Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 3 / 10C / Solution

Use the right-handed rotation matrix which sends the positive $x$ direction to the positive $y$ direction:The Cartesian second-rank tensor transformation law givesHere isotropic tensor means invariant under every proper rotation, as is needed for the rank-three assertion involving the Levi-Civita symbol. A passive coordinate rotation through the same angle uses $R^{\mathsf{T}}$ instead; the invariance conclusions are identical.

Setting $S^{\prime }=S$ forces $S_{11}=S_{22}$, $S_{12}=-S_{21}$, and $S_{13}=S_{23}=S_{31}=S_{32}=0$. No symmetry assumption on $S$ has been made: it now has the formInvariance under a right-handed quarter-turn about the $x$ axis next gives $S_{22}=S_{33}$ and $S_{12}=-S_{13}$, hence $a=c$ and $b=0$. Conversely $R(\lambda I)R^{\mathsf{T}}=\lambda I$ for every orthogonal matrix. The general isotropic second-rank tensor is therefore $S_{ij}=\lambda {\delta }_{ij}$.

An isotropic vector must be fixed by the half-turn about the $z$ axis, so its first two components vanish. The half-turn about the $x$ axis then forces its third component to vanish. Thus the only isotropic vector is zero. The general isotropic third-rank tensor under proper rotations is $A_{ijk}=\lambda {ϵ}_{ijk}$. If invariance under the whole orthogonal group were required of an ordinary rank-three tensor, inversion $R=-I$ would send $A$ to $-A$, forcing $A=0$ instead.