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\heading{E2351:Tension on a chain molecule} 
\auname{Yair Judkovsky}


{\bf The problem:}

\Dn

${N}$ monomeric units are arranged along a straight line to
form a chain molecule. Each unit can be either in a state
${\alpha}$ (with length ${a}$ and energy ${E_{a}}$ ) or in a
state ${\beta}$ (with length ${b}$ and energy ${E_{b}}$ ).

\Dn

(1) Write down the function ${Z_{G}({\beta},f)}$.

(2) Derive the relation between the length ${L}$ of the chain molecule
     and the tension ${f}$ applied at the ends of the molecule.

(3) Find the compressibility ${\chi_{T}=(\partial L/\partial f)_{T}}$.

(4)  Describe the dependence of L and ${\chi_{T}}$ on 
     ${(fa/T) \right)}$. explain.

(5) Write down the partition function ${Z({\beta},L)}$. What is the probability 
      function of L for ${ f=0}$ ?

\Dn\Dn


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{\bf The solution:}

\Dn

     (1)

\Dn

      Let us use ${n}$ for the number of monomers in the state ${\alpha}$, and ${(N-n)}$ for the number of monomers in the state ${\beta}$. In order to derive the relation between ${f}$ and ${L}$, we shall define the grand-hamiltonian
 \Dn     

    ${H_{G}=H+fL=nE_{a}+(N-n)E_{b}+fL}$ 

\Dn

     The physical interpertation of this definition is putting a piston at the end of the polymer applying tension ${f}$ on it. 
     Mathematically, the transition from ${H}$ to ${H_{G}}$ changes the identity of the parameter determining the behavior of the system from ${L}$ to ${f}$ (Legendre transform). Adding the term ${fL}$ to ${H}$ translates into multiplying ${Z}$ by ${e^{-{\beta}fL}}$, and thus the Legendre transform on ${H}$ corresponds to the Laplace transform on the partition function ${Z}$.

\Dn



\Dn

   Putting  ${L=na+(N-n)b}$ yields

\Dn
\Dn

    ${H_{G}=nE_{a}+(N-n)E_{b}+naf+(N-n)bf}$

\Dn 
\Dn

    With this Hamiltonian, we shall calculate the function ${Z_{G}}$, from which we can derive the ${L-f}$ relation:

\Dn

    ${Z_{G}({\beta},f)={\sum}e^{-{\beta}H_{G}}={\sum}{N \choose n}e^{-{\beta}(NE_{b}+Nbf+n(E_{a}-E_{b})+n(af-bf))}={\sum}{\frac{N!}{(N-n)!n!}}e^{-{\beta}(NE_{b}+Nbf+n(E_{a}-E_{b})+n(af-bf))}}$
\Dn
${=e^{-{\beta}N(fb+E_{b})}(e^{-{\beta}(E_{a}-E_{b}+fa-fb)}+1)^{N}}=(e^{-{\beta}(E_{a}+fa)}+e^{-{\beta}(E_{b}+fb)})^{N} }$

\Dn
\Dn

  (2)  

\Dn

     Equipped with the ${f}$-dependent function ${Z_{G}}$, we shall use the free energy function to obtain the connection between ${L}$ and ${f}$:

\Dn
\Dn

    ${F({\beta},f)=-\frac{\ln{Z_{G}}}{\beta}=-\frac{N}{\beta}\ln({e^{-{\beta}(E_{a}+fa)}+e^{-{\beta}(E_{b}+fb)}})}$

\Dn\Dn

    ${L=-(\partial F/\partial f)=-(\frac{N}{\beta})(\frac       {-{\beta}ae^{-{\beta}(E_{a}+fa)}-{\beta}be^{-{\beta}(E_{b}+fb)}}     {e^{-{\beta}(E_{a}+fa)}+e^{-{\beta}(E_{b}+fb)}})}$

\Dn\Dn

    ${L=N{\frac{a+be^{-{\beta}f(b-a)}e^{-{\beta}(E_{b}-E_{a})}}{1+e^{-{\beta}f(b-a)}e^{-{\beta}(E_{b}-E_{a})}}}}$

    

%%%%%%%%%%%%%%
\Dn\Dn
\Dn

(3) 

\Dn

We shall calculate the compressibility directly by differentiating. After some algebra, we get:

\Dn

${\chi_{T}=(\partial L/\partial f)_{T}=-{\frac{N(b-a)^{2}}{4T}}{\frac{1}{\cosh^{2}(\frac{1}{2}{\beta}(E_{b}-E_{a})+\frac{1}{2}{\beta}fa(\frac{b}{a}-1))}}}$



%%%%%%%%%%%%%
\Dn\Dn
\Dn

(4)

\Dn

  Defining

\Dn

  ${    {\alpha}=\frac{b-a}{a}  }$  ;{     }   ${       {\Delta}={\beta}(E_{b}-E_{a})  }$   ; {     }          ${     y=\frac{fa}{T}      }$;

\Dn

   We  get 

\Dn\Dn

   ${L(y)=N{\frac{a+be^{-{\Delta}-{\alpha}y}}{1+e^{-\Delta-{\alpha}y}}}$

\Dn
\Dn

   ${\chi_{T}(y)=-\frac{Na^{2}{\alpha}^{2}}{4T}\frac{1}{\cosh^{2}(\frac{1}{2}{\alpha}y+\frac{1}{2}{\Delta})}}$

\Dn\Dn

Without the loss of generality, we shall assume that ${b>a}$, and thus ${\alpha>0}$. ${L(y)}$ goes to ${Na}$ as ${y}$ goes to positive infinity, and goes to ${Nb}$ as ${y}$ goes to negative infinity.

The physical intuition - a large positive tension ${f}$ will "shrink" the polymer to its minimum length ${(Na)}$ while a large negative tension will "strech" the polymer to its maximum length ${(Nb)}$. Those are the asymptotic lines of ${L(y)}$.

\Dn
\Dn

The derivative ${\chi_{T}(y)}$ is always negative, as ${L}$ always decreases with ${y}$, and it has one global minimum - at ${y=-\frac{\Delta}{\alpha}=-{\beta}a{\frac{E_{b}-E_{a}}{b-a}}}$, which is determined by the energy gap between the two monomer states and by the lengths ratio ${\frac{b}{a}}$.


%%%%%%%%%%%%%
\Dn
\Dn
\Dn
(5)
\Dn
 
We will now return to the ordinary partition function ${Z}$, derived from the original hamiltonian ${H=nE_{a}+(N-n)E_{b}}$. 
\Dn
Once the length ${L}$ is determined, the energy is determined:
\Dn

${H=\frac{Nb-L}{b-a}E_{a}+\frac{L-Na}{b-a}E_{b}}$         

\Dn

(notice that both the fractions are non-negative, by our assumption ${b>a}$).

By taking into account the degeneration, we get

\Dn
\Dn

${Z(\beta,L)=\frac{N!}{(N-n)!n!}e^{-\beta(\frac{Nb-L}{b-a}E_{a}+\frac{L-Na}{b-a}E_{b})}}$         

\Dn
(Which is no other than one summand from the function ${Z_{G}}$ calculated in section 1 without the ${fL}$ term)

\Dn
\Dn

Where ${n=\frac{Nb-L}{b-a}}$

\Dn

For ${f=0}$, each monomer will be totally independent of the other monomers, and thus will have a probability of ${\frac{1}{Z_{1}}e^{-{\beta}E_{a}}}$ to be in the state ${\alpha}$, and a probability of ${\frac{1}{Z_{1}}e^{-{\beta}E_{b}}}$ to be in the state ${\beta}$, when the one-monomer partition function is defined as

\Dn
\Dn

${Z_{1}=e^{-{\beta}E_{a}}+e^{-{\beta}E_{b}}}$.
\Dn

In such a situation, where the ${N}$ components of the system are independent, the partition function ${Z}$ "factorizes" into an N-exponent of a one-component partition function: ${Z_{N}=({Z_{1}})^{N}}$.

\Dn
\Dn

For convenience, we shall define 

\Dn
\Dn

${q=\frac{1}{Z_{1}}e^{-{\beta}E_{a}}}$ 

\Dn

${1-q=\frac{1}{Z_{1}}e^{-{\beta}E_{b}}}$

\Dn
\Dn

The expectation value and standard deviation of the length of one monomer are given by

\Dn
\Dn            

${<L_{1}>=qa+(1-q)b}$

\Dn

${<L_{1}^{2}>=qa^{2}+(1-q)b^{2}}$

\Dn

${Var(L_{1})=<L_{1}^{2}>-<L_{1}>^{2}=q(1-q)(b-a)^{2}}$

\Dn

${{\sigma_{1}}=\sqrt{q(1-q)}(b-a)}$

\Dn
\Dn

By the central limit theorem, for ${N>>1}$, summing on ${N}$ independent variables with a common expectation value and a common standard deviation will sum up (in the limit of infinity) to give a normal distribution, characterized by the expectation value ${N<L_{1}>}$ and by the standard deviation ${\sqrt{N}{\sigma_{1}}}$.

\Dn
\Dn

Thus, the probability function of ${L}$ will be given by

\Dn
\Dn

${P(L)=\frac{1}{\sqrt{2{\pi}N{\sigma_{1}}^{2}}}e^{-\frac{(L-N<L_{1}>)^{2}}{2N{\sigma_{1}}^{2}}}}$

\Dn
\Dn

We can see here, that up to a factor ${Z({\beta},L)}$ is the probability to find the polymer in the length ${L}$ (as pointed in the lecture notes, 3.6). Using the Gaussian approximation (according to the central limit theorem) allowed us to avoid a calculation of the probability function using the Stirling formula.

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\end{document} 
