Plus and Cross Polarizations

A gravitational wave is a ripple in the geometry of spacetime, and like any wave it has a definite polarization — a specification of which way the field points as it oscillates. For light the answer is two transverse directions; for gravity the field is not a vector but a symmetric rank-2 tensor, the metric perturbation $h_{\mu\nu}$ , and the counting is subtler. Ten components sound like ten possibilities, but general relativity is a gauge theory: most are coordinate artifacts. After the Lorenz gauge and the residual transverse-traceless (TT) conditions are spent, only two genuine degrees of freedom survive — the plus ( $+$ ) and cross ( $\times$ ) polarizations. This article is about what they are, why there are exactly two, and what they do to matter. Two facts organize everything. First, the physical content of a polarization is read not from the metric but from the tidal field it carries: a passing wave drives a ring of free test masses into an oscillating ellipse through geodesic deviation, and the $+$ and $\times$ patterns are the two independent ways that ellipse can be oriented — along the axes, or turned by $45°$ . Second, that $45°$ — not the $90°$ of electromagnetism — is a fingerprint: the field transforms with two factors of a rotation, so a turn of $\psi$ rotates the polarization by $2\psi$ . That doubling is the operational meaning of spin 2, and why the graviton is a massless spin-2 particle with helicities $\pm 2$ . Everything downstream — antenna patterns, source inclination, tests of alternative gravity — flows from these two modes.

Key takeaways

In the TT gauge a wave along $z$ leaves only a symmetric, transverse, traceless $2\times2$ block in the $x$ – $y$ plane, with two independent entries: the amplitudes $h_+,h_\times$ .
The $+$ mode stretches along $x$ and squeezes along $y$ (reversing a half-period later); the $\times$ mode does the same along axes rotated by $45°$ .
The physical effect is read from geodesic deviation: a ring of free masses becomes an ellipse, and because the strain is traceless its area is conserved to first order in $h$ .
A rotation by $\psi$ about the propagation axis mixes $h_+$ and $h_\times$ through $2\psi$ — the defining transformation of a spin-2 field, with helicity eigenstates $h_+\mp i h_\times$ .
The numbers are tiny: GW150914's peak strain $h\approx1.0\times10^{-21}$ moves a LIGO $4\ \mathrm{km}$ arm by $\tfrac12 hL = 2\times10^{-18}\ \mathrm{m}$ , a few thousandths of a proton radius.
Key idea: In the TT gauge a wave along $z$ leaves only a symmetric, transverse, traceless $2\times2$ block in the $x$ – $y$ plane; its two independent components are the plus and cross polarizations, whose $45°$ relation encodes spin-2.

Where this lives today

The two-polarization structure is now a measurement to exploit. The global network — two LIGO detectors, Virgo, and KAGRA, each oriented differently — separates $h_+$ and $h_\times$ because each instrument's $\cos2\psi$ antenna pattern samples a different mixture; this pins down a source's sky location and inclination, and inclination-versus-distance breaking drives "standard siren" cosmology, as in GW170817. The same network turns the spin-2 hypothesis into a null test: general relativity permits only the two tensor modes, whereas scalar-tensor and modified-gravity theories predict extra scalar "breathing" or longitudinal polarizations, and analyses so far find the data consistent with pure tensor content. In the nanohertz band, pulsar timing arrays (NANOGrav, EPTA, PPTA, the International PTA) target supermassive black-hole binaries; the predicted Hellings–Downs correlation between pulsars integrates each pulsar's response to $+$ and $\times$ over the sky, so the quadrupolar shape of that curve — seen at growing significance since 2023 — is a direct imprint of the spin-2 content. The space-based LISA mission will read the same two modes from massive black-hole mergers.

The physical picture

Forget the metric and watch the matter. A gravitational wave announces itself only through tides — the relative acceleration of nearby free-falling bodies — because a single freely-falling particle feels nothing (the equivalence principle). So the picture of a polarization is a movie of a ring of free test masses, weightless in the transverse plane, as the wave sweeps through. The $+$ wave pulls the ring into a horizontal ellipse, relaxes it to a circle a quarter-period later, then stretches it vertical; the $\times$ wave does the identical dance rotated by $45°$ , along the diagonals. Because the strain is traceless, whatever it stretches along one axis it squeezes equally along the other, so the enclosed area does not change at first order. The first figure is that movie across one period.

Figure

Plus and cross polarizations deform a ring of free test masses in opposite ways, a quarter period apart. The semi-axes obey a = r(1 + ½ h cos ωt) and b = r(1 − ½ h cos ωt), so the enclosed area stays πr² to first order. The × pattern is the + pattern turned by 45°, not 90° — the fingerprint of spin 2.

Why does a $45°$ turn, not a $90°$ turn, carry the $+$ pattern onto the $\times$ pattern? Because the perturbation is a rank-2 tensor, a rotation acts on it twice, once per index, so the physical response rotates at double the geometric angle: a $45°$ turn is a full $90°$ in polarization space, converting pure $+$ into pure $\times$ . The factor between the geometric angle and the invariance angle is the spin, and here it is $2$ . The second figure translates this into experimental currency — the sub-nuclear length change an interferometer must resolve.

Figure

One LIGO arm of length L = 4 km, stretched by an overhead plus wave of strain h = 1.0×10⁻²¹, moves its mirror by δL = ½ h L = 2×10⁻¹⁸ m while the perpendicular arm shrinks by the same amount — a differential ΔL = h L = 4×10⁻¹⁸ m, about 1/200 of a proton radius.

The core equations

The starting point is the general TT-gauge plane wave. Once all gauge freedom is spent, a wave along $+z$ has its perturbation entirely in the transverse plane, built from two basis tensors weighted by two amplitudes — the entire polarization content of the field.

h_{ij}^{\rm TT}(t,z) = \big(h_+\, e^{+}_{ij} + h_\times\, e^{\times}_{ij}\big)\, e^{-i\omega(t - z/c)} \qquad (1)

where: $h_{ij}^{\rm TT}$ — transverse-traceless perturbation (only the $x$ – $y$ block nonzero); $h_+,\,h_\times$ — complex amplitudes; $e^{+}_{ij},\,e^{\times}_{ij}$ — fixed basis tensors; $\omega$ — angular frequency; $c$ — speed of light. Reading it: the field is a two-dimensional vector $(h_+,h_\times)$ on a plane wave — everything physical is how those numbers sit and rotate. The two basis tensors are the only symmetric, transverse, traceless $2\times2$ tensors there are. One is diagonal, one off-diagonal, and together they span the space of transverse strains.

e^{+}_{ij} = \hat x_i \hat x_j - \hat y_i \hat y_j, \qquad e^{\times}_{ij} = \hat x_i \hat y_j + \hat y_i \hat x_j \qquad (2)

where: $\hat x_i,\hat y_i$ — transverse unit vectors; $e^{+}_{ij}$ — plus tensor, $\mathrm{diag}(1,-1)$ ; $e^{\times}_{ij}$ — cross tensor, symmetric off-diagonal. Reading it: each is manifestly traceless, so neither changes an area — the geometric root of the $\times$ -versus- $+$ distinction. In the real-part convention the whole wave is one matrix times a cosine. Keep this form in your head — you read the tidal distortion straight off it.

h_{ij}^{\rm TT}(t,z) = \begin{pmatrix} h_+ & h_\times & 0 \\ h_\times & -h_+ & 0 \\ 0 & 0 & 0 \end{pmatrix} \cos\!\big[\omega(t - z/c)\big] \qquad (3)

where: $h_+,h_\times$ — real strain amplitudes; the $0$ entries enforce transversality (nothing along $z$ ). Reading it: the diagonal $\pm h_+$ is stretch-one-axis / squeeze-the-other; the off-diagonal $h_\times$ is the same distortion turned $45°$ . Physics enters through geodesic deviation, converting the metric into a measurable acceleration between two nearby free masses. In the TT gauge their coordinates do not move, but their proper separation does, governed by the Riemann tensor's electric part — here just the second time derivative of the strain.

\ddot{\xi}^{\,i} = -R^{i}{}_{0j0}\,\xi^{j} = \tfrac{1}{2}\,\ddot{h}^{\rm TT}_{ij}\,\xi^{j}, \qquad R_{i0j0} = -\tfrac{1}{2}\,\ddot h^{\rm TT}_{ij} \qquad (4)

where: $\xi^{i}$ — separation vector between two nearby test masses; $R^{i}{}_{0j0}$ — electric (tidal) components of the linearized Riemann tensor; $\ddot h^{\rm TT}_{ij}$ — second time derivative of the strain. Reading it: the wave is a tidal field oscillating at $\omega$ ; the factor $\tfrac12$ is the one in the arm-length response $\Delta L/L=\tfrac12 h$ . Integrating (4) once, for masses at rest and small strain, gives the deformed ring directly. A circle of radius $r$ becomes an ellipse whose semi-axes breathe in antiphase.

x(\theta,t) = \Big(1+\tfrac{h_+}{2}\cos\omega t\Big) r\cos\theta, \qquad y(\theta,t) = \Big(1-\tfrac{h_+}{2}\cos\omega t\Big) r\sin\theta \qquad (5)

where: $r$ — ring radius; $\theta$ — angular label of a bead; the two prefactors are the instantaneous semi-axes $a,b$ . Reading it: $a\cdot b = r^2\big(1-\tfrac{h_+^2}{4}\cos^2\omega t\big)\approx r^2$ , so the area is fixed to first order — tracelessness made visible. The spin-2 character is a statement about how $(h_+,h_\times)$ transforms under a rotation by $\psi$ about the propagation axis. Because the strain carries two indices, the amplitudes rotate at twice the angle.

\begin{pmatrix}h'_+\\ h'_\times\end{pmatrix}=\begin{pmatrix}\cos 2\psi & \sin 2\psi\\ -\sin 2\psi & \cos 2\psi\end{pmatrix}\begin{pmatrix}h_+\\ h_\times\end{pmatrix} \qquad (6)

where: $\psi$ — rotation angle in the transverse plane; the primed amplitudes are what a detector rotated by $\psi$ registers. Reading it: the $2\psi$ (not $\psi$ ) is the defining transformation of a helicity- $\pm2$ field — this equation is where "spin 2" is earned. Diagonalizing (6) picks out the two states that merely acquire a phase under rotation: the circular, or helicity, basis. These are the irreducible states of the field.

h_{R} = h_+ - i\,h_\times, \qquad h_{L} = h_+ + i\,h_\times, \qquad h_{R,L}\to e^{\mp 2i\psi}\,h_{R,L} \qquad (7)

where: $h_R$ — right-handed amplitude, helicity $+2$ ; $h_L$ — left-handed amplitude, helicity $-2$ . Reading it: under a rotation by $\psi$ each picks up a pure phase $e^{\mp2i\psi}$ — the hallmark of a definite-helicity state, and the classical shadow of the massless spin-2 graviton.

Derivation ACounting the two physical polarizations

The claim that a gravitational wave has exactly two polarizations is not an assumption but a theorem of gauge freedom, and this derivation earns it by bookkeeping. We start with the ten components of a symmetric metric perturbation, spend four on the Lorenz gauge and four more on the residual transformations it leaves free, and read off the two survivors as the plus and cross amplitudes. The tally $10-4-4=2$ says gravity in four dimensions radiates in exactly two tensor modes.

Step 1. A symmetric rank-2 tensor $h_{\mu\nu}=h_{\nu\mu}$ in four dimensions has $4\cdot5/2=10$ independent components, the raw field for a weak perturbation $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ . Why: symmetry is the only constraint before any gauge is chosen, so ten is the honest starting count.

Step 2. Impose the Lorenz gauge $\partial^\mu\bar h_{\mu\nu}=0$ on the trace-reversed field $\bar h_{\mu\nu}=h_{\mu\nu}-\tfrac12\eta_{\mu\nu}h$ ; this is four conditions (one per index $\nu$ ) and reduces the field equation to the flat wave equation $\Box\bar h_{\mu\nu}=0$ . Why: four conditions remove four components, leaving six, and make the field a genuine wave.

Step 3. Lorenz does not exhaust the gauge freedom: any further shift $x^\mu\to x^\mu+\xi^\mu$ with $\Box\xi^\mu=0$ preserves it, acting as $h_{\mu\nu}\to h_{\mu\nu}-\partial_\mu\xi_\nu-\partial_\nu\xi_\mu$ — four harmonic functions $\xi^\mu$ . Why: these residual transformations are pure coordinate relabelings, so the four components they adjust are not real degrees of freedom.

Step 4. Spend the four residual functions to enforce $h_{0\mu}^{\rm TT}=0$ , $h^{i}{}_{i}=0$ , $\partial^{i}h_{ij}=0$ ; for a wave along $z$ these kill every component with a $z$ index. Why: this transverse-traceless choice is the gauge in which coordinates track the freely-falling frame and only observable strain remains.

Step 5. What survives is a symmetric, traceless $2\times2$ block, $h_{ij}^{\rm TT}=\begin{pmatrix}h_+&h_\times\\ h_\times&-h_+\end{pmatrix}$ , with two entries. The count $N_{\rm phys}=10-4_{\rm Lorenz}-4_{\rm residual}=2$ is complete. Why: two is exactly what a massless spin-2 field must have — helicities $+2$ and $-2$ , no longitudinal or scalar mode.

Go deeper · Derivation B · 1 minDistortion of a test-mass ring and area conservation

Derivation BDistortion of a test-mass ring and area conservation

The goal here is to turn the abstract strain matrix into the concrete motion of a ring of free masses, and to prove the claim that gives tracelessness its physical teeth: to first order in the strain, the wave conserves the ring's enclosed area. We integrate the geodesic deviation equation for a pure plus wave, obtain the ellipse, and multiply its semi-axes to show the area is fixed — the picture every textbook draws and every detector exploits.

Step 1. Take a pure plus wave $h_{ij}^{\rm TT}=h_+\,e^+_{ij}\cos\omega t$ of small amplitude $h$ and feed it into $\ddot\xi^{\,i}=\tfrac12\ddot h^{\rm TT}_{ij}\xi^{j}$ , replacing $\xi^{j}$ on the right by its unperturbed value $\xi^{j}_0$ . Why: the acceleration is already $O(h)$ , so using the $O(h)$ position there would be a second-order error.

Step 2. The diagonal $e^+$ splits the motion: $\ddot\xi^{x}=\tfrac12\ddot h_+\,\xi^{x}_0$ and $\ddot\xi^{y}=-\tfrac12\ddot h_+\,\xi^{y}_0$ , with opposite signs, each a driven oscillation of amplitude set by $h$ . Why: those opposite signs are the whole plus mode — stretch along $x$ is squeeze along $y$ , locked together.

Step 3. Integrate twice, fixing constants so the ring is a circle of radius $r$ when $h=0$ : $\xi^{x}=\big(1+\tfrac{h}{2}\cos\omega t\big)x_0$ , $\xi^{y}=\big(1-\tfrac{h}{2}\cos\omega t\big)y_0$ — a point at angle $\theta$ traces the ellipse of equation (5). Why: the two integrations turn acceleration into the slow breathing displacement seen in the figure.

Step 4. The semi-axes $a=r\big(1+\tfrac{h}{2}\cos\omega t\big)$ , $b=r\big(1-\tfrac{h}{2}\cos\omega t\big)$ give area $A=\pi a b=\pi r^2\big(1-\tfrac{h^2}{4}\cos^2\omega t\big)$ , so $A=\pi r^2$ to first order. Why: the cross term cancels because the axes move in antiphase — tracelessness made physical, distinguishing a real TT wave from a spurious breathing mode.

Go deeper · Derivation C · 1 minRotation behaviour and the spin-2 signature

Derivation CRotation behaviour and the spin-2 signature

This derivation extracts the deepest structural fact about the two polarizations: how they behave under a rotation of the observer about the line of sight. We rotate the transverse frame by an angle $\psi$ , transform the strain as the rank-2 tensor it is, and find that the amplitudes mix through $2\psi$ rather than $\psi$ . From there the helicity eigenstates fall out as the combinations that merely acquire a phase — the classical fingerprint of the spin-2 graviton, obtained with no quantum input at all.

Step 1. Write the transverse rotation $R(\psi)=\begin{pmatrix}\cos\psi&-\sin\psi\\ \sin\psi&\cos\psi\end{pmatrix}$ ; the strain, carrying two lower indices, transforms as $h'=R\,h\,R^{T}$ . Why: a rank-2 tensor picks up one rotation factor per index, and this double action doubles the angle.

Step 2. Direct multiplication gives $R\,e^+\,R^{T}=\cos2\psi\,e^+ + \sin2\psi\,e^\times$ and $R\,e^\times\,R^{T}=-\sin2\psi\,e^+ + \cos2\psi\,e^\times$ , the double angles arising from $\cos^2\psi-\sin^2\psi$ and $2\sin\psi\cos\psi$ . Why: the algebra collapses to double angles automatically — forced by the tensor rank, not assumed.

Step 3. Hence the mixing $h'_+=h_+\cos2\psi+h_\times\sin2\psi$ , $h'_\times=-h_+\sin2\psi+h_\times\cos2\psi$ , equation (6). A $45°$ rotation gives $2\psi=90°$ and $(h_+,h_\times)\to(h_\times,-h_+)$ : pure plus becomes pure cross. Why: this is the $45°$ -not- $90°$ rule, and why a single detector aligned to one mode is blind to the other.

Step 4. Diagonalize by forming $h_{R,L}=h_+\mp i\,h_\times$ ; these obey $h_{R,L}\to e^{\mp2i\psi}h_{R,L}$ . A state transforming as $e^{-i s\psi}$ has helicity $s$ , so here $s=\pm2$ . Why: helicity $\pm2$ defines a massless spin-2 field — the classical algebra already knows the graviton is spin 2, no quantization required.

Limiting cases

Case	Result	Meaning
Pure plus	$h_{ij}^{\rm TT}=h_+\begin{pmatrix}1&0\\0&-1\end{pmatrix}$	stretches along $x$ , squeezes along $y$ ; ellipse aligned with the coordinate axes
Pure cross	$h_{ij}^{\rm TT}=h_\times\begin{pmatrix}0&1\\1&0\end{pmatrix}$	the same deformation turned by $45°$ ; ellipse axes lie along $y=\pm x$
Circular / helicity	$h_\times=\pm i\,h_+$ , so $h_{R,L}=h_+\mp i h_\times$	equal magnitudes $90°$ out of phase; the ellipse rotates rigidly — right/left circular, helicity $\pm2$
Longitudinal	$h_{zj}^{\rm TT}=0$	no motion along propagation; GR waves are strictly transverse, exactly two states
Linearized / weak field	$h\ll1$ , area $=\pi r^2+O(h^2)$	tracelessness conserves area to first order; volume effects appear at $O(h^2)$

Common trapThe plus and cross patterns are rotated by 90°, like the two polarizations of light

They are not — they are rotated by $45°$ , and confusing the two imports the wrong (spin-1) intuition from electromagnetism. Light's two linear polarizations sit $90°$ apart because the field is a vector, carrying one rotation index; gravity's field is a rank-2 tensor, so a rotation by $\psi$ turns the polarization by $2\psi$ , and the physically distinct states sit $90°$ apart in polarization space — only $45°$ of real-world angle. Rotating a $+$ wave by $90°$ maps it back to itself (the sign flip is just a half-period time shift), while a $45°$ rotation turns it entirely into $\times$ . This factor of two is why a gravitational wave has helicity $\pm2$ , not $\pm1$ , and why a detector rotated by $45°$ swaps rather than kills its sensitivity.

Worked examples

1 · How far does a LIGO mirror actually move? (fully worked)

GW150914 reached a peak strain $h\approx1.0\times10^{-21}$ . LIGO's arms are $L=4\ \mathrm{km}$ . For a plus wave from overhead aligned with the arms, find the displacement of one end mirror, the differential arm-length change, and compare it to a proton.

Step 1. The fractional stretch of an arm along a stretch axis is $\Delta L/L=\tfrac12 h$ — the same factor $\tfrac12$ from the deviation equation (4). So one mirror moves by $\delta L=\tfrac12 hL=\tfrac12(1.0\times10^{-21})(4\times10^{3}\ \mathrm{m})=\mathbf{2\times10^{-18}\ m}$ .

Step 2. A plus wave squeezes the perpendicular arm by the same amount at the same instant, $\delta L_y=-2\times10^{-18}\ \mathrm{m}$ . A Michelson reads the difference of the arm lengths, so the signal is $\Delta L=\delta L_x-\delta L_y=hL=(1.0\times10^{-21})(4\times10^{3})=\mathbf{4\times10^{-18}\ m}$ — the differential doubles the single-arm number.

Step 3. Put it on scale: a proton's charge radius is $r_p\approx0.84\ \mathrm{fm}=8.4\times10^{-16}\ \mathrm{m}$ , so $\Delta L/r_p=(4\times10^{-18})/(8.4\times10^{-16})\approx4.8\times10^{-3}$ . The mirrors move by about one two-hundredth of a proton radius, which is why the measurement demands strain sensitivity near $10^{-23}/\sqrt{\mathrm{Hz}}$ and stabilized km-scale optical cavities.

2 · A detector rotated off the polarization axis (fully worked)

A pure plus wave of amplitude $h_0$ arrives from overhead. An L-shaped detector has its arms rotated by $\psi$ relative to the wave's plus axis. What plus-equivalent strain does it register at $\psi=0°$ , $22.5°$ , $45°$ ?

Step 1. In the detector frame the strain it responds to is the rotated plus component, equation (6): $h'_+=h_0\cos2\psi$ (the cross term vanishes since $h_\times=0$ ), so the response scales as $\cos2\psi$ .

Step 2. At $\psi=0°$ : $h'_+=h_0\cos0=\mathbf{h_0}$ — full response, arms aligned with the stretch. At $\psi=22.5°$ : $2\psi=45°$ , $h'_+=h_0\cos45°=\mathbf{0.707\,h_0}$ , and $h'_\times=-h_0\sin45°=-0.707\,h_0$ — an equal mix of plus and cross in the detector frame.

Step 3. At $\psi=45°$ : $2\psi=90°$ , $h'_+=h_0\cos90°=\mathbf{0}$ . Aligned with the diagonals, the detector sees the wave as pure cross and an interferometer sensing only $h'_+$ is completely blind to it. This $\cos2\psi$ blind spot — not the $\cos\psi$ of a dipole antenna — is the observable signature of spin 2, and why a network of differently oriented detectors is needed to reconstruct both polarizations.

3 · Inclination and circular polarization of a binary

A compact binary radiates $h_+\propto\tfrac12(1+\cos^2\iota)$ and $h_\times\propto\cos\iota$ (in quadrature), with $\iota$ the angle between the line of sight and the orbital angular momentum.

Step 1. Face-on, $\iota=0$ : $\tfrac12(1+\cos^2 0)=1$ and $\cos0=1$ , so $h_+$ and $h_\times$ have equal amplitude with a $90°$ phase offset — exactly $h_\times=\pm i h_+$ , a circularly polarized wave, helicity $\pm2$ . Edge-on, $\iota=90°$ : $\cos90°=0$ , so $h_\times=0$ and only $h_+$ survives — linear polarization at half the amplitude.

Step 2. The ratio $h_\times/h_+=2\cos\iota/(1+\cos^2\iota)$ encodes inclination directly, from $1$ (circular, face-on) to $0$ (linear, edge-on). Because inclination and distance both scale the observed amplitude, the polarization ratio partly breaks that degeneracy — why reconstructing $h_+$ and $h_\times$ separately matters for standard-siren cosmology.

Physics of the portrait

For forty years after Einstein wrote them down, it was unclear whether gravitational waves were physical or merely ripples in the coordinate system — a doubt Einstein himself shared. The resolution came not from a new equation but from the right operational question: what would a wave actually do to a detector? That question, and its answer through geodesic deviation, belongs to Felix Pirani.

Origins · 1 minFelix Pirani and how this was discovered

Felix Pirani

1928–2015 · the geometry of gravitational radiation

Felix Arnold Edward Pirani was a Canadian-British physicist who, more than anyone, made gravitational radiation an observable rather than a philosophical puzzle. In a decisive 1956 paper, On the physical significance of the Riemann tensor, he argued that the true content of a gravitational field is invariant — encoded in the Riemann curvature — and that its physical effect must be read from the geodesic deviation of nearby test particles, not from coordinate-dependent metric components. This reframing cut through the long "are the waves real?" controversy that had entangled Einstein and Rosen: whatever coordinates you choose, a passing wave produces a genuine, gauge-invariant relative acceleration between free masses, and that is what the plus and cross patterns describe. His insight set the stage for the 1957 Chapel Hill conference, where Feynman's "sticky bead" argument — a bead sliding on a rod, heated by the wave's tidal force — turned the same idea into a vivid proof that the waves carry energy. With Hermann Bondi and Ivor Robinson, Pirani helped build the modern covariant theory of radiation. The ring-of-masses picture at the heart of this article, and every interferometer that reads a strain, descend directly from his argument that the observable is the tide.

History

1916

— Einstein predicts gravitational waves from the linearized field equations

1937

— Einstein and Rosen argue the waves may be mere coordinate artifacts, then reverse course

1956

— Pirani shows a wave moves free test masses through gauge-invariant geodesic deviation

1957

— Feynman's sticky-bead argument at Chapel Hill demonstrates the waves carry energy

2015

— LIGO detects GW150914, confirming two transverse tensor polarizations

Going further

Texts. Misner, Thorne & Wheeler, Gravitation, §35.6 and §37 (polarization and the tidal action of a plane wave) · Wald, General Relativity, §4.4 · Carroll, Spacetime and Geometry, §7.2 · Hartle, Gravity, ch. 16 · Schutz, A First Course in General Relativity, §9.2 · Maggiore, Gravitational Waves vol. 1, §1.2–1.3 and §2.1.

Provenance. The degree-of-freedom count $10-4-4=2$ and the TT reduction follow Wald §4.4 and Carroll §7.2; the ring-distortion equations and the area-conservation result reproduce MTW §37 and Hartle ch. 16; the $2\psi$ rotation law and the helicity eigenstates $h_+\mp i h_\times$ follow Maggiore §2.1. The worked LIGO numbers use the reported GW150914 peak strain $h\approx1.0\times10^{-21}$ , arm length $L=4\ \mathrm{km}$ , and proton charge radius $r_p\approx0.84\ \mathrm{fm}$ ; all other relations ( $\Delta L/L=\tfrac12 h$ , area to $O(h^2)$ , the inclination factors $\tfrac12(1+\cos^2\iota)$ and $\cos\iota$ ) are standard textbook results. No numbers are fabricated.

Prerequisites. Transverse-Traceless Gauge · Gravitational Wave Equation · Metric Perturbation and Trace-Reversed Field · Geodesic Deviation and the Tidal Action of Waves. Leads to. Helicity and the Spin-2 Nature of Gravitons · The Quadrupole Formula · Gravitational-Wave Luminosity and Energy Flux · The Isaacson Stress-Energy of Gravitational Waves.

Ask the tutor: Ask me to draw the ring-of-masses movie for an arbitrary mix of $h_+$ and $h_\times$ , or to work out the full $\cos2\psi$ antenna pattern for a detector pointed anywhere on the sky.

Why does a gravitational wave have two polarizations while light also has two, if their spins differ?

It is a coincidence of counting, not of physics. A massless field of spin $s$ has exactly two helicity states, $+s$ and $-s$ , in four dimensions — the longitudinal and lower-helicity modes are removed by gauge invariance. Light is spin 1, so its states are helicity $\pm1$ ; gravity is spin 2, so its states are helicity $\pm2$ . The number "two" is the same, but the transformation differs: light's linear polarizations sit $90°$ apart in real space, gravity's sit $45°$ apart. The shared count hides a genuine difference in spin, visible the moment you rotate the detector.

If the test masses do not move in TT coordinates, in what sense does anything happen?

In the TT gauge the coordinates are glued to the free-falling masses, so their coordinate positions are constant by construction — but coordinates are not distances. The measurable quantity is the proper separation, obtained by integrating the metric along the ruler between the masses, and it oscillates as $\big(1\pm\tfrac12 h\cos\omega t\big)$ times its unperturbed value. Equivalently, in the proper detector frame the masses genuinely accelerate relative to one another under the tidal force of equation (4). Both descriptions agree on the one gauge-invariant fact — the changing distance — which is exactly what an interferometer's light travel time measures.

What is the difference between the plus and cross modes, physically, if they look so similar?

They are the same distortion viewed along different axes, but that difference is physical once you fix a frame or a detector. Relative to given axes, the plus mode acts along those axes while the cross mode acts along the diagonals $45°$ away, and a specific detector or source orientation singles out a frame in which $+$ and $\times$ are distinct, independently measurable amplitudes. Only free rotation of the whole apparatus would make them interchangeable — and even then the swap costs a $45°$ turn, the spin-2 content.

Are there really no longitudinal or "breathing" gravitational waves in general relativity?

Not in general relativity. The gauge freedom that removes eight of the ten metric components also removes any longitudinal ( $z$ -directed) motion and any isotropic "breathing" of the ring, leaving only the two transverse-traceless modes. Some alternatives — scalar-tensor gravity, massive gravity, certain vector-tensor models — do predict extra polarizations: a scalar breathing mode that expands the ring uniformly, or longitudinal modes. This makes polarization a clean discriminator: detect a breathing mode and general relativity is in trouble. Current LIGO-Virgo data are consistent with pure tensor ( $+$ and $\times$ ) content.

How can a traceless strain do any work on the detector if it conserves area?

Area conservation is a first-order statement about the shape of the ring, not about energy. The wave still accelerates each mass, and when the masses feel any friction or coupling — Feynman's sticky bead sliding on its rod, or a readout resisting a mirror's motion — the tidal force does real work. That is the point of the sticky-bead argument: even though the strain is volume-preserving, the relative motion it drives is real and can be tapped. Energy transfer is force times displacement, nonzero regardless of the area returning to its original value each cycle.

Do the two polarizations carry energy and angular momentum independently?

Yes. Each mode carries an energy flux set by the Isaacson stress-energy, proportional to the time-averages of $\dot h_+^2$ and $\dot h_\times^2$ , so a source's total luminosity adds the two contributions. Angular momentum is where the helicity basis shines: the circular combinations $h_R$ and $h_L$ carry spin angular momentum $\pm2\hbar$ per graviton, so a net circular polarization — as radiated by a face-on binary — carries a net flux of angular momentum along the line of sight. This is the classical origin of the graviton being spin 2, measurable through the polarization asymmetry of the signal.