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Take $P_0>0$, real $\alpha$, and the real branch $T\ge T_0$, with physical temperature $T>0$. In hydrostatic equilibrium the pressure decreases with height $z$:Thus an atmospheric thermal inversion means $dT/dz>0$, equivalently $dT/dP<0$. For $\alpha \ne 0$, the logarithm of the given relation yieldsThe sign restriction must be retained after squaring. In particular, the inverted branch has $0<P\le P_0$, whereas the non-inverted branch has $P\ge P_0$. By differentiation, on the interior of either branch,Consequently,At $T=T_0$, the inverse temperature gradient vanishes at the branch endpoint, rather than producing a finite isothermal interval. If $\alpha =0$, the relation becomes $P=P_0$ independently of $T$; it is not a nontrivial vertical atmospheric pressure-temperature profile under hydrostatic equilibrium. These distinctions prevent the two sides of the squared parabola from being interpreted as one permissible atmosphere.

The functional form is not specific to exoplanets. With positive pressure normalization and physical temperature, a square-root exponential atmospheric profile can approximate a monotone interval of the terrestrial atmosphere. An outward-cooling interval, such as part of the troposphere or mesosphere, requires $\alpha >0$. An upward-warming interval, such as part of the stratosphere heated by ultraviolet absorption or the thermosphere heated by high-energy radiation, requires $\alpha <0$.

The local atmospheric lapse rate follows from the ideal gas relation $P=\rho k_{B}T/(\mu m_H)$ and hydrostatic equilibrium:For dry terrestrial air, approximately $\gamma =1.4$ and $\mu \simeq 29$ give the dry-adiabatic lapse rate $g/c_p\simeq 9.8\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$. The familiar mean tropospheric value near $6.5\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ is less steep; a local fit must satisfy $\nabla \lesssim 2/7$ to be dry-convectively stable. Moist convection needs the moist parcel thermodynamics instead, and the terrestrial atmosphere is not uniformly dry or chemically homogeneous at all heights.

The squared-logarithm shape cannot reproduce an exactly constant nonzero lapse rate over an arbitrary thick region, all the alternating atmospheric layers, or a finite exactly isothermal region. It is a local parametrization with a fixed sign of the temperature gradient; it has no terrestrial universality.

In the collisionless exosphere, an atom escapes if its outward trajectory has positive total mechanical energy. Neglect tides and stellar forces and use Newtonian gravity at exobase radius $r_e$:With thermal speed $v_{\mathrm{th}}=\sqrt{2k_B{T}_e/m}$, the Jeans escape parameter isA thermal distribution always has an escaping tail; Jeans escape is exponentially suppressed for ${\lambda }_e\gg 1$, with Jeans escape flux proportional to $(1+{\lambda }_e)e^{-{\lambda }_e}$. Efficient escape requires ${\lambda }_e$ of order a few or smaller, with an order-unity energetic estimate $k_B{T}_e\sim G{M}_{p}m/r_e$.

Assume a Neptune-like mass and radius, $r_e\simeq R_p$, and atomic hydrogen. Using the supplied rounded constants givesHenceUsing the mean kinetic energy $3k_{B}T/2$ instead gives an order-unity coefficient and $T\simeq 2.3\times 10^4\mathrm{K}$. An expanded exobase has weaker binding and lowers the estimate by $R_p/r_e$. A comet-like tail can also be shaped by radiation pressure and stellar-wind interactions; it does not by itself measure $T_e$ or prove that a hydrostatic Jeans model is valid. At ${\lambda }_e\sim 1$, $H/r_e\sim 1$ and hydrostatic equilibrium fails as a global description: substantial mass loss must usually be treated as hydrodynamic atmospheric escape.

The exobase is defined by mean free path $\ell \sim H$, not by a universal pressure. For a neutral hydrostatic gas with collision cross-section ${\sigma }_c$,For example, explicitly assuming ${\sigma }_c\sim 10^{-19}\text{–}10^{-20}{\mathrm{m}}^2$ gives $P_e\sim 2\times 10^{-7}\text{–}2\times 10^{-6}\mathrm{Pa}$, or $2\times 10^{-12}\text{–}2\times 10^{-11}\mathrm{bar}$. These are representative extremely dilute neutral-exobase pressures, with orders of magnitude varying with composition, cross-sections and expansion. The supplied constants contain no collision information, so they cannot uniquely determine an exobase pressure; ionization or a non-hydrostatic density profile changes this estimate.

Assume an isothermal terminator, constant gravity and mean molecular weight, and use the stated nightside atmospheric scale height as the terminator value. Hydrostatic equilibrium gives $P(z)=P_{\mathrm{ref}}{e}^{-(z-z_{\mathrm{ref}})/H}$. A band radius three atmospheric scale heights above a measured continuum at $0.1\mathrm{bar}$ therefore corresponds toThis is the band's characteristic pressure, not automatically the top of an exoplanet cloud deck. The distinction matters because an opaque cloud generally raises the continuum and removes the lower part of a molecular feature; it does not cap the height of a strong molecular line above the cloud.

If the quoted $0.1\mathrm{bar}$ is the actual observed continuum and the cloud dominates its opacity, the consistent estimate is insteadIf $0.1\mathrm{bar}$ refers to a clear-atmosphere reference continuum, an additional clear band height is necessary. Let that height be $NH$. A cloud at height $z_c$ leaves a residual feature of $3H$ when $NH-z_c=3H$. Thus the cloud pressure degeneracy of a transmission feature givesFor example, a $5\mathrm{mbar}$ cloud estimate requires the extra assumption that the clear band spans six scale heights, so the cloud raises the continuum by three. Identifying the cloud with the level three scale heights above the original continuum gives the same numerical value, but is not justified by the observed band height alone. Without a clear-band opacity or equivalent reference radius, there is no unique cloud-top pressure. A strong dayside emission feature also does not supply that missing terminator information.

For a quasi-static self-gravitating object of fixed mass, negligible surface pressure, and no accretion, nuclear source or additional deep heating, conservation of energy gives the intrinsic planetary luminosityHere $U$ is internal energy, and $\Omega$ is negative gravitational potential energy. Gravitational contraction makes $\Omega$ more negative and releases power $L_{\mathrm{grav}}\equiv -d\Omega /dt>0$. Some of that release raises the internal energy; it cannot all be radiated while the object maintains hydrostatic equilibrium.

To quantify the split assume thermal gas pressure with constant specific-heat ratio $\gamma >4/3$ and negligible rotation or magnetic support. This excludes the dynamically unstable gas-supported regime. The stellar virial theorem givesIt follows thatFor a monatomic ideal gas, $\gamma =5/3$, so $L_{\mathrm{int}}=L_{\mathrm{grav}}/2$: half the released gravitational energy heats the gas and half escapes. If the density profile remains homologous, write $\Omega =-qG{M}^2/R$, with $q>0$ a fixed structure constant. Thenup to the structure and virial coefficients, where ${\tau }_{\mathrm{KH}}$ is the Kelvin-Helmholtz cooling time.

For a diatomic gas with $\gamma =7/5$ the idealized fraction is $1/6$, already demonstrating that the half-factor is not universal. Partial electron degeneracy pressure, dissociation, changing structure, accretion and deep heating require the full energy equation. The general link between luminosity and Kelvin-Helmholtz contraction is $L_{\mathrm{int}}=-d(U+\Omega )/dt$, with any genuine source terms added explicitly.

Assume a Newtonian, spherically symmetric, nonrotating body in hydrostatic equilibrium, supported by a polytropic equation of state $P=K{\rho }^{1+1/n}$ with constant $K>0$ and $n>0$. The enclosed mass and hydrostatic pressure support equation areEliminate $m$ by first writing $r^2{\rho }^{-1}dP/dr=-Gm$ and then differentiating:Let ${\rho }_c$ be central mass density, and introduce the Lane-Emden variables for a stellar polytropewhereThen $(1/\rho )dP/dr=(n+1)K{\rho }_c^{1/n}d\theta /dr$. Substitution cancels the dimensional factors and yields the Lane-Emden equationThe central conditions enforce the chosen central mass density and regular spherical symmetry. The local regular expansion is $\theta =1-{\xi }^2/6+n{\xi }^4/120+\cdots$. If a finite first zero ${\xi }_1$ exists and surface pressure is negligible, the physical radius is $R=a{\xi }_1$, and the Lane-Emden mass formula is $M=4\pi a^3{\rho }_c[-{\xi }_1^2{\theta }^{\prime }({\xi }_1)]$. For $0<n<5$ the standard isolated solution has a finite surface; $n=5$ has infinite extent, so a finite surface must not be assumed for all indices. Irradiation, composition stratification and non-polytropic equations of state require more general structure equations.

For a quasistatic spherical star with fixed total mass, the stellar structure equations in radius areHere $s$ is the specific entropy of a moving mass shell, and the entropy term accounts for release or storage of thermal and gravitational energy. In thermal equilibrium it vanishes. An equation of state, opacity, energy-generation law and composition evolution close the equations. The absence of mass loss fixes the outer mass coordinate; it does not require the star to be in thermal equilibrium.

The radiative temperature gradient needed to carry the full luminosity isHere the actual temperature gradient is $\nabla =d\log T/d\log P$. For uniform composition the Schwarzschild criterion says that a radiative layer is stable if ${\nabla }_{\mathrm{rad}}<{\nabla }_{\mathrm{ad}}$, with marginal stability at equality, where the adiabatic temperature gradient is $(\partial \log T/\partial \log P{)}_s$. Use $\nabla ={\nabla }_{\mathrm{rad}}$ there. If ${\nabla }_{\mathrm{rad}}>{\nabla }_{\mathrm{ad}}$, convection carries some of the flux. Efficient convection gives $\nabla \simeq {\nabla }_{\mathrm{ad}}$; inefficient surface convection needs a transport prescription and can be superadiabatic. For composition gradients, use the Ledoux criterion, with threshold ${\nabla }_{\mathrm{ad}}+\varphi {\nabla }_{{\mu }_{\mathrm{mol}}}/{\delta }_{\mathrm{th}}$ instead. Here ${\nabla }_{{\mu }_{\mathrm{mol}}}=d\log {\mu }_{\mathrm{mol}}/d\log P$, $\varphi =(\partial \log \rho /\partial \log {\mu }_{\mathrm{mol}}{)}_{P,T}$ and ${\delta }_{\mathrm{th}}=-(\partial \log \rho /\partial \log T{)}_{P,{\mu }_{\mathrm{mol}}}$, using the mean molecular weight ${\mu }_{\mathrm{mol}}$.

Above the thin burning shell, take constant luminosity $L$, constant mean molecular weight ${\mu }_{\mathrm{mol}}$, ideal-gas pressure $P=\mathcal{R}T\rho$ with $\mathcal{R}=k_B/({\mu }_{\mathrm{mol}}{m}_u)$, and the Kramers opacity law $\kappa ={\kappa }_0\rho T^{-7/2}$. Then the temperature equation isTo implement the printed assumption that $P,\rho$ and $M_r$ are all power laws while retaining mass conservation, write $M_r\propto r^m$, $\rho \propto r^b$, $P\propto r^p$ and $T\propto r^d$. The mass equation gives $m=b+3$; the ideal gas law gives $d=p-b$; hydrostatic equilibrium gives $d=m-1$. The diffusion equation gives $d-1=2b-2-13d/2$. Solving this linear system gives the self-gravitating Kramers power-law envelope:The coefficients can also be matched consistently. If $r_c$ is the core boundary and $M_c$ its enclosed mass,Radiative transfer fixes the constant luminosity throughwhose right-hand side is independent of radius for these exponents. Thus this is a solution of all four envelope equations, not just a dimensional estimate.

Matching the boundary temperature to the isothermal core and extrapolating the idealized power law to the specified photosphere givesFor reference $M_R/M_c=(R/r_c{)}^{1/11}=2500^{1/10}\simeq 2.19$, so the envelope mass is not negligible in this particular solution. The given $M_c\simeq M_{\odot }$ fixes the core-radius normalization $r_c=(11/42)G{M}_c/(\mathcal{R}10^7\mathrm{K})$; a numerical absolute radius needs the unspecified molecular weight. The exponent ratio also gives $\nabla =d/p=5/21<2/5$, so the fully ionized monatomic version of this idealization is convectively stable.

A different common approximation neglects the envelope's self-gravity and sets $M_r\simeq M_c$ in the force equation. It cannot obey the exact mass equation with a nonzero density and exactly constant $M_r$. Under that additional approximation, the same calculation instead givesThis is the constant-core-mass power-law opacity radiative envelope limit, not the full power-law solution with changing enclosed mass. Stating which approximation is used resolves the otherwise different numerical radius ratios.

A compact nearly isothermal core, a thin luminosity-producing shell and a much larger cool envelope describe a shell-burning red giant. The full power-law profile is a deliberately simple model. The Kramers opacity law and an ionized ideal gas are not reliable at a $4000\mathrm{K}$ photosphere; partial ionization, other opacity sources and a convective envelope usually modify real cool giants. The quoted radius ratios are extrapolations within the stipulated model.

Use a gas-pressure-supported, optically thick alpha disk, with $\mathcal{R}=k/({\mu }_m{m}_p)$ and positive parameters. Here $\sigma$ is the Stefan–Boltzmann constant in the Stefan–Boltzmann law, and $F_{+}$ and $F_{-}$ below are heating and cooling flux estimates per disk face. Hydrostatic equilibrium and the ideal gas relation give $\mathcal{R}T\sim {\Omega }^2{H}^2$, while $\rho \sim \Sigma /H$. Integrating viscous heating over one semi-thickness and estimating radiative diffusion giveHere the stipulated negative hydrogen ion opacity exponent is $\kappa ={\kappa }_0{\rho }^{1/3}{T}^{10}$, which is a local approximation distinct from other empirical opacity fits. Substituting $T\sim {\Omega }^2{H}^2/\mathcal{R}$ and equating heating and cooling givesDropping numerical constants of order one is essential here. This is the negative-hydrogen-opacity alpha-disk thickness scaling. Its angular thickness estimate isFor a central mass $M$, Keplerian rotation gives $\Omega =(GM/r^3{)}^{1/2}$, so $H/r\propto r^{53/82}\Sigma (r{)}^{-7/41}$. The isothermal sound speed $c_s^2=\mathcal{R}T$ isAn adiabatic sound speed differs by the constant factor $\sqrt{\gamma }$ if the adiabatic index is fixed. For $\Sigma \propto r^{-p}$, constant $\alpha$, and fixed composition, the disk aspect ratio grows as $r^{(53+14p)/82}$, so there is disk flaring when $p>-53/14$; $c_s\propto r^{(6+7p)/41}$. The problem supplies no numerical parameters or specified $\Sigma (r)$, so no unique numerical $H/r$ or unconditional radial power can be inferred.

If the additional standard closure of a constant inward accretion rate well outside a zero-torque edge is intended, $\bar{\nu }\sim \alpha \Omega H^2$ and $\bar{\nu }\Sigma \simeq \dot{M}/(3\pi )$ imply $\Sigma \propto (\alpha \Omega H^2{)}^{-1}$. Eliminating it gives $H^9\propto {\Omega }^{-38/3}$ at fixed $\dot{M},\alpha$, hence $H/r\propto r^{10/9}$ and $c_s\propto r^{11/18}$. These stronger powers use that extra steady-flow assumption. All these local scalings cease to apply if the thin disk or the stated opacity regime fails.

In the vertical equations, $\mu$ denotes dynamic viscosity, while ${\mu }_m$ is the mean molecular weight in units of the proton mass $m_p$. The symbol $k$ denotes the Boltzmann constant. For the intended hydrostatic equilibrium calculation, put $P=K{\rho }^{\gamma }$ with $K>0$, $\gamma >1$, and set $\mathcal{R}=k/({\mu }_m{m}_p)$. Dividing the hydrostatic equation by $\rho$ and integrating from the symmetric midplane givesThe specific enthalpy vanishes at the free surface. Hence the polytropic vertical structure in stellar gravity isOn $|z|<H$, the ideal gas relation gives $T=T_m(1-z^2/H^2)$ and $P=P_m(1-z^2/H^2{)}^{\gamma /(\gamma -1)}$; the vacuum extension sets $\rho =P=0$ outside. The formula needs $\gamma >1$ for a finite positive semi-thickness and assumes the harmonic vertical gravity remains valid through this height.

There is a genuine inconsistency if all four printed structure equations are imposed literally with $F=0$ and finite positive opacity: radiative diffusion then requires $dT/dz=0$, whereas the polytropic temperature above has a nonzero gradient away from the midplane. The intended result is valid for the hydrostatic polytropic equation of state with the radiative closure replaced or neglected, for example an adiabatic opaque limit. It is not an exact simultaneous solution of the stated finite-opacity radiative system. With that radiative system and a nonvacuum ideal gas, $\gamma =1$ instead gives an isothermal solutionwhich has no finite free surface. For $\gamma \ne 1$, constant $T$ and constant $K$ would force constant density, contradicting hydrostatic balance on a nontrivial vertical interval. This is the zero-radiative-flux obstruction for a polytropic ideal gas.

Assume the vertical dynamical timescale of a disk is shorter than the thermal timescale of an accretion disk, which is itself shorter than the viscous timescale. This ordering is controlled in a thin disk with $\alpha \ll 1$: $t_{\mathrm{dyn}}\sim {\Omega }^{-1}$, $t_{\mathrm{th}}\sim (\alpha \Omega {)}^{-1}$ and $t_{\nu }\sim (r/H{)}^2{t}_{\mathrm{th}}$. Therefore compare heating and cooling at fixed surface density of a disk, radius, $\alpha$ and composition. Vertical hydrostatic equilibrium adjusts on the vertical dynamical timescale of a disk, giving $H\propto T^{1/2}$ and $\rho \propto T^{-1/2}$. Consequently the stipulated negative hydrogen ion opacity varies as $\kappa \propto T^{10-1/6}=T^{59/6}$. The rates per disk face then scale asThus heating increases while cooling decreases after a positive temperature perturbation. At equilibrium $Q^{+}=Q^{-}=Q_0$, a small change gives $\delta (Q^{+}-Q^{-})=(41/6)(Q_0/T)\delta T$, which drives the perturbation further from equilibrium for positive effective heat capacity. A negative temperature perturbation likewise increases the net cooling. This proves thermal instability from negative hydrogen ion opacity within the local closure; it is the temperature-slope test of thermal stability of an accretion disk. Holding volume density rather than surface density of a disk fixed would give a different cooling exponent, but still the same qualitative instability; the fixed-column comparison is the relevant thin-disk thermal test.

For a thin spherical shell, the radial pressure force per unit volume is $-dP/dr$. By the shell theorem, the inward gravitational acceleration is $GM(r)/r^2$, because exterior spherical shells contribute no force. Balancing force densities in hydrostatic equilibrium givesThe second equation is mass conservation; it closes the pressure-support equation once an equation of state is supplied.

Assume an isothermal atmosphere in hydrostatic equilibrium, with constant gravity $g$, mean molecular weight $\mu$, and an ideal gas equation of state. Use $h$ for the atmospheric scale height, to distinguish it from the total observable height $H$. Thenso the cloud-top pressure in an isothermal atmosphere isAn opaque, approximately grey exoplanet cloud deck can flatten the exoplanet transmission spectrum if the remaining gas above it has little wavelength-dependent extinction. If its top nearly reaches the observable atmospheric extent, $H_c\simeq H$, this gives $P_c\simeq P_s{e}^{-H/h}$. More generally, a top at $N$ atmospheric scale heights has $P_c/P_s=e^{-N}$; illustrative values $N=5$ and $10$ give $6.7\times 10^{-3}$ and $4.5\times 10^{-5}$.

A flat exoplanet transmission spectrum and $P_s$ alone do not determine a unique cloud-top pressure. The temperature, gravity, mean molecular weight and cloud altitude, or equivalent opacity information, are also needed. A large mean molecular weight can itself suppress spectral features, so the exoplanet cloud deck interpretation is conditional.

This is a semi-grey irradiated atmosphere in radiative equilibrium. The quantities have these meanings:

The model assumes a static plane-parallel atmosphere, separate grey stellar and thermal bands with constant opacities, local thermodynamic equilibrium for thermal emission, negligible scattering, and transport by radiative transfer rather than convection. The Eddington closure approximation sets the thermal angular-moment ratio to $1/3$, and the upper boundary supplies the usual $2/3$ term. Treating the incoming radiation with representative direction cosine $1/\sqrt{3}$ gives its attenuation $e^{-\sqrt{3}\gamma \tau }$. Intrinsic flux enters from below, while the imposed external stellar flux is absorbed from above.

With constant gravity and hydrostatic equilibrium, $\tau \simeq {\kappa }_{\mathrm{IR}}P/g$, allowing conversion to an atmospheric pressure-temperature profile. This last relation additionally assumes constant thermal opacity. The grey treatment describes the energy balance approximately; it does not resolve individual molecular absorption lines.

Take $z$ to increase inward, so the positive coefficient $A$ describes an inward-increasing temperature. Constant gravity and hydrostatic equilibrium give $dP/dz=\rho g$, henceMatching the radiative temperature gradient to the adiabatic temperature gradient gives the formal local boundary relationThe same result follows from radiative diffusion: for constant upward thermal flux $F$ and Rosseland mean opacity $\kappa$, $A=3\kappa F/(16{\sigma }_{\mathrm{SB}})$.

There is an important limitation to treating $A$ as constant over the entire radiative layer. Integration from an irradiated outer boundary givesFor a diatomic ideal gas, ${\nabla }_{\mathrm{ad}}=2/7>1/4$, so this profile cannot actually reach a radiative-convective boundary. Formally, imposing a constant ${\nabla }_{\mathrm{ad}}$ would givewhich is positive only for $0<{\nabla }_{\mathrm{ad}}<1/4$. A finite boundary for a normal molecular atmosphere requires additional opacity, flux, or thermodynamic variation. The local matching formula is usable near a real boundary, but constant $A$ is not a complete global model of it. This is the convective stability of a constant-opacity irradiated atmosphere.

For the intended order-of-magnitude scaling, suppose the local values of $g/A$ and ${\nabla }_{\mathrm{ad}}$ are comparable for Jupiter and a hot Jupiter, and assume $T_b$ scales with planetary equilibrium temperature. Equal absorbed-flux factors around the same stellar luminosity give $T_b\propto a^{-1/2}$. Taking $a_J=5.2\mathrm{AU}$ and a representative close-in orbit $a_h=0.05\mathrm{AU}$ yieldsThis illustrates how irradiation can push a boundary much deeper. It is a conditional estimate calibrated from the supplied $0.1\mathrm{bar}$ reference, not a self-consistent prediction of the globally constant-$A$ model. Different intrinsic cooling flux, opacity, gravity, or atmospheric metallicity of a giant planet can substantially alter it.

Consider a spherical stellar polytrope with $P=K{\rho }^{1+1/n}$, constant $K$, $0<n<5$, zero surface pressure and finite radius $R$. Its specific enthalpy, measured from zero density, is $h={\int }_0^{P}d{P}^{\prime }/\rho =(n+1)P/\rho$. Hydrostatic equilibrium gives $h+\Phi =\Phi (R)=-GM/R$. Integrating over mass therefore yieldsThe static stellar virial theorem gives $3\int PdV=-\Omega$, and eliminating the pressure integral proves the gravitational energy of a stellar polytrope:For an ideal gas with constant specific-heat ratio, the accompanying internal energy isTo express both energies using $\gamma$ alone one additionally identifies the model as an adiabatic stellar polytrope, so $n=1/(\gamma -1)$. ThenFor $\gamma =5/3$ these become $\Omega =-6G{M}^2/(7R)$ and $U=3G{M}^2/(7R)$. A general polytropic index need not be determined by the thermodynamic specific-heat ratio; without the adiabatic identification both $n$ and $\gamma$ remain.

Take $\gamma >1$ for the ideal gas. At fixed cylindrical radius, put $\mathcal{R}=k/(\mu m_p)$, where $k$ is the Boltzmann constant, $\mu$ the mean molecular weight and $m_p$ the proton mass. Also write $a=9\alpha \Omega /4$. Vertical hydrostatic equilibrium, thermal equilibrium and the ideal gas law requireConsequently, with $K=A/a$,Integrating from the midplane yieldsThe polytropic vertical structure in stellar gravity therefore has, for $|z|\le H$,whereThis is a polytropic equation of state $P\propto {\rho }^{1+1/\beta }$ with polytropic index $\beta$. For $|z|>H$, set $\rho =P=0$; temperature has no physical gas value in that vacuum. Here $H$ is the finite semi-thickness, rather than an exponential disk scale height.

Consider an element displaced upward by $\xi$ in hydrostatic equilibrium. Assume it quickly reaches the ambient pressure, but moves adiabatically and retains its composition. Put $\nabla =d\log T/d\log P$, ${\nabla }_{\mu }=d\log \mu /d\log P$, and let ${\nabla }_{\mathrm{ad}}$ be the adiabatic temperature gradient. The equation of state has differentialwhere $\delta =-(\partial \log \rho /\partial \log T{)}_{P,\mu }$ and $\varphi =(\partial \log \rho /\partial \log \mu {)}_{P,T}$.

At the new pressure, the parcel-to-environment density difference isWith positive pressure scale height $H_P=-(d\log P/dr{)}^{-1}$, $\Delta \log P=-\xi /H_P$. The buoyancy acceleration is therefore $\ddot{\xi }=-N^2\xi$, with stellar buoyancy frequencyA negative $N^2$ amplifies the displacement. For a radiative background, $\nabla ={\nabla }_{\mathrm{rad}}$, so the Ledoux criterion for instability isFor uniform composition, ${\nabla }_{\mu }=0$, giving the Schwarzschild criterionEquality is marginal stability. A mean molecular weight increasing inward has ${\nabla }_{\mu }>0$ and supplies a stabilizing composition term when $\varphi ,\delta >0$.

The same polytrope of index one interior equation is ${\nabla }^2\rho +(2\pi G/K)\rho =0$. A positive separated solution with zero density on all six cube faces is the cubic polytropic interiorIts Laplacian is $-3{\pi }^2\rho /L^2$, so the required side length isWith $\Phi =C-2K\rho$ and $P=K{\rho }^2$, it satisfies the interior hydrostatic equilibrium and Poisson equation for Newtonian gravity. Integrating each sine factor yieldsThis formal interior solution is not an isolated physical cubic star. The cubic polytrope fails isolated gravitational matching: at a vertex, the product of sines has $\nabla \rho =0$, so the interior potential predicts zero gravitational acceleration. At the vertex $(0,0,0)$, however, the field generated by its own positive mass isand each component is strictly positive. There can be no continuous matching to the isolated external field without additional forces or mass sources. Thus solving the interior density equation and imposing zero face values is insufficient. Fluid stars also have no rigid structure to maintain sharp cubic faces, and observed stellar shapes are approximately spherical or rotationally flattened rather than cubic.

For $P=K{\rho }^2$, hydrostatic equilibrium gives $2K\nabla \rho =-\nabla \Phi$. The Poisson equation for Newtonian gravity therefore gives the Helmholtz equationFor spherical symmetry, the regular polytrope of index one solution isThe singular $\cos (kr)/r$ solution is excluded by finite central mass density. Taking the first zero as the stellar surface yieldsThe mass integral isDividing by the volume $4\pi R^3/3$ gives