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I know that $\nabla \times f(r) \vec r = \nabla f(r) \times \vec r + f(r) \left ( \nabla \times \vec r \right )$. I figured that the rightmost expression is $0$. How do I prove that $\nabla f(r) \times \vec r = 0$ ?

The actual question in the book is to evaluate the curl, and the answer key says that the answer is 0. I've already expanded the vector product but do not see how it becomes 0.

Haresh
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  • It seems like something is missing from the problem. To show $\nabla f(r) \times \vec{r} = 0$ we would need either that the two vectors are parallel or that one of them has zero magnitude. The latter possibility is not excluded by the Question, but neither does it find much justification. – hardmath Jul 23 '14 at 13:03
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    @hardmath: $f$ is a function of $r$ alone, so its constant surfaces are concentric spheres. – Semiclassical Jul 23 '14 at 13:13

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Your function $f(r)$ is a function of $r$ alone. That should tell you something about the direction of its gradient compared to that of $\vec{r}$, and about their cross product.

Semiclassical
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