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In units $c=1$, a slowly moving isolated source has trace-free mass quadrupole moment $Q^{ij}=\int T^{00}(x^i{x}^j-{\delta }^{ij}|\mathbf{x}{|}^2/3)d^{3}x$. The retarded linearized gravity solution and the stress-energy quadrupole identity give the displayed physical spatial strain, with $\mathbf{n}=\mathbf{x}/r$ and the transverse-traceless projector $\Lambda$. If the spatial metric is written $g_{ij}=-{\delta }_{ij}-2E_{ij}$ for metric signature $(+---)$, its trace-free Lorenz-gauge coefficient before the transverse projection is $E_{ij}=G{\ddot{Q}}_{ij}(t-r)/r$. Thus $H^{\mathrm{TT}}=2E^{\mathrm{TT}}$. Far distance alone does not permit dropping source-dependent retardation: the quadrupole approximation also requires the source size to be small compared with the radiation wavelength.

Let ${\gamma }_{\alpha \beta }={\delta }_{\alpha \beta }$ be the positive Euclidean spatial background metric. Write a tensor-sector metric perturbation as $h_{00}=h_{0\alpha }=0$, $h_{\alpha \beta }=-2E_{\alpha \beta }$, with ${\gamma }^{\alpha \beta }{E}_{\alpha \beta }=0$ and ${\partial }^{\alpha }{E}_{\alpha \beta }=0$. Spatial indices here are raised with $\gamma$, and $\square ={\partial }_t^2-\Delta$.

Substitute the linearized Levi-Civita connection into the cover's curvature definition. To first order,In the transverse-traceless tensor sector $h=0$ and ${\partial }^c{h}_{ac}=0$, so $\delta C=0$ andThe tensor part of the Linearized Einstein equations therefore givesIn a calculation strictly confined to the tensor sector, the source on this equation means its tensor projection; it is already trace-free there.

For a generic compact source, trace-free is not the same as transverse-traceless. For example, the static compactly supported spatial stress $T^{ij}={\partial }^i{\partial }^j\chi -{\gamma }^{ij}\Delta \chi$, with smooth compactly supported $\chi$, is divergence-free, but its trace-free part has divergence $2{\partial }^j\Delta \chi /3$, generally nonzero. There is a precise way to reproduce the unprojected trace-free formula used in the question while retaining this distinction. In Lorenz gauge in linearized gravity, the trace-reversed metric perturbation ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$ obeysDefine its spatial trace-free half-strain byThen its wave equation has exactly the displayed trace-free source $\delta {\hat{T}}^{\alpha \beta }=\delta T^{\alpha \beta }-{\gamma }^{\alpha \beta }\delta T^{\mu }{}_{\mu ,\mathrm{spatial}}/3$. The physical tensor radiation is obtained afterwards with the transverse-traceless projector. Without that projection, this $E$ is a Lorenz-gauge trace-free coefficient, not necessarily a transverse tensor mode. At leading far-zone order, the same directional projector is applied to both sides of the quadrupole field formula below.

The flat contracted Bianchi identity implies stress-energy conservation, ${\partial }_b\delta T^{ab}=0$. LetCompact support removes boundary terms. Differentiate and integrate by parts once:A second time derivative, another integration by parts and symmetry of the stress-energy tensor giveTake the spatial trace-free part. The mass quadrupole moment satisfies the stress-energy quadrupole identityInserting this into the assumed leading retarded quadrupole field givesThis is the requested spatial half-strain convention; the measured transverse-traceless strain is $H_{\alpha \beta }^{\mathrm{TT}}=2E_{\alpha \beta }^{\mathrm{TT}}$. The formula selects the retarded source field and does not include an independently added homogeneous gravitational wave. Dropping the source-dependent part of retardation also needs the long-wavelength source approximation, besides $R/r\ll 1$.

For a slowly moving weakly self-gravitating source, the virial theorem or the dynamical balance $v^2/R\sim GM/R^2$ gives $v^2\sim ϵ=GM/R$. A varying nonspherical mass quadrupole moment has characteristic size $M{R}^2$ and dynamical time $\tau \sim R/v$, so $\ddot{Q}\sim M{R}^2/{\tau }^2\sim M{v}^2$. Consequently the quadrupole strain scaling of a weak self-gravitating source isThis is a characteristic scale, not a nonzero lower bound: a stationary or exactly spherical source does not radiate. Slow motion gives $R/\tau \sim \sqrt{ϵ}\ll 1$, justifying the long-wavelength approximation; for a radiation-zone strain one also takes $r\gg \tau$.

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.

The same invariant interval can be written as $g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }={\tilde{g}}_{\alpha \beta }d{\tilde{x}}^{\alpha }d{\tilde{x}}^{\beta }$. Applying the chain rule to the differentials proves the metric transformation lawFor first-order scalar cosmological perturbations and ${\xi }^{\mu }=(T,{\partial }^{i}L)$, compare perturbations at the same background coordinate label. Expanding the Jacobians and the shifted background gives the linear metric gauge-transformation lawFor the background ${\bar{g}}_{00}=-a^2$, ${\bar{g}}_{ij}=a^2{\delta }_{ij}$, the Lie derivative components areHere primes denote conformal time and $\mathcal{H}$ is the conformal Hubble parameter. These expressions establish all four alternatives, including any chosen pair.

The $00$ component is $\delta g_{00}=-2a^{2}A$, immediately giving the lapse function perturbation transformation. The $0i$ component is $a^2{\partial }_{i}B$, giving the shift vector scalar-potential transformation. The trace of the spatial perturbation is $6a^{2}C$, so tracing the spatial Lie derivative gives the transformation of $C$. Its trace-free scalar part is $2a^2({\partial }_i{\partial }_j-{\delta }_{ij}{\nabla }^2/3)E$, giving the transformation of $E$. ThusAs usual, identifying scalar potentials from their derivatives uses the standard boundary conditions, or nonzero Fourier modes, to remove homogeneous ambiguities.

At first order, tensor cosmological perturbations are spatial transverse-traceless tensors. The coordinate-generated spatial perturbation consists of a trace term and symmetrized derivatives of the displacement. In Fourier space, the latter terms carry a factor $k_i$ or $k_j$. The transverse-traceless projector removes these longitudinal terms and the trace, including the derivative of a transverse vector displacement. Therefore the tensor perturbation is gauge invariant at linear order around the homogeneous background. This is a first-order statement, not a claim of automatic invariance at arbitrary perturbative order.

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.

Remove the constant trace from the second mass moment tensor to obtain the mass quadrupole moment $Q_{ij}=I_{ij}-{\delta }_{ij}{I}_{kk}/3$. Since $I_{kk}$ is constant, ${\stackrel{...}{Q}}_{ij}={\stackrel{...}{I}}_{ij}$. In units $G=c=1$, the quadrupole formula givesIts normalization can also be seen from the gravitational-wave energy flux: $h_{ij}^{\mathrm{TT}}=2{\Lambda }_{ij,kl}{\ddot{Q}}_{kl}/r$, where $\Lambda$ is the transverse-traceless projector. With $S_{ij}={\stackrel{...}{Q}}_{ij}$ and $P_{ij}={\delta }_{ij}-n_i{n}_j$, the projection is $S_{ij}^{\mathrm{TT}}=P_{ik}{P}_{jl}{S}_{kl}-P_{ij}{P}_{kl}{S}_{kl}/2$. For a symmetric trace-free $S$,The isotropic tensor integrals $\int n_i{n}_{j}d\Omega =4\pi {\delta }_{ij}/3$ and $\int n_i{n}_j{n}_k{n}_{l}d\Omega =4\pi ({\delta }_{ij}{\delta }_{kl}+{\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk})/15$ therefore give $\int |S^{\mathrm{TT}}{|}^{2}d\Omega =8\pi S_{ij}{S}_{ij}/5$. Integrating the flux yields $⟨P⟩=(8\pi {)}^{-1}\int ⟨|S^{\mathrm{TT}}{|}^2⟩d\Omega =⟨S_{ij}{S}_{ij}⟩/5$, as above.

Now differentiate the components from part (i). With $\Delta =I_1-I_2$,The off-diagonal component occurs twice in the contraction. ThusIt is already time independent, so averaging givesThis is the leading gravitational radiation from a rotating triaxial body; restoring units multiplies it by $G/c^5$. The source is treated as rotating uniformly over an averaging interval, with radiation reaction negligible at this order.

The outgoing solution obtained from the retarded fundamental solution of the wave equation isIn the radiation zone, put $r=|\mathbf{x}|$, $\mathbf{n}=\mathbf{x}/r$, and retain the leading $1/r$ term at retarded time $u=t-r$. The trace term disappears after applying the transverse-traceless projectorsoTwice using stress-energy conservation, ${\partial }_a{T}^{ab}=0$, and integrating by parts givesThe projector removes the trace, so in terms of the mass quadrupole momentthe far field isThe gravitational-wave energy flux isFor a trace-free symmetric tensor $A_{ij}$, the isotropic tensor integral over the observation direction givesConsequently the standard quadrupole formula, in the units $G=c=1$ used by the paper, isThus the printed coefficient $4\pi /5$ is inconsistent with the stated definition of $Q_{ij}$ and the standard wave-energy normalization; it appears to be a typographical error.

For the planet, choose its circular orbit in the $xy$-plane and write $\omega =2\pi /\tau$. With the star treated as fixed,and henceDirect differentiation givessoRestoring units multiplies this by $G/c^5$. If the printed $4\pi /5$ coefficient is followed literally, the answer is instead $4\pi$ times larger,For two bodies of comparable mass, $M$ is replaced by the reduced mass and $R$ by their separation.

The equal-time canonical commutation relation for the transverse-traceless graviton and its canonical momentum is the transverse-traceless projector. At zero momentum the supplied polarization completeness relation giveswhere symmetry and tracelessness of ${\omega }_{ij}$ remove the trace term. Thus the charge generates precisely the transformation in part i:This is the soft, field-independent part of the Noether charge associated with the large diffeomorphism.

For a compact slowly moving source in the weak-field approximation, with no incoming homogeneous wave and $G=c=1$, the retarded Linearized Einstein equations give this leading far-zone trace-reversed metric perturbation. The key consequence of stress-energy conservation is ${\ddot{I}}_{ij}=2\int T_{ij}{d}^{3}x$. The physical gravitational wave is obtained by the transverse-traceless projector.

A transverse-traceless spatial tensor is orthogonal to its Fourier mode wavevector in each index and has zero trace. The transverse-traceless projector extracts this part. Trace terms and symmetrized coordinate-displacement gradients have no such part, explaining the linear gauge invariance of tensor cosmological perturbations.