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\begin{document} 
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\exnumber{2360}
\heading{The zipper model for DNA molecule} 
\auname{Tamar Simhon}


{\bf The problem:}

\Dn

The DNA molecule forms a double stranded helix with hydrogen bonds stabilizing the double helix.
Under certain conditions the two strands get separated resulting in a sharp ”phase transition” (in
the thermodynamic limit). One can describe the DNA molecule using the zipper model. 
A zipper has N links; each link has a state in which it is closed with energy 0 and a
state in which it is open with energy $\epsilon$, the zipper can unzip only from one side such that the $p$'th link might open only if all of
the links to its unziped side are open. The last link $p=N$ cannot be open.
Each open link has $g$ orientations corresponding to the rotational freedom about the bond.\\
Assume a large number of links.

\begin{figure}[h]
	\includegraphics[scale=0.7]{zipper}
	\centering
\end{figure}

\Dn

(1) Define $x=ge^{-\beta\epsilon}$ and find the canonical partition function $Z(\beta,x)$.\\
(2) Find the average number of open links $<p>$ as a function of $x$.\\
(3) Find the linear approximation for $<p>$.\\
(4) Approximate $\frac{<p>}{N}$ for large $x$.\\
(5) Describe the dependence of $\frac{<p>}{N}$ 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


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{\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}


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(2) The probability to find an open link is $\frac{x^p}{Z}$, thus the average number of open links 

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

\begin{equation}
	<p>=-\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
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(3) Expand (2) around $x=1$ is equivalent to expand it around $\eta=0$ where $\eta=x-1$, thus we can write

\begin{equation}
	<p>=\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}
	<p>\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}
	<p>=\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}
	<p>=\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}
	<p>=\frac{N}{2}+\frac{1}{12}N^2\eta
\end{equation}

Finally,

\begin{equation}
\boxed{
	<p>\approx\frac{N}{2}\Big[ 1+\frac{N}{6}(x-1)\Big]
}
\end{equation}



\Dn


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(4) Using (2) we get

\begin{equation}
	\frac{<p>}{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{<p>}{N}\approx 1
}
\end{equation}

Which means that all the links are open.\\
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(5) Looking at the linear approximation we found for $<p>$ at equation (9) and taking the limits $N\rightarrow\infty$ and $x\rightarrow 0$ we get $\frac{<p>}{N}=0$.\\
Subtituting $x=1$ at (9) we get $\frac{<p>}{N}=\frac{1}{2}$.\\
Considering equation (11) we get that $\frac{<p>}{N}$ behaves like heaviside function

\begin{equation}
	\frac{<p>}{N}=\begin{cases}
	0& x<1\\
	\frac{1}{2} & x=1\\
	1& x>1
	\end{cases}=H(x-1)
\end{equation}







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(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<p>
\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 $<p>\sim N$, and we know that $N\gg \ln(N)$ for large $N$.\\

Thus,

\begin{equation}
\boxed{
	S(x)\Big\rvert_{x=1}\approx\beta\epsilon<p>
}
\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}

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\Dn

(7) The entropy, which is a first derivative of the free energy, is discontinuous, thus this is a first order phase transition.










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