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B Merger of 2 black holes

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  1. May 2, 2017 #1
    LIGO is reputed to have detected gravitational waves from the merger of 2 black holes (BHs).

    For an external observer of such an event, each BH would appear to approach the event horizon of the other, but never cross it in a finite time......is this correct so far? But the event horizons are roughly spheres, so this approach would appear to force at least a small intrusion of one or both BHs into a 4th spatial dimension, the merged object appearing as a topological 3-sphere to the external observer.

    By analogy, in a 2-D universe, the merger of 2 disk-shaped BHs at their circular event horizons would
    force the merged object into a 3rd spatial dimension, forming a 2-sphere (a balloon).

    Am I imagining this scenario correctly, or if not, why not?
     
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  3. May 2, 2017 #2

    Orodruin

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    No. You need to be much more careful with what you mean by "finite time". Your language seems to suggest some assumption of an absolute time, which does not exist. Furthermore, a black hole merger is very far from the static solution to the EFEs that describes a Schwarzschild black hole.
     
  4. May 2, 2017 #3
    Hmmm.....admittedly I am not a trained cosmologist, but I thought I had used the same care which I have seen Susskind use.
    That is, I am sitting at some point (without assuming any absolute coordinates) and at some other location I observe what
    appears to be 2 black holes in the process of merging. At each stage of the merger, time is measured on my timepiece.
    Also, I do not know what EFEs are. Am I resigned to abandon this question?

    Tom McFarland
     
  5. May 2, 2017 #4

    kimbyd

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    What happens is the two black holes merge and form a single, larger black hole. The entire process is over in a few seconds once the black holes touch.
     
  6. May 2, 2017 #5
    If the merger involves a BH and a chunk of normal matter S, like a star, then the external observer will see S approach the BH, moving ever more slowly as S nears the event horizon, but the external observer will never see S cross (or touch) the event horizon. If we now replace S with another BH, I would have (naïvely?) expected similar behavior, but now with more symmetry. Some kind of merged object results, with more mass, but if an external observer never sees the 2 event horizons cross, then what is the nature of the merged object? Will the math answer the question?.......I do not have the needed math skills. However, an external observer would see two BHs with disjoint interiors but with event horizons which were very close together. Such a merged object is a 3-sphere and can only reside in 4-space, which string theory says must be there, but is degenerate.

    Cheers, Tom McFarland
     
  7. May 2, 2017 #6

    kimbyd

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    There is one major difference between a black hole and any other object: the black hole has no proper surface. The event horizon isn't anything physical. It's just a location past which nothing can return. You can't actually observe the event horizon in any meaningful sense: you can only observe matter fall into it (or rather, the effects of that happening).

    So when two black holes merge, the horizons touch, quickly turning into a sort of barbell shape that then very rapidly vibrates until it settles into a new event horizon that's larger than either of the two original black holes' horizons.
     
  8. May 2, 2017 #7
    Put a couple of drops of liquid Mercury on a surface that is slightly curved, so they are bound to shortly collide.
    It's like that.
     
  9. May 2, 2017 #8
    Kimbyd, you make a good point that I did not notice.......the event horizon is not the equivalent of a physical object.

    However, I would like to argue that an external observer still will not see two event horizons "touch" because the observed timing of movement near the two event horizons will be increasingly dilated (stretched) the closer the horizons get. For the external observer, the two event horizons ought to appear to hover near to each other, but this can only happen in 4-space. I am having trouble imagining what such a 4-dimensional configuration would look like to a 3-D external observer. Would you like to try?

    Cheers, Tom McFarland
     
  10. May 3, 2017 #9

    Nugatory

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    That is one of the most common and pervasive misunderstandings about falling into a black hole. It is true that the external observer will never see the infalling object reach and pass the event horizon; it will be redshifted to invisibility so will simply disappear instead.

    However, the external observer will eventually (and fairly quickly, at that) observe that the mass of the black hole has increased by the mass of the infalling object, with commensurate increase in the Schwarzschild radius of the black hole. In most presentations of black holes, this effect is completely ignored because the infalling mass is assumed to be negligible compared with the mass of the black hole (exercise: how much does the Schwarzschild radius of a ten-solar-mass black hole change when you drop a 100-kilogram astrophysicist into it?)... But that's not the case when we're dropping one black hole into another.
     
  11. May 3, 2017 #10
    Nugatory:

    You write "the external observer will eventually (and fairly quickly, at that) observe that the mass of the black hole has increased by the mass of the infalling object"

    Are you claiming that information about the change in mass can travel faster than the (redshifted) light which an external observer uses to observe the change?
     
  12. May 3, 2017 #11

    Vanadium 50

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    Redshifted light still travels at c.
     
  13. May 3, 2017 #12
    My point was that time dilation near the event horizons will give an external observer (using light) the vision of two event horizons drawing ever closer, ever more slowly, and never touching in a finite time.

    Yes,the external observer sees the red-shifted light traveling with speed c, but time dilation near the horizon only allows light from that location to arrive at the external observer much more slowly. Nugatory seems to claim that information on BH mass can somehow arrive faster than it would if that same information had been carried by light ??
     
  14. May 3, 2017 #13

    PeterDonis

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    Moderator's note: Moved to relativity forum.
     
  15. May 3, 2017 #14

    PeterDonis

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    No. The information about the change in mass is not coming from the black hole event horizons. It is coming from the past, from the objects that originally formed the hole.

    This is actually the case for ordinary massive objects, but we normally don't notice it because our normal experience is with situations that are so close to static that the light speed travel time doesn't make a dfiference. For example, the effect of the Sun's mass on the Earth's orbit is not due to the Sun "now"; it's due to the intersection of the Sun with the Earth's past light cone. But that's only 500 seconds ago (by Earth clocks), and the changes in that time are very small anyway (though there are observable effects due to this, such as the perihelion shift of the orbits of the planets, which I believe has now been measured for all the planets out to Mars). And the gravitational redshift due to the Sun's mass has negligible effect on all of this.

    In the case of a black hole, things are very different. Because of the extreme redshift of light coming from just above the hole's horizon, and the accompanying time delay, the gravity you would feel "now" from the hole is coming from very, very far in the past--from the collapsing matter that formed the hole, just before it reached the event horizon. A little bit in the future from now, you will be feeling the gravity from that collapsing matter a little bit closer to the horizon--and so on indefinitely into the future. But you will never feel gravity from the collapsing matter inside the horizon, for the very reason you give: gravity can't travel faster than light. (Note that this is still a heuristic picture, and has a number of simplifications in it, but it's the best I can do in a "B" level thread.)

    Similar remarks apply to the case of two holes merging.
     
  16. May 3, 2017 #15

    kimbyd

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    I don't think this is accurate at all. I'm not really sure precisely what the merger would look like, but I do know the beginning and end states of the merger.

    First, it may be useful to listen to this, which is the waveform that was detected from the merger seen two years ago:

    (Note: the description below the video is pretty comprehensive, and should clear up any questions about it you might have)

    That "blip" at the end of the wave is what happens when the event horizons touch (I don't know the precise moment that the touch happens, it might be right at the peak or a little before, but probably not after). After that touch, the gravity waves become undetectable within a fraction of a second. After that time, the resulting black hole has settled into a nearly-spherical shape. If the touch itself cannot be observed, how would it be possible to observe the nearly-spherical resulting black hole? There has to be some sort of transition between those two states.
     
  17. May 3, 2017 #16
    Kimbyd and Peter Donis:

    I would like to thank you gentlemen for your patience in expressing your thoughts in language I can understand. I am a bit disappointed in losing the case for the merger of 2 BHs being a 3 sphere in 4-space, but honestly, I think you folks have a stronger argument. I see no way to further argue in favor of a 3-sphere as the topology for 2 merged BHs, in spite of it being such a beautiful concept.

    Goodbye for now, Tom McFarland
     
  18. May 4, 2017 #17

    pervect

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    There are two different of gravitational waves reported in the literature, that I'm aware of. There are numerous papers on both, some of the peer-reviewed papers would be https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.061102 and https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.241103. So yes, they've been detected - twice. And the gravitational waves have been attributed to black holes, and not other compact massive objects, based on the characteristics of the recived signal and using General Relativity as a model.

    "Never cross it" is rather a subjective concept, and is probably not entirely accurate depending on one's interpretation of "never." For instance, Schwarzschild coordinates never assign a time coordinate to the event of an infalling test particle reaching the event horizon of a black hole, putting the time coordinate at infinity. Other coordinates would assign finite time coordinates to this event, and hypothetical clocks on test particles falling into the event horizon would have finite readings.

    The binary inspiral case is more complicated in detail than a test particle falling into the Schwarzschild black hole, requiring numerical simultaions to analyze, but the same basic idea is there.

    The issues associated with "never" can be avoided by looking at the General relativistic predictions for a signal emitted from a binary inspiral. Regardless of coordinates use, everyone predicts that the received signal will initially increase in amplitude, as the two black holes approach each other, then reach a peak, then start to die off.

    The logical necessity of the die-off follows from the fact that no signal can escape the event horizon of a black hole. So if we use a viewpoint where we assign finite coordinates to the event (which is helpful in analyzing it!), using this viewpoint we can say that the event "happens", but the signals from the actual crossing of the event horizon can't reach us. But signals are emitted before the event horizon crossing takes place, and we can analyze and measure those.
     
  19. May 4, 2017 #18

    timmdeeg

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    Lets consider the very short time to reach spherical symmetry of the new black hole after the event horizons of the two black holes have just "touched" each other.
    I'm reasoning whether it makes a difference regarding the redshift of gravitational waves if they are emitted locally close to the event horizon or by the extended system consisting of two black holes. The radiation stops after spherical symmetry is reached. Is the radiation emitted in said short time period highly redshifted?
     
  20. May 4, 2017 #19

    PeterDonis

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    I haven't actually addressed this point explicitly. Even leaving aside what has already been said, your argument is based on a misconception. The event horizon of a single black hole, as a surface in spacetime, is not a 2-sphere. It's a 3-surface composed of an infinite series of 2-spheres, and the third "dimension" of the surface is null--heuristically, a curve that links the series of 2-spheres together is a null curve. You can think of this surface as a "hypercylinder" (and visualize it as a cylinder, a stack of circles representing the 2-spheres), as long as you keep in mind that the "axial" dimension of the cylinder is null (lightlike). The description of the horizon as a 2-sphere depends on slicing up the cylinder into a stack of 2-spheres and picking out only one of them. But the horizon, as a surface in spacetime, is not just one of those slices; it's the whole cylinder.

    When you have two black holes merging, there aren't two event horizons becoming one. There is one event horizon, but its "shape" in spacetime (using the heuristic visualization I gave above) is now like a pair of trousers instead of a cylinder. The description "two horizons merge into one" depends on slicing up the pair of trousers into a stack of individual "layers"; then some layers (the ones "before" the merger) consist of two 2-spheres, and other layers (the ones "after" the merger) consist of only one. But the horizon, as a surface in spacetime, is not any single slice; it's the whole pair of trousers.
     
  21. May 4, 2017 #20

    kimbyd

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    Not a man.
     
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