Where $ F_{\mu \nu}= \partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$, $A$ being the four-potential.
I'm interested in whether or not it's possible to show from this: $$ \frac{1}{\epsilon_0 \mu_0} E_{xx} = E_{tt}$$
Lower index meaning partial derivative. Sticking to (1+1) dimensions for simplicity.
This really does need all three space dimensions. I'll set $\mu_0=\epsilon_0=1$ to tidy up; feel free to restore them. Since $\partial_\mu F^{\mu\nu}=0$, $\nabla\cdot E=0$ and $\nabla\times B=\dot{E}$. Since $\partial_\mu\tilde{F}^{\mu\nu}=0$, $\nabla\cdot B=0$ and $\nabla\times E=-\dot{B}$. So$$\ddot{E}=\nabla\times\dot{B}=-\nabla\times\nabla\times E=\Delta E.$$
