シュテファン＝ボルツマンの法則

${\displaystyle I=\sigma T^{4}}$

${\displaystyle I=\epsilon \sigma T^{4}}$

${\displaystyle L={\frac {1}{\pi }}I={\frac {\sigma }{\pi }}T^{4}}$

${\displaystyle u={\frac {4}{c}}I={\frac {4\sigma }{c}}T^{4}}$

${\displaystyle p={\frac {u}{3}}}$

${\displaystyle \left({\frac {\partial U}{\partial V}}\right)_{T}=T\left({\frac {\partial p}{\partial T}}\right)_{V}-p}$

${\displaystyle {\frac {du}{dT}}={\frac {4u}{T}}}$

${\displaystyle u\propto T^{4}}$

${\displaystyle I_{\lambda }(\lambda ,T)={\frac {c_{1}}{\lambda ^{5}}}f(c_{2}/\lambda T)}$

${\displaystyle I_{\nu }(\nu ,T)={\frac {c_{1}}{c^{4}}}\nu ^{3}f{\Big (}(c_{2}/c)\nu /T{\Big )}}$

${\displaystyle I(T)=\int _{0}^{\infty }I_{\nu }(\nu ,T)\,d\nu ={\frac {c_{1}}{c^{4}}}\int _{0}^{\infty }\nu ^{3}f{\Big (}(c_{2}/c)\nu /T{\Big )}\,d\nu ={\frac {c_{1}T^{4}}{{c_{2}}^{4}}}\int _{0}^{\infty }x^{3}f(x)\,dx}$

${\displaystyle \sigma ={\frac {c_{1}}{{c_{2}}^{4}}}\int _{0}^{\infty }x^{3}f(x)\,dx}$

${\displaystyle I(\nu ,T)={\frac {2h}{c^{2}}}{\frac {\nu ^{3}}{\mathrm {e} ^{h\nu /kT}-1}}}$

${\displaystyle f(x)={\frac {1}{\mathrm {e} ^{x}-1}}}$

${\displaystyle c_{1}=2\pi hc^{2},~c_{2}={\frac {hc}{k}}}$

${\displaystyle \sigma ={\frac {2\pi k^{4}}{c^{2}h^{3}}}\int _{0}^{\infty }{\frac {x^{3}\ }{\mathrm {e} ^{x}-1}}dx}$

${\displaystyle \zeta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }{\frac {u^{s-1}}{\mathrm {e} ^{u}-1}}du}$

${\displaystyle \int _{0}^{\infty }{\frac {x^{3}}{\mathrm {e} ^{x}-1}}dx=\Gamma (4)\zeta (4)=6\cdot {\frac {\pi ^{4}}{90}}={\frac {\pi ^{4}}{15}}}$

${\displaystyle \sigma ={\frac {2\pi ^{5}k^{4}}{15c^{2}h^{3}}}}$

${\displaystyle f(x)=\mathrm {e} ^{-x}}$

${\displaystyle \int _{0}^{\infty }x^{3}\,\mathrm {e} ^{-x}dx=\Gamma (4)=6}$

${\displaystyle \sigma _{\text{Wien}}={\frac {12\pi k^{4}}{c^{2}h^{3}}}={\frac {\sigma }{\zeta (4)}}={\frac {\sigma }{1.0823\ldots }}}$

${\displaystyle f(x)={\frac {1}{x}}}$

${\displaystyle \int _{0}^{\infty }x^{2}dx}$

${\displaystyle L_{\text{s}}=4\pi {R_{\text{s}}}^{2}\times \sigma T^{4}}$

${\displaystyle I=a^{2}E}$

${\displaystyle L_{\text{s}}=4\pi I=4\pi a^{2}E}$

${\displaystyle T={\sqrt[{4}]{{\frac {E}{\sigma }}{\frac {a^{2}}{{R_{\text{s}}}^{2}}}}}}$

${\displaystyle T\simeq 5780\ {\text{K}}}$