Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 6

For a constant orthogonal Cartesian change of coordinates $x_i^{\prime }=Q_{ij}{x}_j$, the tensor component transformation law for order $n$ isRepeated indices are summed. In Cartesian orthonormal coordinates the metric identifies upper and lower indices, so the same orthogonal matrix acts on every index.

An isotropic tensor has the same numerical components after every proper orthogonal change of axes: $Q^{\otimes n}T=T$ for every $Q$ with $Q^{T}Q=I$ and $\det Q=1$. For example, ${\delta }_{ij}$ is isotropic because $Q_{ik}{Q}_{jk}={\delta }_{ij}$. Some conventions require invariance under all orthogonal transformations, including reflections; this stronger convention must be distinguished for odd-order tensors, notably the Levi-Civita symbol, which is invariant under proper rotations but changes sign under its ordinary three-index transformation for a reflection.

Under the same Cartesian change, a vector field has $u_i^{\prime }(x^{\prime })=Q_{ip}{u}_p(x)$ with $x=Q^T{x}^{\prime }$. The chain rule gives $\partial /\partial x_j^{\prime }=Q_{jq}\partial /\partial x_q$. Since $Q$ is constant,This is exactly the order-two tensor component transformation law. Thus the Cartesian vector gradient is a Cartesian second-rank tensor, as in the velocity gradient tensor when $u$ is a velocity field.

The restriction to Cartesian transformations matters. In a general position-dependent basis, differentiating the basis contributes extra terms, so the tensorial derivative is the covariant derivative, not an arbitrary array of partial derivatives.

An axial vector transforms under an orthogonal matrix $Q$ asFor a proper rotation, $\det Q=1$, so this is the usual polar vector rule. Under an orientation-reversing change, the extra sign distinguishes the two. In particular, spatial inversion leaves an axial vector unchanged while reversing a polar vector.

Let $C_i=(\nabla \times u{)}_i={ϵ}_{ijk}{\partial }_j{u}_k$. The determinant identity for the Levi-Civita symbol isIt follows by multiplying the three-index determinant identity by $Q_{ir}$ and using orthogonality. The derivative transformation from part (ii) then givesHence the curl of a polar vector field is an axial vector field. The Levi-Civita determinant factor is exactly the extra sign required under reflections.

Write $A_{ij}={\partial }_j{u}_i$. The trace fixes the isotropic part, and splitting the remaining matrix into its symmetric and antisymmetric parts gives the scalar, symmetric-traceless and axial tensor decomposition:Here $s_{ij}=s_{ji}$ and $s_{ii}=0$. The antisymmetric part is $a_{ij}=({\partial }_j{u}_i-{\partial }_i{u}_j)/2$. Its associated axial vector isIndeed the contraction identity ${ϵ}_{ijk}{ϵ}_{k\ell m}={\delta }_{i\ell }{\delta }_{jm}-{\delta }_{im}{\delta }_{j\ell }$ gives ${ϵ}_{ijk}{\omega }_k=({\partial }_j{u}_i-{\partial }_i{u}_j)/2$. Adding the three parts reconstructs $A_{ij}$.

The minus sign is required by the index order $A_{ij}={\partial }_j{u}_i$. The symmetric gradient and curl decomposition written instead for ${\partial }_i{u}_j$ has the opposite axial sign. The divergence supplies $3p$, the trace-free symmetric part describes shape change, and the antisymmetric part describes local rotation.

Direct differentiation givesThe curl is $(0,0,b-a)$, so the preceding convention givesPut $h=(a+b)/2$. The normalized principal axes and eigenvalues of the symmetric tensor $s$ areThe planar axes lie along the two angle bisectors of the coordinate axes. If $h\ne 0$, one planar direction extends and the perpendicular direction contracts, with no change along the third axis. The labels need not be ordered by size when $h<0$. If $a+b=0$, then $s=0$, all principal values vanish and any orthonormal frame is a principal frame; the flow is pure local rotation in the plane.