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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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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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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 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 2:19 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



LEJ Brouwer says...
Quote:  Daryl McCullough wrote:
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.

Perhaps, but the conclusion that there is no interior solution is false.
Quote:  For example I also conclude that the event horizon coincides with point
mass

No, that's false.
Quote:  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.

But the exterior does not cover the entire space. There *is* no
point mass in the region r > 0. The Einstein tensor G is identically
0 in that region, indicating no mass at all.
Quote:  I do *not* deny that the interior solution naively solves the field
equations, or that it is nonstatic.

There is nothing naive about it. It solves Einstein's equations
in the interior.
Quote:  However, the interior solution is incompatible with the fact that
the event horizon coincides with the spatial origin,

That's false. No static spherically symmetric solution extends all
the way to the origin except in the case of no singularity at r=0.

Daryl McCullough
Ithaca, NY 

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

Posted: Tue Jul 18, 2006 3:43 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



LEJ Brouwer wrote:
Quote: 
However, the interior solution is
incompatible with the fact that the event horizon coincides with the
spatial origin,

Where does this nonsense come from?

Jan Bielawski 

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Sue... science forum Guru
Joined: 08 May 2005
Posts: 2684

Posted: Tue Jul 18, 2006 6:59 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



LEJ Brouwer wrote:
Quote:  Tom Roberts wrote:
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).

Tom: << 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.>>
Quote: 
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.
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.

Tom's statement above is one of more sensible ones in this thread
from a qualitative POV. I would not dismiss it out of hand.
The 'rabbit' he can pull out the hat is a spherical region where
long range (1.r^2) Coulomblike messegners which couple
across sparse regions, can transition to an induction mode
(Lorenz) suitable for penetrating (r^1) denser regions.
Where Tom separates static and nonstatic regions
there is a plausible 'quantum' scale mechanism to
accomplish the task.
http://www.mypage.bluewin.ch/Bizarre/GRAV.htm
http://www.mypage.bluewin.ch/Bizarre/
http://www.citebase.org/cgibin/citations?id=oai:arXiv.org:physics/0107015
You would indeed be throwing the baby out to dismiss a
mechanism, for which a current experiment shows some
congruence.
http://www.esa.int/SPECIALS/GSP/SEM0L6OVGJE_0.html
Apart from that,
arXiv:physics/0607102 v1 12 Jul 2006 is quite impressive
up to about page 14, where it seems to fall through the
'looking glass' with implausible 'spooks' that
*appear*, to me at least, inconsistant with a good bit
of experimental data from the field of high energy
physics.
"Fragmentation Functions at ZEUS"
http://ppewww.ph.gla.ac.uk/preprints/1997/10/ggo_preprint/ggo_preprint.html
Sue...


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

Posted: Wed Jul 19, 2006 1:00 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote:  LEJ Brouwer says...
Daryl McCullough wrote:
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.
Perhaps, but the conclusion that there is no interior solution is false.
For example I also conclude that the event horizon coincides with point
mass
No, that's false.
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.
But the exterior does not cover the entire space. There *is* no
point mass in the region r > 0. The Einstein tensor G is identically
0 in that region, indicating no mass at all.
I do *not* deny that the interior solution naively solves the field
equations, or that it is nonstatic.
There is nothing naive about it. It solves Einstein's equations
in the interior.
However, the interior solution is incompatible with the fact that
the event horizon coincides with the spatial origin,
That's false. No static spherically symmetric solution extends all
the way to the origin except in the case of no singularity at r=0.

Daryl McCullough
Ithaca, NY

Are these merely declarations of faith, or are you going to point out
the errors in the proof I have given?
 Sabbir 

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

Posted: Wed Jul 19, 2006 3:15 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



LEJ Brouwer says...
Quote:  That's false. No static spherically symmetric solution extends all
the way to the origin except in the case of no singularity at r=0.
Are these merely declarations of faith, or are you going to point out
the errors in the proof I have given?

I already did. Your proof assumed a static metric (independent of
time). You can derived from that assumption that r>2m. So you have
an implication:
If the metric components are static, then r > 2m.
which is just the contrapositive of
If r <= 2m, then the metric components are not static.
That's truethe metric components are not static in the
interior. How does that imply the interior doesn't exist?
As far as your claim that the mass is concentrated at the event
horizon, that violates Einstein's field equations. Einstein's
field equations say
G_uv = k T_uv
where G_uv is the curvature tensor, and T_uv is the energymomentum
tensor. The Schwarzchild solution has G = 0 *everywhere* for r>0.
So you can't locate the mass at r=2m without violating the field
equations.

Daryl McCullough
Ithaca, NY 

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

Posted: Wed Jul 19, 2006 10:12 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote:  LEJ Brouwer says...
That's false. No static spherically symmetric solution extends all
the way to the origin except in the case of no singularity at r=0.
Are these merely declarations of faith, or are you going to point out
the errors in the proof I have given?
I already did. Your proof assumed a static metric (independent of
time). You can derived from that assumption that r>2m. So you have
an implication:
If the metric components are static, then r > 2m.

The metric components are static in the exterior solution.
Quote:  which is just the contrapositive of
If r <= 2m, then the metric components are not static.
That's truethe metric components are not static in the
interior. How does that imply the interior doesn't exist?

So what you are essentially claiming is that an infalling particle
passes into the (nonexistent) interior of the point particle only to
hit the point particle again at a later time?
Quote:  As far as your claim that the mass is concentrated at the event
horizon, that violates Einstein's field equations. Einstein's
field equations say
G_uv = k T_uv
where G_uv is the curvature tensor, and T_uv is the energymomentum
tensor. The Schwarzchild solution has G = 0 *everywhere* for r>0.
So you can't locate the mass at r=2m without violating the field
equations.

That's a fair comment. I already mentioned earlier that upon formation
of the event horizon during spherical gravitational collapse (which
occurs at time t=Infty according to an external observer), a
topological phase transition must occur so that the event horizon acts
as an impenetrable physical barrier which separates an interior
region/manifold (not the Schwarzschild interior) which contains the
mass, from the (Schwarzschild) exterior region which does not. The
interior region will appear to be concentrated at a point from the
point of view of an external observer, even though to an internal
observer it will have finite volume.
Quote:  
Daryl McCullough
Ithaca, NY 


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

Posted: Wed Jul 19, 2006 11:46 am Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



LEJ Brouwer says...
Quote:  Daryl McCullough wrote:
I already did. Your proof assumed a static metric (independent of
time). You can derived from that assumption that r>2m. So you have
an implication:
If the metric components are static, then r > 2m.
The metric components are static in the exterior solution.

Right. And r > 2m for the exterior solution.
Quote:  which is just the contrapositive of
If r <= 2m, then the metric components are not static.
That's truethe metric components are not static in the
interior. How does that imply the interior doesn't exist?
So what you are essentially claiming is that an infalling particle
passes into the (nonexistent) interior of the point particle

The point particle is not *in* the exterior solution.
There is no point particle anywhere in the region r > 2m.
There is no point particle anywhere in the region r > 0.
Both the interior and exterior solutions are *vacuum*
solutions. No mass *anywhere*, except possibly at the origin.
Quote:  That's a fair comment. I already mentioned earlier that upon formation
of the event horizon during spherical gravitational collapse (which
occurs at time t=Infty according to an external observer), a
topological phase transition must occur so that the event horizon acts
as an impenetrable physical barrier which separates an interior
region/manifold (not the Schwarzschild interior) which contains the
mass, from the (Schwarzschild) exterior region which does not. The
interior region will appear to be concentrated at a point from the
point of view of an external observer, even though to an internal
observer it will have finite volume.

That's a different topic. The Schwarzchild solution is for an
eternal black hole (because it is a static solution), so doesn't
actually describe the collapse of a star into a black hole.
So that explains why you have been saying the
mass is concentrated at the event horizon. You
are thinking of the mass of a star undergoing
spherical collapse. In spherical collapse,
the Schwarzchild solution holds outside the
surface of the star, but a more complicated
solution holds inside.
However, you can calculate the path of a particle
on the surface of the star as it falls inward, and
you can show that (for a sufficiently massive star)
it will cross the line r=2m in a finite amount of proper
time.

Daryl McCullough
Ithaca, NY 

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

Posted: Wed Jul 19, 2006 2:35 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote:  LEJ Brouwer says...
Daryl McCullough wrote:
I already did. Your proof assumed a static metric (independent of
time). You can derived from that assumption that r>2m. So you have
an implication:
If the metric components are static, then r > 2m.
The metric components are static in the exterior solution.
Right. And r > 2m for the exterior solution.
which is just the contrapositive of
If r <= 2m, then the metric components are not static.
That's truethe metric components are not static in the
interior. How does that imply the interior doesn't exist?
So what you are essentially claiming is that an infalling particle
passes into the (nonexistent) interior of the point particle
The point particle is not *in* the exterior solution.
There is no point particle anywhere in the region r > 2m.

Correct.
Quote:  There is no point particle anywhere in the region r > 0.

Wrong.
Quote:  Both the interior and exterior solutions are *vacuum*
solutions. No mass *anywhere*, except possibly at the origin.

....which I have already shown you is at r=2m and coincides with the
point mass.
Quote: 
That's a fair comment. I already mentioned earlier that upon formation
of the event horizon during spherical gravitational collapse (which
occurs at time t=Infty according to an external observer), a
topological phase transition must occur so that the event horizon acts
as an impenetrable physical barrier which separates an interior
region/manifold (not the Schwarzschild interior) which contains the
mass, from the (Schwarzschild) exterior region which does not. The
interior region will appear to be concentrated at a point from the
point of view of an external observer, even though to an internal
observer it will have finite volume.
That's a different topic. The Schwarzchild solution is for an
eternal black hole (because it is a static solution), so doesn't
actually describe the collapse of a star into a black hole.

Sure, but the result of collapse is still relevant, particularly as an
infalling particle experiences hitting the horizon in finite time.
Quote:  So that explains why you have been saying the
mass is concentrated at the event horizon.

Well, no, not really. The point mass coincides with the event horizon
in the Schwarzschild solution, and that's that. This is quite
independent of the dynamics of gravitational collapse.
Quote:  You
are thinking of the mass of a star undergoing
spherical collapse. In spherical collapse,
the Schwarzchild solution holds outside the
surface of the star, but a more complicated
solution holds inside.

I am not thinking about stars at all. I am more concerned with
gravitational waves collapsing to from primordial black holes. In any
case, stellar collapse has other problems which prevent black hole
formation  see the work of Mitra et al, but that is not my concern
here. I mention spherical collapse because there is no such thing
physically as a real point particle (i.e. with infinite mass density),
so when the event horizon forms, it must be due to such a collapse
scenario. The main point to note is that the result of collapse is
something that looks to an external observer like a pointlike mass
which coincides with the (impenetrable) event horizon.
Quote:  However, you can calculate the path of a particle
on the surface of the star as it falls inward, and
you can show that (for a sufficiently massive star)
it will cross the line r=2m in a finite amount of proper
time.

That's another discussion for another time.
Quote:  
Daryl McCullough
Ithaca, NY

Thanks,
Sabbir. 

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

Posted: Wed Jul 19, 2006 6:31 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



LEJ Brouwer says...
Quote:  Daryl McCullough wrote:
There is no point particle anywhere in the region r > 0.
Wrong.

Not according to the Einstein field equations. The
Schwarzchild solution is a *vacuum* solution.
Quote:  Both the interior and exterior solutions are *vacuum*
solutions. No mass *anywhere*, except possibly at the origin.
...which I have already shown you is at r=2m and coincides with the
point mass.

By "origin" I mean r=0, not r=2m. There is no mass at r=2m.
Quote:  That's a different topic. The Schwarzchild solution is for an
eternal black hole (because it is a static solution), so doesn't
actually describe the collapse of a star into a black hole.
Sure, but the result of collapse is still relevant, particularly as an
infalling particle experiences hitting the horizon in finite time.
So that explains why you have been saying the
mass is concentrated at the event horizon.
Well, no, not really. The point mass coincides with the event horizon
in the Schwarzschild solution, and that's that.

No, it doesn't. The event horizon is a region
in which G = 0, which by the field equations
implies that there is no mass there.
Your statement that the mass is located at the event horizon
is *nonsense* for the Schwarzchild metric. If we're talking
about the exterior Schwarzchild metric, then we are talking
about a *vacuum* solutionno mass. If there is mass anywhere
outside r=0, then we aren't talking about Schwarzchild.
Quote:  This is quite independent of the dynamics of gravitational collapse.

If you are talking about mass residing anywhere outside r=0,
then you are not talking about the Schwarzchild solution. If
you are not talking about the Schwarzchild solution, then you
are not talking about a static solution. If you are not talking
about a static solution, then that means that collapse is involved.
Quote:  You
are thinking of the mass of a star undergoing
spherical collapse. In spherical collapse,
the Schwarzchild solution holds outside the
surface of the star, but a more complicated
solution holds inside.
I am not thinking about stars at all. I am more concerned with
gravitational waves collapsing to from primordial black holes.

Well, it doesn't really matter much the nature of the energy
used to create a black hole.
Quote:  I mention spherical collapse because there is no such thing
physically as a real point particle (i.e. with infinite mass density),
so when the event horizon forms, it must be due to such a collapse
scenario. The main point to note is that the result of collapse is
something that looks to an external observer like a pointlike mass
which coincides with the (impenetrable) event horizon.

What do you mean by "impenetrable" here? Anyway, I agree. If all
the mass/energy is spherically distributed, and there is no mass
outside some radius R, then outside radius R, the Schwarzchild solution
will obtain. We agre on that point. The question is what are things
like for r < R?
In particular, where in the world are you getting your idea
that "the point mass resides at r=2m"? That makes no sense
at all.

Daryl McCullough
Ithaca, NY 

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

Posted: Wed Jul 19, 2006 8:19 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote: 
In particular, where in the world are you getting your idea
that "the point mass resides at r=2m"? That makes no sense
at all.

I asked the same question before  no answer besides the canonical
"read the papers" (they are all terribly written, BTW). He stopped
talking about "singular maps removing a singularity" being somehow
wrong but the r=2M singularity just won't go away.
This quote from Richard Feynman describes the situation perfectly. He
writes to his wife Gweneth from a Warsaw conference in 1962:
"I get into arguments [...] whenever anyone [...] starts to tell me
about his "work." The "work" is always:
(1) completely ununderstandable,
(2) vague and indefinite,
(3) something correct that is obvious and selfevident, but worked out
by a long and difficult analysis, and presented as an important
discovery, or
(4) a claim based on the stupidity of the author that some obvious and
correct fact, accepted and checked for years, is, in fact, false (these
are the worst: no argument will convince the idiot),"
(1) and (2) describe the papers quoted by LEJ perfectly. (4) is, well,
nothing new, obviously.

Jan Bielawski 

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

Posted: Wed Jul 19, 2006 8:38 pm Post subject:
Re: Misinterpretation of the radial parameter in the Schwarzschild solution?



Daryl McCullough wrote:
Quote:  LEJ Brouwer says...
Daryl McCullough wrote:
There is no point particle anywhere in the region r > 0.
Wrong.
Not according to the Einstein field equations. The
Schwarzchild solution is a *vacuum* solution.
Both the interior and exterior solutions are *vacuum*
solutions. No mass *anywhere*, except possibly at the origin.
...which I have already shown you is at r=2m and coincides with the
point mass.
By "origin" I mean r=0, not r=2m. There is no mass at r=2m.
That's a different topic. The Schwarzchild solution is for an
eternal black hole (because it is a static solution), so doesn't
actually describe the collapse of a star into a black hole.
Sure, but the result of collapse is still relevant, particularly as an
infalling particle experiences hitting the horizon in finite time.
So that explains why you have been saying the
mass is concentrated at the event horizon.
Well, no, not really. The point mass coincides with the event horizon
in the Schwarzschild solution, and that's that.
No, it doesn't. The event horizon is a region
in which G = 0, which by the field equations
implies that there is no mass there.
Your statement that the mass is located at the event horizon
is *nonsense* for the Schwarzchild metric. If we're talking
about the exterior Schwarzchild metric, then we are talking
about a *vacuum* solutionno mass. If there is mass anywhere
outside r=0, then we aren't talking about Schwarzchild.
This is quite independent of the dynamics of gravitational collapse.
If you are talking about mass residing anywhere outside r=0,
then you are not talking about the Schwarzchild solution. If
you are not talking about the Schwarzchild solution, then you
are not talking about a static solution. If you are not talking
about a static solution, then that means that collapse is involved.
You
are thinking of the mass of a star undergoing
spherical collapse. In spherical collapse,
the Schwarzchild solution holds outside the
surface of the star, but a more complicated
solution holds inside.
I am not thinking about stars at all. I am more concerned with
gravitational waves collapsing to from primordial black holes.
Well, it doesn't really matter much the nature of the energy
used to create a black hole.
I mention spherical collapse because there is no such thing
physically as a real point particle (i.e. with infinite mass density),
so when the event horizon forms, it must be due to such a collapse
scenario. The main point to note is that the result of collapse is
something that looks to an external observer like a pointlike mass
which coincides with the (impenetrable) event horizon.
What do you mean by "impenetrable" here? Anyway, I agree. If all
the mass/energy is spherically distributed, and there is no mass
outside some radius R, then outside radius R, the Schwarzchild solution
will obtain. We agre on that point. The question is what are things
like for r < R?

That's the subject of my next paper.
Quote:  In particular, where in the world are you getting your idea
that "the point mass resides at r=2m"? That makes no sense
at all.

Daryl McCullough
Ithaca, NY

"I refer the honourable gentleman to the answers I gave just a moment
ago..." because you are ignoring them and putting words into my mouth.
 Sabbir. 

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