Solved Problem on Black Body Radiation
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Use the relationship \( \begin{gather} R(\nu )\,d\nu =\frac{c}{4}\rho (\nu )\,d\nu \end{gather} \) between spectral radiance and energy density, and Planck's Radiation Law to derive Stefan's law. That is, show that
\[
\begin{gather}
R_{T}=\int_{{0}}^{{\infty}}{\frac{2\pi h}{c^{2}}\frac{\nu^{3}\;d\nu}{\operatorname{e}^{h\nu /{k T}}-1}}=\sigma T^{4}
\end{gather}
\]
where
\(
\begin{gather}
\sigma =\dfrac{2\pi ^{5}k^{4}}{15c^{2}h^{3}}
\end{gather}
\)
hint \( \begin{gather} {\Large\int}_{{0}}^{{\infty}}{\dfrac{q^{3}\;dq}{\operatorname{e}^{q}-1}}=\dfrac{\pi ^{4}}{15} \end{gather} \)
Problem data:
- Relationship between spectral radiance and energy density: \( R(\nu )\,d\nu =\dfrac{c}{4}\rho (\nu )\,d\nu \).
Planck's Radiation Law for energy density is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\rho (\nu )\,d\nu =\frac{8\pi \nu^{2}}{c^{3}}\frac{h\nu}{\operatorname{e}^{h\nu /{k T}}-1}\,d\nu} \tag{I}
\end{gather}
\]
substituting expression (I) into the relationship between spectral radiance and energy density given in the
problem, the radiance with a frequency between (ν and ν+dν) will be
\[
\begin{gather}
R(\nu)\,d\nu =\frac{c}{4}\frac{8\pi \nu^{2}}{c^{3}}\frac{h\nu}{\operatorname{e}^{h\nu /{k T}}-1}\,d\nu \\[5pt]
R(\nu)\,d\nu =\frac{2\pi h}{c^{2}}\frac{\nu^{3}\,d\nu}{\operatorname{e}^{h\nu /{k T}}-1}
\end{gather}
\]
The integral of the spectral radiance R(ν) over all frequencies will be the total radiated energy
\[
\begin{gather}
R_{T}=\int_{{0}}^{{\infty}}{R(\nu)\,d\nu}\\[5pt]
R_{T}=\int_{{0}}^{{\infty}}{\frac{2\pi h}{c^{2}}\frac{\nu^{3}\,d\nu }{\operatorname{e}^{h\nu /{k T}}-1}}
\end{gather}
\]
Integration of
\( \displaystyle \int_{{0}}^{{\infty}}{\frac{2\pi h}{c^{2}}\frac{\nu^{3}\,d\nu}{\operatorname{e}^{h\nu /{k T}}-1}} \)
substituting
for \( \nu =0 \)
we have \( q=\dfrac{h .0}{kT}\Rightarrow q=0 \)
for \( \nu =\infty \)
we have \( q=\dfrac{h.\infty }{kT}\Rightarrow q=\infty \)
substituting these values in the integral
substituting
\[
\begin{gather}
q=\frac{h\nu }{kT}\Rightarrow \nu =\frac{kT}{h}\;q
\end{gather}
\]
differentiating ν with respect to q
\[
\begin{gather}
\frac{d\nu}{dq}=\frac{kT}{h}\Rightarrow d\nu=\frac{kT}{h}\,dq
\end{gather}
\]
cubing ν
\[
\begin{gather}
\nu^{3}=\left(\frac{kT}{h}\,q\right)^{3}\Rightarrow \nu^{3}=\frac{k^{3}T^{3}}{h^{3}}\,q^{3}
\end{gather}
\]
changing the limits of integration
for \( \nu =0 \)
we have \( q=\dfrac{h .0}{kT}\Rightarrow q=0 \)
for \( \nu =\infty \)
we have \( q=\dfrac{h.\infty }{kT}\Rightarrow q=\infty \)
substituting these values in the integral
\[
\begin{gather}
\int_{{0}}^{{\infty}}{\frac{2\pi h}{c^{2}}\frac{k^{3}T^{3}}{h^{3}}\frac{q^{3}}{\operatorname{e}^{q}-1}\frac{kT}{h}\,dq}=\int_{{0}}^{{\infty}}{\frac{2\pi}{c^{2}}\frac{k^{4}T^{4}}{h^{3}}\frac{q^{3}\,dq}{\operatorname{e}^{q}-1}}
\end{gather}
\]
as the term
\( \dfrac{2\pi}{c^{2}}\dfrac{k^{4}T^{4}}{h^{3}} \)
is constant, it can be moved out of the integral
\[
\begin{gather}
\frac{2\pi}{c^{2}}\frac{k^{4}T^{4}}{h^{3}}\int_{{0}}^{{\infty}}{\frac{q^{3}\,dq}{\operatorname{e}^{q}-1}}
\end{gather}
\]
using the hint given in the problem
\( {\Large\int}_{{0}}^{{\infty}}{\dfrac{q^{3}\;dq}{\operatorname{e}^{q}-1}}=\dfrac{\pi^{4}}{15} \)
\[
\begin{gather}
\int_{{0}}^{{\infty }}{\frac{2\pi h}{c^{2}}\frac{\nu^{3}\,d\nu}{\operatorname{e}^{h\nu /{k T}}-1}}=\frac{2\pi}{c^{2}}\frac{k^{4}T^{4}}{h^{3}}\frac{\pi^{4}}{15}=\frac{2\pi^{5}}{15\,c^{2}}\frac{k^{4}T^{4}}{h^{3}}
\end{gather}
\]
\[
\begin{gather}
R_{T}=\frac{2\pi^{5}}{15\,c^{2}}\frac{k^{4}T^{4}}{h^{3}}
\end{gather}
\]
setting
\( \sigma \equiv \frac{2\pi^{5}}{15\,c^{2}}\frac{k^{4}}{h^{3}} \),
we have Stefan's Law
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{R_{T}=\sigma T^{4}} \tag{Q.E.D.}
\end{gather}
\]
Note: Q.E.D is the abbreviation of the Latin expression "Quod Erat Demonstrandum" which
means "Which was to be Demonstrated".
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .