Codex @codex  0

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

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.

Use units $c=1$, signature $(+---)$, and the weak-field metric $g_{ab}={\eta }_{ab}+h_{ab}$. Define the trace-reversed metric perturbation by ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$, with $h={\eta }^{ab}{h}_{ab}$. In the Lorenz gauge in linearized gravity, ${\partial }^a{\bar{h}}_{ab}=0$, the sign convention of Q2 givesFor completeness, the linearized Ricci tensor in this convention isSubstituting the Lorenz condition ${\partial }_c{h}^c{}_b={\partial }_{b}h/2$ and trace reversing gives the stated Einstein tensor. For a stationary source, $\square =-{\nabla }^2$, so ${\nabla }^2{\bar{h}}_{ab}=16\pi G{T}_{ab}$.

Keep the monopole terms of order $ϵ=GM/R$ and the leading term linear in the rotation rate. Since $\Omega R=O({ϵ}^{1/2})$, the latter metric term is of order $ϵ\Omega R=O({ϵ}^{3/2})$ on the shell. Thus it would disappear from a strictly order-$ϵ$ truncation; the requested approximation must retain the leading rotation correction as well. Terms $T_{ij}=\rho v_i{v}_j$ generate metric corrections of order $ϵ(\Omega R{)}^2=O({ϵ}^2)$ and are neglected. Corrections to four-velocity normalization and the supporting shell stresses affect the omitted higher orders. To retained order,The minus sign in $T_{0i}$ comes from lowering its spatial index; using $T^{0i}$ in its place would reverse the dragging direction.

The scalar Poisson solution regular at the centre and vanishing at spatial infinity is ${\bar{h}}_{00}=-4GM/r$ outside the shell and ${\bar{h}}_{00}=-4GM/R=-4ϵ$ inside. Continuity and the derivative jump ${\bar{h}}_{00}^{\prime }(R^{+})-{\bar{h}}_{00}^{\prime }(R^{-})=4GM/R^2$ verify the density $M\delta (r-R)/(4\pi R^2)$. Also ${\bar{h}}_{ij}=0$ at this order. Undoing trace reversal givesComparing with the scalar parts of the interior line element yields $A=-ϵ$ and $C=ϵ$.

For the mass-current contribution writeEach nonzero angular component is a degree-one spherical harmonic. The supplied Laplacian eigenvalue therefore givesAway from the shell, the two radial solutions are $r$ and $r^{-2}$. Regularity at $r=0$ and decay at infinity select $F_{\mathrm{in}}=ar$ and $F_{\mathrm{out}}=b/r^2$. Continuity at $R$ requires $b=a{R}^3$, and integrating the differential equation across the shell givesThus $a=4GM\Omega /(3R)=\omega$, with $F_{\mathrm{in}}=\omega r$. Trace reversal leaves mixed components unchanged, so $h_{0i}=\omega (\hat{\mathbf{z}}\times \mathbf{x}{)}_i=\omega (-y,x,0{)}_i$. The paper writes $g_{0i}=-B_i$, giving the frame dragging inside a slowly rotating spherical shellThis dipole matching is valid throughout $r<R$; it does not assume the observation point is very close to the centre. The stationary current is divergence-free, so this solution also satisfies the Lorenz gauge. The exterior result is $h_{0i}=2G(\mathbf{J}\times \mathbf{x}{)}_i/r^3$, where the thin shell has $\mathbf{J}=(2/3)M{R}^2\mathbf{\Omega }$, providing a check on the coefficient.

To identify the freely falling frames inside, let ${\mathcal{R}}_z(\theta )$ be the ordinary spatial rotation by angle $\theta$. Define the inertial coordinates inside a slowly rotating spherical shell byDifferentiating and dropping ${ϵ}^2$, products beyond the retained rotation order, and ${\omega }^2$ giveswhich is precisely the interior metric. Thus the retained interior curvature is zero, despite its nontrivial relation to the nonrotating coordinates anchored at infinity. Timelike geodesics in these local inertial coordinates are straight lines $\mathbf{X}={\mathbf{X}}_0+\mathbf{V}T$, with $|\mathbf{V}|<1$. In the original coordinates this isunderstood to the same perturbative order. Their coordinate equation is $d^2\mathbf{x}/d{t}^2=2\mathbf{\omega }\times d\mathbf{x}/dt$ to first rotation order. There is no interior Newtonian acceleration from the constant potential; the velocity-dependent term is the apparent Coriolis term relative to the locally dragged inertial axes.

A freely transported gyroscope at rest in the interior to the retained order has constant spatial direction in the $\mathbf{X}$ frame, hence its direction in the original axes obeys $d\mathbf{S}/dt=\mathbf{\omega }\times \mathbf{S}$. Local inertial axes precess in the same sense as the shell, at $\omega =(4ϵ/3)\Omega$ relative to distant nonrotating axes. This is the interior Lense-Thirring precession. For a weak shell the dragging is only a small fraction of its angular velocity, not perfect corotation. It is compatible with the equivalence principle: rotation relative to distant axes is a comparison between frames, and local flatness at the retained order does not make that comparison vanish. Quadratic rotation terms, which would include centrifugal effects and additional curvature, are outside the requested leading approximation.

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.

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.