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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.

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You might want o check the Wikipedia article on Vanderwaals forces: en.wikipedia.org/wiki/Van_der_Waals_force –  Johannes May 1 '13 at 16:48
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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! –  John Rennie May 1 '13 at 17:30
    
@JohnRennie OK, thanks. I was thinking that the derivation would involve a dipole-dipole interaction, but I guess it is more complicated than that. –  Andrew May 1 '13 at 19:57
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@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. –  gatsu May 2 '13 at 11:00

1 Answer 1

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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 $\vec{u}= \vec{O_1 O_2}/||\vec{O_1O_2}||$.

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}

This quantity is super difficult to compute. However, if the dipoles are sufficiently far appart, 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 saclar product of two independent vectors is zero). The only remaining term is of order 2 in $\beta U$ and decays with as $1/r^6$. 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 term 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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Re:your last remark: retarded Van der Waals = Casimir –  Slaviks May 4 '13 at 18:28
    
Good effort but I noticed you've introduced some notations and not really made a full statement about what they are. Could you fix it? Also why is you arrow like that over both the $\vec{O_1 O_2}$ expressions? –  Magpie May 5 '13 at 15:12
    
$\vec{O_1O_2}$ is the vector going from $O_1$ to $O_2$ i.e. joining the two dipoles and $d\Omega_i$ is the solid angle element associated with the direction of dipole $\vec{p}_i$ and $\beta = 1/k_B T$. –  gatsu May 6 '13 at 13:00

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