I've read a lot of conflicting answers in these forums. However, today saw the awesome announcement of gravitational waves. Two black holes merged: http://www.slate.com/blogs/bad_astronomy/2016/02/11/gravitational_waves_finally_detected_at_ligo.html

Not only that, they merged FAST. In 1/5th of a second revolving around each other 250 times a second. The entire event was quicker than a heartbeat. Moreover, we observed this happening as distant outsiders. So now we can say for sure:

  1. Objects approaching the event horizon DO NOT appear to slow down
  2. Black holes CAN merge in a finite (and quick) amount of time
  3. And all this is wrt a frame of reference far, far away

To quote the NYTimes article:

One of them was 36 times as massive as the sun, the other 29. As they approached the end, at half the speed of light, they were circling each other 250 times a second.

And then the ringing stopped as the two holes coalesced into a single black hole, a trapdoor in space with the equivalent mass of 62 suns. All in a fifth of a second, Earth time.

However, everything I've read so far has let me to believe that an outside observer should never be able to measure the collision happening in a finite time. So what exactly is happening here? I must have read at lest 5 different versions of this so far everywhere in these forums over the past several years.

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    $\begingroup$ At the end the signal becomes weaker very fast. It only asymptotically tends to zero, but LIGO has a finite sensitivity, after a second or so the signal drops below the noise floor. $\endgroup$ Feb 11, 2016 at 18:15
  • $\begingroup$ @CountIblis So this means we can't really make the statement "The black holes have merged" yet? From our perspective they're still circling around each other? I guess it would also mean that when they say "...as the two holes coalesced into a single black hole", it's not accurate - is that right? $\endgroup$ Feb 11, 2016 at 18:19
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    $\begingroup$ Yes, but as John explains in his answer, it is still "effectively true". You can compare this with many other things, like saying that your position is such and such and you move at such and such velocity, when strictly speaking you can't have a definite speed and velocity at the same time. Or the fact that phase transition cannot happen with a finite amount of substance (if you have a finite number of molecules, then the thermodynamic functions will be analytic, they won't have discontinuous derivatives). $\endgroup$ Feb 11, 2016 at 18:40

3 Answers 3


This presumably stems from the fact that in the coordinate system of an external observer nothing can ever cross the event horizon of a black hole.

This is perfectly true, but if you were watching an object fall onto a stellar mass black hole it would red shift to invisibility in a few microseconds and it would look to you just like it crossed the horizon. More precisely, no matter how sensitive your measuring equipment there would be a time after which you could no longer detect that the object had not crossed the horizon, and for any physically reasonable equipment this time is extremely short.

The same principle applies to the merging black holes. We have two objects that can't actually be real black holes because in any finite universe we know real black holes cannot exist. However they are experimentally indistinguishable from real black holes. As these two objects approach each other the spacetime geometry changes and approaches that of a single rotating black hole - the Kerr metric. We know the geometry can never actually become Kerr because that would take an infinite time. However the geometry approaches the Kerr geometry so quickly that after a fifth of a second it is experimentally indistinguishable from the Kerr geometry.

Whether the black holes have merged or not depends on exactly what you mean by merged. They are certainly no longer two separate objects, and that happens in a short time and is observable. In this sense it seems reasonable to me to describe tham as having merged. If you insist the merger is complete only when the transition to the Kerr geometry is complete then this will take an infinite time so they will never merge.

tl;dr - in any sensible meaning of the term merge the two black holes do indeed merge in a finite, and very short, time.

  • $\begingroup$ Thanks for that! I think it almost makes sense to me now :) $\endgroup$ Feb 11, 2016 at 18:37
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    $\begingroup$ @BhagwadJalPark If it helps, when we do the calculations with a diffuse clump of matter rather than a point of matter, we see that the far-away viewer see that as the clump approaches the black hole, the matter inside the clump causes the event horizon to expand, which firmly "eats" the matter. In this way the surface of a black hole is not just a redshifted image of everything that it has absorbed, but stuff does indeed "fall in". $\endgroup$
    – CR Drost
    Feb 11, 2016 at 18:48
  • $\begingroup$ Have your upvote. May the HNQ be with you :) $\endgroup$
    – user10851
    Feb 11, 2016 at 18:50
  • $\begingroup$ But are the singularities merging? Or will they orbit each other in the black hole forever? $\endgroup$
    – Jossi
    Feb 14, 2016 at 17:20
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    $\begingroup$ @Jossi there are no singularities because as this answer says, black hole do not exit. Re-read. $\endgroup$
    – Anixx
    Nov 9, 2019 at 22:57

Just like photons don't age but still move, black hole horizons don't age but move and can thus merge with other black holes.

The formation process of a black hole takes an infinite amount of time according to external observers, but that doesn't mean it needs to remain in its initial position. The frozen horizon moves through space and emits gravitational radiation to communicate its new position to its environment. Nothing besides Hawking radiation escapes the inside of a black hole, including gravitons.


The rotation of an object around the black hole does not slow down (in terms of frequency). It is not affected by time dilation. The object gets sticked on the surface of black hole, even accelerates its rotation.


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