Re: nabla is vectorial?

From: Ken Pledger (Ken.Pledger_at_mcs.vuw.ac.nz)
Date: 01/26/05 

Date: Thu, 27 Jan 2005 09:36:31 +1300

In article <6c8b737a.0501251153.7e7f07cd@posting.google.com>,
 marnodoh@libero.it (marnodoh) wrote:

> nabla operator can be considered a vector or a scalar (can it be
> represented only with a number, or does it need even direction and
> point of application?). And what about Laplace operator: can it be
> considered at the same way of nabla, cause it is derived by it
> (Laplace=nabla*nabla)? Thank you.

      Nabla is a clever gadget which has some features of both a
differential operator and a vector.

Nabla scalar = a vector (the gradient)
as you expect from (vector times scalar).

Nabla dot vector = a scalar (the divergence)
as you expect from (vector dot vector).

Nabla cross vector = a vector (the curl or rotation)
as you expect from (vector cross vector).

Nabla squared = nabla dot nabla
               = a scalar differential operator (the Laplacian)
as you expect from a dot product.

Nabla cross nabla is an operator which maps every smooth scalar or
vector field to zero,
as you expect from (vector cross itself).

      The analogy extends even to the expansion of vector triple
products. For any vectors a, b, c in R^3,

a x (b x c) = (a.c)b - (a.b)c

             = b(a.c) - (a.b)c for convenience below.

Replacing a and b by nabla gives the correct formula

nabla x (nabla x c) = nabla(nabla.c) - (nabla^2)c

i.e. "curl curl equals grad div minus del squared".

      HTH

            Ken Pledger.

Quantcast