Since now two neutron stars have been detected merging via gravitational waves, I was wondering what is the current detection threshold that the LIGO detectors can achieve.

Considering that the first observed objects were two black holes with a combined mass of more than 60 solar masses and they now detected two neutron stars with a combined mass of only about 3 solar masses I was wondering what was the threshold that these detectors can actually detect.

Obviously there are much larger stars out there which orbit each other, but their size and distance from each other make gravitational waves too difficult to detect. So what masses and at what distances can we expect to be detected in the future?


2 Answers 2


I'm afraid this is not straightforward

The amplitude of the gravitational wave strain signal from a merging compact binary (neutrons star or black hole) is $$h \sim 10^{-22} \left(\frac{M}{2.8M_{\odot}}\right)^{5/3}\left(\frac{0.01{\rm s}}{P}\right)^{2/3}\left(\frac{100 {\rm Mpc}}{d}\right),$$ where $M$ is the total mass of the system in solar masses, $P$ is the instantaneous orbital period in seconds and $d$ is the distance in 100s of Mpc. $h \sim 10^{-22}$ is a reasonable number for the sensitivity of LIGO to gravitational wave strain where it is most sensitive (at frequencies of 30-300 Hz).

So you can see that to increase the observability you can increase the mass, decrease the period or decrease the distance.

But here are the complications. LIGO is only sensitive between about 30-300 Hz and the GW frequencies are twice the orbital frequency. Thus you cannot shorten the period to something very small because it would fall outside the LIGO frequency range and you also cannot increase the mass to something too much bigger than the black holes that have been already seen because they merge before they can attain high enough orbital frequencies to be seen. (The frequency at merger is $\propto M^{-1}$).

A further complication is that the evolution of the signals is more rapid at lower masses. That is - the rate of change of frequency and amplitude increase rapidly with total mass. That is why the recent neutron star merger was detectable for 100s by LIGO, whereas the more massive black hole mergers could only be seen for about 1 second. But what this means is that you have fewer cycles of the black hole signal that can be "added up" to improve the signal to noise, which means that higher mass sources are less detectable than a simple application of the formula I gave above would suggest. A further complication is that there is a geometric factor depending on how the source and detectors are orientated with respect to each other.

OK, these are complications, but the formula can still be used as an approximation. So if we take the GW170817 signal, the total mass was about $2.8M_{\odot}$, the source was at 40 Mpc, so at frequencies of 200 Hz (corresponding to a period of 0.01 s) you might have expected a strain signal of about $3\times 10^{-22}$. This gave a very readily detectable signal. The discovery paper (Abbot et al. 2017) says the "horizon" for detection was approximately 218 Mpc for LIGO-Livingston and 107 Mic for LIGO-Hanford. As the source was much closer than these numbers then it is unsurprising that the detection was strong.

Taking the formula above and a fixed orbital period of 0.01 s, we can see that the horizon distance will scale as $\sim M^{5/3}$. So a $10 M_{\odot} + 10 M_{\odot}$ black hole binary might be seen out to $218 \times (20/2.8)^{5/3} = 5.7$ Gpc (this will be an overestimate by a factor of a few because of the issue of the rapidity of the evolution towards merger that I discussed above).

A more through and technical discussion can be read here, although this is a couple of years out of date and LIGO's reach has been extended by about a factor of five since these calculations were done.

  • $\begingroup$ Maybe this should be a separate question, but why doesn't the gravitational wave "brightness" go down by the square of the distance? $\endgroup$
    – antlersoft
    Oct 17, 2017 at 15:16
  • $\begingroup$ @antlersoft Because you are measuring the amplitude of the wave, not the power. $\endgroup$
    – ProfRob
    Oct 17, 2017 at 15:53

Figure 1 of this paper shows the horizon distance (distance to which a circularly polarised overhead signal would be detected at SNR 8) for larger mass systems up to total mass of 1000 solar masses, assuming a search with compact binary coalescence templates. For higher masses the signal amplitude is generally larger, but they merge at lower frequencies so the signals are generally shorter-lived in the sensitive band of the detectors. As they're shorter they also, unfortunately, look a lot more like classes of instrumental glitches, so if they're not that strong (just above a threshold of roughly SNR 8) the background level can be large and lead to lower significance of any candidates.


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