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\begin{document}
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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 ${\frac{b}{a}}$.
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\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
${=qa+(1-q)b}$
\Dn
${=qa^{2}+(1-q)b^{2}}$
\Dn
${Var(L_{1})=-^{2}=q(1-q)(b-a)^{2}}$
\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 ${N}$ and by the standard deviation ${\sqrt{N}{\sigma_{1}}}$.
\Dn
\Dn
Thus, the probability function of ${L}$ will be given by
\Dn
\Dn
${P(L)=\frac{1}{\sqrt{2{\pi}N{\sigma_{1}}^{2}}}e^{-\frac{(L-N)^{2}}{2N{\sigma_{1}}^{2}}}}$
\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}