Algr said:
Your text with the "pair of trousers" describes the two event horizons merging, but does not explain why they would behave that way.
They behave that way because that's how the Einstein Field Equation says they behave. Strictly speaking, we would need to do a detailed numerical solution to get the exact shape of the "pair of trousers"; there is no closed-form analytical solution that describes this scenario the way the Schwarzschild metric describes a single black hole. What I'm describing is simply a simplified, heuristic version of what you would get if you did the detailed numerical solution.
If that doesn't count as "explaining why", then you haven't explained why the light cones tilt in the single black hole case either, because that tilting comes from solving the Einstein Field Equation too.
Algr said:
Suppose the black holes has a mass such that in the absence of other objects, their event horizons would be one million miles radius. The two radii are one foot apart at their closest points.
You are continuing to make the mistake of thinking of the horizon as a "place". It isn't; it's an outgoing null surface, meaning it's a surface formed by radially outgoing light rays. Light rays can never "stay in the same place". Don't be misled by the fact that the horizon, at least for a single black hole, has a constant ##r## coordinate; the ##r## coordinate is not the same as "where" the horizon is--the latter concept has no meaning. (Outside the horizon, ##r## can be interpreted as "where" something is, but you still have to take into account that "space" in this chart is non-Euclidean.)
Also, the event horizon does not have a radius; it has a circumference and an area. The ##r## coordinate is sometimes referred to as the "horizon radius", but that's sloppy terminology; ##r## is really the circumference of the horizon divided by ##2 \pi##, or the square root of the area divided by ##4 \pi##. (Note that this is for a single black hole; the relationships are more complicated for the case of merging black holes that we're discussing, but the fundamental fact remains that the horizon does not have a radius.) There is no such thing as a "distance" from the horizon to the singularity; if you're at the horizon, the singularity is in the future, not at a different place in space.
Algr said:
Is there really a gravitational equation that would show gravity as inescapable at a million miles away, but zero six inches further?
No. To the extent that your picture of the situation makes sense (which is problematic to begin with given the caveats I stated above), the equation shows that gravity is inescapable at each horizon, and almost inescapable at the midpoint between. The fact that there is no "acceleration due to gravity" at that midpoint does not mean it is just like empty space at infinity; there is still a very large gravitational redshift for any light emitted outwards towards infinity at that point. You continue to make this mistake as well--thinking that "zero acceleration" means "zero redshift"--even though it's been pointed out to you several times.
Algr said:
Could you go over your "Apparent" and "Actual" horizons again? I think that one of these is the horizon that I am talking about, but I don't follow why there is a different one.
The apparent horizon is where radially outgoing light no longer moves outward; if a flash of light is emitted spherically in all directions radially outward from the apparent horizon, the area of the sphere does not increase with time--it stays the same.
The absolute horizon is the boundary of the region that cannot send light signals to infinity (this region is the black hole). For a single black hole whose mass stays the same forever (i.e., nothing ever falls in), the absolute horizon coincides with the apparent horizon. But for more complicated cases, where things fall into a single hole or two holes merge, that's not the case. For the case we're discussing, of a black hole merger, the absolute horizon will, in general, be outside the apparent horizon: that is, light emitted outward towards infinity from the absolute horizon might still move outward when it is emitted. But sooner or later that light, if it's emitted from the absolute horizon, will get caught and pulled back by the gravity of the merging (or merged) black holes--it will never reach infinity.
One other thing to keep in mind is that a spacetime with two black holes merging is not spherically symmetric, so defining the "outward" direction--the direction you need to emit light for it to have a chance of reaching infinity--requires some thought for the region in between the holes, which is the region you're concentrating on. If you think of the "pair of trousers", with time vertical, and concentrate on the region in between the legs, obviously light emitted to the left or right won't escape--it will fall into one of the holes. Light emitted "towards" you (meaning perpendicular to the line connecting the two merging holes) has a chance of escaping: but if the holes are close enough together, it still might not, because it can't get out of the region between the holes fast enough--their combined gravity pulls it back. This region is where the "legs" of the trousers merge. All of this can't be described by a simple diagram showing "space", as you tried to do before--you have to include time, and you have to include at least two "space" dimensions as well, because of the lack of spherical symmetry. In short, the intuitions you may have from considering the single black hole case, which is spherically symmetric, might lead you astray in this case.
Algr said:
The horizons DON'T merge because the region of null gravity between two black holes always lies outside the event horizon.
No, it doesn't. If you're going to just keep re-stating your incorrect statements without even addressing the responses that have been made to them, there's no point in continuing discussion. Do you have any response to the multiple posts I and others have made showing why this claim of yours is incorrect?