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The Motion of the Perihelion of Mercury



 
 
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  #21  
Old December 30th 08, 07:45 AM posted to sci.physics,sci.physics.relativity,sci.astro
Eric Gisse
external usenet poster
 
Posts: 1,465
Default The Motion of the Perihelion of Mercury

On Dec 29, 9:41*pm, Koobee Wublee wrote:
On Dec 29, 9:36 pm, Eric Gisse wrote:

On Dec 29, 7:58 pm, Koobee Wublee wrote:
Since the Schwarzschild metric is merely one of the infinite number of
solutions to the Einstein field equations that are static, spherically
symmetric, and asymptotically flat, other solutions do not predict the
same 43”.


Name one that is not related to Schwarzschild through a coordinate
transformation.


Short memory? *You have been told that the following and the
Schwarzschild metric are ones among an infinite solutions to the
Einstein field equations that are static, spherically symmetric, and
asymptotically flat.

ds^2 = c^2 T dt^2 / (1 + 2 K / r) – (1 + 2 K / r) dr^2 – (r + K)^2
dO^2

Where

** *K, T = Constants
** *dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2


My memory is fine. Yours, though, is quite ****ed.

Back in July of 2007 I gave the explicit coordinate transformation
between your "different" solution and Schwarzschild:
http://groups.google.com/group/sci.p...4?dmode=source

Which you promptly ignored / forgot.

You then repeated the same idiocy in May of this year:
http://groups.google.com/group/sci.p...2?dmode=source

Funny how you keep repeating the same idiotic and wrong things over,
and over, and over, and over...

Besides, you didn't show that this "different solution" makes a
different prediction. You are unable to do anything but copy and paste
out of textbooks, as you can't even do a simple area calculation from
a given metric.


Again, notice this solution does not manifest black holes. *shrug


Oh, is this another one of your "by inspection" routines? Like how you
think you can see there is curvature "by inspection"?


With inability to learn, that explains why you remain a multi-year
super-senior today? *Apparently, that free money the state of Alaska
provides must go a long way for you.


I'm not the one who can't follow a basic derivation of Birkhoff's
theorem.

http://groups.google.com/group/sci.p...199f8c2bf4c127

I'm not the one who can't follow through the simplest steps of
deriving the field equations.

http://groups.google.com/group/sci.p...a48e445c49364e

I'm not the one who thinks you can determine curvature by inspection.

http://groups.google.com/group/sci.p...0?dmode=source

I'm not the one who does not believe in differential equation
uniqueness theorems.

http://groups.google.com/group/sci.p...0?dmode=source

I'm not the one who has repeatedly claimed that you can introduce
curvature with a coordinate transformation.

http://groups.google.com/group/sci.p...d?dmode=source
http://groups.google.com/group/sci.p...d?dmode=source

I'm not the one who doesn't understand basic notation, what a tensor
is, what proper time is, etc etc and ETC.

Inability to learn INDEED.


[snip perennial whining crap as usual]




  #22  
Old December 30th 08, 10:02 PM posted to sci.physics,sci.physics.relativity,sci.astro
Koobee Wublee
external usenet poster
 
Posts: 815
Default The Motion of the Perihelion of Mercury

On Dec 29, 11:45 pm, Eric Gisse wrote:
On Dec 29, 9:41 pm, Koobee Wublee wrote:


Short memory? You have been told that the following and the
Schwarzschild metric are ones among an infinite solutions to the
Einstein field equations that are static, spherically symmetric, and
asymptotically flat.


ds^2 = c^2 T dt^2 / (1 + 2 K / r) – (1 + 2 K / r) dr^2 – (r + K)^2
dO^2


Where


** K, T = Constants
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2


My memory is fine. Yours, though, is quite ****ed.

Back in July of 2007 I gave the explicit coordinate transformation
between your "different" solution and Schwarzschild:
http://groups.google.com/group/sci.p...sg/5e4cf198adb...


You still don’t get it. All tensors are matrices. There are indeed
an infinite number of solutions to the field equations. These are
basic mathematical axioms. shrug

Which you promptly ignored / forgot.

You then repeated the same idiocy in May of this year:
http://groups.google.com/group/sci.p...sg/92d926cce89...


Hmmm... What I have written down is correct. shrug

Funny how you keep repeating the same idiotic and wrong things over,
and over, and over, and over...

Besides, you didn't show that this "different solution" makes a
different prediction. You are unable to do anything but copy and paste
out of textbooks, as you can't even do a simple area calculation from
a given metric.


Don’t blame your low intellects on me. Try do that to your parents.
shrug

Again, notice this solution does not manifest black holes. shrug


Oh, is this another one of your "by inspection" routines? Like how you
think you can see there is curvature "by inspection"?


The equation above does not manifest black holes. You are still
hopelessly lost as usual. shrug

With inability to learn, that explains why you remain a multi-year
super-senior today? Apparently, that free money the state of Alaska
provides must go a long way for you.


I'm not the one who can't follow a basic derivation of Birkhoff's
theorem.

http://groups.google.com/group/sci.p...rowse_frm/thre...


You are the one who believes in the nonsense of Birkhoff’s theorem.
shrug

I'm not the one who can't follow through the simplest steps of
deriving the field equations.


You are the one who cannot follow the mathematics involved. shrug

http://groups.google.com/group/sci.p...rowse_frm/thre...

I'm not the one who thinks you can determine curvature by inspection.


Well, I am, and you should be too. No one can determine curvature
just by inspection. shrug

http://groups.google.com/group/sci.p...sg/65e5819e6e0...

I'm not the one who does not believe in differential equation
uniqueness theorems.


The differential equations represented by the field equations yield an
infinite number of solutions. Just how many time do I have to tell
you before your imbecile brain finally get it?

http://groups.google.com/group/sci.p...ea4c5a50?dmode...

I'm not the one who has repeatedly claimed that you can introduce
curvature with a coordinate transformation.


I am not, either. shrug

http://groups.google.com/group/sci.p...sg/ce55cde7fd5...
http://groups.google.com/group/sci.p...sg/a45d3eba38f...

I'm not the one who doesn't understand basic notation, what a tensor
is, what proper time is, etc etc and ETC.


What is that all about?

Inability to learn INDEED.


I think you are under the influence of narcotics. shrug

That explains why you remain a multi-year super-senior.
  #23  
Old December 31st 08, 02:35 PM posted to sci.physics,sci.physics.relativity,sci.astro,fr.sci.physique,fr.sci.astrophysique
[email protected]
external usenet poster
 
Posts: 1
Default The Motion of the Perihelion of Mercury

On 29 déc, 07:40, Pentcho Valev wrote:
On Dec 29, 12:49*am, wrote:





THE MOTION OF THE PERIHELION OF MERCURY
In his general relativity calculation of the motion of the perihelion
of Mercury Albert Einstein had only taken into account the
gravitational actions between the Sun and the Mercury, which he also
assumed as two points.


What will be, according to the theory of general relativity, the value
of the motion of the perihelion of Mercury if the gravitational
actions of all the planets in the solar system are taken into account
and also it is taken into account that the Sun is a little oblate?


Have any done these calculations?


Best regards
Louis Nielsen
Denmark


As fas as I know, the only person dealing explicitly and honestly with
this is the French astrophysicist Jean-Marc Bonnet-Bidaud. Einstein
has made his calculations on the assumption that the mass of the sun
is perfectly spherical, and if it is not, the confirmation of
relativity becomes in fact a refutation:

http://astronomy.ifrance.com/pages/g.../einstein.html
"Le deuxième test classique donne en revanche des inquiétudes.
Historiquement, pourtant, l'explication de l'avance du périhélie de
Mercure, proposé par Einstein lui-même, donna ses lettres de noblesse
à la relativité générale. Il s'agissait de comprendra pourquoi le
périhélie de Mercure ( le point de son orbite le plus proche du
soleil ) se déplaçait de 574 s d'arc par siècle. Certes, sur ces 574
s, 531 s'expliquaient par les perturbations gravitationnels dues aux
autres planètes. Mais restait 43 s, le fameux effet "périhélique"
inexpliqué par les lois de Newton. Le calcul relativiste d'Einstein
donna 42,98 s ! L'accord et si parfait qu'il ne laisse la place à
aucune discussion. Or depuis 1966, le soleil est soupçonné ne pas être
rigoureusement sphérique mais légèrement aplati à l'équateur. Une très
légère dissymétries qui suffirait à faire avancer le périhélie de
quelques secondes d'arc. Du coup, la preuve se transformerait en
réfutation puisque les 42,88 s du calcul d'Einstein ne pourrait pas
expliquer le mouvement réel de Mercure."

More explanation he

http://www.cieletespaceradio.fr/inde...0-histoire-des...
les-preuves-de-la-relativite
(ECOUTEZ!)

Pentcho Valev
- Masquer le texte des messages précédents -

- Afficher le texte des messages précédents -


helo,

Il n'y a aucune erreur dans la théorie d'Einstein, l'avance du
périhélie est correcte, lire l'article :
"NAP applied to gravitation and the implications for Einstein’s theory
of special and general relativity." de la théorie NAP qui confirme ce
résultat.
La rondeur du soleil n'a rien à voir acec ce phénomène.
l'article se trouve sur le site:
www.new-atomic-physics.com

Amicalement
ACE

  #24  
Old December 31st 08, 04:20 PM posted to sci.physics,sci.physics.relativity,sci.astro
George Hammond[_2_]
external usenet poster
 
Posts: 22
Default The Motion of the Perihelion of Mercury

On Mon, 29 Dec 2008 22:41:28 -0800 (PST), Koobee Wublee
wrote:

On Dec 29, 9:36 pm, Eric Gisse wrote:
On Dec 29, 7:58 pm, Koobee Wublee wrote:


Since the Schwarzschild metric is merely one of the infinite number of
solutions to the Einstein field equations that are static, spherically
symmetric, and asymptotically flat, other solutions do not predict the
same 43”.


Name one that is not related to Schwarzschild through a coordinate
transformation.


Short memory? You have been told that the following and the
Schwarzschild metric are ones among an infinite solutions to the
Einstein field equations that are static, spherically symmetric, and
asymptotically flat.

ds^2 = c^2 T dt^2 / (1 + 2 K / r) – (1 + 2 K / r) dr^2 – (r + K)^2
dO^2

Where

** K, T = Constants
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

[Hammond]
It may be "static, spherically symmetric, and
asymptotically flat" but I doubt that it satifies R_uv=0
since Schwarzchild proved that his solution is the ONLY
"static, spherically symmetric, and asymptotically flat"
solution that does! The schwarzchild solution is known to
be the ONLY solution to the spherical mass body problem.
=====================================
HAMMOND'S PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
GOD=G_uv (a folk song on mp3)
http://interrobang.jwgh.org/songs/hammond.mp3
=====================================

  #25  
Old December 31st 08, 07:00 PM posted to sci.physics,sci.physics.relativity,sci.astro
George Hammond[_2_]
external usenet poster
 
Posts: 22
Default The Motion of the Perihelion of Mercury

On Wed, 31 Dec 2008 11:20:24 -0500, George Hammond
wrote:

On Mon, 29 Dec 2008 22:41:28 -0800 (PST), Koobee Wublee
wrote:

On Dec 29, 9:36 pm, Eric Gisse wrote:
On Dec 29, 7:58 pm, Koobee Wublee wrote:


Since the Schwarzschild metric is merely one of the infinite number of
solutions to the Einstein field equations that are static, spherically
symmetric, and asymptotically flat, other solutions do not predict the
same 43”.

Name one that is not related to Schwarzschild through a coordinate
transformation.


Short memory? You have been told that the following and the
Schwarzschild metric are ones among an infinite solutions to the
Einstein field equations that are static, spherically symmetric, and
asymptotically flat.

ds^2 = c^2 T dt^2 / (1 + 2 K / r) – (1 + 2 K / r) dr^2 – (r + K)^2
dO^2

Where

** K, T = Constants
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

[Hammond]
It may be "static, spherically symmetric, and
asymptotically flat" but I doubt that it satifies R_uv=0
since Schwarzchild proved that his solution is the ONLY
"static, spherically symmetric, and asymptotically flat"
solution that does! The schwarzchild solution is known to
be the ONLY solution to the spherical mass body problem.

[Hammond]
P.S....The fact that Schwarzchild's solution is the ONLY
spherically symmetric solution to the EFE is known as
"Birkhoff's Theorem".
=====================================
HAMMOND'S PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
GOD=G_uv (a folk song on mp3)
http://interrobang.jwgh.org/songs/hammond.mp3
=====================================

  #26  
Old December 31st 08, 07:52 PM posted to sci.physics,sci.physics.relativity,sci.astro
George Hammond[_2_]
external usenet poster
 
Posts: 22
Default The Motion of the Perihelion of Mercury

On Wed, 31 Dec 2008 14:00:03 -0500, George Hammond
wrote:

On Wed, 31 Dec 2008 11:20:24 -0500, George Hammond
wrote:

On Mon, 29 Dec 2008 22:41:28 -0800 (PST), Koobee Wublee
wrote:

On Dec 29, 9:36 pm, Eric Gisse wrote:
On Dec 29, 7:58 pm, Koobee Wublee wrote:

Since the Schwarzschild metric is merely one of the infinite number of
solutions to the Einstein field equations that are static, spherically
symmetric, and asymptotically flat, other solutions do not predict the
same 43”.

Name one that is not related to Schwarzschild through a coordinate
transformation.

Short memory? You have been told that the following and the
Schwarzschild metric are ones among an infinite solutions to the
Einstein field equations that are static, spherically symmetric, and
asymptotically flat.

ds^2 = c^2 T dt^2 / (1 + 2 K / r) – (1 + 2 K / r) dr^2 – (r + K)^2
dO^2

Where

** K, T = Constants
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

[Hammond]
It may be "static, spherically symmetric, and
asymptotically flat" but I doubt that it satifies R_uv=0
since Schwarzchild proved that his solution is the ONLY
"static, spherically symmetric, and asymptotically flat"
solution that does! The schwarzchild solution is known to
be the ONLY solution to the spherical mass body problem.

[Hammond]
P.S....The fact that Schwarzchild's solution is the ONLY
spherically symmetric solution to the EFE is known as
"Birkhoff's Theorem".

[Hammond]
P.P.S..... Jorg Jebsen a Norwegian physicist actually
discovered and published Birkhoff's theorem two years
earlier but because he died in poverty of tuberculous in
Italy at the age of 34; when the famous mathematician George
Birkhoff later rediscovered the theorem without knowing of
Jebson's work it was named after him. Alas poor Jebson.
=====================================
HAMMOND'S PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
GOD=G_uv (a folk song on mp3)
http://interrobang.jwgh.org/songs/hammond.mp3
=====================================

  #27  
Old January 1st 09, 05:29 AM posted to sci.physics,sci.physics.relativity,sci.astro
Koobee Wublee
external usenet poster
 
Posts: 815
Default The Motion of the Perihelion of Mercury

On Dec 31, 8:20 am, George Hammond wrote:
Koobee Wublee wrote:


Short memory? You have been told that the following and the
Schwarzschild metric are ones among an infinite solutions to the
Einstein field equations that are static, spherically symmetric, and
asymptotically flat.


ds^2 = c^2 T dt^2 / (1 + 2 K / r) – (1 + 2 K / r) dr^2 – (r + K)^2
dO^2


Where


** K, T = Constants
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2


It may be "static, spherically symmetric, and
asymptotically flat" but I doubt that it satifies R_uv=0


So, you are not sure if that solution above does not satisfy R_uv =
0. Well, Gisse plugged it into his software program and had verified
so a year ago. There are actually infinite solutions to the field
equations that are static (time invariant), spherically symmetric, and
asymptotically flat (approaching flat spacetime at r = infinity).
Through Koobee Wublee’s theorem or the theorem of Generality below,
you can find any solution you wish the universe to be including the
accelerated expanding universe that still behaves like Newtonian at
relatively smaller distances.

ds^2 = c^2 T dt^2 (1 + 2 K / u) – (1 + 2 K / u) (du/dr)^2 dr^2 – (u +
K)^2 dO^2

Where

** u(r) = Any function of r

For example,

1. If (u = r), then you have the solution above.

2. If (u = r^2 / K), you do not get the Newtonian inverse square law
for gravitation.

3. If (u = (r^3 + K^3)^(1/3) – K), you get Schwarzschild’s original
solution.

4. If (u = K / (K / r + r^2 / L^2)), you get the accelerated
expanding universe at cosmological scales and Newtonian physics at
astronomical scales.

5. If (u = r – K), you get the Schwarzschild metric discovered by
Hilbert.

since Schwarzchild proved that his solution is the ONLY
"static, spherically symmetric, and asymptotically flat"
solution that does!


Notice all the examples above are static and spherically symmetric.
All are asymptotically flat except (4). Thus, Birkhoff’s theorem is
proven utter nonsense by example. shrug

The schwarzchild solution is known to
be the ONLY solution to the spherical mass body problem.


Nonsense!

On Dec 31, 11:00 am, George Hammond wrote:

P.S....The fact that Schwarzchild's solution is the ONLY
spherically symmetric solution to the EFE is known as
"Birkhoff's Theorem".


Nonsense!

On Dec 31, 11:52 am, George Hammond wrote:

P.P.S..... Jorg Jebsen a Norwegian physicist actually
discovered and published Birkhoff's theorem two years
earlier but because he died in poverty of tuberculous in
Italy at the age of 34; when the famous mathematician George
Birkhoff later rediscovered the theorem without knowing of
Jebson's work it was named after him. Alas poor Jebson.


Well, either Jebson and Birkhoff are proven to be very shallow minded
mathematicians, or Koobee Wublee is a great genius able to see through
these nonsense. Well, I will leave it up to you to decide. As you
know, yours truly is still a very humble scholar. You, on the other
hand, need to stick to what you do best. That is preaching to the
already religious SR/GR/Einstein worshippers. shrug


  #28  
Old January 1st 09, 06:00 AM posted to sci.physics,sci.physics.relativity,sci.astro
Eric Gisse
external usenet poster
 
Posts: 1,465
Default The Motion of the Perihelion of Mercury

On Dec 31, 8:29*pm, Koobee Wublee wrote:
On Dec 31, 8:20 am, George Hammond wrote:



Koobee Wublee wrote:
Short memory? *You have been told that the following and the
Schwarzschild metric are ones among an infinite solutions to the
Einstein field equations that are static, spherically symmetric, and
asymptotically flat.


ds^2 = c^2 T dt^2 / (1 + 2 K / r) – (1 + 2 K / r) dr^2 – (r + K)^2
dO^2


Where


** *K, T = Constants
** *dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2


* *It may be "static, spherically symmetric, and
asymptotically flat" but I doubt that it satifies R_uv=0


So, you are not sure if that solution above does not satisfy R_uv =
0. *Well, Gisse plugged it into his software program and had verified
so a year ago. *There are actually infinite solutions to the field
equations that are static (time invariant), spherically symmetric, and
asymptotically flat (approaching flat spacetime at r = infinity).


And all of them describe the same manifold - as told to you a year ago
with the explicit construction of the coordinate transformation
between your "different" manifold and Schwarzschild.


Through Koobee Wublee’s theorem or the theorem of Generality below,
you can find any solution you wish the universe to be including the
accelerated expanding universe that still behaves like Newtonian at
relatively smaller distances.

ds^2 = c^2 T dt^2 (1 + 2 K / u) – (1 + 2 K / u) (du/dr)^2 dr^2 – (u +
K)^2 dO^2

Where

** *u(r) = Any function of r

For example,

1. *If (u = r), then you have the solution above.

2. *If (u = r^2 / K), you do not get the Newtonian inverse square law
for gravitation.

3. *If (u = (r^3 + K^3)^(1/3) – K), you get Schwarzschild’s original
solution.

4. *If (u = K / (K / r + r^2 / L^2)), you get the accelerated
expanding universe at cosmological scales and Newtonian physics at
astronomical scales.

5. *If (u = r – K), you get the Schwarzschild metric discovered by
Hilbert.


....and all of them can be converted into the other with simple
coordinate transformations rendering your argument idiotic.


since Schwarzchild proved that his solution is the ONLY
"static, spherically symmetric, and asymptotically flat"
solution that does!


Notice all the examples above are static and spherically symmetric.
All are asymptotically flat except (4). *Thus, Birkhoff’s theorem is
proven utter nonsense by example. *shrug

The schwarzchild solution is known to
be the ONLY solution to the spherical mass body problem.


Nonsense!

On Dec 31, 11:00 am, George Hammond wrote:

* *P.S....The fact that Schwarzchild's solution is the ONLY
spherically symmetric solution to the EFE is known as
"Birkhoff's Theorem".


Nonsense!

On Dec 31, 11:52 am, George Hammond wrote:

P.P.S..... Jorg Jebsen a Norwegian physicist actually
discovered and published Birkhoff's theorem two years
earlier but because he died in poverty of tuberculous in
Italy at the age of 34; when the famous mathematician George
Birkhoff later rediscovered the theorem without knowing of
Jebson's work it was named after him. *Alas poor Jebson.


Well, either Jebson and Birkhoff are proven to be very shallow minded
mathematicians, or Koobee Wublee is a great genius able to see through
these nonsense. *Well, I will leave it up to you to decide. *As you
know, yours truly is still a very humble scholar. *You, on the other
hand, need to stick to what you do best. *That is preaching to the
already religious SR/GR/Einstein worshippers. *shrug


  #29  
Old January 1st 09, 06:46 AM posted to sci.physics,sci.physics.relativity,sci.astro
Koobee Wublee
external usenet poster
 
Posts: 815
Default The Motion of the Perihelion of Mercury

On Dec 31, 10:00 pm, Eric Gisse wrote:
On Dec 31, 8:29 pm, Koobee Wublee wrote:


Nonsense! There is no coordinate transformation. You don’t
understand the mathematics involved. Go back to be a multi-year super-
senior, and get lost.

So, you are not sure if that solution above does not satisfy R_uv =
0. Well, Gisse plugged it into his software program and had verified
so a year ago. There are actually infinite solutions to the field
equations that are static (time invariant), spherically symmetric, and
asymptotically flat (approaching flat spacetime at r = infinity).


And all of them describe the same manifold - as told to you a year ago
with the explicit construction of the coordinate transformation
between your "different" manifold and Schwarzschild.

Through Koobee Wublee’s theorem or the theorem of Generality below,
you can find any solution you wish the universe to be including the
accelerated expanding universe that still behaves like Newtonian at
relatively smaller distances.


ds^2 = c^2 T dt^2 (1 + 2 K / u) – (1 + 2 K / u) (du/dr)^2 dr^2 – (u +
K)^2 dO^2


Where


** u(r) = Any function of r


For example,


1. If (u = r), then you have the solution above.


2. If (u = r^2 / K), you do not get the Newtonian inverse square law
for gravitation.


3. If (u = (r^3 + K^3)^(1/3) – K), you get Schwarzschild’s original
solution.


4. If (u = K / (K / r + r^2 / L^2)), you get the accelerated
expanding universe at cosmological scales and Newtonian physics at
astronomical scales.


5. If (u = r – K), you get the Schwarzschild metric discovered by
Hilbert.


...and all of them can be converted into the other with simple
coordinate transformations rendering your argument idiotic.

since Schwarzchild proved that his solution is the ONLY
"static, spherically symmetric, and asymptotically flat"
solution that does!


Notice all the examples above are static and spherically symmetric.
All are asymptotically flat except (4). Thus, Birkhoff’s theorem is
proven utter nonsense by example. shrug


The schwarzchild solution is known to
be the ONLY solution to the spherical mass body problem.


Nonsense!


On Dec 31, 11:00 am, George Hammond wrote:


P.S....The fact that Schwarzchild's solution is the ONLY
spherically symmetric solution to the EFE is known as
"Birkhoff's Theorem".


Nonsense!


On Dec 31, 11:52 am, George Hammond wrote:


P.P.S..... Jorg Jebsen a Norwegian physicist actually
discovered and published Birkhoff's theorem two years
earlier but because he died in poverty of tuberculous in
Italy at the age of 34; when the famous mathematician George
Birkhoff later rediscovered the theorem without knowing of
Jebson's work it was named after him. Alas poor Jebson.


Well, either Jebson and Birkhoff are proven to be very shallow minded
mathematicians, or Koobee Wublee is a great genius able to see through
these nonsense. Well, I will leave it up to you to decide. As you
know, yours truly is still a very humble scholar. You, on the other
hand, need to stick to what you do best. That is preaching to the
already religious SR/GR/Einstein worshippers. shrug

  #30  
Old January 1st 09, 07:13 AM posted to sci.physics,sci.physics.relativity,sci.astro
Eric Gisse
external usenet poster
 
Posts: 1,465
Default The Motion of the Perihelion of Mercury

On Dec 31, 9:46*pm, Koobee Wublee wrote:
On Dec 31, 10:00 pm, Eric Gisse wrote:

On Dec 31, 8:29 pm, Koobee Wublee wrote:


Nonsense! *There is no coordinate transformation. *You don’t
understand the mathematics involved. *Go back to be a multi-year super-
senior, and get lost.


Liar.

r(R) = 2*R^2/(2*R-G*M)

The coordinate transformation is _right there_. Why don't you check
it?

[snip]
 




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