Coulomb's law from Maxwell's equations.

Question

I'm trying to find general sufficient additional conditions to derive Coulomb equation for the electric field generated by a steady point charge in free space from Maxwell equations in said conditions.

I know that a way to do this is assuming that the solution of Maxwell equations must have spherical symmetry due to the disposition of charges (a single point charge infact) having such a degree of symmetry.

The thing is, how can I prove that the solution of maxwell equations has to be symmetrical if I don't know in advance the equation of the electric field? And if the answer is "because the space is isotropic", how can I mathematically write such a condition?

Infact, if I consider Maxwell's equations (I write just the first two for the case we are considering) I get

\begin{align} \int_{\partial \Omega} E \cdot dS &= \delta(\Omega)q \\ \int_\gamma E \cdot ds =& 0 \quad \text{if} \quad \partial\gamma=\emptyset \end{align}

($\delta$ is $1$ if the carge is inside of $\Omega$, $0$ if it is outside and $q$ is the value of the point charge) so if $E$ solves them, also $E+K$ does for every constant field $K$.

I get that space should not "choose" any direction for $K$, but how to mathematically state it? There is clearly something missing here.

I thought of adding as a condition that

\begin{equation} \lim_{|x|\to +\infty} E = 0 \end{equation}

and this infact eliminates the possibility of adding a costant field, but I think I'm far from proving uniqueness of the solution.

Also another thing that bugged me, is the fact that it is not always the case that $\lim_{|x|\to +\infty} E = 0$, since if we impose this condition for the solution in the case of a homogeneus charged infinite plane, we don't get a solution at all.

Answer 1

For the spherically symmetric case, you can constrain it by forcing $|E|$ to be a function of only distance $|r|$ from the point charge.

Answer 2

I'm answering my own question because I think that maybe I've found a way to put space isotropy in mathematical therms, and infact (in this case) it solves all my problems, so I'd really like an opinion on this.

I think this could be extended with little effort to moving distributions, when I will have studied the subject in greater detail I think I will be able to update this answer.

I'm therefore assuming constitutive relation of the form:

\begin{equation} D = D(x, E, H), \; B=B(x, E ,H), \; J = 0 \end{equation} and the measure with sign $Q$ independent of time.

The condition I think I'm searcing is:

With $A$ being the domain (open connected) we are solving the problem in, let $f: A \to A$ of the form $f(x) = x_0 + \Lambda(x-x_0)$ bijective, where $\Lambda$ is linear and $\det(\Lambda) = \pm1$. If $Q(\Omega) = Q(f(\Omega))$ for every $Q$-measurable set, and we have connected closed bounded 2 and 3 manifolds $\Omega$ and $S$ rispectively for wich integrals are defined, then the solution $(E,H)$ to Maxwell's equations

\begin{align} &\oint_{\partial \Omega} D \cdot dS = Q(\Omega) \\ \oint_{\partial S}& E \cdot ds = - \frac{d}{dt} \int_S D \cdot dS \\ \frac{d}{dt}&\int_S D \cdot dS = \oint_{\partial S} H \cdot ds \\ &\oint_S D \cdot dS = 0 \end{align}

is such that $(E, H)(t,f(x)) = (\Lambda E, \Lambda B)(t,x).$

EDIT: I think I have to require also $D(f(x), \Lambda E, \Lambda H)= \Lambda D(x, E,H), \; B(f(x), \Lambda E, \Lambda H)= \Lambda B(x, E,H)$. END EDIT

For example in the case of a point charge in free space fixed in the origin, we get constitutive relations

\begin{equation} D = \varepsilon_0 E, \; B=\mu_0H, \; J = 0 \end{equation}

and $Q(\Omega) = q\delta_0(\Omega) = q\delta_0(f(\Omega))$ whenever $f$ is a reflection or a rotation, hence with little effort we get spherical symmetry of both $E$ and $B$, and then $E(t,x)=\frac{q}{4\pi\varepsilon_0|x|^3}x, B(t,x)=0$, as desired.

Let me know if you think this could be good.