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(My book uses the notation "ergosphere" as the hypersurface of static limit, "ergoregion" as the hypervolume within.)

Studying the Kerr BH, I've come to the part about horizons and singularities, and I think I understand the ring singularity, the two horizons, and the (outer) ergosphere, as shown here:

However, I also read about an "inner ergosphere" that I'm not finding in my notes, and I'm not understanding it. Does crossing it cause another "switch" between the timelike and spacelike Killing vectors? What shape does it have? Where is it located?

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  • $\begingroup$ Why is the ergosphere a hypersurface? The event horizon isn't one. $\endgroup$ Commented Sep 24, 2020 at 9:06
  • $\begingroup$ my book uses the notation "ergosphere" as the hypersurface of static limit, "ergoregion" as the hypervolume within. Why is the ergosphere a hypersurface (of which dimension?) and the volume a hypervolume? Is the time involved? If so, how? If the volume is a 4d hypervolume then hypersurface isn't the associated surface 3d? $\endgroup$ Commented Sep 24, 2020 at 10:38
  • $\begingroup$ "ergosphere" as the hypersurface of static limit, "ergoregion" as the hypervolume within What's meant by the static limit? $\endgroup$ Commented Sep 24, 2020 at 14:22
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    $\begingroup$ @ChiralAnomaly I thought the word "hyper" referred to manifolds (surfaces or volumes, which are not equivalent to (anti) Riemann manifolds, as you surely know know)) not perceptible to our eye. I.e., for which only a mathematical description can be given. So I learned again something. Thanx! $\endgroup$ Commented Sep 24, 2020 at 14:28

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If you are learning about the Kerr geometry I strongly recommend Matt Visser's paper The Kerr spacetime: A brief introduction as he manages to pack in all the essential information and still keep it readable.

The location of the ergosphere is derived by considering the trajectory of an observer at constant $r$, $\theta$ and $\phi$, and requiring that this trajectory be time like. With a bit of fiddling this comes down to the condition that $g_{00} < 0$ and using the Boyer Lindquist coordinates this gives us:

$$ r^2 − 2mr + a^2 \cos^2\theta < 0 \tag{1} $$

The ergosphere is where the left side above is equal to zero, and since this is a quadratic in $r$ solving it gives us two ergosphere boundaries:

$$ r_e^\pm = m \pm \sqrt{m^2 - a^2 \cos^2\theta } $$

The radius $r_e^+$ is the outer ergosphere and $r_e^-$ is the inner ergosphere. It is the region in between these two radii where it is impossible to remain at constant $r$, $\theta$ and $\phi$. Note that:

$$ r_e^+ \ge r^+ \ge r^- \ge r_e^- $$

i.e. the outer ergosphere lies outside the outer horizon and the inner ergosphere lies inside the inner horizon. So if you start far from the black hole you can hover at constant $\phi$, then as you reach the outer ergosphere you find this is no longer possible. As you keep going inwards you pass through both horizons then eventually reach a radius where you can once again hover at constant $\phi$, and that's the inner ergosphere.

Or you could look at this the other way round: starting from the centre and moving outwards you can hover at constant $\phi$ until you reach the inner ergosphere and above this you cannot hover at constant $\phi$ until you've passed through the outer ergosphere.

Finally I should add the obligatory caveat that the Kerr metric is an ideal case and is probably unstable inside the outer event horizon. A real rotating black hole would probably have a different, and currently unknown, internal geometry.

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    $\begingroup$ I would add that specifically the region $r \leq r^-$ is definitely unstable in general relativity and thus we have no real idea what happens at $r_e^-$. Things in the region $r \in (r^-,r^+)$ probably look a lot like Kerr unless you firewallin' $\endgroup$
    – Void
    Commented Sep 24, 2020 at 13:01
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    $\begingroup$ Can you please help me, what is the definition of the inner and outer event horizons, how do they differ? $\endgroup$ Commented Sep 24, 2020 at 15:39
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I only now realize you are talking about a Kerr BH instead of a normal one. I wasn't very awake yet, I guess when I wrote the answer.

Of course, in this case, you have to use the Kerr metric. I think this is useful reading insofar the Killing vectors for a Kerr BH are considered.

See also this Wikipedia article (I think you have already looked there though) or this one.

Here's a pic:

enter image description here

From the ergosphere, work can be extracted. Hence the word "ergo", which means "work" in Greece.

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    $\begingroup$ Please, if downvoting, explain the reason for downvoting. $\endgroup$ Commented Sep 24, 2020 at 10:46
  • $\begingroup$ 1- you're right: I often use the word "hypersurface" improperly, as "generalized surface". I am of course meaning just a surface here. 2- I asked about the inner ergosphere, and you're explaining the outer. 3- inside the (outer) ergosphere the timelike Killing vector does become spacelike. It is for this reason that one can extract work from the Kerr BH with the Penrose process. $\endgroup$ Commented Sep 24, 2020 at 12:28
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    $\begingroup$ -1 for a) not answering the question asked, b) confused rambling/complaining about hyper. (FYI, "hyper" does not mean 4d) $\endgroup$
    – TimRias
    Commented Sep 24, 2020 at 13:11
  • $\begingroup$ @MauroGiliberti !0How cold I know that? The just say what you mean 2)The photosphere lies within the ergosphere. I'm talking nowhere about the outside of the ergosphere. Please show me.#0You're right. I thought you were talking about the process which occurs after entering the BH at the event horizon. $\endgroup$ Commented Sep 24, 2020 at 14:07
  • $\begingroup$ @mmeent a)Then I must edit. Thanx! 2) I thought a hypersurface, (Riemann manifold) is a surface we can't imagine ourselves and can only be described mathematically. So, higher than 3d. $\endgroup$ Commented Sep 24, 2020 at 14:48

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