# Can Gauss' and Ampere's Laws be written in terms of the divergence of an energy four-vector?

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 $\partial_{\mu}\mathcal{E}^{\mu}=0$.

• Are you not familiar with the special relativistic form of the EM fields, the faraday bivector $F$? Of how Maxwell's equations in vacuum can be written $\nabla \wedge F = -\mu_0 J$ and $\nabla \cdot F = 0$? Nov 7, 2014 at 6:19

The covariant formulation of EM is precisely this. The formulation as a gauge theory also does this. ($c = 1$ in the following)

Given the $E$- and $B$-fields as spatial three-vectors in some frame, we construct the antisymmetric field strength tensor as (roman indices are spatial indices, summation over repeated indices implied)

$$F^{0i} := E^i \; \text{and} \; F^{ij} := \epsilon^{ijk}B^k$$

Using the behaviour of the fields under Lorentz transformations, one can indeed see that this $F_{\mu\nu}$ is a proper tensor on Minkowski space. Similarly, we construct a four-current out of the classical current $\vec j$ and the charge density $\rho$ by

$$j^i = j^i \; \text{and} \; j^0 = \rho$$

$$\partial_\mu F^{\mu\nu} = \mu_0j^\nu \; \text{and} \; \partial_\mu \epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma} = 0$$
This is not yet the most abstract way to present this. Using the language of differential forms, we have forms $F = F_{\mu\nu}\mathrm{d}x^\mu \wedge \mathrm{d}x^\nu$ and $j = j_\mu \mathrm{d}x^\mu$. Maxwell's laws now look, in their most concise form, like this:
$$\mathrm{d}F = 0 \; \text{and} \; \mathrm{d}\star F = \star j$$
where the star is the Hodge dual. One advantage of this language is that one may conclude that there is a 1-form $A$ with $F = \mathrm{d}A$ on any contractible subset of our space. More precisely, it exists if the second deRham cohomology vanishes.
The gauge theory description of electrodynamics starts from gluing together these $A$ which always exist locally to get a globally defined gauge potential $A$.