Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 3 / 12A (Vector Calculus)

A rank-$k$ Cartesian tensor has components $T_{i_1\dots i_k}$ transforming under an orthogonal change of basis $R$ as $T{\prime }_{i_1\dots i_k}=R_{i_1{j}_1}\cdots R_{i_k{j}_k}{T}_{j_1\dots j_k}$.

Contracting one index from each tensor means summing them, for example $C_{i_1\dots i_{p-1}{j}_1\dots j_{q-1}}=A_{i_1\dots i_{p-1}r}{B}_{r{j}_1\dots j_{q-1}}$. Its rank is $p+q-2$. Substitution of the transformation laws for $p=q=3$ and $R_{rs}{R}_{rt}={\delta }_{st}$ proves the tensor law.

Isotropic means invariant under every proper rotation. The general ranks $0,1,2,3$ are respectively a scalar $a$, zero, $b{\delta }_{ij}$, and $c{\epsilon }_{ijk}$ (for full orthogonal invariance the last must also vanish).

Rotational invariance makes this an isotropic vector, hence it is $0$; oddness gives the same result.

It equals $C{\delta }_{ij}$. Taking the trace gives $3C=4\pi a^4{e}^{-a^2}$, hence $C=4\pi a^4{e}^{-a^2}/3$.

This isotropic rank-three integral is $0$ by oddness under $x\mapsto -x$.

Expanding the integrand, the terms linear in $x$ integrate to zero. Thereforeso these are $\alpha$ and $\beta$.