Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 357 / 1 / c

All time components of the perturbation vanish. For $\mu =t$ this immediately givesFor a spatial index $\mu =i$,by the stated property. Thus the perturbation is transverse.

Its Minkowski trace is purely spatial:Adding the displayed components and collecting the coefficients of the independent functions $A$, $B$, and $C$ makes each coefficient vanish separately, soTogether with $h_{t\mu }=0$ and ${\partial }^{\nu }{h}_{\mu \nu }=0$, this proves that the wave is in transverse-traceless gauge.

On the positive $z$ axis, $x=y=0$ and $r=z$. Taking this limit in the spatial components giveswith $A,B,C$ evaluated at retarded time $t-r$. A wave propagating toward the observer along the $z$ direction has polarization matrixConsequently the observed gravitational wave polarization amplitudes areThe $B$ mode and the orthogonal combination of $A,C$ do not contribute on this symmetry axis. The observer therefore sees a purely plus-polarized wave; a rotation of the transverse $x,y$ axes would represent the same physical polarization as the corresponding spin-two mixture of plus and cross.