detection time gate centre at zero.

A detection time gate need not collect the whole pulse. Integrating the same profile only over $-T_g/2\le t\le T_g/2$ gives the Gaussian pulse time-gate energy fractionwhere $\mathrm{erf}$ is the error function. A gate as wide as the power full width at half maximum collects only $\mathrm{erf}(\sqrt{\ln 2})\simeq 0.760968$, or 76.10% of the pulse energy. The pulse has nonzero Gaussian tails outside that interval. A real readout's impulse response can extend the required integration window further.

Alignment adds a parameter. Let the pulse centre be $t_0$, the gate centre be $t_c$, and write the Gaussian gate centring error as $d=t_c-t_0$. Put $\sigma =\tau /(2\sqrt{2\ln 2})$. Integrating the normalized Gaussian pulse profile over the gate givesAt $d=0$ the two error function terms reduce to the centred formula above. With $T_g=\tau$ and $d=\tau /2$, the captured fraction drops to $0.490734$. The pulse centre then lies on a gate edge. Merely keeping the gate width fixed does not keep its collected energy fixed.

The gate works best when centred. With the centred energy-normalized pulse profile $p(s)=P(t_0+s)/U$ and $a=T_g/2$, differentiation of $F(d)={\int }_{d-a}^{d+a}p(s)ds$ givesA Gaussian function is an even function about its centre and decreases with distance from that centre. Thus $F^{\prime }(d)\le 0$ for $d>0$, and symmetry gives the opposite sign for $d<0$. The same argument proves that a centred interval maximizes mass for an even unimodal density; unimodality supplies the required monotonicity.

A centred gate of width $2\tau$ captures $0.981468$ of the energy. Solving $\mathrm{erf}(\sqrt{\ln 2}{T}_g/\tau )=0.99$ gives $T_g\simeq 2.18771\tau$ for 99% capture in this ideal model. The requirement changes if the centring error is nonzero.

Pulse timing jitter can therefore create variation in gated energy even when every full pulse has the same $U$. A wider gate reduces this loss but also changes admitted background and the chance of including neighbouring pulses. Measure the timing and detector response associated with the chosen gate.

For a detector modeled as a linear time-invariant system, its recorded waveform is a convolution of the optical input with its impulse response. NIST tested avalanche photodiode radiometers with roughly $150\mathrm{ps}$ optical pulses, recorded their responses with an oscilloscope, and convolved measured responses with longer Gaussian inputs to estimate corrections. The detector and amplifier were characterized together.^[ref] A gate applied to that electrical waveform must use the readout model rather than assuming the optical fraction is unchanged.

For $H=-d^2/d{x}^2+x^2+ϵx^4$, use the normalized real wavefunction ${\psi }_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, $\alpha >0$. The Gaussian integral gives $⟨x^2⟩=1/(2\alpha )$ and $⟨x^4⟩=3/(4{\alpha }^2)$, while the kinetic contribution is $\alpha /2$. Thus the Rayleigh quotient is the displayed expression. For positive $ϵ$, its stationary width satisfies ${\alpha }^3-\alpha -3ϵ=0$. At $ϵ=2$ the unique positive solution is $\alpha =2$, giving $E_0\le 13/8$ by the Rayleigh-Ritz variational principle. For $ϵ<0$, taking $\alpha \downarrow 0$ makes the negative quartic contribution dominate, so the energy is unbounded below even within this normalized family. The positive-coupling bound and negative-coupling instability concern different signs of the same variational expression.

A Gaussian variational approximation chooses a quadratic trial kernel and minimizes a variational upper bound on the interacting free energy. The optimized kernel often equals the bare kernel plus a self-consistent, wavevector-independent mass shift.

For $H=p^2/(2m)+k{x}^4$ with $k>0$, the normalized even trial wavefunction $f_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$ has Rayleigh quotientThe Gaussian moment recurrence gives the potential term, and integration by parts gives the kinetic term. Its minimum is at ${\alpha }^3=6mk/{\hslash }^2$, proving the displayed upper bound by the Rayleigh-Ritz variational principle.

An odd trial family $g_{\beta }=(4{\beta }^3/\pi {)}^{1/4}x{e}^{-\beta x^2/2}$ is orthogonal to the exact even ground state. The Sturm oscillation theorem identifies the lowest odd state as the first excited state, so minimizing within this family bounds $E_1$:Orthogonality to an approximate ground trial alone would not justify this excited-state bound; odd parity ensures orthogonality to the exact ground state. Both bounds exhibit the characteristic quartic-oscillator energy scale $({\hslash }^{4}k/m^2{)}^{1/3}$.

Use the quantum harmonic oscillator in units with reduced Planck constant $\hslash =1$, mass $m=1$, and angular frequency $\omega =1$. Its Hamiltonian operator and trial Gaussian wave packet areThis is the same Gaussian wave packet used for the Heisenberg uncertainty relation: $|{\psi }_{\sigma }{|}^2$ is a normal distribution with mean zero and variance ${\sigma }^2$. The Gaussian integral gives wavefunction normalization and $⟨x^2⟩={\sigma }^2$. These smooth, rapidly decaying trials lie in the one-dimensional harmonic oscillator form domain and even in the operator domain.

Since ${\psi }_{\sigma }^{\prime }=-x{\psi }_{\sigma }/(2{\sigma }^2)$, integration by parts with vanishing boundary terms gives the kinetic energy and potential energy:Narrow trials cost kinetic energy, while broad trials cost potential energy. The exact differenceshows that the unique optimal width is ${\sigma }^2=1/2$, with $E=1/2$. At $\sigma =1/2$ and $\sigma =1$, the trial energies are both $5/8=0.625$; the kinetic energy and potential energy contributions exchange places.

Minimizing within this trial family alone establishes $E_0\le 1/2$. To establish exactness, use the Heisenberg uncertainty relation to obtain $E_{\psi }\ge 1/2$ for every admissible quantum state, as in its existing oscillator example. Together the bounds give $E_0=1/2$. Direct substitution also yields $H{\psi }_{1/\sqrt{2}}={\psi }_{1/\sqrt{2}}/2$, agreeing with the Gaussian eigenstate for a quadratic potential. This is a useful distinction: the variational method supplies the upper bound, while the Heisenberg uncertainty relation supplies the complementary lower bound.

A gear is a toothed mechanical element that transmits motion and torque through contact with another gear or a rack and pinion. The teeth maintain a prescribed relation between the motions: unlike a smooth friction drive, the ideal gear ratio is set by geometry and tooth counts. The gear pitch circle supplies an imaginary rolling geometry for describing that relation; it is not the circle through the tooth tips.^[ref]

For two external gears on fixed parallel axes, let $N_1,N_2$ be their tooth counts and let signed angular velocities ${\omega }_1,{\omega }_2$ use the same orientation. Each tooth on one gear must pass a tooth space on the other, so counting tooth passages givesThus a 20-tooth input driving a 40-tooth output makes the output turn half as fast in the opposite direction. An input of $1200$ revolutions per minute gives an output of $600$ revolutions per minute in magnitude. These signs describe an external pair; a rack and pinion instead converts rotation into translation.

The useful trade is between angular speed and torque. In an ideal steady transmission, conservation of energy equates input and output power, $|{\tau }_1{\omega }_1|=|{\tau }_2{\omega }_2|$. The same 20-tooth/40-tooth pair therefore turns an input torque of $2\mathrm{N}\mathrm{m}$ into an output torque of $4\mathrm{N}\mathrm{m}$ while halving angular speed. This does not create energy: more torque accompanies less angular displacement per input revolution. Real gears dissipate energy through friction, and backlash permits a small lost motion on reversal; tooth strength and bearings, rather than the gear ratio alone, determine allowable load.

Gear friction work per engagement is the mechanical energy dissipated by sliding during one tooth pair's active engagement. It is the time integral of that pair's gear sliding-friction loss:The result has units of joules, whereas instantaneous loss has units of watts. It excludes other gearbox losses such as support friction. Manufacturer efficiency guidance distinguishes meshing and support losses and emphasizes their dependence on operating conditions.^[ref]

For a fixed-axis external spur gear pair, let signed contact coordinate $s$ run from $-L_a$ to $L_r$ through the gear pitch point. Gear sliding velocity has magnitude $(|{\omega }_1|+|{\omega }_2|)|s|$. At constant shaft speeds, gear base-roll speed gives $ds/dt=u_b>0$ in this chosen direction. If $\mu$ and the force carried by this pair are constant,With hypothetical $\mu =0.05$, $F_n=100\mathrm{N}$, angular-speed sum75rad/s, $L_a=5.0586\mathrm{mm}$, $L_r=4.5960\mathrm{mm}$ and $u_b=0.93969m/s$, this gives about9.32mJ. Doubling both shaft speeds doubles instantaneous power but halves gear contact duration, leaving this idealized work per engagement unchanged. Real friction and load can themselves change with speed.

Gear tooth load sharing is essential when several pairs engage simultaneously. A constant total mesh load is not a constant full load on each active pair. If identical pair events recur at gear mesh frequency $f_m$, averaging their correctly modeled losses over many events gives $\overline{P}=f_m{E}_f$. The event's force history must already include load sharing; assigning the whole mesh load independently to every overlapping pair would overcount transmitted load and loss.

Gear sliding-friction loss is the rate at which relative tooth-flank sliding dissipates mechanical energy. In a simple Coulomb friction model with normal contact load $F_n$, coefficient of friction $\mu$ and gear sliding velocity $v_s$, the nonnegative dissipated power isThis is a local contact model, rather than the entire gearbox loss. Manufacturer guidance distinguishes gear-meshing losses from losses in rotating supports and warns that overall gear efficiency depends on load and speed.^[ref]

To derive the sign, let $v_s=v_1-v_2$ along the common tangent. The friction force on surface1 is $f_1=-\mu F_n\mathrm{sgn}(v_s)$ and that on surface2 is $-f_1$. Their combined mechanical power is $f_1{v}_1-f_1{v}_2=-\mu F_n|v_s|$. The corresponding positive quantity is the rate of mechanical energy loss. At zero sliding the ideal Coulomb sliding-loss term vanishes, even though another resistance mechanism can remain.

For a fixed-axis external spur gear pair with $|{\omega }_1|+|{\omega }_2|=75rad/s$, $|s|=4\mathrm{mm}$ gives $|v_s|=0.30m/s$. Hypothetical $\mu =0.05$ and $F_n=100\mathrm{N}$ give $P_f=1.5\mathrm{W}$. Holding these load and friction assumptions fixed while doubling both shaft speeds doubles the instantaneous loss at that geometric point.

Gear friction work per engagement integrates this varying loss through the active gear path of contact. If one pair carries a constant $F_n$, the contact coordinate runs from $-L_a$ to $L_r$ at constant speed $\dot{s}=u_b>0$, givingFor $L_a=5.0586\mathrm{mm}$, $L_r=4.5960\mathrm{mm}$ and $u_b=0.93969m/s$, the preceding loads and speeds give about $9.32\mathrm{mJ}$. Real tooth pairs share varying loads; an overall loss calculation must sum the applicable contacts and integrate their actual loads, sliding and friction model.

Generalized forces are the coefficients of virtual work in a set of generalized coordinates. A translational coordinate has a force as its coefficient, while an angular coordinate has a torque. A generalized force conjugate to $q_i$ has units of energy divided by the units of $q_i$. Forces not already included in the Lagrangian enter the Euler-Lagrange equations as $d(\partial L/\partial {\dot{q}}_i)/dt-\partial L/\partial q_i=Q_i$.

Each Killing vector field supplies a conserved quantity for freely moving test particles: it turns a geometric symmetry into a first integral of the geodesic equation. Let $\gamma (\lambda )$ be a geodesic with affine parameter $\lambda$, tangent vector $U^{\mu }=d{x}^{\mu }/d\lambda$, and ${\nabla }_{U}U=0$. Wherever a Killing vector field $K$ is defined along the geodesic,The statement applies to Riemannian manifolds and to timelike geodesics, null geodesics and spacelike geodesics in a Lorentzian manifold. It requires an affine parameter; for a massive test particle, proper time is one choice. See Tong, section 4.3.2, equation (4.32).

For a worked example in the Euclidean plane, take $\gamma (t)=(1+t,2-t)$ with affine parameter $t$ and velocity $U=(1,-1)$. The Euclidean Killing vector fields for translation and rotation giveThe rotational field along the geodesic is $K_{\mathrm{rot}}(\gamma (t))=(t-2,1+t)$: it changes from $(-2,1)$ at $t=0$ to $(-1,2)$ at $t=1$, yet its pairing with $U$ stays $-3$. With toy mass $m=1$ and $t$ interpreted as time, the first constant is the $x$ component of linear momentum, and the second is the signed $z$ component of angular momentum about the origin. Their numerical values use toy units. The geodesic is straight; the rotation flow orbits are circles.

This conservation law follows from metric compatibility and the Killing equation:The geodesic equation kills the acceleration term, and the symmetric product $U^{\mu }{U}^{\nu }$ selects the vanishing symmetric derivative. This tensor contraction is the whole proof outline; see Carroll, equation (5.43).

In a stationary spacetime, a chosen time-translation Killing vector field $K$ gives Killing energy. With metric signature $(-+++)$, $c=1$, and a massive test particle's four-momentum $p^{\mu }=m{U}^{\mu }$, the physical normalization isHere $\Phi$ generates the axial rotational symmetry when one exists. In coordinates with $K={\partial }_t$ and $\Phi ={\partial }_{\varphi }$, these are $E=-p_t$ and $L_z=p_{\varphi }$. Both are conserved quantities along the geodesic, even if the metric has a time-angle cross term. A future timelike $K$ makes $E$ positive for a future massive particle; the scale of $E$ depends on the normalization of $K$. Not every spacetime has a global timelike Killing vector field, and a stationary field can become spacelike in an ergoregion of a stationary spacetime. Conservation on the field's domain does not imply a universal global positive energy; see Tong, sections 4.3.1-4.3.2 on the timelike-domain qualification and conserved charges.

Set the inertial cylindrical angle to $\Phi =\varphi +\omega t$. The Minkowski metric becomes $d{s}^2=(c^2-{\omega }^2{r}^2)d{t}^2-2\omega r^{2}dtd\varphi -d{r}^2-r^{2}d{\varphi }^2-d{z}^2$. For a massive geodesic parametrized by proper time, its cyclic coordinates give conserved covariant four-momentum components $p_t=m[(c^2-{\omega }^2{r}^2)\dot{t}-\omega r^2\dot{\varphi }]$, $p_{\varphi }=-m{r}^2(\dot{\varphi }+\omega \dot{t})$ and $p_z=-m\dot{z}$. Their combination gives $\dot{t}=(p_t-\omega p_{\varphi })/(m{c}^2)$. The radial Euler-Lagrange equation gives the displayed coordinate acceleration, although the free worldline is straight in inertial coordinates. The conserved inertial energy is $p_t-\omega p_{\varphi }$, and the inertial axial angular momentum is $-p_{\varphi }$.

Change in wave amplitude as the same propagating energy is distributed across changing wavefront area. It is distinct from intrinsic material absorption even though both can reduce observed amplitude with distance.

Integrating the hydrostatic pressure support equation with the radiative-zero white-dwarf envelope law yields $T=4GM(1/r-1/R)/(17\mathcal{R})$. Matching $T=T_c$ at $r_b$ gives $s=R-r_b$ as displayed. A strongly degenerate bulk core has thermal energy small compared with its Fermi energy, making $q\ll 1$ and hence $s\ll R$. The inequality $T_c\gg T_{\mathrm{eff}}$ alone does not establish geometrical thinness.

In a dilute, homogeneous population with number density $n$, an effective encounter cross section $\Sigma =\pi R_{\mathrm{enc}}^2$ gives the rate per target galaxyHere $R_{\mathrm{enc}}$ is the maximum encounter impact parameter, not automatically one galaxy's visible radius. If the cross section depends on relative speed, average the product instead: ${\Gamma }_{\mathrm{enc}}=n⟨\Sigma (v_{\mathrm{rel}})v_{\mathrm{rel}}⟩$. Gravitational focusing can create such speed dependence.

With independent encounters at a constant rate, a Poisson process givesThe approximation ${\Gamma }_{\mathrm{enc}}T$ is valid only for rare encounters. This is an encounter probability; it is not automatically a galaxy merger probability. Passing galaxies can remain unbound, and a bound pair can take additional time to coalesce.

As a simple additional model, suppose each encounter independently produces a merger within the relevant window with a fixed conditional probability $p$. The probability generating function of the Poisson count givesso the corresponding merger probability is $1-e^{-p{\Gamma }_{\mathrm{enc}}T}$. A realistic $p$ depends on encounter geometry, relative energy, angular momentum, gas and the allowed delay. It cannot be inferred from the geometric encounter rate alone.

For an entire volume $V$ of identical potential targets, the total distinct encounter-pair rate is $nV{\Gamma }_{\mathrm{enc}}/2$, because summing the per-target rate counts each unordered pair twice. This pair correction is separate from the conditional chance of merging.

Consider a layer governed by the linear shallow water equations of mean depth $H$ on an f-plane with nonzero constant Coriolis parameter $f$. Initially the surface is flat, the cross-strip velocity is zero, and the along-strip velocity is $u_0(y)=U{1}_{|y|<a}$. The two edges are opposite vortex sheets. Let $c=\sqrt{gH}$ and use the positive Rossby deformation radius $L_D=c/|f|$.

The conserved linearized shallow-water potential-vorticity anomaly is $q=v_x-u_y-f\eta /H$. Initially it isThe bounded final geostrophic balance has $u_f=-g{\eta }_f^{\prime }/f$ and henceThe decaying Green function of this one-dimensional operator is $-L_D{e}^{-|y|/L_D}/2$. Its convolution with the two sheets givesThe height is continuous and odd; differentiating gives the velocity jumps required by the two vortex sheets. Reversing $f$ reverses this height profile while leaving its positive decay length unchanged. The formula with $c/f$ as a decay length therefore requires the explicit restriction $f>0$.

Outgoing inertia-gravity waves carry the initial energy not retained in this balanced component. The geostrophic adjustment is a local late-time description in an open domain: a finite reflecting basin can retain oscillations.