0
$\begingroup$
  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?

$\endgroup$
  • 3
    $\begingroup$ the formula E(x)=- \frac{\partial V}{\partial x} is only valid for electrostatics, so when $\partial _t$ = 0. As soon as you introduce the vector potential $\mathbf{A}$, you cannot think of the electric field as simply the gradient of the potential. $\endgroup$ – SuperCiocia Jun 21 '14 at 22:22
  • $\begingroup$ Related: physics.stackexchange.com/q/53020/2451 and links therein. $\endgroup$ – Qmechanic Nov 11 '14 at 17:10
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ That means it is not a single surface but an array surface? or a volume? $\endgroup$ – Saprativ Saha Jun 22 '14 at 5:10
  • $\begingroup$ My initial reaction to your question was that yes, you would need some kind of higher-dimensional surface, but in the description I give you do not. Since $d\vec{A}/dt$ is still just a vector, the curve with normal vector $d\vec{A}/dt$ would encode enough information for you to know the electric field. It wouldn't tell you what the entire vector $\vec{A}$ is, so you are missing information in that sense. $\endgroup$ – levitopher Jun 22 '14 at 19:14
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.