$ 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\Dn %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% {\bf The solution:} \Dn (1) The partition function of one open link is $e^{-\beta\varepsilon}+e^{-\beta\varepsilon}+...+e^{-\beta\varepsilon}=ge^{-\beta\varepsilon} = x$. For $p$ open links we multiply the $p$ partition functions and get $g^pe^{-\beta\varepsilon p}=x^p$. So the partition function of the system would be: \[ Z(\beta,x) = \displaystyle\sum\limits_{p=0}^{N-1} x^p = \frac{x^N-1}{x-1} \] (2) The average number of open links is: \[

= \displaystyle\sum\limits_{p=0}^{N-1} Prob(p)\cdot p = \displaystyle\sum\limits_{p=0}^{N-1} \frac{x^p}{Z}\cdot p = \frac{1}{Z}x \frac{\partial}{\partial x} \displaystyle\sum\limits_{p=0}^{N-1} x^p = x\frac{\partial}{\partial x}\ln (Z) = \frac{Nx^N}{x^N-1}- \frac{x}{x-1} \] (3) \[

= \frac{Nx^{N+1} - Nx^N - x^{N+1} + x}{x^{N+1} - x^N - x + 1} \] Now we expand the term for $

$ to a series at $x=1$. We expand the numerator and the denominator separately to a third order and then divide them. We start with the numerator: \[ Nx^{N+1} - Nx^N - x^{N+1} + x \approx \frac{1}{2}(x-1)^2[N^2-N] + \frac{1}{6}(x-1)^3[2N^3-3N^2+N] \] The series expansion for the denominator: \[ x^{N+1} - x^N - x + 1 \approx \frac{1}{2}(x-1)^2N + \frac{1}{6}(x-1)^3[3N^2-3N] \] So we get: \[

\approx \frac{\frac{1}{2}(x-1)^2[N^2-N] + \frac{1}{6}(x-1)^3[2N^3-3N^2+N]}{\frac{1}{2}(x-1)^2N + \frac{1}{6}(x-1)^3[3N^2-3N]} = \frac{\frac{1}{2}(N-N^2)+\frac{1}{6}(x-1)(2N^2-3N+1)}{1+\frac{1}{6}(x-1)(3N-3)} \] Now we expand the term $\frac{1}{1+\frac{1}{6}(x-1)(3N-3)}$ to a series at $x=1$: \[ \frac{1}{1+\frac{1}{6}(x-1)(3N-3)} \approx 1- \frac{1}{6}(3N-3)(x-1) \] So we have: \[

\approx [\frac{1}{2}(N-N^2)+\frac{1}{6}(x-1)(2N^2-3N+1)] \cdot [1- \frac{1}{6}(3N-3)(x-1)] \] And by taking $

$ to first order and taking $N$ to be large we get: \[

\approx \frac{1}{2}N[1+\frac{1}{6}N(x-1)] \] (4) \[ \frac{

}{N}=\frac{x^N}{x^N-1}-\frac{1}{N} \cdot \frac{x}{x-1} = \frac{1}{1-\frac{1}{x^N}} - \frac{1}{N} \cdot \frac{1}{1-\frac{1}{x}} \] And for large $x$ and large $N$ we get: \[ \frac{

}{N} \approx 1 \] (5) We can see that for $x=0$ we get $\frac{

}{N} \approx 0$. The linear approximation gives us $\frac{

}{N}= \frac{1}{2}$ for $x=1$. \Dn The dependence of $\frac{

}{N}$ on $x$ converges to a Heaviside step function as $N$ becomes larger, so for $N \rightarrow \infty$ the dependence is: \[ \frac{

}{N} = \left\{ \begin{array}{rcl} 0 & \mbox{for} & x<1 \\ \frac{1}{2} & \mbox{for} & x=1 \\ 1 & \mbox{for} & x>1 \end{array}\right. \equiv H(x-1) \] (6) We first find the entropy: \[ S(x)=-\frac{\partial F(\beta,x)}{\partial T}= -\beta^2 \frac{\partial}{\partial\beta}(\frac{1}{\beta}\ln(Z)) = \ln(Z) + \frac{1}{\beta}\frac{\partial x}{\partial\beta}\frac{\partial}{\partial x}\ln(Z) = \ln(Z) + \beta \varepsilon x\frac{\partial}{\partial x}\ln(Z) = \ln(Z) + \beta \varepsilon

\] By using L'Hopital's rule we calculate the limit of $\ln(Z)$ at $x=1$ and get $\ln(Z) \approx \ln(N)$. By assuming that $\beta \varepsilon $ is not zero we get $\ln(N) << \beta \varepsilon \cdot

= \beta \varepsilon \frac{N}{2}$. So the entropy at $x=1$ is: \[ S(x=1) \approx \beta \varepsilon

\] And in the neighborhood of the point $x=1$ and for $N\rightarrow\infty$ we get: \[ \frac{S(x)}{\beta \varepsilon N} \approx \frac{

}{N} = H(x-1) \] \Dn We Now calculate the heat capacity: \[ C_V(x) = T\frac{\partial S(x)}{\partial T} = \beta\varepsilon x \frac{\partial S(x)}{\partial x} \Rightarrow \frac{C_V(x)}{\beta \varepsilon N} = x\frac{\partial}{\partial x}(\frac{S(x)}{N}) \] In the limit $N \rightarrow \infty$ and in the neighborhood of $x=1$ the entropy is a step function. So the heat capacity in that limit is: \[ \frac{C_V(x)}{\beta \varepsilon N} = \delta(x-1) \Rightarrow \frac{C_V(x=1)}{\beta\varepsilon N} = \infty \] (7) The entropy which is the first derivation of the Helmholtz free energy is a step function, so the phase transition is of the first order. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \end{document}