The Motion of the Perihelion of Mercury
Lou...@edu.herlufsholm.dk
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
Androcles
<Lou...@edu.herlufsholm.dk> wrote in message
news:1fed2745-f00c-49c2...@x16g2000prn.googlegroups.com...
> 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?
Different for each orbit as Venus, Earth and Jupiter advance.
> Have any done these calculations?
Le Verrier, who had no computer.
Pentcho Valev
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/gdes_theories/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 here:
http://www.cieletespaceradio.fr/index.php/2008/05/26/390-histoire-des-sciences-
les-preuves-de-la-relativite
(ECOUTEZ!)
Pentcho Valev
pva...@yahoo.com
Peter Webb
<Lou...@edu.herlufsholm.dk> wrote in message
news:1fed2745-f00c-49c2...@x16g2000prn.googlegroups.com...
AFAIR, the best experimental evidence we have is good to only a couple of
decimal places. I think we can dispense with the planets pretty quickly.
Venus weighs 1/500,000 of Sun, and that's the nearest one. Effects from the
Sun being oblate you would have to imagine are at least 2nd or 3rd order,
and its not very oblate at all.
Long and short is that measuring these effects would be experimentally
impossible, I bet.
Koobee Wublee
> 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.
In an actual observation of Mercury’s orbital advance, there are
5,600” (in arc-seconds) per century of observed perihelion advance.
Among these, 5,025” are due to the 22,000-year precession of earth’s
orbital around the second. 532” were accounted for through inclusion
of other planets. That leaves (5,600” – 5,015” = 43”) unaccounted
for.
I suspect this 5,600” per century of perihelion advance is not very
accurate in the first place. I want to see error bars associated with
this experiment. Tell me if that is too much to ask.
> 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?
The 43” was calculated based on Paul Gerber’s work. Other
mathematical methods do not yield the same result. <shrug>
> Have any done these calculations?
There are at least 12 such calculations to predict Mercury’s orbital
advance in which the spacetime with the Schwarzschild metric is just
one of them according to Gerber’s method. <shrug>
Eric Gisse
> THE MOTION OF THE PERIHELION OF MERCURY
No need to repeat the title twice, we can read.
> 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.
Entirely correct.
>
> 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?
Unknown to me, but I haven't done a literature search.
The classical analysis for Mercury's perihelion precession is almost
completely explained using perturbation theory, from the effects of
other planets. There is no reason - except for it being hard - that
the analysis can't be replicated while taking into account effects
from the other planets.
However, the effect between the Sun and Mercury is sufficient to
explain what is observed. There's a breakdown in MTW, by the way.
Eric Gisse
[...]
> I suspect this 5,600” per century of perihelion advance is not very
> accurate in the first place. I want to see error bars associated with
> this experiment. Tell me if that is too much to ask.
Is reading the literature too much to ask?
>
> > 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?
>
> The 43” was calculated based on Paul Gerber’s work. Other
> mathematical methods do not yield the same result. <shrug>
No, it was not "based on Paul Gerber's work". All Gerber did was guess
the form of a velocity-dependent potential that would give the same
effects. The actual analysis was based on the works of Le Verrier.
Do you have a literature reference for the assertion that other
methods "do not yield the same result", or is this more of your
typical nonsense that has no scholarly backing?
[...]
Ian Parker
>
> > Have any done these calculations?
>
> > Best regards
> > Louis Nielsen
>
http://www.smad.com/analysis/hpop.htm
You can actually download the program if you want to. Answer is
emphatically yes. Inceased accuracy is in fact serving to confirm GTR.
- Ian Parker
Uncle Al
Robert Dicke. Einstein was correct.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
Strich.9
>
> Einstein was correct.
>
Uncle Al recites his prayers daily.
Lou...@edu.herlufsholm.dk
> Loui...@edu.herlufsholm.dk 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?
>
> Robert Dicke. Einstein was correct.
>
> --
> Uncle Al http://www.mazepath.com/uncleal/
Uncle Al,
If you are right, please give references to Dicke’s
calculations.
As far as I know Robert Henry Dicke (1916-1997) has not made
calculations about the perihelion motion of Mercury, based on the
general theory of relativity, where he had taken into account the
gravitational actions from all the planets and other matter and energy
in the solar system.
BRANS-DICKE SCALAR-TENSOR THEORY
Robert Dicke has together with Carl Henry Brans in 1961 developed the
Brans-Dicke scalar-tensor theory of gravitation in which the
gravitational ‘constant’ is a variable scalar-function.
Dr J R Stockton
ooglegroups.com>, Sun, 28 Dec 2008 14:49:58, Lou...@edu.herlufsholm.dk
posted:
See <http://en.wikipedia.org/wiki/Perihelion_precession_of_Mercury#Perih
elion_precession_of_Mercury> for example.
The "Einstein effect" is 43"/century; the figures for other effects are
: solar oblateness 0.025", other planets 530", co-ordinate system 500",
approximately.
Wikipedia has articles on it in French, Spanish, Portuguese, but not
Danish - presumably because all Danes read English perfectly. The
Iberians have little; but one should read <http://fr.wikipedia.org/wiki/
Tests_exp%C3%A9rimentaux_de_la_relativit%C3%A9_g%C3%A9n%C3%A9rale> which
is quite independent of that cited above.
--
(c) John Stockton, nr London, UK. ?@merlyn.demon.co.uk Turnpike v6.05 MIME.
Web <URL:http://www.merlyn.demon.co.uk/> - FAQqish topics, acronyms & links;
Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.
No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News.
Androcles
"Dr J R Stockton" <j...@merlyn.demon.co.uk> wrote in message
news:FnRPvBeV...@invalid.uk.co.demon.merlyn.invalid...
> In sci.astro message <1fed2745-f00c-49c2-8c22-3ffcabb0a4d7@x16g2000prn.g
> ooglegroups.com>, Sun, 28 Dec 2008 14:49:58, Lou...@edu.herlufsholm.dk
> posted:
>>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?
>
> See <http://en.wikipedia.org/wiki/Perihelion_precession_of_Mercury#Perih
> elion_precession_of_Mercury> for example.
Oh, you found something about it in wackypedia. How clever of you.
Wish I'd thought of that. Fucking idiot!
PD
> On 29 Dec., 16:35, Uncle Al <Uncle...@hate.spam.net> wrote:
>
>
>
> > Loui...@edu.herlufsholm.dk 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?
>
> > Robert Dicke. Einstein was correct.
>
> > --
>
> Uncle Al,
> If you are right, please give references to Dicke’s
> calculations.
>
> As far as I know Robert Henry Dicke (1916-1997) has not made
> calculations about the perihelion motion of Mercury, based on the
> general theory of relativity, where he had taken into account the
> gravitational actions from all the planets and other matter and energy
> in the solar system.
>
> BRANS-DICKE SCALAR-TENSOR THEORY
> Robert Dicke has together with Carl Henry Brans in 1961 developed the
> Brans-Dicke scalar-tensor theory of gravitation in which the
> gravitational ‘constant’ is a variable scalar-function.
>
> Best regards
> Louis Nielsen
> Denmark
He did a little more than that. You could do well to look up more of
his publication list.
While not complete, this link contains a few items of interest to this
discussion:
http://www.nap.edu/html/biomems/rdicke.html
Koobee Wublee
> On Dec 28, 10:07 pm, Koobee Wublee wrote:
> > I suspect this 5,600” per century of perihelion advance is not very
> > accurate in the first place. I want to see error bars associated with
> > this experiment. Tell me if that is too much to ask.
>
> Is reading the literature too much to ask?
What literature?
> > The 43” was calculated based on Paul Gerber’s work. Other
> > mathematical methods do not yield the same result. <shrug>
>
> No, it was not "based on Paul Gerber's work".
Gerber pioneered that particular way of deriving differential
equations. Try reading the literature for a change. <shrug>
> All Gerber did was guess
> the form of a velocity-dependent potential that would give the same
> effects.
Gerber had his reasons. Try reading the literature. Tell me if that
is too much to ask.
> The actual analysis was based on the works of Le Verrier.
Le Verrier was an observer equivalent to an experimenter just like
Professor Roberts. Le Verrier did not do any detailed analyses in the
same level as Gerber. Try to read the literature. <shrug>
> Do you have a literature reference for the assertion that other
> methods "do not yield the same result", or is this more of your
> typical nonsense that has no scholarly backing?
Of course. How much do you want to pay for that?
Koobee Wublee
> On Dec 28, 2:49 pm, Loui...@edu.herlufsholm.dk 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.
>
> In an actual observation of Mercury’s orbital advance, there are
> 5,600” (in arc-seconds) per century of observed perihelion advance.
> Among these, 5,025” are due to the 22,000-year precession of earth’s
> orbital around the second. 532” were accounted for through inclusion
> of other planets. That leaves (5,600” – 5,025” – 532” = 43”)
> unaccounted
> for.
>
> I suspect this 5,600” per century of perihelion advance is not very
> accurate in the first place. I want to see error bars associated with
> this experiment. Tell me if that is too much to ask.
>
> > 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?
>
> The 43” was calculated based on Paul Gerber’s work. Other
> mathematical methods do not yield the same result. <shrug>
>
> > Have any done these calculations?
>
> There are at least 12 such calculations to predict Mercury’s orbital
> advance in which the spacetime with the Schwarzschild metric is just
> one of them according to Gerber’s method. <shrug>
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”.
Professor Roberts, the experimental physicist, has emphasized so much
to demand an error bar to each observation, and yet he remains silent
on Le Verrier’s observation of these 5,600”. Not to mention these
5,025” and 532”. I would have to conclude it is a case of his own
personal bias towards the faith in the nonsense called the general
theory of relativity. <shrug>
Eric Gisse
> On Dec 29, 12:30 am, Eric Gisse wrote:
>
> > On Dec 28, 10:07 pm, Koobee Wublee wrote:
> > > I suspect this 5,600” per century of perihelion advance is not very
> > > accurate in the first place. I want to see error bars associated with
> > > this experiment. Tell me if that is too much to ask.
>
> > Is reading the literature too much to ask?
>
> What literature?
Is there a reason you are unable to perform a basic literature search?
>
> > > The 43” was calculated based on Paul Gerber’s work. Other
> > > mathematical methods do not yield the same result. <shrug>
>
> > No, it was not "based on Paul Gerber's work".
>
> Gerber pioneered that particular way of deriving differential
> equations. Try reading the literature for a change. <shrug>
Since the actual analysis is based on classical perturbation theory
and not Gerber's work, I'm going to go with "you are full of shit".
http://www.mathpages.com/home/kmath527/kmath527.htm
Go read.
[snip rest]
Eric Gisse
[...]
>
> > There are at least 12 such calculations to predict Mercury’s orbital
> > advance in which the spacetime with the Schwarzschild metric is just
> > one of them according to Gerber’s method. <shrug>
>
> 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.
Remember that you've failed this abundantly simple challenge every
time previous, so I doubt this time will be different.
[snip]
Koobee Wublee
> 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
Again, notice this solution does not manifest black holes. <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.
> [snip perennial whining crap as usual]
Koobee Wublee
> On Dec 29, 7:57 pm, Koobee Wublee < wrote:
> > > On Dec 28, 10:07 pm, Koobee Wublee wrote:
> > > > I suspect this 5,600” per century of perihelion advance is not very
> > > > accurate in the first place. I want to see error bars associated with
> > > > this experiment. Tell me if that is too much to ask.
>
> > > Is reading the literature too much to ask?
>
> > What literature?
>
> Is there a reason you are unable to perform a basic literature search?
Is there a reason why you cannot even comprehend the literature you
have dug up with? You remind me of Bill Hubba. That explains why you
remain a multi-year super-senior today. <shrug>
> > Gerber pioneered that particular way of deriving differential
> > equations. Try reading the literature for a change. <shrug>
>
> Since the actual analysis is based on classical perturbation theory
> and not Gerber's work, I'm going to go with "you are full of shit".
>
> http://www.mathpages.com/home/kmath527/kmath527.htm
Equations 6 to 7 are what Gerber did. Learn to understand the
literature. <shrug>
> [snip the rest of whining crap]
Eric Gisse
> On Dec 29, 9:36 pm, Eric Gisse <jowr...@gmail.com> 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 fucked.
Back in July of 2007 I gave the explicit coordinate transformation
between your "different" solution and Schwarzschild:
http://groups.google.com/group/sci.physics.relativity/msg/5e4cf198adbd8234?dmode=source
Which you promptly ignored / forgot.
You then repeated the same idiocy in May of this year:
http://groups.google.com/group/sci.physics.relativity/msg/92d926cce8951cd2?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.physics.relativity/browse_frm/thread/67199f8c2bf4c127
I'm not the one who can't follow through the simplest steps of
deriving the field equations.
http://groups.google.com/group/sci.physics.relativity/browse_frm/thread/bda48e445c49364e
I'm not the one who thinks you can determine curvature by inspection.
http://groups.google.com/group/sci.physics.relativity/msg/65e5819e6e0a3ee0?dmode=source
I'm not the one who does not believe in differential equation
uniqueness theorems.
http://groups.google.com/group/sci.physics/msg/8b886186ea4c5a50?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.physics.relativity/msg/ce55cde7fd58464d?dmode=source
http://groups.google.com/group/sci.physics.relativity/msg/a45d3eba38faed5d?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.
Koobee Wublee
> 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 fucked.
>
> Back in July of 2007 I gave the explicit coordinate transformation
> between your "different" solution and Schwarzschild:
> http://groups.google.com/group/sci.physics.relativity/msg/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.physics.relativity/msg/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.physics.relativity/browse_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.physics.relativity/browse_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.physics.relativity/msg/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.physics/msg/8b886186ea4c5a50?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.physics.relativity/msg/ce55cde7fd5...
http://groups.google.com/group/sci.physics.relativity/msg/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.
Adams...@wanadoo.fr
> les-preuves-de-la-relativite
> (ECOUTEZ!)
>
> Pentcho Valev
> pva...@yahoo.com- 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
George Hammond
<koobee...@gmail.com> wrote:
>On Dec 29, 9:36 pm, Eric Gisse <jowr...@gmail.com> 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
=====================================
George Hammond
[Hammond]
P.S....The fact that Schwarzchild's solution is the ONLY
spherically symmetric solution to the EFE is known as
"Birkhoff's Theorem".
George Hammond
<Nowh...@notspam.com> wrote:
[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.
Koobee Wublee
> 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>
Eric Gisse
> 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.
Koobee Wublee
> 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.
Eric Gisse
> 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]
Sam Wormley
Somebody let you out of your playpen tonight Koobee?
Koobee Wublee
> Koobee Wublee wrote:
> [...]
>
> > 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>
>
> [...]
>
> > 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>
>
> Somebody let you out of your playpen tonight Koobee?
Yes. In case, if you are still hiding under a rock, tonight is a new
year’s eve.
Koobee Wublee
> On Dec 31, 9:46 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.
<shrug>
> r(R) = 2*R^2/(2*R-G*M)
>
> The coordinate transformation is _right there_. Why don't you check
> it?
No, it is not. There is no merit to suggest a coordinate
transformation. You are just so ignorant. <shrug>
> [snip]
You are just an Einstein worshippers’ prostitute. <shrug>
Pentcho Valev
> > (ECOUTEZ!)
>
> 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
C'est parce que Albert le Divin a decouvert la vérité suivante:
"Imagination is more important than knowledge." Albert Einstein
Vous imaginez que "La rondeur du soleil n'a rien à voir avec ce
phénomène" et cela devient beaucoup plus important que le savoir selon
lequel la distribution de la masse du soleil (spherique ou pas) est un
facteur cricual.
Pentcho Valev
pva...@yahoo.com
Eric Gisse
> 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.
http://img58.imageshack.us/img58/8527/idiotcm5.png
It is not a valid solution of the vacuum field equations. Feel free to
rationalize why you are right even though you are wrong. Again.
Idiot.
[snip rest]
Eric Gisse
> On Dec 31, 11:13 pm, Eric Gisse wrote:
>
> > On Dec 31, 9:46 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.
>
> <shrug>
>
> > r(R) = 2*R^2/(2*R-G*M)
You never did check those previous two times, either.
>
> > The coordinate transformation is _right there_. Why don't you check
> > it?
>
> No, it is not. There is no merit to suggest a coordinate
> transformation. You are just so ignorant. <shrug>
You didn't even look. If you had looked, you would have noticed I was
pointing to the wrong line element.
Arrogant stupidity saves the day again.
Your "solution" is not a solution.
http://img58.imageshack.us/img58/8527/idiotcm5.png
George Hammond
[Hammond]
It's obvious Kooby is a Hype since he's claiming
Birkhoff's Theorem is "wrong" when the entire field
confirmed it 75 years ago... and since it explains why a
pulsating star cannot emit gravitational waves it must have
sent another thousand LIGO physicists back to check it again
more recently.
You seemed to be convinced Kooby was simply making a
(radial) coordinate transformation and doesn't actually know
this can't affect the vanishing of R_uv... which sounds very
likely .... on the other hand I just guessed that his metric
probably didn't solve R_uv=0, even though he says it does.
His claim of an "infinite number of solutions" certainly
sounds like an infinite numbers of coordinate
transformations, on the other hand the URL you cite above
appears to show that Ricci isn't actually zero for his
metric as he claims. Since he says it is, could this be a
programming glitch and actually you were right the first
time?
I personally still suspect you're right about his
"solutions" being merely coordinate transformations and he
doesn't know it ...but...which explanation of "Koober's
Folly" do you think is right at this point?
By the way, I'm not an expert on "Koobology", but as the
world's leading "PSYCHOPHYSICIST" I would diagnose Kooby as
what Wikipedia defines as a "putz".....e.g. "sham contempt
fueled by high levels of ironic wonder at the simple power
of ham fisted intimidation". Unfortunately in this case he
has been neatly snared by Birkhoff!
Eric Gisse
He's made the same stupid claim about a dozen different
representations of Schwarzschild. I got confused and thought he was
talking about a different one.
There is no reason to believe it is a programming glitch. The program
has a codebase that is rather old, and the computation is
straightforward.
> I personally still suspect you're right about his
> "solutions" being merely coordinate transformations and he
> doesn't know it ...but...which explanation of "Koober's
> Folly" do you think is right at this point?
They almost universally are coordinate transformations from
Schwarzschild. He hasn't even been clever enough to chain a few
transformations together to make something unintelligible but still
Schwarzschild.
The most likely explanation is he fucked up when he tried to copy from
another source.
Ian Parker
> > > (ECOUTEZ!)
>
> > 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
>
> C'est parce que Albert le Divin a decouvert la vérité suivante:
>
> "Imagination is more important than knowledge." Albert Einstein
>
> Vous imaginez que "La rondeur du soleil n'a rien à voir avec ce
> phénomène" et cela devient beaucoup plus important que le savoir selon
> lequel la distribution de la masse du soleil (spherique ou pas) est un
> facteur cricual.
>
planetary rotation. In this case Newtonian theory predicts attraction
from a point at the center of the Sun. No shape has nothing to do with
it. It is GTR and the Schwartzchild radius.
- Ian Parker
Koobee Wublee
> On Dec 31, 10:37 pm, Koobee Wublee wrote:
> Your "solution" is not a solution.
>
> http://img58.imageshack.us/img58/8527/idiotcm5.png
In the following post, I gave you the following solution to the field
equations that obeys Newtonian law of gravity, but this one exhibits
half of the event horizon than the Schwarzschild metric.
ds^2 = c^2 (1 – 2 K / r)^2 dt^2 – dr^2 / (1 – K / r)^4 – r^2 dO^2 / (1
– K / r)^2
Where
** K = G M / c^2 / 2, HALF OF THE EVENT HORIZON
Reference post:
http://groups.google.com/group/sci.physics.relativity/msg/e3162cddce87191f
With this spacetime, the event horizon occurs at (2 K) which if (G M /
c^2)
A couple posts later, you verified that the above spacetime does
indeed satisfy R_uv = 0 by saying:
“A quick re-roll into grtensor showed that you are, in fact, correct.
It does satisfy R_uv = 0.”
Reference post:
http://groups.google.com/group/sci.physics.relativity/msg/6c66880d4c24fdc6
Now, the solution we have been talking about is much simpler than the
one above. If I can derive the above solution from Koobee Wublee’s
theorem or the theorem of Generality, just how much more difficult can
I derive the following?
ds^2 = c^2 T dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2
Where
** K = 2 G M / c^2
> http://img58.imageshack.us/img58/8527/idiotcm5.png
And just like that last time, you don’t even know how to enter the
inputs correctly. What you have entered is wrong. You need to
replace the 2 instances of (2 K) with K. <shrug>
Koobee Wublee
> It's obvious Kooby is a Hype since he's claiming
> Birkhoff's Theorem is "wrong" when the entire field
> confirmed it 75 years ago...
So, someone believes in a particular of not your liking, and you
believe in the nonsense of Birkhoff’s theorem. Isn’t this all too
familiar? You are doing physics by applying religious teachings.
<shrug>
> and since it explains why a
> pulsating star cannot emit gravitational waves it must have
> sent another thousand LIGO physicists back to check it again
> more recently.
You obviously do not understand the mathematics that gives rise to the
gravitational waves. Hint: Gravitational waves have nothing to do
with Birkhoff’s theorem. <shrug>
> You seemed to be convinced Kooby was simply making a
> (radial) coordinate transformation and doesn't actually know
> this can't affect the vanishing of R_uv... which sounds very
> likely .... on the other hand I just guessed that his metric
> probably didn't solve R_uv=0, even though he says it does.
You are making a conclusion way too soon. <shrug>
You need to understood Koobee Wublee’s theorem or the theorem of
Generality. The derivation is not that difficult if you have the
diligence. <shrug>
> His claim of an "infinite number of solutions" certainly
> sounds like an infinite numbers of coordinate
> transformations,
Sounds like? Once again, this proves that the great reverend Hammond
is clueless. <shrug>
> on the other hand the URL you cite above
> appears to show that Ricci isn't actually zero for his
> metric as he claims. Since he says it is, could this be a
> programming glitch and actually you were right the first
> time?
How about an operator error?
> I personally still suspect you're right about his
> "solutions" being merely coordinate transformations and he
> doesn't know it ...but...which explanation of "Koober's
> Folly" do you think is right at this point?
I love it. The great reverend Hammond does physics with gut feelings.
> By the way, I'm not an expert on "Koobology",
Hmmm... Just what is Koobology?
> but as the
> world's leading "PSYCHOPHYSICIST" I would diagnose Kooby as
> what Wikipedia defines as a "putz".....e.g. "sham contempt
> fueled by high levels of ironic wonder at the simple power
> of ham fisted intimidation". Unfortunately in this case he
> has been neatly snared by Birkhoff!
Now, you are getting very incoherent. Having a hang over?
Eric Gisse
> On Jan 1, 12:32 am, Eric Gisse wrote:
>
> > On Dec 31, 10:37 pm, Koobee Wublee wrote:
> > Your "solution" is not a solution.
>
> >http://img58.imageshack.us/img58/8527/idiotcm5.png
>
> In the following post, I gave you the following solution to the field
> equations that obeys Newtonian law of gravity, but this one exhibits
> half of the event horizon than the Schwarzschild metric.
>
> ds^2 = c^2 (1 – 2 K / r)^2 dt^2 – dr^2 / (1 – K / r)^4 – r^2 dO^2 / (1
> – K / r)^2
>
> Where
>
> ** K = G M / c^2 / 2, HALF OF THE EVENT HORIZON
Since when did any of the constants say where the event horizon is?
>
> Reference post:
>
> http://groups.google.com/group/sci.physics.relativity/msg/e3162cddce8...
>
> With this spacetime, the event horizon occurs at (2 K) which if (G M /
> c^2)
No, the event horizon does not occur there. Do the math, arrogant
idiot.
http://img353.imageshack.us/img353/8273/idiot2ok4.png
LOOK AT THE MATH. Every one of them go to zero or blow up at r = K.
That's where the event horizon is.
As given to you in a post you just referenced, the coordinate
transformation between your "different" metric and Schwarzschild is
given by r(R) = 2*R^2/(2*R-K) where r is the Schwarzschild radial
coordinate and R is the idiot radial coordinate.
Put in R = K, and you get r = 2K. Imagine that - the singularity is
located in the same fucking place and it is the SAME MANIFOLD.
Proven.
>
> A couple posts later, you verified that the above spacetime does
> indeed satisfy R_uv = 0 by saying:
>
> “A quick re-roll into grtensor showed that you are, in fact, correct.
> It does satisfy R_uv = 0.”
>
> Reference post:
>
> http://groups.google.com/group/sci.physics.relativity/msg/6c66880d4c2...
>
> Now, the solution we have been talking about is much simpler than the
> one above. If I can derive the above solution from Koobee Wublee’s
> theorem or the theorem of Generality, just how much more difficult can
> I derive the following?
You aren't deriving jack shit. You are transforming coordinates from
Schwarzschild to something else. Your claims are asinine, childish,
and never backed up by anything more than a repetition of the claim.
>
> ds^2 = c^2 T dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2
>
> Where
>
> ** K = 2 G M / c^2
>
> >http://img58.imageshack.us/img58/8527/idiotcm5.png
>
> And just like that last time, you don’t even know how to enter the
> inputs correctly. What you have entered is wrong. You need to
> replace the 2 instances of (2 K) with K. <shrug>
http://groups.google.com/group/sci.physics.relativity/msg/71f5a677e6fb7519?dmode=source
That's what you wrote, dipshit. I even said later to George that you
were probably transcribing it wrong - and imagine that, you did.
But let's discuss idiot metric #3. I'm numbering them because it is
hard to keep track.
http://img135.imageshack.us/img135/2813/idiot3yw9.png
It _does_ satisfy R_uv = 0. So therefore it is related to
Schwarzschild by a coordinate transformation.
The coordinate transformation between Schwarzschild and idiot3 is
given rather simply by r(R) = r + K.
Another "different" coordinate system proven to be a part of the same
object. How many times do we have to do this before you grow a clue?
Koobee Wublee
> On Jan 1, 1:13 pm, Koobee Wublee wrote:
> > In the following post, I gave you the following solution to the field
> > equations that obeys Newtonian law of gravity, but this one exhibits
> > half of the event horizon than the Schwarzschild metric.
>
> > ds^2 = c^2 (1 – 2 K / r)^2 dt^2 – dr^2 / (1 – K / r)^4 – r^2 dO^2 / (1
> > – K / r)^2
>
> > Where
>
> > ** K = G M / c^2 / 2, HALF OF THE EVENT HORIZON
>
> Since when did any of the constants say where the event horizon is?
Hmmm... If you don’t know, you need to consult one of the textbooks
you have been sitting on. It is actually very simple. <shrug>
> > Reference post:
>
> >http://groups.google.com/group/sci.physics.relativity/msg/e3162cddce8...
>
> > With this spacetime, the event horizon occurs at (2 K) which if (G M /
> > c^2)
>
> No, the event horizon does not occur there. Do the math, arrogant
> idiot.
>
> http://img353.imageshack.us/img353/8273/idiot2ok4.png
>
> LOOK AT THE MATH. Every one of them go to zero or blow up at r = K.
> That's where the event horizon is.
Oh, big deal. G M / c^2 / 2 versus G M c^2 / 4. The bottom line is
that the event horizon of this spacetime which does satisfy as a
static, spherically symmetric, and asymptotically flat solution to the
field equations is not (2 G M / c^2). <shrug>
> As given to you in a post you just referenced, the coordinate
> transformation between your "different" metric and Schwarzschild is
> given by r(R) = 2*R^2/(2*R-K) where r is the Schwarzschild radial
> coordinate and R is the idiot radial coordinate.
There is no coordinate transformation. How many times do I have to
tell you that?
> Put in R = K, and you get r = 2K. Imagine that - the singularity is
> located in the same fucking place and it is the SAME MANIFOLD.
> Proven.
What is this whining crap agan?
> > A couple posts later, you verified that the above spacetime does
> > indeed satisfy R_uv = 0 by saying:
>
> > “A quick re-roll into grtensor showed that you are, in fact, correct.
> > It does satisfy R_uv = 0.”
>
> > Reference post:
>
> >http://groups.google.com/group/sci.physics.relativity/msg/6c66880d4c2...
>
> > Now, the solution we have been talking about is much simpler than the
> > one above. If I can derive the above solution from Koobee Wublee’s
> > theorem or the theorem of Generality, just how much more difficult can
> > I derive the following?
>
> You aren't deriving jack shit. You are transforming coordinates from
> Schwarzschild to something else. Your claims are asinine, childish,
> and never backed up by anything more than a repetition of the claim.
You are merely ignorant. You have been checkmated many times over but
are either too stupid or too hypocritical to accept what I have told
you many times over. <shrug> You are a small man just like that so-
called professor Andersen from an obscure college in Norway. <shrug>
> > ds^2 = c^2 T dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2
>
> > Where
>
> > ** K = 2 G M / c^2
>
> http://img58.imageshack.us/img58/8527/idiotcm5.png
>
> > And just like that last time, you don’t even know how to enter the
> > inputs correctly. What you have entered is wrong. You need to
> > replace the 2 instances of (2 K) with K. <shrug>
>
> http://groups.google.com/group/sci.physics.relativity/msg/71f5a677e6f...
>
> That's what you wrote, dipshit. I even said later to George that you
> were probably transcribing it wrong - and imagine that, you did.
Oh, there is no need to blow your top off over a couple of innocent
typos. I have written the same spacetime over and over and over and
over again before. You should know better. <shrug>
> But let's discuss idiot metric #3. I'm numbering them because it is
> hard to keep track.
>
> http://img135.imageshack.us/img135/2813/idiot3yw9.png
>
> It _does_ satisfy R_uv = 0.
Yes, I told you so. After all, you are going to give the great
reverend Hammond a heart attack after he has already decided to
BELIEVED IN otherwise --- all thanks to your “operator errors”.
> So therefore it is related to
> Schwarzschild by a coordinate transformation.
It is only related to the Schwarzschild metric through being also a
solution that is static, spherically symmetric, and asymptotically
flat to the field equations. <shrug>
> The coordinate transformation between Schwarzschild and idiot3 is
> given rather simply by r(R) = r + K.
Again, there is no transformation of coordinate. However, applying
Koobee Wublee’s theorem or the theorem of Generality, you can select
the following to arrive at the Schwarzschild metric. <shrug>
u = r – K.
> Another "different" coordinate system proven to be a part of the same
> object. How many times do we have to do this before you grow a clue?
Oh, quit behaving like a sore loser. Bow, kneel, and kowtow to Koobee
Wublee, and accept the mathematical nonsense in the bullshit called
GR.
Yes, Einstein was nobody. He was a nitwit, a plagiarist, and a liar.
<shrug>
Eric Gisse
> On Jan 1, 10:56 pm, Eric Gisse wrote:
>
> > On Jan 1, 1:13 pm, Koobee Wublee wrote:
> > > In the following post, I gave you the following solution to the field
> > > equations that obeys Newtonian law of gravity, but this one exhibits
> > > half of the event horizon than the Schwarzschild metric.
>
> > > ds^2 = c^2 (1 – 2 K / r)^2 dt^2 – dr^2 / (1 – K / r)^4 – r^2 dO^2 / (1
> > > – K / r)^2
>
> > > Where
>
> > > ** K = G M / c^2 / 2, HALF OF THE EVENT HORIZON
>
> > Since when did any of the constants say where the event horizon is?
>
> Hmmm... If you don’t know, you need to consult one of the textbooks
> you have been sitting on. It is actually very simple. <shrug>
The objects in a particular coordinate representation of the metric do
not define the location of the event horizon. Idiot.
>
> > > Reference post:
>
> > >http://groups.google.com/group/sci.physics.relativity/msg/e3162cddce8...
>
> > > With this spacetime, the event horizon occurs at (2 K) which if (G M /
> > > c^2)
>
> > No, the event horizon does not occur there. Do the math, arrogant
> > idiot.
>
> >http://img353.imageshack.us/img353/8273/idiot2ok4.png
>
> > LOOK AT THE MATH. Every one of them go to zero or blow up at r = K.
> > That's where the event horizon is.
>
> Oh, big deal. G M / c^2 / 2 versus G M c^2 / 4. The bottom line is
> that the event horizon of this spacetime which does satisfy as a
> static, spherically symmetric, and asymptotically flat solution to the
> field equations is not (2 G M / c^2). <shrug>
Except you are proven wrong. The event horizon is at r = K - as proven
- and r = K corresponds to R = 2K in the Schwarzschild metric.
Repeating the assertion doesn't make you right, just stupid.
> > As given to you in a post you just referenced, the coordinate
> > transformation between your "different" metric and Schwarzschild is
> > given by r(R) = 2*R^2/(2*R-K) where r is the Schwarzschild radial
> > coordinate and R is the idiot radial coordinate.
>
> There is no coordinate transformation. How many times do I have to
> tell you that?
Asserting they are different does not make them different. I proved
they are the same - suck it up, and try again.
We can repeat this idiocy for a thousand years and you will be wrong
_every single time_ because of the fundamental nature of tensors.
Recasting them in a different coordinate system changes nothing but I
encourage you keep trying.
Either that or learn - but if you could learn you wouldn't be here
repeating stupidities, now would you?
>
> > Put in R = K, and you get r = 2K. Imagine that - the singularity is
> > located in the same fucking place and it is the SAME MANIFOLD.
> > Proven.
>
> What is this whining crap agan?
Do you not know what a coordinate transformation is?
I have four metrics:
ds^2 = dx^2 + dy^2 + dz^2
ds^2 = dr^2 + r^2(d\theta^2 + sin^2(\theta)d\phi^2)
ds^2 = dr^2 + r^2 d\theta^2 + dz^2
ds^2 = da^2 + db^2 + dz^2
They all describe the same thing, yet "look different". But I can find
a coordinate transformation relating all of them to eachother, so they
are the same thing.
How is this any different?
[snip idiocy]
You can't refute one single thing I have written. Amazing.
> > The coordinate transformation between Schwarzschild and idiot3 is
> > given rather simply by r(R) = r + K.
>
> Again, there is no transformation of coordinate. However, applying
> Koobee Wublee’s theorem or the theorem of Generality, you can select
> the following to arrive at the Schwarzschild metric. <shrug>
>
> u = r – K.
That's called a "coordinate transformation".
Thanks for proving my point.
>
> > Another "different" coordinate system proven to be a part of the same
> > object. How many times do we have to do this before you grow a clue?
>
> Oh, quit behaving like a sore loser. Bow, kneel, and kowtow to Koobee
> Wublee, and accept the mathematical nonsense in the bullshit called
> GR.
The shield of stupidity protects your huge but fragile ego.
George Hammond
[Hammond]
Yes, R_uv for the above Metric does vanish, but the
reason is, as you have informed us, that making the simple
substitution r=R-K with K=2m immediately transforms the
Kooby metric into the Schwarzchild Metric!
Kooby is trying to refute Birkhoff's theorem which as
anyone with a lick of common sense knows is utterly futile.
In addition a number of times he has transcribed the
metric incorrectly which immediately causes R_uv=/=0 as
required by the vacuum equations.
The quest for a Kooby Metric which will disprove
Schwarzchild-Birkhoff has therefore suffered a final,
disasterous and truly ignominious defeat!
And this certainly puts to the lie his scurrilous remark
that "other solutions to the EFE" "do not predict the 43"
advance of Mercury".
Koobee Wublee
> On Jan 1, 10:48 pm, Koobee Wublee wrote:
> > Hmmm... If you don’t know, you need to consult one of the textbooks
> > you have been sitting on. It is actually very simple. <shrug>
>
> [Gisse’s whining crap snipped]
>
> I have four metrics:
>
> ds^2 = dx^2 + dy^2 + dz^2
> ds^2 = dr^2 + r^2(d\theta^2 + sin^2(\theta)d\phi^2)
> ds^2 = dr^2 + r^2 d\theta^2 + dz^2
> ds^2 = da^2 + db^2 + dz^2
>
> They all describe the same thing, yet "look different". But I can find
> a coordinate transformation relating all of them to eachother, so they
> are the same thing.
Your notation is screwed up. The r of the 2nd is not the same as r of
the 3rd. The phi of the 2nd is the same as theta of the 3rd.
Otherwise, they can all represent the same geometry. <shrug>
This is not what I have been talking about. It looks like very few
have understood what I am saying. That is why I feel very lonely to
be the few who after Riemann and perhaps Christoffel that have
understood the curvature business very well. <shrug>
What I am talking about is that the following geometries are different
given the same dr, dLongitude, dLatitude.
** ds1^2 = g1_11 dr^2 + g1_22 r^2 dO^2
** ds2^2 = g2_11 dr^2 + g2_22 r^2 dO^2
Where
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2
> [Snipped more Gisse’s whining crap]
>
> > Again, there is no transformation of coordinate. However, applying
> > Koobee Wublee’s theorem or the theorem of Generality, you can select
> > the following to arrive at the Schwarzschild metric. <shrug>
>
> > u = r – K.
>
> That's called a "coordinate transformation".
What coordinate transformation? u is any function of r. This is very
basic mathematics. <shrug>
> Thanks for proving my point.
You are still dreaming and jacking off in your mathemagical Realm. In
doing so, probably wishing Einstein can join you. <shrug>
Let me write down the two very different spacetime in a more concise
way.
** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
** ds2^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2
THE GEOMETRY ds1 IS NOT THE SAME AS ds2. THAT IS BECAUSE THE METRIC
OF ds1 IS NOT THE SAME AS ds2, BUT THE COORDINATE SYSTEM DESCRIBING
BOTH ds1 and ds2 IS EXACTLY IDENTICAL.
Can you dig this, punk? I don’t think you possess any minute
intelligence to get after all these years, and I really don’t care.
<shrug>
With any matheMagical tricks, show me how you can coordinate-transform
from one equation to the other and what justify one solution is more
unique than the others. Remember that the coordinate system is the
same with both equations.
> > Oh, quit behaving like a sore loser. Bow, kneel, and kowtow to Koobee
> > Wublee, and accept the mathematical nonsense in the bullshit called
> > GR.
>
> The shield of stupidity protects your huge but fragile ego.
Cheap shot once again.
Koobee Wublee
> Eric Gisse wrote:
> Yes, R_uv for the above Metric does vanish,
So, the great reverend is wishy washy. Did you not just said through
you gut feelings that solution cannot possibly cause the Ricci tensor
to vanish?
> but the
> reason is, as you have informed us,
Oh, excuses to justify your faith. <shrug>
> that making the simple
> substitution r=R-K with K=2m immediately transforms the
> Kooby metric into the Schwarzchild Metric!
There is no coordinate transformation. It is all your ego giving you
excuses to believe in that nonsense. <shrug> It is very funny that
you would allow Gisse, a multi-year super-senior without any college
degree at the University of Alaska, Fairbanks to convince you to
continue your faith. <shrug>
> Kooby is trying to refute Birkhoff's theorem which as
> anyone with a lick of common sense knows is utterly futile.
Ahahaha...
> In addition a number of times he has transcribed the
> metric incorrectly which immediately causes R_uv=/=0 as
> required by the vacuum equations.
Come on. I have written that spacetime correctly for many posts. It
was merely a typo. <shrug>
> The quest for a Kooby Metric which will disprove
> Schwarzchild-Birkhoff has therefore suffered a final,
> disasterous and truly ignominious defeat!
So, you are practicing to write the next bible. <shrug>
> And this certainly puts to the lie his scurrilous remark
> that "other solutions to the EFE" "do not predict the 43"
> advance of Mercury".
Yes, I have not tried to figure out if that spacetime does satisfy Le
Verrier’s claim in Mercury’s orbital advance using Gerber’s
mathematical methodology, but who cares since there is no
justification on which solution is the so-called “Rock of the
Gibraltar”. <shrug>
Eric Gisse
> On Jan 2, 12:03 am, Eric Gisse wrote:
>
>
>
> > On Jan 1, 10:48 pm, Koobee Wublee wrote:
> > > Hmmm... If you don’t know, you need to consult one of the textbooks
> > > you have been sitting on. It is actually very simple. <shrug>
>
> > [Gisse’s whining crap snipped]
>
> > I have four metrics:
>
> > ds^2 = dx^2 + dy^2 + dz^2
> > ds^2 = dr^2 + r^2(d\theta^2 + sin^2(\theta)d\phi^2)
> > ds^2 = dr^2 + r^2 d\theta^2 + dz^2
> > ds^2 = da^2 + db^2 + dz^2
>
> > They all describe the same thing, yet "look different". But I can find
> > a coordinate transformation relating all of them to eachother, so they
> > are the same thing.
>
> Your notation is screwed up. The r of the 2nd is not the same as r of
> the 3rd. The phi of the 2nd is the same as theta of the 3rd.
> Otherwise, they can all represent the same geometry. <shrug>
See? You are a cunt hair's distance away from understanding. THE
LABELS DO NOT MATTER. The COORDINATES do not matter. They all
represent the _same_ geometry.
>
> This is not what I have been talking about. It looks like very few
> have understood what I am saying. That is why I feel very lonely to
> be the few who after Riemann and perhaps Christoffel that have
> understood the curvature business very well. <shrug>
So where did you properly learn this "curvature business"? All I'm
asking is for the book you learned it from. Is that asking so much?
>
> What I am talking about is that the following geometries are different
> given the same dr, dLongitude, dLatitude.
>
> ** ds1^2 = g1_11 dr^2 + g1_22 r^2 dO^2
> ** ds2^2 = g2_11 dr^2 + g2_22 r^2 dO^2
>
> Where
>
> ** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2
Except the coordinate labels are not the same in both coordinate
systems. Calculate the surface area of a sphere - don't just write it
down, CALCULATE it.
>
> > [Snipped more Gisse’s whining crap]
>
> > > Again, there is no transformation of coordinate. However, applying
> > > Koobee Wublee’s theorem or the theorem of Generality, you can select
> > > the following to arrive at the Schwarzschild metric. <shrug>
>
> > > u = r – K.
>
> > That's called a "coordinate transformation".
>
> What coordinate transformation? u is any function of r. This is very
> basic mathematics. <shrug>
Yes, and it is called a "coordinate transformation". You start out in
the (t,u,\theta,\phi) coordinate system and end up in the (t,r,\theta,
\phi) coordinate system.
>
> > Thanks for proving my point.
>
> You are still dreaming and jacking off in your mathemagical Realm. In
> doing so, probably wishing Einstein can join you. <shrug>
>
> Let me write down the two very different spacetime in a more concise
> way.
>
> ** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
> ** ds2^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2
>
> THE GEOMETRY ds1 IS NOT THE SAME AS ds2. THAT IS BECAUSE THE METRIC
> OF ds1 IS NOT THE SAME AS ds2, BUT THE COORDINATE SYSTEM DESCRIBING
> BOTH ds1 and ds2 IS EXACTLY IDENTICAL.
If they are "exactly identical" they should give the same values for
the surface area of the sphere, as #1 is in Schwarzschild isotropic
coordinates which has spheres of constant (r,t) of surface area 4 pi
r^2.
Let us see you calculate the surface area of #2.
I would also like you calculate the Ricci scalars of both. You'll
notice something interesting.
>
> Can you dig this, punk? I don’t think you possess any minute
> intelligence to get after all these years, and I really don’t care.
> <shrug>
>
> With any matheMagical tricks, show me how you can coordinate-transform
> from one equation to the other and what justify one solution is more
> unique than the others. Remember that the coordinate system is the
> same with both equations.
Except the coordinate systems are different - the second one is
displaced from the origin by an amount K, which you can see
immediately by looking at the coefficients of the angular components.
Or by calculating the surface area. Or by calculating the Ricci
scalar. Or by looking at the determinant of the metric.
I cannot claim one solution is more unique than the other, which is
the whole point of my explanation. I just prefer Schwarzschild in
isotropic coordinates. I can easily convert from Schwarzschild and
every /OTHER/ "different" solution to yours - just chain the
transformations.
Eric Gisse
> On Jan 2, 6:18 am, George Hammond wrote:
>
> > Eric Gisse wrote:
> > Yes, R_uv for the above Metric does vanish,
>
> So, the great reverend is wishy washy. Did you not just said through
> you gut feelings that solution cannot possibly cause the Ricci tensor
> to vanish?
You wrote two different metrics. He was right both times, as was I.
[snip]
George Hammond
[Hammond]
Correct, he admits to "making a typo" which is why there
are two different versions of the Kooby Metric floating
around.
In the original one, you discovered and I have
independently confirmed that it is simply a radial
coordinate transformation of Schwarzchild given by r=R-K
with K=2m. So the original Kooby Metric is nothing but the
Schwarzchild Metric in disguise.
In the second case his "Typo Version" of the Kooby Metric
is not a coordinate transformation of Schwarzchild and
naturally therefore R_uv does not vanish as you discovered
and hence it is not a valid solution of the EFE's.
Ergo... Kooby's legal leg is wobbling and all his
statements to the contrary have been PROVEN to be
nonsensical and wrong! At this point he is reduced to
mindless repetition (echolalliation).
Bear in mind, as a mere 25 year old you still think he's
a sincere Physics buff.... but at my age and being a veteran
"Psychophysicist" I've seen enough of his type to know he's
an annoyed, childishly jealous and spiteful personality who
only pretends to be sincere for the purposes of dogging the
uninitiated. He's an ornery prick and he would do well to
read my scientific proof of God where he might be able to
really make an original contribution to a real applied
Physics discovery.... since he seems to have nothing better
to do than play with Psychophysics for entertainment.
Raghar
> 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?
Well van Fandern did this analysis:
http://www.metaresearch.org/cosmology/gravity/PAFfolder/PerihelionAdvanceFormula-combined.asp
Also there are flyby anomalies, which are possibly more accurate
http://www2.phys.canterbury.ac.nz/editorial/Anderson2008.pdf
>Its latitude dependence suggests
>that the Earth’s rotation may be generating an effect much
>larger than the frame dragging effect of General Relativity,
>the Lense-Thirring effect
Eric Gisse
> On Dec 28 2008, 11:49 pm, Loui...@edu.herlufsholm.dk 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?
>
>
TvF writes crank nonsense, and it doesn't address the original
question.
> Also there are flyby anomalies, which are possibly more accuratehttp://www2.phys.canterbury.ac.nz/editorial/Anderson2008.pdf
Completely irrelevant to the original question.
Koobee Wublee
> On Jan 2, 10:42 pm, Koobee Wublee wrote:
> > On Jan 2, 12:03 am, Eric Gisse wrote:
> > > ds^2 = dx^2 + dy^2 + dz^2
> > > ds^2 = dr^2 + r^2(d\theta^2 + sin^2(\theta)d\phi^2)
> > > ds^2 = dr^2 + r^2 d\theta^2 + dz^2
> > > ds^2 = da^2 + db^2 + dz^2
>
> > > They all describe the same thing, yet "look different". But I can find
> > > a coordinate transformation relating all of them to eachother, so they
> > > are the same thing.
>
> > Your notation is screwed up. The r of the 2nd is not the same as r of
> > the 3rd. The phi of the 2nd is the same as theta of the 3rd.
> > Otherwise, they can all represent the same geometry. <shrug>
>
> See? You are a cunt hair's distance away from understanding.
What is that again?
> THE LABELS DO NOT MATTER.
Wrong! The labels do matter. <shrug>
> The COORDINATES do not matter. They all
> represent the _same_ geometry.
Wrong again. To describe a geometry, you need both the labels
(coordinate) and how the labels are connected (metric). <shrug>
> > This is not what I have been talking about. It looks like very few
> > have understood what I am saying. That is why I feel very lonely to
> > be the few who after Riemann and perhaps Christoffel that have
> > understood the curvature business very well. <shrug>
>
> So where did you properly learn this "curvature business"? All I'm
> asking is for the book you learned it from. Is that asking so much?
Just my intelligence which you lack. <shrug>
> > What I am talking about is that the following geometries are different
> > given the same dr, dLongitude, dLatitude.
>
> > ** ds1^2 = g1_11 dr^2 + g1_22 r^2 dO^2
> > ** ds2^2 = g2_11 dr^2 + g2_22 r^2 dO^2
>
> > Where
>
> > ** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2
>
> Except the coordinate labels are not the same in both coordinate
> systems.
What do you mean they are not the same? I have chosen them to be the
same. <shrug>
> Calculate the surface area of a sphere - don't just write it
> down, CALCULATE it.
I did. Just look up on my older post.
> > What coordinate transformation? u is any function of r. This is very
> > basic mathematics. <shrug>
>
> Yes, and it is called a "coordinate transformation". You start out in
> the (t,u,\theta,\phi) coordinate system and end up in the (t,r,\theta,
> \phi) coordinate system.
I did not start out in anything. I merely presented these two
independent geometries of spacetime without showing any favor of one
to the other.
> > You are still dreaming and jacking off in your mathemagical Realm. In
> > doing so, probably wishing Einstein can join you. <shrug>
>
> > Let me write down the two very different spacetime in a more concise
> > way.
>
> > ** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
> > ** ds2^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2
>
> > THE GEOMETRY ds1 IS NOT THE SAME AS ds2. THAT IS BECAUSE THE METRIC
> > OF ds1 IS NOT THE SAME AS ds2, BUT THE COORDINATE SYSTEM DESCRIBING
> > BOTH ds1 and ds2 IS EXACTLY IDENTICAL.
>
> If they are "exactly identical" they should give the same values for
> the surface area of the sphere, as #1 is in Schwarzschild isotropic
> coordinates which has spheres of constant (r,t) of surface area 4 pi
> r^2.
They are not identical. <shrug>
> Let us see you calculate the surface area of #2.
>
> I would also like you calculate the Ricci scalars of both. You'll
> notice something interesting.
Calculating the area does not make any difference. To any observer
observing the geometry of any curved spacetime, the answer is always
going to be 4 pi r^2. <shrug>
> > Can you dig this, punk? I don’t think you possess any minute
> > intelligence to get after all these years, and I really don’t care.
> > <shrug>
>
> > With any matheMagical tricks, show me how you can coordinate-transform
> > from one equation to the other and what justify one solution is more
> > unique than the others. Remember that the coordinate system is the
> > same with both equations.
>
> Except the coordinate systems are different
Hey, I am the one presenting the problem. Both ds1 and ds2 share the
same coordinate system. They are all solved from the set of field
equations with the same coordinate system. <shrug>
> - the second one is
> displaced from the origin by an amount K, which you can see
> immediately by looking at the coefficients of the angular components.
> Or by calculating the surface area. Or by calculating the Ricci
> scalar. Or by looking at the determinant of the metric.
Nonsense! There is no displacement of anything. There is no
coordinate transformation. <shrug>
> I cannot claim one solution is more unique than the other, which is
> the whole point of my explanation. I just prefer Schwarzschild in
> isotropic coordinates. I can easily convert from Schwarzschild and
> every /OTHER/ "different" solution to yours - just chain the
> transformations.
Hold your bullsh*t! Both ds1 and ds2 are spherically symmetric in
isotropic coordinate system. Again, for the n’th time, these
solutions are static, spherically symmetric, and asymptotically flat.
I HAVE SHOWN YOU SOLUTIONS THAT ARE STATIC, SPHERICALLY SYMMETRIC,
ABUT NOT ASYMPTOTICALLY FLAT. That proves Birkhoff’s theorem false by
example. <shrug>
Your ranting is because you refuse the God you worship is a nitwit, a
plagiarist, and a liar. Yes, Einstein was nobody. He was a nitwit, a
plagiarist, and a liar. Despite the nitwit, the plagiarist, and the
liar did not come up with all these nonsense, whatever is credited to
this nitwit, this plagiarist, and this liar is utter nonsense in the
first place. <shrug>
Koobee Wublee
> Correct, he admits to "making a typo" which is why there
> are two different versions of the Kooby Metric floating
> around.
Oh, there are more than that. In the following post, you have been
entertained with a solution that is static, spherically symmetric, but
not asymptotically flat.
http://groups.google.com/group/sci.physics.relativity/msg/14f866a91526793d?hl=en
Of course, allow me to correct my typo. The spacetime I was talking
about is:
ds^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2
> In the original one, you discovered and I have
> independently confirmed that it is simply a radial
> coordinate transformation of Schwarzchild given by r=R-K
> with K=2m. So the original Kooby Metric is nothing but the
> Schwarzchild Metric in disguise.
Oh, it looks like you have been eating the same sh*t with Gisse.
<shrug>
> In the second case his "Typo Version" of the Kooby Metric
> is not a coordinate transformation of Schwarzchild and
> naturally therefore R_uv does not vanish as you discovered
> and hence it is not a valid solution of the EFE's.
<shrug>
> [Rest of whining crap snipped]
So, the great reverend Hammond is also a nitwit. He had no problem to
go to bed with Einstein Dingleberries to save his grace. <shrug>
George Hammond
<koobee...@gmail.com> wrote:
>......... That proves Birkhoff’s theorem false by
>example. <shrug>
>
[Hammond]
"proves Birkhoff's theorem false by example".....aaahahaha
ha hah hah aaaahaha ha ha ha ha......
>
>Your ranting is because you refuse the God you worship is a nitwit, a
>plagiarist, and a liar. Yes, Einstein was nobody. He was a nitwit, a
>plagiarist, and a liar. Despite the nitwit, the plagiarist, and the
>liar did not come up with all these nonsense, whatever is credited to
>this nitwit, this plagiarist, and this liar is utter nonsense in the
>first place. <shrug>
>
[Hammond]
Oh here we go ... turns out you're not just a moron,
you're a jive ass Natzi too....aaahahaha ha hah hah aaaahaha
ha ha ha......
George Hammond
<koobee...@gmail.com> wrote:
>
<snip>
>
>[Koobee Wublee]
>......... That proves Birkhoff’s theorem false by
>example. <shrug>
>
[Hammond]
"proves Birkhoff's theorem false by example".....aaahahaha
ha hah hah aaaahaha ha ha ha ha......
Eric Gisse
> On Jan 3, 3:02 am, Eric Gisse wrote:
>
>
>
> > On Jan 2, 10:42 pm, Koobee Wublee wrote:
> > > On Jan 2, 12:03 am, Eric Gisse wrote:
> > > > ds^2 = dx^2 + dy^2 + dz^2
> > > > ds^2 = dr^2 + r^2(d\theta^2 + sin^2(\theta)d\phi^2)
> > > > ds^2 = dr^2 + r^2 d\theta^2 + dz^2
> > > > ds^2 = da^2 + db^2 + dz^2
>
> > > > They all describe the same thing, yet "look different". But I can find
> > > > a coordinate transformation relating all of them to eachother, so they
> > > > are the same thing.
>
> > > Your notation is screwed up. The r of the 2nd is not the same as r of
> > > the 3rd. The phi of the 2nd is the same as theta of the 3rd.
> > > Otherwise, they can all represent the same geometry. <shrug>
>
> > See? You are a cunt hair's distance away from understanding.
>
> What is that again?
>
> > THE LABELS DO NOT MATTER.
>
> Wrong! The labels do matter. <shrug>
Prove it. Write something with two different labels, and PROVE that
you can measure something to be different.
I have an idea - let's see your proof that "other solutions" that
satisfy the conditions of Birkhoff's theorem produce something other
than 43 arcsec/centry for Mercury's perihelion precession. You did,
after all, assert that to be the case.
>
> > The COORDINATES do not matter. They all
> > represent the _same_ geometry.
>
> Wrong again. To describe a geometry, you need both the labels
> (coordinate) and how the labels are connected (metric). <shrug>
True - to write the metric down, you need coordinates. Except the
coordinates do not matter as I can easily transform from one system to
the next and write the metric down in the new coordinate system.
As has been done repeatedly. As has the explanation to you.
Consider that you are just too stupid to understand, like Androcles
and rbwinn before you.
>
> > > This is not what I have been talking about. It looks like very few
> > > have understood what I am saying. That is why I feel very lonely to
> > > be the few who after Riemann and perhaps Christoffel that have
> > > understood the curvature business very well. <shrug>
>
> > So where did you properly learn this "curvature business"? All I'm
> > asking is for the book you learned it from. Is that asking so much?
>
> Just my intelligence which you lack. <shrug>
So you just expect me to believe you woke up with this knowledge one
morning?
I find it more likely that you arrived at this particular level of
understanding after trying to study modern general relativity
textbooks. Except you failed, and invented all these arguments to
protect your ego.
Why else would you get so defensive and insulting when I ask a simple
question like "where did you learn what you know?".
>
> > > What I am talking about is that the following geometries are different
> > > given the same dr, dLongitude, dLatitude.
>
> > > ** ds1^2 = g1_11 dr^2 + g1_22 r^2 dO^2
> > > ** ds2^2 = g2_11 dr^2 + g2_22 r^2 dO^2
>
> > > Where
>
> > > ** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2
>
> > Except the coordinate labels are not the same in both coordinate
> > systems.
>
> What do you mean they are not the same? I have chosen them to be the
> same. <shrug>
Then calculate the surface area of a sphere using the metric.
>
> > Calculate the surface area of a sphere - don't just write it
> > down, CALCULATE it.
>
> I did. Just look up on my older post.
Nope - you _asserted_ it was 4 pi r^2. You never _derived_ it.
In fact, you seem quite incapable of proving something as simple as
this. Why is the only person who has "truly understood" differential
geometry incapable of deriving the surface area of a sphere?
[...same idiocy as in 2006...]
http://groups.google.com/group/sci.physics.relativity/msg/4897dc5ec4157bec?dmode=source
JanPB explained this to you before with much greater patience than I
can muster. You used the same arguments then as you use now.
> > I cannot claim one solution is more unique than the other, which is
> > the whole point of my explanation. I just prefer Schwarzschild in
> > isotropic coordinates. I can easily convert from Schwarzschild and
> > every /OTHER/ "different" solution to yours - just chain the
> > transformations.
>
> Hold your bullsh*t! Both ds1 and ds2 are spherically symmetric in
> isotropic coordinate system. Again, for the n’th time, these
> solutions are static, spherically symmetric, and asymptotically flat.
...and the same. How quickly you forget that this has all been
explained to you in detail before. With the same metrics no less.
http://groups.google.com/group/sci.physics.relativity/msg/4503cf7f2586e0a7?dmode=source
> I HAVE SHOWN YOU SOLUTIONS THAT ARE STATIC, SPHERICALLY SYMMETRIC,
> ABUT NOT ASYMPTOTICALLY FLAT. That proves Birkhoff’s theorem false by
> example. <shrug>
Except you didn't prove the solution wasn't asymptotically flat
because you never calculated something like CURVATURE which would
actually prove your point. Plus I showed you the coordinate
transformation to navigate between your "different solution" and
Schwarzschild.
>
> Your ranting is because you refuse the God you worship [...]
Funny how you are always the one to bring up God in these little
conversations.
Eric Gisse
[snip]
> Bear in mind, as a mere 25 year old you still think he's
> a sincere Physics buff....
NO, I DO NOT. He is an arrogant idiot who has been repeatedly shown he
can't do something as simple as calculate the surface area of a
sphere, all the way to being unable to understand a derivation of the
field equations.
[snip]
Eric Gisse
> On Jan 3, 9:09 am, George Hammond wrote:
>
> > Correct, he admits to "making a typo" which is why there
> > are two different versions of the Kooby Metric floating
> > around.
>
> Oh, there are more than that. In the following post, you have been
> entertained with a solution that is static, spherically symmetric, but
> not asymptotically flat.
How do you know it is not asymptotically flat when you did not even
consider calculating curvature components?
[snip]
George Hammond
<jow...@gmail.com> wrote:
>
>>
>> > > What I am talking about is that the following geometries are different
>> > > given the same dr, dLongitude, dLatitude.
>>
>> > > ** ds1^2 = g1_11 dr^2 + g1_22 r^2 dO^2
>> > > ** ds2^2 = g2_11 dr^2 + g2_22 r^2 dO^2
>>
>> > > Where
>>
>> > > ** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2
>>
>> > Except the coordinate labels are not the same in both coordinate
>> > systems.
>>
>> What do you mean they are not the same? I have chosen them to be the
>> same. <shrug>
>
>Then calculate the surface area of a sphere using the metric.
>
>>
>> > Calculate the surface area of a sphere - don't just write it
>> > down, CALCULATE it.
>>
>> I did. Just look up on my older post.
>
>Nope - you _asserted_ it was 4 pi r^2. You never _derived_ it.
>
[Hammond]
Spherical surface area in flat space is obtained by
integrating the surface element cos(theta)d(theta)d(phi) in
sperical coordinates over phi and theta for constant r which
results in the ancient formula Area=4 pi r^2.
In curved space of course the metric must be used if the
phi and theta basis vectors are not orthogonal. However in
the Schwarzchild and Kooby metrics there is only radial
space warping and no angularspace warping (radially
symmetric metric), therefore the surface element is still a
simple rectangle and not a parallogram as in the general
case.
Since integrating phi from 0->2pi and theta from 0->pi
over the angular part of the flat space metric or even the
Schwarzchild Metric:
r^2 cos(theta)d(theta)d(phi)
results in 4 pi r^2i,
we see that the area of an ordinary sphere in flat space, or
even a sphere in Schwarzchild space is 4 pi r^2 for a fixed
value of r.
In the case of the Kooby Metric however, the angular part
is not multiplied by r^2, it is multiplied by (R+K)^2
therefore the same surface integration results in an area
equal to:
4 pi (R+K)^2
for a fixed value of R; the new Kooby radial coordinate.
Notice of course that substituting the coordinate
transformation R=r-K transforms thisresult back to the well
known Schwarzchild result area=4 pi r^2 again.
Koobee Wublee
> On Jan 3, 6:29 pm, Koobee Wublee wrote:
> > On Jan 3, 3:02 am, Eric Gisse wrote:
> > > THE LABELS DO NOT MATTER.
>
> > Wrong! The labels do matter. <shrug>
>
> Prove it. Write something with two different labels, and PROVE that
> you can measure something to be different.
I have said that many times over. To describe the geometry, one must
choose a coordinate system first and then define how the set of
coordinates is related to the geometry. <shrug>
> I have an idea - let's see your proof that "other solutions" that
> satisfy the conditions of Birkhoff's theorem produce something other
> than 43 arcsec/centry for Mercury's perihelion precession. You did,
> after all, assert that to be the case.
Well, it is time consuming. I have lacked any motivation to do so
after discovering GR is a total nonsense based on voodoo mathematics.
<shrug>
> > > The COORDINATES do not matter. They all
> > > represent the _same_ geometry.
>
> > Wrong again. To describe a geometry, you need both the labels
> > (coordinate) and how the labels are connected (metric). <shrug>
>
> True - to write the metric down, you need coordinates.
Oh, wow! That is an improvement after all these years. <shrug>
> Except the
> coordinates do not matter as I can easily transform from one system to
> the next and write the metric down in the new coordinate system.
Again, you are stuck in your matheMagical wonderland of coordinate
transformation. There is no such coordinate transformation. You are
schizophrenic. <shrug>
> As has been done repeatedly. As has the explanation to you.
There is no credible coordinate transformation done besides total
nonsense. <shrug>
> Consider that you are just too stupid to understand, like Androcles
> and rbwinn before you.
The Einstein Dingleberries have no problems bouncing these crackpots
up and down the tennis courts, but Koobee Wublee remains
unchallenged. Your whining crap does not count. <shrug>
> > > So where did you properly learn this "curvature business"? All I'm
> > > asking is for the book you learned it from. Is that asking so much?
>
> > Just my intelligence which you lack. <shrug>
>
> So you just expect me to believe you woke up with this knowledge one
> morning?
Well, it is up to you. I have also studies a lot about the nonsense
which you eagerly swallowed. <shrug>
> I find it more likely that you arrived at this particular level of
> understanding after trying to study modern general relativity
> textbooks. Except you failed, and invented all these arguments to
> protect your ego.
There are flaws in GR. I have discovered them, and I have advanced my
studies. In the meantime, you are stuck in what they have told you.
That is why you remain a multi-year super-senior. <shrug>
> Why else would you get so defensive and insulting when I ask a simple
> question like "where did you learn what you know?".
I am not getting defensive. I just don’t think my educational
background is any of your business. <shrug>
> > > Except the coordinate labels are not the same in both coordinate
> > > systems.
>
> > What do you mean they are not the same? I have chosen them to be the
> > same. <shrug>
>
> Then calculate the surface area of a sphere using the metric.
I have told you the answer more than once. To an observer, the
surface area of a sphere is always (4 pi r^2) regardless how curved up
space is. <shrug>
> > > Calculate the surface area of a sphere - don't just write it
> > > down, CALCULATE it.
>
> > I did. Just look up on my older post.
>
> Nope - you _asserted_ it was 4 pi r^2. You never _derived_ it.
>
> [Remaining whining crap snipped]
You are just trolling me. <shrug>
> JanPB explained this to you before with much greater patience than I
> can muster. You used the same arguments then as you use now.
Speaking of the one who calls me mentally unstable, and yet he would
have no problem of swimming naked in a public lake. Mr. Bielawski is
an Einstein Dingleberry who swallows everything the collective group
of Einstein Dingleberries decides what physics should be without any
scientific methodology. You are also one of them. You have also
become a henchman trying to force that crap down on anyone’s throat.
<shrug>
> > > I cannot claim one solution is more unique than the other, which is
> > > the whole point of my explanation. I just prefer Schwarzschild in
> > > isotropic coordinates. I can easily convert from Schwarzschild and
> > > every /OTHER/ "different" solution to yours - just chain the
> > > transformations.
>
> > Hold your bullsh*t! Both ds1 and ds2 are spherically symmetric in
> > isotropic coordinate system. Again, for the n’th time, these
> > solutions are static, spherically symmetric, and asymptotically flat.
>
> [snipped whining crap]
So, you have no answers as usual besides whining nonsense.
> > I HAVE SHOWN YOU SOLUTIONS THAT ARE STATIC, SPHERICALLY SYMMETRIC,
> > ABUT NOT ASYMPTOTICALLY FLAT. That proves Birkhoff’s theorem false by
> > example. <shrug>
>
> Except you didn't prove the solution wasn't asymptotically flat
> because you never calculated something like CURVATURE which would
> actually prove your point. Plus I showed you the coordinate
> transformation to navigate between your "different solution" and
> Schwarzschild.
There is no need to talk about this further. It is obvious that you
do not understand what asymptotically flat means. <shrug>
> > Your ranting is because you refuse the God you worship [...]
>
> Funny how you are always the one to bring up God in these little
> conversations.
That is because you have decided Einstein as your God. <shrug>
Your education is totally based on faith. <shrug>
Sam Wormley
Grade: F-
Koobee Wublee
> Spherical surface area in flat space is obtained by
> integrating the surface element cos(theta)d(theta)d(phi) in
> sperical coordinates over phi and theta for constant r which
> results in the ancient formula Area=4 pi r^2.
Well, is it really that obvious?
> In curved space of course the metric must be used if the
> phi and theta basis vectors are not orthogonal. However in
> the Schwarzchild and Kooby metrics there is only radial
> space warping and no angularspace warping (radially
> symmetric metric), therefore the surface element is still a
> simple rectangle and not a parallogram as in the general
> case.
Well, you have to decide if the question applies to an observer
observing a sphere in curved space or a group of physicists trying to
detect the actual (local) surface area of a sphere situated in known
curved space.
In the case of an observer observing a sphere in curved space, he will
always observe flat space. Thus, the surface area is always (4 pi
r^2) regardless of how much space is curved.
In the case of a group of physicists trying to detect the actual
(local) surface area of a sphere situated in known curved space, you
have to accurately determine what the metric is first, and that is no
trivial task. <shrug>
> [Rest of Hammond whining crap mercifully snipped]
Koobee Wublee
> On Jan 3, 6:37 pm, Koobee Wublee wrote:
> > Oh, there are more than that. In the following post, you have been
> > entertained with a solution that is static, spherically symmetric, but
> > not asymptotically flat.
>
> How do you know it is not asymptotically flat when you did not even
> consider calculating curvature components?
You have to understand what asymptotically flat means. <shrug>
Koobee Wublee
> Koobee Wublee wrote:
> >......... That proves Birkhoff’s theorem false by
> >example. <shrug>
>
> "proves Birkhoff's theorem false by example".....aaahahaha
> ha hah hah aaaahaha ha ha ha ha......
Is that a denial? Do you not understand the disproval of a hypothesis
by example?
For example, if someone claims all oceanic creatures are not
mammalian, and I can show dolphins having habitats in the oceans
around the world are mammals, then that claim must be wrong.
Reverends still cannot get. I guess that would separate reverends
from scientists. <shrug>
Koobee Wublee
> Grade: F-
That is funny. If a very mentally challenged individual such as
yourself gave me a grade of F, without any hesitation I would throw
out and burn any papers related to this subject. <shrug>
Thanks for the laughs. Ahahaha...
Sam Wormley
You got the F- because you FAILED to address any of Eric's points.
Koobee Wublee
> Koobee Wublee wrote:
> > That is funny. If a very mentally challenged individual such as
> > yourself gave me a grade of F, without any hesitation I would throw
> > out and burn any papers related to this subject. <shrug>
>
> > Thanks for the laughs. Ahahaha...
>
> You got the F- because you FAILED to address any of Eric's points.
Why are you trying to justify nonsense? Who gives a f*ck?
Again, thanks for the laughes. Ahahaha...
Eric Gisse
> On Jan 4, 12:29 am, Eric Gisse wrote:
>
> > On Jan 3, 6:29 pm, Koobee Wublee wrote:
> > > On Jan 3, 3:02 am, Eric Gisse wrote:
> > > > THE LABELS DO NOT MATTER.
>
> > > Wrong! The labels do matter. <shrug>
>
> > Prove it. Write something with two different labels, and PROVE that
> > you can measure something to be different.
>
> I have said that many times over. To describe the geometry, one must
> choose a coordinate system first and then define how the set of
> coordinates is related to the geometry. <shrug>
The coordinates are not related to the geometry. The coordinates -
like crackers - do not matter. MY argument is further buttressed by
your inability to calculate anything, like surface area of the amount
Mercury's perihelion would precess.
I do not see why you can't open a tensor analysis book written any
time in the last hundred and thirty or so years and look at the basic
proofs given.
>
> > I have an idea - let's see your proof that "other solutions" that
> > satisfy the conditions of Birkhoff's theorem produce something other
> > than 43 arcsec/centry for Mercury's perihelion precession. You did,
> > after all, assert that to be the case.
>
> Well, it is time consuming. I have lacked any motivation to do so
> after discovering GR is a total nonsense based on voodoo mathematics.
> <shrug>
But you said it gives a different answer, which implies you have done
the calculation before.
However what you are saying implies you never did the calculation. So
you have no way of knowing whether the answer is different.
>
> > > > The COORDINATES do not matter. They all
> > > > represent the _same_ geometry.
>
> > > Wrong again. To describe a geometry, you need both the labels
> > > (coordinate) and how the labels are connected (metric). <shrug>
>
> > True - to write the metric down, you need coordinates.
>
> Oh, wow! That is an improvement after all these years. <shrug>
>
> > Except the
> > coordinates do not matter as I can easily transform from one system to
> > the next and write the metric down in the new coordinate system.
>
> Again, you are stuck in your matheMagical wonderland of coordinate
> transformation. There is no such coordinate transformation. You are
> schizophrenic. <shrug>
People have been telling you what the coordinate transformation is
since 2006, and the transformation itself has been known for most of a
century. Is there any reason you do not understand?
You even wrote down the transformation. Or do you somehow think
writing one coordinate in terms of other coordinates is not a
coordinate transformation?
[snip]
As usual, you slip back into the old routine of slinging insults
because you have been challenged and have nothing else to say. It is
rather sad, but predictable.
Eric Gisse
Again, how do you know it is asymptotically flat when you didn't
calculate anything?
Koobee Wublee
> On Jan 4, 10:12 pm, Koobee Wublee wrote:
> > Eric Gisse <jowr...@gmail.com> wrote:
> > > How do you know it is not asymptotically flat when you did not even
> > > consider calculating curvature components?
>
> > You have to understand what asymptotically flat means. <shrug>
>
> Again, how do you know it is asymptotically flat when you didn't
> calculate anything?
Again, you have to understand what asymptotically flat means, mutli-
year super-senior. <shrug>
Koobee Wublee
> > I have said that many times over. To describe the geometry, one must
> > choose a coordinate system first and then define how the set of
> > coordinates is related to the geometry. <shrug>
>
> The coordinates are not related to the geometry.
>
> [Gisse’s diarrhea snipped]
To describe any geometry, you must choose a set of coordinate system
first. Without any coordinate system, you cannot describe any
geometry. Try to understand this very basic concept.
> Or do you somehow think
> writing one coordinate in terms of other coordinates is not a
> coordinate transformation?
The accusation is broad. Any answer can easily be construed as
malevolence by you.
Thus, I just want you to show me how the following spacetimes ds1 to
ds2 are the same.
** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
** ds2^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2
Since both solutions came from the same set of field equations with
coordinate system chosen right from the very beginning (Christoffel
symbols), the coordinate system in both is the same.
> [More crap snipped]
More diarrhea coming from Gisse.
Eric Gisse
Whether I know it or not does not matter. You claim that your
"different" metrics aren't related by a coordinate transformation, and
setting aside the fact that is wrong, that means you need to actually
_prove_ asymptotic flatness by computing all of the components of the
Riemann curvature tensor R^a_bcd and prove they go to zero as r -->
\infty.
The fact you never calculate anything says a lot about your knowledge
of the subject.
George Hammond
<koobee...@gmail.com> wrote:
>On Jan 4, 11:25 am, George Hammond wrote:
>
>> Spherical surface area in flat space is obtained by
>> integrating the surface element r^2 cos(theta)d(theta)d(phi) in
>> sperical coordinates over phi and theta for constant r which
>> results in the ancient formula Area=4 pi r^2.
>
>[Koobee]
>Well, is it really that obvious?
>
[Hammond]
Is the Pope a Catholic? Of course it's that obvious.
Moreover, exactly the same statement holds true of
Schwarzchild space. A sphere in Schwarzchild space also has
a surface area of 4 pi r^2 for exactly the same reason.
However, for the Kooby metric a new radial coordinate has
been invented via the coordinate transformation r=R+K so
that now a sphere has the surface area 4 pi (R+K)^2
This is of course true because the Kooby Metric is
nothing more than the Schwarzchild Metric with a radial
coordinate transformation r=R+K inserted in it.
.................
So, the long and neverending length of this thread over
something so trivial tells us that the real spactacle we are
discussing is NOT curvature of an objective physical
spacetime (e.g. Gravity), but what we are actually
discussing is the phenomena of an anomolous curvature of
someone's subjective personal spacetime (e.g. God). In this
case the only anomalous curvature being evident is the
curvature of "Koobee's personal perceptual spacetime"...
meaning he has exhibited a notorous and fascinating "God
anomally", not a "Gravity anomally".
That God is a curvature of subjective (individual)
apacetime is throughly explained and proven on my website
below:
Eric Gisse
> On Jan 5, 1:17 am, Eric Gisse wrote:
>
> > Koobee Wublee wrote:
> > > I have said that many times over. To describe the geometry, one must
> > > choose a coordinate system first and then define how the set of
> > > coordinates is related to the geometry. <shrug>
>
> > The coordinates are not related to the geometry.
>
> > [Gisse’s diarrhea snipped]
>
> To describe any geometry, you must choose a set of coordinate system
> first. Without any coordinate system, you cannot describe any
> geometry. Try to understand this very basic concept.
Except the choice of coordinates is irrelevant, as shown repeatedly by
multiple coordinate representations of Euclidean geometry,
Schwarzschild, and Minkowski space. It'd be far simpler to write out
the proof but it has been shown to you before and you simply don't
understand.
To this day you can NOT explain what the difference is between Euclid
written in 4 different coordinate systems and Schwarzschild in 4
different coordinate systems.
Remember the original claim you made?
First you claim the "other solutions" don't predict the 43 arcsec/
century that Schwarzschild does:
Then you rattle off a metric that only has a shifted radial coordinate
with respect to the Schwarzschild isotropic coordinate system.
Finally you explain that you haven't even done the fucking
calculation.
Now you are once again arguing about something rather basic, and are
hitting your head against those same basic concepts. You still don't
know how to compute the surface area of a sphere from the metric - you
just keep repeating the mantra about "observed surface area", while
totally missing the point. You still don't understand what a
coordinate transformation is. You still don't understand what the fuck
a tensor is. Funny how you are making the same mistakes literally year
after goddamn year.
>
> > Or do you somehow think
> > writing one coordinate in terms of other coordinates is not a
> > coordinate transformation?
>
> The accusation is broad. Any answer can easily be construed as
> malevolence by you.
How is it "broad" ? You explicitly wrote the coordinate
transformation. You might not want to call it that, but that's what
it is.
What do YOU call transforming a metric from the one set of coordinates
to another if not a coordinate transformation? Do you even know what a
coordinate transformation is?
>
> Thus, I just want you to show me how the following spacetimes ds1 to
> ds2 are the same.
>
> ** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
>
> ** ds2^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2
Use R = r + K for the second set, and you have equality between ds1^2
and ds2^2.
>
> Since both solutions came from the same set of field equations with
> coordinate system chosen right from the very beginning (Christoffel
> symbols), the coordinate system in both is the same.
The coordinates are CLEARLY not the same - see the r vs r+k attached
to dO^2? That's DIFFERENT. As in "not the same".
>
> > [More crap snipped]
>
> More diarrhea coming from Gisse.
As usual, you slip back into the old routine of slinging insults
George Hammond
[Hammond]
Holy Curvature almighty God!!!! How many times do you
have to repeat it?
ds1 is just the plain vanilla Schwarzchild Metric. If you
make the substitution r=r+K simple algebra immediately gives
you ds2 !!!!
PD
> On Jan 5, 1:17 am, Eric Gisse wrote:
>
> > Koobee Wublee wrote:
> > > I have said that many times over. To describe the geometry, one must
> > > choose a coordinate system first and then define how the set of
> > > coordinates is related to the geometry. <shrug>
>
> > The coordinates are not related to the geometry.
>
> > [Gisse’s diarrhea snipped]
>
> To describe any geometry, you must choose a set of coordinate system
> first. Without any coordinate system, you cannot describe any
> geometry. Try to understand this very basic concept.
Good heavens, this sounds like the vapor lock of an aerospace engineer
or something.
To describe the geometry of a wing, you need to choose a coordinate
system and a system of units first, and only then can you write down
the geometry of the wing. For example, you can choose to work in MKS
units where the origin is at the horizontal extreme of the wingtip.
Of course, you can choose a completely different coordinate system and
a completely different system of units to describe the geometry of the
same wing, and of course your description will be completely
different.
But the *geometry of the wing* is completely unchanged in so doing.
Geeawwww, what a nimrod.
PD
Koobee Wublee
> On Jan 5, 12:05 pm, Koobee Wublee wrote:
> > On Jan 5, 1:17 am, Eric Gisse wrote:
> > > Again, how do you know it is asymptotically flat when you didn't
> > > calculate anything?
>
> > Again, you have to understand what asymptotically flat means, mutli-
> > year super-senior. <shrug>
>
> Whether I know it or not does not matter. [More diarrhea snipped]
Whether you know what asymptotically flat means or not, it still
remains your problem. <shrug>
Koobee Wublee
> On Jan 5, 3:06 pm, Koobee Wublee wrote:
> > To describe any geometry, you must choose a set of coordinate system
> > first. Without any coordinate system, you cannot describe any
> > geometry. Try to understand this very basic concept.
>
> Good heavens, this sounds like the vapor lock of an aerospace engineer
> or something.
It sounds like you had all the experience of an aerospace engineer.
Were you one or something?
> To describe the geometry of a wing, you need to choose a coordinate
> system and a system of units first, and only then can you write down
> the geometry of the wing.
Of course, you can also draw a wing like a kindergarten kid which does
not mean jack sh*t. <shrug>
> For example, you can choose to work in MKS
> units where the origin is at the horizontal extreme of the wingtip.
Well, if that is what you want, why not?
> Of course, you can choose a completely different coordinate system and
> a completely different system of units to describe the geometry of the
> same wing,
Yes, that is correct. Since you have brought up MKS, have you not
heard of the English measurement units such as inches, feet, miles,
etc.?
> and of course your description will be completely
> different.
That is correct. For example, 1 inch = 25.4 mm. <shrug>
> But the *geometry of the wing* is completely unchanged in so doing.
That is correct. Again, 1 inch = 25.4mm. <shrug>
> Geeawwww, what a nimrod.
It sounds like the aerospace engineers whom you look down upon know a
lot more about physics than you do. What a nimrod! <shrug>
Koobee Wublee
> Koobee Wublee wrote:
> >To describe any geometry, you must choose a set of coordinate system
> >first. Without any coordinate system, you cannot describe any
> >geometry. Try to understand this very basic concept.
> Holy Curvature almighty God!!!! How many times do you
> have to repeat it?
It looks like the great reverend Hammond has just run out of his
scriptures. Oh, well. It is time for you to check yourself into a
mental institution. <shrug>
> >Thus, I just want you to show me how the following spacetimes ds1 [and]
> >ds2 are the same.
>
> >** ds1^2 = c^2 (1 K / r) dt^2 dr^2 / (1 K / r) r^2 dO^2
>
> >** ds2^2 = c^2 dt^2 / (1 + K / r) (1 + K / r) dr^2 (r + K)^2 dO^2
>
> >Since both solutions came from the same set of field equations with
> >coordinate system chosen right from the very beginning (Christoffel
> >symbols), the coordinate system in both is the same.
> ds1 is just the plain vanilla Schwarzchild Metric.
Well, why is that a surprise for you?
> If you
> make the substitution r=r+K simple algebra immediately gives
> you ds2 !!!!
If (r = r + K) and (K != 0), the (r = 0). It sounds like you have not
passed grade school mathematics. <shrug>
Koobee Wublee
> On Jan 5, 12:06 pm, Koobee Wublee wrote:
> > On Jan 5, 1:17 am, Eric Gisse wrote:
> > To describe any geometry, you must choose a set of coordinate system
> > first. Without any coordinate system, you cannot describe any
> > geometry. Try to understand this very basic concept.
>
> Except the choice of coordinates is irrelevant, [crap snipped]
Nonsense as usual.
> [More diarrhea from Gisse as prophesized]
> > Thus, I just want you to show me how the following spacetimes ds1 to
> > ds2 are the same.
>
> > ** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
>
> > ** ds2^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2
>
> Use R = r + K for the second set, and you have equality between ds1^2
> and ds2^2.
Again, nonsense as usual. R is irrelevant in the above two
equations. <shrug>
> > Since both solutions came from the same set of field equations with
> > coordinate system chosen right from the very beginning (Christoffel
> > symbols), the coordinate system in both is the same.
>
> The coordinates are CLEARLY not the same - see the r vs r+k attached
> to dO^2? That's DIFFERENT. As in "not the same".
Once again, total nonsense on your part, and that is the third strike.
You have to show me how these two geometries are identical before we
continue further. If not, don’t cry about me not justifying your
ignorance, multi-year super-senior.
> [More whining crap snipped]
George Hammond
[Hammond]
Strictly speaking "asymptotically flat" means the 4th
order Riemann curvature tensor goes to zero at r=infinity.
In simple cases this can be determined by inspecting the
metric.... if it approaches the Minkowski metric for r=oo
then the space is asymptotically flat. For instance the
Schwarzchild metric approaches the Minkowski Metric in
spherical coordinates for r-->oo.
Koobee Wublee
> Koobee Wublee wrote:
> >Whether you know what asymptotically flat means or not, it still
> >remains your problem. <shrug>
>
> Strictly speaking "asymptotically flat" means the 4th
> order Riemann curvature tensor goes to zero at r=infinity.
> In simple cases this can be determined by inspecting the
> metric.... if it approaches the Minkowski metric for r=oo
> then the space is asymptotically flat. For instance the
> Schwarzchild metric approaches the Minkowski Metric in
> spherical coordinates for r-->oo.
It actually does not involve the Rieman curature tensor. As long as
the geometry approaches flat space at r = infinity, it is considered
asymptotically flat. Both spacetimes I have described below satisfy
this criterion. <shrug>
George Hammond
<koobee...@gmail.com> wrote:
>On Jan 5, 2:18 pm, George Hammond wrote:
>> Koobee Wublee wrote:
>
>> >To describe any geometry, you must choose a set of coordinate system
>> >first. Without any coordinate system, you cannot describe any
>> >geometry. Try to understand this very basic concept.
>
>> Holy Curvature almighty God!!!! How many times do you
>> have to repeat it?
>
>It looks like the great reverend Hammond has just run out of his
>scriptures. Oh, well. It is time for you to check yourself into a
>mental institution. <shrug>
>
>> >Thus, I just want you to show me how the following spacetimes ds1 [and]
>> >ds2 are the same.
>>
>> >** ds1^2 = c^2 (1 K / r) dt^2 dr^2 / (1 K / r) r^2 dO^2
>>
>> >** ds2^2 = c^2 dt^2 / (1 + K / r) (1 + K / r) dr^2 (r + K)^2 dO^2
>>
>> >Since both solutions came from the same set of field equations with
>> >coordinate system chosen right from the very beginning (Christoffel
>> >symbols), the coordinate system in both is the same.
>
>> ds1 is just the plain vanilla Schwarzchild Metric.
>
>Well, why is that a surprise for you?
>
[Hammond]
Never said it was.
>
>> If you
>> make the substitution r=r+K simple algebra immediately gives
>> you ds2 !!!!
>
>If (r = r + K) and (K != 0), the (r = 0).
>
[Hammond]
I said "SUBSTITUTE" r+K for r, Tricycle.
>
> It sounds like you have not
>passed grade school mathematics. <shrug>
>
[Hammond]
You couldn't pass grammar Pearl Diver, never mind math.
Koobee Wublee
> Koobee Wublee wrote:
> >If (r = r + K) and (K != 0), the (r = 0).
>
> I said "SUBSTITUTE" r+K for r, Tricycle.
No, you can say “SUBSTITUTE” (r + K) for R. <shrug>
In that case, the two said geometries of spacetime are:
** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
** ds2^2 = c^2 (1 – K / R) dt^2 – dR^2 / (1 – K / R) – R^2 dO^2
In that case, they are still very different where (R = r + K).
<shrug>
> > It sounds like you have not
> >passed grade school mathematics. <shrug>
>
> You couldn't pass grammar Pearl Diver, never mind math.
Even my pre-kindergarten kids know R is not the same as r. Either my
kids are geniuses or you are extra-ordinarily mentally challenged.
<shrug>
George Hammond
<koobee...@gmail.com> wrote:
>On Jan 5, 10:37 pm, George Hammond <Nowhe...@notspam.com> wrote:
>> Koobee Wublee wrote:
>
>> >Whether you know what asymptotically flat means or not, it still
>> >remains your problem. <shrug>
>>
>> Strictly speaking "asymptotically flat" means the 4th
>> order Riemann curvature tensor goes to zero at r=infinity.
>> In simple cases this can be determined by inspecting the
>> metric.... if it approaches the Minkowski metric for r=oo
>> then the space is asymptotically flat. For instance the
>> Schwarzchild metric approaches the Minkowski Metric in
>> spherical coordinates for r-->oo.
>
>It actually does not involve the Rieman curature tensor.
>
[Hammond]
Nope... it actually does.
>
> As long as
>the geometry approaches flat space at r = infinity, it is considered
>asymptotically flat.
>
[Hammond]
That's a tautology not a definition.
>
>
> Both spacetimes I have described below satisfy
>this criterion. <shrug>
>
[Hammond]
Of course they do since they are both identically the
Schwarzchild Metric. Shouldn't you be getting back to the
shop?
>
>** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
>
>** ds2^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2
>
[Hammond]
I'll give you a clud Kooby.... really smart people don't
challenge Relativity.... they discuss how Relativity
explains God, even Einstein couldn't figure that out.
Koobee Wublee
> Koobee Wuble wrote:
> >It actually does not involve the Rieman curature tensor.
>
> Nope... it actually does.
No, it does not. Show me where your God said so. <shrug>
> > As long as
> >the geometry approaches flat space at r = infinity, it is considered
> >asymptotically flat.
>
> That's a tautology not a definition.
Hmmm... That is no tautology but a true description of what
“asymptotically flat” means. <shrug>
> > Both spacetimes I have described below satisfy
> >this criterion. <shrug>
>
> Of course they do since they are both identically the
> Schwarzchild Metric.
It proves you have flunked grade school mathematics. <shrug>
> Shouldn't you be getting back to the shop?
After you.
> >** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
>
> >** ds2^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2
>
> I'll give you a clud Kooby....
From a candidate of mental patients, that is going to be interesting.
> really smart people don't challenge Relativity....
Try “really stupid people are incapable of challenging relativities
(both SR and GR)”. <shrug>
> they discuss how Relativity explains God,
Hmmm... Like yourself, a candidate for mental hospitals. <shrug>
> even Einstein couldn't figure that out.
Of course not, Einstein was nobody. Einstein was a nitwit, a
plagiarist, and a lair. Why do you expect Einstein to figure out even
how to match a pair of socks. It would be way too challenging for a
nitwit, a plagiarist, and a liar. <shrug>
George Hammond
<koobee...@gmail.com> wrote:
>On Jan 5, 11:00 pm, George Hammond wrote:
>> Koobee Wublee wrote:
>
>> >If (r = r + K) and (K != 0), the (r = 0).
>>
>> I said "SUBSTITUTE" r+K for r, Tricycle.
>
>No, you can say “SUBSTITUTE” (r + K) for R. <shrug>
>
>In that case, the two said geometries of spacetime are:
>
>** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
>
>** ds2^2 = c^2 (1 – K / R) dt^2 – dR^2 / (1 – K / R) – R^2 dO^2
>
>In that case, they are still very different where (R = r + K).
>
[Hammond]
The COORDINATES are different, but the METRIC is the same.
The metric is the Schwarzchild metric in BOTH cases.
You don't know the difference.
>
<snip crackpot syrup>
Koobee Wublee
> Koobee Wublee wrote:
> >No, you can say “SUBSTITUTE” (r + K) for R. <shrug>
>
> >In that case, the two said geometries of spacetime are:
>
> >** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
>
> >** ds2^2 = c^2 (1 – K / R) dt^2 – dR^2 / (1 – K / R) – R^2 dO^2
>
> >In that case, they are still very different where (R = r + K).
>
> [Hammond]
> The COORDINATES are different, but the METRIC is the same.
No, the coordinates are different:
** (r, theta, phi)
** (R, theta, phi)
Where
** R = r + k
> The metric is the Schwarzchild metric in BOTH cases.
> You don't know the difference.
No, the metrics are also different. This is basic mathematics. Try
to understand mathematics in grade school. <shrug>
> [snipped diarrhea of Einstein Dingleberries]
George Hammond
<koobee...@gmail.com> wrote:
>On Jan 5, 11:33 pm, George Hammond wrote:
>> Koobee Wuble wrote:
>
>It proves you have flunked grade school mathematics. <shrug>
>
[Hammond]
Nope... I have a BS and MS in Physics from 2 accredited
universities in MA and my CV is clearly posted on my
website, URL below.
>
>
>Of course not, Einstein was nobody. Einstein was a nitwit, a
>plagiarist, and a lair. Why do you expect Einstein to figure out even
>how to match a pair of socks. It would be way too challenging for a
>nitwit, a plagiarist, and a liar. <shrug>
>
[Hammond]
hmm.... did you say "Einstein couldn't match a pair of
socks"... you wouldn't be referring to the discovery that
Relativity explains God would you?
I say that because obviously both Relativity and God have
got you confounded judging from the tone of your anti
Einstein rant.
George Hammond
<koobee...@gmail.com> wrote:
>On Jan 5, 11:48 pm, George Hammond wrote:
>> Koobee Wublee wrote:
>
>> >No, you can say “SUBSTITUTE” (r + K) for R. <shrug>
>>
>> >In that case, the two said geometries of spacetime are:
>>
>> >** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
>>
>> >** ds2^2 = c^2 (1 – K / R) dt^2 – dR^2 / (1 – K / R) – R^2 dO^2
>>
>> >In that case, they are still very different where (R = r + K).
>>
>> [Hammond]
>> The COORDINATES are different, but the METRIC is the same.
>
>No, the coordinates are different:
>
[Hammond]
That's what I just said, you're repeating yourself Koobic.
>
>** (r, theta, phi)
>** (R, theta, phi)
>
>Where
>
>** R = r + k
>
>> The metric is the Schwarzchild metric in BOTH cases.
>> You don't know the difference.
>
>No, the metrics are also different.
>
>
[Hammond]
NO.... the metrics are NOT different. They both describe
the SAME physical space with the same metrical distances.
The space described by (r,theta, phi) is IDENTICAL to the
space described by (R,theta,phi) because the METRICS are
identical, irregardless of what K is...a Titleist golf ball
is exactly the same diameter and shape whether it is
described by R or r...in fact both spaces are identically
Schwarchild space. Obviously you have no credentials in
Physics.
=====================================
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
=====================================
This is basic mathematics. Try
Eric Gisse
[snip]
> You have to show me how these two geometries are identical before we
> continue further. If not, don’t cry about me not justifying your
> ignorance, multi-year super-senior.
It has been shown to you repeatedly in the last two and a half years.
However, you simply do not understand. Which is not surprising because
of how little you actually do understand.
>
> > [More whining crap snipped]
>
>
Eric Gisse
> On Jan 5, 10:37 pm, George Hammond <Nowhe...@notspam.com> wrote:
>
> > Koobee Wublee wrote:
> > >Whether you know what asymptotically flat means or not, it still
> > >remains your problem. <shrug>
>
> > Strictly speaking "asymptotically flat" means the 4th
> > order Riemann curvature tensor goes to zero at r=infinity.
> > In simple cases this can be determined by inspecting the
> > metric.... if it approaches the Minkowski metric for r=oo
> > then the space is asymptotically flat. For instance the
> > Schwarzchild metric approaches the Minkowski Metric in
> > spherical coordinates for r-->oo.
>
> It actually does not involve the Rieman curature tensor.
Do you at all find it strange that every time you argue about
something, it is you saying one thing and several other people saying
the same something else?
[snip]
Koobee Wublee
> Koobee Wublee wrote:
> > ** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
>
> > ** ds2^2 = c^2 (1 – K / R) dt^2 – dR^2 / (1 – K / R) – R^2 dO^2
>
> > In that case, they are still very different where (R = r + K).
>
> >** (r, theta, phi)
> >** (R, theta, phi)
>
> >Where
>
> >** R = r + k
> NO.... the metrics are NOT different. They both describe
> the SAME physical space with the same metrical distances.
OK, I will go for that:
** The coordinates are different.
** The metrics are the same.
> The space described by (r,theta, phi) is IDENTICAL to the
> space described by (R,theta,phi) because the METRICS are
> identical, irregardless of what K is...
This is where your logic is failing you. The coordinate systems are
not the same.
For example, we want to measure two displacements of some sort. In
one instance, we use inches, and the other one, we use millimeters.
For analogy, let’s say both instances we measure the same number 10
(the same metric). Since the first instance, the number 10 describes
inches. Therefore, this length is 10 inches. Since the second
instance, the number 10 describes mm. Thus, the other length is 10
mm. Obviously, both lengths are not the same after reporting the same
metric (10 in this example) and different coordinate system (inches
versus mm).
> a Titleist golf ball
> is exactly the same diameter and shape whether it is
> described by R or r...
Of course, the geometry of a golf ball is invariant. <shrug>
> in fact both spaces are identically
> Schwarchild space.
Logically, this cannot be. <shrug>
> Obviously you have no credentials in
> Physics.
This is not about physics but geometry. Obvious, you have not passed
the first course in geometry. Grade school stuff. <shrug>
This shows all Einstein Dingleberries discuss science with faiths.
There is no logic in them. <shrug>
Eric Gisse
[snip]
> > in fact both spaces are identically
> > Schwarchild space.
>
> Logically, this cannot be. <shrug>
Why not?
ds^2 = dR^2 + (R+K)^2 (d\theta^2 + sin^2(\theta) d\phi^2)
ds^2 = dr^2 + r^2 (d\theta^2 + sin^2(\theta) d\phi^2)
Do these or do these not describe the same manifold?
[snip]
PD
> On Jan 6, 12:26 am, George Hammond wrote:
> > NO.... the metrics are NOT different. They both describe
> > the SAME physical space with the same metrical distances.
>
> OK, I will go for that:
>
> ** The coordinates are different.
> ** The metrics are the same.
>
> > The space described by (r,theta, phi) is IDENTICAL to the
> > space described by (R,theta,phi) because the METRICS are
> > identical, irregardless of what K is...
>
> This is where your logic is failing you. The coordinate systems are
> not the same.
>
The above is what I love about KW.
"The metrics are NOT different. They both describe the SAME physical
space with the same metrical distances."
"OK, I will go for that that."
"The space described by (r,theta,phi) is IDENTICAL to the space
described by (R,theta,phi) because the METRICS are identical.."
"This is where your logic is failing you. The coordinate systems are
not the same."
Indeed, a logical failure is at work here.
PD
Eric Gisse
He writes out the transformation, and then screams that it is not a
transformation.
One cannot help but wonder what he is trying to do here.
PD
>
> For example, we want to measure two displacements of some sort. In
> one instance, we use inches, and the other one, we use millimeters.
> For analogy, let’s say both instances we measure the same number 10
> (the same metric). Since the first instance, the number 10 describes
> inches. Therefore, this length is 10 inches. Since the second
> instance, the number 10 describes mm. Thus, the other length is 10
> mm. Obviously, both lengths are not the same after reporting the same
> metric (10 in this example) and different coordinate system (inches
> versus mm).
>
False analog. The proper one is that you want to measure ONE
displacement of some sort. In one coordinate system, that displacement
is 10 inches. In another coordinate system, that displacement is 254
mm. However, despite the different values of coordinates in the two
coordinate systems, it is a *common* geometry being measured.
PD
George Hammond
<koobee...@gmail.com> wrote:
>On Jan 5, 11:48 pm, George Hammond wrote:
>> Koobee Wublee wrote:
>
>> >No, you can say “SUBSTITUTE” (r + K) for R. <shrug>
>>
>> >In that case, the two said geometries of spacetime are:
>>
>> >** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
>>
>> >** ds2^2 = c^2 (1 – K / R) dt^2 – dR^2 / (1 – K / R) – R^2 dO^2
>>
>> >In that case, they are still very different where (R = r + K).
>>
>> [Hammond]
>> The COORDINATES are different, but the METRIC is the same.
>
>No, the coordinates are different:
>
>** (r, theta, phi)
>** (R, theta, phi)
>
>Where
>
>** R = r + k
>
>> The metric is the Schwarzchild metric in BOTH cases.
>> You don't know the difference.
>
>No, the metrics are also different. This is basic mathematics.
>
[Hammond]
You don't know what you're talking about.
Whether you use (r,theta,phi) or (x,y,z) makes absolutely
no difference to the diameter, circumference, area or volume
of a Titleist Golf Ball in Schwarchild Space, Flat Space,
curved space or any other space, because the golf ball does
not change location, orientation or shape by virtue of a
coordinate transformation. A coordinate transformation has
NO PHYSICAL EFFECT on the space being described or the
objects in it.
Neither does it matter whether you use a "Kooby
Coordinate Transformation" R=r+k since the Metric still has
the identical SAME geometric properties and will give the
same diameter, circumference, area and volume of a physical
object in that space!
Again, you don't know what you're talking about.
George Hammond
[Hammond, post script]
While changing the coordinates has absolutely no physical
effect on the golf ball, changing the metric DOES. For
instance if you add a few terms, invert a few terms, square
a few terms etc... you will suddenly notice the golf ball
becoming lop sided, or grow in size, or the dimples on one
side might blow up while becoming smaller on the other side
etc.
changing the coordinates does nothing.... changing the
metric will physically ALTER the shape of the golf ball!