Since surface area of the remnant black hole must be more that the sum of the binary surface then the maximum energy released via gravitational waves is

$$ ΔE = [M_1 + M_2 – \sqrt{M_1^2 + M_2^2}]c^2 $$

That means that when ratio of the initial masses is very large then almost all mass of smaller black hole may turn into gravity waves.

I went through "SXS Gravitational Waveform Database" - it has 8 simulations of mergers with ratio above 9. For all of them less then 1% of total mass is converted into gravitational waves while typical LIGO merge releases several percents. This implies that in typical super massive BH and stellar mass BH merge there is almost no emission.

How does typical (most likely in the wild) and extreme (with maximum loss) case of high mass ratio merge looks like?


2 Answers 2


If I understand "Gravitational wave snapshots of generic extreme mass ratio inspirals" (thanks to @andrew for the tip) correctly then typically almost all smaller mass body is converted in waves.

EMRIs simulations table

, where μ is the mass of the smaller body and E is wave energy.

But for head-on collision (plunging orbit) conversion ratio will be very small (see "Gravitational radiation from plunging orbits").

  • 1
    $\begingroup$ The $E$ in that table is the energy of the orbit, not the energy of the emitted gravitational waves. $\endgroup$
    – TimRias
    Jan 5 at 0:13

To lowest order approximation the dynamics of a binary with a large mass-ratio can be approximated as the smaller object following a geodesic in the metric generated by the larger object, and the system slowly (adiabatically) evolving due to the emission of energy (and angular momentum) in the form of gravitational waves.

This adiabatic "inspiral" process continues until the system reaches the last stable orbit (LSO) after which it transitions to a plunging geodesic.

The total energy converted to gravitational waves during the inspiral phase is simply the difference between the initial energy of the system and the energy of the LSO. The lowest possible energy LSO is a circular orbit, the innermost stable circular orbit or ISCO. For a binary without spin the energy of the ISCO is

$$E_{\rm isco} = \frac{2\sqrt{2}}{3} m_2 c^2 \approx 0.943 m_2 c^2,$$

where $m_2$ is the mass or the lighter (secondary) object. During the inspiral phase a non-spinning binary can convert at most 5.7% of the rest mass of the lighter object to energy. For spinning binaries (which have ISCOs with much lower energy) this can increase substantially to 42.3% (for a maximally spinning primary).

The plunge phase is relatively short for large marge ratio binaries, leaving little opportunity to convert energy into gravitational waves. The energy lost in this phase is proportional to $ \frac{m_2^2}{m_1}c^2$. If $m_1 \gg m_2$ this is always much smaller than the energy convert in the inspiral phase.

In fact, the next largest contribution to the energy convert to gravitational waves comes from the "transition" phase as the system transitions from the inspiral to the plunging phase in which an amount of energy portional to $m_2 \left(\frac{m_2}{m_1}\right)^{4/5}$ is converted to gravitational waves see this paper by Ori&Thorne, which is still much smaller than the energy lost during the inspiral.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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