Show that the current in a cylindrical conductor is uniformly distributed across its cross section Using relevant equations for E and J, show that the current in a steady current I in a cylindrical conductor with uniform conductivity $\sigma$ is uniformly distributed across its cross-section. 
I think the relevant equations are the divergence of the E field from the Maxwell equations and $\sigma$E=J but calculating the divergence of J using the symmetry of the problem doesn't seem to work at all. I think I might be confused about the definition of the variables or using something I should be. 
 A: By the steady state assumption and cylindrical symmetry, both the potential $V$ and the current per unit area $j_z$ are functions only of the radial distance $\rho$ and the distance along the cylinder $z$, and not of the angle $\phi$.  And since there can be (in the steady state) no current perpendicular to the center of the cylinder,  the potential cannot have a non-zero $\rho$ derivative; thus
$V(\rho,z) = V(z)$.
Now let the current $j_z$ integrated over in a ring of radius $r$ to $r+\delta r$ 
be $I(z,r,\delta_r)$. Ohm's law $\delta V_{ab} = j_ab R_ab$ applied to points in the ring at $z$ and the ring at $z+\delta z$ tells us that 
$$
j_z(z, \rho) \sigma \delta z= V(z+\delta z,\rho) - V(z,\rho)
$$
where $\sigma$ is the resistivity of the wire.  This can be integrated over the area $A$ of the ring:
$$
I(z,r,\delta r) = \sigma \int_{\rho = r}^{r+\delta r} j_z(z,\rho) d\rho  = A [V(z+\delta z,\rho) - V(z,\rho) ]
$$
Taking the limit as $\delta z \to 0$,
$$
\frac{\partial}{\partial z}  \sigma \int_{\rho = r}^{r+\delta r}j_z(z,\rho) d\rho = \frac{\partial V(r,z)}{\partial z}
$$
Now comes the first key point: In the ring from $r - \Delta r$ to $r$, with $r - \Delta r$ chosen such that the (cross section) area from $r - \Delta r$ to $r$ is the same as the area from $r$ to $r + \delta r$, there is no radial current between the two rings so $V(r,z)$ cannot be different for the two rings.  And this is true for the two rings at $z+\delta z$ as well, so the $z$ derivative of $V$ has to be the same whether you are looking at ring one or ring two.
The second key point is that 
that above relation holds for the second ring as well:
$$
\frac{\partial}{\partial z}  \sigma \int_{\rho = r-\Delta r}^{r}j_z(z,\rho) d\rho = \frac{\partial V(r,z)}{\partial z}
$$
So
$$
\frac{\partial}{\partial z}  \sigma \int_{\rho = r-\Delta r}^{r}j_z(z,\rho) d\rho =\frac{\partial}{\partial z}  \sigma \int_{\rho = r}^{r+\delta r}j_z(z,\rho) d\rho \\
I(z,r-\Delta r)=I(z,r,\delta r) \\
\frac{d}{dr} I(z,r-\Delta r) = 0
$$
and this says that the current through a circular ring cross section of the wire must be  dependant only on the area of the ring.
Finally, one can obtain a cross section of arbitrary shape by superposing rings and parts of rings.  By circular symmetry the current passing through an arc of a ring of angle $\theta$ will be  $I(z,r-\Delta r) \frac{\theta}{2\pi}$.  And the area of such a shape is obtained by integrating the area of each ring times that same $\frac{\theta}{2\pi}$.  So the current equallity for equal area holds for arbitrary cross-section shapes.
