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\heading{E3010: Heat capacity of ideal Bose gas} 
\auname{Snir Cohen}


{\bf The problem:}

\Dn

Consider a volume $\mathsf{V}$ that contains $N$ mass $\mass$ bosons.
The gas is in a thermal equilibrium at temperature $T$\,.

\Dn

(1) Write an explicit expression for the condensation temperature $T_c$\,.

(2) Calculate the chemical potential, the energy and the pressure in the Boltzmann approximation $T \gg T_c $\,.

(3) Calculate the chemical potential, the energy and the pressure in the regime $T < T_c$\,.

(4) Calculate  $C_v$  for $ T < T_c$\,.

(5) Calculate $C_v$ for $T = T_c$\,.

(6) Calculate $C_v$ for $T \gg T_c$\,.

(7) Express the ratio $C_p/C_v$ using the polylogarithmic functions. Explain why $C_p \! \rightarrow \! \infty$ in the condensed phase?

(8) Find the $\gamma$ in the adiabatic equation of state. Note that in general it does not equal $C_p/C_v$\,.


\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:} 
{\LARGE Note an impoved version in  [\href{http://physweb.bgu.ac.il/COURSES/StatMechCohen/ExercisesPool/EXERCISES/ex_3009_sol_Y16.pdf}{ex3009}].}

\Dn

(1) The condensation temperature:
  \begin{equation}
		T_c =\frac{2\pi}{m} \left(\frac{N}{\zeta\left(\frac{3}{2}\right) V}\right)^{\frac{2}{3}} \;\;,\;\; \zeta\left(\frac{3}{2}\right)\approx 2.612
	\end{equation}
	
See Pathria, 2nd edition, p.161.

\Dn

(2) In Boltzmann approximation $T \gg T_c$\, and also $g_{\nu}(z)\approx z$\,, therefore:

\begin{equation}
    \frac{N}{V} = \frac{1}{\lambda_{T}^3} \,z \;\;,\;\; z=\eexp{\beta\mu} \;\;,\;\; \lambda_T=\left(\frac{2\pi}{mT}\right)^{\frac{1}{2}}
 \end{equation}         

\begin{equation}
		\frac{N}{V} \lambda_{T}^3=\eexp{\mu/T} \; \Rightarrow \; \mu = T \ln{\left( \frac{N}{V} \lambda_{T}^3 \right) }
\end{equation}

\begin{equation}
    \frac{E}{V} = \frac{3}{2} \frac{T}{\lambda_{T}^3} \,z = \frac{3}{2} \frac{N}{V} T \; \Rightarrow \; E=\frac{3}{2} NT
\end{equation}         

\begin{equation}
   P = \frac{2}{3} \left( \frac{E}{V} \right) = \frac{T}{\lambda_{T}^3}\,z = \frac{N}{V} T
\end{equation} 

These are the classical results for an ideal gas.        

\Dn

(3) For $T < T_c$\,:

\begin{equation}
  \mu = 0 \;\;,\;\; z = 1
 \end{equation}         
\begin{equation}
    E = \frac{3}{2} \frac{T}{\lambda_{T}^3} V g_{5/2}(1) = \frac{3}{2} \zeta\left(\frac{5}{2}\right)  \left(\frac{m}{2\pi}\right)^{\frac{3}{2}} V T^{\frac{5}{2}} 
    \;\;,\;\; \zeta\left(\frac{5}{2}\right)\approx 1.341
\end{equation}         

\begin{equation}
	P = \frac{2}{3} \left( \frac{E}{V} \right) = \zeta\left(\frac{5}{2}\right) \left(\frac{m}{2\pi}\right)^{\frac{3}{2}} T^{\frac{5}{2}} 
\end{equation}

\Dn

%%%%%%%%%%%%%


(4) Heat capacity for $T < T_c$\,:

\begin{equation}
	C_v = \left( \frac{\partial E}{\partial T} \right)_{V,N} = \frac{15}{4} \zeta\left(\frac{5}{2}\right) \left(\frac{m}{2\pi}\right)^{\frac{3}{2}} V  T^{\frac{3}{2}}
\end{equation} 

From now on, all dervivatives with respect to $T$ will be calculated keeping $V$ and $N$ constant, unless noted otherwise.

\Dn

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(5)	Since the expression we got for the energy in (7) is valid for $T \leq T_c$\,, at $T = T_c$\,: 

\begin{equation}
 C_v  = \frac{15}{4} \zeta\left(\frac{5}{2}\right) \left(\frac{m}{2\pi}\right)^{\frac{3}{2}} V T_{c}^{\frac{3}{2}} 
 = \frac{15}{4} N \frac{\zeta\left(\frac{5}{2}\right)}{\zeta\left(\frac{3}{2}\right)} \approx 1.925N 
\end{equation} 

%%%%%%%%%%%%%

(6) For $T \gg T_c$\,:
\begin{equation}
	E = \frac{3}{2} N T \Rightarrow C_v = \frac{3}{2} N
\end{equation}

%%%%%%%%%%%%%%%%%%%%%%%

(7) Let us find $C_v$ for $T > T_c$\,:
\begin{equation}
	\frac{N}{V} = \frac{1}{\lambda_{T}^3} g_{3/2}(z) \;\;,\;\;
	E = \frac{3}{2} V \frac{T}{\lambda_{T}^3} g_{5/2}(z) = \frac{3}{2} N T \frac{g_{5/2}(z)}{g_{3/2}(z)}
\end{equation}

\begin{equation}
	\frac{C_v}{N} = \frac{\partial}{\partial T} \left[\frac{3}{2} T \frac{g_{5/2}(z)}{g_{3/2}(z)} \right]
\end{equation}  

Next, we need to find the derivatives of the polylogarithmic functions with respect to $T$\,:

\begin{equation}
	\frac{\partial}{\partial T} g_{3/2}(z) = \frac{\partial}{\partial T} \left( \frac{N}{V} \lambda_{T}^3 \right)
	= - \frac{3}{2} \frac{N}{V} \left( \frac{2\pi}{mT} \right)^{\frac{3}{2}} \cdot \frac{1}{T}
	= - \frac{3}{2T} g_{3/2}(z)
\end{equation}

Using the recurrence relation (Pathria, 2nd edition, Appendix: D.10)

\begin{equation}
	z \frac{\partial}{\partial z} g_{\nu}(z) = g_{\nu - 1}(z) \,,
\end{equation}

we get:
\begin{equation}
	z \frac{\partial}{\partial z} g_{3/2}(z) = g_{1/2}(z)
\end{equation}

Dividing (14) by (16):
\begin{equation}
	\frac{\partial z}{\partial T} = - \frac{3z}{2T} \frac{g_{3/2}(z)}{g_{1/2}(z)} \;\;,\;\;
	\frac{\partial}{\partial z} = \frac{\partial}{\frac{\partial z}{\partial T} \partial T} = - \frac{2T}{3z} \frac{g_{1/2}(z)}{g_{3/2}(z)} \frac{\partial}{\partial T}
\end{equation}

Using (15) and (17) we get:
\begin{equation}
	z \frac{\partial}{\partial z} g_{5/2}(z) = z \left(- \frac{2T}{3z} \right) \frac{g_{1/2}(z)}{g_{3/2}(z)} \frac{\partial}{\partial T} g_{5/2}(z) = g_{3/2}(z)
\end{equation}

\begin{equation}
	\frac{\partial}{\partial T} g_{5/2}(z) = - \frac{3}{2T} \frac{g_{3/2}^2(z)}{g_{1/2}(z)}
\end{equation}

Using (14) and (19) in the calculation of (13) and after some algebra we finally get:
\begin{equation}
	\frac{C_v}{N} = \frac{15}{4} \frac{g_{5/2}(z)}{g_{3/2}(z)} - \frac{9}{4} \frac{g_{3/2}(z)}{g_{1/2}(z)}
\end{equation}

We find $C_p$ in a similar way:
\begin{equation}
	C_p = \left( \frac{\partial \left( E + PV \right)}{\partial T} \right)_{P,N}
	= \left( \frac{\partial E}{\partial T} \right)_{P,N} + P \left( \frac{\partial V}{\partial T} \right)_{P,N}
\end{equation}

From now on all derivatives with respect to $T$ will be calculated keeping $P$ and $N$ constant,
meaning we cannot use the expressions we got for the derivatives of the polylogarithmic functions and we need to do the calculations again.
From (12):
\begin{equation}
	\frac{\partial E}{\partial T} = \frac{\partial}{\partial T} \left[ \frac{3}{2} N T \frac{g_{5/2}(z)}{g_{3/2}(z)} \right]
\end{equation}  

\begin{equation}
	P = \frac{T}{\lambda_{T}^3} g_{5/2}(z)
\end{equation}

\begin{equation}
	\frac{\partial}{\partial T} g_{5/2}(z) = \frac{\partial}{\partial T} \left( P \frac{\lambda_{T}^3}{T} \right)
	= - \frac{5}{2} P \left( \frac{2\pi}{mT} \right)^{\frac{3}{2}} \frac{1}{T} \cdot \frac{1}{T}
	= - \frac{5}{2T} g_{5/2}(z)
\end{equation}

Recalling that
\begin{equation}
	z \frac{\partial}{\partial z} g_{5/2}(z) = g_{3/2}(z) \,,
\end{equation}

we divide (24) by (25) and get:
\begin{equation}
	\frac{\partial z}{\partial T} = - \frac{5z}{2T} \frac{g_{5/2}(z)}{g_{3/2}(z)} \;\;,\;\;
	\frac{\partial}{\partial z} = \frac{\partial}{\frac{\partial z}{\partial T} \partial T}
	= - \frac{2T}{5z} \frac{g_{3/2}(z)}{g_{5/2}(z)} \frac{\partial}{\partial T} 
\end{equation}

Using (16) and (26) we get: 

\begin{equation}
	z \frac{\partial}{\partial z} g_{3/2}(z)
	= z \left( - \frac{2T}{5z} \right) \frac{g_{3/2}(z)}{g_{5/2}(z)} \frac{\partial}{\partial T} g_{3/2}(z) = g_{1/2}(z)
\end{equation}

\begin{equation}
	\frac{\partial}{\partial T} g_{3/2}(z) = - \frac{5}{2T} \frac{g_{5/2}(z) g_{1/2}(z)}{g_{3/2}(z)}
\end{equation}

\newpage

Using (24) and (28) in the calculation of (22) and arranging we obtain:

\begin{equation}
	\frac{\partial E}{\partial T}
	= N \frac{g_{5/2}(z)}{g_{3/2}(z)} \left[ -\frac{9}{4} + \frac{15}{4} \frac{g_{5/2}(z) g_{1/2}(z)}{g_{3/2}^2(z)} \right]
\end{equation}

From (12) we get an expression for $V$:
\begin{equation}
	V = N \frac{\lambda_{T}^3}{g_{3/2}(z)}
\end{equation}

To make the calculation of the derivative of $V$ with respect to $T$ easier:
\begin{equation}
	\frac{\partial}{\partial T} \lambda_{T}^3 = - \frac{3}{2} \left( \frac{2\pi}{mT} \right)^{\frac{3}{2}} \cdot \frac{1}{T}
	= - \frac{3}{2T} \lambda_{T}^3
\end{equation}

Using (28) and (31) we calculate the derivative and get:

\begin{equation}
	\frac{\partial V}{\partial T} = N \left[ - \frac{3}{2T} \lambda_{T}^3 \frac{1}{g_{3/2}(z)} + \frac{5}{2T} \lambda_{T}^3 \frac{g_{5/2}(z) g_{1/2}(z)}{g_{3/2}^3(z)} \right]
\end{equation}

Recalling that $P = \frac{T}{\lambda_{T}^3} g_{5/2}(z)$\,:

\begin{equation}
	P \left( \frac{\partial V}{\partial T} \right) = N \frac{g_{5/2}(z)}{g_{3/2}(z)}
	\left[ - \frac{3}{2} + \frac{5}{2} \frac{g_{5/2}(z) g_{1/2}(z)}{g_{3/2}^2(z)} \right]
\end{equation}

Using (29) and (33) in (21) we get:
\begin{equation}
	\frac{C_p}{N} = \frac{g_{5/2}(z)}{g_{3/2}(z)} \left[ - \frac{15}{4} + \frac{25}{4} \frac{g_{5/2}(z) g_{1/2}(z)}{g_{3/2}^2(z)} \right]
\end{equation}

Finally: 
\begin{equation}
	\frac{C_p}{C_v} = \frac{\frac{C_p}{N}}{\frac{C_v}{N}} = \frac{5}{3} \frac{g_{5/2}(z) g_{1/2}(z)}{g_{3/2}^2(z)}
\end{equation} 

In condensation $C_p \rightarrow \infty$ $(g_{1/2} (z=1)$ diverges).
We look at the definition of $C_p$\,:

\begin{equation}
	C_p \equiv T \left(\frac{\partial S}{\partial T} \right)_{P,N} = \left(\frac{\dbar Q}{dT} \right)_{P,N}
\end{equation}

In condensed state $P = \const \cdot T^{\frac{5}{2}}$, so by keeping $P$ constant $\Rightarrow dT = 0$\,.
The temperature will not change, no matter how much heat will enter the system, meaning the volume will grow. 

\Dn

%%%%%%%%%%%%%%%%%%%%%%%%

(8) We need to find the entropy. Let us look at:
\begin{equation}
	E - TS + PV \equiv \mu N \Rightarrow \frac{S}{N} = \frac{E + PV}{NT} - \frac{\mu}{T}
\end{equation}

\newpage 

For $T \leq T_c$\,, $\mu = 0$ and from (7) and (8):

\begin{equation}
	E = \frac{3}{2} \frac{T}{\lambda_{T}^3} V \zeta\left( \frac{5}{2} \right) \;\;,\;\;
	P = \frac{T}{\lambda_{T}^3} \zeta\left( \frac{5}{2} \right)
\end{equation}

Hence
\begin{equation}
	\frac{S}{N} = \frac{5}{2} \frac{1}{\lambda_{T}^3} \frac{V}{N} \zeta\left( \frac{5}{2} \right)
\end{equation}

For $T > T_c$\,:

\begin{equation}
	z = \eexp{\mu / T} \Rightarrow \mu = T \ln{z}
\end{equation}
\begin{equation}
	E = \frac{3}{2} \frac{T}{\lambda_{T}^3} V g_{5/2}(z) \;\;,\;\; P = \frac{T}{\lambda_{T}^3} g_{5/2}(z)
\end{equation}

Putting it all together we get:
\begin{equation}
	\frac{S}{N} = \frac{5}{2} \frac{V}{N} \frac{g_{5/2}(z)}{\lambda_{T}^3} - \ln{z}
\end{equation}

Using (12) we get a final expression:
\begin{equation}
	\frac{S}{N} = \frac{5}{2} \frac{g_{5/2}(z)}{g_{3/2}(z)} - \ln{z}
\end{equation}

A reversible adiabatic process implies that $S$ and $N$ are constants. For the last expression it implies that $z$ is constant,
Therefore $g_{3/2}(z)$ is constant and
\begin{equation}
	g_{3/2}(z) = \lambda_{T}^3 \frac{N}{V} = \text{const}
\end{equation}

Equation (39) implies the same. Now,

\begin{equation}
	\left(	\frac{2\pi}{mT} \right)^{\frac{3}{2}} \frac{N}{V} = \text{const} \Rightarrow T^{\frac{3}{2}} V = \text{const}
\end{equation}

For both regions, using the expressions for $P$ we get:
\begin{equation}
	\frac{P}{T^{\frac{5}{2}}} = \text{const}
\end{equation}

From (45):
\begin{equation}
	T = \text{const} \cdot \frac{1}{V^{2/3}}
\end{equation}

\begin{equation}
	\frac{P}{\left( V^{-\frac{2}{3}} \right)^{\frac{5}{2}}} = \text{const} \Rightarrow P V^{\frac{5}{3}} = \text{const}
\end{equation}

Hence $\gamma = \frac{5}{3} $ .

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