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\exnumber{2230}
\heading{Harmonic oscillators, Photons} 
\auname{Ira Wolfson}


{\bf The problem:}

\Dn

Find the state equations of photon gas in 1D/2D/3D cavity
within the framework of the canonical formalism, 
regarding the electromagnetic modes as a collection 
of harmonic oscillators.   

\Dn

Note: additional exercises on photon gas and blackbody radiation 
can be found in the context of quantum gases.  
Formally, photon gas is like Bose gas with chemical potential ${\mu=0}$.
Note that the same type of calculation appears 
in Debye model ("acoustic" phonons instead of "transverse" photons).

\Dn
\begin{itemize}
\item[(1)] Write the Partition function for a single photon.
\item[(2)] Find the occupation function for each $\omega$ mode.
\item[(3)] Find the degeneracy function $g(\omega)$.\\
Note: do not forget dependency on the dimension of the problem.
\item[(4)] Find the total Energy of a photon gas.
\item[(5)] Find the free Energy of a photon gas.
\item[(6)] Find an expression for the pressure of the photon gas.
\end{itemize}
  


\Dn


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{\bf The solution:}

\Dn

\subsubsection*{1 - Partition function} 
In this exercise we are supposed to treat a photon as a single harmonic oscillator ranging in frequencies.\\
The harmonic oscillator Hamiltonian is:
\begin{eqnarray}
	\hat{\mathcal{H}}=\hbar\omega	\left(\hat{n}+\frac{1}{2}\right)
\end{eqnarray} 
And we shall gauge the energy so the ground level energy contribution is zero:
\begin{eqnarray}
	\hat{\mathcal{H}}=\hbar\omega \hat{n}
\end{eqnarray}
Using the canonical ensemble we find the partition function for a single "photon" to be:
\begin{eqnarray}
Z_{\omega,1}=\sum_{n=0}^{\infty}\exp\left(-\beta\hbar\omega \hat{n}\right)=\left(1-\exp\left(-\beta\hbar\omega\right)\right)^{-1}
\end{eqnarray}

\subsubsection*{2 - Occupation function}
\begin{eqnarray}
	\mathit{f}(\omega) = <\hat{n}> = -\frac{1}{\hbar\omega}\frac{\partial \left(\ln(Z_{\omega,1}\right)}{\partial \beta}=\frac{1}{\exp\left(\beta\hbar\omega\right)-1}
\end{eqnarray}
Notice this is the same treatment for Bose gas, with chemical potential $\mu=0$, and the argument is by convention $\omega$ instead of $\varepsilon$.

\subsubsection*{3 - The degeneracy function $g(\omega)$}
We want to count the number of k-modes possible for $\omega$ frequency:
\begin{eqnarray}
	\omega=\frac{k\pi c}{L}
\end{eqnarray}
Thus:
\begin{eqnarray}
	N=\int_{k=0}^{\frac{\omega L}{\pi c}}d^D k =\frac{\Omega^{D}}{2^D}\int k^{D-1}dk=\left.\frac{\Omega^{D}k^{D}}{2^D\cdot D}\right|_{0}^{\frac{\omega L}{\pi c}}=\frac{\Omega^{D}\left(\frac{ L}{\pi	c}\right)^{D}}{2^D\cdot D}\omega^{D}
\end{eqnarray}
Now
\begin{eqnarray}
	g(\omega)=\frac{\partial N}{\partial \omega} \Rightarrow g(\omega)=\left\{\begin{array}{lr}
	\left(\frac{L}{\pi c}\right)\hspace{50pt}&1D\\
	\frac{\pi}{2}\left(\frac{L}{\pi c}\right)^{2}\omega&2D\\
	\frac{\pi}{2}\left(\frac{L}{\pi c}\right)^{3}\omega^{2}&3D
	\end{array}\right.
\end{eqnarray}
\subsubsection*{4 - total energy of photon gas}
\begin{itemize}
%%%%%%%%%%%%%%%%%%%%%%%%%%1D%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item[(1D)] In one dimension we have no degeneracy so the total energy is simply the number of oscillators (or photons) multiplied by the average oscillator energy:
	\begin{eqnarray}
		E=N\cdot <\epsilon> = \frac{N\hbar\omega}{e^{\frac{\hbar\omega}{T}}-1}
	\end{eqnarray}
%%%%%%%%%%%%%%%%%%%%%%%%%%%2D%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	
	\item[(2D)] 
	
	\begin{eqnarray}
		E_{total}=P\int_{0}^{\infty}g(\omega)\hbar\omega f(\omega) d(\omega)=P\cdot \frac{\pi}{2}\left(\frac{L}{\pi c}\right)^{2}\int_{0}^{\infty}\frac{\hbar\omega^{2}\;d\omega}{e^{\frac{\hbar \omega}{T}}-1}=\\
		\nonumber =P\cdot \frac{\pi}{2}\left(\frac{L}{\pi c}\right)^{2}\cdot\hbar\left(\frac{T}{\hbar}\right)^{3}\int_{0}^{\infty}\frac{x^{2}dx}{e^{x}-1}=\frac{P}{\pi}\zeta(3)\left(\frac{L}{\hbar c}\right)^{2}T^{3}\approx \frac{2.404}{\pi}\left(\frac{L}{\hbar c}\right)^{2}T^{3}
	\end{eqnarray}
	Where P is the polarization number, and for photons it's 2\\
	So all in all we have:
	\begin{eqnarray}
		U=\frac{2.404}{\pi}\left(\frac{L}{\hbar c}\right)^{2}T^{3}\\
		\rho(U) =\frac{2.404}{\pi}\left(\frac{1}{\hbar c}\right)^{2}T^{3}
	\end{eqnarray}
	Where here $\rho(U)$ is the energy density per unit area.
%%%%%%%%%%%%%%%%%%%%%%%%%3D%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item[(3D)]
\begin{eqnarray}
		E_{total}=P\int_{0}^{\infty}g(\omega)\hbar\omega f(\omega)=P\hbar \cdot \frac{\pi}{2}\left(\frac{L}{\pi c}\right)^{3}\int_{0}^{\infty}\frac{\omega^{3}\;d\omega}{e^{\frac{\hbar\omega}{T}}-1}=\\
		\nonumber =P\hbar \cdot \frac{\pi}{2}\left(\frac{L}{\pi c}\right)^{3}\frac{T^{4}}{\hbar^{4}}\int_{0}^{\infty}\frac{x^{3}dx}{e^{x}-1}=\frac{P}{2}\frac{\pi^{2}}{15}\left(\frac{L}{\hbar c}\right)^{3}T^{4}
\end{eqnarray}
Again with $P=2$ we get:
\begin{eqnarray}
	U=\frac{\pi^{2}}{15}\left(\frac{L}{\hbar c}\right)^{3}T^{4}	\\
	\rho(U)=\frac{\pi^{2}}{15}\left(\frac{1}{\hbar c}\right)^{3}T^{4}
\end{eqnarray}
\end{itemize}
\subsubsection*{5 - Free Energy}
We've already done most of the work needed to find the free energy function, as we've found $g(\omega)$, and the partition function for a single mode, so now, we just have to apply our previous knowledge:
\begin{itemize}
	\item[(1D)]
	\begin{eqnarray}
		Z_{w}=\frac{1}{1-e^{-\hbar\omega\beta}}\\
		Z_{all-modes}=\prod_{\omega}\frac{1}{1-e^{-\hbar\omega\beta}}\\
		F\equiv -T\ln(Z)=T\sum_{\omega}\ln\left(1-e^{-\frac{\hbar\omega}{T}}\right)\rightarrow T\int_{0}^{\infty}g(\omega)\ln\left(1-e^{-\frac{\hbar\omega}{T}}\right)d\omega=\\
		\nonumber = \frac{LT}{\pi c}\left[\cancelto{0}{ \left.\omega\cdot\ln\left(1-e^{-\frac{\hbar\omega}{T}}\right)\right|_{0}^{\infty}}+\frac{\hbar}{T}\int_{0}^{\infty}\frac{\omega\;d\omega}{e^{\frac{\hbar\omega}{T}}-1} \right]=\frac{LT^{2}}{\hbar\pi c}\int_{0}^{\infty}\frac{xdx}{e^{x}-1}=\frac{\pi}{6}\left(\frac{L}{\hbar c}\right)T^{2}
	\end{eqnarray}
	\item[(2D)]
	\begin{eqnarray}
		F=T\sum_{\omega}\ln\left(1-e^{\frac{-\hbar\omega}{T}}\right)\rightarrow T\int_{0}^{\infty}g(\omega)\ln\left(1-e^{\frac{-\hbar\omega}{T}}\right)d\omega=\\
		\nonumber = T\frac{\pi}{2}\left(\frac{L}{\pi c}\right)^{2}\int_{0}^{\infty}\omega \ln\left(1-e^{\frac{-\hbar\omega}{T}}\right)d\omega=T\frac{\pi}{2}\left(\frac{L}{\pi c}\right)^{2}\frac{1}{2}\left[\cancelto{0}{ \left.\omega^{2}\cdot\ln\left(1-e^{-\frac{\hbar\omega}{T}}\right)\right|_{0}^{\infty}}+\frac{\hbar}{T}\frac{\omega^{2}\;d\omega}{e^{\frac{\hbar\omega}{T}}-1}\right]=\\
		= \nonumber T\frac{\pi}{4}\left(\frac{L}{\pi c}\right)^{2}\left(\frac{T}{\hbar}\right)^{2}\int_{0}^{\infty}\frac{x^{2}dx}{e^{x}-1}=\frac{\zeta(3)}{2\pi}\left(\frac{L}{\hbar c}\right)^{2}T^{3}
	\end{eqnarray}
	Taking polarization into account we get:
	\begin{eqnarray}
		F=\frac{\zeta(3)}{\pi}\left(\frac{L}{\hbar c}\right)^{2}T^{3}
	\end{eqnarray}
	\item[(3D)]
	\begin{eqnarray}
		F=...=T\frac{\pi}{2}\left(\frac{L}{\pi c}\right)^{3}\int_{0}^{\omega}\omega^{2} \ln\left(1-e^{\frac{-\hbar\omega}{T}}\right)d\omega=\\
		\nonumber =T\frac{\pi}{6}\left(\frac{L}{\pi c}\right)^{3}\left[\cancelto{0}{\left.\omega^{3} \ln\left(1-e^{\frac{-\hbar\omega}{T}}\right)\right|_{0}^{\infty}}+\frac{\hbar}{T}\int_{0}^{\infty}\frac{\omega^{3}\;d\omega}{e^{\frac{\hbar\omega}{T}}-1}\right]=T\frac{\pi}{6}\left(\frac{L}{\pi c}\right)^{3}\left(\frac{T}{\hbar}\right)^{3}\int_{0}^{\infty}\frac{x^{3}dx}{e^{x}-1}=\\
		\nonumber =\frac{\pi^{2}}{90}\left(\frac{L}{\hbar c}\right)^{3}T^{4}
	\end{eqnarray}
	Taking into account polarization number we get:
	\begin{equation}
		F=\frac{\pi^{2}}{45}\left(\frac{L}{\hbar c}\right)^{3}T^{4}
	\end{equation}
\end{itemize}
\subsubsection*{6 - State equation}
Since we know the d-volume is the conjugate quantity of the d-pressure we can directly assign:
\begin{equation}
	"P"=\frac{\partial F}{\partial "V"}
\end{equation}
Where, for instance the 1 dimensional pressure is simply the force exerted by at the ends of the infinitesimal length unit (this is similar to tension, only operating outwards and not inwards)
\begin{itemize}
	\item[(1D)]
	\begin{equation}
		f=\frac{\partial F}{\partial L}=\frac{F}{L}=\frac{\pi}{6}\left(\frac{1}{\hbar c}\right)T^{2}
	\end{equation}
	\item[(2D)]
	\begin{equation}
		P_{\lambda}=\frac{\partial F}{\partial L^{2}}=\frac{F}{L^{2}}=F=\frac{\zeta(3)}{\pi}\left(\frac{1}{\hbar c}\right)^{2}T^{3}
	\end{equation}
	\item[(3D)]
	\begin{equation}
		P=\frac{\partial F}{\partial V}=\frac{\partial F}{\partial L^{3}}=\frac{F}{L^{3}}=\frac{\pi^{2}}{45}\left(\frac{1}{\hbar c}\right)^{3}T^{4}=\frac{1}{3}\rho(U)
	\end{equation}
	The last equation yields the known relation (in cosmology) between pressure and energy density 
	\begin{eqnarray}
		P=w\cdot \rho
	\end{eqnarray}
	where $\frac{1}{3}$ is indeed the relation for radiation.
\end{itemize} 
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\end{document} 
