Losing a curl in vector identity

Question

Considering

$$\nabla (\frac{1}{2} \vec{A} \cdot \vec{A}) $$

The full identity has some cross product terms. But, when I try and work from this expression by itself with the summation convention I get the following (wrong) derivation

$$= \partial_i(\frac{1}{2} A_j A_j)$$ $$= \frac{1}{2}(A_j\partial_i A_j + A_j\partial_i A_j)$$ $$=\vec{A} \cdot \nabla \vec{A}$$

Clearly, the cross terms have not popped up (and I very much know the standard method of starting with the cross terms) but why have they gone? Where have I implicitly assumed $\nabla \times \vec{A} = 0$?

Is that last line right? $A_j\partial_i A_j$ not $A_j\partial_j A_i$ — dm63, Apr 05 '20 at 19:56

score 1 · Answer 1 · answered Apr 05 '20 at 20:40

1

You do not assume that $\nabla\times\vec A=0$. But you have $$\vec A\times(\nabla\times\vec A)=0$$For more detailed derivation of this result, see this question

answered Apr 05 '20 at 20:40

Andrei

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Losing a curl in vector identity

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