= Stefan–Boltzmann law
{c}
{title2=$F=\epsilon\sigma T^4$}
{title2=$M_b=\sigma T^4$}
{wiki}
The <Stefan–Boltzmann law> gives the total hemispherical <radiant exitance> of an ideal <blackbody> at absolute <temperature> $T$:
$$
M_b(T)=\sigma T^4,\qquad [M_b]=\mathrm{W\,m^{-2}}.
$$
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> $P_{\rm emit}=AM_b$. A real surface instead has $M=\epsilon(T)\sigma T^4$, where $\epsilon(T)$ is its <total hemispherical emissivity>. The older symbol $F$ in the formula above denotes this emitted <radiant exitance>.<Thermodynamics and Propulsion>[§19.1, Ideal Radiators: blackbody emitted power per area, units, spectrum and cavity]{ref}
For an ideal <blackbody>, doubling absolute <temperature> multiplies <radiant exitance> by $16$. With the approximate <Stefan-Boltzmann constant> $5.670374419\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$, the <radiant exitances> at $300\,\mathrm K$ and $600\,\mathrm K$ are approximately $459.300$ and $7348.805\,\mathrm{W\,m^{-2}}$. 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.
\Image[/stefan-boltzmann-gross-and-net-power.png]
{title=The fourth-power rise of <radiant exitance> and the subtraction needed for <net radiative heat exchange>}
{description=Original model curves. The left panel is total hemispherical <blackbody> <radiant exitance>. The right panel uses <emissivity> $0.8$, emitting <area> $0.01\,\mathrm{m^2}$ and a black enclosure at $300\,\mathrm K$; gross emitted and absorbed <powers> remain positive at equal <temperatures>, while their difference vanishes. The point at $500\,\mathrm K$ is the stipulated example in <net radiative heat exchange>, not a measurement.}
A short derivation integrates <Planck's law> in its <frequency> form. For a <blackbody>, direction-independent <spectral radiance> is $B_\nu(T)=2h\nu^3/[c^2(e^{h\nu/(k_BT)}-1)]$, where $h$ is the <Planck constant>, $k_B$ the <Boltzmann constant> and $c$ the <speed of light>. The outward <hemisphere> contributes $\int\cos\theta\,d\Omega=\pi$, so
$$
M_b=\pi\int_0^\infty B_\nu(T)\,d\nu
=\frac{2\pi k_B^4T^4}{h^3c^2}\int_0^\infty\frac{x^3}{e^x-1}\,dx,
\qquad x=\frac{h\nu}{k_BT}.
$$
Expanding $(e^x-1)^{-1}=\sum_{n=1}^\infty e^{-nx}$, nonnegative terms allow termwise <integration>. Repeated <integration by parts> gives $\int_0^\infty x^3e^{-nx}\,dx=6/n^4$; 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>.
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