Non-Rigid Rotor Stabilized by a Potential

Rotor stabilized by Lennard-Jones type potential

Rather than making the spacing between the disks rigid, we can use a potential to keep the rotor disks together. For this we will use a modified Lennard Jones potential..

The normal LJ potential is given by

$V (r) = 4 ε ({(\frac{σ}{r})}^{12} - {(\frac{σ}{r})}^{6})$

(0.1)

where the minimum potential is -ε and σ is usually chosen as a nominal separation between disks.

We will change this expression by assigning variables to the power of he σ/r ratios.

$V (r) = 4 ε ({(\frac{σ}{r})}^{a} - {(\frac{σ}{r})}^{b})$

(0.2)

The value of r at the minimum potential is computed by setting the derivative of V(r) equal to zero and is

$r_{V \min} = {(\frac{a}{b})}^{\frac{1}{a - b}} σ$

(0.3)

This is also r where the force between particles goes to zero. We can use totally similar math to compute the spacing where maximum attractive force occurs. First defining a power ratio, Pr, using the second derivative of the potential we have:

$P_{r} = \frac{a (a + 1)}{b (b + 1)}$

(0.4)

we find that the spacing where maximum attractive force occurs is:

$r_{F \max} = σ P_{r}^{\frac{1}{a - b}}$

(0.5)

and this leads to the value of the maximum attractive force:

$F_{\max} = \frac{4 ε}{r_{F \max}} (\frac{a}{P_{r}^{2}} - \frac{b}{P_{r}})$

(0.6)

Using F_max we can compute the maximum rotational speed that the diatomic molecule can have, and remain bound, for a given value of ε:

$m \frac{v^{2}}{\frac{r_{F \max}}{2}} < = F_{\max}$

(0.7)

$v < = \sqrt{\frac{F_{\max} r_{F \max}}{2 m}}$

(0.8)

where m is the mass of a single atom.

We can choose an initial separation for the atoms such that their separation stays constant by solving for r₀ in the following equation:

$\begin{array}{l} F_{L J} (r_{0}) + \frac{2 m v^{2}}{r_{0}} = 0 \\ \frac{4 ε (a {(\frac{σ}{r_{0}})}^{a} - b {(\frac{σ}{r_{0}})}^{b})}{r_{0}} + \frac{2 m v^{2}}{r_{0}} = 0 \end{array}$

(0.9)

Equation (0.9) can always be solved by a Newton-Raphson numerical method or it can be solved algebraically for the special case a=2b:

If we let $u = {(\frac{σ}{r_{0}})}^{b}$ then we can rewrite equation (0.9)

$2 b u^{2} - b u + 2 \frac{m v^{2}}{4 ε} = 0$

(0.10)

and equation (0.10) is recognized as a simple quadratic equation. The solution for u is

$u = \frac{b + \sqrt{b^{2} - 16 b \frac{m v^{2}}{4 ε}}}{4 b}$

(0.11)

Again, for the special case a=2b we can write a better expression for v_max

$v_{\max} = \sqrt{\frac{b ε}{4 m}}$

(0.12)

When the rotor is hit by another particle, it could easily be unbound unless the value of ε for the rotor binding is much larger than the value for the other particle. It seems prudent to choose the interaction potential for this kind of collision to be:

$ε_{R A} = \sqrt{ε_{R} ε_{A}}$

(0.13)

where ε_A is the value of ε for the atom and ε_R is the value for the rotor. This should tend to maintain the binding of the rotor components.