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On a contractible Cartesian patch of a flat background, zero linearized Riemann curvature operator says that the connection 1-forms ${\Gamma }^{(1)a}{}_{bc}d{x}^c$ are closed. The Poincaré lemma writes them as derivatives of a matrix-valued potential. Symmetry of the two lower affine connection indices then supplies a second potential, giving ${\Gamma }^{(1)a}{}_{bc}={\partial }_b{\partial }_c{\xi }^a$. Linearized metric compatibility implies that $h_{ab}-2{\partial }_{(a}{\xi }_{b)}$ is constant. Its symmetric constant part is absorbed by an affine vector field. Thus the metric perturbation is locally a coordinate change. Global topology and imposed boundary conditions can obstruct the corresponding global gauge choice.

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.

The phase isBecause the transverse-traceless gauge perturbation has no component alongonly the third term of the linearized Riemann curvature operator contributes to the required contraction:The derivatives and transverse polarization contraction areTherefore the Newman--Penrose scalar Psi4 isIts real and imaginary parts encode the plus and cross gravitational wave polarizations, up to the stated Riemann curvature tensor and complex null tetrad conventions.

At $ϵ=0$ the metric is the constant Minkowski metric, whose Christoffel symbols and curvature vanish, soInsert $R=ϵR^{(1)}+O({ϵ}^2)$ into the Penrose wave equation. Every curvature-square term is $O({ϵ}^2)$, while the covariant wave operator reduces at first order to the flat d'Alembert operator. ThusThe linearized Riemann curvature operator is unchanged by $h_{\mu \nu }\mapsto h_{\mu \nu }+{\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$, because the resulting third derivatives cancel pairwise. The equation therefore requires no gauge choice for $h_{\mu \nu }$.

After the trace gauge, $h_{ab}={\bar{h}}_{ab}=A_{ab}{e}^{ik\cdot x}$. The linearized Riemann curvature operator is built from terms containing two factors of $k$ and one factor of $A$:Contracting with $k^{[c}{v}^{d]}$ gives zero by $k^2=0$ and $k^a{A}_{ab}=0$, so $\mathcal{R}(k\wedge v)=0$. If ${\omega }_{cd}{k}^d=0$, every term in $R_{ab}^{(1)}{}^{cd}{\omega }_{cd}$ also contains such a contraction and vanishes.

To count the kernel, use the null basis $k,\ell ,e_1,e_2$. The forms $k\wedge v$ span the three-dimensional spaceThe condition ${\omega }_{ab}{k}^b=0$ defines the three-dimensional spaceTheir intersection has dimension two, so their sum is a four-dimensional subspace of $\ker \mathcal{R}$. Since $\dim {\Lambda }^2=6$, the rank-nullity theorem gives