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A collision or transient molecular pair can possess a dipole even when the isolated molecules do not. The resulting opacity can be important in a dense molecular hydrogen atmosphere. Pair absorption per unit length scales approximately as the product of the two partner number densities, so its mass absorption coefficient is generally density-dependent. It supplies continuum absorption between ordinary spectral lines.

Here $k$ is a mass absorption coefficient, because $k\rho$ has units of inverse length. The Beer-Lambert law is $dF/dz=-k\rho F$, giving $F(z)=F_0{e}^{-k\rho z}$. A horizontal slab of area $A$ and thickness $dz$ absorbs power $A[F(z)-F(z+dz)]$. Dividing by its volume and taking the limit gives

In the non-scattering grey atmosphere approximation, $I$ is the frequency-integrated specific intensity along a ray, $s$ is distance along that ray, $\rho$ is the mass density, and $\kappa$ is the mass absorption coefficient. The local thermodynamic equilibrium source is the frequency-integrated Planck function $B={\sigma }_{\mathrm{SB}}{T}^4/\pi$. Across a thin element $ds$, absorption removes $\kappa \rho Ids$ and thermal emission adds $\kappa \rho Bds$. Their balance gives $dI/ds=-\kappa \rho (I-B)$.

Take the upward vertical as the polar axis, put $\mu =\cos \theta$, and define downward optical depth byAlong a ray, $dz/ds=\mu$ and $d\tau /ds=-\kappa \rho \mu$. The chain rule therefore gives $\boxed{\mu I_{\tau }=I-B}$. In a plane-parallel, azimuthally symmetric field, the solid angle element is $d\omega =d\phi d\mu$, soLocal radiative equilibrium requires zero net radiative heating of the material: absorption $4\pi \kappa \rho J$ balances emission $4\pi \kappa \rho B$. Where the absorption coefficient is nonzero this gives $B=J$, and substitution gives the required angular radiative transfer equation. It does not require the specific intensity itself to be isotropic.

At the top of an atmosphere with no external incoming long-wave radiation, impose $\boxed{I(0,\mu )=0(-1\le \mu <0)}$. These are inward directions; prescribing the same zero value for outward directions would remove the emergent radiation as well. The upward radiative flux and its derivative areThus $\boxed{\Phi \text{ is constant with height}}$: local radiative equilibrium expresses energy conservation through successive atmospheric layers.

Assume $R_p/R_{\star }=0.10$, $T_{\star }=5800\mathrm{K}$, uniform disk-averaged thermal emission, a nearly blackbody stellar continuum, negligible reflected light at $20\mu \mathrm{m}$, and a flat response over the narrow band. If the eclipse depth is normalized to the out-of-eclipse star-plus-planet flux, the planet/star flux ratio is $r=0.005/(1-0.005)=0.005025$; using $r=0.005$ changes the estimate by less than one percent. The exoplanet secondary eclipse then givesUsing the Planck law at $\lambda =20\mu \mathrm{m}$ gives $T_1\simeq 3.1\times 10^3\mathrm{K}$, or about $3000\mathrm{K}$ at the precision justified by the radius assumption. The Rayleigh-Jeans law instead gives $T_1\simeq 0.5T_{\star }\simeq 2900\mathrm{K}$, a useful rough check. This is a brightness temperature, not the bolometric planetary equilibrium temperature. With zero Bond albedo and global redistribution, the latter would be only about $1250\mathrm{K}$ at $0.05\mathrm{AU}$; a thermal window can sample hotter, deeper gas.

A continuum photon normally emerges from an optical depth of order unity. In hydrostatic equilibrium,The emission-pressure degeneracy of a brightness temperature means that pressure cannot be uniquely recovered from the measured brightness temperature without $g$ and an opacity model. For an explicit nominal estimate, take $g=20\mathrm{m}{\mathrm{s}}^{-2}$ and an effective continuum mass absorption coefficient ${\kappa }_{20}=2\times 10^{-4}{\mathrm{m}}^2\mathrm{k}{\mathrm{g}}^{-1}$. Then $P_{\tau \sim 1}\sim g/{\kappa }_{20}=10^5\mathrm{Pa}$, or approximately one bar. In a hydrogen-rich atmosphere, collision-induced absorption by transient ${\mathrm{H}}_2$ pairs supplies continuum opacity even without ordinary molecular lines; its density dependence makes the constant-opacity calculation an illustrative estimate. With a normal deep atmospheric pressure-temperature profile, temperature increases at larger pressures, eventually following an interior adiabatic temperature gradient.

Two mechanisms for $T_1\ne T_2$ are wavelength-dependent sampling of a vertically varying temperature, and wavelength-dependent weighting of a horizontally nonuniform dayside. In the first, carbon monoxide-rich opacity near $4.5\mu \mathrm{m}$ can sample cooler gas around $0.1$ bar while the $20\mu \mathrm{m}$ continuum samples hotter gas around one bar. In the second, hot and cool areas contribute different proportions of the total flux in different bands because the Planck function depends nonlinearly on temperature. The vertical mechanism alone produces the illustrated pair of temperatures; a single isothermal, horizontally uniform blackbody would have the same brightness temperature in both bands.

If the J-band observation is thermal emission, its low-opacity window normally samples at least as deeply as the continuum at $20\mu \mathrm{m}$. On the illustrated non-inverted profile, $T_3\gtrsim T_1>T_2$. The ordering is conditional: molecular-line absence alone does not specify the relative continuum opacities, and an inversion or reflected-light contamination can change the inference. The exoplanet emission spectrum measures weighted contribution regions rather than three exact thermometers at known pressures.

Interstellar dust both hides and reveals galaxies. Absorption and scattering remove short-wavelength photons from a line of sight, causing interstellar extinction, reddening, anisotropic scattering, and polarization. These effects obscure embedded star formation, bias luminosities, colours, stellar masses, and star-formation rates, and can make an edge-on or dusty galaxy appear older and fainter. Dust also converts absorbed ultraviolet and optical power into far-infrared and submillimetre thermal radiation. That reradiation exposes otherwise hidden star formation, constrains dust temperature and mass, and helps map cold molecular material; polarization traces magnetic-field orientation. Grain surfaces also catalyse molecular-hydrogen formation, while photoelectric emission from grains heats interstellar gas.

One spherical grain has emitting area $4\pi a^2$. Since isotropic blackbody intensity $B_{\nu }$ gives surface flux $\pi B_{\nu }$, its spectral luminosity isFor $N=M_d/m_d$ optically thin grains at distance $D$, the inverse-square law givesDefining the grain mass absorption coefficientthe optically thin dust-mass estimator is

For a constant source function and no incident background, the radiative transfer equation integrates toIf the cloud subtends solid angle $\Omega$, uniform intensity givesThe physical projected area is $A=\Omega D^2$. Its grain column density is $N/A=M_d/(m_d\Omega D^2)$, so the dust optical depth isEliminating ${\tau }_{\nu }$ gives the finite-optical-depth resultIt requires $F_{\nu }<\Omega B_{\nu }$, as demanded by the blackbody brightness limit. When ${\tau }_{\nu }\ll 1$, $1-e^{-{\tau }_{\nu }}={\tau }_{\nu }+O({\tau }_{\nu }^2)$, so $F_{\nu }\simeq \Omega B_{\nu }{\tau }_{\nu }={\kappa }_{\nu }{M}_d{B}_{\nu }/D^2$ and the result reduces to the optically thin estimator.

Assume the added gas mixes throughout the convective column and retain constant mass absorption coefficient for the original gas. The original absorber column mass is unchanged, so its total effective optical depth $D={\tau }_{\infty }$ is unchanged. Under transparent-gas addition at fixed absorber mass, the absorber mass fraction becomes $q=p_s/p_a$, and consequently $t=Dp/p_a$. The new radiating pressure is $p_{\mathrm{rad},a}\simeq p_a/D$, even though the ratio of surface to radiating pressure stays $D$.

The mixture's mass-specific constants areThese pressure weights represent column mass fractions, not mole fractions. Repeating the adiabatic argument givesAt fixed $D>1$, this is warmer than the original surface precisely ifThe equivalence follows by cross multiplication of the positive heat capacities; it holds for any positive added mass. If the two ratios agree, adding a transparent gas leaves this model's surface temperature unchanged. Increased total pressure alone is not a new opacity in a model that fixes absorber mass and mass absorption coefficient. Pressure broadening or collision-induced absorption could change the answer in a more realistic atmosphere, but neither is part of the stated constant-opacity model.

Take $S$ as the integrated line mass absorption coefficient, and $q$ as the absorber mass fraction, assumed constant; for a pure absorbing gas set $q=1$. All spectral intervals below use the paper's wavenumber coordinate. The normalized Lorentz line shape isIf instead an absorption coefficient per length is wanted, multiply by absorber density. Neglecting temperature dependence, pressure broadening gives $\Delta (p)={\Delta }_{s}p/p_s$. In the line wings, with $d=|\nu -{\nu }_0|\gg \Delta (p)$,Define the line-opacity parameter and characteristic wavenumber widthAt an emitting level ${\tau }_{\nu }\sim 1$, the wing estimate implies $d/\Delta (p)\sim \sqrt{C}$. Thus the strong-line limit relevant to outgoing radiation isIt ensures that the opaque-to-transparent transition lies in the Lorentz wings, rather than the narrow central core, and the characteristic absorption width is much larger than ${\Delta }_s$. At a grazing ray angle the criterion is even easier to satisfy because of the larger optical depth. This is a criterion for the radiatively important wings, not a claim that collision broadening remains physical at arbitrarily small pressure.

The pressure-linear grey opacity model interprets $\kappa$ as a mass absorption coefficient. If it is instead a molecular absorption cross section, replace $\kappa$ below by that cross section divided by molecular mass. Assume a plane-parallel atmosphere, hydrostatic balance, local thermodynamic equilibrium, and thermal absorption and emission without scattering of light. With height $z$ increasing upwards,Integrating upwards from the surface givesThusThe vertical optical depth is dimensionless. The atmosphere is optically thick when ${\tau }_{\infty }\gg 1$ and optically thin when ${\tau }_{\infty }\ll 1$. A ray with zenith-angle cosine $\mu$ sees optical depth ${\tau }_{\infty }/\mu$, so even a vertically thin layer can be opaque along a sufficiently grazing ray.

A grey atmosphere model with mass absorption coefficient proportional to pressure. Under hydrostatic balance, $\kappa ={\kappa }_{s}p/p_s$ gives downward optical depth $t={\kappa }_s{p}^2/(2g{p}_s)$. For $T=T_s(p/p_s{)}^a$, a non-scattering optically thick column emits hemispheric radiant flux $F\sim \sigma T_s^4{t}_s^{-2a}\Gamma (2a+1)/(a+1)$. The angular factor follows from integrating ray attenuation over the upward hemisphere; a vertical-ray approximation omits it. The extension of the emitting-depth integral to infinity requires fixed $a$ as $t_s\to \infty$.