Concerning the recent detection of gravitational waves produced by colliding black holes, it has been reported that a significant percentage of the combined mass was lost in the resulting production of the gravitational waves.

So evidently in addition to Hawking radiation, black holes can also lose mass in collisions with other black holes.

Is there a theoretical limit to how much mass, as a percentage, two black holes can lose in a collision as gravitational waves? Could so much mass be lost that the resulting object would no longer have enough gravity to be a black hole?

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    $\begingroup$ The total area of the event horizons always increases. So this should give a bound. For a non-spinning black hole, the area of the event horizon is proportional to the square of the mass, but of course, black holes can acquire or lose spin when they collide, so the actual bound this gives may take a little work. $\endgroup$ Commented Jul 18, 2016 at 19:50
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    $\begingroup$ This great question yields two excellent, simple and clear answers. Many thanks for asking it: this thread is one of those gems that make PhysicsSE so worthwhile $\endgroup$ Commented Jul 19, 2016 at 1:14

3 Answers 3


Suppose you have two black holes of the same mass $M$ and $m = GM/c^2$. The radius of each black hole is then $r = 2m$, and the horizon area is $A = 4\pi r^2$ $ = 16\pi m^2$. Two constraints are imposed. The first is that the type-D solutions have timelike Killing vectors, which are isometries that conserve mass-energy, and with the merger the gravitational radiation is in an asymptotically flat region where we can again localize mass-energy. So the initial mass $2M$ is the total energy. The entropy of the two black holes is a measure of the information they contain and that too is constant. So the horizon area of the resulting black hole is the sum of the two horizon areas, $A_f = 2A$ $ = 32\pi m^2$, that has $\sqrt{2}M$ the mass of the two initial black holes. Now with mass-energy conservation $$ E_t = 2M = \sqrt{2}M + E_{g-wave} $$ and the mass-energy of the gravitational radiation is $.59M$. That is a lot of mass-energy!

This is the upper bound for the generation of gravitational radiation from mass. The assumption here is that the total entropy of the two black holes equals the entropy of the final black hole. Physically this happens if all the curvature exterior to the merging black holes does not result in mass-energy falling into the final black hole. There would be back scatter of gravitational radiation, much as one has to be concerned about the near field EM wave near an antenna that can couple back on it. The final entropy of the merged black hole will in fact be larger, but of course not larger than the mass-squared determined area of the two black holes. This means $1.41m~\le~m_{tot}~\le~2m$.

To estimate this requires numerical methods. Larry Smarr pioneered a lot of this. So far estimates run around $5\%$ of the total mass of the black holes is converted to gravity waves. in this LIGO paper two black holes of mass $39M_{sol}$ and $32M_{sol}$ is computed to have coalesced into a final black hole of $68M_{sol}$, which radiated $3M_{sol}$ is gravitational radiation and accounts for $4.2\%$ of the initial mass. This is about in line with most numerical studies. Consequently a lot of the spacetime curvature generated by these mergers falls back into the final black hole. In terms of area the initial horizon area is $4066M^2_{sol}$ and the final horizon area is $4624M^2_{sol}$, which is an additional area $558M^2_{sol}$ of horizon area with $S~=~k A/4L_p^2$ as the entropy.

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    $\begingroup$ But isn't $A=16 \pi m^2$ only valid for Schwartzchild black holes, and not for Kerr black holes, which have angular momentum? $\endgroup$ Commented Jul 18, 2016 at 23:28
  • $\begingroup$ Admittedly this is Schwarzschild. To work this with the Kerr metric would have required a much longer entry here. $\endgroup$ Commented Jul 19, 2016 at 11:48
  • $\begingroup$ Currently featured: Under what circumstances might gravitational waves impart linear momentum to an object? (e.g. Quasar 3C 186); still hoping for a clearer answer on momentum transfer. $\endgroup$
    – uhoh
    Commented Mar 28, 2017 at 15:11
  • $\begingroup$ How is the radius of each black hole twice the Schwarzschild radius? That's a blatant contradiction. $\endgroup$
    – Melab
    Commented Jul 1 at 5:07
  • $\begingroup$ I did not say that. $\endgroup$ Commented Jul 2 at 17:48

The total surface area of the event horizons never decreases.

We won't consider charged black holes, since in real life, black holes never have very large charge. However, they can have very large angular momentum, as LIGO showed.

The horizon area of a rotating and uncharged black hole (the Kerr metric) is $$ 8 \pi M \left(M+ \sqrt{M^2 - a^2}\right),$$ where $a = J/M$, $J$ being the angular momentum.

So for an uncharged black hole of mass $M$, the area of the event horizon is somewhere between $8\pi M^2$ and $16\pi M^2$, where the angular momentum is respectively $M^2$ (the maximum $J$) and $0$ for these two cases.

Thus, you might be able to collide two extremal rotating black holes of mass $M$, and obtain a non-rotating hole of mass $M$, meaning that you lose half of the total mass in the collision. This scenario is where the largest percentage of combined mass is lost, because $$ \frac{\sqrt{x+y}}{\sqrt{2x} + \sqrt{2y}} $$ is minimized when $x = y$. I don't know whether such a collision is possible.

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    $\begingroup$ The non rotating limit was first published by Hawking in 1971, in Phys Rev Letters, in journals.aps.org/prl/export/10.1103/… $\endgroup$
    – Bob Bee
    Commented Jul 19, 2016 at 5:00
  • $\begingroup$ @BobBee: if I'm reading it correctly, the proof in Hawking's 1971 paper doesn't actually require that the black holes are non-rotating. (But he does specifically apply his result to the case where the initial black holes are non-rotating, which is perhaps what you meant.) $\endgroup$ Commented Jul 19, 2016 at 14:02
  • $\begingroup$ @MichaelSeifert In 1971 the abstract read "It is shown that there is an upper bound to the energy of the gravitational radiation emitted when one collapsed object captures another. In the case of two objects with equal masses m and zero intrinsic angular momenta, this upper bound is (2−2 √ )m". That is approx. 29% max. He certainly also got the right approach generally. In the Houches Lectures he used the equations for the Kerr Newman horizons and got 65% (approx.) max, and a little less than 50% for rotating uncharged. I was not sure if that had specifically been published before. $\endgroup$
    – Bob Bee
    Commented Jul 20, 2016 at 0:16
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    $\begingroup$ @BobBee: Although the abstract doesn't mention it, Hawking's 1971 paper treats the case of Kerr solutions as well (though admittedly not Kerr-Newman). Eq. (2) of that paper is a statement of non-decrease of horizon area applied to Kerr black holes; he also notes that "the highest limit on $\epsilon$ [the efficiency] is $\frac{1}{2}$ which occurs if $m_1 = m_2 = a_1 = a_2$, $a_3 = 0$" (where $a_3$ is the spin of the final black hole.) This is exactly the limit described in the above answer. $\endgroup$ Commented Jul 20, 2016 at 14:11
  • $\begingroup$ Ok, I thought the 50% first time was Les Houches. Thanks for correcting it. I'm editing the answer. $\endgroup$
    – Bob Bee
    Commented Jul 20, 2016 at 22:33

See also a separate related answer by Patrick Gupta, second answer to Did merging Black Holes in GW150914 give up entropy and information to the gravitational waves, since they lost 3 solar masses? (the question was faulty). He calculated the final horizon area for the rotating case, with entropy 1.57 times the original entropy, so entropy did grow, and the second law of BH thermodynamics held up. Using the Kerr solution and equations for the horizon area, as Peter Shor did, is important to do because unless it is a heads on collision (very unlikely) and there was no individual rotations to begin with, the final black hole is very likely to have a significant angular momentum. It is interesting, and should not be surprising, that the observed merger led to the high angular momentum (a=.67) observed in the final black hole.

Worth noting that Hawking derived first the limits for both rotating (edited, as indicated correctly by Michael Seifert in a comment below) and non rotating bodies in 1971 Phys Rev Let, and published more in 1972 for both rotating, non-rotating, and charged bodies (though he may not have been the first on the latter) in the Les Houches Summer School Lectures on Black Hole in 1972. For rotating Kerr black holes the max is just 50%, and for rotating and charged Kerr Newman black holes it's approx. 65% max.

The many numerical and PN calculations done (by Smarr and others), as stated by Lawrence Crowell in his nice answer, over the years and before the LIGO finding eventually led to an understanding that 5% or so was a more likely number in many cases (not sure if those include charged, those are not likely to occur astrophysically).

It's worth noting also that for LIGO they could not get a measurement/estimate of the initial BH rotations, if any, and estimated it would have made only a small difference (smaller than the mass uncertainties estimates) on the final estimate for the radiated grav energy. In later observations they expect to see more earlier and maybe get any pre merger rotation rates.


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