# 普朗克黑体辐射定律

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

${\displaystyle I_{\lambda }(\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}{\frac {1}{e^{\frac {hc}{\lambda kT}}-1}}}$

${\displaystyle I_{\nu }(\nu ,T)\,d\nu =-I_{\lambda }(\lambda ,T)\,d\lambda }$

${\displaystyle d\nu =d\left({\frac {c}{\lambda }}\right)=c\,d\left({\frac {1}{\lambda }}\right)=-{\frac {c}{\lambda ^{2}}}\,d\lambda }$

## 概述

${\displaystyle \lambda ={c \over \nu }.}$

${\displaystyle u_{\nu }(\nu ,T)={4\pi \over c}I_{\nu }(\nu ,T)={\frac {8\pi h\nu ^{3}}{c^{3}}}~{\frac {1}{e^{\frac {h\nu }{kT}}-1}},}$

${\displaystyle u_{\lambda }(\lambda ,T)={8\pi hc \over \lambda ^{5}}{1 \over e^{\frac {hc}{\lambda kT}}-1},}$

${\displaystyle I\,}$ 辐射率，在单位时间内从单位表面积和单位立体角内以单位频率间隔或单位波长间隔辐射出的能量 焦耳·秒-1·米-2·球面度 -1·赫兹-1，或焦耳·秒-1·米-2·球面度- 1·米-1 尔格·秒-1·厘米-2·赫兹-1·球面度-1
${\displaystyle \nu \,}$ 频率 赫兹 (Hz) 赫兹
${\displaystyle \lambda \,}$ 波长 (m) 厘米（cm）
${\displaystyle T\,}$ 黑体的温度 开尔文 (K) 开尔文
${\displaystyle h\,}$ 普朗克常数 焦耳·秒 (J·s) 尔格·秒（erg·s）
${\displaystyle c\,}$ 光速 米/秒 (m/s) 厘米／秒（cm/s）
${\displaystyle e\,}$ 自然对数的底，2.718281... 1 1
${\displaystyle k\,}$ 玻尔兹曼常数 焦耳／开尔文 (J/K) 尔格／开尔文 (erg/K)

## 推导

${\displaystyle \lambda _{i}={\frac {2L}{n_{i}}},}$

${\displaystyle E_{n_{1},n_{2},n_{3}}\left(r\right)=\left(r+{\frac {1}{2}}\right){\frac {hc}{2L}}{\sqrt {n_{1}^{2}+n_{2}^{2}+n_{3}^{2}}}.\qquad {\mbox{(1)}}}$

${\displaystyle P_{r}={\frac {e^{-\beta E\left(r\right)}}{Z\left(\beta \right)}}.}$

${\displaystyle \beta \ {\stackrel {\mathrm {def} }{=}}\ 1/\left(kT\right).}$

${\displaystyle Z\left(\beta \right)=\sum _{r=0}^{\infty }e^{-\beta E\left(r\right)}={\frac {e^{-{\frac {\beta \varepsilon }{2}}}}{1-e^{-\beta \varepsilon }}}.}$

${\displaystyle \varepsilon \ {\stackrel {\mathrm {def} }{=}}\ {\frac {hc}{2L}}{\sqrt {n_{1}^{2}+n_{2}^{2}+n_{3}^{2}}},}$

${\displaystyle \left\langle E\right\rangle =-{\frac {d\ln Z}{d\beta }}={\frac {\varepsilon }{2}}+{\frac {\varepsilon }{e^{\beta \varepsilon }-1}}.}$

${\displaystyle U=\int _{0}^{\infty }{\frac {\varepsilon }{e^{\beta \varepsilon }-1}}g(\varepsilon )\,d\varepsilon .\qquad {\mbox{(2)}}}$

${\displaystyle \varepsilon \ {\stackrel {\mathrm {def} }{=}}\ {\frac {hc}{2L}}n,}$

${\displaystyle n={\sqrt {n_{1}^{2}+n_{2}^{2}+n_{3}^{2}}}.}$

${\displaystyle g(\varepsilon )\,d\varepsilon =2{\frac {1}{8}}4\pi n^{2}\,dn={\frac {8\pi L^{3}}{h^{3}c^{3}}}\varepsilon ^{2}\,d\varepsilon .}$

${\displaystyle U=L^{3}{\frac {8\pi }{h^{3}c^{3}}}\int _{0}^{\infty }{\frac {\varepsilon ^{3}}{e^{\beta \varepsilon }-1}}\,d\varepsilon .\qquad {\mbox{(3)}}}$

${\displaystyle {\frac {U}{L^{3}}}=\int _{0}^{\infty }u(\nu ,T)\,d\nu ,}$

${\displaystyle u(\nu ,T)={8\pi h\nu ^{3} \over c^{3}}{1 \over e^{h\nu /kT}-1}.}$

${\displaystyle {\frac {U}{L^{3}}}=\int _{0}^{\infty }u(\lambda ,T)\,d\lambda ,}$

${\displaystyle u(\lambda ,T)={8\pi hc \over \lambda ^{5}}{1 \over e^{hc/\lambda kT}-1}.}$

${\displaystyle \varepsilon =kTx,}$

${\displaystyle u(T)={\frac {8\pi (kT)^{4}}{(hc)^{3}}}J,}$

${\displaystyle J=\int _{0}^{\infty }{\frac {x^{3}}{e^{x}-1}}\,dx={\frac {\pi ^{4}}{15}}.}$

${\displaystyle {U \over V}={\frac {8\pi ^{5}(kT)^{4}}{15(hc)^{3}}},}$

${\displaystyle I(\nu ,T)={\frac {u(\nu ,T)\,c}{4\pi }},}$

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

## 历史

${\displaystyle E=h\nu .\,}$

## 附录

${\displaystyle J=\int _{0}^{\infty }{\frac {x^{3}}{e^{x}-1}}\,dx}$

${\displaystyle \int _{0}^{\infty }{\frac {x^{n}}{e^{x}-1}}\,dx=\int _{0}^{\infty }{\frac {x^{n}e^{-x}}{1-e^{-x}}}\,dx}$

${\displaystyle {\frac {1}{1-e^{-x}}}=\sum _{k=0}^{\infty }e^{-kx}.}$

${\displaystyle \int _{0}^{\infty }x^{n}e^{-x}\sum _{k=0}^{\infty }e^{-kx}\,dx.}$

${\displaystyle \int _{0}^{\infty }x^{n}\sum _{k=1}^{\infty }e^{-kx}\,dx.}$

${\displaystyle \int _{0}^{\infty }{\frac {u^{n}}{k^{n}}}\sum _{k=1}^{\infty }e^{-u}{\frac {du}{k}}}$

${\displaystyle \int _{0}^{\infty }u^{n}\sum _{k=1}^{\infty }{\frac {1}{k^{n+1}}}e^{-u}du.}$

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{n+1}}}\int _{0}^{\infty }u^{n}e^{-u}\,du.}$

${\displaystyle \int _{0}^{\infty }{\frac {x^{n}}{e^{x}-1}}\,dx=\zeta (n+1)\Gamma {\left(n+1\right)}.}$

${\displaystyle \int _{0}^{\infty }{\frac {x^{n-1}}{e^{x}-1}}\,dx=\zeta {\left(n\right)}\Gamma {\left(n\right)}.}$

${\displaystyle J=\zeta {\left(4\right)}\Gamma {\left(4\right)}={\frac {\pi ^{4}}{90}}\times 6={\frac {\pi ^{4}}{15}}.}$

${\displaystyle \zeta {\left(4\right)}=\sum _{n=1}^{\infty }{\frac {1}{n^{4}}}=\pi ^{4}/90}$

${\displaystyle \Gamma (n+1)=n!}$。（参见黎曼ζ函数Γ函数的有关性质）。

## 参考文献

1. ^ Planck 1914，第42頁
2. ^ Rybicki & Lightman 1979，p.1）
3. Brehm, J.J. and Mullin, W.J., "Introduction to the Structure of Matter: A Course in Modern Physics," (Wiley, New York, 1989) ISBN 0-471-60531-X.
4. Planck, Max, "On the Law of Distribution of Energy in the Normal Spectrum 互联网档案馆存檔，存档日期2008-04-18.". Annalen der Physik, vol. 4, p. 553 ff (1901)
5. ^ 关于究竟是什么动机致使普朗克建立了量子化的能量这一历史争论，请参看
Kuhn, Thomas. Black-Body Theory and the Quantum Discontinuity: 1894-1912. Clarendon Press, Oxford. 1978. ISBN 0-226-45800-8.
Galison, Peter. Kuhn and the Quantum Controversy. British Journal for the Philosophy of Science. 1981, 32 (1): 71–85.
6. ^ Kragh, Helge Max Planck: The reluctant revolutionary页面存档备份，存于互联网档案馆） Physics World, December 2000.

## 延伸阅读

• Peter C. Milonni. The Quantum Vacuum. Academic Press. 1994.