Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 6

A second-order Cartesian tensor assigns components in each orthonormal coordinate system according toHere $R$ is the matrix taking old vector components to new ones, and repeated indices are summed. This is the tensor component transformation law; an arbitrary array of nine numbers is not a tensor unless its components in different frames obey this rule. No symmetry $T_{ij}=T_{ji}$ is implied by being a second-order tensor.

Let $R$ be an orthogonal change of axes. Since $A$ is a second-order tensor, $A_{ij}^{\prime }=R_{ik}{R}_{j\ell }{A}_{k\ell }$. The scalar-contraction assumption givesfor every tensor $A$. In one fixed frame its components can be any array: in particular, choose one component at a time to be one, all others zero. Equality of the coefficients then forcesMultiplying by $R$ and $R^T$ yieldsThis is exactly the tensor component transformation law. It is the full-contraction version of the quotient theorem for Cartesian tensors. The words “for every” are essential: testing a single $A$, such as the zero tensor or the identity, would not determine all components.

Use the quarter-turn rotation matrixFor a general second-order tensor, direct multiplication givesRequiring $RT{R}^T=T$ gives $T_{11}=T_{22}$, $T_{12}=-T_{21}$, $T_{13}=-T_{23}$ together with $T_{23}=T_{13}$, and $T_{31}=-T_{32}$ together with $T_{32}=T_{31}$. Thus all mixed axial-transverse entries vanish andFor an arbitrary rotation angle $\theta$, the transverse rotation is $Q_{\theta }=\cos \theta I+\sin \theta J$, where $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. The transverse block of $T$ is $\alpha I-\omega J$, so it commutes with $Q_{\theta }$. Therefore $Q_{\theta }(\alpha I-\omega J)Q_{\theta }^T=\alpha I-\omega J$, while the axial component $\beta$ is unchanged.

Hence invariance holds under every rotation about the axis, not just the quarter-turn. This is the general axisymmetric second-order tensor; a symmetric tensor would additionally require $\omega =0$, but that restriction was not assumed.

For $i\ne j$, the inertia tensor component is $I_{ij}=-\rho {\int }_V{x}_i{x}_{j}dV$. Reflect either coordinate appearing once: the cylindrical domain and density are unchanged, while the integrand changes sign. Thus every off-diagonal component vanishes. Rotational symmetry also implies $I_{11}=I_{22}$.

Use cylindrical coordinates $x_1=r\cos \theta$, $x_2=r\sin \theta$, $x_3=z$, with volume element $rdrd\theta dz$. The axial moment of inertia isThe two contributions to a transverse component areThereforeThe axial value $M{R}^2/2$ and transverse value $M(3R^2+h^2)/12$ for $R=1$, $h=4$ give an independent check on both diagonal entries.