Nabla Operator

[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)"?

[dextercioby]

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

\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.

[Raparicio]

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

[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.