Mediation of Collisions by Lennard-Jones Potential

Introduction

The physics in Joule Thompson cooling is a sharp departure from the physics of ideal gases that has been used in the Gas Physics chapter. In Joule Thompson cooling, we must invoke potentials between otherwise free gas atoms and allow energy exchange between the kinetic energy and the potential energy as the separations of the atoms changes. Often the algebraic form of this potential energy is chosen to be the Lennard-Jones (LJ) potential. The following section will document how the LJ potential can be used to describe the motion of hard spheres.

Calculations

In order to get macroscopic cooling of a gas, the average separation of the gas atoms has to increase. An increase of separation reduces the attractive force between any two atoms and reduces the negative potential energy from which this force derives.

Since the potential energy has increased, in order to have constant total energy, the kinetic energy of the two atoms must decrease. And, since temperature is a measure of the translational kinetic energy of atoms, this results in a reduction of temperature or cooling of the gas.

An example of a potential energy expression is the Lennard-Jones potential:

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

(1.1)

where 4ε is the depth of the potential well at r=σ, σ is the separation of the atoms when $ϕ$ =0.

To further explore $ϕ$ , we compute r_0s for which the slope of $ϕ$ and the force is zero:

$\begin{array}{l} \frac{d ϕ}{d r} = - \frac{4 ε}{r} (12 {(\frac{σ}{r_{0 s}})}^{12} - 6 {(\frac{σ}{r_{0 s}})}^{6}) = 0 \\ r_{0 s} = \sqrt[6]{2} σ \end{array}$

(1.2)

Inserting r_0s into equation (1.1) we get the minimum value of $ϕ$ :

$ϕ (r_{0 s}) = 4 ε ({(\frac{σ}{\sqrt[6]{2} σ})}^{12} - {(\frac{σ}{\sqrt[6]{2} σ})}^{6}) = 4 ε (\frac{1}{4} - \frac{1}{2}) = - ε$

(1.3)

In order to use equation (1.1) in our animation we need to know the direction of the force and its value. The force is the negative derivative of the potential:

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

(1.4)

where vector r always points from the atom whose motion we want to compute toward the other atom. Therefore, when r>σ, the force will be attractive since the second term has a negative sign in front. Another way of expressing the significance of the sign is a negative force will be attractive (decreasing r) and a positive force will be repulsive (increasing r).

Of course we have a huge number of atoms in our animation and we usually handle the interactions by assuming hard sphere collisions. We can include the LJ potential's attractive forces and exclude the repulsive forces by ignoring the cases where, for example, r<2s. This way we can still use hard sphere math for the actual collisions but retain the effects of the attractive forces on cooling by including the LJ potential at higher r.

A particular atom, A, will have its momentum, p, affected by the forces from all of the other atoms:

$δ p_{A} = δ t \sum_{i \neq A} F_{i A}$

(1.5)

where i is the sum over all other atoms. Of course most of the Force terms in the sum will be cancelled because the atoms are in all directions with respect to atom A.

How is δp used in the animation?

Once we've computed the value of δp, how can we use it to compute the change in motion of the atom? Obviously we can use Newton's law of motion

$m δ v = δ p$

(1.6)

where m is the mass of the atom and v is its velocity. But suppose the force is the same for the present time element as the previous? Do we continue to accelerate the atom even if its potential energy has not changed? The answer should be no since the atom's total energy should not change and therefore its speed should not change. However, the atom's motion will bring it to a new r and that will cause its potential energy to change. And the larger the speed, the greater the change in potential energy.