I have an Allien genius who wants to learn GR
Mike
presented with the GR equations. He knows nothing about Newtonian
physics, he oes not know what force means for earthlinks, he has never
seen F = dp/dt, or F =ma or, F = GMm/r^2
He is asked to find out what are the equations of motion in the weak
field limit.
Will he be able to derive Newton's laws of motion and universal law of
graviation in the weak field limit from the GR equations?
Mike
PD
Yes.
Mike
The one I spoke to told me NO, a flat NO.
Mike
PD
OK, I don't know what you expect me to do with that.
Now, I'm sure you know you're asking a hypothetical question, since
there are no real aliens in this position and there are no real
earthly physicists in this position.
>
> Mike
JanPB
Yes. There is no wiggle room there.
--
Jan Bielawski
Mike
ET says otherwise.
Mike
>
> --
> Jan Bielawski
JanPB
Ask him/her/it to write down for you where the wiggle room is. Then
post it here.
--
Jan Bielawski
Koobee Wublee
> > Suppose this allien mathematical genius arrives to earth and he is
> > presented with the GR equations. He knows nothing about Newtonian
> > physics, he oes not know what force means for earthlinks, he has never
> > seen F = dp/dt, or F =ma or, F = GMm/r^2
>
> > He is asked to find out what are the equations of motion in the weak
> > field limit.
>
> > Will he be able to derive Newton's laws of motion and universal law of
> > graviation in the weak field limit from the GR equations?
>
> Yes. There is no wiggle room there.
>
> Ask him/her/it to write down for you where the wiggle room is. Then
> post it here.
There are an infinite number of solutions to the field equations. In
vacuum, one such solution can be the following.
ds^2 = G c^2 dt^2 / (1 + r / K) – K^4 (1 + r / K) dr^2 / r^4
– K^2 (1 + K / r)^2 dO^2
Where
** G = Dimensionless constant
** K = Constant of length
** dO^2 = r^2 cos^2(Latitude) dLongitude^2 + r^2 dLatitude^2
It is static and spherically symmetric. However, it is not
asymptotically flat. Nevertheless, it is still a valid solution.
Does it degenerate into Newtonian law of gravity? No.
Therefore, Mike is correct on this one. Both Professor Draper and Mr.
Bielawski are very wrong. It is time for both gentlemen to propose a
graceful retreat.
Eric Gisse
Yes, gracefully retreat from the cranks who confuse persistence of
lies with facts.
Mike
To me and to many people I know, a crank and an imbecile, is a person
who while it insists GR is a correct theory of gravitation, he thinks
that a = GM/r^2 is and ordinary differential equation. Such a crank
exists and is called Eric Gisse as these posts proves beyond any
doubt:
http://groups.google.gr/group/sci.physics.relativity/msg/01b8793b2b1f235b?dmode=source
Eric Gisse's excuse for not being able to sove a simple mechanics
problem: "One of these days I'll stop posting on groups when I'm sleep
deprived."
http://groups.google.gr/group/sci.physics.relativity/msg/cc76e4770a6265e8?dmode=source
Now, who is the crank here?
Mike
>
> - Show quoted text -
Eric Gisse
>
> Eric Gisse's excuse for not being able to sove a simple mechanics
> problem: "One of these days I'll stop posting on groups when I'm sleep
> deprived."
>
>
> Now, who is the crank here?
The one who can't admit he made a mistake.
Do you know if radiation pressure exists yet?
Juan R.
But expertises on relativity like R. Wald and Eric Poisson say NO.
--
http://canonicalscience.org/en/miscellaneouszone/guidelines.txt
Juan R.
But expertises on relativity like R. Wald and Eric Poisson say NO.
The weak field limit geodesic is a = 0.
--
http://canonicalscience.org/en/miscellaneouszone/guidelines.txt
Juan R.
> Therefore, Mike is correct on this one. Both Professor Draper and Mr.
> Bielawski are very wrong. It is time for both gentlemen to propose a
> graceful retreat.
Mike is correct. GR does not reduce to NG in the weak field limit.
During the thread on Newtonian limit difficulties of GR I have discussed
this point with an expert on curved spacetime equations of motion, Eric
Poisson [1].
Eric confirms my finding that a = 0 in the linear regime of GR:
(\blockquote
Since the energy-momentum tensor is already of first-order, in the
linearized theory the conservation equations must be written down with
the Minkowski metric, and this implies that the matter cannot have
gravitational interactions. Or as you point out, particles would have
to move on straight lines.
)
Textbooks and lecture notes take the weak field limit in a wrong way.
E.g. Carroll takes the fake limit
Z[a] = - L[\Gamma] Z[v v]
instead the correct one
L[a] = - L[\Gamma v v] = - L[\Gamma] Z[v v]
Prof. Carlip adds a couple of more mistakes and falsifications to his
'derivation'.
Indeed GR has not Newtonian limit in weak field or not.
[1] http://relativity.livingreviews.org/Articles/lrr-2004-6/
--
http://canonicalscience.org/en/miscellaneouszone/guidelines.txt
Eric Gisse
No. The explicit computation of the connection coefficients disproves
that nonsense.
The only inconsistent part of the weak field limit is the covariant
derivative of the stress energy tensor. Since it is assumed that the
quadratic and higher terms as well as the _derivatives_ of those terms
that are quadratic and higher, the covariant equation div . T = 0
reduces to @^a T_ab = 0. That means, to first order, that gravitation
does not affect matter. Except that's only true to first order - there
is a discussion of this in MTW, Carroll, and Wald.
I'd give page numbers if it were not for the fact this is well-
traveled and well-misunderstood ground by Juan R.
>
> --http://canonicalscience.org/en/miscellaneouszone/guidelines.txt
Eric Gisse
<juanREM...@canonicalscience.com> wrote:
> Koobee Wublee wrote on Mon, 05 May 2008 23:23:02 -0700:
>
> > Therefore, Mike is correct on this one. Both Professor Draper and Mr.
> > Bielawski are very wrong. It is time for both gentlemen to propose a
> > graceful retreat.
>
> Mike is correct. GR does not reduce to NG in the weak field limit.
>
> During the thread on Newtonian limit difficulties of GR I have discussed
> this point with an expert on curved spacetime equations of motion, Eric
> Poisson [1].
>
> Eric confirms my finding that a = 0 in the linear regime of GR:
>
> (\blockquote
> Since the energy-momentum tensor is already of first-order, in the
> linearized theory the conservation equations must be written down with
> the Minkowski metric, and this implies that the matter cannot have
> gravitational interactions. Or as you point out, particles would have
> to move on straight lines.
> )
>
This is well known - freely falling particles move on geodesics,
gravity isn't a force, and the conservation law for the stress-energy
tensor takes a specific form in the weak field limit. What are you
complaining about?
> Textbooks and lecture notes take the weak field limit in a wrong way.
> E.g. Carroll takes the fake limit
>
> Z[a] = - L[\Gamma] Z[v v]
>
> instead the correct one
>
> L[a] = - L[\Gamma v v] = - L[\Gamma] Z[v v]
Could you perhaps point out where he actually _takes_ the claimed
limit, why it is wrong, and why yours is correct?
Albertito
<juanREM...@canonicalscience.com> wrote:
> Koobee Wublee wrote on Mon, 05 May 2008 23:23:02 -0700:
>
> > Therefore, Mike is correct on this one. Both Professor Draper and Mr.
> > Bielawski are very wrong. It is time for both gentlemen to propose a
> > graceful retreat.
>
> Mike is correct. GR does not reduce to NG in the weak field limit.
>
If GR doesn't reduce to NG in the weak field limit, then
GR is wrong. How do you dare to claim GR is wrong, crank? :-)
Juan R.
Textbooks and lecture notes often make the claim that geodesic equation
of motion (simplified notation)
a = - \Gamma v v
reduces to Newtonian equation
a = - \grad \phi
when the metric is g_ab = \eta_ab + h_ab with |h| << 1
As proved the correct geodesic equation of motion for that metric is
a = 0
I also proved that equations of motion FTG [1] and relativistic AAAD
theories reduces to
a = - \grad \phi
with correct *spatial* metric \gamma_ij.
Thus the problem is only on a *geometric* description of gravity.
GR is not wrong but incomplete. It is completed in a better non-
geometrical theory.
[1] http://www.amazon.com/Feynman-Lectures-Gravitation-Frontiers-Physics/
dp/0201627345
--
http://canonicalscience.org/en/miscellaneouszone/guidelines.txt
Albertito
It is a pun to claim that GR is not wrong but incomplete.
If something is incomplete, it may yield wrong answers
to fine-tuned questions. If something can't predict the
correct required fine-tuned answer, then it is wrong.
But, anyone daring to claim GR is either wrong or
incomplete, will be fatally dubbed as a crank, and
will be relegated to a ghetto ->The Crackpotery District
:-)
Juan R.
> crank? :-)
Cranks sociology :-)
When by first time i expressed my doubts that GR reduces to NG in the
weak field limit i was said by several "experts" (aka trolls and
ignorants :-) on this newsgroup that "i did not know how to take a linear
limit", that "i was a crank", that "i didn't understand", etc.
I reconsidered the problem once again this year. I wrote a paper on
Newtonian limits of GR and discussed with different people expert in
gravitation and GR, including one author is specialized in Newtonian
limits of GR (e.g he introduced the no-holonomy constraint in NC
spacetime).
Nobody found a serious mistake (typos and others yes!).
Thanks to my work on gravitation i was formally invited by organization
of a next conference in gravity (URSS) where i would talk about my recent
research on Newtonian limits and non-geometrical gravity.
Unfortunately i rejected invitation in last moment because i have got
health problems (probably due to bad food) and cannot do so long travels.
Some weeks ago i received another formal invitation from other
organization to participate in another Conference on cosmology (USA). I
rejected by same motive [#].
Probably i was invited because my work proves that textbooks are not
right at this point.
Or maybe because my approach solves the unphysical cosmological boundary
of island universes.
Or maybe because i have traced the origin of the problems of GR to the
geometrical formulation.
Or maybe because i was arrogant enough to analyze the several mistakes
that genius as Steve Carlip are doing.
In any case, a few days ago another recognized expertise has confirmed
that in the weak field limit the GR correct equation of motion is
a = 0
or as i said "bodies move in straight lines"...
[#] I will write about this unlucky episode in the Canonical Science
Today blog and will apologize again by being forced to reject invitations.
--
http://canonicalscience.org/en/miscellaneouszone/guidelines.txt
Albertito
Congratulations for all those invitations, but
care about your health trying to eat good food.
" los gallegos teneis muy buenos alimentos, coño,
no me lo explico :-) "
Mike
<juanREM...@canonicalscience.com> wrote:
> Koobee Wublee wrote on Mon, 05 May 2008 23:23:02 -0700:
>
> > Therefore, Mike is correct on this one. Both Professor Draper and Mr.
> > Bielawski are very wrong. It is time for both gentlemen to propose a
> > graceful retreat.
>
> Mike is correct. GR does not reduce to NG in the weak field limit.
Although I have not researched the issue extensively as you have done,
my understanding is that one has to assume a specific form of the
metric to make the linearized equations converge to NG because there
are virtually infinite solutions. Maybe you may want to corerct me
here if I am wrong.
The form of the metric chosen though, implies NG at the weak field
limit. Then to come back and assert that GR converges to NG, is a
petitio principii, the worse of all falalcies but one found in many
places in both SR and GR.
It is also found in NM books on some subjects but this is beyond the
subject for now.
Mike
Mike
Dig hard...
you are a troll. You are supposed to be in school, yet you instantly
reply to 99% of the posts in these groups.
Mike
>
>
>
>
>
> > Mike
>
> > > - Show quoted text -- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
Eric Gisse
Does radiation pressure exist or not, Mike?
Koobee Wublee
> There are an infinite number of solutions to the field equations. In
> vacuum, one such solution can be the following.
>
> ds^2 = G c^2 dt^2 / (1 + r / K) – K^4 (1 + r / K) dr^2 / r^4
> – K^2 (1 + K / r)^2 dO^2
>
> Where
>
> ** G = Dimensionless constant
> ** K = Constant of length
> ** dO^2 = r^2 cos^2(Latitude) dLongitude^2 + r^2 dLatitude^2
>
> It is static and spherically symmetric. However, it is not
> asymptotically flat. Nevertheless, it is still a valid solution.
>
> Does it degenerate into Newtonian law of gravity? No.
>
> Therefore, Mike is correct on this one. Both Professor Draper and Mr.
> Bielawski are very wrong. It is time for both gentlemen to propose a
> graceful retreat.
Another solution that is static, spherically symmetric, and (this
time) asymptotically flat is the following.
ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2(1 + K^2 / r^2) dr^2 / K^2
– r^4 dO^2 / K^2
Does is degenerate into Newtonian law of gravity? No, because it
follows an inverse-cubed law instead of the inverse squared law. The
Einstein field equations represent an utter nonsense. They suit for
the ones to promote mysticism as wisdom. <shrug>
Eric Gisse
Its' the same solution as Schwarzschild, idiot. I proved it to you
previously. Have you forgotten _already_ ?!
http://groups.google.com/group/sci.physics.relativity/msg/5e4cf198adbd8234?dmode=source
JanPB
> On May 5, 11:23 pm, Koobee Wublee wrote:
>
>
>
> > There are an infinite number of solutions to the field equations. In
> > vacuum, one such solution can be the following.
>
> > ds^2 = G c^2 dt^2 / (1 + r / K) – K^4 (1 + r / K) dr^2 / r^4
> > – K^2 (1 + K / r)^2 dO^2
>
> > Where
>
> > ** G = Dimensionless constant
> > ** K = Constant of length
> > ** dO^2 = r^2 cos^2(Latitude) dLongitude^2 + r^2 dLatitude^2
>
> > It is static and spherically symmetric. However, it is not
> > asymptotically flat. Nevertheless, it is still a valid solution.
>
> > Does it degenerate into Newtonian law of gravity? No.
>
> > Therefore, Mike is correct on this one. Both Professor Draper and Mr.
> > Bielawski are very wrong. It is time for both gentlemen to propose a
> > graceful retreat.
>
> Another solution that is static, spherically symmetric, and (this
> time) asymptotically flat is the following.
>
> ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2(1 + K^2 / r^2) dr^2 / K^2
> – r^4 dO^2 / K^2
It's the same solution as Schwarzschild's, you merely changed the
numbers labelling the spheres of symmetry.
> Does is degenerate into Newtonian law of gravity? No, because it
> follows an inverse-cubed law instead of the inverse squared law.
I think you meant "inverse-Koobeed" (you knew I'd write that!) Anyway,
since you reshuffled the numbers labelling the events, obviously you
get a different expression wrt those labels. Using the same method, I
can define an electrostatic force that's inversely proportional to the
17th power of the coordinate. This sort of manipulations is just name-
changing and as such it obviously has no influence on the physics.
> The
> Einstein field equations represent an utter nonsense. They suit for
> the ones to promote mysticism as wisdom. <shrug>
You've said that before. It's wrong.
--
Jan Bielawski
Koobee Wublee
> On May 6, 2:24 pm, Koobee Wublee wrote:
> > Another solution that is static, spherically symmetric, and (this
> > time) asymptotically flat is the following.
>
> > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > – r^4 dO^2 / K^2
>
> It's the same solution as Schwarzschild's, you merely changed the
> numbers labelling the spheres of symmetry.
The Schwarzschild metric has the following form.
ds^2 = c^2 dt^2 (1 + K / r) – dr^2 / (1 + K / r) – r^2 dO^2
For the queer of England to continue living in that fat castle in the
air, he/she/it needs to talk himself/herself/itself into believing
both of the above equations are exactly the same.
Well, both are solutions to the field equations using the spherically
symmetric polar coordinate system and nothing else. There are no
coordinate transformations involved. They are just independent
solutions to the field equations. <shrug>
As I said before, for a mere $99,999, I will design the universe of
your choice from the field equations. I think that is a small price
for a universe.
> > Does it degenerate into Newtonian law of gravity? No, because it
> > follows an inverse-cubed law instead of the inverse squared law.
>
> I think you meant "inverse-Koobeed"
Not at all. <shrug>
> (you knew I'd write that!)
No. <shrug>
> Anyway,
> since you reshuffled the numbers labelling the events,
Hardly. <shrug>
> obviously you
> get a different expression wrt those labels.
We are talking about mathematics and not black magic. <shrug>
> Using the same method, I
> can define an electrostatic force that's inversely proportional to the
> 17th power of the coordinate.
You are incoherent as usual. <shrug>
> This sort of manipulations is just name-
> changing and as such it obviously has no influence on the physics.
There are no manipulations. <shrug> You are totally delusional.
<shrug> Please seek psychiatric help.
> > The
> > Einstein field equations represent an utter nonsense. They suit for
> > the ones to promote mysticism as wisdom. <shrug>
>
> You've said that before. It's wrong.
Please check yourself in a mental asylum. It will do you good. You
will have companies also claiming to be queers of England.
Koobee Wublee
> On May 6, 1:24 pm, Koobee Wublee wrote:
> > Another solution that is static, spherically symmetric, and (this
> > time) asymptotically flat is the following.
>
> > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2(1 + K^2 / r^2) dr^2 / K^2
> > – r^4 dO^2 / K^2
>
> > Does it degenerate into Newtonian law of gravity? No, because it
> > follows an inverse-cubed law instead of the inverse squared law. The
> > Einstein field equations represent an utter nonsense. They suit for
> > the ones to promote mysticism as wisdom. <shrug>
>
> Its' the same solution as Schwarzschild, idiot.
It is not the Schwarzschild metric. Any sane and intelligent person
can tell you that.
> I proved it to you
> previously. Have you forgotten _already_ ?!
>
> http://groups.google.com/group/sci.physics.relativity/msg/5e4cf198adb...
Ahahaha... You continue to show why you remain a multi-year super-
senior at the very prestigious University of Alaska majoring in basket
weaving. <shrug>
Eric Gisse
> On May 6, 3:18 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On May 6, 1:24 pm, Koobee Wublee wrote:
> > > Another solution that is static, spherically symmetric, and (this
> > > time) asymptotically flat is the following.
>
> > > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2(1 + K^2 / r^2) dr^2 / K^2
> > > – r^4 dO^2 / K^2
>
> > > Does it degenerate into Newtonian law of gravity? No, because it
> > > follows an inverse-cubed law instead of the inverse squared law. The
> > > Einstein field equations represent an utter nonsense. They suit for
> > > the ones to promote mysticism as wisdom. <shrug>
>
> > Its' the same solution as Schwarzschild, idiot.
>
> It is not the Schwarzschild metric. Any sane and intelligent person
> can tell you that.
If there is a coordinate mapping between two supposedly different
metrics, they are the same.
Koobee Wublee
> On May 6, 8:59 pm, Koobee Wublee wrote:
> > It is not the Schwarzschild metric. Any sane and intelligent person
> > can tell you that.
>
> If there is a coordinate mapping between two supposedly different
> metrics, they are the same.
No. <shrug>
What you are saying is wrong. It is not of any mathematical axiom.
You are merely confused with the similarities in all solutions to a
set of differential equations. <shrug>
Upon proper treatment, all solutions to the field equations are
independent although somewhat related to each other. This
relationship among all these solutions is what you call
transformation. However, at the stage where all these field equations
are laid out to be solved of their solutions, the choice of coordinate
system is cast in concrete. You cannot change the coordinate system
any more. Thus, each solution can only be valid with the already
established choice of coordinate system. In this case, it is the
common spherically symmetric polar coordinate system. You have been
told this many times over. It is time for you to understand this very
basic mathematical relationship. <shrug>
Eric Gisse
> On May 6, 10:27 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On May 6, 8:59 pm, Koobee Wublee wrote:
> > > It is not the Schwarzschild metric. Any sane and intelligent person
> > > can tell you that.
>
> > If there is a coordinate mapping between two supposedly different
> > metrics, they are the same.
>
> No. <shrug>
Yes. Or do you somehow think the coordinate labels determine the
physics?
>
> What you are saying is wrong. It is not of any mathematical axiom.
> You are merely confused with the similarities in all solutions to a
> set of differential equations. <shrug>
>
> Upon proper treatment, all solutions to the field equations are
> independent although somewhat related to each other. This
> relationship among all these solutions is what you call
> transformation. However, at the stage where all these field equations
> are laid out to be solved of their solutions, the choice of coordinate
> system is cast in concrete. You cannot change the coordinate system
> any more. Thus, each solution can only be valid with the already
> established choice of coordinate system. In this case, it is the
> common spherically symmetric polar coordinate system. You have been
> told this many times over. It is time for you to understand this very
> basic mathematical relationship. <shrug>
Once again, I ask you to explicitly apply a coordinate transformation
to G_uv = kT_uv and PROVE that a transformed solution is no longer a
solution.
Do you not know how, or do you somehow think that proves nothing?
Juan R.
> my understanding is that one has to assume a specific form of the metric
> to make the linearized equations converge to NG because there are
> virtually infinite solutions.
The issue is much more complex. Different authors use different false
assumptions on the metrics, on the geodesic equation, etc.
About metrics. Some authors falsify the metric. E.g. Ivanenko et al. 2005
take the metric coefficients
g_00 = -(1 + 2 \phi); g_i0 = 0; g_ij = \delta_ij
for the derivation of the 'Newtonian' limit. Note that the coefficient
g_ij = \delta_ij
is falsified because using the geometric coefficient
g_ij = -\delta_ij / g_00
would give the wrong result during the limit.
The problem here is not selecting /a priori/ a specific form for the
metric, but that the metric violates geometric requirement
g_ij = -\delta_ij / g_00
If one take
g_ij = \delta_ij
then the g_00 = 1
and again the result is
a = 0.
GR does not reduce to NG in the linear regime.
--
http://canonicalscience.org/en/miscellaneouszone/guidelines.txt
Eric Gisse
<juanREM...@canonicalscience.com> wrote:
> Mike wrote on Tue, 06 May 2008 06:24:35 -0700:
>
> > my understanding is that one has to assume a specific form of the metric
> > to make the linearized equations converge to NG because there are
> > virtually infinite solutions.
>
> The issue is much more complex. Different authors use different false
> assumptions on the metrics, on the geodesic equation, etc.
>
> About metrics. Some authors falsify the metric. E.g. Ivanenko et al. 2005
> take the metric coefficients
>
> g_00 = -(1 + 2 \phi); g_i0 = 0; g_ij = \delta_ij
>
> for the derivation of the 'Newtonian' limit. Note that the coefficient
>
> g_ij = \delta_ij
>
> is falsified because using the geometric coefficient
>
> g_ij = -\delta_ij / g_00
>
> would give the wrong result during the limit.
That form of the metric _appears from the limit_, not the other way
around.
>
> The problem here is not selecting /a priori/ a specific form for the
> metric, but that the metric violates geometric requirement
>
> g_ij = -\delta_ij / g_00
What geometric requirement is this?
>
> If one take
>
> g_ij = \delta_ij
>
> then the g_00 = 1
>
> and again the result is
>
> a = 0.
Well yea if you assume the metric coeffficients are all equal to +/-
1.
Koobee Wublee
> On May 6, 10:10 pm, Koobee Wublee wrote:
> Yes. Or do you somehow think the coordinate labels determine the
> physics?
Choice of coordinate does not change physics. This has been what I
have been saying all along. You are merely a malicious troll. That
is why you remain a multi-year super-senior even at the University of
Alaska. <shrug>
Sink this in. We have no more to discuss if you do not understand the
following very basic points.
** The geometry must be invariant.
** The choice of coordinate is observer dependent.
** The metric which together with the coordinate system describes the
geometry.
Therefore, the metric must be coordinate dependent.
> > Upon proper treatment, all solutions to the field equations are
> > independent although somewhat related to each other. This
> > relationship among all these solutions is what you call
> > transformation. However, at the stage where all these field equations
> > are laid out to be solved of their solutions, the choice of coordinate
> > system is cast in concrete. You cannot change the coordinate system
> > any more. Thus, each solution can only be valid with the already
> > established choice of coordinate system. In this case, it is the
> > common spherically symmetric polar coordinate system. You have been
> > told this many times over. It is time for you to understand this very
> > basic mathematical relationship. <shrug>
>
> Once again, I ask you to explicitly apply a coordinate transformation
> to G_uv = kT_uv and PROVE that a transformed solution is no longer a
> solution.
Once again, I have told you there is no such transformation. <shrug>
Each solution to the field equations is independent of each other
using the same set of coordinate system.
> Do you not know how, or do you somehow think that proves nothing?
Can you dig that, multi-year super-senior?
Eric Gisse
[snip]
>
> Once again, I have told you there is no such transformation. <shrug>
> Each solution to the field equations is independent of each other
> using the same set of coordinate system.
Then prove it.
Koobee Wublee
> On May 7, 11:43 am, Koobee Wublee wrote:
> > Once again, I have told you there is no such transformation. <shrug>
> > Each solution to the field equations is independent of each other
> > using the same set of coordinate system.
>
> Then prove it.
The troll sign comes out of you once again. As a multi-year super-
senior, is this not getting old?
I have already presented unchallengeable arguments to my claim. You
need to address my claim instead of resorting to these trollish
behaviors. <shrug>
JanPB
> On May 6, 8:28 pm, JanPB <film...@gmail.com> wrote:
>
> > On May 6, 2:24 pm, Koobee Wublee wrote:
> > > Another solution that is static, spherically symmetric, and (this
> > > time) asymptotically flat is the following.
>
> > > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > > – r^4 dO^2 / K^2
>
> > It's the same solution as Schwarzschild's, you merely changed the
> > numbers labelling the spheres of symmetry.
>
> The Schwarzschild metric has the following form.
>
> ds^2 = c^2 dt^2 (1 + K / r) – dr^2 / (1 + K / r) – r^2 dO^2
>
> For the queer of England to continue living in that fat castle in the
> air, he/she/it needs to talk himself/herself/itself into believing
> both of the above equations are exactly the same.
You did not specify whether you wrote both formulas with respect to
the same coordinate system or not.
Assuming you meant the same coordinates in both formulas, then you are
WRONG: the _first_ formula is NOT a solution. You can verify (it's
quite tedious) that in fact it doesn't satisfy Ricci=0. The second one
does.
On the other hand, if you intended the first formula to be obtained by
substituing r in the second, then you are also WRONG: in that case
both formulas describe the same dot product (i.e., the same solution).
It's exactly the same with dr in polar coordinates and x/
sqrt(x^2+y^2)dx + y/sqrt(x^2+y^2)dy in Cartesian -- they are equal
covectors. Different formulas summing up to the same value. Same thing
when these covectors sit inside a formula for ds^2.
> Well, both are solutions to the field equations using the spherically
> symmetric polar coordinate system and nothing else.
So you say "the" spherically symmetric polar, so it looks like you
indeed intend both expressions for ds^2 to be written in the same
coordinate system. In this case your first formula is not a solution.
> There are no
> coordinate transformations involved. They are just independent
> solutions to the field equations. <shrug>
Only the second one is a solution in this case.
> As I said before, for a mere $99,999, I will design the universe of
> your choice from the field equations.
You've just failed.
[...]
--
Jan Bielawski
Koobee Wublee
> On May 6, 9:55 pm, Koobee Wublee wrote:
> > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > – r^4 dO^2 / K^2
>
> > The Schwarzschild metric has the following form.
>
> > ds^2 = c^2 dt^2 (1 + K / r) – dr^2 / (1 + K / r) – r^2 dO^2
>
> > For the queer of England to continue living in that fat castle in the
> > air, he/she/it needs to talk himself/herself/itself into believing
> > both of the above equations are exactly the same.
>
> You did not specify whether you wrote both formulas with respect to
> the same coordinate system or not.
Do I need to? These solutions are solved after the specific forms of
the field equations are laid out. At this moment, the choice of
coordinate system should be well established, and no one can change
the coordinate. <shrug>
> Assuming you meant the same coordinates in both formulas, then you are
> WRONG: the _first_ formula is NOT a solution.
I beg your pardon. Is there a typo on my part?
> You can verify (it's
> quite tedious) that in fact it doesn't satisfy Ricci=0. The second one
> does.
Both should satisfy G_ij = 0, and thus R_ij = 0. These are just cheap
shots of yours. <shrug>
> On the other hand, if you intended the first formula to be obtained by
> substituing r in the second, then you are also WRONG: in that case
> both formulas describe the same dot product (i.e., the same solution).
There you go again worshipping that dot product as if a true God.
<shrug>
> It's exactly the same with dr in polar coordinates and x/
> sqrt(x^2+y^2)dx + y/sqrt(x^2+y^2)dy in Cartesian -- they are equal
> covectors. Different formulas summing up to the same value. Same thing
> when these covectors sit inside a formula for ds^2.
Yes?
> > Well, both are solutions to the field equations using the spherically
> > symmetric polar coordinate system and nothing else.
>
> So you say "the" spherically symmetric polar, so it looks like you
> indeed intend both expressions for ds^2 to be written in the same
> coordinate system. In this case your first formula is not a solution.
Accusation is very cheap. <shrug>
> > There are no
> > coordinate transformations involved. They are just independent
> > solutions to the field equations. <shrug>
>
> Only the second one is a solution in this case.
Yeah, we all know that all thanks to Hilbert. <shrug>
> > As I said before, for a mere $99,999, I will design the universe of
> > your choice from the field equations.
>
> You've just failed.
Oh, you need to pay up then. Shall I take you to small-claim court?
JanPB
> On May 8, 12:06 am, JanPB <film...@gmail.com> wrote:
>
> > On May 6, 9:55 pm, Koobee Wublee wrote:
> > > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > > – r^4 dO^2 / K^2
>
> > > The Schwarzschild metric has the following form.
>
> > > ds^2 = c^2 dt^2 (1 + K / r) – dr^2 / (1 + K / r) – r^2 dO^2
>
> > > For the queer of England to continue living in that fat castle in the
> > > air, he/she/it needs to talk himself/herself/itself into believing
> > > both of the above equations are exactly the same.
>
> > You did not specify whether you wrote both formulas with respect to
> > the same coordinate system or not.
>
> Do I need to?
Yes.
> These solutions are solved after the specific forms of
> the field equations are laid out.
Not "after the specific forms of the field equations are laid out" but
after a coordinate system has been chosen. This choice is only a
computational convenience, not anything fundamental.
> At this moment, the choice of
> coordinate system should be well established, and no one can change
> the coordinate. <shrug>
What you've just said is like: once the choice of inches as units has
been made during the calculation, no one can convert the result to
centimetres.
> > Assuming you meant the same coordinates in both formulas, then you are
> > WRONG: the _first_ formula is NOT a solution.
>
> I beg your pardon. Is there a typo on my part?
It's not a solution if the (t,r,theta,phi) are the same as in the
second (Schwarzschild's) formula.
> > You can verify (it's
> > quite tedious) that in fact it doesn't satisfy Ricci=0. The second one
> > does.
>
> Both should satisfy G_ij = 0, and thus R_ij = 0.
No, the first one doesn't if its (t,r,theta,phi) are the same as those
in the second formula.
> > On the other hand, if you intended the first formula to be obtained by
> > substituing r in the second, then you are also WRONG: in that case
> > both formulas describe the same dot product (i.e., the same solution).
>
> There you go again worshipping that dot product as if a true God.
> <shrug>
Saying "both formulas describe the same dot product" is worshipping
dot product? How?
> > It's exactly the same with dr in polar coordinates and x/
> > sqrt(x^2+y^2)dx + y/sqrt(x^2+y^2)dy in Cartesian -- they are equal
> > covectors. Different formulas summing up to the same value. Same thing
> > when these covectors sit inside a formula for ds^2.
>
> Yes?
Well, if they are equal, then what miracle suddenly makes them unequal
when they are inserted in the expression for ds^2?
> > > Well, both are solutions to the field equations using the spherically
> > > symmetric polar coordinate system and nothing else.
>
> > So you say "the" spherically symmetric polar, so it looks like you
> > indeed intend both expressions for ds^2 to be written in the same
> > coordinate system. In this case your first formula is not a solution.
>
> Accusation is very cheap. <shrug>
So are the two formulas written wrt to one coordinate system or two?
If two, what transformation did you use?
> > > There are no
> > > coordinate transformations involved. They are just independent
> > > solutions to the field equations. <shrug>
>
> > Only the second one is a solution in this case.
>
> Yeah, we all know that all thanks to Hilbert. <shrug>
If you admit that the first one is not a solution then your claim of
multiplicity of solutions goes away.
> > > As I said before, for a mere $99,999, I will design the universe of
> > > your choice from the field equations.
>
> > You've just failed.
>
> Oh, you need to pay up then.
How does that follow?
> Shall I take you to small-claim court?
--
Jan Bielawski
Eric Gisse
Your argument is nothing even close to "unchallengeable" - all you do
is assert some crap about "metric times coordinate" and therefore
state what the fuck ever.
I thought your shit would make sense after a few beers but I was wrong
- what is the correct number because it isn't 5.
You never ever actually prove anything. You either state the result
[and skip the point of the question] or make arguments to your own
nonexistent authority. When asked to make a _proof_ about the field
equations, you waffle and make excuses. Its' rather pathetic actually,
but hey - its' your life.
Koobee Wublee
> On May 8, 12:21 am, Koobee Wublee wrote:
> > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > – r^4 dO^2 / K^2
>
> > The Schwarzschild metric has the following form.
>
> > ds^2 = c^2 dt^2 (1 + K / r) – dr^2 / (1 + K / r) – r^2 dO^2
>
> > For the queer of England to continue living in that fat castle in the
> > air, he/she/it needs to talk himself/herself/itself into believing
> > both of the above equations are exactly the same.
Both coordinate systems are exactly the same. <shrug>
> > These solutions are solved after the specific forms of
> > the field equations are laid out.
>
> Not "after the specific forms of the field equations are laid out" but
> after a coordinate system has been chosen. This choice is only a
> computational convenience, not anything fundamental.
You are still wrong. <shrug>
These solutions are solved after the specific forms ofthe field
equations are laid out.
> > At this moment, the choice of
> > coordinate system should be well established, and no one can change
> > the coordinate. <shrug>
>
> What you've just said is like: once the choice of inches as units has
> been made during the calculation, no one can convert the result to
> centimetres.
No, that is not what I said. Of course, you are free to convert it to
whatever coordinate system of your choice after the solutions are
derived. However, at these solutions only apply to one same
particular set of coordinate system. <shrug>
> > I beg your pardon. Is there a typo on my part?
>
> It's not a solution if the (t,r,theta,phi) are the same as in the
> second (Schwarzschild's) formula.
Have you personally plug it into the field equations and verify that.
Eric Gisse has the software to do so. I am sure if that is not a
solution, someone would have complained long ago. So, the first
equation is still a valid solution to the field equations using the
same coordinate system of (t, r, theta, phi) as the second equation
(the Schwarzschild metric).
> > Both should satisfy G_ij = 0, and thus R_ij = 0.
>
> No, the first one doesn't if its (t,r,theta,phi) are the same as those
> in the second formula.
<shrug>
> > > On the other hand, if you intended the first formula to be obtained by
> > > substituing r in the second, then you are also WRONG: in that case
> > > both formulas describe the same dot product (i.e., the same solution).
>
> > There you go again worshipping that dot product as if a true God.
> > <shrug>
>
> Saying "both formulas describe the same dot product" is worshipping
> dot product? How?
Because you believe in nonsense. <shrug>
Both equations do not describe the same dot product.
> > > It's exactly the same with dr in polar coordinates and x/
> > > sqrt(x^2+y^2)dx + y/sqrt(x^2+y^2)dy in Cartesian -- they are equal
> > > covectors. Different formulas summing up to the same value. Same thing
> > > when these covectors sit inside a formula for ds^2.
>
> > Yes?
>
> Well, if they are equal, then what miracle suddenly makes them unequal
> when they are inserted in the expression for ds^2?
But both equations do not specify the coordinate system being (t, x,
y, z). <shrug>
> > Accusation is very cheap. <shrug>
>
> So are the two formulas written wrt to one coordinate system or two?
> If two, what transformation did you use?
I am telling you both equations employ the same coordinate system.
Now, I see you are resorting back to your trollish behavior. There is
no point to continue any further, and in the future, please don’t
bring up that you have discussed this issue thoroughly with me because
it never did. You always resort to this kind of trollish behavior.
Now, go back to that fat castle in the air that you call GR and
fantasize about yourself being the queer of England, you majesty.
Eric Gisse
[snip]
>
> Have you personally plug it into the field equations and verify that.
> Eric Gisse has the software to do so. I am sure if that is not a
> solution, someone would have complained long ago. So, the first
> equation is still a valid solution to the field equations using the
> same coordinate system of (t, r, theta, phi) as the second equation
> (the Schwarzschild metric).
This is where your misunderstandings flare up. The coordinate labels
do not matter, but the definitions of the coordinates /do/ and if the /
definitions/ or (t,t,theta,phi) are the same as in Schwarzschild they
are different solutions. However if they are just /labels/, then they
are the same solution because I can [and have] found the isomorphism
between the two supposedly different solutions.
This relates to your inability to compute the area of a sphere - since
you blindingly assert '4pir^2' over and over rather than actually
compute the answer, you cannot see not only why the modern
Schwarzschild solution is preferred but what makes it different from
Hilbert's solution and other iterations.
Plus JanPB is _more_ than capable of computing elements of
differential geometry on his own - he does not need a software
package. The software just makes it crazy easy to dispute your
stupidities.
[snip]
Albertito
[snip crap]
> This is where your misunderstandings flare up. The coordinate labels
> do not matter, but the definitions of the coordinates /do/ and if the /
> definitions/ or (t,t,theta,phi) are the same as in Schwarzschild they
> are different solutions. However if they are just /labels/, then they
> are the same solution because I can [and have] found the isomorphism
> between the two supposedly different solutions.
>
You even don't know what a isomorphism is, fuckhead!
Eric Gisse
And you do? Can you define an isomorphism in your own words?
Koobee Wublee
> On May 7, 10:48 pm, Koobee Wublee wrote:
> > The troll sign comes out of you once again. As a multi-year super-
> > senior, is this not getting old?
>
> > I have already presented unchallengeable arguments to my claim. You
> > need to address my claim instead of resorting to these trollish
> > behaviors. <shrug>
>
> Your argument is nothing even close to "unchallengeable" - all you do
> is assert some crap about "metric times coordinate" and therefore
> state what the fuck ever.
Go back to read my older posts, and don’t come back without
understanding them. Now, get lost.
Koobee Wublee
> On May 8, 12:08 pm, Koobee Wublee wrote:
> > Have you personally plug it into the field equations and verify that.
> > Eric Gisse has the software to do so. I am sure if that is not a
> > solution, someone would have complained long ago. So, the first
> > equation is still a valid solution to the field equations using the
> > same coordinate system of (t, r, theta, phi) as the second equation
> > (the Schwarzschild metric).
>
> [crap snipped]
I say ‘get lost’.
Eric Gisse
I have a better idea - I'll continue to respond to your stupidities
and call them out as such, then laugh as you ineffectively defend
yourself with your shield of stupidity.
Eric Gisse
cry?
Koobee Wublee
> On May 8, 3:36 pm, Koobee Wublee wrote:
> > Go back to read my older posts, and don’t come back without
> > understanding them. Now, get lost.
>
> I have a better idea - I'll continue to respond to your stupidities
> and call them out as such, then laugh as you ineffectively defend
> yourself with your shield of stupidity.
Well, I have even a better idea. You can choose to ignore you stupid
and trollish comments. Next time, whenever you try to bring up how I
did not address your stupid and trollish comments, you will know why.
That is assuming you possess some rudimentary reasoning. If not, it
is no sweat. That is why you remain a multi-year super-senior at the
JanPB
> On May 8, 1:23 am, JanPB <film...@gmail.com> wrote:
>
> > On May 8, 12:21 am, Koobee Wublee wrote:
> > > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > > – r^4 dO^2 / K^2
>
> > > The Schwarzschild metric has the following form.
>
> > > ds^2 = c^2 dt^2 (1 + K / r) – dr^2 / (1 + K / r) – r^2 dO^2
>
> > > For the queer of England to continue living in that fat castle in the
> > > air, he/she/it needs to talk himself/herself/itself into believing
> > > both of the above equations are exactly the same.
>
> Both coordinate systems are exactly the same. <shrug>
Then the first formula is not a solution. Calculate its Ricci tensor
and you'll see it's nonzero.
(I'll skip the rest.)
--
Jan Bielawski
Koobee Wublee
> cry?
Yes, crying won’t do you any good. You remain a multi-year super-
senior. <shrug>
Koobee Wublee
> On May 8, 1:08 pm, Koobee Wublee wrote:
> > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > – r^4 dO^2 / K^2
>
> > The Schwarzschild metric has the following form.
>
> > ds^2 = c^2 dt^2 (1 + K / r) – dr^2 / (1 + K / r) – r^2 dO^2
>
> Then the first formula is not a solution. Calculate its Ricci tensor
> and you'll see it's nonzero.
Oh, you are correct. It is my mistake --- a typo. Please allow me to
correct the first equation.
ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– (r^2 + K^2)^2 dO^2 / K^2
The following is also a solution where the gravitational force follows
the inverse cubed law.
ds^2 = G c^2 dt^2 (1 + K^2 / r^2) – 4 r^2 dr^2 / K^2 / (1 + K^2 / r^2)
– r^4 dO^2 / K^2
JanPB
> On May 9, 1:11 am, JanPB <film...@gmail.com> wrote:
>
> > On May 8, 1:08 pm, Koobee Wublee wrote:
> > > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > > – r^4 dO^2 / K^2
>
> > > The Schwarzschild metric has the following form.
>
> > > ds^2 = c^2 dt^2 (1 + K / r) – dr^2 / (1 + K / r) – r^2 dO^2
>
> > Then the first formula is not a solution. Calculate its Ricci tensor
> > and you'll see it's nonzero.
>
> Oh, you are correct. It is my mistake --- a typo. Please allow me to
> correct the first equation.
>
> ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> – (r^2 + K^2)^2 dO^2 / K^2
Its Ricci curvature is still nonzero if r denotes the same coordinate
as in Schwarzschild. Try it. I recommend Cartan moving frame method as
it's much faster than Christoffel symbols.
> The following is also a solution where the gravitational force follows
> the inverse cubed law.
>
> ds^2 = G c^2 dt^2 (1 + K^2 / r^2) – 4 r^2 dr^2 / K^2 / (1 + K^2 / r^2)
> – r^4 dO^2 / K^2
Same thing: if r is the same then Ricci of the above is nonzero. OTOH
if r denotes sqrt(K * Schwarzschild-r) then Ricci of the above equals
0 but in that case it's the same solution as Schwarzschild - you
simply change the labels you attach to the spheres from "r" to "r^2/
K".
--
Jan Bielawski
Koobee Wublee
> On May 9, 12:32 pm, Koobee Wublee wrote:
> > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > – (r^2 + K^2)^2 dO^2 / K^2
>
> Its Ricci curvature is still nonzero if r denotes the same coordinate
> as in Schwarzschild. Try it.
Its Ricci and Einstein tensors both vanish. In doing so, it is also
sharing the same coordinate system with the Schwarzschild metric.
> I recommend Cartan moving frame method as
> it's much faster than Christoffel symbols.
There is no short cut in such type of mathematics. <shrug>
> > The following is also a solution where the gravitational force follows
> > the inverse cubed law.
>
> > ds^2 = G c^2 dt^2 (1 + K^2 / r^2) – 4 r^2 dr^2 / K^2 / (1 + K^2 / r^2)
> > – r^4 dO^2 / K^2
>
> Same thing: if r is the same then Ricci of the above is nonzero.
But mathematically, its Ricci and Einstein tensors also vanish.
<shrug>
> OTOH
> if r denotes sqrt(K * Schwarzschild-r) then Ricci of the above equals
> 0 but in that case it's the same solution as Schwarzschild - you
> simply change the labels you attach to the spheres from "r" to "r^2/
> K".
Perhaps, there is a flaw in your Cartan whatever. As I said, anyone
who possesses mathematical software like Eric Gisse can easily verify
it, and I do not see any complaint from that multi-year super-senior.
<shrug>
Eric Gisse
> On May 10, 3:09 am, JanPB <film...@gmail.com> wrote:
>
> > On May 9, 12:32 pm, Koobee Wublee wrote:
> > > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > > – (r^2 + K^2)^2 dO^2 / K^2
>
> > Its Ricci curvature is still nonzero if r denotes the same coordinate
> > as in Schwarzschild. Try it.
>
> Its Ricci and Einstein tensors both vanish. In doing so, it is also
> sharing the same coordinate system with the Schwarzschild metric.
Then evaluate surface area of a sphere in both coordinate systems.
You'll see something interesting if you actually do the computation.
>
> > I recommend Cartan moving frame method as
> > it's much faster than Christoffel symbols.
>
> There is no short cut in such type of mathematics. <shrug>
Based on your years of not studying the subject?
>
> > > The following is also a solution where the gravitational force follows
> > > the inverse cubed law.
>
> > > ds^2 = G c^2 dt^2 (1 + K^2 / r^2) – 4 r^2 dr^2 / K^2 / (1 + K^2 / r^2)
> > > – r^4 dO^2 / K^2
>
> > Same thing: if r is the same then Ricci of the above is nonzero.
>
> But mathematically, its Ricci and Einstein tensors also vanish.
> <shrug>
Not if they use the Schwarzschild coordinate system.
>
> > OTOH
> > if r denotes sqrt(K * Schwarzschild-r) then Ricci of the above equals
> > 0 but in that case it's the same solution as Schwarzschild - you
> > simply change the labels you attach to the spheres from "r" to "r^2/
> > K".
>
> Perhaps, there is a flaw in your Cartan whatever. As I said, anyone
> who possesses mathematical software like Eric Gisse can easily verify
> it, and I do not see any complaint from that multi-year super-senior.
> <shrug>
That's because we [JanPB, me] understand that the r you use is
different from the r in Schwarzschild even if you don't.
You appear to think that the coordinates determine the physics. You
claim otherwise and you can't explain why Minkowski space in Cartesian
coordinates is different from Minkowski space in spherical
coordinates, but when presented with the Schwarzschild metric in two
different coordinates your GR myopia sets in and you spray ink to
avoid discussing the subject.
JanPB
> On May 10, 3:09 am, JanPB <film...@gmail.com> wrote:
>
> > On May 9, 12:32 pm, Koobee Wublee wrote:
> > > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > > – (r^2 + K^2)^2 dO^2 / K^2
>
> > Its Ricci curvature is still nonzero if r denotes the same coordinate
> > as in Schwarzschild. Try it.
>
> Its Ricci and Einstein tensors both vanish. In doing so, it is also
> sharing the same coordinate system with the Schwarzschild metric.
Calculate the Ricci tensor. It's not zero. The R_theta,theta is
probably the easiest nonzero component to compute.
> > I recommend Cartan moving frame method as
> > it's much faster than Christoffel symbols.
>
> There is no short cut in such type of mathematics. <shrug>
Cartan's method utilizes the skew symmetry of certain combinations of
Christoffel symbols (rather than symmetry of the symbols themselves).
It allows one to refactor the computations in a more efficient way. In
this regard it's a bit like the Euler-Lagrange equation which also
cleverly refactors the rather ugly brute-force general-coordinate form
of "F=ma".
> > > The following is also a solution where the gravitational force follows
> > > the inverse cubed law.
>
> > > ds^2 = G c^2 dt^2 (1 + K^2 / r^2) – 4 r^2 dr^2 / K^2 / (1 + K^2 / r^2)
> > > – r^4 dO^2 / K^2
>
> > Same thing: if r is the same then Ricci of the above is nonzero.
>
> But mathematically, its Ricci and Einstein tensors also vanish.
> <shrug>
Does "mathematically vanish" mean something different than "vanish"?
All I'm saying is that Ricci curvatures of the two metrics you wrote
are nonzero if you think of (t,r,theta,phi) as the coordinate system
of Schwarzschild.
> > OTOH
> > if r denotes sqrt(K * Schwarzschild-r) then Ricci of the above equals
> > 0 but in that case it's the same solution as Schwarzschild - you
> > simply change the labels you attach to the spheres from "r" to "r^2/
> > K".
>
> Perhaps, there is a flaw in your Cartan whatever.
Use Christoffel symbols then. You'll see the terms don't cancel.
> As I said, anyone
> who possesses mathematical software like Eric Gisse can easily verify
> it, and I do not see any complaint from that multi-year super-senior.
> <shrug>
By all means, use a computer if you prefer.
--
Jan Bielawski
Juan R.
> and I do not see any complaint from that multi-year super-senior.
Sorry by not understanding this jargon but what is a "multi-year super-
senior"?
--
http://canonicalscience.org/en/miscellaneouszone/guidelines.html
Mike
Senior is the fourth year of college in US, before diploma. Mutli -
year senior is a struggling one to get through, like Erica.
Mike
Koobee Wublee
> On May 10, 9:42 pm, Koobee Wublee wrote:
> > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > – (r^2 + K^2)^2 dO^2 / K^2
>
> > Its Ricci and Einstein tensors both vanish. In doing so, it is also
> > sharing the same coordinate system with the Schwarzschild metric.
>
> Calculate the Ricci tensor. It's not zero. The R_theta,theta is
> probably the easiest nonzero component to compute.
What do you mean ‘probably’? Remember dO^2 consists of r, theta, and
phi as the Ricci and the Einstein tensors are only valid in four
dimensions. Since you have not computed the Ricci tensor, you have no
right to call the above solution not a null result of the Ricci and
the Einstein tensors. <shrug> There is no point to continue.
Koobee Wublee
> On May 10, 8:42 pm, Koobee Wublee wrote:
> > Its Ricci and Einstein tensors both vanish. In doing so, it is also
> > sharing the same coordinate system with the Schwarzschild metric.
>
> Then evaluate surface area of a sphere in both coordinate systems.
> You'll see something interesting if you actually do the computation.
The multi-year super-senior has nothing to say besides the stupid
question in which I have given the answers already more than once.
> > There is no short cut in such type of mathematics. <shrug>
>
> Based on your years of not studying the subject?
Equivalent years, yes. <shrug>
> > But mathematically, its Ricci and Einstein tensors also vanish.
> > <shrug>
>
> Not if they use the Schwarzschild coordinate system.
Since both metrics use the same coordinate system, you are just
whining about something you do not understand. <shrug>
> > Perhaps, there is a flaw in your Cartan whatever. As I said, anyone
> > who possesses mathematical software like Eric Gisse can easily verify
> > it, and I do not see any complaint from that multi-year super-senior.
> > <shrug>
>
> That's because we [JanPB, me] understand that the r you use is
> different from the r in Schwarzschild even if you don't.
Hahaha, is this the best I can do with an amateur tag-team of JanPB
the queer of England also a part-time film critics of second rated
movies and Gisse the multi-year super-senior who sits on his piles of
books for fun?
> You appear to think that the coordinates determine the physics.
On the contrary, I have been telling you the exact opposite. <shrug>
There is no need to continue.
Eric Gisse
> On May 10, 11:19 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On May 10, 8:42 pm, Koobee Wublee wrote:
> > > Its Ricci and Einstein tensors both vanish. In doing so, it is also
> > > sharing the same coordinate system with the Schwarzschild metric.
>
> > Then evaluate surface area of a sphere in both coordinate systems.
> > You'll see something interesting if you actually do the computation.
>
> The multi-year super-senior has nothing to say besides the stupid
> question in which I have given the answers already more than once.
Then provide the posts where you drive the surface area of spheres of
constant r and t in both supposedly-different metrics.
[snip]
JanPB
> On May 10, 11:57 pm, JanPB <film...@gmail.com> wrote:
>
> > On May 10, 9:42 pm, Koobee Wublee wrote:
> > > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > > – (r^2 + K^2)^2 dO^2 / K^2
>
> > > Its Ricci and Einstein tensors both vanish. In doing so, it is also
> > > sharing the same coordinate system with the Schwarzschild metric.
>
> > Calculate the Ricci tensor. It's not zero. The R_theta,theta is
> > probably the easiest nonzero component to compute.
>
> What do you mean ‘probably’?
I meant I computed only R_theta,theta and didn't bother with the rest.
> Remember dO^2 consists of r, theta, and
> phi as the Ricci and the Einstein tensors are only valid in four
> dimensions. Since you have not computed the Ricci tensor, you have no
> right to call the above solution not a null result of the Ricci and
> the Einstein tensors. <shrug> There is no point to continue.
If Ricci tensor were zero, then R_theta,theta would be zero. It isn't,
hence Ricci is not zero.
--
Jan Bielawski
Koobee Wublee
> On May 11, 9:53 pm, Koobee Wublee < wrote:
> > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > – (r^2 + K^2)^2 dO^2 / K^2
>
> > Its Ricci and Einstein tensors both vanish. In doing so, it is also
> > sharing the same coordinate system with the Schwarzschild metric.
>
> I meant I computed only R_theta,theta and didn't bother with the rest.
>
> > Remember dO^2 consists of r, theta, and
> > phi as the Ricci and the Einstein tensors are only valid in four
> > dimensions. Since you have not computed the Ricci tensor, you have no
> > right to call the above solution not a null result of the Ricci and
> > the Einstein tensors. <shrug> There is no point to continue.
>
> If Ricci tensor were zero, then R_theta,theta would be zero. It isn't,
> hence Ricci is not zero.
Looking for a generic solution that is static and spherically
symmetric, we employ the spherically symmetric coordinate system that
describes a segment of spacetime in general of the following.
ds^2 = c^2 T(r) dt^2 – P(r) dr^2 – Q(r) dO^2
Where
** T(r), P(r), Q(r) = Functions of r only
** dO^2 = cos^2(Phi) dTheta^2 + dPhi^2
Then, the null Einstein tensor (not Ricci) tensor (in free space) is
represented by the following 3 differential equations. You are
supposed to get 4, but 2 of these are identical.
** 1 + dQ/dr dP/dr / (2 P^2) + (dQ/dr)^2 / (4 P Q) – d^2Q/dr^2 / P =
0
** 1 – dQ/dr dT/dr / (1 T P) – (dQ/dr)^2 / (4 P Q) = 0
** - d^2T/dr^2 / T + dT/dr dP/dr / (2 T P) + (dT/dr)^2 / (2 T^2) – dQ/
dr dT/dr / (2 Q T) + dP/dr dQ/dr / (2 P Q) + (dQ/dr)^2 / (2 Q^2) –
d^2Q/dr^2 / Q = 0
There are an infinite number of solutions found. A few examples are
the following in the order of first discovered. These include the
inverse cubed law one at the beginning of the post.
** Schwarzschild’s original solution
ds^2 = G c^2 dt^2 (1 - K / R) – r^4 dr^2 / R^4 / (1 - K / R) – R^2
dO^2
Where
** R = (r^3 + K^3)^(1/3)
** Schwarzschild (Hilbert’s) solution
ds^2 = G c^2 (1 + K / r) dt^2 – dr^2 / (1 + K / r) – r^2 dO^2
** ?’s solution
ds^2 = G c^2 dt^2 / (1 + K / r) – dr^2 (1 + K / r) – (r + K)^2 dO^2
You need to resolve your own mathematical errors. I cannot help you
on that one. <shrug>
The bottom line is that the Newtonian solution is not unique under the
Einstein field equations, and you need to get over with that. <shrug>
It is also time to vacate from that fat castle in the air, your
majesty, the queer of England.
Mike
Of course there are infinite solutions. But they will not accept this.
What they do is select the solution they want to get as a result. Then
work the math to get that solutions. Then claim they got the solution.
Nice, eh?
Mike
>
> ** Schwarzschild’s original solution
>
> ds^2 = G c^2 dt^2 (1 - K / R) – r^4 dr^2 / R^4 / (1 - K / R) – R^2
> dO^2
>
> Where
>
> ** R = (r^3 + K^3)^(1/3)
>
> ** Schwarzschild (Hilbert’s) solution
>
> ds^2 = G c^2 (1 + K / r) dt^2 – dr^2 / (1 + K / r) – r^2 dO^2
>
> ** ?’s solution
>
> ds^2 = G c^2 dt^2 / (1 + K / r) – dr^2 (1 + K / r) – (r + K)^2 dO^2
>
> You need to resolve your own mathematical errors. I cannot help you
> on that one. <shrug>
>
> The bottom line is that the Newtonian solution is not unique under the
> Einstein field equations, and you need to get over with that. <shrug>
> It is also time to vacate from that fat castle in the air, your
> majesty, the queer of England.- Hide quoted text -
>
> - Show quoted text -
Eric Gisse
[...]
You continue to say "infinite number" like that's supposed to have any
meaning. There is an infinite number of coordinate systems that the
Schwarzschild solution can be projected on, like any other metric.
What makes me wonder is why you think this is meaningful or even
relevant to the uniqueness of the solution.
You write down multiple representations of the same thing and expect
us to agree with you that they are different. It doesn't work that way.
Koobee Wublee
> On May 12, 9:55 pm, Koobee Wublee wrote:
> You continue to say "infinite number" like that's supposed to have any
> meaning.
So, you don’t understand what infinity represents.
> There is an infinite number of coordinate systems
Yes. <shrug>
> that the Schwarzschild solution can be projected on,
The Schwarzschild metric only applies to the spherically symmetric
polar coordinate system and nothing else. So, you do not understand
the Schwarzschild metric, either.
> like any other metric.
Other metric as a solution to the Einstein field equations also only
applies to the spherically symmetric polar coordinate system and
nothing else, and you have never understood the field equations.
<shrug>
> What makes me wonder is why you think this is meaningful or even
> relevant to the uniqueness of the solution.
You remain a multi-year super-senior. <shrug>
> You write down multiple representations of the same thing and expect
> us to agree with you that they are different. It doesn't work that way.
You were expecting to graduate this spring. Did it happen? No.
<shrug>
JanPB
> On May 11, 11:43 pm, JanPB <film...@gmail.com> wrote:
>
>
>
> > On May 11, 9:53 pm, Koobee Wublee < wrote:
> > > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > > – (r^2 + K^2)^2 dO^2 / K^2
>
> > > Its Ricci and Einstein tensors both vanish. In doing so, it is also
> > > sharing the same coordinate system with the Schwarzschild metric.
>
> > I meant I computed only R_theta,theta and didn't bother with the rest.
>
> > > Remember dO^2 consists of r, theta, and
> > > phi as the Ricci and the Einstein tensors are only valid in four
> > > dimensions. Since you have not computed the Ricci tensor, you have no
> > > right to call the above solution not a null result of the Ricci and
> > > the Einstein tensors. <shrug> There is no point to continue.
>
> > If Ricci tensor were zero, then R_theta,theta would be zero. It isn't,
> > hence Ricci is not zero.
>
> Looking for a generic solution that is static and spherically
> symmetric, we employ the spherically symmetric coordinate system that
> describes a segment of spacetime in general of the following.
>
> ds^2 = c^2 T(r) dt^2 – P(r) dr^2 – Q(r) dO^2
>
> Where
>
> ** T(r), P(r), Q(r) = Functions of r only
> ** dO^2 = cos^2(Phi) dTheta^2 + dPhi^2
Yes. But the formula you presented:
ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– (r^2 + K^2)^2 dO^2 / K^2
...is not a solution if you take r to be the same as in
Schwarzschild's formula.
> Then, the null Einstein tensor (not Ricci) tensor (in free space) is
> represented by the following 3 differential equations. You are
> supposed to get 4, but 2 of these are identical.
>
> ** 1 + dQ/dr dP/dr / (2 P^2) + (dQ/dr)^2 / (4 P Q) – d^2Q/dr^2 / P =
> 0
>
> ** 1 – dQ/dr dT/dr / (1 T P) – (dQ/dr)^2 / (4 P Q) = 0
>
> ** - d^2T/dr^2 / T + dT/dr dP/dr / (2 T P) + (dT/dr)^2 / (2 T^2) – dQ/
> dr dT/dr / (2 Q T) + dP/dr dQ/dr / (2 P Q) + (dQ/dr)^2 / (2 Q^2) –
> d^2Q/dr^2 / Q = 0
>
> There are an infinite number of solutions found.
No. While your statement is true in the generic case, it turns out
that in those coordinates these equations become ordinary diff.
equations ("ordinary" is a technical term meaning "in one variable").
And it is well-known that ordinary differential equations have unique
solutions - it follows, oddly enough, from a fixed point theorem. It's
a classic result called Picard-Lindeloef theorem.
> A few examples are
> the following in the order of first discovered. These include the
> inverse cubed law one at the beginning of the post.
> ** Schwarzschild’s original solution
>
> ds^2 = G c^2 dt^2 (1 - K / R) – r^4 dr^2 / R^4 / (1 - K / R) – R^2
> dO^2
>
> Where
>
> ** R = (r^3 + K^3)^(1/3)
>
> ** Schwarzschild (Hilbert’s) solution
>
> ds^2 = G c^2 (1 + K / r) dt^2 – dr^2 / (1 + K / r) – r^2 dO^2
>
> ** ?’s solution
>
> ds^2 = G c^2 dt^2 / (1 + K / r) – dr^2 (1 + K / r) – (r + K)^2 dO^2
Same thing. If "r" in these formulas are presumed the same, then the
third one does not satisfy Ricci=0.
OTOH if the radial coordinates in these formulas are related by
substitutions, then they are the same solution, only written in
different bases. Just like dx equals cos(theta)dr - r sin(theta)dtheta
- just written in different bases.
> You need to resolve your own mathematical errors. I cannot help you
> on that one. <shrug>
>
> The bottom line is that the Newtonian solution is not unique under the
> Einstein field equations, and you need to get over with that. <shrug>
> It is also time to vacate from that fat castle in the air, your
> majesty, the queer of England.
That's nice but have you noticed that your claim that the formula:
ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
– (r^2 + K^2)^2 dO^2 / K^2
...with Schwarzschild's "r" was a solution has just been disproved?
--
Jan Bielawski
JanPB
> On May 13, 1:55 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > There are an infinite number of solutions found. A few examples are
> > the following in the order of first discovered. These include the
> > inverse cubed law one at the beginning of the post.
>
> Of course there are infinite solutions. But they will not accept this.
> What they do is select the solution they want to get as a result. Then
> work the math to get that solutions. Then claim they got the solution.
>
> Nice, eh?
Nice but it's baloney. Good for a Hollywood movie. It's much easier to
believe in silly conspiracies than to do honest work.
--
Jan Bielawski
JanPB
> On May 12, 11:16 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On May 12, 9:55 pm, Koobee Wublee wrote:
> > You continue to say "infinite number" like that's supposed to have any
> > meaning.
>
> So, you don’t understand what infinity represents.
>
> > There is an infinite number of coordinate systems
>
> Yes. <shrug>
>
> > that the Schwarzschild solution can be projected on,
>
> The Schwarzschild metric only applies to the spherically symmetric
> polar coordinate system and nothing else.
No, that's false. It's exactly like saying "the vector pointing to the
North Star applies only to the spherically symmetric polar coordinate
system and nothing else".
The same North-Star-pointing vector has different component triplets
in different coordinates. Do you agree? Tensors of rank higher than 1
are exactly the same way.
> So, you do not understand
> the Schwarzschild metric, either.
>
> > like any other metric.
>
> Other metric as a solution to the Einstein field equations also only
> applies to the spherically symmetric polar coordinate system and
> nothing else,
That's false. Would you say that a vector field (say, velocity field
of a fluid) "only applies to" a particular system?
--
Jan Bielawski
Koobee Wublee
> On May 12, 10:55 pm, Koobee Wublee wrote:
> > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > – (r^2 + K^2)^2 dO^2 / K^2
>
> > Its Ricci and Einstein tensors both vanish. In doing so, it is also
> > sharing the same coordinate system with the Schwarzschild metric.
>
> > Looking for a generic solution that is static and spherically
> > symmetric, we employ the spherically symmetric coordinate system that
> > describes a segment of spacetime in general of the following.
>
> > ds^2 = c^2 T(r) dt^2 – P(r) dr^2 – Q(r) dO^2
>
> > Where
>
> > ** T(r), P(r), Q(r) = Functions of r only
> > ** dO^2 = cos^2(Phi) dTheta^2 + dPhi^2
>
> Yes. But the formula you presented:
>
> ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> – (r^2 + K^2)^2 dO^2 / K^2
>
> ...is not a solution if you take r to be the same as in
> Schwarzschild's formula.
Mathematically, what you are saying is just wrong. <shrug>
Well, all these are solutions to the field equations. Closing your
eyes and yelling they are not does not make them not. <shrug> It is
all in the mathematics. <shrug>
> OTOH if the radial coordinates in these formulas are related by
> substitutions, then they are the same solution, only written in
> different bases.
You are full of crap. The coordinate system is the spherically
symmetric polar coordinate system and nothing else. <shrug>
> Just like dx equals cos(theta)dr - r sin(theta)dtheta
> - just written in different bases.
Wrong analogy. As a film critic of second-rated films, you are very
subjective, aren’t you?
> > You need to resolve your own mathematical errors. I cannot help you
> > on that one. <shrug>
>
> > The bottom line is that the Newtonian solution is not unique under the
> > Einstein field equations, and you need to get over with that. <shrug>
> > It is also time to vacate from that fat castle in the air, your
> > majesty, the queer of England.
>
> That's nice but have you noticed that your claim that the formula:
>
> ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> – (r^2 + K^2)^2 dO^2 / K^2
>
> ...with Schwarzschild's "r" was a solution has just been disproved?
I don’t know what you mean by my own claim. I have claimed that all
these solutions are mathematically tested out to be the solutions to
the Einstein field equations under the spherically symmetric polar
coordinate system. <shrug>
It is better for you majesty, the queer of England, to retire into
that soon-to-demolished fat castle in the air, for it will not be
there long. You don’t have to be such scared sh*tless. At least, you
can still go back being a critic for all these second-rated films, no?
PD
> On May 12, 11:16 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On May 12, 9:55 pm, Koobee Wublee wrote:
> > You continue to say "infinite number" like that's supposed to have any
> > meaning.
>
> So, you don’t understand what infinity represents.
>
> > There is an infinite number of coordinate systems
>
> Yes. <shrug>
>
> > that the Schwarzschild solution can be projected on,
>
> The Schwarzschild metric only applies to the spherically symmetric
> polar coordinate system and nothing else. So, you do not understand
> the Schwarzschild metric, either.
No, that's incorrect. The Schwarzchild metric is a *shape*. It is most
conveniently described in one set of coordinates, but it can certainly
be described in more than one coordinate system.
Likewise, the hydrogen atom wavefunctions are most conveniently
written in spherically polar coordinates, so that one can take
algebraic advantage of the properties of Legendre polynomials and
Bessel radial functions, and because the spherically symmetric
potential allows for the quick finding of these solutions by virtue of
the algebraic method called separation of variables. But if you were
to say that it is simply impossible to write the hydrogen atom
wavefunctions in anything other than spherically polar coordinates,
you'd be daft.
To take it one step further just to hammer in the point, writing the
equation of a sphere of radius R is easiest in spherical polar
coordinates, but it is certainly possible to write the equation for
the same sphere in rectangular coordinates, or cylindrical
coordinates, or any other coordinate system you choose.
Eric Gisse
> On May 12, 11:16 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On May 12, 9:55 pm, Koobee Wublee wrote:
> > You continue to say "infinite number" like that's supposed to have any
> > meaning.
>
> So, you don’t understand what infinity represents.
I usually take it to mean "lots", or in a more general sense, "more
than one".
You chose the coordinate form so there is only one solution for that
particular coordinate form. Ever read anything about ODE uniqueness
theorems?
The other solutions are clearly in different coordinate systems.
>
> > There is an infinite number of coordinate systems
>
> Yes. <shrug>
>
> > that the Schwarzschild solution can be projected on,
>
> The Schwarzschild metric only applies to the spherically symmetric
> polar coordinate system and nothing else. So, you do not understand
> the Schwarzschild metric, either.
There are still an infinite number of spherically symmetric coordinate
representations of Schwarzschild - you have shown me at least 3 of
them.
Who says that the metric is only true in a spherically symmetric
coordinate system? I can write the line element in Cartesian, prolate
spherical, or cylindrical coordinates. Or can I not do that for some
reason?
>
> > like any other metric.
>
> Other metric as a solution to the Einstein field equations also only
> applies to the spherically symmetric polar coordinate system and
> nothing else, and you have never understood the field equations.
> <shrug>
Since when does the coordinate system determine the physics? Spherical
symmetry is a property of the manifold, not the coordinate system.
Writing something in spherical coordinates doesn't make it spherically
symmetric - or did you not know what?
Glass houses, kooby.
>
> > What makes me wonder is why you think this is meaningful or even
> > relevant to the uniqueness of the solution.
>
> You remain a multi-year super-senior. <shrug>
>
> > You write down multiple representations of the same thing and expect
> > us to agree with you that they are different. It doesn't work that way.
>
> You were expecting to graduate this spring. Did it happen? No.
> <shrug>
I was also expecting a pony. Where's my pony?
I'll say this once and no more: my personal life is exactly none of
your business.
Koobee Wublee
> You chose the coordinate form so there is only one solution for that
> particular coordinate form. [The rest of crap is mercifully
> snipped.]
You continue to whine and distort what I have said. Go back to sit on
your piles of books.
Koobee Wublee
> On May 13, 1:24 am, Koobee Wublee wrote:
> > The Schwarzschild metric only applies to the spherically symmetric
> > polar coordinate system and nothing else. So, you do not understand
> > the Schwarzschild metric, either.
>
> No, that's incorrect. The Schwarzchild metric is a *shape*. It is most
> conveniently described in one set of coordinates, but it can certainly
> be described in more than one coordinate system.
The spacetime described by the Schwarzschild metric only applied to
the spherically symmetric polar coordinate. You are welcome to
transform to another coordinate system with a different metric. The
segment of spacetime remains the same and thus invariant while the
coordinate is your choice as well as the metric associated with this
coordinate to describe the invariant segment of spacetime.
This concept is so basic, but you have a lot of trouble understanding
that. There is no need to go through the rest of your babbling.
<shrug>
Koobee Wublee
> On May 12, 11:24 pm, Koobee Wublee wrote:
> > The Schwarzschild metric only applies to the spherically symmetric
> > polar coordinate system and nothing else.
>
> No, that's false. It's exactly like saying "the vector pointing to the
> North Star applies only to the spherically symmetric polar coordinate
> system and nothing else".
The spacetime described by the Schwarzschild metric only applied to
the spherically symmetric polar coordinate. You are welcome to
transform to another coordinate system with a different metric. The
segment of spacetime remains the same and thus invariant while the
coordinate is your choice as well as the metric associated with this
coordinate to describe the invariant segment of spacetime. <shrug>
This concept is so basic, but you have a lot of trouble understanding
that. There is no point to go through the north star nonsense.
PD
> On May 13, 12:21 am, PD <TheDraperFam...@gmail.com> wrote:
>
> > On May 13, 1:24 am, Koobee Wublee wrote:
> > > The Schwarzschild metric only applies to the spherically symmetric
> > > polar coordinate system and nothing else. So, you do not understand
> > > the Schwarzschild metric, either.
>
> > No, that's incorrect. The Schwarzchild metric is a *shape*. It is most
> > conveniently described in one set of coordinates, but it can certainly
> > be described in more than one coordinate system.
>
> The spacetime described by the Schwarzschild metric only applied to
> the spherically symmetric polar coordinate.
Repeating the same erroneous statement does not inch it any further
toward being right.
Koobee Wublee
JanPB
> On May 12, 11:51 pm, JanPB <film...@gmail.com> wrote:
>
> > On May 12, 11:24 pm, Koobee Wublee wrote:
> > > The Schwarzschild metric only applies to the spherically symmetric
> > > polar coordinate system and nothing else.
>
> > No, that's false. It's exactly like saying "the vector pointing to the
> > North Star applies only to the spherically symmetric polar coordinate
> > system and nothing else".
>
> The spacetime described by the Schwarzschild metric only applied to
> the spherically symmetric polar coordinate.
Not _spacetime_, the _coordinate representation of the metric_. There
is no such thing as "spacetime" that can be "applied" to a coordinate
system. The concept doesn't exist.
I notice that you didn't comment on the North Star vector example. You
always evade points that disprove your claims and then you post semi-
responses on a slightly altered topic. (A standard strategy around
these parts - it won't work with me as I've been posting to this NG
since its inception and have seen it all (by definition :-) ).
> You are welcome to
> transform to another coordinate system with a different metric.
Metric is like that vector pointing to the North Star - a coordinate-
independent entity. Here I'm standing, here is the North Star, here is
my finger pointing to it. This is a vector. No coordinate system
involved. [No answer, right?] Metric is the same way except it needs
more data to describe it than just magnitude and direction as in the
vector case.
> The
> segment of spacetime remains the same and thus invariant while the
> coordinate is your choice as well as the metric associated with this
> coordinate
...there is no such thing as "metric associated with this coordinate"
just like there is no "vector pointing to the North Star associated
with this coordinate".
> to describe the invariant segment of spacetime. <shrug>
> This concept is so basic, but you have a lot of trouble understanding
> that. There is no point to go through the north star nonsense.
But it's the crux of the matter. For months now you have been avoiding
answering this question: why is vector coordinate independent and
metric suddenly _isn't_?
--
Jan Bielawski
JanPB
> On May 12, 11:32 pm, JanPB <film...@gmail.com> wrote:
>
>
>
> > On May 12, 10:55 pm, Koobee Wublee wrote:
> > > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > > – (r^2 + K^2)^2 dO^2 / K^2
>
> > > Its Ricci and Einstein tensors both vanish. In doing so, it is also
> > > sharing the same coordinate system with the Schwarzschild metric.
>
> > > Looking for a generic solution that is static and spherically
> > > symmetric, we employ the spherically symmetric coordinate system that
> > > describes a segment of spacetime in general of the following.
>
> > > ds^2 = c^2 T(r) dt^2 – P(r) dr^2 – Q(r) dO^2
>
> > > Where
>
> > > ** T(r), P(r), Q(r) = Functions of r only
> > > ** dO^2 = cos^2(Phi) dTheta^2 + dPhi^2
>
> > Yes. But the formula you presented:
>
> > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > – (r^2 + K^2)^2 dO^2 / K^2
>
> > ...is not a solution if you take r to be the same as in
> > Schwarzschild's formula.
>
> Mathematically, what you are saying is just wrong. <shrug>
It's very easy to see who is wrong: I say the Ricci tensor of that is
nonzero. You are saying it is zero. So just calculate the thing and
see for yourself! What are you waiting for?
Not the third one. Calculate its Ricci curvature.
> Closing your
> eyes and yelling they are not does not make them not. <shrug>
I'm doing no such thing - I'm saying their Ricci curvature is nonzero
(an easy although a bit tedious calculation) which means they can't be
solutions.
> It is
> all in the mathematics. <shrug>
Exactly.
> > OTOH if the radial coordinates in these formulas are related by
> > substitutions, then they are the same solution, only written in
> > different bases.
>
> You are full of crap. The coordinate system is the spherically
> symmetric polar coordinate system and nothing else. <shrug>
OK, if you want it this way then none of the formulas you wrote except
Schwarzschild's own is the solution. Calculate their Ricci curvature
and see it's nonzero. Why aren't you doing it? You prefer to live in
your fantasy world?
> > Just like dx equals cos(theta)dr - r sin(theta)dtheta
> > - just written in different bases.
>
> Wrong analogy.
Actually it's better than an analogy. The dx, dr, etc. above are
objects of exactly the same type as dr dt etc. in Riemannian metrics.
> As a film critic of second-rated films, you are very
> subjective, aren’t you?
I am not a film critic. I run (with a friend) a web site devoted to
the Russian filmmaker Andrei Tarkovsky who is about as second-rate as
Dostoyevsky. There are some other things in that department that I'm
involved with. Besides that I don't see how pointing out a result of
definite computation being nonzero can be seen as "subjective".
> > > You need to resolve your own mathematical errors. I cannot help you
> > > on that one. <shrug>
>
> > > The bottom line is that the Newtonian solution is not unique under the
> > > Einstein field equations, and you need to get over with that. <shrug>
> > > It is also time to vacate from that fat castle in the air, your
> > > majesty, the queer of England.
>
> > That's nice but have you noticed that your claim that the formula:
>
> > ds^2 = G c^2 dt^2 / (1 + K^2 / r^2) – 4 r^2 (1 + K^2 / r^2) dr^2 / K^2
> > – (r^2 + K^2)^2 dO^2 / K^2
>
> > ...with Schwarzschild's "r" was a solution has just been disproved?
>
> I don’t know what you mean by my own claim. I have claimed that all
> these solutions are mathematically tested out to be the solutions to
> the Einstein field equations under the spherically symmetric polar
> coordinate system. <shrug>
I'm just saying the above is not a solution because if you assume "r"
is the same then calculating Ricci curvature of the above yields a
nonzero result.
> It is better for you majesty, the queer of England, to retire into
> that soon-to-demolished fat castle in the air, for it will not be
> there long. You don’t have to be such scared sh*tless. At least, you
> can still go back being a critic for all these second-rated films, no?
I'm not scared at all, where did you get this idea? I know this stuff
is as solid as 1+1=2.
--
Jan Bielawski
PD
Not saying anything at all doesn't inch the incorrect statement any
more towards being right either.
Do you think you cannot write down the equation of a sphere in
anything other than spherical polar coordinates?
PD
Mike
> On May 13, 12:21 am, PD <TheDraperFam...@gmail.com> wrote:
>
> > On May 13, 1:24 am, Koobee Wublee wrote:
> > > The Schwarzschild metric only applies to the spherically symmetric
> > > polar coordinate system and nothing else. So, you do not understand
> > > the Schwarzschild metric, either.
>
> > No, that's incorrect. The Schwarzchild metric is a *shape*. It is most
> > conveniently described in one set of coordinates, but it can certainly
> > be described in more than one coordinate system.
>
> The spacetime described by the Schwarzschild metric only applied to
> the spherically symmetric polar coordinate. You are welcome to
> transform to another coordinate system with a different metric.
No, the Schwarzschild metric IS the metric. There is no such thing as
Schwarzschild metric in a coordinate system with a metric.
I think you are pounding on a non-issue. The real issue is a more
important one: in the limit the Schwarzschild metric reduc es to
Minkowski spacetime, which is gravity free. Yet, in physical reality
there is gravity is such limit.
Thus, the solution is empirically falsified.
Mike
Eric Gisse
> On May 13, 5:46 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
>
> > On May 13, 12:21 am, PD <TheDraperFam...@gmail.com> wrote:
>
> > > On May 13, 1:24 am, Koobee Wublee wrote:
> > > > The Schwarzschild metric only applies to the spherically symmetric
> > > > polar coordinate system and nothing else. So, you do not understand
> > > > the Schwarzschild metric, either.
>
> > > No, that's incorrect. The Schwarzchild metric is a *shape*. It is most
> > > conveniently described in one set of coordinates, but it can certainly
> > > be described in more than one coordinate system.
>
> > The spacetime described by the Schwarzschild metric only applied to
> > the spherically symmetric polar coordinate. You are welcome to
> > transform to another coordinate system with a different metric.
>
> No, the Schwarzschild metric IS the metric. There is no such thing as
> Schwarzschild metric in a coordinate system with a metric.
>
> I think you are pounding on a non-issue. The real issue is a more
> important one: in the limit the Schwarzschild metric reduc es to
> Minkowski spacetime, which is gravity free. Yet, in physical reality
> there is gravity is such limit.
They reduce to the same limit, Mike. Expand the metric in a power
series expansion of GM/r for GM / r << 1.
Koobee Wublee
> On May 13, 5:46 pm, Koobee Wublee wrote:
> > The spacetime described by the Schwarzschild metric only applied to
> > the spherically symmetric polar coordinate. You are welcome to
> > transform to another coordinate system with a different metric.
>
> No, the Schwarzschild metric IS the metric. There is no such thing as
> Schwarzschild metric in a coordinate system with a metric.
Mathematically, you are just wrong. For example, describing flat
spacetime using the linear rectangular (Euclidean) and using the
spherically symmetric polar coordinate systems require you to supply
different metric for each choice of coordinate system. <shrug>
> I think you are pounding on a non-issue. The real issue is a more
> important one: in the limit the Schwarzschild metric reduc es to
> Minkowski spacetime, which is gravity free. Yet, in physical reality
> there is gravity is such limit.
Hmmm... You are very confused.
> Thus, the solution is empirically falsified.
<shrug>
Koobee Wublee
> On May 13, 11:52 pm, Koobee Wublee wrote:
> Not saying anything at all doesn't inch the incorrect statement any
> more towards being right either.
Go back to read my last post. You are very confused as usual.
> Do you think you cannot write down the equation of a sphere in
> anything other than spherical polar coordinates?
Yes, I can, but it does not apply to what we are discussing. <shrug>
Koobee Wublee
> On May 13, 2:48 pm, Koobee Wublee wrote:
> > The spacetime described by the Schwarzschild metric only applied to
> > the spherically symmetric polar coordinate.
>
> Not _spacetime_, the _coordinate representation of the metric_. There
> is no such thing as "spacetime" that can be "applied" to a coordinate
> system. The concept doesn't exist.
You have grossly misunderstood what I said, and I cannot tell if it is
due to your incompetence or just being malicious. There is no point
to go on.
Koobee Wublee
> On May 12, 11:54 pm, Koobee Wublee wrote:
> > Mathematically, what you are saying is just wrong. <shrug>
>
> It's very easy to see who is wrong: I say the Ricci tensor of that is
> nonzero. You are saying it is zero. So just calculate the thing and
> see for yourself! What are you waiting for?
I have given you the complete Einstein tensor in the link below. The
solutions I gave all satisfy the null Einstein tensor and thus the
Ricci tensor. Your application of Cartan stuff is grossly
simplistic. There is no short cut in mathematical grunge. <shrug>
http://groups.google.com/group/sci.physics.relativity/msg/a89098444d9ab5b7
Eric Gisse
> On May 14, 7:10 am, Mike <elea...@yahoo.gr> wrote:
>
> > On May 13, 5:46 pm, Koobee Wublee wrote:
> > > The spacetime described by the Schwarzschild metric only applied to
> > > the spherically symmetric polar coordinate. You are welcome to
> > > transform to another coordinate system with a different metric.
>
> > No, the Schwarzschild metric IS the metric. There is no such thing as
> > Schwarzschild metric in a coordinate system with a metric.
>
> Mathematically, you are just wrong. For example, describing flat
> spacetime using the linear rectangular (Euclidean) and using the
> spherically symmetric polar coordinate systems require you to supply
> different metric for each choice of coordinate system. <shrug>
Is there even one person who agrees with you on this?
Mike
About what?
(a) DO yopu think that the Schwarzschild metric gines Minskowski
spaceitme in the limit?
(b) do you think Minkowski space time alone can explain gravity?
(c) Do you think is pretty scary that everyone else is confused but
you?
Mike
Juan R.
> The real issue is a more
> important one: in the limit the Schwarzschild metric reduc es to
> Minkowski spacetime, which is gravity free. Yet, in physical
> reality
> there is gravity is such limit.
Yes, General Relativity has not Newtonian limit and usual derivations
of Newtonian limits are either incorrect or not care about
mathematical consistency.
Absence of Newtonian limit is directly related to the impossibility to
quantize General Relativity. Notice you can quantize Newtonian
gravity, the resulting equation is the famous Schrödinger-Newton
equation
http://en.wikipedia.org/wiki/Schr%C3%B6dinger-Newton_equations
Those equation are already under experimentalists' target!
Notice one finds problems *only* with the geometrical approach to
gravity. Those problems are *absent* in nongeometrical theories of
gravity such as FTG
http://www.amazon.com/Feynman-Lectures-Gravitation-Frontiers-Physics/dp/0201627345
or R-AAAD
http://arxiv.org/pdf/physics/0612019v10
The usual general relativist motto:
(\blockquote
Mass grips space by telling it how to curve area, space grips mass
by telling it how to move.
)
was once believed to be true.
--
My newsserver continues down
Eric Gisse
wrote:
> On May 14, 4:10 pm, Mike <elea...@yahoo.gr> wrote:
>
> > The real issue is a more
> > important one: in the limit the Schwarzschild metric reduc es to
> > Minkowski spacetime, which is gravity free. Yet, in physical
> > reality
> > there is gravity is such limit.
>
> Yes, General Relativity has not Newtonian limit and usual derivations
> of Newtonian limits are either incorrect or not care about
> mathematical consistency.
It really is too bad that you can't just /accept/ that nobody cares.
It is well understood - there is a whole page of discussion in MTW
about this - that linearized GR is inconsistent with itself. It is
also well understood _why_ the inconsistency is there and _why_ it
doesn't matter.
[snip]
Mike
Yep, inconsistencies do not matter in your inconsistent world full of
inconsistent brains like you.
Have you got that ODE thing straight yet?
http://groups.google.gr/group/sci.physics.relativity/msg/01b8793b2b1f235b?dmode=source
Do you think it is consistent thiikning that you call a = GM/r^2 an
Euler ODE?
Mike
>
> [snip]
Mike
wrote:
> On May 14, 4:10 pm, Mike <elea...@yahoo.gr> wrote:
>
> > The real issue is a more
> > important one: in the limit the Schwarzschild metric reduc es to
> > Minkowski spacetime, which is gravity free. Yet, in physical
> > reality
> > there is gravity is such limit.
>
> Yes, General Relativity has not Newtonian limit and usual derivations
> of Newtonian limits are either incorrect or not care about
> mathematical consistency.
>
> Absence of Newtonian limit is directly related to the impossibility to
> quantize General Relativity. Notice you can quantize Newtonian
> gravity, the resulting equation is the famous Schrödinger-Newton
> equation
>
> http://en.wikipedia.org/wiki/Schr%C3%B6dinger-Newton_equations
>
> Those equation are already under experimentalists' target!
That equation hold the key to the future of physics when people
finally get to their senses and abandon SR and GR after realizing the
inconsistencies present in those theories/
Mike
>
> Notice one finds problems *only* with the geometrical approach to
> gravity. Those problems are *absent* in nongeometrical theories of
> gravity such as FTG
>
> http://www.amazon.com/Feynman-Lectures-Gravitation-Frontiers-Physics/...
Juan R.
> That equation hold the key to the future of physics when people
> finally get to their senses and abandon SR and GR after realizing
> the
> inconsistencies present in those theories/
I am not so optimistic.
A subset of 'believers' will continue to support and promote geometric
approach to gravity without caring about how many recent theoretical
or experimental evidence goes against.
Koobee Wublee
> On May 15, 1:08 am, Koobee Wublee wrote:
> (a) DO yopu think that the Schwarzschild metric gines Minskowski
> spaceitme in the limit?
Yes, of course. This should be a no brainer. <shrug>
> (b) do you think Minkowski space time alone can explain gravity?
No. Gravity has nothing to do with a curvature in space. Only
gravitational time dilation manifests gravity according to spacetime.
Since Minkowski sapcetime does not manifest gravitational time
dilation, it cannot explain gravity. PERIOD.
> (c) Do you think is pretty scary that everyone else is confused but
> you?
No, the god everyone else believes in is very incompetent. <shrug>
JanPB
You said, this is a direct quote: "The spacetime described by the
Schwarzschild metric only applied to the spherically symmetric polar
coordinate". Remember always to define your private terms. We cannot
guess what you mean by terms you invent for your own use. As it
stands, and without further definition of terms, the statement I
quoted is nonsensical.
If you use the words "applied to a coordinate system" in some other
manner that allows "spacetime" in this context, then define this
usage.
--
Jan Bielawski
JanPB
> On May 14, 7:10 am, Mike <elea...@yahoo.gr> wrote:
>
> > On May 13, 5:46 pm, Koobee Wublee wrote:
> > > The spacetime described by the Schwarzschild metric only applied to
> > > the spherically symmetric polar coordinate. You are welcome to
> > > transform to another coordinate system with a different metric.
>
> > No, the Schwarzschild metric IS the metric. There is no such thing as
> > Schwarzschild metric in a coordinate system with a metric.
>
> Mathematically, you are just wrong. For example, describing flat
> spacetime using the linear rectangular (Euclidean) and using the
> spherically symmetric polar coordinate systems require you to supply
> different metric for each choice of coordinate system. <shrug>
No, it's the same metric in both cases, e.g. in the plane the
following are equal:
dx^2 + dy^2
and:
dr^2 + r^2 dtheta^2
...where x = r cos(theta) and y = r sin(theta) (polar coordinates).
The above are two different coordinate decompositions of the same
metric. To see that they are equal, just evaluate them on an arbitrary
vector and see you get the same value in both cases.
For example, consider vector v defined in Cartesian coordinates like
this:
v is situated at point (1,1),
v = (1,2)
Then, applying ds^2 = dx^2 + dy^2:
ds^2(v) = v.v = dx(v) * dx(v) + dy(v) * dy(v) = 1^2 + 2^2 = 5
Now calculate using the polar representation of ds^2:
v is situated at (sqrt(2), pi/4),
v = (3/sqrt(2), -1/2)
Then, applying ds^2 = dr^2 + r^2 dtheta^2:
ds^2(v) = v.v = dr(v) * dr(v) + (sqrt(2))^2 * dtheta(v) *
dtheta(v) =
= (3/sqrt(2))^2 + 2 * (-1/2)^2 =
= 9/2 + 2 * 1/4 =
= 9/2 + 1/2 = 10/2 = 5
Same result! Yout hink that's an accident? :-) You'd obtain the
corresponding same results with any vector. It must be so because both
expressions for ds^2 are equal.
Same thing with Schwarzschild, etc.
The term "metric" as used by everyone refers to that mapping that
takes tangent vectors (like v) to their squared lengths (like v.v).
That mapping is coordinate-INdependent. OTOH its _coordinate
representation_ of course depends on coordinates (by definition). But
dx^2+dy^2 and dr^2 + r^2 dtheta^2 are two different coordinate
representations of the same metric (the same mapping taking v to v.v,
or equivalently, a mapping taking vector pairs v,w to v.w - that's
what metric _is_).
--
Jan Bielawski
PD
> On May 14, 10:08 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
>
> > On May 14, 7:10 am, Mike <elea...@yahoo.gr> wrote:
>
> > > On May 13, 5:46 pm, Koobee Wublee wrote:
> > > > The spacetime described by the Schwarzschild metric only applied to
> > > > the spherically symmetric polar coordinate. You are welcome to
> > > > transform to another coordinate system with a different metric.
>
> > > No, the Schwarzschild metric IS the metric. There is no such thing as
> > > Schwarzschild metric in a coordinate system with a metric.
>
> > Mathematically, you are just wrong. For example, describing flat
> > spacetime using the linear rectangular (Euclidean) and using the
> > spherically symmetric polar coordinate systems require you to supply
> > different metric for each choice of coordinate system. <shrug>
>
> No, it's the same metric in both cases, e.g. in the plane the
> following are equal:
>
> dx^2 + dy^2
>
> and:
>
> dr^2 + r^2 dtheta^2
>
> ...where x = r cos(theta) and y = r sin(theta) (polar coordinates).
>
> The above are two different coordinate decompositions of the same
> metric. To see that they are equal, just evaluate them on an arbitrary
> vector and see you get the same value in both cases.
This is where KW gets the small chicken bone stuck in his throat,
completely unable to swallow the distinction between a vector or
tensor and its coordinate decompositions. To him, if you have a
different component representation, you're talking about a physically
different thing, and he pretends to sniff that it is "mathematically
so".
Koobee Wublee
> On May 14, 10:08 pm, Koobee Wublee wrote:
> > Mathematically, you are just wrong. For example, describing flat
> > spacetime using the linear rectangular (Euclidean) and using the
> > spherically symmetric polar coordinate systems require you to supply
> > different metric for each choice of coordinate system. <shrug>
>
> No, it's the same metric in both cases, e.g. in the plane the
> following are equal:
>
> dx^2 + dy^2
>
> and:
>
> dr^2 + r^2 dtheta^2
>
> ...where x = r cos(theta) and y = r sin(theta) (polar coordinates).
>
> The above are two different coordinate decompositions of the same
> metric. To see that they are equal, just evaluate them on an arbitrary
> vector and see you get the same value in both cases.
You are making the same mistake again. What you refer to is the
geometry itself not the metric. The equations above represent the
same geometry, yes. They are equivalent. However, the coordinates
are different, and the metrics are different. The metric cannot
adequately describe the geometry despite your voodoo conjectures of
dot products, and the coordinates itself cannot adequately describe
the geometry. It takes both well specified coordinate systems and the
metrics to fully describe the geometries. <shrug>
We cannot go on without you understand my point of view, and I have
understood yours and pointed the errors in your logic. If you are not
malicious as Eric Gisse is, you need to understand my point of view.
<shrug>