# I Vector Laplacian

1. May 20, 2016

### olgerm

Is on of these pages wrong or I misunderstand it?

I am asking because I want to know what does Δ2 equal to in this and this equation on this https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field#Coulomb_gauge page.

2. May 20, 2016

### Orodruin

Staff Emeritus
The d'Alembertian is a differential operator. You can apply it to a scalar and get a scalar, or you can apply it to a vector and get a vector.

3. May 20, 2016

### olgerm

So since magnetic potential A in this equation is vector $\nabla^2=(\sum_{i=1}^3(\frac{\partial^2 A}{\partial x_i*\partial x_1});\sum_{i=1}^3(\frac{\partial^2 A}{\partial x_i*\partial x_2});\sum_{i=1}^3(\frac{\partial^2 A}{\partial x_i*\partial x_3}))$ not $\nabla^2=\sum_{i=1}^3(\frac{\partial^2 A}{\partial x_i^2})$?
These are Cartesian coordinates.

4. May 20, 2016

### Orodruin

Staff Emeritus
It is unclear what you intend to write with these expressions. In particular, it is unclear what you intend for $A$ to be (the vector, the vector components, etc). I strongly suggest that you do not write vectors on component form (x;y;z), but instead use basis vectors. This will make it much clearer what is intended.

The Laplace operator applied to a vector in Cartesian coordinates is such that the $x$ component of $\nabla^2 \vec A$ is equal to $\nabla^2 A_x$, where $A_x$ is the $x$ component of $\vec A$.

5. May 20, 2016

### olgerm

Last edited: May 20, 2016
6. May 20, 2016

### Orodruin

Staff Emeritus
Neither, see my previous post.

Your first option is just one term in the sum, your second is grad(div(A)) and not the Laplacian of A, your third is a scalar.

7. May 20, 2016

### olgerm

8. May 20, 2016

### Orodruin

Staff Emeritus
9. May 20, 2016

Ok, thanks.