# Nabla Operator

1. Jan 10, 2005

### Raparicio

Dear Friends,

Another question for dummies...

The operator "nabla" can be locates before or after a vector or a tensor. If you take the vector A, "nabla A" is not the same that "A nabla" but, is it possible to obtain "nabla A - A nabla"? ¿And "(A nabla) A - A (nabla A)"?

2. Jan 10, 2005

### dextercioby

I hope u're not insinuating that we (me included) would be "dummies"...

$$\nabla=\sum_{i=1}^{3} \frac{\partial}{\partial x_{i}} \vec{e}_{i}$$(1)
in the cartezian system of coordinates
$$\vec{A}\cdot \nabla=\sum_{i=1}^{3} A_{i}\frac{\partial}{\partial x_{i}}$$ (2)

$$\nabla\cdot\vec{A}=\sum_{i=1}^{3} \frac{\partial A_{i}}{\partial x_{i}}$$ (3)

That's all u need to know.
$$[\nabla,\vec{A}]_{-}=:\nabla\cdot\vec{A}-\vec{A}\cdot \nabla$$
is kinda weird operator which is made from a multiplicative part and a differential part.
I've never seen it in physics in this form.A bit different form can be found in QM with the operators of position and momentum in the coordinate representations.It's basicaly minus the fundamental commutator relations.

Daniel.

Last edited: Jan 10, 2005
3. Jan 10, 2005

### Raparicio

NOoooooooo

No... the "dummie" in mathmatics and physics am I.

4. Jan 10, 2005

### dextercioby

On a second thought about that commutator of operators,in analogy with the QM case,consider it to be applying on a scarar function [itex] \phi(\vec{r}) [/tex]

$$[\nabla,\vec{A}(\vec{r})]_{-}\phi(\vec{r})=:\nabla\cdot [\vec{A}(\vec{r})\phi(\vec{r})]-\vec{A}\cdot \nabla\phi(\vec{r})=[\nabla\cdot \vec{A}(\vec{r})] \phi(\vec{r})$$

Daniel.