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For a perturbation $h_{xx}=2a(t-z)$, $h_{yy}=2b(t-z)$ with all other components zero, the linearized Vacuum Einstein equations give $(a+b{)}^{\prime \prime }=0$. Put $s=a+b$ and choose ${\xi }_x=sx/2$, ${\xi }_y=sy/2$, ${\xi }_t=-s^{\prime }(x^2+y^2)/4$, ${\xi }_z=s^{\prime }(x^2+y^2)/4$. Under the linearized coordinate gauge transformation, the mixed time/transverse and longitudinal/transverse components cancel, while the other time/longitudinal changes are proportional to $s^{\prime \prime }=0$. The remaining components are $h_{xx}=a-b$, $h_{yy}=-(a-b)$, $h_{xy}=0$, a transverse-traceless gauge perturbation of plus polarization. This explicitly removes a nonzero affine trace without assuming it vanishes by boundary conditions. If $(a-b{)}^{\prime \prime }=0$ too, all linearized curvature vanishes.

The source-free Linearized Einstein equations. For a metric perturbation about the Minkowski metric, imposing the Lorenz gauge in linearized gravity reduces them to $\square h_{\mu \nu }=0$ together with ${\partial }^{\mu }{\bar{h}}_{\mu \nu }=0$, where $\bar{h}$ is the trace-reversed metric perturbation. A linearized coordinate gauge transformation preserves $G^{(1)}[h]=0$, while preserving the imposed gauge additionally requires its generator to satisfy the wave equation.

Use the active sign convention for the linearized coordinate gauge transformation, $h_{\mu \nu }\mapsto h_{\mu \nu }+{\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$. The paper's prescribed generator has precisely this sign; using the passive convention would require its negative. Direct partial differentiation givesThe middle terms cancel when the indices are symmetrized. ThereforeTo test both equations (3), put $H(u)=\tilde{h}(u)={\eta }^{\kappa \sigma }{\tilde{h}}_{\kappa \sigma }(u)$, the trace of the original metric perturbation. Since $n$ is a null vector, $\hat{h}=0$ and its trace-reversed metric perturbation equals $\hat{h}$. Differentiating the result of part (a) givesIn the second identity, the term with $n^2$ is zero. ThusIn general $\hat{h}$ does not satisfy (3). On an open interval of $u$, both equations hold exactly when $H^{\prime \prime }=0$ there: the first equation forces this for $u\ne 0$, and smoothness gives it at $u=0$ as well. A linearized coordinate gauge transformation preserves the linearized vacuum Einstein equations, but only a residual gauge symmetry of linearized gravity preserves the imposed Lorenz gauge in linearized gravity.

For an explicit failure, use the null coordinates of part (c) and take only ${\tilde{h}}_{uv}={\tilde{h}}_{vu}=u^2/4$ nonzero. Then $H=u^2$, the original field obeys (3), but $\varphi =uv/2$ gives ${\partial }^{\mu }{\overline{\hat{h}}}_{\mu u}=u$ and $\square {\hat{h}}_{uu}=2$.

Continue with the active sign convention for the linearized coordinate gauge transformation. Choose local functions $A,B,C,F$ of $u$ by integration so thatFor example, for any fixed $u_0$ one may takeA generator, specified as a covector with components in the null coordinates $(u,v,x,y)$, isIndeed, with $\delta h_{AB}={\partial }_A{\chi }_B+{\partial }_B{\chi }_A$, the mixed components cancel:All components without a $u$ index vanish as well, andSince $n^{♭}=du$, only the $uu$ component needs changing. ThusYes, $\mathring{h}$ satisfies both equations (3). Its trace is zero, ${\partial }_v\psi =0$ gives zero divergence, and the harmonic quadratic plane-wave profile givesThe two surviving functions encode the two gravitational wave polarizations; constant and linear transverse terms have been removed by the explicit linearized coordinate gauge transformation.

Use $c=1$, background Minkowski spacetime metric tensor ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, and $g_{ab}={\eta }_{ab}+h_{ab}$ with small metric perturbation. All indices and derivatives in this linear approximation use $\eta$. The linearized inverse metric is ${\eta }^{ab}-h^{ab}$, and the linearized Levi-Civita connection isDefine the trace-reversed metric perturbation ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$, with $h={\eta }^{ab}{h}_{ab}$.

An infinitesimal coordinate change $x^{\prime a}=x^a+{\xi }^a$ gives the linearized coordinate gauge transformationThe metric perturbation therefore has redundant components. The gauge invariance of the linearized Riemann tensor follows because its change contains third derivatives of $\xi$ cancelling by commutation of partial derivatives. Conversely, flat linearized metric perturbations are locally pure gauge on a contractible patch. Thus linearized curvature, rather than a particular coordinate value of $h$, detects the physical disturbance. Gauge freedom can also be restricted by global topology or boundary conditions.

Keeping the paper's curvature sign convention, its linearized Ricci tensor and scalar areThese are the negatives of the alternative frequently used curvature convention. Put $v_b={\partial }^a{\bar{h}}_{ab}$. The Linearized Einstein equations in vacuum areUnder a gauge change $v_b\mapsto v_b-\square {\xi }_b$, so solving $\square {\xi }_b=v_b$ imposes the Lorenz gauge in linearized gravity. The field equations then become $\square {\bar{h}}_{ab}=0$. The remaining residual gauge symmetry of linearized gravity has $\square {\xi }_b=0$; Lorenz gauge does not exhaust the coordinate freedom.

A Fourier mode ${\bar{h}}_{ab}=A_{ab}{e}^{i{k}_c{x}^c}$ obeys $k^a{k}_a=0$ and $k^a{A}_{ab}=0$. Hence the plane gravitational waves in linearized gravity propagate on the background light cones. Four transversality conditions and four residual gauge amplitudes leave $10-4-4=2$ physical degrees of freedom. The explicit plane-wave reduction to transverse-traceless gauge can set $h_{0a}=0$, spatial trace zero and $k^i{h}_{ij}=0$. For a wave in the $z$ direction, write its physical spatial strain as $H_{ij}=-h_{ij}$, so the spatial metric tensor is $-({\delta }_{ij}+H_{ij})$:There are two gravitational wave polarizations, plus polarization and cross polarization. Rotating transverse axes through $\theta$ rotates their amplitude pair through $2\theta$. Nonconstant profiles with nonzero second derivative have nonzero linearized Riemann curvature tensor and cannot be removed as pure gauge.

The TT coordinate separation and measured separation distinction explains a detector's response. Freely falling particles initially at rest can keep fixed TT coordinates, since ${\Gamma }^a{}_{00}=0$, while their proper separation changes. For a short arm along a unit vector $n$,This is the tidal response described by geodesic deviation; coordinate motion by itself is not the measured signal.

Finally, gravitational-wave energy is second order in the perturbation, so it is absent from a first-order vacuum equation. In a short-wavelength averaging regime the averaged stress-energy of transverse gravitational waves isFor a plane wave its energy flux is $(16\pi G{)}^{-1}⟨{\dot{H}}_{+}^2+{\dot{H}}_{\times }^2⟩$. The averaging scale and weak-field assumptions matter: this does not assign a coordinate-independent local gravitational energy density to an arbitrary first-order perturbation.

Raise perturbation indices with the background Minkowski metric. To first order, the inverse metric is $g^{ik}={\eta }^{ik}-ϵh^{ik}+O({ϵ}^2)$ and the Levi-Civita connection isQuadratic connection products in the Riemann curvature tensor are of order ${ϵ}^2$. Substitution into the paper's curvature convention givesThe signs follow the convention on the cover sheet; reversing the curvature definition would reverse this expression.

The gauge freedom is the freedom to identify points of the perturbed spacetime with points of the flat background in slightly different coordinates. Under the passive infinitesimal change $x^{\prime i}=x^i+ϵ{\xi }^i(x)$, the metric transformation law gives the linearized coordinate gauge transformationThus a nonzero perturbation can partly represent a coordinate change rather than physical curvature. Substituting this variation into the displayed curvature formula givesEach third derivative has a partner differing only by the order of its commuting partial derivatives. This explicitly proves gauge invariance of the linearized Riemann tensor. The restriction to first order matters: a coordinate change acts on nonzero background curvature as well, but the background here is flat.

Define the trace-reversed metric perturbation and d'Alembert operator byIn four dimensions $\bar{h}=-h$ and $h_{ik}={\bar{h}}_{ik}-\frac{1}{2}{\eta }_{ik}\bar{h}$. Contract the curvature formula using $R_{ik}=R^m{}_{imk}$:up to $O({ϵ}^2)$. Consequently the Einstein tensor isChoose Lorenz gauge in linearized gravity, ${\partial }^k{\bar{h}}_{ik}=0$. It can be imposed locally by solving $\square {\xi }_i={\partial }^k{\bar{h}}_{ik}$, since the gauge transformation changes this divergence by $-\square {\xi }_i$. The remaining coordinate freedom satisfies $\square {\xi }_i=0$. In this gauge,This is the requested Linearized Einstein equations in the stated sign convention. Their divergence requires ${\partial }^k{T}_{ik}=0$ to this order.

For the photon beam, introduce $u=t-x$ and the constant null covector ${\ell }_i=(1,-1,0,0)$, so ${\ell }^i=(1,1,0,0)$. The stress-energy tensor isIt is trace-free and conserved: ${\ell }^i{\partial }_i\rho (u)=0$. A beam-adapted particular solution, with no independently added homogeneous gravitational field, isThe equality follows because its trace is zero, and the Lorenz condition holds because ${\ell }^k{\partial }_{k}H=0$. The field equation becomes a transverse Poisson equation:For the transversely uniform source in the question, one particular solution is $H=4\pi G\rho (u)(y^2+z^2)$. Thus the sourced components can be chosen asBoundary conditions and homogeneous solutions determine additional freedom; the source does not uniquely specify every component in every gauge. In particular, taking $H$ to depend only on $t-x$ would give $\square H=0$ and cannot solve the equation for nonzero $\rho$. The uniform idealization is not an asymptotically flat finite beam, and its quadratic particular solution is a weak-field description only where $|ϵH|\ll 1$.

Finally consider another photon with tangent proportional to ${\ell }^i$. Its tangent remains null because $h_{ik}{\ell }^i{\ell }^k=0$. Moreover $h_{ik}{\ell }^k=0$, and since $\ell$ is constant, the connection contraction isThe geodesic equation is therefore satisfied by a straight affinely parametrized ray along the beam. Hence co-propagating photons experience no deflection from the beam's gravitational field. This is parallel photon beams in linearized gravity; it does not assert absence of forces on counter-propagating photons or effects from unrelated homogeneous gravitational perturbations.

A linearized coordinate gauge transformation changes the identification of points of the perturbed spacetime with its Minkowski spacetime background. With $x^{\prime i}=x^i+ϵ{\xi }^i$, the same physical metric has perturbationThese changes are coordinate freedom, not extra physical gravitational wave polarizations. Raising indices with $\eta$, define $h={\eta }^{ij}{h}_{ij}$ and the trace-reversed metric perturbation ${\bar{h}}_{ij}=h_{ij}-{\eta }_{ij}h/2$. Then

The first-order Levi-Civita connection is ${\Gamma }^{(1)k}{}_{ij}=\frac{1}{2}{\eta }^{k\ell }({\partial }_i{h}_{j\ell }+{\partial }_j{h}_{i\ell }-{\partial }_{\ell }{h}_{ij})$. Contracting the Riemann curvature tensor in the stated convention givesPut $A_j={\partial }^i{\bar{h}}_{ij}$. The corresponding Einstein tensor isUnder the linearized coordinate gauge transformation, $A_j^{\prime }=A_j-\square {\xi }_j$. Choosing a solution of $\square {\xi }_j=A_j$ imposes Lorenz gauge in linearized gravity, $A_j^{\prime }=0$. Thus the vacuum Einstein field equations reduce toThere remains residual gauge symmetry of linearized gravity with $\square {\xi }_j=0$.

For the transverse profiles, write both symmetric off-diagonal entries as $h_{xy}=h_{yx}=h_{\times }(t-z)$ and the diagonal entries as $h_{xx}=-h_{yy}=h_{+}(t-z)$. The trace-reversed metric perturbation has zero trace and therefore equals $h$. Its divergence vanishes: its only nonzero index directions are $x,y$, while its coefficients depend only on $t-z$. Also $(-{\partial }_t^2+{\partial }_z^2)F(t-z)=0$ for every twice differentiable profile $F$. These are consequently plane gravitational waves in linearized gravity in transverse-traceless gauge, with the two independent gravitational wave polarizations $h_{+}$ and $h_{\times }$.

To calculate the detector response, use a parallel-propagated orthonormal frame along the centre of a short nonrotating stick. To first order its clock reads $t$, and the tidal curvature isFor a sliding-bead gravitational wave detector, the stick constrains transverse motion but supplies no axial restoring force. Projecting geodesic deviation along an $x$-directed stick givesHere $L$ is the proper separation and $L_0=L(t_0)$, with $\dot{L}(t_0)=0$. Integrating with those actual initial conditions yieldsFor beads at rest before an incident wave, both initial profile terms vanish and the familiar result is $\boxed{L(t)=L_0[1+ϵh_{+}(t-z_0)/2]+O({ϵ}^2)}$. The cross polarization produces a transverse tidal force, balanced by the guide, and no axial first-order displacement in this orientation. In transverse-traceless gauge the same plus response can be read from the proper line element, $d\ell =\sqrt{1+ϵh_{+}}dx$; coordinate distances alone are not measured distances. When a cross polarization is present, keeping the guide physically nonrotating may require transverse coordinate adjustments, which do not change the axial answer at this order.

For a $z$-directed stick, $R_{z0z0}=0$ and the wave is transverse. Therefore beads initially at rest retain their proper separation to first order: $L(t)=L_0+O({ϵ}^2)$. The tidal calculation is the usual linear detector limit, with bead spacing small compared with the wavelength; it does not treat a macroscopic rigid body as exactly rigid over arbitrary relativistic distances.

Use linearization around Minkowski spacetime after constant rescaling of $x,y$: set $A=1+\epsilon a(u)$, $B=1+\epsilon b(u)$, $u=t-z$. To first order,Hence $s=a+b$ is affine in $u$, while $h=a-b$ is the freely propagating profile. The affine transverse trace in a diagonal linearized plane wave is pure gauge: it has zero linearized curvature and is locally a coordinate artifact. To see this without imposing extra boundary conditions, use a first-order linearized coordinate gauge transformation $h_{\mu \nu }\mapsto h_{\mu \nu }-{\partial }_{\mu }{\xi }_{\nu }-{\partial }_{\nu }{\xi }_{\mu }$, withThe $tx,ty,zx,zy$ changes cancel and the remaining time/longitudinal changes are proportional to $s^{\prime \prime }=0$. The transverse perturbation becomeswith all time and longitudinal components zero. It is a transverse-traceless gauge plane gravitational wave travelling in the positive $z$-direction, with plus polarization and no cross polarization in these transverse axes.

For freely falling nearby particles, geodesic deviation gives opposite transverse tidal accelerations, since $R_{txtx}=-\epsilon a^{\prime \prime }$ and $R_{tyty}=-\epsilon b^{\prime \prime }=+\epsilon a^{\prime \prime }$ in vacuum. Thus the transverse geodesic deviation eigenvalues have opposite signs; the stretching and compressing roles interchange when the curvature profile changes sign. If $h^{\prime \prime }=0$ as well, there is no linearized tidal wave: that special profile is also locally flat to this order. The exact diagonal metric describes a fixed transverse polarization through its opposite vacuum curvature eigenvalues; the linearized calculation identifies it as plus relative to the chosen axes.