Answer to Question #86541 in Electricity and Magnetism for ritika

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 $\epsilon_{123}=0$. 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

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

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

$$\nabla \cdot \left[ \nabla f \times \left( f \nabla g \right) \right] = \nabla \cdot \left[ f \nabla f \times \nabla g \right] = \frac12 \nabla \cdot \left[ \nabla f^2 \times \nabla g \right] =$$ $$\frac12 \sum_i \partial_i \left[ \nabla f^2 \times \nabla g \right]_i = \frac12 \sum_{ijk} \partial_i \left( \epsilon_{ijk} \partial_j f^2 \partial_k g \right) =$$ $$\frac12 \sum_{ijk} \epsilon_{ijk} \left( \partial_i \partial_j f^2 \partial_k g + \partial_j f^2 \partial_i \partial_k g \right) = 0 \, .$$

The last equality is valid because $\partial_i \partial_j f^2$ is symmetric with respect to {i, j}, and $\partial_i \partial_k g$ is symmetric with respect to {i, k}, whereas $\epsilon_{ijk}$ is totally anti-symmetric.