Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 37B / a

The Lorenz gauge in linearized gravity givesso the symmetric amplitude is transverse:Under this condition the supplied linearized Einstein tensor reduces toThe vacuum Einstein field equations and the plane-wave ansatz then implyThus the wavevector is null and the perturbation is a plane gravitational wave in linearized gravity. Lorenz gauge alone does not require the trace of $H_{\mu \nu }$ to vanish; that is a further gauge choice.

Under an infinitesimal coordinate transformation generated by ${\xi }_{\mu }$, the trace-reversed metric perturbation changes byTaking a divergence givesThe transformation therefore preserves Lorenz gauge precisely when $\square {\xi }_{\nu }=0$. For a plane-wave parameter ${\xi }_{\mu }=a_{\mu }{e}^{ik\cdot x}$, this holds because $k^2=0$. Absorbing a factor of $i$ into an arbitrary constant vector ${ϵ}_{\mu }$, the amplitude transformation isThis four-parameter freedom is the residual gauge symmetry of linearized gravity.