# Geometric perspective of the vector potential

1. Jun 9, 2012

### PerpStudent

I'm struggling with trying to visualize the vector potential as in the identity:

B = ∇⨯A

For starters, how does A relate to, say, a uniform magnetic field, which is quite easy to visualize. Then, how about the magnetic field around a bar magnet -- where is A?
Any help would be appreciated.

2. Jun 9, 2012

### Muphrid

The formula should look similiar to something you're already familiar with:

$$\mu_0 j = \nabla \times B$$

If you can imagine the magnetic field that goes with a specific current density, then the same picture applies to the vector potential that goes with the magnetic field.

3. Jun 10, 2012

### vanhees71

It's not easy to visualize $\vec{A}$ or give it a physical meaning since it is a gauge-dependent quantity. What's physical is the magnetic field, $\vec{B}$ which is given by

$$\vec{B}=\vec{\nabla} \times \vec{A}.$$

For a constant field, it's easy to get the vector potential. So let's consider

$$\vec{B}=B_0 \vec{e}_z.$$

You have quite some freedom to choose the vector potential. You can take always one constraint since it is only defined from $\vec{B}$ up to the gradient of a scalar field. Here, I'd choose the spatial axial gauge

$$A_z=0.$$

Then you have

$$\vec{B}=\begin{pmatrix} 0 \\ 0 \\ B_0 \end{pmatrix} = \vec{\nabla} \times \vec{A}=\begin{pmatrix} -\partial_z A_y \\ \partial_z A_x \\ \partial_x A_y-\partial_y A_x \end{pmatrix}.$$

Obviously our constraint doesn't fix the solutions completely, and you have some more freedom. You can, e.g., set $A_x=0$ and $A_y=B_0 x$. Then you have

$$\vec{A} = B_0 x \vec{e}_y.$$

Around a bar magnet you have to solve the magnetostatic Maxwell equations for a given magnetization of your bar. You find some calculations about this in Sommerfeld's Lectures on Theoretical Physics, vol. III.