Is $\nabla \cdot (\nabla\times \textbf{F})$ = div $\textbf{F} $?
Where $(\nabla\times \textbf{F})$ = rot $\textbf{F}$. If yes, why?
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Thank you for the answer. Not really, I am not searching for the interpretation if the expression equals zero, but if you can calculate the divergence using the curl. – Pi314 Mar 18 '21 at 20:17
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There is no "if" the expression equals zero. You have failed to read the answers to the cited question. – Eric Towers Mar 18 '21 at 20:19
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The question you have suggested is "was wondering if there were a similar $\textbf{explanation}$ for div curl =0." – Pi314 Mar 18 '21 at 20:41
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And the various answers show that div curl F is always zero. – Eric Towers Mar 18 '21 at 20:47
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Where did you get the idea that it would be true to begin with? Did you try it out on a few examples? You should be able to very easily convince yourself that the identity does not hold in general. – Hans Lundmark Mar 18 '21 at 21:12
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@HansLundmark My professor used that expression when solving a question using Gauss's theorem once we had the curl, – Pi314 Mar 18 '21 at 21:16
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Oh, I see. I think the best person to ask in this case would be your professor, who knows what the context is. Maybe he or she actually meant to write something else. – Hans Lundmark Mar 19 '21 at 06:26
1 Answers
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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 $i\leftrightarrow j$.
J.G.
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