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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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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
up vote 6 down vote accepted

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

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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

The dot product of two vectors $v = (v_1, \ldots, v_n), w = (w_1, \ldots, w_n)$ in $\textbf{R}^n$ is defined as $v \cdot w = \sum_{i = 1}^n v_iw_i$.

The expression $\textbf{A} \cdot \bigtriangledown$ is just a mnemonic to remember the definition of the div operator. Say $\textbf{A} = \textbf{A}_1e_1 + \textbf{A}_2e_2 + \textbf{A}_3e_3$ is a vector field on $\textbf{R}^3$. Then, viewing $\bigtriangledown$ as $\left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)$, you get, analogously as with the dot product of $\textbf{R}^n$,

$$ \textbf{A} \cdot \bigtriangledown = \frac{\partial \textbf{A}_1}{\partial x} + \frac{\partial \textbf{A}_2}{\partial y} + \frac{\partial \textbf{A}_3}{\partial z} $$

which is the definition of div $\textbf{A}$.

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This is misleading - the notation $\mathbf{A}\cdot\nabla$ is far more likely to refer to the operator $\sum_i A_i\partial_i$ than to the divergence of $\mathbf{A}$, which would be notated $\nabla\cdot\mathbf{A}$. – Chris Taylor Mar 16 '12 at 16:52
Under what conditions can we say $A\frac{\partial \textbf{}}{\partial x}=\frac{\partial \textbf{}}{\partial x}A$? – Michael Mar 16 '12 at 18:20
Never. One is a differential operator, the other is a vector field. – Robert Israel Mar 16 '12 at 21:17
@Robert You see this kind of thing all the time in theoretical physics, where $A$ is interpreted as an operator whose action is $(Af)(x) = A(x)f(x)$. For example, in statements about commutators, like $[x,\frac{\partial}{\partial x} ] = x \frac{\partial}{\partial x} - \frac{\partial}{\partial x} x = -1$ – Chris Taylor Mar 20 '12 at 12:48

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