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Work to first order in a metric perturbation $g_{ab}={\eta }_{ab}+h_{ab}$, with $\eta =\mathrm{diag}(1,-1,-1,-1)$, raising perturbation indices with $\eta$ and discarding quadratic terms. The Christoffel symbols areUsing the printed curvature convention gives the linearized Riemann curvature operatorFor definiteness contract $R_{bd}=R^a{}_{bad}$. The corresponding first-order Ricci tensor isThis is the reversed sign convention specified in curvature-sign convention in Killing derivative identities; the vacuum equation is unaffected by an overall curvature sign.

An infinitesimal passive coordinate change $x^{\prime a}=x^a+{\xi }^a$ changes the perturbation by $\delta h_{ab}=-{\partial }_a{\xi }_b-{\partial }_b{\xi }_a$. Substitution into $R^{(1)}$ gives zero change, since partial derivatives commute. Equivalently the change in curvature is the Lie derivative of the background curvature, which vanishes in Minkowski spacetime. Thus linearized curvature, and the linearized Weyl tensor in vacuum, are gauge-invariant tidal observables, whereas individual components of $h$ are generally gauge dependent.

The trace-reversed metric perturbation ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$ transforms by $\delta {\bar{h}}_{ab}=-{\partial }_a{\xi }_b-{\partial }_b{\xi }_a+{\eta }_{ab}{\partial }_c{\xi }^c$. Its divergence changes by $-\square {\xi }_b$. Solving $\square {\xi }_b={\partial }^a{\bar{h}}_{ab}$ therefore imposes Lorenz gauge in linearized gravity, ${\partial }^a{\bar{h}}_{ab}=0$. With the stated sign convention the Einstein tensor isso the linearized vacuum field equation becomes $\boxed{\square {\bar{h}}_{ab}=0}$. The residual gauge symmetry of linearized gravity has $\square {\xi }_a=0$. This makes a hyperbolic evolution problem after the gauge condition and its initial consistency constraints have been imposed.

To separate actual gauge invariants from a gauge choice, use gauge-invariant Minkowski perturbations. For spatial indices contracted with ${\delta }_{ij}$, decomposewhere ${\partial }_i{B}_i={\partial }_i{E}_i=0$, ${\partial }_i{h}_{ij}^{\mathrm{TT}}=0$ and $h_{ii}^{\mathrm{TT}}=0$. Spatial falloff or Fourier-mode conventions fix the decomposition. Write ${\xi }^0=\tau$ and ${\xi }^i={\partial }_{i}L+L_i$ with ${\partial }_i{L}_i=0$. Direct substitution in the gauge change givesHence the invariant combinations areA representative with $E=E_i=B=0$ has $h_{00}=2\Phi$, $h_{0i}=V_i$ and $h_{ij}=2\Psi {\delta }_{ij}+h_{ij}^{\mathrm{TT}}$. The explicit vacuum constraints for gauge-invariant Minkowski perturbations follow fromFor a nonzero spatial Fourier mode, or regular decaying fields on all space, $G_{00}=0$ forces $\Psi =0$, then $G_{0i}=0$ forces $V_i=0$, and the spatial trace gives $\Delta \Phi =0$, hence $\Phi =0$. Only $\square h_{ij}^{\mathrm{TT}}=0$ remains. This removal of the scalar and vector sectors does not exclude singular Coulomb perturbations outside a source, boundary-driven harmonic potentials or spatially homogeneous modes; those require separate boundary and gauge conventions.

For a propagating plane mode $h\propto e^{i{k}_a{x}^a}$, the wave equation gives $k^a{k}_a=0$. A transverse-traceless spatial tensor perpendicular to a fixed propagation direction has two independent components. For propagation along $z$, define the physical strain $H_{ij}=-h_{ij}^{\mathrm{TT}}$ in this mostly-minus convention, so the spatial line element is $({\delta }_{ij}+H_{ij})d{x}^{i}d{x}^j$. It has the formThese are the two gravitational wave polarizations, plus polarization and cross polarization. A rotation in the transverse plane rotates the polarization pair through twice the angle; cross is plus rotated through $45$ degrees. Lorenz constraints and residual gauge freedom give the same count: ten symmetric metric components minus four constraints and four gauge freedoms leave two radiative degrees of freedom.

In transverse-traceless gauge, particles initially at rest keep fixed spatial coordinates since ${\Gamma }^a{}_{00}=0$, but their measured separation changes. For a short separation of length $L$ in unit direction $n$, $\delta L/L=H_{ij}{n}^i{n}^j/2$. The gauge-invariant curvature gives the corresponding geodesic deviation:Here the last equation is in a freely falling orthonormal frame, with the printed reversed curvature convention. Coordinate separation and physical separation must not be confused. The linearized vacuum equations describe propagation and tidal measurements; wave energy transfer and an effective wave stress-energy tensor arise at quadratic order, beyond the linear approximation.