The freely jointed chain model without excluded volume has the following gyration radius, $$ \langle R_g^2 \rangle=\frac{1}{6}\frac{N+2}{N+1}Na^2 $$ where $N$ is the number of bonds and $a$ their length. Now that I have added to my simulations a Lennard-Jones potential between non bonded beads (I take $\sigma=1.5a$), $$ V_{LJ}=4\epsilon\left(\left(\frac{\sigma}{r}\right)^{12}-\left(\frac{\sigma}{r}\right)^{6}\right) \ \text{for} \ r<2^{1/6}\sigma $$ I wanted to check the gyration radius I get. I don't hope to find an analytical formula like the previous one but some approximation which works good for low $N$.

Searching on the internet I found this article, which proposes the following equations for the gyration radius, $$ \langle R_g^2 \rangle=\frac{1}{6}Na^2\alpha^2(\bar{u},z) $$ $$ \alpha^{5.643}(\bar{u},z)-(1-\bar{u})\alpha^{0.395}(\bar{u},z)=2.9713z $$ where $\bar{u}=(1/0.1777)(3/(2\pi))^{3/2}w/a$ and $z=(3/(2\pi))^{3/2}w\sqrt{N}/a$. I'm not entirely sure how this equations are derived as they use renormalization group theory which I'm not familiar with and I don't know either what is the value of $w$ in my case (is it related to $\sigma$?). However I believe they are considering a continuum model and in fact for $N=1$ where there are only two beads and therefore excluded volume doesn't modify anything it gives (if I have solved it correctly) a different prediction for $\langle R_g^2 \rangle$ than the first formula.


1 Answer 1


When your joints aren't free
What's the next thing you see?
It's an angle

With joints correlated
Your chain's free-rotated
With an angle

$\overline{\cos \theta}$ was once null
But now the LJ hull
Gives an angle

If your chain's short at all
Long-range terms become small
Says the angle


$$ \int_0^\pi \cos \theta \exp(-\beta U[\theta]) \mathrm{d} \theta \tag{1}$$

to get $\alpha$


$$ \langle R^2 \rangle = n l^2 \left(\frac{1 + \alpha}{1 - \alpha} - \frac{2(1-\alpha^n)}{n(1-\alpha)^2} \right) \tag{2}$$

is what you're after!

(Here $U[\theta]$ is the LJ energy imparted across the angle between two links by the bead-bead interaction:

$$ U[\theta] = U_{LJ} (r (\theta)), r^2(\theta) = 2l^2(1+ \cos \theta) $$

and $l$ is the bond length and $\beta = 1/(k_B T)$ is the Boltzmann factor. The integral (1) thus gives the Boltzmann-weighted average of $\cos \theta$ over the LJ interaction, where $\theta$ is the angle-from-parallel as conventionally used in polymer physics -- that is, $\theta = 0$ when successive links are parallel and fully extended, and $\theta = \pi$ when successive links are anti-parallel. This average will now be positive because successive links can no longer "double back" on themselves, where the equivalent quantity for the free-jointed chain is zero.)

  • $\begingroup$ Thanks for your answer but part of the text seems to be missing or isn't being rendered correctly. Integral (1) is equal to $\alpha$? $\endgroup$ Mar 22 at 12:09
  • $\begingroup$ Yup, $\alpha$ is integral (1)! $\endgroup$ Mar 22 at 15:48
  • $\begingroup$ +1 for the nice poetry in such a mathematical question ! $\endgroup$ Mar 23 at 4:17
  • $\begingroup$ To be honest I hadn't noticed it... Very nice haha $\endgroup$ Mar 23 at 7:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .