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Consider two supermassive black holes of equal mass orbiting about their common centre of mass. Is it the case that a free-fall trajectory along the axis of rotation would be outside of either event horizon at all black hole separation distances > 0 (based on the symmetry of the situation)?

To rephrase this, would you be able to navigate a rocket along a path at right angles to the orbital plane and bisecting the line between the two black holes with no ill effects whatsoever even when the black holes were very close to each other?

Supplementary question: What is the shape of each of the event horizons prior to coalescence?

Note: I’m assuming that tidal forces would be small until the two singularities were extremely close. Note also that the point midway between the two BHs is the L1 Lagrangian point.

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See my comment below--I don't think you can really talk about Lagrangian points for a system involving two moving black holes. – Jerry Schirmer Jan 21 '11 at 13:45
For information I have generated a question about General Relativity from these discussions. < > The comments include a link to an animation of rotating Black Holes coalescing. – Roy Simpson Jan 24 '11 at 13:05
up vote 3 down vote accepted

You can't have a stable configuration of two orbiting black holes at all separations. Even in the case of a test particle orbiting a single non-spinning black hole, the innermost stable orbit for the test particle is located at $r=6M$, with closer orbits plunging into the hole (circular orbits closer than $6M$ are unstable solutions analogous to balancing a rigid pendulum vertically.)

If you have two black holes of appreciable mass, they will radiate away energy, gradually falling in closer and closer to each other, and then at some time before a common horizon is formed (also meaning before the "center of mass" of the system is inside the horizon), the two holes will hit their last stable orbit, and then will very rapidly combine, giving off a large burst of radiation, and an eventual end state of a single black hole.

As a final aside, I should note that a spinning black hole's singularity is a ring, not a point, and that the center of this ring IS locally flat. Since a binary black hole collision will definitely have net angular momentum (even if just from the orbital angular momenta of the holes), this means that the end state will almost certainly be a spinning hole, so it is not necessary that the "center point" ever not be locally flat.

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I think this is the answer I'm looking for. The point about orbital stability is crucial as it answers negatively the question as to whether there are "rocket trajectories" as described, outside the event horizon, at all > 0 BH separations. I kind of assumed that the black holes could get arbitrarily close as they lost energy through gravitational radiation, and that the orbital diameter contraction rate would be "slow" pretty much all the way in. – Nigel Seel Jan 21 '11 at 14:41

As I understand this question the two Black Holes initially have some distance D between their Event Horizons, a region of Space also containing L1. So yes I think that this journey outside the Event Horizons is theoretically possible assuming that they dont coalesce on you whilst on this journey.

One practical problem, and hence risk, likely arises from the non-stability of L1 (at least in Newtonian 3 Body theory - I assume also true in General Relativity). This would mean that a journey which slightly deviated from it, could orbit the nearest BH (in Newtonian theory tangentially escaping the system is more likely). So determining the exact nature of this instability for GR for this setup is crucial.

Once orbiting a Black Hole - even although outside the Event Horizon - the next issue is whether your Spaceship has the energy to escape that orbit. Meanwhile with the two Black Holes converging around you there could still be problems staying out of one of the Event Horizons....

I should point out that whilst researching this question, I have come across the "Interplanetary Transport Network" -

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If the trajectory is effectively "dropping from infinity" then the flyby is pretty fast - with two BH's pulling it in for most of the way. For the reasons stated in my comment on dbrane's answer, I don't think that for "small deviations" from L1 you would be particularly close to the horizon. So if you missed L1 by delta x then I think you'd do a hyperbolic rather than linear flyby. – Nigel Seel Jan 21 '11 at 13:06
Nigel, As you can see from the other answers there are several assumptions in this scenario. Returning to a Newtonian version for simplicity (where L1 is defined) we can also see that L1 is not a stationary point in general. The exact equality of the two masses is a crucial assumption in your formulation. So in general there maybe nowhere to "aim" for from infinity. – Roy Simpson Jan 21 '11 at 14:12
So the question seems to me to be reformulated as "under what conditions of BH mass, separation, rotation" and spaceship energy is this journey possible? – Roy Simpson Jan 21 '11 at 14:17
Yes, the condition of mass equality was there in the question for just that reason. – Nigel Seel Jan 21 '11 at 14:36
The mass of a supermassive black hole is unlikely to be constant due to accretion, so this is important for a practical solution. – Roy Simpson Jan 21 '11 at 16:53

Even before the horizons merge to form a common horizon, the tidal forces may be very large, even for supermassive BHs. For a single black hole, particles originating from outside the BH cannot get closer than something like 1.5 times the horizon radius without getting captured. So a particle or photon that gets slightly closer to one of the two BHs would get captured. Thus, if you try to "thread the needle" you might get split in two.

For a picture of the shapes of the horizons, see Fig 1.

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Good article you cited. I notice that in figure 1 (page 8) the "Apparent Horizons" (AH) of the two BHs are non-overlapping, although distorted, just prior to merger (right-hand diagram). Following coalescence we have a "peanut-shaped" AH which then settles into its spherical shape. Looks like we can still get our pre-coalescence flyby (see my comment to dbrane's answer)! – Nigel Seel Jan 21 '11 at 10:33

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