Van der Waals force

The Van der Waals force is a distance-dependent nonbonded interaction between atoms or molecules, including dispersion and dipolar contributions. In the attractive nonretarded pair model, an interaction energy $U(r)=-\beta /r^6$ gives force magnitude $6\beta /r^7$. Macroscopic integration introduces a Hamaker constant; it does not replace the sign change and molecular cutoff needed at very short distances.

The interaction potential energy associated with a Van der Waals force.

A pairwise-additive approximation integrating microscopic dispersion interactions over body volumes. For two nonretarded homogeneous half-spaces, it gives interaction energy per area $-A_H/(12\pi d^2)$.

The nonretarded attractive normal pressure obtained by differentiating the half-space interaction energy per area: $P=-A_H/(6\pi d^3)$. Total force additionally requires the interacting area.

Named after Fritz London, this is an attractive interaction between fluctuating electric dipoles of atoms or molecules, present even without a permanent dipole. At large separation compared with atomic size in the nonretarded regime, the dipole interaction scales as $r^{-3}$ and its second-order ground-state energy correction is negative and scales as $r^{-6}$. Differentiating the energy gives an attractive force of magnitude $6C_6/r^7$. The short-distance overlap regime and the very-long-distance retarded regime require different descriptions.

For pairwise-additive nonretarded molecular attraction and equal solid number densities $n$, let $A_H={\pi }^{2}B{n}^2$ and let $L>R$ be the distance from a sphere centre to a planar half-space. Integrating the atom–half-space potential energy givesFor $L\gg R$, expansion gives the displayed expression. Equivalently expand $(L+z{)}^{-3}$ over the sphere: its odd moments vanish and $⟨z^2⟩=R^2/5$. This yields the first relative correction $6R^2/(5L^2)$. In the opposite near-contact limit, the gap $h=L-R\ll R$ gives $U\simeq -A_{H}R/(6h)$ and the sphere-plane dispersion force magnitude $A_{H}R/(6h^2)$. The two limits use different distances and must not be confused.

For a sphere of radius $a$ at separation $d\ll a$ from a half-space, pairwise integration of the attractive $r^{-6}$ molecular potential gives $F\simeq Aa/(6d^2)$. Equivalently, integrate the planar disjoining pressure magnitude $A/(6\pi h^3)$ over the local gap $h=d+r^2/(2a)$. Across a third medium the effective Hamaker constant can instead correspond to repulsion; state its sign convention explicitly.