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\begin{document} 
	\exnumber{3009}
	\heading{Entropy and heat capacity of quantum ideal gases} 
	\auname{Yoav Zigdon}
	{\bf The problem:}
Consider an $N$ particle ideal gas confined in volume $V$ at temperature
$T$. Find (a) the entropy $S$ and (b) the heat capacity $C$, highlighting
its dependence on the temperature.
	
	(1) Consider classical gas.
	
	(2) Consider Fermi gas at low temperatures, using leading order
	Sommerfeld expansion.
	
	(3) Consider Bose gas below the condensation temperature.
	
	(4) Consider Bose gas above the condensation temperature.
	
	(5) What is $C_{Bose}/C_{classical}$ at the condensation temperature?
	
(6) For temperatures that are above but very close to the condensation temperature, find an approximation for $C_V$
in terms of elementary functions.	\\ \\
	Hints: In (4) use the Grand-Canonical formalism to express $N$ and
	$E$ as a function of the temperature $T$ and the fugacity $z$. Use the
	equation for $N$ in order to deduce an expression for $\left(\frac{\partial z}{\partial T}\right)_N$ \\ Note that
	the derivative of the polylogarithmic function $L_{\alpha}(z)$ is
	$(1/z)L_{\alpha-1}(z)$.
	Final results should be expressed in terms of $(N,V,T)$,
	but it is allowed to define and use the notations
	$\lambda_T$ and $\epsilon_F$ and $T_c$.
	In item (4) the final result can include
	ratios of polylogarithmic functions,
	with the fugacity $z$ as an implicit variable.
Note that such ratios are all of order unity
throughout the whole temperature range
provided ${\alpha>1}$, while functions
with ${\alpha<1}$ are singular at $z=1$.
	\\
	\\
	{\bf The solution:}
	\\
	\\
	\textbf{\underline{(1) Classical Gas:}} \\ \\
	Defining the thermal wavelength:
	\begin{equation}
	\lambda_T = \left(\frac{2\pi}{mT}\right)^{\frac{1}{2}} 
	\end{equation}
	The partition function:
	\begin{equation}
     Z_N=\frac{1}{N!} \left( \frac{V}{\lambda_T ^3} \right)^N	
	\end{equation}
	The free energy is:
	\begin{equation}
	F=-T \ln \left(\frac{1}{N!} \left(\frac{V}{\lambda_T ^3} \right)^N \right) \approx -NT\ln{\left(\frac{V}{\lambda_T ^3}\right)} +T\left(N \ln {(N)} -N \right)
	\end{equation}
	We have used the Stirling approximation:
	\begin{equation}
	\ln{(N!)} \approx N \ln{(N)}-N
	\end{equation}
	\\
	(a) \textbf{The entropy is given by:} \\
	\begin{equation}
S=-\left(\frac{\partial F}{\partial T} \right)_{N,V}= N \left[\ln{\left(\frac{V}{N \lambda_T ^3}\right)}+\frac{5}{2} \right]
	\end{equation}
	\\ 
	\\
(b)	\textbf{The heat capacity:} 
\\ \\
	\begin{equation}
	C_V=T \left(\frac{\partial S}{\partial T} \right)_{N,V}=\frac{3}{2}N
	\end{equation}
\\ \\ \\
	\textbf{\underline{(2) Fermi gas at low temperature} $T\ll \epsilon_F$.} \\ \\ In order to calculate the grand free energy, lets start by recalling the Sommerfeld expansion of the number of particles and the energy:
	\begin{equation}
	N=V\frac{1}{6\pi^2} (2m)^{\frac{3}{2}} \mu^{\frac{3}{2}} \left( 1+\frac{\pi^2}{8} \left( \frac{T}{\mu}\right)^2 +...\right)
	\end{equation}
	\begin{equation}
	E=V \frac{3}{5} \frac{1}{6\pi^2} (2m)^{\frac{3}{2}} \mu^{\frac{5}{2}} \left(1+\frac{5\pi ^2}{8} \left(\frac{T}{\mu} \right)^2+... \right)
	\end{equation} 
	The pressure is related to the energy by:
	\begin{equation}
	P=\frac{2}{3} \frac{E}{V}
	\end{equation}
	By extensiveness, the grand free energy is: %The grand free energy is related to the free energy by Legendre transform:
	%\begin{equation}
	%F = F_G+\mu N = -P V + \mu N = -\frac{2}{3} E + \mu N
	%\end{equation}
	\begin{equation}
	F_G=-PV=-V\frac{2}{5} \frac{1}{6\pi^2} (2m)^{\frac{3}{2}} \mu^{\frac{5}{2}} \left(1+\frac{5\pi ^2}{8} \left(\frac{T}{\mu} \right)^2+... \right)
	\end{equation}
	%On top of that,
	%\begin{equation}
	%\mu = \epsilon_F \left(1-\frac{\pi^2}{12} \left(\frac{T}{\epsilon_F}\right)^2 +... \right)
	%\end{equation}
	%Some people just pop this equation out of their adorable mind.. :).
	\textbf{(a) The entropy is: (to the first order)}
	\begin{equation}
	S=-\left(\frac{\partial F_G}{\partial T} \right)_{\mu,V} = \frac{\pi^2}{2} V\frac{1}{6\pi^2} (2m)^{\frac{3}{2}} \mu^{\frac{1}{2}} T
	\end{equation}
We would like to express the entropy using the number of particles
because we are interested in a closed system, where the number of
particles is fixed. It is convenient to express the result in terms
of the Fermi energy:
	\begin{equation}
	\mu= \epsilon_F \left( 1- \frac{\pi^2}{12} \left( \frac{T}{\epsilon_F} \right)^2 +... \right)
	\end{equation}
	In leading order the $O(T^2)$ correction to $\mu(T)$ can be ignored and we
	get the leading linear approximation
	\begin{equation}
	S=\frac{\pi^2}{2} N \frac{T}{\epsilon_F}
	\end{equation}
	\textbf{(b) The heat capacity (to the first order)}
	\begin{equation}
	C_V=T\left(\frac{\partial S}{\partial T} \right)_{N,V}=\frac{\pi^2}{2} N \frac{T}{\epsilon_F}
	\end{equation}
	\newpage
	\textbf{\underline{(3)	Bose gas} $T_C>T$}: \\ \\ %In order to find the heat capacity and the entropy, we first need to calculate the grand free energy $F_G$. To do this, let us remember that the pressure is given by:
	In order to find the grand free energy, we need the pressure, which is:
	\begin{equation}
	P= \frac{T}{\lambda_T ^3}\zeta\left(\frac{5}{2} \right)
	\end{equation}
	By extensiveness:
	\begin{equation}
	F_G=-PV = -V \frac{T}{\lambda_T ^3}\zeta\left(\frac{5}{2} \right)
	\end{equation} 	
	\textbf{(a) The entropy:}
	\\
	\\
	\begin{equation}
	S=-\left(\frac{\partial F_G}{\partial T} \right)_{\mu,V}=\frac{5}{2} V \left(\frac{m}{2\pi}\right) ^{\frac{3}{2}} T^{\frac{3}{2}} \zeta \left(\frac{5}{2} \right)
	\end{equation}
	Formally one can regard the entropy $S$ of a Bose gas, as the sum
	over entropies $S_r$ of harmonic oscillators that have frequencies
	${\omega_r=\epsilon_r-\mu}$. Below the condensation temperature the
	zero-frequency mode ${\omega_0=0}$ contributes an infinite offset $S_0$
	to the total entropy. This offset should be excluded if the system is
	closed, because then the occupation of the lower orbital is not an
	independent variable but is dictated by the occupations of the excited
	orbitals. Differently phrased, one may say that in the regime $T<Tc$ the
	Bose gas is formally equivalent to a cavity with photons/phonons.
	 \\ \\
	\textbf{(b) The heat capacity:}
	\\ \\
	\begin{equation}
	C_V=T \left(\frac{\partial S}{\partial T}\right)_V= \frac{15}{4} V \left(\frac{m}{2\pi}\right) ^{\frac{3}{2}} T^{\frac{3}{2}}\zeta\left(\frac{5}{2} \right)
	\end{equation}
	\newpage
	\textbf{\underline{(4) Bose gas}} $T>T_C$. \\ \\%We keep the number of particles fixed, so that:
		%\begin{equation}
	%		N=\frac{V}{\lambda_T ^ 3} L_{\frac{3}{2}}(z)=\text{const}
	%	\end{equation}
	The system is closed, so
	in order to find the grand free energy, we use the expressions for the number of particles and the energy of Bose gas $T>T_C$. 
	\begin{equation}
	N=V \left(\frac{m}{2\pi} \right) ^{\frac{3}{2}} T^{\frac{3}{2}} L_{\frac{3}{2}} (z)=\text{constant}
	\end{equation}
	The energy is:
	\begin{equation}
	E=\frac{3}{2} V \left(\frac{m}{2\pi} \right) ^{\frac{3}{2}} T^{\frac{5}{2}} L_{\frac{5}{2}} (z)
	\end{equation}
			The grand free energy, as before:
		\begin{equation}
		F_G=-PV = -\frac{2}{3} E 
		\end{equation}
		Therefore: 
		\begin{equation}
		F_G=- V \left(\frac{m}{2\pi} \right) ^{\frac{3}{2}} T^{\frac{5}{2}} L_{\frac{5}{2}} (z)
		\end{equation}
		\\
		\\
		\textbf{(a) The entropy is}:
		\begin{equation}
		S=-\left(\frac{\partial F_G}{\partial T} \right)_{\mu,V} = \left(\frac{m}{2\pi} \right)^{\frac{3}{2}} V \left[ \frac{5}{2} T^{\frac{3}{2}} L_{\frac{5}{2}}(z) +T^{\frac{5}{2}} \frac{\partial L_{\frac{5}{2}}(z)}{\partial T}\right] 
		\end{equation}
		In order to calculate $\frac{\partial}{\partial T}L_{\frac{5}{2}}(z)$
	%	\begin{equation}
	%	\frac{\partial}{\partial T}L_{\frac{5}{2}} (z)=L'_{\frac{5}{2}} (z) \frac{\partial z}{\partial T}=\frac{1}{z} L_{\frac{3}{2}} (z) \frac{\partial z}{\partial T}=L_{\frac{3}{2}}(z) \left( -\frac{\mu}{T^2} \right)=-\frac{1}{T} L_{\frac{3}{2}}(z) \ln(z)
	%	\end{equation}
		We use:
		\begin{equation}
		L_{\frac{5}{2}} ' (z)=\frac{1}{z} L_{\frac{3}{2}} (z)
		\end{equation}
		and 
		\begin{equation}
		\left( \frac{\partial z}{\partial T} \right)_{\mu} = -\frac{1}{T} z \ln(z)
		\end{equation}
		And get:
		\begin{equation}
		\frac{\partial}{\partial T}L_{\frac{5}{2}} (z)=-\frac{1}{T} L_{\frac{3}{2}}(z) \ln(z)
		\end{equation}
		%Note that for $T<T_C$,  $z=1$ and thus the last equation vanish. So we multiply the term $\frac{\partial z}{\partial T}$ by $\Theta(T-T_C)$ from now on. \\ 
		Substituting to the entropy expression:
		\begin{equation}
		S=\left(\frac{mT}{2\pi} \right)^{\frac{3}{2}} V L_{\frac{3}{2}}(z) \left[ \frac{5}{2} \frac{L_{\frac{5}{2}}(z)}{L_{\frac{3}{2}}(z)} - \ln(z) \right]	
		\end{equation}
		This equation is true also for $T<T_C$ since $\mu=0$ and hence $z=1$.
		Identifying the prefactor in the last equation as the number of particles,
		\begin{equation}
			S=N \left[ \frac{5}{2} \frac{L_{\frac{5}{2}}(z)}{L_{\frac{3}{2}}(z)} - \ln(z) \right]
		\end{equation}
		In the Boltzmann regime ($z\ll1$) this agree with the classical gas
		result Eq(5).
		\newpage
		\textbf{(b) Heat capacity:}
		\\
		%It is more convenient to calculate the heat capacity using:
		Our system is closed, so the number of particles should stay constant. We calculate the heat capacity using the energy. (Alternatively we could have done it using the entropy).
				\begin{equation}
				C_V=\left(\frac{\partial E}{\partial T}\right)_{N,V}=\frac{15}{4} V \left(\frac{m}{2\pi} \right) ^{\frac{3}{2}} T^{\frac{3}{2}} L_{\frac{5}{2}} (z)+V\frac{3}{2} \left(\frac{m}{2\pi} \right) ^{\frac{3}{2}} T^{\frac{5}{2}} \frac{\partial}{\partial T}L_{\frac{5}{2}} (z)
				\end{equation}
			%Keeping the number of praticles in our bose gas (N) constant:
			%\begin{equation}
	 		%0=\frac{\partial N}{\partial T} \propto \frac{\partial}{\partial %\end{equation} 
	 		Regarding in (19) as an implicit equation for $z$ as a function of the
	 		free variables $N$ and $T$ we get by differentiation:
			\begin{equation}
			\frac{3}{2} T^{\frac{1}{2}}  L_{\frac{3}{2}} (z)+T^{\frac{3}{2}}  L_{\frac{3}{2}}^{'} (z) \frac{\partial z}{\partial T}=0
			\end{equation}
			Now, we know that 
		    \begin{equation}
	     	  L_{\frac{3}{2}} ' (z)=\frac{1}{z} L_{\frac{1}{2}} (z)
			\end{equation}
			Hence.
		\begin{equation}
		\frac{1}{z} \left(\frac{\partial z}{\partial T}\right)_N=-\frac{3}{2T} \frac{L_{\frac{3}{2}} (z)}{L_{\frac{1}{2}} (z)}
		\end{equation}
		\footnote{Note that this derivative in constant number is different from the same derivative in constant chemical potential.}Lets substitute at Eq(29):
		\begin{equation}
		C_V=\frac{15}{4} V \left(\frac{m}{2\pi} \right) ^{\frac{3}{2}} T^{\frac{3}{2}} L_{\frac{5}{2}} (z)+\frac{3}{2} V \left(\frac{m}{2\pi} \right) ^{\frac{3}{2}} T^{\frac{5}{2}} L_{\frac{3}{2}}(z) (-\frac{3}{2T}\frac{L_{\frac{3}{2}(z)}}{L_{\frac{1}{2}(z)}})
		\end{equation} 
		\begin{equation}
		C_V = \left(\frac{mT}{2\pi} \right) ^{\frac{3}{2}}V L_{\frac{3}{2}}(z) \left[    \frac{15}{4} \frac{L_{\frac{5}{2}}(z)}{L_{\frac{3}{2}}(z)} -\frac{9}{4} \frac{L_{\frac{3}{2}}(z)}{L_{\frac{1}{2}}(z)} \right]
		\end{equation}
		Expressing the heat capacity using the number of particles, 
		We finally get:
		\begin{equation}
			C_V = N \left[    \frac{15}{4} \frac{L_{\frac{5}{2}}(z)}{L_{\frac{3}{2}}(z)} -\frac{9}{4} \frac{L_{\frac{3}{2}}(z)}{L_{\frac{1}{2}}(z)} \right]
		\end{equation}
		 In the Boltzman approximation $z \ll 1$ and $L_{\alpha}(z) \approx z$, and by substitution one can see: $C_V=\frac{3}{2}N$ - in agreement with the classical heat capacity. 
%		Now use the equation for N:
%		\begin{equation}
			% %  C_V =N \left[    \frac{15}{4} \frac{L_{\frac{5}{2}}(z)}{L_{\frac{3}{2}}(z)} -\frac{9}{4} \frac{L_{\frac{3}{2}}(z)}{L_{\frac{1}{2}}(z)} \Theta(T-T_C)  \right]
		%\end{equation}
		\newpage
	(5) In the critical temperature:
		\begin{equation}
		C_V=N\frac{15 \zeta(\frac{5}{2})}{4\zeta(\frac{3}{2})}\approx 1.92 N
		\end{equation}
		This result can be deduced using Eq(18) or alternatively Eq(35). In the latter equation, the right term in the right-hand-side vanish since $L_{\frac{1}{2}}(z)$ diverges at $z \rightarrow 1$.
		Thus,
		\begin{equation}
		\frac{C_{Bose}}{C_{Classical}} \approx 1.28
		\end{equation}
		So it is easier to heat a classical gas by 28\%  than Bose quantum gas.\\ 
		\\
		\\
		(6)	We basically need to approximate polylogarithmic raios: $L_{\frac{3}{2}}(z)/L_{\frac{1}{2}}(z), L_{\frac{5}{2}}(z)/L_{\frac{3}{2}}(z) $ near $z=1$ (look at Eq. (35)). \\ For $z \rightarrow  1^+, T \rightarrow T_C, \mu \rightarrow 0^-$ we shall use the formula:
		\begin{equation}
		F_{\alpha<1} (e^{\beta \mu}) = \int_0 ^{\infty} dx \frac{x^{\alpha-1}}{\frac{e^x}{z}-1} =\Gamma(\alpha) L_{\alpha} (z) \approx \frac{1}{1-\alpha} (-\beta \mu)^{-(1-\alpha)}
		\end{equation}
		Setting $\alpha=\frac{1}{2}$ we get that for $T\rightarrow T_C$:
		\begin{equation}
		L_{\frac{1}{2}} (z) \approx \frac{2}{\sqrt{\pi}} \frac{1}{\sqrt{-\beta \mu}}
		\end{equation}
		To get an approximation for $L_{\frac{3}{2}}(z)$, we remember Eq.(31) and integrate:
				\begin{equation}
				L_{\frac{3}{2}} (z) \approx \zeta \left( \frac{3}{2} \right) - \frac{4}{\sqrt{\pi}} \sqrt{-\frac{\mu}{T_C}}
				\end{equation}
				So:
				\begin{equation}
				\frac{L_{\frac{3}{2}}(z)}{L_{\frac{1}{2}}(z)} \approx  \zeta\left( \frac{3}{2} \right)\frac{\sqrt{\pi}}{2} \sqrt{\frac{-\mu}{T_C}}
				\end{equation}
				On top of that we need:
				\begin{equation}
				\frac{L_{\frac{5}{2}}(z)}{L_{\frac{3}{2}}(z)} \approx \frac{\zeta(\frac{5}{2})}{\zeta \left( \frac{3}{2} \right) - \frac{4}{\sqrt{\pi}} \sqrt{-\frac{\mu}{T_C}}} \approx \frac{\zeta(\frac{5}{2})}{\zeta(\frac{3}{2})} \left [ 1+\frac{4}{\sqrt{\pi} \zeta(\frac{3}{2})} \sqrt{\frac{-\mu}{T_C}} \right]
				\end{equation}
				But the latter expressions depend on $\mu$, which we would like to relate to the small parameter:
			\begin{equation}
			t=\frac{T-T_C}{T_C}
			\end{equation}		
			So we approximate the number of particles $N\propto T^{\frac{3}{2}} L_{\frac{3}{2}}(z)$ near the condensation temperature: 
		\begin{equation}
			T^{\frac{3}{2}} =\left( T_C (1+\frac{T-T_C}{T_C}) \right)^{\frac{3}{2}}\approx T_C ^{\frac{3}{2}} \left(1+\frac{3}{2}t \right)
		\end{equation}
		Putting this together with Eq. (40) in the expresion for N:
		\begin{equation}
			N = V\left(\frac{m T_C}{2\pi} \right)^{\frac{3}{2}} (1+\frac{3}{2} t) \left[\zeta \left( \frac{3}{2} \right) - \frac{4}{\sqrt{\pi}} \sqrt{-\frac{\mu}{T_C}} \right]
		\end{equation}
		After a little algebra,
		\begin{equation}
			\sqrt{-\frac{\mu}{T_C}} \approx \frac{3 \sqrt{\pi} \zeta\left(\frac{3}{2} \right)}{8}t
		\end{equation}
		 	Now we are ready to get the heat capacity:
	\begin{equation}
	C_V=N \left( \frac{15}{4} \frac{\zeta(\frac{5}{2})}{\zeta(\frac{3}{2}}+\left (\frac{3\zeta(\frac{5}{2})}{2\zeta(\frac{3}{2})}-\frac{27 \pi  
		\zeta(\frac{3}{2})^2}{64} \right) t \right) \approx N (1.92-8.27t)
	\end{equation}
		 Comparing with item (3) - the heat capacity is continuous. 
\end{document}