Re: More on the controversy about the Schwarzschild radius and black holes.

On May 14, 9:03 pm, stevendaryl3...@xxxxxxxxx (Daryl McCullough)
wrote:
LEJ Brouwer says...
On May 5, 5:49 am, stevendaryl3...@xxxxxxxxx (Daryl McCullough) wrote:
Imagine sitting on Earth, and surrounded by a spherical shell of
distant stars. Now, imagine attaching powerful rockets to each star
in the shell so that at the same moment (according to the coordinate
system in which the Earth is at rest), the stars are all sent hurtling
towards the earth. The rockets are timed so that all the stars will
arrive simultaneously.

Now, what you can do is add up the masses of all the stars and compute
the Schwarzschild radius for that mass. At some point, the stars will
pass that radius. After that, you and the Earth and all those stars
are inside a black hole.

There still isn't any point mass anywhere except in the *future*.
The predictions of General Relativity are that all the mass inside
the spherical shell will eventually be concentrated into a single
point. That's the singularity, and it exists in the *future*, not
in any particular spatial location.

From the point of view of someone outside the spherical shell of
stars, the geometry will approach that of the Schwarzschild exterior
solution.

[Stuff deleted]

There therefore appear to be three separate regions in the scenario
above:

(i) The exterior of the horizon, which is just an exterior
Schwarzschild solution.

Okay.

(ii) The spherical region in which I sit, which also has a
Schwarzschild exterior-like metric, but with a spherical bounding
surface which shields me from infalling stars, and also prevents
anything from leaving.

I don't think that's correct. Maybe Steve Carlip can explain
to us what this region is like, but it certainly is not like
the Schwarzschild exterior solution. For one thing,
the metric is time-dependent. If you wait long enough, the
curvature will go to infinity (as the stars come crashing
into you).

You are assuming the answer to prove it. Suppose I am the planet Earth
and all the stars start converging upon me. Whatever the metric it is
in my vicinity, it has a spherical symmetry, and will be approximately
Schwarzschild. There will be no net gravitational field acting on me
due to the stars due to the symmetry of the problem. When the event
horizon forms, nothing happens locally to change this sphercal
symmetry. We have already determined that the Schwarzschild interior
solution has an SO(3) symmetry, but it does not have the central
spherical symmetry. Therefore the interior solution cannot coincide
with the metric around me when the horizon forms. Rather, I will
remain quite oblivious of the formation of the horizon, and whatever
the metric was before horizon formation will not change greatly after
it. If it is true as claimed that the infalling stars fall into the
Schwarzschild interior, then that interior cannot coincide with my
little patch of the universe (which I labelled 'A' in my diagram), and
thus must correspond to an independent patch 'C' which only forms upon
formation of the event horizon.


(iii) The interior Schwarzschild solution, complete with temporal
singularity, into which the stars disappear upon crossing the horizon.

I believe that this is the region between the event horizon and
the spherical shell of stars. Anyone who starts off outside the
sphere of stars and falls radially inward will eventually enter
this region.

But once the stars cross the event horizon, these are not spherical
shells in the usual sense, but 'temporal' spheres, which hit the
singularity at a given time relative to passing through the event
horizon. There is no way that this can be identified with the
spherically symmetric region in which I sit. Although the assumption
that the stars continue to fall 'radially inwards' upon crossing the
horizon seems natural, it seems to me to be invalid as the
Schwarzschild interior metric does not describe a space with spatial
spherically symmetry.

In a more realistic collapse scenario where we have a continuous
spherically symmetric distribution of radially infalling matter, this
is what I think would happen:

As the matter collapses inwards, there will eventually be a high
enough mass density to form an event horizon. Any matter external to
the horizon will subsequently be 'diverted' towards the singularity
lying in its future worldline, but this interior region is not the
same as the region which contains the matter which was responsible for
the formation of the horizon.

I think that's right, although the singularity will claim
everyone in freefall, both those initially inside and initially
outside the event horizon.

What mechanism do you propose for those initially inside to fall into
the singularity, when they are physically isolated from each other?
Again this conclusion is due to the invalid assumption that the metric
of those initially inside and the patch described by the Schwarzschild
interior can be identified. They cannot.


Rather, the matter which was interior to
the event horizon has a spherically symmetric distribution, and
becomes isolated from the exterior, and also from the Schwarzschild
interior containing the temporal singularity. Matter can neither enter
nor escape from this region, so it has unusual boundary conditions,
but it remains otherwise well-behaved and is without any kind of
singularity.

I don't think that's correct. If you are inside an inward-rushing
sphere of matter, then you will eventually get crushed by it.

There is no inward-rushing sphere in the sense that you mean - as we
have established, the interior Schwarzschild solution does not
describe a sphere converging on a spatial point, but a sphere
converging on a time relative to the time at which the horizon was
crossed. This implies that the singularity has nothing to do with the
original origin of the problem.


Also, I'm not sure about the claim that this region being
isolated from the exterior. Nothing in this region can
affect the exterior, but I don't see why something in
the exterior couldn't affect this region.

Anything falling in from the exterior goes into region 'C', and not
into region 'A', so nothing from region 'B' can pass into region 'A'.
It is possible, I guess that stuff in region A could end up in region
C if it manages to reach the horizon, but I suspect that the whole of
region A becomes physically disconnected from the full Schwarzschild
solution described by regions B+C as soon as the horizon forms. As I
said, I do not know what the boundary condition on region A would be,
but I am pretty sure that the boundary is impenetrable either way.

I therefore disagree with your claim that all the mass inside the
spherical shell will eventually be concentrated into a single point.

What's to prevent that from happening?

This doesn't happen even in the usual case - I think we have all been
brainwashed into believing that the Schwarzschild singularity is a
spatial point when clearly it is not. As I said, it is not even a
temporal point, though in the rather artificial scenario you describe
where all stars form the event horizon at precisely the same time,
then that might appear to be the case there.

As I mentioned before, and you confirmed, the singularity is not a
'point'. The singularity exists in the (relative) *future*, not at any
particular spatial location or particular time. The mass inside the
spherical shell does not eventually become concentrated into a single
point - rather it lives to evolve happily in a little spherically
symmetric world of its own.

Yes, but a spherically symmetric world that gets smaller all the time.

These are not spatial spheres in the usual sense, so the picture you
paint here is quite misleading.


Here is an (admittedly rather poor attempt at) representation of this
scenario:

C
x
^
|
|
A ----o<---- B

The radially infalling matter is in region B. The trapped region which
arises once the horizon (indicated by 'o') has formed is labelled A,
which contains the matter responsible for creating the horizon in the
first place. The matter crossing the horizon from B falls towards the
singularity at 'x' which lies in region C, corresponding to the
Schwarzschild interior solution. The singularity lies at a fixed time
relative to the time at which an infalling particle crosses the
horizon at 'o'. The region A becomes physically isolated from regions
B and C once the horizon has formed. As I have said before, my guess
is that at the singularity, particles are reflected back in space and
time towards the event horizon and back into the region B, so x acts
like a spacetime reflecting mirror. At a microscopic level, the
interior region A corresponds to an elementary particle, and it
appears to an external observer (i.e. in region B) that particle-
antiparticle annihilations are taking place at the event horizon,
though physically what is happening is that a single particle is being
reflected off of the singularity at 'x' which is acting as a spacetime
mirror.

Let me know if you agree with the above picture. I am not a betting
man, but I will bet that you do not. :)

Well, you have to actually look at the math to know whether one
region is causally disconnected from another region. In particular,
you need to see if there are timelike geodesics connecting the two
regions. If so, then it's possible for objects to travel from one
region to the other.

The picture with the regions I describe makes sense to me. The picture
you describe where you attempt to make region A coincide with the
Schwarzschild interior (which I label 'C') is not borne out by the
form of the interior metric, which is incompatible with the spatial
symmetry in region A.

- Sabbir.

.

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