Dance Of The Spaghetti

The displacement $\vec{d}$ between the start and end of a dancing spaghetti noodle can be written as the sum of displacements between the end points of the small, contiguous rods. If we write the displacement vector between hinge $i-1$ and hinge $i$ as $\vec{r}_i$ , we have $\vec{d} = \sum_i\vec{r}_i$ .

The square of this is then $\begin{aligned} \vec{d}^2 &= \left(\vec{r}_1 + \vec{r}_2 + \ldots + \vec{r}_N\right)\cdot\left(\vec{r}_1 + \vec{r}_2 + \ldots + \vec{r}_N\right) \\ &= \left(r_1^2 + r_2^2 + \ldots + r_N^2\right) + \left(\vec{r}_1\cdot\vec{r}_2 + \vec{r}_1\cdot\vec{r}_3+\ldots + \vec{r}_{N-1}\cdot\vec{r}_N\right) \\ &= N\delta l^2 + \delta l^2\sum_{i\neq j} \cos\theta_{ij} \end{aligned}$ where $\theta_{ij}$ is the angle between displacement vectors $i$ and $j$ .

We must average this expression over all possible realizations to find the mean squared distance $\langle d^2\rangle$ . As there is no correlation between adjacent displacement vectors (and no long range correlation either), we have $\begin{aligned} \langle d^2 \rangle &= N\delta l^2 + \delta l^2 \sum_{ij}\overbrace{\langle\cos\theta_{ij}\rangle}^{\textrm{zero}} \\ &= N\delta l^2 \end{aligned}$ So that the root mean squared distance is simply $\sqrt{N}\delta l$ .