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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 $\mathbf{m}$.

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

$$\mathbf{F} = \mathbf{\nabla}(\mathbf{m} \cdot \mathbf{B}).$$

I need to prove to myself that this can be reduced to
$$\mathbf{F} = (\mathbf{m} \cdot \mathbf{\nabla})\mathbf{B}.$$

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:

$$\mathbf{F} = \mathbf{\nabla}(\mathbf{m} \cdot \mathbf{B}) = (\mathbf{m} \cdot \mathbf{\nabla})\mathbf{B} + (\mathbf{B} \cdot \mathbf{\nabla})\mathbf{m} + \mathbf{m} \times (\mathbf{\nabla} \times \mathbf{B}) + \mathbf{B} \times (\mathbf{\nabla} \times \mathbf{m}).$$

  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 $\mathbf{B} \cdot \mathbf{\nabla} = \mathbf{\nabla} \cdot \mathbf{B} = 0$ because magnetic monopoles don't exist?
  3. On page 374 of Andrew Zangwill's Modern Electrodynamics (2013), he writes, "When the sources of $\mathbf{B}$ are far away so $\mathbf{\nabla} \times \mathbf{B} = 0$, blah blah blah." This is the thing I'm most confused about. How do we know that $\mathbf{\nabla} \times \mathbf{B} = 0$? 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.
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    $\begingroup$ To add to the clarification in the accepted answer. $\mathbf{B}\cdot\nabla$ is an operator itself, $B_i\partial_i$, that can be used on other quantites. This is why it is not the same as $\nabla\cdot\mathbf{B}=\partial_iB_i=0$ $\endgroup$ – Aaron Stevens Oct 20 '18 at 2:44
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Ampere's law says that $$\nabla \times \boldsymbol{B} = \mu_0 \boldsymbol{J} + \epsilon_0\mu_0 \frac{\partial}{\partial t} \boldsymbol{E}$$

so "far away from sources" means that the current density $\boldsymbol{J}$ 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 $\mathbf{m\cdot B}$ is not the dot product of two vector fields. In particular, the spatial derivatives of $\mathbf{m}$ are not defined.

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

$$[\nabla(\mathbf{m\cdot B})]_i = \partial_i m_j B_j = m_j\partial_i B_j$$

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

$$m_j\partial_iB_j = (m_j\partial_iB_j - m_j\partial_jB_i) + m_j\partial_jB_i$$ $$ = (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}) m_j \partial _l B_m + m_j\partial_j B_i$$ $$ =\epsilon_{ijk} \epsilon_{klm} m_j\partial_lB_m + m_j\partial_j B_i$$ $$ =\epsilon_{ijk} m_j (\epsilon_{klm}\partial_l B_m) + m_j \partial_j B_i$$ $$ = [ \mathbf{m}\times (\nabla \times \mathbf{B}) + (\mathbf{m}\cdot \nabla)\mathbf{B}]_i$$ and so if $\mathbf{m}$ and $\mathbf{B}$ are a constant vector and a vector field respectively, the applicable vector identity is

$$\nabla(\mathbf{m\cdot B}) = \mathbf{m} \times (\nabla \times \mathbf{B}) + (\mathbf{m} \cdot \nabla)\mathbf{B}$$

This is, of course, what we would get if we treated $\mathbf{m}$ as a spatially constant vector field, so you could wave your hands and say that $\nabla \mathbf{m}$ and $\nabla \times \mathbf{m}$ 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 $div(\mathbf{B}) = \nabla \cdot \mathbf B$ is nothing more than a useful mnemonic device.

More specifically, $div(\mathbf{B}) = \nabla \cdot \mathbf{B}$ is a scalar field which happens to be equal to $0$ everywhere. On the other hand, $$\mathbf{B}\cdot \nabla = B_i\partial_i = B_x\frac{\partial}{\partial x} + B_y \frac{\partial}{\partial y} + B_z \frac{\partial}{\partial z}$$ is itself a differential operator, which you could apply to either scalar or vector fields:

$$(\mathbf{B}\cdot \nabla)f = \mathbf{B}\cdot (\nabla f)$$ or $$(\mathbf{B}\cdot \nabla)\mathbf{ A} = \big[\mathbf{B}\cdot (\nabla A_x)\big]\hat e_x + \big[\mathbf{B}\cdot (\nabla A_y)\big]\hat e_y+\big[\mathbf{B}\cdot (\nabla A_z)\big]\hat e_z$$

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  • $\begingroup$ You're right I remove my comment. $\endgroup$ – my2cts Oct 20 '18 at 14:16
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    $\begingroup$ @Bunji If you use the deltas to get rid of the dummy indices $l$ and $m$ you should see that the second line agrees with the first. It isn't an obvious algebraic trick, but it is correct, and it leads to the identity we're looking for. $\endgroup$ – J. Murray Oct 20 '18 at 17:40
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    $\begingroup$ You should expand the last paragraph to show that $\mathbf B\cdot\nabla$ is itself a new operator; it'll clear up more confusion than just saying what you have. $\endgroup$ – Kyle Kanos Oct 22 '18 at 10:13
  • $\begingroup$ @KyleKanos Good idea, done. $\endgroup$ – J. Murray Oct 22 '18 at 12:00

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