There is a problem in Griffiths "Introduction to Electrodynamics" whose available "solutions" online all seem to miss the point entirely. The exact problem I'm talking about is problem 5.12 (4th edition), or problem 5.38 (3rd edition).
I will recapitulate the problem here:
Since parallel currents attract each other, it may have occurred to you that the current within a single wire, entirely due to the mobile electrons, must be concentrated in a very thin stream along the wire. Yet, in practice, the current typically distributes itself quite uniformly over the wire. How do you account for this?
Okay, so the problem so far seems to be asking us to show that indeed, contrary to our initial guess, the mobile electrons spread out uniformly across the cross-section of the wire. But then the problem goes on to say the following:
If the uniformly distributed positive charges (charge-density $\rho_+$) are "nailed down", and the negative charges (charge-density $\rho_-$) move at speed $v$ (and none of these depends on the distance from the axis), show that $\rho_-=-\rho_+\gamma^2$, where $\gamma=1/\sqrt{1-(v/c)^2}$.
What?! In the boldfaced-text, Griffiths (seemingly) defeated the purpose of this problem!
My question is, can it be shown classically that, according to Maxwell's equations (and symmetry principles), the electron charge density and/or the velocity of those electrons is independent of the radial distance from the central-axis of the wire? I have tried to show it, but I can only end up with the following equation:
$$\frac{v(s)}{c^2}\int_0^s \rho(s')v(s')s'ds'=\int_0^s(\rho_++\rho_-(s'))s'ds'$$
From here, if I assume $\rho_-(s)$ and $v(s)$ are constant, then the result $\rho_-=-\rho_+\gamma^2$ easily follows.
NOTE: This is not a homework problem for a class of mine, but even if it was I think question is still valid.
EDIT:
If we assume the velocity is independent of the radial distance from the center, then we can show that this implies $\rho_-(s)=\rho_-$ (constant).
$$ \begin{align*} \frac{v(s)}{c^2}\int_0^s \rho(s')v s'ds'&=\int_0^s(\rho_++\rho_-(s'))s'ds'\\ \implies0&=\int_0^s\left(\rho_++\left(1-\frac{v^2}{c^2}\right)\rho_-(s')\right)2\pi s' ds'~~~~\textrm{(for all }s\textrm{)} \\ \implies 0&= \rho_++\left(1-\frac{v^2}{c^2}\right)\rho_-(s)\\ &\implies \boxed{\rho_-(s)=-\gamma^2\rho_+ }~~~~\textrm{(constant!)} \end{align*}$$