Linearized gravity

Linearized gravity writes $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$ with $|h_{\mu \nu }|\ll 1$ and discards terms quadratic in $h$ and its derivatives.

For signature $(+---)$, decompose a metric perturbation into $h_{00}=2A$, $h_{0i}={\partial }_{i}B+B_i$ and $h_{ij}=2C{\delta }_{ij}+2({\partial }_i{\partial }_j-{\delta }_{ij}\Delta /3)E+{\partial }_i{E}_j+{\partial }_j{E}_i+h_{ij}^{\mathrm{TT}}$, where the spatial vectors are divergence-free and $h^{\mathrm{TT}}$ is a transverse-traceless tensor. Under a passive coordinate shift ${\xi }^0=\tau$, ${\xi }^i={\partial }_{i}L+L_i$, with ${\partial }_i{L}_i=0$, the changes are $\delta A=-\dot{\tau }$, $\delta B=\dot{L}-\tau$, $\delta C=\Delta L/3$, $\delta E=L$, $\delta B_i={\dot{L}}_i$, and $\delta E_i=L_i$. The displayed combinations and $h^{\mathrm{TT}}$ are therefore gauge invariant. The decomposition is unique after appropriate spatial decay or mode restrictions; homogeneous and boundary modes require separate conventions.

With the reversed convention in curvature-sign convention in Killing derivative identities, choose the representative $h_{00}=2\Phi$, $h_{0i}=V_i$, $h_{ij}=2\Psi {\delta }_{ij}+h_{ij}^{\mathrm{TT}}$. The linearized Einstein components are $G_{00}^{(1)}=-2\Delta \Psi$, $G_{0i}^{(1)}=-2{\partial }_i\dot{\Psi }-\Delta V_i/2$ and $G_{ij}^{(1)}=({\partial }_i{\partial }_j-{\delta }_{ij}\Delta )(\Phi -\Psi )-2{\delta }_{ij}\ddot{\Psi }-({\partial }_i{\dot{V}}_j+{\partial }_j{\dot{V}}_i)/2+\square h_{ij}^{\mathrm{TT}}/2$. On all space with regular decaying fields, or for nonzero spatial Fourier modes, the vacuum constraints remove the scalar and vector parts, leaving two radiative gravitational wave polarizations. Singular Coulomb fields, spatial boundaries and homogeneous modes are not covered by that removal argument.

For static scalar perturbations with no mixed time-space or spatial shear terms, the displayed metric tensor uses signature $(+---)$. With the reversed convention in curvature-sign convention in Killing derivative identities, its first-order Einstein tensor is $\delta G_{00}=2\Delta C$, $\delta G_{0i}=0$, and $\delta G_{ij}={\partial }_i{\partial }_j(A+C)-{\delta }_{ij}\Delta (A+C)$. These follow by substituting the metric into the first-order Christoffel symbols and Ricci tensor, discarding products of perturbations. A static source with the dust stress-energy tensor obeys $\Delta C=-4\pi G\rho$ and the decaying spatial equation makes $A=-C$. For a point mass the Newtonian potential of a point mass gives $A=-GM/r$, $C=GM/r$. The spatial potential is necessary to determine light propagation; a time potential alone is insufficient.

For a static spherical point source, the weak-field solution in isotropic radius $\rho$ has $g_{tt}=1-2GM/\rho$ and spatial metric $-(1+2GM/\rho )(d{\rho }^2+{\rho }^{2}d{\Omega }^2)$. The isotropic-to-areal weak-field radius change gives the displayed metric with $r_S=2GM$. Its validity requires $r_S/r\ll 1$. Treating it as an exact metric near $r\sim r_S$ can turn omitted terms into large corrections; it does not establish horizons or strong-field photon dynamics.

The first-order null-orbit equation $(u^{\prime }{)}^2=r_S^2{E}^2/h^2-u^2(1-u)$ formally has a circular solution at $u=2/3$, with $|h/E|=3\sqrt{3}{r}_S/2$. Its second-order equation gives a perturbation $\delta u^{\prime \prime }=\delta u$, hence instability. However, $u=2/3$ is outside the weak-field expansion. The exact Schwarzschild spacetime has a corresponding unstable photon sphere, but that physical result needs the exact geometry; extrapolating the linearized metric is not a controlled proof.

The area of a symmetry sphere in the weak-field isotropic metric is $4\pi {\rho }^2(1+2GM/\rho )$. Equating it to $4\pi r^2$ defines the areal radius and gives the displayed change. To first order $dr=d\rho$, and replacing $1/\rho$ by $1/r$ changes the perturbation only at second order. This explains why the isotropic angular correction disappears in the linearized Schwarzschild metric without discarding spatial curvature.

Expand the Einstein tensor about the Minkowski metric. The quadratic term in the metric perturbation defines $t_{\mu \nu }=-G_{\mu \nu }^{(2)}[h,h]/(8\pi )$, a symmetric tensor relative to the background. For a first-order vacuum solution, the expanded contracted Bianchi identity gives ${\partial }^{\mu }{t}_{\mu \nu }=0$. Equivalently, a second-order correction $j$ obeys $G^{(1)}[j]=8\pi t$, whose flat divergence vanishes identically. This conservation is an on-shell statement. The local $t$ depends on perturbative gauge and the background split; conservation alone does not make it a gauge-independent local gravitational stress-energy tensor.

A massless Fierz-Pauli action is the quadratic flat-background action for a symmetric metric perturbation that yields the vacuum Linearized Einstein equations. One normalization, up to a total derivative, isVarying the symmetric field with compact support gives the displayed linearized Einstein tensor coefficient. Its linearized coordinate-gauge symmetry is consistent with the contracted Bianchi identity. It is the massless gravitational action, distinct from imposing the massive Fierz-Pauli equations with five polarizations.

An infinitesimal coordinate change generated by $\xi$ changes a metric perturbation about Minkowski spacetime by $h_{\mu \nu }\mapsto h_{\mu \nu }-{\partial }_{\mu }{\xi }_{\nu }-{\partial }_{\nu }{\xi }_{\mu }$, in the passive sign convention. For a passive coordinate change $x^{\mu }\mapsto x^{\mu }+{\xi }^{\mu }$, the first-order change is minus the Lie derivative of a tensor field of the background metric, equivalently the pullback by the inverse flow. The gauge invariance of the linearized Riemann tensor follows because the added terms have commuting third derivatives, which cancel in the curvature antisymmetrizations. Gauge choices alter the metric components while preserving linearized tidal effects.

For a perturbation $h_{xx}=2a(t-z)$, $h_{yy}=2b(t-z)$ with all other components zero, the linearized Vacuum Einstein equations give $(a+b{)}^{\prime \prime }=0$. Put $s=a+b$ and choose ${\xi }_x=sx/2$, ${\xi }_y=sy/2$, ${\xi }_t=-s^{\prime }(x^2+y^2)/4$, ${\xi }_z=s^{\prime }(x^2+y^2)/4$. Under the linearized coordinate gauge transformation, the mixed time/transverse and longitudinal/transverse components cancel, while the other time/longitudinal changes are proportional to $s^{\prime \prime }=0$. The remaining components are $h_{xx}=a-b$, $h_{yy}=-(a-b)$, $h_{xy}=0$, a transverse-traceless gauge perturbation of plus polarization. This explicitly removes a nonzero affine trace without assuming it vanishes by boundary conditions. If $(a-b{)}^{\prime \prime }=0$ too, all linearized curvature vanishes.

The weak-field approximation requires every dimensionless component of the metric perturbation to have magnitude much smaller than one, so terms quadratic in the perturbation may be neglected.

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

The source-free Linearized Einstein equations. For a metric perturbation about the Minkowski metric, imposing the Lorenz gauge in linearized gravity reduces them to $\square h_{\mu \nu }=0$ together with ${\partial }^{\mu }{\bar{h}}_{\mu \nu }=0$, where $\bar{h}$ is the trace-reversed metric perturbation. A linearized coordinate gauge transformation preserves $G^{(1)}[h]=0$, while preserving the imposed gauge additionally requires its generator to satisfy the wave equation.

For signature $(+---)$ and the curvature convention giving $\square {\bar{h}}_{ik}=-16\pi G{ϵ}^{-1}{T}_{ik}$, take $u=t-x$, ${\ell }_i=(1,-1,0,0)$ and $T_{ik}=ϵ\rho (u){\ell }_i{\ell }_k$. A trace-free Lorenz-gauge particular solution has the displayed form. The vector ${\ell }^i=(1,1,0,0)$ is null, and ${\Gamma }^i{}_{jk}{\ell }^j{\ell }^k=0$, so co-propagating test photons are undeflected by this beam field. For a transversely uniform source, $H=4\pi G\rho (u)(y^2+z^2)$ is one particular solution. A function of $u$ alone has vanishing wave operator and cannot source nonzero $\rho$; homogeneous gravitational perturbations and boundary conditions remain separate data.

The retarded source solution of the Linearized Einstein equations is selected by excluding added incoming or freely specified homogeneous gravitational waves. A homogeneous wave can satisfy Lorenz gauge in linearized gravity while having no source mass quadrupole moment. Thus a formula for the retarded source field is not an identity for every solution of the sourced wave equation.

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.

Dropping the position-dependent source delay in the leading far-zone retarded field requires source size $d$ to be small compared with its variation time $\tau$, in units $c=1$. The retarded quadrupole field also requires $d/r\ll 1$. For a typical nonrelativistic bound source, $\tau \sim d/v$ implies this long-wavelength condition; an arbitrarily rapidly oscillating small displacement is not controlled merely by its instantaneous speed.

If $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$, then $g^{\mu \nu }={\eta }^{\mu \nu }-h^{\mu \nu }+O(h^2)$, with indices on $h$ raised using the background metric.

About a Cartesian flat background,

The linearized Riemann tensor defines an operator on two-forms byUnlike the metric perturbation itself, this operator is invariant under linearized gauge transformations.

Under $h_{\mu \nu }\mapsto h_{\mu \nu }+{\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$, every contribution to the linearized Riemann tensor contains three commuting derivatives of $\xi$ and cancels pairwise. Linearized curvature is therefore gauge invariant about flat spacetime.

On a contractible Cartesian patch of a flat background, zero linearized Riemann curvature operator says that the connection 1-forms ${\Gamma }^{(1)a}{}_{bc}d{x}^c$ are closed. The Poincaré lemma writes them as derivatives of a matrix-valued potential. Symmetry of the two lower affine connection indices then supplies a second potential, giving ${\Gamma }^{(1)a}{}_{bc}={\partial }_b{\partial }_c{\xi }^a$. Linearized metric compatibility implies that $h_{ab}-2{\partial }_{(a}{\xi }_{b)}$ is constant. Its symmetric constant part is absorbed by an affine vector field. Thus the metric perturbation is locally a coordinate change. Global topology and imposed boundary conditions can obstruct the corresponding global gauge choice.

In four dimensions, the trace-reversed perturbation isTrace reversal simplifies the linearized Einstein equations.

The Lorenz condition ${\partial }^{\mu }{\bar{h}}_{\mu \nu }=0$ reduces the vacuum field equation to $\square {\bar{h}}_{\mu \nu }=0$. A plane-wave amplitude therefore satisfies $k^{\mu }{H}_{\mu \nu }=0$ and $k^{\mu }{k}_{\mu }=0$.

An infinitesimal coordinate change generated by ${\xi }_{\mu }$ changesIt preserves Lorenz gauge exactly when $\square {\xi }_{\mu }=0$.

Under the convention $\delta h_{\mu \nu }={\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$, choose ${\xi }^{\mu }=b^{\mu }\sin (k\cdot x)$ with constant $b^{\mu }$ and null $k^{\mu }$. The trace-reversed metric perturbation has amplitudeIt satisfies $k_{\mu }{ϵ}_{\mathrm{gauge}}^{\mu \nu }=0$ and $\square {\xi }^{\mu }=0$, so it is an allowed residual gauge symmetry of linearized gravity. Its linearized Riemann curvature operator is zero: the four wavevector factors in each second-derivative term cancel in the antisymmetric curvature combinations. Nonzero transverse-traceless curvature distinguishes the two physical gravitational wave polarizations from these coordinate modes.^[ref]

A vacuum mode ${\bar{h}}_{\mu \nu }=H_{\mu \nu }{e}^{ik\cdot x}$ has a null wavevector and a transverse symmetric amplitude. Residual gauge freedom can further reduce it to two physical transverse-traceless polarizations.

Transverse-traceless gauge sets the time and longitudinal components and the spatial trace of a vacuum gravitational-wave perturbation to zero.

In a weak plus-polarized gravitational wave, initially stationary test particles have constant spatial coordinates in transverse-traceless gauge because ${\Gamma }^i{}_{00}=0$. Their physical separations nevertheless change. For a small transverse ring, $X=(1+h_{+}/2)x$ and $Y=(1-h_{+}/2)y$, so its principal semiaxes oscillate as $a(1\pm h_{+}/2)$. The radial response has the displayed angular dependence. These are local distance measurements in the long-wavelength limit; finite-arm light-travel times require the full detector response.

For a plane gravitational wave in linearized gravity with $k_{\mu }=(-\omega ,0,0,\omega )$, Lorenz gauge in linearized gravity gives $H_{3\nu }=-H_{0\nu }$. Under $\delta H_{\mu \nu }=i(k_{\mu }{X}_{\nu }+k_{\nu }{X}_{\mu }-{\eta }_{\mu \nu }{k}^{\rho }{X}_{\rho })$, choose $X_i=H_{0i}/(i\omega )$ for $i=1,2$, $X_3-X_0=i{H}_{00}/\omega$, and $X_0+X_3=(H_{11}+H_{22})/(2i\omega )$. This residual gauge symmetry of linearized gravity removes all time and longitudinal entries and the transverse trace. The remaining amplitudes $H_{+},H_{\times }$ are the two gravitational wave polarizations. The formula uses the plus-sign convention for $\delta h=2{\partial }_{(\mu }{\xi }_{\nu )}$; reversing the definition of $\xi$ reverses the gauge change.

In transverse-traceless gauge, particles initially at rest have fixed spatial coordinates because ${\Gamma }^{\mu }{}_{00}=0$. Their separation measured in a parallel-propagated orthonormal frame changes: to first order $S^{\hat{A}}=S^A+h_{AB}{S}^B/2$. Hence the physical geodesic deviation is ${\ddot{\xi }}^{\hat{A}}={\ddot{h}}_{AB}{L}^B/2$, whereas the coordinate displacement obeys ${\ddot{\xi }}_{\mathrm{coord}}^A=0$. The extra ${\partial }_0{\Gamma }^A{}_{0B}{L}^B$ term in the coordinate expansion of the second covariant derivative reconciles the two results. This applies to infinitesimal separations and linear metric perturbations.

At first order, geodesic deviation gives ${\ddot{\xi }}_i={\ddot{h}}_{ij}{X}^j/2$ for a small detector in a plane gravitational wave in linearized gravity. For physical initial displacement zero and initial velocity $V_i$, integration gives ${\xi }_i=[h_{ij}(t)-h_{ij}(0)]X^j/2+[V_i-{\dot{h}}_{ij}(0)X^j/2]t$. A bounded periodic response selects $V_i={\dot{h}}_{ij}(0)X^j/2$. Prescribed positions alone do not select that response when the wave has no wave-free past.

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.

A short rigid nonrotating guide constrains two freely sliding beads to one axis $n$. In its parallel-propagated orthonormal frame, the axial geodesic deviation is $\ddot{L}=-R_{n0n0}L$. For a transverse gravitational wave, $R_{n0n0}=-ϵ{\ddot{h}}_{nn}/2$, so a wave incident after the beads are at rest changes the measured separation by $ϵh_{nn}{L}_0/2$ to first order. Transverse reaction forces do not supply an axial restoring force. Alignment with the propagation direction gives no first-order length change.

A gravitational wave in four-dimensional general relativity has two independent linear polarizations, conventionally called plus and cross. A rotation by angle $\vartheta$ in the transverse plane rotates their pair by $2\vartheta$.

Equal-amplitude plus polarization and cross polarization with a quarter-cycle phase difference form circularly polarized gravitational radiation. The principal transverse strain axes rotate by half the phase angle. An observer on the rotation axis of a circular binary sees this polarization; the sign distinguishes the handedness relative to a stated propagation direction.

A gravitational wave carrying the plus component relative to a chosen pair of transverse axes.

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.

A gravitational wave of plus polarization has opposite transverse strain or geodesic deviation eigenvalues along two fixed perpendicular axes. For a sinusoidal periodic function profile, the stretching and compressing roles alternate every half-period.

The cross polarization is the plus pattern rotated through $45$ degrees in the transverse plane.

The transverse-traceless projector maps a symmetric tensor to the part orthogonal to a chosen wavevector and with zero trace. It is the identity on physical gravitational-wave polarizations.

A polarization completeness relation expresses a projector as a sum of outer products of an orthonormal polarization basis. For transverse-traceless tensors, ${\sum }_s{ϵ}_{ij}^s{ϵ}_{mn}^{s\ast }$ is the transverse-traceless projector up to the chosen normalization.

To first order around flat spacetime,and

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.

For a weak, static field produced by nonrelativistic matter, $T_{00}=\rho$ is the dominant stress-energy component. The field equations and the slow-particle geodesic equation implywhere the Poisson equation is ${\nabla }^2\Phi =4\pi G\rho$.

For smooth spatial $\varphi$ and $g(c)=-c^2{e}^{2\varphi /c^2}d{t}^2+{\delta }_{ij}d{x}^{i}d{x}^j$, the only nonzero Christoffel symbols are ${\Gamma }^t{}_{ti}={\Gamma }^t{}_{it}=c^{-2}{\partial }_i\varphi$ and ${\Gamma }^i{}_{tt}=e^{2\varphi /c^2}{\partial }_i\varphi$. Consequently the Levi-Civita connection converges smoothly on compact sets as $c\to \infty$ although the metric itself diverges in these coordinates. The limiting torsion-free connection has only ${\Gamma }^i{}_{tt}={\partial }_i\varphi$ and gives $d^2{x}^i/d{t}^2=-{\partial }_i\varphi$ when $t$ is an affine parameter. With $R^a{}_{bcd}={\partial }_c{\Gamma }^a{}_{db}-{\partial }_d{\Gamma }^a{}_{cb}+\cdots$, its Ricci tensor is $(\Delta \varphi )dt\otimes dt$: quadratic connection terms vanish and $R^i{}_{tjt}={\partial }_j{\partial }_i\varphi$. Thus zero Ricci tensor is the Laplace equation, while full flatness would require the entire Hessian to vanish.

For a point mass in a positive cosmological constant, the inward Newtonian acceleration $GM/r^2$ balances the outward cosmological acceleration $\Lambda r/3$ at the static radius $r_{\mathrm{s}}=(3GM/\Lambda {)}^{1/3}$.

For a point mass and $r>0$, writeThe vacuum scalar-curvature equation gives $r{f}^{\prime }+3f=0$, so $f=C/r^3$. The remaining vacuum equations fix $C=2GM$ in the corresponding Schwarzschild gauge.

Far from a slowly rotating isolated body of mass $m$ and angular momentum $\mathbf{J}$, the scalar part of the metric is the linearized point-mass field while $g_{0i}$ is the dipole field of $\mathbf{J}$. For rotation about the z-axis, the cross term is $4J{r}^{-3}(ydx-xdy)dt$ in a common convention.