# gyration radius of short freely jointed chain with excluded volume

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.

What's the next thing you see?
It's an angle

With joints correlated
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

Integrate

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

to get $$\alpha$$

Then

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

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.)

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