Nabla operator and working with it

[jink]

While using the ∇ operator, most of the times we can treat it as a vector. I came across a few formulae(basically product rules)..

∇×([B]A[/B]×[B]B[/B])=([B]B[/B].∇)[B]A[/B]-([B]A[/B].∇)[B]B[/B]+[B]A[/B](∇.[B]B[/B])-[B]B[/B](∇.[B]A[/B])

where A and B are vectors

I wanted to know if there is any direct way of deriving it. By direct I mean assuming the basic vector identity

[B]C[/B]×([B]B[/B]×[B]A[/B])=[B]B[/B]([B]C[/B].[B]A[/B])-[B]A[/B]([B]C[/B].[B]B[/B])

. Any other better way is also fine. I derived it by splitting each of the vectors A, B, and ∇(although this is not truly a vector) in their orthogonal components and then doing the appropriate cross products.

 

[Marioeden]

The best way to derive these results is to use summation convention. Everything drops out quite nicely.

The vector identity you've stated is also derived in the same way but you'll notice the subtleties of the fact that you're dealing with an operator when doing the derivation.

 

[jink]

The best way to derive these results is to use summation convention

Thanks for the idea. I have never really used summation notation, always expanded terms and got my way through. Will try summation from now on.

but you'll notice the subtleties of the fact that you're dealing with an operator when doing the derivation

I realized one thing that ∇.A=A.∇ doesn't hold in this case while it holds with vectors.

The vector identity you've stated is also derived in the same way

yeah that's fine, I just wanted to think if there is any result which we can exploit using the fact that ∇ behaves as a vector(well, almost). I was just worried that incase I forget the long identity, and I want to use it, I won't be able to derive it quickly.

 

[SteamKing]

I realized one thing that ∇.A=A.∇ doesn't hold in this case while it holds with vectors.That's because Del isn't really a vector; it's a set of operators which work on vectors.

http://en.wikipedia.org/wiki/Del

 

[jink]

Yeah right.