= Solution
A <linearized coordinate gauge transformation> changes the identification of points of the perturbed <spacetime> with its <Minkowski spacetime> background. With $x'^i=x^i+\epsilon\xi^i$, the same physical metric has perturbation
$$
h'_{ij}=h_{ij}-\partial_i\xi_j-\partial_j\xi_i.
$$
These changes are coordinate freedom, not extra physical <gravitational wave polarizations>. Raising indices with $\eta$, define $h=\eta^{ij}h_{ij}$ and the <trace-reversed metric perturbation> $\bar h_{ij}=h_{ij}-\eta_{ij}h/2$. Then
$$
\bar h'_{ij}=\bar h_{ij}-\partial_i\xi_j-\partial_j\xi_i+\eta_{ij}\partial_k\xi^k.
$$
The first-order <Levi-Civita connection> is $\Gamma^{(1)k}{}_{ij}=\tfrac12\eta^{k\ell}(\partial_ih_{j\ell}+\partial_jh_{i\ell}-\partial_\ell h_{ij})$. Contracting the <Riemann curvature tensor> in the stated convention gives
$$
R^{(1)}_{ij}=\tfrac12\left(\partial_i\partial_kh^k{}_j+\partial_j\partial_kh^k{}_i-\Box h_{ij}-\partial_i\partial_jh\right),
\qquad R^{(1)}=\partial_i\partial_jh^{ij}-\Box h.
$$
Put $A_j=\partial^i\bar h_{ij}$. The corresponding <Einstein tensor> is
$$
G^{(1)}_{ij}=-\tfrac12\Box\bar h_{ij}
+\tfrac12\left(\partial_iA_j+\partial_jA_i-\eta_{ij}\partial^kA_k\right).
$$
Under the <linearized coordinate gauge transformation>, $A'_j=A_j-\Box\xi_j$. Choosing a solution of $\Box\xi_j=A_j$ imposes <Lorenz gauge in linearized gravity>, $A'_j=0$. Thus the vacuum <Einstein field equations> reduce to
$$
\boxed{\partial^i\bar h_{ij}=0,\qquad\Box\bar h_{ij}=0.}
$$
There remains <residual gauge symmetry of linearized gravity> with $\Box\xi_j=0$.
For the transverse profiles, write both symmetric off-diagonal entries as $h_{xy}=h_{yx}=h_\times(t-z)$ and the diagonal entries as $h_{xx}=-h_{yy}=h_+(t-z)$. The <trace-reversed metric perturbation> has zero trace and therefore equals $h$. Its divergence vanishes: its only nonzero index directions are $x,y$, while its coefficients depend only on $t-z$. Also $(-\partial_t^2+\partial_z^2)F(t-z)=0$ for every twice differentiable profile $F$. These are consequently <plane gravitational waves in linearized gravity> in <transverse-traceless gauge>, with the two independent <gravitational wave polarizations> $h_+$ and $h_\times$.
To calculate the detector response, use a <parallel-propagated orthonormal frame> along the centre of a short nonrotating stick. To first order its clock reads $t$, and the tidal curvature is
$$
R_{A0B0}=-\frac\epsilon2\ddot h_{AB}+O(\epsilon^2),\qquad A,B\in\{x,y\}.
$$
For a <sliding-bead gravitational wave detector>, the stick constrains transverse motion but supplies no axial restoring force. Projecting <geodesic deviation> along an $x$-directed stick gives
$$
\ddot L=\frac\epsilon2\ddot h_+(t-z_0)L_0+O(\epsilon^2).
$$
Here $L$ is the proper separation and $L_0=L(t_0)$, with $\dot L(t_0)=0$. Integrating with those actual initial conditions yields
$$
\boxed{L(t)=L_0\left[1+\frac\epsilon2\left(h_+(t-z_0)-h_+(t_0-z_0)
-(t-t_0)\dot h_+(t_0-z_0)\right)\right]+O(\epsilon^2).}
$$
For beads at rest before an incident wave, both initial profile terms vanish and the familiar result is $\boxed{L(t)=L_0[1+\epsilon h_+(t-z_0)/2]+O(\epsilon^2)}$. The cross polarization produces a transverse tidal force, balanced by the guide, and no axial first-order displacement in this orientation. In <transverse-traceless gauge> the same plus response can be read from the proper line element, $d\ell=\sqrt{1+\epsilon h_+}\,dx$; coordinate distances alone are not measured distances. When a cross polarization is present, keeping the guide physically nonrotating may require transverse coordinate adjustments, which do not change the axial answer at this order.
For a $z$-directed stick, $R_{z0z0}=0$ and the wave is transverse. Therefore \b[beads initially at rest retain their proper separation to first order]: $L(t)=L_0+O(\epsilon^2)$. The tidal calculation is the usual linear detector limit, with bead spacing small compared with the wavelength; it does not treat a macroscopic rigid body as exactly rigid over arbitrary relativistic distances.
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