Question about falling into a Kerr black hole.

[intuitionist1]

Hello, I have a question about rotating (Kerr) black holes. Suppose a
particle falls through the ring singularity of a rotating black hole
and emerges on the second (spacetime)sheet. Now suppose that the
particle subsequently falls through the ring singularity of another
black hole on that second sheet, which could be a significant distance
away from the original black hole.

Will that particle then re-emerge on the original sheet, or an
entirely new sheet? (My understanding, which may be incorrect, is that
the time orientation on the second sheet is opposite to that on the
first sheet, as is the mass - i.e. the emergent particle will appear
to have both opposite mass and be travelling backwards in time).

[Anon E. Mouse]

There are not two sheets. At the schwartzsheild radius the local patch
inverts, it goes vertical at the radius and inverts below that radius.
To and outside observer time might seem reversed, except of course the
outside observer can not observe inside the radius.

This type of inversion can not be smoothly projected onto a flat 2d
map of space-time such as a euclidean representation. Perhaps this is
where the two sheet idea came from.

A possibly better projection would be a false color "double exposure"
with the spherical volume of the inverted space-time projected onto
the image of the space-time plane normal to the outside observer.
Thus, the the space and time that we reasonably know do exist, but can
not see, are represented as a false color, negative image of the space-
time we can see. A ring, but a ring that represents a region of space-
time with a very high mean mass density, but otherwise normal region
of space-time we can not see because it is gravitationally bound in
our frame of reference.

An observer inside the radius observing a particle inside the radius
would see a normal relatively normal motion of the particle.

Emerging from a black hole is predicted by relativity to be at least a
significant technical challenge if not a out right impossibility.


AAG

[Norbert Dragon]

* Anon E. Mouse writes:

> At the schwartzsheild radius ...

Karl Schwarzschild could not have inherited a more fitting name.

German "Schwarz" means "black", "Schild" is "shield" (and is
pronounced similarly).

The Schwarzschild solution shields its interior from observation
and makes it appear black, it is a black hole.

Authors which hyphenate Schwarzs--child seem to be unaware that
Schwarz-schild is a self-documenting name.

--
Superstition brings bad luck

www.itp.uni-hannover.de/~dragon

[intuit...@gmail.com]

What about the case of the fast Kerr solution, where there is a naked
ring singularity? Would there still be obstructions to external
observation in that case, or obstructions to "escaping" from the black
hole? Also, if the time direction is reversed in the "inverted"
spacetime, then would it not be correct to say that the particle would
(if it were being observed from the external spacetime) appear to be
falling into the black hole, rather than escaping from one - though from
the point of view of the particle which has fallen through, it would
indeed be escaping from the black hole? Can I repose my question in this
case, where there are no event horizons? Thanks

[fil...@gmail.com]

On Wednesday, April 18, 2012 9:39:11 AM UTC-7, intuitionist1 wrote:
> Hello, I have a question about rotating (Kerr) black holes. Suppose a
> particle falls through the ring singularity of a rotating black hole
> and emerges on the second (spacetime)sheet.

I wouldn't call it "second" because it's an analytic continuation of "our"
sheet. Also, a particle free falling toward the ring won't fall through it. An
extra push is necessary, like a rocket engine or something. E.g. a particle
merely free-falling along the axis will bounce back (the singularity is
repulsive) and fly back thorugh the horizons - not the same horizons it went
through originally, of course, but through the two horizons surrounding a white
hole in some _yet another_ analytic continuation of the Kerr spacetime. Most of
the time people remember about the extension beyond the ring but they forget
about that other extension involving two more horizons.

It's a bit odd that the ring singularity is repulsive (it is actually very
difficult to hit it, a free-faling particle must be flying entirely in the
equatorial plane for this) - it's only the region between the horizons which is
a one-way street. Schwarzschild's black hole is actually the same, except its
one-way street region extends _all the way_ to the repulsive singular set, so
the singularity is unavoidable there. Schwarzschild is a limiting case of Kerr.

> Now suppose that the
> particle subsequently falls through the ring singularity of another
> black hole on that second sheet, which could be a significant distance
> away from the original black hole.
>
> Will that particle then re-emerge on the original sheet, or an
> entirely new sheet?

Generically a new "sheet" since you cannot cross back the horizons you fell
through. But it may be possible to cook up topologically some weird spacetime
in which you could go back.

> (My understanding, which may be incorrect, is that
> the time orientation on the second sheet is opposite to that on the
> first sheet, as is the mass - i.e. the emergent particle will appear
> to have both opposite mass and be travelling backwards in time).

No, everything is the same. The fact that the ring is repulsive sometimes
tricks people into thinking the region of Kerr spacetime with r < 0 is
"reversed" in some sense. This is not the case. Gravity there is still
attractive, it just contains a repulsive object in it (a naked white hole).

--
Jan