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I want to prove $$\vec{A}\times(\nabla\times\vec{A})=\frac{1}{2}\nabla(A \cdot A)-(\vec{A}\cdot\nabla)\vec{A}$$

This looks strikingly similar to the BAC CAB formula

$$\vec{A}\times(\vec{B}\times\vec{C})=\vec{B}(\vec{A}\cdot\vec{C})-\vec{C}(\vec{A}\cdot\vec{B})$$

However, because there is some differentiation involved, I understand I cannot simply move the 'Del' operator around willy nilly. I'm thinking I need to use (Product Rule + Levi Civita + Clairaut's Theorem) to prove this, but when I get to the step:

$$[\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}]A_{j}\partial_lA_m$$

Up until this I have have NOT seperated the $\partial_l(A_m)$ but I'm unsure how to proceed.

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1 Answer 1

up vote 3 down vote accepted

Hint: You are very close. Note, for example, that $$\delta_{il}\delta_{jm}A_{j}\partial_l A_m = A_j \partial_i A_j = \frac{1}{2} \partial_i A_j^2.$$

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