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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.

In terms of the trace-reversed metric perturbation and in the Lorenz gauge in linearized gravity, the linearized Einstein equations on Minkowski spacetime are

Keep only first order in the metric perturbation and use the Minkowski metric to raise and lower indices; set $G=c=1$. The Levi-Civita connection hasThe quadratic products of Christoffel symbols in the Riemann curvature tensor can be dropped at this order. Contracting the remaining derivatives giveswhere $\square ={\eta }^{\rho \sigma }{\partial }_{\rho }{\partial }_{\sigma }=-{\partial }_t^2+{\nabla }^2$ is the d'Alembert operator. In four dimensions the trace-reversed metric perturbation satisfies $\bar{h}=-h$ and $h_{\mu \nu }={\bar{h}}_{\mu \nu }-\frac{1}{2}{\eta }_{\mu \nu }\bar{h}$. Writing the Linearized Einstein equations in terms of it givesIn Lorenz gauge in linearized gravity, the first, second and fourth terms vanish. Therefore $G_{\mu \nu }^{(1)}=8\pi T_{\mu \nu }$ becomesThe divergence of this equation also gives ${\partial }^{\mu }{T}_{\mu \nu }=0$, the leading-order stress-energy conservation required by the contracted Bianchi identity.

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 $\pi /4$.

Use $G=c=1$, select the retarded source field with no added homogeneous radiation, and work to leading order in the weak-field approximation and long-wavelength source approximation. The retarded fundamental solution of the Linearized Einstein equations in Lorenz gauge in linearized gravity isFor $r\gg d$, the denominator is $r$ to leading order. The delay is $t-r+\mathbf{n}\cdot {\mathbf{x}}^{\prime }+O(d^2/r)$, where $\mathbf{n}=\mathbf{x}/r$. In the usual slowly evolving source regime, its characteristic time $\tau$ obeys $d/\tau \ll 1$, so the source-size part of the delay is negligible:The stress-energy conservation equation ${\partial }_{\mu }{T}^{\mu \nu }=0$ is required at the approximation order used; it follows from the divergence of the gauge-fixed field equation. Compact support removes the integration by parts surface terms. Since $T^{00}=T_{00}$ and $T^{ij}=T_{ij}$ in this signature,Combining these identities gives the retarded quadrupole fieldFor an ordinary nonrelativistic bound source, $\tau \sim d/v$ makes $d/\tau \sim v\ll 1$. More generally the size relative to the variation timescale must also be small; speed alone does not exclude a rapidly varying small-amplitude motion.

The radiation boundary condition for linearized gravity is essential for a statement about the full field. Without it, the displayed implication is false: take $T_{\mu \nu }=0$ and add ${\bar{h}}_{11}=A\cos (k(t-z))$, ${\bar{h}}_{22}=-{\bar{h}}_{11}$, all other components zero. This weak plane gravitational wave satisfies both the homogeneous wave equation and Lorenz gauge in linearized gravity, while every $I_{ij}$ is zero. The proved formula is for the retarded source contribution, with the standard slow-source qualification. Its transverse–traceless projection gives the physical radiative strain.

Use geometrized units $G=c=1$ and signature $(-,+,+,+)$. The original PDF starts with the d'Alembert operator $\square {\bar{h}}_{\mu \nu }=-16\pi T_{\mu \nu }$; the local TeX incorrectly transcribes its derivative indices. The harmonic condition is the Lorenz gauge in linearized gravity.

Choose the retarded solution, excluding an incoming homogeneous wave. For a spatially localized source, the Linearized Einstein equations giveLet the source size be $D$ and the characteristic angular frequency be $\omega$. Assume the weak-field approximation, nonrelativistic source velocities, and $D\ll {\omega }^{-1}\ll r$ for a radiative far-zone measurement. Then $|\mathbf{x}-\mathbf{y}|=r-\mathbf{n}\cdot \mathbf{y}+O(D^2/r)$. At leading order in $D/r$ and $\omega D$, one may replace the denominator by $r$ and the retarded time throughout the source by $t-r$, obtaining

To identify this integral, use stress-energy conservation ${\partial }_{\mu }{T}^{\mu \nu }=0$ for a symmetric source tensor. Compact support or sufficient decay permits integration by parts with no boundary flux. Because $T^{00}=T_{00}$ in this signature, the second mass moment tensor satisfiesSpatial indices are raised with ${\delta }_{ij}$, so $T^{ij}=T_{ij}$. Substitution gives the retarded quadrupole field:The assumptions are an isolated conserved source, retarded boundary conditions, weak gravity, a source small compared with the wavelength, and observation far from it. The physical radiative field follows by the transverse-traceless projector applied to this trace-reversed metric perturbation.

Write $g_{\mu \nu }={\eta }_{\mu \nu }+ϵh_{\mu \nu }$. The linearized inverse metric is $g^{\mu \nu }={\eta }^{\mu \nu }-ϵh^{\mu \nu }$, and all quadratic Christoffel products are $O({ϵ}^2)$. The wave-coordinate condition linearizes toUsing the supplied Ricci formula then gives $G_{\mu \nu }^{(1)}=-\frac{1}{2}\square {\bar{h}}_{\mu \nu }$. Consequently the Linearized Einstein equations in Lorenz gauge in linearized gravity areThis Lorenz condition is the first-order form of the harmonic coordinate equations.

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.

Because $R=O(ϵ)$ andthe Taylor series gives $f(R)=R+O({ϵ}^3)$ and $f^{\prime }(R)=1+O({ϵ}^2)$. The extra derivative terms in the f(R) equation are therefore at least second order. At first order the theory reduces to the Linearized Einstein equationsIn harmonic coordinates, equivalently the Lorenz gauge in linearized gravity ${\partial }_{\mu }{\bar{h}}^{\mu }{}_{\nu }=0$, the trace-reversed metric perturbationsatisfiesThus Linearized f(R) gravity about Minkowski spacetime has the same weak-field predictions as general relativity under these assumptions. Differences can first appear through nonlinear curvature terms or in backgrounds about which $f^{\prime \prime }(R)$ is nonzero.

Write $g_{\mu \nu }={\eta }_{\mu \nu }+ϵh_{\mu \nu }$ and retain first-order terms. The quadratic Christoffel products in the supplied Ricci formula drop out. The wave-coordinate condition becomes the Lorenz gauge in linearized gravityThe linearized Ricci tensor and scalar then satisfySubstitution into $G_{\mu \nu }=8\pi T_{\mu \nu }$ givesIf the paper denotes the trace-reversed variable itself by $h_{\mu \nu }$ in its displayed equation, this is exactly that convention.