# Gradient of a Vector Dot Product

1. Nov 20, 2008

### pcalhoun

Hello,

I was messing around with subscript summation notation problems, and I ended up trying to determine a vector identity for the following expresion:

$$\overline{\nabla}(\overline{A}\cdot\overline{B})$$

Here are my steps for as far as I got:

$$\hat{e}_{i}\frac{\partial}{\partial x_{i}}(A_{j}\hat{e}_{j}\cdot B_{k}\hat{e}_{k})$$
$$\hat{e}_{i}\frac{\partial}{\partial x_{i}}(A_{j}B_{k}\delta_{jk})$$
$$\hat{e}_{i}\frac{\partial}{\partial x_{i}}(A_{j}B_{j})$$
$$\hat{e}_{i}(A_{j}\frac{\partial B_{j}}{\partial x_{i}} +B_{j}\frac{\partial A_{j}}{\partial x_{i}} )$$

After these steps, I could not clearly see any ways to continue to manipulate this expression.
Not knowing whether an identity actually existed for this expression, I turned to wikipedia and they suprisingly had the solution (which was more complicated than I thought it would have been.)
Regardless, I wasn't sure what steps could be taken to arrive at the solution.

Thanks,
pcalhoun

2. Nov 20, 2008

### D H

Staff Emeritus
The expression in Wikipedia,

$$\boldsymbol{\nabla}(\boldsymbol{A}\cdot\boldsymbol{B}) = (\boldsymbol{A}\cdot\boldsymbol{\nabla})\boldsymbol{B} + (\boldsymbol{B}\cdot\boldsymbol{\nabla})\boldsymbol{A} + \boldsymbol{A}\times(\boldsymbol{\nabla}\times\boldsymbol{B}) + \boldsymbol{B}\times(\boldsymbol{\nabla}\times\boldsymbol{A})$$

is, IMHO, much worse than your simple expression. The wiki expression is a computation nightmare: much, much more expensive and subject to loss of accuracy. In short, yech.

The way to proceed is to creatively add zero to the right hand side of your simple expression.

3. Nov 21, 2008

### pcalhoun

With a little luck, I found an equivalent expression (assuming no mistakes) dealing with dyadic/outer products of Del and the two vectors:

$$(\overline{ \nabla } \overline{ A } ) \cdot \overline{B} + (\overline{ \nabla } \overline{ B } ) \cdot \overline{ A }$$

Anyways, thanks for the help D H.

Cheers,
pcalhoun

4. Feb 1, 2011