Potential Energy – Calculating the Interaction Between Two Dipoles: A Comprehensive Guide

Question

We know that the potential energy of interaction between two dipoles, which make an angle $\theta_1$ and $\theta_2$ respectively with the line joining them and an angle $\phi$ between the planes containing them is$$U=\frac{\mu_1\mu_2}{4\pi\epsilon_0r^3}(\sin\theta_1\sin\theta_2\cos\phi-2\cos\theta_1\cos\theta_2).$$
I know this follows from the coordinate-free formula for interaction, which is
$$U=\frac{1}{4\pi\epsilon_0r^3}\left[\vec{\mu}_1\cdot\vec{\mu}_2-3\left(\vec{\mu}_1\cdot\hat{r}\right)\left(\vec{\mu}_2\cdot\hat{r}\right)\right],$$
$r$ being the distance between the dipoles and $\hat{r}$ the unit vector along this distance. However, I don't see how the latter immediately implies the former equation. Could someone help to derive the equation containing angles from the coordinate independent form?

Answer 1

Define the $z$-axis so $\hat r = \hat z$, and define $x$ so $\vec{\mu}_{1}$ is in the $xz$-plane. Then $$ \vec \mu_1 = \mu_1(\sin{\theta_1}, 0, \cos{\theta_1}) $$ $$ \vec \mu_2 = \mu_2(\sin{\theta_2}\cos{\phi}, \sin{\theta_2}\sin{\phi}, \cos{\theta_2}),$$ so $$ \vec \mu_1\cdot\vec \mu_2 = \mu_1\mu_2( \sin{\theta_1}\sin{\theta_2}\cos{\phi} + \cos{\theta_1}\cos{\theta_2}).$$ Meanwhile, $$ \hat r \cdot \vec \mu_i = \mu_i\cos{\theta_i}.$$ Then subtract $3\times(\hat{r} \cdot \vec{\mu}_{1})(\hat{r} \cdot \vec{\mu}_{2})$. That turns the $+1$ into a $-2$ multiplying the $\cos\theta$ products.