8.1 The Lennard-Jones pair potential

A liquid or solid is a continuum at macroscopic scales, but its bulk modulus, tensile strength, and surface tension all trace back to the forces between individual molecules. This chapter builds those forces from the ground up and shows how they set macroscopic mechanics. The starting point is the interaction between a single pair of neutral molecules.

Two competing effects

A neutral, closed-shell molecule such as argon feels two opposing interactions with a neighbour, dominant at different ranges:

Short-range repulsion. When the electron clouds of two molecules begin to overlap, the Pauli exclusion principle forces electrons into higher-energy states. The energy cost climbs steeply as the molecules are pushed together — a near-vertical wall.
Long-range attraction. Even without permanent charges, fluctuating dipoles on the two molecules correlate and attract. This is the London dispersion (van der Waals) force, whose origin the next lesson derives; its energy falls off as $-1/r^6$ .

The 12-6 form

A compact two-parameter potential capturing both is the Lennard-Jones potential:

U(r) \;=\; 4\varepsilon\!\left[\left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^{6}\right].

where

$U(r)$: pair interaction energy J
$r$: centre-to-centre separation m
$\varepsilon$: depth of the potential well J
$\sigma$: separation at which $U=0$ m

The $1/r^6$ term is the physically-grounded London attraction; the $1/r^{12}$ term is a convenient stand-in for the repulsive wall, chosen — as the next lesson recounts — largely because $12 = 2\times 6$ makes the algebra clean. The parameter $\sigma$ sets the length at which the potential crosses zero, and $\varepsilon$ the depth of the attractive well. For argon, $\sigma = 0.34\,\text{nm}$ and $\varepsilon/k_B = 120\,\text{K}$ .

r1.200 σ

U(r)-0.891 ε

F(r)-2.212 ε/σ

r_eq = 2^1/61.122 σ

r_inf = (26/7)^1/61.244 σ

max attraction |F|2.396 ε/σ

The repulsive (1/r¹²) term dominates at short range — pushing molecules apart when they overlap — and the attractive (−1/r⁶) term dominates at long range, pulling them together. The two balance at r_eq: the equilibrium spacing of two molecules at zero force. Pull a pair past r_inf and the restoring force decreases; beyond it, you've started breaking the bond.

Two characteristic radii

Differentiating the potential locates the two radii that govern the mechanics.

▶ Equilibrium and inflection radii Derivation

The force is $F = -dU/dr$ . Setting $dU/dr = 0$ ,

\frac{dU}{dr} = 4\varepsilon\!\left[-12\,\frac{\sigma^{12}}{r^{13}} + 6\,\frac{\sigma^6}{r^7}\right] = 0 \;\;\Longrightarrow\;\; r_\text{eq} = 2^{1/6}\sigma \approx 1.122\,\sigma.

This is the well bottom — where the force vanishes and a pair sits at zero temperature. Setting $d^2U/dr^2 = 0$ locates the inflection point,

r_\text{inf} = (26/7)^{1/6}\sigma \approx 1.244\,\sigma,

where the attractive force is strongest.

The two radii mark the two regimes. Inside $r_\text{eq}$ the interaction is stiff and repulsive — this is what resists compression. Between $r_\text{eq}$ and $r_\text{inf}$ the restoring force grows with separation, as a spring’s does. Beyond $r_\text{inf}$ the restoring force weakens with further stretching: the bond is past its strength limit and on its way to breaking. The curvature at $r_\text{eq}$ becomes the material’s stiffness, the subject of lesson 8.3; the maximum attraction near $r_\text{inf}$ becomes its tensile limit, the subject of lesson 8.6.