# I Two black holes colliding - visual

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1. Aug 18, 2016

### mgkii

There's lots of other questions on the forum about 2 black holes, but I think this is different - and I can't get my head around which outcome is consistent with GR.

Black holes here are simplistic - non-rotating and let's assume with a tiny accretion disk; just enough grains of matter to allow us to see the black hole, but small enough to ignore when we are talking about the effects below.

So in the basic case of an object (usually an unfortunate individual) falling into a black hole, what we see from the outside is that object slowing as it approaches the event horizon, never actually passing the event horizon and eventually pausing at the horizon, and then slowly fading from view.

So here's my question. I have two black holes (as per my conditions above) and they start at some random distance from each other but are stationary with respect to each other - i.e. not orbiting, no rotations. etc. As they move together under their own gravitational influence, what will I see when they collide?

Thought 1: The simple "unfortunate individual" case is still a two body problem, so extending the thinking would suggest that the accretion disks would meet, and I would never actually see the two holes combine.

Though 2: Thought 1 is nonsense! I would see something akin to the models I've seen of two galaxies colliding - the two black holes will pass through each other, dance around a bit and end up as a single black hole, with the matter forming the accretion disks simply being scattered and then re-accreted during the process.

Any views?

2. Aug 18, 2016

### BvU

There's a simulation here

3. Aug 18, 2016

### mgkii

Thanks @BvU

The simulation is really interesting, but the simulation has the two holes in orbit which I think would account for the long period in which they stay appart - i.e. this isn't necessarily an apparent effect an external observer would see due to GR, but a real effect due to the orbits.

In the case of the unfortunate person falling straight into the black hole (as opposed to orbiting in the way the accretion disk does), the external observer see's them slow down but their experience would be that they sail right in. My question is an extension of this - if you removed the orbital component, then would the black holes collision appear to slow down at the boundary, or would you see something else?

Thanks
Matt

4. Aug 19, 2016

### tionis

What you would see is that when the black holes get close to each other, the horizon swells up to encompass both of them.

5. Aug 19, 2016

### Staff: Mentor

You won't actually see this, for the same reason that you won't actually see something fall through a single black hole's horizon. Light at or inside the horizon can't escape. But you are correct that the two holes' horizons will merge together to form a single horizon.

You can't see black holes the way you see ordinary objects, because light can't escape from them.

6. Aug 20, 2016

### tionis

Well, maybe not directly but astrophysical BHs can be seen by other methods. Suppose we were within observational distance of two black holes about to merge: couldn't we infer by the distorted light from the background what was about to happen?

7. Aug 20, 2016

### MattRob

You'd probably see the exactly what you saw in that simulation (not being cheeky - just really meant just that: I don't think the two black holes' angular momentum had much to do with how they acted on "swallowing" eachother. It would probably look much the same, albiet extremely sped up, if they fell directly into each other).

As for the person-slowly-falling-in conundrum, that comes about because of, well, reasons to do with the Schwarzchild solution:

The Schwarzchild is a simple, static spacetime solution with no spacetime curvature except the curvature induced from the black hole itself.

As you probably know, gravity is merely the result of objects falling along geodesics - the shortest line through curved spacetime. ie, they're "trying" to go along a straight line through spacetime as inertia demands, but spacetime itself is curved, thus the "straight line" they go along is this geodesic, the path they free-fall along under gravity's influence. So a geodesic is just that - an inertial trajectory of a point particle through spacetime.

In the Schwarzchild, geodesics do this: They approach the horizon asymptotically, never reaching it. Thus goes the saying that things never appear to cross an Event Horizon.

But of course, a black hole isn't massless, thus the Schwarzchild doesn't accurately describe a second black hole, only the first one (and an idealized one at that, too). For this, it's a bit more nuanced. I think what it comes down to, is that the horizon swells up to "swallow" the infalling massive object, as tionis pointed out, but as PeterDonis pointed out, you wouldn't directly see light from this event since nothing escapes a Horizon. You'd merely see some sort of light distortion that you recognize as a black hole "swallow" something up - again, essentially what the animation showed.

I'm being careful to not say you "see" it per sae, because after all, you don't see a black hole - there's nothing there to see - you just see the light distortion from around it, and you recognize the black hole is there, and so in more practical terms you "see" it that way, even though of course, you never get a single photon out of it.

8. Aug 20, 2016

### tionis

You can always try to see the horizon itself, in other words try to map out the region from which light can't escape, but you would have to wait forever. But because the redshift is so strong, you can get a very accurate estimate

9. Aug 20, 2016

### Battlemage!

As the event horizon becomes much bigger, would you see the objects near the two black holes suddenly freeze (as they fall in)? What I mean is, would you just get this large sphere of orbiting matter just freeze as the new, larger event horizon grows too close to them?

10. Aug 20, 2016

### tionis

Emitted light from an object near the event horizon drops by 17 orders of magnitude in a millisecond for a solar mass BH.

11. Aug 20, 2016

### Battlemage!

So lets assume you have a timelapse recorder can separate frames into a comparable interval. Would you see them slow down and freeze if you looked at each frame one by one?

By the way, is it incorrect that if you saw someone falling into a black hole, you'd see them freeze before they ever got to the event horizon? If so, that's the same thing I'm talking about. As the two black holes collapse, the event horizon gets bigger, so would objects orbiting do the same thing as your friend would appear to do if she fell into a black hole, as if they were falling toward it (when in reality the event horizon is just getting bigger)?

12. Aug 20, 2016

### Staff: Mentor

No, they don't. They fall right through the horizon.

No, that is because the light emitted outward from an object falling towards the horizon takes longer and longer to get out to a distant observer as the object gets closer to the horizon (and light emitted by the object at or below the horizon never gets out at all). That is because of the effect of spacetime curvature on outgoing light; it is not because of any "slowing down" of the object itself.

13. Aug 20, 2016

### Staff: Mentor

You would see the distortion change as the holes got closer together and merged, yes.

14. Aug 20, 2016

### Staff: Mentor

It is incorrect. You would never actually see them "freeze" at all, strictly speaking. In the idealized case where you could observe light of arbitrarily long wavelength, you would see them get closer and closer to the horizon, appearing to move slower and slower, but you would never see them actually stop and you would never see them actually reach the horizon.

In the more realistic case where you could not observe light that was of longer wavelength than some finite value, you would see the object get dimmer and dimmer and then disappear; but it would still be moving inward and would still be a finite distance above the horizon when you last saw it.

15. Aug 20, 2016

### Staff: Mentor

No. You would see them either slow down more and more, or disappear, depending on whether you consider the idealized case or the more realistic case described in my previous post just now.

16. Aug 20, 2016

### Battlemage!

Ah, so they would fade out of view. That makes sense.

Which means in this case, as the event horizon gets larger due to the collision, a lot of previously visible stuff would then fade from view as they got closer to the new event horizon, correct?

17. Aug 20, 2016

### Staff: Mentor

Yes.

18. Aug 20, 2016

### MattRob

True, but I should have been more specific - I'm referring to geodesics as the trajectories of free-falling, massless point-particles in Schwarzchild coordinates.

In Schwarzchild coordinates - is what I meant - a massless point particle (since only massless particles can exist in a Schwarzchild solution, since by definition introducing a new mass would no longer be Schwarchild, but a solution with more curvature than just that of the particular black hole) asmyptotically approaches r = 2M but never reaches nor crosses it. An artifact of the coordinates chosen, yes, but that is nonetheless how free-falling particles behave in the Schwarzchild coordinates in a Schwarzchild geometry, and is probably where all the talk of "it slows down as it approaches the horizon" comes from. A coordinate-dependent "fact," but sort of what you'd see, nonetheless. Not necessarily the "true" frame-independent geodesic, but nonetheless, how a geodesic would appear in these particular coordinates.

Really it comes down to what you mean by "slow down". I'm speaking in a coordinate-dependent sense, while you seem to be speaking more fundamentally and inertially?

Both have plenty of validity. Inertially may be more fundamental and important in many respects, but you can't completely ignore coordinate phenomenon, either, since it's coordinate acceleration that breaks a glass if I drop it. Or more relevantly - it's coordinate phenomenon that I say it takes a neutrino 100,000 + years to cross the Milky Way (understood that I'm using a Minkowski description of space), despite it taking only seconds in its own frame, and an arbitrary amount of time if I'm allowed to choose any accelerated frame.

19. Aug 20, 2016

### Staff: Mentor

Yes, I'm aware of that, and I am doing the same.

First, you have misdescribed what a "test particle" is. A test particle does not have to be massless; massless particles travel on null worldlines, and it is perfectly possible to have test particles that travel on timelike worldlines. The correct statement is that test particles must have negligible stress-energy, so that they do not affect the spacetime geometry. But one can meet this criterion by assuming a particle with nonzero rest mass (so it travels on a timelike worldline) and making that mass so small compared to the mass of the hole that it is negligible. (Or, if you really want to be a purist, you can take limits as the rest mass of the particle goes to zero, while holding the particle's worldline constant, so it stays timelike.)

With the above caveat, what you say is technically correct, but it doesn't justify the statement you made before about geodesics. Geodesics are independent of coordinates, and you can't make blanket statements about geodesics based on how they appear to behave in particular coordinates. In this case, a better way of stating what is going on would be: in Schwarzschild coordinates, along the worldline of a test particle free-falling into the hole, the derivative $dr/dt$ of the radial coordinate $r$ of the particle with respect to coordinate time goes to zero as $t \rightarrow \infty$, and in that same limit, $r \rightarrow 2M$. But that is because the coordinates become singular at $r = 2M$; it's a property of the coordinates, not the geodesic.

To see what happens to the geodesic itself, you need to look at $dr / d\tau$, the change in $r$ with respect to the particle's proper time along the worldline. That does not go to zero as $r \rightarrow 2M$.

No, how a portion of the geodesic would appear in these particular coordinates. That portion is not the entire geodesic.

Not for the purposes you are trying to use them.

No, it isn't. I can choose coordinates in which the glass's coordinate acceleration is zero; but it still breaks.

Yes, which means that number is not physically meaningful; nobody will ever actually measure it.

I strongly suggest that you take a step back and think very carefully about the claims you are making about coordinates. Basically you are trying to say that coordinates can cause physical events. That's wrong, and will lead you into misunderstandings. Only invariants--things that are independent of coordinates--have physical meaning. That is one of the key lessons of relativity.

20. Aug 21, 2016

### MattRob

Okay - I guess that's a misnomer - it sort of has mass but doesn't - I wasn't meaning to imply it was null-like, which is, as you mentioned, what I accidentally implied by saying "massless". The point of "massless" was to state how it made no contribution to the stress-energy tensor, and no contribution to the spacetime geometry, but forgot that "massless" also implies null-like trajectory.

You can choose a coordinate system where neither the glass nor the floor accelerate, yet still impact each other with a "high" relative velocity?

That was the point I was getting at - motions induced by the geometry, as measured by coordinates, are useful and important.

Though of course, certain types of coordinates can be misleading. I think I may have miscommunicated my intent - I didn't mean to cite the Schwarzchild to support the claim that objects never cross the horizon, only to mention that in these coordinates, things never cross the horizon by coordinate time $t$, though of course, they actually do cross the horizon in their own local frames and in a more global, invariant sense.

Not exactly what I'm saying, as much as I'm saying that coordinates can reveal interesting information. Coordinates don't cause events but they can report them and certainly make work much easier in handling certain problems, and my main point was that one thing that coordinates did cause is the oft-repeated misunderstanding that "things never fall into a black hole", which of course, isn't an invariant truth, but is conditionally true in Schwarzchild coordinates when speaking of coordinate time $t$ as opposed to invariant time $τ$.

So, it really becomes more a matter of how you look at it - in terms of coordinate time, it never falls in, but the misunderstanding being that that statement is often misinterpreted to be one that is always true. Of course locally, and in different coordinates, it falls right on through.

Hehe, to be a bit more light about it - whether things fall in or not depends on if you're a solipsist or not, and if you ever fall in.

21. Aug 21, 2016

### Staff: Mentor

No. But you can switch the coordinate acceleration from one to the other by choosing coordinates. So "coordinate acceleration" can't be what's causing anything.

No. Motions induced by the geometry are useful and important, period. The "as measured by coordinates" part should not be in there; you don't need to "measure" the motions with coordinates. Coordinates are conveniences in our models; they are not physics.

No. Coordinates can organize interesting information for human convenience; but they don't "reveal" the information. The information is there whether you choose coordinates or not. What "reveals" information is actual physical measurements; but you can make those without choosing any coordinates at all. Don't be misled by the fact that, for our own convenience, we almost always describe measurements using coordinates. As I've said, that's a human convenience, not physics.

Even this attempt to make a limited statement is problematic, because "never" implies something absolute, independent of coordinates. (Maybe it doesn't to you, but it does to the hundreds of posters that have started threads here based on that misunderstanding.) The way I would state it is that Schwarzschild coordinates only cover a limited region of the spacetime, and that region does not include the event where an infalling object crosses the horizon (or any events below the horizon).

22. Aug 21, 2016

### tionis

If the Schwarszchild solution doesn't cover the black hole itself meaning it doesn't describe what is going on beyond the horizon, then is it describing our spacetime and not a BH?

23. Aug 21, 2016

### Staff: Mentor

The Schwarzschild solution (i.e., the spacetime geometry corresponding to that particular solution of the Einstein Field Equation) does. Schwarzschild coordinates do not. Coordinates are not the same as the solution (geometry).

(More precisely, exterior Schwarzschild coordinates do not. There are Schwarzschild coordinates that work inside the horizon, but they are a different coordinate chart, disconnected from the one that covers the region outside the horizon.)

24. Aug 21, 2016

### tionis

I will try to explain my thoughts, but it's difficult 'cause I am like Einstein in that I get flooded with mental imagery, but unlike him, don't know what it means or if it has validity. So, the OP wants to collide two black holes but didn't specify the infalling BH size. My question is: from the background of the stationary black hole, how does the other black hole looks like? If the size of the infalling BH is small, then the bigger BH would view the smaller one as a point particle according to its scale, right? while the smaller BH sees the larger BH not making radical changes to the local physics, right? Now, if the two BH were of the same size, what spacetime would the two BHs see? I suppose it is a question of scales.

How does our spacetime looks to the large BH? From our perspective, we see a Schwarzchild spacetime, but what type of spacetime does the black hole sees around itself when looking at us, the planets, the universe? What type of solution would that BH assign to us? Make sense?

25. Aug 21, 2016

### Staff: Mentor

When they're far apart, each one just looks like a black hole to the other one. When they are close together, about to merge, it's not that simple, and I don't know that anyone really has a simple description of how things would appear to an observer stuck close to the process.

If by "size" you mean "mass", then yes; a BH with a small enough mass would not significantly affect the spacetime geometry of the bigger BH until they were basically merged anyway.

There is no known analytical solution for this case; you have to solve it numerically.

Spacetime geometry is the same no matter whose "perspective" we choose to look at it. If the spacetime in the solar system is (to a good approximation) a piece of the Schwarzschild geometry, then that is what it is.