Pre-Quantum Electrodynamics

Similarly to the electrostatic dipole, the magnetic field of a dipole can be written in coordinate-free form. For this: we define a useful object, the

Using this: we can for example rewrite the cross product as \[ ({\bf A} \times {\bf B})_i = \epsilon_{ijk} A_j B_k. \] Important identity:

\[ \epsilon_{ijk} \epsilon_{ilm} = \delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl} \tag{sumLeviCivita}\label{sumLeviCivita} \] The curl of a vector is thus \[ ({\boldsymbol \nabla} \times {\bf A})_i = \epsilon_{ijk} \partial_j A_k. \]

We can now easily compute the magnetic field of a magnetic dipole (we write the dipole moment \(m\) as \(M\) to make notation clear) \[ (B_{di})_i = \epsilon_{ijk} \partial_j \left( \frac{\mu_0}{4\pi} \frac{{\bf M} \times {\bf r}}{r^3} \right)_k = \frac{\mu_0}{4\pi} \epsilon_{ijk} \partial_j \left( \epsilon_{klm} M_l \frac{r_m}{r^3} \right) = \frac{\mu_0}{4\pi} \epsilon_{ijk} \epsilon_{klm} M_l \partial_j \frac{r_m}{r^3} \] But \[ \partial_j \frac{r_m}{r^3} = \partial_j \frac{r_m}{[\sum_l r_l^2]^{3/2}} = \frac{\delta_{jm}}{r^3} - 3 \frac{r_j r_m}{r^5} \] and \(\epsilon_{ijk} \epsilon_{klm} = \epsilon_{ijk} \epsilon_{lmk} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}\), so we get

\begin{align} ({\bf B}_{di})_i &= \frac{\mu_0}{4\pi} (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}) M_l \left(\frac{\delta_{jm}}{r^3} - 3 \frac{r_j r_m}{r^5}\right) \nonumber \\ &= \frac{\mu_0}{4\pi} \left( \frac{M_i}{r^3} (3 - 3) - M_j (\frac{\delta_{ij}}{r^3} - 3 \frac{r_i r_j}{r^5}) \right) \nonumber \\ &= \frac{\mu_0}{4\pi} \left( 3 \frac{ r_i (M_j r_j)}{r^5} - \frac{M_i}{r^3} \right) \end{align}

Putting back the vector notation, and the \(m\) notation for the magnetic dipole (instead of \(M\)), we obtain

\[ {\bf B}_{di} ({\bf r}) = \frac{\mu_0}{4\pi} \frac{1}{r^3} [3({\bf m} \cdot \hat{\bf r}) \hat{\bf r} - {\bf m}] \tag{B_di}\label{B_di} \]

As an instructive exercise, you can rederive all derivative and product rules using Levi-Civita.