Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 3 / 9C (Vector Calculus)

A Cartesian second-rank tensor has components in every orthonormal Cartesian frame that transform according towhen vector components transform as $x^{\prime }=Qx$ for an orthogonal matrix $Q$. Repeated indices are summed. If $A=B$ in one frame, their transformed difference is $Q(A-B)Q^T=0$, so their equality holds in every such frame, at the same physical point.

Now impose the scalar contraction hypothesis on $C$. For every tensor $A$, invariance in the two frames saysAt a fixed point, choose the components of $A$ to be each matrix unit in turn; each is an admissible tensor once its components in other frames are transformed. Thus every coefficient of $A_{pq}$ agrees, yielding $Q^T{C}^{\prime }Q=C$, orTherefore $C$ is necessarily a Cartesian second-rank tensor. This is the scalar contraction test for a Cartesian tensor. Testing all tensors, rather than only symmetric tensors, is essential to the proof.

Let $R_{\theta }$ be rotation about the third axis, so invariance means $R_{\theta }T{R}_{\theta }^T=T$. At $\theta =\pi$, its diagonal signs are $(-1,-1,1)$; each mixed entry $T_{13},T_{23},T_{31},T_{32}$ changes sign and must vanish.

For the remaining planar block $M=\left(\begin{array}{cc}p& q\\ r& s\end{array}\right)$, the quarter-turn $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ givesHence $s=p$ and $r=-q$. Writing the constants as $\alpha ,\omega ,\beta$ givesConversely, $M=\alpha I-\omega J$ commutes with every planar rotation, so every displayed matrix is invariant under every rotation about the axis. This is an oriented-axis invariant Cartesian tensor. Symmetry was not assumed, so the antisymmetric parameter $\omega$ need not vanish.

For $i\ne j$, the inertia component is $I_{ij}=-\rho {\int }_V{x}_i{x}_{j}dV$. Reflecting just one of these coordinates preserves the cylinder and its uniform density but changes the integrand's sign. Odd-integrand cancellation by reflection therefore makes every off-diagonal entry zero. Rotations about the axis also give $I_{11}=I_{22}$.

Using cylindrical coordinates $x_1=r\cos \theta$, $x_2=r\sin \theta$, $x_3=z$, with $dV=rdrd\theta dz$, the required integrals areThus the inertia tensor isThe mass is $M=4\pi \rho$, so these entries are also $19M/12,19M/12,M/2$, respectively.