While I was studying fluid mechanics and doing some vector calculus. I wondered if the following statement is true or false.
Given that $A$ is a smooth vector field and given that $V\times ( \nabla \times V)=0$. We must have $\nabla \times V=0$
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A related question is found here https://math.stackexchange.com/questions/1090374/is-there-a-vector-field-that-is-equal-to-its-own-curl – Mohamed hamdy merzban Jun 10 '20 at 14:36
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Tip: Consider the field $(z,x,y)$ at point $(1,1,1)$.
Vasily Mitch
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Good. This proves a vector and its curl can be collinear at a point. But the question still open. Can a vector and its curl be collinear at every point in R3. – Mohamed hamdy merzban Jun 10 '20 at 12:07