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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.

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.

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.

Let $I^{ij}(t)=\int T^{00}(t,\mathbf{x})x^i{x}^j{d}^{3}x$, using spatial indices $i,j$. Stress-energy conservation, symmetry of the stress-energy tensor and integration by parts giveand a second derivative givesAll boundary terms vanish by compact support. This proves the stress-energy quadrupole identity,The support is understood to remain bounded over the time interval used for differentiating these integrals.

For the radiation calculation set $c=1$, write $g_{ab}={\eta }_{ab}+h_{ab}$ and use the trace-reversed metric perturbation ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$ in Lorenz gauge in linearized gravity. With the curvature and field-equation signs in this paper, the linear equation isThe retarded solution with no incoming radiation is consequentlyLet $r=|\mathbf{x}|$ be the distance from the source's centre of mass, and let $d$ be its size. In the radiation zone, $r\gg d$. The leading multipole expansion additionally assumes slow internal motion, or $d$ small compared with the radiation wavelength. These assumptions allow $|\mathbf{x}-{\mathbf{x}}^{\prime }|\simeq r$ in the denominator and the common retarded time $u=t-r$ at leading quadrupole order. ThusDefine the trace-free mass quadrupole moment byTo make the requested normalization explicit, write the trace-free spatial metric coefficient as $h_{ij}^{\mathrm{TF}}=-2E_{ij}$. Trace reversal only adds a spatial trace, so the preceding equations give the far-zone quadrupole strain coefficientThe supplied scan does not contain the referenced perturbation handout. The definition above fixes the normalization of $E$ directly. If $E$ denotes the physical transverse-traceless tensor in that handout, the right-hand side is also to be transversely projected for the observation direction. Trace-free alone does not imply transverse. With $\mathbf{n}=\mathbf{x}/r$, putThe transverse-traceless projector gives $E_{ij}^{\mathrm{TT}}={\Lambda }_{ij,kl}{E}_{kl}$ and the physical spatial strain is $H_{ij}^{\mathrm{TT}}=2E_{ij}^{\mathrm{TT}}$. This also states the result without relying on omitted handout notation. The retarded construction and direction-dependent transverse projection are described in Carroll's general relativity lectures, with the appropriate changes of metric-sign convention.

For the equal-mass binary star, its centre of mass is at the origin and the separation is $2R$. The gravitational force on either star is $G{m}^2/(2R{)}^2$, while its centripetal force is $mR{\omega }^2$. HenceAt leading Newtonian order, $T^{00}$ is the sum of the two point-mass densities. Its moments areso the trace-free mass quadrupole moment isThe Newtonian bound system is treated in the usual leading weak-field quadrupole approximation; an exactly conserved effective source also includes the interaction stresses. Accelerated point-mass dust alone would not satisfy the conservation hypothesis.

Differentiating twice and evaluating at $u=t-r$ givesFor an observer on the $z$ axis this is already transverse and traceless. Its alternating plus polarization and cross polarization have a phase difference of $\pi /2$, giving circularly polarized gravitational radiation at twice the orbital frequency. The amplitude decays as $1/r$, as expected for outgoing radiation in three spatial dimensions.

More generally take the line of sight in the $xz$ plane at inclination $\iota$ to the orbital angular-momentum axis. In the transverse basis ${\mathbf{e}}_{\theta }=(\cos \iota ,0,-\sin \iota )$, ${\mathbf{e}}_{\varphi }=(0,1,0)$, define $A=4Gm{R}^2{\omega }^2/r$. Projection gives the physical gravitational wave polarizationsThus a generic view is elliptically polarized, the face-on view is circularly polarized and the edge-on view is linearly polarized. This is circular-binary quadrupole radiation; an overall sign or transverse-axis rotation only changes the polarization convention.

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.

With $u=t-r$, direct integration givesTherefore the time-dependent spatial field isThe gravitational-wave frequency is $2\omega$. On the positive $z$-axis this matrix is transverse and traceless. It is a rotating combination of the plus polarization and cross polarization, with amplitude proportional to $1/z$, exactly as for a plane wave propagating in the $z$ direction.

An independent physical solution is the cross polarizationIt is symmetric, traceless and transverse to $k^{\mu }=(\omega ,0,0,\omega )$. Unlike the pure coordinate mode, it has nonzero linearized Riemann curvature operator: for example, in the convention $R_{0i0j}^{(1)}=-{\partial }_t^2{h}_{ij}/2$,Changing the curvature sign convention reverses this component but cannot make it vanish. Gauge invariance of the linearized Riemann tensor therefore proves that this nonzero mode cannot be removed by a coordinate transformation.

The principal stretch directions are rotated by $45$ degrees relative to the plus polarization. In the same proper-length convention as part (c), its first-order angular response is $\delta \ell /\Delta =-\frac{1}{2}{h}_{\times }\sin (2\theta )\cos [\omega (t-z_0)]$. The two independent transverse patterns are the two propagating gravitational degrees of freedom.