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In the weak, non-crossing, leading-eccentricity model the exterior resonant coefficient has positive sign: write its disturbing function as $R_{\mathrm{res}}=B_q{e}_c^q\cos {\varphi }_{bc}$, with $B_q>0$ for the orders considered. The physical feedback can be seen for first order. Before an outer conjunction near apoapsis, the faster inner planet pulls the outer one backwards; after passing it pulls forwards. On the outward side, the nearer, stronger earlier pull wins, reducing the outer semi-major axis and increasing its mean motion. Thus ${\varphi }_{bc}<\pi$ is driven upward. On the inward side, the later forward pull wins and drives ${\varphi }_{bc}>\pi$ downward. The paired impulses cancel at the symmetric centre.

For higher order, sum the feedback over all $q$ conjunction phases. The first surviving angular Fourier harmonic is $\sin {\varphi }_{bc}$; the lower Fourier harmonics cancel over the equally spaced directions. A local Hamiltonian reduction with canonical momentum $J$ hasThe negative curvature of the Kepler orbit Hamiltonian gives the displayed sign. Linear stability analysis at $\pi$ gives a restoring acceleration. The symmetric libration centres of an eccentricity resonance are thereforeThe corresponding conjunction phases relative to periapsis are $\{\pi \}$, $\{\pi /2,3\pi /2\}$ and $\{\pi /3,\pi ,5\pi /3\}$. They suppress repeatedly close passages and provide resonance protection.

These are expected centres of the specified leading-harmonic model, not centres fixed solely by the integer order for arbitrary orbital eccentricity and masses. Asymmetric resonant-argument libration can occur when higher Fourier harmonics matter. For example, if $R=A\cos \varphi +B\cos 2\varphi$, $A>0$, $B>A/4$, the symmetric $\pi$ equilibrium loses stability and stable centres satisfy $\cos \varphi =-A/(4B)$. Thus further dynamical information is needed beyond the PDF's unrestricted eccentric orbit to assert unique centres. Exterior asymmetric branches are documented by Winter and Murray, “Resonance and chaos. II. Exterior resonances and asymmetric libration”.

Let $f(\mathbf{p})d^{3}p$ be particle number per unit volume in a momentum element, including the spin-state count in $f$. This is the local momentum part of a phase-space distribution function, normalized by the number density $n=\int f(\mathbf{p})d^{3}p$. For an isotropic distribution with particle energy $E(p)=\sqrt{m^2{c}^4+p^2{c}^2}$, the kinetic pressure of an isotropic gas is the average momentum flux:The kinetic energy density, explicitly excluding rest mass, isFor $p\ll mc$, $E-m{c}^2=p^2/(2m)$ and $\mathbf{v}=\mathbf{p}/m$, giving $P=2u_{\mathrm{kin}}/3$. For $p\gg mc$, $E\simeq pc$ and $v\simeq c$, giving $P\simeq u_{\mathrm{kin}}/3$. ThusNeither relation assumes a Maxwellian distribution. At intermediate momenta neither constant ratio is exact, and anisotropic distributions require a pressure tensor rather than this scalar pressure.

For a classical Maxwell-Boltzmann distribution, $f=n(2\pi m{k}_{B}T{)}^{-3/2}\exp [-p^2/(2m{k}_{B}T)]$. Three Gaussian component integrals give $⟨p^2⟩=3m{k}_{B}T$, so the ideal gas hasIn a fully ionized mixture, sum over independent species to obtain $P_{\mathrm{gas}}=\rho k_{B}T/({\mu }_{\mathrm{mol}}{m}_u)$, where the mean molecular weight counts ions and free electrons. This excludes partial-ionization and interaction corrections. A dilute classical ultrarelativistic gas still has $P=n{k}_{B}T$ but $u_{\mathrm{kin}}\simeq 3n{k}_{B}T$; its momentum distribution is proportional to $e^{-pc/(k_{B}T)}$ instead of the nonrelativistic Gaussian.

For a thermal photon gas, the two polarization states and zero photon chemical potential give the Planck photon distributionUse $x=pc/(k_{B}T)$ and ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation constant and photon equation of state are thereforeLikewise $n_{\gamma }=2\zeta (3)(k_{B}T/\hslash c{)}^3/{\pi }^2$. Unlike a gas with a fixed particle number, photons do not have pressure proportional to baryonic mass density.

For fully degenerate electrons, the Fermi-Dirac distribution becomes a filled momentum sphere with two spin states. Counting them gives the Fermi momentumHere ${\mu }_e$ is the mean molecular weight per electron. The equation of state of a cold electron gas follows by integrating momentum flux up to $p_F$:Its two limits areThe corresponding kinetic energy densities are $3n_e[p_F^2/(2m_e)]/5$ and $3n_e{p}_{F}c/4$, respectively. Thus in the high-density relativistic limit the electron pressure scales as ${\rho }^{4/3}$ and is nearly independent of temperature. This is the ideal noninteracting, fixed-composition electron result, not a universal equation of state at nuclear densities where captures, interactions and the composition change.

The boundaries on a stellar equation-of-state regime diagram concern the electron component. Define the electron relativistic density threshold and thermal electron relativistic threshold byA degenerate electron gas is nonrelativistic well below ${\rho }_{\ast }$ and ultrarelativistic well above it: this is an approximately vertical division on a log-density plot. A nondegenerate electron gas instead becomes thermally relativistic near $T_{\ast }$: this is an approximately horizontal division. The regimes have broad crossovers rather than a discontinuity at either line.

The exact electron Fermi temperature, subtracting electron rest energy, isFor electrons with negligible thermal pairs, strong degeneracy requires $T\ll T_F$, whereas $T\gg T_F$ gives a nondegenerate gas. In the pair-rich regime the actual electron and Positron distributions must instead be treated with their chemical potentials, as discussed below. The two limiting degeneracy boundaries areThese slopes $2/3$ and $1/3$ explain the bent degeneracy boundary on the logarithmic graph. A thermal-wavelength test $n_e[2\pi {\hslash }^2/(m_e{k}_{B}T){]}^{3/2}\sim 2$ gives the same nonrelativistic density/temperature scaling, with an order-one definition of the crossover.

The radiation-to-gas pressure boundary in the nondegenerate fully ionized regime isRadiation dominates above this line. Once electrons are strongly degenerate, compare radiation pressure with $P_e(\rho )$ instead of continuing the ideal-gas comparison into that regime. The radiation-to-degeneracy pressure boundary is $T=[3P_e(\rho )/a_{\mathrm{rad}}{]}^{1/4}$, with logarithmic slopes $5/12$ and $1/3$ in the nonrelativistic and ultrarelativistic limits. Degeneracy of the electrons and dominance of their pressure are distinct criteria.

For the pair curve, distinguish baryonic net electrons from thermally created electrons and Positrons. Chemical equilibrium with photons requires opposite electron/Positron chemical potentials when the one-particle energy includes rest energy, as in the distributions below. At low temperature and low degeneracy the zero-chemical-potential density per charge species isCharge neutrality gives $n_{-}-n_{+}=n_b=\rho /({\mu }_e{m}_u)$, while the nondegenerate equilibrium product is $n_{-}{n}_{+}=n_0^2$. HencePairs become important when $n_0$ is comparable with or exceeds $n_b$, approximately the electron-positron thermal pair abundance curve ${\rho }_{\mathrm{pair}}(T)={\mu }_e{m}_u{n}_0(T)$. Below its density at a fixed temperature, pairs dominate over the net charge electrons. The exponential makes the low-temperature portion steep in a log-log plot. For ${\mu }_e=2$, the full zero-potential integral gives approximate pair-marker densities $1.80\times 10^3\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $10^9\mathrm{K}$ and $7.52\times 10^5\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $3\times 10^9\mathrm{K}$. At relativistic temperature use the full zero-chemical-potential Fermi-Dirac distribution instead; it gives $n_0\to [3\zeta (3)/(2{\pi }^2)](k_{B}T/\hslash c{)}^3$, so the high-temperature pair curve approaches slope three in log-density versus log-temperature. Strong net-electron degeneracy suppresses Positrons and requires the full chemical-potential-dependent distribution.

The original diagram uses fully ionized helium, ${\mu }_e=2$ and ${\mu }_{\mathrm{mol}}=4/3$, only to set numerical locations. It plots the full $T_F$ curve and the zero-chemical-potential pair integral, rather than extending their asymptotes into the crossover. The pair boundary is deliberately an approximate abundance marker, not a phase transition; at high temperature the nonrelativistic ion approximation and fixed-composition model also have limits. For $T\gg T_{\ast }$ with abundant nondegenerate pairs, the combined electron/Positron energy density is $7a_{\mathrm{rad}}{T}^4/4$, so photons plus pairs have $u=11a_{\mathrm{rad}}{T}^4/4$ and $P=u/3$. Real stellar matter adds Coulomb effects, partial ionization, nuclear reactions and, at sufficiently high density, nuclear-matter physics beyond the ideal regime map.

Suppress tildes in the normalized equation and let $X=x-vt$. A Gross–Pitaevskii solitary wave stationary in these translating coordinates obeysThe boundary fixes the bulk complex argument and describes a localized disturbance with zero net vortex winding number; a vortex pair is possible. No extra plane-wave factor is needed when only the coordinates are changed. A full Galilean transformation of the field would additionally change the bulk flow and its complex argument convention.

Subtract the uniform background from the grand potential. A convenient dimensionless energy and the renormalized momentum of a condensate areThe subtraction in $\mathbf{p}$ removes the troublesome background complex argument gradient. In particular, for the usual subsonic localized far field $\psi -1=O(r^{-1})$, $\nabla \psi =O(r^{-2})$, and $1-|\psi {|}^2=O(r^{-2})$, the energy density is $O(r^{-4})$ and the momentum integrand is $O(r^{-3})$, both integrable in two dimensions. Merely writing $\psi \to 1$ without sufficient decay is not by itself a convergence proof. The common solitary-wave branch is subsonic: linearizing this normalization gives ${\omega }^2=k^2/2+k^4/4$ and speed of sound $1/\sqrt{2}$.

Integration by parts, with decaying variations and their boundary terms vanishing, givesThus the translating equation is precisely $\delta (E-v{p}_x)=0$. Along a differentiable family ${\psi }_{\lambda }$ with fixed bulk normalization, evaluate this stationarity on ${\partial }_{\lambda }\psi$ to obtainwhere $d{p}_x/d\lambda \ne 0$. This energy–momentum slope of a solitary wave does not require differentiating the Lagrange multiplier $v$ inside the field variation. At a turning point the parametrized identity remains true, while $E$ need not be a single-valued function of $p_x$ locally. The chosen factors make $i{\psi }_t=\delta E/\delta {\psi }^{\ast }$, consistent with the normalized equation; changing the normalization of $E$ alone would change the claimed slope.

The Maxwell equations with vacuum permittivity and permeability give $\rho ={ϵ}_0\nabla \cdot E$ and $J={\mu }_0^{-1}\nabla \times B-{ϵ}_0{\partial }_{t}E$. The current identity is the Ampère-Maxwell equation, and Faraday's law gives ${\partial }_{t}B=-\nabla \times E$. With $g={ϵ}_{0}E\times B$,The identity $(\nabla \times B)\times B=(B\cdot \nabla )B-\frac{1}{2}\nabla B^2$, and its electric analogue, turn this into the divergence of the Maxwell stress tensor $T$. Thus the tensor in the question is its negative:The vector $g$ is electromagnetic momentum density, and ${\sigma }_{ij}$ is the outward flux of its $i$th component across a surface normal in the $j$ direction. In static conditions the total force on matter is $F_i=-\oint {\sigma }_{ij}{n}_{j}dS$; generally subtract the time derivative of the enclosed electromagnetic field momentum as well.

For the two infinite sheets, superposition and the electromagnetic field jumps give $E=(\sigma /{ϵ}_0)e_x$, $B={\mu }_{0}K{e}_y$ between them and zero outside. Substitution yields exactlyA thin pillbox enclosing the $x=0$ sheet has only its $+x$ face in the interior electromagnetic field. ThereforeDirectly, omit the sheet's self-electromagnetic field: the other sheet supplies $E_{\mathrm{other}}=\sigma e_x/(2{ϵ}_0)$ and $B_{\mathrm{other}}={\mu }_{0}K{e}_y/2$. The Lorentz force per area is $\sigma E_{\mathrm{other}}+K{e}_z\times B_{\mathrm{other}}$, giving the same result. The electric contribution is attractive and the magnetic contribution repulsive.

If $K=\sigma v$ for a single moving charge population, use ${\mu }_0{ϵ}_0=c^{-2}$ to obtain $F/A={\sigma }^2(1-v^2/c^2)e_x/(2{ϵ}_0)$. For physical $|v|<c$ this is attractive, so motion of these charges cannot make the force repulsive. Independently imposed currents need not satisfy that restriction on $K/\sigma$.

For covariant quantization of the bosonic string,All other commutators between independent canonical variables vanish. The covariant Fock vacuum obeys ${\alpha }_k^m|0;p⟩=0$ for every $k>0$ and every $m$. Normal ordering moves these annihilating string oscillators to the right and givesThen ${\hat{N}}_{\mathrm{cov}}|0;p⟩=0$. Its commutator $[{\hat{N}}_{\mathrm{cov}},{\alpha }_{-k}^m]=k{\alpha }_{-k}^m$ grades a Fock state basis by $N=\sum k{n}_{km}$, although its inner product is indefinite because of the timelike oscillator.

The quantum Virasoro algebra has central charge $D$:The Virasoro central extension comes from commutators needed to order infinite oscillator sums; it is a quantum effect absent from the classical Poisson brackets. Replacing a classical Poisson bracket by a commutator cannot recover that term without a regularized ordering calculation.

If a state were annihilated by every nonzero ${\hat{L}}_j$, the $j=1,k=-1$ commutator would imply ${\hat{L}}_0|\psi ⟩=0$. The $j=2,k=-2$ commutator would then imply $(D/2)|\psi ⟩=0$. For $D>0$, only the zero vector is annihilated by all nonzero Virasoro constraints. This explains why only positive modes annihilate a physical string state.

The vacuum is a physical string state when $p^2=2\pi Ta$. At level one all states have the form $|A;p⟩=A_m{\alpha }_{-1}^m|0;p⟩$. Since $[{\hat{L}}_j,{\alpha }_k^m]=-k{\alpha }_{j+k}^m$, the positive-mode Virasoro constraints giveThus the complete level-one conditions and norm arewith the common vacuum normalization suppressed.

For $a>1$, choose spacelike momentum $p^m=(0,k,0,\dots )$, $k^2=2\pi T(a-1)$. A purely timelike polarization has $A\cdot p=0$ and norm $-1$. Hence a negative-norm physical string state exists when $a>1$.

For $a<1$, the mass-shell condition gives $M^2=2\pi T(1-a)>0$. In a rest frame, $A\cdot p=0$ forces $A_0=0$, leaving $D-1$ positive-norm vector-particle polarizations, those of a massive vector.

For $a=1$ and nonzero null momentum, $A\cdot p=0$ leaves a null direction $A_m\propto p_m$. The corresponding null string state is $p\cdot {\alpha }_{-1}|0;p⟩=\sqrt{\pi T}{\hat{L}}_{-1}|0;p⟩$. It is orthogonal to every physical string state because ${\hat{L}}_1$ annihilates them. Quotienting by this gauge redundancy, $A_m\sim A_m+\zeta p_m$, leaves $D-2$ positive vector-particle polarizations. Thus the level-one spectrum agrees with light-cone gauge in string theory at $a=1$. This level-one argument alone does not establish consistency or absence of negative norms at all higher levels.

In the displayed version of light-cone gauge in string theory, the oscillators have $D-2$ transverse components andThe center-of-mass Poisson brackets remain $\{x^m,p_n\}={\delta }_n^m$, and the transverse oscillator Poisson brackets are $\{{\alpha }_j^a,{\alpha }_k^b\}=-ij{\delta }^{ab}{\delta }_{j+k,0}$. Other independent brackets vanish. With $\hslash =1$, canonical quantization gives the canonical commutation relationsThe oscillator Fock vacuum at momentum $p$ is defined by ${\alpha }_k^a|0;p⟩=0$ for $k>0$. It is the ground state of one string, rather than the empty spacetime vacuum. Set $a_k^a={\alpha }_k^a/\sqrt{k}$ for $k>0$. The normal ordering prescription gives the string level operatorA Fock state basis is obtained by applying ${\prod }_{k,a}((a_k^a{)}^{†}{)}^{n_{ka}}/\sqrt{n_{ka}!}$ to $|0;p⟩$, with string level operator eigenvalue $N={\sum }_{k,a}k{n}_{ka}$. In particular, its level-one states areThey transform as the transverse vector of the little group rotation subgroup $SO(D-2)$, precisely the vector-particle polarizations of a massless vector. A massive vector would instead require $D-1$ vector-particle polarizations. This conclusion uses a quantization compatible with the Lorentz group of the bosonic string theory.

The classical mass constraint alone has no quantum zero-point energy shift. Its quantum version includes the normal-ordering constant of a string $a$:Masslessness at level one fixes $a=1$, soThus the ground state is a tachyon. Without the quantum ordering shift, the displayed classical constraint would give $M_1^2=2\pi T$ and would not support the stated massless interpretation. In the usual transverse vacuum regularization, $a=(D-2)/24$; consistency with $a=1$ also gives the critical dimension of string theory $D=26$.

The real canonical variables $x^m,p_m$ describe the center-of-mass position and total momentum of the open string. The nonzero string oscillators are complex Fourier modes with ${\alpha }_{-k}^m=({\alpha }_k^m{)}^{\ast }$; the Lagrange multipliers obey ${\lambda }_{-j}={\lambda }_j^{\ast }$. The Virasoro constraints satisfy $L_{-j}=L_j^{\ast }$, so the multiplier term is real.

For each $k>0$, the oscillator kinetic term differs from the manifestly real expressionby $\frac{i}{2k}\frac{d}{dt}({\alpha }_k\cdot {\alpha }_{-k})$. Thus the written action is real up to a boundary term, which does not change its symplectic form or bulk dynamics. Adding the corresponding endpoint term makes reality exact.

The independent nonzero Poisson brackets areAll brackets between these independent center-of-mass and oscillator variables vanish. The dependent zero mode is ${\alpha }_0^m=p^m/\sqrt{\pi T}$, so, if it is used, $\{x^m,{\alpha }_0^n\}={\eta }^{mn}/\sqrt{\pi T}$.

The Fourier coefficients of the Virasoro constraints areTheir Poisson brackets areThey form the classical Witt algebra, with no Virasoro central extension. In particular they are first-class constraints, closing on the constraint surface, and generate the remaining worldsheet diffeomorphisms rather than independent physical excitations.

A collective coordinate describes a position, orientation, or another parameter of a family of static solitons. For a family $\Phi (\mathbf{x};q)$, promote $q$ to a slowly varying $q(t)$ and substitute into the field action. In a scalar theory with unit kinetic coefficient, this gives the collective-coordinate effective LagrangianFor a gauge-theory soliton, one also solves the Gauss law constraint in gauge theory constraint and projects out pure gauge transformations; arbitrary variations of gauge representatives do not define the physical metric. Tangent vectors to an exactly equal-energy family are zero modes in field theory. If $V=E_0$ is constant, the Euler-Lagrange equations of this moduli-space approximation arethe geodesic equations of its Riemannian metric. The approximation neglects radiation and deformation modes; it describes motion sufficiently slow that these omitted degrees of freedom remain unexcited to the required accuracy.

In collective-coordinate quantization, take the wavefunction measure $\sqrt{\det g}{d}^{n}q$ and the minimal scalar Hamiltonian operatorThe Laplace-Beltrami operator supplies coordinate-invariant kinetic energy. Global identifications and statistics must be imposed on the wavefunctions; the classical metric alone does not choose them. Curvature-ordering terms and loop corrections are additional quantum input.

For the phi-four kink, use the normalization and profile of Question 1. Substituting $\varphi (x,t)={\varphi }_K(x-X(t))$ givesThe metric is constant because of translation invariance. This proves the translational dynamics of a phi-four kink: classically the centre moves at constant velocity, and quantum mechanicallyPlane waves label the continuous translational momentum, with no position-dependent potential. Uniform-motion Lorentz invariance upgrades the dispersion to $\sqrt{M^2+p^2}$; the displayed collective Lagrangian is its small-velocity expansion. Small perturbations also include an internal shape mode and continuum radiation, which this single collective coordinate omits. The fluctuation operator of a phi-four kink in this normalization isthe translational eigenfunction has ${\omega }^2=0$, the shape mode has ${\omega }^2=3c^2$, and continuum modes in the spectrum start at ${\omega }^2=4c^2$. Thus the free-coordinate states describe the kink's low translational energies, not its full excitation spectrum or quantum mass correction.

For two Abelian Higgs vortices at critical coupling, the static energy is $2M_v$ and the Abelian Higgs vortex moduli space has four real dimensions. Let $z_1,z_2$ be their positions, $Z=(z_1+z_2)/2$, and $q=z_1-z_2$. The centre of mass decouples; the relative metric is rotationally symmetric and can be writtenAt large separation, $F(\rho )\to M_v/2$, recovering two free particles. Although there is no static separation potential, the nonconstant metric produces velocity-dependent interaction. Coincidence is smooth in the relative coordinate for two identical vortices $w=q^2$, not in the double-valued $q$. Smoothness gives $F(\rho )\sim C{\rho }^2$ for $C>0$ near zero. A head-on geodesic continues through $w=0$ to the opposite real ray, so $q$ changes its line by $\pi /2$: the vortices scatter through a right angle. This geometric argument does not require an explicit formula for $F$.

With ordinary bosonic exchange statistics, relative wavefunctions are single-valued in $w$ and smooth at coincidence. In the separated polar coordinate they obey $\Psi (\rho ,\theta +\pi )=\Psi (\rho ,\theta )$, with even integer angular labels. Their kinetic operator isThe apparent singularity at $\rho =0$ must be resolved with the smooth $w$ coordinate and regularity, rather than arbitrary boundary conditions on a punctured cone. The free centre-of-mass motion and the asymptotically free relative geometry give quantum scattering states; a flat static energy does not imply that the metric is flat or that scattering is absent. This is not a prediction of a discrete family of static two-vortex bound separations. The smooth collision geometry is developed in David Tong's arxiv.org/abs/hep-th/0509216.

For a Skyrmion of baryon number one, the Skyrmion hedgehog ansatz iswhere $\mathbf{\sigma }$ are the Pauli matrices. Include a centre $\mathbf{X}$ and an orientation $A\in SU(2)$ through $U=A{U}_0(\mathbf{x}-\mathbf{X})A^{-1}$. Hedgehog symmetry identifies spatial rotations with opposite internal rotations, so there are three independent orientation coordinates, not six. Since $A$ and $-A$ give the same field, the physical orientation space is $SO(3)$, with SU(2) group as its double cover. Write $A^{-1}\dot{A}=i\mathbf{\omega }\cdot \mathbf{\sigma }/2$. The leading collective Lagrangian has the formwhere $\Lambda >0$ is the rotational moment of inertia obtained by integrating the profile's field kinetic energy.

For the fermionic quantization appropriate to baryons, the Finkelstein-Rubinstein constraints on the double cover impose $\Psi (-A)=-\Psi (A)$. In SU(2) representations, the central element $-1$ acts by $(-1{)}^{2j}$, so $j$ must be half-integer. Left and right group actions supply isospin and spin angular momentum; hedgehog symmetry makes their magnitudes equal. The rotational quantization of a unit Skyrmion therefore givesThe $j=1/2$ level has four spin-isospin states and models the nucleon doublet, proton and neutron, each with two spin states. The $j=3/2$ level has sixteen states and models the Delta baryon quartet, each with four spin states. The rotor predicts a splitting $3{\hslash }^2/(2\Lambda )$. Without the fermionic sign, single-valued functions on $SO(3)$ would instead allow integer $j$, which is a different quantization. High rotor levels can couple to deformation and pion radiation; this semiclassical approximation does not establish that its entire formal tower consists of stable particles.

The outward radiative flux is $F=L/(4\pi r^2)$. A photon of energy $E$ carries momentum $E/c$, so the momentum transferred per unit volume and time to matter is $\kappa \rho F/c$, when $\kappa$ is the appropriate transport opacity. Momentum conservation makes the radiation stress decrease outward by this amount. In the thin atmosphere, where the radiation stress is treated by its scalar radiation pressure,This is the first moment of the radiative transfer equation. The grey atmosphere approximation uses a frequency-independent transport opacity; in an optically thick diffusion regime it becomes the Rosseland mean opacity. Near the surface the plane-parallel atmosphere approximation removes geometric stress-divergence corrections.

Let $p$ denote the fluid pressure perturbation divided by the reference mass density, and let $b=-g{\rho }^{\prime }/{\rho }_0$ be the buoyancy perturbation. The buoyancy frequency satisfies $N^2=-g{\hat{\rho }}_z/{\rho }_0$. Linearization requires small boundary slope $k|{\eta }_0|\ll 1$, small displacements compared with the vertical scales of the disturbance and background, and advection small compared with the oscillatory acceleration. For a propagating internal gravity wave this includes $|m{\eta }_0|\ll 1$; boundary slope alone is insufficient when $|m|\gg k$. The Boussinesq approximation also requires small relative mass density differences over the region of interest. These conditions must hold for the resulting disturbance, including any amplification by resonance.

The inviscid Linearized Boussinesq equations and linear kinematic boundary condition areWrite the velocity field as $(u,w)=\mathrm{Re}\{(U(z),W(z))e^{i(kx-\omega t)}\}$. Incompressibility and horizontal momentum balance giveThe vertical momentum balance therefore yieldsFor a real vertical wavenumber $m$, this gives the dispersion relationWhen $0<\omega <N$, define $\ell =k\sqrt{N^2/{\omega }^2-1}$. The radiation condition selects $m=-\ell$, because energy must leave the boundary upward. The boundary-forced internal gravity wave isIts wavevector is $\mathbf{K}=(k,m)$, its phase velocity is ${\mathbf{c}}_p=\omega \mathbf{K}/(k^2+m^2)$, and its group velocity isThus the phase velocity points down and right, while the group velocity points up and right. They are perpendicular. Internal-wave polarization makes the particle motion an oscillation along the group velocity direction, with zero first-order mean transport. The constant-phase lines of an internal gravity wave are also parallel to the group velocity. Their inclination $\vartheta$ above the horizontal satisfies $\sin \vartheta =\omega /N$.

When $\omega >N$, set $\kappa =k\sqrt{1-N^2/{\omega }^2}$. Boundedness at infinity selects the evanescent waveThe horizontal and vertical velocity components are in quadrature: fluid particles describe small ellipses, and the response decays over ${\kappa }^{-1}$. There is no upward time-averaged energy flux, because $P$ and $W$ are in quadrature. A real vertical group velocity is not defined for this evanescent wave. The pattern travels horizontally with phase velocity $\omega /k$.

For $N=0$ the response is the evanescent wave with $\kappa =k$; the ratios $\omega /N$ should not be used. At the cutoff $\omega =N>0$, the bounded harmonic solution has $W=-i\omega {\eta }_0$, $U=0$: it neither decays nor has nonzero upward group velocity. It is the limiting cutoff response, rather than a localized radiating disturbance. A uniform nonzero buoyancy frequency in an infinitely deep Boussinesq approximation is itself a local idealization of the background mass density.

The arrows distinguish the phase velocity, group velocity, and oscillatory particle motion of the internal gravity wave. The right panel shows the decay envelope and particle ellipses of the evanescent wave.

Use the shallow-water approximation, a hydrostatic vertically mixed layer, a deep motionless ambient, and no ambient fluid entrainment. Particle loss changes reduced gravity but changes total layer volume only at the discarded dilute-particle order. Likewise thermal expansion changes the equation of state while the leading Boussinesq approximation retains volume conservation. DefineDepth integration of mass conservation, horizontal momentum balance, particle transport, and the heating law givesHere the integrated excess hydrostatic pressure is $b{h}^2/2$. There is no particle source from the bed. Bed drag and the particle contribution to inertia are neglected at this order. The equivalent variable-buoyancy shallow water equations areThe term $h{b}_x/2$ is necessary when heating or particle loss makes the reduced gravity vary horizontally.

For $v=(h,u,\varphi ,\theta {)}^T$, the quasilinear system $v_t+A(v)v_x=S(v)$ hasIts characteristic polynomial factors asThus the characteristic curves have slopesOn the two repeated contact characteristic curves, the particle and heating laws are $D\varphi =-W_s\varphi /h$ and $D\theta =\beta \varphi$. Writing $c_g=\sqrt{bh}$ and $D_{\pm }={\partial }_t+(u\pm c_g){\partial }_x$, the gravity-wave compatibility relations areFor $h>0$ and $b>0$, gravity-wave right eigenvectors can be taken as $(1,\pm c_g/h,0,0{)}^T$. The contact eigenspace has dimension two: $\delta u=0$ and $b\delta h+(h/2)\delta b=0$, with independent $\delta \varphi ,\delta \theta$. Therefore $A$ has a full set of real eigenvectors.