Stefan–Boltzmann law ($F=ϵ\sigma T^4$, $M_b=\sigma T^4$)

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 small, uniform surface at temperature $T$, enclosed by much larger black walls all at temperature $T_s$, the outward net heat transfer by thermal radiation isAssume an opaque grey body that is a diffuse thermal emitter, with the same emissivity $ϵ$ and radiometric absorptance at the surface's current temperature. Its emitting area $A$ sees only the enclosure: the view factor to the walls is one, with no self-viewing surface, opening or intervening obstruction. The intervening medium must neither absorb nor emit appreciable thermal radiation; a vacuum supplies a useful idealization. The enclosure is maintained at uniform $T_s$. The small-body assumption makes that reservoir description physically reasonable; genuinely black isothermal walls supply the incoming irradiance even without an infinite area ratio. Positive ${\dot{Q}}_{\mathrm{rad}}$ means net loss by the surface; negative means net gain. This expression counts thermal radiation alone, not all heat transfer.^[ref]

The reasoning is a two-way energy balance. Black walls supply irradiance $E=\sigma T_s^4$, so the surface absorbs $P_{\mathrm{abs}}=ϵAE$ and emits $P_{\mathrm{emit}}=ϵA\sigma T^4$. Subtract these to obtain ${\dot{Q}}_{\mathrm{rad}}$. For an opaque surface the remaining fraction $1-ϵ$ of incoming radiant flux is reflected; counting emitted plus reflected outgoing radiant flux minus total incoming radiant flux gives the same result. Kirchhoff's law of thermal radiation justifies equal spectral absorption and emission factors; the grey body assumption lets one factor be taken outside both spectral integrals.

Choose $ϵ=0.8$, $A=0.01{\mathrm{m}}^2$, $T=500\mathrm{K}$ and $T_s=300\mathrm{K}$. These are stipulated model inputs, not measured material properties. Using $\sigma \approx 5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$ givesThe incoming irradiance is $459.300\mathrm{W}{\mathrm{m}}^{-2}$, corresponding to $4.5930\mathrm{W}$ incident on the surface; only $80\%$ is absorbed. At $T=T_s=300\mathrm{K}$, emitted and absorbed powers are both $3.6744\mathrm{W}$, so radiative equilibrium has zero net exchange despite continuing emission. Changing only emissivity to $0.2$ reduces both one-way powers and the net loss by a factor of four, giving $6.1694\mathrm{W}$ net. A low-emissivity coating therefore suppresses both radiative cooling to colder walls and radiative heating from hotter walls.

Near equal temperatures, write $T=T_s+\Delta T$. Taylor's theorem givesFor constant emissivity, the radiative heat transfer coefficient $h_{\mathrm{rad}}$ is approximately $4.8992\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$ at $T_s=300\mathrm{K}$ and $ϵ=0.8$. This local linear approximation is useful for small temperature differences; the $200\mathrm{K}$ difference in the worked example needs the full fourth-power expression.

A cooling experiment requires a wider first law of thermodynamics balance. In a uniform-temperature model without a heater,where $C(T)$ is the sample's heat capacity and the last two terms are signed outward convection and thermal conduction losses. A vacuum suppresses convection but does not eliminate thermal conduction through supports or sensor leads. A cooling slope therefore cannot establish emissivity unless these paths and heat capacity are independently constrained. Multiple wall temperatures, incomplete view factors, wavelength-selective materials or appreciable intervening absorption require a more complete radiative transfer calculation.

The view factor $F_{i\to j}$ is the fraction of uniformly distributed, diffuse outgoing radiant flux from surface $i$ that directly reaches surface $j$. It depends on geometry, including obstruction, rather than on temperature or emissivity. All destinations in a closed enclosure have fractions summing to one. For a convex body fully enclosed by walls, $F_{\mathrm{body}\to \mathrm{walls}}=1$; the reverse view factor is generally smaller. Reciprocity gives $A_i{F}_{i\to j}=A_j{F}_{j\to i}$. Two unobstructed spheres with outer radius ten times the inner radius have $F_{\mathrm{inner}\to \mathrm{outer}}=1$ and $F_{\mathrm{outer}\to \mathrm{inner}}=1/100$. Do not multiply both directions by an assumed identical view factor.^[ref]

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.

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.

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.

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.

Directional emissivity is the ratio of emitted radiance to blackbody radiance at the same temperature in a specified outgoing direction. It can be total over wavelength or additionally spectrally resolved. Normal-direction emissivity is not automatically total hemispherical emissivity: the latter integrates all directions with the projected area weighting. A diffuse thermal emitter removes this directional dependence of emitted radiance.

A grey body has spectral emissivity independent of wavelength over the range in which the approximation is used. Under Kirchhoff's law of thermal radiation, the corresponding radiometric absorptance is also wavelength-independent. This allows the same factor to weight emission at $T$ and black-enclosure irradiation at a different $T_s$ in net radiative heat exchange. The factor can still depend on temperature; treating it as constant over a cooling curve is an additional assumption. Wavelength independence alone does not imply direction independence.^[ref]

A diffuse thermal emitter has emitted radiance independent of outgoing direction over the outward hemisphere. Its emitted radiant intensity from a planar patch still varies with viewing angle through projected area, as described by Lambert's cosine law. Diffuse emission does not mean equal power per solid angle from the patch. This approximation does not by itself impose a grey body spectrum or specify how incident radiation is reflected.

Oberlin College's documented demonstration heats a cube internally with a bulb and compares polished aluminium, dull aluminium, black-painted and white-painted faces at the same temperature. The apparatus includes a thermopile and a digital meter. Its preparation specifies 20 minutes of preheating and heater setting 6; these are instructions, not measured outcomes. The thermopile supplies the electrical readout for the relative thermal radiation comparison. Cover it between readings: otherwise detector warming changes its reference temperature. The account reports little difference between painted black and white faces and a larger contrast between shiny and painted faces. The comparison motivates a surface-dependent emissivity and shows why visible color alone is insufficient.^[ref]

Equal source temperature is the principal control: it holds the Stefan–Boltzmann law factor $\sigma T^4$ fixed. For a sharper comparison, also hold detector distance, viewing angle, viewed area and surrounding temperatures fixed, and record a covered-detector baseline. These are additional diagnostic controls. Under those assumptions, a changed readout can reveal changed directional emissivity averaged over the detector's wavelength response. The demonstration supplies neither an absolute calibration nor a scan over all directions and wavelengths; it therefore does not directly measure total hemispherical emissivity or establish the fourth-power law. Comparing cooling rates would additionally require separating convection, thermal conduction and heat capacity effects, as in net radiative heat exchange.

A body is in radiative equilibrium when absorbed radiative power equals emitted radiative power. A surface absorbing flux $F$ and emitting as a grey body has equilibrium temperature $T=(F/(ϵ\sigma ){)}^{1/4}$.