I'm reading a paper (Ni et al, J Comp Phys (2007)) where the author introduced the divergence form of the Coriolis force. $$ \omega \times \mathbf{u} = \mathbf{u} \cdot \nabla(\omega \times \mathbf{r}) \tag{1} $$ where $ \omega, \mathbf{u}\ $and$\ \mathbf{r}$ are the rotating speed vector of the reference frame relative to the absolute inertia frame, the velocity vector and the distance vector. From this the author derives the formula for the Lorentz force where the magnetic field is not constant: $$ \mathbf{J}\times \mathbf{B} = -\mathbf{J}\ \cdot\ \nabla(\mathbf{B}\ \times \ \mathbf{r})\ +\ (J \cdot \ \nabla\mathbf{B})\ \times \mathbf r\tag{2} $$ with $ \bf{J},\ \bf{B} $ and $\bf{r}$ are the current density, the magnetic field and the distance vector.

I would like to know how to derive relation (1) and (2).


1 Answer 1


You can understand this looking at the single components (i.e. using "index notation" and the Einstein sum convention -- I hope you understand this).

About (1), start with the r.h.s.. The $m$-th component of $\mathbf{u} \cdot \nabla(\omega \times \mathbf{r} )$ is:

$$u_i\nabla_i(\omega_j r_k\epsilon_{jkm})$$

note that $\omega$ does not depend on space and $\epsilon$ is a constant. So this is the same as

$$u_i (\nabla_i r_k) \omega_j \epsilon_{jkm}$$

now use the fact that $\nabla_i r_k = \delta_{ik}$ which means that the derivative in the direction $i$ of the $k$-th component of $r$ is 1 if $i=k$ and 0 otherwise. Then our quantity is $$ u_i \delta_{ik}\omega_j\epsilon_{jkm} = u_i\omega_j\epsilon_{jim}$$ which is nothing else but the $m$-th component of $\omega \times \bf{u}$. Whence (1) is true.

For (2), the transcription in index notation of the r.h.s. is

$$ -J_i\nabla_i(B_jr_k\epsilon_{jkm}) + J_i(\nabla_iB_j)r_k\epsilon_{jkm}$$ expanding the derivative we get $$ -J_i(\nabla_iB_j)r_k\epsilon_{jkm} -J_i(\nabla_ir_k)B_j\epsilon_{jkm} + J_i(\nabla_iB_j)r_k\epsilon_{jkm} = $$ $$ = -J_i(\nabla_iB_j)r_k\epsilon_{jkm} -J_i\delta_{ik}B_j\epsilon_{jkm} + J_i(\nabla_iB_j)r_k\epsilon_{jkm} = $$ $$ = -J_iB_j\epsilon_{jim} = J_iB_j\epsilon_{ijm}$$ which is the cross-product.

  • $\begingroup$ I understand this may look quite scary, feel free to ask if you're not used to it. $\endgroup$
    – Martino
    Apr 25, 2017 at 15:52
  • 2
    $\begingroup$ It's something of a nitpick in this case but if you're talking about "abstract index notation" then it is hard to talk about either "Einstein sum convention" or "components" and some of your indices should be raised. It's probably best to just delete the word "abstract" so that you've got actual, concrete indices with respect to some orthogonal Euclidean coordinates: then you indeed don't need to draw any distinction between vectors and covectors and you can leave all of your indices lowered. $\endgroup$
    – CR Drost
    Apr 25, 2017 at 16:00

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