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Display information for equation id:math.2565.20 on revision:2565

* Page found: Das Photonengas im Strahlungshohlraum (eq math.2565.20)

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Das Photonengas im Strahlungshohlraum Eq: math.2569.20

Hash: 75d31515ac7f2b894fcacc28d9157052

TeX (original user input):

\begin{align}

& U\left( T \right):=V\int_{{}}^{{}}{{}}d\nu \frac{1}{V}D\left( \nu  \right)h\nu \left\langle {{N}_{\nu }} \right\rangle =\frac{8\pi hV}{{{c}^{3}}}\int_{0}^{\infty }{{}}d\nu \frac{{{\nu }^{3}}}{{{e}^{\frac{h\nu }{kT}}}-1} \\

& =\frac{8\pi V}{{{\left( ch \right)}^{3}}}{{\left( kT \right)}^{4}}\int_{0}^{\infty }{{}}dx\frac{{{x}^{3}}}{{{e}^{x}}-1} \\

& \int_{0}^{\infty }{{}}dx\frac{{{x}^{3}}}{{{e}^{x}}-1}=\frac{{{\pi }^{4}}}{15} \\

& U\left( T \right)=\frac{8{{\pi }^{5}}V}{15{{\left( ch \right)}^{3}}}{{\left( kT \right)}^{4}} \\

\end{align}

TeX (checked):

{\begin{aligned}&U\left(T\right):=V\int _{}^{}{}d\nu {\frac {1}{V}}D\left(\nu \right)h\nu \left\langle {{N}_{\nu }}\right\rangle ={\frac {8\pi hV}{{c}^{3}}}\int _{0}^{\infty }{}d\nu {\frac {{\nu }^{3}}{{{e}^{\frac {h\nu }{kT}}}-1}}\\&={\frac {8\pi V}{{\left(ch\right)}^{3}}}{{\left(kT\right)}^{4}}\int _{0}^{\infty }{}dx{\frac {{x}^{3}}{{{e}^{x}}-1}}\\&\int _{0}^{\infty }{}dx{\frac {{x}^{3}}{{{e}^{x}}-1}}={\frac {{\pi }^{4}}{15}}\\&U\left(T\right)={\frac {8{{\pi }^{5}}V}{15{{\left(ch\right)}^{3}}}}{{\left(kT\right)}^{4}}\\\end{aligned}}

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\begin{aligned} U (T) : = V \int_{}^{} d ν \frac{1}{V} D (ν) h ν ⟨ N_{ν} ⟩ = \frac{8 π h V}{c^{3}} \int_{0}^{\infty} d ν \frac{ν^{3}}{e^{\frac{h ν}{k T}} - 1} \\ = \frac{8 π V}{{(c h)}^{3}} {(k T)}^{4} \int_{0}^{\infty} d x \frac{x^{3}}{e^{x} - 1} \\ \int_{0}^{\infty} d x \frac{x^{3}}{e^{x} - 1} = \frac{π^{4}}{15} \\ U (T) = \frac{8 π^{5} V}{15 {(c h)}^{3}} {(k T)}^{4} \end{aligned}

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Identifiers

$U$

$T$

$V$

$ν$

$V$

$D$

$ν$

$h$

$ν$

$N_{ν}$

$π$

$h$

$V$

$c$

$ν$

$ν$

$e$

$h$

$ν$

$k$

$T$

$π$

$V$

$c$

$h$

$k$

$T$

$x$

$x$

$e$

$x$

$x$

$x$

$e$

$x$

$π$

$U$

$T$

$π$

$V$

$c$

$h$

$k$

$T$

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