Interferometer antenna patterns
Contract polarization tensors with a detector geometry, define sky/basis conventions and distinguish angular sensitivity from network localization.
On this page
The question for today
What you’ll understand
- Compute the leading response of a right-angle interferometer.
- Rotate polarization components and antenna coefficients consistently.
- Distinguish sky/polarization averages, source inclination and arrival-time localization.
An interferometer measures one linear combination of the two tensor strains, determined by its arm directions and the wave's sky position and polarization basis. Its antenna pattern is broad rather than a narrow telescope beam, and some directions or polarization combinations have little response. A network gains information by combining different projections and arrival times. These geometric effects are central to distance and localization inference, but no single antenna coefficient is a universal conversion between source amplitude and measured strain.
The detector tensor defines the projection
Use an ideal equal-arm right-angle detector whose arms lie along x and y, in the long-wavelength free-mass approximation. Its tensor is one half the difference of the two arm outer products. Contracting this tensor with each transverse polarization tensor defines the corresponding antenna coefficient. The measured strain is the sum of those coefficients times the incoming polarization components, with propagation delays and optical transfer effects added when required.
Let the source line of sight have polar angle theta from positive z and azimuth phi from the x arm. The wave propagates toward the detector along minus that line of sight. At zero polarization-basis angle choose transverse axes p equal to the usual spherical theta vector and q equal to minus the spherical phi vector. This explicitly fixes the cross sign and makes the transverse axes right handed with the propagation direction.
Patterns rotate with the polarization basis
With the chosen base axes the plus coefficient is one half of one plus cosine-squared theta times cosine twice phi, and the cross coefficient is cosine theta times sine twice phi. A passive polarization-basis rotation mixes them with the same component rotation used earlier for the strains. Their scalar contraction remains invariant when both are transformed. Changing only the source's physical polarization relative to a fixed detector is a different operation.
For an overhead source aligned with the arm basis the plus response can reach one. A source in the detector plane along an arm bisector has both coefficients zero in this ideal model. Other zeros affect only one component or a particular combination. Averaging uniformly over sky and polarization-basis angle gives mean square one fifth for each coefficient, but this does not include a binary's inclination average. Optimal, root-mean-square and population-averaged amplitudes therefore require different stated conventions.
Networks constrain projections and delays
Separated detectors sample the same wave with different antenna coefficients and geometric delays. A measured delay on one baseline constrains a direction cosine and hence a ring on the sky, rather than a unique point. Additional baselines, amplitude ratios, polarization consistency and the time evolution of detector orientation can reduce that ambiguity. Noise, calibration and finite bandwidth control how accurately those constraints can be extracted.
A single observed amplitude mixes luminosity distance with inclination, polarization and sky projection. Even a network can retain substantial degeneracies when its responses are similar or the signal is weak. The long-wavelength pattern also omits frequency-dependent optical response and arm-transfer effects. A full likelihood combines calibrated strain, noise and a propagated waveform; neither an antenna map nor an arrival-time ring alone supplies a complete astrophysical position or distance estimate.
Work it through
Change basis without changing a detector prediction
Take θ=60°, φ=0 and ψ=0. At an instant h+=2×10⁻²¹ and h×=0. Find the antenna coefficients and detected strain. Then passively rotate the polarization basis by 45° and transform both the strain and response components.
Cosθ=1/2 and cos2φ=1, so a=(1+1/4)/2=5/8 and b=0. At ψ=0 the coefficients are F+=5/8 and F×=0.
The detector strain is (5/8)(2×10⁻²¹)=1.25×10⁻²¹. This differs from the source plus component because the response is not optimal for this line of sight.
The passive 45° rotation sends the response pair to (0,−5/8) and the strain pair to (0,−2×10⁻²¹). Their contraction is again +1.25×10⁻²¹. Rotating only one pair would incorrectly predict a changed physical signal.
The detected strain is 1.25×10⁻²¹ in either basis. The apparent transfer from plus to cross is a representation change, not a change of source orientation or detector hardware.
The living notebook
Explore the model
Interactive lab · waves
A detector combination can cancel coherently
As an analogy for two projected coherent strain contributions, Reset and set both controls to Phase difference 0° and Amplitude ratio 1; predict resultant amplitude 2. For a fresh comparison, Reset and set both controls to Phase difference 180° and Amplitude ratio 1; predict resultant amplitude 0. Both components remain present even when their selected sum cancels.
- Sum at x = 0
- 0.866
- Resultant amplitude
- 1.732
- Phase difference
- 60°
Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). 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 forThe equal-frequency wave lab illustrates one coherent projection cancellation. It has no sky angles, arm tensor, arrival-time baselines or detector antenna calculation, and it does not infer that the incoming gravitational-wave energy vanished.
Pause & reason
Check your understanding
Your turn
Try it yourself
Two ideal detectors have baseline length B=3000 km. Use the source-direction convention . If Δt=−6.0 ms and c=3.0×10⁸ m/s, find the angle between source direction and baseline. Does this one delay locate a unique sky point?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The maximum delay magnitude is B/c. Divide the measured signed delay by that scale before taking an inverse cosine.
Compare with the solution
The baseline light time is 3.0×10⁶/(3.0×10⁸)=0.010 s=10 ms. Thus cosα=−Δt/(B/c)=0.6 and α≈53.13°. One direction cosine specifies a ring around the baseline, not a unique sky point. Additional detector baselines and coherent amplitude/polarization information help localize the source, with finite noise and calibration affecting the actual uncertainty.
Keep the idea
Take it with you
- A detector tensor projects the two physical strains into a geometry-dependent scalar output.
- Antenna and strain components must rotate together under passive polarization-basis changes.
- Sky averages, source inclination and network delay localization provide different information and retain different degeneracies.
Read further
Original sources and deeper explanations to follow at your own pace.
Optional progress, saved in this browser when available. Revisit any lesson at any time.
Lesson not marked complete.