Can we use a magnetic vector potential in the case of time varying $E$-fields?

Question

Most discussions of the magnetic vector potential defined through $\mathbf{B}=\nabla\times\mathbf{A}$ are only for working with static electric fields (for example, Griffiths:

If we instead require that

$$\nabla\times\mathbf{B}=\nabla\times(\nabla\times\mathbf{A}) = \nabla(\nabla\cdot\mathbf{A})\ -\nabla^2\mathbf{A} = \mu_0\mathbf{J} +\mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}$$

using the full form of Ampere's law, then we can still have the freedom to choose an $\mathbf{A}$ with zero divergence, giving us a form of Poisson's equation:

$$\nabla^2\mathbf{A} = -\mu_0\mathbf{J} -\mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}$$

Which can be solved in principle using analogies to the electric scalar potential $V$.

Why is this form of the magnetic vector potential in its full generality seldom seen? Is there a contradiction which prevents its use?

Answer 1

Yes it can be used,

Given,$ \nabla × A = B$

Then

$\nabla ×E = -\frac{\partial B}{\partial t}$

$\nabla × E = -\frac{\partial (\nabla × A) }{\partial t}$

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

$\nabla × E + \nabla × \frac{\partial A }{\partial t} = 0$

$\nabla × (E + \frac{\partial A }{\partial t})= 0$

Because the curl of this quantity is zero, it can be written as the gradient of a scalar function ( or the negative of a gradient of a scalar function, which is used to match the definition of electrostatic potential)

$E + \frac{\partial A }{\partial t} = -\nabla V$

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

From here we can substitute the definitions of E into the other maxwell equations

This will obtain 2 equations that interlink A and V in terms of the source terms $\rho$ and J.

using the "lorenz" gauge choice ( not lorentz) we can decouple these and solve easier( or using the coulomb gauge)

This is called the potential formulation of maxwells equations