Re: Question about analysis of Schwarzschild solution on Fo. of Phys. 1988, 18, 6
Sorcerer
"Juan R." <juanrgo...@canonicalscience.com> wrote in message news:1167572792....@48g2000cwx.googlegroups.com...
| It is well-known the role that classical Schwarzschild solution play in
| the classical tests of GR as Mercury perihelion anomaly.
The advance of Mercury's perihelion is accounted for by Newtonian
celestial mechanics, Einstein's 3 significant figure slide rule not
withstanding.
The only anomaly is the idiot Einstein and his stupid GR.
100 years, 415 orbits, 360 degrees per orbit, 3600 arc
seconds per degree.
43 arc seconds/ (415*360*3600) =
7.9949427338985571917298824929347e-8
"Amateurs look at data, professionals look at errorbars." - Tom Fuckwit Roberts.
Eric Gisse
Juan R. wrote:
> It is well-known the role that classical Schwarzschild solution play in
> the classical tests of GR as Mercury perihelion anomaly.
>
> the Schw solution requires at the same time
>
> T_1{}^1 = T_0{}^0
>
> AND
>
> T_0{}^0 as the only non-zero component of T_a{}^b
The Schwarzschild solution is the *vacuum* solution where T_ab = 0. All
the components of T should be zero _by assumption_.
>
> He cites Narlikar and Padmanabhan. Found. of Physics 1988, 18, 6.
>
> Someone with a copy of that paper? Some public rebuttal published or
> so?
>
> Without a copy of the work i cannot understand from where they claim
> two different values for the T-00.
Surfer
<juanrgo...@canonicalscience.com> wrote:
>It is well-known the role that classical Schwarzschild solution play in
>the classical tests of GR as Mercury perihelion anomaly.
>
>Recently i was thought that solution is a sheer imposibility because
>the Schw solution requires at the same time
>
>T_1{}^1 = T_0{}^0
>
>AND
>
>T_0{}^0 as the only non-zero component of T_a{}^b
>
>He cites Narlikar and Padmanabhan. Found. of Physics 1988, 18, 6.
>
>Someone with a copy of that paper? Some public rebuttal published or
>so?
>
>Without a copy of the work i cannot understand from where they claim
>two different values for the T-00.
>
It may help to look at the original derivations by Schwarzschild.
"On the gravitational field of a mass point according to Einstein's
theory."
http://arxiv.org/abs/physics/9905030
On the gravitational field of a sphere of incompressible fluid
according to Einstein's theory
http://arxiv.org/abs/physics/9912033
An exact solution appears as expression 14 on Page 6 of the first
paper. It includes the following term R:
R = (r^3 + alpha^3)^1/3
However when alpha << r, it is possible to make use of the
APPROXIMATION R = r.
When textbooks present the Schwartzschild metric, they generally
present an approximate metric based on R=r, rather than
Schwarzschild's original metric.
The second paper contains an interesting statement at the bottom of
page 8.
"...there is a limit to the concentration, above which a sphere of
incompressible fluid can not exist."
This seems suggestive of gravitational collapse (eg into a black hole)
This is interesting, because the foreword to the first paper says that
the exact metric "leaves no room for the science fiction of the black
holes".
However, the exact metric was derived by assuming that all the mass is
concentrated at a point. Since this assumption is not physically
realistic, the exact metric is exact in a mathematical sense only, and
not in the sense of exactly matching reality.
So the approximate metric might match reality just as well or better.
JanPB
>
> It may help to look at the original derivations by Schwarzschild.
>
> "On the gravitational field of a mass point according to Einstein's
> theory."
> http://arxiv.org/abs/physics/9905030
>
> On the gravitational field of a sphere of incompressible fluid
> according to Einstein's theory
> http://arxiv.org/abs/physics/9912033
>
> An exact solution appears as expression 14 on Page 6 of the first
> paper. It includes the following term R:
>
> R = (r^3 + alpha^3)^1/3
>
> However when alpha << r, it is possible to make use of the
> APPROXIMATION R = r.
>
> When textbooks present the Schwartzschild metric, they generally
> present an approximate metric based on R=r, rather than
> Schwarzschild's original metric.
No, they present the same metric as (14). The reason expressions like
(r^3+alpha^3)^1/3 do not appear in them is that they start with a
different radial coordinate (determined by the areas of the spheres of
symmetry) - Schwarzschild started with the standard polar r instead.
--
Jan Bielawski
shuba
> He cites Narlikar and Padmanabhan. Found. of Physics 1988, 18, 6.
>
> Someone with a copy of that paper? Some public rebuttal published or
> so?
>
> Without a copy of the work i cannot understand from where they claim
> two different values for the T-00.
So in the four months since this same post was rejected through
moderation at s.p.research, Juan R., who claims to be a serious
researcher, has been unable to locate anyone with access to that
particular journal who is willing to provide a copy. Funny!
The paper can be purchased for $30 without a subscription.
http://www.springerlink.com/content/1572-9516/?k=Narlikar+and+Padmanabhan
Perhaps Juan R. can ask for a $1 donation from each "researcher"
at the "center for canonical science" which would bring down the
personal cost for him to, oh around $30.
---Tim Shuba---
Koobee Wublee
> On 31 Dec 2006 05:46:32 -0800, "Juan R."
> "On the gravitational field of a mass point according to Einstein's
> theory."http://arxiv.org/abs/physics/9905030
>
> On the gravitational field of a sphere of incompressible fluid
> according to Einstein's theoryhttp://arxiv.org/abs/physics/9912033
>
> An exact solution appears as expression 14 on Page 6 of the first
> paper. It includes the following term R:
>
> R = (r^3 + alpha^3)^1/3
>
> However when alpha << r, it is possible to make use of the
> APPROXIMATION R = r.
>
> When textbooks present the Schwartzschild metric, they generally
> present an approximate metric based on R=r, rather than
> Schwarzschild's original metric.
>
> The second paper contains an interesting statement at the bottom of
> page 8.
>
> "...there is a limit to the concentration, above which a sphere of
> incompressible fluid can not exist."
>
> This seems suggestive of gravitational collapse (eg into a black hole)
>
> This is interesting, because the foreword to the first paper says that
> the exact metric "leaves no room for the science fiction of the black
> holes".
>
> However, the exact metric was derived by assuming that all the mass is
> concentrated at a point. Since this assumption is not physically
> realistic, the exact metric is exact in a mathematical sense only, and
> not in the sense of exactly matching reality.
>
> So the approximate metric might match reality just as well or better.
In late 1915, Schwarzschild received the field equations. To solve
these equations in which the solution is the metric, he had to wade
through the vastly complicated Ricci tensor to do so. However, if he
could transform the coordinate system into something that the
corresponding metric would yield a determinant of -1, the complexity of
the Ricci tensor would reduce by half. That is from this new
transformed coordinate system with the reduced Ricci tensor that easily
yielded for him his original solution in 1915. It actually looks like
the Schwarzschild metric as a function of R. However, transforming it
back to the polar coordinate system, that results in the
Schwarzschild's original metric and not the Schwarzschild metric.
Schwarzschild's original metric and others do not manifest black
holes. Schwarzschild was able to find a solution so soon because there
are actually an infinite number of solutions to the field equations.
The Schwarzschild metric itself did not come about until Hilbert
introduced it a year or two later. Thus, Einstein did not have any
approximate solution to Schwarzschild's original metric. In fact,
before the field equations, it is impossible even to approximate any
thing about the curvature in spacetime. Einstein's 1915 or
pre-field-equation solution to Mercury's orbital anomaly was a total
BS.
JanPB
It's the same metric. Metric doesn't change under coordinate
transformations.
> Schwarzschild's original metric and others do not manifest black
> holes.
False.
> Schwarzschild was able to find a solution so soon because there
> are actually an infinite number of solutions to the field equations.
False.
> The Schwarzschild metric itself did not come about until Hilbert
> introduced it a year or two later.
False.
> Thus, Einstein did not have any
> approximate solution to Schwarzschild's original metric. In fact,
> before the field equations, it is impossible even to approximate any
> thing about the curvature in spacetime. Einstein's 1915 or
> pre-field-equation solution to Mercury's orbital anomaly was a total
> BS.
False.
--
Jan Bielawski
JanPB
> Koobee Wublee wrote:
[...]
>
> > Schwarzschild's original metric and others do not manifest black
> > holes.
>
> False.
I want to qualify this: Schwarzschild cut off the radial coordinate at
what we today call the horizon. That's why the domain of his formula
did not include the central singularity (he thought the horizon was a
genuine singularity). This is simply a trivial restriction of the
domain of a unique solution - it does not introduce any new
"singularity-free" metric.
--
Jan Bielawski
Koobee Wublee
> Koobee Wublee wrote:
> > In late 1915, Schwarzschild received the field equations. To solve
> > these equations in which the solution is the metric, he had to wade
> > through the vastly complicated Ricci tensor to do so. However, if he
> > could transform the coordinate system into something that the
> > corresponding metric would yield a determinant of -1, the complexity of
> > the Ricci tensor would reduce by half. That is from this new
> > transformed coordinate system with the reduced Ricci tensor that easily
> > yielded for him his original solution in 1915. It actually looks like
> > the Schwarzschild metric as a function of R. However, transforming it
> > back to the polar coordinate system, that results in the
> > Schwarzschild's original metric and not the Schwarzschild metric.
>
> It's the same metric. Metric doesn't change under coordinate
> transformations.
I notice you do not label the metric as a tensor any more. Are you
wishy-washy?
The metric is merely a matrix which changes under coordinate
transformation.
> > Schwarzschild's original metric and others do not manifest black
> > holes.
>
> False.
>
> > Schwarzschild was able to find a solution so soon because there
> > are actually an infinite number of solutions to the field equations.
>
> False.
>
> > The Schwarzschild metric itself did not come about until Hilbert
> > introduced it a year or two later.
>
> False.
>
> > Thus, Einstein did not have any
> > approximate solution to Schwarzschild's original metric. In fact,
> > before the field equations, it is impossible even to approximate any
> > thing about the curvature in spacetime. Einstein's 1915 or
> > pre-field-equation solution to Mercury's orbital anomaly was a total
> > BS.
>
> False.
I have shown you mathematically how the field equations can yield an
infinite number of solutions. Your hand-waving defense was to claim
the metric being tensors which are the same under any coordinate
transformations. Now, I have shattered your BS claim that the metric
are invariant. The metric is not a tensor. Your silly responses are
typical of a sore loser.
> > > Schwarzschild's original metric and others do not manifest black
> > > holes.
>
> > False.
>
> I want to qualify this: Schwarzschild cut off the radial coordinate at
> what we today call the horizon. That's why the domain of his formula
> did not include the central singularity (he thought the horizon was a
> genuine singularity). This is simply a trivial restriction of the
> domain of a unique solution - it does not introduce any new
> "singularity-free" metric.
This is total BS.
Schwarzschild's original solution is valid. It is one of the many
independent solutions out there. The mathematics does not indicate a
singularity.
Why are you not calling the metric a tensor any more? Are you going to
suggest the metric being a mere matrix is invariant under any
coordinate transformation?
Eric Gisse
Koobee Wublee wrote:
[...]
> The metric is not a tensor.
[...]
WRONG.
Surfer
However the adoption of a different radial coordinate changes the
boundary condition at r = 0.
For example reading from:
http://www.mathpages.com/rr/s8-07/8-07.htm
[NB I have represented rho by p ]
"Nevertheless, following along with Schwarzschild's thought, he
obviously needs to require that the equality r^3 + p = a^3 be
satisfied only when r = 0, which implies p = a^3. Consequently he
argues that the expression (r3 + p )^1/3 should not be reduced to r.
Instead, he defines the parameter R = (r3 + p )^1/3, in terms of which
the metric has the familiar form (1). Of course, if we put p = 0 then
R = r and equation (1) reduces to the usual form of the
Schwarzschild/Droste solution. However, with p = a^3 we appear to have
a physically distinct result, free of any coordinate singularity
except at r = 0, which corresponds to the location R = a . The
question then arises as to whether this is actually a physically
distinct solution from the usual one."
The author then presents a case as to why the two solutions should be
considered the same. But as this is not a proof, it seems to me to
leave room for doubt.
So I feel, if the familiar form is preferred, it would be more
appropriate to simply argue for its merits.
Would if really matter if Schwarzschild's concept of the metric
differed from the modern concept?
-- Surfer
JanPB
> On Jan 1, 4:28 pm, "JanPB" <film...@gmail.com> wrote:
> > Koobee Wublee wrote:
>
> > > In late 1915, Schwarzschild received the field equations. To solve
> > > these equations in which the solution is the metric, he had to wade
> > > through the vastly complicated Ricci tensor to do so. However, if he
> > > could transform the coordinate system into something that the
> > > corresponding metric would yield a determinant of -1, the complexity of
> > > the Ricci tensor would reduce by half. That is from this new
> > > transformed coordinate system with the reduced Ricci tensor that easily
> > > yielded for him his original solution in 1915. It actually looks like
> > > the Schwarzschild metric as a function of R. However, transforming it
> > > back to the polar coordinate system, that results in the
> > > Schwarzschild's original metric and not the Schwarzschild metric.
> >
> > It's the same metric. Metric doesn't change under coordinate
> > transformations.
>
> I notice you do not label the metric as a tensor any more. Are you
> wishy-washy?
Metric is a tensor. I use both terms interchangeably.
> The metric is merely a matrix which changes under coordinate
> transformation.
That's the same as tensor.
Yes.
> It is one of the many
> independent solutions out there.
They are just different restrictions of one and the same solution.
> The mathematics does not indicate a singularity.
In his restricted domain there is no singularity, correct. But that's
simply because he sets the values of the radial coordinate to be
greater than the radius of what is now called the horizon.
> Why are you not calling the metric a tensor any more?
Pure stylistic convention. Metric, metric tensor - same thing.
> Are you going to
> suggest the metric being a mere matrix is invariant under any
> coordinate transformation?
No.
--
Jan Bielawski
Koobee Wublee
> For example reading from:http://www.mathpages.com/rr/s8-07/8-07.htm
>
> [NB I have represented rho by p ]
>
> "Nevertheless, following along with Schwarzschild's thought, he
> obviously needs to require that the equality r^3 + p = a^3 be
> satisfied only when r = 0, which implies p = a^3. Consequently he
> argues that the expression (r3 + p )^1/3 should not be reduced to r.
> Instead, he defines the parameter R = (r3 + p )^1/3, in terms of which
> the metric has the familiar form (1). Of course, if we put p = 0 then
> R = r and equation (1) reduces to the usual form of the
> Schwarzschild/Droste solution. However, with p = a^3 we appear to have
> a physically distinct result, free of any coordinate singularity
> except at r = 0, which corresponds to the location R = a . The
> question then arises as to whether this is actually a physically
> distinct solution from the usual one."
>
> The author then presents a case as to why the two solutions should be
> considered the same. But as this is not a proof, it seems to me to
> leave room for doubt.
Yes, all that is BS.
> So I feel, if the familiar form is preferred, it would be more
> appropriate to simply argue for its merits.
Once again, your instinct is correct. From their point of view, they
have no idea which metric is the real one while others are mere
transformations from that metric.
> Would if really matter if Schwarzschild's concept of the metric
> differed from the modern concept?
Yes, the fate of these monsters such as black holes rest at the
decision.
Eric Gisse
Koobee Wublee wrote:
> On Jan 1, 9:42 pm, Surfer <sur...@no.spam.net> wrote:
>
> > For example reading from:http://www.mathpages.com/rr/s8-07/8-07.htm
> >
> > [NB I have represented rho by p ]
> >
> > "Nevertheless, following along with Schwarzschild's thought, he
> > obviously needs to require that the equality r^3 + p = a^3 be
> > satisfied only when r = 0, which implies p = a^3. Consequently he
> > argues that the expression (r3 + p )^1/3 should not be reduced to r.
> > Instead, he defines the parameter R = (r3 + p )^1/3, in terms of which
> > the metric has the familiar form (1). Of course, if we put p = 0 then
> > R = r and equation (1) reduces to the usual form of the
> > Schwarzschild/Droste solution. However, with p = a^3 we appear to have
> > a physically distinct result, free of any coordinate singularity
> > except at r = 0, which corresponds to the location R = a . The
> > question then arises as to whether this is actually a physically
> > distinct solution from the usual one."
> >
> > The author then presents a case as to why the two solutions should be
> > considered the same. But as this is not a proof, it seems to me to
> > leave room for doubt.
>
> Yes, all that is BS.
...and why do you feel you are qualified to make this judgement?
You haven't exactly made that clear, now have you? Everything you say
is at odds with modern understandings of general relativity, tensor
analysis, and differential geometry.
>
> > So I feel, if the familiar form is preferred, it would be more
> > appropriate to simply argue for its merits.
>
> Once again, your instinct is correct. From their point of view, they
> have no idea which metric is the real one while others are mere
> transformations from that metric.
...and why do you feel you are qualified to make this judgement?
Do you still confuse metrics and transformation matricies?
>
> > Would if really matter if Schwarzschild's concept of the metric
> > differed from the modern concept?
>
> Yes, the fate of these monsters such as black holes rest at the
> decision.
...and why do you feel you are qualified to make this judgement?
How come you think four generations of physicists have missed what you
have 'discovered' ?
Koobee Wublee
On Jan 1, 11:49 pm, "Eric Gisse" <jowr...@gmail.com> wrote:
> Koobee Wublee wrote:
> > On Jan 1, 9:42 pm, Surfer <sur...@no.spam.net> wrote:
>
> > > For example reading from:http://www.mathpages.com/rr/s8-07/8-07.htm
>
> > > [NB I have represented rho by p ]
>
> > > "Nevertheless, following along with Schwarzschild's thought, he
> > > obviously needs to require that the equality r^3 + p = a^3 be
> > > satisfied only when r = 0, which implies p = a^3. Consequently he
> > > argues that the expression (r3 + p )^1/3 should not be reduced to r.
> > > Instead, he defines the parameter R = (r3 + p )^1/3, in terms of which
> > > the metric has the familiar form (1). Of course, if we put p = 0 then
> > > R = r and equation (1) reduces to the usual form of the
> > > Schwarzschild/Droste solution. However, with p = a^3 we appear to have
> > > a physically distinct result, free of any coordinate singularity
> > > except at r = 0, which corresponds to the location R = a . The
> > > question then arises as to whether this is actually a physically
> > > distinct solution from the usual one."
>
> > > The author then presents a case as to why the two solutions should be
> > > considered the same. But as this is not a proof, it seems to me to
> > > leave room for doubt.
>
> > Yes, all that is BS.
>
> ...and why do you feel you are qualified to make this judgement?
Read my posts.
> You haven't exactly made that clear, now have you?
Yes, I have.
> Everything you say
> is at odds with modern understandings of general relativity, tensor
> analysis, and differential geometry.
That is correct.
> > > So I feel, if the familiar form is preferred, it would be more
> > > appropriate to simply argue for its merits.
>
> > Once again, your instinct is correct. From their point of view, they
> > have no idea which metric is the real one while others are mere
> > transformations from that metric.
>
> ...and why do you feel you are qualified to make this judgement?
Again, read my posts.
> Do you still confuse metrics and transformation matricies?
Not at all. Physicists since Riemann or maybe Christoffel have.
By the way, have you finally understood the calculus of variations yet?
> > > Would if really matter if Schwarzschild's concept of the metric
> > > differed from the modern concept?
>
> > Yes, the fate of these monsters such as black holes rest at the
> > decision.
>
> ...and why do you feel you are qualified to make this judgement?
Once again, you've guessed it. Read my post.
> How come you think four generations of physicists have missed what you
> have 'discovered' ?
Why don't you go ask them?
Eric Gisse
Koobee Wublee wrote:
> On Jan 1, 11:49 pm, "Eric Gisse" <jowr...@gmail.com> wrote:
> > Koobee Wublee wrote:
> > > On Jan 1, 9:42 pm, Surfer <sur...@no.spam.net> wrote:
> >
> > > > For example reading from:http://www.mathpages.com/rr/s8-07/8-07.htm
> >
> > > > [NB I have represented rho by p ]
> >
> > > > "Nevertheless, following along with Schwarzschild's thought, he
> > > > obviously needs to require that the equality r^3 + p = a^3 be
> > > > satisfied only when r = 0, which implies p = a^3. Consequently he
> > > > argues that the expression (r3 + p )^1/3 should not be reduced to r.
> > > > Instead, he defines the parameter R = (r3 + p )^1/3, in terms of which
> > > > the metric has the familiar form (1). Of course, if we put p = 0 then
> > > > R = r and equation (1) reduces to the usual form of the
> > > > Schwarzschild/Droste solution. However, with p = a^3 we appear to have
> > > > a physically distinct result, free of any coordinate singularity
> > > > except at r = 0, which corresponds to the location R = a . The
> > > > question then arises as to whether this is actually a physically
> > > > distinct solution from the usual one."
> >
> > > > The author then presents a case as to why the two solutions should be
> > > > considered the same. But as this is not a proof, it seems to me to
> > > > leave room for doubt.
> >
> > > Yes, all that is BS.
> >
> > ...and why do you feel you are qualified to make this judgement?
>
> Read my posts.
>
[...]
I have. You display a stunning amount of confusion about simple
subjects and get the not-so-simple subjects wrong *EVERY TIME*.
Do you still believe the area of a sphere in Schwarzschild geometry is
4*pi*r^2 despite the metric telling you differently? That argument was
fun. You simply ignored the explicit calculation and said "NUH-UH!".
Do you still believe you can introduce curvature through a coordinate
transformation? That was a fun argument. You never did do the one thing
I asked: calculate the curvature scalar after your amazing
transformation.
Do you still believe a tensor is a coordinate-dependent entity? An
oldie but goodie, you continually seem to not understand what it means
to be a tensor.
Do you still believe one can create a new metric via an invertable
coordinate transformation? The concept of the "fundamental form" seems
to elude you.
How can someone who believed and most likely still does believe these
things actually be qualified to critique ANYTHING more complicated than
high school algebra? How can someone who says such things seriously
actually expect to not be laughed at when he says he is the only one to
truly understand differential geometry since Riemann?
JanPB
> On 31 Dec 2006 11:26:00 -0800, "JanPB" <fil...@gmail.com> wrote:
>
> >Surfer wrote:
> >>
> >> It may help to look at the original derivations by Schwarzschild.
> >>
> >> "On the gravitational field of a mass point according to Einstein's
> >> theory."
> >> http://arxiv.org/abs/physics/9905030
> >>
> >> On the gravitational field of a sphere of incompressible fluid
> >> according to Einstein's theory
> >> http://arxiv.org/abs/physics/9912033
> >>
> >> An exact solution appears as expression 14 on Page 6 of the first
> >> paper. It includes the following term R:
> >>
> >> R = (r^3 + alpha^3)^1/3
> >>
> >> However when alpha << r, it is possible to make use of the
> >> APPROXIMATION R = r.
> >>
> >> When textbooks present the Schwartzschild metric, they generally
> >> present an approximate metric based on R=r, rather than
> >> Schwarzschild's original metric.
> >
> >No, they present the same metric as (14). The reason expressions like
> >(r^3+alpha^3)^1/3 do not appear in them is that they start with a
> >different radial coordinate (determined by the areas of the spheres of
> >symmetry) - Schwarzschild started with the standard polar r instead.
> >
> However the adoption of a different radial coordinate changes the
> boundary condition at r = 0.
Sure - relabelling a manifold alters the numbers assigned to
boundaries, singularities, etc.
> For example reading from:
> http://www.mathpages.com/rr/s8-07/8-07.htm
>
> [NB I have represented rho by p ]
>
> "Nevertheless, following along with Schwarzschild's thought, he
> obviously needs to require that the equality r^3 + p = a^3 be
> satisfied only when r = 0, which implies p = a^3. Consequently he
> argues that the expression (r3 + p )^1/3 should not be reduced to r.
> Instead, he defines the parameter R = (r3 + p )^1/3, in terms of which
> the metric has the familiar form (1). Of course, if we put p = 0 then
> R = r and equation (1) reduces to the usual form of the
> Schwarzschild/Droste solution. However, with p = a^3 we appear to have
> a physically distinct result, free of any coordinate singularity
> except at r = 0, which corresponds to the location R = a . The
> question then arises as to whether this is actually a physically
> distinct solution from the usual one."
>
> The author then presents a case as to why the two solutions should be
> considered the same. But as this is not a proof, it seems to me to
> leave room for doubt.
I don't know why the author writes "with p = a^3 we appear to have a
physically distinct result, free of any coordinate singularity" - all
one has to do is keep track of the various coordinate changes (which
includes keeping track of their domains) and it is then obvious what
the meaning of the constant rho is. In brief, Schwarzschild got the
following solution in his coordinates (x1,x2,x3,x4):
ds^2= f4 dx4^2 - f1 dx1^2 -f2 (dx2^2/(1-x2^2)) - f3 (1-x2^2) dx3^2,
where:
f4 = 1 - alpha/(3x1 + rho)^1/3
f2 = f3 = (3x1 + rho)^2/3
f1 = (3x1 + rho)^(-4/3) / (1 - alpha(3x1 + rho)^(-1/3))
and alpha and rho are constant and the domain is:
x1 > 0 (it came from x1 = r^3/3, with r > 0),
AND x1 =/= (alpha^3 - rho)/3 AND x1 =/= -rho/3,
(where the metric coefficients become zero or singular),
-1 < x2 < 1 (from x2 = -cos(theta), with 0 < theta < pi),
0 < x3 < 2pi (from x3 = phi),
x4 - any real number (from x4 = t).
Notice the excluded values for x1 do not necessarily imply points
excluded from the manifold - they can be just coordinate problems since
we obviously don't know a priori that our initial chart covers the
whole manifold (even a simple 2D sphere cannot be covered by a single
chart).
Now comes the claim:
For every value of the constant rho the manifold portion with the above
metric is isometric to an open submanifold of what is commonly shown
nowadays in textbooks as "the Schwarzschild solution" (with possible
extra region corresponding to the negative values of alpha (the mass)).
Proof - very easy, Schwarzschild pretty much did most of the work in
his paper: given a rho, define the following change of variables (i.e.,
a diffeomorphism) from (x1,x2,x3,x4) to (t,R,theta,phi):
t = x4,
R = (3x1 + rho)^1/3
theta = arccos(-x2)
phi = x3
This is a diffeomorphism for all values in the domain of (x1,x2,x3,x4)
listed above: notice that the change of variables fails to be a
diffeomorphism when its Jacobian is zero which happens at x1=-rho/3 but
this value is already not in the domain of (x1,x2,x3,x4).
So we transform the line element (the metric's expression in
coordinates) to these new variables and we get:
ds^2 = (1 - alpha/R) dt^2 - 1/(1 - alpha/R) dR^2 -
R^2 (dtheta^2 + sin^2(theta) dphi^2)
where the domain in the (t,R,theta,phi)-space is:
t - any real number,
R > rho^1/3, AND R =/= 0 AND R =/= alpha,
0 < theta < pi
0 < phi < 2pi.
And this is the standard textbook Schwarzschild metric with the extra
"variability" in the form of a mere cutoff value for the radial
coordinate R, with the cutoff actually equal to rho^1/3.
QED.
Schwarzschild argued for this cutoff to be equal to alpha:
rho^1/3 = alpha, or rho=alpha^3,
...because he assumed the singularity was an actual manifold
singularity (an excluded point) and the only such point was the point
x1=0 which implied rho had to be alpha^3.
This cut his solution off just past the horizon, so in his domain there
are no singularities. But that's just restriction of a function.
What's perhaps confusing to some is that in the (t,R,theta,phi)-space
we have the cutoff R=rho^1/3 which corresponds to r=0 in the original
polar coordinates. This looks as if a "point" was in fact a "sphere".
But this is not the case - the crux of the matter is that the points
R=rho^1/3 and the point r=0 are NOT in the manifold. So from the
(t,R,theta,phi)-space we have a closed ball R<=rho^1/3 removed and from
the (t,r,theta,phi)-space we have the point r=0 removed - and these two
complements are _diffeomorphic_ (in fact, isometric, which of course is
the point).
So: instead of cutting off at R=rho^1/3=alpha, one can cut off at
R=rho^1/3=0 which extends the domain of the (unique) solution all the
way to the central excluded point. In principle one could continue in
an abstract manner even further and cut off R at negative values of
rho^1/3. These manifolds are again isometric to various open subsets of
a unique Schwarzschild-like solution with the constant alpha _negative_
(negative mass - whatever that would mean physically). This portion of
the solution has no horizon as (1+alpha/R) is only singular at R=0 - a
naked singularity.
At the end of the day we have only one solution whose maximal extension
is given by e.g. the Kruskal-Szekeres coordinate system (ignoring the
negative mass business).
--
Jan Bielawski
Don Mortel
>> the metric has the familiar form (1). Of course, if we put p = 0 then
>> R = r and equation (1) reduces to the usual form of the
>> Schwarzschild/Droste solution. However, with p = a^3 we appear to have
>> a physically distinct result, free of any coordinate singularity
>> except at r = 0, which corresponds to the location R = a . The
>> question then arises as to whether this is actually a physically
>> distinct solution from the usual one."
>> The author then presents a case as to why the two solutions should be
>> considered the same. But as this is not a proof, it seems to me to
>> leave room for doubt.
>
>I don't know why the author writes "with p = a^3 we appear to have a
>physically distinct result, free of any coordinate singularity" - all
I think the key word is "appear", when it says "we APPEAR to have a
physically distinct solution", because the next sentence is "The
question then arises as to whether this is ACTUALLY a physically
distinct solution from the usual one", and the web page then goes on
to explain that it is NOT (physically distinct).
Of course, we now know that it couldn't possibly be physically
distinct, since, by Birkhoff's theorem, there is only one spherically
symmetrical solution of the Einstein field equations (aside from
insignificant differences due to the arbitrary choice of coordinate
systems). So, the previous poster was in error when he said "But as
this is not a proof, it seems to me to leave room for doubt." It IS a
perfectly rigorous proof, and there is no room for doubt.
Koobee Wublee
> On 3 Jan 2007 01:48:31 -0800, "JanPB" <film...@gmail.com> wrote:
> Of course, we now know that it couldn't possibly be physically
> distinct, since, by Birkhoff's theorem, there is only one spherically
> symmetrical solution of the Einstein field equations (aside from
> insignificant differences due to the arbitrary choice of coordinate
> systems). So, the previous poster was in error when he said "But as
> this is not a proof, it seems to me to leave room for doubt." It IS a
> perfectly rigorous proof, and there is no room for doubt.
Proof of Birkhoff's theorem is convoluted. It assumes Schwarzschild
metric is the only solution in the first place. Birkhoff's theorem
is not only wrong but serves no purpose.
So far, we have two ridiculous claims. These two claims are not
substantiated, false, and silly.
** Because of Birkoff's theorem, the Schwarzschild metric is a
unique solution to the field equations.
Birkhoff's theorem is just wrong.
There is no credible proof.
** The metric is a tensor. Thus, the Schwarzschild metric is a unique
solution to the field equations.
The metric is merely an interpretation to
actual physical appearance using one's own
choice of coordinate system. Of course, the
physical appearance is invariant. The
coordinate system depends on the observer.
Thus, the metric must be dependent on the
Observer's choice of coordinate system.
Koobee Wublee
> Surfer wrote:
> Sure - relabelling a manifold alters the numbers assigned to
> boundaries, singularities, etc.
True.
However, with the introduction of a new metric while keeping the
coordinate system the same, the manifold has to change.
The mathematics showing this is the following.
ds^2 = g_ij dq^i dq^j
Where
** ds^2 = the manifold
** g_ij = the metric
** dq^i dq^j = the coordinate
The metric is not the manifold itself. It is an interpretation to the
manifold using the choice of coordinate system.
> > For example reading from:
> >http://www.mathpages.com/rr/s8-07/8-07.htm
>
> I don't know why the author writes "with p = a^3 we appear to have a
> physically distinct result, free of any coordinate singularity"
Because that is exactly what the mathematics shows.
This is total BS and hand-waving in the most ridiculous scale. Any of
the "variability" solution can be the metric itself. How can you
prove the Schwarzschild metric (textbook one) is a "variability"
version?
> Schwarzschild argued for this cutoff to be equal to alpha:
>
> rho^1/3 = alpha, or rho=alpha^3,
>
> ...because he assumed the singularity was an actual manifold
> singularity (an excluded point) and the only such point was the point
> x1=0 which implied rho had to be alpha^3.
Schwarzschild had no assumptions on anything. Singularities are not
even mentioned in both of his papers. You are fabricating history to
justify your silly and wrong interpretations.
JanPB
> On Jan 3, 1:48 am, "JanPB" <film...@gmail.com> wrote:
> > Surfer wrote:
>
> > Sure - relabelling a manifold alters the numbers assigned to
> > boundaries, singularities, etc.
>
> True.
>
> However, with the introduction of a new metric while keeping the
> coordinate system the same, the manifold has to change.
Point is, there is no new metric. Until you clear up in your mind what
metric is you'll be running in circles and seeing new "metrics" where
there are none.
> The mathematics showing this is the following.
>
> ds^2 = g_ij dq^i dq^j
>
> Where
>
> ** ds^2 = the manifold
> ** g_ij = the metric
> ** dq^i dq^j = the coordinate
>
> The metric is not the manifold itself.
No, of course not. It's a certain rank-2 tensor on it. In this post you
don't explain what your symbols mean but on the other thread you do,
and I'll address it there.
[...]
I provided the exact description and you provide some English prose.
You don't understand the problem. This was obvious from day one. What I
wrote above is in fact far more precise than any textbook I've seen.
> Any of
> the "variability" solution can be the metric itself. How can you
> prove the Schwarzschild metric (textbook one) is a "variability"
> version?
Anyone else can parse this sentence? What does "Any of the
"variability" solution can be the metric itself" mean? "What is
"variability solution"?
> > Schwarzschild argued for this cutoff to be equal to alpha:
> >
> > rho^1/3 = alpha, or rho=alpha^3,
> >
> > ...because he assumed the singularity was an actual manifold
> > singularity (an excluded point) and the only such point was the point
> > x1=0 which implied rho had to be alpha^3.
>
> Schwarzschild had no assumptions on anything.
Except on the 3rd page (bottom) he says: "...three functions of x1 have
to fulfil the following conditions: [...] continuity of the f, except
for x1=0".
> Singularities are not
> even mentioned in both of his papers.
Of course not, he calls them "discontinuities". When he sets:
rho = alpha^3
...he does so because f1 has a "discontinuity" when:
3x1 = alpha^3 - rho.
At this point he assumes that the only discontinuity can be at the
origin x1=0 (he misses the possibility the coordinate system itself is
simply discontinuous there), so that he has:
0 = alpha^3 - rho.
> You are fabricating history to
> justify your silly and wrong interpretations.
I'm just following the paper 100% exactly - something that Crothers,
Abrams, et al. never did (they never understood it).
--
Jan Bielawski
JanPB
> On Jan 3, 6:51 am, dmor...@no-spam.com (Don Mortel) wrote:
> > On 3 Jan 2007 01:48:31 -0800, "JanPB" <film...@gmail.com> wrote:
>
> > Of course, we now know that it couldn't possibly be physically
> > distinct, since, by Birkhoff's theorem, there is only one spherically
> > symmetrical solution of the Einstein field equations (aside from
> > insignificant differences due to the arbitrary choice of coordinate
> > systems). So, the previous poster was in error when he said "But as
> > this is not a proof, it seems to me to leave room for doubt." It IS a
> > perfectly rigorous proof, and there is no room for doubt.
>
> Proof of Birkhoff's theorem is convoluted. It assumes Schwarzschild
> metric is the only solution in the first place. Birkhoff's theorem
> is not only wrong but serves no purpose.
Complete nonsense. Birkhoff's theorem follows from a derivation which
assumes only spherical symmetry, vacuum, and the usual boundary
conditions. When one writes the Einstein equation in this case, only
four of the Ricci tensor components are distinct and nontrivial: R_tt,
R_rr, R_{theta,theta} (equal to R_{phi,phi}), and R_tr.
It is the last Ricci component that gives you the theorem because the
corresponding equation R_tr=0 says:
dG/dt = 0
...where G is the metric coefficient in front of the dr^2. So the
equation says that G is independent of t (a function of r only). Then
the next two equations:
R_tt = 0
R_rr = 0
...imply that F*G=constant (denoting by F the metric coefficient in
front of the dt^2), hence F is independent of t as well.
Uniqueness of this solution follows from the fact that all four Ricci
tensor equations happen to be ODEs so the theorems on existence and
uniqueness of solutions apply here.
> So far, we have two ridiculous claims. These two claims are not
> substantiated, false, and silly.
>
> ** Because of Birkoff's theorem, the Schwarzschild metric is a
> unique solution to the field equations.
>
> Birkhoff's theorem is just wrong.
> There is no credible proof.
Nonsense. Anybody can say "2+2=5". It doesn't make it true.
> ** The metric is a tensor. Thus, the Schwarzschild metric is a unique
> solution to the field equations.
>
> The metric is merely an interpretation to
> actual physical appearance using one's own
> choice of coordinate system.
This is all English prose. In order to substantiate your claims you
have to show, exactly: what other mathematical object is necessary
besides ds^2 (what you call "invariant segment of displacement") in
order to describe the gravitational field.
[...]
--
Jan Bielawski
Eric Gisse
Koobee Wublee wrote:
> On Jan 3, 6:51 am, dmor...@no-spam.com (Don Mortel) wrote:
> > On 3 Jan 2007 01:48:31 -0800, "JanPB" <film...@gmail.com> wrote:
>
> > Of course, we now know that it couldn't possibly be physically
> > distinct, since, by Birkhoff's theorem, there is only one spherically
> > symmetrical solution of the Einstein field equations (aside from
> > insignificant differences due to the arbitrary choice of coordinate
> > systems). So, the previous poster was in error when he said "But as
> > this is not a proof, it seems to me to leave room for doubt." It IS a
> > perfectly rigorous proof, and there is no room for doubt.
>
> Proof of Birkhoff's theorem is convoluted. It assumes Schwarzschild
> metric is the only solution in the first place. Birkhoff's theorem
> is not only wrong but serves no purpose.
You confuse the RESULT of a theorem and the PROOF of a theorem.
The proof of Birkhoff's theorem does *not* assume Schwarzschild's
solution. Have you actually looked at the proof?
>
> So far, we have two ridiculous claims. These two claims are not
> substantiated, false, and silly.
Yet you are the only one who thinks this. That should tell you
something.
>
> ** Because of Birkoff's theorem, the Schwarzschild metric is a
> unique solution to the field equations.
>
> Birkhoff's theorem is just wrong.
> There is no credible proof.
Just because you do not understand the proof does not mean the proof is
not credible. Would you like some references?
>
> ** The metric is a tensor. Thus, the Schwarzschild metric is a unique
> solution to the field equations.
>
> The metric is merely an interpretation to
> actual physical appearance using one's own
> choice of coordinate system. Of course, the
> physical appearance is invariant. The
> coordinate system depends on the observer.
> Thus, the metric must be dependent on the
> Observer's choice of coordinate system.
You fail differential geometry. Again.
Koobee Wublee
> Koobee Wublee wrote:
> > Proof of Birkhoff's theorem is convoluted. It assumes Schwarzschild
> > metric is the only solution in the first place. Birkhoff's theorem
> > is not only wrong but serves no purpose.
>
> Complete nonsense. Birkhoff's theorem follows from a derivation which
> assumes only spherical symmetry, vacuum, and the usual boundary
> conditions. When one writes the Einstein equation in this case, only
> four of the Ricci tensor components are distinct and nontrivial: R_tt,
> R_rr, R_{theta,theta} (equal to R_{phi,phi}), and R_tr.
>
> It is the last Ricci component that gives you the theorem because the
> corresponding equation R_tr=0 says:
>
> dG/dt = 0
>
> ...where G is the metric coefficient in front of the dr^2. So the
> equation says that G is independent of t (a function of r only). Then
> the next two equations:
>
> R_tt = 0
> R_rr = 0
>
> ...imply that F*G=constant (denoting by F the metric coefficient in
> front of the dt^2), hence F is independent of t as well.
>
> Uniqueness of this solution follows from the fact that all four Ricci
> tensor equations happen to be ODEs so the theorems on existence and
> uniqueness of solutions apply here.
This is how the field equations are solved by setting the Ricci tensor
to be zero in free space. You just prove my point that the proof to
the Birkhoff's theorem is convoluted.
> > However, with the introduction of a new metric while keeping the
> > coordinate system the same, the manifold has to change.
>
> Point is, there is no new metric. Until you clear up in your mind what
> metric is you'll be running in circles and seeing new "metrics" where
> there are none.
Each of the infinite numbers of solutions to the field equations using
the same coordinate is each a metric. Each different metric with the
same coordinate describes a different spacetime.
> > The mathematics showing this is the following.
>
> > ds^2 = g_ij dq^i dq^j
>
> > Where
>
> > ** ds^2 = the manifold
> > ** g_ij = the metric
> > ** dq^i dq^j = the coordinate
>
> > The metric is not the manifold itself.
>
> No, of course not. It's a certain rank-2 tensor on it. In this post you
> don't explain what your symbols mean but on the other thread you do,
> and I'll address it there.
The metric cannot be invariant just by examining the mathematics above.
If you claim tensors are invariant, then the metric is not a tensor.
<shrug>
> ds^2 = (1 - alpha/R) dt^2 - 1/(1 - alpha/R) dR^2 -
> R^2 (dtheta^2 + sin^2(theta) dphi^2)
> where the domain in the (t,R,theta,phi)-space is:
>
> t - any real number,
> R > rho^1/3, AND R =/= 0 AND R =/= alpha,
> 0 < theta < pi
> 0 < phi < 2pi.
>
> > This is total BS and hand-waving in the most ridiculous scale.
>
> I provided the exact description and you provide some English prose.
> You don't understand the problem. This was obvious from day one. What I
> wrote above is in fact far more precise than any textbook I've seen.
You provided your opinion. The equation above does not have these
boundary conditions you impose on it. Your boundary conditions are
totally unjustified.
> > Any of
> > the "variability" solution can be the metric itself. How can you
> > prove the Schwarzschild metric (textbook one) is a "variability"
> > version?Anyone else can parse this sentence? What does "Any of the
> "variability" solution can be the metric itself" mean? "What is
> "variability solution"?
>
> Schwarzschild argued for this cutoff to be equal to alpha:
>
> rho^1/3 = alpha, or rho=alpha^3,
>
> ...because he assumed the singularity was an actual manifold
> singularity (an excluded point) and the only such point was the point
> x1=0 which implied rho had to be alpha^3.
>
> > Schwarzschild had no assumptions on anything.
>
> Except on the 3rd page (bottom) he says: "...three functions of x1 have
> to fulfil the following conditions: [...] continuity of the f, except
> for x1=0".
3) and 4) are very obvious. 1) and 2) are the case because he
transformed the line element into a new metric with a new coordinate
system such that the determinant of the new metric is -1. Now, the
solution to the field equations becomes the new metric. So, after this
solution is found, he had to transform this new metric back to the
original polar coordinate system. This is very basic mathematics.
The transformed metric or the solution turned out to be the [textbook]
Schwarzschild metric as a function of R. However, the associated
coordinate is not the polar coordinate. Therefore, to get back to the
polar coordinate, the [textbook] Schwarzschild metric as a function of
R has to be transformed back into its corresponding metric as a
function of r. The metric is not a Schwarzschild metric, and it does
not manifest a black hole.
> > Singularities are not
> > even mentioned in both of his papers.
>
> Of course not, he calls them "discontinuities". When he sets:
>
> rho = alpha^3
>
> ...he does so because f1 has a "discontinuity" when:
>
> 3x1 = alpha^3 - rho.
>
> At this point he assumes that the only discontinuity can be at the
> origin x1=0 (he misses the possibility the coordinate system itself is
> simply discontinuous there), so that he has:
>
> 0 = alpha^3 - rho.
The sentence right above 13) specifically says the discontinuity is
right at zero. There is no singularity (other than 0). You are trying
to justify for a singularity in Schwarzschild original derivation, and
you are not going to find any --- no black hole.
> > You are fabricating history to
> > justify your silly and wrong interpretations.
>
> I'm just following the paper 100% exactly - something that Crothers,
> Abrams, et al. never did (they never understood it).
The bottom line is that you don't understand Schwarzschild's
derivation. You are the one who has 90 years to catch up.
Eric Gisse
Durrr...what?
How is that even an argument? Do you think the Ricci tensor is nonzero
in free space?
[rest of idiocy snipped]
Koobee Wublee
> Koobee Wublee wrote:
> > This is how the field equations are solved by setting the Ricci tensor
> > to be zero in free space. You just prove my point that the proof to
> > the Birkhoff's theorem is convoluted.
>
> Durrr...what?
Knowledge indigestion? Have you learned the calculus of variations
yet? It has only been half a year.
> How is that even an argument? Do you think the Ricci tensor is nonzero
> in free space?
No, the Ricci is always zero in free space. You are utterly confused.
<shrug>
Eric Gisse
Koobee Wublee wrote:
> On Jan 3, 5:40 pm, "Eric Gisse" <jowr...@gmail.com> wrote:
> > Koobee Wublee wrote:
>
> > > This is how the field equations are solved by setting the Ricci tensor
> > > to be zero in free space. You just prove my point that the proof to
> > > the Birkhoff's theorem is convoluted.
> >
> > Durrr...what?
>
> Knowledge indigestion? Have you learned the calculus of variations
> yet? It has only been half a year.
Do you still believe the area of a sphere in Schwarzschild geometry is
4*pi*r^2 despite the metric telling you differently? That argument was
fun. You simply ignored the explicit calculation and said "NUH-UH!".
Do you still believe you can introduce curvature through a coordinate
transformation? That was a fun argument. You never did do the one thing
I asked: calculate the curvature scalar after your amazing
transformation.
Do you still believe a tensor is a coordinate-dependent entity? An
oldie but goodie, you continually seem to not understand what it means
to be a tensor.
Do you still believe one can create a new metric via an invertable
coordinate transformation? The concept of the "fundamental form" seems
to elude you.
How can someone who believed and most likely still does believe these
things actually be qualified to critique ANYTHING more complicated than
high school algebra? How can someone who says such things seriously
actually expect to not be laughed at when he says he is the only one to
truly understand differential geometry since Riemann?
>
> > How is that even an argument? Do you think the Ricci tensor is nonzero
> > in free space?
>
> No, the Ricci is always zero in free space. You are utterly confused.
> <shrug>
WHAT THE FUCK IS YOUR ARGUMENT THEN?
JanPB
OK, fine, whatever. My point was only that Birkhoff's theorem did not
assume Schwarzschild's solution was unique.
> > > However, with the introduction of a new metric while keeping the
> > > coordinate system the same, the manifold has to change.
> >
> > Point is, there is no new metric. Until you clear up in your mind what
> > metric is you'll be running in circles and seeing new "metrics" where
> > there are none.
>
> Each of the infinite numbers of solutions to the field equations using
> the same coordinate is each a metric. Each different metric with the
> same coordinate describes a different spacetime.
Agreed as far as changing the components of a metric in a fixed
coordinate system changes the metric.
Disagreed with "infinite numbers of solutions to the field equations".
[...]
>
> > ds^2 = (1 - alpha/R) dt^2 - 1/(1 - alpha/R) dR^2 -
> > R^2 (dtheta^2 + sin^2(theta) dphi^2)
>
> > where the domain in the (t,R,theta,phi)-space is:
> >
> > t - any real number,
> > R > rho^1/3, AND R =/= 0 AND R =/= alpha,
> > 0 < theta < pi
> > 0 < phi < 2pi.
> >
> > > This is total BS and hand-waving in the most ridiculous scale.
> >
> > I provided the exact description and you provide some English prose.
> > You don't understand the problem. This was obvious from day one. What I
> > wrote above is in fact far more precise than any textbook I've seen.
>
> You provided your opinion.
Since when mathematics is an opinion? What I wrote was simply the
description of the mathematical objects under consideration: two
coordinate systems, their resp. domains, and the expression for ds^2 in
those two coordinates. None of it is debatable - it's just a
description of the situation.
> The equation above does not have these
> boundary conditions you impose on it. Your boundary conditions are
> totally unjustified.
They come from the standard spherical coordinates in space. They are
exactly the same as Schwarzschild's (although he wasn't as explicit
with the domains toward the end as I was).
Start with spherical coordinates and arbitrary time variable. The
domains are:
t - any real number
r > 0
0 < theta < pi
0 < phi < 2pi
...and then simply keep track of the domains as you switch to new
coordinate systems. There are several coordinate changes chained one
after another in Schwarzschild's paper, all you have to do is follow
them.
> > > Any of
> > > the "variability" solution can be the metric itself. How can you
> > > prove the Schwarzschild metric (textbook one) is a "variability"
> > > version?Anyone else can parse this sentence? What does "Any of the
> > "variability" solution can be the metric itself" mean? "What is
> > "variability solution"?
> >
> > Schwarzschild argued for this cutoff to be equal to alpha:
> >
> > rho^1/3 = alpha, or rho=alpha^3,
> >
> > ...because he assumed the singularity was an actual manifold
> > singularity (an excluded point) and the only such point was the point
> > x1=0 which implied rho had to be alpha^3.
> >
> > > Schwarzschild had no assumptions on anything.
> >
> > Except on the 3rd page (bottom) he says: "...three functions of x1 have
> > to fulfil the following conditions: [...] continuity of the f, except
> > for x1=0".
>
> 3) and 4) are very obvious. 1) and 2) are the case because he
> transformed the line element into a new metric with a new coordinate
> system such that the determinant of the new metric is -1.
It's the same metric in new coordinates.
> Now, the
> solution to the field equations becomes the new metric. So, after this
> solution is found, he had to transform this new metric back to the
> original polar coordinate system. This is very basic mathematics.
No, he didn't _have_ to transform back - the metric is fine as is. It
only looks and feels far better when written in more or less standard
polars rather then those odd (x1,x2,x3,x4).
> The transformed metric or the solution turned out to be the [textbook]
> Schwarzschild metric as a function of R. However, the associated
> coordinate is not the polar coordinate. Therefore, to get back to the
> polar coordinate, the [textbook] Schwarzschild metric as a function of
> R has to be transformed back into its corresponding metric as a
> function of r. The metric is not a Schwarzschild metric, and it does
> not manifest a black hole.
It is an open submanifold of the standard Schwarzschild solution
(because its ds^2 is the same). It does not manifest a black hole
simply because the entire interior region including the horizon has
been ignored. When one transforms from R back to r, the exterior region
(space minus a closed ball) gets mapped onto the polar coordinate space
minus the origin. This is a perfectly valid isometry.
And no, it doesn't mean that "the quantity r=alpha [the horizon] [...]
is actually a point" either (as Crothers naively writes in his April
2005 article on p. 70).
> > > Singularities are not
> > > even mentioned in both of his papers.
> >
> > Of course not, he calls them "discontinuities". When he sets:
> >
> > rho = alpha^3
> >
> > ...he does so because f1 has a "discontinuity" when:
> >
> > 3x1 = alpha^3 - rho.
> >
> > At this point he assumes that the only discontinuity can be at the
> > origin x1=0 (he misses the possibility the coordinate system itself is
> > simply discontinuous there), so that he has:
> >
> > 0 = alpha^3 - rho.
>
> The sentence right above 13) specifically says the discontinuity is
> right at zero.
I know - he means a discontinuity of a solution. This is correct since
the physical setup is vacuum everywhere except possibly at the origin.
But this still leaves open the possibility of discontinuity of the
coordinates.
> There is no singularity (other than 0). You are trying
> to justify for a singularity in Schwarzschild original derivation, and
> you are not going to find any --- no black hole.
I'm simply stating the obvious: when one starts with a coordinate
system on an unknown manifold (like Schwarzschild does at the very
beginning), one does not know a priori that this system covers the
entire manifold. The points not covered by the coordinates but lying on
the boundary of their domain can result in weird limiting behaviour as
one approaches the domain boundary. In the case of Schwarzschild's f1
the weirdness showed up as blowing f1 up to infinity at
x1=1/3*(alpha^3-rho). Such weird limits are simply properties of the
coordinate system.
For example, f(x) = 1/x is a good coordinate system on the real line
minus the origin. But this doesn't mean there is anything funny going
on at the origin of the real line. It just means the coordinate system
f has a domain at whose boundary it blows up.
Same thing happens with Schwarzschild's coordinate system: it's not
defined not only at the origin but also at the horizon. By cutting the
entire interior out Schwarzschild "removed" the singularities: the real
one and the coordinate one.
> > > You are fabricating history to
> > > justify your silly and wrong interpretations.
> >
> > I'm just following the paper 100% exactly - something that Crothers,
> > Abrams, et al. never did (they never understood it).
>
> The bottom line is that you don't understand Schwarzschild's
> derivation. You are the one who has 90 years to catch up.
Well, you keep talking and I'll supply the mathematics proving you
wrong.
--
Jan Bielawski
Surfer
<koobee...@gmail.com> wrote:
It is interesting though, to look at the last few sentences of:
"On the gravitational field of a sphere of incompressible fluid
according to Einstein's theory"
http://arxiv.org/abs/physics/9912033
Here Schwarzschild says:
At the center of the sphere (X = 0) velocity of light and pressure
become infinite when cosXa = 1/3 , and the fall velocity becomes
sqrt(8/9) of the (naturally measured) velocity of light. Hence there
is a limit to the concentration, above which a sphere of
incompressible fluid can not exist. If one would apply our equations
to values cosXa < 1/3, one would get discontinuities already outside
the center of the sphere. One can however find solutions of the
problem for larger Xa, which are continuous at least outside the
center of the sphere, if one goes over to the case of either lamda > 0
or lamba < 0, and satisfies the condition K = 0 (Eq. 27). On the road
of these solutions, that are clearly not physically meaningful, since
they give infinite pressure at the center, one can go over to the
limit case of a mass concentrated to one point, and retrieves then the
relation rho = alpha^3 which, according to the previous study, holds
for the mass point. It is further noticed here that one can
speak of a mass point only as far as one avails of the variable r,
that otherwise in a surprising way plays no role for the geometry and
for the motion inside our gravitational field.
For an observer measuring from outside it follows from (40) that a
sphere of given gravitational mass alpha/2k^2 can
not have a radius measured from outside smaller than:
Po = alpha
For a sphere of incompressible fluid the limit will be 9/(8 * alpha).
(For the Sun alpha is equal to 3 km, for a mass of 1 gram is equal to
1.5 * 10^-28 cm.)"
So at the very least, he predicted the following:
1) The Schwarzschild radius.
2) Strange happenings within this radius.
This could be seen as suggestive of the concept of black holes.
Or, if not of black holes, then some alternative new concept would
seem to be needed.
-- Surfer
Koobee Wublee
> Koobee Wublee wrote:
> [R_ij = 0]
>
> > This is how the field equations are solved by setting the Ricci tensor
> > to be zero in free space. You just prove my point that the proof to
> > the Birkhoff's theorem is convoluted.
>
> OK, fine, whatever. My point was only that Birkhoff's theorem did not
> assume Schwarzschild's solution was unique.
I still have not seen the proof.
> > Each of the infinite numbers of solutions to the field equations using
> > the same coordinate is each a metric. Each different metric with the
> > same coordinate describes a different spacetime.
>
> Agreed as far as changing the components of a metric in a fixed
> coordinate system changes the metric.
Excellent, the toilet paper serving as the shield that you try to
protect your matrix is finally removed.
> > 3) and 4) are very obvious. 1) and 2) are the case because he
> > transformed the line element into a new metric with a new coordinate
> > system such that the determinant of the new metric is -1.
>
> It's the same metric in new coordinates.
You are contradicting yourself again. In order to describe an
invariant geometry, you need a new metric whenever you are using a new
coordinate system.
> > Now, the
> > solution to the field equations becomes the new metric. So, after this
> > solution is found, he had to transform this new metric back to the
> > original polar coordinate system. This is very basic mathematics.
>
> No, he didn't _have_ to transform back - the metric is fine as is. It
> only looks and feels far better when written in more or less standard
> polars rather then those odd (x1,x2,x3,x4).
You are still contradicting yourself. For example, we have the
following quadratic equation.
x^2 - 6 x + 8 = 0
There are many ways to solve it. Let use a substitution first.
y = x / 2
Thus, the quadratic equation is transformed into the following.
y^2 - 3 y + 2 = 0
The solutions are
y = 1, and y = 2
Since x is not y, therefore we have to transform y back to x. In this
case, we have arrived at the correct solutions to the original
quadratic equation.
x = 2, and x = 4
That is why Schwarzschild had to change the coordinate back. This
should be the most basic of the mathematics! Why do you have so much
trouble understanding that?
> > The transformed metric or the solution turned out to be the [textbook]
> > Schwarzschild metric as a function of R. However, the associated
> > coordinate is not the polar coordinate. Therefore, to get back to the
> > polar coordinate, the [textbook] Schwarzschild metric as a function of
> > R has to be transformed back into its corresponding metric as a
> > function of r. The metric is not a Schwarzschild metric, and it does
> > not manifest a black hole.
>
> It is an open submanifold of the standard Schwarzschild solution
> (because its ds^2 is the same). It does not manifest a black hole
> simply because the entire interior region including the horizon has
> been ignored. When one transforms from R back to r, the exterior region
> (space minus a closed ball) gets mapped onto the polar coordinate space
> minus the origin. This is a perfectly valid isometry.
You are assuming too much again. The interior region is rightfully not
addressed. There is no need. You are too influenced with the
Schwarzschild metric to see ghosts like these. You are way too
opinionated. Hindsight is always 20/20.
> And no, it doesn't mean that "the quantity r=alpha [the horizon] [...]
> is actually a point" either (as Crothers naively writes in his April
> 2005 article on p. 70).
According to basic mathematics, when you specify (r = alpha), you are
describing a point in one dimension, a line in two dimensions, a
surface in 3 dimensions.
Koobee Wublee
> On 1 Jan 2007 23:37:14 -0800, "Koobee Wublee"
> Would if really matter if Schwarzschild's concept of the metric
What Schwarzschild described is very different from a black hole.
Eric Gisse
Koobee Wublee wrote:
> JanPB wrote:
> > Koobee Wublee wrote:
>
> > [R_ij = 0]
> >
> > > This is how the field equations are solved by setting the Ricci tensor
> > > to be zero in free space. You just prove my point that the proof to
> > > the Birkhoff's theorem is convoluted.
> >
> > OK, fine, whatever. My point was only that Birkhoff's theorem did not
> > assume Schwarzschild's solution was unique.
>
> I still have not seen the proof.
....are you shitting me?
You are complaining about a proof you have never even seen?
Are you ever going to respond to my other posts where I ask if you
still believe in stupid shit you have said in the past, or are you
going to continue ignoring them in the hope I will forget how stupid
you are?
[...]
JanPB
> JanPB wrote:
> > Koobee Wublee wrote:
>
> > [R_ij = 0]
> >
> > > This is how the field equations are solved by setting the Ricci tensor
> > > to be zero in free space. You just prove my point that the proof to
> > > the Birkhoff's theorem is convoluted.
> >
> > OK, fine, whatever. My point was only that Birkhoff's theorem did not
> > assume Schwarzschild's solution was unique.
>
> I still have not seen the proof.
The proof is in the derivation which takes several pages of
calculations - I just posted the highlights: the fact that R_tr=0
implies the relevant metric components are independent of t, and the
fact that all four nontrivial Einstein equations happen to be ODEs.
Beyond that you only need to know what metrics on manifolds are.
> > > Each of the infinite numbers of solutions to the field equations using
> > > the same coordinate is each a metric. Each different metric with the
> > > same coordinate describes a different spacetime.
> >
> > Agreed as far as changing the components of a metric in a fixed
> > coordinate system changes the metric.
>
> Excellent, the toilet paper serving as the shield that you try to
> protect your matrix is finally removed.
I think you misread something. I said that if you _fix_ a coordinate
system then varying the matrix of components does change the metric
(i.e., the metric tensor - that's what I always mean when I say
"metric"). This is simply linear algebra. Of course in the Einstein
equation case once you fix a coordinate system, the equations admit
only one solution (parametrised by the constants we talked about)
satisfying given boundary conditions.
That's why I said "agreed _as far as_...". I disagree with the multiple
solutions bit because it's false: solutions to ODEs are unique, it's
THE basic theorem in all of differential equations theory. Entire
science of mechanics relies on it too.
> > > 3) and 4) are very obvious. 1) and 2) are the case because he
> > > transformed the line element into a new metric with a new coordinate
> > > system such that the determinant of the new metric is -1.
> >
> > It's the same metric in new coordinates.
>
> You are contradicting yourself again. In order to describe an
> invariant geometry, you need a new metric whenever you are using a new
> coordinate system.
Again, if you are using your own private terminology according to which
"metric" means "the matrix of coefficients wrt a particular basis" then
we have no argument here. So it looks like we are both saying that the
coefficient matrix of the metric changes with coordinates. That's
correct - the matrix changes according to the usual (X-transpose)*[g]*X
law.
> > > Now, the
> > > solution to the field equations becomes the new metric. So, after this
> > > solution is found, he had to transform this new metric back to the
> > > original polar coordinate system. This is very basic mathematics.
> >
> > No, he didn't _have_ to transform back - the metric is fine as is. It
> > only looks and feels far better when written in more or less standard
> > polars rather then those odd (x1,x2,x3,x4).
>
> You are still contradicting yourself. For example, we have the
> following quadratic equation.
>
> x^2 - 6 x + 8 = 0
>
> There are many ways to solve it. Let use a substitution first.
>
> y = x / 2
>
> Thus, the quadratic equation is transformed into the following.
>
> y^2 - 3 y + 2 = 0
>
> The solutions are
>
> y = 1, and y = 2
>
> Since x is not y, therefore we have to transform y back to x. In this
> case, we have arrived at the correct solutions to the original
> quadratic equation.
>
> x = 2, and x = 4
>
> That is why Schwarzschild had to change the coordinate back. This
> should be the most basic of the mathematics! Why do you have so much
> trouble understanding that?
May I remind you that we are on a manifold here (spacetime) - your
example is an equation on an _unparametrised real line_ and is thus a
wrong analogy. See, if you think of x^2 - 6 x + 8 as a scalar on a
manifold and you think of y=x/2 as a coordinate change, then indeed you
_don't_ need to go back from y to x. Either x or y points to the same
1D manifold point pair which is your solution and you are done. It's up
to you if you prefer the expression of the solution points in terms of
x or y.
But your quadratic is an equation on the real line (no parametrisation,
just the real line) so obviously to pinpoint the solution you need to
express it in the original variable x.
On top of that in our case we have not a scalar but a tensor on a
manifold which means the extra complication in the form of coefficients
becoming coordinate-dependent (which is something scalars don't do).
> > > The transformed metric or the solution turned out to be the [textbook]
> > > Schwarzschild metric as a function of R. However, the associated
> > > coordinate is not the polar coordinate. Therefore, to get back to the
> > > polar coordinate, the [textbook] Schwarzschild metric as a function of
> > > R has to be transformed back into its corresponding metric as a
> > > function of r. The metric is not a Schwarzschild metric, and it does
> > > not manifest a black hole.
> >
> > It is an open submanifold of the standard Schwarzschild solution
> > (because its ds^2 is the same). It does not manifest a black hole
> > simply because the entire interior region including the horizon has
> > been ignored. When one transforms from R back to r, the exterior region
> > (space minus a closed ball) gets mapped onto the polar coordinate space
> > minus the origin. This is a perfectly valid isometry.
>
> You are assuming too much again.
I'm not "assuming" anything - I just followed all the relevant objects
exactly. They lead one down a path towards the conclusion I posted: one
solution parametrised by the mass and the cutoff value of the radial
coordinate. There is no other interpretation of it, it's just like any
other mathematical argument: a implies b implies c, etc. At each step I
had absolutely no choice but to push the argument certain way.
> The interior region is rightfully not
> addressed. There is no need. You are too influenced with the
> Schwarzschild metric to see ghosts like these. You are way too
> opinionated. Hindsight is always 20/20.
I'm going to stop posting to this thread if you keep just adding more
prose like this.
> > And no, it doesn't mean that "the quantity r=alpha [the horizon] [...]
> > is actually a point" either (as Crothers naively writes in his April
> > 2005 article on p. 70).
>
> According to basic mathematics, when you specify (r = alpha), you are
> describing a point in one dimension, a line in two dimensions, a
> surface in 3 dimensions.
Yes, but Crothers says r=alpha is a point in 3 dimensions.
--
Jan Bielawski
Koobee Wublee
> Koobee Wublee wrote:
> > I still have not seen the proof [of Birkhoff's theorem].
>
> The proof is in the derivation which takes several pages of
> calculations - I just posted the highlights: the fact that R_tr=0
> implies the relevant metric components are independent of t, and the
> fact that all four nontrivial Einstein equations happen to be ODEs.
I am not objecting to the static part of the derivation. It follows
exactly the Schwarzschild metric. What I am objecting to is the claim
of the unique solution to the field equations.
> Beyond that you only need to know what metrics on manifolds are.
>
> > Excellent, the toilet paper serving as the shield that you try to
> > protect your matrix is finally removed.
>
> I think you misread something. I said that if you _fix_ a coordinate
> system then varying the matrix of components does change the metric
> (i.e., the metric tensor - that's what I always mean when I say
> "metric"). This is simply linear algebra. Of course in the Einstein
> equation case once you fix a coordinate system, the equations admit
> only one solution (parametrised by the constants we talked about)
> satisfying given boundary conditions.
Mathematically, the field equations yield an infinite number of
solutions. How can you justify all solutions are indeed the same after
demoting the metric from being a divine tensor to be a mere common
matrix. Are you not able to understand what the solutions to ODEs are
in general?
> That's why I said "agreed _as far as_...". I disagree with the multiple
> solutions bit because it's false: solutions to ODEs are unique, it's
> THE basic theorem in all of differential equations theory. Entire
> science of mechanics relies on it too.
There is no theorem that dictates a set of ODE in general to yield only
one unique solution.
> > You are contradicting yourself again. In order to describe an
> > invariant geometry, you need a new metric whenever you are using a new
> > coordinate system.
>
> Again, if you are using your own private terminology according to which
> "metric" means "the matrix of coefficients wrt a particular basis" then
> we have no argument here. So it looks like we are both saying that the
> coefficient matrix of the metric changes with coordinates. That's
> correct - the matrix changes according to the usual (X-transpose)*[g]*X
> law.
Let's try this again. When you transform the whole set of ODE to a
different coordinate system, the metric, of course, changes. However,
the geometry still does not change because it is invariant. The
solutions you obtained from the transformed ODE still account to
infinite. Each solution (or each metric) as interpreted by the new
transformed coordinate does describe a different geometry based on the
transformed coordinates. You have not changed the scope of the issue
by trying transforming coordinate system.
Your argument is so lame. There are no differences in perspectives
between these naked ODEs and the quadratic equation.
The transformed metric or the solution turned out to be the [textbook]
Schwarzschild metric as a function of R. However, the associated
coordinate is not the polar coordinate. Therefore, to get back to the
polar coordinate, the [textbook] Schwarzschild metric as a function of
R has to be transformed back into its corresponding metric as a
function of r. The metric is not a Schwarzschild metric, and it does
not manifest a black hole.
>
> It is an open submanifold of the standard Schwarzschild solution
> (because its ds^2 is the same). It does not manifest a black hole
> simply because the entire interior region including the horizon has
> been ignored. When one transforms from R back to r, the exterior region
> (space minus a closed ball) gets mapped onto the polar coordinate space
> minus the origin. This is a perfectly valid isometry.
>
> > You are assuming too much again.
>
> I'm not "assuming" anything - I just followed all the relevant objects
> exactly. They lead one down a path towards the conclusion I posted: one
> solution parametrised by the mass and the cutoff value of the radial
> coordinate. There is no other interpretation of it, it's just like any
> other mathematical argument: a implies b implies c, etc. At each step I
> had absolutely no choice but to push the argument certain way.
You are way too influenced by the event horizon manifested by the
Schwarzschild metric. By looking at Schwarzschild's original
solution as naked as it is, it really does not justify your claim.
Specifically, the choice of coordinate system is always the simple
polar coordinate. He transformed it to one that yields the determinant
of the also transformed metric to be -1. He had to transform it back
to simple polar coordinate.
> > The interior region is rightfully not
> > addressed. There is no need. You are too influenced with the
> > Schwarzschild metric to see ghosts like these. You are way too
> > opinionated. Hindsight is always 20/20.
>
> I'm going to stop posting to this thread if you keep just adding more
> prose like this.
That is fine. Until next time. Hopefully, you will be wiser. At
least, in this round of exchanges, you have finally agreed the metric
is not invariant. That is called progress even if you have hard time
swallowing that.
> > According to basic mathematics, when you specify (r = alpha), you are
> > describing a point in one dimension, a line in two dimensions, a
> > surface in 3 dimensions.
>
> Yes, but Crothers says r=alpha is a point in 3 dimensions.
You really do not like Mr. Crothers, do you? That is between you and
he.
JanPB
I didn't "demote" it - tensor _is_ a set of matrices, one matrix per
coordinate system, linked by the law:
(X-transpose) * [matrix wrt coordinates q] * X = [matrix wrt
coordinates p]
...where X is the Jacobian matrix from coordinates p to coordinates q.
For example, Euclidean metric in the 2D plane is:
the matrix [1 0] wrt the rectangular coordinates xy,
[0 1]
AND
the matrix [1 0 ] wrt the polar coordinates r,theta,
[0 r^2]
AND
infinity of other matrices corresponding to the infinity of other
coordinate systems in 2D plane via the "(X-transpose)*[g]*X" formula.
One geometry, infinitely many matrices of g_ij's corresponding to
infinitely many coordinates, all linked by the above formula.
For example, the Jacobian matrix X from the polar r,theta to the
rectangular xy is:
X = [ cos(theta) -r sin(theta) ]
[ sin(theta) r cos(theta) ]
...and you can verify that indeed:
(X-transpose) * [1 0] * X = [1 0 ]
[0 1] [0 r^2]
So the answer to your question is trivial:
1. get hold of the Einstein equation,
2. fix a spherically symmetric coordinate system - this fixes the
corresponding metric matrix although we don't know yet what its entries
are,
3. write Einstein's equation in this coordinate system,
4. solve it - this yields the matrix entries mentioned in step 2 above.
These entries are unique by the ODE theory,
5. thus obtain the matrix with respect to _any other coordinate system
we can think of_ via the formula:
(X-transpose * [the solution matrix obtained in step 4] * X
This infinite set of matrices _is the single metric on the manifold_.
Any representative matrix from this set describes how to calculate _one
and the same dot product_ (the above formula is precisely what makes
this definition of "dot product" invariant).
Hence all the geometric invariants are uniquely determined, hence the
gravitational physics of the Schwarzschild solution is uniquely
determined.
> Are you not able to understand what the solutions to ODEs are
> in general?
>
> > That's why I said "agreed _as far as_...". I disagree with the multiple
> > solutions bit because it's false: solutions to ODEs are unique, it's
> > THE basic theorem in all of differential equations theory. Entire
> > science of mechanics relies on it too.
>
> There is no theorem that dictates a set of ODE in general to yield only
> one unique solution.
Yes, there is. It's called Picard-Lindeloef theorem and it says that as
long as a very general condition is satisfied (mainly a Lipschitz
condition) any system of ODEs has a unique solution once the initial
data is specified.
> > > You are contradicting yourself again. In order to describe an
> > > invariant geometry, you need a new metric whenever you are using a new
> > > coordinate system.
> >
> > Again, if you are using your own private terminology according to which
> > "metric" means "the matrix of coefficients wrt a particular basis" then
> > we have no argument here. So it looks like we are both saying that the
> > coefficient matrix of the metric changes with coordinates. That's
> > correct - the matrix changes according to the usual (X-transpose)*[g]*X
> > law.
>
> Let's try this again. When you transform the whole set of ODE to a
> different coordinate system, the metric, of course, changes.
If by "metric" you mean the matrix of g_ij then - yes.
> However,
> the geometry still does not change because it is invariant.
OK.
> The
> solutions you obtained from the transformed ODE still account to
> infinite. Each solution (or each metric) as interpreted by the new
> transformed coordinate does describe a different geometry based on the
> transformed coordinates.
No, it's the same geometry. The consistency of tensor calculus
guarantees that the "new" solution matrix you obtain from the
transformed "new" ODEs will automatically be linked with the "old"
solution obtained by solving the original "old" ODEs by the usual
formula [g-new]=(X-transpose)*[g-old]*X, where X is the Jacobian from
the "new" to "old" coordinates.
So the geometry hasn't changed.
There is a huge difference between an equation on a manifold whose
points have identities independently of any parametrisation and an
equation on the parametrising space whose points are _fixed_ n-tuples
of numbers.
> The transformed metric or the solution turned out to be the [textbook]
> Schwarzschild metric as a function of R. However, the associated
> coordinate is not the polar coordinate.
It makes no difference.
> Therefore, to get back to the
> polar coordinate, the [textbook] Schwarzschild metric as a function of
> R has to be transformed back into its corresponding metric as a
> function of r.
Yes, in order to write this metric tensor in polar coordinates one must
transform it to polar coordinates - obviously.
> The metric is not a Schwarzschild metric, and it does
> not manifest a black hole.
That's only because it's a subset.
> > It is an open submanifold of the standard Schwarzschild solution
> > (because its ds^2 is the same). It does not manifest a black hole
> > simply because the entire interior region including the horizon has
> > been ignored. When one transforms from R back to r, the exterior region
> > (space minus a closed ball) gets mapped onto the polar coordinate space
> > minus the origin. This is a perfectly valid isometry.
> >
> > > You are assuming too much again.
> >
> > I'm not "assuming" anything - I just followed all the relevant objects
> > exactly. They lead one down a path towards the conclusion I posted: one
> > solution parametrised by the mass and the cutoff value of the radial
> > coordinate. There is no other interpretation of it, it's just like any
> > other mathematical argument: a implies b implies c, etc. At each step I
> > had absolutely no choice but to push the argument certain way.
>
> You are way too influenced by the event horizon manifested by the
> Schwarzschild metric. By looking at Schwarzschild's original
> solution as naked as it is, it really does not justify your claim.
> Specifically, the choice of coordinate system is always the simple
> polar coordinate. He transformed it to one that yields the determinant
> of the also transformed metric to be -1. He had to transform it back
> to simple polar coordinate.
That's fine but his solution has this second constant rho which only
serves as the lower bound for R. If you choose to set this lower bound
above the horizon then what's left obviously has no singularities. But
this is not any new solution, it's just a restriction.
> > > The interior region is rightfully not
> > > addressed. There is no need. You are too influenced with the
> > > Schwarzschild metric to see ghosts like these. You are way too
> > > opinionated. Hindsight is always 20/20.
> >
> > I'm going to stop posting to this thread if you keep just adding more
> > prose like this.
>
> That is fine. Until next time. Hopefully, you will be wiser. At
> least, in this round of exchanges, you have finally agreed the metric
> is not invariant.
Not sure where you got it from. All I have ever said was that the
metric tensor was invariant (all tensors are invariant in fact - just
like vectors and scalars which are special cases of tensors) and that
the metric tensor can be represented in a coordinate system by a
matrix, with the usual formula linking such matrix representations
across different coordinate systems.
> That is called progress even if you have hard time
> swallowing that.
You must have misread something.
> > > According to basic mathematics, when you specify (r = alpha), you are
> > > describing a point in one dimension, a line in two dimensions, a
> > > surface in 3 dimensions.
> >
> > Yes, but Crothers says r=alpha is a point in 3 dimensions.
>
> You really do not like Mr. Crothers, do you? That is between you and
> he.
I don't like people who have the front to pontificate condescendingly
while doing nothing to fix their ignorance of the basics. The following
quote from Crothers' paper is typical: "It is plain from the foregoing
that the Kruskal-Szekeres extension is meaningless, that the
'Schwarzschild radius' is meaningless, that the orthodox conception of
gravitational collapse is incorrect, and that the black hole is not
consistent at all with General Relativity". Just that and with a
perfectly straight face.
--
Jan Bielawski
Surfer
>Koobee Wublee wrote:
>>
>> The transformed metric or the solution turned out to be the [textbook]
>> Schwarzschild metric as a function of R. However, the associated
>> coordinate is not the polar coordinate. Therefore, to get back to the
>> polar coordinate, the [textbook] Schwarzschild metric as a function of
>> R has to be transformed back into its corresponding metric as a
>> function of r. The metric is not a Schwarzschild metric, and it does
>> not manifest a black hole.
>
>It is an open submanifold of the standard Schwarzschild solution
>(because its ds^2 is the same). It does not manifest a black hole
>simply because the entire interior region including the horizon has
>been ignored. When one transforms from R back to r, the exterior region
>(space minus a closed ball) gets mapped onto the polar coordinate space
>minus the origin. This is a perfectly valid isometry.
>
After reading your earlier responses and also looking more carefully
at Schwarzschild's papers, I believe you are completely correct about
this.
Thanks,
Surfer
Surfer
<koobee...@gmail.com> wrote:
He might not have had time to explore such possibilities.
Since he was at the Russian front his papers were probably written
under pressing conditions.
However, suppose that a sphere of fluid gradually gained volume eg
like a star growing larger due to attacting debris and gas.
Its radius would increase in proportion to the cube root of the volume
(even more slowly if the fluid was compressible) but the
Schwarzschild radius would increase in proportion to the volume, so
the Schwarzschild radius would eventually become larger than the
radius of the sphere.
When that point was reached, the solution for a sphere of fluid would
suddenly have to be replaced by the solution for a mass point, which
implies the sphere suffers a dramatic collapse.
Wouldn't this be equivalent or very similar to forming a black hole?
Also on the last page he says:
<<<
"It is further noticed here that one can speak of a mass point only as
far as one avails of the variable r, that otherwise in a surprising
way plays no role for the geometry"
>>>
But if r plays "no role in the geometry", how can we be sure that r=0
is actually at the centre of the system?
In fact it is not.
Near the bottom of page 7 he writes:
<<<<
According to the expression (36) of the line element outside the
sphere this Po is clearly identical with the value Ra = (ra^3 +
rho)^1/3 that the variable R assumes at the surface of the sphere.
>>>>
So it would seem that the true centre would be at R=0.
But at this point we have
R=0=(r^3 + rho)^1/3
which means that at the centre of the system
r = - rho^1/3
So that is where the point mass has to be located in order to be at
the geometrical center of the system.
So Schwarzschild did not have the boundary condition for the mass
point quite right and it is the familiar form of the Schwarzschild
solution which is correct.
It seems that by placing the mass at r=0, Schwarzschild inadvertently
solved for mass distributed on a spherical shell of radius,
R = rho^1/3.
But beyond this radius, the metric is the same as for a point mass, so
he got the right solution in spite of this oversight.
-- Surfer
Koobee Wublee
> Koobee Wublee wrote:
> > Mathematically, the field equations yield an infinite number of
> > solutions. How can you justify all solutions are indeed the same after
> > demoting the metric from being a divine tensor to be a mere common
> > matrix.
>
> I didn't "demote" it - tensor _is_ a set of matrices, one matrix per
> coordinate system, linked by the law:
You have already agreed that the metric is not a tensor any more.
> So the answer to your question is trivial:
>
> 1. get hold of the Einstein equation,
> 2. fix a spherically symmetric coordinate system - this fixes the
> corresponding metric matrix although we don't know yet what its entries
> are,
> 3. write Einstein's equation in this coordinate system,
> 4. solve it - this yields the matrix entries mentioned in step 2 above.
> These entries are unique by the ODE theory,
At this stage, the solutions (or the metric matrices) are only valid to
this spherically symmetric coordinate system. They are not valid for
any other type of coordinate system. This is so because your step 2
above. Agreed?
> If by "metric" you mean the matrix of g_ij then - yes.
No, both g_ij and [g].
> > Therefore, to get back to the
> > polar coordinate, the [textbook] Schwarzschild metric as a function of
> > R has to be transformed back into its corresponding metric as a
> > function of r.
>
> Yes, in order to write this metric tensor in polar coordinates one must
> transform it to polar coordinates - obviously.
You have missed the point. That metric is only valid in R. To qualify
a metric valid in r, it has to be transformed back into the coordinate
of r.
> > The metric is not a Schwarzschild metric, and it does
> > not manifest a black hole.
>
> That's only because it's a subset.
How do you justify that the [textbook] Schwarzschild metric is not a
subset of Schwarzschild's original solution?
> > You are way too influenced by the event horizon manifested by the
> > Schwarzschild metric. By looking at Schwarzschild's original
> > solution as naked as it is, it really does not justify your claim.
> > Specifically, the choice of coordinate system is always the simple
> > polar coordinate. He transformed it to one that yields the determinant
> > of the also transformed metric to be -1. He had to transform it back
> > to simple polar coordinate.
>
> That's fine but his solution has this second constant rho which only
> serves as the lower bound for R. If you choose to set this lower bound
> above the horizon then what's left obviously has no singularities. But
> this is not any new solution, it's just a restriction.
How do you determine what these constants in Schwarzschild's original
solution are?
> Not sure where you got it from. All I have ever said was that the
> metric tensor was invariant (all tensors are invariant in fact - just
> like vectors and scalars which are special cases of tensors) and that
> the metric tensor can be represented in a coordinate system by a
> matrix, with the usual formula linking such matrix representations
> across different coordinate systems.
This is not what you were saying. You clearly agreed that the metric
is not a tensor anymore.
> > That is called progress even if you have hard time
> > swallowing that.
>
> You must have misread something.
You just have lousy memory.
> > You really do not like Mr. Crothers, do you? That is between you and
> > he.
>
> I don't like people who have the front to pontificate condescendingly
> while doing nothing to fix their ignorance of the basics.
You yourself are acting exactly like you have described above. <shrug>
> The following
> quote from Crothers' paper is typical: "It is plain from the foregoing
> that the Kruskal-Szekeres extension is meaningless, that the
> 'Schwarzschild radius' is meaningless, that the orthodox conception of
> gravitational collapse is incorrect, and that the black hole is not
> consistent at all with General Relativity". Just that and with a
> perfectly straight face.
Mr. Crothers is correct. He should say that with a straight face. If
you embrace voodoo mathematics, it is not his fault.
Koobee Wublee
> Koobee Wublee wrote:
> >What Schwarzschild described is very different from a black hole.
>
> He might not have had time to explore such possibilities.
> Since he was at the Russian front his papers were probably written
> under pressing conditions.
He served as a meteorologist in the eastern front. He was not doing
combat. His death was due to an infection of some sort not due to
combat injuries.
> http://arxiv.org/abs/physics/9912033
Let me be more blunt. This paper deals with solutions in matter and
not in free space. Thus, Schwarzschild metric does not apply.
Eric Gisse
Koobee Wublee wrote:
[...]
Where were you taught your general relativity and differential geometry?
JanPB
> JanPB wrote:
> > Koobee Wublee wrote:
>
> > > Mathematically, the field equations yield an infinite number of
> > > solutions. How can you justify all solutions are indeed the same after
> > > demoting the metric from being a divine tensor to be a mere common
> > > matrix.
> >
> > I didn't "demote" it - tensor _is_ a set of matrices, one matrix per
> > coordinate system, linked by the law:
>
> You have already agreed that the metric is not a tensor any more.
No - reread what I wrote: "tensor _is_ a set of matrices, one matrix
per coordinate system, linked by the [usual] law". Metric is a tensor -
it's the basis of Riemannian geometry.
> > So the answer to your question is trivial:
> >
> > 1. get hold of the Einstein equation,
> > 2. fix a spherically symmetric coordinate system - this fixes the
> > corresponding metric matrix although we don't know yet what its entries
> > are,
> > 3. write Einstein's equation in this coordinate system,
> > 4. solve it - this yields the matrix entries mentioned in step 2 above.
> > These entries are unique by the ODE theory,
>
> At this stage, the solutions (or the metric matrices) are only valid to
> this spherically symmetric coordinate system. They are not valid for
> any other type of coordinate system. This is so because your step 2
> above. Agreed?
Sure. But you can transform those matrices to obtain the corresponding
metric matrices wrt to any other coordinate system of your choice via
the "(X-transpose)*[g]*X" thing.
> > If by "metric" you mean the matrix of g_ij then - yes.
>
> No, both g_ij and [g].
I use [g] to denote the matrix of g_ij's. You seem to be now using [g]
to denote something other than g_ij's - what is it?
> > > Therefore, to get back to the
> > > polar coordinate, the [textbook] Schwarzschild metric as a function of
> > > R has to be transformed back into its corresponding metric as a
> > > function of r.
> >
> > Yes, in order to write this metric tensor in polar coordinates one must
> > transform it to polar coordinates - obviously.
>
> You have missed the point. That metric is only valid in R.
Yes, the matrix of g_ij is meant to be used in that particular
coordinate system. Of course this automatically defines the
corresponding metric matrices in all other coordinate systems.
> To qualify
> a metric valid in r, it has to be transformed back into the coordinate
> of r.
Sure. Perhaps I misunderstood you: I thought you said Schwarzschild had
to pull the metric back to r because the expression with respect to R
was invalid somehow. So I said that the R coordinate was as good as the
r coordinate.
> > > The metric is not a Schwarzschild metric, and it does
> > > not manifest a black hole.
> >
> > That's only because it's a subset.
>
> How do you justify that the [textbook] Schwarzschild metric is not a
> subset of Schwarzschild's original solution?
To avoid more confusion regarding the negative mass solution let me
state exactly the situation (I'm using r for Schwarzschild radial
coordinate):
* the textbook Schwarzschild metric tensor is defined over the domain
r>0 (plus the usual domains for the other coordinates),
* Schwarzschild's paper defines the same metric tensor over the domain
r>rho^1/3 (plus the usual domains for the other coordinates),
...where rho is an arbitrary constant.
So depending on how one sets this constant we either have
Schwarzschild's original metric contained in the textbook metric
(rho>0) or equal to it (rho=0) or containing it (rho<0).
Of course there is nothing preventing any of the textbooks to allow
negative numbers for the radial coordinate - the resulting metric still
satisfies Einstein's equations - but since it's isometric to the
standard Schwarzschild with _negative alpha_ (the mass) (via changing
the coordinate r to -r), it is not usually considered as a physical
spacetime. It also has a naked singularity at the origin (no horizon).
Finally, the portion of Schwarzschild's solution corresponding to
negative r is (topologically) disconnected from the portion with
positive r.
> > > You are way too influenced by the event horizon manifested by the
> > > Schwarzschild metric. By looking at Schwarzschild's original
> > > solution as naked as it is, it really does not justify your claim.
> > > Specifically, the choice of coordinate system is always the simple
> > > polar coordinate. He transformed it to one that yields the determinant
> > > of the also transformed metric to be -1. He had to transform it back
> > > to simple polar coordinate.
> >
> > That's fine but his solution has this second constant rho which only
> > serves as the lower bound for R. If you choose to set this lower bound
> > above the horizon then what's left obviously has no singularities. But
> > this is not any new solution, it's just a restriction.
>
> How do you determine what these constants in Schwarzschild's original
> solution are?
alpha can be determined from comparing with the Newtonian case. rho is
an arbitrary parameter which happens to cut the radial coordinate from
below, so it doesn't really "parametrise" the solution in any other
sense than FAPP tautological. IOW one can set rho so as to make the
solution a maximally topologically connected component. This happens
for rho=0.
> > Not sure where you got it from. All I have ever said was that the
> > metric tensor was invariant (all tensors are invariant in fact - just
> > like vectors and scalars which are special cases of tensors) and that
> > the metric tensor can be represented in a coordinate system by a
> > matrix, with the usual formula linking such matrix representations
> > across different coordinate systems.
>
> This is not what you were saying. You clearly agreed that the metric
> is not a tensor anymore.
Sorry, a misunderstanding then. There is no way I could have ever
"given up" what is absolutely fundamental in differential geometry.
[...]
> > > You really do not like Mr. Crothers, do you? That is between you and
> > > he.
> >
> > I don't like people who have the front to pontificate condescendingly
> > while doing nothing to fix their ignorance of the basics.
>
> You yourself are acting exactly like you have described above. <shrug>
I don't publish articles about stuff I don't understand in which I
declare virtually all previous research worthless.
> > The following
> > quote from Crothers' paper is typical: "It is plain from the foregoing
> > that the Kruskal-Szekeres extension is meaningless, that the
> > 'Schwarzschild radius' is meaningless, that the orthodox conception of
> > gravitational collapse is incorrect, and that the black hole is not
> > consistent at all with General Relativity". Just that and with a
> > perfectly straight face.
>
> Mr. Crothers is correct. He should say that with a straight face. If
> you embrace voodoo mathematics, it is not his fault.
He is provably wrong.
--
Jan Bielawski
Russell
> JanPB wrote:
> > Koobee Wublee wrote:
>
> > > Mathematically, the field equations yield an infinite number of
> > > solutions. How can you justify all solutions are indeed the same after
> > > demoting the metric from being a divine tensor to be a mere common
> > > matrix.
> >
> > I didn't "demote" it - tensor _is_ a set of matrices, one matrix per
> > coordinate system, linked by the law:
>
> You have already agreed that the metric is not a tensor any more.
I assume you mean that Jan, by making his *above*
statement, has according to you just asserted it is not
a tensor.
(Otherwise, I'd challenge you to cite a previous post in
which Jan asserted anything like what you claim. Surely
I've never seen one.)
If my reading of you is right, then I'd say that at best you're
engaging in a deconstruction of the word "is" that rivals
that of a recent U.S. President. Whatever definition of
tensor you might use, there is certainly an isomorphism
to the entity that Jan described above. In math, that's
good enough for "is".
[snip]
> > > The metric is not a Schwarzschild metric, and it does
> > > not manifest a black hole.
> >
> > That's only because it's a subset.
>
> How do you justify that the [textbook] Schwarzschild metric is not a
> subset of Schwarzschild's original solution?
Eh? Why in the world should subset or superset status
depend on historical order?
Hmm, I'll answer that -- if both solutions are correct, and
if we believe in scientific progress, then the earlier ought to
be a subset of the later.
[snip]
Russell
> Koobee Wublee wrote:
> > JanPB wrote:
> > > Koobee Wublee wrote:
> >
> > > > Mathematically, the field equations yield an infinite number of
> > > > solutions. How can you justify all solutions are indeed the same after
> > > > demoting the metric from being a divine tensor to be a mere common
> > > > matrix.
Heh, guess I didn't read that final phrase very well.
> > >
> > > I didn't "demote" it - tensor _is_ a set of matrices, one matrix per
> > > coordinate system, linked by the law:
> >
> > You have already agreed that the metric is not a tensor any more.
>
> I assume you mean
Note to self: beware posting late at night.
that Jan, by making his *above*
> statement, has according to you just asserted it is not
> a tensor.
>
> (Otherwise, I'd challenge you to cite a previous post in
> which Jan asserted anything like what you claim. Surely
> I've never seen one.)
Well then, KW, consider yourself so challenged.
Tom Roberts
>> http://arxiv.org/abs/physics/9912033
> Let me be more blunt. This paper deals with solutions in matter and
> not in free space. Thus, Schwarzschild metric does not apply.
It is well known that Schwarzschild discovered _TWO_ different solutions
to the EFE. One in vacuum and one in matter with uniform density. This
paper is quite clearly discussing his second.
Admittedly, when mentioning "the Schwarzschild solution" people usually
mean the first (vacuum) one. But both are known by his name.
Tom Roberts
Surfer
<koobee...@gmail.com> wrote:
>JanPB wrote:
>
>> The following
>> quote from Crothers' paper is typical: "It is plain from the foregoing
>> that the Kruskal-Szekeres extension is meaningless, that the
>> 'Schwarzschild radius' is meaningless, that the orthodox conception of
>> gravitational collapse is incorrect, and that the black hole is not
>> consistent at all with General Relativity". Just that and with a
>> perfectly straight face.
>
>Mr. Crothers is correct.
>
He may be correct about some things, but compare his claim
"the 'Schwarzschild radius' is meaningless"
with the following comments by Schwarzschild.
From:
On the gravitational field of a sphere of incompressible fluid
according to Einstein's theory
http://arxiv.org/abs/physics/9912033
Near bottom of page 8
<<<
At the center of the sphere (X = 0) velocity of light and pressure
become infinite when cosXa = 1/3...
>>>
What do you imagine might happen to matter if it was subjected to
"infinite pressure"?
Next, near the bottom of page 9, Schwarzschild gives us the radius at
which this would occur.
<<<
For an observer measuring from outside it follows from (40) that a
sphere of given gravitational mass alpha/2k^2 can
not have a radius measured from outside smaller than:
Po = alpha
>>>
I believe this is the Schwarzschild radius, which implies this radius
is far from meaningless.
For the sun, Schwarzschild calculates Po to be 3 km which agrees with
the value at
http://en.wikipedia.org/wiki/Schwarzschild_radius
Regards,
Surfer
Koobee Wublee
> Koobee Wublee wrote:
> No - reread what I wrote: "tensor _is_ a set of matrices, one matrix
> per coordinate system, linked by the [usual] law". Metric is a tensor -
> it's the basis of Riemannian geometry.
Did you not mention a tensor is invariant, independent of coordinate
system?
> 1. get hold of the Einstein equation,
> 2. fix a spherically symmetric coordinate system - this fixes the
> corresponding metric matrix although we don't know yet what its entries
> are,
> 3. write Einstein's equation in this coordinate system,
> 4. solve it - this yields the matrix entries mentioned in step 2 above.
> These entries are unique by the ODE theory,
>
> > At this stage, the solutions (or the metric matrices) are only valid to
> > this spherically symmetric coordinate system. They are not valid for
> > any other type of coordinate system. This is so because your step 2
> > above. Agreed?
>
> Sure.
Since you have great tendency to speak with a forked tongue, please
acknowledge once again that the solutions to the field equations are
only valid at the particular choice of coordinate system.
> But you can transform those matrices to obtain the corresponding
> metric matrices wrt to any other coordinate system of your choice via
> the "(X-transpose)*[g]*X" thing.
Yes, I do agree with that. However, if you do mean you have agreed any
solutions obtained by the field equations for this particular set of
coordinate system, each solution served as a metric to this particular
set of coordinate system should describe an independent geometry. The
math is very simply presented below.
ds1^2 = g1 * dq^2
ds2^2 = g2 * dq^2
ds3^2 = g3 * dq^2
...
Where
** g1, g2, g3... = Solutions (matrices) to field equations
** dq^2 = Coordinate choice of these field equations
** * = Inner product of 2 4x4 matrices
> [...]
The rest is irrelevant if we do not agree on the above.
Koobee Wublee
> >Mr. Crothers is correct.
>
> He may be correct about some things, but compare his claim
> "the 'Schwarzschild radius' is meaningless"
> with the following comments by Schwarzschild.
The field equations based on a particular choice of coordinate system
yield an infinity number of solutions. Most of them do not address the
Schwarzschild radius. Only the simpler ones do. Are you going to ask
me what constitute simper ones?
> From:
> On the gravitational field of a sphere of incompressible fluid
> according to Einstein's theory
> http://arxiv.org/abs/physics/9912033
>
> Near bottom of page 8
> <<<
> At the center of the sphere (X = 0) velocity of light and pressure
> become infinite when cosXa = 1/3...
> >>>
>
> What do you imagine might happen to matter if it was subjected to
> "infinite pressure"?
>
> Next, near the bottom of page 9, Schwarzschild gives us the radius at
> which this would occur.
> <<<
> For an observer measuring from outside it follows from (40) that a
> sphere of given gravitational mass alpha/2k^2 can
> not have a radius measured from outside smaller than:
> Po = alpha
> >>>
>
> I believe this is the Schwarzschild radius, which implies this radius
> is far from meaningless.
>
> For the sun, Schwarzschild calculates Po to be 3 km which agrees with
> the value at
> http://en.wikipedia.org/wiki/Schwarzschild_radius
I am sorry that I am not interested in a particular interior solution
and all possible interpretations to this solution at this moment due to
the limited bandwidth of my analytic channel. However, the
Schwarzschild radius is strictly an artifact of the vacuum solution.
Clearly, Schwarzschild's original solution did not allow any room for
such a silly concept of a Schwarzschild radius. If you want to beat
the Schwarzschild radius to death, the following paper by Schwarzschild
describes a particular solution in free space.
Eric Gisse
JanPB
> Surfer wrote:
> > On 5 Jan 2007 22:05:12 -0800, "Koobee Wublee"
> > <koobee...@gmail.com> wrote:
>
> > >Mr. Crothers is correct.
> >
> > He may be correct about some things, but compare his claim
> > "the 'Schwarzschild radius' is meaningless"
> > with the following comments by Schwarzschild.
>
> The field equations based on a particular choice of coordinate system
> yield an infinity number of solutions.
After all this time you still haven't explained how it's mathematically
possible to have a "particular choice of coordinate system" and
"infinity number of solutions".
--
Jan Bielawski
Eric Gisse
ODE uniqueness theorems do not exist in his reality.
Would you start hammering him about where he was educated?
I don't think KW likes to talk to me anymore because I ask him
complicated questions like "where were you taught this shit?" and "do
you STILL think <x>?" where <x> was something fantastically stupid he
said not that long ago.
>
> --
> Jan Bielawski
Koobee Wublee
> Koobee Wublee wrote:
> > The field equations based on a particular choice of coordinate system
> > yield an infinity number of solutions.
>
> After all this time you still haven't explained how it's mathematically
> possible to have a "particular choice of coordinate system" and
> "infinity number of solutions".
You just choose your own measurement or choice of coordinate system.
For example, in this post of yours, you specifically demanded the
spherically symmetric coordinate system.
http://groups.google.com/group/sci.physics.relativity/msg/77c1f71289a9709e?hl=en&
In another post of yours below which you have agreed on a particular
set of field equations only valid for your own choice of coordinate
system, if there are multiple solutions, each solution (or the metric)
must be addressing the same coordinate system. In doing so, the
combination of each metric and this choice of coordinate system must
describe a unique set of geometry. Thus, there must be an infinite
number of solutions to a particular set of field equations based on one
choice of coordinate system.
http://groups.google.com/group/sci.physics.relativity/msg/6585d894984cc314?hl=en&
The castle in the air called General Relativity must come down in doing
so, and we have just scratched the surface.
JanPB
> JanPB wrote:
> > Koobee Wublee wrote:
> > > Surfer wrote:
> > > > On 5 Jan 2007 22:05:12 -0800, "Koobee Wublee"
> > > > <koobee...@gmail.com> wrote:
> > >
> > > > >Mr. Crothers is correct.
> > > >
> > > > He may be correct about some things, but compare his claim
> > > > "the 'Schwarzschild radius' is meaningless"
> > > > with the following comments by Schwarzschild.
> > >
> > > The field equations based on a particular choice of coordinate system
> > > yield an infinity number of solutions.
> >
> > After all this time you still haven't explained how it's mathematically
> > possible to have a "particular choice of coordinate system" and
> > "infinity number of solutions".
>
> ODE uniqueness theorems do not exist in his reality.
>
> Would you start hammering him about where he was educated?
I don't think I ever asked this question anybody around here - all that
matters is the subject matter :-)
--
Jan Bielawski
JanPB
> JanPB wrote:
> > Koobee Wublee wrote:
>
> > > The field equations based on a particular choice of coordinate system
> > > yield an infinity number of solutions.
> >
> > After all this time you still haven't explained how it's mathematically
> > possible to have a "particular choice of coordinate system" and
> > "infinity number of solutions".
>
> You just choose your own measurement or choice of coordinate system.
> For example, in this post of yours, you specifically demanded the
> spherically symmetric coordinate system.
>
> http://groups.google.com/group/sci.physics.relativity/msg/77c1f71289a9709e?hl=en&
>
> In another post of yours below which you have agreed on a particular
> set of field equations only valid for your own choice of coordinate
> system, if there are multiple solutions, each solution (or the metric)
> must be addressing the same coordinate system.
IF there are multiple solutions.
> In doing so, the
> combination of each metric and this choice of coordinate system must
> describe a unique set of geometry.
> Thus, there must be an infinite
> number of solutions to a particular set of field equations based on one
> choice of coordinate system.
> http://groups.google.com/group/sci.physics.relativity/msg/6585d894984cc314?hl=en&
What do you mean "thus"? You only said "if there are multiple
solutions...". You haven't shown there are any. In fact, there can't by
multiple solutions because all the PDEs are gone.
> The castle in the air called General Relativity must come down in doing
> so, and we have just scratched the surface.
Sounds very nice, like all dreamlands.
--
Jan Bielawski
Koobee Wublee
> Koobee Wublee wrote:
> IF there are multiple solutions.
>
> [...]
>
> What do you mean "thus"? You only said "if there are multiple
> solutions...". You haven't shown there are any. In fact, there can't by
> multiple solutions because all the PDEs are gone.
There are indeed multiple solutions. The Schwarzschild's original
solution and the following two spacetime are examples.
ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
All satisfy the field equations in spherically symmetric polar
coordinate system.
In the meantime, my previous concerns are still not discussed by you.
In this post of yours, you specifically demanded the spherically
symmetric coordinate system.
http://groups.google.com/group/sci.physics.relativity/msg/77c1f71289a9709e?hl=en&
In another post of yours below which you have agreed on a particular
set of field equations only valid for your own choice of coordinate
system, if there are multiple solutions, each solution (or the metric)
must be addressing the same coordinate system. In doing so, the
combination of each metric and this choice of coordinate system must
describe a unique set of geometry. Thus, there must be an infinite
number of solutions to a particular set of field equations based on one
choice of coordinate system.
http://groups.google.com/group/sci.physics.relativity/msg/6585d894984cc314?hl=en&
This has turned into a discussion of faith, and I have no interest to
discussion how you defend your faith. You are the same as Reverend
Hammond except you are still an excellent voodoo "mathemagician".
JanPB
> JanPB wrote:
> > Koobee Wublee wrote:
>
> > IF there are multiple solutions.
> >
> > [...]
> >
> > What do you mean "thus"? You only said "if there are multiple
> > solutions...". You haven't shown there are any. In fact, there can't by
> > multiple solutions because all the PDEs are gone.
>
> There are indeed multiple solutions. The Schwarzschild's original
> solution and the following two spacetime are examples.
>
> ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>
> ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
These are identical: same metric tensor written wrt two different
coordinate systems (aka. same geometry aka. same Riemannian manifold
aka. same spacetime). They only look different just like the vector
field:
X = d/dx (Cartesian)
and:
X = cos(theta) d/dr - sin(theta)/r d/dtheta (polar)
...look different but are one and the same vector field (unit length,
pointing right).
Or consider this covector field:
w = dx (Cartesian)
and:
w = cos(theta) dr -r sin(theta) dtheta (polar)
...again, same covectors.
Same with tensors.
> All satisfy the field equations in spherically symmetric polar
> coordinate system.
Sure, they are equal.
> In the meantime, my previous concerns are still not discussed by you.
> In this post of yours, you specifically demanded the spherically
> symmetric coordinate system.
>
> http://groups.google.com/group/sci.physics.relativity/msg/77c1f71289a9709e?hl=en&
I'm not sure which portion of that post you are referring to but in
general - yes, one is free to pick any convenient coordinate system and
work with tensor components expressed in terms of that system.
> In another post of yours below which you have agreed on a particular
> set of field equations only valid for your own choice of coordinate
> system, if there are multiple solutions, each solution (or the metric)
> must be addressing the same coordinate system.
Sure, if in a _fixed_ coordinate chart you get _multiple_ solution
functions for the g_ij's then you _may_ get multiple solutions. You
must check - in this case the multiplicity of the solutions is
represented by their parametrisation by rho and alpha. The alpha bit is
clear and by inspection one verifies that once alpha is fixed the
remaining parametrisation by rho corresponds to different restrictions
of the domain of one and the same tensor (aka. same geometry, aka. same
spacetime).
> In doing so, the
> combination of each metric and this choice of coordinate system must
> describe a unique set of geometry. Thus, there must be an infinite
> number of solutions to a particular set of field equations based on one
> choice of coordinate system.
>
> http://groups.google.com/group/sci.physics.relativity/msg/6585d894984cc314?hl=en&
Not necessarily - one may get a bunch of isometric spacetimes (or a
bunch of different restrictions of some spacetime).
> This has turned into a discussion of faith, and I have no interest to
> discussion how you defend your faith. You are the same as Reverend
> Hammond except you are still an excellent voodoo "mathemagician".
It only seems like faith to those who did not study this in depth. If
you understand this stuff _really_ then everything is 100% verifiable
and transparent "all the way to the empty set" :-) That feeling of
haziness or "faith" is a sure sign of the lack of understanding.
I have provided a proof that the solution is unique. Such proofs
unfortunately require a bit of prerequisites. That's the standard
conundrum on this NG.
--
Jan Bielawski
Koobee Wublee
> There are indeed multiple solutions. The Schwarzschild's original
> solution and the following two spacetime are examples.
>
> ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2
>
> ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
>
> All satisfy the field equations in spherically symmetric polar
> coordinate system.
>
> In the meantime, my previous concerns are still not discussed by you.
> In this post of yours, you specifically demanded the spherically
> symmetric coordinate system.
>
> http://groups.google.com/group/sci.physics.relativity/msg/77c1f71289a9709e?hl=en&
>
> In another post of yours below which you have agreed on a particular
> set of field equations only valid for your own choice of coordinate
> system, if there are multiple solutions, each solution (or the metric)
> must be addressing the same coordinate system. In doing so, the
> combination of each metric and this choice of coordinate system must
> describe a unique set of geometry. Thus, there must be an infinite
> number of solutions to a particular set of field equations based on one
> choice of coordinate system.
>
> http://groups.google.com/group/sci.physics.relativity/msg/6585d894984cc314?hl=en&
>
> This has turned into a discussion of faith, and I have no interest to
> discussion how you defend your faith. You are the same as Reverend
> Hammond except you are still an excellent voodoo "mathemagician".
Reverend Bielawski the voodoo mathemagician,
You are still full of sh*t. You have not even tried to address my
comments and concerns above. Your BS cannot stand in a moderated forum
of discussion if they allow our discussions to move over there.