Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 6

For Cartesian tensors, let $R$ be the orthogonal matrix taking old vector components to new ones, $v_i^{\prime }=R_{ij}{v}_j$. A rank-$n$ tensor transforms by one copy of $R$ on each index:Repeated indices are summed according to Einstein notation. In general coordinates a contravariant tensor index instead carries the corresponding Jacobian matrix $\partial x^{\prime i}/\partial x^j$, and a covariant tensor index carries its inverse; in orthonormal bases of Cartesian axes their component laws coincide with the displayed form.

An isotropic tensor has identical components after every proper orthogonal transformation, so $T^{\prime }=T$ for every $R$ with $R^{T}R=I$ and $\det R=1$. Requiring reflection invariance as well is a stronger convention for odd tensor ranks. The Kronecker delta is invariant under all orthogonal transformations, while the Levi-Civita symbol is invariant under proper rotations and acquires a determinant sign under reflections.

First separate the symmetric tensor and antisymmetric tensor parts:The trace of the antisymmetric part vanishes. In three dimensions, remove one third of the trace from the symmetric part, and identify the remaining antisymmetric part with an axial vector. The scalar, symmetric-traceless and axial tensor decomposition isHere $S_{ji}=S_{ij}$ and $S_{ii}=P_{ii}-P_{kk}=0$. The contraction of two Levi-Civita symbols verifies the reconstruction:ThereforeThe pieces contain $1$, $5$ and $3$ independent components respectively, accounting for all nine entries. Taking the trace, then the traceless symmetric part, then the antisymmetric part shows that the decomposition is unique.

Contract the given isotropic fourth-rank tensor with the strain array. The stress tensor isWrite $p_{ij}=p{\delta }_{ij}+s_{ij}+{ϵ}_{ijk}{a}_k$, where $s_{ij}$ is symmetric and traceless. Then $p_{kk}=3p$ and $p_{ji}=p{\delta }_{ij}+s_{ij}-{ϵ}_{ijk}{a}_k$. Substitution givesBy uniqueness of the scalar, symmetric-traceless and axial tensor decomposition,This is isotropic elasticity on tensor components: the three tensor sectors do not mix. If the strain is symmetric, $a_k=0$, hence $A_k=0$ and $\boxed{P_{ij}=P_{ji}}$. No equality between $\mu$ and $\nu$ is needed for that conclusion.