Why do we need Gauss' laws for electricity and magnetism?

Question

The source of an electromagnetic field is a distribution of electric charge, $\rho$, and a current, with current density $\mathbf{J}$. Considering only Faraday's law and Ampere-Maxwell's law: $$ \nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}\qquad\text{and}\qquad\nabla\times\mathbf{B}=\mu_0\mathbf{J}+\frac{1}{c^2}\frac{\partial\mathbf{E}}{\partial t}\tag{1} $$ In an isolated system the total charge cannot change. Thus, we have the continuity equation that is related to conservation of charge: $$ \frac{\partial\rho}{\partial t}=-\nabla\cdot\mathbf{J}\tag{2} $$ From these three equations, if we take the divergence of both equations in $(1)$, and using $(2)$ in the Ampere-Maxwell's law, we can get the two Gauss' laws for electricity and magnetism: $$ \nabla\cdot\mathbf{B}=0\qquad\text{and}\qquad\nabla\cdot\mathbf{E}=\frac{\rho}{\varepsilon_0}\tag{3} $$

Therefore, the assumption of $(1)$ and $(2)$ implies $(3)$. At first glance, it could be said that we only need these three equations. Also, conservation of charge looks like a stronger condition than the two Gauss' laws (it's a conservation law!), but, as the article in Wikipedia says, ignoring Gauss' laws can lead to problems in numerical calculations. This is in conflict with the above discussion, because all the information should be in the first three equations.

So, the question is, what is the information content of the two Gauss' laws? I mean, apart of showing us the sources of electric and magnetic field, there has to be something underlying that requires the divergence of the fields. If no, then, what is the reason of the inherently spurious results in the numerical calculations referred?

(Also, I don't know what type of calculation is referred in the article.)

Answer 1

I don't agree that you get that you obtain the Gauss law using the method proposed. What you obtain instead is $$\frac{\partial\nabla\cdot\mathbf{B}}{\partial t} = 0,\\ \frac{1}{c^2}\frac{\partial\nabla\cdot\mathbf{E}}{\partial t} + \mu_0\nabla\cdot\mathbf{J}= \frac{1}{c^2}\frac{\partial\nabla\cdot\mathbf{E}}{\partial t} - \mu_0\frac{\partial\rho}{\partial t}=0.$$ These equations give you only the rate of change of $\nabla\cdot\mathbf{B}$ and $\nabla\cdot\mathbf{E}$, but not their value, which needs to be defined by time integration and gives you the answer up to a position-dependent constant (whose time derivative is zero). E.g., the Gauss law for the electricity is given now by $$\nabla\cdot\mathbf{E}(\mathbf{r},t) = \frac{1}{\epsilon_0}\rho(\mathbf{r},t) +C(\mathbf{r}).$$ So we do need an additional constraint to specify function $C(\mathbf{r})$, i.e. the Gauss law, which in these terms can be written as: $$C(\mathbf{r}) =0.$$

Answer 2

There is a paper linked to the cited statement at wikipedia. In short the system is actually not overdetermined. The authors report that numerical methods, which ignore the divergence-free conditions lead to inaccurate solutions. They show that they are needed to guarantee the uniqueness of the solutions (you have to take account for the boundary conditions).

Answer 3

This just an explicit example to @vadim's answer: Pick a function $f(\vec x)$, constant in time, such that $\Delta f =5$. Set $\vec B=\vec\nabla f$, $\vec E=\vec J=0$, $\rho=17$. Then Eqns. (1) and (2) are satisfied, buth both equations in (3) are not.