Solved Problem on Black Body Radiation

Show that the constant between spectral radiance R(ν) and energy density ρ(ν) is c/4.


The energy per unit of volume with frequencies between (ν, ν+) 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 will be
\[ \begin{gather} \frac{E}{V}=\rho (\nu)\,d\nu \,d\Omega \tag{I} \end{gather} \]
Figure 1
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)
\[ \begin{gather} E=\frac{A\,\cos \theta \,c\,\Delta t\,\rho (\nu)\,d\nu \, \sin \theta \,d\theta \,d\varphi}{4\pi} \tag{VI} \end{gather} \]
Figure 3

The spectral radiance is defined as radiation energy with frequencies between (ν, ν+) 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), 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
\[ \begin{array}{l} u=\sin \theta \\ \dfrac{du}{d\theta}=\cos\theta \Rightarrow d\theta=\dfrac{du}{\cos\theta } \end{array} \]
changing the limits of integration

for \( \theta =0 \)
we have \( u=\sin 0\Rightarrow u=0 \)

for \( \theta =\frac{\pi}{2} \)
we have \( u=\sin \frac{\pi}{2}\Rightarrow u=1 \)
\[ \begin{split} \int_{0}^{\frac{\pi}{2}}{\cos\theta\sin \theta \,d\theta} &\Rightarrow \int_{0}^{1}{\cos\theta \,u}\,\frac{du}{\cos\theta}\Rightarrow \int_{0}^{1}{u\,du}\Rightarrow\\[5pt] &\Rightarrow \left.\frac{u^{2}}{2}\,\right|_{\,0}^{\,1}\Rightarrow \frac{1^{2}}{2}-\frac{0^{2}}{2}\frac{1}{2}-0=\frac{1}{2} \end{split} \]
\[ \begin{gather} R(\nu )\,d\nu =\frac{c\rho (\nu )\,d\nu }{\cancel{4}\cancel{\pi}}\,\cancel{2}\cancel{\pi} \,\frac{1}{2}\\[5pt] R(\nu )\,d\nu =\frac{c}{4}\,\rho (\nu )\,d\nu \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {R(\nu)=\frac{c}{4}\,\rho (\nu)} \end{gather} \]