Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 3 / 12A (Vector Calculus)

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.

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.

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 ${\alpha }^2(\alpha +\gamma )\ne 0$.