On May 6, 12:33*am, Mike wrote:
On May 6, 3:49*am, Eric Gisse wrote:



On May 5, 10:23*pm, Koobee Wublee wrote:

On May 5, JanPB wrote:

Mike wrote:
Suppose this allien mathematical genius arrives to earth and he is
presented with the GR equations. He knows nothing about Newtonian
physics, he oes not know what force means for earthlinks, he has never
seen F = dp/dt, or F =ma or, F = GMm/r^2

He is asked to find out what are the equations of motion in the weak
field limit.

Will he be able to derive Newton's laws of motion and universal law of
graviation in the weak field limit from the GR equations?

Yes. There is no wiggle room there.

Ask him/her/it to write down for you where the wiggle room is. Then
post it here.

There are an infinite number of solutions to the field equations. *In
vacuum, one such solution can be the following.

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

Where

** *G = Dimensionless constant
** *K = Constant of length
** *dO^2 = r^2 cos^2(Latitude) dLongitude^2 + r^2 dLatitude^2

It is static and spherically symmetric. *However, it is not
asymptotically flat. *Nevertheless, it is still a valid solution.

Does it degenerate into Newtonian law of gravity? *No.

Therefore, Mike is correct on this one. *Both Professor Draper and Mr.

On May 6, 5:34*am, "Juan R." González-Álvarez
wrote:
Koobee Wublee wrote on Mon, 05 May 2008 23:23:02 -0700:

Therefore, Mike is correct on this one. *Both Professor Draper and Mr.
Bielawski are very wrong. *It is time for both gentlemen to propose a
graceful retreat.

Mike is correct. GR does not reduce to NG in the weak field limit.

Although I have not researched the issue extensively as you have done,
my understanding is that one has to assume a specific form of the
metric to make the linearized equations converge to NG because there
are virtually infinite solutions. Maybe you may want to corerct me
here if I am wrong.

The form of the metric chosen though, implies NG at the weak field
limit. Then to come back and assert that GR converges to NG, is a
petitio principii, the worse of all falalcies but one found in many
places in both SR and GR.

It is also found in NM books on some subjects but this is beyond the
subject for now.

Mike






During the thread on Newtonian limit difficulties of GR I have discussed
this point with an expert on curved spacetime equations of motion, Eric
Poisson [1].

Eric confirms my finding that a = 0 in the linear regime of GR:

(\blockquote
*Since the energy-momentum tensor is already of first-order, in the
*linearized theory the conservation equations must be written down with
*the Minkowski metric, and this implies that the matter cannot have
*gravitational interactions. Or as you point out, particles would have
*to move on straight lines.
)

Textbooks and lecture notes take the weak field limit in a wrong way.
E.g. Carroll takes the fake limit

Z[a] = - L[\Gamma] Z[v v]

instead the correct one

L[a] = - L[\Gamma v v] = - L[\Gamma] Z[v v]

Prof. Carlip adds a couple of more mistakes and falsifications to his
'derivation'.

Indeed GR has not Newtonian limit in weak field or not.

[1] *http://relativity.livingreviews.org/...es/lrr-2004-6/

--http://canonicalscience.org/en/miscellaneouszone/guidelines.txt

On May 6, 4:52*am, Eric Gisse wrote:
On May 6, 12:33*am, Mike wrote:





On May 6, 3:49*am, Eric Gisse wrote:

On May 5, 10:23*pm, Koobee Wublee wrote:

On May 5, JanPB wrote:

Mike wrote:
Suppose this allien mathematical genius arrives to earth and he is
presented with the GR equations. He knows nothing about Newtonian
physics, he oes not know what force means for earthlinks, he has never
seen F = dp/dt, or F =ma or, F = GMm/r^2

He is asked to find out what are the equations of motion in the weak
field limit.

Will he be able to derive Newton's laws of motion and universal law of
graviation in the weak field limit from the GR equations?

Yes. There is no wiggle room there.

Ask him/her/it to write down for you where the wiggle room is. Then
post it here.

There are an infinite number of solutions to the field equations. *In
vacuum, one such solution can be the following.

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

Where

** *G = Dimensionless constant
** *K = Constant of length
** *dO^2 = r^2 cos^2(Latitude) dLongitude^2 + r^2 dLatitude^2

It is static and spherically symmetric. *However, it is not
asymptotically flat. *Nevertheless, it is still a valid solution.

Does it degenerate into Newtonian law of gravity? *No.

Therefore, Mike is correct on this one. *Both Professor Draper and Mr.
Bielawski are very wrong. *It is time for both gentlemen to propose a
graceful retreat.

Yes, gracefully retreat from the cranks who confuse persistence of
lies with facts.- Hide quoted text -

To me and to many people I know, a crank and an imbecile, is a person
who while it insists GR is a correct theory of gravitation, he thinks
that a = GM/r^2 is and ordinary differential equation. Such a crank
exists and is called Eric Gisse as these posts proves beyond any
doubt:

http://groups.google.gr/group/sci.ph...g/01b8793b2b1f...

Eric Gisse's excuse for not being able to sove a simple mechanics
problem: "One of these days I'll stop posting on groups when I'm sleep
deprived."

http://groups.google.gr/group/sci.ph...g/cc76e4770a62...

Now, who is the crank here?

The one who can't admit he made a mistake.

Do you know if radiation pressure exists yet?

Dig hard...

you are a troll. You are supposed to be in school, yet you instantly
reply to 99% of the posts in these groups.

Mike









Mike

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On May 6, 5:25*am, Mike wrote:
On May 6, 4:52*am, Eric Gisse wrote:



On May 6, 12:33*am, Mike wrote:

On May 6, 3:49*am, Eric Gisse wrote:

On May 5, 10:23*pm, Koobee Wublee wrote:

On May 5, JanPB wrote:

Mike wrote:
Suppose this allien mathematical genius arrives to earth and he is
presented with the GR equations. He knows nothing about Newtonian
physics, he oes not know what force means for earthlinks, he has never
seen F = dp/dt, or F =ma or, F = GMm/r^2

He is asked to find out what are the equations of motion in the weak
field limit.

Will he be able to derive Newton's laws of motion and universal law of
graviation in the weak field limit from the GR equations?

Yes. There is no wiggle room there.

Ask him/her/it to write down for you where the wiggle room is. Then
post it here.

There are an infinite number of solutions to the field equations. *In
vacuum, one such solution can be the following.

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

Where

** *G = Dimensionless constant
** *K = Constant of length
** *dO^2 = r^2 cos^2(Latitude) dLongitude^2 + r^2 dLatitude^2

It is static and spherically symmetric. *However, it is not
asymptotically flat. *Nevertheless, it is still a valid solution..

Does it degenerate into Newtonian law of gravity? *No.

Therefore, Mike is correct on this one. *Both Professor Draper and Mr.
Bielawski are very wrong. *It is time for both gentlemen to propose a
graceful retreat.

Yes, gracefully retreat from the cranks who confuse persistence of
lies with facts.- Hide quoted text -

To me and to many people I know, a crank and an imbecile, is a person
who while it insists GR is a correct theory of gravitation, he thinks
that a = GM/r^2 is and ordinary differential equation. Such a crank
exists and is called Eric Gisse as these posts proves beyond any
doubt:

http://groups.google.gr/group/sci.ph...g/01b8793b2b1f....

Eric Gisse's excuse for not being able to sove a simple mechanics
problem: "One of these days I'll stop posting on groups when I'm sleep
deprived."

http://groups.google.gr/group/sci.ph...g/cc76e4770a62....

Now, who is the crank here?

The one who can't admit he made a mistake.

Do you know if radiation pressure exists yet?

Dig hard...

you are a troll. You are supposed to be in school, yet you instantly
reply to 99% of the posts in these groups.

Does radiation pressure exist or not, Mike?


Mike



Mike

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- Show quoted text -

On May 5, 11:23 pm, Koobee Wublee wrote:

There are an infinite number of solutions to the field equations. In
vacuum, one such solution can be the following.

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

Where

** G = Dimensionless constant
** K = Constant of length
** dO^2 = r^2 cos^2(Latitude) dLongitude^2 + r^2 dLatitude^2

It is static and spherically symmetric. However, it is not
asymptotically flat. Nevertheless, it is still a valid solution.

Does it degenerate into Newtonian law of gravity? No.

Therefore, Mike is correct on this one. Both Professor Draper and Mr.
Bielawski are very wrong. It is time for both gentlemen to propose a
graceful retreat.

Another solution that is static, spherically symmetric, and (this
time) asymptotically flat is the following.

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

Does is degenerate into Newtonian law of gravity? No, because it
follows an inverse-cubed law instead of the inverse squared law. The
Einstein field equations represent an utter nonsense. They suit for
the ones to promote mysticism as wisdom. shrug

On May 6, 2:24 pm, Koobee Wublee wrote:
On May 5, 11:23 pm, Koobee Wublee wrote:



There are an infinite number of solutions to the field equations. In
vacuum, one such solution can be the following.

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

Where

** G = Dimensionless constant
** K = Constant of length
** dO^2 = r^2 cos^2(Latitude) dLongitude^2 + r^2 dLatitude^2

It is static and spherically symmetric. However, it is not
asymptotically flat. Nevertheless, it is still a valid solution.

Does it degenerate into Newtonian law of gravity? No.

Therefore, Mike is correct on this one. Both Professor Draper and Mr.
Bielawski are very wrong. It is time for both gentlemen to propose a
graceful retreat.

Another solution that is static, spherically symmetric, and (this
time) asymptotically flat is the following.

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

It's the same solution as Schwarzschild's, you merely changed the
numbers labelling the spheres of symmetry.

Does is degenerate into Newtonian law of gravity? No, because it
follows an inverse-cubed law instead of the inverse squared law.

I think you meant "inverse-Koobeed" (you knew I'd write that!) Anyway,
since you reshuffled the numbers labelling the events, obviously you
get a different expression wrt those labels. Using the same method, I
can define an electrostatic force that's inversely proportional to the
17th power of the coordinate. This sort of manipulations is just name-
changing and as such it obviously has no influence on the physics.

The
Einstein field equations represent an utter nonsense. They suit for
the ones to promote mysticism as wisdom. shrug

You've said that before. It's wrong.

--
Jan Bielawski

On May 6, 8:28 pm, JanPB wrote:
On May 6, 2:24 pm, Koobee Wublee wrote:

Another solution that is static, spherically symmetric, and (this
time) asymptotically flat is the following.

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

It's the same solution as Schwarzschild's, you merely changed the
numbers labelling the spheres of symmetry.

The Schwarzschild metric has the following form.

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

For the queer of England to continue living in that fat castle in the
air, he/she/it needs to talk himself/herself/itself into believing
both of the above equations are exactly the same.

Well, both are solutions to the field equations using the spherically
symmetric polar coordinate system and nothing else. There are no
coordinate transformations involved. They are just independent
solutions to the field equations. shrug

As I said before, for a mere $99,999, I will design the universe of
your choice from the field equations. I think that is a small price
for a universe.

Does it degenerate into Newtonian law of gravity? No, because it
follows an inverse-cubed law instead of the inverse squared law.

I think you meant "inverse-Koobeed"

Not at all. shrug

(you knew I'd write that!)

No. shrug

Anyway,
since you reshuffled the numbers labelling the events,

Hardly. shrug

obviously you
get a different expression wrt those labels.

We are talking about mathematics and not black magic. shrug

Using the same method, I
can define an electrostatic force that's inversely proportional to the
17th power of the coordinate.

You are incoherent as usual. shrug

This sort of manipulations is just name-
changing and as such it obviously has no influence on the physics.

There are no manipulations. shrug You are totally delusional.
shrug Please seek psychiatric help.

The
Einstein field equations represent an utter nonsense. They suit for
the ones to promote mysticism as wisdom. shrug

You've said that before. It's wrong.

Please check yourself in a mental asylum. It will do you good. You
will have companies also claiming to be queers of England.

On May 6, 3:18 pm, Eric Gisse wrote:
On May 6, 1:24 pm, Koobee Wublee wrote:

Another solution that is static, spherically symmetric, and (this
time) asymptotically flat is the following.

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

Does it degenerate into Newtonian law of gravity? No, because it
follows an inverse-cubed law instead of the inverse squared law. The
Einstein field equations represent an utter nonsense. They suit for
the ones to promote mysticism as wisdom. shrug

Its' the same solution as Schwarzschild, idiot.

It is not the Schwarzschild metric. Any sane and intelligent person
can tell you that.

I proved it to you
previously. Have you forgotten _already_ ?!

http://groups.google.com/group/sci.p...sg/5e4cf198adb...

Ahahaha... You continue to show why you remain a multi-year super-
senior at the very prestigious University of Alaska majoring in basket
weaving. shrug

On May 6, 8:59*pm, Koobee Wublee wrote:
On May 6, 3:18 pm, Eric Gisse wrote:

On May 6, 1:24 pm, Koobee Wublee wrote:
Another solution that is static, spherically symmetric, and (this
time) asymptotically flat is the following.

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

Does it degenerate into Newtonian law of gravity? *No, because it
follows an inverse-cubed law instead of the inverse squared law. *The
Einstein field equations represent an utter nonsense. *They suit for
the ones to promote mysticism as wisdom. *shrug

Its' the same solution as Schwarzschild, idiot.

It is not the Schwarzschild metric. *Any sane and intelligent person
can tell you that.

If there is a coordinate mapping between two supposedly different
metrics, they are the same.


I proved it to you
previously. Have you forgotten _already_ ?!

http://groups.google.com/group/sci.p...sg/5e4cf198adb...

Ahahaha... *You continue to show why you remain a multi-year super-
senior at the very prestigious University of Alaska majoring in basket
weaving. *shrug

On May 6, 11:25 pm, Eric Gisse wrote:
On May 6, 10:10 pm, Koobee Wublee wrote:

Yes. Or do you somehow think the coordinate labels determine the
physics?

Choice of coordinate does not change physics. This has been what I
have been saying all along. You are merely a malicious troll. That
is why you remain a multi-year super-senior even at the University of
Alaska. shrug

Sink this in. We have no more to discuss if you do not understand the
following very basic points.

** The geometry must be invariant.

** The choice of coordinate is observer dependent.

** The metric which together with the coordinate system describes the
geometry.

Therefore, the metric must be coordinate dependent.

Upon proper treatment, all solutions to the field equations are
independent although somewhat related to each other. This
relationship among all these solutions is what you call
transformation. However, at the stage where all these field equations
are laid out to be solved of their solutions, the choice of coordinate
system is cast in concrete. You cannot change the coordinate system
any more. Thus, each solution can only be valid with the already
established choice of coordinate system. In this case, it is the
common spherically symmetric polar coordinate system. You have been
told this many times over. It is time for you to understand this very
basic mathematical relationship. shrug

Once again, I ask you to explicitly apply a coordinate transformation
to G_uv = kT_uv and PROVE that a transformed solution is no longer a
solution.

Once again, I have told you there is no such transformation. shrug
Each solution to the field equations is independent of each other
using the same set of coordinate system.

Do you not know how, or do you somehow think that proves nothing?

Can you dig that, multi-year super-senior?