How to get the square of this expression?

51 Views Asked by At

I would like to open brackets of the following expression:

$$(\nabla+\textbf{A})^2$$ where $\textbf{A}=(2r^2,0,r sin(\theta))$.

I do: $(\nabla+\textbf{A})^2=\Delta+(\nabla\textbf{A})+(\textbf{A}\nabla)+A^2$

The expression $(\nabla\textbf{A})$ is divergence of vector $A$ and it is simplified to $8r$, but how to simplify expression $(\textbf{A}\nabla)$?

Help me please with this task

1

There are 1 best solutions below

0
On

In cartesian coordinates, the expression you are looking for is $A_i\partial_i$. It represents an derivation in the "direction" of the vector field A. There is not much more you can do to it until you apply this new opperator onto something.