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 m.
It is well known that the force on a magnetic dipole moment in a magnetic field is given by
F=∇(m⋅B).
I need to prove to myself that this can be reduced to
F=(m⋅∇)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:
F=∇(m⋅B)=(m⋅∇)B+(B⋅∇)m+m×(∇×B)+B×(∇×m).
- The first term is good -- it can stay!
- For the second term, can I use the commutative property of the dot product to say that B⋅∇=∇⋅B=0 because magnetic monopoles don't exist?
- On page 374 of Andrew Zangwill's Modern Electrodynamics (2013), he writes, "When the sources of B are far away so ∇×B=0, blah blah blah." This is the thing I'm most confused about. How do we know that ∇×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?)
- I think the fourth term is simple -- since a dipole moment is just 1 vector, the curl is always zero.
Answer
Ampere's law says that ∇×B=μ0J+ϵ0μ0∂∂tE
so "far away from sources" means that the current density 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 m⋅B is not the dot product of two vector fields. In particular, the spatial derivatives of m are not defined.
Instead, we can use index notation to get the actual identity we're looking for. Note that
[∇(m⋅B)]i=∂imjBj=mj∂iBj
This doesn't have an immediately obvious vector form, but we can do the following sorcery (which, full disclosure, I did backwards):
mj∂iBj=(mj∂iBj−mj∂jBi)+mj∂jBi
∇(m⋅B)=m×(∇×B)+(m⋅∇)B
This is, of course, what we would get if we treated m as a spatially constant vector field, so you could wave your hands and say that ∇m and ∇×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(B)=∇⋅B is nothing more than a useful mnemonic device.
More specifically, div(B)=∇⋅B is a scalar field which happens to be equal to 0 everywhere. On the other hand, B⋅∇=Bi∂i=Bx∂∂x+By∂∂y+Bz∂∂z
(B⋅∇)f=B⋅(∇f)
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