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\exnumber{5811}
\heading{Mechanical model for symmetry breaking}
\auname{Moamen Jbara}
% \label{name}  to label an equation

{\bf The problem:}\\

%Baruch's  A14

 An airtight piston of mass ${M}$ is free to move inside a cylindrical tube of cross sectional area ${a}$. The tube is bent into a semicircular shape of radius
${R}$. On each side of the piston there is an ideal gas of ${N}$
atoms at a temperature ${T}$.
The angular position of the piston is ${\varphi}$ (see figure). The gravitation field of Earth exerts a force Mg on the piston, while its effect on the gas particles can be neglected.
The partition function of the system can be written as d${\varphi}$ integral over exp[-A(${\varphi}$)]. The variable ${\varphi}$ is regarded as the "order parameter" of the system. A small difference ${\Delta N}$ in the occupation of the two sides is regarded as the conjugate field. The susceptibility is defined via the relation ${\langle{\varphi}\rangle}$${\approx}$ ${\chi}$${\Delta N}$.\\

(1) Write an explicit expression for A($\phi$).\\
(2) Find the coefficients in the expansion : $A(\phi) = (a/2)\phi^{2} + (u/4)\phi^{4} - h\phi$.\\
(3) Deduce what is the critical temperature $T_c$.\\
(4) Using Gaussian approximation find what is ${\chi}$ for T\textgreater$T_c$.\\ 
(5) Using Gaussian approximation find what is ${\chi}$ for T\textless$T_c$.\\
(6) Sketch a plot of ${\chi}$ versus T indicating by dashed lines the Gaussian approximations and by solid line the expected exact result. Write what is the range ${\Delta T}$ around $T_c$ where the Gaussian approximation fails.\\
(7) What is the way to take the "thermodynamic limit" such as to have a phase transition at finite temperature ?\\
(8) In reality, as the temperature is lowered, droplets condense on the walls of the left (larger) chamber. What do you expect to find in the right chamber (gas? liquid? both?).\\

Guidlines: In items (4) and (5) simplify the result assuming T\textasciitilde$T_c$ and express it in terms of $T_c$ and T-$T_c$.
The final answer should include one term only. Care about numerical prefactors - their correctness indicates that the algebra is done properly. In item (7) you are requested to identify the parameter that shouls be taken to infinity in order to get a "phase transition". Please specify what are the other parameters that should be kept constant while taking this limit.\\  

\Dn\Dn


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{\bf The solution:}

\Dn
(1) We know that the free energy of the system equals the sum of the free energy of the two gases and the energy of the piston so we can write :\\
\begin{equation}
A(\phi) = N_1 ln \left( \frac{N_1\lambda_T^3}{aR(\frac{\pi}{2}-\phi)}\right)-N_1+N_2ln\left(\frac{N_2\lambda_T^3}{aR(\frac{\pi}{2}+\phi)}\right)-N_2+\beta MgRcos(\phi)      \\\\
\end{equation}

(2) We define the following :\\
\begin{equation}
N_1 = N - (1/2)\Delta N \\
\end{equation}
\begin{equation}
N_2 = N + (1/2)\Delta N \\
\end{equation}
Taylor expansion:\\
\begin{equation} 
cos(\phi) = 1 - \frac{\phi^{2}}{2} + \frac{\phi^{4}}{24}+... \\
\end{equation}
\begin{equation} 
ln(1+x) = x-x^2+x^3-x^4+..\\
\end{equation}

We insert equations (3),(4),(5) in equation (2) and after some algebra we get the following :
\begin{equation}
A(\phi) = (N - \Delta N)ln\left(\frac{1-\frac{\Delta N}{N}}{1-\frac{2\phi}{\pi}}\right) + (N + \Delta N)ln\left(\frac{1+\frac{\Delta N}{N}}{1+\frac{2\phi}{\pi}}\right)+\beta MgR(-\frac{\phi^{2}}{2} + \frac{\phi^{4}}{24})
\end{equation} 
We omit constants that are dependent on $\Delta N$ :\\
\begin{equation}
A(\phi) = -(N - \Delta N)ln\left(1-\frac{2\phi}{\pi}\right) - (N + \Delta N)ln \left(1+\frac{2\phi}{\pi}\right)+\beta MgR(-\frac{\phi^{2}}{2} + \frac{\phi^{4}}{24})
\end{equation} 

Now we use equation (6) and after little algebra :
\begin{equation}
A(\phi) = (1/2)\left(\frac{8N}{\pi^{2}}-\beta MgR\right)\phi^{2} + (1/4)\left(\frac{32}{\pi^{4}}+\frac{\beta MgR}{6}\right)\phi^{4} - \left(\frac{4\Delta N}{\pi}\right)\phi
\end{equation}

Its easy to see that : 
\begin{equation}
a = \frac{8N}{\pi^{2}}-\beta MgR = \left(\frac{8N}{\pi^{2}}\right)\left(\frac{T-T_c}{T}\right) \\
u = \frac{32N}{\pi^{4}}+\frac{\beta MgR}{6}\\ 
h = \frac{4\Delta N}{\pi}\\
\end{equation}

(3) We can calculate $T_c$ be equating the coeffecient "a" to zero :
\begin{equation}
T_c = \frac{MgR\pi^2}{8N}
\end{equation}

(4) For T\textgreater$T_c$ , a\textgreater0 and u=0 :\\
\begin{equation} 
Z = \int d\phi e^{-A(\phi)} = \int d\phi e^{-(a/2)\phi^{2} - h\phi}\\
\end{equation}
The minimum of $A(\phi)$ is when derivative of A is zero, $\phi$ = -$\frac{h}{a}$.\\
 If there is a difference in the number of molecules then we can write :\\
 \begin{equation}
 h = \frac{2}{\pi}\Delta N
  \end{equation} 
We use the last equation and get : \\
\begin{equation} 
Z =  \int d\phi e^{-(a/2)(\phi+\frac{h}{a})^2 + (a/2)(\frac{h}{a})^2} = \sqrt{\frac{2\pi}{a}} e^{\frac{h^2}{2a}}
 \end{equation}
 
\begin{equation}
ln(Z)= \frac{h^2}{2a} + (1/2)ln\left(\frac{2\pi}{a}\right)
 \end{equation} 
 \begin{equation}
\langle\phi\rangle = \frac{dlnZ}{d\Delta N} = \left(\frac{2}{a\pi}\right)\Delta N 
 \end{equation} 
 From the last equation we can conclude that :
 \begin{equation}
\chi = \frac{2}{\pi a} = \frac{\pi}{4N}\left(\frac{T_c}{T-T_c}\right)
 \end{equation} 

(5) For T\textless$T_c$ , a\textless0 and we choose the field h equal zero. There are two gaussians that we have to take in account becuase they contribute the most to the propability :
 \begin{equation}
A(\phi) = (a/2)\phi^{2}+(u/4)\phi^{4}
 \end{equation} 
 In order to calculate the peak's values, we demand : 
  \begin{equation}
 \dfrac{dA}{d\phi}=0 \Rightarrow a\phi - u{\phi}^3 = 0  \Rightarrow \phi=\pm \sqrt{\dfrac{\lvert a \rvert}{u}} 
  \end{equation}
  As we said before the main contribution come from two gaussians that we had calculated there mean values above.
    \begin{equation}
    A(\phi_{extremum}) = -\frac{{\lvert a \rvert}^2}{2u} + \frac{u}{4}\frac{{\lvert a \rvert}^2}{u^2} \mp h\sqrt{\frac{\lvert a \rvert}{u}} = -\frac{{\lvert a \rvert}^2}{4u} \mp h\sqrt{\frac{\lvert a \rvert}{u}}
    \end{equation}
 
We use the gaussian approximation and assume that the distance between the two gaussians is much larger than their width (we keep the fourth order in order to get a symmetry brekaing) : 
we calculate Z :
\begin{equation}
Z =  \int d\phi e^{-(a/2)\phi^2 - (u/4)\phi^4} = \int d\phi_{right} + \int d\phi_{left} = 2\sqrt{\frac{2\pi}{a}}e^{-\frac{{\lvert a \rvert}^2}{4u}}cosh\left({h\frac{{\lvert a \rvert}}{u}}\right)
\end{equation}
\begin{equation}
\langle\phi\rangle = \frac{dlnZ}{d\Delta N} = tanh\left(h\frac{{\lvert a \rvert}}{u}\right)\sqrt{\frac{\lvert a \rvert}{u}} = \frac{\lvert a \rvert}{u}\frac{2}{\pi}\Delta N = \left(\frac{12\pi}{24 + \pi^{2}}\right)\left(\frac{\lvert T- T_c \rvert}{T_c}\right)\Delta N
\end{equation}
\begin{equation}
\chi = \left(\frac{12\pi}{24 + \pi^{2}}\right)\left(\frac{\lvert T- T_c \rvert}{T_c}\right)
\end{equation}
(6) See the Figure below.\\
\begin{figure}[p]
	\centering
	\includegraphics[width=0.8\textwidth]{IMG_20160830_125500.jpg}
	\caption{A plot for $\chi$ versus T}
	\label{fig:awesome_image}
\end{figure}


(7) In order to have a phase transition at finite temperature we have to approach the thermodynamic limit by taking N to infinity while keeping $T_c$ constant. The point here that at finite N the gaussian approximation is right only at range that is far away from $T_c$ (as we had seen in (6)).\\
When N starts to increase the gaussian approximation will be better, until we reach the thermodynamic limit at which the approximation will be right for every temprature.
At the thermodynamic limit $\chi$ will the have values according to equations (16) and (22), zero for T\textgreater$T_c$ and a constant for T\textless$T_c$ . \\\\
(8)  As the temperature is lowered, droplets will condense also on the walls of the right chamber becuase :\\
\begin{equation}
P_{1}>P_{2}\\
 V_{2}>V_{1}
\end{equation}
So we expect to find liquid in the right chamber.
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\end{document} 
