I've been looking at the formulae for magnetic dipole moments, and keep coming across something like this:

$$\mathbf{m}=\int \mathbf{r}\times\mathbf{J}dV$$

Which is something I would be perfectly happy with, if anything actually explained what $\mathbf{r}$ was. I've been told that this is the position vector of the moment, but not with respect to what - and I would assume that the origin would make a difference to the dipole moment. It would also beg the question of where the dipole moment actually is. Should the formula, to be more specific, actually say:

$$\mathbf{m}(\mathbf{r})=\int \mathbf{r}\times\left(\mathbf{J}\left(\mathbf{r'}\right)\right)d^3\mathbf{r'}$$

Basically I'm just not certain on what the $\mathbf{r}$ actually refers to. Any help would be appreciated!


A more explicit formula is $$ {\bf m} = \int {\bf r}\times {\bf j}({\bf r})\, d^3{\bf r} $$ so ${\bf r}$ is the quantity being integrated over. Consequently ${\bf m}$ is independent of ${\bf r}$. You might be happier if, instead, we wrote

$$ {\bf m} = \int ({\bf r}-{\bf r}_0)\times {\bf j}({\bf r})\, d^3{\bf r} $$ so that ${\bf r}_0$ is the point about which the moment is being taken. However, since ${\nabla}\cdot {\bf j}=0$ for all cases of interest, we have (by using Stokes' theorem) $$ \int {\bf j}({\bf r})\,d^3 {\bf r}= 0 $$ so the moment is independent of the choice of ${\bf r}_0$.

Added comment: Emilio says that the last step is not obvious, so here it is: As ${\bf j}$ is zero outside some volume $\Omega$ with boundary $\partial\Omega$ $$ 0= \int_{\partial\Omega} j^\mu x^\nu dS^\mu= \int_\Omega \partial_\mu (x^\nu j^\mu)= \int x^\nu \partial_\mu j^\mu d^3x + \int (\partial_\mu x^\nu) j^\mu d^3 x. $$
As the first term in the last equality is zero, we have
$$ 0= \int (\partial_\mu x^\nu) j^\mu d^3 x= \delta^\nu_\mu \int j^\mu d^3x = \int j^\nu d^3 x $$

  • $\begingroup$ The use of Stokes' theorem here isn't obvious at all. It would be if it was a surface integral, b it that's not the case here. $\endgroup$ – Emilio Pisanty Sep 6 '18 at 18:23
  • $\begingroup$ Thanks! I just wasn't entirely certain. I had a hunch that it might be independent of origin, but had absolutely no idea how to prove it! $\endgroup$ – DoublyNegative Sep 6 '18 at 18:29
  • $\begingroup$ @Emilio Pisanty I'll add the derivation $\endgroup$ – mike stone Sep 6 '18 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.