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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.

For a short pair of stationary test particles in transverse-traceless gauge, spatial coordinates stay fixed while the measured proper length changes. In background signature $+---$, let $h_{xx}=-h_{yy}=H_{+}$ and $h_{xy}=H_{\times }$. A coordinate separation of length $\Delta$ at angle $\theta$ hasThe plus polarization and cross polarization differ by a $45$-degree rotation of their principal axes. With background signature $-+++$ and the same numerical spatial perturbation, the sign reverses because the positive spatial metric is then ${\delta }_{ij}+h_{ij}$ rather than ${\delta }_{ij}-h_{ij}$. Consistently redefining the signed strain leaves the physical effect unchanged.

In a weak plus-polarized gravitational wave, initially stationary test particles have constant spatial coordinates in transverse-traceless gauge because ${\Gamma }^i{}_{00}=0$. Their physical separations nevertheless change. For a small transverse ring, $X=(1+h_{+}/2)x$ and $Y=(1-h_{+}/2)y$, so its principal semiaxes oscillate as $a(1\pm h_{+}/2)$. The radial response has the displayed angular dependence. These are local distance measurements in the long-wavelength limit; finite-arm light-travel times require the full detector response.

In the radiation zone, the time-averaged energy flux of a weak wave in transverse-traceless gauge isCombining this with the far-zone quadrupole field and integrating over directions gives the quadrupole formula.

For a nonzero plane-wave trace-reversed metric perturbation,The wave equation therefore requires a null wave vector. The Lorenz gauge in linearized gravity supplies transversality:The amplitude is symmetric because it represents a metric perturbation; its real part is the physical wave. The two displayed conditions do not by themselves make it spatially transverse and traceless: the additional reduction to transverse-traceless gauge uses the remaining residual gauge symmetry of linearized gravity.

Take the real physical wave and write $h(t,z)=A\cos [k(z-ct)]$. The specified amplitude has zero trace, so its metric perturbation equals its trace-reversed metric perturbation. It is in transverse-traceless gauge:to first order. Assume the freely falling particles are initially at rest in these coordinates. Since $g_{00}=-1$ and $g_{0i}=0$ in coordinates $(ct,x,y,z)$,The geodesic equations therefore admitUniqueness preserves these initial conditions. Thus the coordinate locations remain fixed; this conclusion does not apply to arbitrary initially moving particles.

Coordinate position is not measured separation. In the central observer's local orthonormal frame in spacetime, and in the long-wavelength limit appropriate to a small detector, the transverse physical separations areto first order. In particular the two principal semiaxes of the ring areFor a point at angular position $\phi$ the radial displacement isThe freely falling ring in transverse-traceless gauge alternates between ellipses elongated in the $x$ and $y$ directions. Its area is unchanged to first order. This is the plus gravitational wave polarization; the axes exchange their roles after half a period.

An observer at the center detects these changing distances, for example by exchanging light signals with the particles. Equivalently geodesic deviation gives the local tidal accelerationto the same order. There is physical relative motion and nonzero Riemann curvature tensor, even though the selected coordinate positions are constant. For a detector comparable to a wavelength, its light-travel-time response must be included rather than treating all distances as simultaneously measured local lengths.

Use metric signature $-+++$ and transverse-traceless gauge, $g_{ij}={\delta }_{ij}+h_{ij}^{\mathrm{TT}}$. To resolve the source's origin ambiguity, put the detector centre at $z=0$ and the source centre at $z=-r_0$, so the gravitational wave travels in the positive $z$ direction. Reversing that direction replaces $z$ by $-z$. Work in the weak-field approximation, with $\omega l\ll c$, $r_0\gg l,c/\omega$, and detector size $a\ll c/\omega ,r_0$. The radiation reaction is neglected, so the rod maintains the prescribed steady motion.

At the retarded source time $u=t-(r_0+z)/c$, take the source positions as ${\mathbf{x}}_{\pm }=\pm l(\cos (\omega u+\beta ),\sin (\omega u+\beta ),0)$. The arbitrary constant $\beta$ specifies the initial source phase. The second mass moment tensor $I_{ij}={\sum }_{A}M{x}_{Ai}{x}_{Aj}$ obeysThe quadrupole formula $h_{ij}^{\mathrm{TT}}=2G{\ddot{I}}_{ij}^{\mathrm{TT}}/(c^4{r}_0)$ projects away the transverse trace, which is already constant here. Differentiating twice therefore givesAll time and longitudinal metric perturbation components vanish in transverse-traceless gauge. The gravitational-wave frequency is $2\omega$ in angular units; $M$ is each source mass, not their sum, and the detector mass $m$ does not enter the strain. The plane wave approximation holds the amplitude fixed at $r_0$; retaining spherical spreading replaces its denominator by $r_0+z$ near the axis. The retarded quadrupole formula and the distinction between source and wave frequency are derived in ^[ref].

Displacement must be specified relative to an observer. In transverse-traceless gauge the free particles can have fixed coordinates, while their measured proper lengths oscillate. In a local, freely falling orthonormal frame, the geodesic deviation equation for an initial separation $\mathbf{X}=(X,Y,0)$ is, to first order in $H$,Its general solution with physical displacement $\mathbf{\xi }(0)=0$ and initial physical velocity $\mathbf{V}$ isThe source has been running indefinitely, so there is no wave-free initial state. Its initial conditions for an eternal monochromatic gravitational wave need more than positions to fix $\mathbf{V}$. The usual bounded, periodic response selects $V_i={\dot{h}}_{ij}(0,0)X^j/2$; if the particles instead start at physical rest at $t=0$, set $V_i=0$ and retain the linear term. Both solve the same linear geodesic deviation problem. The distinction between fixed TT coordinates and measured separation is explained in ^[ref].

Choose $\beta$ so the wave phase at the detector is $\Phi (t,0)=2\omega t$. For the bounded response the four initial placements then giveThese are displacements from the specified positions at $t=0$; positions are $\mathbf{X}+\mathbf{\xi }$. Equivalently, relative to mean positions the usual oscillatory displacement is ${\mathbf{\xi }}_{\mathrm{osc}}=h\mathbf{X}/2$ to first order. The response repeats after $\pi /\omega$.

Both plus polarization and cross polarization occur with equal amplitude and a quarter-cycle phase difference. Their superposition is circular gravitational-wave polarization, with handedness reversed when the viewing direction or source rotation is reversed. The principal stretching axis turns through half the wave phase. There is no leading longitudinal motion.

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.

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.

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 $\pi /4$.

Put $A,B\in \{1,2\}$ and write $h_{AB}=H_{AB}{e}^{ik\cdot x}$, where $H_{11}=H_{+}$, $H_{22}=-H_{+}$ and $H_{12}=H_{21}=H_{\times }$. Physical fields are the real parts. Use $c=1$ here, and take the initial separation at equal time, so $L^0=0$. For propagation in the positive $x^3$ direction one may set $k_{\mu }=(-\omega ,0,0,\omega )$; only $k_0^2={\omega }^2$ enters the answer.

The linearized Levi-Civita connection in this transverse-traceless gauge has the potentially nonzero componentswith all others zero to this order. In particular ${\Gamma }^{\mu }{}_{00}=0$, so a particle initially at rest has fixed spatial coordinates in transverse-traceless gauge, and $t$ equals its proper time. Expanding the Riemann curvature tensor givesThe plus sign in the last expression agrees with $R(T,S)T$ in the geodesic deviation convention above.

For the separation measured in a parallel-propagated orthonormal frame, write $S^{\hat{\mu }}=L^{\hat{\mu }}+{\xi }^{\hat{\mu }}$. The frame removes the connection terms from the time derivative, so the geodesic deviation equation reduces to ${\ddot{\xi }}^{\hat{A}}=R^{\hat{A}}{}_{\hat{0}\hat{0}\hat{B}}{L}^{\hat{B}}={\ddot{h}}_{AB}{L}^{\hat{B}}/2$ at first order. The observable transverse relative acceleration isThese are the plus polarization and cross polarization tidal patterns; there is no longitudinal acceleration. The curvature is evaluated along the reference worldline, and replacing $S$ by $L$ on the right changes only second-order terms. This is the infinitesimal-separation approximation intrinsic to geodesic deviation.

There is an important component distinction if the symbols $L^{\mu }+{\xi }^{\mu }$ are instead interpreted literally in the transverse-traceless gauge coordinate basis. Expanding the left side, rather than replacing a covariant derivative by an ordinary derivative, givesThus in TT coordinatesWith zero initial coordinate displacement and velocity, ${\xi }_{\mathrm{coord}}=0$ throughout. This does not remove the physical gravitational wave strain. To first order an orthonormal coframe in spacetime along the reference worldline is ${\vartheta }^{\hat{A}}=({\delta }^A{}_B+h_{AB}/2)d{x}^B$; its dual frame is parallel propagated because ${\partial }_0({\delta }^A{}_B-h_{AB}/2)+{\Gamma }^A{}_{0B}=0$. Henceafter choosing the initial flat-frame reference. The TT coordinate separation and measured separation distinction accounts exactly for the apparently different accelerations. In a region initially free of the wave, the measured displacement is ${\xi }^{\hat{A}}=h_{AB}{L}^{\hat{B}}/2$ with the corresponding initial conditions.

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.

Use $G=c=1$ and the physical real part of the printed plane wave. Its null wavevector obeys $k^{\alpha }{k}_{\alpha }=0$, so each component satisfies $\square h_{\alpha \beta }=0$. Direct differentiation of the defined stress tensor givesThe trace is proportional to $k^{\nu }{k}_{\nu }$ for a single plane wave and hence vanishes. ThereforeThis is local conservation of energy and momentum conservation; spatial integration gives conserved total charges when the boundary flux vanishes. For a finite-amplitude complex phasor, one takes the real field before quadratic products and may time-average: $⟨{\tau }_{\mu \nu }⟩=k_{\mu }{k}_{\nu }{H}^{\alpha \beta }{\overline{H}}_{\alpha \beta }/(64\pi )$ in transverse-traceless gauge. Its energy density is nonnegative and its flux travels in the positive $z$ direction.

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.

A spatial rotation and rescaling put the future null vector in the form $k^{\mu }=(1,0,0,1)$, so the profile depends on $z-t$. Transversality gives four linear relations among the ten symmetric components. The four residual gauge functions satisfying $\square {\xi }^{\mu }=0$ remove the time and longitudinal components; the remaining trace can be removed by the residual transformation indicated in the question. The resulting transverse-traceless gauge isThe two arbitrary functions are the plus and cross gravitational-wave polarizations. They are the two physical degrees of freedom left after the four gauge conditions and four residual coordinate freedoms are removed.

All time components of the perturbation vanish. For $\mu =t$ this immediately givesFor a spatial index $\mu =i$,by the stated property. Thus the perturbation is transverse.

Its Minkowski trace is purely spatial:Adding the displayed components and collecting the coefficients of the independent functions $A$, $B$, and $C$ makes each coefficient vanish separately, soTogether with $h_{t\mu }=0$ and ${\partial }^{\nu }{h}_{\mu \nu }=0$, this proves that the wave is in transverse-traceless gauge.