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Black-body radiation is the electromagnetic radiation in thermal equilibrium with an ideal blackbody. It can be described as a photon gas: a cavity whose walls are all at the same temperature contains radiation with the universal Planck's law spectrum. A small opening in such a cavity approximates a blackbody emitter because incoming radiation undergoes repeated absorption inside. The radiation's equilibrium spectrum depends on temperature, rather than the wall material. ^[ref]

There are three useful quantities to keep distinct: directional spectral radiance $B_{\nu }$ or $B_{\lambda }$, total energy density $u$ in the cavity, and outward radiant exitance $j^{\ast }$ from its opening. Planck's law supplies the first; integrals over frequency and directions supply the others. For isotropy, every direction has the same $B_{\nu }$, so the full $4\pi$ solid angle givesHere $\hslash =h/(2\pi )$ is the reduced Planck constant, and the radiation constant is $a_{\mathrm{rad}}\approx 7.565733250\times 10^{-16}\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-4}$. The explicit Boltzmann constant makes this an SI formula; adopting natural units with $k_B=1$ recovers $u={\pi }^2{T}^4/(15{\hslash }^3{c}^3)$.

For a short integral outline, substitute $x=h\nu /(k_{B}T)$ and useThe expansion is a geometric series with nonnegative terms; repeated integration by parts evaluates each integral, and $\zeta (4)={\pi }^4/90$ is a value of the Riemann zeta function. This gives the coefficient above.

The photon gas also exerts radiation pressure, $p=u/3$. The factor $1/3$ comes from the angular average of the squared directional component of photon momentum, $⟨{\cos }^2\theta ⟩=1/3$. The thermodynamic derivation of blackbody radiation pressure gives the same result. In contrast, outgoing power involves one projected-direction factor over an outward hemisphere, yielding radiant exitanceThis is the Stefan–Boltzmann law with unit emissivity; $\sigma$ is the Stefan-Boltzmann constant. The factor $\pi$ follows from Lambert's cosine law, rather than the $4\pi$ full-sphere factor used for energy density.

For example, an ideal $6000\mathrm{K}$ cavity has $u\approx 0.980519\mathrm{J}{\mathrm{m}}^{-3}$, radiation pressure $p\approx 0.326840\mathrm{Pa}$, and outward radiant exitance $j^{\ast }\approx 7.34881\times 10^7\mathrm{W}{\mathrm{m}}^{-2}$. A stipulated $3000\mathrm{K}$ cavity has one-sixteenth each value because all three scale as $T^4$. These describe stored energy, mechanical pressure, and outgoing power per area, respectively. In a closed equilibrium cavity, equal opposing radiation balances the directional power flow even though $u$ and $p$ are nonzero.

The usefulness of black-body radiation is this common reference: Planck's law, Wien's displacement law, and the Stefan–Boltzmann law consistently connect spectral measurements, temperature, and total emission. The three illustrative curves in Planck's law are ideal equilibrium spectra; applying them to a material requires considering its emissivity and whether thermal equilibrium is a suitable approximation.

Emissivity compares thermal emission from a surface with emission from an ideal blackbody at the same absolute temperature, using the same wavelength and angular averaging convention. For ordinary passive surfaces in the far field at local thermal equilibrium, the ratio lies between zero and one. A blackbody has emissivity one. Unless a spectral or directional qualification is given, the $ϵ$ in $M=ϵ\sigma T^4$ is total hemispherical emissivity: a ratio of emitted radiant exitances, not of all outgoing radiant flux including reflected radiation.^[ref]

Emissivity belongs to a specified surface state, not just a bulk material name. Coating, oxidation and surface finish can change it; wavelength, viewing direction and temperature also matter. A surface that looks white in visible light need not have low infrared emissivity. This distinction is useful when choosing a radiator coating or interpreting a radiation-based temperature measurement: a narrow-band detector needs the appropriate spectral emissivity, and its signal can include reflected surrounding radiation. The grey body and diffuse thermal emitter assumptions are separate simplifications, addressing wavelength and direction respectively.

Irradiance measures incident radiant flux per actual receiving area. Its units $\mathrm{W}{\mathrm{m}}^{-2}$ match radiant exitance, but it counts the incoming side of the radiation exchange. Measure incoming angles from the unit normal pointing toward the incident hemisphere. ThenFor $2\mathrm{W}$ distributed uniformly over $0.010{\mathrm{m}}^2$, $E=200\mathrm{W}{\mathrm{m}}^{-2}$. If the receiving plane is tilted $60^{\circ }$ in the same broad uniform parallel beam, the incident power per actual area is halved to $100\mathrm{W}{\mathrm{m}}^{-2}$. This receiving-surface cosine factor is geometric and does not assume Lambertian reflectance. Together, irradiance and radiant exitance describe how radiation supplies and removes energy at a surface, while radiance preserves the directional information.

Lambertian reflectance is the ideal diffuse-reflection model with direction-independent outgoing radiance, the frequency-integrated specific intensity. If a fraction $0\le \rho \le 1$ of incident radiant flux is reflected, a surface receiving irradiance $E$ has reflected radiance $L=\rho E/\pi$. The reflected outward power per area is $\pi L=\rho E$ by the same hemispherical integral used for radiant exitance. Its radiant intensity still acquires the projected area factor in Lambert's cosine law.

For $E=100\mathrm{W}{\mathrm{m}}^{-2}$ and $\rho =0.80$, the reflected radiance is $80/\pi \approx 25.4648\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$. A patch of area $0.020{\mathrm{m}}^2$ reflects $1.6\mathrm{W}$ in total, and its radiant intensities at $0^{\circ }$ and $60^{\circ }$ are respectively $1.6/\pi$ and $0.8/\pi \mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$. The radiance stays fixed as the view changes, provided the illumination stays fixed. Changing the incident angle can change irradiance and therefore the reflected radiance.

For a circular stellar disk whose specific intensity is independent of azimuth, its mean specific intensity over projected area isThe polar coordinates element is $d{A}_{\mathrm{proj}}=2\pi rdr$ after angular integration, and $rdr=-R_{\star }^2\mu d\mu$. Substitution of the linear limb-darkening law givesFor a distant observer at distance $D\gg R_{\star }$ receiving light normally to the detecting plane, the observed radiative flux isIt is spectral irradiance, with units $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{H}{\mathrm{z}}^{-1}$ when $I_{\nu }$ is per hertz. The local outward radiative flux per actual surface area is instead $F_{\nu }=2\pi {\int }_0^1{I}_{\nu }(\mu )\mu d\mu =\pi {\overline{I}}_{\nu }$, a spectral radiant exitance. In a spherical model with the same stellar atmosphere everywhere, the luminosity is $L_{\nu }=4\pi R_{\star }^2{F}_{\nu }$ and $f_{\nu }=L_{\nu }/(4\pi D^2)$, consistent with the inverse-square law. Compare disks at fixed $I_c$ or fixed total observed radiative flux explicitly: increasing $u$ lowers the total in the first comparison, while the second requires a larger $I_c$.

Use spectral radiant exitance per unit wavelength. The Planck law isA hotter blackbody radiates more at every wavelength, its wavelength peak moves to shorter wavelengths according to Wien's displacement law, and its integrated flux rises as $\sigma T^4$ by the Stefan–Boltzmann law. The wavelength convention matters: a spectrum plotted per unit frequency has a different peak transformation.

For planetary equilibrium temperature, assume an isotropic blackbody star, a circular orbit, negligible internal heating, unit surface thermal emissivity and efficient global heat redistribution so the surface has one temperature. The received stellar irradiance is $F_{\ast }=\sigma T_{\ast }^4(R_{\ast }/a{)}^2$. A planet of radius $R_p$ intercepts $\pi R_p^2{F}_{\ast }$ but radiates over $4\pi R_p^2$, giving the projected-area factor in planetary energy balanceWithout the uniform-temperature assumption, this calculation gives $⟨T^4{⟩}^{1/4}$, not a unique arithmetic mean of surface temperature.

With Bond albedo $A=0.05$, the absorbed fraction is $1-A$. Hencea decrease of approximately $3.87\mathrm{K}$. This albedo weights incident stellar light and does not imply thermal emissivity $0.95$.

Define the planetary effective emissivity by outgoing top-of-atmosphere flux $F_{\mathrm{out}}={ϵ}_{\mathrm{eff}}\sigma T_s^4$. Maintaining the same absorbed stellar power with $T_s=312\mathrm{K}$ givesThe greenhouse effect allows thermal radiation to escape from colder layers, so the outgoing flux is less than a black surface at $T_s$ would emit. Downward atmospheric emission contributes to the surface energy balance; this does not violate the top-of-atmosphere energy balance.^[ref]

A single-layer greenhouse model makes the distinction quantitative. Take a shortwave-transparent, isothermal layer with infrared absorptivity and emissivity ${ϵ}_a$, above a surface black in the thermal infrared. Writing $Q=(1-A)F_{\ast }/4$, its surface and atmospheric balances areFor ${ϵ}_a>0$, $T_a^4=T_s^4/2$. Radiation escaping to space is $(1-{ϵ}_a)\sigma T_s^4+{ϵ}_a\sigma T_a^4$, soThus $0.849$ is the planet's effective emissivity; $0.301$ is the layer's infrared emissivity within this particular model. They are not interchangeable. Real atmospheres require wavelength-dependent absorption, a vertical temperature profile and nonradiative heat transport; the measured temperatures alone do not uniquely retrieve an atmospheric optical profile.

A blackbody is an ideal body that absorbs all incident electromagnetic radiation, at every frequency and direction. Its absorptivity is one, so Kirchhoff's law of thermal radiation makes its equilibrium emissivity one as well. At a specified temperature its thermal spectrum is universal: it depends on temperature, not material or shape. This does not mean it looks black when hot; it can glow brightly while remaining a perfect absorber.

A small hole in a nearly isothermal opaque cavity approximates a blackbody, because radiation entering the hole undergoes many interactions before it could escape. In thermal equilibrium the cavity contains an approximately isotropic Planck spectrum. The emitted radiance follows Planck's law, its integrated outward radiant exitance follows the Stefan–Boltzmann law, and its wavelength peak follows Wien's displacement law. A real surface can have wavelength- and direction-dependent emissivity below this ideal limit.

A blackbody absorbs all incident electromagnetic radiation, at every frequency, direction and polarization: its absorptivity is one. In thermal equilibrium, Kirchhoff's law of thermal radiation makes its emissivity one as well. Its emitted spectrum depends only on its absolute temperature, through Planck's law, rather than on the material of its surface.

A small aperture in a large isothermal cavity approximates a blackbody because radiation entering it undergoes many reflections and is very unlikely to escape before absorption. The equilibrium radiation field is isotropic and unpolarized. The total emitted radiant exitance is $\sigma T^4$ by the Stefan–Boltzmann law; the peak of the wavelength spectrum shifts to shorter wavelengths as temperature increases according to Wien's displacement law. An ideal absorber can therefore glow brightly when hot: “black” describes absorption, not the absence of thermal emission.

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.

Radiance resolves both position and direction: it is radiant flux per projected area and per solid angle. For a surface element of area $dA$, and a small outgoing cone $d\Omega$ at angle $0\le \theta <\pi /2$ to its outward unit normal,Here $\mathbf{x}$ locates the patch and $\mathbf{s}$ is a unit vector along the outgoing direction. The factor $dA\cos \theta$ is the patch's apparent area viewed along that direction. The same local definition applies to transmitted or received radiation, choosing the positive projected area for the crossing direction. ^[ref].

For a planar patch of area $A$ with spatially uniform, direction-independent radiance $L$, integrating over the patch givesTake $L=10\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$ and $A=0.020{\mathrm{m}}^2$. The radiant intensities are $I(0)=0.20\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$ and $I(60^{\circ })=0.10\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$. A sufficiently narrow $0.010\mathrm{sr}$ cone therefore carries approximately $2.0\mathrm{mW}$ or $1.0\mathrm{mW}$, respectively. Both views have the same radiance $L$: the oblique view presents half the projected area. This is Lambert's cosine law, not a loss of radiance with viewing angle.

Radiance is useful because it separates a source's local directional output from the geometry of an observation. Integrating over outgoing directions gives radiant exitance; integrating incoming radiance at a receiver gives irradiance. In the stationary vacuum setting of vacuum conservation of specific intensity, a resolved uniform source retains its radiance along each light ray, while the radiant flux received from the whole source can decrease as its apparent solid angle shrinks.

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.

In geometrical optics, suppose radiation has a nonnegative radiance $L(\mathbf{x},\mathbf{s})$ whose directional contributions add as power. Here $\mathbf{s}$ is the actual propagation unit vector, $d\Omega$ is a solid angle element, and the collecting oriented surface $A$ has chosen unit normal $\mathbf{n}(\mathbf{x})$. Assume the following integrals exist. The outward and inward radiant fluxes areBoth quantities are nonnegative; their difference is the signed net radiant flux. The dot product supplies the cosine of the angle with the unit normal, hence the projected area. For incoming radiation the outward unit normal gives a negative dot product; the minus sign makes the collected power positive. Measuring the incoming angle from a source-facing unit normal instead gives the familiar positive $\cos \theta$. The units check is $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\times {\mathrm{m}}^2\times \mathrm{sr}=\mathrm{W}$.

Equivalently, the incoming surface integral is ${\Phi }_{-}={\int }_{A}EdA$ with local irradiance $E$; the emitted part of ${\Phi }_{+}$ is ${\int }_{A}MdA$ with radiant exitance $M$. For a small flat patch and a narrow accepted cone, nearly constant radiance and angle give $\Phi \simeq LA\cos \theta \Delta \Omega$. This approximation requires uniformity across the patch and cone. Direction-independent outgoing radiance over a whole hemisphere gives the different exact result ${\Phi }_{+}=\pi LA$, by Lambert's cosine law and the spherical coordinates calculation in radiant exitance. Replacing the cosine-weighted integral by the hemisphere's $2\pi$ solid angle would double the answer.^[ref]

These radiance formulas presume a description in additive directional power; they do not reconstruct arbitrary coherent wave interference from separate ray contributions. For general electromagnetic fields, the signed instantaneous power crossing an oriented surface is ${\int }_A\mathbf{S}\cdot \mathbf{n}dA$, using the Poynting vector $\mathbf{S}$; optical measurements commonly average over many wave periods. A zero net radiant flux can conceal equal nonzero incoming and outgoing radiant fluxes.

Radiometry measures the energy and power carried by electromagnetic radiation, including light. It asks how much radiation a source supplies, where it travels, and how much reaches a receiving surface. The measurements are physical rather than weighted by the sensitivity of the human eye: equal radiant fluxes at different wavelengths need not look equally bright. ^[ref].

The five quantities below answer different questions about the same radiation. The International System of Units makes the distinctions visible: a watt is a joule per second, while a steradian measures solid angle. ^[ref].

Resolving wavelength or frequency adds a further “per unit interval”, as in spectral radiance. The notation is consistent with NIST's table of spectral quantities. In particular, radiant flux is power, whereas the existing astronomical radiative flux is power per area. ^[ref].

This bookkeeping makes an optical camera's collected signal calculable from a source's radiance and the geometry of the receiving aperture. It also lets radiation measurements from different instruments be compared. For example, NIST describes separate calibrations of spectral radiance and spectral irradiance for instruments used to observe Earth. ^[ref].

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.

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.