Question

Show that for two scalar fields f and g:
∇.[∇f ×( f ∇g)]= 0

Answer 1

This is most easily proved by using the Kronecker totally anti-symmetric symbol $\epsilon_{ijk}$ such that $\epsilon_{123} = 1$. Given two vectors $a$ and $b$ with the Cartesian components, respectively, $a_i$ and $b_i$, the Cartesian components of their vector product are given by

$$\left[ \bm a \times \bm b \right]_i = \sum_{ij} \epsilon_{ijk} a_j b_k$$

and their scalar product is $a \cdot b = \sum_i a_i b_i$. Applying these formulas, and using the simple relation $f {\nabla} f = \frac12 {\nabla} f^2$, we have

$${\nabla} \cdot \left [ {\nabla} f \times \left ( f {\nabla} g \right ) \right ] = {\nabla} \cdot \left [ f {\nabla} f \times { \nabla} g \right ] = \frac{1}{2} { \nabla} \cdot \left [{ \nabla} f^2 \times { \nabla} g \right ] =$$ $$\frac{1}{2} \sum_i \partial_i \left [ {\nabla} f^2 \times { \nabla} g \right ]_i = \frac{1}{2} \sum_{ijk} \partial_i \left ( \epsilon_{ijk} \partial_j f^2 \partial_k g \right ) =$$ $$\frac{1}{2} \sum_{ ijk } \epsilon_{ ijk } \left( \partial_i \partial_j f^2 \partial_k g + \partial_j f^2 \partial_i \partial_k g \right ) = 0\, .$$

The last equality is valid because is symmetric with respect to {i, j}, and is symmetric with respect to {i, k}, whereas is totally anti-symmetric.