We've all heard the news about the detection by gravitational waves of two black holes, one 29 solar masses and the other 36 solar masses, spiraling into each other to create a single black hole of 62 solar masses.

For me, the loss of three solar masses into the gravitational waves over a fraction of a second is the most amazing aspect of this. I can understand how the waves were created (and mass changed into broadcast energy) by the spiraling motion of the black holes as they did their final do-si-do, and it's clear that losing that energy was a necessary part of their joining; otherwise they would have orbited forever.

However, does the amount of mass lost depend on just how the two holes merged? For instance, if they had rammed each other head-on, would they still have somehow lost that amount of energy? I expect that less energy would have been diverted into the resultant black hole's spinning, but I don't know how that would have changed the resulting mass.


2 Answers 2


Yes. The amount of energy radiated as gravitational waves will depend on the details of the two black holes before the merger. The answers to questions like:

  • Was the orbit circular or elliptical?
  • Were they spinning?
  • Were the spins of the black holes aligned with the orbital plane?

will affect the energy of the gravitational waves. The most import detail in terms of energy radiated is the mass ratio of the two precursor black holes.

The final mass of the system $M_\mathrm{fin}$ is always less than the original total mass ($m_1 + m_2$), since some of the mass energy gets converted to gravitational waves, $M_\mathrm{rad}$.

$$M_\mathrm{fin} + M_\mathrm{rad} = m_1 + m_2 $$

Because of something akin to the second law of thermodynamics the final black hole must be bigger than the biggest original black hole. Basically, you can't radiate so much energy in gravitational waves that a black hole shrinks.

$$M_\mathrm{fin} > m_1 \quad \mathrm{and} \quad M_\mathrm{fin} > m_2$$

We can define the fraction of mass radiated aways as:

$$ e = \frac{M_\mathrm{rad}}{m_1 + m_2} $$

This is sometimes called the efficiency of radiation.

If the black holes have about the same mass (as they did in the LIGO detection), about 5% of the total mass will be radiated away. This is the most efficient possibility.

On the other extreme imagine the case where one black hole is way more massive than the other: maybe 1 million solar masses and 1 solar mass. In order to follow the two rules stated above, $M_\mathrm{fin}$ is less than 1 million and one solar masses and greater than 1 million solar masses. In this case the efficiency would be about $e=10^{-6}$ or 0.0001%. Extreme mass ratios produce the weakest gravitational waves.

  • 2
    $\begingroup$ +1, but a minor quibble. You seem to be aware of this effect, but it's not made clear in your answer. The most efficient possibility is apparently when the black holes have high spins aligned with the orbital angular momentum. In this case, you can get more than 11% of the total mass radiated away. See, e.g., this paper for a simulation that radiates 11.3%. $\endgroup$
    – Mike
    Feb 22, 2016 at 18:29

In short, yes. Lubos Motl wrote a very nice article on his blog explaining all one needs to know about the gravitational waves. I want to focus on one of the parts he mentions - "quasinormal modes". Quasinormal modes explain how the black hole settles down to a final, stable configuration and it very much depends on the black hole in states i.e. momentum, mass etc. These QNM's are a description of something called the "ringdown" which we all saw in the data i.e. the ringdown to the Kerr solution. During ringdown, there are still gravitational waves being emitted which are detectable by LIGO. So if the ringdown mechanism is different, the certainly the gravitational waves produced are different.

I'm sorry that I cannot give you more details of calculations since this is very numerical and difficult and I don't have them at hand.


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