# Is it possible to prove that the curl of a gradient equals zero in this way?

If $(\nabla\times\nabla\Phi)_i = \epsilon_{ijk}\partial_j\partial_k\Phi$, where Einstein summation is being used to find the $i$th component...

Using Clairaut's theorem $\partial_{i}\partial_{j}\Phi = \partial_{j}\partial_{i}\Phi$, so $$\epsilon_{ijk}\partial_j\partial_k\Phi = \epsilon_{ijk}\partial_k\partial_j\Phi = -\epsilon_{ikj}\partial_k\partial_j\Phi$$

Now here is where I am confused. If the positive is equal to the negative, the value must be zero. Have I adequately shown this?

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Just a note that $\epsilon$ or $\varepsilon$ is the symbol to use for the Levi-Civita symbol, not $\in$ (which indicates set membership). – David Z Oct 8 '12 at 23:21

## 1 Answer

Yes, that's fine. You could write out each component individually if you want to assure yourself.

A more-intuitive argument would be to prove that line integrals of gradients are path-independent, and therefore that the circulation of a gradient around any closed loop is zero. The curl is a limit of such a circulation, and so the curl must be zero.

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