Lecture 4

Elasticity (Force-Extension Relations)

Consider a polymer where one end is fixed while the other end is free.

Now we apply a force, $\vec{f}$, to the free end of the polymer. This will cause it to stretch in that direction and so the chain links will want to align themselves parallel to the direction of the force. We can associate the following change in energy with this behavior:
$$\Delta E_i = \vec{f} \cdot \vec{u}_i$$
Consider a state $s$ from the set of all possible states. e.g. $s \in \{\vec{u}_i\}_i^N$

• Assume that the FJC is in a temperature bath $T$ and so its Hamiltonian is the following: [#1]
  $$H(s) = \sum_{i=1}^N -\vec{f}\cdot\vec{u}_i = -fb\sum_{i=1}^N \cos\theta$$
• Let $x(s)$ be length of the FJC projected along the direction of $\vec{f}$.
  $$\rightarrow x(s) = b\cos\theta \\ \Rightarrow H(s) = -fx(s)$$
  
  • Note that there was originally no energy cost for an orientation in the FJC.

What is the average length of $x$?

The average can be calculated by summing over all possible $x(s)$ weighted with the probability of being in the particular state $s$. We use Boltzmann factors and the partition function to express the probabilities.
$$\bar{x} = \frac{\sum_s x(s) e^{-\beta H(s)}}{\sum_s e^{-\beta H(s)}} \\ Z = \sum_s e^{-\beta H(s)} \\ \beta = \frac{1}{kT} \\ \bar{x} = \frac{\sum_s x(s) e^{-\beta f x(s)}}{\sum_s e^{-\beta f x(s)}}$$
Notice that in the last expression, the numerator can expressed as the derivative of an exponential.
$$\rightarrow \bar{x} = \frac{\sum_s \partial_{\beta f} e^{-\beta f x(s)}}{\sum_s e^{-\beta f x(s)}} = \frac{\partial_{\beta f} Z}{Z}$$

Now lets calculate $Z$.
$$Z = \sum_s e^{-\beta H(s)} = \int d\Omega_1 ... d\Omega_N e^{\beta f b \sum_i \cos\theta_i} \\ \rightarrow Z = \left[\int d\Omega e^{\beta f b \cos\theta} \right]^N$$
We can express $Z$ as a product of a single integral since although although different states will have different orientation, the integral over the solid angle will be the same.

• Let $Z_1$ be defined as the following:
  $$Z_1 \equiv \int d\Omega e^{\beta f b \cos\theta} \\ \Rightarrow Z = Z_1^N$$
  Evaluating the integral for $Z_1$ gives:
  $$Z_1 = \frac{4\pi}{\beta f b}\sinh(\beta f b)$$

• Now we can calculate $\bar{x}$:
  $$\bar{x} = \frac{\partial_{\beta f} Z_1^N}{Z_1^N} = \frac{N \partial_{\beta f} Z_1}{Z_1} \\ \Rightarrow \bar{x} = Nb\coth(\beta f b) - \frac{N}{\beta f} = NB \left[\coth(\beta f b) - \frac{1}{\beta f b} \right]$$
  The Langevin function $\mathcal{L}(x)$ is defined as $\mathcal{L}(x) = \coth x - \frac{1}{x}$.
  $$\rightarrow \bar{x} = Nb\mathcal{L}(\beta fb)$$

What does this mean for the Polymer?

• Limits:
  
  • Small force: $f<<\frac{1}{\beta b}$
    
    • note: Taylor Expansion of $\coth y \approx \frac{1}{y} + \frac{y}{3}$.
      $$\bar{x} \approx Nb\frac{\beta fb}{3} \\ \Rightarrow f = \frac{3kT}{Nb^2} \bar{x}$$
      This force has the same form as Hook’s Law! Hence, for small forces the polymer behaves like a spring with spring constant $k_{\text{eff}} \equiv \frac{3kT}{Nb^2}$.
  • Large force: $f>>\frac{1}{\beta b}$
    $$\bar{x} \approx Nb$$
    
    • The polymer is completely stretched out.

Is the result from the small force limit reasonable?

We can check whether our result with different approaches to the problem to see if we get the same thing.

• Assume polymer does behave like a spring. Using the spring potential
  energy compare the resulting expression for the probability
  distribution $P(x)$ with the result derived in a previous lecture.
  With this method we get the same $k_{\text{eff}}$.
  $$V_{\text{eff}} = \frac{1}{2}k_{\text{eff}}^2$$
  Therefore the probability distribution looks like $P(x) \propto e^{-V_{\text{eff}}/kT}$.
  Recall, from a previous lecture, $P(|\vec{r}|=x) \propto e^{-x^2/2\sigma^2}$.
  $$\Rightarrow k_{\text{eff}} = \frac{3kT}{Nb^2}$$
• Another check is from the free energy. Calculate the free energy $F$ and minimize it with respect to $x$. Again, we get the same $k_{\text{eff}}$.

$$F=E-TS \ \text{;} \ E=-fx \ \text{;} \ \Omega(x) \propto P(x) \propto e^{-x^2/2\sigma^2} \\ \Rightarrow S=k\frac{-x^2}{2\sigma^2} + \text{const.} \\ F = -fx + kT\frac{x^2}{2\sigma^2} \\ 0 = \frac{\partial F}{\partial x} = -f + \frac{x}{\sigma^2} \Rightarrow x = \frac{f\sigma^2}{kT} \\ \Rightarrow k_{\text{eff}} = \frac{3kT}{Nb^2}$$

Worm-like Chain

A worm-like chain is model for polymer which prescribes an energy for misalignment. There is energy cost for chain links not being parallel with one another. This is expressed in the following form:
$$\Delta E_i = \frac{-J}{b^2}\vec{u}_i\vec{u}_{i+1}$$
$J$ is a parameter with units of energy and quantifies the penalty. The worm-like chain’s Hamiltonian can be expressed as the following:
$$H(s) = \frac{-J}{b^2} \sum_{i=1}^N \vec{u}_i\cdot\vec{u}_{i+1} = -J \sum_i^N \cos\theta_i$$
- Notice that this $H(s)$ has the same for as that of the stretched described above, so the partition function will also have the same form.
$$Z = Z_1^N \ \text{;} \ \ Z_1 = \frac{4\pi}{\beta J} \sinh(\beta J)$$

MSD

Note, for simplicity I will write $\left<r^2\right>$ instead of $\left<\vec{r}^2\right>$ when discussing the MSD.
$$\left<r^2\right>=\left<\left(\sum_i \vec{u}_i \right)^2\right>=\sum_{ij} \left<\vec{u}_i \cdot \vec{u}_j\right>$$

• Now we need to calculate $\left<\vec{u}_i \cdot \vec{u}_j\right>$.
  
  • For the nearest-neighbors:
    $$\left<\vec{u}_i \cdot \vec{u}_{i+1}\right> = b^2 \left<\cos\theta_i\right> =b^2\frac{\partial_{\beta J}Z_1}{Z_1} \\ \left<\vec{u}_i \cdot \vec{u}_{i+1}\right> = b^2 \mathcal{L}(x)$$
  • For the non-neighboring chains we get a recusion relation:
    $$\left<\vec{u}_i \cdot \vec{u}_{i+n}\right> = b^2 \left<\cos\theta_1 ... \cos\theta_{i+n} \right> = b^2[\left<\cos\theta\right>]^n = b^2[\mathcal{L}(x)]^n \\ \text{let} \ \ b^2 e^{-nb/\xi} \equiv b^2[\mathcal{L}(x)]^n \\ \Rightarrow \xi = \frac{b}{\ln(1/\mathcal{L}(x))}$$
    
    • Compare this to the persistence length we derived for the FRC:
      $$\xi_{FRC} = \frac{b}{\ln(\cos\beta)}$$

Lets focus back our attention on $\left<r^2\right>$, we will come back to persistence length in the next section. Using the calculations for $\left<\vec{u}_i \cdot \vec{u}_j\right>$ we derived above we can express $\left<r^2\right>$ as a series of Langevin functions.
$$\frac{\left<r^2\right>}{b^2} = \sum_{i=1}^N \left[ \sum_{j=1}^{i-1}\mathcal{L}^{i-j} + \mathcal{L}^0 + \sum_{j=i+1}^N \mathcal{L}^{j-i} \right]$$
In the large $N$ limit both summation terms in the brackets look like infinite sums.
$$\approx \sum_{k=1}^{\infty} \mathcal{L}^k = \frac{\mathcal{L}}{1-\mathcal{L}} \\ \Rightarrow \left<r^2\right> = Nb^2 \left(1+ \frac{2\mathcal{L}}{1-\mathcal{L}}\right) = Nb^2 \left(\frac{1+\mathcal{L}}{1-\mathcal{L}}\right)$$

• Again, compare this to the result derived for the FRC:

$$\left<r^2\right>_{FRC}=Nb^2\left(\frac{1+\cos\beta}{1-\cos\beta}\right)$$

Stiff Worm-like chain Polymers

The worm-like chain is considered stiff if its energy penalty $J$ is big. We quantify this by the following:
$$J>>kT \Leftrightarrow \beta J>>1 \\ \mathcal{L}(\beta J) \approx 1 - \frac{1}{\beta J} \\ \Rightarrow \xi \approx \frac{b}{\ln\left(\frac{1}{1-1/\beta J}\right)} \approx \frac{1}{\ln(1+\frac{1}{\beta J})} \approx \frac{1}{1/ \beta J} \\ \xi = \frac{J}{kT} b$$
Notice that the persistence length is larger than the chain length, $\xi > b$.

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lecture3notes

Lecture 3

Polymers

• In previous lectures we discussed Freely Jointed Chains (FJC) and Freely Rotating Chains (FRC) and derived some properties relating to them.
• We calculated some of the following properties for the FJC and FRC in 3D. Now we will calculate these properties specifically for the 1D FJC
  
  • note that $N$ is the number of links, and $b$ is the chain length
  • $b_{eff}$ is the effective chain length for the FRC such that $b_{eff}=b\sqrt{\frac{1+\cos\beta}{1-\cos\beta}}$.

Freely Roating Chains

3D FJC
mean-square end-to-end distance, MSD, $<r^2> = Nb^2$
end-to-end distribution, $\ P(\vec{r}) = \frac{1}{(2\pi\sigma_x^2)^{3/2}} e^{-r^2/2\sigma^2}$

Freely Rotating Chain (FRC)

3D FRC
MSD, $<r^2> = Nb_{eff}^2$
$\frac{<\vec{u}_m\vec{u}_n>}{b^2}=e^{-b|m-n|/\xi}$

• $\xi = \frac{b}{\ln(\cos\beta)}$ is defined as the persistence length
• Stiffness is characterized by persistence length $\xi$. The bigger $\xi$ gets the stiffer the polymer is.
• In the limit of $\beta<<1$, $\xi \approx \frac{2b}{\beta^2}$

1D FJC

• These can be modeled using a stochastic, dynamical process (like a random walk)
• Consider a chain made of link of length b which can be in the + or - direction.
  
  • $x_i$ is the length after i steps are taken
    $\Rightarrow x_i = x_{i-1} + u_i$
  • $u_i$ is a random number, $u_i \in \{-b,+b\}$
  • $x_0=0$, $x_1=0+u_1$, $x_2=u_1+u_2$ and so on…
    $\Rightarrow x_N = \sum_{i=1}^{N}u_i$
  • let $x = x_N$
  • The chain links $u_i$ have the following properties
    $$\left<\vec{u}_i\right>=0 \\ \left<\vec{u}_i \cdot \vec{u}_j\right> = 0, b^2 \\ \left<x\right> = \left<\sum_{i=1}^{N}u_i\right>=\sum_{i=1}^{N}\left<u_i\right>=0 \\ \left<x^2\right>=\left<\sum_{i=1}^{N}u_i\sum_{j=1}^{N}u_j\right> = \sum_{ij}\left<u_iu_j\right>=Nb^2$$
  • The MSD is the same as in the 3D case!

End-to-End Distribution

• Using the Central Limit Theorem we deduce that the distribution must be Gaussian.
  $$P(x) = \frac{1}{\sqrt{2\pi\sigma_x}}e^{-x^2/2\sigma_x^2} \\ \sigma_x^2 = \left<x^2\right>-\left<x\right>^2=Nb^2$$

Aside: Coin Flips

• A coin flip replicates the statistics of the 1D FJC since just as there are only two possible ways for the chain link to go, there are only two possible outcomes for a coin flip.
• In coin flips, the probabilities of outcomes obey a binomial distribution.
• Ex: The probability of getting m heads out of N flips is the following:
  $$P(m) = { N \choose m}p^mq^{N-m}$$
  
  • $p$ is the probability of getting heads, $q$ is the probability of getting tails.
• In the case of the 1D FJC, $p=q=1/2$ and $m-(N-m)=x/b$. Where $m$ is the number of chains going in the positive direction and $N-m$ is the number of chain links going in the negative direction.
  
  • Note, in the limit $N>>1$ the binomial distribution becomes a Gaussian distribution.

Connecting 1D & 3D

• Now consider $\vec{u}_i$ as a 3d vector. Let $u_xi$ be its projection along the x-axis.
  $$\Rightarrow x = \sum_{i=1}^N u_xi \\ \left<x\right>=0, \left<x^2\right>=N\left<u_x^2\right> \\ P(x)=\frac{1}{\sqrt{2\pi\sigma_x^2}}e^{-x^2/2\sigma_x^2}$$
• We know that $\left<\vec{u}^2\right>=b^2$, and $\left<\vec{u}^2\right>=\left<u_x^2+u_y^2+u_z^2\right> = \left<u_x^2\right>+\left<u_y^2\right>+\left<u_z^2\right>$. Since this is a random walk in all three dimensions all of the variances should be equal.
  $$\Rightarrow \left<u_x^2\right>=\left<u_y^2\right>=\left<u_z^2\right> = \frac{1}{3}b^2 \\ \Rightarrow \sigma_x^2 = Nb^2/3$$
• Let $P(\vec{r})$ be the probability distribution in 3D space. Since movement in each of the three dimensions are independent of one another then this probability can be written as a product.
  $$P(\vec{r}) = P(x,y,z) = P(x)P(y)P(z) = \frac{1}{(2\pi\sigma_x^2)^{3/2}}e^{-(x^2+y^2+z^2)/2\sigma_x^2}$$

What is the most likely value of $r=|\vec{r}|$?

• In this case it is easier to consider $P(\vec{r})=P(r,\theta,\phi)$.
  $$P(r) = \int dA P(r,\theta,\phi) = \int_0^\pi\int_0^{2\pi}d\theta d\phi r^2\sin(\theta) P(r,\theta,\phi) \\ \rightarrow P(r) = \frac{4\pi r^2}{(2\pi\sigma_x^2)^{3/2}}e^{-r^2/2\sigma_x^2} \\ \Rightarrow \left<r\right>=b\sqrt{\frac{8N}{3\pi}}$$

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