# Why does the dipole-dipole interaction scale with $r^{-6}$?

I often see the formula $$V \propto \frac{-C}{r^6}$$ for the interaction energy of two electric dipoles. However there interaction energy of an electric dipole in an electric field generated by an electric dipole obviously is

$$E_{pot} = -p_2 \cdot E_1 = \frac{1}{4\pi \epsilon_0 \epsilon}\left(\frac{p_1 p_2}{r^3} -\frac{3(p_1r)(p_2r)}{r^5}\right) \propto r^{-3}$$ so where is this $$r^{-6}$$ dependency coming from?

• I'd guess you're thinking of the London dispersion force, but this is not the interaction energy between two fixed dipoles. May 4, 2021 at 10:18

The $$1/r^6$$ potential you're thinking of is likely the London force. This can be seen as the interaction energy of two dipoles, but they're not static dipoles.
• For static dipoles, as you correctly point out, the interaction energy scales as $$1/r^3$$.
• The London force, on the other hand, describes the interaction between two induced dipoles. That is, you have two atoms (or molecules) which are each neutral and have no permanent dipole of their own. However, they still have an internal charge structure, so they can polarize each other if they are close enough $$-$$ they can each induce a dipole in the other one $$-$$ and these dipoles can then interact.
This is, of course, a weak effect: it is very hard to induce a dipole in the partner system if you don't have a permanent dipole of your own! This is why the London force scales so unfavourably (as $$1/r^6$$) when the systems stop being close to each other.