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I'm having trouble proving $$\nabla\times(\nabla f)=0$$ using index notation. I have started with: $$(\hat{e_i}\partial_i)\times(\hat{e_j}\partial_j f)=\partial_i\partial_jf(\hat{e_i}\times\hat{e_j})=\epsilon_{ijk}(\partial_i\partial_j f)\hat{e_k}$$ I know I have to use the fact that $\partial_i\partial_j=\partial_j\partial_i$ but I'm not sure how to proceed.

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2 Answers 2

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The point is that the quantity $M_{ijk}=\epsilon_{ijk}\partial_i\partial_j$ is antisymmetric in the indices $ij$, $$M_{ijk}=-M_{jik}$$

So when you sum over $i$ and $j$, you will get zero because $M_{ijk}$ will cancel $M_{jik}$ for every triple $ijk$.

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    $\begingroup$ I would specify, to avoid confusion, that you don't use the summation convention in the definition of $M_{ijk}$ (note that OP uses this in his/her expression). $\endgroup$
    – Winther
    Oct 10, 2017 at 22:05
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You have that $\nabla f = (\partial_x f, \partial_y f, \partial_z f)$

and $curl f = (\partial_y f_3 - \partial_z f_2, \partial_z f_1 - \partial_x f_3, \partial_x f_2 - \partial_y f_1) $. Where $f_i =$ i:th element in the vector.

So $curl \nabla f = (\partial_{yz} f - \partial_{zy} f, \partial_{zx} - \partial_{xz}, \partial_{xy} - \partial_{yx} )$.

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