2
$\begingroup$

Let's assume we have two black holes with equal mass. They move towards each other, each heading towards the center of the other black hole. Both black holes rotate with equal speed. What happens to a particle that is placed in the middle of both black holes in respect to the approaching ergospheres?

If they rotate in opposite directions, do the ergospheres add up, shooting the particle outwards? If they rotate in the same direction, do the ergospheres cancel each other out, not moving the particle at all? Does the ergosphere of one black hole affect the other black hole?

What happens if the two black holes do not have equal mass? Does the ergosphere of the heavier black hole "win" and force the particle to move in its direction? Do the ergospheres influence the other black hole in this case?

Please excuse my possible layman terms, physics is not my area of expertise.

$\endgroup$
7
  • 2
    $\begingroup$ There are no simple equations to describe this; you need to do a full numerical simulation like the ones here: einsteintoolkit.org/gallery/bbh/index.html. Note the computing requirements for just one set of initial conditions! $\endgroup$
    – m4r35n357
    Jul 31, 2021 at 17:50
  • $\begingroup$ @m4r35n357 My intuition tells me that this should be much simpler because of the symmetry. However, it is quite possible that my intuition is completely off here. In that case I would like to understand why it's that hard of a problem. $\endgroup$ Aug 1, 2021 at 11:27
  • $\begingroup$ Unfortunately the symmetry you speak of is not the sort of symmetry that simplifies any equations. It is "that hard a problem" because the mathematics is hard. Here is what you need to solve: en.wikipedia.org/wiki/ADM_formalism. Your first step is to find out what all the pieces mean ;) $\endgroup$
    – m4r35n357
    Aug 1, 2021 at 16:34
  • 1
    $\begingroup$ I would guess the spacetime near the collision is so complicated and the evolution so non-linear that the notion of an ergosphere probably doesn't even make sense in this regime, and that matter so close to the black holes will either fall in during or get ejected from the system. $\endgroup$
    – Andrew
    Aug 2, 2021 at 20:35
  • 2
    $\begingroup$ Most numerical relativity simulations dont use the ADM formalism anymore, rather the BSSN or evolutionary types @m4r35n357 $\endgroup$
    – Johnny
    Aug 6, 2021 at 9:22

1 Answer 1

1
+100
$\begingroup$

Firstly, the ergosphere of black holes is considered an indirect rotational quality, its just the effect of the massive rotation speed morphing the area of space outside the event horizon.

Geometrical Theory:

(2) If they rotate in opposite directions, do the ergospheres add up, shooting the particle outwards?

(1) If they rotate in the same direction, do the ergospheres cancel each other out, not moving the particle at all?

Until the particle is not under any influence of a ergophere rotation, the objects pulled into a black hole is due to its gravitational pull (which extends way further than the ergosphere).

From what I suspect, the particle will stay in the middle for both situations until the above statement is true,

assuming the particle can sustain pull forces from both the black holes combined (realistically not possible), the black hole have the same mass and are equidistant from the particles (as you have mentioned) and they have the same gravitational pull.

Now, once the black the particle enters the ergospheres of the black holes (assuming the both the black hole move at the exact same speed and the particle enters the ergospheres at the same time):

In condition (1):

Condition 1

The particle will stay constant / still. This is because, referencing the image above the opposite rotational forces of black holes A, B on the particle, will cancel out and with the balanced gravitational forces, the particle simply stays still.

In condition (2):

enter image description here

The object will be ejected out in the direction relative to the added rotational direction of the black holes. Each force has two components; horizontal and vertical, Fx and Fy as seen in the image above. Due to the direction of x component of the rotations of the black holes, they will cancel out, leaving only the y. Since the y component is in the same direction for both black holes and at a equal magnitude (quantity), its only obvious that the particle should move straight in the y and relative to the rotation direction of the black holes (downwards in the image) and it will eventually be ejected.

Other possibilities include:

enter image description here

The particles rotates around the two black holes, following the path above

OR

below

enter image description here

Experimental / Real Life:

To stay true to reality, the above theory will not work due to a simple reason: When the black holes come near enough one another, the will orbit each other, which gradually decays, leading them to collide and become a bigger black hole, which in-turn leads to two scenarios:

either the force and factors affecting the particle cause it to be ejected out into space,

OR,

your particle gets sucked into the black hole :(.

So why did I do the geometrical theory above?

  1. Its always fun to play around with geometry and theory and what it would be like excluding any other variables and problems.
  2. If I just said the black hole ate up the particle, does not really seem to explain a lot + plus its cooler to explain geometry.

Hope this helps :) !!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.