Molecular Simulation/The Lennard-Jones Potential

The Lennard-Jones potential describes the interactions of two neutral particles using a relatively simple mathematical model. Two neutral molecules feel both attractive and repulsive forces based on their relative proximity and polarizability. The sum of these forces gives rise to the Lennard-Jones potential, as seen below:

$$\mathcal{V}\left( r \right) = 4\varepsilon \left[ \left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^{6} \right] = \varepsilon \left[ \left(\frac{R_{min}}{r}\right)^{12} - 2\left(\frac{R_{min}}{r}\right)^{6} \right]$$

where ε is the potential well depth, σ is the distance where the potential equals zero (also double the van der Waals radius of the atom), and Rmin is the distance where the potential reaches a minimum, i.e. the equilibrium position of the two particles.

The relationship between the $$\sigma$$ and $$R_{\min}$$ is  $$R_{min}=\sqrt[6]{2}\sigma$$

Pauli Repulsion
The first half of the Lennard-Jones potential is Pauli-Repulsion. This occurs when two closed shell atoms of molecules come in close proximity to each other and their electron density distributions overlap. This causes high inter-electron repulsion and extremely short distances, inter-nuclei repulsion. This repulsion follows an exponential distribution of electron density:

$$\mathcal{V}_{rep}\left( r \right) = Ae^{-cr}$$

where A and c are constants and r is the intermolecular distance. In a liquid, however, it is very unlikely that two particles will be at highly repulsive distances, and thus a simplified expression can be used by assuming the potential has a r-12 dependence (note that this high exponent means that the energy of repulsion drops off extremely fast as molecules separate). The resulting simple polynomial is as follows:

$$\mathcal{V}\left( r \right) = \frac{C_{12}}{r^{12}}$$

Where the C12 coefficient is defined as:

$$C_{12} = 4\varepsilon\sigma^{12} = \varepsilon R_{min}^{12}$$

$$\Phi_{12}(r) = A \exp \left(-Br\right) - \frac{C}{r^6}$$

Buckingham

London Dispersion
The Second half of the Lennard-Jones potential is known as London dispersion, or induced Dipole-Dipole interaction. While a particular molecule may not normally have a dipole moment, at any one instant in time its electrons may be asymmetrically distributed, giving an instantaneous dipole. The strength of these instantaneous dipoles and thus the strength of the attractive force depends on the polarizability and ionization potential of the molecules. The ionization potential measures how strongly outer electrons are held to the atoms. The more polarizable the molecule, the more its electron density can distort, creating larger instantaneous dipoles. Much like with Pauli Repulsion, this force is dependent on a coefficient, C6, and also decays as the molecules move further apart. In this case, the dependence is r-6

$$\mathcal{V}\left( r \right)= \frac{-C_6}{r^6}$$

$$C_6 = \frac{3}{2} \alpha^'_1 \alpha^'_2 \frac{I_1I_2}{I_1 + I_2}$$

Where α' is the polarizability, usually given as a volume and I is the ionization energy, usually given as electron volts. Lastly, the C6 constant can also be expressed by the variables as seen in the Lennard-Jones equation:

$$C_{6} = 4\varepsilon\sigma^{6} = 2\varepsilon R_{min}^{6}$$

Combination Rules


In the case of two separate molecules interacting, a combination rule called the Lorentz-Berthelot combination rule can be imposed to create new σ and ε values. These values are arithmetic and geometric means respectively. For example, an Ar-Xe L-J plot will have intermediate σ and ε values between Ar-Ar and Xe-Xe. An example of this combination rule can be seen in the figure to the right.

$$ \sigma_{AB} = \frac{\sigma_{AA} + \sigma_{BB}}{2}$$

$$ \varepsilon_{AB} = \sqrt{\varepsilon_{AA} \varepsilon_{BB}} $$

Example
Calculate the intermolecular potential between two Argon (Ar) atoms separated by a distance of 4.0 Å (use ϵ=0.997 kJ/mol and σ=3.40 Å).

$$\mathcal{V}\left( r \right) = 4 \varepsilon \left[ \left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^{6} \right]$$

$$\mathcal{V}\left( r \right) = 4 (0.997~\text{kJ/mol}) \left[ \left(\frac{3.40}{4.00}\right)^{12} - \left(\frac{3.40}{4.00}\right)^{6} \right]$$

$$\mathcal{V}\left( r \right) = 3.988(0.142242-0.377150) $$

$$\mathcal{V}\left( r \right) = -0.94~\text{kJ/mol} $$