On Jul 23, 10:08 pm, Koobee Wublee wrote:
On Jul 21, 5:05 pm, Chalky wrote:

On Jul 21, 6:41 pm, Tom Roberts wrote:
Yes, of course. This becomes obvious when applying GR. Indeed, the
ability to model gravity using a potential phi inherently requires weak
fields and an approximation.

Quite. Nevertheless, GM / R in the exact Schwarzchild solution
certainly looks and smells like Newtonian potential energy, even
though we are advised to avoid stepping in it.

Each solution or the metric to the Einstein field equations is unique
and describes an independent universe from all other solutions. One
such particular solution that Hilbert had discovered is the famous
Schwarzschild metric

Really? I thought Schwarzschild discovered the Schwarzschild metric
only months after Einstein published his gravitational field equation.
Are you claiming Hilbert did it all first?

Of course, the (2 G M /c^2 / r) term smells like twice the Newtonian
potential.

We seem to have our wires crossed here. I was referring to the
redshift which is (1 - 2GM/R) -!/2

This only acts like twice the Newtonian potential at the relativistic
event horizon, which is exactly the same as what we find by applying
the General Principle to the rotating disk.

Why should I be surprised by that?

On Jul 23, 10:08 pm, Ian Parker wrote:

Something is wrong though. Of we move a mass to the rim energy will
increase.

No, the disk will slow down. (Spinning skater, ballerina
etc....Remember?)

Thus for an observer at the axis, if a sufficiently massive part of
the disk is moving radially outwards, this will cause experienced
'gravitational field' to decrease in magnitude near the origin.

If you think about it, something similar would happen if you moved a
sufficiently massive fraction of the Earth away from the Earth too.

Chalky wrote:
On Jul 23, 10:08 pm, Ian Parker wrote:

Something is wrong though. Of we move a mass to the rim energy will
increase.

No, the disk will slow down. (Spinning skater, ballerina
etc....Remember?)

Correct,
The motion creates a path that moves outward at a spiral,
the further out it gets the slower rotations will occur even if at the
same "path speed" but in reality it is constantly slowing
unless constant energy is applied.
If no constant energy is supplied, the energy will slowly
be decreasing.


Thus for an observer at the axis, if a sufficiently massive part of
the disk is moving radially outwards, this will cause experienced
'gravitational field' to decrease in magnitude near the origin.

If you think about it, something similar would happen if you moved a
sufficiently massive fraction of the Earth away from the Earth too.

Again, Correct.
The gravitational potential strength will decrease when the distance
increases.


--
James M Driscoll Jr
Spaceman

[Moderator's note: Post might be garbled due to unnecessary MIME stuff.
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On Jul 23, 10:08pm, Ian Parker wrote:
On 22 Jul, 01:05, Chalky wrote:

The
disc is not gravitational,

No, but an observer located on it can consider himself at rest in a
gravitational field, under the General Principle, just as the linearly
accelerating observer could too, as explicitly spelled out by the
General Postulate.

No he doesn't.

Who doesn't what? You are not making sense.

If we are in a gravitational field we can describe it
by Tensors.

We can describe it any way we like, that works.

We do not observe the Universe revolving round us.

Of course we DO! (what planet are you on?)

We INFER that this is because the planet is rotating (and orbiting
the Sun). This is because we OBSERVE the Sun and the stars revolving
around us every 24 hours, and the seasons cycling every year.

You do in point of fact get the same answer however you look at it. If
you apply special relativity you get a time dilatrion factor of (1-
v*v/c*c) If you integrate your centrifugal force, you will get (1-v*v/
c*c).

Assuming ???? means sqrt, yes. That is the whole point

Something is wrong though. Of we move a mass to the rim energy will
increase.

Actually, if the a mass is an appreciable fraction of the disk's mass,
the disk will slow down, instead,

But, yes, the situation is not identical to the gravitational fields
_induced by mass_

You seem to be following the Kierkegaard argument that once you have
used a ladder to get to a higher point, you can throw away the ladder.
You seem to be going one step beyond though by now denying that the
ladder ever existed.

On Jul 23, 10:08 pm, Ian Parker wrote:
On 22 Jul, 01:05, Chalky wrote:

On Jul 21, 6:41 pm, Ian Parker wrote:

This seems to me to be "Special" and not "General" Relativity.

It is at the interface between the 2

The
disc is not gravitational,

No, but an observer located on it can consider himself at rest in a
gravitational field, under the General Principle, just as the linearly
accelerating observer could too, as explicitly spelled out by the
General Postulate.

No he doesn't.

Jim Akerlund has carefully typed out the relevant text (by Einstein)
at

http://groups.google.com/group/sci.p...0c7aacfc6038e#

I suggest you read it very carefully. As I have already pointed out,
your argument contradicts Einstein.

[Moderator's note: Post might be garbled due to unnecessary mime stuff.
Since not all email gateways are 8-bit clean, and even if they are there
is not a standard 8-bit character set, MIME encoding if there are 8-bit
characters are OK and I usually change them back to 8-bit characters
which, as far as I know, show up correctly for everyone reading them.
However, encoding 7-bit characters, whether or not there are any 8-bit
characters in the post, is just plain stupid. Sorry, but that's the
only way to put it. This is a text-based newsgroup, and there is no
reason to send anything but 7-bit ASCII if that is what the text is to
begin with. In the future, we might have to start deleting posts which
have too much MIME garbage. Find out how to send plain text and do so.
-P.H.]

On Jul 24, 12:25 pm, Chalky wrote:
On Jul 23, 10:08 pm, Koobee Wublee wrote:

Each solution or the metric to the Einstein field equations is unique
and describes an independent universe from all other solutions. One
such particular solution that Hilbert had discovered is the famous
Schwarzschild metric

Really? I thought Schwarzschild discovered the Schwarzschild metric
only months after Einstein published his gravitational field equation.
Are you claiming Hilbert did it all first?

Yes, really. See Schwarzschild's own writing below.

http://arxiv.org/abs/physics/9905030

In summary, below spacetime represents Schwarzschild's original
solution which is not to be confused as the Schwarzschild metric.

ds^2 ppp c^2 (1 =96 K / u) dt^2 =96 dr^2 (du/dr)^2 / (1 =96 K / u) =96 u^2 dO^2

Where

** K ppp Integration constant
** u^3 ppp r^3 + K^3
** dO^2 ppp cos^2(Latitude) dLongitude^2 + dLatitude^2

Notice Schwarzschild's original solution does not manifest any black
holes. It was Hilbert a year or two later that finally wrote down the
Schwarzschild metric below.

ds^2 ppp c^2 (1 =96 K / r) dt^2 =96 dr^2 / (1 =96 K / r) =96 r^2 dO^2

As you know the Schwarzschild metric does manifest black holes.
Another simple solution to the field equations is described as follows
where it also does not manifest black holes as the original
Schwarzschild's solution.

ds^2 ppp c^2 dt^2 / (1 + K / r) =96 dr^2 (1 + K / r) =96 (r + K)^2 dO^2

All these three solutions satisfy the following:

** Newtonian gravity as the limiting case
** Static
** Spherically symmetric
** Asymptotically flat

Thus, they are all valid solutions. The most acceptable solution is
the Scwharzschild metric discovered by Hilbert that manifest black
holes.

Of course, the (2 G M /c^2 / r) term smells like twice the Newtonian
potential.

We seem to have our wires crossed here. I was referring to the
redshift which is (1 - 2GM/R) -!/2

http://groups.google.com/group/sci.p...3d816f?hlpppen

This only acts like twice the Newtonian potential at the relativistic
event horizon, which is exactly the same as what we find by applying
the General Principle to the rotating disk.

http://groups.google.com/group/sci.p...dc9470?hlpppen

Why should I be surprised by that?

Since you raised the issue, I thought you were surprised. That is
all. shrug

On Jul 25, 9:52 pm, Spaceman
wrote:
Chalky wrote:
On Jul 23, 10:08 pm, Ian Parker wrote:

Something is wrong though. Of we move a mass to the rim energy will
increase.

No, the disk will slow down. (Spinning skater, ballerina
etc....Remember?)

Correct,
The motion creates a path that moves outward at a spiral,
the further out it gets the slower rotations will occur even if at the
same "path speed" but in reality it is constantly slowing
unless constant energy is applied.
If no constant energy is supplied, the energy will slowly
be decreasing.

Thus for an observer at the axis, if a sufficiently massive part of
the disk is moving radially outwards, this will cause experienced
'gravitational field' to decrease in magnitude near the origin.

If you think about it, something similar would happen if you moved a
sufficiently massive fraction of the Earth away from the Earth too.

Again, Correct.
The gravitational potential strength will decrease when the distance
increases.


Yes, you are right. The gravitational potential difference *does*
decrease in both situations, irrespective of whether the potential
difference encourages or discourages that radial motion.

Thanks also for responding via the moderated newsgroup instead of via
the rogue sub-thread created by Kubee Wooblie removing SPResearch from
the loop.

It is unfortunate that Eric Gisse (for example) has not also tried
responding in this way, since this does tend to filter out a lot of
undesirable noise (and invariably supresses flames).

On Jul 27, 7:51 pm, Koobee Wublee wrote:
On Jul 24, 12:25 pm, Chalky wrote:

We seem to have our wires crossed here. I was referring to the
redshift which is (1 - 2GM/R)^ -1/2

http://groups.google.com/group/sci.p...sg/7dd4e0a10f3...

Your central argument in this link is that the rim is blue shifted not
red shifted (as was Ian Parker's), and, by implication, that
Einstein's special relativistic time dilation formula is upside down.

Perhaps you should show a rigorous derivation of this alleged
relativistic time contraction, and an explain why, despite this,
Einstein's formula had already been verified observationally by the
early 70s using accurate clocks on aircraft flights (MTW). This would
be more sensible than challenging me to prove that red is blue.

This only acts like twice the Newtonian potential at the relativistic
event horizon, which is exactly the same as what we find by applying
the General Principle to the rotating disk.

http://groups.google.com/group/sci.p...sg/f42dddd7a3d...

This link does at least present your prior formulae uncorrupted.

Why should I be surprised by that?

Since you raised the issue, I thought you were surprised. That is
all. shrug

When I first raised this question I did not know that the Schwarzchild
redshift formula was effectively identical to the rotating disk
formula if you interpret GM/R as Newtonian potential.

I am relieved rather than surprised that the General Principle works
that well, in this situation.