= Solution
{c}
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 $H$,
$$
\ddot\xi_i=\frac12\ddot h_{ij}(t,0)X^j,\qquad \ddot\xi_z=0.
$$
Its general solution with physical displacement $\boldsymbol\xi(0)=0$ and initial physical <velocity> $\mathbf V$ is
$$
\boxed{\xi_i(t)=\frac12[h_{ij}(t,0)-h_{ij}(0,0)]X^j
+\left[V_i-\frac12\dot h_{ij}(0,0)X^j\right]t.}
$$
The 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 <General Relativity by David Tong>[Section 5.2.2, response of free particles to a wave]{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 give
$$
\begin{array}{c|ccc}
\mathbf X&\xi_x&\xi_y&\xi_z\\ \hline
(\sigma a,0,0)&\dfrac{\sigma aH}{2}(1-\cos2\omega t)&-\dfrac{\sigma aH}{2}\sin2\omega t&0\\[2pt]
(0,\sigma a,0)&-\dfrac{\sigma aH}{2}\sin2\omega t&\dfrac{\sigma aH}{2}(\cos2\omega t-1)&0
\end{array}\qquad (\sigma=\pm1).
$$
These are displacements from the specified positions at $t=0$; positions are $\mathbf X+\boldsymbol\xi$. Equivalently, relative to mean positions the usual oscillatory displacement is $\boldsymbol\xi_{\rm osc}=h\mathbf X/2$ 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.
\Image[/past-exam-of-the-general-relativity-course-of-the-university-of-oxford/2021-circular-gravitational-wave-cycle.png]
{title=One full circularly polarized gravitational-wave cycle, with the four particles at their specified initial positions at phase zero}
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