Linearized Gravity and the Wave Equation

Derivation

Linearized Gravity and the Wave Equation

D-397 Home PU-403 Threads waves · fields · light Depends on Field Equations from the Einstein-Hilbert Action

Statement

Starting from the exact Einstein field equations, we expand the metric as a small perturbation on flat spacetime, \( g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu} \) with \( |h_{\mu\nu}| \ll 1 \), keep only terms linear in \( h_{\mu\nu} \), impose the Lorenz (harmonic) gauge \( \partial^{\mu}\bar h_{\mu\nu}=0 \) on the trace-reversed perturbation \( \bar h_{\mu\nu} \), and obtain \( \Box\,\bar h_{\mu\nu} = -\frac{16\pi G}{c^{4}}\,T_{\mu\nu} \). In vacuum this reduces to the homogeneous wave equation \( \Box\,\bar h_{\mu\nu}=0 \); exhausting the residual gauge freedom fixes the transverse-traceless (TT) gauge, in which the two physical polarizations \( h_{+},h_{\times} \) propagate as transverse plane waves at exactly the speed of light \( c \).

Why it matters

This is the theoretical bridge between Einstein's static field equations and the dynamical prediction that spacetime carries radiation. It tells us gravitational waves exist, travel at \( c \), are transverse, and come in exactly two polarizations — every one of which was confirmed by the LIGO/Virgo detections beginning in 2015.

The linearized theory is also the workhorse for detector modelling, post-Newtonian expansions, and the weak-field limit that must reproduce Newtonian gravity. It shows that the metric \( h_{\mu\nu} \) itself is the radiative field, with \( G/c^{4} \) setting the (extraordinarily small) coupling to matter that makes gravitational radiation so hard to produce and detect.

Assumptions

Weak field: a global near-inertial chart exists with \( |h_{\mu\nu}|\ll 1 \).If dropped, second-order terms \( \mathcal{O}(h^{2}) \) are no longer negligible, the equations become nonlinear, gravitational-wave energy (which is quadratic in \( h \)) must be kept, and the clean superposition of plane waves fails.

Background is exactly flat Minkowski, \( \eta_{\mu\nu}=\mathrm{diag}(-1,+1,+1,+1) \), with \( \partial_\alpha\eta_{\mu\nu}=0 \).If the background is curved, \( \partial_\alpha \) must be promoted to the background covariant derivative and background curvature couples to \( h_{\mu\nu} \); indices are then raised and lowered with the background metric, not \( \eta \).

Coordinates are chosen so that \( h_{\mu\nu} \) is a genuine perturbation, not a gauge artefact.If dropped, a diffeomorphism \( x^{\mu}\to x^{\mu}+\xi^{\mu} \) can generate a spurious \( h_{\mu\nu}=-\partial_\mu\xi_\nu-\partial_\nu\xi_\mu \) that carries no curvature; physical statements must be made in a fixed gauge (here TT) or through gauge invariants like the linearized Riemann tensor.

Slow secular growth is ignored over the region of interest (linearity is uniform, not just pointwise).If dropped, small nonlinearities accumulate over many wavelengths (as in the self-interaction that makes GR nonlinear), and the linear solution drifts from the exact one.

Derivation

\[ g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu},\qquad g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}+\mathcal{O}(h^{2}) \]

Definition of the perturbation; the inverse follows from \( g^{\mu\alpha}g_{\alpha\nu}=\delta^{\mu}_{\nu} \) to first order. Indices are raised and lowered with \( \eta \) from here on. A

\[ \Gamma^{\lambda}_{\mu\nu}=\tfrac{1}{2}\eta^{\lambda\rho}\left(\partial_\mu h_{\rho\nu}+\partial_\nu h_{\rho\mu}-\partial_\rho h_{\mu\nu}\right)+\mathcal{O}(h^{2}) \]

Christoffel symbols to first order: \( \partial_\alpha\eta=0 \) kills the background piece, leaving only derivatives of \( h \). Because \( \Gamma=\mathcal{O}(h) \), any \( \Gamma\Gamma \) term in the Riemann tensor is \( \mathcal{O}(h^{2}) \) and is dropped. A

\[ R^{(1)}_{\mu\nu}=\tfrac{1}{2}\left(\partial^{\alpha}\partial_{\mu}h_{\alpha\nu}+\partial^{\alpha}\partial_{\nu}h_{\alpha\mu}-\partial_\mu\partial_\nu h-\Box h_{\mu\nu}\right) \]

Linearized Ricci tensor from \( R_{\mu\nu}=\partial_\lambda\Gamma^{\lambda}_{\mu\nu}-\partial_\nu\Gamma^{\lambda}_{\mu\lambda}+\mathcal{O}(h^2) \), with \( h\equiv\eta^{\mu\nu}h_{\mu\nu} \) the trace and \( \Box\equiv\partial^{\alpha}\partial_{\alpha}=-\tfrac{1}{c^{2}}\partial_t^{2}+\nabla^{2} \). B

\[ R^{(1)}=\eta^{\mu\nu}R^{(1)}_{\mu\nu}=\partial^{\alpha}\partial^{\beta}h_{\alpha\beta}-\Box h,\qquad G^{(1)}_{\mu\nu}=R^{(1)}_{\mu\nu}-\tfrac{1}{2}\eta_{\mu\nu}R^{(1)} \]

Contract Step 3 with \( \eta^{\mu\nu} \) for the Ricci scalar, then assemble the linearized Einstein tensor. The trace reversal that defines \( G \) is what motivates the next substitution. B

\[ \bar h_{\mu\nu}\equiv h_{\mu\nu}-\tfrac{1}{2}\eta_{\mu\nu}h,\qquad \bar h\equiv\eta^{\mu\nu}\bar h_{\mu\nu}=h-2h=-h,\qquad h_{\mu\nu}=\bar h_{\mu\nu}-\tfrac{1}{2}\eta_{\mu\nu}\bar h \]

Introduce the trace-reversed perturbation. In four dimensions \( \eta^{\mu\nu}\eta_{\mu\nu}=4 \), so the trace flips sign; the inversion shows the map is its own inverse up to that sign. B

\[ G^{(1)}_{\mu\nu}=-\tfrac{1}{2}\left(\Box\,\bar h_{\mu\nu}+\eta_{\mu\nu}\,\partial^{\alpha}\partial^{\beta}\bar h_{\alpha\beta}-\partial^{\alpha}\partial_{\nu}\bar h_{\mu\alpha}-\partial^{\alpha}\partial_{\mu}\bar h_{\nu\alpha}\right) \]

Substitute Step 5 into Step 4 and collect. Every term in \( G^{(1)}_{\mu\nu} \) is now expressed through \( \bar h_{\mu\nu} \); the leading \( \Box\bar h_{\mu\nu} \) is the wave operator we are chasing, the other three are gauge-dependent. C

\[ x^{\mu}\to x^{\mu}+\xi^{\mu}\ \Rightarrow\ \bar h_{\mu\nu}\to\bar h_{\mu\nu}-\partial_\mu\xi_\nu-\partial_\nu\xi_\mu+\eta_{\mu\nu}\partial_\alpha\xi^{\alpha},\qquad \partial^{\mu}\bar h_{\mu\nu}=0 \]

Under an infinitesimal diffeomorphism \( \partial^{\mu}\bar h_{\mu\nu}\to\partial^{\mu}\bar h_{\mu\nu}-\Box\xi_\nu \), so choosing \( \xi_\nu \) to solve \( \Box\xi_\nu=\partial^{\mu}\bar h_{\mu\nu} \) enforces the Lorenz (harmonic / de Donder) gauge. Such \( \xi_\nu \) always exists because \( \Box \) is invertible with retarded boundary conditions. C

\[ \partial^{\mu}\bar h_{\mu\nu}=0\ \Rightarrow\ \partial^{\alpha}\partial^{\beta}\bar h_{\alpha\beta}=0,\ \ \partial^{\alpha}\partial_{\nu}\bar h_{\mu\alpha}=0,\ \ \partial^{\alpha}\partial_{\mu}\bar h_{\nu\alpha}=0 \]

The Lorenz condition annihilates the last three terms of Step 6 (each contains a factor \( \partial^{\alpha}\bar h_{\alpha\cdot} \)). B

\[ G^{(1)}_{\mu\nu}=-\tfrac{1}{2}\Box\,\bar h_{\mu\nu}=\frac{8\pi G}{c^{4}}\,T_{\mu\nu}\quad\Longrightarrow\quad \boxed{\ \Box\,\bar h_{\mu\nu}=-\frac{16\pi G}{c^{4}}\,T_{\mu\nu}\ } \]

Insert Step 8 into Step 6, then set it equal to the matter side of the Einstein equation \( G_{\mu\nu}=\frac{8\pi G}{c^{4}}T_{\mu\nu} \) (linearized: \( T_{\mu\nu} \) is evaluated on the flat background). This is a sourced wave equation, one for each component. A

\[ T_{\mu\nu}=0\ \Rightarrow\ \Box\,\bar h_{\mu\nu}=0,\qquad \bar h_{\mu\nu}=A_{\mu\nu}\,e^{\,i k_\alpha x^{\alpha}},\ \ k^{\alpha}k_{\alpha}=0 \]

In vacuum the source vanishes. A plane-wave ansatz turns \( \Box\to -k^{\alpha}k_{\alpha} \); a nontrivial amplitude requires the null dispersion relation \( k^{\alpha}k_{\alpha}=-\omega^{2}/c^{2}+|\mathbf{k}|^{2}=0 \), i.e. \( \omega=c|\mathbf{k}| \): phase speed exactly \( c \). B

\[ \Box\xi_\mu=0:\quad h\to h,\ \ \text{choose }\xi^{\mu}\text{ to set }\ \bar h=0\ (\Rightarrow h_{\mu\nu}=\bar h_{\mu\nu}),\ \ h_{0\mu}=0,\ \ k^{i}h_{ij}=0 \]

Lorenz gauge leaves residual freedom \( \xi^{\mu} \) with \( \Box\xi^{\mu}=0 \) — four functions. They are spent to make \( \bar h_{\mu\nu} \) traceless, purely spatial, and transverse to \( \mathbf{k} \). This is the transverse-traceless (TT) gauge; only physical degrees of freedom survive. C

\[ h^{\mathrm{TT}}_{ij}=\begin{pmatrix} h_{+} & h_{\times} & 0\\ h_{\times} & -h_{+} & 0\\ 0 & 0 & 0\end{pmatrix}\cos\!\big[\omega(t-z/c)\big]\quad(\text{wave along }z) \]

Imposing tracelessness, transversality, and \( h_{0\mu}=0 \) on a wave travelling in \( +z \) leaves exactly two independent amplitudes: the plus \( (h_{+}) \) and cross \( (h_{\times}) \) polarizations. The \( 10 \) components of a symmetric \( 4\times4 \) tensor are reduced by \( 4 \) Lorenz + \( 4 \) residual gauge conditions to \( 2 \) physical ones. A

Result

\[ \Box\,\bar h_{\mu\nu}=-\frac{16\pi G}{c^{4}}\,T_{\mu\nu}\qquad\xrightarrow{\ \text{vacuum}\ }\qquad \Box\,\bar h_{\mu\nu}=0,\quad \omega=c|\mathbf{k}| \]

Reading. In the weak-field, Lorenz-gauge limit each component of the trace-reversed metric perturbation obeys an ordinary flat-space wave equation sourced by the local energy–momentum. Where matter is absent the perturbation propagates as a free wave whose null wavevector forces the phase and group speed to equal \( c \). Fixing the residual gauge (TT) strips the ten components down to the two transverse, traceless polarizations \( h_{+} \) and \( h_{\times} \) that a distant observer can measure as a quadrupolar stretch-and-squeeze of proper distances.

Units check. \( h_{\mu\nu} \) is dimensionless, so \( \Box\bar h \) has units \( \mathrm{m^{-2}} \). On the right, \( G/c^{4} \) carries \( \frac{\mathrm{m^{3}\,kg^{-1}\,s^{-2}}}{\mathrm{m^{4}\,s^{-4}}}=\mathrm{m^{-1}\,kg^{-1}\,s^{2}} \), and \( T_{\mu\nu} \) (an energy density) carries \( \mathrm{J\,m^{-3}}=\mathrm{kg\,m^{-1}\,s^{-2}} \). Their product is \( \mathrm{m^{-1}\,kg^{-1}\,s^{2}}\times\mathrm{kg\,m^{-1}\,s^{-2}}=\mathrm{m^{-2}} \), matching the left side.

Limiting cases

Static, slow source. Dropping \( \partial_t^2 \) gives \( \nabla^{2}\bar h_{00}=-\frac{16\pi G}{c^{4}}T_{00} \); with \( T_{00}=\rho c^{2} \) and \( h_{00}=-2\Phi/c^{2} \) this reproduces the Newtonian Poisson equation \( \nabla^{2}\Phi=4\pi G\rho \).
Vacuum plane wave. \( T_{\mu\nu}=0 \) yields \( \Box\bar h_{\mu\nu}=0 \) with \( \omega=c|\mathbf{k}| \): non-dispersive propagation at \( c \).
Single polarization. Setting \( h_{\times}=0 \) leaves the pure "+" mode — a ring of test masses oscillates along the \( x \)- and \( y \)-axes in antiphase.
Zero amplitude. \( h_{+}=h_{\times}=0 \) returns exact flat Minkowski spacetime, \( R^{(1)}_{\mu\nu}=0 \).

Breaks when

Strong field, \( |h_{\mu\nu}|\sim1 \). Near black-hole horizons or neutron-star surfaces the quadratic terms dominate; gravity self-gravitates, the equations are genuinely nonlinear, and linear superposition and the fixed background both fail.
Near the source / wave-zone energy transport. Gravitational-wave energy and momentum are second order in \( h \) (the Isaacson stress tensor \( \sim\langle\partial h\,\partial h\rangle \)); a strictly linear treatment carries no energy and cannot describe radiation reaction or orbital inspiral.
Curved or cosmological background. On an expanding FRW or Schwarzschild background \( \partial_\alpha\eta_{\mu\nu}\neq0 \); the flat \( \Box \) must become the covariant wave operator with curvature couplings, and the simple TT decomposition no longer diagonalizes the equations.
Gauge not fully fixed. If the Lorenz condition is not imposed, the three extra terms in Step 6 survive and \( \bar h_{\mu\nu} \) does not obey a wave equation at all — pure coordinate waves masquerade as physical ones.

Failure modes

Sign error in the trace reversal. Writing \( \bar h=+h \) instead of \( \bar h=-h \) (forgetting \( \eta^{\mu\nu}\eta_{\mu\nu}=4 \)) corrupts every subsequent contraction; the factor \( 16\pi \) comes out wrong.
Confusing \( h_{\mu\nu} \) with \( \bar h_{\mu\nu} \). The wave equation is for the trace-reversed field; imposing the Lorenz condition on \( h_{\mu\nu} \) rather than \( \bar h_{\mu\nu} \) leaves stray trace terms.
Treating coordinate waves as physical. A "wave" that is pure gauge, \( h_{\mu\nu}=-\partial_\mu\xi_\nu-\partial_\nu\xi_\mu \), has \( R^{(1)}_{\mu\nu\alpha\beta}=0 \); students mistake it for radiation. Always check the linearized Riemann tensor or work in TT gauge.
Wrong \( \Box \) sign from signature. With \( (-+++) \), \( \Box=-\tfrac{1}{c^{2}}\partial_t^{2}+\nabla^{2} \); flipping to \( (+---) \) without also flipping the field equation's sign scrambles the dispersion relation.
Counting six polarizations. Forgetting the residual gauge freedom (Step 11) leaves apparent extra modes; only two survive in GR.
Using \( \tfrac{1}{2}h_{+}L \) vs \( h_{+}L \) for detector strain. The fractional change of a single arm along a polarization axis is \( \Delta L/L=\tfrac{1}{2}h_{+} \); dropping the \( \tfrac{1}{2} \) doubles the predicted displacement.

Discussion

The central conceptual move is that gravity, linearized about flat space, behaves like any other relativistic field: its potential \( \bar h_{\mu\nu} \) satisfies a sourced wave equation formally identical to electromagnetism's \( \Box A_\mu=-\mu_0 J_\mu \) in Lorenz gauge. The analogy is deep but not perfect — the source is a rank-2 tensor \( T_{\mu\nu} \) rather than a vector current, so the lowest radiating multipole is the quadrupole (mass conservation and momentum conservation kill the monopole and dipole), and the field has spin 2 rather than spin 1, giving two tensor polarizations at \( 45^{\circ} \) rather than transverse vector polarizations.

That the waves travel at exactly \( c \) is a direct consequence of the graviton being massless, encoded in the null condition \( k^{\alpha}k_{\alpha}=0 \). Any graviton mass would add a term \( \propto m^{2}\bar h_{\mu\nu} \), making the dispersion relation \( \omega^{2}=c^{2}|\mathbf{k}|^{2}+m^{2}c^{4}/\hbar^{2} \) and delaying low-frequency components; the near-simultaneous arrival of GW170817 and its gamma-ray counterpart bounds any such deviation to better than one part in \( 10^{15} \).

The TT gauge makes the physical content manifest: only \( h_{+} \) and \( h_{\times} \) survive, and they act tidally. Proper distances between freely falling test masses oscillate — a transverse ring is squeezed along one axis while stretched along the perpendicular axis, then vice versa a half-period later. This is precisely what a laser interferometer measures as a differential arm-length change \( \Delta L/L\sim h\sim10^{-21} \), the smallest fractional length ever measured.

A subtlety that separates linearized gravity from a mere field theory on flat space is that the theory is only self-consistent to first order: the very energy carried by the waves gravitates, and this back-reaction, quadratic in \( h \), is what completes the nonlinear Einstein equations. The Isaacson effective stress–energy tensor \( t_{\mu\nu}=\frac{c^{4}}{32\pi G}\langle\partial_\mu h^{\mathrm{TT}}_{\alpha\beta}\,\partial_\nu h_{\mathrm{TT}}^{\alpha\beta}\rangle \) is gauge invariant only after averaging over several wavelengths, reflecting the fact that gravitational energy cannot be localized to a point — a hallmark of the equivalence principle, which lets one transform away the field locally.

Common misconceptions. (i) Gravitational waves are not "ripples in a medium" — there is no aether; \( h_{\mu\nu} \) is the metric of spacetime itself. (ii) The wave does not change coordinate distances in TT gauge — the coordinates comove with free masses — it changes proper distances; both descriptions agree on the measurable interferometer signal. (iii) Two polarizations does not mean "vertical and horizontal": the "+" and "×" modes differ by a \( 45^{\circ} \) rotation, reflecting the spin-2 (helicity \( \pm2 \)) nature of the graviton.

Worked examples

Example 1 — dispersion and wavelength of a LIGO-band wave.

\[ \Box\,\bar h_{\mu\nu}=0,\qquad \bar h_{\mu\nu}=A_{\mu\nu}\cos\!\big(k z-\omega t\big) \]

Vacuum wave equation with a plane wave travelling along \( z \). A

\[ \Box\to -\frac{1}{c^{2}}(-\omega)^{2}+(k)^{2}=0\ \Rightarrow\ k=\frac{\omega}{c},\qquad \lambda=\frac{2\pi}{k}=\frac{c}{f} \]

Apply \( \Box \) with signature \( (-+++) \); the null condition gives the dispersion relation and hence the wavelength. Symbols first. B

\[ f=150\ \mathrm{Hz}\ \Rightarrow\ \omega=2\pi f=942\ \mathrm{rad\,s^{-1}},\quad k=\frac{942}{2.998\times10^{8}}=3.14\times10^{-6}\ \mathrm{m^{-1}} \]

Insert a representative LIGO-band frequency and \( c=2.998\times10^{8}\ \mathrm{m\,s^{-1}} \). A

\[ \lambda=\frac{c}{f}=\frac{2.998\times10^{8}}{150}=2.0\times10^{6}\ \mathrm{m} \]

Compute the wavelength directly. A

\[ \lambda=2.0\times10^{3}\ \mathrm{km},\qquad v_{\text{phase}}=\frac{\omega}{k}=c \]

Reading. A \( 150\ \mathrm{Hz} \) gravitational wave has a wavelength of about \( 2000\ \mathrm{km} \) — vastly larger than the \( 4\ \mathrm{km} \) detector, which is why LIGO responds to the wave essentially instantaneously across its arms. The phase speed is exactly \( c \).

Units check. \( c/f=(\mathrm{m\,s^{-1}})/(\mathrm{s^{-1}})=\mathrm{m} \). Correct.

Example 2 — interferometer arm-length change.

\[ ds^{2}=\big(1+h^{\mathrm{TT}}_{xx}\big)dx^{2}+\ldots,\qquad L_{\text{proper}}=\int_0^{L}\sqrt{1+h_{+}\cos\omega t}\;dx \]

Proper length of an arm along \( x \) in the field of a "+" polarized wave (arm short compared with \( \lambda \), so \( h_{+} \) is uniform along it). B

\[ L_{\text{proper}}\approx L\left(1+\tfrac{1}{2}h_{+}\cos\omega t\right)\ \Rightarrow\ \frac{\Delta L}{L}=\tfrac{1}{2}h_{+} \]

Expand \( \sqrt{1+x}\approx1+\tfrac{1}{2}x \) for \( |h_{+}|\ll1 \). Symbols before numbers. A

\[ h_{+}=1.0\times10^{-21},\quad L=4.0\ \mathrm{km}=4.0\times10^{3}\ \mathrm{m} \]

Insert a typical detected strain amplitude and the LIGO arm length. A

\[ \Delta L=\tfrac{1}{2}h_{+}L=\tfrac{1}{2}\,(1.0\times10^{-21})(4.0\times10^{3})=2.0\times10^{-18}\ \mathrm{m} \]

Numeric evaluation of the peak arm displacement. A

\[ \Delta L=2.0\times10^{-18}\ \mathrm{m}\ \approx\ \tfrac{1}{1000}\ \text{of a proton radius} \]

Reading. Even for a "loud" event, one arm of a \( 4\ \mathrm{km} \) interferometer changes length by only \( \sim10^{-18}\ \mathrm{m} \). The differential signal between the two perpendicular arms doubles this, but it still explains why gravitational-wave detection demanded decades of instrument development.

Units check. \( h_{+} \) dimensionless, so \( \tfrac{1}{2}h_{+}L \) has units of \( \mathrm{m} \). Correct.

Problems

Show that for a wave travelling in the \( +z \) direction the Lorenz gauge condition \( \partial^{\mu}\bar h_{\mu\nu}=0 \) applied to \( \bar h_{\mu\nu}=A_{\mu\nu}e^{ik_\alpha x^\alpha} \) implies \( k^{\mu}A_{\mu\nu}=0 \).
Solution
\( \partial^{\mu}\bar h_{\mu\nu}=\partial^{\mu}\!\big(A_{\mu\nu}e^{ik_\alpha x^\alpha}\big)=iA_{\mu\nu}k^{\mu}e^{ik_\alpha x^\alpha} \). For this to vanish for all \( x \), the non-zero exponential must be multiplied by zero, so \( k^{\mu}A_{\mu\nu}=0 \). Combined with \( k^{\alpha}k_{\alpha}=0 \), transversality of the amplitude to the null wavevector is established — the algebraic backbone of the TT reduction.
For static pressureless dust the only large stress–energy component is \( T_{00}=\rho c^{2} \). Solve the static field equation for \( \bar h_{00} \), reconstruct \( h_{00} \), and (identifying \( h_{00}=-2\Phi/c^{2} \)) recover the Newtonian Poisson equation.
Solution
Work on the trace-reversed field, where each component decouples. With only \( T_{00}\neq0 \), only \( \bar h_{00}\neq0 \). Static limit \( \Box\to\nabla^{2} \): \( \nabla^{2}\bar h_{00}=-\frac{16\pi G}{c^{4}}T_{00}=-\frac{16\pi G}{c^{4}}\rho c^{2}=-\frac{16\pi G}{c^{2}}\rho \). Now undo the trace reversal, \( h_{\mu\nu}=\bar h_{\mu\nu}-\tfrac12\eta_{\mu\nu}\bar h \). Since \( \bar h_{00} \) is the only nonzero component, \( \bar h=\eta^{00}\bar h_{00}=-\bar h_{00} \), so \( h_{00}=\bar h_{00}-\tfrac12\eta_{00}\bar h=\bar h_{00}-\tfrac12(-1)(-\bar h_{00})=\tfrac12\bar h_{00} \) (and \( h_{ij}=\tfrac12\delta_{ij}\bar h_{00}\neq0 \) — the spatial metric is perturbed too, which is why "only \( h_{00} \)" is wrong). Hence \( \bar h_{00}=2h_{00}=-4\Phi/c^{2} \), and \( \nabla^{2}(-4\Phi/c^{2})=-\frac{16\pi G}{c^{2}}\rho \) gives \( \boxed{\nabla^{2}\Phi=4\pi G\rho} \). The factor \( 4 \) that would otherwise appear is exactly cancelled by the trace-reversal, confirming the \( 16\pi \) in the field equation.
A gravitational wave has \( f=60\ \mathrm{Hz} \). Find \( \omega \), \( k \), and \( \lambda \).
Solution
\( \omega=2\pi f=2\pi(60)=377\ \mathrm{rad\,s^{-1}} \). \( k=\omega/c=377/(2.998\times10^{8})=1.26\times10^{-6}\ \mathrm{m^{-1}} \). \( \lambda=c/f=2.998\times10^{8}/60=5.0\times10^{6}\ \mathrm{m}=5.0\times10^{3}\ \mathrm{km} \).
For a "×" polarized wave along \( z \), \( h^{\mathrm{TT}}_{xy}=h_{\times}\cos[\omega(t-z/c)] \) with \( h_{\times}=5\times10^{-22} \), estimate the peak differential displacement of two test masses separated by \( L=3\ \mathrm{km} \) along the \( x \)-axis.
Solution
The "×" mode has off-diagonal \( h_{xy} \); a mass pair along \( x \) experiences \( \Delta L/L=\tfrac12 h_{xy} \) only for a separation vector with a \( y \)-component. For a pair purely along \( x \), the diagonal \( h_{xx}=0 \) for the pure cross mode, so the along-\( x \) stretch vanishes — the response is maximal for masses at \( 45^{\circ} \). Rotating axes by \( 45^{\circ} \), the cross mode becomes a plus mode of amplitude \( h_{\times} \); then \( \Delta L=\tfrac12 h_{\times}L=\tfrac12(5\times10^{-22})(3\times10^{3})=7.5\times10^{-19}\ \mathrm{m} \). This highlights the \( 45^{\circ} \) geometry of the cross polarization.
Verify that the pure-gauge perturbation \( h_{\mu\nu}=-\partial_\mu\xi_\nu-\partial_\nu\xi_\mu \) with \( \xi_\mu=C_\mu e^{ik_\alpha x^\alpha} \), \( k^\alpha k_\alpha=0 \), satisfies \( \Box h_{\mu\nu}=0 \) yet carries no physical curvature.
Solution
\( \Box h_{\mu\nu}=-\partial_\mu(\Box\xi_\nu)-\partial_\nu(\Box\xi_\mu) \), and \( \Box\xi_\mu=\Box(C_\mu e^{ik x})=-k^\alpha k_\alpha C_\mu e^{ikx}=0 \) since \( k \) is null; hence \( \Box h_{\mu\nu}=0 \). But the linearized Riemann tensor \( R^{(1)}_{\mu\nu\alpha\beta}=\tfrac12(\partial_\alpha\partial_\nu h_{\mu\beta}+\partial_\beta\partial_\mu h_{\nu\alpha}-\partial_\alpha\partial_\mu h_{\nu\beta}-\partial_\beta\partial_\nu h_{\mu\alpha}) \) is gauge invariant and vanishes identically when \( h_{\mu\nu} \) is pure gauge (substitute and note every term cancels in pairs by symmetry of mixed partials). So this "wave" is a coordinate artefact: it solves the wave equation but produces no tidal forces. Moral — only curvature, or equivalently TT-gauge amplitudes, is physical.