How to prove the curl curl of a vector?

[shedrick94]

I've got ∇×(∇×R)=∇(∇.R)-∇²R [call it eq.1]

However I have the identity ∇×(A×B)=A(∇.B)-B(∇⋅A)+ (B⋅∇)A-(A⋅∇)B [call it eq.2]

Substituting in A=∇ and R=B into eq.2 we get ∇×(∇×R)=∇(∇.R)-R(∇⋅∇)+ (R⋅∇)∇-(∇⋅∇)R

which i work out to be ∇×(∇×R)=∇(∇.R)-R(∇⋅∇)+ (R⋅∇)∇-∇²R

Basically I don't understand what happens to the two terms -R(∇⋅∇)+ (R⋅∇)∇ from eq.2 when we get to eq.1, why do they disappear?

 

[BvU]

Substituting in A=∇

Can't be done:$$ \vec A \times \vec B = - \vec B \times \vec A $$ but $$ \vec \nabla \times \vec A \ne - \vec A \times \vec \nabla $$the left hand side is a vector, the righthand side an operation

 

[DrDu]

The disappearing terms are differential operators. As long as they don't act on anything, they will be zero.

 

[shedrick94]

The disappearing terms are differential operators. As long as they don't act on anything, they will be zero.

Why is this though?

 

[BvU]

basically then you might put a factor 1 behind the expression and $\nabla 1=0$, but as I indicated in #2, the equals sign doesn't hold, just like ${\delta f\over \delta x }\ne f{\delta 1\over x} = 0$