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For a plane gravitational wave in linearized gravity with $k_{\mu }=(-\omega ,0,0,\omega )$, Lorenz gauge in linearized gravity gives $H_{3\nu }=-H_{0\nu }$. Under $\delta H_{\mu \nu }=i(k_{\mu }{X}_{\nu }+k_{\nu }{X}_{\mu }-{\eta }_{\mu \nu }{k}^{\rho }{X}_{\rho })$, choose $X_i=H_{0i}/(i\omega )$ for $i=1,2$, $X_3-X_0=i{H}_{00}/\omega$, and $X_0+X_3=(H_{11}+H_{22})/(2i\omega )$. This residual gauge symmetry of linearized gravity removes all time and longitudinal entries and the transverse trace. The remaining amplitudes $H_{+},H_{\times }$ are the two gravitational wave polarizations. The formula uses the plus-sign convention for $\delta h=2{\partial }_{(\mu }{\xi }_{\nu )}$; reversing the definition of $\xi$ reverses the gauge change.

For a nonzero, nonconstant vacuum plane gravitational wave in linearized gravity with real wave covector $k$, the Linearized Einstein equations and Lorenz gauge in linearized gravity requireThe zero-amplitude solution places no restriction on $k$; a constant mode $k=0$ is not a propagating wave. Under the sign convention for the infinitesimal change specified here, the trace-reversed metric perturbation changes byThus the residual gauge symmetry of linearized gravity is characterized by $\square {\xi }_{\nu }=0$. The proposed plane-wave form obeys this for every constant complex $X_{\nu }$, because $k$ is null. Changing the sign used to name $\xi$ would reverse both gauge formulas, with no physical consequence.

Rotate the spatial axes so the wave propagates along $+z$, choosing $k_{\mu }=(-\omega ,0,0,\omega )$ with $\omega >0$. The Lorenz gauge in linearized gravity says $H_{3\nu }=-H_{0\nu }$; in particular $H_{03}=-H_{00}$, $H_{33}=H_{00}$ and $H_{3i}=-H_{0i}$ for $i=1,2$. The amplitude change isChooseIndeed, $\delta H_{00}=i\omega (X_3-X_0)$ and $\delta H_{0i}=-i\omega X_i$ cancel the time components, while $\delta H_{11}=\delta H_{22}=-i\omega (X_0+X_3)$ cancels the transverse trace. The preserved Lorenz gauge in linearized gravity then cancels all longitudinal components. This is an explicit plane-wave reduction to transverse-traceless gauge, yieldingThe trace-reversed metric perturbation has zero trace in this transverse-traceless gauge, so $h^{\prime }={\bar{h}}^{\prime }$. The null dispersion relation is ${\omega }^2=|\mathbf{k}{|}^2$, giving the speed of light. Only components perpendicular to the direction of propagation remain, and the two independent amplitudes give the plus and cross gravitational wave polarizations.

These are physical transverse tidal distortions, rather than just a convenient display of the metric perturbation: in transverse-traceless gauge, $R_{0i0j}^{(1)}=-\frac{1}{2}{\partial }_t^2{h}_{ij}^{\prime }$, so a freely falling detector has ${\ddot{S}}^i=\frac{1}{2}{\ddot{h}}_{ij}^{\prime }{S}^j$ at first order. There is no longitudinal tidal acceleration. The plus polarization stretches one transverse axis while compressing the other; the cross polarization does the same along axes rotated by $\pi /4$.

The quadratic gravitational effective stress-energy tensor depends on the split between the fixed background and the metric perturbation. A residual gauge symmetry of linearized gravity changes $h$ without changing the first-order tidal field, but generally changes $G^{(2)}[h,h]$ and hence the local value of $t$. At second order the compensating change of $j$ restores the same physical geometry; assigning only the quadratic term to an energy density loses this compensation.

Moreover, a stress-energy superpotential can change the local energy expression while preserving its divergence and, with suitable boundary behavior, its integrated charges. Symmetry and flat conservation do not make $t$ a unique, gauge-independent local gravitational energy density. Averaging in an appropriate short-wavelength regime can lead to a useful physical gravitational-wave energy flux, but that is a qualified approximation, not a cure for the local definition requested here.

Under an infinitesimal coordinate transformation generated by ${\xi }_{\mu }$, the trace-reversed metric perturbation changes byTaking a divergence givesThe transformation therefore preserves Lorenz gauge precisely when $\square {\xi }_{\nu }=0$. For a plane-wave parameter ${\xi }_{\mu }=a_{\mu }{e}^{ik\cdot x}$, this holds because $k^2=0$. Absorbing a factor of $i$ into an arbitrary constant vector ${ϵ}_{\mu }$, the amplitude transformation isThis four-parameter freedom is the residual gauge symmetry of linearized gravity.

Under $h_{\mu \nu }^{\prime }=h_{\mu \nu }+{\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$, trace reversal givesTaking a divergence yieldsApplying $\square$ to the transformed field likewise produces only derivatives of $\square \xi$. Therefore $\square {\xi }^{\mu }=0$ preserves both the gauge condition and the sourced wave equation. These are the residual gauge transformations.

An infinitesimal coordinate change generated by ${\xi }_a$ acts on the trace-reversed metric perturbation asIts divergence changes bySolving $\square {\xi }_b={\partial }^a{\bar{h}}_{ab}$ therefore imposes Lorenz gauge in linearized gravity, ${\partial }^a{\bar{h}}_{ab}=0$. Transformations satisfying $\square {\xi }_b=0$ remain and form the residual gauge symmetry of linearized gravity.