Lets try to re-create maxwell equations
🐌 Guess: here we guess that the following solution is true and we will verify it later
$$ \boxed{\partial _\mu F^{\mu \nu}=\mu_0 j^\nu} $$
For $\nu =0$ we have
$$ \begin{aligned} \partial _\mu F^{\mu 0} &=0+\frac{\partial }{\partial x^1} \left ( \frac{E_1}{c} \right )+ \frac{\partial }{\partial x^2} \left ( \frac{E_2}{c} \right )+\frac{\partial }{\partial x^3} \left ( \frac{E_3}{c} \right ) \\ &=\vec \nabla\cdot \left ( \frac{\vec E}{c} \right )=\mu_0 j^0=\mu_0 c \rho
\end{aligned} $$
Using $\mu_0 c^2= \frac{1}{\epsilon_0}$ we recognize Gauss’s law
$$ \vec \nabla \cdot \vec E= \frac{\rho}{\epsilon_0} $$
For $\nu=1,2,3$
$$ \begin{aligned} \partial _\mu F^{\mu i} & = \frac 1c \frac{\partial}{\partial t} \left ( -\frac{E_1}{c} \right )+0 + \frac{\partial }{\partial x^2} (B_3)+\frac{\partial}{\partial x^3} (-B_2) \\ &=\left [ -\frac{1}{c^2} \frac{\partial \vec E}{\partial t} + \vec \nabla \times \vec B \right ]1=\mu_0 j^1 \\ \partial\mu F^{\mu2} &=\frac 1c \frac{\partial}{\partial t} \left ( -\frac{E_2}{c}\right )+\frac{\partial}{\partial x^1} (-B_3) + \frac{\partial}{\partial x^3} (B_1) \\ &=\left [ -\frac 1 {c^2} \frac{\partial \vec E}{\partial t} + \vec \nabla \times \vec B \right ]2=\mu_0 j^2 \\ \partial\mu F^{\mu3} &=\frac 1c \frac{\partial}{\partial t} \left ( -\frac{E_3}{c}\right )+\frac{\partial}{\partial x^1} (B_2) + \frac{\partial}{\partial x^2} (-B_1) \\ &=\left [ -\frac 1 {c^2} \frac{\partial \vec E}{\partial t} + \vec \nabla \times \vec B \right ]_3=\mu_0 j^3 \\
\end{aligned} $$
We can rearrange this so that we recognize Faraday’s correction to Ampere law
$$ \vec \nabla \times \vec B =\mu_0 \vec j + \epsilon_0 \mu_0 \frac{\partial \vec E}{\partial t} $$
Going back to our equation and using the Lorenz gauge ie $\partial_\mu A^\mu =0$ we write
$$ \begin{aligned} \mu_0 j^\nu &= \partial _\mu F^{\mu \nu} = \partial \mu (\partial ^\mu A^\nu - \partial ^\nu A^\mu ) \\ &=\partial \mu \partial ^\mu A^\nu - \partial ^\nu \underbrace{(\partial\mu A^\mu )}{=0} \\ &=\square ^2 A^\nu
\end{aligned} $$
which we recognize as the wave equation
Lets try to uncover the other 2 Maxwell equations using
$$ \boxed{\partial ^\mu F^ {\nu \lambda}+\partial ^{\lambda } F^{\mu \nu} + \partial ^{\nu} F^{\lambda \mu}=0} $$
For $\mu,\nu,\lambda=1,2,3$
$$ \begin{aligned} 0&= \partial ^1 F^{23} + \partial^3 F^{12} + \partial ^2 F^{31} \\ &=-\left [ \frac{\partial }{\partial x^1} (-B_1)+\frac{\partial}{\partial x^3}(-B_3) + \frac{\partial}{\partial x^2} (-B_2) \right ]=\vec \nabla \cdot \vec B
\end{aligned} $$
we recognise the no magnetic monopole condition
$$ \vec \nabla \cdot \vec B =0 $$
For $\mu,\nu,\lambda = 0,i,j$ we have
$$ \begin{aligned} 0&= \partial ^0 F^{ij} + \partial^j F^{0i} + \partial ^i F^{j0} \\ &=\frac{1}{c}\frac{\partial }{\partial t}(-\epsilon_{ijk} B_k)+\left ( -\frac{\partial}{\partial x^j} \right )\left ( - \frac{E_i}{c} \right )+\left ( -\frac{\partial}{\partial x^i} \right )\left ( \frac{E_j}{c} \right ) \\ &=\frac{-\epsilon_{ijk}}{c}\frac{\partial B_k}{\partial t} -\frac{\epsilon_{ijk}}{c} [\vec \nabla \times \vec E]_k \\ &=-\frac{\partial _{ijk}}{c} \left [ \frac{\partial \vec B}{\partial t}+\vec \nabla \times \vec E \right ]_k
\end{aligned} $$
doing the same for $\nu=0$ and $\lambda =0$ gives us Faraday’s law
$$ \frac{\partial \vec B}{\partial t} + \vec \nabla \times \vec E=0 $$
Lets try to get something analogous to $F=ma$ but for electromagnetism