Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If time stops at the event horizon, can we ever detect two black holes merging? In other words, if you are a short distance away, would you encounter a spherically symmetric gravitational field, or a dipole field?

share|cite|improve this question
Time measured by observer wrt himself doesn't stop at event horizon. – adobriyan Jul 4 '11 at 12:04
Dipole fields are impossible with gravity. Quadrupole is the "lowest" possoble. – Georg Nov 25 '11 at 11:42 (archived version if link doesn't work)

The thing to remember whenever dealing with spacetime weirdness surrounding extremely dense objects is that what happens, that is, what the objects in question actually experience, can be completely different from what appears to happen to a distant observer.

From the point of view of the singularities at the core of each black hole, they "swirl down the drain," no problem. That is, each one spirals in faster and faster, until they merge. Things do get a little weird once the smaller singularity is inside the event horizon of the larger one, but any small object falling into a big black hole will experience the same thing. For one thing, because of the extreme curvature of spacetime (so curved it is, in fact, closed- that's kind of the point of a black hole) every direction that the smaller black hole observes is towards the other black hole. That's one way to view the inescapability of a black hole.

Now, from the point of view of a distant observer, as the event horizons of the black holes get closer and closer, things do seem to get really weird, but that's only because time as experienced by objects in the immediate neighborhood of the merger gets more and more stretched out compared to time as experienced by the distant observer. I can't stress enough that the weirdness ONLY arises when comparing two different points of view. Neither the distant observer nor the suicidal one going along for the ride with the black holes (Call him Dr. Strangelove) experiences anything other than seemingly normal time*.

According to the distant observer, the spiraling black holes will drag spacetime around with them. You can imagine a grid of lines drawn on a sheet of rubber or taffy. Because spacetime has to be "smooth" in the calculus sense, the non-constant motion of the black holes around each must be transmitted outwards, sort of like waves on your rubber grid. This is called gravitational radiation.

Things will appear strange to the distant observer as the event horizons approach each other. Since Dr. Strangelove's time appears to stretch all the way out to infinity, it is tempting to imagine that the distant observer will see a normal image of him, just waving in slow motion like the bullet time special effects from the Matrix series of movies. That misses an important part! Since all aspects of Dr. Strangelove's time are seeming to slow down relative to us, the photons he emits as he waves backwards get farther and farther apart, and also get stretched out so that initially blue photons become green, then yellow, orange, red, infrared, radio, etc. So as Dr. Strangelove approaches the black hole, the most pronounced weirdness we see is not actually the slowing down, directly, but rather fading out, getting dimmer and redder.

Likewise, if we could see the spacetime around the merger directly, we would see the ripples fade out to "red" as well. In a finite time, even according to the distant observer, we would see all the action and details fade out to red while slowing down, leaving only the single, merged black hole.

*OK, technically, if Dr. Strangelove is a finite size, then his head and feet will have different experiences, but that's beyond the scope of this inquiry.

EDIT: By the way, the gravitational field can have mono- and dipole moments, but gravitational radiation is quadrupole

share|cite|improve this answer
Frustrated grad student's favorite phrase: "Beyond the scope of this inquiry." – Andrew Jul 4 '11 at 16:10
I once had the honour of taking the place of Dr. Strangelove in a classroom thought-excursion to the vicinity of a black hole. I guess a significant fraction of all physicists have had that honour though... – Emilio Pisanty Jun 3 '12 at 16:04
The singularities do not swirl down the drain! This is just not true. There is no calculation which gives this, they do not move like point particles. The view that the singularity is the particle at the center of the black hole is completely misguided and comes up time after time. Please, no more. – Ron Maimon Jun 4 '12 at 5:24
As Ron Maimon said, I'd be curious if there's any basis for the claim about the singularities other than personal intuition--exact solutions like this one only deal with spacetime outside the horizon, and this source says that computational models require "special procedures (known under the generic name of singularity avoidance) devised to lay the numerical grid only over the well-behaved region of the spacetime and away from the singular points." – Hypnosifl Jan 13 '15 at 20:36
@Hypnosifl IIRC the case is that the numerical grid does not (can not?) exist at the singularity, which is what defines it as a singularity in the first place. – Asher Jul 29 '15 at 16:13

A binary system of black holes - assuming no infalling matter - would be detectable by the gravitational waves it emits. As they spiral in to each other, the waves will get shorter and more intense (I think), and eventually stop altogether when they merge. It's not exactly a dipole field (though I know what you mean), since both poles are the same 'sign'. IIRC, you can distinguish the difference between one massive star and a relatively equal binary system from a sufficiently close distance. From further away, of course, everything looks like a point source.

share|cite|improve this answer
+1 observation biased to reference frame relativity – Argus Jun 3 '12 at 2:12

(I tried several times to "Add Comment" this to the post by @Andrew but it never worked)

Presumably, an observer outside the event horizon could always see Dr. Strangelove by shining a light on him. It would be blue shifted on the way down, and red shifted on the way up. Assuming Dr. S. reflects gamma rays, the light would be unshifted after the round trip.

My confusion stems from what we would be able to detect during a merger. People are building devices to look for the gravity waves from merging black holes. Are these waves from what is happening just before the black holes merge, or a product of the merger itself?

Here is a thought experiment. Two black holes are circling each other with their event horizons a short distance apart. An observer (for movie consistency, let's call him Col. "Bat" Guano) is orbiting in the same plane a little farther out. His orbit is not circular since the gravitational field of the two black holes is not spherically symmetric. For a single black hole it is.

The black holes now get a little closer together and merge*. Bat is now either in a circular orbit, or he is not. If he is, then the merger (as measured by the universe outside the event horizon) happened at a particular moment in time. It isn't smeared out over an infinite period of time.

I agree with @Ben, if you are far away, the difference in the field is hard to detect. But it seems like near a supermassive black hole there should be a measurable difference between having millions of solar masses in a ring at the "equator" of the black hole versus all concentrated at the center.

*As the black holes get closer together, does Bat see the rotational speed increase due to conservation of angular momentum, or slow down due to general relativistic effects closer to the event horizon? I have to stop now, my head hurts...

share|cite|improve this answer
... is now either in a circular orbit, or he is not. ... at a particular moment in time. -- assuming global uniform time at any point of reference. That's barely relativistic. – ulidtko Jun 17 '13 at 10:40

As a distant observer we can watch the shadow of the black holes forming in front of the background stars. According to a nice little paper by Daisuke Nitta, Takeshi Chiba, and Naoshi Sugiyama ("Shadows of Colliding Black Holes, 2011") the answer is yes. To a distant observer in a finite span of time two black holes form a shadow that is indistinguishable from a single black hole's. I would "conjecture" that in a similar time frame the gravitational field around the pair would be same as that of a single black hole too. The question of black holes "actually" merging is not answerable in the sense that we can never know what is actually going on in our universe; we have to contend with limited observables...

share|cite|improve this answer

protected by Qmechanic Aug 13 '15 at 10:57

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?