Chapter 1 - Problem 1, 2

Problem 1
Prove the relation (1-1) between the energy density in a cavity and the emissive power. [Hint: To do so, look at the figure. The shaded volume element is of magnitude $r^2 dr \sin\theta~d\theta~ d\phi = dV$, where $r$ is the distance to the origin (at the aperture of area $dA$), $\theta$ is the angle with the vertical, and $\phi$ is the azimuthal angle about the perpendicular axis through the opening. The energy contained in the volume element is $dV$ multiplied by the energy density. The radiation is isotropic, so that what emerges is given by the solid angle $dA \cos\theta/4\pi r^2$ multiplied by the energy. This is to be integrated over the angles $\phi$ and $\theta$ and, if the flow of radiation in time $\Delta t$ is wanted, over $dr$ from 0 to $c~\Delta t$ - the distance from which the radiation will escape in the given time interval.]

Solution:
The energy contained in a volume $dV$ is
\[U(v,T)dV=U(v,T)r^2 dr sin \theta d\theta d\phi\]
when the geometry is that shown in the figure. The energy from this source that emerges through a hole of area $dA$ is
$\begin{aligned}
dE(v,T)&=\int_0^{c\Delta t} dr\int_0^{\pi/2} d\theta~d\phi U(v,T) \sin\theta \cos\theta \frac{dA}{4\pi}\\
&=\frac{dA}{4\pi} 2\pi c~\Delta t~U(v,T) \int_0^{\pi/2} d\theta~ \sin\theta \cos\theta\\
&=\frac{1}{4} c~\Delta t ~dA~ U(v,T)
\end{aligned}$
By definition of the emissivity, this is equal to $E~\Delta t~ dA$. Hence
\[E(v,T)=\frac{c}{4}U(v,T)\]

Problem 2
Given (1-6), calculate the energy density in a wavelength interval $\Delta\lambda$. Use your expression to calculate the value of $\lambda = \lambda_{\rm max}$, for which this density is maximal. Show that $\lambda_{\rm max}$ is of the form $b/T$, calculate $b$, and use 6000K as an estimate of the sun's surface temperature to calculate $\lambda_{\rm max}$ for solar radiation. [Hint: In calculating $b$ you will need the solution $x$ of the equation $(5-x)=5e^{-x}$.]

Solution
We have
\[w(\lambda,T)=U(v,T)~|dv/d\lambda|=U\left(\frac{c}{\lambda}\right)\frac{c}{\lambda^2}=\frac{8\pi hc}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1}\]
This density will be maximal when $dw(\lambda,T)/d\lambda=0$. What we need is
\[\frac{d}{d\lambda}\left(\frac{1}{\lambda^5}\frac{1}{e^{A/\lambda}-1}\right)=\left[-5\frac{1}{\lambda^6}-\frac{1}{\lambda^5}\frac{e^{A/\lambda}}{e^{A/\lambda}-1}\left(-\frac{A}{\lambda^2}\right)}\right]\frac{1}{e^{A/\lambda}-1}=0\]
Where $$A=\frac{hc}{kT}$$. The above implies that with $x=A/\lambda$, we must have
\[5-x=5e^{-x}\]
A solution of this is $x=4.965$ so that
\[\lambda_{\rm max}T=\frac{hc}{4.965k}=2.898\times 10^{-3}~\mbox{m}\]
In example 1.1 we were given an estimate of the sun's surface temperature as 6000 K. From this we get
\[\lambda_{\rm max}^{\rm sun}}=\frac{28.98\times 10^{-4}~\mbox{mK}}{6\times 10^3~\mbox{K}}=4.83 \times 10^{-7}~\mbox{m}=483~\mbox{nm}\]