# How do you calculate the Kuhn length of a polymer?

Gold Member
2019 Award
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!

Last edited:
• berkeman

fresh_42
Mentor
I think it is easier to see geometrical situation, if you write ##\sqrt{\dfrac{1+cos \theta}{1-\cos \theta}}=\left|\cot \left(\dfrac{\theta}{2}\right)\right|##.

The closest I came was: "It can be shown..." (sic!) "... that ##E(\cos \theta)=e^{-l/l_p}##, where ##l_p## is the persistence length, defined as ...".

In any case, it will need a stochastic argument, why the formula "... where the rods are no longer freely jointed, is actually ..." holds. Wiki unfortunately says "clarification needed". It feels like "assume friction". As far as I could see, the deduction uses the law of large numbers. Maybe you get to the desired result if you use the central limit theorem instead. Or you understand this on page 21 (29):
and again without details on page 10 (25):
https://www.theorie.physik.uni-muen...06/softmatter/talks/Peter_Jensen-Polymers.pdf

• etotheipi
Gold Member
2019 Award
Awesome, that's exactly it! Assume that bonds between adjacent steps have mean ##\langle \theta \rangle = \gamma##, and thus ##\langle \cos{\theta} \rangle## for steps separated by ##j-i## is ##(\cos{\gamma)}^{j-i}##, so\begin{align*} \langle \vec{R}_n^2 \rangle = l^2 \sum_{i=1}^n \sum_{j=1}^n \langle \cos{\theta_{ij}}\rangle &= l^2 \left (\sum_{i=j} \langle \cos{\theta_{ij}} \rangle + 2\sum_{i < j} \langle \cos{\theta_{ij}} \rangle \right) \\ &= l^2(n + 2\sum_{i < j}(\cos{\gamma})^{j-i}) = nl^2 \left(1+ 2\frac{\cos{\gamma}}{1-\cos{\gamma}} \right) \end{align*}and simplify for the result. Thanks!

Are you doing materials science?

• etotheipi
Gold Member
2019 Award
Are you doing materials science?
Yes! Are you?

Yes! Are you?
No, just chemistry.