- This exercise focuses on the Flory characteristic ratio from Rubinstein’s textbook on Polymer Physics.
- Definition of Flory Characteristic Ratio:
$C_\infty:=\lim_{n\to\infty}\frac{\langle \mathbf{R}^2 \rangle}{nb^2}$
where $\langle \mathbf{R}^2 \rangle$ represents the mean square end-to-end vector, $b$ is the Kuhn length of the polymer, and $n$ is the corresponding chain length. - Calculation of $\langle \mathbf{R}^2\rangle$:
Let $\mathbf{r}_i$ denote the bond vector between the $i$th and $(i-1)$th particle, and let $\mathbf{r}_1=\mathbf{R}_1-\mathbf{R}_0\equiv0$. For an $n$-bead chain:
$$\begin{align}\langle \mathbf{R}^2 \rangle&=\left\langle\left(\sum_{i=1}^n \mathbf{r}_i^2\right)\cdot\left(\sum_{j=1}^n \mathbf{r}_j^2\right)\right\rangle\\
&=\sum_i\sum_j\langle \mathbf{r}_i\cdot\mathbf{r}_j\rangle\end{align}$$ - Calculation of $\langle\mathbf{r}_i\cdot\mathbf{r}_j\rangle$:
Consider a local coordinate frame where $\mathbf{r}_i=(b,0,0)^T$. Then, $\mathbf{r}_{i+1}$ can be viewed as a rotation of $\mathbf{r}_i$ by $\theta$ about the $z$-axis (bond angle) and by $\phi_{i}$ about the $x$-axis (torsion). The rotation matrix is:
$$\mathbf{T}_i:=\left(
\begin{array}{ccc}
\cos (\theta ) & -\sin (\theta ) & 0 \\
\sin (\theta ) \cos (\phi_i) & \cos (\theta ) \cos (\phi_i) & -\sin (\phi_i) \\
\sin (\theta ) \sin (\phi_i ) & \cos (\theta ) \sin (\phi_i ) & \cos (\phi_i ) \\
\end{array}\right)$$
or $\pi-\theta$, $\pi – \phi$ based on the definitions of bond angle $\theta$ and torsion angle $\phi$. In each reference frame $R_i$, the bond vector is $(b,0,0)^T$, and in $R_i$, $\mathbf{r}_{i+1}=\mathbf{T}_i\cdot\mathbf{r}_i=(b \cos (\theta ),b \sin (\theta ) \cos (\phi_i ),b \sin (\theta ) \sin (\phi_i ))^T$.
Now, consider the simple case of $\mathbf{r}_i\cdot\mathbf{r}_{i+2}$. $\mathbf{r}_{i+2}$ is calculated in the $(i+1)$th reference frame $R_{i+1}$ where $\mathbf{r}_{i+1}$ is $(b,0,0)^T$ and $\mathbf{r}_{i+2}=\mathbf{T}_{i+1}\cdot\mathbf{r}_{i+1}$. To perform the inner product, we need to rotate one of the vectors into the other frame. This involves rotating $\mathbf{r}_i=(b,0,0)^T$ from $R_i$ to $R_{i+1}$ or vice versa.
Note that $\mathbf{r}_{i+1}$ in the $R_i$ frame is $(b \cos (\theta ),b \sin (\theta ) \cos (\phi_i ),b \sin (\theta ) \sin (\phi_i ))^T$ and $(b,0,0)^T$ in the $R_{i+1}$ frame. Therefore, to transform a vector from $R_i$ to $R_{i+1}$, we need a transformation matrix that converts $(b \cos (\theta ),b \sin (\theta ) \cos (\phi_i ),b \sin (\theta ) \sin (\phi_i ))^T$ to $(b,0,0)^T$. This matrix is $\mathbf{T}_i^{-1}$, or $\mathbf{T}_i^T$, because in frame $R_i$, we obtain $\mathbf{r}_{i+1}$ using $\mathbf{T}_i\cdot(b,0,0)^T$, and $\mathbf{T}_i$ is an orthonormal rotation matrix.
By transforming $\mathbf{r}_i$ into $R_{i+1}$, the inner product $\mathbf{r}_i\cdot\mathbf{r}_{i+2}$ becomes:
$$\langle\mathbf{T}_i^T\mathbf{r}_i,\mathbf{T}_{i+1}\mathbf{r}_{i+1}\rangle=$$
$$\langle\mathbf{r}_i,\mathbf{T}_i\mathbf{T}_{i+1}\mathbf{r}_{i+1}\rangle=(b,0,0)\mathbf{T}_i\mathbf{T}_{i+1}(b,0,0)^T$$
By induction, $\mathbf{r}_i\cdot\mathbf{r}_j=(b,0,0)\mathbf{T}_i\mathbf{T}_{i+1}\mathbf{T}_{i+2}\cdots\mathbf{T}_{j-1}(b,0,0)^T$. Since $\theta$ is the same for all bonds and $\phi_{1,2,\dots,n}$ are independently distributed, let $\mathbf{T}$ represent the average transformation matrix of ${\mathbf{T}_i}$:
$$\begin{align}\mathbf{T}&:=\frac{\int_{-\pi}^{\pi}\mathbf{T}_i(\phi)\exp{(-U(\phi)/k_BT)}\mathrm{d}\phi}{\int_{-\pi}^{\pi}\exp{(-U(\phi)/k_BT)}\mathrm{d}\phi}\\
&=\left(
\begin{array}{ccc}
\cos (\theta ) & -\sin (\theta ) & 0 \\
\sin (\theta ) \langle\cos (\phi) \rangle & \cos (\theta )\langle \cos (\phi)\rangle &0 \\
0& 0& \langle\cos (\phi)\rangle \\
\end{array}\right)\end{align}$$
Therefore, we have $\langle\mathbf{r}_i\cdot\mathbf{r}_j\rangle=(b,0,0)\mathbf{T}^{|j-i|}(b,0,0^T)=b^2(\mathbf{T}^{|j-i|})_{11}$. Every $\sin(\phi)$ term vanishes in the integral because $U$ is even. - Flory Characteristic Ratio:
$$\begin{align}C_\infty&=\lim_{n\to\infty}\frac{1}{nb^2}\sum_i\sum_j \mathbf{r}_i\cdot\mathbf{r}_j\\
&=\lim_{n\to\infty}\frac{1}{n}(\sum_i\sum_j T^{|j-i|})_{11}\\
&=\lim_{n\to\infty}\left(-\frac{1}{n}\frac{-2 \mathbf{T}^{n+1}+\mathbf{T}^2 n+2 \mathbf{T}-n\mathbf{I}}{(\mathbf{I}-\mathbf{T})^2}\right)_{11}\\
&=\left(\frac{\mathbf{I}+\mathbf{T}}{\mathbf{I}-\mathbf{T}}\right)_{11}\\
&=\frac{1+\cos(\theta)}{1-\cos(\theta)}\frac{1-\langle\cos(\phi)\rangle}{1+\langle\cos(\phi)\rangle}\end{align}$$
Here, $\frac{\bullet}{n}=0$ and $\mathbf{T}^n=0$ as $n\to\infty$ because $\mathbf{T}$ is constant, and the correlation between two beads approaches zero as the distance between them increases.
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