1
$\begingroup$

The magnetic vector potential $\textbf{A}$ can be defined up to a gradient of a field. Adding or subtracting such gradient should not change the physics of the problem. The same reasoning is applied for the time derivative of $\textbf{A}$.

Imagine a simple vector potential such that $$\frac{\partial\textbf{A}}{\partial t} = (0, -z, y)$$

Inside a finite conductor (e.g., a cube), such field originates a charge accumulation at the boundaries, that creates a total electric field

$$\textbf{E} = -\nabla \phi -\frac{\partial\textbf{A}}{\partial t} $$

where the gradient of the potential is due to the charge accumulation.

So.. When $\textbf{A}$ is shifted, there shouldn't be any change, but what I see doing FEM simulations is that $\phi$ and $\nabla \phi$ change radically, producing the same total electric field in the two cases. A second vector potential is such that: $$\frac{\partial\textbf{A}}{\partial t} = (0, -z-1, y-1)$$

Looking at the electric field equation, that makes sense. But if the gradient of the potential is due to the charge accumulation, then there are two different charge densities for the two presented situations.

Since I believe the physics is right, there should be an error in my reasoning. Can someone please point where?

$\endgroup$
  • 3
    $\begingroup$ You wrote: "The new vector potential is" but what you really wrote is its time derivative. Please either provide the actual potential (by integrating over time?) or correct phrasing. Anyway, gauge transformation does modify $\phi$ for explicitly time-dependent transforms. $\endgroup$ – user23660 Oct 16 '13 at 16:27
3
$\begingroup$

Hint: OP's gauge transformation of the magnetic vector potential $$\vec{A}^{\prime}~=~\vec{A} +\nabla \Lambda$$ can e.g. be described by a time-dependent gauge parameter $$\Lambda~=~ -(y+z)t. $$

But this means that the electric scalar potential $$\phi^{\prime}~=~\phi -\partial_t \Lambda$$ transforms as well.

$\endgroup$
  • $\begingroup$ good point! I had thought it was enough to guarantee the same gauge (which for the mentioned example was div(A)=0 ). Now it feels obvious.. Thank you! $\endgroup$ – serigado Oct 17 '13 at 10:51

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.