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

LEJ Brouwer says...

What the mathematics has shown is that the interior solution does not
have spherically symmetry about a point, and it is quite unclear as to
what physical matter configuration could give rise to the properties
described by the interior solution - it is certainly not due to a
point mass. Could you explain what matter configuration the interior
solution describes?

This has been discussed by Steven Carlip. Here's an intuitive answer:
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.

The exterior solution does have a spatial spherical symmetry - and
this symmetry only holds as far as the event horizon. The
Schwarzschild coordinates are special because one can transform to
coordinates in which the metric is locally Minkowski by simply
rescaling r and t. Their physical intepretation is therefore clear.

I'm not sure what you mean by that. You can always transform
coordinates to get coordinates that are locally Minkowskian.

The fact that Kruskal and related coordinate systems cannot be
transformed to Schwarzschild (or locally flat) coordinates at the
event horizon indicates that the event horizon itself corresponds to
the 'edge' of the solution manifold.

No, it doesn't. Once again, look at this analogous case.
Suppose we have a spacetime metric that looks like this

ds^2 = X/2m dT^2 - 2m/X dX^2

This 2D spacetime metric has an "event horizon" at X=0, in the
sense that

(1) An object in freefall at rest at X > 0 will asymptotically
approach X = 0 as T --> infinity.
(2) An object that starts off in the region X < 0 can never
reach the region X > 0.
(3) The force F(X) required to keep an object stationary at a
constant value of X goes to infinity as X --> 0.

On the other hand, this metric is easily seen to be just
the Minkowski metric written in unusual accelerated coordinates:
If you let x = square-root(8m X) cosh(T/4m) and let
t = square-root(8m X) sinh(T/4m), then the line element
in terms of x and t is just

ds^2 = dt^2 - dx^2

To see this: dx = square-root(2m/X) cosh(T/4m) dX
+ square-root(X/2m) sinh(T/4m) dT
dt = square-root(2m/X) sinh(T/4m) dX
+ square-root(X/2m) cosh(T/4m) dT

dt^2 - dx^2 = -(2m/X) dX^2 + X/2m dT^2

The event horizon has disappeared by changing the coordinates,
showing that it's a *coordinate* singularity, not a curvature
singularity.

The event horizon for Schwarzschild coordinates has *exactly*
the same character. To see this, start with the full Schwarzschild
metric:

ds^2 = (r-2m)/r dt^2 - r/(r-2m) dr^2 - r^2 dOmega^2

Now, for radially infalling objects, dOmega = 0, so it
reduces to

ds^2 = (r-2m)/r dt^2 - r/(r-2m) dr^2

Now, change coordinates from r to X = r-2m:

ds^2 = X/(X+2m) dt^2 - (X+2m)/X dX^2

Now, when the particle is very close to the event horizon, so
X << 2m, we can approximate X+2m by just 2m, and this becomes,
approximately

ds^2 = X/2m dt^2 - 2m/X dX^2

The four regions of Kruskal spacetime have close analogies
in this 2D metric:

Region I, or the exterior solution of
the Schwarzschild solution, corresponds to the
region x > 0, -x < t < +x.

Region II, or the interior solution, corresponds
to the region t > 0, -t < x < +t.

Region III, or the mirror exterior solution, corresponds
to the region x < 0, -|x| < t < |x|.

Region IV, or the "white hole" solution, corresponds
to the region t < 0, -|t| < x < |t|

The separation between the regions is the
"event horizon", which is defined by X = 0,
which corresponds to the locus of points
in which x^2 = t^2.

Note: any object initially at "rest" in region IV will
eventually "fall" into Region I or Region III. So it's
like a white hole, repelling matter out of it. Any object
initially at rest in region I or region III will eventually
"fall" into region II. So it's like a black hole, drawing
all matter into it.

To suggest that an infalling particle will somehow pass beyond
this edge into the interior manifold which does not even share
the same spherical symmetry as the exterior defies reason.

To suggest *otherwise* defies reason. There is no *reason* for
an infalling particle *not* to pass through the event horizon.
From the point of view of the infalling particle, there is nothing
special about the event horizon that could possibly *prevent* it
from falling through.

If the horizon were not an edge of the manifold, then
there should be a one-to-one (and certainly not one-to-four)
transformation from Kruskal coordinates to locally Minkowski
coordinates (up to a Lorentz transformation).

There is no problem transforming Kruskal coordinates to
locally Minkowski coordinates. I think what you are talking
about is the fact that there are four regions of the Kruskal
spacetime.

On the other hand, what does happen at the event horizon needs to be
investigated, bearing in mind that the exterior solution has two
sheets. My guess is that a particle is reflected off the event
horizon onto the second with space and time directions reversed (i.e.
a radially infalling particle will be reflected outwards travelling
backwards in time - i.e. it will look like a radially infalling
antiparticle, so that the overall process looks like a particle-
antiparticle annihilation occuring at the horizon).

You don't have to *guess*. You can look at what the spacetime
looks like in the neighborhood of the event horizon.

You can't transform from Schwarzschild coordinates to local
Minkowsky coordinates at the event horizon because the Schwarzschild
coordinates are singular there. But I don't see why that's a problem.

Because the Schwarzschild coordinates are easy to interpret
physically.

No, I would say they are definitely *not* easy to interpret,
except in the region in which r is much greater than the event
horizon, in which case it looks like Minkowsky space. Close
to the event horizon, the meaning of Schwarzschild coordinates
is very difficult to interpret.

The fact that they break down at the horizon indicates
that the horizon is the edge of the solution manifold.

No, it doesn't. Once again, look at the metric

ds^2 = X/2m dT^2 - 2m/X dX^2

The fact that this metric breaks down at X=0 doesn't
tell us *anything*, because this is *just* Minkowsky
spacetime written using accelerated coordinates.

Near the event horizon, Schwarzschild coordinates are
accelerated coordinates, just like X and T in the
above 2D case. Kruskal coordinates are the corresponding
well-behaved coordinates.

It's rare that you can describe curved spacetime using a single
nonsingular coordinate system. The more general case requires
several overlapping patches to describe the whole manifold.

Yes, but that is not quite analogous to the situation here.

Yes, it certainly is. That's exactly what's going on here.
The interior solution is a perfectly good solution of the
Einstein field equations. The exterior solution is a perfectly
valid solution. It's just a matter of gluing them together.
The "glue" has to be a coordinate system that is continuous
across the event horizon. Kruskal coordinates work for that
purpose. Or we can use local Minkowski coordinates.

The passage of time and the measurement of distance are physically
meaningful

Yes, but t is *not* the physically measurable time, and r is
not a physically measurable distance. There is nothing physical
about the Schwarzschild coordinates.

and the mapping from Kruskal coordinates to locally
Minkowski coordinates is not possible for Kruskal or Schwarzschild or
any other at the horizon.

That's plainly not true. In terms of Kruskal coordinates, the
metric looks like this:

ds^2 = 32m^3/r exp(-r/2m) (dT^2 + dR^2) - r^2 dOmega^2

where r is an implicit function of T and R.

At the event horizon r=2m, so the metric reduces to

ds^2 = 16 m^2 exp(-1) (dT^2 + dR^2) - 4 m^2 dOmega^2

There is no problem rescaling this to get Minkowski coordinates.

I'm not sure why you're saying "worse". There is nothing
bad about anything you've said so far.

Surely you must be joking, Mr McCullough?

No, I'm not. You really haven't given any reason to be
suspicious of the interior solution *other* than the fact
that it is contrary to your intuitions. From the point of
view of a freefalling test particle, there is nothing
weird going on anywhere in the Schwarzschild spacetime
*except* at the singularity at r=0. There are weird
*global* properties, but nothing weird locally. And
all laws of physics are local. There is nothing local
at the event horizon to prevent someone from falling
through.

It's just that things
aren't as simple as they are in good old Minkowsky spacetime.

[I had already conceded that I made a sign error regarding the
signature of the metric, so I am not going to discuss it]. I do not
understand how you cannot be concerned when you have conceded that the
interior solution does not have a central spherical symmetry.

There is a pointlike singularity in the interior solution, but
it occurs in the *future*, rather than at any particular point
in space.

All of the textbooks and the understanding of modern general
relativists is just wrong.

We had it right pre-1960, and misinterpreted the Kruskal
extension - which is not an extension but rather the sewing of two
physically distinct manifolds with boundary having an SO(3) symmetry.

Look, the whole point of Riemannian geometry is that a general
manifold is created by sewing together pieces that locally look
like sections of good old Euclidean. General Relativity says,
similarly, that every spacetime manifold can be created by
sewing together pieces that locally look like a section of
Minkowski spacetime. Any such sewing job is a legitimate
solution to General Relativity, provided that the field equations
are satisfied.

The Kruskal solution is a perfectly good solution (except possibly
at r=0). In contrast, the Schwarzschild solution is *not* a perfectly
good solution. That's because the Schwarzschild coordinates are
singular at r=2m, even though the curvature tensor is perfectly
well-defined there.

The fact that we continue to accepted the standard picture without
question (and rebuke anyone who dares to question it)

Why do you say that people are accepting it without question?
The reason people changed their minds about it was because
they *studied* it from many different angles, using many different
methods and many different coordinate systems, and they discovered
that they were *wrong* to think of a black hole as a "frozen star".

In contrast, the people who are saying otherwise (like you)
*haven't* studied Kruskal coordinates, haven't studied differential
geometry, haven't studied Rindler coordinates (with their apparent
"event horizon" in ordinary Minkowski spacetime).

is merely a reflection of our sheep-like mentality
and our unfortunate lack of critical thought.

There has been plenty of critical thought about black holes,
and it's still going on. People still don't know how black
holes interact with quantum mechanics. And certainly, the
various approaches to unifying quantum mechanics and General
Relativity may very well show that GR is *wrong* about black
holes. However, there is nothing in your arguments that say
anything about quantum effects. *If* you were arguing that
quantum effects change the picture, then I would not be
arguing against it. You haven't been arguing that. You've
been arguing about *coordinates*. Your argument assumes that
there is something special about Schwarzschild coordinates.
There isn't anything special about them. Any physically
meaningful claim *must* be expressable in whatever coordinate
system you please. The idea that an object slows to a halt
at the event horizon or worse reflects off it may have a
certain plausibility when considering Schwarzschild coordinates,
but it is completely mysterious and unmotivated looked at in
any other coordinate system.

I don't think that *you* have put very much critical thought
into it. You seem completely unable or unwilling to get past
Schwarzschild coordinates. If there is something weird happening
at the event horizon then *demonstrate* it using Kruskal coordinates.

--
Daryl McCullough
Ithaca, NY

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