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Let $g=Gm/r^2>0$, $H_P=-dr/d\log P>0$, $\nabla =d\log T/d\log P$, and ${\nabla }_{\mu }=d\log \mu /d\log P$. Consider a small radial fluid displacement $\xi$ whose sound-crossing time is short enough to keep its pressure equal to its surroundings. Its composition is frozen and its heat exchange negligible. This is the local stellar convective stability test.

For gas-pressure-dominated ideal gas matter, $\rho \propto \mu P/T$. The ambient gradient is thereforeThe displaced element conserves specific entropy and mean molecular weight, giving $(d\log \rho /dr{)}_{\mathrm{parcel}}=-(1-{\nabla }_{\mathrm{ad}})/H_P$, where ${\nabla }_{\mathrm{ad}}=({\Gamma }_2-1)/{\Gamma }_2$. The parcel-minus-environment mass density difference isIts buoyancy acceleration is minus $g$ times this difference. ThusPositive stellar buoyancy frequency squared gives a restoring force. The gas-pressure-dominated Ledoux criterion is thereforeEquality is marginal in this ideal adiabatic test. For monatomic gas ${\nabla }_{\mathrm{ad}}=2/5$. An inward increase of mean molecular weight has ${\nabla }_{\mu }>0$ and stabilizes the layer; uniform composition recovers the Schwarzschild criterion. For a radiative layer one substitutes the stellar radiative temperature gradient, ${\nabla }_{\mathrm{rad}}=3\kappa LP/(16\pi acGm{T}^4)$.

More generally define $\delta =-(\partial \log \rho /\partial \log T{)}_{P,\mu }$ and $\phi =(\partial \log \rho /\partial \log \mu {)}_{P,T}$. The same displacement argument gives $N^2=(g/H_P)[\delta ({\nabla }_{\mathrm{ad}}-\nabla )+\phi {\nabla }_{\mu }]$ and stability for $\nabla <{\nabla }_{\mathrm{ad}}+(\phi /\delta ){\nabla }_{\mu }$. Both coefficients equal one in the gas-dominated ideal-gas limit used above. This is local dynamical stability against convection; it is distinct from global radial collapse and from instabilities requiring heat or composition diffusion.

Interpret the quoted band amplitude as an absolute transit-depth increment $\Delta D=0.0003$, or $300$ parts per million. For a thin atmospheric annulus of extra height $\Delta z$, the atmospheric spectral-feature amplitude isLet the water band span $N_H$ atmospheric scale heights, so $\Delta z=N_{H}H$. For a nearly isothermal atmosphere with constant gravity and mean molecular weight $\mu$, $H=k_{B}T/(\mu m_{u}g)$. The temperature from a transmission-band amplitude is thereforeHere $T_0,H_0,{\mu }_0,g_0$ refer to the chosen Jupiter reference layer. Assuming $g=g_J\simeq 25\mathrm{m}{\mathrm{s}}^{-2}$, an H/He exoplanet atmosphere with $\mu \simeq 2.3$, and a clear strong band with $N_H\simeq 5$, gives $H\simeq 210\mathrm{km}$ and $T\simeq 1.4\times 10^3\mathrm{K}$. Equivalently, a Jovian reference with $T_0\simeq 165\mathrm{K}$ has $H_0\simeq 24\mathrm{km}$, giving $H/H_0\simeq 8.7$.

The assumptions are hydrostatic, approximately isothermal limb gas, constant gravity, negligible cloud truncation, H/He composition, and an effective opacity contrast yielding about five scale heights. In an ideal clear limb, $N_H\simeq \log ({\kappa }_{\mathrm{band}}/{\kappa }_{\mathrm{cont}})$ follows from the slant optical depth of an isothermal atmosphere. The data fix $N_{H}H$, not $H$ separately: the temperature is conditional on the band opacity contrast and composition. For $N_H=3$–$7$ with the same composition and gravity, the estimate is about $1000$–$2400\mathrm{K}$.

The usual major carbon/oxygen reservoirs in a hydrogen-rich hot Jupiter are water, carbon monoxide and methane, with their relative importance set by thermochemical equilibrium. This is not a universal ranking for every temperature and composition: nitrogen molecules or carbon dioxide can exceed a strongly depleted member of this trio.

At about one bar, the useful net reaction isThe rightward reaction is exothermic. Cooler gas favors methane and water; warming favors carbon monoxide and suppresses methane. The CO/CH4 crossover is of order $10^3\mathrm{K}$ and shifts with pressure, elemental inventory and metallicity; it is not a universal temperature. Water remains an important oxygen reservoir for oxygen-rich compositions but can dissociate at sufficiently high temperature.

Increasing the atmospheric metallicity of a giant planet raises the available carbon and oxygen. In a dilute, H2-dominated regime, major CO and H2O abundances roughly increase with the enrichment factor, and cool-regime CH4 does likewise. CO2 can rise faster, approximately quadratically in suitable warm regimes. At very high enrichment, the H2 fraction and mean molecular weight also change, invalidating simple linear scalings.

At high temperature and atmospheric carbon-to-oxygen ratio below unity, CO binds much of the carbon, leaving excess oxygen for H2O. As C/O approaches or exceeds one, CO consumes nearly all available oxygen and H2O is strongly depleted; excess carbon enhances CH4, hydrogen cyanide and acetylene. Cooler CH4-dominated chemistry uses less oxygen in CO and can leave more water even at the same elemental ratio. These trends are carbon partition and atmospheric water abundance, and assume equilibrium rather than vertical or horizontal quenching.

Cloud particles can scatter or absorb, so not every atmospheric haze makes a high-albedo planet. Weak molecular bands alone also admit low abundance or high mean molecular weight; a consistent spectral combination is stronger evidence than a single signature.

Assume a hydrogen-rich gas with approximately solar elemental ratios and local thermochemical equilibrium when identifying the limiting compositions. At $T\lesssim 500\mathrm{K}$, a representative set of dominant molecules is ${\mathrm{H}}_2,{\mathrm{H}}_{2}O,{\mathrm{CH}}_4,{\mathrm{NH}}_3$. In the deeper $T\gtrsim 1300\mathrm{K}$ regime, it is ${\mathrm{H}}_2,{\mathrm{H}}_{2}O,\mathrm{CO},{\mathrm{N}}_2$. Helium is an abundant atom, not a molecule. The carbon and nitrogen switches, carbon monoxide–methane quenching and nitrogen–ammonia quenching when frozen by mixing, are represented byLow temperature and high pressure favor the right-hand sides, while hotter gas favors carbon monoxide and ${\mathrm{N}}_2$. Exact boundaries depend on pressure and composition; ${\mathrm{CO}}_2$ can become important at high metallicity. If mixing is strong, the cool upper atmosphere need not retain its local-equilibrium four-species ordering.

To preserve the hot-region reactant $A_1$ aloft, require transport to beat its conversion at the quench region. With a mixing length $\ell$ and vertical eddy diffusivity $K_{zz}$,For the simplest use of the supplied atmospheric scale height reference, assume the same temperature and mean molecular weight as the reference atmosphere and take $\ell =H$. Since $H\propto 1/g$, $H\simeq 30/0.4=75\mathrm{km}$, giving $K_{zz}\gtrsim 5.6\times 10^4{\mathrm{m}}^2{\mathrm{s}}^{-1}=5.6\times 10^8\mathrm{c}{\mathrm{m}}^2{\mathrm{s}}^{-1}$.

The hotter quench layer requires a temperature correction if the reference is ordinary Jupiter; the temperature of that scale-height reference was not specified. An explicit estimate using $T_q=1300\mathrm{K}$, dimensionless mean molecular weight $\mu =2.3$, and $g\simeq 0.4\times 25=10\mathrm{m}{\mathrm{s}}^{-2}$ givesThus a scale-height-based estimate using the hot layer is of order $10^{10}\mathrm{c}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. Both estimates state their assumptions: the reference scaling alone does not include the temperature ratio. Choosing $\ell =0.1H$ lowers the threshold by a factor of $100$, and a full quench calculation needs the reaction timescale along the profile. The dependence of a quench diffusivity on mixing length shows why the stated reaction timescale supplies a mixing constraint, not a unique measured coefficient.

A super-Earth is a planet more massive than Earth but below the usual ice-giant scale, conventionally about $1$--$10M_{\oplus }$. The term does not guarantee an Earth-like or rocky composition; water-rich planets and objects with small H/He envelopes can lie in this range.

The most direct constraints are transit radius and a mass from radial velocities or transit-timing variations, giving $\bar{\rho }=3M/(4\pi R^3)$. Comparing these with interior models constrains the possible iron core, silicate mantle, water layer and gas envelope, but many mixtures share one mass and radius.

Additional observables include atmospheric spectra/mean molecular weight and escape signatures, which test whether a low-density envelope is H/He or heavier volatiles; stellar Fe/Mg/Si abundances as priors on refractory composition; and age, irradiation and orbital history, which control thermal expansion and envelope survival. Where measurable, a Love number or related apsidal response constrains central concentration. Mass and radius constrain families of interiors, not a unique composition; atmospheric and dynamical information can reduce this exoplanet interior-composition degeneracy.

Use stellar homology: dimensionless profiles and composition are fixed, so proportionality constants do not change along the model sequence. Scaling mass conservation and the stellar hydrostatic equation giveswhere the last relation uses the ideal gas equation of state with fixed mean molecular weight. The specific stellar energy-generation rate is $ϵ\propto \rho T^{\eta }$, soBy stellar radiative diffusion, $T_{\ast }/R\sim {\kappa }_{\ast }{\rho }_{\ast }L/R^2{T}_{\ast }^3$, henceEquate the generated and transported luminosity in thermal equilibrium. The homology relations for radiative stars giveThis division presumes a nonzero denominator. If it vanishes, the scaling balance instead imposes a condition on $M$ or leaves the radius scaling undetermined, depending on the numerator. For later use, the corresponding mass-luminosity relation exponent is

The two measurements sample different atmospheric regions and different radiative transfer geometries. An exoplanet emission spectrum probes the dayside; an exoplanet transmission spectrum probes the morning and evening limbs of the day-night terminator. Several alternatives to an exoplanet cloud deck can therefore suppress the transmission feature.

Temperature, mean molecular weight, limb chemistry, continuum opacity and the dayside temperature gradient are alternatives or degeneracies to test. In particular, the three-scale-height amplitude alone cannot establish different CO abundances between the hemispheres.

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.

Three complementary methods available in the setting of this 2017 paper are:

Transmission probes the limb, eclipses probe the dayside, and direct imaging isolates spatially resolved planetary light. Radial velocities alone measure orbital motion rather than atmospheric composition; high-resolution planetary Doppler spectroscopy can instead detect atmospheric spectral lines if planetary light is isolated statistically.

At fixed mass, three broad controls of a super-Earth-sized body's observed radius are:

Bulk composition, envelope fraction/composition, and thermal/irradiation history are three independent controls. If mass is not fixed, the mass itself is an additional major variable. The label super-Earth does not ensure a rocky composition or an Earth-like atmosphere, and a radius alone does not determine which of these effects dominates.

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.

Assume a stationary spherical outflow with speed $v>0$, mass-loss rate $\dot{M}>0$, an optically thick thermal radiation field and negligible radiation inertia compared with matter inertia. The stellar wind continuity equation givesUse the printed gas fraction $\beta =P_{\mathrm{gas}}/P$, with $P=P_{\mathrm{gas}}+P_{\mathrm{rad}}$, and assume constant mean molecular weight ${\mu }_{\mathrm{mol}}$. Put $\mathcal{R}=k_B/({\mu }_{\mathrm{mol}}{m}_u)$ and $a_g^2=\mathcal{R}T$ for the isothermal sound speed squared. Then $P_{\mathrm{gas}}=\rho a_g^2$ and $P_{\mathrm{rad}}=a_{\mathrm{rad}}{T}^4/3$. The momentum equation including the radiation-pressure gradient isDo not also add a separate radiative force here: the diffusion radiation-pressure gradient already represents that force.

Using radiative diffusion in a star,Write $g=G{M}_r/r^2$ and $\Gamma =L_r/L_{\mathrm{crit}}=\kappa L_r/(4\pi cG{M}_r)$, the local ratio to the Eddington luminosity. Substituting continuity into the gas-pressure gradient gives the gas and radiation pressure stellar wind equationThe denominator changes sign at the sonic point $v=a_g$. A smooth transonic branch must have its right-hand numerator vanish there too. In a subsonic flow the denominator is negative, so acceleration requires a negative numerator; for a supersonic flow it requires a positive numerator. A regular subsonic-to-supersonic branch must therefore pass through a compatible common zero, with a finite slope obtained by differentiating the equations. Boundary conditions and that regularity requirement normally select the mass flux. Other branches can remain subsonic, remain supersonic or encounter a singular gradient. In the isothermal nonradiating limit the numerator is $2a_g^2/r-GM/r^2$, recovering the Parker wind equation and critical radius $GM/(2a_g^2)$.

To expose the role of $\beta$, rewrite the diffusion temperature gradient asHenceAt the stipulated local value $\Gamma =1$, with $0<\beta <1$, its numerator isThe gas-pressure contribution is not removed by setting the luminosity equal to the Eddington value. Instead the negative temperature gradient makes the thermal term in the numerator positive as well. A subsonic outflow then has $dv/dr<0$, and no common numerator/denominator zero is available for smooth sonic passage. This is the sonic-point obstruction at the Eddington luminosity.

Thus a subsonically launched wind cannot accelerate smoothly to supersonic speed under these diffusion assumptions at $L_r/L_{\mathrm{crit}}=1$. It is not a claim that every initially supersonic solution decelerates; those have positive denominator and accelerate in this equation. Even in the radiation-dominated limit $\beta \to 0$, the spherical thermal term remains positive for $a_g^2>0$. The singular pressureless limit is not a finite-temperature sonic crossing. Finally $a_g$ is the thermal sound scale of this diffusion-controlled stationary equation, not the full adiabatic sound speed of a tightly trapped gas-plus-radiation perturbation. Optically thin winds or different heating/driving mechanisms require another transport model.

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.

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 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.