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

For a 3D FJC:

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

Freely Rotating Chain (FRC)

For a 3D FRC:

• The MSD is \(<r^2> = Nb_{eff}^2\).
• The effective chain length is given by the following equality.

\[\frac{<\vec{u}_m\vec{u}_n>}{b^2}=e^{-b|m-n|/\xi} \ .\]

• \(\xi = \frac{b}{\ln(\cos\beta)}\) 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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