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With the curvature sign convention in which $\delta R_{bc}={\nabla }_c\delta {\Gamma }^a{}_{ba}-{\nabla }_a\delta {\Gamma }^a{}_{bc}$, the inverse-metric variation of $\int \sqrt{-g}\Phi R$ is $\int \sqrt{-g}{E}_{ab}\delta g^{ab}$ plus a boundary term, for a scalar multiplier held fixed in that variation. The Palatini identity expresses the curvature variation as divergences of $\delta \Gamma$. Two integrations by parts move the derivatives onto $\Phi$, yielding the displayed terms. For constant $\Phi$ they vanish. The opposite curvature convention instead has $E_{ab}=\Phi G_{ab}+g_{ab}\square \Phi -{\nabla }_a{\nabla }_b\Phi$.

Use the curvature sign conventionwhich has the derivative ordering printed in the source for $R^i{}_{0j0}$. A Christoffel symbol is of order $\mu /r^2$, so its derivative is of order $\mu /r^3$, whereas a product of two connections is of order ${\mu }^2/r^4$. Their ratio is $\mu /r\ll 1$. The metric is stationary, hence the temporal derivative term vanishes. ConsequentlyHere $g_{00}=-1+2\mu /r$, and ${\partial }_i{\partial }_j(1/r)=(3x_i{x}_j-r^2{\delta }_{ij})/r^5$. The weak-field tidal curvature in a stationary metric is

Printed sign inconsistency. The source's positive half-Hessian and its final negative-prefactor expression are the opposites of this result. A concrete check at $(r,0,0)$ is ${\Gamma }^1{}_{00}\simeq \mu /r^2$, giving ${\partial }_1{\Gamma }^1{}_{00}=-2\mu /r^3$, while the printed final expression gives $+2\mu /r^3$. Thus both printed equalities cannot hold with its stated derivative ordering. Reversing the definition of the Riemann curvature tensor would give the printed final sign, but would also require reversing the curvature term in the geodesic deviation equation. The convention used here and in the next part gives the physical radial stretching and transverse compression.

For the Levi-Civita connection, the all-lowered Riemann curvature tensor has the algebraic symmetriesThe first Bianchi identity is the cyclic identityThese statements concern the Riemann curvature tensor at each point and are unchanged by reversing its overall curvature sign convention.

Use $x^0=ct$ and metric signature $(-,+,+,+)$. With the source's curvature sign convention, the Ricci tensor isIts time-time component is thereforeFor a slowly moving, pressure-negligible source, the stress-energy tensor is approximated by relativistic dust: $T^{\mu \nu }=\rho U^{\mu }{U}^{\nu }$, where $\rho$ is mass density and $U^{\mu }$ is the four-velocity. In its approximate rest local inertial frame, $U^{\mu }\simeq (c,0,0,0)$ and $U_0\simeq -c$. Thusand the trace-reversed stress-energy tensor satisfies

Independently, introduce the metric perturbation $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$, with $|h_{\mu \nu }|\ll 1$ and constant Minkowski metric ${\eta }_{\mu \nu }$. The Christoffel symbols are first order in $h$, so their products are discarded in the weak-field approximation. With a comma denoting a partial derivative, the linearized Riemann curvature tensor isTaking the tensor contraction gives, to this order,In the static Newtonian limit, all time partial derivatives vanish. The constant ${\eta }_{00}$ also has zero partial derivatives, leavingHere ${\nabla }^2={\sum }_i{\partial }_i^2$ is the spatial Laplace operator in the approximately Cartesian background coordinates. This sign follows the paper's curvature sign convention.

For the Newtonian gravitational potential $\Phi$, the weak static metric tensor has $g_{00}\simeq -1-2\Phi /c^2$. The Newtonian limit derived above consequently givesMatching the Poisson equation for Newtonian gravity, ${\nabla }^2\Phi =4\pi G\rho$, fixesHere $G$ is the gravitational constant and $c$ is the speed of light. This negative coupling agrees with the source formula sheet. The familiar positive coupling uses the opposite curvature sign convention for the Ricci tensor; the physical Poisson equation for Newtonian gravity is the same.

A Killing vector field $K$ generates local isometries: its flow of a vector field ${\Phi }_s$ satisfies ${\Phi }_s^{\ast }g=g$. Equivalently, the Lie derivative of the metric tensor vanishes. For the Levi-Civita connection, metric compatibility converts the coordinate expression into the Killing equation:The Levi-Civita connection is an implicit extra assumption in this question. The printed assumption of vanishing torsion tensor alone does not imply metric compatibility. For example, on Minkowski spacetime, take the affine connection whose only nonzero Christoffel symbol is ${\Gamma }^0{}_{00}=1$. The vector field $K={\partial }_0$ preserves the Minkowski metric, but $K_0=-1$ gives ${\nabla }_0{K}_0=1$, contradicting the Killing equation. The following calculation, and the covariant formulas in part (b), use the Levi-Civita connection; this is the role of metric compatibility in Killing identities.

For the stated curvature sign convention, the Ricci identity on a covector has a minus sign:Write $A_{\mu \nu \rho }={\nabla }_{\mu }{\nabla }_{\nu }{K}_{\rho }$. Differentiating the Killing equation gives $A_{\mu \nu \rho }=-A_{\mu \rho \nu }$. Alternately using this relation and the Ricci identity givesThe first Bianchi identity and antisymmetry of the last two indices of the Riemann curvature tensor reduce the three curvature terms to $2R^{\sigma }{}_{\mu \nu \rho }{K}_{\sigma }$. Consequently the second covariant derivative of a Killing vector is

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.

Although connection coefficients are not tensor components, their variation $C^a{}_{bc}=\delta {\Gamma }^a{}_{bc}$ is tensorial: the difference of affine connections is a tensor. Varying the Levi-Civita connection at fixed coordinates givesIn the paper's curvature sign convention, varying the derivative and quadratic-connection terms and grouping them into covariant derivatives gives the Palatini identityOne may verify the grouping in normal coordinates at a point, where the affine connection vanishes and the expression is just the difference of two partial derivatives; both sides are tensors, so it holds in every coordinate frame. Reversing the definition of the Riemann curvature tensor reverses this identity's right-hand side.

For completeness, put $q_{ab}=\delta g_{ab}$ and $q=g^{ab}{q}_{ab}$. The contracted variation uses $C^a{}_{ba}={\nabla }_{b}q/2$ and $g^{bc}{C}^a{}_{bc}={\nabla }_b{q}^{ab}-{\nabla }^{a}q/2$. After two integrations by parts,up to the discarded boundary term. Substituting $q^{ab}=-\delta g^{ab}$ gives the differentiated-scalar terms in the metric variation of a scalar-curvature coupling.

The Brans-Dicke theory couples a scalar field $\Phi$ to the Ricci scalar. Assume $\Phi \ne 0$, constant $\omega$, matter independent of $\Phi$, and variations of compact support so boundary terms may be discarded. Use the paper's curvature sign convention and signature $(+---)$. Set $X=g^{ab}{\partial }_a\Phi {\partial }_b\Phi$ and $\square \Phi ={\nabla }_a{\nabla }^a\Phi$. Define the stress-energy tensor consistently byThis convention fixes the sign of the matter term.

Vary with respect to $g^{ab}$. The variation of metric volume density contributes $-g_{ab}/2$ times each scalar Lagrangian, while $\delta R=R_{ab}\delta g^{ab}+g^{ab}\delta R_{ab}$. Applying the Palatini identity followed by two rounds of integration by parts gives the metric variation of a scalar-curvature coupling,The differentiated $\Phi$ terms remain because the curvature multiplier is not constant. Varying the kinetic term at fixed $\Phi$ givesCombining these terms with the matter variation, the Euler-Lagrange equation isRaising both free indices reproduces the contravariant form, because ${\nabla }_a{g}_{bc}=0$.

The independent scalar variation givesor $R+\omega {\Phi }^{-2}X-2\omega {\Phi }^{-1}\square \Phi =0$. The trace of the metric tensor equation in four dimensions isMultiplying the scalar equation by $\Phi$ eliminates the same combination $\Phi R+\omega {\Phi }^{-1}X$. Thus the scalar equation of Brans-Dicke theory isAt the degenerate coupling of Brans-Dicke theory, $\omega =-3/2$, the combined equations impose $T=0$ instead; division by $3+2\omega$ is unavailable. If matter depended directly on $\Phi$, its scalar variation would add a source and the displayed scalar equation would change.

The following lettered sections justify the supplied variation identities; they are supporting identities in the PDF, rather than three independent field-equation problems.

A tidal force is a difference of gravitational acceleration across an extended region. In Newtonian gravity, $g_i=-{\partial }_i\Phi$. Two neighboring freely falling particles separated by ${\xi }^j$ have, to first order,For a point mass, $\Phi =-GM/r$. In a local orthonormal radial/transverse frame,A small freely falling cloud stretches radially and compresses transversely. Outside the source the three eigenvalues sum to zero, consistently with ${\nabla }^2\Phi =0$; within matter the trace is $-4\pi G\rho$. These are accelerations of separations, not the much larger common acceleration of the cloud's centre.

In general relativity, freely falling test particles follow timelike geodesics. A freely falling frame can eliminate the connection at one event, but not the Riemann curvature tensor throughout a finite laboratory. Let $u$ be the geodesic tangent and $\xi$ a connecting vector for neighboring geodesics with $[u,\xi ]=0$. In the explicitly chosen curvature conventionthe geodesic deviation equation isIt follows by commuting ${\nabla }_u{\nabla }_{\xi }u$, using ${\nabla }_{u}u=0$ and zero torsion. In a parallel-propagated orthonormal frame, its spatial components directly describe measured relative acceleration. The weak-field, slow-motion limit recovers the Newtonian Hessian above. Reversing the curvature sign convention reverses the displayed curvature side, without changing the physical stretching. The equivalence principle removes locally uniform gravity, while geodesic deviation measures the residual tidal field. The connection between parallel transport, curvature and neighboring geodesics is developed systematically in the standard geometric construction.^[ref]