Can a detector tell a lensed chirp from an ordinary one?

This chapter assumes the exact amplification factor of a point-mass lens. You also need least-squares fitting, complex inner products, and the idea that detector noise is random with a known spectrum.

Noise and the inner product

A detector's noise is described by its one-sided power spectral density $S_n(f)$, in units of $1/\mathrm{Hz}$: the noise power per unit frequency. Signals are compared with a noise-weighted inner product,

so frequencies where the detector is quiet count more. The optimal signal-to-noise ratio of a signal $h$ is $\rho=\sqrt{\langle h,h\rangle}$. We use this form, Eq. (4.3) of Ajith (2011), and his Eq. (4.7) fit to the Advanced LIGO design noise curve, with $f_{\rm low}=20$ Hz. This is a design curve, not the noise of any observing run. Ajith 2011, Sec. IV

The chirp

Two black holes of $10\,M_\odot$ each spiral together. To leading order the frequency-domain signal from the stationary-phase approximation is

where $\mathcal A$ is a constant amplitude and $\mathcal M_c=(m_1m_2)^{3/5}/(m_1+m_2)^{1/5}$ is the chirp mass, written in seconds by multiplying by $G/c^3$. This is the leading term of Eqs. (41) and (42) in Cutler and Flanagan. Cutler and Flanagan 1994, Sec. III

Worked example. For $10+10\,M_\odot$, $\mathcal M_c=8.71\,M_\odot$. We stop the signal at the innermost stable circular orbit, $f_{\rm ISCO}=c^3/(6^{3/2}\pi GM)=219.9$ Hz for total mass $M=20\,M_\odot$. The time from 20 Hz to merger is $(5/256)\,\mathcal M_c^{-5/3}(\pi f)^{-8/3}=5.96$ s. Merger and ringdown are left out, so treat this as a teaching waveform.

Match: how alike two chirps are

Lensing multiplies the signal by $F(f)$. Compare the lensed chirp $h_{\rm L}=F\,h$ with the unlensed chirp $h_{\rm U}=h$ through the match,

The maximum over arrival time $t_0$ and phase $\phi_0$ lets the unlensed chirp slide and rotate to fit as well as it can. A match of 1 means the lens left only an overall amplitude change and a shift; anything less means the shape changed. In practice one fast Fourier transform gives the overlap at every $t_0$ at once.

From match to evidence

Suppose the data are exactly the lensed signal, with no noise. Fit it with the unlensed template of free amplitude $A$, after the best shift. The squared residual is

It is smallest at $A=\langle h_{\rm L},\hat h_{\rm U}\rangle=\rho_{\rm L}\,\mathrm{match}$, where the residual is $\rho_{\rm L}^2-\rho_{\rm L}^2\mathrm{match}^2$. A lensed template fits exactly and leaves zero. So

The Lab calls $\Delta\chi^2$ the lens signature strength. Because it scales with $\rho_{\rm L}^2$, halving the signal-to-noise ratio divides it by four.

The Lab presets in numbers

All values below use an unlensed signal-to-noise ratio of 20 and $10+10\,M_\odot$ black holes. The lens magnifies the signal, so $\rho_{\rm L}$ differs from 20.

Two rows teach the most. The second copy is equally loud in the first two rows, but the waves are too long for the $15\,M_\odot$ lens to split, so the fringes never form and the match stays at 0.998. The faint copy has the longest delay and still gives the weakest signature after the long-wave case, because a 17 percent echo changes the shape very little.

What noise does

Real data contain noise, so the two best fits also fit some of the noise. The lensed family has two extra parameters, lens mass and $y$, and many fringe phases to try, so it always wins a little. To measure how much, the recompute guide injects chirps into synthetic Gaussian noise drawn from the same design curve and runs a template search. Its results at an unlensed signal-to-noise ratio of 20, in $\Delta\ln L$:

• No lens, 120 trials: median 2.49, 90th percentile 4.35, maximum 7.97. A threshold of 4 is crossed in 11.7 percent of trials.
• Two paths at $y=0.5$, 20 trials: median 110.2. Without noise the value is 115.2. Every trial passes.
• Faint copy at $y=2$, 40 trials: median 4.27. Without noise the value is 6.03, matching half of the 12.1 in the table. Only 52.5 percent of trials pass.

The faint case and the noise-only case overlap. Lowering the threshold would catch more faint copies and more noise; raising it would lose both. The only real escape is a louder signal.

Why the Lab's number is optimistic

The Lab keeps the unlensed template at the true black-hole masses and only shifts and rotates it. A real search also varies masses, spins and the merger, and some of the lens wiggle can be absorbed that way. Real noise also has short bursts, called glitches. In the third LIGO–Virgo–KAGRA observing run, the event GW200208_130117 gave a mild preference for a microlensed waveform, and the authors suggest that a short noise fluctuation produced a dip that mimics a beating pattern. LIGO–Virgo–KAGRA 2023, App. B An isolated point lens is also only a first approximation. Most stellar-mass lenses would sit inside a galaxy, whose own field changes the pattern. The paper guide follows these issues through the published search.

Lab tasks

1. In the gravitational-wave lens Lab, start from the preset Two paths, clear fringes. Open Source and loudness, halve the loudness without a lens from 20 to 10, and check that the lens signature falls by a factor of four.
2. Choose A faint second copy. Under Source and loudness, raise the loudness without a lens until the lens signature passes 16, twice the largest noise-only $\Delta\ln L$ in the tutorial. Record the value.
3. In the same group, raise the component masses. Watch the band edge move left on the amplification plot, and explain why the signature changes.

Exercises

1. Find the chirp mass and the cutoff frequency for $15+15\,M_\odot$.
2. Derive $\Delta\chi^2=\rho_{\rm L}^2(1-\mathrm{match}^2)$ by minimizing $\chi^2(A)$.
3. The faint copy gives $\Delta\chi^2=12.1$ at an unlensed signal-to-noise ratio of 20. Ignoring noise, at what signal-to-noise ratio does it reach 16?
4. The two-path preset has a shorter delay than the faint copy, yet a far larger signature. Explain why using the amplitude ratio.

Next: recompute point-lens amplification and test a lens search.