Monday, February 12, 2018

electromagnetism - When does nablatimesB=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 m.


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


F=(mB).


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=(mB)=(m)B+(B)m+m×(×B)+B×(×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 B=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 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?)

  4. 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+ϵ0μ0tE


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 mB 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


[(mB)]i=imjBj=mjiBj


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


mjiBj=(mjiBjmjjBi)+mjjBi

=(δilδjmδimδjl)mjlBm+mjjBi
=ϵijkϵklmmjlBm+mjjBi
=ϵijkmj(ϵklmlBm)+mjjBi
=[m×(×B)+(m)B]i
and so if m and B are a constant vector and a vector field respectively, the applicable vector identity is


(mB)=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=Bii=Bxx+Byy+Bzz

is itself a differential operator, which you could apply to either scalar or vector fields:


(B)f=B(f)

or (B)A=[B(Ax)]ˆex+[B(Ay)]ˆey+[B(Az)]ˆez



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