Coriolis force in the divergence form

Question

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).

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.