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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: $\nabla \times (\mathbf{A} \times \mathbf{B}) = \mathbf{A} (\nabla \cdot \mathbf{B}) - \mathbf{B} (\nabla \cdot \mathbf{A}) + (\mathbf{B} \cdot \nabla) \mathbf{A} - (\mathbf{A} \cdot \nabla) \mathbf{B}$

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  • $\begingroup$ Where in that section do you see $\vec A\cdot \nabla$? I only see $\nabla\cdot F$, which is just an alternative form for "$\mathrm{div}\,F$". $\endgroup$ – Marc van Leeuwen Mar 16 '12 at 15:23
  • $\begingroup$ The 3rd and 4th addends the in the second identity for vector cross product. $\endgroup$ – Michael Mar 16 '12 at 15:44
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$$\vec A \cdot \nabla = \sum_{i=1}^3 A_i \frac{\partial}{\partial x_i}$$

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  • $\begingroup$ Does this mean $(\vec A \cdot \nabla)\vec B = \sum_{i=1}^3 A_i (\frac{\partial}{\partial x_i}\vec B)$? $\endgroup$ – Michael Mar 16 '12 at 18:16
  • $\begingroup$ Yes, that's right. $\endgroup$ – Robert Israel Mar 16 '12 at 21:15
  • $\begingroup$ Thank you Robert and Shabat Shalom $\endgroup$ – Michael Mar 17 '12 at 11:18
  • $\begingroup$ 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 $\mathbf{a}\cdot \boldsymbol{\nabla}\neq \boldsymbol{\nabla}\cdot \mathbf{a}$?? $\endgroup$ – pluton Oct 24 '17 at 3:25

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