Question in Curl of a cross product.

• Mar 21st 2011, 06:00 PM
yungman
Question in Curl of a cross product.
$\nabla X (\vec A X \vec B) \;=\; (\vec B \cdot \nabla)\vec A - \vec B(\nabla \cdot \vec A) -(\vec A \cdot \nabla)\vec B + \vec A ( \nabla \cdot \vec B)$.

What is $\vec A \cdot \nabla$?
• Mar 21st 2011, 06:15 PM
Prove It
Have they told you what $\displaystyle \mathbf{A}$ and $\displaystyle \mathbf{B}$ are equal to?
• Mar 21st 2011, 06:17 PM
yungman
Quote:

Originally Posted by Prove It
Have they told you what $\displaystyle \mathbf{A}$ and $\displaystyle \mathbf{B}$ are equal to?

Nop, this is just a general formula of the product of two vectors.
• Mar 22nd 2011, 02:02 AM
Ackbeet
Quote:

Originally Posted by yungman
$\nabla X (\vec A X \vec B) \;=\; (\vec B \cdot \nabla)\vec A - \vec B(\nabla \cdot \vec A) -(\vec A \cdot \nabla)\vec B + \vec A ( \nabla \cdot \vec B)$.

What is $\vec A \cdot \nabla$?

It's the vector $\vec{A}$ dotted with the gradient operator. The result of this dot product is a scalar operator. So, for example, the expression

$\displaystyle(\vec{A}\cdot\nabla)\vec{B}=\left(\su m_{j=1}^{3}A_{j}\,\dfrac{\partial}{\partial x_{j}}\right)\vec{B}=\left\langle\sum_{j=1}^{3}A_{ j}\,\dfrac{\partial B_{1}}{\partial x_{j}},\sum_{j=1}^{3}A_{j}\,\dfrac{\partial B_{2}}{\partial x_{j}},\sum_{j=1}^{3}A_{j}\,\dfrac{\partial B_{3}}{\partial x_{j}}\right\rangle.$

Does that make sense?