Why is the London dispersion force inversely related to distance but not to the second power but the sixth power? Why is there not an inverse square relationship as there is for Coulombic forces?

Is there something non-Coulombic about London dispersion forces?


2 Answers 2


One somewhat simplified way of looking at it is that the London dispersion forces are a dipole-dipole interactions. The interaction between two dipoles depends upon the relative orientation of the two dipoles. Some positions are attractive and some are replusive, but averaged over all the positions (and considering that lower energy positions are favored by Boltzmann statistics), the interaction is attractive with potential energy having an 1/$r^6$ dependence.

See for example Stephan Franzen's lecture

and especially the corporate Intermolecular Van Der Waals page

as well as Fritz London's The General Theory of Molecular Forces Trans. Faraday Soc., 1937, vol. 33, pages 8-26.

for more on the 1/$r^6$ dependence of dipoles.

However, a proper quantum mechanical treatment shows that London dispersion interaction is not always proportional to 1/$r^6$

For example, between two hydrogen atoms in 1s states, the potential energy is proportional to 1/$r^6$ but

between a hydrogen atom in a 1s state and a hydrogen atom in a 2p state, energy is proportional to 1/$r^3$

See: Complement $C_{XI}$ of Quantum Mechanics vol. 2 by Cohen-Tannoudji et al.

Also, in 1948 Verwey and Overbeek demonstrated experimentally that the London dispersion interaction is even weaker than 1/$r^6$ at long distance (say hundreds of Angstroms or more).

Casimir and Polder soon thereafter explained with quantum electrodynamics (QED) that the dependence should be 1/$r^7$ at relatively long distances.

For a nice historical overview and QED prespective see Some QED vacuum effects: van der waals forces in The Quantum Vacuum by Peter W. Milonni.

  • $\begingroup$ @DavePhD: "but averaged over all the positions (and considering that lower energy positions are favored by Boltzmann statistics), the interaction is attractive" why then might one of my profs ask questions starting with "Based on vdW repulsions, there are how many different bond angles in ...)? $\endgroup$
    – Dissenter
    Jan 14, 2015 at 21:35
  • $\begingroup$ @Dissenter I think he/she is referring to the "12" part of the Lennard Jones 6-12 potential, which has nothing to do with dipole-dipole interaction, it is due to the Pauli exclusion principle (like electron degeneracy pressure). So once the molecules are close enough that electron wavefunctions overlap, the 1/r^6 attractive transitions to 1/r^12 repulsive $\endgroup$
    – DavePhD
    Jan 14, 2015 at 23:04

London dispersion forces (or van der Waals) can be rationalized (classically) as a relationship between induced dipole moments.

Some electrostatic (Coulombic) relationships:

  • Two Point Charges: $V = \frac{q_1 q_2}{4\pi\epsilon_0 r}$ (i.e., $r^{-1}$)
  • Point Charge and Permanent Dipole: $V = \frac{\mu_1 q_2}{4\pi\epsilon_0 r^2}$ (i.e., $r^{-2}$)
  • Two Permanent Dipoles: $V = \frac{\mu_1 \mu_2}{4\pi\epsilon_0 r^3}(1 - 3 \cos^2 \theta)$ (i.e., $r^{-3}$)

The last assumes that the polar molecules are stationary with angle $\theta$ between the dipole moment vectors. If the molecules rotate then it falls off to $1/r^6$.

In general, the interaction of a $2^n$-pole (dipole, quadrupole, octupole, etc.) with $2^m$-pole is $V \propto \frac{1}{r^{n+m-1}}$.

So why the $1/r^6$ for dispersion? Here's a quick justification. Let's take the system of a permanent dipole that's creating a induced dipole moment in another molecule:

$\mu_2^* = \alpha_2 E$

where $\alpha_2$ is the polarizability in the second molecule and E is the electric field generated by the polar molecule:

$E = \mu \sqrt{1 + 3\cos^2 \theta} / 4\pi\epsilon_0r^3$

The potential energy is given by:

$V = \alpha_2 E^2 / 2$

$V \propto r^{-6}$

Since there are multiple dispersion interactions (two dipole moments averaged over all rotations, permanent dipole-induced dipole, hydrogen bonding, and induced dipole-induced dipole moment) the general potential energy is classically approximated as:

$V = \frac{C}{r^6}$

  • $\begingroup$ what is "$\theta$" here? the third result seems wrong. $\endgroup$
    – RE60K
    Jan 14, 2015 at 14:26
  • $\begingroup$ That's the general result regardless of the angle theta between two dipole vectors. $\endgroup$ Jan 14, 2015 at 14:28
  • $\begingroup$ As to the difference in constants with induced dipoles, I checked these with Atkins' textbook before posting. IIRC, it depends on the different ways of defining polarizability. I'll check with a colleague who breathes electrostatics. $\endgroup$ Jan 14, 2015 at 14:31
  • $\begingroup$ @GeoffHutchison For permanent dipole- induced dipole $E = \mu\sqrt {1 + 3 \cos^2 \theta}/4\pi\epsilon_0 r^3$ and $V=-\alpha E^2/2$ Then according to London 1937 (explaining Debye's theory), considering all orientations, the $\cos^2 \theta$ term is 1/3. $\endgroup$
    – DavePhD
    Jan 14, 2015 at 15:09
  • $\begingroup$ @DavePhD if you see the still in my answer and see the surroundings, the illustration 27 says $E=kp/r^3 . \sqrt{(1+3\cos^2\theta)}$ $\endgroup$
    – RE60K
    Jan 14, 2015 at 15:15

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