# Vector identity proof

I am told that ∇×(∇×A)=∇(∇⋅A)−△A\nabla\times(\nabla\times A)=\nabla(\nabla\cdot A)-\triangle A.

I’m then asked to use this to show that ∇×(∇×(∇×A))=−△(∇×A).\nabla\times(\nabla\times(\nabla\times A))=-\triangle(\nabla\times A).

So ∇×(∇×(∇×A))=∇×(∇(∇⋅A)−△A)=∇×(∇(∇⋅A)−∇⋅∇A))=ϵijk∂j(∂k∂mAm−∂k∂kA)\begin{align*}\nabla\times(\nabla\times(\nabla\times A))
&=\nabla\times(\nabla(\nabla\cdot A)-\triangle A)\\
&=\nabla\times(\nabla(\nabla\cdot A)-\nabla\cdot\nabla A))\\
&=\epsilon_{ijk}\partial_j(\partial_k\partial_mA_m-\partial_k\partial_kA)
\end{align*}

It doesn’t look like this is taking me anywhere though. What should I be doing?

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You are probably asked to use other formulas for the cross product that you already have, instead of using the definition
– b00n heT
2 days ago

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2

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You should keep the inner ∇×A\nabla \times A term and calculate from the ‘outside’. Then using the stated identity:
∇×(∇×(∇×A))=∇(∇⋅(∇×A))−Δ(∇×A)=−Δ(∇×A) \nabla\times (\nabla \times (\nabla\times A)) =\nabla (\nabla \cdot (\nabla\times A)) – \Delta (\nabla \times A) =- \Delta (\nabla \times A)
(in the first term one has ∇⋅(∇×A)=0\nabla \cdot (\nabla\times A)=0 ).

… or keep ∇×∇×→A\nabla \times \nabla \times \vec A and note that ∇×∇Φ=0\nabla \times \nabla \Phi=0 and ∇×∇2→A=∇2∇×→A\nabla \times \nabla^2 \vec A=\nabla^2 \nabla \times \vec A.
– Dr. MV
2 days ago

@H. H. Rugh I’m not entirely sure what you mean here. Could you explain further?
– user332597
2 days ago

HINTS:

∇×∇Φ=0\nabla \times \nabla \Phi=0

and

∇×∇2→A=∇2∇×→A\nabla \times \nabla^2 \vec A=\nabla^2 \nabla \times \vec A