The Motion of the Perihelion of Mercury

On Jan 4, 11:25 am, George Hammond wrote:

Spherical surface area in flat space is obtained by
integrating the surface element cos(theta)d(theta)d(phi) in
sperical coordinates over phi and theta for constant r which
results in the ancient formula Area=4 pi r^2.

Well, is it really that obvious?

In curved space of course the metric must be used if the
phi and theta basis vectors are not orthogonal. However in
the Schwarzchild and Kooby metrics there is only radial
space warping and no angularspace warping (radially
symmetric metric), therefore the surface element is still a
simple rectangle and not a parallogram as in the general
case.

Well, you have to decide if the question applies to an observer
observing a sphere in curved space or a group of physicists trying to
detect the actual (local) surface area of a sphere situated in known
curved space.

In the case of an observer observing a sphere in curved space, he will
always observe flat space. Thus, the surface area is always (4 pi
r^2) regardless of how much space is curved.

In the case of a group of physicists trying to detect the actual
(local) surface area of a sphere situated in known curved space, you
have to accurately determine what the metric is first, and that is no
trivial task. shrug

[Rest of Hammond whining crap mercifully snipped]

On Jan 4, 12:32 am, Eric Gisse wrote:
On Jan 3, 6:37 pm, Koobee Wublee wrote:

Oh, there are more than that. In the following post, you have been
entertained with a solution that is static, spherically symmetric, but
not asymptotically flat.

How do you know it is not asymptotically flat when you did not even
consider calculating curvature components?

You have to understand what asymptotically flat means. shrug

On Jan 3, 9:55 pm, George Hammond wrote:
Koobee Wublee wrote:

......... That proves Birkhoff’s theorem false by
example. shrug

"proves Birkhoff's theorem false by example".....aaahahaha
ha hah hah aaaahaha ha ha ha ha......

Is that a denial? Do you not understand the disproval of a hypothesis
by example?

For example, if someone claims all oceanic creatures are not
mammalian, and I can show dolphins having habitats in the oceans
around the world are mammals, then that claim must be wrong.
Reverends still cannot get. I guess that would separate reverends
from scientists. shrug

On Jan 4, 11:09 pm, Sam Wormley wrote:

Grade: F-

That is funny. If a very mentally challenged individual such as
yourself gave me a grade of F, without any hesitation I would throw
out and burn any papers related to this subject. shrug

Thanks for the laughs. Ahahaha...

On Jan 4, 11:33 pm, Sam Wormley wrote:
Koobee Wublee wrote:

That is funny. If a very mentally challenged individual such as
yourself gave me a grade of F, without any hesitation I would throw
out and burn any papers related to this subject. shrug

Thanks for the laughs. Ahahaha...

You got the F- because you FAILED to address any of Eric's points.

Why are you trying to justify nonsense? Who gives a f*ck?

Again, thanks for the laughes. Ahahaha...

On Jan 4, 10:03*pm, Koobee Wublee wrote:
On Jan 4, 12:29 am, Eric Gisse wrote:

On Jan 3, 6:29 pm, Koobee Wublee wrote:
On Jan 3, 3:02 am, Eric Gisse wrote:
THE LABELS DO NOT MATTER.

Wrong! *The labels do matter. *shrug

Prove it. Write something with two different labels, and PROVE that
you can measure something to be different.

I have said that many times over. *To describe the geometry, one must
choose a coordinate system first and then define how the set of
coordinates is related to the geometry. *shrug

The coordinates are not related to the geometry. The coordinates -
like crackers - do not matter. MY argument is further buttressed by
your inability to calculate anything, like surface area of the amount
Mercury's perihelion would precess.

I do not see why you can't open a tensor analysis book written any
time in the last hundred and thirty or so years and look at the basic
proofs given.



I have an idea - let's see your proof that "other solutions" that
satisfy the conditions of Birkhoff's theorem produce something other
than 43 arcsec/centry for Mercury's perihelion precession. You did,
after all, assert that to be the case.

Well, it is time consuming. I have lacked any motivation to do so
after discovering GR is a total nonsense based on voodoo mathematics.
shrug

But you said it gives a different answer, which implies you have done
the calculation before.

However what you are saying implies you never did the calculation. So
you have no way of knowing whether the answer is different.


The COORDINATES do not matter. They all
represent the _same_ geometry.

Wrong again. *To describe a geometry, you need both the labels
(coordinate) and how the labels are connected (metric). *shrug

True - to write the metric down, you need coordinates.

Oh, wow! *That is an improvement after all these years. *shrug

Except the
coordinates do not matter as I can easily transform from one system to
the next and write the metric down in the new coordinate system.

Again, you are stuck in your matheMagical wonderland of coordinate
transformation. *There is no such coordinate transformation. *You are
schizophrenic. *shrug

People have been telling you what the coordinate transformation is
since 2006, and the transformation itself has been known for most of a
century. Is there any reason you do not understand?

You even wrote down the transformation. Or do you somehow think
writing one coordinate in terms of other coordinates is not a
coordinate transformation?

[snip]

As usual, you slip back into the old routine of slinging insults
because you have been challenged and have nothing else to say. It is
rather sad, but predictable.

On Jan 4, 10:12*pm, Koobee Wublee wrote:
On Jan 4, 12:32 am, Eric Gisse wrote:

On Jan 3, 6:37 pm, Koobee Wublee wrote:
Oh, there are more than that. *In the following post, you have been
entertained with a solution that is static, spherically symmetric, but
not asymptotically flat.

How do you know it is not asymptotically flat when you did not even
consider calculating curvature components?

You have to understand what asymptotically flat means. *shrug

Again, how do you know it is asymptotically flat when you didn't
calculate anything?

On Jan 5, 1:17 am, Eric Gisse wrote:
On Jan 4, 10:12 pm, Koobee Wublee wrote:
Eric Gisse wrote:

How do you know it is not asymptotically flat when you did not even
consider calculating curvature components?

You have to understand what asymptotically flat means. shrug

Again, how do you know it is asymptotically flat when you didn't
calculate anything?

Again, you have to understand what asymptotically flat means, mutli-
year super-senior. shrug

On Jan 5, 1:17 am, Eric Gisse wrote:
Koobee Wublee wrote:

I have said that many times over. To describe the geometry, one must
choose a coordinate system first and then define how the set of
coordinates is related to the geometry. shrug

The coordinates are not related to the geometry.

[Gisse’s diarrhea snipped]

To describe any geometry, you must choose a set of coordinate system
first. Without any coordinate system, you cannot describe any
geometry. Try to understand this very basic concept.

Or do you somehow think
writing one coordinate in terms of other coordinates is not a
coordinate transformation?

The accusation is broad. Any answer can easily be construed as
malevolence by you.

Thus, I just want you to show me how the following spacetimes ds1 to
ds2 are the same.

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

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

Since both solutions came from the same set of field equations with
coordinate system chosen right from the very beginning (Christoffel
symbols), the coordinate system in both is the same.

[More crap snipped]

More diarrhea coming from Gisse.

On Jan 5, 12:05*pm, Koobee Wublee wrote:
On Jan 5, 1:17 am, Eric Gisse wrote:

On Jan 4, 10:12 pm, Koobee Wublee wrote:
Eric Gisse wrote:
How do you know it is not asymptotically flat when you did not even
consider calculating curvature components?

You have to understand what asymptotically flat means. *shrug

Again, how do you know it is asymptotically flat when you didn't
calculate anything?

Again, you have to understand what asymptotically flat means, mutli-
year super-senior. *shrug

Whether I know it or not does not matter. You claim that your
"different" metrics aren't related by a coordinate transformation, and
setting aside the fact that is wrong, that means you need to actually
_prove_ asymptotic flatness by computing all of the components of the
Riemann curvature tensor R^a_bcd and prove they go to zero as r --
\infty.

The fact you never calculate anything says a lot about your knowledge
of the subject.