$\nabla . (ϕ a) = ϕ \nabla . a + a . \nabla ϕ$

By Dan Coomber on Wednesday, April 17, 2002 - 03:07 pm:

Can anyone help with proving the following vector identities? I have found many texts stating them, but none with proofs:

With suitably smooth scalar field

ϕ

and suitably smooth vector fields

a

and

b

$\nabla . (ϕ a) = ϕ \nabla . a + a . \nabla ϕ$

and

$\nabla . (a \times b) = b . (\nabla \times a) - a . (\nabla \times b)$

Many thanks...

By David Loeffler on Wednesday, April 17, 2002 - 09:52 pm:

Use suffix notation. Write

a = {a_{i}}

, where

i = 1

, 2, 3. Then

$\nabla . (ϕ a) = \sum_{i} \partial / \partial x_{i} (ϕ a_{i})$

$= \sum_{i} (ϕ \partial a_{i} / \partial x_{i} + a_{i} \partial ϕ / \partial x_{i})$

$= ϕ \nabla . a + a . (\nabla ϕ)$

The curl one is similar.

David