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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 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.

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 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.

Thermal radiation is electromagnetic radiation emitted by matter because of its temperature. Matter in thermal equilibrium emits a spectrum governed by its absorptivity, approaching the Planck law for a blackbody.

Total hemispherical emissivity integrates over all emitted wavelengths and the outward hemisphere. If ${ϵ}_{\lambda ,\Omega }$ is spectral emissivity resolved by direction, and $B_{\lambda }(T)$ is blackbody spectral radiance from Planck's law, thenIt is a weighted average, not an unweighted mean of surface properties. Even if the wavelength-resolved material response stays unchanged, the Planck's law weighting changes with temperature, so the total ratio can change. Likewise, total radiometric absorptance averaged over radiation from hotter surroundings generally differs from this emission-weighted ratio. Kirchhoff's law of thermal radiation equates matched spectral and directional quantities; it does not equate arbitrary broadband averages.

For a blackbody, the wavelength at which $B_{\lambda }(T)$ is largest obeyswhere $b$ is Wien displacement constant.

Consider an idealized planet that is a sphere of radius $R$, illuminated by parallel sunlight of irradiance $S$, measured on a plane perpendicular to the rays. Its intercepting projected area is $\pi R^2$, whereas its total surface area is $4\pi R^2$. Therefore the incoming power averaged per unit total surface area iswhere $\alpha$ is the broadband Bond albedo. The factor four is a sphere geometry result, not an atmospheric transmission factor. Local sunlight still varies over the illuminated hemisphere and vanishes on the dark side.

Now stipulate a single spatially uniform temperature $T$, no atmosphere, no internal heating, and emissivity one for outgoing thermal radiation. This is a thermal blackbody model using the Stefan–Boltzmann law; the prescribed solar-band reflection is treated separately, so the body is not a perfect absorber at all wavelengths. Let $C>0$ be the effective heat capacity per unit surface area, in $\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$. Conservation of energy gives the ordinary differential equationHere $F$ is net heating per unit surface area, in $\mathrm{W}{\mathrm{m}}^{-2}$, and $\sigma$ is the Stefan-Boltzmann constant. Use $\sigma =5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, rounded from the exact constant reported by NIST. ^[ref] The uniform temperature is a model assumption: generally the surface average of $T^4$ differs from the fourth power of the surface-average temperature.

For fixed Bond albedo $\alpha$, radiative equilibrium requires $F(T_{\ast })=0$, givingWith the illustrative solar constant $S=1360\mathrm{W}{\mathrm{m}}^{-2}$, the global incoming average is $340\mathrm{W}{\mathrm{m}}^{-2}$:

Doubling albedo here lowers the equilibrium temperature by $33.231\mathrm{K}$, rather than halving it: the absorbed fraction changes from $0.7$ to $0.4$, and thermal radiation scales as $T^4$. These stipulated Bond albedos are whole-planet examples, not measurements of bare water and ice. The model predicts neither Earth's modern surface temperature nor its response to actual climate change; it omits the greenhouse effect, spatial heat transfer and seasons.

Each fixed-albedo equilibrium is a stable equilibrium: for $T_{\ast }>0$, the derivative is $F^{\prime }(T_{\ast })=-4\sigma T_{\ast }^3<0$. A small temperature increase produces net cooling, and a small decrease produces net heating. This is the simplest expression of the stabilizing Planck feedback. Allowing albedo to fall with temperature adds a competing term, as in ice-ocean albedo feedback.