In the first 20 minutes of this video, Susskind derives the continuity equation for charge conservation:
$$\dot{\rho}+\nabla\cdot\vec{J}=0$$
(Where $\vec{J}=\frac{\partial\dot{q}^m}{\partial A^m} \:; \:\:\:m=1,2,3 \:\:$ is the current density, or as Susskind calls it, just the current.)
(And $m=1,2,3$ indexes the x,y,z space components)
And then shows that this continuity equation can be written more compactly as the divergence of a current four-vector:
$$\dot{\rho}+\nabla\cdot\vec{J}=\partial_{\mu}\mathcal{J}^{\mu}=0\:; \:\:\:\mu=0,1,2,3 \:\:$$
(Where $\mu=0$ indexes the time component. For the rest of this post, a curly letter indicates a four vector, whereas a non-curly letter with an arrow over it indicates an ordinary vector with three space components.)
That is, he demonstrates that $\dot{\rho}$ can be thought of as the time component of the divergence of the current.
My main question is: if $\rho$ is a time component (i.e. differentiates w.r.t. to time), then doesn't Gauss' Law also have a more compact expression as the divergence of an Energy four-vector? Here is the reasoning:
When the continuity equation is derived by taking the divergence of Ampere's Law we see that $\nabla\cdot \dot{E}=\dot{\rho}$. Well, if $\rho$ is a time component, then this must mean that $\dot{E}=\rho$. From this it follows that Gauss' Law can be written as the divergence of the Energy four-vector:
$$\nabla\cdot \vec{E}=\rho \:;\:\:\: \vec{E}=(E^1,E^2,E^3)\rightarrow \partial_{\mu}\mathcal{E}^{\mu}=0\:; \:\:\:\mu=0,1,2,3 \:\:$$
And as a corollary does it follow that the divergence of the current four-vector equals the Laplacian of the Energy four-vector?
$$\partial_{\mu}^2\mathcal{E}^{\mu}=\partial_{\mu}\mathcal{J}^{\mu}$$
Reasoning: If $\rho$ is a time component, then the divergence of Gauss' Law gives:
$$\nabla^2\vec{E}=\dot{\rho}$$
...which, since $\dot{E}=\mathcal{E}^{0}=\rho$, can also be written
$$\partial_{\mu}^2\mathcal{E}^{\mu}=0$$
But by Susskind's compact expression of the continuity equation we know that $\partial_{\mu}\mathcal{J}^{\mu}=0$. Therefore the two can be equated: $\partial_{\mu}^2\mathcal{E}^{\mu}=\partial_{\mu}\mathcal{J}^{\mu}$.
And then finally, if this corollary is true, then doesn't it also follow that Ampere's Law has a more compact expression as:
$$\partial_{\mu}\mathcal{E}^{\mu}=\nabla\times\vec{B}$$
Reasoning: By integrating both sides of the corollary you get:
$$\partial_{\mu}\mathcal{E}^{\mu}=\mathcal{J}^{\mu}$$
But the RHS of this can also be written:
$$\mathcal{J}^{\mu}=\rho+\vec{J}$$
...which, since $\rho=\dot{E}$, is also:
$$\mathcal{J}^{\mu}=\dot{E}+\vec{J}$$
But the RHS of this is just Ampere's Law ($\nabla\times\vec{B}=\dot{E}+\vec{J}$). Therefore Ampere's Law has the following compact expression in terms of the divergence of the energy four-vector:
$$\partial_{\mu}\mathcal{E}^{\mu}=\nabla\times\vec{B}$$
From which a desirable property (I think) follows automatically--that the magnetic field is irrotational ($\nabla\times \vec{B}=0 $)--since the compact expression for Gauss' Law is $\partial_{\mu}\mathcal{E}^{\mu}=0$.