Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 6

For a constant orthogonal Cartesian coordinate change $x_i^{\prime }=Q_{ij}{x}_j$, the tensor component transformation law givesRepeated indices are summed. Each index of an ordinary Cartesian tensor contributes one factor of $Q$. A pseudotensor would have an additional determinant factor under an orientation-reversing transformation; the rules here concern ordinary tensors.

Define the outer product $T_{ij}=u_i{v}_j$. Transforming each vector givesThus $u_i{v}_j$ is a rank-two tensor, since it obeys the two-index tensor component transformation law. This is the component form of the tensor product of two vectors; its two indices are independent, not contracted.

Adding and subtracting the transpose uniquely separates a Cartesian second-rank tensor into its symmetric tensor and antisymmetric tensor parts:Then $S_{ji}=S_{ij}$, $A_{ji}=-A_{ij}$ and $S_{ij}+A_{ij}=B_{ij}$. Both parts transform as tensors because transposition commutes with the orthogonal similarity $B^{\prime }=QB{Q}^T$. Uniqueness follows by taking the symmetric and antisymmetric parts of any such decomposition.

The tensor contraction of a symmetric tensor with an antisymmetric one vanishes. Rename the two summed indices and use their opposite exchange symmetries:Hence $R_{ij}{A}_{ij}=0$. Using the decomposition from part (iii) givesThis is orthogonality of the real symmetric and antisymmetric matrix subspaces under the Frobenius inner product. It is a summed scalar identity, not an assertion about the unsummed products of individual entries.

Let $B$ be the rank-two tensor and $v$ the vector. The components of their tensor product are $T_{ijk}=B_{ij}{v}_k$. Applying the transformation rules givesThe product therefore obeys the three-index tensor component transformation law, proving that $B_{ij}{v}_k$ is a rank-three tensor. Tensor-product order adds because no pair of indices is summed out.

Start from the three-index transformation law and contract the first two Cartesian indices:Orthogonality supplies ${\sum }_i{Q}_{ia}{Q}_{ib}={\delta }_{ab}$, where $\delta$ is the Kronecker delta. The remaining factor is exactly the vector transformation rule. Therefore $C_k$ is a vector.

This tensor contraction removes two indices by pairing them through the Euclidean metric. In general nonorthonormal coordinates, contraction must use the appropriate metric or pair a covariant and a contravariant index; the bare repeated-lower-index sum relies on Cartesian orthonormal coordinates.

Write the curl and cross product with the Levi-Civita symbol, then use its contraction identity:Expanding with the product rule givesRecognize the divergence in the first and third terms, and the directional derivatives in the other two. The curl of a cross product identity is thereforeThe derivatives act on both vector fields. No assumption of vanishing divergence or spatially constant fields has been made.