Relationship between Kuhn length and persistence length What is the persistence length of poly(N-isopropyl  acrylamide)? Is there a definite relation between kuhn length and persistence length? I know that the persistence length is considered to be one half of the Kuhn length, but to what extent is this statement true?
 A: The persistence length is not in general half of the Kuhn length.
The Kuhn monomer length $b$ is defined as 
$$
b = \frac{\langle R^2 \rangle}{R_{max}}
$$
where $\langle R^2 \rangle$ is the root mean squared end-to-end distance and $R_{max}$ is the maximum length of the chain.
The persistence length $l_p$ is defined as the length over which the correlation between the bond vectors decays. This definition is quite vague, so let's see some examples.
In general, we have
$$
\langle R^2 \rangle = \langle \vec R \cdot \vec R \rangle = \sum_{i,j}^n \langle \vec r_i \cdot \vec r_j \rangle
$$
where $n$ is the number of bonds in the chain. If we make the assumption that the bond length is constant and equal to $l$, we have $\vec r_i \cdot \vec r_j = l^2 \cos \theta_{ij}$, and therefore
$$
\langle R^2 \rangle = l^2 \sum_{i,j}^n \langle \cos \theta_{ij} \rangle
$$
The value of $\langle \cos \theta_{ij} \rangle$ depends on the model that we choose. The simplest models are the ideal chain models, in which it is assumed that there is no correlation between distant bonds:
$$
\lim_{|i-j|\to \infty} \langle \cos \theta_{ij} \rangle = 0
$$
Let's see what are the values of $b$ and $l_p$ for some ideal chain models.


*

*Freely jointed chain
Here the assumption is the simplest one: no correlation between bond vectors, i.e.
$$
\langle \cos \theta_{ij} \rangle = 0 \ \ \ \ \text{for} \ \ i\neq j
$$
We have in this case
$$
\langle R^2 \rangle = n l^2\\
R_{max} = n l
$$
and therefore $b=l$. The persistence length is equal to the bond length, $l_p=l$. Therefore the Kuhn monomer length coincides with the persistence length:
$$
l_p=b
$$


*

*Freely rotating chain
Here the assumption is
$$
\langle \cos \theta_{ij} \rangle = (\cos \theta_p)^{|i-j|}
$$
which means that we are assuming that all bond angles are equal, but oriented in random directions in 3d space (see picture below, taken from Polymer Physics by Rubinstein and Colby)

In this case, can define a persistence length in the following way:
$$
(\cos \theta_p)^{|i-j|} = \exp[|i-j| \ln (\cos \theta)] = \exp \left( - \frac{|i-j|}{s_p} \right) = \exp \left( - \frac{l |i-j|}{l_p} \right) 
$$
i.e.
$$
l_p = - \frac{l}{\cos \theta}
$$
The Kuhn monomer length is a bit cumbersome to calculate. Provided that $\theta$ is not too close to $0$, we can approximate $\langle R^2\rangle$ as
$$
\langle R^2 \rangle \approx nl^2 \frac{1+\cos \theta}{1-\cos \theta} 
$$
and since we have
$$
R_{max} = n l \cos (\theta/2)
$$
we obtain
$$
b = l \frac{1+\cos \theta}{\cos (\theta/2)-\cos \theta \cos (\theta/2)}
$$
If you plot the two quantities, you will see that $b$ is always larger than $l_p$:



*

*Worm-like chain
If we assume, in the freely rotating chain model, that the bond angle $\theta$ is very small, we can do a Taylor expansion of the various functions, getting
$$
l_p \approx \frac {2l}{\theta^2}
$$
$$
b \approx \frac{4 l}{\theta^2}
$$
and therefore, in this particular case, you indeed have $b \approx 2 l_p$.
If you want to calculate the persistence length of poly(N-isopropyl acrylamide), you should therefore try to understand which model describes better its properties. If the freely rotating chain models seems adequate and if the typical bond angle is small enough, than you can make the approximation $b \approx 2 l_p$.
