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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!

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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
$$ 0= \int (\partial_\mu x^\nu) j^\mu d^3 x= \delta^\nu_\mu \int j^\mu d^3x = \int j^\nu d^3 x $$

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  • $\begingroup$ 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

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