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


Question: Setting up form table checkboxes
Answer: Drupal has a built-in tableselect element type for situations just like this, it would be a lot easier to use that instead:

$disputes = array(
  1 =&gt; array('Joe', 'Smith'),
  2 =&gt; array('Susan', 'Doe'),
  3 =&gt; array('John', 'Doe')
);

$header = array(t('First Name'), t('Last Name'));

$form['checkboxes'] = array(
  '#type' =&gt; 'tableselect',
  '#header' =&gt; $header,
  '#options' =&gt; $disputes,
  '#empty' =&gt; 'No users were found.'
);


The above code produces this table:



In your form submit function you can get the selected user ids with code like this:

$selected_uids = array_filter($form_state['values']['checkboxes']);


EDIT

The above is for Drupal 7, if you're using Drupal 6 you can use the same code but you need to install the Elements module.