What is the Dual Nature of Nabla in Vector Differential Operators?

[abrowaqas]

I didn't get the concept of dual or hybrid nature of nabla? I-e vector differential operator .. Is it means that nabla can produce a vector from scalar field (gradient) and scalar from vector field(divergence) ? What's the concept of Nabla's Dual nature ? Please explain..

 

[tiny-tim]

hi abrowaqas!

∇ behaves just like an ordinary vector …

the vector itself is ∇f (where f is a scalar)​

and you can dot-product it, or cross-product it, with a vector A …

∇.A (divA) or ∇xA (curlA) ​

 

[HallsofIvy]

If you multiply a vector by a scalar, you get a vector. If you multiply a vector by a vector (dot product) you get a scalar. If you multiply a vector by a vector (cross product), you get vector.

If that vector is "nabla" those three types of "multiplication" become
\nabla f(x,y,z)= grad f(x,y,z)
\nabla\cdot\vec{f}(x,y,z)= div \vec{f}(x, y, z)
\nabla\times\vec{f}(x,y,z)= curl \vec{f}(x, y, z)

 

[abrowaqas]

Thanks for explanation.. i got it... i have given exam last last friday. i have written the same answer to the question stated below

Elaborate the Hybrid Nature of Operator NABLA ?

what this link say about dual nature of nabla? can anybody explain?

http://books.google.com.pk/books?id...a=X&ei=p3g2T8DXJpSIhQe9jKn7AQ&ved=0CDwQ6AEwAg

 

[tiny-tim]

what this link say about dual nature of nabla? can anybody explain?

http://books.google.com.pk/books?id...a=X&ei=p3g2T8DXJpSIhQe9jKn7AQ&ved=0CDwQ6AEwAg

oh, that's completely different from what i thought you were asking about!

that book (by bernard maxum) is saying that ∇ is both a vector and a derivative …

in general (not cartesian) coordinates, the simple "dot" and "cross" procedure doesn't work …

that's what it means by the "dual nature" of ∇ ​