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I Kerr Black Hole Questions

  1. Jul 1, 2016 #1
    I have two questions regarding Kerr black holes, which I am hoping some of you might be able to shed some light on for me.

    1. What is the physical significance / meaning of the inner ergosurface, the one beyond the inner horizon ? If considered as a boundary surface, what would it separate from what ?

    2. With regards to the hypothetical closed time-like curves beyond the Cauchy horizon, how would a test particle need to free-fall in order to describe such a geodesic ? Would it have to fall through the Cauchy horizon in a specific manner to "enter" a CTC, or does it always happen when you cross this boundary ?

    I am aware of the instability issue in the Kerr geometry; the above questions are meant only to help me better understand the geometric structure of this metric.
     
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  3. Jul 2, 2016 #2
    Hi Markus,

    My understanding is that Roy Kerr himself and other professors, like David Wiltshire, believe that the interior geometry is just mathematical science fiction:

    ''The surface beyond the inner horizon is not an ergosurface, but rather the inner ring singularity. I.e., where curvature goes infinite and our geometrical interpretation breaks down.''

    ''Mathematically, a CTC is a point particle going round and round in time. It does does not make sense to think about a real particle "getting there" precisely because of the instability issue; the inner Cauchy horizon is unstable. Therefore, personally I think the inner horizon does not exist in the sense of classical GR. Quantum gravity has to become relevant. The whole interior geometry is simply unphysical. So if you want to imagine the "geometric structure" then you just accept "particles going around in time forever" mathematically. You have to suspend physics and not ask about particles "falling in". As far as I am concerned, the existence of CTCs is like having a map with "here be dragons"... As far as actual physics goes it is all just fantasy.''
     
  4. Jul 2, 2016 #3
    Thank you @tionis, I understand what you are saying, and agree on the instability issue. However, I don't think I can agree with your statement that the inner ergo surface is the same as the ring singularity - mathematically these seem to be quite distinct, unless I am missing something. Consider :

    https___arxiv_org_pdf_0706_0622_pdf.jpg

    ( This is taken from https://arxiv.org/pdf/0706.0622.pdf ). I am fairly clear on the meaning of the other surfaces ( I hope ), but what they call the "inner ergo surface" here eludes me, and I can find no real information about it.
     
  5. Jul 2, 2016 #4

    PeterDonis

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    Look at it by analogy with the outer ergosurface. Outside the outer ergosurface, objects can stay "at rest" (zero angular velocity) at the same altitude above the hole. But between the outer ergosurface and the outer horizon, objects that want to maintain a constant altitude are forced to revolve around the hole at some nonzero angular velocity in the same sense as the hole's rotation. (Also, the range of possible angular velocities gets narrower and narrower, until very close to the horizon any object must be revolving with an angular velocity very close to that of the hole itself. This is often referred to as "frame dragging".)

    The inner ergosurface works the same way: between the inner horizon and the inner ergosurface, objects that want to maintain "altitude" (i.e., distance from the inner horizon, though of course the inner horizon is now "above" them rather than "below" them) are forced to revolve around the hole at nonzero angular velocity, and the closer they are to the inner horizon, the closer the angular velocity has to be to the angular velocity of the hole. But inside the inner ergosurface (i.e., further away from the inner horizon), objects can stay at a constant "altitude" with zero angular velocity.

    They are, but there is still a connection between them: note from your image that on the "equatorial plane" ##z = 0##, the inner ergosurface touches the ring singularity. That means that on the equatorial plane, inside the inner horizon, there are no observers with zero angular velocity--the inner ergoregion (the region between the ergosurface and the horizon, where observers are forced to have nonzero angular velocity) goes all the way down to the singularity at ##r = 0##.

    It can't, if by "free-fall" you mean "free-fall in from outside the hole". The CTCs are geodesics, but they are geodesics that are entirely confined to the region beyond the Cauchy horizon. So there is no way for a test particle to describe such a geodesic if it free-falls in from outside. The test particle would have to be accelerated at some point to change its trajectory from an infalling geodesic (one that comes in from outside the hole) to one of the CTC geodesics.
     
  6. Jul 3, 2016 #5
    That's exactly what I was looking for, thank you :)

    Very interesting, I wasn't aware of this. So basically you'd need a rocket with thrusters, instead of just a free fall particle. Do you happen to know precisely where and how that acceleration would have to take place ? I'm not trying to get at anything in particular, I am really just being curious to learn more.
     
  7. Jul 3, 2016 #6

    PeterDonis

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    Not really, no. I'm not really clear on what the CTC trajectories look like anyway.
     
  8. Jul 3, 2016 #7

    PeterDonis

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    This page on Andrew Hamilton's site on black holes has a very brief discussion and diagram of where the CTC region is in Kerr spacetime:

    http://jila.colorado.edu/~ajsh/insidebh/waterfall.html

    Note that he also mentions an "antiverse", the region you go to if you pass through the ring singularity at ##r = 0##, where the ##r## coordinate is negative.
     
  9. Jul 4, 2016 #8
    Last edited by a moderator: May 8, 2017
  10. Jul 4, 2016 #9
    Sorry, Markus. It was a mistake. You are correct, there is an inner ergosphere and it is not the ring singularity. Excellent questions, BTW. The number of people in the world who have thought about this cannot be more than a few dozen maybe.
     
  11. Jul 6, 2016 #10
    @PeterDonis
    What is meant by "unstable" in this case?
     
  12. Jul 6, 2016 #11

    George Jones

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    There is a weak curvature singularity at the inner Cauchy horizon. Research indicates, however, that this weak curvature singularity might be survivable. For example, if a curvature singularity blows up like a Dirac delta function, then integration produces only a finite contribution to the tidal deformation of an object, which, if the object is robust enough, it can withstand.
     
  13. Jul 6, 2016 #12
    @George Jones Thanks, something like only moving my brain a couple of inches with respect to the rest of my body or being hit by a tiny unstoppable bullet? My impression/memory is that all of these named "surfaces" are basically 2D surfaces separating regions of 3 space and the solution is static. Thus all of the defined manifold sections have time-like extensions that are invariant with respect to free falling observers; i.e. time-like geodesics. Is this right? Or is the invariance with respect to some global reference frame; it's been some time. Does the Cauchy surface locally have a space-like normal direction? Or is the normal time-like; like the region around a Schwarzschild center: all time-like "roads lead to Rome"?? Actually a 2D surface in four space seems like it should have both a time-like and space-like normal.
    I apologize for any silly questions. I have just bought the book mentioned above and will read it. I have meant to get a better understanding of the Kerr solutions and Godel's global solution for some time.
    BTW: I have acquired a Sage program that allows manipulation of Kerr solutions; I haven't tried it but the code looks great.
     
    Last edited: Jul 6, 2016
  14. Jul 6, 2016 #13

    PeterDonis

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    No, this is not correct.

    First, a technical point: even outside the outer horizon (and inside the inner horizon--see below), Kerr spacetime is only stationary, not static. These terms have particular technical definitions in GR, which I won't go into because I don't think they're really necessary for this discussion; they both share the intuitive property that there is a sense in which "space" is not changing with "time".

    Now for the real issue: Kerr spacetime is only stationary outside the outer (event) horizon and inside the inner (Cauchy) horizon. In between the two horizons, Kerr spacetime is not even stationary. That means that there is no way to view this region as having a "space" that does not change with "time". Which also means that there is no way to view the boundaries of this region--the two horizons--as surfaces that just separate different regions of "space".

    No. Only the stationary regions do. See above.

    Neither. Both of the horizons (outer event horizon and inner Cauchy horizon) are null surfaces. That is a big part of the reason why they cannot be viewed as separating "regions of space".

    No. Unlike the case in Schwarzschild spacetime, observers who fall inside the inner horizon of Kerr spacetime--at least in the idealized maximally extended version--are not forced to hit the singularity.
     
  15. Jul 6, 2016 #14
    @PeterDonis
    Thanks!! I had forgotten the fine points/truth; perhaps I never did see the things you mentioned. I will read:)
    Actually, I think a reference page correcting my mistakes would be informative; unless it's already around. I don't like to make an example of myself but it's easy to get turned around in four dimensions and a hyperbolic type metric.
     
  16. Jul 8, 2016 #15
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