$$ \partial _\mu \partial ^\mu = \frac{1}{c^2}\frac{\partial ^2}{\partial t^2} -\nabla^2 = \square^2 $$
🗒️ Note: since $\square^2$ is a product of covariant and contravariant it is Lorentz invariant
🧠 Remember: we have the following wave equations
$$ \begin{aligned} \square ^2 \left ( \frac{V}{c} \right )&=\frac{\rho}{c\epsilon_0} = \mu_0 (\rho c) \\ \square ^2 \vec A&= \mu_0 \vec j
\end{aligned} $$
👈 LHS: $\square^2$ Lorentz invariant times components of a $4$-vector
👉 RHS: $\mu_0$ constant times components of $j^\mu$
Thus we define the potential 4-vector $A^\mu$
$$ A^\mu \equiv \left ( \frac{V}{c}, \vec A \right ) $$
This allows us to write the wave equation as
$$ \boxed{\square ^2 A^\mu = \mu_0 j^\mu} $$
🗒️ Note: $\square^2,\mu_0,j^\mu$ are Lorentz invariant, constants or 4-vector meaning that $A^\mu$ is a 4-vector
⚙️ Properties:
- Lorentz transformation: $A'^\mu = \Lambda ^\mu _\nu A^\nu$
- Magnitude: $A^\mu A_\mu = \frac{V^2}{c^2} - |\vec{A}|^2$
🧠 Remember: the Lorenz gauge condition
$$ \frac{1}{c} \frac{\partial }{\partial t} \left [ \frac{V}{c} \right ]+ \vec \nabla \cdot \vec A = 0 $$
Which we can write in 4-vector form simply as
$$ \boxed{\partial _\mu A^\mu = 0} $$
🗒️ Notes: