Solved Problem on Black Body Radiation
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Show that the constant between spectral radiance R(ν) and energy density ρ(ν) is c/4.
Solution
The energy per unit of volume with frequencies between (ν, ν+dν) is given by
\[
\begin{gather}
\rho (\nu)\,d\nu =\frac{E}{V}
\end{gather}
\]
Considering the emission of a point P inside the body (Figure 1 cutaway diagram), the energy emitted
from this point through a solid angle dΩ will be
\[
\begin{gather}
\frac{E}{V}=\rho (\nu)\,d\nu \,d\Omega \tag{I}
\end{gather}
\]
The solid angle is defined as
\[
\begin{gather}
d\Omega =\sin \theta \,d\theta \,d\varphi \tag{II}
\end{gather}
\]
As radiation is isotropic, it is emitted in all directions, the fraction of this energy that propagates
towards the body hole per unit of solid angle will be (Figure 2)
\[
\begin{gather}
\frac{E}{V}=\frac{\text{ângulo sólido atravessado pela radiação}}{\text{ângulo sólido total}}=\rho (\nu )\,d\nu \frac{d\Omega }{\Omega _{T}} \tag{III}
\end{gather}
\]
Figure 2
Note: A total solid angle is equal to 4π steradian, corresponding to a unitary radius
sphere centered on the point where radiation spreads.
substituting the expression (II) into expression (III), we get the energy in a volume
\[
\begin{gather}
E=\frac{\rho (\nu)\,d\nu \sin \theta \,d\theta \,d\varphi }{4\pi}\,V \tag{IV}
\end{gather}
\]
The area of the hole is equal to A, and the projection of this area in the direction of radiation
from P is equal to Acos θ (Figure 3-A). Since radiation propagates with the speed of
light c, in a time interval Δt, it moves a distance cΔt, this is
the height of the cylinder containing the radiation coming from P, and its volume will be
\[
\begin{gather}
V=\underbrace{A\,\cos \theta}_{\text{ area of the base }} \,\underbrace{c\,\Delta t}_{\text{ height}} \tag{V}
\end{gather}
\]
substituting the expression (V) into expression (IV)
\[
\begin{gather}
E=\frac{A\,\cos \theta \,c\,\Delta t\,\rho (\nu)\,d\nu \, \sin \theta \,d\theta \,d\varphi}{4\pi} \tag{VI}
\end{gather}
\]
The spectral radiance is defined as radiation energy with frequencies between (ν, ν+dν) per unit of area per unit of time
\[
\begin{gather}
R(\nu )\,d\nu =\frac{E}{A\Delta t} \tag{VII}
\end{gather}
\]
substituting the expression (VI) into the expression (VII)
\[
\begin{gather}
R(\nu )\,d\nu =\frac{1}{A\Delta t}\frac{A\cos \theta \,c\,\Delta t\,\rho (\nu )\,d\nu \sin \theta \,d\theta \,d\varphi }{4\pi }
\end{gather}
\]
Radiation reaches the hole from all directions (Figure 3-B, with the hole expanded), integrating on the
angles φ, from 0 to 2π, one turning around the hole, and θ from 0 to
\( \frac{\pi}{2} \),
from the z-axis to the xy plane of the hole
\[
\begin{gather}
R(\nu)\,d\nu =\frac{1}{A\Delta t}\int_{{0}}^{{2\pi}}\int_{{0}}^{{\frac{\pi}{2}}}{\frac{A\, \cos \theta \,c\, \Delta t\,\rho (\nu)\,d\nu \sin \theta \,d\theta \,d\varphi}{4\pi}}
\end{gather}
\]
as the area of hole A, the speed of light c, the time interval Δt, the energy
density ρ(ν)dν, and the denominator 4π do not depend on the integration
variables θ and φ they can be moved out of the integral so that we can rewrite the integral as
\[
\begin{gather}
R(\nu)\,d\nu =\frac{1}{A\Delta t}\frac{A\,c\,\Delta t\,\rho (\nu)\,d\nu}{4\pi}\int_{{0}}^{{2\pi}}\,d\varphi \,\int_{{0}}^{{\frac{\pi}{2}}}{{\,\cos \theta \sin \theta \;d\theta}}\\[5pt]
R(\nu)\,d\nu =\frac{\,c\,\rho (\nu)\,d\nu }{4\pi}\int_{{0}}^{{2\pi}}\,d\varphi\,\int_{{0}}^{{\frac{\pi}{2}}}{{\,\cos \theta \sin \theta \,d\theta}}
\end{gather}
\]
Integration of \( {\large\int}_{{0}}^{{2\pi }}\;d\varphi \)
\[
\begin{gather}
\int_{{0}}^{{2\pi }}\,d\varphi =\left.\varphi \,\right|_{\,0}^{\,2\pi}=2\pi -0=2\pi
\end{gather}
\]
Integration of \( {\large\int}_{{0}}^{{\frac{\pi}{2}}}{{\,\cos\theta \sin \theta \,d\theta}} \)
Changing the variable
for \( \theta =0 \)
we have \( u=\sin 0\Rightarrow u=0 \)
for \( \theta =\frac{\pi}{2} \)
we have \( u=\sin \frac{\pi}{2}\Rightarrow u=1 \)
Changing the variable
\[
\begin{array}{l}
u=\sin \theta \\
\dfrac{du}{d\theta}=\cos\theta \Rightarrow d\theta=\dfrac{du}{\cos\theta }
\end{array}
\]
changing the limits of integration
for \( \theta =0 \)
we have \( u=\sin 0\Rightarrow u=0 \)
for \( \theta =\frac{\pi}{2} \)
we have \( u=\sin \frac{\pi}{2}\Rightarrow u=1 \)
\[
\begin{split}
\int_{0}^{\frac{\pi}{2}}{\cos\theta\sin \theta \,d\theta} &\Rightarrow \int_{0}^{1}{\cos\theta \,u}\,\frac{du}{\cos\theta}\Rightarrow \int_{0}^{1}{u\,du}\Rightarrow\\[5pt]
&\Rightarrow \left.\frac{u^{2}}{2}\,\right|_{\,0}^{\,1}\Rightarrow \frac{1^{2}}{2}-\frac{0^{2}}{2}\frac{1}{2}-0=\frac{1}{2}
\end{split}
\]
\[
\begin{gather}
R(\nu )\,d\nu =\frac{c\rho (\nu )\,d\nu }{\cancel{4}\cancel{\pi}}\,\cancel{2}\cancel{\pi} \,\frac{1}{2}\\[5pt]
R(\nu )\,d\nu =\frac{c}{4}\,\rho (\nu )\,d\nu
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{R(\nu)=\frac{c}{4}\,\rho (\nu)}
\end{gather}
\]
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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 .