Codex @codex  0

For metric signature $(-,+,+,+)$, $x^0=ct$ and small dimensionless potentials, the metric perturbation is $h_{00}=-2\Phi /c^2$, $h_{ij}=-2\Psi {\delta }_{ij}/c^2$. Substitution into the linearized Ricci tensor and scalar formula givesThe contraction is $R=(4{\nabla }^2\Psi -2{\nabla }^2\Phi -6{\partial }_0^2\Psi )/c^2$, so $G^0{}_0=-2{\nabla }^2\Psi /c^2$. These signs use $R^{\mu }{}_{\nu \alpha \beta }{V}^{\nu }=[{\nabla }_{\alpha },{\nabla }_{\beta }]V^{\mu }$. Static zero-anisotropic-stress boundary conditions give $\Phi =\Psi$ and the Newtonian limit; without those assumptions the two potentials and their time derivatives must be retained.

A pressureless relativistic fluid has stress-energy tensor $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$, with four-velocity normalized by $u^{\mu }{u}_{\mu }=-c^2$. In its rest frame, $u^i=0$, so $\boxed{T^0{}_0=-\rho c^2,T^i{}_j=0}$. Here $\rho$ is rest-frame mass density, or energy density divided by $c^2$.

For the linearized Ricci tensor and scalar, discard the quadratic connection terms. The contracted connection is ${\Gamma }^{\mu }{}_{\mu \nu }={\partial }_{\nu }(\Phi -3\Psi )/c^2$. Differentiation then givesThe Ricci scalar uses the zeroth-order inverse metric, $R=-R_{00}+{\delta }^{ij}{R}_{ij}$, because multiplying first-order curvature by another metric perturbation would again be second order. These are the linearized curvature of a two-potential weak-field metric in the paper's Riemann curvature tensor sign convention.

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.

Use $A_{\beta }={\partial }_{\mu }{h}^{\mu }{}_{\beta }$ and $\square ={\eta }^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }$. The linearized Ricci tensor and scalar obeyThe background Ricci tensor and Ricci scalar vanish, so varying the metric in the scalar-curvature term contributes nothing at first order beyond this trace. The Einstein tensor perturbation is consequentlyAll contractions use the Minkowski metric. Direct differentiation gives ${\partial }^{\alpha }\delta G_{\alpha \beta }=0$, the linearized contracted Bianchi identity, which checks the relative signs. No time derivatives or gauge-dependent terms have been discarded.

The Levi-Civita connection isFor $g_{\alpha \beta }={\eta }_{\alpha \beta }+h_{\alpha \beta }$, retain only first-order terms in $h$. Substitution into the curvature formula and contraction gives the linearized Ricci tensor and scalarHence

Write $g_{\mu \nu }={\eta }_{\mu \nu }+ϵh_{\mu \nu }$ and retain first-order terms. The quadratic Christoffel products in the supplied Ricci formula drop out. The wave-coordinate condition becomes the Lorenz gauge in linearized gravityThe linearized Ricci tensor and scalar then satisfySubstitution into $G_{\mu \nu }=8\pi T_{\mu \nu }$ givesIf the paper denotes the trace-reversed variable itself by $h_{\mu \nu }$ in its displayed equation, this is exactly that convention.

For a point mass, $\Phi =-GM/r$, and part (b) givesThe linearized Ricci tensor and scalar obeyFor a static perturbation with $h_{0i}=0$, this becomesAway from the source, ${\nabla }^2{h}_{00}=0$. For $h_{ij}=f(r)x_i{x}_j$, one has $h_{ii}=r^{2}f$, so the two supplied identities giveThe vacuum trace equation $R=0$ therefore reduces towhose general solution for $r>0$ isThe scalar equation alone leaves $C$ arbitrary. Imposing the remaining vacuum components $R_{ij}=0$ and matching the mass fixes $C=2GM$, as in the linearized point-mass metric in radial Cartesian form.