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Today's Phys. Rev Letter GW170104: Observation of a 50-Solar-Mass Binary Black Hole Coalescence at Redshift 0.2 shows the analysis of the third confirmed detection of a black hole merger by the LIGO collaboration. According to the paper this even may differ from the previous two:

The black hole spins are best constrained through measurement of the effective inspiral spin parameter, a mass-weighted combination of the spin components perpendicular to the orbital plane, $\chi_{eff} = -0.12^{+0.21}_{-0.30}$. This result implies that spin configurations with both component spins positively aligned with the orbital angular momentum are disfavored.

You can see a simulation in this LIGO/Caltech video (which I found here), showing what I believe is precession of the tilted spin axis of one of the objects. https://youtu.be/S2vp7iVWrkE

If I understand correctly, while there is significant pre-processing in frequency space, the final data for each site is really only the one-dimensional strain versus time plot, and all extracted parameters from the event come from the maximum-likelihood fitting of theoretically generated strain waveforms to these two plots for the two LIGO sites. See the several good answers to How were the solar masses and distance of the GW150914 merger event calculated from the signal? for further information.

A great deal of work has gone in to the statistical analysis of these events, both for low latency detection (alerts to the community for possible coincident optical detection) and more careful analysis later. The fact that $\chi_{eff}$ is so many error bars away from unity suggest there is a good chance the spins were not aligned to the orbit. In Section IV. at the bottom of the first column on page 221101-4 it says for example:

Considering the two spins together, the probability that both tilt angles are less than 90° is 0.05

Question (Edited): How does the alignment of the spins manifest itself in the waveform shown in the LIGO data? Is it a subtle shift in phase near the end, or a modulation of amplitude?

Are there any examples of a simulated waveform for very different values of $\chi_{eff}$ that show the difference?

enter image description here

above: Enlarged, cropped section of Figure 1, shown below.

enter image description here

enter image description here

above: Figure 1.

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    $\begingroup$ I don't know the answer to this, but I suspect it may be some hairy best-fit thing: they've run simulations of all sorts of mergers and they then match the observed waveform against the simulated ones (perhaps in some nice distance-in-vector-space way) and find the nearest. If that's right then the only way to answer the question would be to look at a lot of simulated event waveforms: I guess those are available. $\endgroup$
    – user107153
    Commented Jun 3, 2017 at 14:09
  • $\begingroup$ @tfb ok, you could call what I call "...maximum-likelihood fitting of theoretically generated strain waveforms to these two plots for the two LIGO sites." a "best fitting thing" if you like. I'd also assume there are sensitivity analyses for the parameters, and "show-me" plots of fits with and without parallel spins, but I don't know where to find these. $\endgroup$
    – uhoh
    Commented Jun 3, 2017 at 14:14
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    $\begingroup$ Spin-orbital coupling shows up already at 1,5 PN order and it essentially allows you to decrease or increase the orbital period at the given orbital separation, and thus the frequency-amplitude relation for the GW emission. But using this to measure the spin parameter is a complex issue because it assumes you know the distance of the binary and binary masses. Perhaps someone more qualified can give you a more precise answer, but I am afraid the answer is really more or less "you have to fit the whole wave-form". $\endgroup$
    – Void
    Commented Jun 3, 2017 at 14:43
  • $\begingroup$ @Void Thanks! That simple explanation is excatly the kind of answer I'm looking for - how it would "show up" in the data, or what it would look like. Ideally a plot showing two fits, one with $\chi_{eff}$ constrained to all-parallel spins/orbits to one with the constraint relaxed (the current fit) would be great - I'm really interested to see just how different the would be from each other. I know the error bars on $\chi_{eff}$ are fairly tight, but sometimes seeing a plot of two alternate fits on top of the data is really helpful. Please consider posting an answer, even sans plot. $\endgroup$
    – uhoh
    Commented Jun 3, 2017 at 15:19
  • $\begingroup$ @uhoh My point was that it may be that the thing simply doesn't show up in any obvious single parameter of the waveform as you seem to be looking for. $\endgroup$
    – user107153
    Commented Jun 3, 2017 at 15:34

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$\chi_{eff} = (m_1S_{1z} + m_2S_{2z})/(m_1+m_2)$ is a measure of how the z-component of the spins is aligned with the orbital angular momentum. The term which affects the evolution of binary after quadruple (Newtonian) term depends on a combination of $\chi_{eff}$ and mass ratio. Newtonian term depends on the chirp mass of the binary. So you can measure chirp mass fairly accurately but you can only measure a combination of $\chi_{eff}$ and mass ratio. Coming on to how aligned projection of spin effects the evolution of binary here is a nice plot showing it. If both the spins are aligned with the angular momentum then binary takes longer to merge and vice-versa if they are anti-aligned. The business with precession is very different. If the spins are not aligned with the angular momentum then the waveform modulates. Off-course it depends on from where are you looking at the binary. If you are looking from the edge the modulation is most. Something like this. Currently LIGO searches are not including precession when they search for gravitational waves. A face-on binary will show no precession so no one is caring for it right now (off-course this is purely search sensitivity based argument - an edge-on binary is seen to distances much smaller than face-on binary) but observation of prescessing binaries will be great.

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  • $\begingroup$ Thank you, this is exactly the kind of answer I was hoping for! I appreciate you taking the time to explain that the antiparallel and perpendicular cases would evolve very differently. It's up to you but I think your answer will be more robust long term if you include one or two images here, as a precaution against inevitable link rot. $\endgroup$
    – uhoh
    Commented Jun 7, 2017 at 10:44

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