# What was the rate of Black Hole - Black Hole mergers expected to be detected by LIGO prior to GW150914?

My question is a more detailed version of the one found here, which elicited some good information but the question was never really answered.

From table 4 in a 2010 paper we see the estimated rate of BH-BH mergers ranged from $10^{-4}$ to $0.3 \rm \,Mpc^{-3} Myr^{-1}$. This corresponds to $10^{-10}$ to $3\times10^{-7} \rm\,Mpc^{-3} yr^{-1}$. They write:

For BH–BH inspirals, horizon distances of ...2187 Mpc are assumed. These distances correspond to a choice of ... 10M for BH mass.

In the left panel of figure 4 found in this 2016 paper by LIGO there is a plot of BH mass vs horizon distance. For 2015-2016 sensitivity levels and 10M mass, the horizon distance is 300 Mpc. So we need to multiply the above numbers by $300^3=2.7\times10^{7}$. This gives a range of $2.7\times10^{-3}$ to $8.1$ detectable mergers per year, given the prior assumptions that a typical signal would come from black holes with ~10 solar masses. Also, in the discovery paper we find:

We present the analysis of 16 days of coincident observations between the two LIGO detectors from September 12 to October 20, 2015.

Therefore the duration of the experiment was $16/365$ years. This gives an expected rate ranging from $1.18\times10^{-4}$ to $3.55\times10^{-1}$ mergers during the experiment. As far as I can tell, this can be taken as an upper bound, since not all of these events will really be detected (due to non-optimal location, coincident noise, etc).

Are there other factors to take into account?

• This calculation does not appear to take into account the directional nature of GW wavefronts? It calculates how many and how big, but doesn't appear to make provision from the planar nature of the radiation (is that right - that the radiation is first order contained within the plane of the orbits of the black holes?) – GreenAsJade Feb 20 '16 at 1:44
• @GreenAsJade I think that is an interesting question but there is no mention of that factor in anything I have read. So I assume that the waves are expected to propagate with equal amplitude in all three spatial dimensions. – Livid Feb 20 '16 at 1:58
• @Livid I think it's more than an interesting question: I think it's a key consideration. I think so, because I can't see how quadropole polaried radiation can propagate spherically. Is there something I'm missing there: can this actually happen? Because if it can't then this must be a fundamental component of computing detection probabilities. – GreenAsJade Feb 20 '16 at 2:37
• @GreenAsJade: Just for the record, I do agree with you that a naive linear field theory will, at least intuitively, predict that there be some zeros in the radiation pattern, so my initial guess would have been that we are losing some fraction of events that fall below the detectability threshold, however, GR is not linear at the source. I don't think I would take "information targeted at amateurs" any seriously about any level of detail of any physics experiment. There is an awful lot of really poor science reporting out there, even from people who should know better but just don't care. – CuriousOne Feb 20 '16 at 4:40
• The paper referred to clearly states (in the appendix) that they integrate over the range of possible geometries/orientations to get the effectively sampled volume for BH mergers. – Rob Jeffries Feb 20 '16 at 14:02

I think there are some problems.

The figure 4 you refer to has total system mass on the lower x-axis. So the horizon distance for the merger of two 10 solar mass BHs appears to be 800 Mpc. So that increases your naive rate by a factor of 20.

In the right hand plot of the same Fig 4, you see that the authors say that the effective volume for a merger of two 10 solar mass black holes is 0.1 Gpc$^3$.

So we can compare this with the naive volume of $(4\pi/3)\times 0.8^3 = 2.14$ Gpc$^3$.

So they are saying they detect about 20 times fewer mergers than your way of calculating it would suggest, ironically cancelling out with the earlier error, and so suggesting that your headline figures are roughly correct.

Two interesting points arise. First, the naive estimate of the volume in which a detection can be made is way too large. The reason for this is explained in the appendix of the paper. The left hand panel curves are for an equal mass binary, overhead with a face-on orbit. All other geometries will result in a weaker signal by factors of a few. As signal amplitude scales as 1/distance this means a shorter horizon distance by factors of a few. The volume then goes as the cube of this. As the appendix explains, the curves in the right hand plot were produced by integrating over random directions and orientations.

Second, the headline figure in your calculation turns out to be right (because of two cancelling errors), and suggests that the detection of a merger of 10 solar mass BHs was unlikely in 16 days of data. However, the mass dependence is large - GW amplitude goes as $M^2$. The detected source had a total mass of 60 solar masses and could have been seen (RH panel of Fig. 4) in an effective volume that was 20 times bigger than for a 20 solar mass system.

• Thanks. This seems usable as a first approximation. I don't like the idea of considering larger sources without also knowing the distribution. I expect that >60 solar mass sources would have a rate much less than 10 solar masses. Perhaps the effects due to increased horizon distance and decreased rate will approximately cancel? – Livid Feb 21 '16 at 2:51
• Also, in that same appendix: "The actual sensitivity will depend on the exact network configuration, the data quality, and the signal parameters, so the curves in Figure 4 should be viewed only as approximations." I interpret this to mean these estimations are on the high side, does that seem correct? – Livid Feb 21 '16 at 2:53
• As for a guess at the rate, you might say that $n(m) \propto m^{-2.3}$ (Salpeter mass function), which suggests GW detections from big BHs will be more common. – Rob Jeffries Feb 21 '16 at 8:41
• I'm not sure I follow. The total mass ratio is $60M/20M=3$. So we would expect the event rate to to be $3^{-2.3}=0.08$ times lower. From that fig 4 the ratio of horizon distances is $1.15/0.8=1.4375$. Cubing to get the volume ratio gives $1.4375^{3}=2.97$. Together we get $0.08*2.97=0.2376$. The rate of detectable signals for a 60M system would be about 1/4 that of a 20M system. – Livid Feb 21 '16 at 20:29
• Yes, my mistake. Don't know what I was thinking with that 1.15 horizon distance. Thanks again. – Livid Feb 21 '16 at 21:35