2-D Polymer Bundle (Microcanonical Approach)

This is a microcanonical ensemble approach to a simple model of a two dimensional polymer bundle. The professor of the course I took does a lot of research in the area of polymer physics and so set a few problems pertaining to them. They aren’t easily found in textbooks or online either (this website notwithstanding) but are nevertheless quite interesting in themselves. We also approached this same problem from the canonical ensemble which I’ll upload soon.

Question

A polymer is represented by a chain of $N$ segments of length $a$. $N_x^{+}$ segments are oriented in the positive $x$-direction $N_x^{-}$ are oriented in the negative $x$-direction. $N_y^{+}$ and $N_y^{-}$ segments are oriented in the positive and negative $y$-direction, respectively.

1. Assign an energy penalty $\epsilon$ for every polymer segment oriented in $y$-direction. This penalty reflects the confinement in a polymer bundle.
2. For prescribed extensions $L_x$ and $L_y$ and energy $E$, calculate the entropy $S(E,N,L_x,L_y)$ of the chain.
3. Set $L_y=0$ in the following. Derive the mechanical equation of state for the force $F_x(E,N,L_x)$ in the $x$-direction.
4. Derive the thermal equation of state for the temperature $T(E,N,L_x)$.
5. Eliminate $E/\epsilon$ from the two equations of state and show that $$\begin{array}{}\text{(1)}& L_x(T,N,F_x)=aN\frac{\sinh (\frac{F_{x}a}{k_{B}T})}{\cosh (\frac{F_{x}a}{k_{B}T})+\exp (\frac{-\epsilon }{k_{B}T})}\end{array}$$
6. Expand $L_x$ for small force and extract the entropic spring constant.

Solution

1. Energy Penalty

To include an energy penalty for segments oriented in the $y$-direction, we simply write $E=\epsilon [N_y^{+}+N_y^{-}]$.

2. Entropy

To calculate the entropy of the chain, we first need to deduce the number of states. In the one-dimensional case this is simply the result for the one-dimensional random walk: $$\mathrm{\Omega }=\frac{N!}{N^{+}!N^{-}!}$$

Extending this to two dimensions, we obtain

$$\begin{array}{}\text{(2)}& \mathrm{\Omega }=\frac{N!}{N_x^{+}!N_x^{-}!N_y^{+}!N_y^{-}!}\end{array}$$

We find the entropy using the well-known formula $S=k_B\ln \mathrm{\Omega }$ and applying Stirling’s approximation: $$S=k_B[N\ln N-N_x^{+}\ln N_x^{+}-N_x^{-}\ln N_x^{-}-N_y^{+}\ln N_y^{+}-N_y^{-}\ln N_y^{-}]$$

If we define ${\rho }_{x/y}^{\pm }=\frac{N_{x/y}^{\pm }}{N}$, we can write it a bit more succinctly:

$$\begin{array}{}\text{(3)}& S=-k_{B}N[{\rho }_x^{+}\ln {\rho }_x^{+}+{\rho }_x^{-}\ln {\rho }_x^{-}+{\rho }_y^{+}\ln {\rho }_y^{+}+{\rho }_y^{-}\ln {\rho }_y^{-}]\end{array}$$

We set $L_y=0$. Now let us briefly take pause to define some variables for convenience:

For convenience, $\eta :=\frac{E}{N\epsilon }$, $x:=\frac{L_x}{L_0}$, and $L_0:=Na$, where $L_0$ is the “stretched out” (i.e. $N=N_x$) length of the polymer.

We also note that the effective number of steps in the positive $x$-direction is the effective length over the segment length: $N_x^{+}-N_x^{-}=\frac{L_x}{a}$ and the same for $y$: $N_y^{+}-N_y^{-}=\frac{L_y}{a}$.

Combining these variables, it is possible to write $$\frac{N_x^{\pm }}{N}=\frac{1}{2}[1-\eta \pm x]$$

3. Mechanical Equation of State

So, to the question. The equation for the force in the $x$-direction $F_x$ is given by $\frac{-F_x}{T}=\frac{\partial S}{\partial L_x}$, so

$$\begin{array}{}\text{(4)}& \frac{-F_x}{T}=\frac{\partial S}{\partial L_x}=\frac{\partial S}{\partial x}\frac{\partial x}{\partial L_x}=\frac{1}{L_0}\frac{\partial S}{\partial x}\end{array}$$

In order to find the derivative $\frac{\partial S}{\partial x}$ we need to rewrite the entropy with the new constraint of $L_y=0$.

$$\begin{array}{}\text{(5)}& \frac{S}{-k_{B}N}=\frac{1}{2}(1-\eta +x)\ln (1-\eta +x)+\frac{1}{2}(1-\eta –x)\ln (1-\eta –x)+2\left(\frac{\eta }{2}\right)\ln \left(\frac{\eta }{2}\right)\end{array}$$

Taking the derivative for $x$ then, we find $$\frac{\partial S}{\partial x}=-\frac{k_{B}N}{2}\ln \left[\frac{1-\eta +x}{1-\eta -x}\right]$$

The mechanical equation of state for the force $F_x$, found by rearranging $\text{(4)}$ and using the result just obtained, is then:

$$\begin{array}{}\text{(6)}& F_x=\frac{k_{B}T}{2a}\ln \left[\frac{1-\eta +\frac{L_x}{L_0}}{1-\eta -\frac{L_x}{L_0}}\right]\end{array}$$

4. Thermal Equation of State

To obtain the thermal equation of state, we use the formula $\frac{1}{T}=\frac{\partial S}{\partial E}$

Using $\text{(5)}$:

$$\frac{2\epsilon }{k_{B}T}=\ln [\frac{1}{2}(1-\eta +x)]+\ln [\frac{1}{2}(1-\eta -x)]-2\ln \frac{\eta }{2}$$

which we can write much more neatly as:

$$\begin{array}{}\text{(7)}& \frac{2\epsilon }{k_{B}T}=\ln \left(\frac{1-\eta +x}{\eta }\right)+\ln \left(\frac{1-\eta ,–,x}{\eta }\right)\end{array}$$

5. Calculating the Extension

The equation we are to derive is quite an long process as the equations get somewhat bulky. In fact, the same result can be obtained much simpler (just a few lines) in the canonical ensemble approach to this problem. The steps here are only an outline as a guide.

Start by solving $\text{(6)}$ for $\eta$:

$$\begin{array}{}\text{(8)}& \eta =1-x·\coth \frac{F_{x}a}{k_{B}T}\end{array}$$

If we now sub. this into equation $\text{(7)}$:

$$\frac{2\epsilon }{k_{B}T}=\ln \left[\frac{\frac{L_x}{Na}\coth \left(\frac{F_{x}a}{k_{B}T}\right)+\frac{L_x}{Na}}{1-\frac{L_x}{Na}\coth \left(\frac{F_{x}a}{k_{B}T}\right)}\right]+\ln \left[\frac{\frac{L_x}{Na}\coth \left(\frac{F_{x}a}{k_{B}T}\right)-\frac{L_x}{Na}}{1-\frac{L_x}{Na}\coth \left(\frac{F_{x}a}{k_{B}T}\right)}\right]$$

Simplifying and taking the exponent of both sides:

$$e^{\frac{2\epsilon }{k_{B}T}}=\frac{\frac{L_x^2}{N^2{a}^2}{\coth }^2\left(\frac{F_{x}a}{k_{B}T}\right)–\frac{L_x^2}{N^2{a}^2}}{{[1-\frac{L_x}{Na}\coth \left(\frac{F_{x}a}{k_{B}T}\right)]}^2}$$

Using that ${\coth }^2(x)-1=\frac{1}{{\sinh }^2(x)}$:

$$e^{\frac{\epsilon }{k_{B}T}}=\frac{\frac{L_x}{Na}\sqrt{{\coth }^2\left(\frac{F_{x}a}{k_{B}T}\right)-1}}{1-\frac{L_x}{Na}\coth \left(\frac{F_{x}a}{k_{B}T}\right)}$$

Finally, we solve for $L_x$:

$$\begin{array}{}\text{(9)}& \frac{L_x}{Na}=\frac{\sinh (\frac{F_{x}a}{k_{B}T})}{\cosh (\frac{F_{x}a}{k_{B}T})+\exp (\frac{-\epsilon }{k_{B}T})}\end{array}$$

which is the equation $\text{(1)}$ given.

6. Extracting the Entropic Spring Constant

We need to take the limit of $\text{(1)}$ for small force. For small $x$: $\sinh (x)\to x$ and $\cosh (x)\to 1$.

Therefore $L_x$ becomes

$$L_x\approx aN\frac{\left(\frac{F_{x}a}{k_{B}T}\right)}{1+\exp \left(\frac{-\epsilon }{k_{B}T}\right)}$$

If we re-arrange for $F_x$:

$$F_x=\frac{k_{B}T}{a^{2}N}[1+\exp \left(\frac{-\epsilon }{k_{B}T}\right)]L_x$$

And compare it with Hooke’s Law $F=-kx$:

$$k_{\text{entropy}}=\frac{k_{B}T}{a^{2}N}[1+\exp \left(\frac{-\epsilon }{k_{B}T}\right)].$$