Pre-Quantum Electrodynamics

We've encountered:

\begin{align} (i) &{\boldsymbol \nabla} \cdot {\bf E} = \frac{\rho}{\varepsilon_0}, \hspace{1cm} &\mbox{Gauss}, \nonumber \\ (ii) &{\boldsymbol \nabla} \cdot {\bf B} = 0, \hspace{1cm} &\mbox{anonymous} \nonumber \\ (iii) &{\boldsymbol \nabla} \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}, \hspace{1cm} &\mbox{Faraday}, \nonumber \\ (iv) &{\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J}, \hspace{1cm} &\mbox{Ampère}. \end{align}

Fatal inconsistency: div of curl must always vanish. Check on \((iii)\): \[ {\boldsymbol \nabla} \cdot ({\boldsymbol \nabla} \times {\bf E}) = {\boldsymbol \nabla} \cdot \left( -\frac{\partial {\bf B}}{\partial t} \right) = -\frac{\partial}{\partial t} ({\boldsymbol \nabla} \cdot {\bf B}) = 0. \] But: try same with \((iv)\): \[ {\boldsymbol \nabla} \cdot ({\boldsymbol \nabla} \times {\bf B}) = \mu_0 {\boldsymbol \nabla} \cdot {\bf J} \label{Gr(7.35)} \] LHS must be zero, but RHS is not zero for non-steady currents. Cannot be right !

Other way of seeing that Ampère's law must fail for non-steady currents: suppose we're charging a capacitor. In integral form, \[ \oint {\bf B} \cdot d{\bf l} = \mu_0 I_{\mbox{enc}}. \] But the surface can either cut the charging wire, or not (by going 'around' the capacitor plate). So for non-steady currents, the 'current enclosed by a loop' is ill-defined.