This question is from Physics overflow: question in physicsoverflow.

I am reading LIGO's new discovery of gravitational waves by black hole merger. During the merger, two phases are not hard to manipulate by hand or on PC, the in-spiral and ring-down phases.

During the violent merging before ring-down, the binary contacts each other closely and excite gravitational perturbations around the final Kerr space-time. If I understand correctly, this violent excitation includes frequencies over a large range, possibly well beyond the hundreds Hertz seen by LIGO. They only observed that low frequency in ring-down, only because this mode damps the slowest.

Now comes the question: How much in fraction (or amplitude) are the modes excited during merging? If some high frequency mode get excited with a very large amplitude, shouldn't they also show up during the early ring-down in data, even though they damps faster?

Thank you very much!

  • $\begingroup$ I expect your question could be answered by an expert (not me) analysis of the total amount of gravitational energy radiated. $\endgroup$ – Lewis Miller May 3 '16 at 15:01
  • $\begingroup$ If you are really interested you can freely access the event data here: losc.ligo.org/about and calculate a PSD. $\endgroup$ – docscience May 3 '16 at 16:14

After a binary merger the resulting black hole will experience a ringdown. The ringdown is characterized by a spectrum of quasi-normal modes (QNM) which, according to GR (specifically black hole perturbation theory), depend only on the final black hole mass, angular momentum, and charge. This is effectively the no-hair theorem.

QNMs are oscillations that decay. Regular old normal modes don't decay.

Which QNMs are excited depends on the details of how the black hole is perturbed. This is similar to QNMs in a musical instrument. The timbre you hear when plucking a guitar string depends on where you physically pluck the string. Plucking in the middle will result in a bassy tone, while plucking at the edge a trebly tone. In all cases the dominate mode is the fundamental for that string.

Unlike musical overtones which are characterized by a single index. The black hole QNMs are related to spherical harmonics and are characterized by two indices $\ell$ and $m$. The fundamental mode is the quadrupole $\ell=m=2$.

The excitation spectrum of BH ringdown modes depends on the spins and masses of the two objects. Due to the non-linear nature of the GR two body problem, there is no simple method to predict how much one mode will be excited in general. There are several approximations that work in special cases. A very simple model assumes non-spinning initial black holes so the excitation spectrum is a function of the mass ratio of the black holes, $q$.

$$ q = \frac{M_a}{M_b}\ge 1, \,\,\, i.e. \,\, M_a \ge M_b $$

For equal mass mergers, $q=1$, almost all of the excitation is in the fundamental mode. As the mass ratio increases there is more and more excitations in the higher order modes. Although, the total excitation energy decreases. A large mass ratio means a little black hole is falling into a big one, so it doesn't perturb it much.

The two LIGO detections GW150914 and GW151226 are both in the low mass ratio regime. So very little energy is radiated at higher order modes.

Kamaretsos, et al followed the non-spinning method to determine the relative amplitude between modes for different mass ratios using numerical simulations. For instance, they found that for $q=2$ the fundamental (2,2) mode has about 73% of the total energy. The next most excited (3,3) had about 12%. Then comes (2,1) with about 9% and (4,4) with about 2%...

The first LIGO detection, GW150914, was very close to $q=1$. The more recent LIGO detection, GW151226, is closer to $q=2$. In either case no one expected to detect higher order modes of the ringdown.

This has been an active area of research for a while now. If you are interested in digging deeper, you could check out:

  • $\begingroup$ Thanks a lot for this comprehensive answer! I really appreciate it. $\endgroup$ – Ruifeng Dong Jun 28 '16 at 17:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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