Physical representation of magnetic vector potential

Question

1. In electrostatics, for scalar potential $V$, we can represent the equipotential surface as a perpendicular surface of the direction of derivative. such as $$ {\bf E}~=~-\nabla V \tag{1} $$ & the equipotential surface is as a perpendicular of ${\bf E}$.

2. In electromagnetism, for time varying potential $$ {\bf E}~=~ - \nabla V - \frac{\partial {\bf A}}{\partial t}\tag{2} $$ Then how can I represent the vector potential ${\bf A}$ having equal in everywhere as the above case (1). Is it perpendicular to time?

Answer 1

The equipotential surfaces of $V$ give you the direction of the electric field (perpendicular to them) and the magnitude (by how far apart they are), so you could try to do the same thing with the vector potential. At every point in space there is a vector $\vec{A}(x)$ representing the vector potential, and a vector $\partial \vec{A}/\partial t$ representing it's derivative. This vector (the time derivative) should be the normal vector to the "surfaces of equal $\vec{A}$" which you are looking for.

Answer 2

One way to think of a vector field like $\mathbf{E}$ is to separate it into a divergent part and a curling part. Roughly, at a given location in space, the divergent part spreads out from that location and the curling part curls in a closed contour around that location. The divergent part (also called "irrotational") has $\mathbf{\nabla}\times\mathbf{E} = 0$ and the curling part (also called "solenoidal" or "divergenceless") has $\mathbf{\nabla}\cdot\mathbf{E} = 0$.

Taking the curl and divergence of your eq. (2) (and setting $\mathbf{\nabla}\cdot\mathbf{A} = 0$, which is referred to as "choosing the Coulomb gauge"), we see that $-\mathbf{\nabla} V$ corresponds to the divergent part of $\mathbf{E}$ and $-\dfrac{\partial \mathbf{A}}{\partial t}$ corresponds to the curling part.

If $\mathbf{A}$ is changing only in magnitude and not direction, then it points in the same direction as $\dfrac{\partial \mathbf{A}}{\partial t}$ and also in the opposite direction to the curling part of $\mathbf{E}$. In that case, to visualize $\mathbf{A}$, imagine the contours of the curling part of $\mathbf{E}$ and reverse their direction.