# Hindered rotation model for flexible polymers: deriving the Flory characteristic ratio

In the hindered rotation model we assumes constant bond angles $\theta$ and lengths $\ell$, with torsion angles between adjacent monomers being hindered by a potential $U(\phi_i)$. In Rubinstein's book problem 2.9 asks us to derive the Flory characteristic ratio for such a model, which is given as $C_\infty=\frac{1+\cos\theta}{1-\cos\theta}\cdot\frac{1+\langle\cos\phi\rangle}{1-\langle\cos\phi\rangle}$.

I am not sure where to start in working out the correlations between bonds to prove this relation. Starting from $\langle \vec{r_i}\cdot \vec{r_j}\rangle$ it seems from earlier derivations that I am expecting the correlations to be of the form

$\langle \vec{r_i}\cdot \vec{r_j}\rangle = \ell^2\left(\cos^{|i-j|}\theta + \langle \cos\phi\rangle^{|i-j|}\right)$, but I am having a hard time seeing how to show that. Any insights into how to see the correlations geometrically (or an indication that I am on the wrong track entirely) would be greatly appreciated.

Incidentally, it's a bit odd that there are no tags specifically for polymer physics.

The vector $\mathrm{r}_i$ is between the beads $i$ and $i+1$. Define a local coordinate system so that $x$ is along $\mathrm{r}_i$. The coordinate $y$ is defined so that $\mathrm{r}_{i-1}$, $\mathrm{r}_{i}$ are both on the same plane, and $z$ is normal to this plane. Thus we can write

\begin{align} \mathrm{r}_{i-1} &= (\ell\cos\theta, -\ell\sin\theta, 0)_i^T \\ \mathrm{r}_i &= (\ell, 0, 0)_i^T \\ \mathrm{r}_{i+1} &= (\ell\cos\theta, -\ell\sin\theta\cos\varphi_{i+1}, \ell\sin\theta\sin\varphi_{i+1})_i^T \end{align}

Or after working out the full coordinate transformation matrix from the above information, $$\begin{pmatrix}x \\ y \\ z \end{pmatrix}_i = A_i\begin{pmatrix}x' \\ y' \\ z' \end{pmatrix}_{i+1}$$ where $$A_i = \begin{pmatrix} \cos\theta & -\sin \theta & 0 \\ -\sin\theta\cos\varphi_{i+1} & -\cos\theta\cos\varphi_{i+1} & -\sin\varphi_{i+1} \\ \sin\theta\sin\varphi_{i+1} & \cos\theta\sin\varphi_{i+1} & -\cos\varphi_{i+1} \end{pmatrix}$$

Now \begin{align} \langle R^2 \rangle &= \left \langle \left(\sum_{i=1}^n \mathbf{r}_i\right) \cdot \left(\sum_{j=1}^n \mathbf{r}_j\right) \right\rangle = \sum_{i=1}^n \sum_{j=1}^n \langle \mathbf{r}_i \cdot \mathbf{r}_j\rangle \\ &= \sum_{i=1}^n \left( \sum_{j=1}^{i-1} \langle \mathbf{r}_i \cdot \mathbf{r}_j\rangle + \langle \| \mathbf{r}_i \|^2\rangle + \sum_{j=i+1}^{n} \langle \mathbf{r}_i \cdot \mathbf{r}_j\rangle \right) \\ &= n\ell^2 + 2\sum_{i=1}^n \sum_{j=i+1}^{n} \langle \mathbf{r}_i \cdot \mathbf{r}_j\rangle \end{align}

Using the transition matrix and noting that $j>i$, \begin{align} \langle \mathbf{r}_i \cdot \mathbf{r}_j\rangle &= \ell^2 \langle(1, 0, 0) A_i \cdots A_{j-1} (1, 0, 0)^T\rangle = \ell^2 \langle A_i \cdots A_{j-1}\rangle_{11} \\ &=\ell^2 (\langle A\rangle^{j-i})_{11} \end{align} Where $A$ is any of the matrices $A_i$, for example we could set $A = A_1$.

Thus $$\langle R^2 \rangle = n\ell^2 + 2\ell^2 \sum_{i=1}^n \sum_{j=i+1}^{n} (\langle A\rangle^{j-i})_{11} = n\ell^2 + 2\ell^2 \left(\sum_{i=1}^n (n-i) \langle A\rangle^{i}\right)_{11}$$

So towards the characteristic ratio: \begin{align} \frac{\langle R^2 \rangle}{n\ell^2} &= 1 + \frac{2}{n} \left(n\langle A\rangle(I-\langle A\rangle^{n})(I-\langle A\rangle)^{-1}\right. \\ & \qquad \left. - \langle A\rangle(I-(n+1)\langle A\rangle^{n} + n\langle A\rangle^{n+1})(I-\langle A\rangle)^{-2}\right)_{11} \\ & = \left((I+\langle A\rangle)(I-\langle A\rangle)^{-1} - \frac{2\langle A \rangle}{n}(I-\langle A\rangle^{n})(I-\langle A\rangle)^{-2}\right)_{11} \end{align}

Correlation between two beads tends to zero as distance goes to infinity, i.e. $\lim_{n\to\infty}\langle A\rangle^{n} = 0$, so $$C_\infty = \left((I+\langle A\rangle)(I-\langle A\rangle)^{-1}\right)_{11}$$

If the potential is symmetric, by the virtue of sine being odd $\langle \sin\varphi\rangle = 0$, and we have $$\langle A\rangle = \begin{pmatrix} \cos\theta & -\sin \theta & 0 \\ -\sin\theta\langle\cos\varphi\rangle & -\cos\theta\langle\cos\varphi\rangle & 0 \\ 0 & 0 & -\langle\cos\varphi\rangle \end{pmatrix}$$

Remembering that for a 3x3 matrix $T$ $$T^{-1} = \frac{1}{\det(T)}\begin{pmatrix}\det\begin{pmatrix} T_{22} & T_{23} \\ T_{32} & T_{33} \end{pmatrix} & \det\begin{pmatrix} T_{13} & T_{12} \\ T_{33} & T_{32} \end{pmatrix} & \det\begin{pmatrix} T_{12} & T_{13} \\ T_{22} & T_{23} \end{pmatrix} \\ \det\begin{pmatrix} T_{23} & T_{21} \\ T_{33} & T_{31} \end{pmatrix} & \det\begin{pmatrix} T_{11} & T_{13} \\ T_{31} & T_{33} \end{pmatrix} & \det\begin{pmatrix} T_{13} & T_{11} \\ T_{23} & T_{21} \end{pmatrix} \\ \det\begin{pmatrix} T_{21} & T_{22} \\ T_{31} & T_{32} \end{pmatrix} & \det\begin{pmatrix} T_{12} & T_{11} \\ T_{32} & T_{31} \end{pmatrix} & \det\begin{pmatrix} T_{11} & T_{12} \\ T_{21} & T_{22} \end{pmatrix} \end{pmatrix}$$ we have $$C_\infty = \frac{(1+\cos\theta)(1+\cos\theta\langle\cos\varphi\rangle) + \sin\theta \sin\theta\langle\cos\varphi\rangle}{(1-\cos\theta)(1+\cos\theta\langle\cos\varphi\rangle) - \sin\theta \sin\theta\langle\cos\varphi\rangle}$$

Which gives us the final relation $$C_\infty = \frac{(1+\cos\theta)(1+\langle\cos\varphi\rangle)}{(1-\cos\theta)(1-\langle\cos\varphi\rangle)}$$