Confused with Electric field and current density directions The typical relationship equation $\mathbf{J}=\sigma \mathbf{E}$ implies that the electric field and the current density have the same direction. However, in a coaxial cable the Efield has a radial direction due to the opposite charge in the two conductors and the current density $\mathbf{J}$ has the direction along the wire, thus they are perpendicular. Why is this the case? Doesn't the Efield and $\mathbf{J}$ direction always coincide?
 A: 
However, in a coaxial cable the Efield has a radial direction due to the opposite charge in the two conductors and the current density J has the direction along the wire, thus they are perpendicular.

You're comparing the $\vec{J}$ in the conductive part of the coaxial structure with the $\vec{E}$ in the dielectric part of the structure.
If you used a crappy dielectric that allowed substantial leakage current, then leakage current would flow between the outer and inner conductors and the $\vec{J}$ in the dielectric would indeed be in the same direction as the $\vec{E}$.
When exciting the coax with an AC signal the $\vec{J}$ in the conductors doesn't necessarily correspond (either in magnitude or direction) to the $\vec{E}$ because not only electric field but also the magnetic field is significant in this system.
As others have pointed out, even in actual quasi-electrostatic scenarios in conductive materials, $\vec{J}$ and $\vec{E}$ don't necessarily point in the same direction and $\sigma$ might have to be considered as a tensor rather than a scalar. But that doesn't relate to the example you asked about.
A: No $\boldsymbol{\mathrm{E}}$ and $\boldsymbol{\mathrm{J}}$ do not always point in the same direction. The $\boldsymbol{\sigma}$ is in general a tensor (the conductivity tensor), which you can think of as a matrix relating two vectors (with certain symmetry properties).
There's a good overview on the associated Wikipedia page.
A: The relation
\begin{equation}
\mathbf{J} = \sigma \mathbf{E} 
\end{equation}
is a simplification. More generally the conductivity $\sigma$ is a matrix (more precisely a tensor), hence the relation (using Einstein notation) is
\begin{equation}
\mathbf{J}_i = \sigma_{ij} \mathbf{E}_j 
\end{equation}
For the case of a coaxial cable, as you said, you have a component of the electric field in the radial direction, which, however, does not produce any current because of the insulator between the inner and outer conductor. On the other hand, you also have a component of the field along the cable, assuming that you are measuring a non-zero signal.  The consequent voltage drop in turn produces the current along the cable.
A: $\vec J=\sigma\vec E$ holds (approximately) in materials.  In the case of your example, the $\vec E$ field related to $\vec J$ would be the field resulting from the voltage difference that drives the current flowing in your wire: the size of the current density would be proportional to this external applied $\vec E$ resulting from the potential difference.
In a coaxial cable, one usually computes $\vec E$ for static (or quasistatic) charge distributions, and in this setup $\vec E$ is perpendicular (as it would in wire with constant static charge distribution).  In fact, in a (perfect) conductor, there is no $\vec E$ inside a perfect conductor in the static limit, so the idea that you'd have an uniform current density $\vec J$ related to $\vec E$ inside a conductor shows this is not an electrostatics problem.
The $\vec E$ between the conductors in a coax cable has nothing to do with the current density and depends only on the (quasistatic) charge density on the surface of the conductor.
