The Lennard-Jones potential has the form:

$$U(r) = 4\epsilon\left[ \left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^{6} \right]$$

The (attractive) $r^{-6}$ term describes the dispersion force (attraction at long range). Why is the power $-6$ rather than something else?

I think that it has to do with dipole-dipole interactions, but I am having difficulty working out the physics. Can you please help me?

I know that, at a distance $z$ away, the on-axis electric potential due to a dipole at the origin and aligned on the z-axis is given by:

$$V = \frac{1}{4\pi\epsilon_0} \frac{p}{z^2}$$

where $p$ is the (magnitude of the) dipole moment.

But if there is an atom nearby, the dipole will induce a dipole on the atom. The induced dipole will in turn induce a dipole on the first dipole. Is it possible to classically derive the $r^{-6}$ dependence of the dispersion force? Thanks for your time.

  • 2
    $\begingroup$ You might want o check the Wikipedia article on Vanderwaals forces: en.wikipedia.org/wiki/Van_der_Waals_force $\endgroup$
    – Johannes
    May 1, 2013 at 16:48
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    $\begingroup$ The $r^{-6}$ comes from the London dispersion force (en.wikipedia.org/wiki/London_dispersion_force) and the derivation of this is complicated. Googling for "derivation of london dispersion force" finds several promising looking articles, but they won't be easy reading! $\endgroup$ May 1, 2013 at 17:30
  • $\begingroup$ @JohnRennie OK, thanks. I was thinking that the derivation would involve a dipole-dipole interaction, but I guess it is more complicated than that. $\endgroup$
    – Andrew
    May 1, 2013 at 19:57
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    $\begingroup$ @Andrew you are right it is essentially a dipole-dipole interaction but when averaging over all possible orientations the power of the decay changes from 2 to 6. $\endgroup$
    – gatsu
    May 2, 2013 at 11:00

1 Answer 1


The "simplest" classical explanation I know is the van der Waals interaction described by Keesom between two permanent dipoles. Let us consider two permanent dipoles $\vec{p}_1$ (located at $O_1$) and $\vec{p}_2$ located at $O_2$. Their potential energy of interaction is: \begin{equation} U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) = -\vec{p}_1\cdot \vec{E}_2 = -\vec{p}_2 \cdot \vec{E}_1 = -\frac{[3(\vec{p}_1\cdot\vec{u})(\vec{p}_2\cdot\vec{u})-\vec{p}_1\cdot \vec{p}_2]}{4\pi \epsilon_0 ||\vec{O_1 O_2}||^3} \end{equation} where $\vec{u}= \vec{O_1 O_2}/||\vec{O_1O_2}||$ and $\vec{O_1 O_2}$ is a vector going from $O_1$ to $O_2$.

Now, if the system is subject to thermal fluctuations then the dipoles can change their orientations according to a Boltzmann weight. If the time scale associated to the motion of the particles is much longer than the one associated to the dipole orientations then for each distance $r = ||\vec{O_1O_2}||$ on can trace over all the possible orientations.

The effective interaction between the dipoles is then characterized by the free energy of the system i.e. by:

\begin{equation} \mathcal{F}(r|p_1,p_2) \equiv -k_B T \ln \int \frac{d\Omega_1 d\Omega_2}{(4\pi)^2} \:e^{-\beta U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2})} \end{equation}

where $d\Omega_i$ is the solid angle element associated with the direction of dipole $\vec{p_i}$ and $\beta=1/k_B T$.

This quantity is super difficult to compute. However, if the dipoles are sufficiently far apart, their interaction energy is small and one may expand the exponential in powers of $\beta U$: \begin{eqnarray} \mathcal{F}(r|p_1,p_2) \approx -k_B T \ln \int \frac{d\Omega_1 d\Omega_2}{(4\pi)^2} && \times\\ && \:\left(1 \right.\\ && -\beta U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) \\ && \left.+\frac{1}{2}\beta^2 U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2})^2\right) \end{eqnarray}

The first integral $\int \frac{d\Omega_1 d\Omega_2}{(4\pi)^2} 1 $ is trivial and equal to $1$. The second integral of the potential is zero by symmetry (each dipole is as likely to be at either a positive or negative angle from $\vec{u}$ and the integral of a scalar product of two independent vectors is zero). The only remaining term is of order 2 in $\beta U$ which decays with $1/r^6$. At the end of the day we end up with something like: \begin{equation} \mathcal{F}(r|p_1,p_2) \approx -k_B T \ln \left(1+\beta^2\frac{C}{r^6} \right) \approx -\frac{C}{k_B T}\frac{1}{r^6} \end{equation} where

\begin{equation} C = \int \frac{d\Omega_1 d\Omega_2}{(4\pi)^2}\left( \frac{[3(\vec{p}_1\cdot\vec{u})(\vec{p}_2\cdot\vec{u})-\vec{p}_1\cdot \vec{p}_2]}{4\pi \epsilon_0}\right)^2 \end{equation} is a number not important in terms of physical insight.

What matters here is that the average value of the dipole-dipole interaction is zero while the first non zero term not to be zero is owing to fluctuations. This is roughly the rule with van der Waals like interactions which is that they are fluctuation driven (the same is true in the quantum case).

Remark: the previous derivation assumes that the interactions travel at infinite speed which is not true. When the limitation of the speed of light is introduced then at large distances, the van der Waals interaction decays as $1/r^7$.

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    $\begingroup$ +1 Great answer. I have a small doubt : in the equation for the free energy what happens to the internal energy part. I see that you have just indicated the entropy part. $\endgroup$
    Sep 16, 2015 at 5:16
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    $\begingroup$ @WYSIWYG: what do you mean by "entropy part"? I have used the usual definition that the free energy is the log of the partition function; or are you referring to something else? $\endgroup$
    – gatsu
    Sep 17, 2015 at 8:04
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    $\begingroup$ I am sorry but you are mistaken. I am taking the log of the partition function i.e. $\ln Q = \ln \sum_i e^{-\beta E_i}$ which is minus the free energy. To see this do as follows: $-\ln Q = -\ln Q -\sum_i p_i \ln p_i + \sum_i p_i \ln p_i$, where the second equality is a tautology. Now, do not touch at the second Shannon entropy term and focus on the first one. We now relate $p_i$ and $Q$ via $p_i = e^{-\beta E_i}/Q$. Upon inserting this expression into the log of the first entropy we get $\sum_i p_i \beta E_i + \ln Q$ which gives $-kT \ln Q = F = \langle E \rangle -TS$, which what I used. $\endgroup$
    – gatsu
    Sep 18, 2015 at 10:50

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