Antenna Patterns and Polarization Response

An interferometer does not measure plus and cross strain separately. It measures one direction- and basis-dependent linear combination whose coefficients are the antenna patterns. For a right-angle Michelson in the small-antenna limit, those patterns follow entirely from the detector tensor and the transverse polarization screen. Their zeros, maxima, and rotation law explain both the reach and the degeneracies of a single detector.

Required background. The Wave Equation and Characteristics supplies plane-wave propagation and the transverse screen.

Helpful background. Gravitational Radiation and Waveforms fixes the plus-cross convention, while Probability and Inference for Gravity supplies the averaging language used for sky and polarization populations.

A declared sky and screen convention

Let $\boldsymbol n$ point from the detector toward the source, so the wave propagates along $-\boldsymbol n$. In detector Cartesian axes,

$$\boldsymbol n =(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta).$$

Choose the right-handed transverse screen

$$\boldsymbol p =(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta), \qquad \boldsymbol q=(-\sin\phi,\cos\phi,0),$$

and define

$$e^+_{ij}=p_ip_j-q_iq_j, \qquad e^\times_{ij}=p_iq_j+q_ip_j.$$

These tensors obey

$$e^A_{ij}e^{B,ij}=2\delta^{AB}, \qquad n^ie^A_{ij}=0, \qquad \delta^{ij}e^A_{ij}=0.$$

The normalizations matter: changing $e^A:e^B$ changes the numerical definitions of $h_A$ and $F_A$ even though their product can remain unchanged. Here $\boldsymbol p\times\boldsymbol q=\boldsymbol n$, so the screen is oriented with the source direction rather than the propagation direction. The transverse-traceless field and its two tensor polarizations are derived in Flanagan and Hughes 2005, §2.1, Eqs. (2.18)–(2.25), PDF.

For orthogonal arms along $x$ and $y$, the ideal detector tensor is

$$D^{ij}=\frac12 \begin{pmatrix} 1&0&0\\ 0&-1&0\\ 0&0&0 \end{pmatrix}.$$

The Michelson-response derivation then gives

$$s(t)=F_+(\theta,\phi,\psi)h_+(t) +F_\times(\theta,\phi,\psi)h_\times(t).$$

Antenna patterns

First set the polarization angle $\psi=0$ for the screen above. Direct contraction yields

$$F_+^{(0)} =\frac12(1+\cos^2\theta)\cos2\phi, \qquad F_\times^{(0)}=-\cos\theta\sin2\phi.$$

Rotate the screen through $\psi$ by

$$\boldsymbol p_\psi=\boldsymbol p\cos\psi+\boldsymbol q\sin\psi, \qquad \boldsymbol q_\psi=-\boldsymbol p\sin\psi+\boldsymbol q\cos\psi.$$

Because the polarization is spin two,

$$\begin{pmatrix} F_+\\ F_\times \end{pmatrix}_{\!\psi} = \begin{pmatrix} \cos2\psi&\sin2\psi\\ -\sin2\psi&\cos2\psi \end{pmatrix} \begin{pmatrix} F_+^{(0)}\\ F_\times^{(0)} \end{pmatrix}.$$

Equivalently,

$$\begin{aligned} F_+ &=\frac12(1+\cos^2\theta)\cos2\phi\cos2\psi -\cos\theta\sin2\phi\sin2\psi,\\ F_\times &=-\frac12(1+\cos^2\theta)\cos2\phi\sin2\psi -\cos\theta\sin2\phi\cos2\psi. \end{aligned}$$

The strain components transform by the same orthogonal matrix: $h_A=\tfrac12 e_A^{ij}h_{ij}$ implies $\boldsymbol h_\psi=\mathbf R(2\psi)\boldsymbol h_0$. Consequently $\boldsymbol F_\psi^T\boldsymbol h_\psi=\boldsymbol F_0^T\boldsymbol h_0$. This is a change of basis for one fixed physical wave; rotating the source polarization while holding the detector basis fixed is a different operation and can change the output.

Other sources may define $\boldsymbol n$ as the propagation direction, reverse $\boldsymbol q$, or rotate the screen oppositely. Those choices can change signs in the displayed formulas. The invariant test is the contraction $D^{ij}h_{ij}$ after the source and detector conventions have both been translated. The response-tensor construction and its convention warnings are developed in Romano and Cornish 2017, §5.2, especially Eqs. (5.9)–(5.13) and (5.29)–(5.32), PDF.

Maxima, nulls, and averages

The polarization-independent response power is

$$\mathcal R^2(\theta,\phi) =F_+^2+F_\times^2 =\frac14(1+\cos^2\theta)^2\cos^22\phi +\cos^2\theta\sin^22\phi.$$

It is independent of $\psi$, as it must be: relabeling the screen cannot change an unpolarized response.

For a source directly above or below the detector, $\lvert\cos\theta\rvert=1$ and $\mathcal R=1$. The named plus and cross coefficients still depend on the screen, but their quadrature sum does not. Four exact small-antenna nulls lie in the detector plane along the arm bisectors:

$$\theta=\frac\pi2, \qquad \phi=\frac\pi4,\frac{3\pi}4,\frac{5\pi}4,\frac{7\pi}4.$$

A null is a property of this channel and geometry, not proof that no gravitational wave passed through the site.

For sources uniformly distributed over the sphere and a uniformly distributed polarization angle,

$$\left\langle F_+^2\right\rangle =\left\langle F_\times^2\right\rangle =\frac15, \qquad \left\langle\mathcal R^2\right\rangle=\frac25.$$

These are power averages, not amplitude averages. They do not include a source inclination distribution, distance selection, detector duty cycle, a frequency-dependent response, or a search threshold.

What one detector cannot identify

At a fixed time and frequency a single output supplies one complex number,

$$\widetilde s=F_+\widetilde h_++F_\times\widetilde h_\times.$$

Even with known sky position, two unknown polarization amplitudes cannot generally be recovered from one equation. For a compact binary, polarization, inclination, distance, phase, and sky position can produce related amplitude-phase changes. Earth’s rotation modulates a long signal and supplies additional projections over time, but a short transient does not receive that benefit. Independent detector tensors are therefore essential for polarization reconstruction and localization.

The small-antenna functions also cease to be complete when the phase changes along an arm. Then $F_A$ becomes complex and frequency dependent. The exact path-integrated response is derived on Finite-Arm and Frequency-Dependent Response.

Observational boundary and checks

This page makes no current polarization measurement. To use an antenna factor in a likelihood, an analysis must bind the detector geometry and orientation, Earth-reference frame and time standard, sky-coordinate convention, polarization convention, complex response and calibration version, selected data segment, and frequency range. Current public geometry and data interfaces are taken from the GWOSC documentation available on 2026-08-18; a later detector or metadata release triggers a review.

Three independent checks are especially useful:

• $F_+^2+F_\times^2$ must be unchanged by $\psi$.
• Normal incidence must give unit quadrature response for a right-angle interferometer.
• Numerical sky integration must recover $\langle F_+^2\rangle=\langle F_\times^2\rangle=1/5$ before any population or selection weighting is added.

Common pitfalls

Calling an antenna null a field null. The wave is present even when one detector tensor is orthogonal to its polarization tensor. A differently oriented detector may respond strongly.

Mixing source and propagation directions. Replacing $\boldsymbol n$ by $-\boldsymbol n$ without translating the screen can flip cross-polarization and arrival-time signs.

Averaging amplitudes instead of power. Signed antenna amplitudes average to zero over the sky. Sensitivity calculations involve $F_A^2$, a signal model, noise weighting, and selection.

Exercises

1. Locate the blind directions. Starting from $\mathcal R^2$, prove that the right-angle small-antenna response vanishes only in the detector plane along the arm bisectors.

Solution

Both terms in

$$\mathcal R^2 =\frac14(1+\cos^2\theta)^2\cos^22\phi +\cos^2\theta\sin^22\phi$$

are nonnegative. The second can vanish with nonzero $\sin2\phi$ only when $\cos\theta=0$, so $\theta=\pi/2$. The first then requires $\cos2\phi=0$, giving $\phi=\pi/4+k\pi/2$. Conversely, those four directions make both terms zero.

2. Recover the sky average. Average $\mathcal R^2$ uniformly over the sphere and show that it equals $2/5$.

Solution

Let $\mu=\cos\theta$. Averaging first over $\phi$ uses $\langle\cos^22\phi\rangle_\phi=\langle\sin^22\phi\rangle_\phi=1/2$:

$$\left\langle\mathcal R^2\right\rangle_\phi =\frac18(1+\mu^2)^2+\frac12\mu^2 =\frac18(1+6\mu^2+\mu^4).$$

Uniform solid angle makes $\mu$ uniform on $[-1,1]$, so $\langle\mu^2\rangle=1/3$ and $\langle\mu^4\rangle=1/5$. Therefore

$$\left\langle\mathcal R^2\right\rangle =\frac18\left(1+2+\frac15\right)=\frac25.$$

A uniform polarization rotation shares this power equally between plus and cross, giving $1/5$ for each.

References

• Flanagan, Éanna É., and Scott A. Hughes. “The Basics of Gravitational Wave Theory.” New Journal of Physics 7 (2005): 204. DOI. Open PDF.
• Gravitational Wave Open Science Center. “API Documentation.” Version available 2026-08-18. Official documentation.
• Romano, Joseph D., and Neil J. Cornish. “Detection Methods for Stochastic Gravitational-Wave Backgrounds: A Unified Treatment.” Living Reviews in Relativity 20 (2017): 2. DOI. Open PDF.