Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

2 Answers 2

up vote 5 down vote accepted

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

share|improve this answer
    
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
add comment

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 $\textbf{A}$.

share|improve this answer
2  
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
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.