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.

  • $\begingroup$ At which point are you having trouble? the beginning, or the very the very last step. I suspect a vector operator identity would do the trick, rewrite the last equation using something like div(cross(r, A)), expand and reduce terms. $\endgroup$
    – user196418
    Aug 28, 2019 at 14:19
  • $\begingroup$ @ggcg the problem is the last step, namely identifying $\iiint d \tau {\bf J}_d ( {\bf r} \cdot {\nabla_0} ) {\bf \cdot} {\bf A} ({\bf 0})$ with $\iiint d \tau \frac{1}{2} {\bf J}_d \times {\bf r} \cdot {\nabla_0} \times {\bf A} ({\bf 0})$. I will take up your suggestion. $\endgroup$
    – jim
    Aug 28, 2019 at 14:32
  • $\begingroup$ Yeah, I'm not sure which identity will work but I think it makes sense to use div(something) that will allow you to use the divergence theorem to argue that something vanishes on a surface at infinity. This and other constraints based on the approximations made might help. $\endgroup$
    – user196418
    Aug 28, 2019 at 15:06

1 Answer 1


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.


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