Scalar and Vector Potentials
Since \(\nabla\cdot\vec{B}=0\) always, write \(\vec{B}=\nabla\times\vec{A}\).
Faraday's law \(\nabla\times\vec{E}=-\partial\vec{B}/\partial t\) then gives:
\(\nabla\times(\vec{E}+\partial\vec{A}/\partial t)=\mathbf{0}\), so:
Fields in terms of potentials\begin{equation}
\vec{E} = -\nabla V - \frac{\partial\vec{A}}{\partial t},\qquad
\vec{B} = \nabla\times\vec{A}
\end{equation}
Only M2 and M3 are automatically satisfied. M1 and M4 give:
\begin{align}
\nabla^2V + \frac{\partial(\nabla\cdot\vec{A})}{\partial t}
&= -\frac{\rho}{\eps} \\[4pt]
\nabla^2\vec{A} - \muz\eps\frac{\partial^2\vec{A}}{\partial t^2}
- \nabla\!\left(\nabla\cdot\vec{A}+\muz\eps\frac{\partial V}{\partial t}\right)
&= -\muz\vec{J}
\end{align}
