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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.

Choose an orthonormal basis with $\mathbf{n}={\mathbf{e}}_3$. A half-turn about $\mathbf{n}$ kills all mixed axial-transverse components. Commutation with a quarter-turn makes the transverse block a scalar identity matrix plus an antisymmetric matrix. The resulting oriented-axis invariant Cartesian tensor, in coordinate-free form, iswith real constants $a,d,h$. The Kronecker delta, the outer product $n_i{n}_j$, and the displayed Levi-Civita symbol contraction are invariant under every proper rotation about $\mathbf{n}$, so this form is also sufficient. In the chosen basis,The antisymmetric part must be retained unless further symmetry is imposed.

Decompose the electric field as $\mathbf{E}={\mathbf{E}}_{\perp }+{\mathbf{E}}_{\parallel }$, where ${\mathbf{E}}_{\parallel }=(\mathbf{n}\cdot \mathbf{E})\mathbf{n}$. The electric current density isFor a nonzero real electric field, the alignment of current and field in an axisymmetric conductor gives the complete classification of collinearity:

For completeness, if ${\mathbf{E}}_{\perp }\ne 0$ and $\mathbf{J}=\lambda \mathbf{E}$, taking the dot product with $\mathbf{n}\times {\mathbf{E}}_{\perp }$ forces $h=0$. The transverse equation then gives $\lambda =a$, while a nonzero axial component gives $\lambda =d$. This proves there are no missing directions. Zero current is included as the zero scalar multiple; if “parallel” requires a direction for $\mathbf{J}$, exclude cases where that scalar is zero. For a strictly anisotropic crystal the last possibility is absent.