Past exam of the general relativity course of the University of Oxford / 2021 / 2 / c / Solution

Displacement must be specified relative to an observer. In transverse-traceless gauge the free particles can have fixed coordinates, while their measured proper lengths oscillate. In a local, freely falling orthonormal frame, the geodesic deviation equation for an initial separation $\mathbf{X}=(X,Y,0)$ is, to first order in $H$,Its general solution with physical displacement $\mathbf{\xi }(0)=0$ and initial physical velocity $\mathbf{V}$ isThe source has been running indefinitely, so there is no wave-free initial state. Its initial conditions for an eternal monochromatic gravitational wave need more than positions to fix $\mathbf{V}$. The usual bounded, periodic response selects $V_i={\dot{h}}_{ij}(0,0)X^j/2$; if the particles instead start at physical rest at $t=0$, set $V_i=0$ and retain the linear term. Both solve the same linear geodesic deviation problem. The distinction between fixed TT coordinates and measured separation is explained in ^[ref].

Choose $\beta$ so the wave phase at the detector is $\Phi (t,0)=2\omega t$. For the bounded response the four initial placements then giveThese are displacements from the specified positions at $t=0$; positions are $\mathbf{X}+\mathbf{\xi }$. Equivalently, relative to mean positions the usual oscillatory displacement is ${\mathbf{\xi }}_{\mathrm{osc}}=h\mathbf{X}/2$ to first order. The response repeats after $\pi /\omega$.

Both plus polarization and cross polarization occur with equal amplitude and a quarter-cycle phase difference. Their superposition is circular gravitational-wave polarization, with handedness reversed when the viewing direction or source rotation is reversed. The principal stretching axis turns through half the wave phase. There is no leading longitudinal motion.