Pre-Quantum Electrodynamics

\begin{align} (i)~~ &{\boldsymbol \nabla} \cdot {\bf E} = \frac{\rho}{\varepsilon_0}, &(iii)~~ {\boldsymbol \nabla} \times {\bf E} + \frac{\partial {\bf B}}{\partial t} &= 0, \nonumber \\ (ii)~~ &{\boldsymbol \nabla} \cdot {\bf B} = 0, &(iv)~~ {\boldsymbol \nabla} \times {\bf B} - \mu_0 \varepsilon_0 \frac{\partial {\bf E}}{\partial t} &= \mu_0 {\boldsymbol J}. \end{align}

Solving these for generale time-dependent sources \(\rho({\boldsymbol r}, t)\) and \({\boldsymbol J} ({\boldsymbol r}, t)\) is not an easy task.

Useful strategy: represent fields in terms of potentials.

Easiest:

Putting this into Faraday's law gives \[ {\boldsymbol \nabla} \times \left({\boldsymbol E} + \frac{\partial {\boldsymbol A}}{\partial t} \right) = 0 \] so this can be written as the gradient of a scalar (by choice: \(-{\boldsymbol \nabla} V\)) so we get

Using this potential representation for \({\boldsymbol E}\) and \({\boldsymbol B}\) automatically fulfills the two homogeneous Maxwell equations. For the inhomogeneous equations, substituting (\ref{eq:E_from_Potentials}) into Gauss's law gives

whereas Amp{\`ere}-Maxwell becomes \[ {\boldsymbol \nabla} \times \left({\boldsymbol \nabla} {\boldsymbol A}\right) = \mu_0 {\boldsymbol J} - \mu_0 \varepsilon_0 {\boldsymbol \nabla} \left(\frac{\partial V}{\partial t}\right) - \mu_0 \varepsilon_0 \frac{\partial^2 {\boldsymbol A}}{\partial t^2} \] which becomes after simple rearrangement and use of the identity \({\boldsymbol \nabla} \times \left({\boldsymbol \nabla} \times {\boldsymbol A}\right) = {\boldsymbol \nabla} ({\boldsymbol \nabla} \cdot {\boldsymbol A}) - {\boldsymbol \nabla}^2 {\boldsymbol A}\),