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An absorption line is a deficit of observed spectral specific intensity relative to the surrounding continuum. In a photosphere with temperature falling outward, increased line opacity makes photons emerge from cooler layers than the continuum. For an absorption-dominated LTE source, the Planck function then supplies a smaller line specific intensity. A foreground layer with source below the incident continuum gives another direct transfer description.

For a thermalized radiative transfer source function $S_{\lambda }=B_{\lambda }(T)$ constant along a ray, the specific intensity after optical-depth interval $\Delta \tau$ is $I_{\lambda }=B_{\lambda }+(I_{\mathrm{in}}-B_{\lambda })e^{-\Delta \tau /\mu }$ for $\mu >0$. As $\Delta \tau \to \infty$, it tends to the Planck function, independently of a bounded boundary input. A temperature gradient is unnecessary for blackbody emission. In a scattering medium, local thermodynamic equilibrium alone does not guarantee $S_{\lambda }=B_{\lambda }$; thermalization must be justified.

Consider a hypothetical star whose emitted specific intensity is exactly that of a uniform $5000\mathrm{K}$ blackbody. Assume transparent propagation with negligible Doppler shift and interstellar extinction, so geometric dilution multiplies every spectral interval by the same factor. The stipulated temperature givesUsing the converted frequency peak as though it maximized $B_{\lambda }$ would return $b/(c/{\nu }_{\max })\simeq 2841.26\mathrm{K}$, a substantial error caused solely by mixing spectral conventions.

Now stipulate a photodetector with constant quantum efficiency $\eta$ over the relevant wavelengths, collecting equal narrow wavelength bins. For fixed accepted projected area times solid angle $G$, with dilution included in $G$, its expected photon count rate density isThis follows by dividing the received power by the photon energy $hc/\lambda$. It peaks where $x=4(1-e^{-x})$, at $x=3.9206903949\dots$, giving $0.73394057\mu \mathrm{m}$ for the same model star. Constant quantum efficiency is not constant energy spectral responsivity: the latter includes the wavelength-dependent photon energy conversion. For example, $G=10^{-12}{\mathrm{m}}^2\mathrm{sr}$, $\eta =0.5$ and a $1\mathrm{nm}$ bin centered at $0.600\mu \mathrm{m}$ give approximately $1.93\times 10^{10}$ detected photons per second in the narrow-bin approximation. These are calculated model values, not observations of a named star.

A more general linear detector channel has outputwhere $S_{i,0}$ is the background reading and $R_i$ includes spectral selection, optical transmission and detector spectral responsivity. A raw channel maximum need not locate either Planck function maximum. Recovering a temperature requires an appropriate calibrated response and an emission model. For a real stellar atmosphere, opacity and radiative transfer can reshape the spectrum; interstellar extinction can reshape it again. A single peak does not establish a unique physical temperature without those assumptions.

At fixed $T>0$, substitute $x=hc/(\lambda k_{B}T)>0$ into the Planck function:Here $h$ is the Planck constant, $k_B$ the Boltzmann constant, and $c$ the speed of light. Since $dx/d\lambda =-x/\lambda \ne 0$, an interior maximum in wavelength is also a stationary point in $x$. A logarithmic derivative givesThe required positive root is $x_{\lambda }=4.9651142317\dots$. The rearranged equation also has $x=0$, but that endpoint is not a peak. For $g(x)=5(1-e^{-x})-x$, $g^{\prime }(x)=5e^{-x}-1$ changes sign once, at $x=\log 5$. Thus $g$ first increases and then decreases to negative values, crossing zero exactly once for $x>0$. Since $x^5/(e^x-1)$ tends to zero at both endpoints, this stationary point is the unique global maximum. Solving the definition of $x$ for wavelength givesThe inverse-temperature dependence follows before any numerical root-finding. More generally $B_{\lambda }(T)=T^{5}F(\lambda T)$ for one fixed function $F$, so spectra at all temperatures have the same rescaled shape. This is stronger than a relation between two peak positions. Replacing the denominator by $e^x$ in the Wien approximation would instead give $x=5$ and a peak wavelength about $0.70\%$ too short.

A brightness temperature equates a measured intensity with a Planck function; it does not determine its contribution pressure. In hydrostatic equilibrium, ${\tau }_{\lambda }(P)={\int }_0^P{\kappa }_{\lambda }d{P}^{\prime }/g$. Estimating $P\sim g/{\kappa }_{\lambda }$ therefore needs gravity and opacity, as well as a model of the contribution function. Different pressures can produce the same band brightness.

The Far Infrared Absolute Spectrophotometer measured the spectrum of the cosmic microwave background by comparing the sky with a calibrated reference. Its differential Fourier transform spectroscopy and blackbody calibration tested whether the radiation followed one Planck function, rather than only measuring differences between sky directions.

The key observable is a spectral residual after fitting a temperature and accounting for instrumental and foreground contributions. A small residual constrains departures from a blackbody spectrum and thus energy exchange in the early universe. This absolute spectral test is distinct from the directional map supplied by the Differential Microwave Radiometer.^[ref]

At a fixed temperature, the frequency maximizing $B_{\nu }$ and the wavelength maximizing $B_{\lambda }$ in Planck's law are different ways of describing the same black-body radiation. They are maxima of different densities. The inverse relation between frequency and wavelength does not preserve those maxima because its interval-conversion factor varies with wavelength.

For a short extremum calculation, substitute $x=h\nu /(k_{B}T)$ into $B_{\nu }$, leaving a positive constant times $x^3/(e^x-1)$. Substitute $x=hc/(\lambda k_{B}T)$ into $B_{\lambda }$, leaving a positive constant times $x^5/(e^x-1)$. Setting each derivative to zero givesSelect the positive roots: $x_{\nu }=2.821439372\dots$ and $x_{\lambda }=4.965114232\dots$. Although $x=0$ solves the rearranged equations, it is an endpoint, not either peak. For $m=3$ or $5$, the function $m(1-e^{-x})-x$ initially increases, its derivative $m{e}^{-x}-1$ crosses zero once, and it eventually becomes negative. Hence it has exactly one positive root. Each density tends to zero at both ends, so this root gives its unique maximum. The bisection method brackets these roots in $(1,3)$ and $(1,5)$.

Consequently Wien's displacement law givesThese constants agree with NIST's tabulation. ^[ref] The wavelength peaks at $3000$, $4500$, and $6000\mathrm{K}$ are $965.924$, $643.949$, and $482.962\mathrm{nm}$. Doubling temperature halves this peak wavelength.

At $6000\mathrm{K}$, the frequency peak is ${\nu }_{\max }\approx 352.736\mathrm{THz}$, whose converted wavelength is $c/{\nu }_{\max }\approx 0.849907\mu \mathrm{m}$. The $B_{\lambda }$ peak is instead $0.482962\mu \mathrm{m}$, corresponding to about $620.737\mathrm{THz}$. These numbers are compatible: equal-width frequency bins and equal-width wavelength bins partition the same radiation differently. Specifying which density was maximized is essential when using a peak to infer temperature.

Limb darkening is the decrease of emergent specific intensity toward the apparent edge of a stellar disc. A transit depth then depends on the intensity sampled by the planet as well as its projected area; the uniform-disc area ratio is only an approximation.

For a spherical star of radius $R_{\star }$, let $r$ be distance from the centre of its projected disk and let $\theta$ be the angle between the outward unit normal and the line of sight. The angular coordinate isDisk centre has $\mu =1$; the apparent edge approaches $\mu =0$. A resolved image measures specific intensity in each direction. Limb darkening means that this specific intensity itself falls toward the edge; the projected area factor used when collecting radiant flux is a separate geometric effect.

A simple explanation assumes a horizontally uniform plane-parallel atmosphere, a sufficiently deep optically thick medium, negligible scattering of light, and thermal absorption and emission in local thermodynamic equilibrium. Let ${\tau }_{\nu }$ be inward vertical optical depth at fixed frequency $\nu$. An oblique light ray has line of sight optical depth ${\tau }_{\nu }/\mu$, so it encounters the same attenuation at a smaller vertical optical depth. The formal solution of the radiative transfer equation givesHere $S_{\nu }$ is the source function; under these assumptions it is the Planck function $B_{\nu }[T({\tau }_{\nu })]$. If temperature increases inward, an oblique view preferentially samples cooler layers and hence lower specific intensity. ^[ref]

The weight $e^{-{\tau }_{\nu }/\mu }/\mu$ is an exponential distribution in vertical optical depth, with expected value $\mu$ and standard deviation $\mu$. Thus “seeing down to optical depth one” refers to the line of sight, and represents a broad sampling of layers rather than a sharp emitting surface. For a source function that is linear throughout the atmosphere, the Eddington-Barbier relation is exact:This follows by integrating the constant and linear terms separately. A constant source function instead gives constant specific intensity, as in the blackbody limit of an isothermal atmosphere. An outward-rising temperature or important scattering of light changes the argument. Close enough to the actual edge, a plane-parallel atmosphere also loses validity because of curvature. Neither one limb-darkening coefficient nor a universal dependence on wavelength follows from this simple model.

A material element is in local radiative equilibrium when its absorbed and emitted radiative powers balance, so radiation causes no net local heating. In a non-scattering grey atmosphere in local thermodynamic equilibrium, this reduces to equality of the angular mean specific intensity $J$ and the Planck function $B$. Integrating the radiative transfer equation over solid angle then shows that the radiative flux is constant through a plane-parallel atmosphere. The radiation can remain anisotropic; equilibrium constrains its angular mean, not every ray separately.

Local thermodynamic equilibrium assumes matter populations follow thermal distributions at the local temperature: excitation follows the Boltzmann distribution and ionization the Saha equation. It permits a temperature gradient; it does not require the radiation field itself to be a blackbody.

For true absorption and its inverse thermal emission, $j_{\nu }^{\mathrm{abs}}/{\alpha }_{\nu }^{\mathrm{abs}}=B_{\nu }(T)$, the Planck function. The total source function need not equal $B_{\nu }$ when scattering contributes. Distinguishing matter equilibrium from radiation equilibrium is essential near a star's escaping surface.^[ref]

When radiative rates compete with or dominate collisions, level populations need statistical equilibrium coupled to radiative transfer, rather than thermal population laws at the local temperature. Radiation from other depths can make the local radiation field differ from the Planck function. This changes ionization and spectral-line strengths, so a geometrically thin atmosphere may still require non-LTE abundance modelling.

A non-scattering isothermal atmospheric layer in local thermodynamic equilibrium, of temperature $T_a$ and slant optical depth $t_{\lambda }$, lies above a blackbody background at $T_b$. The specific intensity isThus an increase in opacity shifts the specific intensity toward the layer's Planck function. If a line is more opaque than its continuum, it appears in absorption for $T_a<T_b$, in emission for $T_a>T_b$, and has no contrast for $T_a=T_b$. This is a local explanation, not a claim that all features diagnose one uniform layer.

Multiplying an absorber abundance by $a$ with fixed cross sections and physical temperature profile multiplies optical depth by $a$. The Planck function as a function of the new coordinate must therefore be composed with division by $a$. A term proportional to ${\tau }^n$ acquires coefficient $a^{-n}$. Holding its old optical-depth coefficient fixed would instead change the physical temperature profile. This is the immediate radiative comparison before thermal adjustment.

Use the Einstein coefficients convention in which $B_{ij}$ multiplies spectral radiation energy density $u_{\nu }$, with $u_{\nu }=c^{-1}\int I_{\nu }d\Omega$. A normalized spectral line profile obeys $\int \phi (\nu )d\nu =1$. The spontaneous emission probability per upper-state atom per unit time is $A_{21}$; the stimulated emission and upward absorption rates per atom are respectively $B_{21}\int u_{\nu }\phi (\nu )d\nu$ and $B_{12}\int u_{\nu }\phi (\nu )d\nu$.

For isotropic spontaneous emission, $n_2{A}_{21}$ transitions occur per unit volume and time, each supplying photon energy $h{\nu }_{12}$. Dividing among $4\pi$ steradians givesFor a small incident beam, its contribution to $u_{\nu }$ is $I_{\nu }d\Omega /c$. Absorption removes and stimulated emission adds photons in the incident direction. Matching the net loss to ${\alpha }_{\nu }{I}_{\nu }d\Omega$ therefore givesReplacing ${\nu }_{12}$ by $\nu$ across the narrow spectral line profile gives the printed expressions. The negative sign is essential: an inverted population can amplify the incident specific intensity. Thermal equilibrium and the Planck function imply $g_1{B}_{12}=g_2{B}_{21}$ and $A_{21}/B_{21}=8\pi h{\nu }_{12}^3/c^3$, where $g_i$ is the state degeneracy. A convention using mean intensity instead changes both $B$ coefficients by $4\pi /c$.

Use the energy-density convention of part (a), a normalized common spectral line profile $\int \varphi (\nu )d\nu =1$, and isotropic spontaneous emission. For a narrow spectral line centered at ${\nu }_0$,The factor $1/c$ follows from $u_{\nu }=4\pi J_{\nu }/c$: angle-integrated absorption removes $c{\kappa }_{\nu }{u}_{\nu }$ of energy per volume and time. The emission coefficient is per solid angle, whereas the absorption coefficient has units of inverse length.

The Boltzmann distribution in thermal equilibrium gives $n_2/n_1=(g_2/g_1)e^{-h{\nu }_0/(k_{B}T)}$. Dividing the two expressions and using the relations between Einstein coefficients,Thus the source function equals the Planck function at the transition frequency, and in the usual narrow-line approximationWith an exactly common finite-width $\varphi$ and constant line-center Einstein coefficients, their ratio is constant across the profile; exact equality to a varying $B_{\nu }$ at every frequency is only a narrow-line approximation. Exact thermal radiative transfer requires frequency-resolved detailed balance, which supplies the corresponding small profile corrections. This distinction does not alter the intended equilibrium result.

In local thermodynamic equilibrium, write the one-ray radiative transfer equation asLarge total optical depth removes the boundary contribution, but an additional local condition is essential: the temperature and opacity vary little over an absorption length ${\kappa }_{\nu }^{-1}$. Iterating once in that small gradient givesThe integrated Planck function and its derivative areUsing the specified Rosseland mean opacity, with ${\kappa }_R$ measured here in inverse length, the source's definition $F_{\mathrm{ray}}=\int I_{\nu }d\nu$ therefore givesThe printed leading term $\sigma T^4$ lacks a factor $1/\pi$ under that definition. An isothermal optically thick medium is a concrete counterexample: $I_{\nu }=B_{\nu }$ gives $F_{\mathrm{ray}}=\sigma T^4/\pi$ exactly, while the printed formula gives $\sigma T^4$.

A physical radiative flux requires an angular integral. For direction cosine $\mu$, the same gradient expansion gives $I_{\nu }(\mu )=B_{\nu }-\mu {\kappa }_{\nu }^{-1}{\partial }_z{B}_{\nu }$. Integration over all directions gives the net radiative diffusion fluxThe isotropic equilibrium contribution cancels between opposite directions. For comparison the outward-hemisphere flux alone is $\sigma T^4-8\sigma T^3(dT/dz)/(3{\kappa }_R)$ at this order. None of these consistent definitions yields the printed mixture of coefficients.

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.

The Rosseland mean opacity is the transport opacity appropriate to an optically thick region close to local thermodynamic equilibrium. If ${\kappa }_{\nu }$ is monochromatic transport opacity and $B_{\nu }(T)$ the Planck function, thenIt is a weighted harmonic mean: low-opacity frequency windows carry much of the diffusive flux. One must combine the monochromatic processes before averaging, rather than adding their separately averaged opacities.

In hot fully ionized material, electron-scattering opacity supplies an almost density- and temperature-independent contribution at fixed composition. In the nonrelativistic Thomson scattering limit, ${\kappa }_{\mathrm{es}}={\sigma }_T{n}_e/\rho \simeq 0.2(1+X)\mathrm{c}{\mathrm{m}}^2{\mathrm{g}}^{-1}$. Free-free opacity, the absorption counterpart of thermal bremsstrahlung, comes from photon absorption during encounters between free electrons and ions. Its approximate mean obeys Kramers' opacity law, ${\kappa }_{\mathrm{ff}}\propto \rho T^{-7/2}$, with composition and quantum correction factors.

When ions retain bound electrons, bound-free absorption is photoionization; edges, occupations and ionization fractions strongly affect its frequency dependence. Bound-bound absorption produces absorption lines. Numerous metal lines can close low-opacity windows and produce important opacity features in partially ionized layers. Line populations depend on excitation as well as ionization, while Doppler broadening, pressure broadening and overlapping lines change the transport windows. Stimulated inverse processes must be included consistently with detailed balance and the local thermal radiation field.

In cooler hydrogen-rich atmospheres, negative hydrogen ion opacity from both bound-free and free-free processes is often important. Its strong temperature dependence helps determine convective surface relations. In still cooler layers, molecular bands, and at sufficiently low temperatures dust absorption/scattering, may be important. These processes cannot be represented by one Kramers power law over the whole density-temperature plane.

An opacity calculation therefore needs the chemical abundances, the equation of state, electron density and bound-state populations. Saha ionization equation and thermal excitation populations give useful dilute LTE estimates; high-density pressure ionization, nonideal effects and altered level populations require corrections. Composition matters even when one writes only $\kappa (\rho ,T)$: hydrogen, helium and the metal abundance pattern are additional fixed inputs.

In dense degenerate cores, thermal conduction by electrons can dominate energy transport. An equivalent conductive opacity may be introduced so that $1/{\kappa }_{\mathrm{eff}}=1/{\kappa }_R+1/{\kappa }_{\mathrm{cond}}$, because the two diffusive fluxes add; conduction is not itself a photon absorption process. Very hot regimes also require departures from the simple nonrelativistic scattering approximation. Near the optically thin photosphere, a Rosseland diffusion description alone is inadequate, and frequency-dependent radiative transfer with appropriate boundary conditions is required. These regime changes are why realistic stellar opacity tables, rather than one global power law, enter stellar structure.

Write $T=\overline{T}(1+\Theta )$, with $\Theta =\Delta T/\overline{T}$ at linear order. Since $\overline{T}\propto a^{-1}$, the background Planck function depends only on comoving momentum. Expanding its temperature-shifted argument givesUsing $T$ rather than $\overline{T}$ in the perturbation denominator changes only higher-order terms.

The photon brightness perturbation is the energy-weighted angular perturbationIntegration by parts gives $\int q^4{f}_0^{\prime }dq=-4\int q^3{f}_{0}dq$; the boundary term vanishes at both endpoints for a Planck spectrum. Therefore $\Delta =4\Theta$. Multiplying the transport equation by $q^3$ and integrating yields the synchronous photon brightness equationThe PDF's definition is $4\Delta T/T$; the TeX's denominator $\hslash$ is a transcription error.

Expand $\Delta$ in Legendre moments in $\mu$. The monopole is the density/temperature perturbation, and the dipole is the bulk velocity. Before decoupling, rapid Thomson scattering isotropizes photons in the electron rest frame. In the tight-coupling approximation, the scattering rate is much larger than $k$ and $\mathcal{H}$, so the quadrupole and higher moments are suppressed by the mean-free-path parameter, with the quadrupole typically of order $k/\dot{\kappa }$ times the dipole. To leading order only monopole and dipole are needed. This suppression uses the collision term restored before decoupling; it does not follow from the collisionless equation by itself. Finite quadrupole corrections generate polarization and diffusion damping.

Use the notation $I_{\nu }(\mu ,\tau )$ throughout, with optical depth increasing inward and $0<\mu \le 1$ for an outward ray. This removes the change in argument order used in the printed surface-intensity notation. Multiplying the radiative transfer equation by its integrating factor givesIntegrating from the surface to a deeper boundary $b$ yieldsFor a semi-infinite atmosphere with no exponentially growing homogeneous term, the boundary term vanishes as $b\to \infty$. The formal solution of the radiative transfer equation is thereforeThe usual deep-atmosphere boundary condition is necessary; a first-order differential equation alone does not fix the emerging specific intensity.

For a linear source function, the zeroth and first exponential moments give $I_{\nu }(\mu ,0)=a_0+a_1\mu$. Equivalently $I_{\nu }(\mu ,0)=S_{\nu }(\tau =\mu )$, the Eddington-Barbier relation. If the source rises inward, rays near the limb sample cooler, shallower layers, producing limb darkening.

For the finite polynomial, put $u=t/\mu$ and integrate each term. Repeated integration by parts gives ${\int }_0^{\infty }{u}^j{e}^{-u}du=j!$, so the polynomial source function moments for emergent intensity giveThe limiting grazing-ray specific intensity is $a_0$. If the displayed polynomial is only a local Taylor approximation to a more general source, this formula applies to that approximation and a source-function remainder also contributes; a finite local expansion is not automatically an exact description at every optical depth.

For a constant source in an overlying layer of optical thickness $\tau$, the radial ray has the exact transfer relationExpanding the exponentials gives the constant source transfer through a thin layer:This finite-layer result does not require the specific intensity below the layer to equal its source function. The sign of $S_{\nu }-I_{\nu }(1,\tau )$ already distinguishes emission from absorption.

An absorption line forms when a bound-bound transition increases opacity at selected frequencies. In a photosphere whose temperature decreases outward, optical depth of order one at the line frequency occurs higher than at a nearby continuum frequency. In an absorption-dominated region in local thermodynamic equilibrium, the source is the Planck function; the cooler line-forming layer then emits less specific intensity than the deeper continuum-forming layer. The result is a dark spectral feature. Scattering and departures from LTE modify the source, so this temperature-gradient picture is a useful mechanism rather than a universal identity for every line.

Two distinct situations can produce emission lines. First, a heated chromosphere has an outward temperature rise, so an optically thick line can sample gas hotter than the continuum photosphere beneath it. Secondly, a hot optically thin circumstellar envelope or stellar wind emits line photons through collisional excitation or recombination, with no comparably bright background along all rays; its added line flux can exceed the underlying continuum. Winds of hot luminous stars can also give a mixture of emission and absorption across a Doppler-broadened line profile. These examples explain the change in specific intensity through either an enhanced source function or additional emitting material.