Misinterpretation of the radial parameter in the Schwarzschild solution?
LEJ Brouwer
response has settled this issue, and refuse to allow further
discussion. I disagree, so I am continuing the thread here, and would
like to invite Steve to respond if he wishes - Sabbir.
=============================================
The following papers claim that
there is an error in the interpretation of the radial coordinate 'r' in
the standard Schwarzschild metric:
L. S. Abrams, "Black holes: The legacy of Hilbert's error", Can. J.
Phys. 23 (1923) 43, http://arxiv.org/abs/gr-qc/0102055
S. Antoci, "David Hilbert and the origin of the 'Schwarzschild
solution'", http://arxiv.org/abs/physics/0310104
S. J. Crothers, "On the general solution to Einstein's vacuum field and
its implications for relativistic degeneracy", Prog. Phys. 1 (2006)
68-73.
In particular they show, in a rather simple fashion, that the event
horizon is at radius zero, coinciding with the position the point mass
itself, and actually appears pointlike to an external observer.
They claim that the reason that the original misinterpretation occurred
is because Hilbert incorrectly assumed a priori that the 'r' which
appears in the metric must be the radial coordinate (in fact, it need
only parametrise the radii to ensure a spherically symmetric solution).
The careful analysis of Abrams et al shows that the point mass actually
resides at r=2m, which therefore corresponds to the true origin, so
that there is in fact no 'interior' solution. In particular, they
mention that Schwarzschild's original paper never allowed for an
interior solution either.
If the event horizon is at the origin, and there is no interior
solution, then this tends to raise the question, "well, where does a
radially infalling particle actually go?". Does it just bounce off the
'brick wall' (or rather, 'brick point')? (I do not agree with the above
papers that the Kruskal extension is invalid - it is absolutely
necessary to have a consistent well-defined timelike direction).
Have we really all been making this silly mathematical error, and is
our present understanding of the simplest classical black hole way off
the mark?
Cheers,
Sabbir.
-----------------------------------------------------------------------
LEJ Brouwer <intuitioni...@yahoo.com> wrote:
[...]
> The reason I am interested is because the following papers claim that
> there is an error in the interpretation of the radial coordinate 'r' in
> the standard Schwarzschild metric:
> L. S. Abrams, "Black holes: The legacy of Hilbert's error", Can. J.
> Phys. 23 (1923) 43, http://arxiv.org/abs/gr-qc/0102055
> S. Antoci, "David Hilbert and the origin of the 'Schwarzschild
> solution'", http://arxiv.org/abs/physics/0310104
> S. J. Crothers, "On the general solution to Einstein's vacuum field and
> its implications for relativistic degeneracy", Prog. Phys. 1 (2006)
> 68-73.
> In particular they show, in a rather simple fashion, that the event
> horizon is at radius zero, coinciding with the position the point mass
> itself, and actually appears pointlike to an external observer.
These papers are complete nonsense. In particular, the authors seem
not to understand the basic fact that physics does not depend on what
coordinates one chooses to use.
It is trivially true that if one changes coordinates in the standard
Schwarzschild solution from r to r-2m, then the horizon is at r=0.
It is also trivially true that this does not change the spacetime
geometry -- the horizon is still a lightlike surface, with an area at
fixed time of 4m^2. Choosing a coordinate that makes the horizon
look like a point doesn't make it a point -- it just means that you've
made a dumb coordinate choice.
> They claim that the reason that the original misinterpretation occurred
> is because Hilbert incorrectly assumed a priori that the 'r' which
> appears in the metric must be the radial coordinate (in fact, it need
> only parametrise the radii to ensure a spherically symmetric solution).
It is radial in the sense that the set of points at constant r and t
is a two-sphere of area 4pi r^2. It is not "radial distance," but
no one has claimed that it is.
> The careful analysis of Abrams et al shows that the point mass actually
> resides at r=2m, which therefore corresponds to the true origin, so
> that there is in fact no 'interior' solution.
This analysis is not "careful" -- it's mathematically awful. How can
a "point mass" reside at a two-sphere of finite area? What sense does
it make to say that a mass resides at a position at which the Ricci
tensor is zero?
Abrams makes an elementary mistake. He concludes that r=2m (in
standard
Schwarzschild coordinates) is singular because the "radius" of a circle
around this "point" goes to zero as r->2m while its "circumference"
does not. But this is not a singularity -- it's just a reflection
of the fact that r=2m is a two-sphere, not a point.
> If the event horizon is at the origin, and there is no interior
> solution, then this tends to raise the question, "well, where does a
> radially infalling particle actually go?". Does it just bounce off the
> 'brick wall' (or rather, 'brick point')?
To answer this, you just compute the motion. You find that it falls
right past the "origin," with nothing peculiar happening there. (Of
course, you can insist on using bad coordinates, but that's your own
fault...).
> Have we really all been making this silly mathematical error, and is
> our present understanding of the simplest classical black hole way off
> the mark?
No.
Steve Carlip
---------------------------------------------------------------------------------------------------
carlip...@physics.ucdavis.edu wrote:
> LEJ Brouwer <intuit...@yahoo.com> wrote:
> > In particular they show, in a rather simple fashion, that the event
> > horizon is at radius zero, coinciding with the position the point mass
> > itself, and actually appears pointlike to an external observer.
>
> These papers are complete nonsense. In particular, the authors seem
> not to understand the basic fact that physics does not depend on what
> coordinates one chooses to use.
>
> It is trivially true that if one changes coordinates in the standard
> Schwarzschild solution from r to r-2m, then the horizon is at r=0.
> It is also trivially true that this does not change the spacetime
> geometry -- the horizon is still a lightlike surface, with an area at
> fixed time of 4m^2. Choosing a coordinate that makes the horizon
> look like a point doesn't make it a point -- it just means that you've
> made a dumb coordinate choice.
Yes, it is true that the choice of coordinates does not change the
physics (including, I might add, the impossibility of a timelike vector
suddenly becoming spacelike, and vice versa). However, you have
overlooked that there is a constraint on the range of values r can take
(namely, if r=r0 is the position of the point mass in one's chosen
coordinates, then one cannot use coordinates with r<r0). As the papers
make clear, for the standard Hilbert coordinate choice, we must have r
> 2m, and in the shifted coordinates r -> r-2m, this constraint becomes r > 0. 'r' isn't the same as the radial coordinate in either case, but at least the latter coordinate choice makes it harder to make the elementary error of believing that there exists an interior solution. So I don't agree with you that this
choice is 'dumb'. The only thing that is 'dumb' here is to try to use
values of r which are not within the physically allowable range.
And yes, it might naively seem strange that the EH should have area
despite being at 'zero' distance from the point mass, but that is only
because the way the radius and area are defined and calculated are by
using different components of the metric (g_11 for the radius and g_22,
g_33 for the area). There is no reason why intuition regarding
Euclidean metrics should apply to the non-Euclidean case.
> > They claim that the reason that the original misinterpretation occurred
> > is because Hilbert incorrectly assumed a priori that the 'r' which
> > appears in the metric must be the radial coordinate (in fact, it need
> > only parametrise the radii to ensure a spherically symmetric solution).
>
> It is radial in the sense that the set of points at constant r and t
> is a two-sphere of area 4pi r^2. It is not "radial distance," but
> no one has claimed that it is.
Again, one should not expect Euclidean identities to hold in a
non-Euclidean space, where the 'radial' distance and 'area' no longer
have the usual connection.
> > The careful analysis of Abrams et al shows that the point mass actually
> > resides at r=2m, which therefore corresponds to the true origin, so
> > that there is in fact no 'interior' solution.
>
> This analysis is not "careful" -- it's mathematically awful. How can
> a "point mass" reside at a two-sphere of finite area? What sense does
> it make to say that a mass resides at a position at which the Ricci
> tensor is zero?
When the event horizon is formed (and note that this happens t=infinity
so that no external observer sees it) a topological phase transition
must occur which separates the space external to the event horizon
where the Ricci tensor is zero which _does not_ contain the mass, from
the space contained within the event horizon which _does_ contain the
mass. Note that although the interior space _appears_ to be contained
within a point from the perspective of an observer in the exterior
space, this is merely an artefact of the topological disconnectedness
of the two spaces. The interior space in general will have finite
volume and is nonsingular.
> Abrams makes an elementary mistake. He concludes that r=2m (in standard
> Schwarzschild coordinates) is singular because the "radius" of a circle
> around this "point" goes to zero as r->2m while its "circumference"
> does not. But this is not a singularity -- it's just a reflection
> of the fact that r=2m is a two-sphere, not a point.
Agreed, and as Crother points out, one can define 'radius' in different
ways depend on whether on is measuring distance, circumference, area,
or some other typically radially dependent parameter. Once again these
are different because we are working with a non-Euclidean metric.
> > If the event horizon is at the origin, and there is no interior
> > solution, then this tends to raise the question, "well, where does a
> > radially infalling particle actually go?". Does it just bounce off the
> > 'brick wall' (or rather, 'brick point')?
>
> To answer this, you just compute the motion. You find that it falls
> right past the "origin," with nothing peculiar happening there. (Of
> course, you can insist on using bad coordinates, but that's your own
> fault...).
What is usually done is a bad coordinate transformation is applied to a
bad set of coordinates in order to obtain a supposedly 'good' set of
coordinates. This is an invalid mathematical procedure and is analogous
to dividing infinity by infinity. The proper time taken to reach the
horizon is vaguely believable, but the proper time taken to get from
the event horizon to the supposed singularity is not - particularly so
given that the interior solution does not exist.
> > Have we really all been making this silly mathematical error, and is
> > our present understanding of the simplest classical black hole way off
> > the mark?
>
> No.
Well admittedly there does appear to be a small handful of people who
do not make this mistake, but you and your colleagues are clearly not
amongst them, and I predict that this silly mathematical error will be
looked back upon as one of the most embarassing blunders of twentieth
century physics.
> Steve Carlip
- Sabbir.
Phineas T Puddleduck
On 6/7/06 11:58, in article
1152183483....@j8g2000cwa.googlegroups.com, "LEJ Brouwer"
<intuit...@yahoo.com> wrote:
> The moderators of s.p.research have decided that Steve Carlip's
> response has settled this issue, and refuse to allow further
> discussion. I disagree, so I am continuing the thread here, and would
> like to invite Steve to respond if he wishes - Sabbir.
> =============================================
Take it to email if you've been spnaked.
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Koobee Wublee
news:1152183483....@j8g2000cwa.googlegroups.com...
> The moderators of s.p.research have decided that Steve Carlip's
> response has settled this issue, and refuse to allow further
> discussion. I disagree, so I am continuing the thread here, and would
> like to invite Steve to respond if he wishes - Sabbir.
> =============================================
>
> The following papers claim that
> there is an error in the interpretation of the radial coordinate 'r' in
> the standard Schwarzschild metric:
>
> L. S. Abrams, "Black holes: The legacy of Hilbert's error", Can. J.
> Phys. 23 (1923) 43, http://arxiv.org/abs/gr-qc/0102055
>
> S. Antoci, "David Hilbert and the origin of the 'Schwarzschild
> solution'", http://arxiv.org/abs/physics/0310104
>
> S. J. Crothers, "On the general solution to Einstein's vacuum field and
> its implications for relativistic degeneracy", Prog. Phys. 1 (2006)
> 68-73.
Einstein Field Equations are
R_ij - R g_ij / 2 = constant Lm g _ij
Where
** i, j have variations from 0 to 3 each.
** R_ij = Ricci tensor
** R = g^uv R_uv
** g_ij = metric = curvature correction factors
** Lm = Lagrangian of mass energy
Lm = constant rho
Where
** rho = mass density if inside an object
** rho = (- 1 / r^3) if in free space near an object
To make things much simpler, we allow only one spatial and one temporal
dimensions and also only diagonal metric. The field equations above
simplify to the following after many steps of still very tedious
caculations. Concentrating with solutions in free space and matching
with the Newtonian result, we have
- @^2g_00/@r^2 / (2 g_00 g_11) - @g_00/@r @(1/(g_00 g_11))/@r / 4 = 2 G
M / r^3 / c^2
Where
** @/@r = partial derivative operator with respect to r
Set the following.
g_00 g_11 = -1
Then, the equation becomes
@^2g_00/@r^2 = - 4 G M / r^3 / c^2
Solving that very simple differential equation above, we get
g_00 = k1 + k2 r - 2 G M / r / c^2
Where
** k1, k2 = integration constants
Applying the boundary condition at (g_00 = 1) where (r = infinity), you
get
g_00 = 1 - 2 G M / r / c^2
This means Hilbert is correct. However, if you decide that the
Newtonian gravitational law is only an approximation which it mostly
likely is anyway, the Poisson equation in free space has the following
form.
@^2g_00/@r^2 = - 4 G M / r^3 / c^2 + (smaller components)
Where
** 4 G M / r^3 / c^2 >> (smaller components)
This will open up an infinite number of solutions in which
Schwarzschild's original solution is just one of them. The
Schwarzschild metric today is valid only if we take the Newtonian
result exactly as it is.
An interesting scenario is when setting (g_11 = -1) or flat space, the
field equation becomes
@^2g_00/@r^2 + (@g_00/@r)^2 / (4 g_00) = - 2 G M g_00 / r^3 / c^2
To solve the differential equation above, it appears to be quite
challenging. I have a feeling that it would not even lead to anything
remotely logical in nature. I will attempt to solve it later on when I
go out jogging with my dog using Taylor series approximations. If you
are interested, I can post my finding.
shuba
> The moderators of s.p.research have decided that Steve Carlip's
> response has settled this issue, and refuse to allow further
> discussion. I disagree, so I am continuing the thread here, and would
> like to invite Steve to respond if he wishes - Sabbir.
Ah, this is the same stunt Sabbir pulled in 2001, back before
he decided to use the handle of a dead mathematician. Usenet
theatre at its finest!
http://groups.google.com/group/sci.physics/msg/79a4b9362363fd93
> L. S. Abrams, "Black holes: The legacy of Hilbert's error", Can. J.
> Phys. 23 (1923) 43, http://arxiv.org/abs/gr-qc/0102055
Judging from the citebase citations to that paper, this appears
to (once again) have everything to do with the discredited ideas
of a certain Abhas Mitra, just as in 2001. It's good this was posted,
as otherwise some people might have been led to believe that
Sabbir's purpose in bringing up the articles on s.p.research was
to solicit informed opinion.
Mitra has been ripped apart by Carlip, Hillman, Baez, and many
others. The monomaniacal posturing of Sabbir Rahman just
enhances the effect. In the words of Baez, in a recent (2004)
foray into this newsgroup (responding to Sabbir and Mitra),
"Dignity, eh? If I had any "dignity" I wouldn't be posting here."
http://groups.google.com/group/sci.physics.relativity/msg/ca9aef3604acf620
---Tim Shuba---
Tom Roberts
> Have we really all been making this silly mathematical error, and is
> our present understanding of the simplest classical black hole way off
> the mark?
No. If what you (those papers) claim were true, than any pair of
timelike objects approaching r=2M from different directions would
approach arbitrarily close to each other. This does not happen, and the
distance between such infalling geodesics does not go to zero for all
possible pairs of geodesics.
More remarkably, for the approach to r=0 such objects can have an
arbitrarily large distance between them.
The whole confusion is about coordinates that behave badly at r=2M. The
Schwarzschild coordinates are INVALID there, and that destroys the whole
"argument". Use coordinate-independent notions (such as distance between
geodesics, curvature tensors, etc.) and the truth comes out.
Steve Carlip and the moderators got it right.
Tom Roberts
LEJ Brouwer
shuba wrote:
> Sabbir Rahman wrote:
>
> > The moderators of s.p.research have decided that Steve Carlip's
> > response has settled this issue, and refuse to allow further
> > discussion. I disagree, so I am continuing the thread here, and would
> > like to invite Steve to respond if he wishes - Sabbir.
>
> Ah, this is the same stunt Sabbir pulled in 2001, back before
> he decided to use the handle of a dead mathematician. Usenet
> theatre at its finest!
Indeed, a 'stunt' which made pretty clear to all the double standards
of the moderators of that group.
> http://groups.google.com/group/sci.physics/msg/79a4b9362363fd93
>
> > L. S. Abrams, "Black holes: The legacy of Hilbert's error", Can. J.
> > Phys. 23 (1923) 43, http://arxiv.org/abs/gr-qc/0102055
>
> Judging from the citebase citations to that paper, this appears
> to (once again) have everything to do with the discredited ideas
> of a certain Abhas Mitra, just as in 2001. It's good this was posted,
> as otherwise some people might have been led to believe that
> Sabbir's purpose in bringing up the articles on s.p.research was
> to solicit informed opinion.
>
> Mitra has been ripped apart by Carlip, Hillman, Baez, and many
> others. The monomaniacal posturing of Sabbir Rahman just
> enhances the effect. In the words of Baez, in a recent (2004)
> foray into this newsgroup (responding to Sabbir and Mitra),
>
> "Dignity, eh? If I had any "dignity" I wouldn't be posting here."
> http://groups.google.com/group/sci.physics.relativity/msg/ca9aef3604acf620
>
>
> ---Tim Shuba---
Tim,
It is true that Abhas was subjected to an unreasonable amount of abuse
from a couple of the individuals you mention (Steve Carlip was not one
of them), but he certainly did not lose the technical battle, and
responded satisfactorily to all objections made. Maybe you did not have
the patience to read the entire thread?
As for John Baez's lame effort, his "proof" of Mitra's error amounted
to the dubious claim that lim f(x)/g(x) -> Infty if g(x) -> 0,
carelessly overlooking the fact that f(x) -> 0 in the same limit. This
was also the reason for the subsequent thread entitled "John Baez
teaches calculus".
As for Abhas's supposedly "discredited ideas", his work has being
published in several established journals, and has most recently been
accepted by Phys Rev D and MNRAS:
He is currently visiting the Max Planck Institute in Heidelberg.
I find it curious that you feel justified in judging the technical
merits of a paper on the arxiv by the number of citations alone. This
is remarkably similar to the method of Hillman and Baez, who also have
the curious habit of reading a paper only *after* they have ripped its
contents apart. So might I suggest that you also try reading Abrams
before passing judgment?
Note that Abrams original paper actually predates Mitra's work by many
years (the correct journal reference is actually Can. J. Phys. 67
(1989) 919 - apologies for the incorrect reference given above).
- Sabbir.
LEJ Brouwer
Tom Roberts wrote:
> LEJ Brouwer wrote:
> > Have we really all been making this silly mathematical error, and is
> > our present understanding of the simplest classical black hole way off
> > the mark?
>
> No. If what you (those papers) claim were true, than any pair of
> timelike objects approaching r=2M from different directions would
> approach arbitrarily close to each other. This does not happen, and the
> distance between such infalling geodesics does not go to zero for all
> possible pairs of geodesics.
No. I think we both agreed that this was an unusual situation where
apparently the event horizon has finite area (hence the separated
geodesics) but zero radial distance from the pointlike mass. Note that
we are dealing with a non-Euclidean metric here, so the radial distance
from the origin and the radius associated with the square root of the
area need not be the same. Note also that the metric becomes
ill-defined in this limit (the EH forms at t=Infty according to an
external observer), and that is why I proposed the existence of a phase
transition occuring at the formation of the event horizon, so that this
singular limit is not precisely realised.
>
> More remarkably, for the approach to r=0 such objects can have an
> arbitrarily large distance between them.
It is not remarkable because there is no interior solution for r<2m.
> The whole confusion is about coordinates that behave badly at r=2M. The
> Schwarzschild coordinates are INVALID there, and that destroys the whole
> "argument". Use coordinate-independent notions (such as distance between
> geodesics, curvature tensors, etc.) and the truth comes out.
No it's not. The confusion is due to the use of radial parameters
outside of their physically valid range.
> Steve Carlip and the moderators got it right.
No they didn't.
> Tom Roberts
- Sabbir.
Ben Rudiak-Gould
> I think we both agreed that this was an unusual situation where
> apparently the event horizon has finite area (hence the separated
> geodesics) but zero radial distance from the pointlike mass.
The radial "distance" from the center is timelike, which is not the same as
being zero.
Do you agree that the event horizon is at r = 2m (not r = 0) in
Schwarzschild coordinates? I.e. is this whole debate over how one ought to
define "radius"?
> Note also that the metric becomes ill-defined in this limit (the EH forms
> at t=Infty according to an external observer),
Event horizons don't form. They aren't dynamical objects. They can't be
defined in terms of any local property of the geometry or the metric.
-- Ben
LEJ Brouwer
> LEJ Brouwer wrote:
> > I think we both agreed that this was an unusual situation where
> > apparently the event horizon has finite area (hence the separated
> > geodesics) but zero radial distance from the pointlike mass.
>
> The radial "distance" from the center is timelike, which is not the same as
> being zero.
>
> Do you agree that the event horizon is at r = 2m (not r = 0) in
> Schwarzschild coordinates? I.e. is this whole debate over how one ought to
> define "radius"?
Yes, the event horizon is at r=2m in Schwarzschild coordinates, and
this coincides with the position of the mass, i.e. r=2m _is_ the
origin/centre, not r=0. Values of r<2m are physically meaningless, and
there is no 'interior' solution at all.
>
> > Note also that the metric becomes ill-defined in this limit (the EH forms
> > at t=Infty according to an external observer),
>
> Event horizons don't form. They aren't dynamical objects. They can't be
> defined in terms of any local property of the geometry or the metric.
Not so - the norm of the 4-acceleration on a test particle is a local,
invariant, intrinsic quantity that diverges on the event horizon. See
sections 3 and 4 of,
Antoci and Liebscher, "Reinstating Schwarzschild's original manifold
and its singularity", http://arxiv.org/abs/gr-qc/0406090
> -- Ben
- Sabbir
Tom Roberts
> Yes, the event horizon is at r=2m in Schwarzschild coordinates, and
> this coincides with the position of the mass, i.e. r=2m _is_ the
> origin/centre, not r=0. Values of r<2m are physically meaningless, and
> there is no 'interior' solution at all.
You are confusing your coordinate chart with the manifold itself. Yes,
indeed, the exterior Schw. coordinates are invalid for r<2M.
But as is well known, coordinates are human artifacts and do not
determine the geometry of the manifold (or any other properties).
Transform to other coordinates, such as Kruskal coordinates, and it
becomes QUITE CLEAR that the manifold does not "end" at r=2M as you
claim. In particular, there are infalling timelike geodesic paths that
go right through the surface r=2M and continue all the way to a limit
point at r=0. Moreover, Ricci=0 at every point of every such geodesic
(i.e. there is no energy density anywhere for r>0).
Note that the locus r=0 must be deleted from the manifold. Note I did
not say "point", as this is most definitely not a point. While one might
naively expect it to be a 1-d timelike locus (all possible values of t),
in fact it is a 2-d spacelike locus and that disconnect is at base why
it must be deleted from the manifold (please do remember that for r<2M
the Schw. coordinate t is spacelike).
> Not so - the norm of the 4-acceleration on a test particle is a local,
> invariant, intrinsic quantity that diverges on the event horizon.
That is plain and simply not true. An infalling timelike geodesic path
can go right through the horizon, and it has zero 4-acceleration
everywhere (because it is a timelike geodesic).
The abstract of gr-qc/0406090 discusses a test particle at rest relative
to the black hole, and points out that such worldlines have divergent
4-accelerations for positions approaching the horizon. I suspect you
think that is universal. Not so.
[That paper considers this a "defect", but in fact it is
quite reasonable and well known.]
You, and the papers you quote, have very serious misunderstandings about
the basics of GR.
> See
> sections 3 and 4 of,
> Antoci and Liebscher, "Reinstating Schwarzschild's original manifold
> and its singularity", http://arxiv.org/abs/gr-qc/0406090
I'll look at it when I get a chance....
Tom Roberts
LEJ Brouwer
Tom Roberts wrote:
> LEJ Brouwer wrote:
> > Yes, the event horizon is at r=2m in Schwarzschild coordinates, and
> > this coincides with the position of the mass, i.e. r=2m _is_ the
> > origin/centre, not r=0. Values of r<2m are physically meaningless, and
> > there is no 'interior' solution at all.
>
> You are confusing your coordinate chart with the manifold itself. Yes,
> indeed, the exterior Schw. coordinates are invalid for r<2M.
>
> But as is well known, coordinates are human artifacts and do not
> determine the geometry of the manifold (or any other properties).
> Transform to other coordinates, such as Kruskal coordinates, and it
> becomes QUITE CLEAR that the manifold does not "end" at r=2M as you
> claim. In particular, there are infalling timelike geodesic paths that
> go right through the surface r=2M and continue all the way to a limit
> point at r=0. Moreover, Ricci=0 at every point of every such geodesic
> (i.e. there is no energy density anywhere for r>0).
Tom, I don't claim that the manifold ends at r=2M. Rather, I agree with
you totally and accept the Kruskal extension and that the exterior
solution is double-sheeted. I actually DISAGREE with Abrams et alii on
this point. I think they made a mistake. However, they are _correct_
that the the coordinates r<2M are invalid (note that no point on the
exterior Kruskal-extended solution has r<2M). It is a shame that there
are mistakes in the works I refer to (it seems that no work on this
surprisingly tricky subject has been perfect), just as there have been
clear mistakes in Hilbert's work and those who followed him. Antoci's
paper on Hilbert's derivation makes this point very clearly.
[Note that the very fact that the Schwarzschild solution had to be
extended by Kruskal and Szekeres means that the Schwarzschild/Hilbert
solution is *not* the most general spherically symmetric solution to
the problem - the reason that _this_ mistake was made was because of
the prior assumption, which is subtly hidden in the way that the
problem is usually formulated, and which again depends upon the
misinterpretation of r as the radial coordinate as opposed to an
apriori unknown parametristation of the radius, that there can only be
single-sheeted solution. It was not until Synge's derivation in 1974,
that the most general solution (i.e. the Kruskal-extened one) was
derived from first principles.]
As in all matters, we should be objective about this, and take what is
right and leave what is wrong. Until we are willing to do this and
accept that some of the things we have 'learnt' in the past may in fact
be mistaken, we are going to have great difficulty making progress in
our understanding.
>
> Note that the locus r=0 must be deleted from the manifold. Note I did
> not say "point", as this is most definitely not a point. While one might
> naively expect it to be a 1-d timelike locus (all possible values of t),
> in fact it is a 2-d spacelike locus and that disconnect is at base why
> it must be deleted from the manifold (please do remember that for r<2M
> the Schw. coordinate t is spacelike).
Yes, r=0 is a singularity and is physically unreasonable. Indeed
pointlike masses themselves are physically unreasonable. In a more
realistic scenario of gravitational collapse we will have, say, a
spherical distribution of matter collapsing until an event horizon is
formed at time t=Infty. To an external observer (though I guess there
won't be any around to watch at t=Infty), all of the mass contained
within the event horizon will appear to be contained at a point,
admittedly with apparently finite area. My proposal as to what must
happen to explain the lack of metric/curvature singularities at the
event horizon is that a topological phase transition occurs as a result
of which the mass contained within the event horizon becomes physically
separated from the exterior solution (including its Kruskal extension).
The only consistent picture that comes to mind is that any particles
falling in radially towards the event horizon will not experience any
singularity, but will reappear apparently travelling radially outwards
and backwards in time on the other sheet to an external observer. An
infalling particle will actually look like a particle-antiparticle
annihilation event, though the infalling particle itself will be
totally oblivious of how its trajectory appears to external observers.
As for the mass in the interior of the event horizon, it will not
actually be concentrated at a point but will occupy a finite volume
within which it will continue to gravitate normally, as an isolated
system from the exterior manifold. [This interior manifold should also
be time non-orientable like the exterior manifold.]
> > Not so - the norm of the 4-acceleration on a test particle is a local,
> > invariant, intrinsic quantity that diverges on the event horizon.
>
> That is plain and simply not true. An infalling timelike geodesic path
> can go right through the horizon, and it has zero 4-acceleration
> everywhere (because it is a timelike geodesic).
How can spacelike and timelike directions suddenly swap places if there
is nothing unusual going on at the event horizon? This is completely
unphysically, yet is accepted in a completely matter-of-fact way in all
standard textbooks. Anyone considering this objectively should
immediately smell a rat here.
> The abstract of gr-qc/0406090 discusses a test particle at rest relative
> to the black hole, and points out that such worldlines have divergent
> 4-accelerations for positions approaching the horizon. I suspect you
> think that is universal. Not so.
>
> [That paper considers this a "defect", but in fact it is
> quite reasonable and well known.]
>
>
> You, and the papers you quote, have very serious misunderstandings about
> the basics of GR.
I will need to take a look at the paper again as it is a while since I
last read it, but even if I were to concede this last point, I will not
concede the main point I am making which is the fact that r<2M has no
meaning for the Schwarzschild coordinates, so that the usual interior
solutions are invalid.
I am not certainly not God (wa aoudhu bika min dhalik), and make no
personal claims to infallibility, but the mistake that Hilbert made,
which everyone after him followed, seems *very* clear, and so one could
say that Hilbert and all of his followers also have 'serious
misunderstandings about the basics of GR'. But I am not going to say
that because 'to err is human', and sometimes simple mistakes do
propagate for a long time without being noticed. There is a problem
here which needs to be addressed - petty name-calling and pretending
that nothing is wrong is not going to change that.
> > See
> > sections 3 and 4 of,
> > Antoci and Liebscher, "Reinstating Schwarzschild's original manifold
> > and its singularity", http://arxiv.org/abs/gr-qc/0406090
>
> I'll look at it when I get a chance....
Okay, I appreciate it.
> Tom Roberts
Best wishes,
Sabbir.
JanPB
>
> [Note that the very fact that the Schwarzschild solution had to be
> extended by Kruskal and Szekeres means that the Schwarzschild/Hilbert
> solution is *not* the most general spherically symmetric solution to
> the problem - the reason that _this_ mistake was made
It's not a mistake, Schwarzschild's coordinate system is simply a chart
covering a subset covered by Kruskal and Szekeres. One counterintuitive
aspect of the Schwarzschild is that what appears as a "line" r=2M is
really a "point" in 1+1 (really a 2-sphere, or the Einstein-Rosen
"neck") in a manner similar to the Mercator representation of earth's
north pole (say) as a line. On top of that, Schwarzschild omits the
most important portion of the horizon through which matter and light
signals fall(!) So these are very counterintuitive coordinates.
> was because of
> the prior assumption, which is subtly hidden in the way that the
> problem is usually formulated, and which again depends upon the
> misinterpretation of r as the radial coordinate as opposed to an
> apriori unknown parametristation of the radius,
I'm not sure why you keep repeating that - the usual derivation only
assumes the radial coordinate is related to the sphere surface area via
the usual 4 pi r^2. It's just a matheatical assumption, nobody ever
claimed this coordinate was supposed to have a meaning besides that.
--
Jan Bielawski
LEJ Brouwer
JanPB wrote:
> LEJ Brouwer wrote:
> >
> > [Note that the very fact that the Schwarzschild solution had to be
> > extended by Kruskal and Szekeres means that the Schwarzschild/Hilbert
> > solution is *not* the most general spherically symmetric solution to
> > the problem - the reason that _this_ mistake was made
>
> It's not a mistake, Schwarzschild's coordinate system is simply a chart
> covering a subset covered by Kruskal and Szekeres. One counterintuitive
> aspect of the Schwarzschild is that what appears as a "line" r=2M is
> really a "point" in 1+1 (really a 2-sphere, or the Einstein-Rosen
> "neck") in a manner similar to the Mercator representation of earth's
> north pole (say) as a line. On top of that, Schwarzschild omits the
> most important portion of the horizon through which matter and light
> signals fall(!) So these are very counterintuitive coordinates.
One of the points that I have been trying to make is that matter and
light do not fall through the event horizon, because there is (a) no
interior solution since r<2M is not allowed, and (b) timelike and
spacelike directions cannot just swap places, which is what would have
to happen if matter were to fall past the horizon. In any case, it is
invalid to apply a singular transformation to a singular coordinate
choice to produce a 'nonsingular' coordinate system. As I have already
said, that is like dividing infinity by infinity and believing the
answer.
>
> > was because of
> > the prior assumption, which is subtly hidden in the way that the
> > problem is usually formulated, and which again depends upon the
> > misinterpretation of r as the radial coordinate as opposed to an
> > apriori unknown parametristation of the radius,
>
> I'm not sure why you keep repeating that - the usual derivation only
> assumes the radial coordinate is related to the sphere surface area via
> the usual 4 pi r^2. It's just a matheatical assumption, nobody ever
> claimed this coordinate was supposed to have a meaning besides that.
The use of the metric in the form ds^2 = A(r) dt^2 - B(r) dr^2 - r^2
dA^2 contains the (incorrect) hidden assumption that 0<r<Infty, and
that there is a one-to-one mapping between values of r and 2-spheres.
This immediately disallows solutions like that of Kruskal where the
solution manifold 'folds back' on itself, so that there is more than
one 2-sphere for a given r. And that is why the standard solution (i.e.
the Schwarzschild solution) needs to be analytically extended (though
the process by which this is done, as I have hinted at is rather
dubious). A correct definition of a spherically symmetric metric was
given by Synge in 1974, and as a result he was able to derive the
Kruskal-extended solution directly (though he still misinterprets the
radial coordinates):
J. L. Synge, "Model universes with spherical symmetry", Ann. di Mat.
Pura ed App., 98 (1974) 239-255.
> --
> Jan Bielawski
- Sabbir.
JanPB
> JanPB wrote:
> >
> > It's not a mistake, Schwarzschild's coordinate system is simply a chart
> > covering a subset covered by Kruskal and Szekeres. One counterintuitive
> > aspect of the Schwarzschild is that what appears as a "line" r=2M is
> > really a "point" in 1+1 (really a 2-sphere, or the Einstein-Rosen
> > "neck") in a manner similar to the Mercator representation of earth's
> > north pole (say) as a line. On top of that, Schwarzschild omits the
> > most important portion of the horizon through which matter and light
> > signals fall(!) So these are very counterintuitive coordinates.
>
> One of the points that I have been trying to make is that matter and
> light do not fall through the event horizon, because there is (a) no
> interior solution since r<2M is not allowed,
What do you mean "not allowed"? True, the metric for r<2M does not
satisfy the constraint you set up for yourself (namely, it's not
static) and it may contain other hidden assumptions (like r<infty) but
so what? It's still a solution to Einstein's equation and the
requirement for being static, etc., was just a simplifying (i.e., ad
hoc, random, man-made) assumption which in the end just turned out to
be an overkill. So you ended up with a more general solution covering a
larger region (the extra region - the interior - is disconnected, in
the topological sense, from the static exterior).
> and (b) timelike and
> spacelike directions cannot just swap places, which is what would have
> to happen if matter were to fall past the horizon.
This is a mathematical feature of the Schwarzschild representation of
the metric. Nothing really swaps places, it's just that the
mathematical process of solving the relevant differential equations
mirrors one another over the two components of the disconnected domain
so the *letters* r and t end up denoting different things for r<2M
and r>2M due to what is basically an abuse of notation. One reason for
the confusion is that this sort of "switch" in the signature is rarely
seen elsewhere so people tend to see something physical going on here.
The domain of this chart is topologically disconnected and the equation
describing a particular representation of the metric over such domain
is free to switch the letters denoting the coordinates over each
connected component as it pleases, without violating continuity. If it
bothers you, just switch the letters r and t in the interior
Schwarzschild formula. (It would be nice if GR textbooks did it, it
also would be nice if pop-sci books stopped incessantly harping on this
nonexistent "switch" - it's about as bad as the dreadful rubber sheet
so-called model.)
> In any case, it is
> invalid to apply a singular transformation to a singular coordinate
> choice to produce a 'nonsingular' coordinate system.
Why? Imagine your manifold is the real line minus the origin (zero).
You have 1/x there as a coordinate chart. It's singular at 0. Someone
then comes in and says: "OK, here I have the entire real line
(including zero) with the identity chart. Now if I temporarily ignore
the zero in my manifold, then it will be diffeomorphic to yours - in
fact 1/x is the diffeomorphism. But my manifold has another chart - the
identity - which covers the whole thing! So here I applied a singular
transformation to a singular coordinate system to produce a nonsingular
one."
_This is differential geometry 101_
> As I have already
> said, that is like dividing infinity by infinity and believing the
> answer.
No, these guys who wrote those papers need to study the basics more.
Spivak is back in print in a nicely TeX-reset 3rd edition so there is
no excuse not to read it (first two volumes)! This is all really basic
and must be clearly understood before climbing that GR mountain.
> > > was because of
> > > the prior assumption, which is subtly hidden in the way that the
> > > problem is usually formulated, and which again depends upon the
> > > misinterpretation of r as the radial coordinate as opposed to an
> > > apriori unknown parametristation of the radius,
> >
> > I'm not sure why you keep repeating that - the usual derivation only
> > assumes the radial coordinate is related to the sphere surface area via
> > the usual 4 pi r^2. It's just a matheatical assumption, nobody ever
> > claimed this coordinate was supposed to have a meaning besides that.
>
> The use of the metric in the form ds^2 = A(r) dt^2 - B(r) dr^2 - r^2
> dA^2 contains the (incorrect) hidden assumption that 0<r<Infty, and
> that there is a one-to-one mapping between values of r and 2-spheres.
> This immediately disallows solutions like that of Kruskal where the
> solution manifold 'folds back' on itself, so that there is more than
> one 2-sphere for a given r.
Yes, so what? You've constrained yourself to seek only solutions of
certain type, over certain domain, that's what you got (more or less,
you actually got a bit more: the non-static interior). This is not
"incorrect", it's a man-made self-limitation and it resulted in a
coordinate chart valid over a domain which - it turns out - is a subset
of some other domain over which another coordinate representation *of
the same metric* is valid.
This is again the basics - see Spivak volume 1.
> And that is why the standard solution (i.e.
> the Schwarzschild solution) needs to be analytically extended (though
> the process by which this is done, as I have hinted at is rather
> dubious).
No, it's logically and mathematically very plain and boring. Even to
call analytical extension a "process" is a bit of a stretch because it
suggests some definite procedure is applied whose validity might then
be questioned. In fact one merely gets hold of *another* solution
(another coordinate representation of the metric) and then notices that
its domain is larger than the original domain. Such new solution is
then referred to as an "extension", as if some sort of action of
"extending" has taken place. This is just reification, scientists like
to talk like that.
> A correct definition of a spherically symmetric metric was
> given by Synge in 1974, and as a result he was able to derive the
> Kruskal-extended solution directly (though he still misinterprets the
> radial coordinates):
>
> J. L. Synge, "Model universes with spherical symmetry", Ann. di Mat.
> Pura ed App., 98 (1974) 239-255.
That's all fine. It just doesn't mean what you seem to think it means.
It's all far less sinister :-)
--
Jan Bielawski
LEJ Brouwer
called "Flogging the Xprint" on sci.physics.research:
tes...@um.bot wrote:
> Sabbir Rahman ("LEJ Brouwer") asked about a group of arXiv eprints which
> (as I think he knows) have previously been castigated in this newsgroup
> for committing elementary student errors.
I referenced preprints by Abrams, Antoci and Crothers. I did a search
in sci.physics.research for each of these authors, and none of their
papers have been discussed at all, let alone been 'castigated'. Given
that your first statement is blatantly untrue, this does not leave much
hope for the objectivity of the rest of your post.
> Steve Carlip briefly explained the most fundamental of these errors, but
> unfortunately the comment by Mark Hopkins muddies the waters by going off
> in a different direction (and by stating as established "truth" a
> controversial assertion).
I have responded to these assertions on sci.physics.relativity. The
moderators can explain why I am not being allowed to reply on
s.p.research.
> I just want to make sure that everyone understands the only really
> important point here: the papers Rahman mentioned belong to a group of
> papers which are founded upon serious but elementary misconceptions. I
> call these "Xprints" in homage to a Monty Python skit.
Yes, "bicycle repair man" was one of my favourites.
> No bits were actually harmed in producing this post :-/
>
> === The papers ===
>
> It turns out there is a sizable group of arXiv eprints
>
> http://arxiv.org/find/gr-qc/1/au:+Antoci_S/0/1/0/all/0/1
> http://www.arxiv.org/find/gr-qc/1/au:+Liebscher_D/0/1/0/all/0/1
> http://arxiv.org/find/gr-qc/1/au:+Mihich_L/0/1/0/all/0/1
> http://arxiv.org/find/grp_physics/1/au:+Loinger_A/0/1/0/all/0/1
> http://arxiv.org/find/grp_physics/1/au:+Abbassi_A_H/0/1/0/all/0/1
>
> http://arxiv.org/find/grp_physics/1/au:+Zakir_Zahid/0/1/0/all/0/1
> (I now see that his 1999 eprints have been withdrawn)
>
> http://arxiv.org/find/grp_physics/1/au:+Abrams_L/0/1/0/all/0/1
> http://arxiv.org/find/grp_physics/1/au:+Mitra_Abhas/0/1/0/all/0/1
>
> and even published papers, including several by Leonard S. Abrams and
> Stephen J. Crothers, which
Well, I am referring to specific papers by Abrams, Antoci and Crothers,
so I am not sure why you are dragging a whole bunch of other authors
into the discussion.
> 1. claim that all the standard gtr textbooks misrepresent and misinterpret
> the Schwarzschild vacuum solution, following a "mistake" they generally
> attribute to Hilbert 1917,
>
> 2. repeat a number of -elementary- mistakes, all involving fundamental
> misconceptions concerning the roles played by atlases, pullbacks, and
> diffeomorphisms in manifold theory, and consequently,
>
> 3. confuse the concept of coordinate singularity with the concept of a
> singularity in some physical field (or in a geometric quantity such as the
> metric or curvature tensor).
>
> These papers often use absurdly overcomplicated notation which can obscure
> the elementary nature of their most fundamental errors. (Some of these
> papers also commit more sophisticated errors--- but here, I just want to
> make sure everyone understands the -simple- stuff!)
So you have just made a long list of claims about these papers, without
proving any of them claims. Shall we just assume that your claims
represent incontrovertible facts, as if handed down from God, or are
you actually going to justify your claims? Also, I don't know about the
other authors you reference, but the papers I referred to seem pretty
clear in terms of notation used.
> One of the above cited authors, Salvatore Antoci, has also translated some
> relevant historical papers into English and posted them to the arXiv
> (unfortunately adding misleading "editorial comments"):
Unfortunately you omit to mention which 'editorial comments' you find
to be misleading or why.
> Karl Schwarzschild,
> On the gravitational field of a mass point according to Einstein's theory
> http://www.arxiv.org/abs/physics/9905030
>
> Karl Schwarzschild,
> On the gravitational field of a sphere of incompressible fluid according
> to Einstein's theory
> http://www.arxiv.org/abs/physics/9912033
>
> Marcel Brillouin,
> The singular points of Einstein's Universe,
> http://www.arxiv.org/abs/physics/0002009
>
> === The problems ===
>
> Recall that in the now standard "Scharzschild exterior chart" for the
> Schwarzschild vacuum, the line element expressing the Schwarzschild vacuum
> solution takes the form
>
> ds^2 = -(1-2m/r) dt^2 + 1/(1-2m/r) dr^2 + r^2 dOmega^2,
>
> 2m < r < infty, 0 < theta < pi, -pi < phi < pi
>
> I have often argued that physicists must learn the habit of -always-
> stating coordinate ranges, since -failing- to do so can easily mislead
> students, while -following- this precept can prevent embarrassing errors.
Well, essentially that's what I have been arguing all along.
> For example, in (2) of gr-qc/0310104, the stated range 0 < r < infty is
> -incorrect- because of the coordinate singularity at r=2m. In fact, the
> range 0 < r < 2m gives Schwarschild coordinates for a perfectly valid
> -interior- patch, but this patch is -disjoint- from the -exterior- patch
> above!
I see, in one breath you mention that the solution is only valid for 2m
< r < infty, and in the next breath you say that, 'in fact' it is valid
for 0 < r < 2m as well. Talk about inconsistent logic. Using the words
'in fact' prior to making a claim does not constitute a proof of that
claim.
> This is not just a quibble because this oversight is in fact an
> integral part of a mistake made in this paper, as we shall see.
The only oversight here is your own.
In fact, the range 0 < r < 2m gives coordinates for a patch that does
not physically exist. Since you haven't bothered reading the papers I
mentioned to find out why, and are intent on misleading everyone else
who hasn't, I shall explain once again why 0 < r < 2m does not
correspond to any physical region of spacetime.
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/sqrt{C}) dt^2 - (sqrt{C} / (sqrt{C} - 2m)) (C'^2/4C)
dr^2 - C dOmega^2 (eqn 7 of Crothers).
Now, if the coordinate associated with the point mass is r=sqrt{C}=r0
(which may be chosen such that r0=0), then we can define the 'proper
radius' from the mass at r0 to any r>r0 to be the integral from ro to r
of sqrt{-grr}:,
R = int sqrt{-grr} dr from r0 to r (eqn 11 of Crothers)
Given the boundary condition that R->0 as r->r0, this implies the
condition C(r0)=4m^2, and that R is given by:
R(r) = sqrt{sqrt{C}*(sqrt{C} - 2m)} + 2m ln
|(sqrt{sqrt{C}}+sqrt{(sqrt{C} - 2m)} / sqrt{2m}| (eqn 14 of Crothers)
where we must have C(r0)=4m^2, i.e. r*(r0) = 2m, so that the radial
position of the point mass in the Schwarzschild solution must be at r*
= 2m (which also coincides with the event horizon). Hilbert's error was
to confuse r* with r.
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.
It's as simple as that. So there is NO solution for r* < 2m, because r*
< 2m DOES NOT PHYSICALLY EXIST.
If you do not understand this proof, then I suggest you read through
Crother's paper again, bearing in mind Antoci's clear explanation of
the incorrect assumption Hilbert made on setting r = sqrt{C}.
> Recall also that the "Schwarzschild radial coordinate" r has a simple
> geometric interpretation, which was apparently first pointed out by
> Hilbert 1917, namely: the locus t=t0, r=r0 is a geometric sphere with line
> element
>
> dsigma^2 = r0^2 (dtheta^2 + sin(theta)^2 dphi^2)
>
> and with surface area 4 pi r0^2.
That's fine, but it does not affect the 'proper radius' argument
outlined above.
> Now, in his 1923 paper (posted to the arXiv by Antoci), Brillouin makes
> the simple substitution
>
> R = r-2m,
>
> 2m < r < infty
>
> This defines a diffeomorphism (2m,infty) -> (0,infty), i.e. a coordinate
> transformation. This substitution brings the line element into the form
>
> ds^2 = -1/(1+2m/R) dt^2 + (1+2m/R) dR^2 + (R+2m)^2 dOmega^2,
>
> 0 < R < infty, 0 < theta < pi, -pi < phi < phi
>
> (In modern language, we have pulled back the metric tensor under our
> diffeomorphism.) This is a new coordinate chart covering precisely the
> same region as the old chart. Geoemetrically speaking, this expression
> gives the -same- metric tensor (placed on some underlying smooth manifold,
> if you like) as before--- it has just been expressed in a new chart.
> Actually, this new chart doesn't really improve anything, because it also
> has a coordinate singularity at the event horizon; it simply relabels the
> locus r=2m as the locus R=0. Unfortunately, Brillouin failed to recognize
> this.
As I already mentioned in response to Steve Carlip, it does improve
things in the sense that we are less likely to assume that -2m < r < 0
is physical, when it is not. Hilbert and indeed pretty much everyone
else made the incorrect association of r with the proper radius, and
this choice of coordinates makes such an error less easy to make as we
are unlikely to associate -2m < r < 0 with the proper radius.
> In addition, Brillouin incorrectly used the range 0 < r < infty in the
> original chart,
And so did Hilbert.
> and he then became confused by the fact that the
> coordinate vector field @/@t is timelike "outside" and spacelike "inside"
> the horizon, while @/@r is spacelike "outside" but timelike "inside".
He was absolutely correct to be confused. The confusion disappears when
we realise that 0 < r < 2m is physically invalid.
> This
> observation, which unfortunately is often repeated verbatim by modern
> physicists--- who ought to know better--- doesn't make sense as stated
> since these are in fact -two disjoint charts- neither of which can be
> extended as they stand through the event horizon (to do that you have to
> adopt a more suitable chart, such as the charts introduced by Painleve
> 1921, Eddington 1922, or LeMaitre 1933, all of which are well defined at
> r=2m, and in fact -overlap- both the interior and exterior regions, which
> is: they can be used to -extend- from the exterior to the interior, or
> vice versa).
Of course this is all rubbish as we have just shown that there is no
'interior region'
> Brillouin felt that this means that the event horizon is some kind of
> "impassable physical barrier". But of course, according to gtr this is
> wrong, as all the modern textbooks explain.
You mean, as all the modern textbooks 'fudge'. Brillouin was absolutely
right for the reasons mentioned above.
> (The notion of frame fields, a way of "decorating" the Lorentzian manifold
> by a structure which can be drawn, and which also has an immediate
> physical meaning in terms of the physical experience of some class of
> observers, renders such issues transparent. Due to lack of time, this
> important notion is rarely taught in introductory gtr courses, but see the
> new book by Eric Poisson.)
>
> Even worse, Brillouin apparently thought that the locus t=t0, R=0 is a
> geometric -point-. In fact, it is a -sphere- with surface area A = 8m^2,
> as is carefully explained in standard textbooks like Misner, Thorne, &
> Wheeler, Gravitation, 1973. Several contemporary authors have repeated
> this second error of Brillouin, some even arguing that since "the point"
> R=0 "has no interior" (sic), the interior region must not exist!
Adding an exclamation mark does not make the statement any less true.
Brillouin was correct because the locus is at proper radius (as defined
above, or according to any other sensible measure of radial distance
from the origin) is zero. It is also true that the locus has finite
area, and this apparent inconsistency is a reflection of the
singularity in the metric at that point.
> The papers by Antoci et al., Loinger, and Abrams all use a much more
> confusing notation but make essentially the same basic errors, although
> they are a bit harder to spot.
That's complete rubbish.
> These authors make much of the fact that Schwarzschild's original paper
> used a more general coordinate chart than is found in modern textbooks.
> The implicit argument seems to be that since these physics textbooks
> oversimplify the historical details, they must be "lying" about the
> physics too! ;-/
So you say that the papers accuse the textbooks of 'lying'? Where? In
any case, Crothers lists a set of solutions which is even more general
than Schwarzschild's - see eqn 17 of his paper, and following that
summarises very neatly how earlier descriptions correspond to varous
parameter choices. I think that the fact that he was able to do this is
highly praiseworthy.
> To obtain this more general chart from the now standard Schwarzschild
> exterior chart, put
>
> rho = (a^3 + r^3)^(1/3)
>
I think you mean rho = (a^3 + r^3)^(2/3) here?
> where a is a second parameter (the first being the mass parameter m).
a=2m by definition. a is not a 'second parameter'. Unless you mean a is
the same as my r0 above - in which case the above equation is wrong.
Reference please?
> It
> should be immmediately apparent that this second parameter merely -adjusts
> the radial coordinate-, but does not affect any physics. (To be truly
> fussy I should write "rho_a" since changing "a" gives a -different- radial
> coordinate--- but not, of course, a different manifold!)
I have lost you now - what work are you referring to, or are you just
making this up as you go along?
> But in gr-qc/0102084, Antoci et al. incorrectly state (I have slightly
> changed the notation to agree with that used here): "for different values
> of a... the solutions are geometrically and physically different". This
> claim exhibits the same fundamental misconception about diffeomorphisms
> and coordinate charts in manifold theory as Brillouin 1923.
Although I can't find the statement in the paper you refer to, the
statement is correct as 'a' is proportional to the mass 'm' as
mentioned above. Maybe you are confusing notation again?
> If we carry out the coordinate transformation, the line element now takes
> the form
>
> ds^2 = -(1-2m/(rho^3+a^3)^(1/3)) dt^2
>
> + rho^4/(rho^3+a^3)/((rho^3+a^3)^(1/3) -2m) drho^2
>
> + (rho^3+a^3)^(2/3) dOmega^2,
>
> (8m^3-a^3)^(1/3) < rho < infty
>
> As before, this is the -same- metric tensor, namely the metric tensor
> defining the Schwarzschild vacuum solution; it has simply been
> re-expressed in new coordinates. Unfortunately, many physicists
> carelessly speak of "the new metric" in cases like this, which can easily
> confuse students--- or authors like Antoci!
>
> Hilbert noticed that if we set a = 0, the somewhat complicated line
> element we just obtained simplifies considerably, giving the now standard
> "Schwarzschild exterior chart". Even better, with a = 0, rho = r now
> acquires the memorable geometric interpretation mentioned above.
>
> OTH, if we set a = 2m, we obtain a chart valid on the range 0 < rho <
> infty. Here, the locus rho = 0 corresponds to the event horizon. This
> chart is the one favored by Antoci et al., but to paraphrase Carlip's
> comment, using this chart is a -really bad idea-, because it -greatly-
> increases the complexity of the components of the metric and other
> quantities, while yielding no compensatory advantage whatever!
Again, I disagree - see my comments above.
> For example, the magnitude of the acceleration of static test particles is
>
> m/(rho^3 + 8 m^3)^(1/2)/(rho^3 + 8 m^3)^(1/3)-2m))^(1/2)
>
> (Expanding this in powers of 1/rho confirms that the parameter m has the
> same interpretation in terms of the mass of the central object as it has
> in the standard Schwarzschild exterior chart.) And the tidal tensor as
> measured by static observers is
>
> m/(rho^3+a^3) diag(-2,1,1)
>
> Here, the meaning of rho is that the surface area of the sphere t=t0,
> rho=rho0 is
>
> A = 4 pi (rho0^3 + 8 m^3)^(2/3)
>
> Compare these expressions with (respectively):
>
> m/r/sqrt(1-2m/r)
>
> m/r^3 diag(-2,1,1)
>
> A = 4 pi r0^2
>
> This shows why the mainstream has been -wise- to adopt the coordinate
> normalization suggested by Hilbert!
>
> (I say "normalization" because the only remaining coordinate freedom in
> the standard chart corresponds to the time translation symmetry and the
> spherical symmetry which were assumed in deriving the solution.)
>
> But authors like Antoci and Abrams insist that Hilbert made some kind of
> mistake and that all the textbooks are wrong :-/
>
> Several of these authors also claim that there is something wrong with
> Eddington's 1922 extension of the exterior chart past the event horizon.
> Recall the Carter-Penrose conformal diagram:
>
> future
> singularity
>
> 888888888888 i^+ ("future timelike infinity")
> /\ /\
> / \ future / \
> / \ int. / \
> / \ / \ scri^+ ("future null infinity")
> / second \ / first \
> / exterior \/ exterior \
> \ region /\ region / i^0 ("spatial infinity", r = infty)
> \ / \ /
> \ / \ /
> \ / past \ / scri^- ("past null infinity")
> \ / int. \ /
> \/ \/
> 888888888888 i^- ("past timelike infinity")
>
> past
> singularity
>
> Eddington's extension covers the region
>
> 888888888888
> /\**********/\
> / \********/**\
> / \******/****\
> / \****/******\
> / \**/********\
> / \/**********\
> \ /\**********/
> \ / \********/
> \ / \******/
> \ / \****/
> \ / \**/
> \/ \/
> 888888888888
>
> Some of these careless/confused authors note that the coordinate
> transformation (a diffeomorphism defined on the exterior region) between
> the Schwarzschild and Eddington charts is -not defined- on the event
> horizon. They claim that this implies that the extension is spurious!
>
> But this is analogous to considering a locally flat chart covering the
> left half plane
>
> ds^2 = dxi^2/xi^2 + dy^2,
>
> 0 < xi < infty, -infty < y < infty
>
> and claiming that this cannot be extended to the usual Euclidean plane
> because the diffeomorphism x = 1/xi is well-defined only on the right half
> plane. Such a claim would of course be nonsense, and would exhibit a
> fundamental misconception concerning the role of diffeomorphisms in
> manifold theory. In this example, note also that 0 < xi < infty and
> -infty < xi < 0 give -two- nonoverlapping charts; it would be quite wrong
> to give the range as -infty < xi < infty! But this is precisely analogous
> to one of the elementary errors which is committed by Antoci et al., as
> noted above!
>
> So the proper response to these authors is that -of course- the coordinate
> transformations in question are not defined on the horizon! That is why
> we say this locus represents a "coordinate singularity"! :-/
>
> Abhas Mitra makes an odd computational error in the Kruskal-Szekeres
> chart. His error is esssentially a goof using elementary calculus to
> compute a limit of an implicitly defined function. Since we extensively
> discussed this error in this very newgroup, many years hence, I will just
> say that it can be helpful to use the Lambert W function to express the
> Kruskal-Szekeres line element -explicitly-, rather than use an implicit
> formulation as in Misner, Thorne, & Wheeler and other textbooks.
>
> The Lambert W function is the special function given by solving w = z
> exp(z) for z. This is a multivalued function with a branch point at -1/e,
> but the principal branch is both real valued and single valued on
> (-1/e,infty), which is just the range we need to write down the
> Schwarzschild metric tensor on the maximal analytic extension in terms of
> the Kruskal-Szekeres chart.
>
> (BTW, the KS chart is not quite a -global chart-, since it still has
> coordinate singularites at the "international date line" of each of our
> nested spheres, but it comes close.)
>
> Mitra claims that the tangent vector to the world line of a freely and
> radially infalling test particle becomes null on the horizon. Here too,
> frame fields are useful in verifying that this is incorrect, because we
> can -draw- the frame fields and then we "see" the result: a (correct)
> computation of the Kruskal-Szekeres components of the timelike unit vector
> field in question shows that these infalling timelike geodesics are
> perfectly well behaved at the horizon.
None of the above is related to the point I am trying to make. For one
thing, I do not reject the Kruskal extension to the exterior solution
(and as I said, I DISAGREE with Abrams et alii on this point) - I
reject the physical existence of an interior solution, and secondly, I
am not defending Abhas Mitra here - he is quite capable of defending
himself, and if you would like to point out his mistakes to him, I am
sure he will be glad to respond to you in person.
>
> === The Lessons ===
>
> What can we learn from reviewing this protracted parade of goofs?
That the physics establishment is more than capable of screwing up en
masse.
- Sabbir.
LEJ Brouwer
JanPB wrote:
> LEJ Brouwer wrote:
> > JanPB wrote:
>
> What do you mean "not allowed"? True, the metric for r<2M does not
> satisfy the constraint you set up for yourself (namely, it's not
> static) and it may contain other hidden assumptions (like r<infty) but
> so what? It's still a solution to Einstein's equation and the
> requirement for being static, etc., was just a simplifying (i.e., ad
> hoc, random, man-made) assumption which in the end just turned out to
> be an overkill. So you ended up with a more general solution covering a
> larger region (the extra region - the interior - is disconnected, in
> the topological sense, from the static exterior).
I refer you to the reply I gave to T.Essel's post. It is clearly not a
solution to the Einstein equation if the interior region yuo refer to
does not even physically exist.
> > and (b) timelike and
> > spacelike directions cannot just swap places, which is what would have
> > to happen if matter were to fall past the horizon.
>
> This is a mathematical feature of the Schwarzschild representation of
> the metric. Nothing really swaps places, it's just that the
> mathematical process of solving the relevant differential equations
> mirrors one another over the two components of the disconnected domain
> so the *letters* r and t end up denoting different things for r<2M
> and r>2M due to what is basically an abuse of notation.
I see. And I take it that you will also remember to swap these letters
in the Einstein field equations? Why not go ahead and also swap M and
theta? In fact, why not change 'ds^2' to 'sd^2' too, just for the hell
of it? They are just letters after all.
- Sabbir.
Koobee Wublee
> [...]
>> Recall that in the now standard "Scharzschild exterior chart" for the
>> Schwarzschild vacuum, the line element expressing the Schwarzschild vacuum
>> solution takes the form
>>
>> ds^2 = -(1-2m/r) dt^2 + 1/(1-2m/r) dr^2 + r^2 dOmega^2,
>>
>> 2m < r < infty, 0 < theta < pi, -pi < phi < pi
>>
>> I have often argued that physicists must learn the habit of -always-
>> stating coordinate ranges, since -failing- to do so can easily mislead
>> students, while -following- this precept can prevent embarrassing errors.
>
> Well, essentially that's what I have been arguing all along.
> [...]
Schwarzschild metric is only valid in free space where the mass density
is zero. The characterize where (r <= 2 G M / c^2), you have to find
another metric that satisfy the field equations below.
R_ij - R g_ij / 2 = 8 pi G Lm g_ij / c^4, where in this case (Lm > 0).
While forming a black hole, as you have correctly pointed out, it can
only form at eternal time. In doing so, the above equation must be
valid at some point to give another solution different from the
Schwarzschild metric, and very possibly it will prevent the formation
of a black hole or pushing the event horizon down to zero.
JanPB
> JanPB wrote:
> > LEJ Brouwer wrote:
> > > JanPB wrote:
> >
> > What do you mean "not allowed"? True, the metric for r<2M does not
> > satisfy the constraint you set up for yourself (namely, it's not
> > static) and it may contain other hidden assumptions (like r<infty) but
> > so what? It's still a solution to Einstein's equation and the
> > requirement for being static, etc., was just a simplifying (i.e., ad
> > hoc, random, man-made) assumption which in the end just turned out to
> > be an overkill. So you ended up with a more general solution covering a
> > larger region (the extra region - the interior - is disconnected, in
> > the topological sense, from the static exterior).
>
> I refer you to the reply I gave to T.Essel's post.
I can't see it.
> It is clearly not a
> solution to the Einstein equation if the interior region yuo refer to
> does not even physically exist.
*If*. You haven't shown this. If 2+2=5 then all sorts of marvellous
results follow.
> > > and (b) timelike and
> > > spacelike directions cannot just swap places, which is what would have
> > > to happen if matter were to fall past the horizon.
> >
> > This is a mathematical feature of the Schwarzschild representation of
> > the metric. Nothing really swaps places, it's just that the
> > mathematical process of solving the relevant differential equations
> > mirrors one another over the two components of the disconnected domain
> > so the *letters* r and t end up denoting different things for r<2M
> > and r>2M due to what is basically an abuse of notation.
>
> I see. And I take it that you will also remember to swap these letters
> in the Einstein field equations?
> Why not go ahead and also swap M and
> theta? In fact, why not change 'ds^2' to 'sd^2' too, just for the hell
> of it? They are just letters after all.
No, this is incorrect. Very briefly the logic of finding this solution
goes like this. You restrict yourself to spherically symmetric static
metrics and seek solutions of this type. This forces certain
constraints on the coefficients of the metric. You denote the timelike
coordinate by t and the spacelike coordinate by r. You obtain a
solution which is only valid for r>2M (because of the assumption of t
timelike and r spacelike).
_Then_ you look at it and suddenly you notice that for r<2M the formula
still describes _a_ metric of Lorentz signature. This metric is not the
one obtained from the derivation, so to make sure it satisfies the
Einstein equation, you have to check it. Of course this is trivial
because the r<2M case is sufficiently close in purely mathematical
terms to the metric with r>2M that it's obvious its Ricci tensor is
also zero.
But note that the meaning of each coordinate in the new (r<2M) case has
nothing a priori to do with the coordinates that _happen_ to be denoted
by the same letters in the r>2M case - that's because you simply
mechanically "extended" your solution to a region not covered by the
assumptions of your derivation, where by "extended" one means simply
_mechanical copying_ of the original (r>2M) formula, retaining the
original Latin and Greek symbols and all - out of pure human laziness.
This is the abuse of notation I referred to previously.
There is no a priori physical connection between the two sets of Latin
and Greek symbols - they are the same only because we didn't feel like
using new symbols for the new metric.
The meaning of the coordinate symbols in the new metric must be deduced
from the form of the coefficients, not from the _names_ of the
coordinates! In particular, what is denoted by r is time and what is
denoted by t is a spatial coordinate because of the signs.
--
Jan Bielawski
Sue...
Yes... you are correct.Where the appearance of a distant moving
clock is to be corrected then the term 'proper time' has a literal
application to the coordinate. (Special Relativity)
When a space-time is being defined to represent energy density
as volume, then the temporal component becomes imaginary.
(General Relativity and the Swartzchild solution)
The second URL below explains in nearly identical terms.
http://farside.ph.utexas.edu/teaching/em/lectures/node114.html
http://www.aoc.nrao.edu/~smyers/courses/astro12/speedoflight.html
Sue...
>
> --
> Jan Bielawski
Koobee Wublee
> The moderators of s.p.research have decided that Steve Carlip's
> response has settled this issue, and refuse to allow further
> discussion. I disagree, so I am continuing the thread here, and would
> like to invite Steve to respond if he wishes - Sabbir.
Both Schwarzschild and Hilbert made errors. Refer to the following
article written by Schwarzschild himself.
http://arxiv.org/abs/physics/9905030
Equation (4) is only valid when the metric has a determinant of -1.
Subsequent derivation of the Schwarzschild metric has been derived
based on this error. In spherical polar cordinate, the determinant of
the metric is not -1. The retangular coordinate should not be -1
either in which Schwarzschild made another mistake in his equation
right above equation (6). Equating the spherical polar coordinate with
the rectangular one, we have
dx^2 + dy^2 + dz^2 = dr^2 + r^2 cos^2V dH^2 + r^2 dV^2
Where
** r^2 = x^2 + y^2 + z^2
** H = longitude
** V = latitude
We can also write
dx^2 + dy^2 + dz^2 = (x dx + y dy + z dz)^2 / r^2 + ((y^2 + z^2) dx^2 +
(z^2 + x^2) dy^2 + (x^2 + y^2) dz^2 - 2 x y d xdy - 2 y z dy dz - 2 z x
dz dx) / r^2
Where
** dr^2 = (x dx + y dy + z dz)^2 / r^2
** r^2 (cos^2V dH^2 + dV^2) = ((y^2 + z^2) dx^2 + (z^2 + x^2) dy^2 +
(x^2 + y^2) dz^2 - 2 x y dx dy - 2 y z dy dz - 2 z x dz dx) / r^2
This is basic geometry so far. This is flat space so far. Now, adding
the curvature in the radial direction as well as the angular direction,
we have
F dr^2 + G r^2 (cos^2V dH^2 + dV^2) = (F^2 x^2 + G y^2 + G z^2) dx^2 +
(G^2 x^2 + F y^2 + G z^2) dy^2 + (G^2 x^2 + G y^2 + F z^2) dx^2 + 2 ( F
- G) (x y dx dy + y z dy dz + z x dz dx)
Common sense says G = 1, F = 1 at r = infinity.
The spatial components of the metric in spherical polar coordinate is
** g_rr = F
** g_HH = G r^2 cos^2V
** g_VV = G r^2
** g_rH = g_rV = g_HV = 0
The spatial components of the metric in rectangular coordinate is
** g_xx = (F^2 x^2 + G y^2 + G z^2) / r^2
** g_yy = (G^2 x^2 + F y^2 + G z^2) / r^2
** g_xx = (G^2 x^2 + G y^2 + F z^2) / r^2
** g_xy = 2 (F - G) x y / r^2
** g_yz = 2 (F - G) y z / r^2
** g_zx = 2 (F - G) z x / r^2
This is not the same as what Schwarzschild wrote down. Both metrics
don't have determinants equivalent to -1. Using the more complicated
Ricci Tensor, there is no way anyone can arrive at the Schwarzschild
metric with Ricci Tensor equals to zero in free space. Ricci Tensor is
wrong. Einstein's Field Equations are total nonsense.
There are only 9 non-zero Christoffel Symbols of the 2nd kind in
spherical polar coordinate. It is a start. However, it is getting
late for me. If there is anyone interested, I can post more of the
mechanical derivations.
JanPB
> "LEJ Brouwer" <intuit...@yahoo.com> wrote in message
> news:1152183483....@j8g2000cwa.googlegroups.com...
>
> > The moderators of s.p.research have decided that Steve Carlip's
> > response has settled this issue, and refuse to allow further
> > discussion. I disagree, so I am continuing the thread here, and would
> > like to invite Steve to respond if he wishes - Sabbir.
>
> Both Schwarzschild and Hilbert made errors. Refer to the following
> article written by Schwarzschild himself.
>
> http://arxiv.org/abs/physics/9905030
>
> Equation (4) is only valid when the metric has a determinant of -1.
> Subsequent derivation of the Schwarzschild metric has been derived
> based on this error.
It's not an error, just a historical and - as Einstein realised before
November 1915 - unnecessary restriction on allowable coordinate
changes.
> [...]
> This is not the same as what Schwarzschild wrote down. Both metrics
> don't have determinants equivalent to -1. Using the more complicated
> Ricci Tensor, there is no way anyone can arrive at the Schwarzschild
> metric with Ricci Tensor equals to zero in free space. Ricci Tensor is
> wrong. Einstein's Field Equations are total nonsense.
Study this stuff some more, and seriously, then you won't have to waste
your time on nonexisting problems.
--
Jan Bielawski
Koobee Wublee
news:1152727769.7...@75g2000cwc.googlegroups.com...
>> Both Schwarzschild and Hilbert made errors. Refer to the following
>> article written by Schwarzschild himself.
>>
>> http://arxiv.org/abs/physics/9905030
>>
>> Equation (4) is only valid when the metric has a determinant of -1.
>> Subsequent derivation of the Schwarzschild metric has been derived
>> based on this error.
>
> It's not an error, just a historical and - as Einstein realised before
> November 1915 - unnecessary restriction on allowable coordinate
> changes.
It is not an error for Einstein because he changed the rules of
mathematics. However, it remains a problem for everyone else.
>> [...]
>> This is not the same as what Schwarzschild wrote down. Both metrics
>> don't have determinants equivalent to -1. Using the more complicated
>> Ricci Tensor, there is no way anyone can arrive at the Schwarzschild
>> metric with Ricci Tensor equals to zero in free space. Ricci Tensor is
>> wrong. Einstein's Field Equations are total nonsense.
>
> Study this stuff some more, and seriously, then you won't have to waste
> your time on nonexisting problems.
Assuming the angular distortion is unity just like the Schwarzschild
metric. The radial distortion from one field equation looks like
g_11 = 1 / (2 + k sqrt(r))
Where
** k = integration constant
Versus
g_11 = 1 / (1 + k U)
Where
** U = G M / r / c^2
I have not even bothered to calculate g_00 from another field equation.
It appears not to be (1 / g_11) though. The other two field equations
are identical because we have assumed no angular distortion in space.
You are right. I have spent a lot of time doing this, and I feel
satisfied. The field equations are utterly nonsense because of the
mathematical errors made by Schwarzschild and then Hilbert according to
Mr. Rahman.
Tom Roberts
> How can spacelike and timelike directions suddenly swap places if there
> is nothing unusual going on at the event horizon?
That does not happen. What does happen, when using the standard Schw.
coordinates, is that the coordinates LABELED r and t swap their
spacelike and timelike characters at the horizon. But that's OK, because
neither set of coordinates is valid at the horizon -- these TWO sets of
coordinates are valid on disjoint regions of the manifold, and how the
labels get assigned is of no physical significance whatsoever.
This is a source of confusion to students today, and was
confusing to experts until the 1960s or so.
> This is completely
> unphysically,
No physical anything happens. Merely the labels of coordinates get shuffled.
> yet is accepted in a completely matter-of-fact way in all
> standard textbooks. Anyone considering this objectively should
> immediately smell a rat here.
Only someone who does not understand what is happening would say such
silly things.
> the main point I am making which is the fact that r<2M has no
> meaning for the Schwarzschild coordinates,
You don't seem to realize that there are _TWO_SET_ of Schwarzschild
coordinates. One set is valid for r>2M and the other is valid for r<2M.
They have no intersection. Yes, the exterior coordinates are not valid
in the region r<2M. <shrug>
> so that the usual interior
> solutions are invalid.
This is not true. They are valid, but are covered by a different chart
in the Schw. coordinates.
Tom Roberts
Tom Roberts
> The meaning of the coordinate symbols in the [interior] metric must be
> deduced from the form of the coefficients, not from the _names_ of the
> coordinates! In particular, what is denoted by r is time and what is
> denoted by t is a spatial coordinate because of the signs.
And to make it even more confusing to the unwary, in the region r<2M the
future-pointing basis vector is -d/dr (note the minus sign) -- that is
what makes the singularity at r=0 be in the future of every timelike
path in this region (such paths enter this region at r=2M, which is
greater than 0).
Tom Roberts
brian a m stuckless
> IOW: this discussion is not in a vacuum, it is in the context of GR
$$ Any SPACE with "discussion" going on in it, clearly isN'T EMPTY.
> Do you "suspend your belief" that 1+1=2? Why ask me to "suspend
> belief" for a concept equally well established?
$$ That BOTH (ALL) ones are identical is an ARBiTRARY *assumption*.
$$ [That 1+1=2, is a totally ARBiTRARY *ANTHROPOLOGiCAL* STANDARD].
$$ [Argument against SI (international) ARBiTRARY STANDARDs apply].
$$ [No MORE arbitrary to ASSUME SI than ASSUME all ONEs are EXACT].
$$ ASSUMPTiON that ALL ONEs are EXACT ..totally a HUMAN convention.
$$ [NATURE has no such LAW saying, "All ONEs _must_ be iDENTiCAL"].
$$ <adjust crotch>
> > And you agreed that:
> > "In general relativity, above an event horizon of a black hole,
> > an object falling freely from REST at infinity passes each
> > altitude at a directly measured velocity equal to the escape
> > velocity there."
>
> Yes. Given my other caveats.
> [#] Relative to the succession of locally inertial frames
> I discussed before. They only exist for r>2M.
> In the region r>2M, escape velocity is always < c [Even for light].
$$ Mathematically in GR, the concept, (escape velocity)^2 is, 2*v1^2.
$$ Mathematically in GUESS iSS, (escape velocity)^2 = 2*(n - 1)*v1^2.
$$ [Where escape velocity, vesc, equals orbit velocity, v1, @ n=3/2].
$$
$$ There is NO iNERTiAL (REST) frame AT iNFiNiTY for a GR Black Hole.
$$ There can be nothing EXTERiOR to a GR Black Hole ..on a TiME-line.
$$
$$ You have STARTED CONTRADiCTiNG YOURSELF ..in the SAME essay, Tom!.
REPLY to:
| You don't seem to realize that there are _TWO_SET_ of Schwarzschild
| coordinates. One set is valid for r>2M and the other is valid for
| r<2M. They have no intersection. Yes, the exterior coordinates are
| not valid in the region r<2M. <shrug> | Tom Roberts.
$$ You have BEGUN to CONTRADiCT YOURSELF ..*per essay*, lately, Tom!.
Re: A Flaw of General Relativity, a New Metric and Cosmological Imps.
LEJ Brouwer
Tom Roberts wrote:
> LEJ Brouwer wrote:
> > How can spacelike and timelike directions suddenly swap places if there
> > is nothing unusual going on at the event horizon?
>
> That does not happen. What does happen, when using the standard Schw.
> coordinates, is that the coordinates LABELED r and t swap their
> spacelike and timelike characters at the horizon. But that's OK, because
> neither set of coordinates is valid at the horizon -- these TWO sets of
> coordinates are valid on disjoint regions of the manifold, and how the
> labels get assigned is of no physical significance whatsoever.
>
> This is a source of confusion to students today, and was
> confusing to experts until the 1960s or so.
>
>
> > This is completely
> > unphysically,
>
> No physical anything happens. Merely the labels of coordinates get shuffled.
As I said before, if you are going to 'merely' shuffle the labels of
the coordinates, then you would have to 'merely' shuffle them in the
Einstein field equations too - which means that what you are now
solving are not the Einstein field equations.
> > yet is accepted in a completely matter-of-fact way in all
> > standard textbooks. Anyone considering this objectively should
> > immediately smell a rat here.
>
> Only someone who does not understand what is happening would say such
> silly things.
Quite the contrary, only someone who does not understand would _not_
say such things. The pre-1960s experts were right to be concerned.
> > the main point I am making which is the fact that r<2M has no
> > meaning for the Schwarzschild coordinates,
>
> You don't seem to realize that there are _TWO_SET_ of Schwarzschild
> coordinates. One set is valid for r>2M and the other is valid for r<2M.
> They have no intersection. Yes, the exterior coordinates are not valid
> in the region r<2M. <shrug>
What you still don't seem to realise is that r<2m does not physically
exist, so there is only ONE valid set of coordinates, i.e. r>2m. I
pointed this out at the start of the thread (in fact this is the whole
point of bringing the matter up), and I even went through the proof of
it in another post. Obviously its easier to just ignore these facts and
mindlessly regurgitate what the textbooks tell you, but perhaps you
could try thinking for yourself for a change. It makes a world of
difference.
> > so that the usual interior
> > solutions are invalid.
>
> This is not true. They are valid, but are covered by a different chart
> in the Schw. coordinates.
>
>
> Tom Roberts
- Sabbir.
Dirk Van de moortel
"Koobee Wublee" <koobee...@gmail.com> wrote in message news:1152766216....@35g2000cwc.googlegroups.com...
> "JanPB" <fil...@gmail.com> wrote in message
> news:1152727769.7...@75g2000cwc.googlegroups.com...
>
>>> Both Schwarzschild and Hilbert made errors. Refer to the following
>>> article written by Schwarzschild himself.
>>>
>>> http://arxiv.org/abs/physics/9905030
>>>
>>> Equation (4) is only valid when the metric has a determinant of -1.
>>> Subsequent derivation of the Schwarzschild metric has been derived
>>> based on this error.
>>
>> It's not an error, just a historical and - as Einstein realised before
>> November 1915 - unnecessary restriction on allowable coordinate
>> changes.
>
> It is not an error for Einstein because he changed the rules of
> mathematics. However, it remains a problem for everyone else.
"He changed the rules of mathematics" :-)
Why do you bother breaking your mind over differential
geometry if you can't even handle baby analytic geometry?
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LorentzTale.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SRBogus.html
Dirk Vdm
Bilge
>Tom Roberts wrote:
>> No physical anything happens. Merely the labels of coordinates
>get shuffled.
>
>As I said before, if you are going to 'merely' shuffle the labels of
>the coordinates, then you would have to 'merely' shuffle them in the
>Einstein field equations too - which means that what you are now
>solving are not the Einstein field equations.
Don't be ridiculous. The metric has one timelike and three spacelike
coordinates. For a spacelike metric, -+++, the entry with the - sign
is timelike, regardless of what label you give it.
[...]
>> Only someone who does not understand what is happening would say such
>> silly things.
>
>Quite the contrary, only someone who does not understand would _not_
>say such things. The pre-1960s experts were right to be concerned.
Sure, but around 1960, the issue was resolved. Anyone living in 2006
ought to know this.
[...]
>> You don't seem to realize that there are _TWO_SET_ of Schwarzschild
>> coordinates. One set is valid for r>2M and the other is valid for r<2M.
>> They have no intersection. Yes, the exterior coordinates are not valid
>> in the region r<2M. <shrug>
>
>What you still don't seem to realise is that r<2m does not physically
>exist, >so there is only ONE valid set of coordinates, i.e. r>2m. I
Since it takes an atlas with at least six charts in the standard basis
to cover a sphere are you also going to claim that only one-sixth of
a sphere exists? Of course, if you use polar coordinates, you only need
at least two charts, so now you could claim that a hemisphere can exist
unless someone claims only the standard coordinates are valid. Gee,
what a conundrum.
>pointed this out at the start of the thread (in fact this is the whole
>point of bringing the matter up), and I even went through the proof of
>it in another post.
If that is your point, why have you not made an attempt to prove that
``there is only ONE valid set of coordinates?'' The fact that there are
a number of charts which do not suffer the problem you have created
by making a poor choice contradicts you.
>Obviously its easier to just ignore these facts and
>mindlessly regurgitate what the textbooks tell you, but perhaps you
>could try thinking for yourself for a change. It makes a world of
>difference.
What ``facts?'' All you have done is assert something which indicates
you have no idea what you are talking about. If your idea of ``thinking
for yourself'' is to accept non-sense just because it is recognized as
non-sense by the majority of physicists, have at it.
LEJ Brouwer
Bilge wrote:
> LEJ Brouwer:
> >Tom Roberts wrote:
>
> >> No physical anything happens. Merely the labels of coordinates
> >get shuffled.
> >
> >As I said before, if you are going to 'merely' shuffle the labels of
> >the coordinates, then you would have to 'merely' shuffle them in the
> >Einstein field equations too - which means that what you are now
> >solving are not the Einstein field equations.
>
> Don't be ridiculous. The metric has one timelike and three spacelike
> coordinates. For a spacelike metric, -+++, the entry with the - sign
> is timelike, regardless of what label you give it.
Good grief. I despair for humanity. If dy/dx = x is solved by y = x^2 /
2 + c, and then I changed the x into a z wrote, y = z^2 / 2 + c, would
I not have to change the original problem to dy/dz = z for the new
solution to be valid? What is so ridiculous about that?
The fact that the entry with the - signature is timelike has absolutely
nothing to do with it. Or are you claiming that if a metric contains a
time-like coordinate, then this is sufficient for it to satisfy the
Einstein field equations? Because that is what it sounds like you are
arguing.
> [...]
> >> Only someone who does not understand what is happening would say such
> >> silly things.
> >
> >Quite the contrary, only someone who does not understand would _not_
> >say such things. The pre-1960s experts were right to be concerned.
>
> Sure, but around 1960, the issue was resolved. Anyone living in 2006
> ought to know this.
No around 1960, the issue was fudged, shoved under the carpet, and
banished from discussion ever again. Anyone living in 2006 is expected
not to look under the carpet in case embarassing creatures which are
supposed to be extinct emerge therefrom.
Now, I have gone through the proof of why r<2m is not physical. Could
you please use your own brain to tell me why that proof is wrong? It is
only a few lines long, and should not tax even your good self unduly.
> [...]
> >> You don't seem to realize that there are _TWO_SET_ of Schwarzschild
> >> coordinates. One set is valid for r>2M and the other is valid for r<2M.
> >> They have no intersection. Yes, the exterior coordinates are not valid
> >> in the region r<2M. <shrug>
> >
> >What you still don't seem to realise is that r<2m does not physically
> >exist, >so there is only ONE valid set of coordinates, i.e. r>2m. I
>
> Since it takes an atlas with at least six charts in the standard basis
> to cover a sphere are you also going to claim that only one-sixth of
> a sphere exists? Of course, if you use polar coordinates, you only need
> at least two charts, so now you could claim that a hemisphere can exist
> unless someone claims only the standard coordinates are valid. Gee,
> what a conundrum.
We are not talking about a sphere here, and I am not arguing about
coordinate charts anyway, so your analogy is utterly worthless. It has
absolutely nothing to do with why r<2m is not physical.
> >pointed this out at the start of the thread (in fact this is the whole
> >point of bringing the matter up), and I even went through the proof of
> >it in another post.
>
> If that is your point, why have you not made an attempt to prove that
> ``there is only ONE valid set of coordinates?'' The fact that there are
> a number of charts which do not suffer the problem you have created
> by making a poor choice contradicts you.
The proof is in the papers I reference in post #1, and I have also
outlined the proof in another post. WHY DON'T YOU TRY READING IT??? If
you understand the proof, you will also understand why your 'number of
charts' argument is completely irrelevant.
> >Obviously its easier to just ignore these facts and
> >mindlessly regurgitate what the textbooks tell you, but perhaps you
> >could try thinking for yourself for a change. It makes a world of
> >difference.
>
> What ``facts?'' All you have done is assert something which indicates
> you have no idea what you are talking about. If your idea of ``thinking
> for yourself'' is to accept non-sense just because it is recognized as
> non-sense by the majority of physicists, have at it.
And this from a person who has not even bothered to go through the
proof to identify what is factual and what is not. Truly pathetic.
LEJ Brouwer
Bilge wrote:
> LEJ Brouwer:
> >Tom Roberts wrote:
>
> >> No physical anything happens. Merely the labels of coordinates
> >get shuffled.
> >
> >As I said before, if you are going to 'merely' shuffle the labels of
> >the coordinates, then you would have to 'merely' shuffle them in the
> >Einstein field equations too - which means that what you are now
> >solving are not the Einstein field equations.
>
> Don't be ridiculous. The metric has one timelike and three spacelike
> coordinates. For a spacelike metric, -+++, the entry with the - sign
> is timelike, regardless of what label you give it.
Good grief. I despair for humanity. If dy/dx = x is solved by y = x^2 /
2 + c, and then I changed the x into a z wrote, y = z^2 / 2 + c, would
I not have to change the original problem to dy/dz = z for the new
solution to be valid? What is so ridiculous about that?
The fact that the entry with the - signature is timelike has absolutely
nothing to do with it. Or are you claiming that if a metric contains a
time-like coordinate, then this is sufficient for it to satisfy the
Einstein field equations? Because that is what it sounds like you are
arguing.
> [...]
> >> Only someone who does not understand what is happening would say such
> >> silly things.
> >
> >Quite the contrary, only someone who does not understand would _not_
> >say such things. The pre-1960s experts were right to be concerned.
>
> Sure, but around 1960, the issue was resolved. Anyone living in 2006
> ought to know this.
No around 1960, the issue was fudged, shoved under the carpet, and
banished from discussion ever again. Anyone living in 2006 is expected
not to look under the carpet in case embarassing creatures which are
supposed to be extinct emerge therefrom.
Now, I have gone through the proof of why r<2m is not physical. Could
you please use your own brain to tell me why that proof is wrong? It is
only a few lines long, and should not tax even your good self unduly.
> [...]
> >> You don't seem to realize that there are _TWO_SET_ of Schwarzschild
> >> coordinates. One set is valid for r>2M and the other is valid for r<2M.
> >> They have no intersection. Yes, the exterior coordinates are not valid
> >> in the region r<2M. <shrug>
> >
> >What you still don't seem to realise is that r<2m does not physically
> >exist, >so there is only ONE valid set of coordinates, i.e. r>2m. I
>
> Since it takes an atlas with at least six charts in the standard basis
> to cover a sphere are you also going to claim that only one-sixth of
> a sphere exists? Of course, if you use polar coordinates, you only need
> at least two charts, so now you could claim that a hemisphere can exist
> unless someone claims only the standard coordinates are valid. Gee,
> what a conundrum.
We are not talking about a sphere here, and I am not arguing about
coordinate charts anyway, so your analogy is utterly worthless. It has
absolutely nothing to do with why r<2m is not physical.
> >pointed this out at the start of the thread (in fact this is the whole
> >point of bringing the matter up), and I even went through the proof of
> >it in another post.
>
> If that is your point, why have you not made an attempt to prove that
> ``there is only ONE valid set of coordinates?'' The fact that there are
> a number of charts which do not suffer the problem you have created
> by making a poor choice contradicts you.
The proof is in the papers I reference in post #1, and I have also
outlined the proof in another post. WHY DON'T YOU TRY READING IT??? If
you understand the proof, you will also understand why your 'number of
charts' argument is completely irrelevant.
> >Obviously its easier to just ignore these facts and
> >mindlessly regurgitate what the textbooks tell you, but perhaps you
> >could try thinking for yourself for a change. It makes a world of
> >difference.
>
> What ``facts?'' All you have done is assert something which indicates
> you have no idea what you are talking about. If your idea of ``thinking
> for yourself'' is to accept non-sense just because it is recognized as
> non-sense by the majority of physicists, have at it.
And this from a person who has not even bothered to go through the
Tom Roberts
> Tom Roberts wrote:
>> No physical anything happens. Merely the labels of coordinates get shuffled.
>
> As I said before, if you are going to 'merely' shuffle the labels of
> the coordinates, then you would have to 'merely' shuffle them in the
> Einstein field equations too - which means that what you are now
> solving are not the Einstein field equations.
This is not true. Re-labeling the coordinates is of no consequence, and
the field equation is invariant under such a shuffling. It does not
matter how you label the individual coordinates, except that it can be
confusing if you re-use a given label for different coordinates, such as
is done in the two sets of Schw. coordinates.
> What you still don't seem to realise is that r<2m does not physically
> exist,
Hmmm. We are discussing the Schwarzschild solution of the Einstein field
equation, not any real, physical system. This is all _theoretical_, and
the manifold in the region r<2M is every bit as much a manifold as that
in r>2M. So is the entire "second half" of the Kruskal extension. <shrug>
> so there is only ONE valid set of coordinates, i.e. r>2m.
That is just plain wrong.
> I
> pointed this out at the start of the thread (in fact this is the whole
> point of bringing the matter up), and I even went through the proof of
> it in another post.
Your "proof" assumed that it is possible to escape from any point in the
manifold. This assumption is not valid for the Schwarzschild spacetime.
<shrug>
Tom Roberts
JanPB
> Bilge wrote:
> > LEJ Brouwer:
> > >Tom Roberts wrote:
> >
> > >> No physical anything happens. Merely the labels of coordinates
> > >get shuffled.
> > >
> > >As I said before, if you are going to 'merely' shuffle the labels of
> > >the coordinates, then you would have to 'merely' shuffle them in the
> > >Einstein field equations too - which means that what you are now
> > >solving are not the Einstein field equations.
> >
> > Don't be ridiculous. The metric has one timelike and three spacelike
> > coordinates. For a spacelike metric, -+++, the entry with the - sign
> > is timelike, regardless of what label you give it.
>
> Good grief. I despair for humanity. If dy/dx = x is solved by y = x^2 /
> 2 + c, and then I changed the x into a z wrote, y = z^2 / 2 + c, would
> I not have to change the original problem to dy/dz = z for the new
> solution to be valid? What is so ridiculous about that?
You missed the point. As I said earlier - you need to get the new
edition of Spivak, take a year off from posting nonsense on Usenet, and
study the basics.
> The proof is in the papers I reference in post #1, and I have also
> outlined the proof in another post. WHY DON'T YOU TRY READING IT???
They are no "proofs", just incompetent ramblings. It's not my fault
that I cannot prove to you that you are wrong as any such proof
requires by definition that you have already understood the errors
you've made.
--
Jan Bielawski
LEJ Brouwer
JanPB wrote:
> LEJ Brouwer wrote:
> > 2 + c, and then I changed the x into a z wrote, y = z^2 / 2 + c, would
> > I not have to change the original problem to dy/dz = z for the new
> > solution to be valid? What is so ridiculous about that?
>
> You missed the point. As I said earlier - you need to get the new
> edition of Spivak, take a year off from posting nonsense on Usenet, and
> study the basics.
I do not feel any great urge to revise elementary differential geometry
simply because you cannot be bothered to check a simple proof.
> > The proof is in the papers I reference in post #1, and I have also
> > outlined the proof in another post. WHY DON'T YOU TRY READING IT???
>
> They are no "proofs", just incompetent ramblings. It's not my fault
> that I cannot prove to you that you are wrong as any such proof
> requires by definition that you have already understood the errors
> you've made.
Yet you know mention no specifics of mathematical errors in these
"incompetent ramblings".
LEJ Brouwer
Tom Roberts wrote:
> LEJ Brouwer wrote:
> > the coordinates, then you would have to 'merely' shuffle them in the
> > Einstein field equations too - which means that what you are now
> > solving are not the Einstein field equations.
>
> This is not true. Re-labeling the coordinates is of no consequence, and
> the field equation is invariant under such a shuffling. It does not
> matter how you label the individual coordinates, except that it can be
> confusing if you re-use a given label for different coordinates, such as
> is done in the two sets of Schw. coordinates.
I must admit I haven't check whether the field equation is invariant. I
would be surprised if it was, though it is not impossible. HOWEVER, if
the r in the field equation is spacelike - how can the r in the
solution be timelike? (And similarly for t). And even if this seems
fine to you, this argument is still a red herring and does not remove
the constraint r>2m.
> > What you still don't seem to realise is that r<2m does not physically
> > exist,
>
> Hmmm. We are discussing the Schwarzschild solution of the Einstein field
> equation, not any real, physical system. This is all _theoretical_, and
> the manifold in the region r<2M is every bit as much a manifold as that
> in r>2M. So is the entire "second half" of the Kruskal extension. <shrug>
Well, it's usually associated with the gravitational field outside a
spherical mass distribution (and typically a point mass). Even if you
disagree that this is a physical system, 2M is still a constant of
integration, and the condition r>2M arises from the _mathematical_ fact
that one cannot have a negative radial distance from the origin.
> > so there is only ONE valid set of coordinates, i.e. r>2m.
>
> That is just plain wrong.
My, what a powerfully convincing argument.
> > pointed this out at the start of the thread (in fact this is the whole
> > point of bringing the matter up), and I even went through the proof of
> > it in another post.
>
> Your "proof" assumed that it is possible to escape from any point in the
> manifold. This assumption is not valid for the Schwarzschild spacetime.
> <shrug>
What do you mean by 'escape' here? The proof involves measuring the
radial distance of any point from the origin and ensuring that this is
not negative.
> Tom Roberts
- Sabbir.
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"LEJ Brouwer" <intuit...@yahoo.com> wrote in message
news:1152850629.1...@s13g2000cwa.googlegroups.com...
>
> Tom Roberts wrote:
>> LEJ Brouwer wrote:
>> > As I said before, if you are going to 'merely'
>> > shuffle the labels of the coordinates, then you
>> > would have to 'merely' shuffle them in the
>> > Einstein field equations too - which means
>> > that what you are now solving are not the
>> > Einstein field equations.
>>
>> This is not true. Re-labeling the coordinates is
>> of no consequence, and the field equation is
>> invariant under such a shuffling. It does not
>> matter how you label the individual coordinates,
>> except that it can be confusing if you re-use a
>> given label for different coordinates, such as
>> is done in the two sets of Schw. coordinates.
>
> I must admit I haven't check whether the field
> equation is invariant. I would be surprised if it
> was, though it is not impossible. HOWEVER,
> if the r in the field equation is spacelike
* outside the horizon *
> - how can the r in the solution be timelike?
* inside the horizon *
That is the problem with the choice of "simple" coordinates.
Choose Kruskal coordinates and there is no such problem.
Consider what you are straining at is equivalent to using the LT
to evaluate something for the range 0 <= v <= oo, noting that
gamma blows up at c, and beyond c things become imaginary.
> (And similarly for t). And even if this seems
> fine to you,
It is just a mathematical *map*, using a certain type of *ruler*,
not some fundamental desciption of "reality".
> this argument is still a red herring
No, it is your fundamental misunderstanding, and until you can
see past this, you will coninue to be frustrated. I ought to
know, as I am still struggling with this too.
> and does not remove the constraint r>2m.
Or what happens to gamma when v -> c.
Maybe I'm helping, maybe I'm not...
David A. Smith
LEJ Brouwer
N:dlzc D:aol T:com (dlzc) wrote:
> Dear LEJ Brouwer:
>
> "LEJ Brouwer" <intuit...@yahoo.com> wrote in message
> news:1152850629.1...@s13g2000cwa.googlegroups.com...
> > equation is invariant. I would be surprised if it
> > was, though it is not impossible. HOWEVER,
> > if the r in the field equation is spacelike
>
> * outside the horizon *
>
> > - how can the r in the solution be timelike?
>
> * inside the horizon *
Actually, I meant it as I said it - in the Einstein field equation, r
is spacelike and t is timelike. It is not possible therefore for r to
be timelike in the solution to the Einstein field equation, which means
that only the exterior solution is correct, and that the interior
solution is wrong.
> That is the problem with the choice of "simple" coordinates.
> Choose Kruskal coordinates and there is no such problem.
> Consider what you are straining at is equivalent to using the LT
> to evaluate something for the range 0 <= v <= oo, noting that
> gamma blows up at c, and beyond c things become imaginary.
Even the Kruskal coordinates has problems - in this case two quadrants
of the solution (i.e. those corresponding to r<2m) are invalid - the
change in coordinates cannot change that. I don't understand your
analogy with the Lorentz transformation.
>
> > (And similarly for t). And even if this seems
> > fine to you,
>
> It is just a mathematical *map*, using a certain type of *ruler*,
> not some fundamental desciption of "reality".
I disagree - the radial distance is a function of r, and the radius
must be nonnegative, so this constrains the possible range of the
parameter.
>
> > this argument is still a red herring
>
> No, it is your fundamental misunderstanding, and until you can
> see past this, you will coninue to be frustrated. I ought to
> know, as I am still struggling with this too.
The only thing that frustrates me is that people have become
conditioned not to recognise the blindingly obvious if it jars with
their preconceptions. I am not surprised that you are struggling with
it, as the standard picture is simply inconsistent.
> > and does not remove the constraint r>2m.
>
> Or what happens to gamma when v -> c.
>
> Maybe I'm helping, maybe I'm not...
>
> David A. Smith
I don't know, but I appreciate the effort.
Best wishes,
Sabbir.
Sue...
I understand why he makes the analogy but I consider it faulty.
Let's examine the real physical phenomena that gives us a
basis to interchange a spatial and temporal axis. It is not
railroading or philosophy about a rigid grid that god hung the
planets on.
<< Note that the time-dependent solutions, (509) and (510),
are the same as the steady-state solutions, (504) and (505),
apart from the weird way in which time appears in the former.
According to Eqs. (509) and (510), if we want to work out the
potentials at position and time then we have to perform
integrals of the charge density and current density over all
space (just like in the steady-state situation). However, when
we calculate the contribution of charges and currents at position
to these integrals we do not use the values at time , instead
we use the values at some earlier time >>
http://farside.ph.utexas.edu/teaching/em/lectures/node50.html
The freedom we take to resolve this little nearfield issue
with an imaginary time doesn't give us carte' blanche with
anything we can rotate into the Lorenz gauge and demonstrate
a principle of invariance. It only shows that an operation is
Lorenz invariant without any consideration what other absurdities
are created or masked.
The most obvious of which is operating on neutral far-field
objects as though they were charged nearfield objects.
At some point SR's basis of retarded potential has to
be distinguished from GR's notions of mass and
energy density equivalenece.
When I look at something based on the Schawrtzchild solution,
I automatically assume an error because I know a change in
energy was expressed as a change in time. Pound-Sinder
is the perfect example.
Sue...
Daryl McCullough
>Actually, I meant it as I said it - in the Einstein field equation, r
>is spacelike and t is timelike. It is not possible therefore for r to
>be timelike in the solution to the Einstein field equation, which means
>that only the exterior solution is correct, and that the interior
>solution is wrong.
Back up here. The Einstein field equations don't say anything about
"r" and "t". They don't say anything about which basis vectors are
spacelike and which are timelike. What they say is that
G_uv = k T_uv
where G_uv is a tensor formed from second derivatives of
the metric tensor g_uv, and T_uv is the stress-energy tensor,
and k is a constant related to Newton's constant G (there
might be a factor of pi or something).
How do you get r is spacelike and t is timelike from that?
--
Daryl McCullough
Ithaca, NY
Tom Roberts
> Tom Roberts wrote:
>> the field equation is invariant under such a shuffling.
>
It is trivial. <shrug>
> I
> would be surprised if it was, though it is not impossible.
Then you sure don't know much about this at all. This is a trivial
consequence of the arbitrariness of coordinates and the tensor nature of
the field equation. <shrug>
> HOWEVER, if
> the r in the field equation is spacelike - how can the r in the
> solution be timelike? (And similarly for t).
Your question does not make sense, because of the nature of the field
equation.
Whether or not a given coordinate is timelike or spacelike depends on
the metric components, and one is solving the field equation for those
components. So it is not possible to say "the r in the field equation is
spacelike" until you have solved the field equation and _looked_ at the
g_rr metric component. For the Schwarzschild manifold, using the usual
Schw. coordinates, r is spacelike in the region r>2M and is timelike in
the region r<2M, and the coordinate singularity at r=2M makes it
impossible to answer for that value. Yes, this is a PUN on the symbol
"r" -- there are _TWO_ disjoint coordinate systems here, using the same
symbols _WITH_DIFFERENT_MEANINGS_. Sloppy mathematicians (aka many
physicists) do not always make such things clear, and indeed until 1960
or so this was not known by anybody; today is 2006 and you have no
excuse for your ignorance. <shrug>
> And even if this seems
> fine to you, this argument is still a red herring and does not remove
> the constraint r>2m.
You remain adamantly confused. You need to _study_. <shrug>
> the condition r>2M arises from the _mathematical_ fact
> that one cannot have a negative radial distance from the origin.
The Schw. r coordinate is most definitely _NOT_ "radial distance from
the origin". And your _ASSUMPTION_ that there is a "radial distance from
the origin" is _FALSE_ in the Schwarzschild manifold. Indeed, the locus
r=0 is deleted from the manifold and there is no "origin" (and that
locus is NOT a point). One could imagine substituting the limit point(s)
of incoming timelike paths from r=2M to the limit r->0, but there is no
definite value for such paths and that "distance" can have any positive
value (for different paths).
You seem to think that the Schwarzschild manifold is "just like we
perceive outside the earth (neglecting the air, etc.)". This is just
plain not so. You need to learn that this is NON-Euclidean geometry.
Indeed, this manifold is not anywhere close to Euclidean near and inside
its event horizon. The only way to understand the geometry is to study
the metric.
Tom Roberts
dda1
Why is that you Pakis never admit to error? Is it your religion? You
afraid that you loose face?
Well, you already lost a lot of face by continuing to post, you have
been exposed as a fraud by several of us already. Why do you think that
you can't publish your stuff in any peer reviewed journal? Eh?
LEJ Brouwer
Hi, I think the semantics are getting a little mixed up here. I am not
claiming that r is spacelike *because* of the Einstein field equation -
I am simply stating the *fact* that r is spacelike in the Einstein
field equation - and indeed in any other equation in which r is a
radial parameter. The radial direction is spacelike by definition and
the temporal direction is timelike by definition. That's all there is
to it. The fact that the radial direction is timelike in the interior
Schwarzschild solution is further evidence that this solution is not a
valid one.
- Sabbir.
LEJ Brouwer
Tom Roberts wrote:
> LEJ Brouwer wrote:
> > Tom Roberts wrote:
> >> Re-labeling the coordinates is of no consequence, and
> >> the field equation is invariant under such a shuffling.
> >
> > I must admit I haven't check whether the field equation is invariant.
>
> It is trivial. <shrug>
Well, maybe all the signs just happen to cancel - frankly I don't
really
> > I
> > would be surprised if it was, though it is not impossible.
>
> Then you sure don't know much about this at all. This is a trivial
> consequence of the arbitrariness of coordinates and the tensor nature of
> the field equation. <shrug>
No, it's because working out curvature components etc is usually pretty
boring and time-consuming and I see no reason to do so when the outcome
will not affect the present discussion in any way.
> > HOWEVER, if
> > the r in the field equation is spacelike - how can the r in the
> > solution be timelike? (And similarly for t).
>
> Your question does not make sense, because of the nature of the field
> equation.
I really don't know what you are talking about.
> Whether or not a given coordinate is timelike or spacelike depends on
> the metric components, and one is solving the field equation for those
> components. So it is not possible to say "the r in the field equation is
> spacelike" until you have solved the field equation and _looked_ at the
> g_rr metric component.
Ignoring for a moment the fact that this entire thread of reasoning is
a separate argument from the one which I began to explain why r<2m is
physically invalid, your statement is completely wrong. If I have to
solve an equation for x where x is subject to some constraint, then the
solution had better satisfy that constraint. As I said elsewhere, the
radial direction is spacelike by definition.
> For the Schwarzschild manifold, using the usual
> Schw. coordinates, r is spacelike in the region r>2M and is timelike in
> the region r<2M, and the coordinate singularity at r=2M makes it
> impossible to answer for that value. Yes, this is a PUN on the symbol
> "r" -- there are _TWO_ disjoint coordinate systems here, using the same
> symbols _WITH_DIFFERENT_MEANINGS_. Sloppy mathematicians (aka many
> physicists) do not always make such things clear, and indeed until 1960
> or so this was not known by anybody; today is 2006 and you have no
> excuse for your ignorance. <shrug>
I am sorry, but you are very very confused. This is probably not your
fault, as the textbooks must say the same things to have even the
semblence of consistency. If r had a single 'meaning' in the equation
to be solved, how can it have two different 'meanings' in the solution?
<shrug>
> > And even if this seems
> > fine to you, this argument is still a red herring and does not remove
> > the constraint r>2m.
>
> You remain adamantly confused. You need to _study_. <shrug>
<shrug>
> > the condition r>2M arises from the _mathematical_ fact
> > that one cannot have a negative radial distance from the origin.
>
> The Schw. r coordinate is most definitely _NOT_ "radial distance from
> the origin".
Ah, so you finally figured that out did you? Well done. Slow, but
you're getting there...
> And your _ASSUMPTION_ that there is a "radial distance from
> the origin" is _FALSE_ in the Schwarzschild manifold. Indeed, the locus
> r=0 is deleted from the manifold and there is no "origin" (and that
> locus is NOT a point).
Ah, you had me fooled for a moment - you haven't figured it out at all,
have you? r is just a parameter. r=0 need not be the origin. In the
Schwarzschild solution, the origin is at r=2m. r=0 does not exist -
because r<2m does not exist - that is why the interior solution does
not exist. I suggest you go back to the start of the thread and try
again. <shrug>
> One could imagine substituting the limit point(s)
> of incoming timelike paths from r=2M to the limit r->0, but there is no
> definite value for such paths and that "distance" can have any positive
> value (for different paths).
>
> You seem to think that the Schwarzschild manifold is "just like we
> perceive outside the earth (neglecting the air, etc.)". This is just
> plain not so. You need to learn that this is NON-Euclidean geometry.
> Indeed, this manifold is not anywhere close to Euclidean near and inside
> its event horizon. The only way to understand the geometry is to study
> the metric.
Umm, I think you will find that I was the one that pointed out that
this point is non-Euclidean - which is why the radial distance and the
area do not satisfy the usual relationship. So there is little point in
lecturing me about it when you are the one that has clearly
misunderstood what is going on. Can I suggest that you review your
differential geometry? <shrug>
> Tom Roberts
I am sorry Tom, but you are totally clueless <shrug>.
LEJ Brouwer
I don't know - maybe it's because you don't like Pakis? :)
dda1
This is why no peer reviewed journal would publish your stuff? This is
why you get kicked of moderated forums? Because we all don't like Pakis?
LEJ Brouwer
Well, YOU certainly don't seem to like Pakis.
A major reason for my 'stuff' not being published could be that I tend
to question knowledge that has become 'established' when the
foundations of that knowledge are dubious. For example:
I claim that there is a luminiferous aether, when it has been
'established' that there isn't one.
I claim that quantum theory has a classical basis, when it has already
been 'established' that quantum theory is more fundamental than
classical mechanics.
I claim that classical black holes have been seriously misunderstood,
even though their nature has already been 'well established'.
I claim that classical electrodynamics is a direct consequence of
general relativity, which is just pretty hard to believe.
I claim that antigravity exists, even though it is generally taken for
granted that it does not.
I claim that general relativity is responsible for the existence of the
standard model gauge group, and that the elementary particles are
solitonic solutions of GR, when clearly such a claim is absurd.
I claim that string theory is a red herring, and a great sapper of
funding, resources and manpower, even though it is the only theory
currently considered to be a possible 'theory of everything'.
I claim that cold dark matter consists of neutrinos, which are
gravitational dipoles responsible for MOND, and which can be predicted
from first principles from GR, which is also pretty hard to believe.
I speculate that there was probably no 'big bang', but that it is more
likely that 'in the beginning' there was a 'big collapse'.
...and so on (I am sure you get the picture).
In general, I don't blindly assume that everything I am told or read in
books is correct, and am willing to question when things are not clear
or do not seem right.
And as it happens, yes, I was born in East Pakistan (now Bangladesh)
and am also a practising Muslim - I believe that there is no God but
Allah, and that the prophet Muhammad (may peace be upon him), is His
messenger, and I believe in all the prophets sent by Allah to guide
mankind, from the first prophet Adam (pbuh), all the way through to
Jesus (pbuh) and Muhammad (pbuh), who was the final prophet. Some
people might not like that.
And I think that's more than enough reason for establishment folk to
discard my work without looking at it. Don't you? Besides, my writing
style and my tendency to make imaginative extrapolations isn't
necessarily to everyone's liking. Not all my ideas turn out to be right
it's true, but I still think it's good to come up with bright new
ideas, and to always question the status quo, as that is ultimately
what leads to progress.
- Sabbir.
Tom Roberts
> I am simply stating the *fact* that r is spacelike in the Einstein
> field equation - and indeed in any other equation in which r is a
> radial parameter.
Where does the symbol "r" obtain this magical power???
You _really_ need to learn the basics. Symbols have no power whatsoever,
they mean merely what we designate them to mean, and in the case of GR,
as I said before, we don't know whether a given corodinate is timelike
or spacelike until we solve the field equation and examint the relevant
metric components. <shrug>
> The radial direction is spacelike by definition
Your definition of words and symbols has no power over the mathematics.
Tom Roberts
LEJ Brouwer
Tom Roberts wrote:
> LEJ Brouwer wrote:
> > I am simply stating the *fact* that r is spacelike in the Einstein
> > field equation - and indeed in any other equation in which r is a
> > radial parameter.
>
> Where does the symbol "r" obtain this magical power???
>
> You _really_ need to learn the basics. Symbols have no power whatsoever,
Some might disagree with you quite strongly. Can I recommend you take a
look at "The Temple of Man" by R A Schwaller de Lubicz to get some feel
for how thoroughly symbolism can permeate an entire civilisation?
> they mean merely what we designate them to mean, and in the case of GR,
> as I said before, we don't know whether a given corodinate is timelike
> or spacelike until we solve the field equation and examint the relevant
> metric components. <shrug>
Well, I wish you luck in your research. You might need it.
> > The radial direction is spacelike by definition
>
> Your definition of words and symbols has no power over the mathematics.
So, you dare question the power of my words and symbols? :-)
> Tom Roberts
- Sabbir.
Sue...
LEJ Brouwer wrote:
> Daryl McCullough wrote:
> > LEJ Brouwer says...
> >
> > >Actually, I meant it as I said it - in the Einstein field equation, r
> > >is spacelike and t is timelike. It is not possible therefore for r to
> > >be timelike in the solution to the Einstein field equation, which means
> > >that only the exterior solution is correct, and that the interior
> > >solution is wrong.
> >
> > Back up here. The Einstein field equations don't say anything about
> > "r" and "t". They don't say anything about which basis vectors are
> > spacelike and which are timelike. What they say is that
> >
> > G_uv = k T_uv
> >
> > where G_uv is a tensor formed from second derivatives of
> > the metric tensor g_uv, and T_uv is the stress-energy tensor,
> > and k is a constant related to Newton's constant G (there
> > might be a factor of pi or something).
> >
> > How do you get r is spacelike and t is timelike from that?
> >
> > --
> > Daryl McCullough
> > Ithaca, NY
>
> Hi, I think the semantics are getting a little mixed up here.
Yes.. they are problematic even in texts that should be
'authoritative'.
> I am not
> claiming that r is spacelike *because* of the Einstein field equation -
> I am simply stating the *fact* that r is spacelike in the Einstein
> field equation - and indeed in any other equation in which r is a
> radial parameter. The radial direction is spacelike by definition and
> the temporal direction is timelike by definition.
The definition doesn't count. We need to go back to the field
equations and confirm that the temporal components are
transformed out before the interval is defined as spatial.
Likewise confirming that the spatial component is transformed
out before 'defining' the interval as temporal. The function
chi(x.t) comes to mind but I'd check here to be sure.
http://arxiv.org/abs/physics/0204034
> That's all there is
> to it. The fact that the radial direction is timelike in the interior
> Schwarzschild solution is further evidence that this solution is not a
> valid one.
That is my gut feeling too. janPB came up with the idea, perhaps
in this thread, to ignore the words and look at the ingredients.
It makes perfect sense to me.
Now if the ingredients are Portland cement, the label says
Self Rising Flour and the customer is "All Smiles Bakery"
...then Houston has a problem. :o)
Sue...
>
> - Sabbir.
JanPB
> dda1 wrote:
> > LEJ Brouwer wrote:
> > > dda1 wrote:
> > > > <snipped.
> > > >
> > > > Why is that you Pakis never admit to error? Is it your religion? You
> > > > afraid that you loose face?
> > > > Well, you already lost a lot of face by continuing to post, you have
> > > > been exposed as a fraud by several of us already. Why do you think that
> > > > you can't publish your stuff in any peer reviewed journal? Eh?
> > >
> > > I don't know - maybe it's because you don't like Pakis? :)
> >
> > This is why no peer reviewed journal would publish your stuff? This is
> > why you get kicked of moderated forums? Because we all don't like Pakis?
>
> Well, YOU certainly don't seem to like Pakis.
>
> A major reason for my 'stuff' not being published could be that I tend
> to question knowledge that has become 'established' when the
> foundations of that knowledge are dubious.
No, no, no, no, please let go of those dreams at once. A far simpler
explanation: your papers are not good enough. There is no need to
search for grand explanations and Hollywoodesque "establishment"
conspiracies when good old plain incompetence suffices. If you don't
know how coordinate labels relate to tensor equations then there is
really nothing to add, is there?
Remember that 101 years ago a certain no-name government bureaucrat in
Switzerland was able to publish what you would call an
anti-establishment paper. So much for not being able to publish because
of discussing unpopular ideas.
--
Jan Bielawski
Daryl McCullough
>Hi, I think the semantics are getting a little mixed up here. I am not
>claiming that r is spacelike *because* of the Einstein field equation -
>I am simply stating the *fact* that r is spacelike in the Einstein
>field equation - and indeed in any other equation in which r is a
>radial parameter. The radial direction is spacelike by definition and
>the temporal direction is timelike by definition.
If the radial direction is spacelike by definition, then that
means that r is *not* a radial coordinate in the interior region.
It's a temporal coordinate in that region. If the time direction
is timelike by definition, then that means that t is *not* a
temporal coordinate in the interior region. It's a spatial
coordinate there.
>That's all there is to it. The fact that the radial direction is
>timelike in the interior Schwarzschild solution is further evidence
>that this solution is not a valid one.
It is not true that the "radial direction is timelike in the interior
Schwarzschild solution". By definition, if a coordinate is timelike,
then it is *temporal*, not *radial*. The fact that the same variable
*name* r is used for a radial coordinate in the region outside the
event horizon, and is used for a temporal coordinate in the region
inside the event horizon has no significance.
If you prefer to restrict the name "r" to radial coordinates, and
restrict the name "t" to temporal coordinates, then you can certainly
do the following:
In the region outside the event horizon, use the metric
ds^2 = -((r-R)/r)dt^2 + (r/(r-R)) dr^2 + r^2 dOmega^2
This region is described by the coordinates (t,r,theta,phi)
where t is a time coordinate that runs from -infinity to +infinity,
r is a radial coordinate that runs from R to +infinity, and theta
and phi are angular coordinates.
In the region inside the event horizion, use the metric
ds^2 = -((R-t)/t) dt^2 + (t/(R-t)) dz^2 + (R-t)^2 dOmega^2
This region is described by the coordinates (t,z,theta,phi)
where t is a time coordinate that runs from -infinity to R,
z is a spatial coordinate that runs from -infinity to +infinity,
and theta and phi are angular coordinates. With these coordinates,
there is a catastrophe awaiting every observer in the region at
the time t=R (the singularity).
So with these new coordinates, t always refers to the temporal
coordinate, and r always refers to a radial coordinate.
What I think you are worried about is the relationship between
the coordinates in the exterior region and coordinate in the
interior region. There really *is* no relationship. They are
two separate regions of spacetime, and we use different coordinates
in the different regions.
I think people are spoiled by cartesian coordinates in Euclidean
space. Those coordinates are global; they can be used to describe
any point anywhere in the manifold. But that's not the common
case. Usually, when the manifold is curved, coordinates are only
valid in a small region, and if you travel outside of that region
you have to switch to a different coordinate system. Someone
falling towards a black hole has to switch coordinate systems
as he passes from one region to another (when he crosses the
event horizon) but that has no more significance than having
to reset your watch on Earth when you cross into a different
time zone.
dda1
LEJ Brouwer wrote:
> I claim that there is a luminiferous aether, when it has been
> 'established' that there isn't one.
Hmmm
1. Aether could not be detected experimentally (all experiments
atttempting to detect it failed to do so)
2. aether is not necessary as an explanation of physical reality (has
been superceeded by SR explanations)
Now, you claim that there is such a "luminiferous aether". Your only
chance is to come up with yet another experiment that detects it. Have
you done that? No? And you wonder why you are thrown out on your ass?
> I claim that quantum theory has a classical basis, when it has already
> been 'established' that quantum theory is more fundamental than
> classical mechanics.
Let's try a simple question: can you show that quantum mechanics is
based on classical mechanics?
Did you study physics or are you the "home schooled" breed?
<rest snipped as increasingly idiotic>
Daryl McCullough
>Not so - the norm of the 4-acceleration on a test particle is a local,
>invariant, intrinsic quantity that diverges on the event horizon.
What is infinite is the acceleration necessary to hover in place
right at the event horizon. What that means is that it is *impossible*
to hover at the event horizon. If you are at the event horizon, then
you're going to fall in.
But that doesn't mean that there is anything singular happening
at the event horizon. Here's an analogy: Suppose I have a spaceship
that is undergoing constant acceleration (as measured by someone
aboard the spaceship). If the proper acceleration is A, then any
object (even a light signal) that starts out a distance of 1/A or
more behind the spaceship will *never* catch up.
To someone in the spaceship, it appears that there is a horizon
a distance of 1/A below him such that anything that crosses that
horizon can never come out again. In his coordinate system, it
requires an *infinite* amount of acceleration for someone to
hover right at that horizon.
However, someone at that horizon doesn't feel anything weird
going on. The same is true of the event horizon in Schwarzschild
coordinates. The horizon seems significant to someone above
the horizon, but has no physical significance right at the horizon.
LEJ Brouwer
Hi Daryl,
Thanks for taking the time to write that detailed explanation. I
understand your (and others') point that different coordinates can be
used on different coordinate patches, so that r and t on one patch need
not be related to r and t on the other patch. However, in this case,
both of these coordinates r and t are the same r and t that appear in
the Einstein field equations - and this surely induces a relationship
between the two coordinate systems (i.e. they must have the same
'meaning' as someone put it, as the meaning they have in the field
equation).
Another issue of concern is that the two solutions (i.e. the interior
and exterior) are not generally considered to be independent of each
other, but rather that a radially infalling particle supposedly moves
smoothly from the exterior solution to the interior solution. This is
the reason for the development of the various coordinate systems which
seem to patch the interior and exterior together. The problem is that
while everything looks fine in the new coordinates, these new
coordinates are not the 'physical' ones (i.e. radius and time) and they
unfortunately serve to mask the fact that what is physically happening
is that the radial direction *is* becoming timelike and the temporal
direction *is* becoming radial upon crossing the horizon. This is
_physically_ impossible, even though the mathematical mapping in the
transformed coordinates is smooth.
The whole matter is really quite non-trivial which is no doubt why
there has been so much confusion about it in the past. Unfortunately I
think the interpretation which survived, i.e. the particle crosses the
event horizon into the interior region, is wrong, for the reasons I
have already mentioned (most importantly because there is no interior
region). If there is no interior region, the only place the particle
can really go after it hits the event horizon is onto the other
external Kruskal sheet (and note that this has to happen smoothly as
there are no curvature singularities etc). But according to an
independent observer the orientation of time is reversed on the second
sheet relative to the first, so an external observer will actually see
what appears to be a particle-antiparticle annihilation event occurring
at t=Infty at the event horizon.
One of the reasons that I am so confident that my interpretation is
correct (besides the mathematical proofs already given by Abrams,
Antoci and Crothers) is that it has allowed me to derive classical
electrodynamics and resolve the dark matter/MOND problem essentially
from first principles in the context of classical GR. The appearance of
electrodynamics and MOND I think is pretty miraculous, and an
indication that we are on the right path. I derived the existence of a
fluid consisting of neutral spinors last year when I was not at all
concerned with MOND or dark matter, and was pretty amazed when Luc
Blanchet's paper came out in May stating that this scenario is
precisely what is required to potentially resolve the MOND/dark matter
problem.
Best wishes,
Sabbir.
LEJ Brouwer
JanPB wrote:
> LEJ Brouwer wrote:
> explanation: your papers are not good enough. There is no need to
> search for grand explanations and Hollywoodesque "establishment"
> conspiracies when good old plain incompetence suffices. If you don't
> know how coordinate labels relate to tensor equations then there is
> really nothing to add, is there?
Hmm, I think you are possibly confusing me with the entire population
of America. I do not claim that there is some kind of conspiracy going
on. It is just a sad state of affairs and this is just how the current
system works. As someone said, if Einstein were around today, he would
probably have a Yahoo or Hotmail email account and his submissions
would not be accepted by the arxiv. Its a shame, but that's just how
things are. Maybe my papers are not 'good enough' for whatever reason
(I have not actually submitted most of my papers for publication) but I
will still stand by my results, because I am confident that the
mathematics is sound. Nothing you have said has convinced me otherwise.
> Remember that 101 years ago a certain no-name government bureaucrat in
> Switzerland was able to publish what you would call an
> anti-establishment paper. So much for not being able to publish because
> of discussing unpopular ideas.
>
> --
> Jan Bielawski
So you seriously believe that things work in the same way now as they
did 101 years ago? Any you accuse me of wishful thinking!
LEJ Brouwer
I have a PhD in theoretical physics (string field theory) from MIT. My
supervisor was Barton Zwiebach.
> <rest snipped as increasingly idiotic>
I rest my case.
- Sabbir.
P.S. Perhaps the easiest prediction of my model to test is the negative
mass of antiparticles.
LEJ Brouwer
Daryl McCullough wrote:
> LEJ Brouwer says...
>
> >Not so - the norm of the 4-acceleration on a test particle is a local,
> >invariant, intrinsic quantity that diverges on the event horizon.
>
> What is infinite is the acceleration necessary to hover in place
> right at the event horizon. What that means is that it is *impossible*
> to hover at the event horizon. If you are at the event horizon, then
> you're going to fall in.
Daryl, this goes back to some papers by Doughty - I'll have to dig them
out again to see exactly what he says. I think his results indicate
that there is something singular about the horizon (which is not
surprising as it coincides with the 'surface' of the point mass). But
your conclusion that you cannot hover at the event horizon is probably
correct. I disagree with you that a test particle is going to fall 'in'
because there is no 'in'. Rather the test particle is going to fall
'out' - into the second exterior sheet. I recommend you take a look at
my paper where I discuss this and its consequences.
>
> But that doesn't mean that there is anything singular happening
> at the event horizon. Here's an analogy: Suppose I have a spaceship
> that is undergoing constant acceleration (as measured by someone
> aboard the spaceship). If the proper acceleration is A, then any
> object (even a light signal) that starts out a distance of 1/A or
> more behind the spaceship will *never* catch up.
>
> To someone in the spaceship, it appears that there is a horizon
> a distance of 1/A below him such that anything that crosses that
> horizon can never come out again. In his coordinate system, it
> requires an *infinite* amount of acceleration for someone to
> hover right at that horizon.
Sure, I agree with you.
> However, someone at that horizon doesn't feel anything weird
> going on. The same is true of the event horizon in Schwarzschild
> coordinates. The horizon seems significant to someone above
> the horizon, but has no physical significance right at the horizon.
Again, I agree with you. I only disagree about what happens next.
dda1
LEJ Brouwer wrote:
Did you study physics or are you the "home schooled" breed?
>
> I have a PhD in theoretical physics (string field theory) from MIT. My
> supervisor was Barton Zwiebach.
>
> > <rest snipped as increasingly idiotic>
>
> I rest my case.
>
> - Sabbir.
>
Looks like a huge waste of US taxpayer money. So let me ask you again:
LEJ Brouwer
dda1 wrote:
> LEJ Brouwer wrote:
>
> Did you study physics or are you the "home schooled" breed?
> >
> > I have a PhD in theoretical physics (string field theory) from MIT. My
> > supervisor was Barton Zwiebach.
> >
> > > <rest snipped as increasingly idiotic>
> >
> > I rest my case.
> >
> > - Sabbir.
> >
> Looks like a huge waste of US taxpayer money. So let me ask you again:
Yeah, even bigger than the occupation of Iraq.
> 1. Aether could not be detected experimentally (all experiments
> atttempting to detect it failed to do so)
>
> 2. aether is not necessary as an explanation of physical reality (has
> been superceeded by SR explanations)
>
> Now, you claim that there is such a "luminiferous aether". Your only
> chance is to come up with yet another experiment that detects it. Have
> you done that? No? And you wonder why you are thrown out on your ass?
My 'luminiferous aether' is none other than the physical vacuum. I
don't need an experiment to detect it because we already know it's
there. My aether does not define an absolute frame of reference that it
can be detected in the way you suggest. Read my paper and maybe you
will understand. I wouldn't put my money on that though.
Well, now you know about me, why don't you tell me a little about
yourself?
- Sabbir.
dda1
LEJ Brouwer wrote:
>
> My 'luminiferous aether' is none other than the physical vacuum. I
> don't need an experiment to detect it because we already know it's
> there.
Another kook. And we paid money for an asshole like you to get a PhD?
You" already know it's there"? This is the standard kook - fare: "it is
because I think so"
"My aether does not define an absolute frame of reference that it
> can be detected in the way you suggest."
I don't suggest anything, asshole, I just asked what proof you have.
Now we all know :NONE.
"Read my paper and maybe you
> will understand. I wouldn't put my money on that though."
What "paper"? Do you have it anywhere on the web? In arxiv? If you
haven't run any experiment is worth shit. You know at least that much
if you passed thry MIT.
What pieces of shit we bring in this country. At least you don't bomb
trains with innocent people.
>
> Well, now you know about me, why don't you tell me a little about
> yourself?
I teach. I publish (in peer reviewed journals) . In my free time I
debunk frauds. Like you.
JanPB
>
> What pieces of shit we bring in this country. At least you don't bomb
> trains with innocent people.
Aw, come on, this is a bit too much!
--
Jan Bielawski
dda1
Here is Chris Hillman ripping Abhas a new asshole:
http://www.lns.cornell.edu/spr/2001-07/msg0034336.html
Now, somehow Abhas got an entry into wiki:
http://en.wikipedia.org/wiki/Abhas_Mitra
....which claims that he was proven right by Stanley Robertson and
Darryl Leiter
This is an interesting thread to follow.....:
LEJ Brouwer
dda1 wrote:
> LEJ Brouwer wrote:
>
> >
> > My 'luminiferous aether' is none other than the physical vacuum. I
> > don't need an experiment to detect it because we already know it's
> > there.
>
> Another kook. And we paid money for an asshole like you to get a PhD?
> You" already know it's there"? This is the standard kook - fare: "it is
> because I think so"
Not all research leads to new theoretical results. There is often value
in obtaining a clearer or more intuitive conceptual picture of what
underlies what is already observed. I will quote from my earlier paper:
"Maxwell and others had struggled to find a mathematical description of
the underlying medium, the `aether', in which electromagnetic waves
were presumed to propagate. Although the continuum we have described is
not precisely equivalent to the notion which the earlier proponents had
had in mind, our analysis does show that a description of
electrodynamics in terms of an underlying continuum is possible. This
is particularly important as the failure to find such a formulation
historically contributed to the origin of the concept of `fields'
postulated not to require such a medium. The field concept may not have
been necessary after all."
Does this make me a 'kook' and an 'asshole'? And yes, MIT did pay me
more than other graduate students as I was awarded the "Karl Taylor
Compton Fellowship" for physics, which paid $18,000 tax free p/a on top
of tuition fees.
> "My aether does not define an absolute frame of reference that it
> > can be detected in the way you suggest."
>
> I don't suggest anything, asshole, I just asked what proof you have.
> Now we all know :NONE.
The model predicts that antigravity has negative mass. Experimental
observations do not 'prove' models to be right or wrong, they provide
'evidence' one way or the other.
> "Read my paper and maybe you
> > will understand. I wouldn't put my money on that though."
>
> What "paper"? Do you have it anywhere on the web? In arxiv?
Try searching in the arxiv under "Sabbir". I would have hoped that you
would have had the intelligence to at least do that.
> If you
> haven't run any experiment is worth shit.
So that pretty much rules our string theory and loop quantum gravity.
> You know at least that much
> if you passed thry MIT.
> What pieces of shit we bring in this country. At least you don't bomb
> trains with innocent people.
Was that a compliment?
> >
> > Well, now you know about me, why don't you tell me a little about
> > yourself?
>
>
> I teach. I publish (in peer reviewed journals) . In my free time I
> debunk frauds. Like you.
I see - so you hate Pakis, are upset about train bombings, you teach
and you publish. You also label me as a 'kook', a 'fraud', an 'asshole'
and a 'piece of shit', but you won't reveal your name. Hardly signs of
noble character.
dda1
LEJ Brouwer wrote:
> dda1 wrote:
> > LEJ Brouwer wrote:
> >
> > >
> > > My 'luminiferous aether' is none other than the physical vacuum. I
> > > don't need an experiment to detect it because we already know it's
> > > there.
> >
> > Another kook. And we paid money for an asshole like you to get a PhD?
> > You" already know it's there"? This is the standard kook - fare: "it is
> > because I think so"
>
> Not all research leads to new theoretical results. There is often value
> in obtaining a clearer or more intuitive conceptual picture of what
> underlies what is already observed. I will quote from my earlier paper:
>
> "Maxwell and others had struggled to find a mathematical description of
> the underlying medium, the `aether', in which electromagnetic waves
> were presumed to propagate. Although the continuum we have described is
> not precisely equivalent to the notion which the earlier proponents had
> had in mind, our analysis does show that a description of
> electrodynamics in terms of an underlying continuum is possible. This
> is particularly important as the failure to find such a formulation
> historically contributed to the origin of the concept of `fields'
> postulated not to require such a medium. The field concept may not have
> been necessary after all."
>
So you back up your gratuitous statements with....more gratuitous
statements?
No wonder no one publishes your shit.
Daryl McCullough
>Thanks for taking the time to write that detailed explanation. I
>understand your (and others') point that different coordinates can be
>used on different coordinate patches, so that r and t on one patch need
>not be related to r and t on the other patch. However, in this case,
>both of these coordinates r and t are the same r and t that appear in
>the Einstein field equations
I don't know what you mean. There *is* no "r" and "t" in the
Einstein field equations. They are *tensor* equations, and there
is no mention of r or t.
>- and this surely induces a relationship between the two coordinate
>systems (i.e. they must have the same 'meaning' as someone put it,
>as the meaning they have in the field equation).
Well, it is false that "r" and "t" appear in the field equations.
The names of coordinates are completely *arbitrary*. Think about
2D Euclidean space. It can be described using the coordinates
x and y, or it can be described using the cooordinates r and theta.
Names of coordinates are *irrelevant*.
The only way to relate coordinates in two different patches is
by having a third patch that overlaps both of them. If you
have patch A with coordinates x_A, y_A, z_A, t_A, and patch
B with coordinates x_B, y_B, z_B, t_B, then you can relate
them by having a third small patch C that overlaps A and B
and giving coordinate transformations L_AC and L_BC such that
L_AC is an invertible transformation between A-coordinates
and C-coordinates (valid on the overlap of A and C)
L_BC is an invertible transformation between B-coordinates
and C-coordinates (valid on the overlap of B and C)
Using overlapping patches, you can relate A-coordinates to
B-coordinates through the use of C-coordinates. But C isn't
unique: there can many different overlapping patches, and
many different coordinate transformations. So there is no
meaning to the claim that coordinate r in patch A is "the
same coordinate" as coordinate r in patch B.
In the case of the Schwarzschild geometry, you can let
A be the region outside the event horizon (not including
the horizon) and let B be the region inside the horizon
(not including the horizon), and let C be the local
frame of an infalling observer. The transformation
L_AC will map a timelike vector of region A to some
timelike vector of region C. The transformation L_BC
will map a timelike vector of region B to some timelike
vector of region C. All perfectly fine. But there is
no mapping from vectors in A to vectors in B, because
they don't overlap. There is no meaning to the question
"Which vector of A corresponds to this vector of B?"
>Another issue of concern is that the two solutions (i.e. the interior
>and exterior) are not generally considered to be independent of each
>other, but rather that a radially infalling particle supposedly moves
>smoothly from the exterior solution to the interior solution.
That's true. The regions are *compatible*, in the sense that
an infalling observer can map to both regions.
>The problem is that while everything looks fine in the new coordinates,
>these new coordinates are not the 'physical' ones (i.e. radius and time)
General Relativity doesn't have any notion of one coordinate system being
more "physical" than another. All coordinate systems are equally good
in any region for which they are nonsingular.
>and they unfortunately serve to mask the fact that what is physically
>happening is that the radial direction *is* becoming timelike and the
>temporal direction *is* becoming radial upon crossing the horizon.
Here's an analogy. Suppose you stand on the North Pole with your
right arm pointing straight out. Your arm is pointing South. Now
move straight ahead, without turning, until you reach the Equator.
Now your arm is pointing West. You haven't turned, but South
has turned into West. That's the way curved manifolds work.
LEJ Brouwer
dda1 wrote:
> No wonder no one publishes your shit.
I wouldn't want to smell one of your publications.
Bilge
>
>Bilge wrote:
>> LEJ Brouwer:
>> >Tom Roberts wrote:
>>
>> >> No physical anything happens. Merely the labels of coordinates
>> >get shuffled.
>> >
>> >As I said before, if you are going to 'merely' shuffle the labels of
>> >the coordinates, then you would have to 'merely' shuffle them in the
>> >Einstein field equations too - which means that what you are now
>> >solving are not the Einstein field equations.
>>
>> Don't be ridiculous. The metric has one timelike and three spacelike
>> coordinates. For a spacelike metric, -+++, the entry with the - sign
>> is timelike, regardless of what label you give it.
>
>Good grief. I despair for humanity. If dy/dx = x is solved by y = x^2 /
>2 + c, and then I changed the x into a z wrote, y = z^2 / 2 + c, would
>I not have to change the original problem to dy/dz = z for the new
>solution to be valid? What is so ridiculous about that?
Are you claiming that those are two different equations which
might have different solutions by virtue of using different labels
for the independent variables?
>
>The fact that the entry with the - signature is timelike has absolutely
>nothing to do with it.
Sure it does, since you are claiming that the interior solution
is invalid ased upon your inability to understand why the radial
coordinate is timelike for r < 2m.
>Or are you claiming that if a metric contains a time-like coordinate,
>then this is sufficient for it to satisfy the Einstein field equations?
>Because that is what it sounds like you are arguing.
No, that is not my argument, since that argument would be manifestly
incorrect. The field equations would be valid, independent of even the
signature. The lorentzian signature just happens to describe the
possible metrics of physical interest for this universe.
[...]
>Now, I have gone through the proof of why r<2m is not physical. Could
>you please use your own brain to tell me why that proof is wrong? It is
>only a few lines long, and should not tax even your good self unduly.
You have spent a few lines proving you don't understand the difference
between a removale singularity and one which is not removable.
>> Since it takes an atlas with at least six charts in the standard basis
>> to cover a sphere are you also going to claim that only one-sixth of
>> a sphere exists? Of course, if you use polar coordinates, you only need
>> at least two charts, so now you could claim that a hemisphere can exist
>> unless someone claims only the standard coordinates are valid. Gee,
>> what a conundrum.
>
>We are not talking about a sphere here, and I am not arguing about
>coordinate charts anyway,
Well. I'll agree that you are trying very hard to avoid it,
since doing so would ruin your argument.
>so your analogy is utterly worthless. It has absolutely nothing to
>do with why r<2m is not physical.
>> >pointed this out at the start of the thread (in fact this is the whole
>> >point of bringing the matter up), and I even went through the proof of
>> >it in another post.
>>
>> If that is your point, why have you not made an attempt to prove that
>> ``there is only ONE valid set of coordinates?'' The fact that there are
>> a number of charts which do not suffer the problem you have created
>> by making a poor choice contradicts you.
>
>The proof is in the papers I reference in post #1, and I have also
>outlined the proof in another post. WHY DON'T YOU TRY READING IT???
Why don't you try not making erroneous assumptions?
[...]
>> What ``facts?'' All you have done is assert something which indicates
>> you have no idea what you are talking about. If your idea of ``thinking
>> for yourself'' is to accept non-sense just because it is recognized as
>> non-sense by the majority of physicists, have at it.
>
>And this from a person who has not even bothered to go through the
>proof to identify what is factual and what is not. Truly pathetic.
Ibid.
LEJ Brouwer
Daryl McCullough wrote:
> LEJ Brouwer says...
>
> >Thanks for taking the time to write that detailed explanation. I
> >understand your (and others') point that different coordinates can be
> >used on different coordinate patches, so that r and t on one patch need
> >not be related to r and t on the other patch. However, in this case,
> >both of these coordinates r and t are the same r and t that appear in
> >the Einstein field equations
>
> I don't know what you mean. There *is* no "r" and "t" in the
> Einstein field equations. They are *tensor* equations, and there
> is no mention of r or t.
Are you just arguing for the sake of it or are you expecting me to take
you seriously?
> >- and this surely induces a relationship between the two coordinate
> >systems (i.e. they must have the same 'meaning' as someone put it,
> >as the meaning they have in the field equation).
>
> Well, it is false that "r" and "t" appear in the field equations.
>
> The names of coordinates are completely *arbitrary*. Think about
> 2D Euclidean space. It can be described using the coordinates
> x and y, or it can be described using the cooordinates r and theta.
> Names of coordinates are *irrelevant*.
Sure, but there meanings are not. x and y, I presume, are Cartesian
coordinates, while r and t are polar. The difference is quite relevant.
> The only way to relate coordinates in two different patches is
> by having a third patch that overlaps both of them. If you
> have patch A with coordinates x_A, y_A, z_A, t_A, and patch
> B with coordinates x_B, y_B, z_B, t_B, then you can relate
> them by having a third small patch C that overlaps A and B
> and giving coordinate transformations L_AC and L_BC such that
>
> L_AC is an invertible transformation between A-coordinates
> and C-coordinates (valid on the overlap of A and C)
>
> L_BC is an invertible transformation between B-coordinates
> and C-coordinates (valid on the overlap of B and C)
>
> Using overlapping patches, you can relate A-coordinates to
> B-coordinates through the use of C-coordinates. But C isn't
> unique: there can many different overlapping patches, and
> many different coordinate transformations. So there is no
> meaning to the claim that coordinate r in patch A is "the
> same coordinate" as coordinate r in patch B.
Yes, but be careful here - if A and B have overlapping patches, there
would be a meaning to the statement that a set of coordinates in A
represents the same point as a set of coordinates in B.
> In the case of the Schwarzschild geometry, you can let
> A be the region outside the event horizon (not including
> the horizon) and let B be the region inside the horizon
> (not including the horizon), and let C be the local
> frame of an infalling observer. The transformation
> L_AC will map a timelike vector of region A to some
> timelike vector of region C. The transformation L_BC
> will map a timelike vector of region B to some timelike
> vector of region C. All perfectly fine. But there is
> no mapping from vectors in A to vectors in B, because
> they don't overlap. There is no meaning to the question
> "Which vector of A corresponds to this vector of B?"
Ah, but we are talking about smooth geodesics, not individual vectors,
in which case your argument falls flat on its face.
> >Another issue of concern is that the two solutions (i.e. the interior
> >and exterior) are not generally considered to be independent of each
> >other, but rather that a radially infalling particle supposedly moves
> >smoothly from the exterior solution to the interior solution.
>
> That's true. The regions are *compatible*, in the sense that
> an infalling observer can map to both regions.
Ignoring the fact momentarily that there is no interior region, the
smooth movement you refer to is only so in the transformed coordinates.
In physical terms, a spacelike vector has suddenly become timelike and
vice versa, which is not only not smooth, it is physically impossible.
> >The problem is that while everything looks fine in the new coordinates,
> >these new coordinates are not the 'physical' ones (i.e. radius and time)
>
> General Relativity doesn't have any notion of one coordinate system being
> more "physical" than another. All coordinate systems are equally good
> in any region for which they are nonsingular.
>
> >and they unfortunately serve to mask the fact that what is physically
> >happening is that the radial direction *is* becoming timelike and the
> >temporal direction *is* becoming radial upon crossing the horizon.
>
> Here's an analogy. Suppose you stand on the North Pole with your
> right arm pointing straight out. Your arm is pointing South. Now
> move straight ahead, without turning, until you reach the Equator.
> Now your arm is pointing West. You haven't turned, but South
> has turned into West. That's the way curved manifolds work.
That's a poor analogy. Here's a better one: You take one step forward,
and your hair turns grey with age. You take one step back, and you
return to childhood. Does this sound ridiculous? Good.
- Sabbir.
Ken S. Tucker
LEJ Brouwer wrote:
...
> The model predicts that antigravity has negative mass. Experimental
> observations do not 'prove' models to be right or wrong, they provide
> 'evidence' one way or the other.
Ah, that is contrary to the Principle of Equivalence,
whereby, the acceleration is independant of the
substance, specifically you can use a massless
point to obtain the geodesic. Use the "elevator".
Provide a link to the paper you want reviewed.
I agree with your ideas about the horizon while
following this thread and understand both sides
of the argument.
Regards
Ken S. Tucker
...
LEJ Brouwer
Tom Roberts wrote:
> LEJ Brouwer wrote:
> >> Re-labeling the coordinates is of no consequence, and
> >> the field equation is invariant under such a shuffling.
> >
> > I must admit I haven't check whether the field equation is invariant.
>
> It is trivial. <shrug>
>
>
> > I
> > would be surprised if it was, though it is not impossible.
>
> Then you sure don't know much about this at all. This is a trivial
> consequence of the arbitrariness of coordinates and the tensor nature of
> the field equation. <shrug>
The field equation definitely is not invariant under exchange of r and
t, as the Schwarzschild metric solving it is clearly not. Therefore
your statement is completely false.
LEJ Brouwer
Ken S. Tucker wrote:
> LEJ Brouwer wrote:
> ...
> > The model predicts that antigravity has negative mass. Experimental
> > observations do not 'prove' models to be right or wrong, they provide
> > 'evidence' one way or the other.
>
> Ah, that is contrary to the Principle of Equivalence,
> whereby, the acceleration is independant of the
> substance, specifically you can use a massless
> point to obtain the geodesic. Use the "elevator".
Yes, the principle of equivalence needs to be generalised to allow for
particles of negative mass.
> Provide a link to the paper you want reviewed.
> I agree with your ideas about the horizon while
> following this thread and understand both sides
> of the argument.
> Regards
> Ken S. Tucker
> ...
The paper is here:
http://arxiv.org/abs/physics/0607102
Thanks,
Sabbir.
Daryl McCullough
>The field equation definitely is not invariant under exchange of r and
>t
I don't understand what you mean by that. There *is* no "r" or "t"
in the field equations. The field equations treat all 4 variables
identically. *Solutions* may have some variables that are different
in nature than others, but that difference does not come from the
field equations. The difference comes in the boundary conditions.
Daryl McCullough
>Daryl McCullough wrote:
>> I don't know what you mean. There *is* no "r" and "t" in the
>> Einstein field equations. They are *tensor* equations, and there
>> is no mention of r or t.
>
>Are you just arguing for the sake of it or are you expecting me to take
>you seriously?
I'm telling you the truth. There is no r or t in the field equations.
Perhaps you should post what you believe to be the field equations,
and show why you think it says anything about r and t.
>> The names of coordinates are completely *arbitrary*. Think about
>> 2D Euclidean space. It can be described using the coordinates
>> x and y, or it can be described using the cooordinates r and theta.
>> Names of coordinates are *irrelevant*.
>
>Sure, but there meanings are not. x and y, I presume, are Cartesian
>coordinates, while r and t are polar. The difference is quite relevant.
That difference shows up as differences in the metric. The GR
field equations treat all coordinate systems equally, and all
coordinates. It doesn't say anything about "r" or "t".
>> The only way to relate coordinates in two different patches is
>> by having a third patch that overlaps both of them. If you
>> have patch A with coordinates x_A, y_A, z_A, t_A, and patch
>> B with coordinates x_B, y_B, z_B, t_B, then you can relate
>> them by having a third small patch C that overlaps A and B
>> and giving coordinate transformations L_AC and L_BC such that
>>
>> L_AC is an invertible transformation between A-coordinates
>> and C-coordinates (valid on the overlap of A and C)
>>
>> L_BC is an invertible transformation between B-coordinates
>> and C-coordinates (valid on the overlap of B and C)
>>
>> Using overlapping patches, you can relate A-coordinates to
>> B-coordinates through the use of C-coordinates. But C isn't
>> unique: there can many different overlapping patches, and
>> many different coordinate transformations. So there is no
>> meaning to the claim that coordinate r in patch A is "the
>> same coordinate" as coordinate r in patch B.
>
>Yes, but be careful here - if A and B have overlapping patches
In the case of the Schwarzschild coordinates, there *is* no
overlap. One set of coordinates is valid in the *exterior*,
one is valid in the *interior*. Neither is valid *at* the
event horizon. So there is no overlap in their regions of
validity.
>> In the case of the Schwarzschild geometry, you can let
>> A be the region outside the event horizon (not including
>> the horizon) and let B be the region inside the horizon
>> (not including the horizon), and let C be the local
>> frame of an infalling observer. The transformation
>> L_AC will map a timelike vector of region A to some
>> timelike vector of region C. The transformation L_BC
>> will map a timelike vector of region B to some timelike
>> vector of region C. All perfectly fine. But there is
>> no mapping from vectors in A to vectors in B, because
>> they don't overlap. There is no meaning to the question
>> "Which vector of A corresponds to this vector of B?"
>
>Ah, but we are talking about smooth geodesics, not individual vectors,
>in which case your argument falls flat on its face.
No, it doesn't. I'm not talking about a geodesic, I'm talking about
a coordinate patch for a region of spacetime such that the freefalling
geodesics are straight lines (approximately). In this patch, we can
use local Minkowsky coordinates, and the freefalling observer has
the simple trajectory
x^0(tau) = c tau
x^1(tau) = x^2(tau) = x^3(tau) = 0
>> That's true. The regions are *compatible*, in the sense that
>> an infalling observer can map to both regions.
>
>Ignoring the fact momentarily that there is no interior region,
In the Schwarzschild spacetime, there certainly is an interior
region.
>the smooth movement you refer to is only so in the transformed coordinates.
>In physical terms, a spacelike vector has suddenly become timelike and
>vice versa, which is not only not smooth, it is physically impossible.
What does "suddenly become" mean? From the point of view of the freefalling
observer, timelike vectors stay timelike as he crosses the horizon. There
is no coordinate system in which a timelike vector becomes a spacelike
vector.
>> Here's an analogy. Suppose you stand on the North Pole with your
>> right arm pointing straight out. Your arm is pointing South. Now
>> move straight ahead, without turning, until you reach the Equator.
>> Now your arm is pointing West. You haven't turned, but South
>> has turned into West. That's the way curved manifolds work.
>
>That's a poor analogy.
It's an *exact* analogy. Timelike is analogous to south, and
spacelike is analogous to west.
>Here's a better one: You take one step forward,
>and your hair turns grey with age. You take one step back, and you
>return to childhood. Does this sound ridiculous? Good.
No, that's a very bad analogy, since nothing like that happens
in the neighborhood of a black hole. An infalling observer is
free to turn on his rocket and move up and down, and nothing
special happens. There are no locally observable effects at the
horizon.
Tom Roberts
> The field equation definitely is not invariant under exchange of r and
> t, as the Schwarzschild metric solving it is clearly not.
You obviously do not know what the field equation is. As I keep saying:
you _REALLY_ need to learn the basics. What you think is the "field
equation" is not -- it is the field equation _projected_ onto specific
coordinates. More on this below.
Here is the Einstein field equation (omit the cosmological constant):
G = T where G is the Einstein curvature tensor, and T is
the energy-momentum tensor, and I have chosen
convenient units (c=1 and (8pi)k=1, where k is Newton's
gravitational constant)
Note that neither r nor t appear there; indeed no coordinates appear at
all, and this equation is completely independent of coordinates. So
exchanging r and t has no effect whatsoever, because the field equation
is completely independent of them.
This equation is valid _throughout_ the manifold (we'll come back to
that). For the Schwarzschild spacetime T=0 (i.e. vacuum), but I'll
retain it for generality.
It is common to project this equation onto an arbitrary (but specific)
coordinate basis, obtaining a set of equations:
G^ij = T^ij [i and j are component indexes ranging 0,1,2,3]
Of course this can be done only within the region of the manifold for
which the specific coordinates used are valid. One can further express
the {G^ij} in terms of the metric components {g_ij} and their
derivatives with respect to the coordinates {x^i}; one then has a set of
nonlinear second-order partial differential equations of the {g_ij}. One
can re-label the coordinates: {x^i} => {t,r,theta,phi} -- and write down
the equations as you are familiar with in terms of such coordinate labels.
The key point is that when written in terms of tensor components and
coordinates, the equation is valid only in the region of the manifold in
which the coordinates are valid. The _actual_ field equation (above),
expressed in terms of the tensor themselves, has no such restriction.
[It should come as no surprise that when projecting a
tensor equation onto a set of coordinates, the resulting
set of equations depends on those coordinates,
_including_ their region of validity.]
In solving for the metric components of the Schwarzschild manifold, one
normally selects spherical coordinates {t,r,theta,phi) at spatial
infinity and solves the set of equations, giving the usual Schw. line
element. Sloppy mathematicians (aka most physicists) don't bother to
point out the limitations of this procedure: it is valid _ONLY_ within
the region for which those coordinates are valid. Indeed, a careful
analysis shows that those coordinates are only valid in the region:
-infinity<t<infinity, r>2M, 0<theta<pi, 0<phi<2pi
This is known as the exterior region.
One can, however, look at the line element and notice that it is also
valid in the region:
-infinity<t<infinity, 0<r<2M, 0<theta<pi, 0<phi<2pi
But in this region r is timelike and t is spacelike, with the future in
the -r direction (hence the singularity at r=0 is in the _future_ of
every point in this region). This is known as the interior region.
All your blathering is simply ignoring the basic mathematical structure
presented above. You _really_ need to learn the basics. There is no
shortcut, and the papers you quote are equally ignorant of the basic
mathematics used in GR and described above. <shrug>
In short, you think the portion of the manifold r<2M "does not exist",
while in reality it is the _COORDINATES_ you use that do not exist (i.e.
are invalid), but the interior region of the manifold is perfectly
valid, albeit rather strange to naive Euclidean sensibilities.
Tom Roberts
Tom Roberts
> Yes, the principle of equivalence needs to be generalised to allow for
> particles of negative mass.
Why? First you need to find some....
Tom Roberts
Phineas T Puddleduck
JanPB <fil...@gmail.com> wrote:
> Remember that 101 years ago a certain no-name government bureaucrat in
> Switzerland was able to publish what you would call an
> anti-establishment paper. So much for not being able to publish because
> of discussing unpopular ideas.
For some people, it seems a lot easier to blame the system then blame
themselves.
--
Relf's Law? -+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
"Bullshit repeated to the limit of infinity asymptotically approaches
the odour of roses."
Corollary -+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Å‚It approaches the asymptote faster, the more ÅšpseduosÄ… you throw in
your formulas.Ë›
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Å‚Gravity is one of the four fundamental interactions. The classical
theory of gravity - Einstein's general relativity - is the subject
of this book.Ë› : Hartle/ Gravity pg 1
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Jaffa cakes. Sweet delicious orange jaffa goodness, and an abject lesson
why parroting information from the web will not teach you cosmology.
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Ken S. Tucker
LEJ Brouwer wrote:
> Ken S. Tucker wrote:
> > LEJ Brouwer wrote:
> > ...
> > > The model predicts that antigravity has negative mass. Experimental
> > > observations do not 'prove' models to be right or wrong, they provide
> > > 'evidence' one way or the other.
> >
> > Ah, that is contrary to the Principle of Equivalence,
> > whereby, the acceleration is independant of the
> > substance, specifically you can use a massless
> > point to obtain the geodesic. Use the "elevator".
>
> Yes, the principle of equivalence needs to be generalised to allow for
> particles of negative mass.
I have no problem regarding an "energy deficit"
like E = (+e)*(-e)/r being regarded as "negative
energy". However the PoE liken's the acceleration
of those parts by observing them from an accel-
erating elevator, by gedanken. IOW's everything
follows the same geodesics, that's damn general!
You'll need to specify how to generalize PoE
so that negative mass differs from +mass,
geodesically.
I'm all EARs, I'll consider your arguments including
refs to your papers. I'll even consider that the
PoE is erroneous if you have an argument to that
effect that is reasonable.
Ken S. Tucker
LEJ Brouwer
Tom Roberts wrote:
> In solving for the metric components of the Schwarzschild manifold, one
> normally selects spherical coordinates {t,r,theta,phi) at spatial
> infinity and solves the set of equations, giving the usual Schw. line
> element.
I thought you said that there was no 'r' or 't' in the Einstein field
equations? Well, now you have introduced them as I knew you would, do
you *still* claim that the Einstein field equations in vacuo are
invariant under swapping r and t?
> In short, you think the portion of the manifold r<2M "does not exist",
> while in reality it is the _COORDINATES_ you use that do not exist (i.e.
> are invalid), but the interior region of the manifold is perfectly
> valid, albeit rather strange to naive Euclidean sensibilities.
No, the manifold *really* does not exist there.
> Tom Roberts
- Sabbir.
LEJ Brouwer
Phineas T Puddleduck wrote:
> In article <1152920675.1...@p79g2000cwp.googlegroups.com>,
> JanPB <fil...@gmail.com> wrote:
>
> > Remember that 101 years ago a certain no-name government bureaucrat in
> > Switzerland was able to publish what you would call an
> > anti-establishment paper. So much for not being able to publish because
> > of discussing unpopular ideas.
>
> For some people, it seems a lot easier to blame the system then blame
> themselves.
On the other hand, for most sheep it seems a lot easier to blame the
people than blame the system.
dda1
The fun part is that you probably read them. I do not think you
understood any of them but heck, this is your problem.
So your so called "papers " are languishing in arxiv, none of them
getting any chance of being published. This must hurt, you wasted all
these years, you are no longer that young and nothing to show for.
Based on the crass errors that were exposed in these two forums, I
think that MIT should revoke your PhD (Pile of Dung).
All you do is you talk back, you never answer the technical challenges,
you argue to death. If you act this way with the journal editors, no
wonder that you never publish. Is this what is eating you? That you get
rejected?
dda1
Sabbir Rahman
I thought that you are dumb but it turns out you are a psycho.
LEJ Brouwer
dda1 wrote:
> LEJ Brouwer wrote:
> > dda1 wrote:
> > > No wonder no one publishes your shit.
> >
> > I wouldn't want to smell one of your publications.
>
> The fun part is that you probably read them.
Little things please little minds after all. What is amusing is that
while you boast about your purported publications you cannot tell us
which papers you supposedly wrote because it would blow your cover and
we would be able to put a face to the foul-mouthed expletives.
> I do not think you
> understood any of them but heck, this is your problem.
So... you claim to have publications, you don't say what they are, then
you claim that I probably did not understand them, and then you say
that this is 'my problem'.
You appear to have the emotional IQ a two-year-old. Perhaps you should
seek professional counsel.
> So your so called "papers " are languishing in arxiv, none of them
> getting any chance of being published. This must hurt, you wasted all
> these years, you are no longer that young and nothing to show for.
Quick, someone bring the Kleenex tissues! I think I'm going to cry!
> Based on the crass errors that were exposed in these two forums, I
> think that MIT should revoke your PhD (Pile of Dung).
>
> All you do is you talk back, you never answer the technical challenges,
> you argue to death. If you act this way with the journal editors, no
> wonder that you never publish. Is this what is eating you? That you get
> rejected?
Have you ever considered switching to psychology? You wouldn't be very
good at it, but you'd probably be better at it than physics.
LEJ Brouwer
dda1 wrote:
> I thought that you are dumb but it turns out you are a psycho.
I'm just curious - were you born in India?
dda1
Sabbir Rahman , the defensive shithead wrote:
<snipped, never answers direct questions>
So, shithead, I asked you a very clear question that you never answered
: what is the experimental foundation of your "luminiferous" aether?
You gave us the runaround for countless posts. this is how you deal
with the editors?
It must be painful to see that you can't get published. With a mind and
a character like this, you will have to wait a loong time, if ever. You
are getting old and your "reputation" with the editors is not winning
you any points (incompetent, argumentative, deceptive, can't accept
constructive criticism).
dda1
No fuckface, I am an African American . There are a lot of non-Indians
that hate your ilk. You are still an inept psycho.
LEJ Brouwer
Ken S. Tucker wrote:
> LEJ Brouwer wrote:
> > Yes, the principle of equivalence needs to be generalised to allow for
> > particles of negative mass.
>
> I have no problem regarding an "energy deficit"
> like E = (+e)*(-e)/r being regarded as "negative
> energy". However the PoE liken's the acceleration
> of those parts by observing them from an accel-
> erating elevator, by gedanken. IOW's everything
> follows the same geodesics, that's damn general!
>
> You'll need to specify how to generalize PoE
> so that negative mass differs from +mass,
> geodesically.
> I'm all EARs, I'll consider your arguments including
> refs to your papers. I'll even consider that the
> PoE is erroneous if you have an argument to that
> effect that is reasonable.
> Ken S. Tucker
As you know, the gravitational force is very weak compared to the
electromagnetic force, which is one of the reasons why
antigravitational effects are difficult to detect in the lab.
The only amendment to the principle of equivalence is that for
antiparticles the inertial mass is equal and opposite to the
gravitational mass. The table in section 5.4 of my paper summarises
this, as well as the relationship between mass, energy and spacetime
curvature.
- Sabbir.
LEJ Brouwer
Have you ever considered switching to psychology? You wouldn't be very
LEJ Brouwer
Have you ever considered switching to psychology? You wouldn't be very
dda1
Yes, I am very good at detecting frauds. Like you. No need for
psychology, just a look at the papers you write.
dda1
Not necessary, shithead. I am very good at detecting frauds. Like you.
All I have to do is to take a look at your papers.
dda1
<all snipped, scumbags have no vote>
Ah, I finally got your endgame , you fucking troll.
-You wanted to attract attention to your latest "masterpiece" that you
have just uploaded in arxiv.
-You were afraid that this would be too crass" hey, look what I just
wrote", so you started this polemic about Abrams and Antoci when all
along it was about you and your newest shitty paper that will languish
10 years in arxiv like all your previous ones.
You got all of us fooled for quite a while but now your intent is
becoming apparent.
Why couldn't you be upfront and ask us to comment on your paper, you
fucking deceptive piece of shit?
What a scumbag!.
LEJ Brouwer
Daryl McCullough wrote:
> LEJ Brouwer says...
> a coordinate patch for a region of spacetime such that the freefalling
> geodesics are straight lines (approximately). In this patch, we can
> use local Minkowsky coordinates, and the freefalling observer has
> the simple trajectory
>
> x^0(tau) = c tau
> x^1(tau) = x^2(tau) = x^3(tau) = 0
Okay, and you would argue that x^0 is associated with 'r' outside the
event horizon, and then with 't' inside the horizon, whereas I am
claiming that x^0 is always associated with 'r' - with 'r' decreasing
until the particle hits the event horizon, and then increasing again as
it bounces out into the second Kruskal sheet.
LEJ Brouwer
:)
JanPB
> Tom Roberts wrote:
> > In solving for the metric components of the Schwarzschild manifold, one
> > normally selects spherical coordinates {t,r,theta,phi) at spatial
> > infinity and solves the set of equations, giving the usual Schw. line
> > element.
>
> I thought you said that there was no 'r' or 't' in the Einstein field
> equations? Well, now you have introduced them as I knew you would, do
> you *still* claim that the Einstein field equations in vacuo are
> invariant under swapping r and t?
Reread what Tom wrote. He addresses this very issue explicitly. Hope
it's just an eyeskip on your part.
> > In short, you think the portion of the manifold r<2M "does not exist",
> > while in reality it is the _COORDINATES_ you use that do not exist (i.e.
> > are invalid), but the interior region of the manifold is perfectly
> > valid, albeit rather strange to naive Euclidean sensibilities.
>
> No, the manifold *really* does not exist there.
It does: the set of all (a,b,c,d) in R^4 such that
-infinity<a<infinity, 0<b<2M, 0<c<pi, 0<d<2pi.
--
Jan Bielawski
JanPB
> Daryl McCullough wrote:
> > LEJ Brouwer says...
> > No, it doesn't. I'm not talking about a geodesic, I'm talking about
> > a coordinate patch for a region of spacetime such that the freefalling
> > geodesics are straight lines (approximately). In this patch, we can
> > use local Minkowsky coordinates, and the freefalling observer has
> > the simple trajectory
> >
> > x^0(tau) = c tau
> > x^1(tau) = x^2(tau) = x^3(tau) = 0
>
> Okay, and you would argue that x^0 is associated with 'r' outside the
> event horizon, and then with 't' inside the horizon, whereas I am
> claiming that x^0 is always associated with 'r'
But there is no x^0 in the sense you are imagining it. There is no
"common" coordinate that would associate with r or t or whatever else
over the two regions. The exterior chart was derived by solving the
Einstein equation. The interior chart OTOH was derived by
reverse-engineering a metric that was _pulled our of thin air_ and just
_happened_ to recycle the letters marking the coordinates in a
confusing manner. The price for laziness is confusion.
--
Jan Bielawski
LEJ Brouwer
And there lies a problem - you are assuming a priori that the topology
of the solution is homotopic to R^4, whereas the topology guaranteeing
spherical symmetry is S^2 x R^2, which is not quite the same (see
Synge's 1974 paper).
In any case, it seems that Tom is just repeating the usual textbook
arguments, and not really considering the work of Abrams and Crothers
which I refer him to. As I keep repeating (and I think I am going to
stop doing so now), their analysis precludes the existence of an
interior solution because the event horizon coincides with the point
mass.
> --
> Jan Bielawski
- Sabbir.
JanPB
> JanPB wrote:
> > LEJ Brouwer wrote:
> > > No, the manifold *really* does not exist there.
> >
> > It does: the set of all (a,b,c,d) in R^4 such that
> > -infinity<a<infinity, 0<b<2M, 0<c<pi, 0<d<2pi.
>
> And there lies a problem - you are assuming a priori that the topology
> of the solution is homotopic to R^4, whereas the topology guaranteeing
> spherical symmetry is S^2 x R^2, which is not quite the same (see
> Synge's 1974 paper).
You must be really dense or perhaps it's just a bad day. The domain of
*any* coordinate chart is diffeomorphic to R^4 (that's a _definition_
of coordinate charts in general). In this case the domain is:
(-infty,+infty) x (0, 2M) x (0, pi) x (0, 2pi)
...diffeomorphic to R^4.
Moreover, you can see that this particular domain is in obvious way a
subset of R^2 x S^2 (as it must) - the missing part arises from the
spatial spherical system not charting properly the points
"corresponding" to c=0,pi and d=0,2pi.
> In any case, it seems that Tom is just repeating the usual textbook
> arguments, and not really considering the work of Abrams and Crothers
> which I refer him to.
Tom is telling the truth. Therefore, he must sound the same as the
"usual textbooks", since they also are telling the truth. What's so
repetitive about that?
> As I keep repeating (and I think I am going to
> stop doing so now), their analysis precludes the existence of an
> interior solution because the event horizon coincides with the point
> mass.
Why? (Don't tell me to read your paper, please).
--
Jan Bielawski
LEJ Brouwer
JanPB wrote:
> LEJ Brouwer wrote:
> > And there lies a problem - you are assuming a priori that the topology
> > of the solution is homotopic to R^4, whereas the topology guaranteeing
> > spherical symmetry is S^2 x R^2, which is not quite the same (see
> > Synge's 1974 paper).
>
> You must be really dense or perhaps it's just a bad day. The domain of
> *any* coordinate chart is diffeomorphic to R^4 (that's a _definition_
> of coordinate charts in general). In this case the domain is:
>
> (-infty,+infty) x (0, 2M) x (0, pi) x (0, 2pi)
>
> ...diffeomorphic to R^4.
>
> Moreover, you can see that this particular domain is in obvious way a
> subset of R^2 x S^2 (as it must) - the missing part arises from the
> spatial spherical system not charting properly the points
> "corresponding" to c=0,pi and d=0,2pi.
I am talking about the global topology of the solution, not its local
geometry which you seem to be referring to. R^4 is a very special case
of S^2 x R^2, and for one thing, excludes the exterior Kruskal
solution, which is why I referred you to Synge's paper. (BTW, it is
getting late here - it's almost 2am in London).
> > In any case, it seems that Tom is just repeating the usual textbook
> > arguments, and not really considering the work of Abrams and Crothers
> > which I refer him to.
>
> Tom is telling the truth. Therefore, he must sound the same as the
> "usual textbooks", since they also are telling the truth. What's so
> repetitive about that?
That's fine, but its ignoring the whole point of the thread.
> > As I keep repeating (and I think I am going to
> > stop doing so now), their analysis precludes the existence of an
> > interior solution because the event horizon coincides with the point
> > mass.
>
> Why? (Don't tell me to read your paper, please).
Well, I did refer you to Crother's paper right at the beginning, and I
did at some point try to lead you through the derivation - you'll
probably have to dig that particular post out of the pile somewhere.
It's not a one line intuitive argument that I can explain in a couple
of paragraphs - I'm afraid that there is really no alternative to going
through the mathematics.
JanPB
> JanPB wrote:
> > LEJ Brouwer wrote:
> > > JanPB wrote:
> > > > LEJ Brouwer wrote:
> > > > > No, the manifold *really* does not exist there.
> > > > It does: the set of all (a,b,c,d) in R^4 such that
> > > > -infinity<a<infinity, 0<b<2M, 0<c<pi, 0<d<2pi.
> > > And there lies a problem - you are assuming a priori that the topology
> > > of the solution is homotopic to R^4, whereas the topology guaranteeing
> > > spherical symmetry is S^2 x R^2, which is not quite the same (see
> > > Synge's 1974 paper).
> >
> > You must be really dense or perhaps it's just a bad day. The domain of
> > *any* coordinate chart is diffeomorphic to R^4 (that's a _definition_
> > of coordinate charts in general). In this case the domain is:
> >
> > (-infty,+infty) x (0, 2M) x (0, pi) x (0, 2pi)
> >
> > ...diffeomorphic to R^4.
> >
> > Moreover, you can see that this particular domain is in obvious way a
> > subset of R^2 x S^2 (as it must) - the missing part arises from the
> > spatial spherical system not charting properly the points
> > "corresponding" to c=0,pi and d=0,2pi.
>
> I am talking about the global topology of the solution, not its local
> geometry which you seem to be referring to.
But you _were_ talking about the _interior solution_ (namely, that it
didn't exist). This is not the global topology. That's why I answered
with the precise set of coordinate values corresponding to the interior
solution.
> R^4 is a very special case
> of S^2 x R^2, and for one thing, excludes the exterior Kruskal
> solution,
OK, make up your mind what it is you want to talk about. Last time I
looked it was: "No, the manifold *really* does not exist there [the
interior]". So my answer was: "it does, here it is: (a,b,c,d) in R^4
such that... (etc.)".
Your answer? (And do not change the subject, pretty pleeeease?)
--
Jan Bielawski
LEJ Brouwer
I did not mean to change the subject. The point is that R^4 always
contains the origin and is single-valued wrt the radial distance from
the origin, whereas S^2 x R^2 is more general in that it need not
contain the origin and can be multivalued in r. One should therefore
not assume a priori that the topology of the solution is R^4. Synge
realised this and that is why he produced his alternative derivation of
the Schwarzschild solution in 1974.
I think the assumption of R^4 topology is one of the reasons why 'r' is
incorrectly characterised as a radial coordinate, when in fact it is
just a parametrisation for the radial coordinate, and hence why there
is such a strong attachment to the values r<2m, even though they are
not physical. Anyway, I hope you will take the time to read Crothers'
proof that r<2m is unphysical - it's less than one page and it
basically starts off with the Schwarzschild solution so there's nothing
particularly mysterious about it.
>
> --
> Jan Bielawski
JanPB
But I'm not assuming that. I simply wrote a chart covering a portion of
the region you claimed didn't exist.
> I think the assumption of R^4 topology is one of the reasons why 'r' is
> incorrectly characterised as a radial coordinate, when in fact it is
> just a parametrisation for the radial coordinate, and hence why there
> is such a strong attachment to the values r<2m, even though they are
> not physical.
Nobody assumes the R^4 topology. Only domains of coordinate charts are
homeomorphic to R^4 (as they must by definition). And what does
"incorrectly characterised as a radial coordinate" mean? r simply
labels spheres around the origin in a particular way.
> Anyway, I hope you will take the time to read Crothers'
> proof that r<2m is unphysical - it's less than one page and it
> basically starts off with the Schwarzschild solution so there's nothing
> particularly mysterious about it.
I don't really feel like going to the library to check something that's
obvously wrong. Would you want to waste time checking a new
revolutionary construction for angle trisection?
--
Jan Bielawski