# Divergence Simplification/Identities

1. Oct 2, 2011

### feedmeister

Quick question…

what does the following simplify to? Can it be written in any other way?

$\nabla\bullet (a \bullet b)b$

where a and b are vectors.

Thanks,

2. Oct 2, 2011

### issacnewton

in general

$$\nabla (\varphi \mathbf{F})=(\nabla \varphi)\bullet\mathbf{F}+\varphi (\nabla \bullet \mathbf{F})$$

let $$\varphi = \mathbf{a}\bullet \mathbf{b}$$

and $$\mathbf{F}=\mathbf{b}$$

3. Oct 2, 2011

### feedmeister

Thanks, but I didn't think that $\nabla\bullet (\mathbf{a} \bullet \mathbf{b})\mathbf{b}$ was the same as $$\nabla (\varphi \mathbf{F})$$... there's still a $$\bullet$$ between the $$\nabla$$ and the rest of the statement.

Can you clarify?

4. Oct 2, 2011

### issacnewton

little mishtake...above should be

$$\nabla \bullet (\varphi \mathbf{F})=(\nabla \varphi)\bullet\mathbf{F}+\varphi (\nabla \bullet \mathbf{F})$$

:tongue:

5. Oct 3, 2011

### feedmeister

Thanks, issacnewton.

When substituting in $$\mathbf{a}\bullet \mathbf{b}$$ and $$\mathbf{b}$$ into the equation, it looks like it'd simplifies further.. but it looks like it'd be ugly. :yuck:

Any good way of simplifying it?

Thanks,

6. Oct 3, 2011