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The Planck function is the equilibrium spectral specific intensity of a blackbody, given by Planck's law. Per unit wavelength it isIt can also be written per unit frequency, with $B_{\nu }d\nu =-B_{\lambda }d\lambda$ under the corresponding change of variable.

Planck's law gives the spectral radiance of an ideal blackbody in thermal equilibrium at absolute temperature $T>0$. It predicts how much power is carried in each small frequency or wavelength interval, per projected area and per solid angle. The whole curve is determined by $T$: no material-specific parameter enters this ideal Planck function.

Writing $\nu$ for frequency in hertz and $\lambda$ for wavelength in metres, the two forms of the Planck function areHere $h$ is the Planck constant, $k_B$ the Boltzmann constant, and $c$ the speed of light in vacuum. Their exact values in the International System of Units are $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. ^[ref]

The units matter: $B_{\nu }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$, whereas $B_{\lambda }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$. Both are spectral radiance, rather than radiant exitance or energy density. To report $B_{\lambda }$ per micrometre, multiply its per-metre value by $10^{-6}$; a micrometre-wide interval is $10^{-6}$ metres wide.

For a computed example, at $T=6000\mathrm{K}$ and $\lambda =0.500\mu \mathrm{m}$, the exponent is $hc/(\lambda k_{B}T)\approx 4.79592$. Substitution into Planck's law gives $B_{\lambda }\approx 3.17569\times 10^{13}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$, or $3.17569\times 10^7\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$ per micrometre. At the same wavelength, the illustrative temperatures $3000\mathrm{K}$ and $4500\mathrm{K}$ give $2.60268\times 10^5$ and $6.37793\times 10^6$ in the latter units. These are ideal predictions for stipulated temperatures.

The key change of variables is $\nu =c/\lambda$. Corresponding positive intervals carry the same power, so $B_{\lambda }d\lambda =B_{\nu }|d\nu |$ andEqual-width wavelength intervals have unequal widths in frequency; the factor $c/{\lambda }^2$ compensates for this. Merely replacing the horizontal-axis labels of $B_{\nu }$ by $c/\nu$ does not produce $B_{\lambda }$. This explains the frequency and wavelength peaks of Planck's law.

A short derivation uses the Boltzmann distribution for a single electromagnetic mode with thermal energies $nh\nu$, $n=0,1,\dots$. Put $q=e^{-h\nu /(k_{B}T)}$. The geometric series gives $Z={\sum }_{n\ge 0}{q}^n=(1-q{)}^{-1}$ and mean thermal energy $h\nu q/(1-q)$. The electromagnetic mode density in a box, including both independent polarization of an electromagnetic wave states of a photon, is $8\pi {\nu }^2/c^3$ per volume per unit frequency. Multiplication gives spectral energy density $u_{\nu }=8\pi h{\nu }^3/[c^3(e^{h\nu /(k_{B}T)}-1)]$; isotropy gives $B_{\nu }=c{u}_{\nu }/(4\pi )$. This counts thermal excitation energy, excluding zero-point energy. ^[ref]

Planck's law connects quantum mechanics to a measurable radiation curve. Its small-$h\nu /(k_{B}T)$ limit is the Rayleigh-Jeans law, $B_{\nu }\approx 2k_{B}T{\nu }^2/c^2$; its large-$h\nu /(k_{B}T)$ limit is the Wien approximation, $B_{\nu }\approx (2h{\nu }^3/c^2)e^{-h\nu /(k_{B}T)}$. The suppression at high frequency makes the total energy density finite. Integrating the curve gives the Stefan–Boltzmann law, while locating its peak gives Wien's displacement law. Thus one formula predicts both the spectral shape and the total emission.

For a planet of Bond albedo $A_B$ orbiting a blackbody star of radius $R_{\ast }$ and temperature $T_{\ast }$, complete global redistribution and negligible internal heat giveUniform dayside-only reradiation replaces $4a^2$ by $2a^2$. The planetary equilibrium temperature describes absorbed stellar power; including an intrinsic cooling flux gives a total effective temperature $T_{\mathrm{eff}}^4=T_{\mathrm{int}}^4+T_{\mathrm{eq}}^4$, distinct from $T_{\mathrm{eq}}$. Neither quantity is necessarily the greenhouse-warmed surface temperature.

For blackbody star and planet observed at one common wavelength sufficiently long for the Rayleigh-Jeans law to hold for both,This follows by substituting $B_{\lambda }=2c{k}_{B}T/{\lambda }^4$ in the thermal eclipse depth. A radiative flux ratio alone does not determine the radius without a planetary brightness temperature and a stellar temperature. Spectral peaks interpreted with Wien's displacement law must refer to $F_{\lambda }$, not $F_{\nu }$ or $\lambda F_{\lambda }$.

Radiant exitance is total outward emitted radiant flux per actual emitting area, measured in $\mathrm{W}{\mathrm{m}}^{-2}$. Integrate radiance over the outward hemisphere, keeping the projected area factor:If the radiance is direction-independent, the spherical coordinates element is $d\Omega =\sin \theta d\theta d\phi$, soThe weighted solid angle is $\pi \mathrm{sr}$, although the whole hemisphere covers $2\pi \mathrm{sr}$. Directions near grazing contribute less because of the projected area. This geometric factor is the bridge between Lambert's cosine law and the total output of a diffuse patch. ^[ref].

For the uniform patch with $L=10\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, radiant exitance is $M=10\pi \mathrm{W}{\mathrm{m}}^{-2}$. Its area $0.020{\mathrm{m}}^2$ gives total radiant flux $\Phi =MA=0.20\pi \mathrm{W}\approx 0.628319\mathrm{W}$ over the outward hemisphere. This total differs from its radiant intensity $I(0)=0.20\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$, which describes only one direction.

For an ideal blackbody, integrate Planck's law over frequency as well: $M=M_e=j^{\ast }=\sigma T^4$ by the Stefan–Boltzmann law. An ideal $6000\mathrm{K}$ emitting patch of area $1\mathrm{m}{\mathrm{m}}^2$ emits about $73.4881\mathrm{W}$ outward. Irradiance counts incoming radiation separately; for purely emitted output the net outward exchange per area is $M-E$. Reflected output must also be included if the surface reflects incident radiation.

A radiative-convective climate column applies a radiative-convective equilibrium model to the vertical thermal structure relevant to climate change. Radiative transfer determines heating and cooling; convective adjustment represents unresolved vertical redistribution when the chosen atmospheric lapse rate is too steep. Radiative equilibrium alone need not give a physically admissible lower-atmospheric profile.

In 1967, Syukuro Manabe and Richard T. Wetherald computed equilibrium with prescribed relative humidity, allowing atmospheric water vapor to change with temperature. Their average-cloud case warmed by $2.36\mathrm{K}$ for carbon dioxide increasing from 300 to 600 parts per million, versus $1.33\mathrm{K}$ with fixed absolute humidity. These are model outputs, not measured warming. The comparison isolates an important water-vapor feedback.^[ref]

An original two-temperature calculation exposes the bookkeeping. Let an opaque surface absorb solar irradiance $Q$ and emit blackbody thermal radiation $\sigma T_s^4$. A single isothermal atmospheric layer transmits all sunlight and absorbs a fraction $ϵ$ of surface infrared emission, emitting $ϵ\sigma T_a^4$ both upward and downward. This is a grey thermal-band emissivity, with $0<ϵ\le 1$. Let $H$ be upward nonradiative heat flux, and let $C_s,C_a>0$ be heat capacities per unit area. Then conservation of energy givesThe two equations sum toHere $L$ is outgoing longwave radiation. Both $H$ and downward atmospheric emission cancel internally: they redistribute energy without creating it. The factor two in atmospheric emission counts two faces. Dropping the downward face would spoil the surface budget. Each right-hand term has units $\mathrm{W}{\mathrm{m}}^{-2}$; each $C$ has units $\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$.

Choose $Q=240\mathrm{W}{\mathrm{m}}^{-2}$, $H=80\mathrm{W}{\mathrm{m}}^{-2}$ and $ϵ=1$. At equilibrium, $L=\sigma T_a^4=240\mathrm{W}{\mathrm{m}}^{-2}$ and $\sigma T_s^4=400\mathrm{W}{\mathrm{m}}^{-2}$, giving $T_a\simeq 255.06\mathrm{K}$ and $T_s\simeq 289.81\mathrm{K}$. Direct substitution gives surface balance $240+240-400-80=0$ and atmospheric balance $400-480+80=0$. These deliberately chosen fluxes illustrate the model; they are not an observed global budget or a reconstruction of the 1967 calculation. Prescribing $H$ illustrates heat transfer accounting but does not close a realistic convection law.

For fixed $Q,H,ϵ$, the positive-temperature equilibrium is locally stable. Put $a=4\sigma T_s^3>0$ and $b=4ϵ\sigma T_a^3>0$. The Jacobian matrix of the temperature equations has trace $-a/C_s-2b/C_a<0$ and determinant $ab(2-ϵ)/(C_s{C}_a)>0$. Thus both eigenvalues have negative real part. Temperature-dependent climate feedbacks alter this test. A single grey layer also omits the optical spectrum, cloud structure and the independent thermal behavior of the troposphere and stratosphere.

For luminosity $L$, mass-energy equivalence gives $|\dot{M}|=L/c^2$. Holding $L$ constant gives a time $t=fM{c}^2/L$ to radiate a mass fraction $f$. A blackbody estimate uses the Stefan–Boltzmann law, $L=4\pi R^2\sigma T^4$. This is a radiative-energy estimate, not a model of a star's evolving nuclear fuel or mass loss in a wind.

A column near pressure $p$ contains mass per area of order $p/g$. Linearizing its blackbody cooling around temperature $T$ gives the estimatewhere $c_p$ is specific heat capacity at constant pressure. This estimate omits details of wavelength-dependent radiative transfer, but captures faster relaxation in hotter and lower-pressure atmospheric layers.

A regulated bath holds its temperature near a chosen setting through feedback control, while stirring reduces internal temperature gradients. Its finite heat capacity buffers disturbances; external heating and cooling replace the ideal heat bath's unlimited capacity. This is useful for thermometer calibration and for maintaining a blackbody calibration source at a known temperature.

Joel B. Fowler's A Third Generation Water Bath Based Blackbody Source reports a NIST test of a stirred water bath containing a copper blackbody cavity. A bottom plate supplied heating and cooling; a double stirrer circulated water. Electronic feedback control regulated heating. An electronic thermometer with a thermistor probe provided the readout. The preparation was to approach a selected setting, wait for stabilization, adjust against the measured temperature and then log continuously.^[ref]

To test spatial uniformity, one thermistor stayed near the cavity tip while another visited ten locations at three heights. The simultaneous reference distinguished spatial differences from temporal drift. An insulated dome reduced thermal conduction along probe casings. Table 1 reports nonzero departures at several settings: circulation did not make every location identical. The measurement uncertainty analysis separately included uniformity, stability, setting, immersion and readout contributions. These tests supported a bounded temperature assignment around the cavity, rather than infinite heat capacity or perfect thermal equilibrium throughout the apparatus.^[ref]

A satellite test of greenhouse spectral change compares outgoing radiance with radiative transfer predictions. The 2001 study compared IRIS on Nimbus 4 (1970) and IMG on ADEOS (1997). Matching April–June ocean regions, masking land, screening clouds and correcting resolution and wavenumber controlled comparability. Spectral radiance became brightness temperature readouts. Regional comparisons and independent calculations tested interpretation. Carbon dioxide and methane features supported changed absorption; residual clouds, calibration and interannual variability limited background inference.^[ref]

This observational comparison tests radiative transfer; it does not by itself measure equilibrium climate sensitivity or validate a complete climate model.

To understand the readout, write an instrument's spectral channel schematically asHere $I$ is spectral radiance in the viewing direction $\hat{\mathbf{n}}$, $K_j$ is a normalized channel response, $b_j$ is calibration offset and $e_j$ is random error. Subtracting two channel readings isolates an atmospheric difference only after their response functions and sampling are comparable. Matching kernels reduces a false difference caused merely by observing a narrow spectral line at different resolutions; it cannot remove every calibration offset.

Outgoing longwave radiation in an energy budget also integrates radiance over upward directions and over the thermal spectrum, with the appropriate projected-angle factor. A brightness temperature converts one spectral radiance to the equivalent blackbody value through Planck's law; it is not automatically surface temperature. This explains why a spectral difference, a broadband radiative flux difference and surface warming are related observables with different units and different inferences.

The Shockley-Queisser limit is the radiative conversion-efficiency bound for an ideal single-junction solar cell, with one collected electron-electron hole pair per absorbed above-gap photon, carrier thermalization, and radiative recombination required by the reverse optical process. Photons below the band gap are not absorbed, while excess photon energy above it is thermalized. Maximizing the electrical power over the operating electric potential difference and then the band gap gives a ceiling of roughly thirty percent for ordinary unconcentrated sunlight; its precise value depends on the incident spectrum, cell temperature and emission geometry. The original calculation by William Shockley and Hans-Joachim Queisser obtained approximately $30\%$ at $1.1\mathrm{eV}$ for a $6000\mathrm{K}$ blackbody source and a cell at $300\mathrm{K}$.^[ref]

A single-layer greenhouse model replaces the vertical structure of an exoplanet atmosphere by one layer transparent to incoming stellar light and having thermal emissivity $ϵ$. Let the surface be a uniform blackbody receiving absorbed stellar irradiance $Q$. In radiative equilibrium, the layer absorbs $ϵ\sigma T_s^4$ and emits $ϵ\sigma T_a^4$ in each direction, so for $ϵ>0$,The second equation follows from surface balance, $\sigma T_s^4=Q+ϵ\sigma T_a^4$. It demonstrates how absorption and downward reemission warm the surface at fixed absorbed power. The limiting $ϵ=1$ surface is $2^{1/4}$ times the planetary equilibrium temperature. The model omits convection, spectral variation and a continuous atmospheric pressure-temperature profile, so the fitted emissivity is not a unique atmospheric composition.

A single-line greenhouse deficit measures the reduction of outgoing thermal radiative flux by one isolated absorption line. Consider a plane-parallel atmosphere in local thermodynamic equilibrium, without scattering, above a blackbody surface. Let pressure $p$ increase from zero at the top to $p_s$ at the surface, and let the surface share the adjacent gas temperature $T_s$.

To specify the angular approximation, use one effective upward direction cosine $0<{\mu }_d\le 1$ and a flux closure $F_{\nu }=\pi I_{\nu }$. Define the corresponding transmission ${\mathfrak{T}}_{\nu }(p)=\exp [-{\tau }_{\nu }(p)/{\mu }_d]$. Integration by parts of the formal radiative transfer solution givesFor a sufficiently narrow absorption line centred at frequency ${\nu }_0$, replace the Planck function and its temperature derivative across the line by their values at ${\nu }_0$. With the line equivalent widththe outgoing radiative flux becomesIf temperature increases downward, $D\ge 0$. An atmospheric thermal inversion can invalidate that sign argument; absorption alone does not guarantee a positive deficit for every imposed thermal profile.

For constant absorber mass fraction $q$, constant gravity $g$ and an opacity ${\kappa }_{\nu }=S{\varphi }_{\nu }$ per absorber mass, assume $\int {\varphi }_{\nu }d\nu =1$ and constant integrated strength $S$. Then ${\tau }_{\nu }=qS{\varphi }_{\nu }p/g$. In the optically thin limit for the specified direction,This leading term is independent of the normalized shape of the absorption spectrum. The angular closure and abundance convention must accompany the coefficient; a full angular integration uses its own flux-weighted transmission.

Spectral emissivity compares emission with an ideal blackbody at the same temperature at a specified wavelength, or frequency under a consistently transformed convention. A hemispherical spectral emissivity uses spectral radiant exitance; a directional spectral emissivity uses spectral radiance. Always state which angular convention applies. A wavelength-selective coating can emit strongly in one band while emitting weakly elsewhere, so one band measurement does not determine total hemispherical emissivity.

Spectral radiance is radiance per unit frequency or wavelength interval. Its astronomical synonym specific intensity is usually written $I_{\nu }=L_{\nu }$ or $I_{\lambda }=L_{\lambda }$; this $I_{\nu }$ is a spectral radiance, distinct from the radiant intensity $I$ in radiometry. For positive small intervals,The two expressions refer to the same spectral slice, with the positive widths related through frequency $\nu$ and wavelength $\lambda$. For frequency measured in hertz, $\mathrm{Hz}={\mathrm{s}}^{-1}$, the units are $[L_{\nu }]=\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$. For wavelength measured in metres, $[L_{\lambda }]=\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$. Integrating over a spectral band gives its radiance, and then multiplying by the collecting projected area and accepted solid angle gives a small uniform beam's radiant flux. A spectral measurement therefore retains information that the band-integrated power alone loses.

For an ideal blackbody, Planck's law gives $L_{\nu }=B_{\nu }(T)$ and $L_{\lambda }=B_{\lambda }(T)$ at temperature $T$, independent of outgoing direction. Its existing frequency and wavelength peaks of Planck's law illustrate why the coordinate matters. The simpler flat-band calculation below isolates that same change of variables without the thermal physics.

The Stefan-Boltzmann constant converts the fourth power of absolute temperature into blackbody radiant exitance:Here $k_B$ is the Boltzmann constant, $h$ the Planck constant and $c$ the speed of light. In the current International System of Units, these three constants have exact defining values, so $\sigma$ is exact as an expression in them and $\pi$. The displayed finite decimal is an approximation, not an exact decimal or a measured uncertainty. The 2022 recommended-constants table lists an ellipsis and an exact uncertainty for $\sigma$.^[ref]

A dimensional check uses $[k_B]=\mathrm{J}{\mathrm{K}}^{-1}$, $[h]=\mathrm{J}\mathrm{s}$ and $[c]=\mathrm{m}{\mathrm{s}}^{-1}$. Consequently $[k_B^4/(h^3{c}^2)]=\mathrm{J}{\mathrm{s}}^{-1}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}=\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, exactly the units needed in the Stefan–Boltzmann law.

For an equilibrium photon gas, the energy density is $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$. The radiation pressure is $p_{\mathrm{rad}}=u/3=4\sigma T^4/(3c)$. These describe the radiation filling an equilibrium cavity; $u$ has units $\mathrm{J}{\mathrm{m}}^{-3}$, whereas the outward radiant exitance through a small opening is $cu/4$. The factor $1/4$ comes from averaging propagation directions and their projected area over a radiation field with isotropy, not from reducing the opening's emissivity.

The Stefan–Boltzmann law gives the total hemispherical radiant exitance of an ideal blackbody at absolute temperature $T$:Here $\sigma$ is the Stefan-Boltzmann constant. Total means integrated over all wavelengths; hemispherical means integrated over all outward directions with the projected area weighting. Thus $M_b$ is emitted power per actual emitting area, rather than directional radiance or spectral radiance. Absolute temperature must be in kelvin; raising a Celsius reading to the fourth power gives the wrong result. A uniform surface of area $A$ emits radiant flux $P_{\mathrm{emit}}=A{M}_b$. A real surface instead has $M=ϵ(T)\sigma T^4$, where $ϵ(T)$ is its total hemispherical emissivity. The older symbol $F$ in the formula above denotes this emitted radiant exitance.^[ref]

For an ideal blackbody, doubling absolute temperature multiplies radiant exitance by $16$. With the approximate Stefan-Boltzmann constant $5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, the radiant exitances at $300\mathrm{K}$ and $600\mathrm{K}$ are approximately $459.300$ and $7348.805\mathrm{W}{\mathrm{m}}^{-2}$. This steep response explains why thermal radiation becomes so consequential at high temperature. It also enables heat transfer through a vacuum, where thermal conduction through a gas and convection are absent.

A short derivation integrates Planck's law in its frequency form. For a blackbody, direction-independent spectral radiance is $B_{\nu }(T)=2h{\nu }^3/[c^2(e^{h\nu /(k_{B}T)}-1)]$, where $h$ is the Planck constant, $k_B$ the Boltzmann constant and $c$ the speed of light. The outward hemisphere contributes $\int \cos \theta d\Omega =\pi$, soExpanding $(e^x-1{)}^{-1}={\sum }_{n=1}^{\infty }{e}^{-nx}$, nonnegative terms allow termwise integration. Repeated integration by parts gives ${\int }_0^{\infty }{x}^3{e}^{-nx}dx=6/n^4$; the Riemann zeta function value $\zeta (4)={\pi }^4/90$ gives the remaining integral as ${\pi }^4/15$. Hence $M_b=\sigma T^4$ with the coefficient in Stefan-Boltzmann constant. The angular integral is essential: omitting it confuses radiance with radiant exitance.

For a Planck function expressed as spectral radiance per frequency interval,$h$ is the Planck constant, $c$ the speed of light and $k_B$ the Boltzmann constant. Differentiating at fixed frequency givesThe proof uses ${\partial }_{T}x=-x/T$ and the chain rule. At low frequency, the multiplier tends to one, recovering the Rayleigh-Jeans law. At high frequency it grows approximately as $x$: the same fractional temperature change has a larger fractional brightness effect on the declining spectral tail.

For an original example choose $T=2.725\mathrm{K}$ and $\delta T=1\mathrm{mK}$. At $100\mathrm{GHz}$, $x=1.76119$, so the first-order fractional residual is $7.80415\times 10^{-4}$. At $300\mathrm{GHz}$, $x=5.28357$, it is $1.94881\times 10^{-3}$. Both spectra are exact blackbodies: this residual is a temperature mismatch, not evidence of a nonthermal spectrum. Fitting temperature before assessing remaining spectral residuals therefore removes a physically distinct degree of freedom. The figure divides by a fixed peak brightness rather than by the rapidly vanishing high-frequency tail.

Instrument overview of FIRAS, describing its differential Michelson interferometer, internal reference and external calibrated blackbody. It reports the absolute cosmic microwave background temperature inferred from the spectral comparison.

For blackbody star and planet, define the planet-star contrastA complete exoplanet secondary eclipse removes the planet's light from the combined out-of-eclipse light. Its exact normalized depth is $d_{\lambda }=r_{\lambda }/(1+r_{\lambda })$. The common approximation $d_{\lambda }\simeq r_{\lambda }$ requires $r_{\lambda }\ll 1$. This thermal model omits reflected light and spatial variations.