Estimating the final spin of a binary black hole coalescence

We present a straightforward approach for estimating the final black hole spin of a binary black hole coalescence with arbitrary initial masses and spins. Making some simple assumptions, we estimate the final angular momentum to be the sum of the individual spins plus the orbital angular momentum of a test particle orbiting at the last stable orbit around a Kerr black hole with a spin parameter of the final black hole. The formula we obtain is able to reproduce with reasonable accuracy the results from available numerical simulations, but, more importantly, it can be used to investigate what configurations might give rise to interesting dynamics. In particular, we discuss scenarios which might give rise to a flip in the direction of the total angular momentum of the system. By studying the dependence of the final spin upon the mass ratio and initial spins, we find that our simple approach suggests that it is not possible to spin-up a black hole to extremal values through merger scenarios irrespective of the mass ratio of the objects involved.

Figure 1

The final spin $a_f/M$ vs $\nu =m_1{m}_2/M^2$ for initial black holes with extreme spin parameters $\chi =1$. Clearly, as the mass ratio approaches $\nu =1/4$ (equal-mass case), the final expected spin parameter decreases.

Figure 2

Final spin $a_f/M$ vs $\nu$ for initial black holes for several spin parameters $\chi =0$, 0.2, 0.4, 0.6, 0.8, 0.85, 0.9, 0.95, 1. For spins $\gtrsim 0.948$ the maximum final spin is achieved as the extreme mass-ratio case is approached while for initial spins $\lesssim 0.948$ the equal-mass case would give rise to the maximum final spin.

Figure 3

Final spin $a_f/M$ vs $\nu$ for initial black holes with initially nonspinning black holes ($\chi =0$) as predicted by our simple model (filled squares), and as obtained numerically in Ref. 8 (red diamonds), by the EOB model combined with nonspinning test-particle predictions in Ref. 6 (green circle) and by numerical relativity combined with nonspinning test-particle expectations in Ref. 9 (blue triangle). The largest final black hole spin is obtained in the equal-mass case.

Figure 4

Final spin $a_f/M$ vs $\chi$ for equal-mass black holes as predicted by our expression (filled squares), numerically obtained (filled diamonds) in Ref. 11 and from the EOB model (filled triangle) used in Ref. 5, but without taking into account the angular momentum released during merger-ringdown (thus, the latter curve overestimates the results and should be considered as an upper limit). Irrespective of the alignment or antialignment of the individual spins, the final black hole spin is aligned with the initial orbital angular momentum.

Figure 5

Relation between the mass ratio $\nu$ and the individual spins of the black holes to achieve a final nonspinning black hole. Only mass ratios $\nu \lesssim 0.183$ ($q\gtrsim 3.15$) will allow for the final spin to be antialigned with the initial orbital angular momentum direction if the individual black holes spin fast enough.

Figure 6

Final spin $a_f/M$ vs $\alpha$ for equal-mass black holes with ${\chi }_1=0.584$ and ${\chi }_2\in [-0.584,0.584]$ as predicted by our simple model (filled squares) and as obtained numerically in Ref. 10 (red diamonds).

Figure 7

Final spin $a_f/M$ vs $\alpha$ for equal-mass black holes with ${\chi }_1=-0.584$ and ${\chi }_2\in [-0.584,0.584]$ as predicted by our simple model (filled squares) and as obtained numerically in Ref. 10 (red diamonds).

Figure 8

At ISCO, the total angular momentum will be the sum of the orbital angular momentum and the spin vectors. The angle ${\theta }_{\mathrm{LS}}$ is assumed to be given and remain fixed during the inspiral prior to ISCO. The inclination angle $\iota$ is solved for along with the magnitude of the total angular momentum.