PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
meda
Here's the proof that the Schwarzschild Radius is wrong.
The formula for time dilation in respect to free space,
according to Relativity Theory (RT), and especially
for General Relativity (GR), is [1]:
factor = 1 - GravPotential / c^2
where
GravPotential = G*M/r
Solving this for r so that it gives a singularity
(--> factor=0), ie. for a radius where the gravitational
attraction is so strong that even light cannot escape
(--> this also means for RT that time freezes there):
factor = 1 - (G*M/r) / c^2 = 0
solving this for r gives
r = G*M/c^2
As can be seen this is different from the
Schwarzschild Radius r=2*G*M/c^2
Ergo:
If one believes in Relativity Theory (0 <= v) then the
Schwarzschild Radius r=2*G*M/c^2 is not the correct radius
of a Black Hole.
The correct radius (ie. the event horizon) of a Black Hole
is rather the Gravitational Radius r=G*M/c^2 .
And: if one does not believe in Relativity Theory, and
rather believes in faster-than-light (FTL) speed for gravitation,
then the correct radius of a Black Hole must be r=sqrt(G^2*M).
And the correct formula for gravitational time dilation in respect
to free space is then:
factor = 1 - g * G
where
g=G*M/r^2
References / See also:
[1] Gravitational Radius r=GM/c^2 in Goldstein et.al. p.538, and
[2] Time dilation using Gravitational Potential according to RT & GR:
John G. Cramer: http://www.npl.washington.edu/AV/altvw38.html
meda
>
> PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
>
> Here's the proof that the Schwarzschild Radius is wrong.
>
> The formula for time dilation in respect to free space,
> according to Relativity Theory (RT), and especially
> for General Relativity (GR), is [1]:
> factor = 1 - GravPotential / c^2
> where
> GravPotential = G*M/r
>
> Solving this for r so that it gives a singularity
> (--> factor=0), ie. for a radius where the gravitational
> attraction is so strong that even light cannot escape
> (--> this also means for RT that time freezes there):
> factor = 1 - (G*M/r) / c^2 = 0
> solving this for r gives
> r = G*M/c^2
>
> As can be seen this is different from the
> Schwarzschild Radius r=2*G*M/c^2
>
> Ergo:
> If one believes in Relativity Theory (0 <= v) then the
correction:
If one believes in Relativity Theory (v <= c) then the
Dirk Van de moortel
"meda" <ad...@topkale.net> wrote in message news:1189083986.5...@y42g2000hsy.googlegroups.com...
> PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
What happened to your q-bit soundex?
Dirk Vdm
Androcles
:
: Here's the proof that the Schwarzschild Radius is wrong.
It's wrong, it's wrong, it's all wrong!
You sounds just like q-bit. Funny how "meda" pops into
existence just when "q-bit" vanishes... not related by any chance,
are you? If so, where's the proof he promised last weekend?
Tom Roberts
> PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
You obviously don't have a clue where the Schw. radius comes from. It
comes from a rigorous solution to the Einstein field equation, not your
sound bite approach to physics that uses simplistic formulas that don't
actually apply to the situation you apply them to.
If you are truly interested in how GR is applied to this, it will
require study and effort on your part. A good starting place is:
Geroch, _General_Relativity_from_A_to_B_.
Tom Roberts
meda
> "meda" <a...@topkale.net> wrote
>
> : PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
> :
> : Here's the proof that the Schwarzschild Radius is wrong.
>
> It's wrong, it's wrong, it's all wrong!
> You sounds just like q-bit. Funny how "meda" pops into
> existence just when "q-bit" vanishes... not related by any chance,
> are you? If so, where's the proof he promised last weekend?
I have been "Jimmy Carterized". My previous news host (aioe.org)
has banned me, censored me, because of jewish complaints against me.
You know, jews are so so mighty, they control everything... No joke.
But do they really believe they can silence me? :-)
Regarding the proof: see this thread and and the other
thread "NEW Gravitational Time Dilation Formula".
They both practically prove that Relativity Theory (RT),
and especially Einstein's General Relativity (GR), is wrong.
Dono
[not worth reprinting]
Welcome back, Shit-bot, err, Shit-bit.
Dirk Van de moortel
"meda" <ad...@topkale.net> wrote in message news:1189085637....@22g2000hsm.googlegroups.com...
> On Sep 6, 3:22 pm, "Androcles" <Engin...@hogwarts.physics> wrote:
>> "meda" <a...@topkale.net> wrote
>>
>> : PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
>> :
>> : Here's the proof that the Schwarzschild Radius is wrong.
>>
>> It's wrong, it's wrong, it's all wrong!
>> You sounds just like q-bit. Funny how "meda" pops into
>> existence just when "q-bit" vanishes... not related by any chance,
>> are you? If so, where's the proof he promised last weekend?
>
> I have been "Jimmy Carterized". My previous news host (aioe.org)
> has banned me, censored me, because of jewish complaints against me.
> You know, jews are so so mighty, they control everything... No joke.
> But do they really believe they can silence me? :-)
Silencing trolls (like you - and like Androcles) can only be achieved
by completely ignoring them. This place was speciallay *created*
to support trolls and keep them alive and well by explicitly *not*
ingoring them.
Silencing idiots is the last thing we have in mind.
You hadn't figured that out by now?
We know that Androcles is, by are you *that* stupid as well?
Dirk Vdm
Androcles
: On Sep 6, 3:22 pm, "Androcles" <Engin...@hogwarts.physics> wrote:
: > "meda" <a...@topkale.net> wrote
: >
: > : PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
: > :
: > : Here's the proof that the Schwarzschild Radius is wrong.
: >
: > It's wrong, it's wrong, it's all wrong!
: > You sounds just like q-bit. Funny how "meda" pops into
: > existence just when "q-bit" vanishes... not related by any chance,
: > are you? If so, where's the proof he promised last weekend?
:
: I have been "Jimmy Carterized".
What, you are a peanut farmer now, or the president of the USA?
You should try to get Ronnie Reaganized or Arny Schwarzeneggerized
instead.
: My previous news host (aioe.org)
: has banned me, censored me, because of jewish complaints against me.
: You know, jews are so so mighty, they control everything... No joke.
: But do they really believe they can silence me? :-)
:
: Regarding the proof: see this thread and and the other
: thread "NEW Gravitational Time Dilation Formula".
: They both practically prove that Relativity Theory (RT),
: and especially Einstein's General Relativity (GR), is wrong.
:
It's wrong, I tell you!
WRONG, WRONG, WRONG!
How many other names do you have after you fucked them all up?
gu...@hotmail.com
> PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
>
> Here's the proof that the Schwarzschild Radius is wrong.
>
If critical mass (which also generates an implosion and then fusion +
explosion) occurs before this radius then how can a black hole occur.
gu...@hotmail.com
SperM.hotmail.com> wrote:
>
> Silencing trolls (like you - and like Androcles) can only be achieved
> by completely ignoring them.
.
gu...@hotmail.com
If critical mass (which also generates implosion and then fusion +
explosion) occurs before this radius then how can a black hole occur?
Tom Roberts
> bombs used this radius to determine implosion otherwise there would be
> no fusion?
This is so outrageously incorrect it is "not even wrong".
Tom Roberts
Eric Gisse
<gu...@hotmail.com> wrote:
>On Sep 6, 9:06 am, meda <a...@topkale.net> wrote:
>> PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
>>
>
>It's been a while since I read such documents but I thought Nuclear
>bombs used this radius to determine implosion otherwise there would be
>no fusion?
Observing this kind of stupidity in nature is difficult.
[...]
meda
> On Sep 6, 9:06 am, meda <a...@topkale.net> wrote:
>
> > PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
> > Here's the proof that the Schwarzschild Radius is wrong.
> >
> If critical mass (which also generates an implosion and then fusion +
> explosion) occurs before this radius then how can a black hole occur.
A so called Black Hole is nothing but a dead sun
shrinked to an extreme size and density, ie. the smallest
possible for it's mass. It is IMO a massive body; it is not hollow,
so one cannot fall into it; falling into a BH is just a tale
of some irresponsible and dumb book writers...
(you know, creating some sensation for marketing... :-)
The only difference to a normal sun or planet is
that its gravitational force is so strong that even light
shined from its surface cannot escape it due to its strong "g" force.
Here are data for a calculated Black Hole (cf. the formula below):
M = 5.9673E30 kg (3 times the mass of our Sun)
According to GR (using Schwarzschild Radius method):
R = 8864.050 m
g = 5,067,669,715,454.84 m/s^2
According to my method:
R = 162998.535 m
g = 14,986,673,179.29 m/s^2
As you can see, the gravity force g is so strong
(compare this to our 1g of 9.81 m/s^2 :-) that it leads to
a downward velocity (ie. attraction) even faster than
the speed of light. That's all... :-)
There's nothing else spectacular about BHs, especially
there is no "hollow body" :-) And the singularity (freezeing
of time) happens at the radius, not in the center.
You can verify this if you solve the formula below.
Time in the center of any celestial object has the same value
like that in free space (ie. factor=1).
And "time travel" (ROTFL! :-) is of course an impossible thing,
Wormholes? Forget it, it's all tales! :-)
As said: that's all :-)
Much hot air in science about Black Holes, isn't it? :-)
JanPB
> PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
>
> Here's the proof that the Schwarzschild Radius is wrong.
>
> The formula for time dilation in respect to free space,
> according to Relativity Theory (RT), and especially
> for General Relativity (GR), is [1]:
> factor = 1 - GravPotential / c^2
> where
> GravPotential = G*M/r
No, this is not what general relativity actually says for the very
simple reason that the notion of gravitational potential doesn't even
exist there. (Grav. potential is used only in approximations to
general relativity - these approximations can be used in certain cases
to simplify the calculations.)
Hence your entire argument stops dead right here. You are using a
Hollywood version of relativity, so you are getting a Hollywood
result.
[snip]
--
Jan Bielawski
meda
We both know that I'm right, and you are wrong.
Above the Schwarzschild Radius was proven wrong,
and everything related to it.
If Mr. Schwarzschild back in 1916 used GR to get this
solution, and yes he indeed did, then he undoubtfully used
a wrong tool which gave him this wrong formula.
That simple and logical is the matter.
But you can prove what you say by showing us how
you calculate the gravitational time dilation.
Can you or can you not? That's the question...
gu...@hotmail.com
I did say it's been a while and which is why I had already changed my
post to what's below:
("IF" I remember, a nuclear bomb's implosion occurs at a much less
denser state known as critical mass, not Schwarzchild's radius which
is a more dense state)
gu...@hotmail.com
H-Bomb (hydrogen nuclear bomb). Both Sun and H-Bomb use hydrogen fuel.
Both have Hydrogen fuel implosion, known as radiation implosion, for H-
Bomb this implosion occurs when a certain density is reached known as
critical mass. For Sun's the implosion leading to black holes occurs
when a certain density known as the Schwarzchild's radius is reached.
JanPB
> "JanPB" <film...@gmail.com> wrote
>
>
>
> > On Sep 6, 6:06 am, meda <a...@topkale.net> wrote:
>
> > > PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
>
> > > Here's the proof that the Schwarzschild Radius is wrong.
>
> > > The formula for time dilation in respect to free space,
> > > according to Relativity Theory (RT), and especially
> > > for General Relativity (GR), is [1]:
> > > factor = 1 - GravPotential / c^2
> > > where
> > > GravPotential = G*M/r
>
> > No, this is not what general relativity actually says for the very
> > simple reason that the notion of gravitational potential doesn't even
> > exist there. (Grav. potential is used only in approximations to
> > general relativity - these approximations can be used in certain cases
> > to simplify the calculations.)
>
> > Hence your entire argument stops dead right here. You are using a
> > Hollywood version of relativity, so you are getting a Hollywood
> > result.
>
> We both know that I'm right, and you are wrong.
I can prove I'm right.
> Above the Schwarzschild Radius was proven wrong,
> and everything related to it.
Your proof is incorrect.
> If Mr. Schwarzschild back in 1916 used GR
It was late 1915 actually.
> to get this
> solution, and yes he indeed did, then he undoubtfully used
> a wrong tool which gave him this wrong formula.
No, he just solved the Einstein equation using a particular coordinate
system. One could use slightly more elegant coordinates but this is
all gravy - point is Schwarzschild got it right. This claim, being a
claim about a mathematical argument, can be verified.
> That simple and logical is the matter.
>
> But you can prove what you say by showing us how
> you calculate the gravitational time dilation.
> Can you or can you not? That's the question...
I went through the derivation of Schwarzschild's solution quite
carefully (much more carefully than GR textbooks), tell me where I'm
wrong in it:
http://www.mastersofcinema.org/jan/sch.pdf
--
Jan Bielawski
Androcles
"JanPB" <fil...@gmail.com> wrote in message
news:1189197516.0...@57g2000hsv.googlegroups.com...
: I went through the derivation of Schwarzschild's solution quite
: carefully
As carefully as you went through Einstein's 1905 paper, or as
carefully as you help change light bulbs?
'we establish by definition that the "time" required by
light to travel from A to B doesn't equal the "time" it requires
to travel from B to A in the stationary system, obviously.' --
Heretic Jan Bielawski, assistant light-bulb changer.
JanPB
> "JanPB" <film...@gmail.com> wrote in message
Cut the rhetoric. Read through the PDF, if you find a mistake - post
it.
--
Jan Bielawski
Androcles
I've already found the mistake that you missed.
The cuckoo malformations are derived from the silly rhetoric
'we establish by definition that the "time" required by
light to travel from A to B equals the "time" it requires
to travel from B to A' because I SAY SO and you have to
agree because I'm the great genius, STOOOPID, don't you
dare question it. -- Rabbi Albert Einstein
Your boastful bullshit that you've ever checked anything carefully
is just your arrogant braggadocio and rhetoric, which is obvious,
obviously.
JanPB
Irrelevant nonsense.
--
Jan Bielawski
Androcles
That's right, Schwarzschild's shit is irrelevant nonsense but you
have no way of checking it, you don't even know how Einstein
hoodwinked you, obviously. You are just a simple-minded troll
with nothing better to do than simple denial. Get back to changing
light-bulbs, your fellow Poles need you.
JanPB
Irrelevant nonsense.
--
Jan Bielawski
Androcles
That's right, Schwarzschild's shit is irrelevant nonsense but you
have no way of checking it, you don't even know how Einstein
hoodwinked you, obviously. You are just a simple-minded troll
with nothing better to do than simple denial. Get back to changing
light-bulbs, your fellow Poles need you, drooling fuckin' idiot.
JanPB
It's all irrelevant nonsense. Reread meda's objections.
--
Jan Bielawski
Androcles
meda aka q-bit aka a dozen other names is jerking your chain.
"Black holes" are irrelevant nonsense, they are a theoretical
construction that doesn't exist in Nature and the crackpot
irrelevant nonsense "theory" is GR.
This is REAL physics as opposed to your head-up-arse theory crap:
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac.htm
http://www.androcles01.pwp.blueyonder.co.uk/Algol/Algol.htm
Daryl McCullough
[From Androcles]
>> : > I've already found the mistake that you missed.
>> : > The cuckoo malformations are derived from the silly rhetoric
>> : > 'we establish by definition that the "time" required by
>> : > light to travel from A to B equals the "time" it requires
>> : > to travel from B to A'
As far as I can tell, Androcles has wasted *decades* of his life
because he misunderstood a single line of a paper of Einstein's.
What a normal, non-insane person would do is to ask for *help*.
The misunderstanding could have been explained to him in 15 minutes.
But for some reason, he'd prefer to spend his entire life raging
against the line that he could never understand.
Very sad.
--
Daryl McCullough
Ithaca, NY
JanPB
That's not the point. The point is he/she/it said Schwarzschild's
radius formula was wrong to which I responded with a derivation of
Schwarzschild's solution. Then you jump in with something completely
irrelevant from SR.
--
Jan Bielawski
Koobee Wublee
> I went through the derivation of Schwarzschild's solution quite
> carefully (much more carefully than GR textbooks), tell me where I'm
> wrong in it:http://www.mastersofcinema.org/jan/sch.pdf
What do you mean by the '/\' operator?
Eric Gisse
Try opening a GR textbook.
Androcles
Irrelevant nonsense.
Androcles
"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:fbt61...@drn.newsguy.com...
: JanPB says...
:
: [From Androcles]
:
: >> : > I've already found the mistake that you missed.
: >> : > The cuckoo malformations are derived from the silly rhetoric
: >> : > 'we establish by definition that the "time" required by
: >> : > light to travel from A to B equals the "time" it requires
: >> : > to travel from B to A'
:
As far as McCullough can tell, McCullough has wasted *decades*
of his life because he didn't bother to read a single line of a
paper of Einstein's upon which the ridiculous cuckoo malformations
are based.
Pathetic.
--
'we establish by definition that the "time" required by
light to travel from A to B equals the "time" it requires
to travel from B to A' because I SAY SO and you have to
agree because I'm the great genius, STOOOPID, don't you
dare question it. -- Rabbi Albert Einstein
http://www.androcles01.pwp.blueyonder.co.uk/Smart/tAB=tBA.gif
'we establish by definition that the "time" required by
light to travel from A to B doesn't equal the "time" it requires
to travel from B to A in the stationary system, obviously.' --
Heretic Jan Bielawski, assistant light-bulb changer.
Ref: news:1188363019....@k79g2000hse.googlegroups.com
"SR is GR with G=0." -- Uncle Stooopid.
The Uncle Stooopid doctrine:
http://sound.westhost.com/counterfeit.jpg
"What can be asserted without evidence can also be dismissed without
evidence." -- Uncle Stooopid.
"Counterfactual assumptions yield nonsense.
If such a thing were actually observed, reliably and reproducibly, then
relativity would immediately need a major overhaul if not a complete
replacement." -- Humpty Roberts.
Rabbi Albert Einstein in 1895 failed an examination that would
have allowed him to study for a diploma as an electrical engineer
at the Eidgenössische Technische Hochschule in Zurich
(couldn't even pass the SATs).
According to Phuckwit Duck it was geography and history that Einstein
failed on, as if Eidgenössische Technische Hochschule would give a
damn. That tells you the lengths these lying bastards will go to to
protect their tin god, but its always a laugh when they slip up.
Trolls, the lot of them.
"This is PHYSICS, not math or logic, and "proof" is completely
irrelevant." -- Humpty Roberts.
meda
> On Sep 6, 5:17 pm, meda <a...@topkale.net> wrote:
> > "JanPB" <film...@gmail.com> wrote
> > > On Sep 6, 6:06 am, meda <a...@topkale.net> wrote:
> >
> > > > PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
> >
> > > > Here's the proof that the Schwarzschild Radius is wrong.
> >
> > > > The formula for time dilation in respect to free space,
> > > > according to Relativity Theory (RT), and especially
> > > > for General Relativity (GR), is [1]:
> > > > factor = 1 - GravPotential / c^2
> > > > where
> > > > GravPotential = G*M/r
> >
> > > No, this is not what general relativity actually says for the very
> > > simple reason that the notion of gravitational potential doesn't even
> > > exist there. (Grav. potential is used only in approximations to
> > > general relativity - these approximations can be used in certain cases
> > > to simplify the calculations.)
> >
> > > Hence your entire argument stops dead right here. You are using a
> > > Hollywood version of relativity, so you are getting a Hollywood
> > > result.
> >
> > We both know that I'm right, and you are wrong.
>
> I can prove I'm right.
>
> > Above the Schwarzschild Radius was proven wrong,
> > and everything related to it.
>
> Your proof is incorrect.
You are not correct because my given proof is perfect.
> > If Mr. Schwarzschild back in 1916 used GR
>
> It was late 1915 actually.
It was 1916 !
> > to get this
> > solution, and yes he indeed did, then he undoubtfully used
> > a wrong tool which gave him this wrong formula.
>
> No, he just solved the Einstein equation using a particular coordinate
> system. One could use slightly more elegant coordinates but this is
> all gravy - point is Schwarzschild got it right.
LOL! :-) I have proved him wrong :-)
> This claim, being a claim about a mathematical argument, can be verified.
My proof can be verified too, and it proves that Schwarzschild Radius
is the wrong radius.
> > That simple and logical is the matter.
> >
> > But you can prove what you say by showing us how
> > you calculate the gravitational time dilation.
> > Can you or can you not? That's the question...
>
> I went through the derivation of Schwarzschild's solution quite
> carefully (much more carefully than GR textbooks), tell me where I'm
> wrong in it:
> http://www.mastersofcinema.org/jan/sch.pdf
> --
> Jan Bielawski
I wonder if you are blind?
The proof that the Schwarzschild Radius is wrong is so simple,
as was shown above. So, your 9 pages of maths is needless and useless,
you just use the usual blending tactics of "high maths" to cover the problem...
These tricks of the Relativists no longer work, boy!
And: some people have asked you questions about your math
and you still have not answered them...
meda
> On Sep 6, 3:06 pm, meda wrote:
> >
> > PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
> >
> > Here's the proof that the Schwarzschild Radius is wrong.
> >
> > The formula for time dilation in respect to free space,
> > according to Relativity Theory (RT), and especially
> > for General Relativity (GR), is [1]:
> > factor = 1 - GravPotential / c^2
> > where
> > GravPotential = G*M/r
> >
> > (--> factor=0), ie. for a radius where the gravitational
> > attraction is so strong that even light cannot escape
> > (--> this also means for RT that time freezes there):
> > factor = 1 - (G*M/r) / c^2 = 0
> > solving this for r gives
> > r = G*M/c^2
> >
> > As can be seen this is different from the
> > Schwarzschild Radius r=2*G*M/c^2
> >
> > Ergo:
> > Schwarzschild Radius r=2*G*M/c^2 is not the correct radius
> > of a Black Hole.
> > The correct radius (ie. the event horizon) of a Black Hole
> > is rather the Gravitational Radius r=G*M/c^2 .
> >
> > And: if one does not believe in Relativity Theory, and
> > rather believes in faster-than-light (FTL) speed for gravitation,
> > then the correct radius of a Black Hole must be r=sqrt(G^2*M).
> > And the correct formula for gravitational time dilation in respect
> > to free space is then:
> > factor = 1 - g * G
> > where
> > g=G*M/r^2
> >
> > References / See also:
> > [1] Gravitational Radius r=GM/c^2 in Goldstein et.al. p.538, and
> > [2] Time dilation using Gravitational Potential according to RT & GR:
> > John G. Cramer: http://www.npl.washington.edu/AV/altvw38.html
As a supporting argument for the above said see also this,
written by John G. Cramer, a Relativist (see link above):
|>
|> However, general relativity does provide us with another way
|> of thinking about the twin paradox. According to general relativity,
|> a clock runs slower in a gravitational potential V by a factor of (1-V/c2).
|> This means that a clock on the surface of the Earth runs slower than
|> an identical clock in gravity free space by a factor of (1 - 6.60 × 10^-10).
|> In one year of running, the earth-based clock would lose 20 milliseconds.
One can find dozens of examples where Relativity used
this formula for gravitational time dilation.
As was clearly demonstrated above, there are even
two good reasons why the Schwarzschild Radius
as the event horizon of a Black Hole is wrong.
I have given even two replacements: one for the Relativists,
and one for the Non-Relativists, both better than the Schwarzschild Radius.
And the latter one is obviously the more correct one of all these three.
Daryl McCullough
>
>"JanPB" <fil...@gmail.com> wrote
>> No, he just solved the Einstein equation using a particular coordinate
>> system. One could use slightly more elegant coordinates but this is
>> all gravy - point is Schwarzschild got it right.
>
>LOL! :-) I have proved him wrong :-)
Yes, you've presented something that you consider a proof.
But the problem is that you are a complete idiot. You have
no idea what you are talking about. You don't have the competence
to do any nontrivial physics problem. So your proof doesn't
mean anything at all.
If you don't know any mathematics, then you can't do a mathematical
proof. If you don't know any physics, then you can't make a coherent
argument about physics. Learn some math. Learn some physics.
Androcles
: meda says...
McCullough speaks from experience, he's another troll
that doesn't know any math or physics either.
T.M. Sommers
> "JanPB" <fil...@gmail.com> wrote
>>On Sep 6, 5:17 pm, meda <a...@topkale.net> wrote:
>>
>>>If Mr. Schwarzschild back in 1916 used GR
>>
>>It was late 1915 actually.
>
> It was 1916 !
The work was done in 1915, but not published until 1916.
--
Thomas M. Sommers -- t...@nj.net -- AB2SB
T.M. Sommers
>
> Pathetic.
You certainly are.
Koobee Wublee
> meda wrote:
> > It was 1916 !
>
> The work was done in 1915, but not published until 1916.
Although this is a minor point in history, nevertheless it is still
important to keep things straight. So, let's see.
Hilbert orally presented the field equations on 11/20/15 at
Goettingen. Five days later on 11/25/15, Einstein did his oral
presentation in Berlin. In the meantime, Schwarzschild was already
sent to the eastern front as an officer trying to predict weather of
some sort. The written version of either presentation was not
available until the end of December. Are you serious suggesting
Schwarzschild only spent a few days deriving his original solution? I
do find that highly improbable.
Also, Schwarzschild's original solution is orders of magnitude more
complex than the Schwarzschild metric. It was Hilbert who discovered
the Schwarzschild metric after Schwarzschild's death due to illness
unrelated to combat.
So, again, Schwarzschild never discovered the Schwarzschild metric.
Instead in 1916, he discovered another solution that is also static,
asymptotically flat, and spherically symmetric, but Schwarzschild's
solution does not manifest black holes.
However, I do think it is very highly probable that Hilbert had
already had the Schwarzschild metric in 1915. In fact, he might had
the field equations already while Einstein was celebrating his
understanding of Newtonian gravity by rediscovering the principle of
equivalence.
T.M. Sommers
> On Sep 8, 12:01 pm, "T.M. Sommers" wrote:
>>meda wrote:
>
>>>It was 1916 !
>>
>>The work was done in 1915, but not published until 1916.
>
> Although this is a minor point in history, nevertheless it is still
> important to keep things straight. So, let's see.
>
> Hilbert orally presented the field equations on 11/20/15 at
> Goettingen. Five days later on 11/25/15, Einstein did his oral
> presentation in Berlin. In the meantime, Schwarzschild was already
> sent to the eastern front as an officer trying to predict weather of
> some sort. The written version of either presentation was not
> available until the end of December. Are you serious suggesting
> Schwarzschild only spent a few days deriving his original solution?
A few weeks, not just a few days. He evidently received an
account of one or both of the talks soon after they happened. He
submitted his papers before the end of the year.
> I
> do find that highly improbable.
Well, he was smarter than you. Or me.
JanPB
The wedge product, i.e. antisymmetrisation of the tensor product. For
1-forms a and b it amounts to:
a /\ b = a x b - b x a
...where "x" is tensor product (there are several unimportant
conventions for the multiplicative factor in front of the RHS - here
it's 1).
--
Jan Bielawski
JanPB
> On Sep 8, 12:01 pm, "T.M. Sommers" wrote:
>
> > meda wrote:
> > > It was 1916 !
>
> > The work was done in 1915, but not published until 1916.
>
> Although this is a minor point in history, nevertheless it is still
> important to keep things straight. So, let's see.
>
> Hilbert orally presented the field equations on 11/20/15 at
> Goettingen. Five days later on 11/25/15, Einstein did his oral
> presentation in Berlin. In the meantime, Schwarzschild was already
> sent to the eastern front as an officer trying to predict weather of
> some sort. The written version of either presentation was not
> available until the end of December. Are you serious suggesting
> Schwarzschild only spent a few days deriving his original solution? I
> do find that highly improbable.
You must be missing some part of the story because Schwarzschild's
letter to Einstein communicating his solution is dated 22 December
1915. He must have heard of the presentations soon after they took
place, obviously.
> Also, Schwarzschild's original solution is orders of magnitude more
> complex than the Schwarzschild metric.
He used a coordinate system which is uncommonly used today.
> It was Hilbert who discovered
> the Schwarzschild metric after Schwarzschild's death due to illness
> unrelated to combat.
No, Schwarzschild wrote down what we would consider the standard form
of the solution already in his first letter to Einstein in 1915.
> So, again, Schwarzschild never discovered the Schwarzschild metric.
> Instead in 1916, he discovered another solution that is also static,
> asymptotically flat, and spherically symmetric, but Schwarzschild's
> solution does not manifest black holes.
We've been through this before. What you wrote above is incorrect.
There can be only one solution in the spherically symmetric vacuum
case in principle, so Schwarzschild's is the one. (That "another
solution" you mention is the standard solution merely rewritten in
different coordinates and its "not manifesting" the singularity is
only due to restricting the domain).
> However, I do think it is very highly probable that Hilbert had
> already had the Schwarzschild metric in 1915. In fact, he might had
> the field equations already while Einstein was celebrating his
> understanding of Newtonian gravity by rediscovering the principle of
> equivalence.
Read the research based on Einstein's and Hilbert's archives. The
story is there, no need to guess (wrongly).
--
Jan Bielawski
JanPB
You have no proof, just some meaningless formula salad.
> > > If Mr. Schwarzschild back in 1916 used GR
>
> > It was late 1915 actually.
>
> It was 1916 !
Schwarzschild's letter to Einstein containing the well-known solution
is dated 22 December 1915. You can read it here:
http://www.mastersofcinema.org/jan/tt.pdf
You can see the famous "Linienelement" smack in the middle of the
page.
> > > to get this
> > > solution, and yes he indeed did, then he undoubtfully used
> > > a wrong tool which gave him this wrong formula.
>
> > No, he just solved the Einstein equation using a particular coordinate
> > system. One could use slightly more elegant coordinates but this is
> > all gravy - point is Schwarzschild got it right.
>
> LOL! :-) I have proved him wrong :-)
You say you proved him wrong. That's a different thing :-)
> > This claim, being a claim about a mathematical argument, can be verified.
>
> My proof can be verified too, and it proves that Schwarzschild Radius
> is the wrong radius.
Your proof is wrong - it assumes a falsehood, for a start
(gravitational potential doesn't exist in GR).
> > > That simple and logical is the matter.
>
> > > But you can prove what you say by showing us how
> > > you calculate the gravitational time dilation.
> > > Can you or can you not? That's the question...
>
> > I went through the derivation of Schwarzschild's solution quite
> > carefully (much more carefully than GR textbooks), tell me where I'm
> > wrong in it:
> >http://www.mastersofcinema.org/jan/sch.pdf
> > --
> > Jan Bielawski
>
> I wonder if you are blind?
> The proof that the Schwarzschild Radius is wrong is so simple,
> as was shown above. So, your 9 pages of maths is needless and useless,
> you just use the usual blending tactics of "high maths" to cover the problem...
> These tricks of the Relativists no longer work, boy!
> And: some people have asked you questions about your math
> and you still have not answered them...
Words, words, words. Anybody can say anything. Where is your proof?
You have none.
--
Jan Bielawski
Koobee Wublee
> On Sep 8, 2:38 pm, Koobee Wublee wrote:
> > Hilbert orally presented the field equations on 11/20/15 at
> > Goettingen. Five days later on 11/25/15, Einstein did his oral
> > presentation in Berlin. In the meantime, Schwarzschild was already
> > sent to the eastern front as an officer trying to predict weather of
> > some sort. The written version of either presentation was not
> > available until the end of December. Are you serious suggesting
> > Schwarzschild only spent a few days deriving his original solution? I
> > do find that highly improbable.
>
> You must be missing some part of the story because Schwarzschild's
> letter to Einstein communicating his solution is dated 22 December
> 1915. He must have heard of the presentations soon after they took
> place, obviously.
Yes, in that letter, there is no mentioning of the field equations
being solved. It was just a congratulatory letter to Einstein. Why
did Schwarzschild do that? Because earlier while Einstein was
struggling to understand Newtonian law of gravity Schwarzschild,
Hilbert, Klein, and Minkowski were already playing with curvature in
spacetime. What they did not know was how to derive the metric or the
solution to the field equations.
> > Also, Schwarzschild's original solution is orders of magnitude more
> > complex than the Schwarzschild metric.
>
> He used a coordinate system which is uncommonly used today.
No, his choice of coordinate system when he started the derivation was
still the common spherically symmetric polar coordinate system.
In order to reduce the mathematical complexity of working with the
Ricci tensor, he chose to transform his spherically symmetric polar
coordinate system into one that allows the metric tensor to have a
determinant of -1. In doing so, the complexity in the Ricci tensor is
marginally reduced. Indeed, I do acknowledge Schwarzschild's genius
in this move.
After solved the Ricci tensor in free space, Schwarzschild had to
transform the coordinate back to the spherically symmetric polar
coordinate system. This original solution of Schwarzschild is then
very different from the Schwarzschild metric.
> > It was Hilbert who discovered
> > the Schwarzschild metric after Schwarzschild's death due to illness
> > unrelated to combat.
>
> No, Schwarzschild wrote down what we would consider the standard form
> of the solution already in his first letter to Einstein in 1915.
All historic records are against you. We have been through this
before.
> > So, again, Schwarzschild never discovered the Schwarzschild metric.
> > Instead in 1916, he discovered another solution that is also static,
> > asymptotically flat, and spherically symmetric, but Schwarzschild's
> > solution does not manifest black holes.
>
> We've been through this before. What you wrote above is incorrect.
> There can be only one solution in the spherically symmetric vacuum
> case in principle, so Schwarzschild's is the one. (That "another
> solution" you mention is the standard solution merely rewritten in
> different coordinates and its "not manifesting" the singularity is
> only due to restricting the domain).
This is not true. The static, asymptotically flat, and spherically
symmetric solutions to the field equations in general take one such
form below.
ds^2 = c^2 dt^2 / (1 + K / R) - (1 + K / R) (dR/dr)^2 dr^2 - (R + K)^2
dO^2
Where
** R(r) = Function of r
** dO^2 = cos^2Phi dTheta^2 + dPhi^2
** K = Integration constant
** dr, dTheta, dPhi = Choiced coordinate system
Only if the following is true, you get the Schwarzschild metric.
** R = r - K
Where
** K = 2 G M / c^2
If the following is true, you get Schwarzschild's original solution
which does not manifest any black holes.
** R = (r^3 + K^3)^(1/3)
If the following is true, you get another solution just as simple as
the Schwarzschild metric but without manifestation of black holes.
** R = r
If the following is true, you get a constantly expanding universe that
also obeys the Schwarzschild metric --- a trait that even the FLRW
metric fails to do so.
** R = r / (1 + r^2 / K / L)
Where
** L = Cosmic constant
If the following is true, you get an accelerated expanding universe.
** R = r / (1 + r^2 / K / L + r^3 / K / L / N)
Where
** L, N = Cosmic constants
Each of these solutions is uniquely independent of the others.
Claiming these solutions being the same is utter nonsense --- a
misunderstanding on your part of failure to understand the metric is
not a tensor but merely a matrix. In addition, the last two metrics
prove the Birkhoff's theorem wrong.
> > However, I do think it is very highly probable that Hilbert had
> > already had the Schwarzschild metric in 1915. In fact, he might had
> > the field equations already while Einstein was celebrating his
> > understanding of Newtonian gravity by rediscovering the principle of
> > equivalence.
>
> Read the research based on Einstein's and Hilbert's archives. The
> story is there, no need to guess (wrongly).
Yes, the archives are all available. You don't have to lie about the
history. <shrug>
JanPB
I'm not going to comment on your posting besides stating that almost
all of it is incorrect (both the mathematics and the history). We've
been through this before and there is little point in repeating it.
More detail on Schwarzschild's derivation is found here (for those who
haven't been around at the previous iteration of Koobee's nonsensical
claims):
http://groups.google.com/group/sci.physics.relativity/msg/068e4e33d3fecf0d
--
Jan Bielawski
meda
> On Sep 8, 1:58 am, "meda" <m...@topkale.net> wrote:
> > "JanPB" <film...@gmail.com> wrote
> > > On Sep 6, 5:17 pm, meda <a...@topkale.net> wrote:
> > > > "JanPB" <film...@gmail.com> wrote
> > > > > On Sep 6, 6:06 am, meda <a...@topkale.net> wrote:
> >
> > > > > > PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
> >
> > > > > > Here's the proof that the Schwarzschild Radius is wrong.
> >
> > > > > > The formula for time dilation in respect to free space,
> > > > > > according to Relativity Theory (RT), and especially
> > > > > > for General Relativity (GR), is [1]:
> > > > > > factor = 1 - GravPotential / c^2
> > > > > > where
> > > > > > GravPotential = G*M/r
> >
> > > > > No, this is not what general relativity actually says for the very
> > > > > simple reason that the notion of gravitational potential doesn't even
> > > > > exist there. (Grav. potential is used only in approximations to
> > > > > general relativity - these approximations can be used in certain cases
> > > > > to simplify the calculations.)
You are lying. Here's the proof of your lying, written by Relativist John G. Cramer himself.
It shows that Gravitational Potential is very well used in GR. So you are lying.
| Time dilation using Gravitational Potential according to RT & GR:
| John G. Cramer: http://www.npl.washington.edu/AV/altvw38.html
| However, general relativity does provide us with another way
| of thinking about the twin paradox. According to general relativity,
| a clock runs slower in a gravitational potential V by a factor of (1-V/c^2).
| This means that a clock on the surface of the Earth runs slower than
| an identical clock in gravity free space by a factor of (1 - 6.60 × 10^-10).
| In one year of running, the earth-based clock would lose 20 milliseconds.
Here gravitational potential V of course means GM/r .
> > > > > Hence your entire argument stops dead right here. You are using a
> > > > > Hollywood version of relativity, so you are getting a Hollywood
> > > > > result.
> >
> > > > We both know that I'm right, and you are wrong.
> >
> > > I can prove I'm right.
> >
> > > > Above the Schwarzschild Radius was proven wrong,
> > > > and everything related to it.
> >
> > > Your proof is incorrect.
> >
> > You are not correct because my given proof is perfect.
>
> You have no proof, just some meaningless formula salad.
BS. Your 9 pages of math is meaningless formula salad,
not my proof. My proof is simple and verifyable by everybody.
> > > > If Mr. Schwarzschild back in 1916 used GR
> >
> > > It was late 1915 actually.
> >
> > It was 1916 !
>
> Schwarzschild's letter to Einstein containing the well-known solution
> is dated 22 December 1915. You can read it here:
> http://www.mastersofcinema.org/jan/tt.pdf
There is nothing about his formula Rs=2GM/c^2.
This letter is about the precession of the perihelion of Mercury.
I know German, and in this letter he even admits that his
method is mere a curve fitting found by trial & error, nothing exact.
And see the following. Also here it is said that his solution is "physically incorrect":
http://de.wikipedia.org/wiki/Schwarzschild-Radius#Nichtrelativistische_Betrachtungen_zum_Schwarzschild-Radius
| "Da diese Herleitung nichtrelativistisch ist, ist sie jedoch
| physikalisch nicht korrekt. Bei der Herleitung der Schwarzschild-Metrik
| betrachtet man die Kugel theoretisch aus unendlicher Entfernung,
| also aus einem Gebiet mit flacher Raumzeit, damit man die Gleichungen
| durch eine newtonsche Näherung vereinfachen kann. Dass die Ergebnisse
| übereinstimmen ist demnach nur teilweise zufällig."
> You can see the famous "Linienelement" smack in the middle of the page.
I doubt you know German, because this letter as a whole has nothing
todo with the Schwarzschild Radius nor has this word "Linienelement"
anything todo related to it.
> > > > to get this
> > > > solution, and yes he indeed did, then he undoubtfully used
> > > > a wrong tool which gave him this wrong formula.
> >
> > > No, he just solved the Einstein equation using a particular coordinate
> > > system. One could use slightly more elegant coordinates but this is
> > > all gravy - point is Schwarzschild got it right.
> >
> > LOL! :-) I have proved him wrong :-)
>
> You say you proved him wrong. That's a different thing :-)
>
> > > This claim, being a claim about a mathematical argument, can be verified.
> >
> > My proof can be verified too, and it proves that Schwarzschild Radius
> > is the wrong radius.
>
> Your proof is wrong - it assumes a falsehood, for a start
> (gravitational potential doesn't exist in GR).
And? It doesn't matter whether it exists in GR or not.
The goal is to determine such a radius where even
light cannot escape from the surface due to the strength of the gravitation.
GR is just a tool, much like NM or any other, and as was
demonstrated by me, the result obtained with
GR is not the correct result. Q.E.D.
> > > > That simple and logical is the matter.
> >
> > > > But you can prove what you say by showing us how
> > > > you calculate the gravitational time dilation.
> > > > Can you or can you not? That's the question...
> >
> > > I went through the derivation of Schwarzschild's solution quite
> > > carefully (much more carefully than GR textbooks), tell me where I'm
> > > wrong in it:
> > >http://www.mastersofcinema.org/jan/sch.pdf
> >
> > I wonder if you are blind?
> > The proof that the Schwarzschild Radius is wrong is so simple,
> > as was shown above. So, your 9 pages of maths is needless and useless,
> > you just use the usual blending tactics of "high maths" to cover the problem...
> > These tricks of the Relativists no longer work, boy!
> > And: some people have asked you questions about your math
> > and you still have not answered them...
>
> Words, words, words. Anybody can say anything. Where is your proof?
> You have none.
Like I said: you must be blind, my proof has been given in this thread.
JanPB
> "JanPB" <film...@gmail.com> wrote
>
>
>
> > On Sep 8, 1:58 am, "meda" <m...@topkale.net> wrote:
> > > "JanPB" <film...@gmail.com> wrote
> > > > On Sep 6, 5:17 pm, meda <a...@topkale.net> wrote:
> > > > > "JanPB" <film...@gmail.com> wrote
> > > > > > On Sep 6, 6:06 am, meda <a...@topkale.net> wrote:
>
> > > > > > > PROOF: Schwarzschild Radius r=2*G*M/c^2 is wrong
>
> > > > > > > Here's the proof that the Schwarzschild Radius is wrong.
>
> > > > > > > The formula for time dilation in respect to free space,
> > > > > > > according to Relativity Theory (RT), and especially
> > > > > > > for General Relativity (GR), is [1]:
> > > > > > > factor = 1 - GravPotential / c^2
> > > > > > > where
> > > > > > > GravPotential = G*M/r
>
> > > > > > No, this is not what general relativity actually says for the very
> > > > > > simple reason that the notion of gravitational potential doesn't even
> > > > > > exist there. (Grav. potential is used only in approximations to
> > > > > > general relativity - these approximations can be used in certain cases
> > > > > > to simplify the calculations.)
>
> You are lying. Here's the proof of your lying, written by Relativist John G. Cramer himself.
> It shows that Gravitational Potential is very well used in GR. So you are lying.
Gravitational potential is "used" in GR only in approximate or very
specific situations. You haven't provided any justification for its
use. In principle, GR uses spacetime metric which is a 2-tensor not
derivable from a scalar.
[snip]
> > > > > > Hence your entire argument stops dead right here. You are using a
> > > > > > Hollywood version of relativity, so you are getting a Hollywood
> > > > > > result.
>
> > > > > We both know that I'm right, and you are wrong.
>
> > > > I can prove I'm right.
>
> > > > > Above the Schwarzschild Radius was proven wrong,
> > > > > and everything related to it.
>
> > > > Your proof is incorrect.
>
> > > You are not correct because my given proof is perfect.
>
> > You have no proof, just some meaningless formula salad.
>
> BS. Your 9 pages of math is meaningless formula salad,
> not my proof. My proof is simple and verifyable by everybody.
These are just words. Find a mistake in my derivation - otherwise you
have no argument.
> > > > > If Mr. Schwarzschild back in 1916 used GR
>
> > > > It was late 1915 actually.
>
> > > It was 1916 !
>
> > Schwarzschild's letter to Einstein containing the well-known solution
> > is dated 22 December 1915. You can read it here:
> >http://www.mastersofcinema.org/jan/tt.pdf
>
> There is nothing about his formula Rs=2GM/c^2.
There is his solution there from which the formula for Rs follows.
> This letter is about the precession of the perihelion of Mercury.
> I know German, and in this letter he even admits that his
> method is mere a curve fitting found by trial & error, nothing exact.
I suggest you improve either your German or your understanding - he
says the opposite. The bits about approximation are about his initial
unsuccessful attempt and about Einstein's inexact method for the
calculation of Mercury perihelion precession, as opposed to his exact
solution which yields numerically an almost identical result since -
as he points out - using his (Schwarzschild's) solution amounts to
using an only slightly altered Einstein's formula.
> And see the following. Also here it is said that his solution is "physically incorrect":http://de.wikipedia.org/wiki/Schwarzschild-Radius#Nichtrelativistisch...
> | "Da diese Herleitung nichtrelativistisch ist, ist sie jedoch
> | physikalisch nicht korrekt. Bei der Herleitung der Schwarzschild-Metrik
> | betrachtet man die Kugel theoretisch aus unendlicher Entfernung,
> | also aus einem Gebiet mit flacher Raumzeit, damit man die Gleichungen
> | durch eine newtonsche Näherung vereinfachen kann. Dass die Ergebnisse
> | übereinstimmen ist demnach nur teilweise zufällig."
Either your German or your reading comprehension is completely out to
lunch. This paragraph concerns derivation of Schwarzschild's radius
formula by _nonrelativistic, Newtonian_ approach: setting escape
velocity to the speed of light. It is well-known that this approach
happens to produce the same formula (by accident). That's why they say
that it (the _derivation_, not the radius formula) is "physikalisch
nicht korrekt".
> > You can see the famous "Linienelement" smack in the middle of the page.
>
> I doubt you know German, because this letter as a whole has nothing
> todo with the Schwarzschild Radius nor has this word "Linienelement"
> anything todo related to it.
It has everything to do with it. The line element contains in it the
radius formula. And it is not your formula. At the time of this letter
neither Schwarzschild nor anybody else had yet realised the importance
of that radius but it is of course sitting right there, since
everything follows from that metric. In the letter the radius is
denoted by gamma. The actual value of gamma (2M in geometric units)
follows from comparing with the metric simulating Newtonian model in
the weak gravity case.
> > > > > to get this
> > > > > solution, and yes he indeed did, then he undoubtfully used
> > > > > a wrong tool which gave him this wrong formula.
>
> > > > No, he just solved the Einstein equation using a particular coordinate
> > > > system. One could use slightly more elegant coordinates but this is
> > > > all gravy - point is Schwarzschild got it right.
>
> > > LOL! :-) I have proved him wrong :-)
>
> > You say you proved him wrong. That's a different thing :-)
>
> > > > This claim, being a claim about a mathematical argument, can be verified.
>
> > > My proof can be verified too, and it proves that Schwarzschild Radius
> > > is the wrong radius.
>
> > Your proof is wrong - it assumes a falsehood, for a start
> > (gravitational potential doesn't exist in GR).
>
> And? It doesn't matter whether it exists in GR or not.
> The goal is to determine such a radius where even
> light cannot escape from the surface due to the strength of the gravitation.
> GR is just a tool, much like NM or any other, and as was
> demonstrated by me, the result obtained with
> GR is not the correct result. Q.E.D.
You should have said from the beginning that you were not going to use
relativity. In this case, fine, you've got a formula for something.
So? Why should anybody care? Unless you show first that the theory you
base your derivation on can reproduce all experimentally confirmed
results of GR, you haven't got anything.
[snip]
(I noticed you removed sci.physics.relativity from the newsgroup
followup list. I'm restoring it. Is this how you discuss losing
positions? By running away from them?)
--
Jan Bielawski
Androcles
[snip]
: These are just words.
That's right, fuckhead.
Koobee Wublee
> On Sep 8, 11:56 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > This is not true. The static, asymptotically flat, and spherically
> > symmetric solutions to the field equations in general take one such
> > form below.
I take this back. The following solution is only static and
spherically symmetric but not necessarily asymptotically flat.
> > ds^2 = c^2 dt^2 / (1 + K / R) - (1 + K / R) (dR/dr)^2 dr^2 - (R + K)^2
> > dO^2
>
> > Where
>
> > ** R(r) = Function of r
> > ** dO^2 = cos^2Phi dTheta^2 + dPhi^2
> > ** K = Integration constant
> > ** dr, dTheta, dPhi = Choiced coordinate system
>
> > Only if the following is true, you get the Schwarzschild metric.
>
> > ** R = r - K
>
> > Where
>
> > ** K = 2 G M / c^2
>
> > If the following is true, you get Schwarzschild's original solution
> > which does not manifest any black holes.
>
> > ** R = (r^3 + K^3)^(1/3)
I have a correction. The original Schwarzschild's metric should be
the following.
** R = (r^3 + K^3)^(1/3) - K
> > If the following is true, you get another solution just as simple as
> > the Schwarzschild metric but without manifestation of black holes.
>
> > ** R = r
>
> > If the following is true, you get a constantly expanding universe that
> > also obeys the Schwarzschild metric --- a trait that even the FLRW
> > metric fails to do so.
>
> > ** R = r / (1 + r^2 / K / L)
>
> > Where
>
> > ** L = Cosmic constant
>
> > If the following is true, you get an accelerated expanding universe.
>
> > ** R = r / (1 + r^2 / K / L + r^3 / K / L / N)
>
> > Where
>
> > ** L, N = Cosmic constants
>
> > Each of these solutions is uniquely independent of the others.
> > Claiming these solutions being the same is utter nonsense --- a
> > misunderstanding on your part of failure to understand the metric is
> > not a tensor but merely a matrix. In addition, the last two metrics
> > prove the Birkhoff's theorem wrong.
>
> I'm not going to comment on your posting besides stating that almost
> all of it is incorrect (both the mathematics and the history). We've
> been through this before and there is little point in repeating it.
Just like before, you cannot explain why you are so fixated on the
Schwarzschild metric as the unique solution. You are just a priest.
You are telling the congregation which God is a preferably
worshipping. In reality according to the rules of mathematics, all
these solutions are independent of each other, and they are all
solutions to the field equations. <shrug> There are more, of course.
Just like before, you have no counter argument against what I have
brought us as the true history. As you said, it is all in the
archives. One just has to do some diligent research. <shrug>
> More detail on Schwarzschild's derivation is found here (for those who
> haven't been around at the previous iteration of Koobee's nonsensical
> claims):
> http://groups.google.com/group/sci.physics.relativity/msg/068e4e33d3f...
Still promoting that pure nonsense. This is physics. It is not about
which God is more powerful. <shrug>
Koobee Wublee
> On Sep 7, 11:16 pm, Koobee Wublee wrote:
> > What do you mean by the '/\' operator?
>
> The wedge product, i.e. antisymmetrisation of the tensor product. For
> 1-forms a and b it amounts to:
>
> a /\ b = a x b - b x a
>
> ...where "x" is tensor product (there are several unimportant
> conventions for the multiplicative factor in front of the RHS - here
> it's 1).
Thanks for clarifying what you meant by the above operator in which in
you original paper below you have not.
http://www.mastersofcinema.org/jan/sch.pdf
What you have discovered is merely one of the infinite solutions to
the field equations. You discovered
ds^2 = c^2 (1 - K / u) dt^2 - (du/dr)^2 dr^2 / (1 - K / u) - u^2 dO^2
The following metric works just fine as a solution to the field
equations.
ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) (du/dr)^2 dr^2 - (u + K)^2
dO^2
Just because the latter is not discovered by you, it does not mean the
latter is a transformation of yours. <shrug>
You are indeed a 3-dimensional creature trapped in a 2-dimensional
world. <shrug>
Eric Gisse
<koobee...@gmail.com> wrote:
>On Sep 9, 2:07 am, JanPB <film...@gmail.com> wrote:
>> On Sep 8, 11:56 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
>
>> > This is not true. The static, asymptotically flat, and spherically
>> > symmetric solutions to the field equations in general take one such
>> > form below.
>
>I take this back. The following solution is only static and
>spherically symmetric but not necessarily asymptotically flat.
You can't change the geometry through a coordinate transformation! How
many times do you need to be told?
Schchwarzschild is asymptotically equivalent to Minkowski space.
There exists a coordinate transformation between the two "different"
manifolds. Thus, this one is asymptotically flat.
[...]
>
>Just like before, you cannot explain why you are so fixated on the
>Schwarzschild metric as the unique solution.
IT IS NOT UNIQUE - that is the WHOLE point. We have been telling this
to you for years now!
The Schwarzschild coordinates are preferred because they reduce to
Minkowski space in the asymptotic limit and because it is a more
compact representation compared to others.
[snip remaining]
Once again you are always the first one to bring up religion..
JanPB
> Just like before, you cannot explain why you are so fixated on the
> Schwarzschild metric as the unique solution. You are just a priest.
> You are telling the congregation which God is a preferably
> worshipping. In reality according to the rules of mathematics, all
> these solutions are independent of each other, and they are all
> solutions to the field equations. <shrug> There are more, of course.
As I said earlier, I'm not going to comment on those incorrect claims
as we've been through all this before. If anyone is interested, just
read the thread containing the article I referred to above:
JanPB
> On Sep 8, 10:22 pm, JanPB wrote:
>
> > On Sep 7, 11:16 pm, Koobee Wublee wrote:
> > > What do you mean by the '/\' operator?
>
> > The wedge product, i.e. antisymmetrisation of the tensor product. For
> > 1-forms a and b it amounts to:
>
> > a /\ b = a x b - b x a
>
> > ...where "x" is tensor product (there are several unimportant
> > conventions for the multiplicative factor in front of the RHS - here
> > it's 1).
>
> Thanks for clarifying what you meant by the above operator in which in
> you original paper below you have not.
>
> http://www.mastersofcinema.org/jan/sch.pdf
It's a standard operator.
> What you have discovered is merely one of the infinite solutions to
> the field equations. You discovered
>
> ds^2 = c^2 (1 - K / u) dt^2 - (du/dr)^2 dr^2 / (1 - K / u) - u^2 dO^2
>
> The following metric works just fine as a solution to the field
> equations.
>
> ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) (du/dr)^2 dr^2 - (u + K)^2
> dO^2
>
> Just because the latter is not discovered by you, it does not mean the
> latter is a transformation of yours. <shrug>
They are the same solution - you get one from the other by a
coordinate change.
> You are indeed a 3-dimensional creature trapped in a 2-dimensional
> world. <shrug>
You need to go over the basics of analysis on manifolds (let alone
differential geometry).
--
Jan Bielawski
Koobee Wublee
> On Sun, 09 Sep 200, Koobee Wublee wrote:
> >I take this back. The following solution is only static and
> >spherically symmetric but not necessarily asymptotically flat.
>
> You can't change the geometry through a coordinate transformation! How
> many times do you need to be told?
I am not changing the geometry. All the mathematics I have shown is
based on the invariance in the geometry. <shrug>
> Schchwarzschild is asymptotically equivalent to Minkowski space.
Yes, but this does not mean all the static and the spherically
symmetric solutions are asymptotically flat. <shrug>
> There exists a coordinate transformation between the two "different"
> manifolds. Thus, this one is asymptotically flat.
Since there are infinite choices of coordinate systems, there exist
infinite coordinate transformations of the same invariant geometry.
However, this is not what I am addressing. I need to establish a
choice of coordinate system before I can proceed to derive the
Christoffel symbols of the second kind. Thus, the choice of
coordinate system must be fixed at this point. Then, we see the
Riemann curvature tensor, the Ricci curvature tensor, the field
equations, and finally the metrics (solutions). In doing so, the
choice of coordinate system must remain the same throughout each
process. Since there are an infinite number of solutions to the same
choice of coordinate system, we must conclude that each metric
(solution) describes a unique universe. Also, you can always find
solutions that are static and spherically symmetric but not
asymptotically flat. Thus, this invalidates Birkhoff's theorem in
which you have bet all your chips on. <shrug>
> >Just like before, you cannot explain why you are so fixated on the
> >Schwarzschild metric as the unique solution.
>
> IT IS NOT UNIQUE - that is the WHOLE point. We have been telling this
> to you for years now!
You have being bullsh*tting for years now. It is time to act yourself
and not one of these henchmen of the Church of Einstein to enforce
your Gestapo-like marshal thoughts. It would not work in an
intellectual society. Do we even reside in an intellectual society
with all you henchmen of the Church of Einstein spreading false
gospels?
> The Schwarzschild coordinates are preferred because they reduce to
> Minkowski space in the asymptotic limit and because it is a more
> compact representation compared to others.
Preferred has no mathematical basis in the discussion. It is purely
humanly emotional decision without any logical bases. <shrug>
> Once again you are always the first one to bring up religion.
I am always the first to point out the religious nature of your
belief. <shrug>
Koobee Wublee
> On Sep 9, 7:55 pm, Koobee Wublee wrote:
> > Just like before, you cannot explain why you are so fixated on the
> > Schwarzschild metric as the unique solution. You are just a priest.
> > You are telling the congregation which God is a preferably
> > worshipping. In reality according to the rules of mathematics, all
> > these solutions are independent of each other, and they are all
> > solutions to the field equations. <shrug> There are more, of course.
>
> As I said earlier, I'm not going to comment on those incorrect claims
> as we've been through all this before. If anyone is interested, just
> read the thread containing the article I referred to above:
> http://groups.google.com/group/sci.physics.relativity/msg/068e4e33d3f...
Just like before, you have shunned away very simple and basic
mathematics. Since you are good at copying others on higher-end
mathematics without understanding the basics, it is best that you
remain silent. You are merely a person good at duplicating nonsense.
<shrug>
Koobee Wublee
> On Sep 9, 8:08 pm, Koobee Wublee wrote:
> > What you have discovered is merely one of the infinite solutions to
> > the field equations. You discovered
>
> > ds^2 = c^2 (1 - K / u) dt^2 - (du/dr)^2 dr^2 / (1 - K / u) - u^2 dO^2
>
> > The following metric works just fine as a solution to the field
> > equations.
>
> > ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) (du/dr)^2 dr^2 - (u + K)^2
> > dO^2
>
> > Just because the latter is not discovered by you, it does not mean the
> > latter is a transformation of yours. <shrug>
>
> They are the same solution - you get one from the other by a
> coordinate change.
I see. You still see this nonsense. You are continuingly twisting
the logic of mathematics.
1. These two equations are drastically different. They cannot be the
same equation.
2. The choice of coordinate system must be established as early as in
the derivation of the Christoffel symbols of the second kind.
Therefore, each solution to the same set of field equations must all
reference to the same choice of coordinate system established as early
as the derivation of the Christoffel symbols of the second kind.
<shrug>
> > You are indeed a 3-dimensional creature trapped in a 2-dimensional
> > world. <shrug>
>
> You need to go over the basics of analysis on manifolds (let alone
> differential geometry).
I have done that. Have you? Please stop spreading the nonsense of
the ones who want you to serve as a henchman to repeat this nonsense.
<shrug>
JanPB
> On Sep 9, 8:50 pm, JanPB <film...@gmail.com> wrote:
>
>
>
> > On Sep 9, 8:08 pm, Koobee Wublee wrote:
> > > What you have discovered is merely one of the infinite solutions to
> > > the field equations. You discovered
>
> > > ds^2 = c^2 (1 - K / u) dt^2 - (du/dr)^2 dr^2 / (1 - K / u) - u^2 dO^2
>
> > > The following metric works just fine as a solution to the field
> > > equations.
>
> > > ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) (du/dr)^2 dr^2 - (u + K)^2
> > > dO^2
>
> > > Just because the latter is not discovered by you, it does not mean the
> > > latter is a transformation of yours. <shrug>
>
> > They are the same solution - you get one from the other by a
> > coordinate change.
>
> I see. You still see this nonsense. You are continuingly twisting
> the logic of mathematics.
>
> 1. These two equations are drastically different. They cannot be the
> same equation.
They are one and the same tensor. Just like one and the same vector
can have different coordinate representations.
> 2. The choice of coordinate system must be established as early as in
> the derivation of the Christoffel symbols of the second kind.
> Therefore, each solution to the same set of field equations must all
> reference to the same choice of coordinate system established as early
> as the derivation of the Christoffel symbols of the second kind.
> <shrug>
Einstein field equation is an equation for an unknown tensor. To solve
it, one usually selects a convenient coordinate system which results
in the field equations producing tensor components wrt that coordinate
system. This determines the sought for tensor. Since it's a coordinate-
independent object, one can transform it every which way after the
fact - it changes nothing except the tensor's components.
It's exactly analogous to covectors:
X = dx in the Cartesians
and
Y = cos(theta) dr - r sin(theta) dtheta in polars.
They are the same: X = Y. Your "two solutions" are exactly like this X
and Y except they are rank-1 tensors not 2.
> > > You are indeed a 3-dimensional creature trapped in a 2-dimensional
> > > world. <shrug>
>
> > You need to go over the basics of analysis on manifolds (let alone
> > differential geometry).
>
> I have done that. Have you? Please stop spreading the nonsense of
> the ones who want you to serve as a henchman to repeat this nonsense.
> <shrug>
You are ignorant of the basics of analysis on manifolds and you need
to go over the basics. Tensors do not depend on coordinates, period.
--
Jan Bielawski
JanPB
I have not shunned away - you are repeating verbatim your incorrect
arguments from last year so I'm providing the relevant link in the
archive. Your comments about my mathematics mean nothing - until you
stop talking mathematical nonsense they'll remain only rhetoric.
--
Jan Bielawski
JanPB
> On Sep 9, 8:27 pm, Eric Gisse wrote:
>
> > On Sun, 09 Sep 200, Koobee Wublee wrote:
> > >I take this back. The following solution is only static and
> > >spherically symmetric but not necessarily asymptotically flat.
>
> > You can't change the geometry through a coordinate transformation! How
> > many times do you need to be told?
>
> I am not changing the geometry. All the mathematics I have shown is
> based on the invariance in the geometry. <shrug>
Eric - remember Koobee has his own private terminology for everything.
> > Schchwarzschild is asymptotically equivalent to Minkowski space.
>
> Yes, but this does not mean all the static and the spherically
> symmetric solutions are asymptotically flat. <shrug>
>
> > There exists a coordinate transformation between the two "different"
> > manifolds. Thus, this one is asymptotically flat.
>
> Since there are infinite choices of coordinate systems, there exist
> infinite coordinate transformations of the same invariant geometry.
> However, this is not what I am addressing. I need to establish a
> choice of coordinate system before I can proceed to derive the
> Christoffel symbols of the second kind. Thus, the choice of
> coordinate system must be fixed at this point. Then, we see the
> Riemann curvature tensor, the Ricci curvature tensor, the field
> equations, and finally the metrics (solutions). In doing so, the
> choice of coordinate system must remain the same throughout each
> process.
You phrase it extraordinarily clumsily and in a typical amateurish
overcomplicated fashion but agreed so far.
> Since there are an infinite number of solutions to the same
> choice of coordinate system,
No, this is false in _this particular case_. In general yes, an
equation may have many solutions but the peculiarity of the
spherically symmetric vacuum problem is that it happens to lead to
equations which are guaranteed to have unique solution by virtue of
the general class of differential equations to which they happen to
belong.
> we must conclude that each metric
> (solution) describes a unique universe.
It would be so if there were multiple solutions corresponding to the
same coordinate system. But there is only one. I repeat that this is
not a generic situation but a peculiarity of the spherically symmetric
vacuum problem.
> Also, you can always find
> solutions that are static and spherically symmetric but not
> asymptotically flat. Thus, this invalidates Birkhoff's theorem in
> which you have bet all your chips on. <shrug>
Name one.
> > >Just like before, you cannot explain why you are so fixated on the
> > >Schwarzschild metric as the unique solution.
>
> > IT IS NOT UNIQUE - that is the WHOLE point. We have been telling this
> > to you for years now!
>
> You have being bullsh*tting for years now. It is time to act yourself
> and not one of these henchmen of the Church of Einstein to enforce
> your Gestapo-like marshal thoughts.
It's a mathematical statement you are making, a false one, and it can
be resolved 100% and without throwing silly tantrums invoking "Church
of Einstein" and similar idiocies.
> It would not work in an
> intellectual society. Do we even reside in an intellectual society
> with all you henchmen of the Church of Einstein spreading false
> gospels?
Again, this is mathematics. Save your metaphors for a different
newsgroup.
[cut]
--
Jan Bielawski
Eric Gisse
<koobee...@gmail.com> wrote:
>On Sep 9, 8:27 pm, Eric Gisse wrote:
>> On Sun, 09 Sep 200, Koobee Wublee wrote:
>
>> >I take this back. The following solution is only static and
>> >spherically symmetric but not necessarily asymptotically flat.
>>
>> You can't change the geometry through a coordinate transformation! How
>> many times do you need to be told?
>
>I am not changing the geometry. All the mathematics I have shown is
>based on the invariance in the geometry. <shrug>
You aren't?
Calculate the half dozen independent Riemann components and let M-->
0. Then tell me what you get.
Do be sure to use the right definition for Riemann this time.
>
>> Schchwarzschild is asymptotically equivalent to Minkowski space.
>
>Yes, but this does not mean all the static and the spherically
>symmetric solutions are asymptotically flat. <shrug>
With the added assumption of a time-like Killing vector at infinity,
it sure does! Welcome to Birkhoff's theorem - enjoy your stay.
>
>> There exists a coordinate transformation between the two "different"
>> manifolds. Thus, this one is asymptotically flat.
>
>Since there are infinite choices of coordinate systems, there exist
>infinite coordinate transformations of the same invariant geometry.
>However, this is not what I am addressing. I need to establish a
>choice of coordinate system before I can proceed to derive the
>Christoffel symbols of the second kind. Thus, the choice of
>coordinate system must be fixed at this point.
Says who?
Do the phrases "chain rule" and "change of variables" not mean
anything to you?
> Then, we see the
>Riemann curvature tensor, the Ricci curvature tensor, the field
>equations, and finally the metrics (solutions). In doing so, the
>choice of coordinate system must remain the same throughout each
>process.
Would this be the same Riemann tensor you wrote down that had the
incorrect index placement in several of the components that you seem
to think can be represented by a matrix, or would this be the actual
Riemann tensor?
Anyways, I'll tell you what. If you can show me even one case in which
changing the coordinate system changes the physical content of a
tensor, I'll admit your right.
> Since there are an infinite number of solutions to the same
>choice of coordinate system, we must conclude that each metric
>(solution) describes a unique universe.
Then why can I always construct a coordinate transformation that
relates these "unique" universes?
I can do it EVERY SINGLE TIME, as I have shown you EVERY SINGLE TIME.
You have been unable to find a solution that I cannot do this to.
> Also, you can always find
>solutions that are static and spherically symmetric but not
>asymptotically flat. Thus, this invalidates Birkhoff's theorem in
>which you have bet all your chips on. <shrug>
If you can find me one solution that conforms to the _explicit_
conditions of Birkhoff's theorem [not just the conditions you think
are relevant] but is not asymptotically flat I'll admit you are right.
[...]
Koobee Wublee
> On Sep 9, 9:29 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > ds^2 = c^2 (1 - K / u) dt^2 - (du/dr)^2 dr^2 / (1 - K / u) - u^2 dO^2
>
> > ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) (du/dr)^2 dr^2 - (u + K)^2
> > dO^2
>
> > 1. These two equations are drastically different. They cannot be the
> > same equation.
>
> They are one and the same tensor. Just like one and the same vector
> can have different coordinate representations.
You are twisting the logic of mathematics once again. The choice of
coordinate system in both equations is the same (t, r, theta, phi)
with a single observer.
> > 2. The choice of coordinate system must be established as early as in
> > the derivation of the Christoffel symbols of the second kind.
> > Therefore, each solution to the same set of field equations must all
> > reference to the same choice of coordinate system established as early
> > as the derivation of the Christoffel symbols of the second kind.
> > <shrug>
>
> Einstein field equation is an equation for an unknown tensor.
Einstein field equations are not only one equation. They represent a
set of 16 equations that the solutions yield exactly the 16 elements
to the metric. You really have trouble with the basics. <shrug>
> To solve
> it, one usually selects a convenient coordinate system which results
> in the field equations producing tensor components wrt that coordinate
> system.
Baloney. The field equations are ordinary differential equations
themselves with the choice of coordinate system already well
established at this stage. By far, the most popular one is the
spherically symmetric polar coordinate.
You are just confused over Schwarzschild's own method of solving the
field equations by transforming the entire field equations into
another set of coordinate system that the corresponding metric in
describing the same, invariant geometry would yield a determinant of
-1. This new metric with this new coordinate system is very different
from the original metric with the spherically symmetric polar
coordinate system where the old metric does not yield a determinant of
-1. After solving the field equations for the new metric with the new
coordinate system, Schwarzschild has to convert the new metric back to
the old metric using the spherically symmetric coordinate system. You
even have failed to understand Schwarzschild's classic paper. <shrug>
Schwarzschild's original solution to the transformed equation turned
out to resemble the Schwarzschild metric. If Schwarzschild were to
discover another solution instead, you will be singing the tunes of
this another solution. You are really full of it. <shrug>
> This determines the sought for tensor. Since it's a coordinate-
> independent object, one can transform it every which way after the
> fact - it changes nothing except the tensor's components.
The metric, the Christoffel symbols of the second kind, the Riemann
tensor, the Ricci tensor, and the Einstein tensor are merely
matrices. They must be matrices to satisfy the invariance in the
geometry as each observer chooses his own set of coordinate system.
You just cannot get any more basic than this. <shrug>
> You are ignorant of the basics of analysis on manifolds and you need
> to go over the basics. Tensors do not depend on coordinates, period.
There are so many problems with GR right from the start. The most
basic is even more embarrassing. Any diligent grade school children
can identify the mathematical relationship below.
Given that
A = B C
If (A = constant and B != 0), then (C = A / B).
This blunder came as early as during Ricci's time when the Riemann
tensor which is merely a matrix was incorrectly identified as a tensor
which means it becomes invariant to any coordinate transformation.
Similarly, the metric is merely a matrix. Ricci deified the metric
into a tensor. Mathematically, this can easily be proven wrong.
d[s] = [Q] d[q] = [Q'] d[q']
Where
** d[s] = Invariant geometry in displacement vector
** [Q], [Q'] = Matrices
** d[q], dq[q'] = Coordinate systems
The above equation squared is
ds^2 = [g] * d[q^2] = [g'] * d[q'^2]
Where
** [g] = [Q]^Transpose [Q]
** [g'] = [Q']^Transpose [Q']
** d[q^2] = d[q] d[q]^Transpose
** d[q'^2] = d[q'] d[q']^Transpose
** [A] * [B] = SUM(SUM(A_ij B_ij)), dot product
** ds^2 = Invariant, still
The metric [g] and the metric [g'] cannot be the same if the choice of
coordinate system [q] is different from [q'].
ds^2 = [g] * d[q^2] = g_ij dq^i dq^j = Invariant
The geometry, ds^2, must be invariant due to obvious reason. The
choice of coordinate system, d[q^2], is observer dependent. This can
only mean the metric, [g], must also be observer dependent. The
metric, the Riemann, and the Ricci tensors cannot be tensors after
all. All solutions to the field equations must be unique and
independent of each one where all solutions must reference to the same
choice of coordinate system in describing vastly different invariant
geometries. What good is the set of field equations that can generate
an infinite numbers of solutions to describe infinitely different and
independent universes? What good is the set of field equations can
either generate a solution that manifests black holes and also ones
that don't?
Koobee Wublee
> On Sep 9, 9:18 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > I am not changing the geometry. All the mathematics I have shown is
> > based on the invariance in the geometry. <shrug>
>
> Eric - remember Koobee has his own private terminology for everything.
Again, I am not changing the geometry. All the mathematics I have
shown is
> > based on the invariance in the geometry. <shrug>
> > Also, you can always find
> > solutions that are static and spherically symmetric but not
> > asymptotically flat. Thus, this invalidates Birkhoff's theorem in
> > which you have bet all your chips on. <shrug>
>
> Name one.
I have shown such two in this post below replying to your comment.
http://groups.google.com/group/sci.physics.relativity/msg/f812a0b372722b40?hl=en&
With the following errata,
The Schwarzschild's original solution is when
** R = (r^3 + K^3)^(1/3) - K
The two solutions that are not asymptotically flat are when
** R = r / (1 + r^2 / K / L)
** R = r / (1 + r^2 / K / L + r^3 / K / L / N)
Where
** ds^2 = c^2 dt^2 / (1 + K / R) - (1 + K / R) (dR/dr)^2 dr^2 - (R +
K)^2 dO^2
Have you not read my postings before replying?
> > Since there are infinite choices of coordinate systems, there exist
> > infinite coordinate transformations of the same invariant geometry.
> > However, this is not what I am addressing. I need to establish a
> > choice of coordinate system before I can proceed to derive the
> > Christoffel symbols of the second kind. Thus, the choice of
> > coordinate system must be fixed at this point. Then, we see the
> > Riemann curvature tensor, the Ricci curvature tensor, the field
> > equations, and finally the metrics (solutions). In doing so, the
> > choice of coordinate system must remain the same throughout each
> > process.
>
> You phrase it extraordinarily clumsily and in a typical amateurish
> overcomplicated fashion but agreed so far.
>
> > Since there are an infinite number of solutions to the same
> > choice of coordinate system,
>
> No, this is false in _this particular case_. In general yes, an
> equation may have many solutions but the peculiarity of the
> spherically symmetric vacuum problem is that it happens to lead to
> equations which are guaranteed to have unique solution by virtue of
> the general class of differential equations to which they happen to
> belong.
Then, you are contradicting with yourself. You have agreed that the
coordinate system must be already established even before the
Christoffel symbols of the second kind are computed. At the solution
level, the choice of coordinate system should be the same as the
original choice chosen before the Christoffel symbols of the second
kind are computed. Since each solution or metric must be only applied
to the same choice of coordinate system, each solution or metric must
represent a different unique and independent geometry. You need to
resolve your own contradiction before we can go on further.
JanPB
[snip everything]
As I said: you've posted it all last year. Bottom line is likewise the
same as 100 years ago: tensors are coordinate-independent entities.
End of story.
--
Jan Bielawski
JanPB
> On Sep 9, 11:41 pm, JanPB <film...@gmail.com> wrote:
>
> > On Sep 9, 9:18 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
[snip]
> > > Since there are an infinite number of solutions to the same
> > > choice of coordinate system,
>
> > No, this is false in _this particular case_. In general yes, an
> > equation may have many solutions but the peculiarity of the
> > spherically symmetric vacuum problem is that it happens to lead to
> > equations which are guaranteed to have unique solution by virtue of
> > the general class of differential equations to which they happen to
> > belong.
>
> Then, you are contradicting with yourself. You have agreed that the
> coordinate system must be already established even before the
> Christoffel symbols of the second kind are computed. At the solution
> level, the choice of coordinate system should be the same as the
> original choice chosen before the Christoffel symbols of the second
> kind are computed.
You still don't understand. It is OK to fix a coordinate system in
order to work with tensor coefficients. But these coefficients in the
end determine a tensor which is coordinate-independent. Given a tensor
(any tensor, like a solution to Einstein's equation) one can re-
express it in any coordinate system. It's just like vectors or
covectors.
> Since each solution or metric must be only applied
> to the same choice of coordinate system,
No, that's false. Tensor is coordinate-independent. Coordinates were
merely used as a computational aid to find the tensor.
> each solution or metric must
> represent a different unique and independent geometry.
False. You need to learn how analysis on manifolds is done. Look at
Spivak's "Calculus on manifolds".
> You need to
> resolve your own contradiction before we can go on further.
I don't think an ASCII bboard is efficient enough to show you your
mistakes given the fact you are completely unwilling to learn anyway.
That's my last word on this thread.
--
Jan Bielawski
Koobee Wublee
> As I said: you've posted it all last year. Bottom line is likewise the
> same as 100 years ago: tensors are coordinate-independent entities.
> End of story.
Just like last year, whenever you were confronted with the scenario
where a grade school kid would have understood easily, you just
angrily called it a quit. You are indeed a henchman to the Church of
Einstein spreading nonsensical gospel in disguise as sound physics.
Go back to watching these second rated films. Wow! Someone can even
make a living on that. You must know how to BS really, really good.
This is a trait that I lack myself and envy you. End of discussion.
Now, get lost.
JanPB
> Just like last year, whenever you were confronted with the scenario
> where a grade school kid would have understood easily, you just
> angrily called it a quit.
I call it quits because there are only so many ways to say "tensors
are coordinate-independent entities". This fact nukes your arguments.
--
Jan Bielawski
Eric Gisse
This transcends relativity. This is a fundamental misunderstanding
that spans all of space and time.
What KW constantly forgets [or does not believe me when I tell him] is
that tensors are entities that are used in more than just relativity.
If I thought it'd matter at all, I'd go through the derivation of the
inertia tensor or the /classical/ E&M stress tensor. Neither of which
have anything to do with relativity.
gu...@hotmail.com
wrote:
1st proper time and now this.
#1. It's called "Maxwell's" stress tensor. E&M are related to "c"
which is related to relativity.
#2. A Lorentz scalar is generated from vectors and tensors. A Lorentz
scalar is related to relativity.
#3. Gravity is a tensor and General Relatitvity makes use of it.
N.O.V.I.C.E
Now hide & give another bunch of stupid replies, the same way you kept
doing for proper time.
Eric Gisse
>On Sep 11, 5:13 am, Eric Gisse <jowr.pi.nos...@gmail-nospam.com>
>wrote:
>> On Tue, 11 Sep 2007 06:05:37 -0000, JanPB <film...@gmail.com> wrote:
>> >On Sep 10, 10:09 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
>>
>> >> Just like last year, whenever you were confronted with the scenario
>> >> where a grade school kid would have understood easily, you just
>> >> angrily called it a quit.
>>
>> >I call it quits because there are only so many ways to say "tensors
>> >are coordinate-independent entities". This fact nukes your arguments.
>>
>> This transcends relativity. This is a fundamental misunderstanding
>> that spans all of space and time.
>>
>> What KW constantly forgets [or does not believe me when I tell him] is
>> that tensors are entities that are used in more than just relativity.
>>
>> If I thought it'd matter at all, I'd go through the derivation of the
>> inertia tensor or the /classical/ E&M stress tensor. Neither of which
>> have anything to do with relativity.
>
>1st proper time and now this.
Oh shut the fuck up about proper time already. You had your victory,
but since that is the only one you have had you have to drag it out
for months on end.
>
>#1. It's called "Maxwell's" stress tensor. E&M are related to "c"
>which is related to relativity.
I know the word, dipshit. The choice was deliberate.
When you have studied /any/ classical electromagnetic theory beyond
the level of "Elements of Physics", we can have this discussion. Need
I remind you about your nice long misunderstanding about what EMF is?
I'll entertain your delusions as much as you want if you can write
down the classical mechanical Lagrangian and then derive the equations
of motion. Until then, you are just a poser child who read Wikipedia
and his "Elements of Physics" textbook who thinks he knows what he is
talking about.
>
>#2. A Lorentz scalar is generated from vectors and tensors. A Lorentz
>scalar is related to relativity.
>
>#3. Gravity is a tensor and General Relatitvity makes use of it.
It'd help if you actually knew what you are talking about. Go read
about classical mechanics and the inertia tensor.
My example was a deliberate choice to highlight misunderstandings
_just like this_. Thanks for playing.
>
>N.O.V.I.C.E
>
>Now hide & give another bunch of stupid replies, the same way you kept
>doing for proper time.
Liar. I admitted the mistake immediately.
You, on the other hand, fuck up EVERY SINGLE TIME you talk and drag
out the mistakes for months on end. Then when someone proves to you
that you are wrong, you either ignore the proof or simply move on to
emit stupidities about /yet another/ subject you have no clue about.
Why don't you tell us more about the Bohr model of the atom? Or are
you done writing stupid shit about that now that people have given you
dozens of explainations and links as to why you are stupid as hell?
Better yet, why not tell us about EMF some more? I notice that you
NEVER AGAIN talked about EMF after I provided you a scan of an
electrodynamics text that dealt with the exact misunderstanding you
were using. Why is that, Peter?
http://groups.google.com/group/sci.physics.relativity/msg/463b18998e84adc1?dmode=source
When will you finally shut the fuck up? You have misunderstood pretty
much all of low level 17th, 18th, 19th, and 20th century physics. Will
you graduate to misunderstanding high level concepts, or would you
like to come full circle and argue about EMF some more?
>
>
>
>
>
>
>
>
gu...@hotmail.com
Dummy says Relativity not related to tensors. Lorentz scalar is based
on tensors.
Lorentz transform contracts time and space dummy....known as the
spacetime TENSOR MANIFOLD.
What an idiot.
Eric Gisse
<gu...@hotmail.com> wrote:
No, that is not what I said. Learn to read.
How do you function with this kind of reading disability?
>Lorentz transform contracts time and space dummy....known as the
>spacetime TENSOR MANIFOLD.
Tensors are not manifolds, idiot.
>
>What an idiot.
Why don't you respond to the rest of my post, idiot?
Why not tell us about EMF some more? I notice that you
Androcles
RUDE IDIOTIC FUCKHEAD!
NO!
Koobee Wublee
> On Sep 10, 9:40 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> > Then, you are contradicting with yourself. You have agreed that the
> > coordinate system must be already established even before the
> > Christoffel symbols of the second kind are computed. At the solution
> > level, the choice of coordinate system should be the same as the
> > original choice chosen before the Christoffel symbols of the second
> > kind are computed.
>
> You still don't understand. It is OK to fix a coordinate system in
> order to work with tensor coefficients. But these coefficients in the
> end determine a tensor which is coordinate-independent.
Yes, I understand perfectly where you come from. I have been trying
to point out where you have been wrong. I really don't care what a
tensor is. However, the metric must vary with the choice of
coordinate system to describe an invariant geometry.
> Given a tensor
> (any tensor, like a solution to Einstein's equation) one can re-
> express it in any coordinate system. It's just like vectors or
> covectors.
At the end of the day, the metric must be associated with the initial
choice of and none other than the coordinate system that you started
to derive the Christoffel's symbols of the second kind with. You
cannot change the coordinate system and call it the same solution. If
so, you are just cheating yourself in mathematics.
> > Since each solution or metric must be only applied
> > to the same choice of coordinate system,
>
> No, that's false. Tensor is coordinate-independent. Coordinates were
> merely used as a computational aid to find the tensor.
Without choosing a set of coordinate system in the very beginning, it
is impossible for you to describe what the geometry is. <shrug>
> > each solution or metric must
> > represent a different unique and independent geometry.
>
> False. You need to learn how analysis on manifolds is done. Look at
> Spivak's "Calculus on manifolds".
This is the most basic stuff. There is no need to consult with higher
mathematics. If the reference agrees with you, and if I can get my
hand on that reference, I am going to toss it into the recycle bin.
<shrug>
> > You need to
> > resolve your own contradiction before we can go on further.
>
> I don't think an ASCII bboard is efficient enough to show you your
> mistakes given the fact you are completely unwilling to learn anyway.
You don't have to show me. Just allow me to show you where your error
is. It is that simple.
> That's my last word on this thread.
We shall see.
Koobee Wublee
You have claimed to be a mathemaTician. In reality, you are a
matheMagician. When pushed to the corner, your argument has always
been "BECAUSE IT IS A TENSOR". This type of argument is totally based
on religion. Your introduction at Masters of Cinema should read
Jan Bielawski is a mathemaGician (differential bullsh*t, gravitational
nonsense) ... Once coming down on discussing these issues, he loves to
end the discussions with religion intonation. As a henchman serving
under the Church of Einstein, he personal led the crusade to make sure
the next generation continue to accept the differential bullsh*t and
gravitational nonsense. <shrug>
Eric Gisse
<koobee...@gmail.com> wrote:
>On Sep 10, 11:05 pm, JanPB <film...@gmail.com> wrote:
>> On Sep 10, 10:09 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
>
>> > Just like last year, whenever you were confronted with the scenario
>> > where a grade school kid would have understood easily, you just
>> > angrily called it a quit.
>>
>> I call it quits because there are only so many ways to say "tensors
>> are coordinate-independent entities". This fact nukes your arguments.
>
>You have claimed to be a mathemaTician. In reality, you are a
>matheMagician. When pushed to the corner, your argument has always
>been "BECAUSE IT IS A TENSOR". This type of argument is totally based
>on religion. Your introduction at Masters of Cinema should read
Study some classical mechanics, and learn about how tensors are used
in that subject.
Eric Gisse
<koobee...@gmail.com> wrote:
[...]
>
>You don't have to show me. Just allow me to show you where your error
>is. It is that simple.
Start off by computing the area in your metrics.
Koobee Wublee
> On Tue, 11 Sep 2007, JanPB <film...@gmail.com> wrote:
> >I call it quits because there are only so many ways to say "tensors
> >are coordinate-independent entities". This fact nukes your arguments.
>
> This transcends relativity. This is a fundamental misunderstanding
> that spans all of space and time.
Yes, indeed. The fundamental misunderstanding led Mr. Bielawski and
you to embrace differential bullsh*t and gravitational nonsense
spanning the entire spacetime continuum in every dimensions. <shrug>
> What KW constantly forgets [or does not believe me when I tell him] is
> that tensors are entities that are used in more than just relativity.
What fat Gisse argues from is his collection of bibles disguising as
textbooks. Taken away these bibles, he is an uneducated buffoon of no
substance. Having no college degree to show for, he has to resort to
trolling tactics to disrupt any scholarly discussions.
> If I thought it'd matter at all, I'd go through the derivation of the
> inertia tensor or the /classical/ E&M stress tensor. Neither of which
> have anything to do with relativity.
Fat Gisse needs to learn the relationship among the following first.
** The geometry (invariant)
** The metric
** The choice of coordinate system (variant)
He is incapable of learning though. All his education is a recitation
in pieces from these pathetic bibles he possesses.
Koobee Wublee
> On Tue, 11 Sep 2007, Koobee Wublee wrote:
> >You don't have to show me. Just allow me to show you where your error
> >is. It is that simple.
>
> Start off by computing the area in your metrics.
Here we go again. You love to lick up the nonsense spit out by that
janitor from Cornell, don't you?
Let me remind you again. Besides you are not even a college graduate,
you are no God. You cannot tell how space or spacetime is curved just
by a direct observation. <shrug>
That janitor from Cornell, Mr. Bielawski, and you all need to go back
and study the most basics.
So, repeat after me.
Since the geometry must be invariant, the metric must vary with the
choice of coordinate system to guarantee the invariance in the
geometry.
Let's try again.
Since the geometry must be invariant, the metric must vary with the
choice of coordinate system to guarantee the invariance in the
geometry.
Again.
Since the geometry must be invariant, the metric must vary with the
choice of coordinate system to guarantee the invariance in the
geometry.
And again.
Since the geometry must be invariant, the metric must vary with the
choice of coordinate system to guarantee the invariance in the
geometry.
Eric Gisse
<koobee...@gmail.com> wrote:
>On Sep 11, 2:13 am, Eric Gisse wrote:
>> On Tue, 11 Sep 2007, JanPB <film...@gmail.com> wrote:
>
>> >I call it quits because there are only so many ways to say "tensors
>> >are coordinate-independent entities". This fact nukes your arguments.
>>
>> This transcends relativity. This is a fundamental misunderstanding
>> that spans all of space and time.
>
>Yes, indeed. The fundamental misunderstanding led Mr. Bielawski and
>you to embrace differential bullsh*t and gravitational nonsense
>spanning the entire spacetime continuum in every dimensions. <shrug>
<fart, shrug>
>
>> What KW constantly forgets [or does not believe me when I tell him] is
>> that tensors are entities that are used in more than just relativity.
>
>What fat Gisse argues from is his collection of bibles disguising as
>textbooks. Taken away these bibles, he is an uneducated buffoon of no
>substance. Having no college degree to show for, he has to resort to
>trolling tactics to disrupt any scholarly discussions.
Tell us about your college degree then, oh anonymous one. Or do you
not have one, as everyone here suspects? I'll graduate in the spring
with a bachelors of science in Physics. In the scale of things that
isn't much, but its much more than you will EVER accomplish in the
field.
Your whining would carry more weight if you had the stones to insult
me and my education under your actual name. Instead you are just yet
another pathetic little man who can only rationalize his poor choices
in life by insulting others who are willing to do the work.
>
>> If I thought it'd matter at all, I'd go through the derivation of the
>> inertia tensor or the /classical/ E&M stress tensor. Neither of which
>> have anything to do with relativity.
>
>Fat Gisse needs to learn the relationship among the following first.
>
>** The geometry (invariant)
>** The metric
>** The choice of coordinate system (variant)
The metric is the geometry since the metric is the only piece of
information you need to determine angles between vectors, distances,
areas, and volumes. Every piece of information that you need to know
is either a function of the metric or one of its' derivatives.
The choice of coordinate system is utterly irrelevant since I can
project the metric upon any coordinate basis that I please. The
understanding behind this sentence has eluded you for years, and will
continue to do so.
>
>He is incapable of learning though. All his education is a recitation
>in pieces from these pathetic bibles he possesses.
Apparently you are either unable or unwilling to accept that my
education is not an exercise in rote memorization.
Eric Gisse
<koobee...@gmail.com> wrote:
>On Sep 11, 9:05 pm, Eric Gisse wrote:
>> On Tue, 11 Sep 2007, Koobee Wublee wrote:
>
>> >You don't have to show me. Just allow me to show you where your error
>> >is. It is that simple.
>>
>> Start off by computing the area in your metrics.
>
>Here we go again. You love to lick up the nonsense spit out by that
>janitor from Cornell, don't you?
It is a good thing you hide under that pseudonym. I'd love to see how
much of a failure you are vs how much of one I think you are.
>
>Let me remind you again. Besides you are not even a college graduate,
>you are no God. You cannot tell how space or spacetime is curved just
>by a direct observation. <shrug>
>
Show me where I claimed I could in the sentence "Start off by
computing the area in your metrics".
Then show me your computation of area. Any computation will do,
actually.
[snip junk]
gu...@hotmail.com
Eric Gisse
>Oh this just so easy Erica, I swear you're just looking for a date.
>
Why don't you respond to the rest of my post, idiot? BTW, I already
have someone I'm interested in.