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LEJ Brouwer science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120

Posted: Tue Jul 18, 2006 2:04 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote:  LEJ Brouwer says...
Are you then claiming that a nonstatic exterior solution exists for
which this constraint does not hold?
Sure, the Kruskal coordinates is an example.

I wasn't using Kruskal coordinates.
Quote:  How can the existence of a nonstatic interior solution be compatible
with the nonexistence of an interior?
I'm saying that your conclusion (that there is no interior) is *false*.
The only thing you can conclude from your argument is that there is no
*static* interior solution.

Not at all  that is only one of a number of things I can conclude. For
example I also conclude that the event horizon coincides with point
mass, and hence the spatial origin. I notice that you did *not*
complain about that.
If you do accept this conclusion, then surely you must accept that this
solution is incompatible with any interior solution, static or not?
Quote:  So the assumption that the metric has
the form
A(r) dt^2  B(r) dr^2  C(r) dOmega^2
with t a timelike coordinate, fails in the interior.
Your argument (or whoever's argument it is) does *not*
prove that there is no interior.

All of this is just a technicality and I am sure you realise it. Even
if I had started with a metric of the form,
A(r,t) dt^2  B(r,t) dr^2  C(r,t) dOmega^2
I would *still* get the same solution for the exterior metric (I am
holding the concise derivation of this fact by Landau & Lifschitz even
as I type), and the rest of the proof would follow completely
unaltered.
I do *not* deny that the interior solution naively solves the field
equations, or that it is nonstatic. However, the interior solution is
incompatible with the fact that the event horizon coincides with the
spatial origin, and therefore cannot be physical (except possibly in
some weird context where there is no accompanying exterior solution).
The argument is due to Leonard Abrams, who passed away quite recently.
Quote:  
Daryl McCullough
Ithaca, NY

 Sabbir. 

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Daryl McCullough science forum Guru
Joined: 24 Mar 2005
Posts: 1167

Posted: Tue Jul 18, 2006 12:50 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



LEJ Brouwer says...
Quote:  Are you then claiming that a nonstatic exterior solution exists for
which this constraint does not hold?

Sure, the Kruskal coordinates is an example.
Quote:  How can the existence of a nonstatic interior solution be compatible
with the nonexistence of an interior?

I'm saying that your conclusion (that there is no interior) is *false*.
The only thing you can conclude from your argument is that there is no
*static* interior solution. So the assumption that the metric has
the form
A(r) dt^2  B(r) dr^2  C(r) dOmega^2
with t a timelike coordinate, fails in the interior.
Your argument (or whoever's argument it is) does *not*
prove that there is no interior.

Daryl McCullough
Ithaca, NY 

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LEJ Brouwer science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120

Posted: Tue Jul 18, 2006 12:30 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Tom Roberts wrote:
Quote:  LEJ Brouwer wrote:
Starting with a general spherically symmetric metric:
ds^2 = A(r)dt^2  B(r)dr^2  C(r) dOmega^2, (eqn 3 of Crothers)
You have implicitly assumed that the manifold is static (i.e. the metric
components are independent of t, your time coordinate). The interior
region of Schw. spacetime is not static. You have thrown the baby out
with the bathwater by making an assumption that is too restrictive.

No I haven't because I only use the exterior solution in my proof, and
that is static. I would get exactly the same exterior solution if I
added the possible tdependence in the metric. Methinks you are
clutching at straws now.
Quote:  The exterior region is static, the interior is not. This has been known
for many decades. <shrug
Tom Roberts

Unless you can pull a rabbit out of the proverbial hat, I am afraid
that the game is over.
Cheers,
Sabbir. 

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Tom Roberts science forum Guru
Joined: 24 Mar 2005
Posts: 1399

Posted: Tue Jul 18, 2006 12:12 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



LEJ Brouwer wrote:
Quote:  Starting with a general spherically symmetric metric:
ds^2 = A(r)dt^2  B(r)dr^2  C(r) dOmega^2, (eqn 3 of Crothers)

You have implicitly assumed that the manifold is static (i.e. the metric
components are independent of t, your time coordinate). The interior
region of Schw. spacetime is not static. You have thrown the baby out
with the bathwater by making an assumption that is too restrictive.
The exterior region is static, the interior is not. This has been known
for many decades. <shrug>
Tom Roberts 

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Tom Roberts science forum Guru
Joined: 24 Mar 2005
Posts: 1399

Posted: Tue Jul 18, 2006 12:04 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote:  The *only* way to know for sure what's a real singularity and
what isn't is by seeing what phenomena are present in *every*
coordinate system.

Hmmm. Much better is to compute in a coordinateindependent way: i.e.
compute an invariant. There are a number of curvature invariants that
can be formed from the Riemann curvature tensor, and if any of them are
unbounded as one approaches some locus, then that locus is singular.
Conversely, if they are all regular in some region, then there is no
curvature singularity in that region.
They are all regular in any region of Schwarzschild spacetime that does
not include the locus r=0. This includes any such region containing the
horizon.
Tom Roberts 

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LEJ Brouwer science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120

Posted: Mon Jul 17, 2006 11:53 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote:  LEJ Brouwer says...
Daryl McCullough wrote:
These complications are a symptom of the fact that the 'interior'
region is not physical.
No, they are not. As you point out, the Schwarzchild coordinates
are perfectly good solutions to the Einstein field equations both
in the interior region and the exterior region. They only fail
at the horizon. There is no more reason to dismiss the interior
solution than there is to dismiss the exterior solution.
There is, because the radial direction 'r' which is by definition
spacelike in the Einstein field equations when expressed in spherical
coordinates, is timelike in the interior Schwarzschild solution.
The Einstein field equations don't say anything about whether
r is spacelike.
But if you don't like using r for a timelike
coordinate, then use different coordinates. Make the substitutions
r > Lt'
t > r'
Then the metric looks like this in the interior
ds^2 = (Lt')/t' dr'^2  (Lt')^2 dOmega^2 + t'/(Lt') dt'^2
This is a metric that is valid in the patch
infinity < t' < L
infinity < r' < +infinity
Everywhere in the patch, t' is timelike, and r' is spacelike.
What's the problem?
Here goes:
Starting with a general spherically symmetric metric:
ds^2 = A(r)dt^2  B(r)dr^2  C(r) dOmega^2, (eqn 3 of Crothers)
No. That isn't the most general spherically symmetric
metric. That is the most general *static* spherically
symmetric metric. The most general metric (with
coordinates chosen so that there are no crossterms
such as dr dt) looks like this:
ds^2 = A(r,t) dt^2  B(r,t) dr^2  C(r,t) dOmega^2
where A,B,C>0, and applying the same transformation r* = sqrt{C} used
by Hilbert (we write r* instead of r to avoid confusion with r above),
the general solution is,
ds^2 = (1  2m/r*) dt^2  (r* / (r*  2m)) dr*^2  r*^2 dOmega^2 (eqn
7 of Crothers).
That's for static solutions. You can also get a solution by
letting A,B and C be functions of t (rather than r) and you
get the interior Schwarzchild solution:
ds^2 =  (t/(t2m)) dt^2 + (12m/t) dr^2  t^2 dOmega^2
= (t/(2mt)) dt^2  (2m/t  1) dr^2  t^2 dOmega^2
or, if you let t' = 2mt
ds^2 = ((2mt')/t') dt'^2  (t'/(2mt')) dr^2  (2mt')^2 dOmega^2
Substituting sqrt{C} for r* this becomes,
ds^2 = (1  2m/sqrt{C}) dt^2  (sqrt{C} / (sqrt{C}  2m)) (C'^2/4C)
dr^2  C dOmega^2 (eqn 7 of Crothers).
[C' = dC/dr here]. This is just the usual Schwarzschild solution, but
note that r* = sqrt{C} does not need to lie in the range [0,Infty] as
Hilbert incorrectly assumed.
Now, if the coordinate associated with the point mass is r = sqrt{C} =
r0 (which may if you like be chosen such that r0 = 0  it makes no
difference), then we can define the 'radial distance' from the mass at
r0 to any r >= r0 to be the integral from r0 to r of sqrt{g_rr},
R = int sqrt{g_rr} dr from r0 to r (eqn 11 of Crothers)
where g_rr is just the radial component of the metric:
g_rr =  (sqrt{C} / (sqrt{C}  2m)) (C'^2/4C) (eqn 12 of Crothers)
That's perfectly fine under the assumption that the metric
is static throughout the region. That *won't* be the case.
There is no static solution that covers the entire manifold.
If you make the solution outside the event horizon static,
then the solution inside the event horizon will *not* be.
...Since the radial distance must be real, this means that
the range of possible values for r* = sqrt{C(r)} is given
by r* >= 2m.
Yes, for a *static* (timeindependent) metric. The interior
is *not* static.

Are you then claiming that a nonstatic exterior solution exists for
which this constraint does not hold? If not, then could you please
explain what exactly your argument is? How can the existence of a
nonstatic interior solution be compatible with the nonexistence of an
interior?
Quote:  
Daryl McCullough
Ithaca, NY

 Sabbir. 

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Daryl McCullough science forum Guru
Joined: 24 Mar 2005
Posts: 1167

Posted: Mon Jul 17, 2006 10:43 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



LEJ Brouwer says...
Quote: 
Daryl McCullough wrote:
These complications are a symptom of the fact that the 'interior'
region is not physical.
No, they are not. As you point out, the Schwarzchild coordinates
are perfectly good solutions to the Einstein field equations both
in the interior region and the exterior region. They only fail
at the horizon. There is no more reason to dismiss the interior
solution than there is to dismiss the exterior solution.
There is, because the radial direction 'r' which is by definition
spacelike in the Einstein field equations when expressed in spherical
coordinates, is timelike in the interior Schwarzschild solution.

The Einstein field equations don't say anything about whether
r is spacelike.
But if you don't like using r for a timelike
coordinate, then use different coordinates. Make the substitutions
r > Lt'
t > r'
Then the metric looks like this in the interior
ds^2 = (Lt')/t' dr'^2  (Lt')^2 dOmega^2 + t'/(Lt') dt'^2
This is a metric that is valid in the patch
infinity < t' < L
infinity < r' < +infinity
Everywhere in the patch, t' is timelike, and r' is spacelike.
What's the problem?
Quote:  Here goes:
Starting with a general spherically symmetric metric:
ds^2 = A(r)dt^2  B(r)dr^2  C(r) dOmega^2, (eqn 3 of Crothers)

No. That isn't the most general spherically symmetric
metric. That is the most general *static* spherically
symmetric metric. The most general metric (with
coordinates chosen so that there are no crossterms
such as dr dt) looks like this:
ds^2 = A(r,t) dt^2  B(r,t) dr^2  C(r,t) dOmega^2
Quote:  where A,B,C>0, and applying the same transformation r* = sqrt{C} used
by Hilbert (we write r* instead of r to avoid confusion with r above),
the general solution is,
ds^2 = (1  2m/r*) dt^2  (r* / (r*  2m)) dr*^2  r*^2 dOmega^2 (eqn
7 of Crothers).

That's for static solutions. You can also get a solution by
letting A,B and C be functions of t (rather than r) and you
get the interior Schwarzchild solution:
ds^2 =  (t/(t2m)) dt^2 + (12m/t) dr^2  t^2 dOmega^2
= (t/(2mt)) dt^2  (2m/t  1) dr^2  t^2 dOmega^2
or, if you let t' = 2mt
ds^2 = ((2mt')/t') dt'^2  (t'/(2mt')) dr^2  (2mt')^2 dOmega^2
Quote:  Substituting sqrt{C} for r* this becomes,
ds^2 = (1  2m/sqrt{C}) dt^2  (sqrt{C} / (sqrt{C}  2m)) (C'^2/4C)
dr^2  C dOmega^2 (eqn 7 of Crothers).
[C' = dC/dr here]. This is just the usual Schwarzschild solution, but
note that r* = sqrt{C} does not need to lie in the range [0,Infty] as
Hilbert incorrectly assumed.
Now, if the coordinate associated with the point mass is r = sqrt{C} =
r0 (which may if you like be chosen such that r0 = 0  it makes no
difference), then we can define the 'radial distance' from the mass at
r0 to any r >= r0 to be the integral from r0 to r of sqrt{g_rr},
R = int sqrt{g_rr} dr from r0 to r (eqn 11 of Crothers)
where g_rr is just the radial component of the metric:
g_rr =  (sqrt{C} / (sqrt{C}  2m)) (C'^2/4C) (eqn 12 of Crothers)

That's perfectly fine under the assumption that the metric
is static throughout the region. That *won't* be the case.
There is no static solution that covers the entire manifold.
If you make the solution outside the event horizon static,
then the solution inside the event horizon will *not* be.
Quote:  ...Since the radial distance must be real, this means that
the range of possible values for r* = sqrt{C(r)} is given
by r* >= 2m.

Yes, for a *static* (timeindependent) metric. The interior
is *not* static.

Daryl McCullough
Ithaca, NY 

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LEJ Brouwer science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120

Posted: Mon Jul 17, 2006 9:37 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote:  LEJ Brouwer says...
Daryl McCullough wrote:
No justification must be used for changing coordinates. General
Relativity was designed to work in any coordinate system.
Even with only piecewise continous coordinate changes?
That's the usual case: A coordinate system is only valid in
a small region, or patch, of spacetime, and a coordinate transformation
between two coordinate systems is only meaningful where the
two patches overlap
when clearly it does not satisfy the usual requirements in differential
geometry of a continuous map between coordinate patches.
That's not true.
Why not?
Because it *does* satisfy the usual requirements in differential
geometry of a continuous map between coordinate patches. There
are three regions that are relevant here:
1. Exterior to the event horizon.
2. At the event horizon.
3. Interior to the event horizon (but r > 0).
The Schwarzchild coordinates are valid in regions 1 and 3,
and the Kruskal coordinates are valid in all three regions.
The map between Kruskal and Schwarzchild is nonsingular in
region 1, and it's nonsingular in region 3. The map isn't
valid in region 2, but that is irrelevant, since the
Schwarzchild coordinates are not valid in region 2, either.

Okay, I am going to eat humble pie here, and admit that I got a bit
carried away with my coordinate bashing and slipped up  sorry for
that.
Quote:  Ultimately it seems that you must have decided that somehow the entire
Kruskal coordinate system is okay,
That's because there are no singularities or other weirdness
in the Kruskal coordinates. Why would they *not* be okay?
Okay there are no singularities or other 'weirdness' in the naive
mathematical sense but they do have serious problems in a physical
sense because the whole locus R^2=T^2 corresponds to the point mass.
Why in the world do you believe that? That's completely wrong.

No it's not (see below).
Quote:  This is not at all obvious just by looking at the Kruskal coordinates
Not only is it not obvious, it is false.

No it's not.
Quote:  themselves which give the _illusion_ that welldefined geodesics exist
entering into a region interior to the horizon.
Once again, you've got things exactly backwards. The Schwarzchild
coordinates give the _illusion_ that there is a problem with
geodesics passing through the event horizon, but there is no
such problem.

No, we are both right (and wrong) in different senses. Yes, there would
be a smooth
mathematical transition from exterior to interior via the maps you to
describe,
were it not for the case that there is no interior region.
Quote:  If you think that there is a problem, show it using the mathematics
of the Kruskal coordinates.

Well, let me do so for Schwarzschild coordinates first, and take things
from there.
Quote:  You seem to be arguing that they are not okay, because they can be
transformed to Schwarzchild coordinates, where there is a singularity
at the event horizon. But you are *simultaneously* arguing that that
transformation is suspect (since it is discontinuous). If you disallow
the transformation, then doesn't your argument for there being a
problem in the Kruskal coordinates disappear? You are being inconsistent
here.
No, I am not really arguing that (though I would be concerned as to why
the interior Kruskal coordinates seem okay but the interior
Schwarzschild coordinates are not)
There is nothing wrong with Schwarzchild coordinates in the
interior region. It is just that for Schwarzchild coordinates,
the interior region and the exterior region are two *disjoint*
patches. They don't overlap at all. In contrast, for Kruskal
coordinate, the interior and exterior regions are part of the
*same* patch.
but most textbooks try to argue
(and indeed you stated this earlier) that there is a nice smooth
mapping of Kruskal geodesics into Schwarzschild geodesics across the
event horizon.
No, I argue that there are two smooth, continuous maps:
1. A map from the exterior region's Schwarzchild coordinates
to the Kruskal coordinates. This map becomes singular at the
horizon, but the horizon is *not* part of the exterior region.
2. A map from the interior region's Schwarzchild coordinates
to the Kruskal coordinates. This map is also singular at the
horizon.
There is no map that works *across* the horizon, because the
Schwarzchild coordinates are not *valid* at the horizon.
The fact that there is not (as you yourself have figured
out) should suggest to you the presence of a serious problem
 after all, the interior Schwarzschild solution also naively
solves the vacuum field equations,
Yes. They are *perfectly* good coordinates in the interior
region. They only fail to be good coordinates at the
horizon.
so why are there so many insurmountable complications
in the mapping from Kruskal coordinates to Schwarzschild
coordinates?
It's been explained to you many times. The map is discontinuous
because the Schwarzchild coordinates are singular at the horizon.
You can't have a continuous map between a singular coordinate
system and a nonsingular coordinate system.

Okay.
Quote:  These complications are a symptom of the fact that the 'interior'
region is not physical.
No, they are not. As you point out, the Schwarzchild coordinates
are perfectly good solutions to the Einstein field equations both
in the interior region and the exterior region. They only fail
at the horizon. There is no more reason to dismiss the interior
solution than there is to dismiss the exterior solution.

There is, because the radial direction 'r' which is by definition
spacelike in the Einstein field equations when expressed in spherical
coordinates, is timelike in the interior Schwarzschild solution.
Quote:  Give an argument for there being some problem at the event horizon
*using* Kruskal coordinates. No fair using a transformation to some
other coordinate system if the transformation itself is singular.
What problems do you see with Kruskal metric?
The Kruskal coordinates are misleading in a way that the Schwarzschild
coordinates are not in that they make one believe that an infalling
particle moves continuously into a region that does not physically
exist
You are making a circular argument. I want you to *prove*, using
the Kruskal coordinates, that there is no interior region. You
are assuming what you are trying to prove.
(i.e the upper and lower quadrants of the Kruskal solution as it
is usually drawn). The proof of the nonexistence of this interior
region in the Schwarzschild coordinates is based upon the fact that the
event horizon coincides with the position of the point mass
That is false. The Schwarzchild solution has *no* mass
at all. It's a *vacuum* solution of the field equations.
There is no mass anywhere in the Schwarzchild spacetime
except the singularity at the origin.
Indeed the proof is actually quite simple to follow. Certainly it
should be possible to translate this proof into Kruskal coordinates,
Then do it.
... but what is the point of doing this...?
To prove that your conclusions are not artifacts of using
an illbehaved coordinate system.

Daryl McCullough
Ithaca, NY

Here goes:
Starting with a general spherically symmetric metric:
ds^2 = A(r)dt^2  B(r)dr^2  C(r) dOmega^2, (eqn 3 of Crothers)
where A,B,C>0, and applying the same transformation r* = sqrt{C} used
by Hilbert (we write r* instead of r to avoid confusion with r above),
the general solution is,
ds^2 = (1  2m/r*) dt^2  (r* / (r*  2m)) dr*^2  r*^2 dOmega^2 (eqn
7 of Crothers).
Substituting sqrt{C} for r* this becomes,
ds^2 = (1  2m/sqrt{C}) dt^2  (sqrt{C} / (sqrt{C}  2m)) (C'^2/4C)
dr^2  C dOmega^2 (eqn 7 of Crothers).
[C' = dC/dr here]. This is just the usual Schwarzschild solution, but
note that r* = sqrt{C} does not need to lie in the range [0,Infty] as
Hilbert incorrectly assumed.
Now, if the coordinate associated with the point mass is r = sqrt{C} =
r0 (which may if you like be chosen such that r0 = 0  it makes no
difference), then we can define the 'radial distance' from the mass at
r0 to any r >= r0 to be the integral from r0 to r of sqrt{g_rr},
R = int sqrt{g_rr} dr from r0 to r (eqn 11 of Crothers)
where g_rr is just the radial component of the metric:
g_rr =  (sqrt{C} / (sqrt{C}  2m)) (C'^2/4C) (eqn 12 of Crothers)
Integrating, this (I use the substitution u = sqrt{C} followed by the
substitution u = 2m sec^2 x), we find,
R = sqrt{sqrt{C}*(sqrt{C}  2m)} + 2m ln (sqrt{sqrt{C}}+sqrt{(sqrt{C}
 2m)}) / sqrt{2m}   sqrt{sqrt{C0}*(sqrt{C0}  2m)} + 2m ln
(sqrt{sqrt{C0}}+sqrt{(sqrt{C0}  2m)}) / sqrt{2m} 
[C0 = C(r0) here], where we have fixed the constant of integration by
applying the boundary condition that R>0 as r>r0. Since the radial
distance must be real, this means that the range of possible values for
r* = sqrt{C(r)} is given by r* >= 2m. In particular, the origin must
correspond to the minimum of this range, i.e r*(r0) = sqtr{C0} = 2m, so
that C0 = 4m^2. But r* = 2m is the position of the event horizon, i.e.
the event horizon coincides with the point particle which is the radial
origin. Moreover, r* < 2m is unphysical.
Plugging sqrt{C0} = 2m into the above equation, we find the following
equation for the radial distance of a point at r from the origin,
R(r) = sqrt{sqrt{C}*(sqrt{C}  2m)} + 2m ln
(sqrt{sqrt{C}}+sqrt{(sqrt{C}  2m)} / sqrt{2m} (eqn 14 of Crothers)
Again. one cannot consider radii with r* < 2m, because these radii make
no physical sense, as they would have to have a smaller radius than
radius corresponding to the position of the point mass which defines
the origin of our space.
There is therefore NO interior solution for r* < 2m, because r* < 2m
does not physically exist.
As I said, it should be possible to redo this calculation in Kruskal
coordinates  this would require applying an appropriate coordinate
transformation in the integral defining the radial distance R. Whatever
this coordinate transformation is, it can simply be inverted again in
order to calculate the integral and apply the boundary conditions, and
the above results will once again be obtained.
Let me know if you agree or disagree with this analysis.
 Sabbir. 

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Jan Bielawski science forum Guru
Joined: 08 May 2005
Posts: 388

Posted: Mon Jul 17, 2006 4:39 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



LEJ Brouwer wrote:
Quote: 
I took a look at the thread you referred to above, and I have to say
that I am surprised  it seems that you yourself have pointed out first
hand some of the numerous problems which occur at the event horizon.
What should be very clear from your own analysis is that the mapping
from Schwarzschild to Kruskal (or vice versa) is discontinuous at the
event horizon  this is what gives rise to the analogue of "dividing
infinity by infinity to get something finite". A great deal of
handwaving and fudging and brushing under the carpet and excusemaking
goes on to try to justify the use of such a transformation, when
clearly it does not satisfy the usual requirements in differential
geometry of a continuous map between coordinate patches.

What you wrote above shows your confusion over _elementary_ manifold
theory (at this level it's not even "differential geometry" yet). I
gave you a 1dimensional analogy some time ago that you haven't
commented upon:
1. Let your manifold be the real line R^1 with the identity chart. No
singularities (Yes/No)
2. Consider the open submanifold M=R^1{0}. It consists of two
disconnected regions (Yes/No)
3. Consider a coordinate chart on M: phi(x)=1/x. These coordinates are
singular at the origin (Yes/No)
4. There is a diffeomorphism between M and the submanifold R^1{0} of
R^1. In the local charts this diffeomorphism is F(x)=1/x. This F is
singular at the origin (Yes/No)
5. The image of the diffeomorphism F is an open submanifold of the
singularityless manifold R^1 (namely, the open subset R^1{0})
(Yes/No)
6. Therefore, the manifold M had no singularity at the origin to begin
with (Yes/No).
Again, I highly recommend the 1st volume of Spivak.
Quote:  Ultimately it seems that you must have decided that somehow the entire
Kruskal coordinate system is okay, and that there is something wrong
with the Schwarzschild coordinates,

Well, there is nothing to "decide", it simply _is_ OK by the definition
of: manifold, coordinate chart, singularities of smooth mappings,
singularities of manifolds.

Jan Bielawski 

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Daryl McCullough science forum Guru
Joined: 24 Mar 2005
Posts: 1167

Posted: Mon Jul 17, 2006 4:26 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



LEJ Brouwer says...
Quote:  Daryl McCullough wrote:
No justification must be used for changing coordinates. General
Relativity was designed to work in any coordinate system.
Even with only piecewise continous coordinate changes?

That's the usual case: A coordinate system is only valid in
a small region, or patch, of spacetime, and a coordinate transformation
between two coordinate systems is only meaningful where the
two patches overlap
Quote:  when clearly it does not satisfy the usual requirements in differential
geometry of a continuous map between coordinate patches.
That's not true.
Why not?

Because it *does* satisfy the usual requirements in differential
geometry of a continuous map between coordinate patches. There
are three regions that are relevant here:
1. Exterior to the event horizon.
2. At the event horizon.
3. Interior to the event horizon (but r > 0).
The Schwarzchild coordinates are valid in regions 1 and 3,
and the Kruskal coordinates are valid in all three regions.
The map between Kruskal and Schwarzchild is nonsingular in
region 1, and it's nonsingular in region 3. The map isn't
valid in region 2, but that is irrelevant, since the
Schwarzchild coordinates are not valid in region 2, either.
Quote:  Is there a new field of mathematics called "nondifferential
geometry" which I am unaware of?

The field is called "differential geometry", and according
to that field of mathematics, there is nothing singular at
the event horizon.
Quote:  Ultimately it seems that you must have decided that somehow the entire
Kruskal coordinate system is okay,
That's because there are no singularities or other weirdness
in the Kruskal coordinates. Why would they *not* be okay?
Okay there are no singularities or other 'weirdness' in the naive
mathematical sense but they do have serious problems in a physical
sense because the whole locus R^2=T^2 corresponds to the point mass.

Why in the world do you believe that? That's completely wrong.
Quote:  This is not at all obvious just by looking at the Kruskal coordinates

Not only is it not obvious, it is false.
Quote:  themselves which give the _illusion_ that welldefined geodesics exist
entering into a region interior to the horizon.

Once again, you've got things exactly backwards. The Schwarzchild
coordinates give the _illusion_ that there is a problem with
geodesics passing through the event horizon, but there is no
such problem.
If you think that there is a problem, show it using the mathematics
of the Kruskal coordinates.
Quote:  You seem to be arguing that they are not okay, because they can be
transformed to Schwarzchild coordinates, where there is a singularity
at the event horizon. But you are *simultaneously* arguing that that
transformation is suspect (since it is discontinuous). If you disallow
the transformation, then doesn't your argument for there being a
problem in the Kruskal coordinates disappear? You are being inconsistent
here.
No, I am not really arguing that (though I would be concerned as to why
the interior Kruskal coordinates seem okay but the interior
Schwarzschild coordinates are not)

There is nothing wrong with Schwarzchild coordinates in the
interior region. It is just that for Schwarzchild coordinates,
the interior region and the exterior region are two *disjoint*
patches. They don't overlap at all. In contrast, for Kruskal
coordinate, the interior and exterior regions are part of the
*same* patch.
Quote:  but most textbooks try to argue
(and indeed you stated this earlier) that there is a nice smooth
mapping of Kruskal geodesics into Schwarzschild geodesics across the
event horizon.

No, I argue that there are two smooth, continuous maps:
1. A map from the exterior region's Schwarzchild coordinates
to the Kruskal coordinates. This map becomes singular at the
horizon, but the horizon is *not* part of the exterior region.
2. A map from the interior region's Schwarzchild coordinates
to the Kruskal coordinates. This map is also singular at the
horizon.
There is no map that works *across* the horizon, because the
Schwarzchild coordinates are not *valid* at the horizon.
Quote:  The fact that there is not (as you yourself have figured
out) should suggest to you the presence of a serious problem
 after all, the interior Schwarzschild solution also naively
solves the vacuum field equations,

Yes. They are *perfectly* good coordinates in the interior
region. They only fail to be good coordinates at the
horizon.
Quote:  so why are there so many insurmountable complications
in the mapping from Kruskal coordinates to Schwarzschild
coordinates?

It's been explained to you many times. The map is discontinuous
because the Schwarzchild coordinates are singular at the horizon.
You can't have a continuous map between a singular coordinate
system and a nonsingular coordinate system.
Quote:  These complications are a symptom of the fact that the 'interior'
region is not physical.

No, they are not. As you point out, the Schwarzchild coordinates
are perfectly good solutions to the Einstein field equations both
in the interior region and the exterior region. They only fail
at the horizon. There is no more reason to dismiss the interior
solution than there is to dismiss the exterior solution.
Quote:  Give an argument for there being some problem at the event horizon
*using* Kruskal coordinates. No fair using a transformation to some
other coordinate system if the transformation itself is singular.
What problems do you see with Kruskal metric?
The Kruskal coordinates are misleading in a way that the Schwarzschild
coordinates are not in that they make one believe that an infalling
particle moves continuously into a region that does not physically
exist

You are making a circular argument. I want you to *prove*, using
the Kruskal coordinates, that there is no interior region. You
are assuming what you are trying to prove.
Quote:  (i.e the upper and lower quadrants of the Kruskal solution as it
is usually drawn). The proof of the nonexistence of this interior
region in the Schwarzschild coordinates is based upon the fact that the
event horizon coincides with the position of the point mass

That is false. The Schwarzchild solution has *no* mass
at all. It's a *vacuum* solution of the field equations.
There is no mass anywhere in the Schwarzchild spacetime
except the singularity at the origin.
Quote:  Indeed the proof is actually quite simple to follow. Certainly it
should be possible to translate this proof into Kruskal coordinates,

Then do it.
Quote:  ... but what is the point of doing this...?

To prove that your conclusions are not artifacts of using
an illbehaved coordinate system.

Daryl McCullough
Ithaca, NY 

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LEJ Brouwer science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120

Posted: Mon Jul 17, 2006 3:23 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote:  LEJ Brouwer says...
I took a look at the thread you referred to above, and I have to say
that I am surprised  it seems that you yourself have pointed out first
hand some of the numerous problems which occur at the event horizon.
My posts were not about problems with the event horizon, they
were about problems relating the Kruskal coordinates to the
Schwarzschild coordinates. That is a question about coordinates,
not about anything physically measurable (you can't measure
the value of the parameter t  you can only measure it up
to an arbitrary constant).
What should be very clear from your own analysis is that the mapping
from Schwarzschild to Kruskal (or vice versa) is discontinuous at the
event horizon  this is what gives rise to the analogue of "dividing
infinity by infinity to get something finite".
Yes, the *relationship* between the two coordinate systems is
discontinuous at the boundary. If you have a nonsingular coordinate
system, and a singular coordinate system, then the relationship
between them has to be discontinuous.
A great deal of handwaving and fudging and brushing under the
carpet and excusemaking goes on to try to justify the use of such
a transformation
No justification must be used for changing coordinates. General
Relativity was designed to work in any coordinate system.

Even with only piecewise continous coordinate changes?
Quote:  when clearly it does not satisfy the usual requirements in differential
geometry of a continuous map between coordinate patches.
That's not true.

Why not? Is there a new field of mathematics called "nondifferential
geometry" which I am unaware of?
Quote:  Ultimately it seems that you must have decided that somehow the entire
Kruskal coordinate system is okay,
That's because there are no singularities or other weirdness
in the Kruskal coordinates. Why would they *not* be okay?

Okay there are no singularities or other 'weirdness' in the naive
mathematical sense but they do have serious problems in a physical
sense because the whole locus R^2=T^2 corresponds to the point mass.
This is not at all obvious just by looking at the Kruskal coordinates
themselves which give the _illusion_ that welldefined geodesics exist
entering into a region interior to the horizon.
Quote:  You seem to be arguing that they are not okay, because they can be
transformed to Schwarzchild coordinates, where there is a singularity
at the event horizon. But you are *simultaneously* arguing that that
transformation is suspect (since it is discontinuous). If you disallow
the transformation, then doesn't your argument for there being a
problem in the Kruskal coordinates disappear? You are being inconsistent
here.

No, I am not really arguing that (though I would be concerned as to why
the interior Kruskal coordinates seem okay but the interior
Schwarzschild coordinates are not)  but most textbooks try to argue
(and indeed you stated this earlier) that there is a nice smooth
mapping of Kruskal geodesics into Schwarzschild geodesics across the
event horizon. The fact that there is not (as you yourself have figured
out) should suggest to you the presence of a serious problem  after
all, the interior Schwarzschild solution also naively solves the vacuum
field equations, so why are there so many insurmountable complications
in the mapping from Kruskal coordinates to Schwarzschild coordinates?
These complications are a symptom of the fact that the 'interior'
region is not physical.
Quote:  Give an argument for there being some problem at the event horizon
*using* Kruskal coordinates. No fair using a transformation to some
other coordinate system if the transformation itself is singular.
What problems do you see with Kruskal metric?

The Kruskal coordinates are misleading in a way that the Schwarzschild
coordinates are not in that they make one believe that an infalling
particle moves continuously into a region that does not physically
exist (i.e the upper and lower quadrants of the Kruskal solution as it
is usually drawn). The proof of the nonexistence of this interior
region in the Schwarzschild coordinates is based upon the fact that the
event horizon coincides with the position of the point mass, and this
proof does not need to make use of the interior solution/metric at all.
Indeed the proof is actually quite simple to follow. Certainly it
should be possible to translate this proof into Kruskal coordinates,
where it will show you that the upper and lower quadrants are not
physical, but what is the point of doing this, when the boundary
condition being applied (i.e. one cannot have negative radial
positions) is more naturally suited to the Schwarzschild solution? If
you don't believe the proof in (exterior) Schwarzschild coordinates,
you are certainly not going to believe it in Kruskal coordinates.
Quote: 

Daryl McCullough
Ithaca, NY

 Sabbir. 

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Daryl McCullough science forum Guru
Joined: 24 Mar 2005
Posts: 1167

Posted: Mon Jul 17, 2006 2:16 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



LEJ Brouwer says...
Quote:  I took a look at the thread you referred to above, and I have to say
that I am surprised  it seems that you yourself have pointed out first
hand some of the numerous problems which occur at the event horizon.

My posts were not about problems with the event horizon, they
were about problems relating the Kruskal coordinates to the
Schwarzschild coordinates. That is a question about coordinates,
not about anything physically measurable (you can't measure
the value of the parameter t  you can only measure it up
to an arbitrary constant).
Quote:  What should be very clear from your own analysis is that the mapping
from Schwarzschild to Kruskal (or vice versa) is discontinuous at the
event horizon  this is what gives rise to the analogue of "dividing
infinity by infinity to get something finite".

Yes, the *relationship* between the two coordinate systems is
discontinuous at the boundary. If you have a nonsingular coordinate
system, and a singular coordinate system, then the relationship
between them has to be discontinuous.
Quote:  A great deal of handwaving and fudging and brushing under the
carpet and excusemaking goes on to try to justify the use of such
a transformation

No justification must be used for changing coordinates. General
Relativity was designed to work in any coordinate system.
Quote:  when clearly it does not satisfy the usual requirements in differential
geometry of a continuous map between coordinate patches.

That's not true.
Quote:  Ultimately it seems that you must have decided that somehow the entire
Kruskal coordinate system is okay,

That's because there are no singularities or other weirdness
in the Kruskal coordinates. Why would they *not* be okay?
You seem to be arguing that they are not okay, because they can be
transformed to Schwarzchild coordinates, where there is a singularity
at the event horizon. But you are *simultaneously* arguing that that
transformation is suspect (since it is discontinuous). If you disallow
the transformation, then doesn't your argument for there being a
problem in the Kruskal coordinates disappear? You are being inconsistent
here.
Give an argument for there being some problem at the event horizon
*using* Kruskal coordinates. No fair using a transformation to some
other coordinate system if the transformation itself is singular.
What problems do you see with Kruskal metric?

Daryl McCullough
Ithaca, NY 

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LEJ Brouwer science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120

Posted: Mon Jul 17, 2006 12:44 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Quote:  ....But as is
well known....

Well, there's no arguing with that then. I give up. You win. 

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LEJ Brouwer science forum Guru Wannabe
Joined: 07 May 2005
Posts: 120

Posted: Mon Jul 17, 2006 12:13 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote:  LEJ Brouwer says...
Daryl McCullough wrote:
Let me try another tack with you. Look at the metric in
Kruskal coordinates (ignoring angular coordinates for simplicity):
ds^2 = 1/r exp(r/L) [ dR^2  dT^2 ]
where r is an implicit function of R and T:
T^2  R^2 = (1  r/L) exp(r/L)
Whoops. That should be 1  r/L) exp(+r/L)
The entire locus R^T = T^2 coincides with the point mass.
Absolutely not. With this geometry, the Einstein tensor
G_uv is zero everywhere except at r=0, (at which point it
is singular).
Well, as I said, all kind of weird and wonderful things happen when you
divide infinity by infinity.
That's irrelevant because there *are* no infinities in
the Kruskal coordinates, except at r=0.
The metric is manifestly singular at the event horizon if you
work in the (r,t) coordinates.
That is *irrelevant*. There are coordinate singularities, and
there are physical singularities. The way to tell the difference
is that a physical singularity doesn't disappear when you change
coordinates, but coordinate do.
The fact that it miraculously becomes nonsingular after
this dubious coordinate transformation shouldn't convince anyone of
anything
It should convince anyone that it wasn't really a singularity.
The *only* way to know for sure what's a real singularity and
what isn't is by seeing what phenomena are present in *every*
coordinate system.
In any case, back in the (r,t) frame (if you dare to invert the
transformation  it can't be done analytically for sure), the
geodesic involves the sudden swapping of spacelike and timelike
directions across the event horizon.
No, it does not. If you paralleltransport any timelike vector
along the infalling geodesic, you will get another timelike
geodesic. Parallel transport across the event horizon relates
timelike vectors on one side to timelike vectors on the other
side, and relates spacelike vectors on one side to spacelike
vectors on the other. What you are saying is just incorrect.
So even if this mathematical wizardry makes everything appear
just fine in the Kruskal coordinates, this is just hiding the
fact that deep down something has gone horribly wrong.
You've got things exactly backwards. The use of Schwarzschild
coordinates hides the fact that *nothing* is going wrong as
you cross the event horizon. There is no experiment that
the infalling observer can perform that will reveal anything
out of the ordinary as he crosses the event horizon. The geometry
is perfectly wellbehaved.
You have to apply the Abrams/Crothers analysis before applying the
miracle map.
You don't have to do any mapping at all. You can *start* with
the Kruskal coordinates. That's a perfectly valid solution to
Einstein's field equations.
It makes everything so much simpler when you realise that
the interior solution does not exist.
That's what you are trying to *prove*.
So there is no mass anywhere except possibly
at r=0. The surface R^2 = T^2 has no mass on it anywhere.
The boundary condition applied would be the same  that
there cannot be any points with radial distance less than
zero from the origin.
What do you mean by "radial distance"? The coordinates
here are R and T. If there is some kind of boundary
condition, then express it in terms of R and T.
It is the integral of sqrt{g_rr} from the position r0 of the point mass
to the position r of the test point in the original (r,t).
What is the significance of that integral?
It would be some pretty ugly expression in the new
coordinates I imagine, but the
mathematics should give the same result (rather you than me to
calculate it and do the integrals given the complex multivalued
relationship between (r,t) and (R,T)).
I went through that exercise once before. It's in Google
archives.
http://groups.google.com/group/sci.physics.research/browse_thread/thread/7aabb45980d5bc08/7ac2ad2ff3ef46b9?lnk=st&q=daryl+event+horizon+complex+group%3Asci.physics.research&rnum=1&hl=en#7ac2ad2ff3ef46b9
The quantity r above is *always* nonnegative. R and T are
allowed to be both positive and negative, because R is not
a radial distance.
That's why I recommend you avoid all these obfuscations and stick to
the original coordinates
You've got it exactly backwards. The obfuscations come from the
Schwarzschild coordinates. To avoid all these obfuscations, you
should stick to Kruskal coordinates.
Use Kruskal coordinates exclusively, and prove to yourself
that there is no singularity at the event horizon R^2 = T^2.
Consider it a homework problem.
Here's the metric:
ds^2 = 1/r exp(r/L) [ dR^2  dT^2 ]
Show me what goes wrong at the locus R^2 = T^2. If you can't
show it in these coordinates, then it *isn't* there.
and prove to yourself in the simplest possible
way that there is no interior solution in the
way described by Abrams and Crothers. Consider it a
homework problem.
Abrams and Crothers are WRONG. The Kruskal coordinates prove
that they are wrong. No actual singularity can be transformed
away by a coordinate transformation. The quantity that goes
to infinity at the event horizon is the proper acceleration of
an observer at rest at the event horizon. The fact that this
quantity is infinite just means that it is *impossible* to
remain at rest at the event horizon.

Daryl McCullough
Ithaca, NY

I took a look at the thread you referred to above, and I have to say
that I am surprised  it seems that you yourself have pointed out first
hand some of the numerous problems which occur at the event horizon.
What should be very clear from your own analysis is that the mapping
from Schwarzschild to Kruskal (or vice versa) is discontinuous at the
event horizon  this is what gives rise to the analogue of "dividing
infinity by infinity to get something finite". A great deal of
handwaving and fudging and brushing under the carpet and excusemaking
goes on to try to justify the use of such a transformation, when
clearly it does not satisfy the usual requirements in differential
geometry of a continuous map between coordinate patches. T. Essel's
apologetics on that thread are hilarious  "Oh, I seem to have messed
up my coordinate labels... I'm an old timer and tired of explaining
this over and over again to newbies... Actually the Kruskal coordinates
are confusing  better to use Painleve or Lemaitre coordinates  or
just take a look at the Penrose diagram  I am sure it could be done
more easily that way, though I am not able to tell you how or even
whether they avoid having the same problems as the Kruskal
coordinates... (etc)".
Ultimately it seems that you must have decided that somehow the entire
Kruskal coordinate system is okay, and that there is something wrong
with the Schwarzschild coordinates, though you don't explain what
lifeshattering experience caused this change of heart. Did the
establishment propaganda machine finally win the day by confusing you
into submission?
 Sabbir. 

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dda1 science forum Guru
Joined: 06 Feb 2006
Posts: 762

Posted: Mon Jul 17, 2006 3:35 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl and Tom
You are very generous in debunking Sabbir's BS but he has thrown you a
red herring.
He is not really interested in debunking Crothers and Abrams.
Sabbir posted this link earlier, pointing directly to HIS paper :
http://groups.google.com/group/sci.physics.relativity/browse_frm/thread/c4b3943a10d29a04/3eeb41b03d225c82?lnk=st&q=&rnum=37#3eeb41b03d225c82
No one "bit" , so he decided to use a subterfuge and to post something
about the "origins" of his paper. What he really wants is to have his
OWN paper debugged but he doesn't have the elementary ethics to ask for
this. You will find the errors of Abrams and Crothers repeated (and
amplified) in Sabbir's paper, freshly posted on arxiv. 

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