What is the meaning of $\vec A \cdot \nabla$?

Looking at the application of divergence in Cartesian coordinates in Wikipedia I was wondering about the meaning of $\vec A \cdot \nabla$?

This dot product is found at the vector cross product identity: $\nabla \times (\mathbf{A} \times \mathbf{B}) = \mathbf{A} (\nabla \cdot \mathbf{B}) - \mathbf{B} (\nabla \cdot \mathbf{A}) + (\mathbf{B} \cdot \nabla) \mathbf{A} - (\mathbf{A} \cdot \nabla) \mathbf{B}$

• Where in that section do you see $\vec A\cdot \nabla$? I only see $\nabla\cdot F$, which is just an alternative form for "$\mathrm{div}\,F$". – Marc van Leeuwen Mar 16 '12 at 15:23
• The 3rd and 4th addends the in the second identity for vector cross product. – Michael Mar 16 '12 at 15:44

1 Answer

$$\vec A \cdot \nabla = \sum_{i=1}^3 A_i \frac{\partial}{\partial x_i}$$

• Does this mean $(\vec A \cdot \nabla)\vec B = \sum_{i=1}^3 A_i (\frac{\partial}{\partial x_i}\vec B)$? – Michael Mar 16 '12 at 18:16
• Yes, that's right. – Robert Israel Mar 16 '12 at 21:15
• Thank you Robert and Shabat Shalom – Michael Mar 17 '12 at 11:18
• So for any vectors $\mathbf{a}$ and $\mathbf{b}$ we have commutativity of the scalar product: $\mathbf{a}\cdot \mathbf{b}=\mathbf{b}\cdot \mathbf{a}$ but not with the del operator since $\mathbf{a}\cdot \boldsymbol{\nabla}\neq \boldsymbol{\nabla}\cdot \mathbf{a}$?? – pluton Oct 24 '17 at 3:25