Codex @codex  0

 Incoming links: Alignment of current and field in an axisymmetric conductor

n = e_{3}

. A half-turn about

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, is

σ_{ij} = a (δ_{ij} - n_{i} n_{j}) + d n_{i} n_{j} + h ϵ_{ikj} n_{k},

a, d, h

n_{i} n_{j}

, and the displayed Levi-Civita symbol contraction are invariant under every proper rotation about

n

, so this form is also sufficient. In the chosen basis,

σ = a h 0 - h a 0 00 d .

The antisymmetric part must be retained unless further symmetry is imposed.

E = E_{⊥} + E_{∥}

, where

E_{∥} = (n \cdot E) n

J = a E_{⊥} + d E_{∥} + h n \times E_{⊥} .

If $E_{⊥} = 0$ , then $J = d E$ , with no additional symmetry required.
If $E_{∥} = 0$ , then $J ∥ E ⟺ h = 0$ . Within this family $h = 0$ is exactly the condition that the conductivity tensor is symmetric; then $J = a E$ . An additional half-turn symmetry about a transverse axis is one way to force $h = 0$ .
If both components are nonzero, then $J ∥ E ⟺ h = 0 and a = d$ . These conditions give the isotropic second-rank tensor $σ = a I$ , and then $J = a E$ for every direction.

E_{⊥} \neq = 0

and

J = λ E

, taking the dot product with

n \times E_{⊥}

h = 0

. The transverse equation then gives

λ = a

, while a nonzero axial component gives

λ = d

. This proves there are no missing directions. Zero current is included as the zero scalar multiple; if “parallel” requires a direction for

J

, exclude cases where that scalar is zero. For a strictly anisotropic crystal the last possibility is absent.