I see that the formula giving the potential (interaction) energy of a dipole and an induced dipole is $$V=-\frac{C}{r^6}$$ where $$C=\frac{\mu_1^2 \alpha'_2}{4 \pi \epsilon_0}$$ and that the formula giving the potential (interaction) energy of an induced dipole and another induced dipole is $$V=-\frac{C}{r^6}$$ where $$C=\frac{3}{2}\alpha'_1\alpha'_2\frac{I_1I_2}{I_1+I_2}$$. Subscripts 1 and 2 represent the dipole or induced dipole, $\mu$ represents the dipole moment of a permanent dipole, and $\alpha$ represents the polarizability of an induced dipole. Also, $r$ represents the separation between the two in each case, and $I$ represents the ionization energy of 1 or 2. I have been unable to figure out how people have derived the interaction energy in cases that involve an induced dipole (and couldn't find it in Atkins or on Google), and was wondering if someone might be able to show me how it is done. Thank you.

  • $\begingroup$ The dipole induced-diple interaction equation does not seem right. Permitivity should have a square since we have it twice--once inducing the dipole, once calculating dipole-dipole interactions. And polarizability should not have a square. Can you double check? $\endgroup$ – Xiaolei Zhu Apr 15 '15 at 17:07

I don't think you need quantum mechanics to understand what's going on in dipole-induced dipole interaction. The basic mechanism is quite simple and just the details of the calculations change by switching to a quantum description.

Polarizable molecule in an external field

So first things first. Let us consider a simple model of polarizable molecule as being a charge with valence $z$ attached to a spring of strength $k$ and zero length at rest (this is valid model in the harmonic approximation of a dipole).

Upon placing such a molecule in an electric field $\vec{E}$, the charge feels a force $ze \vec{E}$. This force pulls in one direction while the spring pulls in the opposite direction and eventually the system reaches a new mechanical equilibrium where

\begin{equation} -k \vec{d} + ze \vec{E} = \vec{0} \end{equation}

We can find then that the spring constant can be related to the size of the induced dipole $ d = || \vec{d}||$, the amount of charge displaced $ze$ and the electric field magnitude $E = ||\vec{E}||$ via:

\begin{equation} k = \frac{ze E}{d} \end{equation}

Now, to induce this dipole, of course the electric field had to work. The amount of work it provided is equal to the potential energy gained by the spring-like molecule i.e.

\begin{equation} W_{induced} = \frac{1}{2}k d^2 = \frac{1}{2} ze E d \end{equation}

Now, we can define the induced dipole $\vec{p}_i$ as being $ze \vec{d}$, it then comes that:

\begin{equation} W_{induced} = \frac{1}{2} \vec{p}_i\cdot \vec{E} \end{equation}

Now, the energy of dipole in an electric field is given by $U_d = -\vec{p}\cdot \vec{E}$. It then comes that the total energy of the dipole induced by an external field is then:

\begin{equation} U_{induced} = U_d + W_{induced} = -\frac{1}{2}\vec{p}_i \cdot \vec{E} \end{equation}

Potential energy of dipole-induced dipole system

Let us consider now that the external field is generated by permanent dipole $\vec{p}_p$ such that the interaction energy is now:

\begin{equation} U_{int} = -\frac{1}{2}\vec{p}_i \cdot \vec{E}_p \equiv \frac{1}{2}\alpha_i ||E_p||^2 \end{equation}

where I have introduced the polarizability $\alpha_i$ of the induced dipole such that $\vec{p}_i = \alpha_i \vec{E}$ in the general case.

Now, the electric field generated by a permanent dipole $\vec{p}_p$ at a point $M$ is

\begin{equation} \vec{E}_p(M) = \frac{1}{4 \pi \varepsilon_0 r^3}\left(3(\vec{p}_p \cdot \vec{u}) \: \vec{u}-\vec{p}_p \right) \end{equation}

upon taking the square of it, we get:

\begin{equation} ||\vec{E}_p(M)|| = \frac{1}{(4 \pi \varepsilon_0 r^3)^2}\left(3(\vec{p}_p \cdot \vec{u}) \: \vec{u}-\vec{p}_p \right) \cdot \left(3(\vec{p}_p \cdot \vec{u}) \: \vec{u}-\vec{p}_p \right) = \frac{p_p^2}{(4 \pi \varepsilon_0 r^3)^2}(3 \cos^2 \theta +1) \end{equation}

where the angle $\theta$ is defined such that $\vec{p}_p \cdot \vec{u} = p_p \cos \theta$.

Finally the total interaction energy reads:

\begin{equation} U_{int}(p_p, \theta, r) = \frac{\alpha_i p_p^2 (3 \cos^2 \theta +1)}{2(4 \pi \varepsilon_0 r^3)^2} \end{equation}

Approximate free energy of the system

The free energy of the system, when in contact with bath at inverse temperature $\beta$ is defined as follows:

\begin{equation} e^{-\beta \mathcal{F}(p_p,r)} \equiv \int d\Omega_p \: e^{-\beta U_{int}(p_p,\theta,r)} \end{equation}

where $d\Omega_p \equiv \sin \theta d\theta d\phi$ is the integral element of the solid angle over the possible orientation of the permanent dipole $\vec{p}_p$.

At sufficiently high temperatures, we can expand the exponential inside the integral and this gives:

\begin{equation} e^{-\beta \mathcal{F}(p_p,r)} = \int d\Omega_p \: [1-\beta U_{int}(p_p,\theta,r) + \mathcal{O}(\beta^2 U_{int}(p_p,\theta,r)^2) ] \end{equation}

this integral gives then:

\begin{equation} e^{-\beta \mathcal{F}(p_p,r)} \approx 4\pi + \frac{4\pi \beta p_p^2 \alpha}{(4\pi \varepsilon_0) r^3)^2} \end{equation}


\begin{equation} \mathcal{F}(p_p,r,\alpha) = -\beta^{-1} \ln \left[ 4\pi + \frac{4\pi \beta p_p^2 \alpha}{(4\pi \varepsilon_0) r^3)^2} + \mathcal{O}(\beta^2U_{int}^2) \right] \approx - \frac{ p_p^2 \: \alpha}{(4\pi \varepsilon_0)^2) r^6} \end{equation}

This interaction is called the Debye van der Waals interaction. So you can also look it up in other textbooks to get more details, especially on the quantum treatment of what I have done here.

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  • $\begingroup$ So, the heart of this really just was that the induced dipole itself depends on the applied electric field of the permanent dipole. Together with the fact that the electric potential for a dipole falls as 1/r^2, and thus, the electric field for a dipole falls as 1/r^3, the 1/r^6 follows pretty naturally. Thanks. I was looking around for explanations and though I can't say I followed all of what you did, especially around the point where you talked about the approximate free energy, this is the best and clearest explanation I could find online. $\endgroup$ – Sophisticated Idiot Aug 26 '15 at 5:37
  • $\begingroup$ If there is a point that you think requires more explanation, please tell me, I will try to edit my answer accordingly when I get the time. $\endgroup$ – gatsu Aug 27 '15 at 8:13

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