The Motion of the Perihelion of Mercury

On Mon, 5 Jan 2009 22:16:57 -0800 (PST), Koobee Wublee
wrote:

On Jan 5, 2:18 pm, George Hammond wrote:
Koobee Wublee wrote:

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.

Holy Curvature almighty God!!!! How many times do you
have to repeat it?

It looks like the great reverend Hammond has just run out of his
scriptures. Oh, well. It is time for you to check yourself into a
mental institution. shrug

Thus, I just want you to show me how the following spacetimes ds1 [and]
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.

ds1 is just the plain vanilla Schwarzchild Metric.

Well, why is that a surprise for you?

[Hammond]
Never said it was.

If you
make the substitution r=r+K simple algebra immediately gives
you ds2 !!!!

If (r = r + K) and (K != 0), the (r = 0).

[Hammond]
I said "SUBSTITUTE" r+K for r, Tricycle.

It sounds like you have not
passed grade school mathematics. shrug

[Hammond]
You couldn't pass grammar Pearl Diver, never mind math.
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On Jan 5, 11:00 pm, George Hammond wrote:
Koobee Wublee wrote:

If (r = r + K) and (K != 0), the (r = 0).

I said "SUBSTITUTE" r+K for r, Tricycle.

No, you can say “SUBSTITUTE” (r + K) for R. shrug

In that case, the two said geometries of spacetime a

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

** ds2^2 = c^2 (1 – K / R) dt^2 – dR^2 / (1 – K / R) – R^2 dO^2

In that case, they are still very different where (R = r + K).
shrug

It sounds like you have not
passed grade school mathematics. shrug

You couldn't pass grammar Pearl Diver, never mind math.

Even my pre-kindergarten kids know R is not the same as r. Either my
kids are geniuses or you are extra-ordinarily mentally challenged.
shrug

On Mon, 5 Jan 2009 22:58:52 -0800 (PST), Koobee Wublee
wrote:

On Jan 5, 10:37 pm, George Hammond wrote:
Koobee Wublee wrote:

Whether you know what asymptotically flat means or not, it still
remains your problem. shrug

Strictly speaking "asymptotically flat" means the 4th
order Riemann curvature tensor goes to zero at r=infinity.
In simple cases this can be determined by inspecting the
metric.... if it approaches the Minkowski metric for r=oo
then the space is asymptotically flat. For instance the
Schwarzchild metric approaches the Minkowski Metric in
spherical coordinates for r--oo.

It actually does not involve the Rieman curature tensor.

[Hammond]
Nope... it actually does.

As long as
the geometry approaches flat space at r = infinity, it is considered
asymptotically flat.

[Hammond]
That's a tautology not a definition.


Both spacetimes I have described below satisfy
this criterion. shrug

[Hammond]
Of course they do since they are both identically the
Schwarzchild Metric. Shouldn't you be getting back to the
shop?

** 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

[Hammond]
I'll give you a clud Kooby.... really smart people don't
challenge Relativity.... they discuss how Relativity
explains God, even Einstein couldn't figure that out.
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GOD=G_uv (a folk song on mp3)
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=====================================

On Jan 5, 11:33 pm, George Hammond wrote:
Koobee Wuble wrote:

It actually does not involve the Rieman curature tensor.

Nope... it actually does.

No, it does not. Show me where your God said so. shrug

As long as
the geometry approaches flat space at r = infinity, it is considered
asymptotically flat.

That's a tautology not a definition.

Hmmm... That is no tautology but a true description of what
“asymptotically flat” means. shrug

Both spacetimes I have described below satisfy
this criterion. shrug

Of course they do since they are both identically the
Schwarzchild Metric.

It proves you have flunked grade school mathematics. shrug

Shouldn't you be getting back to the shop?

After you.

** 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

I'll give you a clud Kooby....

From a candidate of mental patients, that is going to be interesting.

really smart people don't challenge Relativity....

Try “really stupid people are incapable of challenging relativities
(both SR and GR)”. shrug

they discuss how Relativity explains God,

Hmmm... Like yourself, a candidate for mental hospitals. shrug

even Einstein couldn't figure that out.

Of course not, Einstein was nobody. Einstein was a nitwit, a
plagiarist, and a lair. Why do you expect Einstein to figure out even
how to match a pair of socks. It would be way too challenging for a
nitwit, a plagiarist, and a liar. shrug

On Mon, 5 Jan 2009 23:09:29 -0800 (PST), Koobee Wublee
wrote:

On Jan 5, 11:00 pm, George Hammond wrote:
Koobee Wublee wrote:

If (r = r + K) and (K != 0), the (r = 0).

I said "SUBSTITUTE" r+K for r, Tricycle.

No, you can say “SUBSTITUTE” (r + K) for R. shrug

In that case, the two said geometries of spacetime a

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

** ds2^2 = c^2 (1 – K / R) dt^2 – dR^2 / (1 – K / R) – R^2 dO^2

In that case, they are still very different where (R = r + K).

[Hammond]
The COORDINATES are different, but the METRIC is the same.
The metric is the Schwarzchild metric in BOTH cases.
You don't know the difference.

snip crackpot syrup
=====================================
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mirror site:
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GOD=G_uv (a folk song on mp3)
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=====================================

On Jan 5, 11:48 pm, George Hammond wrote:
Koobee Wublee wrote:

No, you can say “SUBSTITUTE” (r + K) for R. shrug

In that case, the two said geometries of spacetime a

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

** ds2^2 = c^2 (1 – K / R) dt^2 – dR^2 / (1 – K / R) – R^2 dO^2

In that case, they are still very different where (R = r + K).

[Hammond]
The COORDINATES are different, but the METRIC is the same.

No, the coordinates are different:

** (r, theta, phi)
** (R, theta, phi)

Where

** R = r + k

The metric is the Schwarzchild metric in BOTH cases.
You don't know the difference.

No, the metrics are also different. This is basic mathematics. Try
to understand mathematics in grade school. shrug

[snipped diarrhea of Einstein Dingleberries]

On Mon, 5 Jan 2009 23:47:52 -0800 (PST), Koobee Wublee
wrote:

On Jan 5, 11:33 pm, George Hammond wrote:
Koobee Wuble wrote:


It proves you have flunked grade school mathematics. shrug

[Hammond]
Nope... I have a BS and MS in Physics from 2 accredited
universities in MA and my CV is clearly posted on my
website, URL below.


Of course not, Einstein was nobody. Einstein was a nitwit, a
plagiarist, and a lair. Why do you expect Einstein to figure out even
how to match a pair of socks. It would be way too challenging for a
nitwit, a plagiarist, and a liar. shrug

[Hammond]
hmm.... did you say "Einstein couldn't match a pair of
socks"... you wouldn't be referring to the discovery that
Relativity explains God would you?
I say that because obviously both Relativity and God have
got you confounded judging from the tone of your anti
Einstein rant.
=====================================
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mirror site:
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GOD=G_uv (a folk song on mp3)
http://interrobang.jwgh.org/songs/hammond.mp3
=====================================

On Mon, 5 Jan 2009 23:54:30 -0800 (PST), Koobee Wublee
wrote:

On Jan 5, 11:48 pm, George Hammond wrote:
Koobee Wublee wrote:

No, you can say “SUBSTITUTE” (r + K) for R. shrug

In that case, the two said geometries of spacetime a

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

** ds2^2 = c^2 (1 – K / R) dt^2 – dR^2 / (1 – K / R) – R^2 dO^2

In that case, they are still very different where (R = r + K).

[Hammond]
The COORDINATES are different, but the METRIC is the same.

No, the coordinates are different:

[Hammond]
That's what I just said, you're repeating yourself Koobic.

** (r, theta, phi)
** (R, theta, phi)

Where

** R = r + k

The metric is the Schwarzchild metric in BOTH cases.
You don't know the difference.

No, the metrics are also different.


[Hammond]
NO.... the metrics are NOT different. They both describe
the SAME physical space with the same metrical distances.
The space described by (r,theta, phi) is IDENTICAL to the
space described by (R,theta,phi) because the METRICS are
identical, irregardless of what K is...a Titleist golf ball
is exactly the same diameter and shape whether it is
described by R or r...in fact both spaces are identically
Schwarchild space. Obviously you have no credentials in
Physics.
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mirror site:
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GOD=G_uv (a folk song on mp3)
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=====================================



This is basic mathematics. Try
to understand mathematics in grade school. shrug

[snipped diarrhea of Einstein Dingleberries]

On Jan 5, 9:23*pm, Koobee Wublee wrote:
[snip]

You have to show me how these two geometries are identical before we
continue further. *If not, don’t cry about me not justifying your
ignorance, multi-year super-senior.

It has been shown to you repeatedly in the last two and a half years.

However, you simply do not understand. Which is not surprising because
of how little you actually do understand.


[More whining crap snipped]

On Jan 5, 9:58*pm, Koobee Wublee wrote:
On Jan 5, 10:37 pm, George Hammond wrote:

Koobee Wublee wrote:
Whether you know what asymptotically flat means or not, it still
remains your problem. *shrug

* *Strictly speaking "asymptotically flat" means the 4th
order Riemann curvature tensor goes to zero at r=infinity.
* *In simple cases this can be determined by inspecting the
metric.... if it approaches the Minkowski metric for r=oo
then the space is asymptotically flat. *For instance the
Schwarzchild metric approaches the Minkowski Metric in
spherical coordinates for r--oo.

It actually does not involve the Rieman curature tensor.

Do you at all find it strange that every time you argue about
something, it is you saying one thing and several other people saying
the same something else?

[snip]