Magnetic dipole in a magnetic field In this  AmJPhys  article Griffiths discusses stationary electric and magnetic dipoles in external electric and magnetic fields.
For the case of a standard magnetic dipole (due to an electric current) at the origin in an external magnetic field ${\bf B}$, Poynting's Theorem gives the interaction energy between an external magnetic field $\bf{ B}$ and the field due to the dipole ${\bf B}_d$ as $$W_{int} = \frac{1}{\mu_0} \iiint {\bf B}_d {\bf \cdot} {\bf B} d \tau.$$
He then proceeds to write the external field in terms of the vector potential, ${\bf B} = \nabla \times {\bf A}$, and the interaction energy can be rewritten as $$W_{int} = \frac{1}{\mu_0} \iiint {\bf B}_d {\bf \cdot} \nabla \times {\bf A} d \tau =  \frac{1}{\mu_0} \iiint \nabla \times {\bf B}_d {\bf \cdot} {\bf A} d \tau,$$
which, after identifying $\nabla \times {\bf B}_d$ with the current density ${\bf J}_d$ (assuming static fields), becomes $$ \iiint {\bf J}_d {\bf \cdot} {\bf A} d \tau.$$
After this it is argued that because the magnetic dipole is localised it is reasonable to expand the vector potential due to the external field in a Taylor series about the origin $${\bf A} ({\bf r}) \approx {\bf A} ({\bf 0}) + {\bf r} \cdot {\nabla_0} {\bf A} ({\bf 0}) + \cdots .$$
The integral involving ${\bf J}_d \cdot {\bf A}({\bf 0})$ vanishes leaving the final expression as $$W_{int} = \iiint d \tau {\bf J}_d {\bf \cdot} ( {\bf r} \cdot {\nabla_0} ) {\bf A} ({\bf 0}) = \iiint d \tau {\bf J}_d ( {\bf r} \cdot {\nabla_0} ) {\bf \cdot} {\bf A} ({\bf 0}).$$
This apparently can be written as $ {\bf m} {\bf \cdot} {\bf B}$ where ${\bf m} = \frac{1}{2} \iiint {\bf r} \times {\bf J} d \tau$ is the magnetic dipole moment and ${\bf B} = \nabla \times {\bf A}$ is the external magnetic field (dropping the subscript "$0$" which tells you the field is evaluated at the origin where the magnetic dipole has been placed). I am having difficultly getting this result and any help, hints would be appreciated.
 A: Think I have figured it out:  In terms of the alternating tensor $\epsilon_{ijk}$ have
\begin{align}
{\bf r} \times {\bf J} {\bf \cdot} {\bf c} \times {\bf d} & = \epsilon_{ijk} x_i J_j {\bf e}_k {\bf \cdot} \epsilon_{lmn} c_l d_m {\bf e}_n \nonumber \\
 & = \epsilon_{ijk} \epsilon_{ijk} {\bf e}_k {\bf \cdot} {\bf e}_n x_i J_j c_l d_m  \nonumber \\
 & = \epsilon_{ijk} \epsilon_{lmn} \delta_{kn} x_i J_j c_l d_m \nonumber  \\
 & = \epsilon_{ijk} \epsilon_{lmk} x_i J_j c_l d_m  \,\,\,\, (1).
\end{align}
Now use the identity
$$\epsilon_{ijk} \epsilon_{lmk} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$$
to write (1) as
\begin{equation}
{\bf r} \times {\bf J} {\bf \cdot} {\bf c} \times {\bf d} = x_i J_j c_i d_j - x_j J_i c_i d_j,
\end{equation}
where we have relabelled some dummy indices for convenience. Now consider
$$\iiint x_i J_j d\tau $$
We have the following identity:
\begin{align}
{\bf \nabla} {\bf \cdot} \left( x_i x_j {\bf J} \right) & = x_i x_j  {\bf \nabla} {\bf \cdot} {\bf J} + {\bf J} {\bf \cdot} {\bf \nabla} x_i x_j \nonumber \\
 & =  x_i x_j  {\bf \nabla} {\bf \cdot} {\bf J} + J_i x_j + J_j x_i
\end{align}
As we are going to consider volume integrals, then making use of the assumption ${\bf \nabla} {\bf \cdot} {\bf J} = 0$ and that surface integrals involving ${\bf J}$ vanish we see that we have
$$\iiint x_j J_i d \tau = - \iiint x_i J_j d \tau$$
so that we have the desired result
\begin{equation}
\iiint x_i J_j c_i d_j d \tau = \frac{1}{2} \iiint {\bf r} \times {\bf J} {\bf \cdot} {\bf c} \times {\bf d} d \tau
\end{equation}
as required.
