Vector Identities formula #6 | Quantum Science Philippines

## Vector Identities formula #6

Prove: $\vec{\nabla}\times(\vec{\nabla}\times\vec{a})=\vec{\nabla}(\vec{\nabla}\cdot\vec{a})-\nabla^2\vec{a}$

let: $\vec{\nabla}=\partial_l\widehat{e_l}$ $\vec{a}=a_i\widehat{e_i}$

Solution: $=\vec{\nabla}\times(\vec{\nabla}\times\vec{a})$ $=\partial_l\widehat{e_l}\times[\partial_l\widehat{e_l}\times a_i\widehat{e_i}]$ $=\partial_l\widehat{e_l}\times[\partial_la_i\in_{lij}(\widehat{e_l}\times\widehat{e_i})]$ $=\partial_l\widehat{e_l}\times[\partial_l a_i\in_{lij}\widehat{e_j}]$ $=\partial_l\partial_la_i\in_{lji}\in_{ljn}(\widehat{e_l}\times\widehat{e_j})$ $=\partial_l\partial_la_i\in_{jil}\in_{jln}\widehat{e_n}$ $=\partial_l\partial_la_i\delta_{il}\delta_{ln}\widehat{e_n}-\partial_l\partial_la_i\delta_{in}\delta_{ll}\widehat{e_n}$

note: $\delta_{ll}=1$ $\delta_{il}=1,i=l$

thus; $=\partial_l\partial_la_i\widehat{e_l}-\partial_l\partial_la_i\widehat{e_i}$ $=\partial_ia_i\partial_l\widehat{e_l}-\partial_l\partial_la_i\widehat{e_i}$ $=(\vec{\nabla}\cdot\vec{a})\vec{\nabla}-(\vec{\nabla}\cdot\vec{\nabla})\vec{a}$ $=\vec{\nabla}(\vec{\nabla}\cdot\vec{a})-\nabla^2\vec{a}$





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