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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.)

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  • $\begingroup$ I think the article is refering to iterative numerical solutions, which of necessity are not exact. It suggests that artefacts of the algorithm can become large. $\endgroup$ Commented Apr 2, 2020 at 9:13
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    $\begingroup$ In the textbook Novozhilov, Electrodynamics, chapter 1, it is shown that either one can take Maxwell's four equations as the starting point, or one can take the continuity equation and the two curl Maxwell equations, however one must ALSO still take the two divergence Maxwell equations as assumptions, however now one only need assume they must hold at one instant of time, and one can use the other equations to then show they will hold at all later times. $\endgroup$
    – bolbteppa
    Commented Apr 2, 2020 at 19:51

3 Answers 3

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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: $$C(\mathbf{r}) =0.$$

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  • $\begingroup$ So, the information of Gauss' laws is not contained in the curls. Then, the overdetermination of Maxwell's equations is not true, isn't it? I read one of the sources of the wiki article and it has the following assumption in the case of the time derivative of $\nabla\cdot\mathbf{B}$: "If ever in its past history the field has vanished, this constant must be zero and, since one may reasonably suppose that the initial generation of the field was at a time not infinitely remote", concluding $\nabla\cdot\mathbf{B}=0$. Are these assumptions "realistic"? $\endgroup$
    – Verktaj
    Commented Apr 2, 2020 at 18:20
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    $\begingroup$ Actually, the Maxwell equations are underdetermined, since they lack the description of how the charge and the currents are affected by the fields, so called material equations. Also, one need not justify these equations - they are postulated phenomenologically, like gravity law or Newton laws. $\endgroup$
    – Roger V.
    Commented Apr 2, 2020 at 19:28
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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).

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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.

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