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Electrical conductivity $\sigma$ measures the local response of current density to electric field. In the rest frame of an isotropic Ohmic conductor, the local form of Ohm's law isLarger positive electrical conductivity gives larger current density at the same electric field. It is a bulk material coefficient, distinct from the geometry-dependent electrical conductance of a whole component and from a sheet's surface conductivity. Electrical resistivity is ${\rho }_{\mathrm{el}}=1/\sigma$ in this scalar model. ^[ref].

The local and whole-wire forms fit together particularly neatly. For uniform current in a straight homogeneous wire, with length $L$, wire cross-sectional area $A$, constant electrical conductivity and terminal electric potential difference $V$, the axial electric field has magnitude $E=V/L$. ThusFor the defined electrical resistance example $\sigma =5\times 10^5(\Omega \mathrm{m}{)}^{-1}$, $L=1\mathrm{m}$, $A=10^{-6}{\mathrm{m}}^2$ and $V=5\mathrm{V}$, the local values are $E=5V/m$ and $j=2.5\times 10^{6}A/{\mathrm{m}}^2$; integrating gives $I=2.5\mathrm{A}$. One material law therefore supplies both a spatial description and a terminal relation.

The scalar form requires isotropy. In an anisotropic material the response can instead be a tensor, $J_i={\sum }_j{\sigma }_{ij}{E}_j$, and current density need not be parallel to electric field; uniaxial conductivity tensor is one existing example. These statements concern the material's rest frame. A moving electrical conductor instead uses the moving-conductor Ohm law.

Decompose the electric field into its orthogonal parts $\mathbf{E}={\mathbf{E}}_{\perp }+{\mathbf{E}}_{\parallel }$, where ${\mathbf{E}}_{\parallel }=n(n\cdot \mathbf{E})$. The uniaxial conductivity tensor acts on them asIf $\mathbf{J}=0$, its perpendicular and parallel components each vanish. Since $\alpha \ne 0$ and $\alpha +\gamma \ne 0$, this forces both field components to be zero. The contrapositive givesEquivalently, the tensor has nonzero eigenvalues $\alpha ,\alpha ,\alpha +\gamma$ and determinant ${\alpha }^2(\alpha +\gamma )\ne 0$.

Writing the uniaxial conductivity tensor in vector form gives the electric current densityChoose the plane through the origin perpendicular to $n$:For an electric field in this plane, the anisotropic term vanishes and $\mathbf{J}=\alpha \mathbf{E}$. Thus every nonzero such field is parallel or antiparallel to its current according to the sign of the nonzero $\alpha$. The zero field gives zero current. This establishes collinearity throughout the plane without assuming that the tensor is isotropic.

A Cartesian isotropic tensor is unchanged by every proper rotation matrix. For a rank-two tensor this means $B^{\prime }=RB{R}^{\mathsf{T}}=B$, or $B_{ij}^{\prime }=R_{ik}{R}_{jl}{B}_{kl}=B_{ij}$. No symmetry of $B$ is assumed. For a quarter-turn about the $z$ axis,Comparing entries yields $B_{11}=B_{22}$, $B_{12}=-B_{21}$, $B_{13}=-B_{23}=-B_{13}$, and $B_{31}=-B_{32}=-B_{31}$. In particular,The remaining matrix has the form $\left(\begin{array}{ccc}b& c& 0\\ -c& b& 0\\ 0& 0& d\end{array}\right)$. A quarter-turn about the $x$ axis givesInvariance now forces $c=0$ and $d=b$, proving that an isotropic second-rank tensor isa scalar multiple of the Kronecker delta.

For the final unheaded conductivity identity, let $P=n{n}^{\mathsf{T}}$, the orthogonal projection onto the unit vector $n$. Then $P^2=P$ and the uniaxial conductivity tensor is $\sigma =\alpha I+\gamma P$. The antisymmetric tensor $D$ sends a vector $v$ to $v\times n$. Applying it twice and using the vector triple product givesConsequentlyThis equals $-\sigma =-\alpha I-\gamma P$ exactly when $(\alpha +\gamma )P=0$. Since $P$ has rank one and is nonzero, the requested parameter isAt this value conductivity vanishes along $n$; it is precisely the singular case excluded in part (ii), so there is no conflict with that part's conclusion.