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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.

For a matched receiver accepting one spatial mode and one polarization of waves, the thermal contribution to the available spectral power of a blackbody isThis expression excludes a zero-point term; it is the thermally emitted contribution. Defining the Rayleigh-Jeans law equivalent antenna temperature by $dP/d\nu =k_B{T}_A$ gives the displayed relation. With $x=h\nu /(k_{B}T)$,Thus $T_A\simeq T$ requires $x\ll 1$. It is an approximation with a frequency-dependent error, rather than a redefinition of thermodynamic temperature.

For the stipulated $T=3\mathrm{K}$ and $\nu =4.080\mathrm{GHz}$, $h\nu /k_B\simeq 0.19581\mathrm{K}$ and $T_A\simeq 2.90316\mathrm{K}$. Replacing $T_A$ by $T$ here would differ by about $3.23\%$. At much higher frequencies, the approximation fails more strongly. A real instrument averages over its measured frequency response and antenna beam pattern, so using a single central frequency adds a narrow-band approximation.

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.

Wien's displacement law says that the wavelength maximizing the Planck function per unit wavelength, $B_{\lambda }(T)$, moves inversely with absolute temperature:Here $b$ is the Wien displacement constant. Heating an ideal blackbody therefore shifts this peak toward shorter wavelengths, while also increasing its emission at every fixed wavelength. The peak identifies where equal narrow wavelength intervals carry the most power; it is not a single emitted wavelength. Planck's law supplies the full curve, and the Stefan–Boltzmann law supplies its integral.

Specify the spectral coordinate before finding a peak. The frequency form of Wien's displacement law maximizes $B_{\nu }$, which is spectral radiance per unit frequency. Its maximum is not $c/{\lambda }_{\max }$. The frequency and wavelength spectral radiance conversion changes the density as well as the horizontal coordinate; the existing frequency and wavelength peaks of Planck's law give a complementary numerical comparison.

In his 11 December 1911 Nobel lecture, Wilhelm Wien explained the displacement principle through a thought experiment: compress black-body radiation between reflecting moving walls. Work against radiation pressure raises the energy density, while reflection changes each frequency by the Doppler effect. The Second law of thermodynamics connects the resulting spectrum to one at a higher temperature. This determines how the spectrum scales between temperatures, without determining its complete shape. The Wien approximation is a separate, restricted spectral formula.^[ref]

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.