# When does $\nabla \times B =0$?

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.
• 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 $\nabla\cdot\mathbf{B}=\partial_iB_i=0$ Commented Oct 20, 2018 at 2:44

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

• You're right I remove my comment. Commented Oct 20, 2018 at 14:16
• @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. Commented Oct 20, 2018 at 17:40
• 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. Commented Oct 22, 2018 at 10:13
• @KyleKanos Good idea, done. Commented Oct 22, 2018 at 12:00