Divergence Simplification/Identities

[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,

 

[issacnewton]

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,

in general

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

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

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

 

[feedmeister]

in general

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

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

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

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?

 

[issacnewton]

in general

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

little mishtake...above should be

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

 

[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.

Any good way of simplifying it?

Thanks,

 

[issacnewton]

well you can use gradient of vector dot product to simplify

http://en.wikipedia.org/wiki/Vector_calculus_identities#Product_rule_for_the_gradient

but things WILL look ugly... you can probably simplify further using levi civita tensor

http://folk.uio.no/patricg/teaching/a112/levi-civita/

https://www.physicsforums.com/showthread.php?t=488903

google "levi civita tensor identities" for further help