I read reference (1),but I am confused about how to introduce a distance vector in Lorentz force like the author does it in equation (11):

$$\rm J\times B = -J\cdot \nabla \left( B\times r\right)+\left(J\cdot \nabla B\right)\times r$$

$\rm J$ is a current, $\rm B$ is magnetic field, $\rm r$ is the distance vector

and equation (8)

$$\rm \omega\times u=u\cdot\nabla\left( \omega\times r\right)$$

$\omega$ is the rotating speed vector of the reference frame relative to the absolute inertial frame, $\rm r$ is the distance vector.

Then, how to introduce a distance vector when decompose a force?

The author says that $\rm -J\cdot\nabla \left(B\times r\right)$ is a globally conservative term while $\rm \left(J\cdot\nabla B\right)\times r$ is a locally conservative term, and he does not explain it.

How can we judge a term is a globally conservative term or a locally conservative term?


  1. Ni, Ming-Jiu, et al. "A current density conservative scheme for incompressible MHD flows at a low magnetic Reynolds number. Part II: On an arbitrary collocated mesh." Journal of Computational Physics 227.1 (2007): 205-228.

I believe your first two questions are both just applications of the general vector calculus identity


where I've taken the liberty of relabelling the coordinate vector from $\mathbf{r}$ to $\mathbf{x}$. The easiest way of seeing this is probably coordinatewise, noting that $[\nabla_\mathbf{x}\mathbf{x}]_{ij}=\frac{\partial x_j}{\partial x_i}=\delta_{ij}$. We then get

$\epsilon_{ijk}(a_m\partial_m b_j)x_k-a_m\partial_m(\epsilon_{ijk}b_j x_k) = \epsilon_{ijk}(a_m\partial_m b_j)x_k - \epsilon_{ijk}(a_m\partial_m b_j)x_k - a_m\epsilon_{ijk}b_j \delta_{km}=\epsilon_{imj}b_ja_m$.

For the case with rotation vector, $\mathbf{\omega}$ is a spatially constant vector (solid body rotation) so that $\mathbf{u}\cdot\nabla \mathbf{\omega}$ vanishes.

Your last question is more open to interpretation, but I believe here globally conservative is being used in the sense of "divergence form", while locally conservative means in the sense of, the force is conservative travelling with the vector $\mathbf{J}$.

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  • 1
    $\begingroup$ To augment this answer, what @origimbo is saying is that the formula is true for all points $\mathbf x$ (or $\mathbf r$). It is not meant to be read in the sense that there is one point $\mathbf r$ for which the formula is true. $\endgroup$ – Wolfgang Bangerth Feb 27 '16 at 21:24

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