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We wish to prove $${\mbox grad(curl f)} = 0$$

$$\nabla \times (\nabla f) = \epsilon_{ijk}\partial_j\partial_kf$$

From here,

$$\nabla \times (\nabla f) = \epsilon_{ijk}\partial_j(f' \frac{r_k}{r})$$

I attempted to differentiate $(f' \frac{r_k}{r})$ with respect to $\partial_j$ however could not get 0.

Any help would be appreciated, thanks.

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    I don't know how you got to the third line. Here's how I would go about it - under the assumption that $f$ has continuous second partial derivatives, we know that the mixed partial derivatives are symmetric. We also know that $\epsilon_{ijk}$ is anti-symmetric. Therefore...? – Andrew D Jan 08 '14 at 22:49

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$\partial_k\partial_jf=\partial_j\partial_k f$, however, $\epsilon_{ijk}=-\epsilon_{ikj}$. So your second summation always gives $0$

MoonKnight
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