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In my lab, I use electromagnets to apply a magnetic gradient force to lots of very small (superparamagnetic) nanoparticles embedded in an elastic medium. I believe that these can be treated as magnetic dipoles, with a dipole moment $mathbfm$.

It is well known that the force on a magnetic dipole moment in a magnetic field is given by

$$mathbfF = mathbfnabla(mathbfm cdot mathbfB).$$

I need to prove to myself that this can be reduced to
$$mathbfF = (mathbfm cdot mathbfnabla)mathbfB.$$

I know that we can rewrite the first equation using one of those vector calculus identities that appears, e.g. on the inside covers of Jackson:

$$mathbfF = mathbfnabla(mathbfm cdot mathbfB) = (mathbfm cdot mathbfnabla)mathbfB + (mathbfB cdot mathbfnabla)mathbfm + mathbfm times (mathbfnabla times mathbfB) + mathbfB times (mathbfnabla times mathbfm).$$

1. The first term is good -- it can stay!


2. For the second term, can I use the commutative property of the dot product to say that $mathbfB cdot mathbfnabla = mathbfnabla cdot mathbfB = 0$ because magnetic monopoles don't exist?


3. On page 374 of Andrew Zangwill's Modern Electrodynamics (2013), he writes, "When the sources of $mathbfB$ are far away so $mathbfnabla times mathbfB = 0$, blah blah blah." This is the thing I'm most confused about. How do we know that $mathbfnabla times mathbfB = 0$? Can someone show me a proof and/or help me understand what the "far away" criterion means in real life? (Far away relative to what?)


4. I think the fourth term is simple -- since a dipole moment is just 1 vector, the curl is always zero.

Ampere's law says that
$$nabla times boldsymbolB = mu_0 boldsymbolJ + epsilon_0mu_0 fracpartialpartial t boldsymbolE$$

so "far away from sources" means that the current density $boldsymbolJ$ can be taken to be zero, and that there are no time-varying electric fields. The latter is actually a general approximation that can often be made for relatively low frequency (including steady-state) phenomena.

As for the other questions, that identity actually does not apply here, because $mathbfmcdot B$ is not the dot product of two vector fields. In particular, the spatial derivatives of $mathbfm$ are not defined.

Instead, we can use index notation to get the actual identity we're looking for. Note that

$$[nabla(mathbfmcdot B)]_i = partial_i m_j B_j = m_jpartial_i B_j$$

This doesn't have an immediately obvious vector form, but we can do the following sorcery (which, full disclosure, I did backwards):

$$m_jpartial_iB_j = (m_jpartial_iB_j - m_jpartial_jB_i) + m_jpartial_jB_i$$
$$ = (delta_ildelta_jm - delta_imdelta_jl) m_j partial _l B_m + m_jpartial_j B_i$$
$$ =epsilon_ijk epsilon_klm m_jpartial_lB_m + m_jpartial_j B_i$$
$$ =epsilon_ijk m_j (epsilon_klmpartial_l B_m) + m_j partial_j B_i$$
$$ = [ mathbfmtimes (nabla times mathbfB) + (mathbfmcdot nabla)mathbfB]_i$$
and so if $mathbfm$ and $mathbfB$ are a constant vector and a vector field respectively, the applicable vector identity is

$$nabla(mathbfmcdot B) = mathbfm times (nabla times mathbfB) + (mathbfm cdot nabla)mathbfB$$

This is, of course, what we would get if we treated $mathbfm$ as a spatially constant vector field, so you could wave your hands and say that $nabla mathbfm$ and $nabla times mathbfm$ are equal to zero. However, you should remember that those expressions are formally not defined.

Lastly, I want to clarify that there is no "commutative property of the dot product" when it comes to the divergence operator, because divergence is not a dot product. It only looks like one (and only in Cartesian coordinates), so $div(mathbfB) = nabla cdot mathbf B$ is nothing more than a useful mnemonic device.