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