Vector Identities formula #6 | Quantum Science Philippines
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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