Lagranian Method Revisited

[Koobee Wublee]

The Euler-Lagrange equations are pretty much what govern the physical laws discovered by the self-claimed scientists. The Lagrangian method (to derive the special equations) was discovered during Euler's time that was slightly before Newton. <shrug>

It starts with the following action as an integral of a Lagrangian (density to this action) with respect to instances (time). <shrug>

** A = Integral(T1, T2)[L dt]

Where

** A = action accumulated from T1 to T2
** T1 = starting instance
** T2 = ending instance
** L = Lagrangian
** dt = discrete instance

For a special path, s, there exists a stationary condition to this action from T1 to T2 along this path. So,

** dA/ds = 0, definition of stationary action (min, max, saddle)

Or

** Integral(T1, T2)[@L/@s dt] = 0

Where

** @ = partial derivative operator

No, you can choose to describe L in an observer's coordinate system, [q], instead of invariant geometry, s. If L is also a function of (ds/dt), the above can be rewritten as follows. <shrug>

** Integral(T1, T2)[(@L/@s+ @L/@(ds/dt)) dt] = 0

Applying integration by parts, we have

** @L/@(ds/dt)[T1, T2] + Integral(T1, T2)[(@L/@s - d(@L/@(ds/dt))/dt) dt] = 0

The fixed points at T1 and T2 are just nonsense. In reality, the mathematics says when (@L/@(ds/dt) at T1) = (@L/@(ds/dt) at T2), the above equation simplify into the equation below. <shrug>

** Integral(T1, T2)[(@L/@s - d(@L/@(ds/dt))/dt) dt] = 0

Where

** @L/@(ds/dt)[T1, T2] = 0

Thus, the Euler-Lagrange equation can be derived as follows where it is true between T1 and T2 that the specific condition described above must be met. It is not the fixed end points as the self-styled physicists have proclaimed. <shrug>

** @L/@s - d(@L/@(ds/dt))/dt = 0

Where

** @L/@(ds/dt) at T1 = @L/@(ds/dt) at T2

So, when dealing with Mercury's perihelion advance, the Euler-Lagrange equation cannot possibly apply since with advancing perihelion, there is no way to establish the condition required of the Euler-Lagrange equation. The whole fixed, coherent derivation of Mercury's perihelion advance is just a SCAM! <shrug>

[Gertrude Kräuter]

Koobee Wublee wrote:

> The Euler-Lagrange equations are pretty much what govern the physical
> laws discovered by the self-claimed scientists. The Lagrangian method
> (to derive the special equations) was discovered during Euler's time
> that was slightly before Newton. <shrug>

Good post. So we can learn something. I wonder why not more people are
posting derivations direct related to Relativity.

[al...@interia.pl]

W dniu poniedziałek, 25 stycznia 2016 09:27:07 UTC+1 użytkownik Koobee Wublee

> So, when dealing with Mercury's perihelion advance, the Euler-Lagrange equation cannot possibly apply since with advancing perihelion, there is no way to establish the condition required of the Euler-Lagrange equation. The whole fixed, coherent derivation of Mercury's perihelion advance is just a SCAM!

Maybe it's somethig... but a direct
computition using the energy-force relation,
gives correct result:

df = -3(v/c)^2, using the additional term of energy: 3/8 v^2/c^4,
and now the precession is just: 3pi k/p... what is observed.

[Dirk Van de moortel]

Op 25-jan-2016 om 17:59 schreef Gertrude Kräuter:

See
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PrivateLagrangian.html

Ten yuears ago he was an idiot already.

Dirk Vdm

[Odd Bodkin]

Spectacular self delusion.

--
Odd Bodkin --- maker of fine toys, tools, tables

[Koobee Wublee]

On January 25, 2016, Dirk Van de moortel aka the sperm lover wrote:
> Koobee Wublee wrote:

> > The Euler-Lagrange equations are pretty much what govern the
> > physical laws discovered by the self-claimed scientists. The
> > Lagrangian method (to derive the special equations) was
> > discovered during Euler's time that was slightly before Newton.
> > <shrug>
>

> > It starts with the following action as an integral of a
> > Lagrangian (density to this action) with respect to instances
> > (time). <shrug>
>
> > ** A = Integral(T1, T2)[L dt]
>
> > Where
>
> > ** A = action accumulated from T1 to T2
> > ** T1 = starting instance
> > ** T2 = ending instance
> > ** L = Lagrangian
> > ** dt = discrete instance
>
> > For a special path, s, there exists a stationary condition to
> > this action from T1 to T2 along this path. So,
>
> > ** dA/ds = 0, definition of stationary action (min, max, saddle)
>
> > Or
>
> > ** Integral(T1, T2)[@L/@s dt] = 0
>
> > Where
>
> > ** @ = partial derivative operator
>

> > Now, you can choose to describe L in an observer's coordinate

> http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/PrivateLagrangian.html
>
> Ten yuears ago he was an idiot already.

Ten years ago, Koobee Wublee had defined the potential energy as positive. So, the sign in front must be reversed. Ten years later, Dirk van de moortel aka the sperm lover still does not get it. <shrug>

[JanPB]

95% nonsense, not worth discussing.

--
Jan

[Koobee Wublee]

On January 25, 2016, JanPB the relativistic moron wrote:
> Koobee Wublee wrote:

> > The Euler-Lagrange equations are pretty much what govern the
> > physical laws discovered by the self-claimed scientists. The
> > Lagrangian method (to derive the special equations) was
> > discovered during Euler's time that was slightly before Newton.
> > <shrug>
>
> > It starts with the following action as an integral of a
> > Lagrangian (density to this action) with respect to instances
> > (time). <shrug>
>
> > ** A = Integral(T1, T2)[L dt]
>
> > Where
>
> > ** A = action accumulated from T1 to T2
> > ** T1 = starting instance
> > ** T2 = ending instance
> > ** L = Lagrangian
> > ** dt = discrete instance
>
> > For a special path, s, there exists a stationary condition to
> > this action from T1 to T2 along this path. So,
>
> > ** dA/ds = 0, definition of stationary action (min, max, saddle)
>
> > Or
>
> > ** Integral(T1, T2)[@L/@s dt] = 0
>
> > Where
>
> > ** @ = partial derivative operator
>

> > Now, you can choose to describe L in an observer's coordinate

95% of what Koobee Wublee has written is what the self-styled physicists have called the calculus of variations. This shows how ignorant Jan the relativistic moron is. <shrug>

[JanPB]

No. It's the LSD version of it :-)

--
Jan

[Koobee Wublee]

> No. It's the LSD version of it :-)

This is a ridiculous comment from an imbecile who knows no math about differential geometry but myths. It is the same imbecile who raises objection when Koobee Wublee declared to use the polar coordinate system to describe a geometry. What an imbecile whether on LSD or not! <shrug>

[JanPB]

You can use polar coordinates but they are not the whole story. Before solving
Einstein's equation one makes more coordinate changes to make the metric diagonal.
(The Einstein equation itself does not necessarily guarantee the diagonal form.)

It is this extra coordinate change that introduces certain new artifacts:

1. a radial coordinate shift so that Schwarzschild's "r=0' no longer refers to the origin
but is a locus of points forming a sphere, see https://drive.google.com/open?id=0B2N1X7SgQnLRV1otWnE2T2F6aEk

2. the metric tensor extends over r=0 (i.e. the singularity there is not real).

--
Jan

[Koobee Wublee]

> Before solving Einstein's equation one makes more coordinate
> changes to make the metric diagonal.

More bullshit! Anyone can solve the field equations after deciding on a coordinate system through computing the Christoffel symbols of the second kind, the Ricci tensor, and the Ricci scalar. In vacuum, the field equations aka the Einstein tensor degenerates into the null Ricci tensor. Thus, the Ricci scalar is not needed. <snrug>

> (The Einstein equation itself does not necessarily guarantee the
> diagonal form.)

More bullshit! Schwarzschild had already solved the null Ricci tensor prior to 1915. At first, according to his paper, he started with the rectangular coordinate system but found the metric to be non-diagonal. So, he wisely transformed the Cartesian coordinate system to the polar coordinate system. In doing, the metric becomes diagonal. So, he only had to deal with 4 differential equations. Demanding spherical symmetric which is not a demand too far stretched as the Einstein dingleberries tried to discredit Schwarzschild's work, these 4 differential equations degenerate into just 3. From that, any solution can be solved by experts in solving these differential equations. <shrug>

> It is this extra coordinate change that introduces certain new
> artifacts:

Jan the relativistic moron is very ignorant. The reason why Schwarzschild decided to transform the polar coordinate system into another is that he was not aware that the null Einstein tensor will degenerate into the null Ricci tensor. So, because of the sqrt(- determinant(the metric)) term in the Lagrangian to the field equations that Hilbert had conveniently pulled out of his ass from, if Schwarzschild is able to find another coordinate system that will result in a determinant of -1, the null Einstein tensor will become the null Ricci tensor. That was the motivation behind Schwarzschild's part. Jan the relativistic moron is an idiot. The imbecile thinks it knows a lot, but it actually is very fvcking ignorant. One-on-one discussion is very difficult to make Jan the relativistic moron understand that Jan the relati1vistic moron is indeed an imbecile. Hopefully, the next generation will pick it up. <shrug>

> [rest of garbage from Jan the relativistic moron snipped]

[JanPB]

On Tuesday, January 26, 2016 at 11:57:10 PM UTC-8, Koobee Wublee wrote:
> On January 26, 2016, JanPB the relativistic moron wrote:
> >
> > Before solving Einstein's equation one makes more coordinate
> > changes to make the metric diagonal.
>
> More bullshit! Anyone can solve the field equations after deciding on a coordinate system through computing the Christoffel symbols of the second kind, the Ricci tensor, and the Ricci scalar.

Sure one can solve the equations but this solving process is based on changing
the coordinates so that the metric becomes diagonal. To my knowledge nobody
solved the Einstein equation without assuming the metric in the diagonal form.

Point is, the field equations in the standard coordinates may or may not
yield the solution in the diagonal form: we don't know because we never solve
them that way.

Here is an example of a non-diagonal metric in the a 2D spacetime:

ds^2 = dT dX

It is in fact the flat Minkowski metric (shouldn't be hard to see) but it's
not diagonal. And without the coordinate change _we don't know_ that the
unchanged Einstein equation does not produce a metric with off-diagonal terms
like the one above.

> In vacuum, the field equations aka the Einstein tensor degenerates into the null Ricci tensor. Thus, the Ricci scalar is not needed. <snrug>

Whatever. Not relevant.

> > (The Einstein equation itself does not necessarily guarantee the
> > diagonal form.)
>
> More bullshit!

You can yell all you want but Einstein's equation is a system of PDEs about
which we don't know whether it produces a diagonal metric or not. Nobody
knows that, nobody solved these equations "straight". Instead, one transforms
the coordinates to get rid of the t-dependence of the g_ij AND to get rid
of the cross terms.

> Schwarzschild had already solved the null Ricci tensor prior to 1915. At first, according to his paper, he started with the rectangular coordinate system but found the metric to be non-diagonal. So, he wisely transformed the Cartesian coordinate system to the polar coordinate system. In doing, the metric becomes diagonal.

It's not so fast: you have to make sure g_ij do not depend on time
(the x4 coordinate) and there are no cross terms g_i4 (i = 1,2,3).
This is not guaranteed to happen without a coordinate change. For example,
the following is another example of the Minkowski flat 2D metric:

ds^2 = -T^2 dT^2 + dX^2

See? The first coefficient is T-dependent. How do you know that the
un-transformed Einstein's equation does not force this sort of thing on you?

> So, he only had to deal with 4 differential equations. Demanding spherical symmetric which is not a demand too far stretched as the Einstein dingleberries tried to discredit Schwarzschild's work, these 4 differential equations degenerate into just 3. From that, any solution can be solved by experts in solving these differential equations. <shrug>

Point is, neither you, nor I, nor Schwarzschild, nor anybody else can
guarantee the diagonal form of the metric in the solution without a coordinate
change _on top_ of the obvious change to the spherical system.

> > It is this extra coordinate change that introduces certain new
> > artifacts:
>
> Jan the relativistic moron is very ignorant. The reason why Schwarzschild decided to transform the polar coordinate system into another is that he was not aware that the null Einstein tensor will degenerate into the null Ricci tensor.

I'm not talking about that. I'm talking about the coordinate change which
he only mentions implicitly:

"Then, according to Mr. Einstein, loc. cit. p. 833, the following conditions
must be fulfilled too:"

...which is then followed by the diagonality statement. In other words, he does
not dwell on it, he just quotes Einstein's paper and moves on.

> So, because of the sqrt(- determinant(the metric)) term in the Lagrangian to the field equations that Hilbert had conveniently pulled out of his ass from, if Schwarzschild is able to find another coordinate system that will result in a determinant of -1, the null Einstein tensor will become the null Ricci tensor. That was the motivation behind Schwarzschild's part. Jan the relativistic moron is an idiot. The imbecile thinks it knows a lot, but it actually is very fvcking ignorant. One-on-one discussion is very difficult to make Jan the relativistic moron understand that Jan the relati1vistic moron is indeed an imbecile. Hopefully, the next generation will pick it up. <shrug>

This has nothing to do with this discussion. Read what I write more carefully
next time so you don't waste time typing irrelevancies in.

--
Jan

[David Waite]

The ones I recently posted aren't diagonal. Neither is Brinkmann's That being said, not anyone can solve them. Though many have replicated solutions in hindsight few have found independent solutions for the first time. KW certainly can't solve them. Also as a note to KW the Ricci-scalar does matter in the region where the massive source is. The Ricci-scalar will be nonzero where the matter isn't comprised of massless particles. So for example you may have a region with charge, and inside that region the Ricci-scalar is not zero, while outside that region you may have an elecromagnetic field yielding a nonzero stress-energy tensor, but that external region has a zero Ricci-scalar because the photon is massless.

[Koobee Wublee]

On January 26, 2016, JanPB the relativistic moron wrote:

> > Anyone can solve the field equations after deciding on a
> > coordinate system through computing the Christoffel symbols of
> > the second kind, the Ricci tensor, and the Ricci scalar.
>
> Sure one can solve the equations but this solving process is based
> on changing the coordinates so that the metric becomes diagonal.

That is the beauty of the spherically symmetric polar coordinate system. In this coordinate system, you can now assume if the solution is diagonal. If no solution found, they you need to do something else. <shrug>

> To my knowledge nobody solved the Einstein equation without
> assuming the metric in the diagonal form.

Jan the relativistic moron is wrong again. Schwarzschild was the first to have solved the null Ricci tensor years before 1915. <shrug>

> Point is, the field equations in the standard coordinates may or
> may not yield the solution in the diagonal form: we don't know
> because we never solve them that way.

Nonsense. You have to first to try using the polar coordinate system because under this coordinate system, the metric becomes diagonal. Schwarzschild proved that although with a tiny mistake that would not proliferate to the later derivation of Schwarzschild's original metric. <shrug>

> Here is an example of a non-diagonal metric in the a 2D spacetime:
>
> ds^2 = dT dX

This metric yields null determinant. You are not allowed to go any further. <shrug>

> > In vacuum, the field equations aka the Einstein tensor degenerates
> > into the null Ricci tensor. Thus, the Ricci scalar is not needed.

> > <shrug>
>
> Whatever. Not relevant.

That is correct. You don't have to computer the Ricci scalar if looking for a vacuum solution. Doesn't Jan the relativistic moron know that null Einstein and the null Ricci tensors are the same. Koobee Wublee shall leave that as a homework exercise for the readers. Prove null Einstein tensor = null Ricci tensor. <shrug>

> [nonsense snipped]

> > Schwarzschild had already solved the null Ricci tensor prior to
> > 1915. At first, according to his paper, he started with the
> > rectangular coordinate system but found the metric to be
> > non-diagonal. So, he wisely transformed the Cartesian coordinate
> > system to the polar coordinate system. In doing, the metric
> > becomes diagonal.
>
> It's not so fast: you have to make sure g_ij do not depend on time
> (the x4 coordinate) and there are no cross terms g_i4 (i = 1,2,3).

Gee! How fvcking stupid can Jan the relativistic moron be? Schwarzschild was looking for a static solution that will degenerate into the Newtonian gravity. <shrug>

[more bullshit snipped]

> > So, he only had to deal with 4 differential equations. Demanding
> > spherical symmetric which is not a demand too far stretched as
> > the Einstein dingleberries tried to discredit Schwarzschild's
> > work, these 4 differential equations degenerate into just 3.
> > From that, any solution can be solved by experts in solving these
> > differential equations. <shrug>
>
> Point is, neither you, nor I, nor Schwarzschild, nor anybody else
> can guarantee the diagonal form of the metric in the solution
> without a coordinate change _on top_ of the obvious change to the
> spherical system.

Nonsense! Jan the relativistic moron obviously has never solved any differential equations. <shrug>

> > The reason why Schwarzschild decided to transform the polar
> > coordinate system into another is that he was not aware that
> > the null Einstein tensor will degenerate into the null Ricci
> > tensor.
>
> I'm not talking about that. I'm talking about the coordinate
> change which he only mentions implicitly:
>
> "Then, according to Mr. Einstein, loc. cit. p. 833, the following
> conditions must be fulfilled too:"
>
> ...which is then followed by the diagonality statement. In other
> words, he does not dwell on it, he just quotes Einstein's paper
> and moves on.

This means nothing. <shrug>

> > So, because of the sqrt(- determinant(the metric)) term in the
> > Lagrangian to the field equations that Hilbert had conveniently
> > pulled out of his ass from, if Schwarzschild is able to find
> > another coordinate system that will result in a determinant
> > of -1, the null Einstein tensor will become the null Ricci
> > tensor. That was the motivation behind Schwarzschild's part.
> > Jan the relativistic moron is an idiot. The imbecile thinks it
> > knows a lot, but it actually is very fvcking ignorant.
> > One-on-one discussion is very difficult to make Jan the
> > relativistic moron understand that Jan the relati1vistic moron
> > is indeed an imbecile. Hopefully, the next generation will
> > pick it up. <shrug>
>
> This has nothing to do with this discussion. Read what I write
> more carefully next time so you don't waste time typing
> irrelevancies in.

Jan the relativistic moron has never solved a differential equation before. <shrug>

[al...@interia.pl]

Another stupid nonsense...

a transform type of;
r = r - a, is just a spimple translation alnog r in the polar coord...

therefore the so-called standard Schwarzchild's solution
is just a translated version of the oryginal solution,
where the r parameter is limited: r > rs.

the region r < 2m, simply doesnt exists in the solution,
and the point r = 2m is just the place of the mass m,
which is in standard polar cood. for r = 0.

[JanPB]

On Wednesday, January 27, 2016 at 9:50:54 AM UTC-8, Koobee Wublee wrote:

> On January 26, 2016, JanPB the relativistic moron [hello?] wrote:
> >
> > Sure one can solve the equations but this solving process is based
> > on changing the coordinates so that the metric becomes diagonal.
>
> That is the beauty of the spherically symmetric polar coordinate system. In this coordinate system, you can now assume if the solution is diagonal.

If you change to the spherical system, all you do is you change the Einstein
equation to the spherical system. You _don't_ get the diagonality yet, simply
because there is no reason for it. The transformed field equations might as
well produce something like the example of the Minkowski (flat) metric I gave:

ds^2 = -T^2 dT^2 + dX^2

You don't know it until you solve the PDEs and in this non-diagonal form
they are FAPP impossible to solve. So you don't know. Fortunately, one
can show that _another_ coordinate transformation can get rid of the time
dependence of the g_ij and the cross terms. Many textbooks are quite sloppy at
this point and simply state that the g_ij can be "assumed" to be t-independent
because of the staticity assumption. This is actually false as the above simple
counterexample demonstrates. In that counterexample you also need _another_
coordinate transformation to get rid of the T-dependence:

T = sqrt(2t)
X = x

> > To my knowledge nobody solved the Einstein equation without
> > assuming the metric in the diagonal form.
>
> Jan the relativistic moron is wrong again. Schwarzschild was the first to have solved the null Ricci tensor years before 1915. <shrug>

Reread what I wrote. I said nobody solved the _original_ equation, meaning
the original (t,x,y,z) coordinates and nothing assumed about g_ij except
the symmetry and staticity. Without the coordinate changes required to
make the metric diagonal, the Einstein equation is an impenetrable mess.

> > Point is, the field equations in the standard coordinates may or
> > may not yield the solution in the diagonal form: we don't know
> > because we never solve them that way.
>
> Nonsense.

Prove it. Show us that the Einstein equation after transforming to
spherical coordinates produces g_ij independent of time and no cross
terms. Good luck.

> You have to first to try using the polar coordinate system because under this coordinate system, the metric becomes diagonal.

It doesn't. It becomes diagonal only after a _second_ coordinate change.
That second coordinate change is the root cause of all the misunderstandings
regarding Schwarzschild's paper. It is this _second_ coordinate change that
introduces the spherical-r coordinate shift (and other changes) so that, for
example, the transformed r (let's denote it by R) acquires the property that
R=0 is not pointlike.

> > Here is an example of a non-diagonal metric in the a 2D spacetime:
> >
> > ds^2 = dT dX
>
> This metric yields null determinant.

No, the determinant is equal to -1/4. (Of course you'll never admit
you've just made a mistake.)

> You are not allowed to go any further. <shrug>

Brush up on your linear algebra.

> > > In vacuum, the field equations aka the Einstein tensor degenerates
> > > into the null Ricci tensor. Thus, the Ricci scalar is not needed.
> > > <shrug>
> >
> > Whatever. Not relevant.
>
> That is correct.

Good.

> > > Schwarzschild had already solved the null Ricci tensor prior to
> > > 1915. At first, according to his paper, he started with the
> > > rectangular coordinate system but found the metric to be
> > > non-diagonal. So, he wisely transformed the Cartesian coordinate
> > > system to the polar coordinate system. In doing, the metric
> > > becomes diagonal.
> >
> > It's not so fast: you have to make sure g_ij do not depend on time
> > (the x4 coordinate) and there are no cross terms g_i4 (i = 1,2,3).
>
> Gee! How fvcking stupid can Jan the relativistic moron be? Schwarzschild was looking for a static solution that will degenerate into the Newtonian gravity. <shrug>

Again, staticity does not imply time independence of the g_ij, this is a
common bit of sloppiness but for the third time, here is a simple
counterexample of a static metric (Minkowski flat, in fact) which nevertheless
has a time-dependent coefficient:

ds^2 = -T^2 dT^2 + dX^2

[...]

> > > The reason why Schwarzschild decided to transform the polar
> > > coordinate system into another is that he was not aware that
> > > the null Einstein tensor will degenerate into the null Ricci
> > > tensor.
> >
> > I'm not talking about that. I'm talking about the coordinate
> > change which he only mentions implicitly:
> >
> > "Then, according to Mr. Einstein, loc. cit. p. 833, the following
> > conditions must be fulfilled too:"
> >
> > ...which is then followed by the diagonality statement. In other
> > words, he does not dwell on it, he just quotes Einstein's paper
> > and moves on.
>
> This means nothing. <shrug>

What do you mean, "nothing"? You mean Einstein did nothing in the loc. cit.?
Do you assume that a scientist writing for other professionals always
writes everything down explicitly for your convenience? You just take
everything at face value without examining anything on your own?

--
Jan

[JanPB]

The extra coordinate change I'm talking about is not just a shift in r.
It also involves a logarithmic-like transformation of the time coordinate
with the logarithm blowing up at the horizon (that's why the coordinate
singularity ends up there).

> the region r < 2m, simply doesnt exists in the solution,
> and the point r = 2m is just the place of the mass m,
> which is in standard polar cood. for r = 0.

False.

--
Jan

[Juan Sebastián]

JanPB wrote:

> ds^2 = -T^2 dT^2 + dX^2
>
> You don't know it until you solve the PDEs and in this non-diagonal form
> they are FAPP impossible to solve. So you don't know. Fortunately, one
> can show that _another_ coordinate transformation can get rid of the
> time dependence of the g_ij and the cross terms. Many textbooks are
> quite sloppy at this point and simply state that the g_ij can be
> "assumed" to be t-independent because of the staticity assumption. This
> is actually false as the above simple counterexample demonstrates. In
> that counterexample you also need _another_ coordinate transformation to
> get rid of the T-dependence:

Are you going to write a book with this stuff unsloppy.

[Juan Sebastián]

Suggestion for the title: "Relativity, when the stuff in it is not sloppy"

[Koobee Wublee]

> Show us that the Einstein equation after transforming to spherical
> coordinates produces g_ij independent of time and no cross terms.
> Good luck.

Once again, obviously, Jan the relativistic moron has never solved any differential equations. To solve such, you have to start with some educated guessing. One such assumption is the static case hoping to find one such solution that will degenerate into Newtonian gravity at weak curvature of spacetime. <shrug>

Besides, the assumption that space and time can possibly intertwine lie what you are so fvcking gung ho over has never shown such a necessary merit of doing so. It is not found in the Minkowski spacetime. Why do you expect it to find it in the real world? The equation describing a segment of spacetime geometry can very well be defined as follows. <shrug>

** ds^2 = c^2 [g]_00 dt^2 - [g]_ij d[q]^i d[q]^j

Where

** i, j = 1, 2, 3 of special dimensions only
** [g] = 3x3 matrix that represents the metric
** [q] = 3x1 (or 1x3) matrix that represents the coordinate system

Koobee Wublee does not have to prove the stupidity of the silly assumptions done by the mystic generations to follow after the founding fathers of GR. So, the burden is up to Jan the relativistic moron to prove time and space can intertwine. Good luck with that. <shrug>

> > You have to first to try using the polar coordinate system because
> > under this coordinate system, the metric becomes diagonal.
>
> It doesn't. It becomes diagonal only after a _second_ coordinate
> change.

No, Schwarzschild started with the Cartesian coordinate system and found the metric cannot possibly be diagonal. Thus, being a person (very unlike Jan the relativistic moron) that had solved many differential equations in the past, instinct just told him rightfully that the polar coordinate system will certainly yield a diagonal metric. Solving such a complex set of differential equations require some trials and errors in which Jan the relativistic has no experiences in doing so. <shrug>

> That second coordinate change is the root cause of all the
> misunderstandings regarding Schwarzschild's paper.

No, there is no mysticism to the second coordinate change. After Nordstrom had suggested the null Ricci tensor as a more precise modeling of gravity after the Laplace equation, Schwarzschild had spent many years afterwards trying to solve these differential equations associated with the null Ricci tensor. After Hilbert, which later on plagiarized by Einstein the nitwit, the plagiarist, and the liar, had presented the field equations with the trace term appended to the Ricci tensor, Schwarzschild, in order to salvage all his works on the null Ricci tensor, chose to find a coordinate system that will yield a determinant of -1. In doing so, the null Einstein tensor with the trace term becomes the null Ricci tensor without the trace term. He did not have to do that since the null Einstein tensor is mathematically the same as the null Ricci tensor, but he did not realize so. After finding a solution with the new and transformed coordinate system, he, as a good mathematician and physicist, did not forget to transform it back to the polar coordinate system. There is no misunderstanding. The only misunderstanding is that Jan the relativistic moron tries to turn this episode into a myth. <shrug>

> No, the determinant is equal to -1/4.

You are correct. Koobee Wublee was thrown astray by your expertise in bringing up irrelevant topics. Although this is not the first time, Koobee Wublee has been learning these hard lessons. <shrug>

> If you change to the spherical system, all you do is you change the
> Einstein equation to the spherical system. You _don't_ get the
> diagonality yet, simply because there is no reason for it. The
> transformed field equations might as well produce something like
> the example of the Minkowski (flat) metric I gave:
>
> ds^2 = -T^2 dT^2 + dX^2

Notice there is no hints of time and space can possibly be intertwined to create non-diagonal metric. However, that is a topic worth of philosophic discussions not really a scientific one. With that said, why doesn't Jan the relativistic moron show how Jan the relativistic moron would lay out these differential equations of the field equations into whatever the God coordinate system that Jan the relativistic moron has in mind. Koobee Wublee thinks this is a fair challenge since Jan the relativistic moron has been tossing divine scriptures from the religion of relativity for quite some time without getting its hands dirty. <shrug>

Just for comparison, Koobee Wublee would follow Schwarzschild's footsteps and declare the polar coordinate system as the choice of coordinate system to describe the metric solved as solution in later differential equations derived from the field equations, and no other coordinate system can possibly fvcking contaminate with this decision made by Koobee Wublee. This means the metric must be diagonal. So, only 4 differential equations are involved. Of course, the extra assumption of spherical symmetry reduces to just 3 differential equations. Then, the Christoffel symbols of the second kind are to be derived from the polar coordinate system and no other fvcking coordinate systems are fvcking allowed. They are followed by the Ricci tensor also described in the polar coordinate system and nothing else. Only then, Koobee Wublee following Schwarzschild's footsteps would be able to attempt to solve these 3 differential equations. Again, the solution can only be applied to the polar coordinate system and no fvcking others. <shrug>

[JanPB]

On Wednesday, January 27, 2016 at 11:49:06 PM UTC-8, Koobee Wublee wrote:

> On January 27, 2016, JanPB the relativistic moron[sic] wrote:
>
> > Show us that the Einstein equation after transforming to spherical
> > coordinates produces g_ij independent of time and no cross terms.
> > Good luck.
>
> Once again, obviously, Jan the relativistic moron has never solved any differential equations. To solve such, you have to start with some educated guessing. One such assumption is the static case hoping to find one such solution that will degenerate into Newtonian gravity at weak curvature of spacetime. <shrug>

Yes, correct so far.

> Besides, the assumption that space and time can possibly intertwine lie what you are so fvcking gung ho over has never shown such a necessary merit of doing so.

Stop! Here is the problem: you simply _assume_ that in a spherically symmetric static vacuum
the coordinate space t=const. and the coordinate time (xyz)=const. are orthogonal. But there
is simply no reason for that. And you cannot deduce this from the Einstein equation either:
without _assuming_ diagonal metric that equation is a very complicated system of PDEs, so
you simply don't know if its solution (assuming you had it) would NOT be analogous to
something like:

ds^2 = dt dx (cross-terms present)

or:

ds^2 = -t^2 dt^2 + dx^2 (a time-dependent g_tt)

...both of which are _static spherically symmetric vacua_ (in fact they are both the flat
Minkowski space or a portion thereof).

In fact, with hindsight, we know that this is precisely what happens: the curvature of
spacetime in the large is such that _insisting_ on [1] coordinate space and coordinate time
orthogonality (i.e. no cross-terms) and insisting on [2] labelling the spheres of symmetry
by (essentially) the square root of their areas FORCES the coordinate spaces to bend
near the horizon _so much_ that they are not able to cross it (they look like graphs of
2M * ln|r - 2M| + (arbitrary constants) on the Eddington-Finkelstein diagram).

You are free to _change_ the coordinates so that there are no cross terms and g_ij are
time-independent but this means forfeiting the standard spherical coordinates in R^3. From
that point on you cannot claim that your system's domain covers the entire manifold anymore,
you cannot claim any singularity of the coefficients g_ij is a real singularity of the
manifold, and you cannot claim any coordinate-values-specific implication for the geometry
like "(0,0,0) is the centre" or "(0,0,0) is pointlike", etc. etc. etc.

BTW, if you are willing to give up the constraint [2] above, then you can get a diagonal
metric without coordinate singularities, that's how the Kruskal-Szekeres does it.

> It is not found in the Minkowski spacetime.

It is:

ds^2 = dt dx

or:

ds^2 = -t^2 dt^2 + dx^2

...or countless others. How do you know that a complicated mess of Christoffel symbols, each
of which is a sum involving g_ij and their derivatives, does _not_ yield a solution of
the above type?? You need to _prove_ it if you want to use it. Plus, you are also assuming
that the solution would label the spheres of symmetry by sqrt(their area/4pi). In other
words, you are assuming the existence of the term r^2*dO^2. How do you know it's not
some _other_ function of r?? In fact, as Kruskal-Szekeres coordinates show, you _do_ need
a rather complicated function f(r) in that term, so it's equal to f(r)*dO^2 there.

> Why do you expect it to find it in the real world?

Because you _don't know_ otherwise. One cannot just assume things without proof. If you
don't know something, you keep the option open. If it turns out after you're done that
the assumption was in fact correct, fine. But you cannot _start_ by assuming stuff willy-nilly.

> The equation describing a segment of spacetime geometry can very well be defined as follows. <shrug>
>
> ** ds^2 = c^2 [g]_00 dt^2 - [g]_ij d[q]^i d[q]^j
>
> Where
>
> ** i, j = 1, 2, 3 of special dimensions only
> ** [g] = 3x3 matrix that represents the metric
> ** [q] = 3x1 (or 1x3) matrix that represents the coordinate system

It can be defined that way after the aforementioned coordinate change. Here you are already
past the step of aligning the original coordinate system with the hypersurfaces that define
the word "static". This alignment is accomplished by a coordinate change.

> Koobee Wublee does not have to prove the stupidity of the silly assumptions done by the mystic generations to follow after the founding fathers of GR. So, the burden is up to Jan the relativistic moron to prove time and space can intertwine. Good luck with that. <shrug>

The burden is on you though. I _don't_ assume something, you _do_ assume something.

> > > You have to first to try using the polar coordinate system because
> > > under this coordinate system, the metric becomes diagonal.
> >
> > It doesn't. It becomes diagonal only after a _second_ coordinate
> > change.
>
> No, Schwarzschild started with the Cartesian coordinate system and found the metric cannot possibly be diagonal. Thus, being a person (very unlike Jan the relativistic moron) that had solved many differential equations in the past, instinct just told him rightfully that the polar coordinate system will certainly yield a diagonal metric.

It doesn't. The change from the Cartesians to the sphericals will simply change one complicated
mess of PDEs into another. You still don't know if that other mess does _not_ produce solutions
of the form I quoted above for the 2D Minkowski flat spacetime.

Schwarzschild quotes Einstein for that step, I haven't read his paper, it would be
of historical interest to see how Einstein describes the process (although his opinion on
the subject is not relevant as mathematics does not depend on _who_ is saying stuff).

> Solving such a complex set of differential equations require some trials and errors in which Jan the relativistic has no experiences in doing so. <shrug>

Except ages ago I posted a derivation of the solution:
https://drive.google.com/file/d/0B2N1X7SgQnLRQ2ctaXA4aTVId28

> > That second coordinate change is the root cause of all the
> > misunderstandings regarding Schwarzschild's paper.
>
> No, there is no mysticism to the second coordinate change.

Nobody says there is any "mysticism" there. Just an easily overlooked step.

> > No, the determinant is equal to -1/4.
>
> You are correct.

Nice of you to say this!

> Koobee Wublee was thrown astray by your expertise in bringing up irrelevant topics.

Meaning: we were discussing whether a static spherically symmetric metric must be diagonal,
and I wrote down a couple of easy counterexamples.

How that is "astray" I shall never know :-)

> > If you change to the spherical system, all you do is you change the
> > Einstein equation to the spherical system. You _don't_ get the
> > diagonality yet, simply because there is no reason for it. The
> > transformed field equations might as well produce something like
> > the example of the Minkowski (flat) metric I gave:
> >
> > ds^2 = -T^2 dT^2 + dX^2
>
> Notice there is no hints of time and space can possibly be intertwined to create non-diagonal metric.

Point is, there are no hints of time and space NOT intertwined. You cannot just assume
stuff - if you don't know something, you have to keep the option open.

> However, that is a topic worth of philosophic discussions not really a scientific one.

This is not a philosophy problem but a mathematical one (or logical).

> With that said, why doesn't Jan the relativistic moron show how Jan the relativistic moron would lay out these differential equations of the field equations into whatever the God coordinate system that Jan the relativistic moron has in mind. Koobee Wublee thinks this is a fair challenge since Jan the relativistic moron has been tossing divine scriptures from the religion of relativity for quite some time without getting its hands dirty. <shrug>

The definition of "static"(*) includes a construction of such coordinate system. This system
is thus not given _explicitly_. The assumption is that such coordinate change _exists_.
Ultimately the justification for this assumption is that the metric found using this
approach does in fact satisfy Ricci=0.

(*) Spacetime is static if there exists a timelike Killing field(**) in it (that's the definition
of "stationary") AND this field is hypersurface-orthogonal. The required coordinate system
is then set up by aligning these hypersurfaces with coordinate spaces of t=const. See e.g.
Wald's "General Relativity" or MTW or many others.

(**) In other words, translating the metric along this field's flow lines does not change it.

> Just for comparison, Koobee Wublee would follow Schwarzschild's footsteps and declare the polar coordinate system as the choice of coordinate system to describe the metric solved as solution in later differential equations derived from the field equations, and no other coordinate system can possibly fvcking contaminate with this decision made by Koobee Wublee.

Perhaps you are following Schwarzschild, point is neither Schwarzschild nor Einstein nor
anybody else can just assume that the polar system implies the Einstein equation magically
will yield solutions in the diagonal form, labelling the spheres of symmetry by sqrt(area/4pi)
to boot. Anything else you care to assume? Why not just assume that the solution is
the flat space and be done with it? This would be an equally arbitrary nonsense assumption.

> This means the metric must be diagonal.

No, it's just wishful thinking.

> So, only 4 differential equations are involved. Of course, the extra assumption of spherical symmetry reduces to just 3 differential equations. Then, the Christoffel symbols of the second kind are to be derived from the polar coordinate system and no other fvcking coordinate systems are fvcking allowed. They are followed by the Ricci tensor also described in the polar coordinate system and nothing else. Only then, Koobee Wublee following Schwarzschild's footsteps would be able to attempt to solve these 3 differential equations. Again, the solution can only be applied to the polar coordinate system and no fvcking others. <shrug>

Footsteps don't matter. Correctness matters.

--
Jan

[Koobee Wublee]

> Stop! Here is the problem:

The problem is that Jan the relativistic moron has been claiming voodoo mathematics, and Jan the relativistic moron has never solved any differential equations before. So, why doesn't Jan the relativistic moron show us how the Schwarzschild metric is derived without using any coordinate system, without computing for the Christoffel symbols, nor without deriving for the Ricci tensor? It just cannot be done. It is only a myth without no proof. <shrug>

As comparison, below shows how such a solution that is static, spherically symmetric, an asymptotically flat can be solved. <shrug>

> > Schwarzschild started with the Cartesian coordinate system and
> > found the metric cannot possibly be diagonal. Thus, being a
> > person (very unlike Jan the relativistic moron) that had solved
> > many differential equations in the past, instinct just told him
> > rightfully that the polar coordinate system will certainly yield

> > a diagonal metric. Solving such a complex set of differential

> > equations require some trials and errors in which Jan the
> > relativistic has no experiences in doing so. <shrug>
>

> > After Nordstrom had suggested the null Ricci tensor as a more
> > precise modeling of gravity after the Laplace equation,
> > Schwarzschild had spent many years afterwards trying to solve
> > these differential equations associated with the null Ricci
> > tensor. After Hilbert, which later on plagiarized by Einstein
> > the nitwit, the plagiarist, and the liar, had presented the
> > field equations with the trace term appended to the Ricci tensor,
> > Schwarzschild, in order to salvage all his works on the null
> > Ricci tensor, chose to find a coordinate system that will yield
> > a determinant of -1. In doing so, the null Einstein tensor with
> > the trace term becomes the null Ricci tensor without the trace
> > term. He did not have to do that since the null Einstein tensor
> > is mathematically the same as the null Ricci tensor, but he did
> > not realize so. After finding a solution with the new and
> > transformed coordinate system, he, as a good mathematician and
> > physicist, did not forget to transform it back to the polar
> > coordinate system. There is no misunderstanding. The only
> > misunderstanding is that Jan the relativistic moron tries to
> > turn this episode into a myth. <shrug>

> > Just for comparison, Koobee Wublee would follow Schwarzschild's
> > footsteps and declare the polar coordinate system as the choice
> > of coordinate system to describe the metric solved as solution
> > in later differential equations derived from the field equations,
> > and no other coordinate system can possibly fvcking contaminate

> > with this decision made by Koobee Wublee. This means the metric
> > must be diagonal. So, only 4 differential equations are involved.

[Tom Roberts]

On 1/27/16 1/27/16 - 3:30 PM, JanPB wrote:
> Again, staticity does not imply time independence of the g_ij, this is a
> common bit of sloppiness but for the third time, here is a simple
> counterexample of a static metric (Minkowski flat, in fact) which nevertheless
> has a time-dependent coefficient:
>
> ds^2 = -T^2 dT^2 + dX^2

Yes. In GR, "static" means the manifold has a timelike Killing vector.

For coordinates in which the metric components are independent of time, the time
coordinate _IS_ a timelike Killing vector. But as has been said many times,
there is no necessity of using such coordinates -- the line element above is for
the manifestly static Minkowski manifold, but using coordinates in which T is
not a timelike Killing vector.


Tom Roberts

[Tom Roberts]

On 1/29/16 1/29/16 - 12:46 PM, Tom Roberts wrote:
> On 1/27/16 1/27/16 - 3:30 PM, JanPB wrote:
>> Again, staticity does not imply time independence of the g_ij, this is a
>> common bit of sloppiness but for the third time, here is a simple
>> counterexample of a static metric (Minkowski flat, in fact) which nevertheless
>> has a time-dependent coefficient:
>>
>> ds^2 = -T^2 dT^2 + dX^2
>
> Yes. In GR, "static" means the manifold has a timelike Killing vector.

I forgot to add: which is orthogonal to a set of spacelike hypersurfaces the
union of which covers the region of interest.

[Juan Sebastián]

Tom Roberts wrote:

> On 1/27/16 1/27/16 - 3:30 PM, JanPB wrote:
>> Again, staticity does not imply time independence of the g_ij, this is
>> a common bit of sloppiness but for the third time, here is a simple
>> counterexample of a static metric (Minkowski flat, in fact) which
>> nevertheless has a time-dependent coefficient:
>>
>> ds^2 = -T^2 dT^2 + dX^2
>
> Yes. In GR, "static" means the manifold has a timelike Killing vector.

You mean it means steady-state.

> For coordinates in which the metric components are independent of time,
> the time coordinate _IS_ a timelike Killing vector. But as has been said
> many times, there is no necessity of using such coordinates -- the line
> element above is for the manifestly static Minkowski manifold, but using
> coordinates in which T is not a timelike Killing vector.

If there is no necessity, then you contradict what JanPB said.

[Juan Sebastián]

Tom Roberts wrote:

>>> ds^2 = -T^2 dT^2 + dX^2
>>
>> Yes. In GR, "static" means the manifold has a timelike Killing vector.
>
> I forgot to add: which is orthogonal to a set of spacelike hypersurfaces
> the union of which covers the region of interest.

The problem you have is that your 3d space manifold already has all three
axes orthogonal. Another vector orthogonal to all is calligraphically
impossible.

[JanPB]

On Friday, January 29, 2016 at 10:56:21 AM UTC-8, tjrob137 wrote:
> On 1/29/16 1/29/16 - 12:46 PM, Tom Roberts wrote:
> > On 1/27/16 1/27/16 - 3:30 PM, JanPB wrote:
> >> Again, staticity does not imply time independence of the g_ij, this is a
> >> common bit of sloppiness but for the third time, here is a simple
> >> counterexample of a static metric (Minkowski flat, in fact) which nevertheless
> >> has a time-dependent coefficient:
> >>
> >> ds^2 = -T^2 dT^2 + dX^2
> >
> > Yes. In GR, "static" means the manifold has a timelike Killing vector.
>
> I forgot to add: which is orthogonal to a set of spacelike hypersurfaces the
> union of which covers the region of interest.

Or yet another way: "...means the manifold has a timelike Killing vector which
is a gradient"(*).

(*) Assuming the usual identification of the gradient covector with
a vector.

--
Jan

[JanPB]

On Friday, January 29, 2016 at 10:58:35 AM UTC-8, Juan Sebastián wrote:

> Tom Roberts wrote:
>
> > For coordinates in which the metric components are independent of time,
> > the time coordinate _IS_ a timelike Killing vector. But as has been said
> > many times, there is no necessity of using such coordinates -- the line
> > element above is for the manifestly static Minkowski manifold, but using
> > coordinates in which T is not a timelike Killing vector.
>
> If there is no necessity, then you contradict what JanPB said.

I think Tom meant this generically: any coordinates are allowed in tensor
equations (but only some of them make the _calculations_ easy).

--
Jan

[JanPB]

On Thursday, January 28, 2016 at 5:45:22 PM UTC-8, Koobee Wublee wrote:

> On January 28, 2016, JanPB the relativistic moron[sic] wrote:
>
> > Stop! Here is the problem:
>
> The problem is that Jan the relativistic moron has been claiming voodoo mathematics, and Jan the relativistic moron has never solved any differential equations before.

Except for example here:
https://drive.google.com/file/d/0B2N1X7SgQnLRQ2ctaXA4aTVId28
etc.

> So, why doesn't Jan the relativistic moron show us how the Schwarzschild metric is derived without using any coordinate system, without computing for the Christoffel symbols, nor without deriving for the Ricci tensor?

Needless to say this is NOT what I said. What I said was simply that in
derivations of Schwarzschild's metric there is a coordinate change involved
which is frequently glossed over.

--
Jan

[Koobee Wublee]

> https://drive.google.com/file/d/0B2N1X7SgQnLRQ2ctaXA4aTVId28

You know what the answer is and very creative fudging is done staring with equations (i), (ii), (iii), and (iv). <shrug>

> Needless to say this is NOT what I said. What I said was simply
> that in derivations of Schwarzschild's metric there is a
> coordinate change involved which is frequently glossed over.

This is still wrong. Koobee Wublee has asked you to solve the differential equations not LSD induced voodoo math. For the readers, below was how Schwarzschild derived the original metric. <shrug>

[JanPB]

Translation: "I don't know how the Cartan calculus works".

> > Needless to say this is NOT what I said. What I said was simply
> > that in derivations of Schwarzschild's metric there is a
> > coordinate change involved which is frequently glossed over.
>
> This is still wrong. Koobee Wublee has asked you to solve the differential equations not LSD induced voodoo math.

That's a different topic. Besides, it's done at the note I linked:
https://drive.google.com/file/d/0B2N1X7SgQnLRQ2ctaXA4aTVId28

> For the readers, below was how Schwarzschild derived the original metric. <shrug>
>
> > > Schwarzschild started with the Cartesian coordinate system and
> > > found the metric cannot possibly be diagonal. Thus, being a
> > > person (very unlike Jan the relativistic moron) that had solved
> > > many differential equations in the past, instinct just told him
> > > rightfully that the polar coordinate system will certainly yield
> > > a diagonal metric.

No, it won't.

--
Jan

[Koobee Wublee]

> I don't know how the Cartan calculus works.

Cartan approach of Ricci tensor doesn't work. Your nonsense is created through liberal usage of LSD as you have eluded earlier. <shrug<>

For the readers, there is just no way to solve the field equations without deciding on a set of coordinate system and stick with it all the way. You are free to transform it to another, but at the end, you must translate it back to the original set of coordinate system you have started with. Then, the Christoffel symbols are derived. Then, the Ricci tensor. Then, solve these differential equations. <shrug>

What Schwarzschild did was the following. <shrug>

> > Schwarzschild started with the Cartesian coordinate system and
> > found the metric cannot possibly be diagonal. Thus, being a
> > person (very unlike Jan the relativistic moron) that had solved
> > many differential equations in the past, instinct just told him
> > rightfully that the polar coordinate system will certainly yield

> > a diagonal metric. Solving such a complex set of differential
> > equations require some trials and errors in which Jan the
> > relativistic has no experiences in doing so. <shrug>
>
> > After Nordstrom had suggested the null Ricci tensor as a more
> > precise modeling of gravity after the Laplace equation,
> > Schwarzschild had spent many years afterwards trying to solve
> > these differential equations associated with the null Ricci
> > tensor. After Hilbert, which later on plagiarized by Einstein
> > the nitwit, the plagiarist, and the liar, had presented the
> > field equations with the trace term appended to the Ricci tensor,
> > Schwarzschild, in order to salvage all his works on the null
> > Ricci tensor, chose to find a coordinate system that will yield
> > a determinant of -1. In doing so, the null Einstein tensor with
> > the trace term becomes the null Ricci tensor without the trace
> > term. He did not have to do that since the null Einstein tensor
> > is mathematically the same as the null Ricci tensor, but he did
> > not realize so. After finding a solution with the new and
> > transformed coordinate system, he, as a good mathematician and
> > physicist, did not forget to transform it back to the polar

> > coordinate system. <shrug>

[JanPB]

On Friday, January 29, 2016 at 3:59:18 PM UTC-8, Koobee Wublee wrote:
> On January 29, 2016, JanPB the relativistic moron wrote:

> Cartan approach of Ricci tensor doesn't work.

Of course it does. It's just decomposing the tensors wrt to an orthonormal
frame rather than wrt to a coordinate frame. There exists yet another approach
due to Newman and Penrose using null frames and spinors. Main advantage of
decomposing wrt an orthonormal frame is that the coefficients so obtained
are skew-symmetric rather than symmetric (as in the coordinate Christoffel
approach) so you end up with far fewer nonzero terms to deal with.

> For the readers, there is just no way to solve the field equations without deciding on a set of coordinate system and stick with it all the way.

Deciding which coordinate system to use is fine. Your problem is that
you insist that the equations, when written in a particular coordinate
system (the spherical one), will automagically produce a solution
in some form you find attractive (diagonal and with the spheres of
symmetry labelled by sqrt(area/4pi)).

Such "approach" leads nowhere. When I get home I'll write down the Einstein
equation in spherical coordinates and ask you to prove that it produces
diagonal metrics with the spheres of symmetry labelled as above. Good luck!

> You are free to transform it to another, but at the end, you must translate it back to the original set of coordinate system you have started with.

You can translate to _any_ coordinate system. Tensors are coordinate-invariant.

--
Jan

[Koobee Wublee]

On January 29, 2016, JanPB the relativistic moron wrote:

> > Cartan approach of Ricci tensor doesn't work.
>
> Of course it does. It's just decomposing the tensors wrt to an
> orthonormal frame rather than wrt to a coordinate frame.

For the readers, constructing the Ricci tensor cannot be that simple as Jan the relativistic moron has claimed so. Historically, you have to start with a coordinate system of your choice, compute the Christoffel symbols through minimizing the path of geodesics according to the Lagrangian method, derive the Riemann tensor by constrainting the divergence of two adjacent points in space, decimate the Riemann tensor into the Ricci tensor. It is a very fvcking tedious process. Koobee Wublee has been through it. Don't let these relativistic morons tell you how simple the Ricci tensor can be constructed. <shrug>

> Deciding which coordinate system to use is fine.

How do you translate your thing not represented by any coordinate system into one with coordinate system? Under differential geometry, only a known coordinate system can be transform into another coordinate system. <shrug>

> Your problem is that you insist that the equations, when written
> in a particular coordinate system (the spherical one), will
> automagically produce a solution in some form you find attractive
> (diagonal and with the spheres of symmetry labelled by
> sqrt(area/4pi)).

Not automatic. When a set of differential equations are presented, it is a matter of solving them. There is no magic unless you are talking about the magic thing of Cartan method. <shrug>

> Such "approach" leads nowhere.

Jan the relativistic moron has never solved any differential equations before to claim such a demeaning statement above. <shrug>

> When I get home I'll write down the Einstein equation in spherical
> coordinates and ask you to prove that it produces diagonal metrics
> with the spheres of symmetry labelled as above. Good luck!

Go ahead and make Koobee Wublee's day. <shrug>

> > You are free to transform it to another, but at the end, you must
> > translate it back to the original set of coordinate system you have
> started with.
>
> You can translate to _any_ coordinate system. Tensors are
> coordinate-invariant.

It is impossible to describe any tensor without a set of coordinate system. To this day, Jan the relativistic moron still has not presented a tensor without using any coordinate system to describe it. <shrug>

[danco...@gmail.com]

On Friday, January 29, 2016 at 1:19:43 PM UTC-8, JanPB wrote:
> In derivations of Schwarzschild's metric there is a coordinate change involved

> which is frequently glossed over.

Are you saying that standard textbook derivations of the Schwarzschild metric are deficient in some way? Precisely what "coordinate change" are you referring to?

[Polikwaptiwa]

Is that picture faked? Because if it is not, then it is saying more than a
million. You don't even need to listen.

https://www.youtube.com/watch?v=IfAj4kfAd_g

[JanPB]

On Friday, January 29, 2016 at 10:24:48 PM UTC-8, Koobee Wublee wrote:

> On January 29, 2016, JanPB the relativistic moron[ta dah] wrote:
>
> > Your problem is that you insist that the equations, when written
> > in a particular coordinate system (the spherical one), will
> > automagically produce a solution in some form you find attractive
> > (diagonal and with the spheres of symmetry labelled by
> > sqrt(area/4pi)).
>
> Not automatic. When a set of differential equations are presented, it is a matter of solving them. There is no magic unless you are talking about the magic thing of Cartan method. <shrug>
>
> > Such "approach" leads nowhere.
>
> Jan the relativistic moron has never solved any differential equations before to claim such a demeaning statement above. <shrug>
>
> > When I get home I'll write down the Einstein equation in spherical
> > coordinates and ask you to prove that it produces diagonal metrics
> > with the spheres of symmetry labelled as above. Good luck!
>
> Go ahead and make Koobee Wublee's day. <shrug>

I went home and am preparing the post. But the issue is even more delicate
than I thought, so I'll take another day or two (I have to attend to some
emergency at work today, sigh). But here is some good news for you: you
are not completely wrong and I was not completely right. On certain issues
we can meet in the middle (details coming up soon).

But the final conclusions remain unchanged: Schwarzschild's solution as
derived in his paper is the same solution as the one from modern textbooks.
In other words the claim of Schwarzschild's paper containing a different
solution (without a black hole) is _false_.

The delicacy I mentioned above also includes the standard thorn: extending
a tensor field over a (removable) singularity. This is actually not always
what it seems. Again, final conclusions remain unchanged.

--
Jan

[al...@interia.pl]

W dniu środa, 27 stycznia 2016 22:33:27 UTC+1 użytkownik JanPB napisał:
> > a transform type of;
> > r = r - a, is just a spimple translation alnog r in the polar coord...
> >
> > therefore the so-called standard Schwarzchild's solution
> > is just a translated version of the oryginal solution,
> > where the r parameter is limited: r > rs.
>
> The extra coordinate change I'm talking about is not just a shift in r.
> It also involves a logarithmic-like transformation of the time coordinate
> with the logarithm blowing up at the horizon (that's why the coordinate
> singularity ends up there).

It blows up in the center, ie. r = 0,
what is shifted to the r = 2m in the standardised form.

The regon r < 0 simply dosnt exists in a any geometry...
you are just a full cretin in the math.

> > the region r < 2m, simply doesnt exists in the solution,
> > and the point r = 2m is just the place of the mass m,
> > which is in standard polar cood. for r = 0.
>
> False.

r -> 0 for m -> 0,
it's just the great fallacy, which confused idiots for 100 years.

[JanPB]

On Sunday, January 31, 2016 at 3:16:29 PM UTC-8, al...@interia.pl wrote:
> W dniu środa, 27 stycznia 2016 22:33:27 UTC+1 użytkownik JanPB napisał:
> > > a transform type of;
> > > r = r - a, is just a spimple translation alnog r in the polar coord...
> > >
> > > therefore the so-called standard Schwarzchild's solution
> > > is just a translated version of the oryginal solution,
> > > where the r parameter is limited: r > rs.
> >
> > The extra coordinate change I'm talking about is not just a shift in r.
> > It also involves a logarithmic-like transformation of the time coordinate
> > with the logarithm blowing up at the horizon (that's why the coordinate
> > singularity ends up there).
>
> It blows up in the center, ie. r = 0,
> what is shifted to the r = 2m in the standardised form.
>
> The regon r < 0 simply dosnt exists in a any geometry...
> you are just a full cretin in the math.

You haven't understood this completely yet. I'll post more on that later.

> > > the region r < 2m, simply doesnt exists in the solution,
> > > and the point r = 2m is just the place of the mass m,
> > > which is in standard polar cood. for r = 0.
> >
> > False.
>
> r -> 0 for m -> 0,
> it's just the great fallacy, which confused idiots for 100 years.

No, you are the confused one. For now just trust me.

--
Jan

[al...@interia.pl]

I have facked the GR long time ago.
It's just a shit... no any physics.

BTW. The metrices in the relativity,
can describe only the light propagation...
no any so-called 'spacetime geomety' exists.

The end... of the GR sztory. :)

[JanPB]

Stop dreaming and act like an adult.

> BTW. The metrices in the relativity,
> can describe only the light propagation...
> no any so-called 'spacetime geomety' exists.
>
> The end... of the GR sztory. :)

Not even wrong.

--
Jan

[Koobee Wublee]

> you are not completely wrong and I was not completely right.

No, Jan, you are still wrong since you have been an arm-chair amateur physicist. Koobee Wublee, on the other hand, had personally derived the field equations. Koobee Wublee understands it is absolutely impossible to write down the set of field equations in a coordinate system other than the polar coordinate system. If you want, Koobee Wublee can write down the null Einstein tensor for you to verify that the following solution, not the only ones, satisfy as solutions. <shrug>

** dS^2 = c^2 (1 + K / R) dt^2 - dr^2 (dR/dr)^2 / (1 + K / R) - R^2 dO^2

Where

** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2
** K = integration constant
** R = any function of r

> On certain issues we can meet in the middle (details coming up soon).

This is not a political compromise but a scientific debate. There can only be right or wrong. Koobee Wublee is not willing to meet you in the middle. <shrug>

> But the final conclusions remain unchanged: Schwarzschild's solution
> as derived in his paper is the same solution as the one from modern
> textbooks.

No, this is not the case. After Nordstrom has introduced the null Ricci tensor as the more general form of the Laplace equation, Schwarzschild had already solved it well before 1915, and the solution is exactly what the Schwarzschild metric is. <shrug>

However, Hilbert had derived the Einstein tensor with this added trace term. Schwarzschild needed to eliminate this trace term in which he did not know that the null Einstein tensor is the same as the null Ricci tensor. So, he needed to transform the polar coordinate system into one that yields a metric with the determinant of -1 to eliminate the trace term and then transform it back to the polar coordinate system. Schwarzschild unknowingly to himself has presented an infinite such type of solutions to the field equations. Unfortunately, this brilliant man died with a few months of deriving the original metric. <shrug>

> In other words the claim of Schwarzschild's paper containing a different
> solution (without a black hole) is _false_.

You will start to understand how right Koobee Wublee has been if you actually derived the field equations. There is no other way. <shrug>

> The delicacy I mentioned above also includes the standard thorn:
> extending a tensor field over a (removable) singularity.

Don't worry about the singularity. Just solve the field equations. <shrug>

> This is actually not always what it seems. Again, final
> conclusions remain unchanged.

Again, you need to experience how the field equations are derived and solved. Then, show Koobee Wublee how you can achieve so without using a set of coordinate system. This is not a delicate matter but cut and dry one. <shrug>

[JanPB]

On Sunday, January 31, 2016 at 9:39:35 PM UTC-8, Koobee Wublee wrote:
> On January 31, 2016, JanPB wrote:
>
> > you are not completely wrong and I was not completely right.
>
> No, Jan, you are still wrong since you have been an arm-chair amateur physicist.

Just wait a day or two, you don't know half of the story :-)

--
Jan

[shan]

Koobee Wublee wrote:

> No, Jan, you are still wrong since you have been an arm-chair amateur
> physicist. Koobee Wublee, on the other hand, had personally derived the
> field equations. Koobee Wublee understands it is absolutely impossible
> to write down the set of field equations in a coordinate system other
> than the polar coordinate system. If you want, Koobee Wublee can write
> down the null Einstein tensor for you to verify that the following

Why should that writing be impossible in other coordinate system apart
from the polar. The polar can always be transformed into any other. I
don't understand.

[Koobee Wublee]

Sorry. Koobee Wublee meant impossibly more difficult. <shrug>

[al...@interia.pl]

W dniu poniedziałek, 1 lutego 2016 03:42:12 UTC+1 użytkownik JanPB napisał:

> > I have facked the GR long time ago.
> > It's just a shit... no any physics.
>
> Stop dreaming and act like an adult.
>
> > BTW. The metrices in the relativity,
> > can describe only the light propagation...
> > no any so-called 'spacetime geomety' exists.
> >
> > The end... of the GR sztory. :)
>
> Not even wrong.

Indeed: full truth can't be wrong in any way, even even.

[Koobee Wublee]

On Sunday, January 31, 2016 at 3:30:00 PM UTC-8, JanPB wrote:

> No, you are the confused one. For now just trust me.

It is more than a week. Where is the set of differential equations that represents the field equations without any coordinate system? Has Jan the relativistic moron given up? Please advise and update. <shrug>

[JanPB]

No, just being busy. The subject is quite subtle and it must involve some
discussion of extending spacetime geometries. This alone is a source of
endless confusion. I'll have something written soon, I'll post it on a
new thread but I'll post a note here as well.

--
Jan

[David Fuller]

No, Schwarzschild started with the Cartesian coordinate system and found the metric cannot possibly be diagonal. Thus, being a person (very unlike Jan the relativistic moron) that had solved many differential equations in the past, instinct just told him rightfully that the polar coordinate system will certainly yield a diagonal metric. Solving such a complex set of differential equations require some trials and errors in which Jan the relativistic has no experiences in doing so. <shrug>

Diagonal.

http://i68.tinypic.com/xfn9rt.jpg

[JanPB]

On Tuesday, February 9, 2016 at 3:28:46 PM UTC-8, David Fuller wrote:
> No, Schwarzschild started with the Cartesian coordinate system and found the metric cannot possibly be diagonal. Thus, being a person (very unlike Jan the relativistic moron) that had solved many differential equations in the past, instinct just told him rightfully that the polar coordinate system will certainly yield a diagonal metric.

No, it doesn't work that way. Diagonal metric can only arise from performing
a coordinate transformation. For example, the flat Minkowski metric can
be written as follows using polar coordinates:

ds^2 = -dt^2 - 2 dt dr + R^2 dO^2

See the mixed term "dt*dr" there? Polar coordinates alone won't give you
a diagonal form.

What Schwarzschild does instead, is EITHER relying on Einstein's paper
for that coordinate change (I haven't checked Einstein's paper yet
to see if he does this) OR is simply positing an Ansatz, just like it's
frequently done when solving PDEs: _assume_ the solution in a certain
form, plug it into the equation, and see if it works. This method is
used all over the place, most notably in the analogous approach called
"separation of variables" where the Ansatz consists of seeking solutions
f(x1, x2,... xN) in the form f1(x1) + f2(x2) +...+ fN(xN).

The only problem Schwarzschild hits IF he uses the second approach is
that contrary to his claim he DOESN'T get the uniqueness then. But my
hunch is he simply quotes Einstein's coordinate change (but as I said
I haven't looked it up yet).

> Solving such a complex set of differential equations require some trials and errors in which Jan the relativistic has no experiences in doing so. <shrug>

Hit the player, the needle is stuck.

--
Jan

[al...@interia.pl]

Nonsense.
The math is just about these so-called invariances in physics,
ie. about completely different things, what you think.

[JanPB]

You probably haven't noticed it, but you've just commented on the content
of an article that hasn't been written yet.

--
Jan

[al...@interia.pl]

W dniu środa, 10 lutego 2016 01:21:30 UTC+1 użytkownik JanPB napisał:

> You probably haven't noticed it, but you've just commented on the content
> of an article that hasn't been written yet.

Very well - any such stupid article can be destroyed in advance. ;)

[oriel36]

The news tomorrow will contain an announcement that scientists have discovered magic moonbeams based on an idea lately dumped on Albert but given the sciency name of 'gravitational waves'.

From my seat it is just a journey into a linguistic oblivion that began fairly recently as mathematicians eventually squeezed out the theorists even when the theorists themselves had misgivings -

"A Langrangian is not a physical thing;it is a mathematical thing - a
kind of differential equation to be exact.But physics and maths are so
closely connected these days that it is hard to separate the numbers
from the things they describe.In fact,a month after [Philip]
Morrison's remarks,Nobel Prize winner Burton Richter of the Stanford
Linear Accelerator Center said something that eerily echoed it: "
Mathematics is a language that is used to describe nature" he said
"But the theorists are beginning to think it is nature.To them the
Langrangians are the reality " Discover Magazine ,1983

[Koobee Wublee]

On Wednesday, February 10, 2016 at 10:32:10 AM UTC-8, oriel36 wrote:

> "A Langrangian is not a physical thing;it is a mathematical thing - a
> kind of differential equation to be exact.But physics and maths are so
> closely connected these days that it is hard to separate the numbers
> from the things they describe.In fact,a month after [Philip]
> Morrison's remarks,Nobel Prize winner Burton Richter of the Stanford
> Linear Accelerator Center said something that eerily echoed it: "
> Mathematics is a language that is used to describe nature" he said
> "But the theorists are beginning to think it is nature.To them the
> Langrangians are the reality " Discover Magazine ,1983

Yes, for example, the Lagrangian below that Hilbert had pulled out of his ass that derives the Einstein field equations offers no correlations to reality, and no self-styled physicists really care about this issue. <shrug>

** L = (k R + rho) sqrt(- det([g]))

Where

** k = constant
** R = Ricci scalar
** rho = mass density
** [g] = matrix that represents the metric
** det([g]) = determinant of [g]

The self-styled physicists managed to create universes from something they are hopeless fvcking clueless about. <shrug?

[al...@interia.pl]

W dniu środa, 10 lutego 2016 19:32:10 UTC+1 użytkownik oriel36 napisał:
> On Wednesday, February 10, 2016 at 6:18:25 PM UTC, al...@interia.pl wrote:
> > W dniu środa, 10 lutego 2016 01:21:30 UTC+1 użytkownik JanPB napisał:
> >
> > > You probably haven't noticed it, but you've just commented on the content
> > > of an article that hasn't been written yet.
> >
> > Very well - any such stupid article can be destroyed in advance. ;)
>
> The news tomorrow will contain an announcement that scientists have discovered magic moonbeams based on an idea lately dumped on Albert but given the sciency name of 'gravitational waves'.
>
> From my seat it is just a journey into a linguistic oblivion that began fairly recently as mathematicians eventually squeezed out the theorists even when the theorists themselves had misgivings -

Stupid people always discover, what they want to discover.

The famous 'half spin' fallacy and the so-called Bell's
(pseudo)theorem are nice examples of such fallacious practice.

[oriel36]

You are from Poland, am I right ?.

A famous countryman of yours discovered that it was possible to infer a Sun-centered system by concentrating on relative speeds between a faster moving Earth and slower moving outer planets where those planets fall temporarily behind in view as the Earth overtakes them -

http://apod.nasa.gov/apod/ap011220.html

The fact is that neither you nor the rest of your Polish countrymen protected this main argument for the Earth's motion and a Sun-centered system and that is a national disgrace.

[al...@interia.pl]

W dniu czwartek, 11 lutego 2016 18:45:25 UTC+1 użytkownik oriel36 napisał:
> On Thursday, February 11, 2016 at 5:39:21 PM UTC, al...@interia.pl wrote:
> > W dniu środa, 10 lutego 2016 19:32:10 UTC+1 użytkownik oriel36 napisał:
> > > On Wednesday, February 10, 2016 at 6:18:25 PM UTC, al...@interia.pl wrote:
> > > > W dniu środa, 10 lutego 2016 01:21:30 UTC+1 użytkownik JanPB napisał:
> > > >
> > > > > You probably haven't noticed it, but you've just commented on the content
> > > > > of an article that hasn't been written yet.
> > > >
> > > > Very well - any such stupid article can be destroyed in advance. ;)
> > >
> > > The news tomorrow will contain an announcement that scientists have discovered magic moonbeams based on an idea lately dumped on Albert but given the sciency name of 'gravitational waves'.
> > >
> > > From my seat it is just a journey into a linguistic oblivion that began fairly recently as mathematicians eventually squeezed out the theorists even when the theorists themselves had misgivings -
> >
> > Stupid people always discover, what they want to discover.
> >
> > The famous 'half spin' fallacy and the so-called Bell's
> > (pseudo)theorem are nice examples of such fallacious practice.
>
> You are from Poland, am I right ?.
>
> A famous countryman of yours discovered that it was possible to infer a Sun-centered system by concentrating on relative speeds between a faster moving Earth and slower moving outer planets where those planets fall temporarily behind in view as the Earth overtakes them -

That fact has been known even in the ancient world already.
But the western stupidity is big - very very big,
so big, that any rational man can't live there.

Therefore Copernicus had to be from the east.

> http://apod.nasa.gov/apod/ap011220.html
> The fact is that neither you nor the rest of your Polish countrymen protected this main argument for the Earth's motion and a Sun-centered system and that is a national disgrace.

Copernicus only in Poland could live and think .. quite freely.
In the west have been burned at the stake, along with their ideas.

[oriel36]

I see one participant scream about science being about models and probably unbeknownst to himself that is what empiricism is rather than science or astronomy. The emergence of modeling came at the expense of astronomy insofar as the great discovery of Copernicus was accounting for the observed behavior of the outer planets as the faster Earth overtook them hence the sun-centered system was inferred from this perspective -

http://apod.nasa.gov/apod/ap011220.html

Kepler cam along and extended this observation to include multiple passes of Mars by the Earth -

https://upload.wikimedia.org/wikipedia/commons/e/e9/Kepler_Mars_retrograde.jpg

"Copernicus, by attributing a single annual motion to the earth,
entirely rids the planets of these extremely intricate coils,
leading the individual planets into their respective orbits
,quite bare and very nearly circular. In the period of time
shown in the diagram, Mars traverses one and the same orbit as many
times as the 'garlands' you see looped towards the
center,with one extra, making nine times, while at the same time the
Earth repeats its circle sixteen times " Kepler

The working principles of planetary dynamics and a Sun-centered system was based on judging observations using a moving Earth hence the modeling is consistent with this perspective .

Along comes Sir Isaac and changes things to suit himself. He thinks (and they still do) that Kepler's representation of retrogrades is geocentric and if you plonk the Sun in the middle of the diagram retrogrades disappear like so -

http://www.conservapedia.com/images/thumb/0/0e/Copernicus_system.gif/300px-Copernicus_system.gif

This is the only way to account for Newton's illegal form of double modeling where he created a relative space and motion for observations seen from Earth as opposed to absolute space and motions for hypothetical observations seen from the Sun -

"For to the earth planetary motions appear sometimes direct, sometimes
stationary, nay, and sometimes retrograde. But from the sun they are
always seen direct,..." Newton


Mathematicians neither know nor care as they exist in a world inside their own heads so there is perhaps no audience for what happened that some of the greatest astronomical works were destroyed for little more than empirical pretense. I know what happened and why it is dangerous to maintain the pretense using indiscriminate modeling with no pedigree .

[Odd Bodkin]

On 2/14/2016 2:47 AM, oriel36 wrote:
> I see one participant scream about science being about models and probably unbeknownst to himself
> that is what empiricism is rather than science or astronomy.

I'm curious what you think science is. Perhaps you could define it or
describe it operationally.

--
Odd Bodkin --- maker of fine toys, tools, tables

[Céline Desirée]

Odd Bodkin wrote:

> On 2/14/2016 2:47 AM, oriel36 wrote:
>> I see one participant scream about science being about models and
>> probably unbeknownst to himself that is what empiricism is rather than
>> science or astronomy.
>
> I'm curious what you think science is. Perhaps you could define it or
> describe it operationally.

No idea what it is. I can tell you what it must be. Searching for truth no
mater where it leads toward, at. How do you like it. What you have today
is not science, but corruption and technology. Technology is not science,
but rather anti-science.

I am saying the above, thinking merely at origin of the word science, the
Latin scientia. Otherwise we can safely drop the science terminology and
call it something else.

[al...@interia.pl]

W dniu niedziela, 14 lutego 2016 22:18:53 UTC+1 użytkownik Odd Bodkin napisał:
> On 2/14/2016 2:47 AM, oriel36 wrote:
> > I see one participant scream about science being about models and probably unbeknownst to himself
> > that is what empiricism is rather than science or astronomy.
>
> I'm curious what you think science is. Perhaps you could define it or
> describe it operationally.

Define the communism or capitalism... babe.

Do you know what is a stupidity?

It's just that... :)

[JanPB]

Just posted it under the heading "Schwarzschild's paper and the uniqueness
of the solution".

--
Jan

[Koobee Wublee]

On Tuesday, February 23, 2016 at 5:06:26 PM UTC-8, JanPB wrote:

> On Monday, February 8, 2016 at 11:20:40, Koobee Wublee wrote:

> > It is more than a week. Where is the set of differential
> > equations that represents the field equations without any
> > coordinate system? Has Jan the relativistic moron given up?
> > Please advise and update. <shrug>
>

> Just posted it under the heading "Schwarzschild's paper and
> the uniqueness of the solution".
>

> https://groups.google.com/d/msg/sci.physics.relativity/h49c7pfteQQ/jrJqiL7XBAAJ

Bullshit! <shrug>

https://groups.google.com/d/msg/sci.physics.relativity/h49c7pfteQQ/5PdRRGfpBAAJ

It has been painfully a long time since Jan the relativistic moron had claimed the moron shall produce differential equations to the null Einstein tensor in the form without any coordinate system and shall solve these differential equations showing the unique solution presented in polar coordinate system is none other than the Schwarzschild metric. Well, instead, that did not happen, and we are bombarded with even more bullshit. <shrug>

Jan Bullshit Bielawski the relativistic moron is another bullshitter just like its crony Paul Bullshit Andersen another relativistic moron. <shrug>

Case closed. <shrug>

[JanPB]

On Tuesday, March 1, 2016 at 12:54:42 PM UTC-8, Koobee Wublee wrote:
> On Tuesday, February 23, 2016 at 5:06:26 PM UTC-8, JanPB wrote:
> > On Monday, February 8, 2016 at 11:20:40, Koobee Wublee wrote:
>
> > > It is more than a week. Where is the set of differential
> > > equations that represents the field equations without any
> > > coordinate system? Has Jan the relativistic moron given up?
> > > Please advise and update. <shrug>
> >
> > Just posted it under the heading "Schwarzschild's paper and
> > the uniqueness of the solution".
> >
> > https://groups.google.com/d/msg/sci.physics.relativity/h49c7pfteQQ/jrJqiL7XBAAJ
>
> Bullshit! <shrug>

No. It's how it is.

> https://groups.google.com/d/msg/sci.physics.relativity/h49c7pfteQQ/5PdRRGfpBAAJ
>
> It has been painfully a long time since Jan the relativistic moron had claimed the moron shall produce differential equations to the null Einstein tensor in the form without any coordinate system and shall solve these differential equations showing the unique solution presented in polar coordinate system is none other than the Schwarzschild metric. Well, instead, that did not happen, and we are bombarded with even more bullshit. <shrug>

I've just proved it in that article. Have you read it?

> Jan Bullshit Bielawski the relativistic moron is another bullshitter just like its crony Paul Bullshit Andersen another relativistic moron. <shrug>
>
> Case closed. <shrug>

Case remains open because all you did here was yelling while providing
no substance.

--
Jan

[Koobee Wublee]

On Tuesday, March 1, 2016 at 1:00:46 PM UTC-8, JanPB wrote:
> Koobee Wublee wrote:

> > https://groups.google.com/d/msg/sci.physics.relativity/h49c7pfteQQ/5PdRRGfpBAAJ
> >
> > It has been painfully a long time since Jan the relativistic moron
> > had claimed the moron shall produce differential equations to the
> > null Einstein tensor in the form without any coordinate system and
> > shall solve these differential equations showing the unique
> > solution presented in polar coordinate system is none other than
> > the Schwarzschild metric. Well, instead, that did not happen, and
> > we are bombarded with even more bullshit. <shrug>
>
> I've just proved it in that article. Have you read it?

You should have known the answer. Are you really that stupid? <shrug>

> > Jan Bullshit Bielawski the relativistic moron is another
> > bullshitter just like its crony Paul Bullshit Andersen another
> > relativistic moron. <shrug>
> >
> > Case closed. <shrug>
>
> Case remains open because all you did here was yelling while
> providing no substance.

Koobee Wublee would not hold any breath for that. <shrug>

[JanPB]

As always.

--
Jan