determining electrostatic field using only symmetries As an exercise, I'm trying to (rigorously) determine as much as possible about the electrostatic field due to a infinite line of charge (along the z-axis) without using Maxwell's equations or any of their Green's functions (i.e. Coulomb's law etc.)
So far I've managed to show that $\textbf{E} = f(r)\hat{r}$ using a reflection on the yz-plane, a reflection on the xy-plane, and azimuthal rotations. The only "axioms" I've had to employ are the Lorentz Force Law (to show that $\textbf{E}$ transforms as a vector) and that $\textbf{E}$ is a function of charge and position.
This makes me wonder if I can go any farther with just symmetries or similar fundamental arguments to show that $f(r) \propto \frac{1}{r} $ without just using Guass' Law and calling it a day.
 A: It's possible to make such an argument without using Gauss's law.
Let's compute electric potential rather than electric field. As we know, in Electrostatics, electric potential is observable: it can be defined as a difference energy divided by charge of a small particle in an electric field. After such symmetric arguments you have made, using Dimensional analysis one can show potential $\phi$ is in this form:
$$\phi=\lambda/\epsilon_0*f(a/r)$$ where $\lambda$ is linear charge density, $a$ is an arbitrary positive constant, and $r$ is the distance between the point and the line. $f$ is a function we want to derive it.
Now let use the fact that the energy of an small particle in an electric field is additive and conserved. Choose two arbitrary possitive numbers called $x$ and $y$. Let $r_1$ and $r_2$ denote these values:
$$r_1=a/x$$ $$r_2=r_1/y$$
Let a small particle goes from point $r_1$ to $r_2$ and use the fact its energy is conserved. So:
$$f(a/r_2)=f(a/r_1)+f(r_1/r_2)$$
Hence for any two possitive numbers we have:
$$f(xy)=f(x)+f(y)$$
As a result, $f(1)=0$. So $a$ is the zero of potential.
Let: $$ x=\exp(t) $$ $$ y=\exp(s)$$ define $h$ as below: $$ h(t):=f[\exp(t)] $$ for all $t>=0$. So: $$h(t+s)=h(t)+h(s)$$ Now let us assume we know electric potential is bounded (finite) in an closed interval. It is absolutely intuitive because we are expected the energy difference between two points with finite distance to be finite. By this assumption, one can show $h$ should be $h(t)=\alpha t$ for all possitive $t$ which $\alpha$ is a constant. It is easy to show this. It is known as Cauchy's functional equation. So: $$ f(x)=\alpha \log(x)$$ for all possitive $x$. Hence the electric potential is:
$$\phi=\alpha\lambda/\epsilon_0 \log(a/r)$$ one can use this to calculate electric field.
