From: "Fabrice P. Laussy" <laussy@nat.fr>
Subject: Re: Nabla operator
Date: Fri, 18 Aug 2000 03:29:47 +0200
Newsgroups: sci.math
Summary: [missing]

MJ Ferguson wrote:

> what exactly does the nabla operator do??????

V being a scalar field and F a vector field, it's a very handy
notation (it's not a vector, but it transforms as a vector, so it can
be used like one, somehow). Tackled formally as a vector which
components are partial derivatives, it allows for instance to
express:-

    * The gradient--that is the field of greatest variations--as nabla
times V
    * The divergence--that is the strength of sources--as the scalar
product of nabla and F
    * The curl--that is the difference in stretching of a field--as
vector product of nabla with F
    * The laplacian--that is the departure at a point from the average
in a neighborhood--as nabla squared times V (or F if laplacian of a
field).

It's convenient for scores of other things. For instance, it allows by
formal vector operations to remind of formulas like curl (curl F) =
grad (div F) - laplacian (F). A trap: it works nicely only in
cartesians coordinates.

You will find a very good discussion of it in a "Feynman's lecture on
physics" of tome II (at the start, may be the second or third
chapter).

F.P.L.