Transverse-traceless polarizations
Identify the two tensor polarizations, their spin-two basis rotation, and the orientation dependence of a leading circular-binary strain.
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The question for today
What you’ll understand
- Write a vacuum plane wave in a normalized plus/cross tensor basis.
- Rotate polarization components without changing the physical tensor.
- Compare source-inclination amplitudes while keeping detector projection separate.
In general relativity a vacuum gravitational plane wave has two independent tensor polarizations. A transverse-traceless representation displays them as plus and cross distortions in the plane perpendicular to propagation. These names depend on a chosen transverse basis; rotating that basis mixes the components with twice the rotation angle. Physical strain tensors and detector predictions remain invariant when components and response tensors are transformed consistently. Source inclination adds another, physically different orientation dependence that should not be confused with a coordinate relabeling.
Two transverse tensors remain in vacuum
Consider a locally plane wave traveling along positive z on a weakly perturbed flat background with the previous signature. Its retarded argument is t minus z over c. In TT gauge the time components vanish, the spatial tensor is transverse, and its spatial trace vanishes. The remaining x-y block contains two real functions. This construction applies to the radiative vacuum field, not to every constrained near-source component.
Normalize each polarization tensor to have squared tensor norm two. The cross tensor therefore includes both xy and yx entries, a detail that later produces a factor of two in energy formulas. The strain functions themselves are dimensionless. A transverse circular motion of coordinate labels is not their definition: they are components of a metric perturbation connected to measurable tidal curvature and light-propagation responses in an appropriate observer frame.
A basis rotation has twice the angle
Rotate the transverse basis axes by an angle psi. The plus tensor is quadratic in those basis vectors, so its transformation uses cosine and sine of twice psi. The cross tensor transforms as its orthogonal partner. Projecting the same physical tensor onto the new basis mixes the strain components by the corresponding two-dimensional rotation. A forty-five-degree basis change consequently interchanges the roles of plus and cross with a sign.
This spin-two behavior differs from rotating an ordinary planar arrow by the same angle. The squared component sum is invariant, and the tensor contraction is twice that sum. A detector's polarization response coefficients must transform alongside the strains, preserving their scalar contraction. Rotating only the strain labels while holding response coefficients in their old basis would create an artificial change in the predicted signal.
Source inclination changes the physical polarization ellipse
At leading quadrupole order a nonprecessing circular binary can be written with plus strain proportional to one half of one plus cosine-squared inclination times a cosine phase, and cross strain proportional to cosine inclination times a sine phase. Let their common scale be A. A face-on source has equal quadrature components of amplitude A; an edge-on source has only plus amplitude A over two. This is a physical change of viewing direction, not a passive basis rotation.
These formulas assume a slowly evolving circular orbit, a dominant quadrupole harmonic and far-zone propagation. Higher harmonics, precession and eccentricity can alter the polarization history. Averaging over source inclination is distinct from averaging a detector's sky pattern, and neither should be silently inserted into an optimally oriented amplitude. A particular interferometer records its projection of the two strains, so a source polarization amplitude is not automatically the measured instrument strain.
Work it through
Relabel a pure plus snapshot
At one retarded instant a wave has h+=2×10⁻²² and h×=0. Rotate the transverse basis by ψ=45° using the displayed convention. Find the new components and compare the component sum of squares and full tensor norm.
Twice the basis angle is 90°, so its cosine vanishes and its sine equals one. The rotated plus component is the old cross component, giving h′+=0.
The rotated cross component is minus the old plus component: h′×=−2×10⁻²². The sign is fixed by the stated passive convention and does not mean that the wave has physically reversed propagation.
Both component sums of squares equal 4×10⁻⁴⁴. Since each basis tensor has norm squared two, the full tensor contraction is 8×10⁻⁴⁴ in either basis. This checks the normalization as well as the rotation.
The snapshot becomes (0,−2×10⁻²²), while its physical tensor and norm remain unchanged. A detector projection must be transformed consistently to give the same measurement.
The living notebook
Explore the model
Interactive lab · vectors
Twice the polarization-basis angle
Represent the pair (cos2ψ,sin2ψ) as a unit planar vector. Reset and set both controls to Angle 0° and Magnitude 1; predict (1,0), corresponding to ψ=0°. For a fresh comparison, Reset and set both controls to Angle 90° and Magnitude 1; predict (0,1), corresponding to ψ=45°. The lab angle is twice the physical basis angle in this analogy.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThis illustrates the coefficients of a spin-two basis transformation. It does not contain a metric tensor, tidal ring, binary source or detector response, and the vector itself is not a gravitational-wave trajectory.
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Check your understanding
Your turn
Try it yourself
Use h+=A(1+cos²ι)cosΦ/2 and h×=A cosι sinΦ at fixed common scale A. For ι=60°, find the two component amplitudes and the cycle average of h+²+hײ as a fraction of its face-on value.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Cosι=1/2. Average cos²Φ and sin²Φ separately to one half; do not add amplitudes before squaring.
Compare with the solution
The plus amplitude is 5A/8 and the cross amplitude A/2. Their cycle-averaged squared sum is A²[(5/8)²+(1/2)²]/2=41A²/128. Face-on components have amplitudes A in quadrature, whose squared sum averages to A². The requested fraction is therefore 41/128≈0.3203. This is a source-inclination comparison at fixed A, without a detector sky average or an assumption that both components are independently observed.
Keep the idea
Take it with you
- TT gauge displays two transverse tensor polarizations in an appropriate vacuum radiation region.
- Passive basis rotations mix components with twice the axis angle and preserve their norm.
- Source inclination, polarization-basis choice and detector orientation are separate ingredients.
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