- #1

etotheipi

My lecturer said that as a first approximation, we can take the polymer chain to consist of ##n##

*freely jointed*rods of length ##l##. That's just going to give$$\langle \vec{R}_n^2 \rangle = \langle \sum_{i=1}^n \vec{r}_i \cdot \sum_{j=1}^n \vec{r}_j \rangle = \langle \sum_{i=1}^n \sum_{j=1}^n \vec{r}_i \cdot \vec{r}_j \rangle = l^2 \sum_{i=1}^n \sum_{j=1}^n \langle \cos{\theta_{ij}}\rangle$$The freely jointed condition gives ##\langle \cos{\theta}_{ij} \rangle = \delta_{ij}##, so the ##\sum_{i=1}^n \sum_{j=1}^n \langle \cos{\theta_{ij}}\rangle = \sum_{i=1}^{n} \delta_{ii} = n## and thus ##\sqrt{\langle \vec{R}_n^2 \rangle} = l\sqrt{n}##. A more accurate equation, where the rods are no longer freely jointed, is actually $$\sqrt{\langle \vec{R}_n^2 \rangle}= l\sqrt{n}\sqrt{\frac{1+\cos{\theta}}{1-\cos{\theta}}} $$I looked around but couldn't find a derivation of this, so I wondered if someone could give me a hint as to proceed with the ##\sum_{i=1}^n \sum_{j=1}^n \langle \cos{\theta_{ij}}\rangle## term in order to obtain the more accurate result? Thank you!
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