Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 3 / 10C (Vector Calculus)

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.

Both the Levi-Civita symbol and $T$ are invariant under proper rotations. Their tensor contractionis consequently an isotropic vector. The introductory proof shows that it must vanish, so ${ϵ}_{ijk}{T}_{ijkl}=0$.

For the displayed combination of Kronecker deltas, the three terms areIn each term the tensor contraction repeats two indices of the Levi-Civita symbol; each term is zero. This verifies the identity for arbitrary $\alpha ,\beta ,\gamma$.

The tensor contraction $B_{kl}={\delta }_{ij}{T}_{ijkl}$ is an isotropic second-rank tensor, because the Kronecker delta and $T$ are invariant under rotations. The classification proved above yields $B_{kl}=\mu {\delta }_{kl}$ for some scalar $\mu$.

For the given three-term tensor, contraction over $i,j$ givesThus $\mu =3\alpha +\beta +\gamma$.

The tensor contraction $C_{mkl}={ϵ}_{ijm}{T}_{ijkl}$ is an isotropic third-rank tensor under proper rotations. It therefore equals $\nu {ϵ}_{mkl}$. Cyclically permuting three indices is even, so ${ϵ}_{mkl}={ϵ}_{klm}$ and ${ϵ}_{ijm}{T}_{ijkl}=\nu {ϵ}_{klm}$.

For the given Kronecker delta expression, the $\alpha$ term vanishes by antisymmetry, while the $\beta$ and $\gamma$ terms give respectively $\beta {ϵ}_{klm}$ and $\gamma {ϵ}_{lkm}=-\gamma {ϵ}_{klm}$. HenceThese computations also illustrate isotropic contractions of a fourth-rank tensor.