Alignment of current and field in an axisymmetric conductor ($\sigma =a(I-n{n}^T)+dn{n}^T+h[n\times ]$)

For a real conductivity tensor invariant under every rotation about a unit vector $n$, the electric current density is $J=a{E}_{\perp }+d{E}_{\parallel }+hn\times E_{\perp }$. Its collinearity with a nonzero real electric field $E$ has three cases. Axial fields are always mapped to an axial current. Purely transverse fields give a collinear current exactly when $h=0$, equivalently when this tensor is symmetric. Fields with nonzero axial and transverse parts require both $h=0$ and $a=d$, making the tensor isotropic. The zero vector is included as a scalar multiple; requiring a nonzero current also requires the relevant scalar conductivity to be nonzero.

To prove necessity, take the dot product of $J=\lambda E$ with $n\times E_{\perp }$. If $E_{\perp }\ne 0$ this gives $h=0$; the transverse and axial equations then respectively give $\lambda =a$ and, when $E_{\parallel }\ne 0$, $\lambda =d$. Reversing these steps proves sufficiency. Rotations about the axis alone permit $h\ne 0$; symmetry under a transverse half-turn would eliminate this antisymmetric term.