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Use the spatial-potential convention of this paper and assume $\bar{\rho }+\bar{P}\ne 0$. A passive time displacement changes the spatial scalar by $\tilde{\Psi }-\Psi =\bar{N}H{\xi }^0$ and changes the energy density byThe two contributions to the uniform-density curvature perturbation therefore cancel:A spatial scalar displacement changes $E$ but not this combination. Hence $\zeta$ is gauge invariant at linear order. Pure cosmological-constant matter, for which $\bar{\rho }+\bar{P}=0$, does not supply a energy density clock and this formula is not defined for that component alone.

Now take the definite constant-$w$ barotropic equation of state and use physical-time differentiation $\mathcal{D}={\bar{N}}^{-1}{\partial }_t$. Put $A=\bar{\rho }+\bar{P}$. Background stress-energy conservation gives $\mathcal{D}A=-3H(1+w)A$. Differentiating $\zeta =-\Psi +\delta \rho /(3A)$, then substituting the two supplied perturbation equations, givesThus the lapse and expansion perturbations cancel without imposing a gauge. In terms of the non-adiabatic pressure perturbation $\delta P_{\mathrm{nad}}=\delta P-w\delta \rho$,For the stipulated barotropic equation of state, $\delta P=w\delta \rho$, so the first term vanishes. For a Fourier mode, $\Delta =-k^2/a^2$. For regular long-wavelength solutions the remaining gradients are suppressed when $k\ll aH$. Consequently $\dot{\zeta }\simeq 0$ on superhorizon scales, with exact conservation in the leading zero-gradient limit. It is not an exact finite-$k$ identity. Entropy perturbations or a singular gradient expansion can invalidate that conclusion; the supplied adiabatic assumption is what removes the unsuppressed pressure source.

This superhorizon conservation of uniform-density curvature is useful because inflationary fluctuations leave the horizon during cosmic inflation and may re-enter only much later. In an adiabatic single-clock evolution, the conserved uniform-density curvature perturbation transports their initial amplitude through reheating and the later radiation in cosmology and matter eras without solving every short-distance process. It then determines the initial conditions for large-scale structure and Cosmic microwave background anisotropy. Extra fluctuating fields or fluids can introduce isocurvature perturbations and convert them into curvature, so conservation must be checked rather than assumed during such transitions.

Retain the homogeneous lapse function $\bar{N}(t)$, and define the physical Hubble parameter $H=\dot{a}/(\bar{N}a)$. At first order the scalar metric perturbations give $\delta g_{00}=-2{\bar{N}}^2\Phi$, $\delta g_{0i}=a^2{B}_{,i}$, and $\delta g_{ij}=-2a^2(\Psi {\delta }_{ij}+E_{,ij})$. Applying the passive coordinate displacement component by component gives the scalar gauge transformations with a background lapseFor example the $0i$ transformation is $\delta {\tilde{g}}_{0i}=a^2{\partial }_{i}B-a^2{\partial }_i\dot{\lambda }+{\bar{N}}^2{\partial }_i{\xi }^0$. The spatial isotropic part acquires $-2a\dot{a}{\xi }^0{\delta }_{ij}$, explaining the positive sign in the transformation of $\Psi$.

Preserving synchronous gauge requires both transformed $\Phi$ and $B$ to remain zero. The first condition is ${\partial }_t(\bar{N}{\xi }^0)=0$, and the second is $\dot{\lambda }={\bar{N}}^2{\xi }^0/a^2$. Integrating them givesThus fixing the lapse function and shift vector perturbations does not exhaust the scalar coordinate freedom. The functions $C,D$ describe residual synchronous-gauge freedom; an integration-origin change in the integral is absorbed into $D$.

To move from synchronous gauge to Newtonian gauge, choose $\lambda =-E_S$ to set $\tilde{E}=0$. Requiring $\tilde{B}=0$ then givesA scalar energy density transforms as $\delta {\rho }_N=\delta {\rho }_S-\dot{\bar{\rho }}{\xi }^0$. Dividing by the background energy density, which changes the contrast only at second order if perturbed in that denominator, yields the synchronous-to-Newtonian density transformation with a lapseThe last form uses background stress-energy conservation, $\dot{\bar{\rho }}=-3\bar{N}H(\bar{\rho }+\bar{P})$. With cosmic time $\bar{N}=1$ this is ${\delta }_N={\delta }_S-3(1+w)a^{2}H{\dot{E}}_S$; with conformal time $\bar{N}=a$ it is ${\delta }_N={\delta }_S-3(1+w)\mathcal{H}{E}_S^{\prime }$, where $\mathcal{H}=a^{\prime }/a$. Keeping these clock conventions distinct is essential to the factors of $a$.

Let $\mathcal{H}=a^{\prime }/a$ and ${\theta }_N=i\mathbf{k}\cdot {\mathbf{v}}_N$. Choose the synchronous gauge rest frame of cold dark matter, so ${\theta }_C=0$. Its continuity equation then gives $h^{\prime }=-2{\delta }_C^{\prime }$. Substitution in the trace Einstein equation givesFor radiation in cosmology the continuity and Euler equations are ${\delta }_R^{\prime }+4{\theta }_R/3+2h^{\prime }/3=0$ and ${\theta }_R^{\prime }=k^2{\delta }_R/4$. Differentiate the first, eliminate ${\theta }_R^{\prime }$ and $h^{\prime \prime }$, and obtainradiation in cosmology pressure supplies the acoustic restoring term while its energy density contributes twice as strongly to the trace gravitational source through $1+3w_R=2$.

Deep in radiation domination, $a\propto \tau$, $\mathcal{H}=1/\tau$, ${\Omega }_R\simeq 1$ and ${\Omega }_C\simeq 0$. At long wavelengths $k\tau \ll 1$, the adiabatic initial conditions require ${\delta }_C=3{\delta }_R/4$. The cold-matter equation reduces toA power ${\tau }^p$ gives $p^2-4=0$. The regular growing solution therefore hasThe other adiabatic solution diverges as ${\tau }^{-2}$ and is discarded for the regular primordial growing mode. The relation between the density contrasts fixes their relative entropy to zero, not their subsequent equality inside the acoustic horizon.

One can also solve the leading radiation-era equations explicitly, which proves that the same regular mode develops logarithmic cold-matter growth after horizon entry. Put $x=k\tau /\sqrt{3}$. The first equation gives ${\delta }_R=(x^2{\delta }_{C,xx}+x{\delta }_{C,x})/3$. Substituting in the second givesDefineIts useful derivative is $F^{\prime }=1/x-2\sin x/x^2+2(1-\cos x)/x^3$. Direct differentiation verifies the fourth-order equation and gives the associated radiation in cosmology solution:At zero, $F=x^2/8+O(x^4)$ and ${\delta }_R=K{x}^2/6+O(x^4)$, so this is precisely the regular adiabatic radiation-era cold-dark-matter transfer solution with $K=24A/k^2$. The other indicial behaviors of the fourth-order equation are $1,x,x^{-2}$; they are either nonadiabatic at leading order or singular and do not add another regular growing adiabatic mode.

For $x\gg 1$, the integral is $\gamma +\ln x-\mathrm{Ci}(x)$, where $\gamma$ is the Euler's constant and $\mathrm{Ci}$ the cosine integral. Its oscillatory tail cancels the explicit leading $\sin x/x$ term, givingThus the requested late sound-horizon behavior isAn additive constant in ${\delta }_C$ is absorbed into the reference ${\tau }_{\ast }$; the regular adiabatic solution fixes the acoustic phase, with $D=0$ in the explicit time-origin convention above. A general acoustic mixture has both terms. The physical expansion parameter is $k\tau /\sqrt{3}$; comparisons with $2\pi /k$ describe the same long- and short-wavelength regimes up to fixed numerical factors.

The contrast grows much more slowly after entry during radiation domination: radiation in cosmology undergoes pressure-supported acoustic oscillations and cold matter only the logarithmic Mészáros effect. Modes entering well before matter-radiation equality consequently have suppressed late amplitudes relative to modes that stay outside the horizon until near equality. After matter domination begins, pressureless growing modes can grow as $a$. This scale-dependent history produces the turnover and small-scale suppression of the matter transfer function and affects the initial conditions for large-scale structure of the universe. The early subhorizon limit requires both $k\tau \gg 1$ and $\tau \ll {\tau }_{\mathrm{eq}}$; it is not applicable to a mode that enters only in the matter domination.

Before decoupling, rapid Thomson scattering makes the photon distribution nearly isotropic in the local photon-fluid rest frame. A local blackbody temperature perturbation changes its energy density by ${\delta }_{\gamma }=4\delta T/T$. A first-order boost by the fluid velocity adds a temperature photon dipole $\mathbf{n}\cdot \mathbf{v}$; it does not generate a photon quadrupole until second order in that boost. Thus the photon brightness perturbation at instantaneous decoupling isHigher angular moments are suppressed by the photon mean-free-path to perturbation-length ratio in the tight-coupling approximation. The idealized equilibrium assumption neglects those finite-mean-free-path corrections, polarization effects and finite thickness of the last-scattering surface. It is not an assertion that collisionless photons remain isotropic after decoupling. The distinction between a fluid treatment and the subsequent hierarchy is discussed in the primary synchronous and Newtonian perturbation treatment.

Let $\mu =\hat{\mathbf{k}}\cdot \mathbf{n}$ and $W(\tau )=e^{ik\mu (\tau -{\tau }_0)}$. Multiplication of the synchronous photon brightness equation by its integrating factor givesIntegration from ${\tau }_{\ast }={\tau }_{\mathrm{dec}}$ to ${\tau }_0$ and division by four yieldFor the real-space observation point ${\mathbf{x}}_0$, the corresponding background ray is $\mathbf{x}(\tau )={\mathbf{x}}_0-\mathbf{n}({\tau }_0-\tau )$. Fourier inversion gives the synchronous Sachs-Wolfe line-of-sight formulaEvery source is evaluated at its appropriate point on the ray, rather than at the observation position. Here $\mathbf{n}$ is photon propagation direction, opposite to the outward observer-to-source sky direction; changing that convention changes the written sign of the Doppler projection consistently.

The first term is the intrinsic photon temperature at last scattering. On wavelengths larger than the sound horizon, it retains primordial adiabatic information; inside that horizon the photon-baryon fluid has acoustic energy density oscillations, producing the acoustic temperature peaks. The second term is the Doppler effect from the fluid's bulk motion at emission. A regular long-wavelength velocity is gradient-suppressed; the Doppler contribution is important near and below the acoustic horizon, with its acoustic phase displaced from that of the energy density oscillation.

The integral is the gravitational energy shift produced by the changing spatial metric tensor along the photon path. In synchronous gauge it includes endpoint gravitational redshifts as well as evolving-potential effects. It must not be identified solely with the Integrated Sachs-Wolfe effect: in Newtonian gauge the same total separates into the ordinary last-scattering Sachs-Wolfe effect, local observer terms, and an integral of time-varying potentials. The ordinary contribution is important on scales outside the sound horizon at decoupling. A late integrated contribution is strongest on large angular scales when dark energy or curvature changes the potentials; evolution around matter-radiation equality also contributes near the first acoustic scales.

The relevant comoving sound scale is $r_s({\tau }_{\ast })={\int }^{{\tau }_{\ast }}{c}_s(\tau )d\tau$, while the projection distance is $D_{\ast }\simeq {\tau }_0-{\tau }_{\ast }$. Modes project roughly to photon temperature multipole $\ell \sim k{D}_{\ast }$, and the acoustic scale is $\ell \sim D_{\ast }/r_s$, with successive peak spacing of order $\pi D_{\ast }/r_s$. At much shorter wavelengths Silk damping and the finite last-scattering thickness suppress energy density and velocity anisotropies; the instantaneous-decoupling approximation alone does not model that cutoff. These scale distinctions explain why the three terms cannot be assigned one common range of important wavelengths.

Set $s=a(t)/a(t_{\mathrm{dec}})$. Free propagation in the homogeneous expanding universe redshifts frequencies by $\nu ={\nu }_{\mathrm{dec}}/s$ and dilutes physical number density by $s^{-3}$. Number conservation in corresponding frequency intervals givesInserting the initial Planck spectrum, the factor $s^{-2}(s\nu {)}^2$ becomes ${\nu }^2$, while its exponent is $hs\nu /(k{T}_{\mathrm{dec}})$. HenceThe shape is preserved even without collisions maintaining thermal equilibrium; it follows from redshift of an initially Planckian distribution.

For the total photon number density put $x=h\nu /(kT)$. For the energy density include the energy $h\nu$ of each photon. Thenwhere $\alpha =(8\pi /c^3)(k/h{)}^3{\int }_0^{\infty }{x}^2/(e^x-1)dx$ and $\xi =(8\pi h/c^3)(k/h{)}^4{\int }_0^{\infty }{x}^3/(e^x-1)dx$. Both integrals are finite positive constants. Thus $n\propto a^{-3}$ and $ϵ\propto a^{-4}$, with the extra power supplied by photon-energy redshift.

For the flat FRW metric, $g^{00}=-1$, $g^{0i}=0$, and $g^{ij}=a^{-2}{\delta }^{ij}$. Consequently the kinetic contraction of the scalar field isInserting this in $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi +g^{\mu \nu }\mathcal{L}$, with ${\partial }^0\varphi =-\dot{\varphi }$ and ${\partial }^i\varphi =a^{-2}{\partial }_i\varphi$, givesThese are contravariant coordinate components of the stress-energy tensor; spatial components in an orthonormal frame carry different powers of $a$. In particular $T^{00}$ is the energy density seen by the comoving observer.

With zero scalar potential and a homogeneous scalar field, the preceding perfect fluid identification givesThis equation of state is that of a stiff fluid. The field equation integrates to $a^3\dot{\varphi }=K$, since ${\partial }_t(a^3\dot{\varphi })=0$. Hence $\rho =K^2/(2a^6)$, agreeing with the cosmological continuity equation for $w=1$. If this nonzero component dominates a flat universe, the Friedmann equation giveson the expanding branch. Integrating once more, $a^3=3\sqrt{4\pi G/3}|K|(t-t_B)$. Choosing the Big Bang at $t_B=0$ yieldsThe homogeneous free scalar as a stiff fluid construction requires $K\ne 0$; a constant zero-potential field has zero energy density and cannot dominate this expansion.

Before electron-positron annihilation in cosmology, photons contribute two helicity states; Electrons and positrons contribute two spin states each. Each of the three light neutrino flavours contributes one active neutrino helicity and one antineutrino helicity. At a common temperature the effective number of relativistic energy degrees of freedom is thereforeUsing $G_F=1.17\times 10^{-5}{\mathrm{GeV}}^{-2}$ and $M_{\mathrm{Pl}}\simeq 2.435\times 10^{18}\mathrm{GeV}$ in the scaling formula alone gives $T_F\sim 2\mathrm{MeV}$. Including the explicit expansion coefficient from (ii), but setting the unknown rate coefficient $A$ to one, gives $T_F\simeq 1.5\mathrm{MeV}$. Thus the warranted conclusion is a freeze-out temperature of order $1\mathrm{MeV}$, above the $0.511\mathrm{MeV}$ Electron mass. The omitted weak-rate coefficient prevents a more precise prediction from this estimate.

Neutrino decoupling occurs before most electron-positron pairs annihilate. Afterwards neutrinos redshift freely, $T_{\nu }a=\mathrm{constant}$, while the photons receive the pair entropy. For the still-coupled photon-pair gas, cosmological entropy conservation gives $g_{\ast s}{T}_{\gamma }^3{a}^3=\mathrm{constant}$. Its entropy degrees of freedom drop from $2+(7/8)4=11/2$ to $2$. Comparing this with the freely cooling neutrinos givesAfter annihilation, photon and neutrino temperatures differ. To write the total energy density as $({\pi }^2/30)g_{\ast }{T}_{\gamma }^4$, the neutrino contribution must include that temperature ratio:This is the paper's post-annihilation ${\mathcal{N}}_{\mathrm{eff}}$. Merely deleting the Electron term while keeping all species at the same temperature would be incorrect. For comparison the entropy count, normalized to $T_{\gamma }^3$, is $g_{\ast s}=2+(7/8)6(4/11)=43/11\simeq 3.91$. These results assume instantaneous neutrino decoupling and relativistic neutrinos; the two counts represent different thermal moments.

During radiation domination, the effective number of relativistic energy degrees of freedom, denoted $g_{\ast }$, gives the energy densityThe paper's ${\mathcal{N}}_{\mathrm{eff}}$ is this $g_{\ast }$; it is not the conventionally normalized effective number of neutrino species. With the specified reduced Planck mass, the Friedmann equation reads $H^2={\rho }_R/(3M_{\mathrm{Pl}}^2)$, whenceWrite the weak interaction conversion rate as $\Gamma =A{G}_F^2{T}^5$. The ratio $\Gamma /H$ scales as $T^3$, so it falls during cooling: conversion eventually becomes too slow to maintain neutron-proton chemical equilibrium. Equating the conversion and expansion rates givesOn suppressing the numerical coefficient, the cosmological weak freeze-out temperature is consequentlyThe one-sixth power reflects a balance between a conversion rate proportional to $T^5$ and a Hubble parameter proportional to $\sqrt{g_{\ast }}{T}^2$.

In the Boussinesq approximation, replace density by a constant reference density ${\rho }_0$ in inertia and incompressible flow, retaining the small density departure only in the gravitational force. Write $\rho =\bar{\rho }(z)+{\rho }^{\prime }$ and subtract the resting hydrostatic pressure. The upward buoyancy perturbation acceleration and squared buoyancy frequency areThus stable density stratification has $N^2>0$. Let $p$ henceforth denote the pressure anomaly divided by ${\rho }_0$. The ideal, nonrotating Boussinesq equations areThe buoyancy equation follows by materially conserving the total buoyancy $b_0(z)+\sigma$, where $b_0^{\prime }=N^2$; it does not assume constant $N$.

For an $x$-periodic flow of period $P$, define $\bar{a}=P^{-1}{\int }_0^{P}adx$. An infinite-domain average with vanishing end contributions works as well. The necessary pressure condition is no mean zonal pressure gradient, $\overline{p_x}=0$: periodic pressure satisfies it, whereas a pressure containing a term proportional to $x$ does not. Use incompressible flow to write the zonal momentum equation in conservative form,Averaging its zonal derivative terms yields the exact Reynolds stress relation

For two-dimensional small disturbances about rest, the Linearized Boussinesq equations are $u_t=-p_x$, $w_t=-p_z+\sigma$, ${\sigma }_t=-N^{2}w$, and $u_x+w_z=0$. Differentiate the two momentum equations to eliminate pressure: $(u_z-w_x{)}_t=-{\sigma }_x$. Differentiating in $x$ and using $u_{xz}=-w_{zz}$ gives ${\nabla }^2{w}_t={\sigma }_{xx}$. One further time derivative givesThe equation remains valid for $N=N(z)$. No vertical derivative of $N$ was taken: ${\partial }_x^2(N^{2}w)=N^2{w}_{xx}$.

For constant $N$ and $k\ne 0$, a plane internal gravity wave hasBoth signs give the prescribed lower-boundary velocity. The radiation condition excludes energy arriving from infinity. At positive frequency, the vertical group velocity isFor a source below the fluid, energy must propagate upward, so the outgoing solution has $m<0$, even though its vertical phase velocity $\omega /m$ is downward.

To obtain this selection causally, let $T=\mu t$ and $Z=\mu z$, and replace the constant amplitude by $a(Z,T)$. The method of multiple scales treats $(z,t,Z,T)$ as independent during differentiation: the carrier derivatives hold $Z,T$ fixed, while $a_T$ holds $Z$ fixed and $a_Z$ holds $T$ fixed. Only after differentiating do we restrict to $T=\mu t$, $Z=\mu z$. Acting on the complex wave, the operators becomeWith $K^2=k^2+m^2$, the wave equation has residualThe zeroth-order term is the dispersion relation; the first-order term vanishes precisely whenIf $a_0(T)$ is the switched-on boundary amplitude, with $a_0(T)=0$ for $T<0$, the outgoing envelope is $a(Z,T)=a_0(T-Z/b)$ for $b>0$. Its characteristics carry the boundary signal into initially undisturbed fluid. If $b<0$, characteristics instead arrive from the upper half-space; zero initial disturbances and no incoming signal cannot establish the proposed boundary-forced wavetrain. This is the internal-wave envelope radiation condition.

The leading incompressible flow polarization gives $u^{\prime }=-(m/k)w^{\prime }$. Hence, to second order in amplitude and leading order in the slow modulation,There is no meridional transport here. Envelope transport gives ${\partial }_t|a{|}^2=-b{\partial }_z|a{|}^2$, so an initially vanishing mean develops asThese are the leading expressions for internal-wave momentum deposition; higher slow-modulation corrections to the stress are beyond the displayed approximation. In particular, a spatially uniform fully established wavetrain has zero stress divergence, but the front has already produced its mean flow.

The mean wave energy density per unit mass follows from ${\sigma }^{\prime }=-i{N}^2{w}^{\prime }/\omega$:Thus $\bar{u}=(k/\omega )\mathcal{E}$. In the original frame the mean kinetic energy is ${\bar{u}}^2/2=O(a^4)$. In a frame translating at $c=\omega /k$, the undisturbed velocity is $-c$, and the change in mean kinetic energy is $-c\bar{u}+O(a^4)=O(a^2)$. Indeed $\mathcal{E}-c\bar{u}=0$ at this order behind the front. Mean-flow energy is therefore essential at the same order as wave energy in the translating frame. This follows also from the frame transformation of energy, $E\mapsto E-cP$ up to the fixed background constant, and is why conservation of momentum cannot be omitted from the moving-frame conservation of energy budget.

Assume a nonrotating, nonmagnetic equilibrium, inviscid adiabatic perturbations, and no dissipation. Let $g={\varphi }^{\prime }(R)>0$, so hydrostatic equilibrium is $p^{\prime }=-\rho g$. Use $\mathbf{\xi }{e}^{i\omega t}$ for the fluid displacement. To avoid a source ambiguity, write Eulerian fluid perturbations as $p_1,{\rho }_1,{\varphi }_1$, and use $\Delta$ for quantities following the displaced material element, as in Eulerian and Lagrangian fluid perturbations. They are related by $\Delta p=p_1+{\xi }_R{p}^{\prime }$ and similarly for mass density and potential.

The printed identity needs two corrections: its compression denominator must be $\gamma p$, not $\gamma \rho$, and the pressure and potential appearing in this form of the identity must be Eulerian fluid perturbations. The printed denominator has the wrong dimensions for an energy density. The claimed Lagrangian interpretation does not produce the displayed buoyancy-compression split; the following derivation specifies the consistent version.

Linearized mass conservation, the adiabatic relation, and the Poisson equation giveConsequently $p_1=-\gamma p\nabla \cdot \mathbf{\xi }-{\xi }_R{p}^{\prime }$. Linearizing the Euler equations for an inviscid fluid about the static star gives the linear adiabatic stellar oscillation equationsRequire regularity at the center, a vanishing Lagrangian surface pressure, and the usual matching to a decaying exterior gravitational potential. Because $p$ and $\rho$ vanish at the surface, the pressure boundary term vanishes for regular displacements; compactly supported trials also suffice below. The perturbed Newtonian gravitational potential solves Laplace's equation outside the star and must be included there when its field energy is computed.

Multiply the momentum equation by $-{\mathbf{\xi }}^{\ast }$ and integrate. The pressure term is $-{\int }_V{p}_1\nabla \cdot {\mathbf{\xi }}^{\ast }dV$. Let $d=\nabla \cdot \mathbf{\xi }$. The adiabatic relation and hydrostatic equilibrium implyThe pressure and equilibrium-gravity terms then combine pointwise aswhere $N^2=g[(1/\gamma )p^{\prime }/p-{\rho }^{\prime }/\rho ]$ is the stellar buoyancy frequency. The potentially troublesome cross terms cancel exactly.

For the perturbed self-gravity term, integration by parts givesThus the corrected stellar displacement energy identity isThe right side is a real quadratic form. With these conservative boundary conditions, the force operator is a self-adjoint operator. Its Rayleigh quotient characterizes squared mode frequencies, and a negative admissible trial quotient proves an unstable spectral component; it does not require the trial displacement itself to be an exact eigenfunction.

For a generic compressive disturbance of scale $\ell$ in a region with slowly varying equilibrium coefficients, estimate ${\rho }_1\sim \rho \xi /\ell$, $p_1\sim \gamma p\xi /\ell$ and ${\varphi }_1\sim 4\pi G{\rho }_1{\ell }^2$. The self-gravity and compression energy densities have orderswith adiabatic sound speed $c_s^2=\gamma p/\rho$. Under the plausible stellar estimate $c_s^2\sim G\bar{\rho }{a}^2$, global modes can have comparable gravity and pressure contributions, while their ratio for small-scale compressive modes is of order $(\rho /\bar{\rho })(\ell /a{)}^2$. Thus perturbations with $\ell \ll c_s/\sqrt{4\pi G\rho }$ admit the Cowling approximation. This estimate compares two nonzero compressive terms; it cannot alone justify neglecting gravity for a trial deliberately chosen to have $p_1=0$.

For an explicit convective trial, choose $c<d<e<b$ and a nonzero smooth bumpLet $Y$ be a normalized spherical harmonic of degree $\ell \ge 1$, let $L=\ell (\ell +1)$, and let ${\nabla }_{\Omega }$ be the unit-sphere gradient. The pressure-free localized stellar convective trial isIt is supported away from the center and surface, so it satisfies every needed mechanical boundary condition. Since ${\Delta }_{\Omega }Y=-LY$,Therefore $p_1=-\gamma p\nabla \cdot \mathbf{\xi }+\rho g{\xi }_R=0$ exactly. Its buoyancy energy is ${\int }_d^e\rho N^2{R}^2|f{|}^{2}dR<0$, whereas its kinetic norm ${\int }_d^e\rho R^2(|f{|}^2+L|h{|}^2)dR$ is positive. In the Cowling approximation this gives a negative Rayleigh quotient, and hence $N^2<0$ in an interior shell implies convective instability. Retaining perturbed self-gravity only adds a nonpositive term, so the same trial proves instability in the full conservative problem too.

There is also a quantitative small-scale check for this very trial. Its mass density amplitude is ${\rho }_1=(\rho N^2/g)fY$, independent of $\ell$. Write ${\rho }_1=s(R)Y$ and ${\varphi }_1=F(R)Y$. The radial Poisson equation givesBy the Cauchy-Schwarz inequality, $A\le 4\pi G(\int R^4|s{|}^{2}dR{)}^{1/2}(A/L{)}^{1/2}$, and so the angular-degree bound on perturbed stellar self-gravity isIt tends to zero with horizontal scale $R/\ell$, while the strictly negative buoyancy integral is fixed. This justifies the small-scale neglect of gravity even though the trial's compression term vanishes.

Let $e=\rho u^2/2+p/(\gamma -1)$ be the total gas energy density, with $\gamma >1$. The one-dimensional continuity and momentum equations areUsing continuity to put the momentum equation into conservative form givesThe internal energy density $e_{\mathrm{int}}=p/(\gamma -1)$ satisfiesby the adiabatic pressure equation. Multiplying the momentum equation by $u$ and using continuity givesAdding these equations combines the pressure work into $-{\partial }_z(pu)$. The conservation law variables and conservation law fluxes are thereforeThe three rows express conservation of mass, momentum and total energy. Their integral conservation laws remain meaningful across a shock wave, where the differential pressure and velocity equations cannot be applied pointwise.