$ as a function of $x$.\\ (3) Find the linear approximation for $

$.\\ (4) Approximate $\frac{

}{N}$ for large $x$.\\ (5) Describe the dependence of $\frac{

}{N}$ on $x$.\\ (6) Find expressions for the entropy $S(x)$ and the heat capacity $C_{V}(x)$ at $x=1$.\\ (7) What is the order of the phase transition? \Dn %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% {\bf The solution:} \Dn (1) Assuming there are $p$ open links, the hamiltonian of the system is $\mathcal{H}=p\epsilon$, taking into consideration the number of orientations, the partition function is \begin{equation} Z(\beta,x)=\sum_{p=0}^{N-1}g^pe^{-\beta p\epsilon}=\sum_{p=0}^{N-1}x^p=\boxed{\frac{x^N-1}{x-1}} \end{equation} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% (2) The probability to find an open link is $\frac{x^p}{Z}$, thus the average number of open links \begin{equation}

=\sum_{p=0}^{N-1}\frac{x^p}{Z}p=\frac{1}{Z}x\frac{\partial}{\partial x}\sum_{p=0}^{N-1}x^p=x\frac{\partial}{\partial x} \ln(Z)=\boxed{\frac{Nx^N}{x^N-1}-\frac{x}{x-1}} \end{equation} Another way is noticing that $

=\frac{U}{\epsilon}$, where $U$ is the energy of the system defined $U=-\frac{\partial}{\partial\beta}\ln(Z)$.\\ Thus, \begin{equation}

=-\frac{1}{\epsilon}\frac{\partial}{\partial\beta}\ln(Z)=\frac{1}{\epsilon}x\epsilon\frac{\partial}{\partial x}\ln(Z)=\frac{Nx^N}{x^N-1}-\frac{x}{x-1} \end{equation} Which is the same as eqation (2). \Dn %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%55 (3) Expand (2) around $x=1$ is equivalent to expand it around $\eta=0$ where $\eta=x-1$, thus we can write \begin{equation}

=\frac{N\eta(1+\eta)^N-(1+\eta)\Big((1+\eta)^N-1\Big)}{\eta\Big((1+\eta)^N-1\Big)} \end{equation} Expanding the term $(1+\eta)^N$ up to third order, \begin{equation}

\approx\frac{N\eta(1+N\eta+\frac{1}{2}N^2\eta^2+\frac{1}{6}N^3\eta^3+O(\eta^4))-(1+\eta)(N\eta+\frac{1}{2}N^2\eta^2+\frac{1}{6}N^3\eta^3+O(\eta^4))}{\eta(N\eta+\frac{1}{2}N^2\eta^2+\frac{1}{6}N^3\eta^3+O(\eta^4))} \end{equation} After some algebra we get \begin{equation}

=\frac{(\frac{N}{2}-1)+\eta(\frac{1}{3}N^2-\frac{1}{2}N)+\eta^2(\frac{1}{6}N^3-\frac{1}{6}N^2)}{1+\frac{1}{2}N\eta+\frac{1}{6}N^2\eta^2} \end{equation} Expanding the denominator $(1+\frac{1}{2}N\eta+\frac{1}{6}N^2\eta^2)^{-1}\approx 1-\frac{1}{2}N\eta+\frac{1}{12}N^2\eta^2$ we get, \begin{equation}

=\Big((\frac{N}{2}-1)+\eta(\frac{1}{3}N^2-\frac{1}{2}N)+\eta^2(\frac{1}{6}N^3-\frac{1}{6}N^2)\Big)\Big( 1-\frac{1}{2}N\eta+\frac{1}{12}N^2\eta^2 \Big) \end{equation} Taking up to first order at $\eta$ and $N$ to be large we get, \begin{equation}

=\frac{N}{2}+\frac{1}{12}N^2\eta \end{equation} Finally, \begin{equation} \boxed{

\approx\frac{N}{2}\Big[ 1+\frac{N}{6}(x-1)\Big] } \end{equation} \Dn %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% (4) Using (2) we get \begin{equation} \frac{

}{N}=\frac{1}{1-\frac{1}{x^N}}-\frac{1}{N}\frac{1}{1-\frac{1}{x}} \end{equation} Thus for large $x$ and large $N$ we get \begin{equation} \boxed{ \frac{

}{N}\approx 1 } \end{equation} Which means that all the links are open.\\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5 (5) Looking at the linear approximation we found for $

$ at equation (9) and taking the limits $N\rightarrow\infty$ and $x\rightarrow 0$ we get $\frac{

}{N}=0$.\\ Subtituting $x=1$ at (9) we get $\frac{

}{N}=\frac{1}{2}$.\\ Considering equation (11) we get that $\frac{

}{N}$ behaves like heaviside function \begin{equation} \frac{

}{N}=\begin{cases} 0& x<1\\ \frac{1}{2} & x=1\\ 1& x>1 \end{cases}=H(x-1) \end{equation} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% (6) The entropy is proportional to the derivative of the free energy with regard to $T$ \begin{equation} S=-\frac{\partial F}{\partial T}=-\beta^2\frac{\partial}{\partial\beta}(\frac{1}{\beta}\ln(Z))=\ln(Z)-\beta\frac{\partial x}{\partial\beta}\frac{\partial}{\partial x}\ln(Z)=\ln(Z)-\beta(-\epsilon x)\frac{\partial}{\partial x}\ln(Z) \end{equation} Thus, \begin{equation} S=\ln(Z)+\beta\epsilon

\end{equation} Taking the limit of $\ln(Z)$ for $x\rightarrow 1$ using L'Hopital's rule we get that $\ln(Z)\approx \ln(N)$. But, according to (12) we see that for this limit $

\sim N$, and we know that $N\gg \ln(N)$ for large $N$.\\ Thus, \begin{equation} \boxed{ S(x)\Big\rvert_{x=1}\approx\beta\epsilon

} \end{equation} The heat capacity is defined, \begin{equation} C_{V}=T\frac{\partial S}{\partial T}=\beta\epsilon x \frac{\partial S}{\partial x}=\beta\epsilon xN\delta(x-1) \end{equation} Thus at the limit $x\rightarrow 1$ \begin{equation} \boxed{ C_{V}\rightarrow\infty } \end{equation} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \Dn (7) The entropy, which is a first derivative of the free energy, is discontinuous, thus this is a first order phase transition. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \end{document}