Pre-Quantum Electrodynamics

• Gr 10.1.3

The Coulomb Gauge is specified by taking \[ {\boldsymbol \nabla} \cdot {\boldsymbol A} = 0 \] in which case Lapphi becomes simply \[ {\boldsymbol \nabla}^2 \phi = -\frac{\rho}{\varepsilon_0} \] i.e. Poisson's equation, whose solution we already know: \[ \phi({\boldsymbol r}, t) = \frac{1}{4\pi \varepsilon_0} \int d\tau' \frac{\rho({\boldsymbol r}^\prime, t)}{|{\boldsymbol r} - {\boldsymbol r}^\prime|} \] Note that this is an equal-time relationship (it does not mean instantaneous action at a distance, since \(\phi\) by itself is not physically measurable).

Although Gauss's law looks nice in the Coulomb gauge, Ampère-Maxwell does not: \[ {\boldsymbol \nabla}^2 {\boldsymbol A} - \mu_0 \varepsilon_0 \frac{\partial^2 {\boldsymbol A}}{\partial t^2} = -\mu_0 {\boldsymbol J} + \mu_0 \varepsilon_0 {\boldsymbol \nabla} \left( \frac{\partial \phi}{\partial t} \right). \]