The Motion of the Perihelion of Mercury

On Jan 1, 1:13*pm, Koobee Wublee wrote:
On Jan 1, 12:32 am, Eric Gisse wrote:

On Dec 31, 10:37 pm, Koobee Wublee wrote:
Your "solution" is not a solution.

http://img58.imageshack.us/img58/8527/idiotcm5.png

In the following post, I gave you the following solution to the field
equations that obeys Newtonian law of gravity, but this one exhibits
half of the event horizon than the Schwarzschild metric.

ds^2 = c^2 (1 – 2 K / r)^2 dt^2 – dr^2 / (1 – K / r)^4 – r^2 dO^2 / (1
– K / r)^2

Where

** *K = G M / c^2 / 2, HALF OF THE EVENT HORIZON

Since when did any of the constants say where the event horizon is?


Reference post:

http://groups.google.com/group/sci.p...sg/e3162cddce8...

With this spacetime, the event horizon occurs at (2 K) which if (G M /
c^2)

No, the event horizon does not occur there. Do the math, arrogant
idiot.

http://img353.imageshack.us/img353/8273/idiot2ok4.png

LOOK AT THE MATH. Every one of them go to zero or blow up at r = K.
That's where the event horizon is.

As given to you in a post you just referenced, the coordinate
transformation between your "different" metric and Schwarzschild is
given by r(R) = 2*R^2/(2*R-K) where r is the Schwarzschild radial
coordinate and R is the idiot radial coordinate.

Put in R = K, and you get r = 2K. Imagine that - the singularity is
located in the same ****ing place and it is the SAME MANIFOLD.
Proven.


A couple posts later, you verified that the above spacetime does
indeed satisfy R_uv = 0 by saying:

“A quick re-roll into grtensor showed that you are, in fact, correct.
It does satisfy R_uv = 0.”

Reference post:

http://groups.google.com/group/sci.p...sg/6c66880d4c2...

Now, the solution we have been talking about is much simpler than the
one above. *If I can derive the above solution from Koobee Wublee’s
theorem or the theorem of Generality, just how much more difficult can
I derive the following?

You aren't deriving jack ****. You are transforming coordinates from
Schwarzschild to something else. Your claims are asinine, childish,
and never backed up by anything more than a repetition of the claim.


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

Where

** *K = 2 G M / c^2

http://img58.imageshack.us/img58/8527/idiotcm5.png

And just like that last time, you don’t even know how to enter the
inputs correctly. *What you have entered is wrong. *You need to
replace the 2 instances of (2 K) with K. *shrug

http://groups.google.com/group/sci.p...9?dmode=source

That's what you wrote, dip****. I even said later to George that you
were probably transcribing it wrong - and imagine that, you did.

But let's discuss idiot metric #3. I'm numbering them because it is
hard to keep track.

http://img135.imageshack.us/img135/2813/idiot3yw9.png

It _does_ satisfy R_uv = 0. So therefore it is related to
Schwarzschild by a coordinate transformation.
The coordinate transformation between Schwarzschild and idiot3 is
given rather simply by r(R) = r + K.

Another "different" coordinate system proven to be a part of the same
object. How many times do we have to do this before you grow a clue?

On Jan 1, 10:56 pm, Eric Gisse wrote:
On Jan 1, 1:13 pm, Koobee Wublee wrote:

In the following post, I gave you the following solution to the field
equations that obeys Newtonian law of gravity, but this one exhibits
half of the event horizon than the Schwarzschild metric.

ds^2 = c^2 (1 – 2 K / r)^2 dt^2 – dr^2 / (1 – K / r)^4 – r^2 dO^2 / (1
– K / r)^2

Where

** K = G M / c^2 / 2, HALF OF THE EVENT HORIZON

Since when did any of the constants say where the event horizon is?

Hmmm... If you don’t know, you need to consult one of the textbooks
you have been sitting on. It is actually very simple. shrug

Reference post:

http://groups.google.com/group/sci.p...sg/e3162cddce8...

With this spacetime, the event horizon occurs at (2 K) which if (G M /
c^2)

No, the event horizon does not occur there. Do the math, arrogant
idiot.

http://img353.imageshack.us/img353/8273/idiot2ok4.png

LOOK AT THE MATH. Every one of them go to zero or blow up at r = K.
That's where the event horizon is.

Oh, big deal. G M / c^2 / 2 versus G M c^2 / 4. The bottom line is
that the event horizon of this spacetime which does satisfy as a
static, spherically symmetric, and asymptotically flat solution to the
field equations is not (2 G M / c^2). shrug

As given to you in a post you just referenced, the coordinate
transformation between your "different" metric and Schwarzschild is
given by r(R) = 2*R^2/(2*R-K) where r is the Schwarzschild radial
coordinate and R is the idiot radial coordinate.

There is no coordinate transformation. How many times do I have to
tell you that?

Put in R = K, and you get r = 2K. Imagine that - the singularity is
located in the same ****ing place and it is the SAME MANIFOLD.
Proven.

What is this whining crap agan?

A couple posts later, you verified that the above spacetime does
indeed satisfy R_uv = 0 by saying:

“A quick re-roll into grtensor showed that you are, in fact, correct.
It does satisfy R_uv = 0.”

Reference post:

http://groups.google.com/group/sci.p...sg/6c66880d4c2...

Now, the solution we have been talking about is much simpler than the
one above. If I can derive the above solution from Koobee Wublee’s
theorem or the theorem of Generality, just how much more difficult can
I derive the following?

You aren't deriving jack ****. You are transforming coordinates from
Schwarzschild to something else. Your claims are asinine, childish,
and never backed up by anything more than a repetition of the claim.

You are merely ignorant. You have been checkmated many times over but
are either too stupid or too hypocritical to accept what I have told
you many times over. shrug You are a small man just like that so-
called professor Andersen from an obscure college in Norway. shrug

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

Where

** K = 2 G M / c^2

http://img58.imageshack.us/img58/8527/idiotcm5.png

And just like that last time, you don’t even know how to enter the
inputs correctly. What you have entered is wrong. You need to
replace the 2 instances of (2 K) with K. shrug

http://groups.google.com/group/sci.p...sg/71f5a677e6f...

That's what you wrote, dip****. I even said later to George that you
were probably transcribing it wrong - and imagine that, you did.

Oh, there is no need to blow your top off over a couple of innocent
typos. I have written the same spacetime over and over and over and
over again before. You should know better. shrug

But let's discuss idiot metric #3. I'm numbering them because it is
hard to keep track.

http://img135.imageshack.us/img135/2813/idiot3yw9.png

It _does_ satisfy R_uv = 0.

Yes, I told you so. After all, you are going to give the great
reverend Hammond a heart attack after he has already decided to
BELIEVED IN otherwise --- all thanks to your “operator errors”.

So therefore it is related to
Schwarzschild by a coordinate transformation.

It is only related to the Schwarzschild metric through being also a
solution that is static, spherically symmetric, and asymptotically
flat to the field equations. shrug

The coordinate transformation between Schwarzschild and idiot3 is
given rather simply by r(R) = r + K.

Again, there is no transformation of coordinate. However, applying
Koobee Wublee’s theorem or the theorem of Generality, you can select
the following to arrive at the Schwarzschild metric. shrug

u = r – K.

Another "different" coordinate system proven to be a part of the same
object. How many times do we have to do this before you grow a clue?

Oh, quit behaving like a sore loser. Bow, kneel, and kowtow to Koobee
Wublee, and accept the mathematical nonsense in the bull**** called
GR.

Yes, Einstein was nobody. He was a nitwit, a plagiarist, and a liar.
shrug

On Jan 1, 10:48*pm, Koobee Wublee wrote:
On Jan 1, 10:56 pm, Eric Gisse wrote:

On Jan 1, 1:13 pm, Koobee Wublee wrote:
In the following post, I gave you the following solution to the field
equations that obeys Newtonian law of gravity, but this one exhibits
half of the event horizon than the Schwarzschild metric.

ds^2 = c^2 (1 – 2 K / r)^2 dt^2 – dr^2 / (1 – K / r)^4 – r^2 dO^2 / (1
– K / r)^2

Where

** *K = G M / c^2 / 2, HALF OF THE EVENT HORIZON

Since when did any of the constants say where the event horizon is?

Hmmm... *If you don’t know, you need to consult one of the textbooks
you have been sitting on. *It is actually very simple. *shrug

The objects in a particular coordinate representation of the metric do
not define the location of the event horizon. Idiot.


Reference post:

http://groups.google.com/group/sci.p...sg/e3162cddce8....

With this spacetime, the event horizon occurs at (2 K) which if (G M /
c^2)

No, the event horizon does not occur there. Do the math, arrogant
idiot.

http://img353.imageshack.us/img353/8273/idiot2ok4.png

LOOK AT THE MATH. Every one of them go to zero or blow up at r = K.
That's where the event horizon is.

Oh, big deal. *G M / c^2 / 2 versus G M c^2 / 4. *The bottom line is
that the event horizon of this spacetime which does satisfy as a
static, spherically symmetric, and asymptotically flat solution to the
field equations is not (2 G M / c^2). *shrug

Except you are proven wrong. The event horizon is at r = K - as proven
- and r = K corresponds to R = 2K in the Schwarzschild metric.

Repeating the assertion doesn't make you right, just stupid.

As given to you in a post you just referenced, the coordinate
transformation between your "different" metric and Schwarzschild is
given by r(R) = 2*R^2/(2*R-K) where r is the Schwarzschild radial
coordinate and R is the idiot radial coordinate.

There is no coordinate transformation. *How many times do I have to
tell you that?

Asserting they are different does not make them different. I proved
they are the same - suck it up, and try again.

We can repeat this idiocy for a thousand years and you will be wrong
_every single time_ because of the fundamental nature of tensors.
Recasting them in a different coordinate system changes nothing but I
encourage you keep trying.

Either that or learn - but if you could learn you wouldn't be here
repeating stupidities, now would you?


Put in R = K, and you get r = 2K. Imagine that - the singularity is
located in the same ****ing place and it is the SAME MANIFOLD.
Proven.

What is this whining crap agan?

Do you not know what a coordinate transformation is?

I have four metrics:

ds^2 = dx^2 + dy^2 + dz^2
ds^2 = dr^2 + r^2(d\theta^2 + sin^2(\theta)d\phi^2)
ds^2 = dr^2 + r^2 d\theta^2 + dz^2
ds^2 = da^2 + db^2 + dz^2

They all describe the same thing, yet "look different". But I can find
a coordinate transformation relating all of them to eachother, so they
are the same thing.

How is this any different?

[snip idiocy]

You can't refute one single thing I have written. Amazing.

The coordinate transformation between Schwarzschild and idiot3 is
given rather simply by r(R) = r + K.

Again, there is no transformation of coordinate. *However, applying
Koobee Wublee’s theorem or the theorem of Generality, you can select
the following to arrive at the Schwarzschild metric. *shrug

u = r – K.

That's called a "coordinate transformation".

Thanks for proving my point.


Another "different" coordinate system proven to be a part of the same
object. How many times do we have to do this before you grow a clue?

Oh, quit behaving like a sore loser. *Bow, kneel, and kowtow to Koobee
Wublee, and accept the mathematical nonsense in the bull**** called
GR.

The shield of stupidity protects your huge but fragile ego.


Yes, Einstein was nobody. *He was a nitwit, a plagiarist, and a liar.
shrug

On Thu, 1 Jan 2009 22:56:25 -0800 (PST), Eric Gisse
wrote:

On Jan 1, 1:13*pm, Koobee Wublee wrote:
On Jan 1, 12:32 am, Eric Gisse wrote:

On Dec 31, 10:37 pm, Koobee Wublee wrote:
Your "solution" is not a solution.

http://img58.imageshack.us/img58/8527/idiotcm5.png

In the following post, I gave you the following solution to the field
equations that obeys Newtonian law of gravity, but this one exhibits
half of the event horizon than the Schwarzschild metric.

ds^2 = c^2 (1 – 2 K / r)^2 dt^2 – dr^2 / (1 – K / r)^4 – r^2 dO^2 / (1
– K / r)^2

Where

** *K = G M / c^2 / 2, HALF OF THE EVENT HORIZON

Since when did any of the constants say where the event horizon is?


Reference post:

http://groups.google.com/group/sci.p...sg/e3162cddce8...

With this spacetime, the event horizon occurs at (2 K) which if (G M /
c^2)

No, the event horizon does not occur there. Do the math, arrogant
idiot.

http://img353.imageshack.us/img353/8273/idiot2ok4.png

LOOK AT THE MATH. Every one of them go to zero or blow up at r = K.
That's where the event horizon is.

As given to you in a post you just referenced, the coordinate
transformation between your "different" metric and Schwarzschild is
given by r(R) = 2*R^2/(2*R-K) where r is the Schwarzschild radial
coordinate and R is the idiot radial coordinate.

Put in R = K, and you get r = 2K. Imagine that - the singularity is
located in the same ****ing place and it is the SAME MANIFOLD.
Proven.


A couple posts later, you verified that the above spacetime does
indeed satisfy R_uv = 0 by saying:

“A quick re-roll into grtensor showed that you are, in fact, correct.
It does satisfy R_uv = 0.”

Reference post:

http://groups.google.com/group/sci.p...sg/6c66880d4c2...

Now, the solution we have been talking about is much simpler than the
one above. *If I can derive the above solution from Koobee Wublee’s
theorem or the theorem of Generality, just how much more difficult can
I derive the following?

You aren't deriving jack ****. You are transforming coordinates from
Schwarzschild to something else. Your claims are asinine, childish,
and never backed up by anything more than a repetition of the claim.


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

Where

** *K = 2 G M / c^2

http://img58.imageshack.us/img58/8527/idiotcm5.png

And just like that last time, you don’t even know how to enter the
inputs correctly. *What you have entered is wrong. *You need to
replace the 2 instances of (2 K) with K. *shrug

http://groups.google.com/group/sci.p...9?dmode=source

That's what you wrote, dip****. I even said later to George that you
were probably transcribing it wrong - and imagine that, you did.

But let's discuss idiot metric #3. I'm numbering them because it is
hard to keep track.

http://img135.imageshack.us/img135/2813/idiot3yw9.png

It _does_ satisfy R_uv = 0. So therefore it is related to
Schwarzschild by a coordinate transformation.
The coordinate transformation between Schwarzschild and idiot3 is
given rather simply by r(R) = r + K.

Another "different" coordinate system proven to be a part of the same
object. How many times do we have to do this before you grow a clue?


[Hammond]
Yes, R_uv for the above Metric does vanish, but the
reason is, as you have informed us, that making the simple
substitution r=R-K with K=2m immediately transforms the
Kooby metric into the Schwarzchild Metric!
Kooby is trying to refute Birkhoff's theorem which as
anyone with a lick of common sense knows is utterly futile.
In addition a number of times he has transcribed the
metric incorrectly which immediately causes R_uv=/=0 as
required by the vacuum equations.
The quest for a Kooby Metric which will disprove
Schwarzchild-Birkhoff has therefore suffered a final,
disasterous and truly ignominious defeat!
And this certainly puts to the lie his scurrilous remark
that "other solutions to the EFE" "do not predict the 43"
advance of Mercury".
=====================================
HAMMOND'S PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
GOD=G_uv (a folk song on mp3)
http://interrobang.jwgh.org/songs/hammond.mp3
=====================================

On Jan 2, 12:03 am, Eric Gisse wrote:
On Jan 1, 10:48 pm, Koobee Wublee wrote:

Hmmm... If you don’t know, you need to consult one of the textbooks
you have been sitting on. It is actually very simple. shrug

[Gisse’s whining crap snipped]

I have four metrics:

ds^2 = dx^2 + dy^2 + dz^2
ds^2 = dr^2 + r^2(d\theta^2 + sin^2(\theta)d\phi^2)
ds^2 = dr^2 + r^2 d\theta^2 + dz^2
ds^2 = da^2 + db^2 + dz^2

They all describe the same thing, yet "look different". But I can find
a coordinate transformation relating all of them to eachother, so they
are the same thing.

Your notation is screwed up. The r of the 2nd is not the same as r of
the 3rd. The phi of the 2nd is the same as theta of the 3rd.
Otherwise, they can all represent the same geometry. shrug

This is not what I have been talking about. It looks like very few
have understood what I am saying. That is why I feel very lonely to
be the few who after Riemann and perhaps Christoffel that have
understood the curvature business very well. shrug

What I am talking about is that the following geometries are different
given the same dr, dLongitude, dLatitude.

** ds1^2 = g1_11 dr^2 + g1_22 r^2 dO^2
** ds2^2 = g2_11 dr^2 + g2_22 r^2 dO^2

Where

** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

[Snipped more Gisse’s whining crap]

Again, there is no transformation of coordinate. However, applying
Koobee Wublee’s theorem or the theorem of Generality, you can select
the following to arrive at the Schwarzschild metric. shrug

u = r – K.

That's called a "coordinate transformation".

What coordinate transformation? u is any function of r. This is very
basic mathematics. shrug

Thanks for proving my point.

You are still dreaming and jacking off in your mathemagical Realm. In
doing so, probably wishing Einstein can join you. shrug

Let me write down the two very different spacetime in a more concise
way.

** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
** ds2^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2

THE GEOMETRY ds1 IS NOT THE SAME AS ds2. THAT IS BECAUSE THE METRIC
OF ds1 IS NOT THE SAME AS ds2, BUT THE COORDINATE SYSTEM DESCRIBING
BOTH ds1 and ds2 IS EXACTLY IDENTICAL.

Can you dig this, punk? I don’t think you possess any minute
intelligence to get after all these years, and I really don’t care.
shrug

With any matheMagical tricks, show me how you can coordinate-transform
from one equation to the other and what justify one solution is more
unique than the others. Remember that the coordinate system is the
same with both equations.

Oh, quit behaving like a sore loser. Bow, kneel, and kowtow to Koobee
Wublee, and accept the mathematical nonsense in the bull**** called
GR.

The shield of stupidity protects your huge but fragile ego.

Cheap shot once again.

Yes, Einstein was nobody. He was a nitwit, a plagiarist, and a liar.
shrug

On Jan 2, 6:18 am, George Hammond wrote:
Eric Gisse wrote:

Yes, R_uv for the above Metric does vanish,

So, the great reverend is wishy washy. Did you not just said through
you gut feelings that solution cannot possibly cause the Ricci tensor
to vanish?

but the
reason is, as you have informed us,

Oh, excuses to justify your faith. shrug

that making the simple
substitution r=R-K with K=2m immediately transforms the
Kooby metric into the Schwarzchild Metric!

There is no coordinate transformation. It is all your ego giving you
excuses to believe in that nonsense. shrug It is very funny that
you would allow Gisse, a multi-year super-senior without any college
degree at the University of Alaska, Fairbanks to convince you to
continue your faith. shrug

Kooby is trying to refute Birkhoff's theorem which as
anyone with a lick of common sense knows is utterly futile.

Ahahaha...

In addition a number of times he has transcribed the
metric incorrectly which immediately causes R_uv=/=0 as
required by the vacuum equations.

Come on. I have written that spacetime correctly for many posts. It
was merely a typo. shrug

The quest for a Kooby Metric which will disprove
Schwarzchild-Birkhoff has therefore suffered a final,
disasterous and truly ignominious defeat!

So, you are practicing to write the next bible. shrug

And this certainly puts to the lie his scurrilous remark
that "other solutions to the EFE" "do not predict the 43"
advance of Mercury".

Yes, I have not tried to figure out if that spacetime does satisfy Le
Verrier’s claim in Mercury’s orbital advance using Gerber’s
mathematical methodology, but who cares since there is no
justification on which solution is the so-called “Rock of the
Gibraltar”. shrug

On Jan 2, 10:42*pm, Koobee Wublee wrote:
On Jan 2, 12:03 am, Eric Gisse wrote:



On Jan 1, 10:48 pm, Koobee Wublee wrote:
Hmmm... *If you don’t know, you need to consult one of the textbooks
you have been sitting on. *It is actually very simple. *shrug

[Gisse’s whining crap snipped]

I have four metrics:

ds^2 = dx^2 + dy^2 + dz^2
ds^2 = dr^2 + r^2(d\theta^2 + sin^2(\theta)d\phi^2)
ds^2 = dr^2 + r^2 d\theta^2 + dz^2
ds^2 = da^2 + db^2 + dz^2

They all describe the same thing, yet "look different". But I can find
a coordinate transformation relating all of them to eachother, so they
are the same thing.

Your notation is screwed up. *The r of the 2nd is not the same as r of
the 3rd. *The phi of the 2nd is the same as theta of the 3rd.
Otherwise, they can all represent the same geometry. *shrug

See? You are a **** hair's distance away from understanding. THE
LABELS DO NOT MATTER. The COORDINATES do not matter. They all
represent the _same_ geometry.


This is not what I have been talking about. *It looks like very few
have understood what I am saying. *That is why I feel very lonely to
be the few who after Riemann and perhaps Christoffel that have
understood the curvature business very well. *shrug

So where did you properly learn this "curvature business"? All I'm
asking is for the book you learned it from. Is that asking so much?


What I am talking about is that the following geometries are different
given the same dr, dLongitude, dLatitude.

** *ds1^2 = g1_11 dr^2 + g1_22 r^2 dO^2
** *ds2^2 = g2_11 dr^2 + g2_22 r^2 dO^2

Where

** *dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2

Except the coordinate labels are not the same in both coordinate
systems. Calculate the surface area of a sphere - don't just write it
down, CALCULATE it.


[Snipped more Gisse’s whining crap]

Again, there is no transformation of coordinate. *However, applying
Koobee Wublee’s theorem or the theorem of Generality, you can select
the following to arrive at the Schwarzschild metric. *shrug

u = r – K.

That's called a "coordinate transformation".

What coordinate transformation? *u is any function of r. *This is very
basic mathematics. *shrug

Yes, and it is called a "coordinate transformation". You start out in
the (t,u,\theta,\phi) coordinate system and end up in the (t,r,\theta,
\phi) coordinate system.


Thanks for proving my point.

You are still dreaming and jacking off in your mathemagical Realm. *In
doing so, probably wishing Einstein can join you. *shrug

Let me write down the two very different spacetime in a more concise
way.

** *ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
** *ds2^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2

THE GEOMETRY ds1 IS NOT THE SAME AS ds2. *THAT IS BECAUSE THE METRIC
OF ds1 IS NOT THE SAME AS ds2, BUT THE COORDINATE SYSTEM DESCRIBING
BOTH ds1 and ds2 IS EXACTLY IDENTICAL.

If they are "exactly identical" they should give the same values for
the surface area of the sphere, as #1 is in Schwarzschild isotropic
coordinates which has spheres of constant (r,t) of surface area 4 pi
r^2.

Let us see you calculate the surface area of #2.

I would also like you calculate the Ricci scalars of both. You'll
notice something interesting.


Can you dig this, punk? *I don’t think you possess any minute
intelligence to get after all these years, and I really don’t care.
shrug

With any matheMagical tricks, show me how you can coordinate-transform
from one equation to the other and what justify one solution is more
unique than the others. *Remember that the coordinate system is the
same with both equations.

Except the coordinate systems are different - the second one is
displaced from the origin by an amount K, which you can see
immediately by looking at the coefficients of the angular components.
Or by calculating the surface area. Or by calculating the Ricci
scalar. Or by looking at the determinant of the metric.

I cannot claim one solution is more unique than the other, which is
the whole point of my explanation. I just prefer Schwarzschild in
isotropic coordinates. I can easily convert from Schwarzschild and
every /OTHER/ "different" solution to yours - just chain the
transformations.


Oh, quit behaving like a sore loser. *Bow, kneel, and kowtow to Koobee
Wublee, and accept the mathematical nonsense in the bull**** called
GR.

The shield of stupidity protects your huge but fragile ego.

Cheap shot once again.

Yes, Einstein was nobody. *He was a nitwit, a plagiarist, and a liar.

On Jan 2, 10:52*pm, Koobee Wublee wrote:
On Jan 2, 6:18 am, George Hammond wrote:

Eric Gisse wrote:
* *Yes, R_uv for the above Metric does vanish,

So, the great reverend is wishy washy. *Did you not just said through
you gut feelings that solution cannot possibly cause the Ricci tensor
to vanish?

You wrote two different metrics. He was right both times, as was I.

[snip]

On Sat, 3 Jan 2009 03:03:13 -0800 (PST), Eric Gisse
wrote:

On Jan 2, 10:52*pm, Koobee Wublee wrote:
On Jan 2, 6:18 am, George Hammond wrote:

Eric Gisse wrote:
* *Yes, R_uv for the above Metric does vanish,

So, the great reverend is wishy washy. *Did you not just said through
you gut feelings that solution cannot possibly cause the Ricci tensor
to vanish?

You wrote two different metrics. He was right both times, as was I.

[snip]

[Hammond]
Correct, he admits to "making a typo" which is why there
are two different versions of the Kooby Metric floating
around.
In the original one, you discovered and I have
independently confirmed that it is simply a radial
coordinate transformation of Schwarzchild given by r=R-K
with K=2m. So the original Kooby Metric is nothing but the
Schwarzchild Metric in disguise.
In the second case his "Typo Version" of the Kooby Metric
is not a coordinate transformation of Schwarzchild and
naturally therefore R_uv does not vanish as you discovered
and hence it is not a valid solution of the EFE's.
Ergo... Kooby's legal leg is wobbling and all his
statements to the contrary have been PROVEN to be
nonsensical and wrong! At this point he is reduced to
mindless repetition (echolalliation).
Bear in mind, as a mere 25 year old you still think he's
a sincere Physics buff.... but at my age and being a veteran
"Psychophysicist" I've seen enough of his type to know he's
an annoyed, childishly jealous and spiteful personality who
only pretends to be sincere for the purposes of dogging the
uninitiated. He's an ornery prick and he would do well to
read my scientific proof of God where he might be able to
really make an original contribution to a real applied
Physics discovery.... since he seems to have nothing better
to do than play with Psychophysics for entertainment.
=====================================
HAMMOND'S PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
GOD=G_uv (a folk song on mp3)
http://interrobang.jwgh.org/songs/hammond.mp3
=====================================

On Dec 28 2008, 11:49*pm, wrote:
THE MOTION OF THE PERIHELION OF MERCURY
In his general relativity calculation of the motion of the perihelion
of Mercury Albert Einstein had only taken into account the
gravitational actions between the Sun and the Mercury, which he also
assumed as two points.

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

Have any done these calculations?

Well van Fandern did this analysis:
http://www.metaresearch.org/cosmolog...a-combined.asp

Also there are flyby anomalies, which are possibly more accurate
http://www2.phys.canterbury.ac.nz/ed...derson2008.pdf

Its latitude dependence suggests
that the Earth’s rotation may be generating an effect much
larger than the frame dragging effect of General Relativity,
the Lense-Thirring effect