Is $\nabla \cdot (\nabla\times \textbf{F})$ = div $\textbf{F} $?
Where $(\nabla\times \textbf{F})$ = rot $\textbf{F}$. If yes, why?
2026-02-23 20:42:10.1771879330
Is $\nabla \cdot (\nabla\times \textbf{F})$ = div $\textbf{F} $?
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In general, no. In fact, $\nabla\cdot(\nabla\times F)=\sum_{ijk}\epsilon_{ijk}\partial_i\partial_j F_k=0$, because $\epsilon_{ijk}$ ($\partial_i\partial_j$) is antisymmetric (symmetric) under $i\leftrightarrow j$.