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\heading{E2341: DNA molecule} 
\auname{Eyal Gavish}


{\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).
We model the described above as a zipper consisting of N parallel links that can be opened from one side of the zipper.
If links $1,2,3,...,p$ are opened, then the energy to open link $p+1$ is $\varepsilon$ and the energy to open link $p+2$ is infinity.
The last link $p=N$ cannot be opened. Each open link has $g$ orientations corresponding to the rotational freedom about the bond. Assume a large number of links $N$.

\Dn

(1) Define $x=ge^{-\varepsilon/T}$ 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\Dn


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

(3)
\[
<p> = \frac{Nx^{N+1} - Nx^N - x^{N+1} + x}{x^{N+1} - x^N - x + 1}
\]

Now we expand the term for $<p>$ 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:
\[
<p> \approx \frac{\frac{1}{2}(x-1)^2[N^2-N] +
\frac{1}{6}(x-1)^3[2N^3-3N^2+N]}{\frac{1}{2}(x-1)^2N + \frac{1}{6}(x-1)^3[3N^2-3N]} = \frac{\frac{1}{2}(N-N^2)+\frac{1}{6}(x-1)(2N^2-3N+1)}{1+\frac{1}{6}(x-1)(3N-3)}
\]
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:
\[
<p> \approx [\frac{1}{2}(N-N^2)+\frac{1}{6}(x-1)(2N^2-3N+1)] \cdot [1- \frac{1}{6}(3N-3)(x-1)]
\]
And by taking $<p>$ to first order and taking $N$ to be large we get:
\[
<p> \approx \frac{1}{2}N[1+\frac{1}{6}N(x-1)]
\]

(4)
\[
\frac{<p>}{N}=\frac{x^N}{x^N-1}-\frac{1}{N} \cdot \frac{x}{x-1} = \frac{1}{1-\frac{1}{x^N}} - \frac{1}{N} \cdot \frac{1}{1-\frac{1}{x}}
\]
And for large $x$ and large $N$ we get:
\[
\frac{<p>}{N} \approx 1
\]

(5)
We can see that for $x=0$ we get $\frac{<p>}{N} \approx 0$. The linear approximation gives us $\frac{<p>}{N}= \frac{1}{2}$ for $x=1$.
\Dn

The dependence of $\frac{<p>}{N}$ on $x$ converges to a Heaviside step function as $N$ becomes larger, so for $N \rightarrow \infty$ the dependence is:
\[
\frac{<p>}{N} = \left\{ \begin{array}{rcl}
0 & \mbox{for}
& x<1 \\ \frac{1}{2} & \mbox{for} & x=1 \\
1 & \mbox{for} & x>1
\end{array}\right. \equiv H(x-1)
\]

(6)
We first find the entropy:
\[
S(x)=-\frac{\partial F(\beta,x)}{\partial T}=  -\beta^2 \frac{\partial}{\partial\beta}(\frac{1}{\beta}\ln(Z)) = \ln(Z) + \frac{1}{\beta}\frac{\partial x}{\partial\beta}\frac{\partial}{\partial x}\ln(Z) = \ln(Z) + \beta \varepsilon  x\frac{\partial}{\partial x}\ln(Z) = \ln(Z) + \beta \varepsilon <p>
\]
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 $\ln(N) << \beta \varepsilon \cdot <p> = \beta \varepsilon \frac{N}{2}$. So the entropy at $x=1$ is:
\[
S(x=1) \approx \beta \varepsilon <p>
\]

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

\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.


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