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\heading{E3010: Heat capacity of ideal Bose gas} 
\auname {Submited by Yigal Tirosh}

{\bf The problem:}

\Dn

 $N$ bosons with spin 1 and mass $m$ are in a box with a volume $V$ .
  The gas is in a thermic equilibrium in a temperature .
\Dn

(1) Write an explicit expression for $Tc$  using $ N$, $m$ and  $V$.

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

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

(4) calculate  $C_v$  in  the  temperature  $T<T_c$

(5) What is the result  for  $ C_v$   in the temperatures T = Tc and in Boltzmann approximation $T  \gg Tc$.

(6)  Show that  $ \frac{C_p}{C_v} =  \frac{5g_\frac{5}{2}(z) g_\frac{1}{2}(z)}{3g^2_\frac{3}{2}(z)}$.

        Why is  $C_p  \rightarrow   \infty $  in the condensed phase?.

(7)   Find $ \gamma$  in the adiabatic equation of state.  Note that in general $ \gamma  \not=  \frac{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)$T  = T_c$
 \begin{equation}
T_c =\frac{1}{2m\pi}  (\frac{N}{3V\zeta(\frac{3}{2})})^\frac{2}{3} =  \frac{1}{2m\pi}  (\frac{N}{3V2.612})^\frac{2}{3}.
\end{equation}

(2) $ T \gg T_c $   { Boltzmann approximation}.
\vspace{5 mm}
\begin{equation}
    \frac{N}{V} = 3\frac{z}{\lambda^3}   \rightarrow   z  = \frac{N\lambda^3}{3V}.
 \end{equation}         
\vspace{5 mm}
\begin{equation}
    E =3\frac{3}{2}V\frac{T}{\lambda^3}z = \frac{9}{2}V\frac{T}{\lambda^3}e^{\beta\mu} = \frac{3}{2}NT.
\end{equation}         
\vspace{5 mm}
\begin{equation}
   P = \frac{2E}{3V} = 3\frac{T}{\lambda^3}e^{\beta\mu} = \frac{NT}{V}.
\end{equation}         
\vspace{5 mm}

(3 ) $T < T_c$.
\begin{equation}
  \mu = 0 .
 \end{equation}         
\begin{equation}
    E =  3 V(2m)^\frac{3}{2}2\pi \Gamma(\frac{5}{2})\zeta(\frac{5}{2})T^\frac{5}{2} = 3 V(2m)^\frac{3}{2}2\pi \frac{3}{4}\sqrt{\pi} 1.341T^\frac{5}{2}.
\end{equation}         
\begin{equation}
       P = \frac{2E}{3V} = 2 (2m)^\frac{3}{2}2\pi\frac{3}{4}\sqrt{\pi} 1.341T^\frac{5}{2}.
\end{equation}         

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


(4)  $T<T_c$
\vspace{5 mm}
      
\begin{equation}
C_v = \frac{\partial E}{\partial T} = 3 V(2m)^\frac{3}{2}2\pi \Gamma(\frac{5}{2})\zeta(\frac{5}{2})T^\frac{3}{2}\frac{5}{2} =3 V(2m)^\frac{3}{2}2\pi\sqrt{\pi} 1.341T^\frac{3} {2}\frac{15}{8}.
\end{equation} 
\begin{equation}
  N_e = N{\frac{T}{T_c}}^\frac{3}{2}
 \end{equation}     

 particles  in  the excited  states  increased  with $T$.
\vspace{5 mm}

  $C_v \propto  N_e(T)$.

 $C_v$  increased  with  $N_e$   until  $N_e(T_c=T) = N$.

\vspace{5 mm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(5)  $T = T_c$.
\begin{equation}
 C_v = \frac{\partial E}{\partial T} = 3 V(2m)^\frac{3}{2}2\pi \Gamma(\frac{5}{2})\zeta(\frac{5}{2})T^\frac{3}{2}_c\frac{5}{2} = 3V(2m)^\frac{3}{2}2\Pi\sqrt{\pi} 1.341T^\frac{3}{2}_c\frac{15}{8} = 1.925NK.
\end{equation} 

 \vspace{5 mm}
        
  $C_v(T=T_c) $  Heat  capacity  have  maximum,   phase  transition .
\vspace{5 mm}
  
  $T>T_c$.
\vspace{5 mm}
\begin{equation}
 C_v =  \frac{\partial }{\partial T}[(T\frac{3g_\frac{5}{2}(z)}{2g_\frac{3}{2}(z)})]_v  =  \frac{15g_\frac{5}{2}}{4g_\frac{3}{2}(z)}  - \frac{9g_\frac{3}{2}(z)}{4g_\frac{1}{2}(z)} .  
\end{equation} 
\vspace{5 mm}

 for  $T = T_c    \rightarrow C_v =  1.925NK$.
 \vspace{5 mm}

$C_v$  grow  smaller  with  $T$ .
 \vspace{5 mm}
 
 $T\gg T_c$.     { Boltzmann approximation}.
\vspace{5 mm}

\begin{equation}
E = \frac{3}{2}NT  \rightarrow  C_v =  \frac{3}{2}N.
\end{equation} 

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

(6)
\begin{equation}    
\frac{C_p}{NK} =  \frac{\partial }{\partial T}[(T\frac{5g_\frac{5}{2}(z)}{2g_\frac{3}{2}(z)})]_p =  [\frac{\partial z}{\partial T}]_p [\frac{\partial F(g)}{\partial z}]_p + F(g)  =
\frac{25}{4}\frac{g^2_\frac{5}{2}(z)g_\frac{1}{2}(z)}{g^3_\frac{3}{2}(z)}  - \frac{15}{4}\frac{g_\frac{5}{2}(z)}{g_\frac{3}{2}(z)}.
\end{equation}
\begin{equation}
  F(g) = \frac{5g_\frac{5}{2}(z)}{2g_\frac{3}{2}(z)}.
\end{equation}
\begin{equation}
 [\frac{\partial z}{\partial T}]_p = \frac{-5z}{2T} \frac{g_\frac{5}{2}(z)}{g_\frac{3}{2}(z)}.
\end{equation}
\vspace{5 mm}

 $C_v $   From  (11)
\begin{equation}
\rightarrow   \frac{C_p}{C_v} =  \frac{5g_\frac{5}{2}(z) g_\frac{1}{2}(z)}{3g^2_\frac{3}{2}(z)}.
\end{equation}
$C_p (T <  T_c)  \rightarrow   \infty$   all  particles  in  the  condensed  phase  and  they  compress  to   zero  volume the  system  have  only  one phase. 

\vspace{5 mm}
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  (7)   Adiabat  of an  ideal Bose  gas.
\vspace{5 mm}
  
 For  $T \leq  T_c$

\vspace{5 mm}

     $S$ - Entropy
\begin{equation}
   \frac{S}{NK} = \frac{5}{2} \frac{V}{\lambda^3}\zeta(\frac{5}{2})  \rightarrow   VT^\frac{3}{2}  =  const.
\end{equation}
\begin{equation}
        P = \frac{KT}{\lambda^3}\zeta(\frac{5}{2})  \rightarrow  \frac{P}{T^\frac{5}{2}} =   const.
\end{equation}

\vspace{5 mm}

 From  (17) and  (18)  
\begin{equation}
 PV^\frac{5}{3}  =  const      \rightarrow   \gamma = \frac{5}{3}.   
\end{equation}

\vspace{5 mm}
 \
 Ideal  Bose  gas  only  for  $T  \gg  T_c $     then       $ \gamma  \simeq  \frac{5}{3}$.      
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\end{document} 
