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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 Lorenz gauge in linearized gravity givesso the symmetric amplitude is transverse:Under this condition the supplied linearized Einstein tensor reduces toThe vacuum Einstein field equations and the plane-wave ansatz then implyThus the wavevector is null and the perturbation is a plane gravitational wave in linearized gravity. Lorenz gauge alone does not require the trace of $H_{\mu \nu }$ to vanish; that is a further gauge choice.

Only the transverse coordinates occur in the perturbation, so its tensor components are unchanged by replacing $(t,z)$ with $(u,v)$. The plane gravitational wave in linearized gravity isEvery component depends on $u$ alone. The Minkowski wave operator in null coordinates therefore givesThus the field satisfies the vacuum Linearized Einstein equations.

Let $s=k_{\rho }{x}^{\rho }$. In vacuum,A nonzero localized profile cannot have $H^{\prime \prime }=0$, because an affine function does not decay at both ends. HenceDecay removes the integration constant, so the second condition is equivalently $k^{\mu }{H}_{\mu \nu }=0$. Thus a nontrivial plane gravitational wave in linearized gravity has a null wavevector and transverse amplitude.