Past exam of the theoretical physics I course of the University of Cambridge / 2016 / Paper 1 / 4 / d / Solution

Contract the two Cartesian indices and use the unit length of each molecular orientation:This is exact for each configuration, with no ensemble average assumption. The nematic order parameter is also a real symmetric second-rank tensor. Under a common rotation it transforms as $S\mapsto RS{R}^{\mathsf{T}}$, so its powers give rotational invariants of a symmetric traceless tensor by the cyclic property of the trace. It is unchanged by independent sign flips of the molecular vectors, explaining why it is adapted to nematic rather than polar order.