- Trajectory Of In-fall Into An Un-Equal Binary Black Hole Pair (Orbit Eccentricity=0) last update 07-Nov-2010
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The above chart shows mass in-fall into a Binary Black Hole with Un-Equal Partners of 5 and 10 Solar masses. In-fall from the +Z direction are white trajectories (enter from 1st & 2nd quadrants) and in-fall from -Z are violet trajectories (enter from 3rd & 4th quadrants). Once the in-fall has achieved or exceeds 'Schrarzschild Energy', all trajectory are colored blue. As usual, the red circular zones around each singularity denotes the Schwarzschild radius for an individual BH of that mass. The space bounded by the red trace that circumscribes blue trajectory field represents the combined Schwarzschild radius of the Binary pair. Notice that this area is shaped somewhat like a lazy (sideways) bowling pin with the head toward -X and the base toward +X.

Keep in mind that for a real Binary, this BH pair is spinning about the Z-axis in the X-Y plane about the center of gravity (spin rate is 295.5 revolutions per second). In addition, notice the small red arrow along the X-axis at about -13 miles; This is the position along 'X' at Z=0 where the mutual force of attraction on M2 is zero. It also marks trajectory inflection (change in curvature). Non-Equal mass Binary pairs have mis- alignment of the Force=0 position compared to the barycenter (CG) position (Z-spin axis).

Notice also that 'trajectory inflection' (2nd derivative=0) denotes the position of balanced attraction (zero net force) of M1 and M0 acting on M2 where ever inflection occurs.

Also be aware that the Binary pair in the 2-D solver analysis above *ARE NOT* spinning ; therefore, the
described trajectories would be different in 3-D space depending on the position of the (M1 & M0) masses in the
X-Y plane when the interloping M2 particle approached and traversed the Schwarzschild volume. Also note that Binary
BH's (for M1=M0, or not) all present opportunity for interloping mass to fall into and escape the Schwarzschild
event horizon. I believe this is true with single black holes as well. But the case is harder to make due to issues
imposed by special relativity on time, distance, and mass in regard to influence on in-fall trajectories.

Suffice to say, I do not believe that all mass entering the event horizon of a single black hole ends up
contributing to the mass of that BH by colliding with the point singularity at the center (if it is, indeed, a
point); and that is my position because I believe gravity behaves as a conservative field inside and out side
of the event horizon. Stated before and repeated here; conserved energy is the ticket to escape the BH event
horizon. Conserved Energy can enable mass do what light cannot *-escape!*

This chart shows the classical traverse time (T_e2e) for a M2 particle mass dropped from 6000 miles at an X-Z angle of 130 degrees along a line through the Binary BH pair CG. For this case, the T_e2e time is 8.76E-5 seconds. Since the Binary BH is spinning at 295.5 RPS, it takes 0.00338 seconds to make one 360 degree revolution. During the Schwarzschild transit time from event horizon to event horizon, the Binary pair will have rotated just 9.32 degrees.

Of course, the actual angle in the X-Y plane of a line through the singularity CG's could be anywhere during M2 transit thus dispersing the M2 particle masses in a volume of space about the Z-axis. This would probably reduce the probability of opposing in-fall particle collisions causing an observable 'jet'.

The argument, however, for the M1=M0 Binary case (of minimal dispersion of in-fall along the +Z and -Z axis) producing 'jets' remains true because no matter where the X-Y M1-M2 angle through the singularities happens to be during transit, the X-Y forces acting on Z-axis in-fall remains equal to zero, thus not dispersing M2 particles away from the Z-axis which fact may enhance probability of opposing particle collisions to produce observable jets along the Z-axis for near equal mass Binary BH's

The animation below is for unequal-equal 14-Sm and 7-Sm BH's in elliptic orbit. The problem is set up with initial X0 and X1 distances of 24.7 and 49.4 miles from the barycenter. The velocity Vy0 was set at 0.12c and the velocity Vy1 set at -0.24c so that total linear system momentum is zero (no system CG drift). The case is set up to provide elliptic orbits that maintain a continuous event horizon at maximum semi-major axis extension.

The above analytical result was computed using algorithm and images from the initial valued solver routine (using Newtonian physics). The Red Boundary is combined event horizon. Blue traces are in-fall trajectories, the ends of which map the event horizon at the point where Schwarzschild energy is achieved on ingress and lost upon egress.

The equation for the orbital period of unequal Mass Elliptical Binary BH's was added to the BH_Equation page. In addition, this equation is integrated into the solver code. For each B-BH case the solver is run for just over one revolution and the M0 and M1 trajectory ellipse(s) semi-major axes (a0 and a1) are captured for use in the UNEQUAL mass B-BH revolutions/second (Rps) equation.

Knowing this period also facilitates setting time between frame grabbing (Tinc) equal to the period/frames. I was doing frame setup by trial and error to get an integer number of frames per revolution, a huge waste of time.

I am going to add the ellipticity computation (b/a)=sqrt(1-e^2) to the solver analysis. Also going to change RTime (CPU time) from (min) to (hrs). It takes about 5 hrs to run 20 picture images (frames) with 33 in-fall trajectories for each frame.

It is interesting to note that inertial acceleration for in-fall into Elliptical Binaries along Z-axis is irregularly pulsing by nature due to the time varying acceleration as a function of the closeness of M0 and M1 to the binary barycenter as the in-fall makes closest approach.