Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The merging black hole binary system GW150914 was detected in only 16 days of aLIGO data at a signal level that appears to be well above the detection threshold at around 5 sigma. There are no further events between 4 and 5 sigma in the same data.

Could this event have been detected by previous LIGO/VIRGO incarnations which observed for much longer, albeit with lower sensitivity? If so, does this indicate that the aLIGO team have struck lucky and that this is a comparatively rare event that may not be repeated for many years?

EDIT: The answers I have agree that LIGO couldn't have seen it, but don't yet completely explain why. GW150914 had a strain that rose from a few $10^{-22}$ to $10^{-21}$ over about 0.2 seconds. This appears to make the characteristic strain maybe a few $10^{-22}$ Hz$^{-1/2}$ and thus would appear from the published sensitivity curves to lie above original LIGO's detection sensitivity at frequencies of $\sim 100$ Hz. Is my estimate of the characteristic strain way off?

share|cite|improve this question
    
Did they say why they picked this particular data set? Can we speculate that there are other datasets from the run that contain more events that will be published later? – CuriousOne Feb 14 at 19:36
    
This just lets me think how unlucky it would have been, if the ligo would have been offline due to its improvement while this event had been occured... >.<' – Zaibis Feb 15 at 14:56
    
@CuriousOne Possibly just because they had to pick a stretch of data so they could start writing the paper. Still, I look forward to seeing what the next 88 days of data contained :) – Will Vousden Feb 16 at 16:18
    
@WillVousden: I am simply asking if there are any rumors about what else is in the rest of the data... :-) – CuriousOne Feb 16 at 20:59
    
@CuriousOne It had crossed my mind as well... I wish I knew! I left the collaboration just after the detection :-( – Will Vousden Feb 16 at 21:11
up vote 19 down vote accepted

To expand on HDE's answer, initial LIGO indeed wouldn't have detected GW150914, but it's not quite as simple as the peak strain being below the curve in the sensitivity plot: the integration time also matters.

These plots can be misleading; the curves they show don't represent a minimum detectable strain. Indeed, the units on the y-axis of these plots are $\mathrm{Hz}^{-1/2}$, while the GW strain is dimensionless, so you can't actually compare them! It's entirely possible to detect a signal that peaks well below the noise curve, as long as it's in-band for sufficiently long.

The curves that you see describing LIGO detector sensitivities conventionally show the amplitude spectral density of the detector noise. Meanwhile, the threshold for a detection is determined by the SNR from matched (Wiener) filtering. Assuming we know the form of the signal $h$ in advance (see caveats below), this is defined in terms of the noise-weighted inner product of $h$ with itself: $$ \mathrm{SNR}^2 = \left<h,h\right> \equiv \int_0^\infty \frac{4|\tilde{h}(f)|^2}{S_n(f)}\,\mathrm{d}f $$ where $S_n(f)$ is the noise power spectral density (i.e., the square of what's shown in the sensitivity plots). The SNR therefore depends on the spectral composition of the signal and its overlap with the detector bandwidth.

If you imagine this in the time domain (Parseval's theorem), the (squared) SNR actually accumulates in proportion to the number of cycles the waveform spends in-band. For a monochromatic source, this is proportional to the integration time. For example, if $\tilde{h}(f) = \delta(f-f_0)h_0$ and, without loss, the noise PSD is a constant $S_n(f_0)$, then the SNR is given by: $$ \mathrm{SNR}^2 = \frac{2}{S_n}\int_{-\infty}^\infty|\tilde{h}(f)|^2\,\mathrm{d}f = \frac{2}{S_n}\int_{-\infty}^\infty|h(t)|^2\,\mathrm{d}t $$ Therefore, since $|h(t)| = h_0$, for a finite observation window $T$, the SNR scales with $\sqrt{T}$: $$ \mathrm{SNR} = \sqrt{\frac{2T}{S_n}}h_0 $$

So, let's approximate GW150914 as a monochromatic source. Reading off the plots in the detection paper, let's say it has a average frequency of $f_0 \approx 60 \,\mathrm{Hz}$, an amplitude of $h_0\approx 5\times10^{-22}$, and a duration of $T \approx 0.2\,\mathrm{sec}$. Then, reading off a strain ASD of $\sqrt{S_n(f_0)} \approx 10^{-22}$ for initial LIGO, we'd get an SNR of around 3, which doesn't meet the standard detection threshold of 8 (also, see the caveats below).

There's a much more complete discussion of detector sensitivity curves in this paper; it's worth a read! A more useful quantity, described in this paper, is the characteristic strain, which attempts to account for the frequency evolution of an inspiral signal such as GW150914, to ease comparison between detector sensitivity and strain amplitude.


Caveats: in practice, it's more complicated than the matched filter model, since the detector noise is annoyingly non-stationary and non-Gaussian. There are more sophisticated search algorithms that use things like signal quality vetoes and $\chi^2$ discriminants that reject spurious responses of the matched filter. There are also search algorithms that don't require a priori knowledge of the signal waveform and can detect unmodelled bursts. It was actually this sort of generic search that detected GW150914; references are available in the detection paper.

Also note that the SNR defined above is the optimal SNR that you get if:

  1. you filter the data stream with the exact signal that you're looking for, and
  2. the noise realisation is zero.

Since the mean of the noise is zero, number 2 above is equivalent to taking the expectation of the SNR over all noise realisations.

In practice, we don't know the precise signal a priori, and some SNR is lost in the approximation. For a candidate waveform $u$, the expected SNR (over all noise realisations) is then given by $$ \mathrm{SNR} = \frac{\left<u,h\right>}{\|u\|} $$

share|cite|improve this answer
1  
Right, to compare with the sensitivity plots you need to multiply the strain by the square root of the time the signal was observed for (roughly)? – Rob Jeffries Feb 14 at 22:45
1  
@RobJeffries To the extent that the signal is approximately monochromatic, yes (and of course you must divide by the noise PSD at the signal's characteristic frequency). That will give you an approximate SNR, which you can compare to with whatever threshold you've picked (conventionally 8) to determine whether the signal would be detected. – Will Vousden Feb 14 at 23:02
1  
(For more complicated/broadband signals, you have to know the waveform $h(f)$ and do the integration above.) – Will Vousden Feb 14 at 23:50
1  
Given that, then the strain of this event was a few $10^{-22}$ up to a peak of $10^{-21}$ seen over about 0.2 secs. So isn't that characteristic strain $>10^{-22}$ Hz$^{-1/2}$? In which case the sensitivity curves I've seen (see HDE226868's answer) suggest that original LIGO would have detected it. I'm guessing then that my estimate of the characteristic strain is too high by factors of a few - why? – Rob Jeffries Feb 15 at 12:10
1  
@RobJeffries See my update for an example calculation of the SNR for GW150914 :) – Will Vousden Feb 15 at 16:10

I have a direct quote from the website:

The event would not have registered in LIGO's first-generation detectors; the fact that it appeared with striking clarity in both L1 and H1 indicates the leap in detector performance that the Advanced LIGO program has produced.

This was a sensitivity issue: in most frequencies, Advanced LIGO is more strain-sensitive than LIGO by a factor of 10.

Hild (2012) gives an overview of first- and second- generation detectors, including this graph:

The detected wave had a peak strain of $\thicksim 10^{-21} ~\mathrm{1/\sqrt{\textrm{Hz}}}$ and was detected at frequencies between $35$ and $250~\textrm{Hz}.$ A portion of this does fall in the original LIGO sensitivity range. However, as Figure 3 of the Advanced LIGO detection paper shows, most falls below it.

Regarding the odds of detection, older estimates said that Advanced LIGO should be able to observe $\thicksim 40$ neutron star "inspiral events", and $\thicksim 30$ black hole binary events of the same type. This would be attributed in part to the noise reduction, making it easier to observe larger areas. The group says that event rate detection would increase by 3000 following the sensitivity upgrades.

share|cite|improve this answer
2  
Agreed original LIGO is a factor 4 less sensitive, but there were intermediate upgrades. I'd also like to know the quantitative details. It does look from published sensitivity curves that it might have been seen... – Rob Jeffries Feb 14 at 19:26
    
@RobJeffries I understand; I'm looking for the particulars. – HDE 226868 Feb 14 at 19:27
1  
@RobJeffries I don't have a reference on hand, but a LIGO person the other day said in a talk this event, which was SNR=24 by some measure, would have been SNR=3 or 4 in old LIGO iirc, below their threshold of 8 for reporting. – Chris White Feb 14 at 20:30
1  
I think that strain number you quote is a bit tricky. The peak strain was $10^{-21}$, but the equivalent strain (which is on the y-axis of your plot) was more like $3\times 10^{-22}$. – Rob Jeffries Feb 14 at 22:41
1  
I meant characteristic strain, not equivalent strain. – Rob Jeffries Feb 14 at 23:58

Despite my hopes, it appears that old LIGO would not have detected GW150914. Sigh...

From the discovery announcement of GW150914, I created a facsimile of the measured strain signal:

h(t)

From that paper and also the summary of Run 6 of original LIGO, in 2009-2010, I digitized amplitude spectral densities of the two:

enter image description here

From these inputs I calculated the characteristic strain and noise curves:

enter image description here

Integrating these up per the very helpful reference identified by Will Vousden, I calculate the following results for signal-to-noise ratio $\rho$ for a matched filter:

\begin{equation*} \begin{array}{lcccc} \text{detector} & \rho & \chi_r^2 & \hat{\rho} & \hat{\rho_c} \\ \text{aLIGO} & 18.73 & 1.44 & 16.69 & 23.6 \\ \text{LIGO run 6} & 4.88 & 1.44 & 4.35 & 6.2 \end{array} \end{equation*}

Here:

  • $\rho$ is the calculated signal-to-noise ratio for the detector.
  • the $\chi_r^2$ statistic measures how closely the signal matches the expected waveform template. This value was not reported, that I could see; I have chosen it to get the correct final result for aLIGO, and then used the same value for LIGO Run 6.
  • $\hat{\rho}$ penalizes $\rho$ based on the value of $\chi_r^2$.
  • $\hat{\rho}_c$ accounts for the signal being observed by both LIGO detectors: the two signal-to-noise ratios are added in quadrature (so here the value is multiplied by $\sqrt{2}$.)

Since the Run 6 signal-to-noise ratio $\rho < 8$, it would have been very difficult to claim a detection: in this review of Run 6's search for massive black hole inspiral/merger signals, Figure 2 shows that signals with these values of $\rho$ and $\chi_r^2$ would be buried in noise events.

Curiously, that same review includes Figure 1, which appears to show a 60 total solar mass event being detectable out to about 470 Mpc (with favorable orientation), which would include most of the the reported distance range of $410_{-180}^{+160}$ Mpc. I cannot explain the discrepancy.

share|cite|improve this answer
    
Belated thanks for this. Yes it appears to be close, but not quite. – Rob Jeffries Apr 25 at 7:51
    
Some awesome work here! – WetSavannaAnimal aka Rod Vance May 17 at 8:48

protected by Qmechanic Feb 15 at 17:12

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.