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Use the usual atmospheric plotting convention: temperature on the horizontal axis, logarithmic pressure increasing downward. To illustrate a physically plausible local range for a hot Jupiter, set $T_0=1000\mathrm{K}$ and $|\alpha |=0.05{\mathrm{K}}^{-1/2}$. The sign is not fixed by the fact that the planet is a hot Jupiter; irradiation and opacity determine whether an atmospheric thermal inversion is present.

The original sketch shows both admissible branches of the square-root exponential atmospheric profile, with separate assumed normalizations $P_0=0.01\mathrm{bar}$ for the outward-cooling example and $P_0=1\mathrm{bar}$ for the inverted example. Each extends from $1000$ to $2600\mathrm{K}$ and has $|\log (P/P_0)|\le 2$. For $\alpha >0$, increasing depth increases both pressure and temperature, so the curve bends rightward downward. For $\alpha <0$, moving upward lowers pressure and raises temperature, so the curve bends rightward upward. Each branch meets $T_0$ with $dT/d\log P=0$; it must not be continued through that endpoint onto the other branch.

For the outward-cooling example, $\alpha \sqrt{T_0}=1.58<3.5$, so a section of the imposed profile is unstable for a homogeneous diatomic ideal gas. It should be interpreted as an illustrative retrieved radiative profile that would be modified by convection if realised, rather than a self-consistent deep convective atmosphere. The inverted branch is stable under the same assumptions.

Assume a homogeneous ideal gas, constant specific-heat ratio $\gamma >1$, and an upward displacement that is an adiabatic process and maintains pressure balance with its surroundings. If the ambient temperature falls faster with decreasing pressure than the parcel's adiabatic temperature gradient, the parcel becomes hotter and less dense than its surroundings; buoyancy amplifies the displacement. This is the Schwarzschild criterion:The square-root exponential atmospheric profile hasAn atmospheric thermal inversion with $\alpha <0$ has $\nabla <0$ and is stable to these ordinary adiabatic displacements. For $\alpha >0$, put $s=\sqrt{T-T_0}$. The condition becomes the quadratic inequalityintersected with $s\ge 0$ and $T_0+s^2>0$. In the usual case $T_0>0$, the maximum gradient occurs at $s=\sqrt{T_0}$, namely $T=2T_0$, and is $1/(\alpha \sqrt{T_0})$. Hence a convectively unstable interval exists precisely whenEquality in the existence condition gives a single neutrally stable point. A molecular-hydrogen-dominated hot Jupiter with rotational modes active but vibrational excitation and dissociation negligible has approximately $\gamma =7/5$, so $A=2/7$. The existence condition is then $\alpha \sqrt{T_0}<7/2$. For $T_0\le 0$ the general inequality above remains valid, but its roots must be intersected with the positive-temperature domain; the positive-$T_0$ maximum formula must not be reused. Composition gradients require the Ledoux criterion, and dissociation or variable heat capacities change $A$. Sustained convection normally adjusts a superadiabatic profile toward an adiabatic temperature gradient.

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.

Three physically distinct mechanisms are as follows.

Thermal-tail escape, a bulk hydrodynamic wind, and nonthermal particle energization are three distinct channels; a detected tail need not distinguish them alone. Jeans escape and hydrodynamic atmospheric escape are both thermal mechanisms in the broad sense, but have different distribution functions and dynamical assumptions.

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 a single-phase ideal gas mixture in a closed parcel, with common temperature $T$, total pressure $P$, and molar species amounts $N_i$. Define $N={\sum }_i{N}_i$ and volume mixing ratios $x_i=N_i/N$, so ${\sum }_i{x}_i=1$. Ideal mixing makes partial pressure $P_i=x_{i}P$.

For a pure ideal species, $d{\mu }_i=-{\bar{S}}_{i}dT+{\bar{V}}_{i}d{P}_i$, with molar volume ${\bar{V}}_i=\mathcal{R}T/P_i$. At fixed temperature, integration from standard pressure $P^{\circ }$ givesThe same logarithmic composition term follows from the ideal mixing entropy $\Delta S_{\mathrm{mix}}=-\mathcal{R}{\sum }_i{N}_i\log x_i$. Extensivity gives $G={\sum }_i{N}_i{\mu }_i$. Hence the Gibbs free energy of an ideal-gas mixture isEquivalently, write each standard chemical term as ${\mu }_i^{\circ }=h_i^{\circ }-T{s}_i^{\circ }$, using molar enthalpy and entropy. For zero abundance the limit $x\log x\to 0$ defines the extensive expression continuously.

The fractions alone do not specify the extensive Gibbs free energy: the total amount $N$ or an equivalent elemental inventory must also be fixed. Moreover, in a chemically closed reacting system $N$ need not remain constant; for example, hydrogenation of carbon monoxide changes the total molecular count while conserving atoms. Minimization is therefore over species amounts subject to elemental constraints, rather than over fractions with an artificially fixed molecular count. Nonideal gases require activities or fugacities; condensates contribute separate phase terms.

Let the closed atmospheric parcel exchange heat and volume work with reservoirs at fixed temperature $T$ and pressure $P$, while reactions rearrange its conserved elemental inventory. For a reversible differential between equilibrium states, the first law of thermodynamics, with chemical work included, isThe Gibbs free energy $G=U+PV-TS$ therefore obeysThe first law of thermodynamics supplies this differential but does not alone choose the direction of a reaction. The Second law of thermodynamics does: for the parcel plus its reservoirs, at fixed $T,P$ and with no non-volume work,Thus spontaneous reactions decrease Gibbs free energy, and stable thermochemical equilibrium is its constrained minimum. For an allowed reaction of extent $d\xi$, let $d{N}_i={\nu }_{i}d\xi$, with positive stoichiometric vector entries for products and negative entries for reactants. At an interior minimum,The condition must hold for every independent allowed reaction, with fixed total atoms of each element and any relevant charge constraint. At a boundary with absent species, only feasible one-sided variations are permitted. These are the hypotheses of constrained Gibbs minimization for chemical equilibrium.

For an ideal gas mixture, ${\mu }_i={\mu }_i^{\circ }(T)+\mathcal{R}T\log (x_{i}P/P^{\circ })$, where $x_i$ is the volume mixing ratio and $\mathcal{R}$ the molar gas constant. ThereforeThis is a dimensionless activity-based equilibrium constant, not a bare product of dimensional pressures.

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.

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.

The three parameter groups are local temperature, local pressure, and elemental composition. The elemental composition includes both atmospheric metallicity of a giant planet and atmospheric carbon-to-oxygen ratio; these are independent compositional controls, so a specification by $T$ and $P$ alone is incomplete. Given these inputs, thermochemical equilibrium minimizes Gibbs free energy with element conservation.

Three useful chemical observables in a hydrogen-rich hot Jupiter are:

CO/CH4, H2O, and CO2 bands provide complementary temperature and elemental-abundance diagnostics, but their observed strengths also depend on the atmospheric pressure-temperature profile, exoplanet clouds and observing geometry. A spectral feature is not by itself a direct equilibrium abundance.