Gravitational Redshift On Rotating Disk (and its implications)
Chalky
rim
= sqrt(1 - [v/c]^2) = sqrt(1 - [wr/c]^2)
= sqrt(1 - integral[ w^2 r]dr/[c]^2)
= sqrt(1 - 2delphi/c]^2)
where w = angular velocity, and del phi is difference in gravitational
potential energy from observer located at middle of disk.
Clearly this is only equal to the traditional gravitational Doppler
shift factor (1 - delphi/c^2), to a first approximation.
Is there any more subtle relativistic effect here that I have missed?
If not, does the general principle thus establish that the following
statement is, consequently, a general law of nature?
"The gravitational Doppler shift factor of 1 - del phi/c^2, is only
generally valid to a first approximation."
Chalky
On Jul 18, 12:07 pm, Chalky <chalkys...@bleachboys.co.uk> wrote:
> On repeating Einstein's rotating disk experiment, I find redshift of
> rim is given by 1(1 + z)
> = sqrt(1 - [v/c]^2) = sqrt(1 - [wr/c]^2)
> = sqrt(1 - 2integral[ w^2 r]dr/[c]^2)
> = sqrt(1 - 2delphi/c]^2)
> where w = angular velocity, and del phi is difference in gravitational
> potential energy from observer located at middle of disk.
>
> Clearly this is only equal to the traditional gravitational Doppler
> shift factor (1 - delphi/c^2), to a first approximation.
>
> Is there any more subtle relativistic effect here that I have missed?
>
> If not, does the general principle thus establish that the following
> statement is, consequently, a general law of nature?
>
> "The gravitational Doppler shift factor of 1 - del phi/c^2, is only
> generally valid to a first approximation."
>From a reference given at spf (http://www.teleconnection.info/rqg/
Gravitation#TheWeakFieldLimit) it seems that the Schwarzchild solution
of EFE also gives an accurate solution of 1 + z = (1 - 2 del phi/c^2)^
- 1/2, thus confirming that this formula IS more accurate, generally.
Comments?
Tom Roberts
> "The gravitational Doppler shift factor of 1 - del phi/c^2, is only
> generally valid to a first approximation."
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.
Tom Roberts
Ian Parker
This seems to me to be "Special" and not "General" Relativity. The
disc is not gravitational, it is constrained to spin by quantum
electrodynamics NOT by gravity/curvature of space. What we must do is
take the instantaneous Lorenz Transformation at a point on the rim.
This is quite a spectacular example. Blue rather than red shift
produces Vacuum UV/Soft X rays, but the principle is the same.
- Ian Parker
Koobee Wublee
> On repeating Einstein's rotating disk experiment,
Einstein the nitwit, the plagiarist, and the liar was totally wrong
about this.
The emitter is rotating around the detector in a perfect circle. It
emits light to be observed by a detector at the center of this
rotating platform. The emitter experiences an acceleration of the
following in terms of g’s.
v^2 / r / g
Where
** v = Rotating speed of the emitter
** r = Rotating radius
** g = Gravitational acceleration
> I find redshift of
> rim
> = sqrt(1 - [v/c]^2) = sqrt(1 - [wr/c]^2)
> = sqrt(1 - integral[ w^2 r]dr/[c]^2)
> = sqrt(1 - 2delphi/c]^2)
> where w = angular velocity, and del phi is difference in gravitational
> potential energy from observer located at middle of disk.
The frequency shift of light is actually blue shifted in according to
the following.
1 / sqrt(1 – B^2) ~ 1 + B^2 / 2
Where
** B c = v
** ~ if (1 >> B)
In the meantime, the gravitational red shift is fudged into the
following by the priests preferred to be called physicists.
Sqrt(1 – 2 U) ~ 1 - U
Where
** U = G M / c^2 / r
** ~ if (1 >> U)
> Clearly this is only equal to the traditional gravitational Doppler [...]
Now, please go back and re-do your analysis. Show the priests how you
can justify (+ B^2 / 2, blue shift) with (- U, red shift).
Ps. This reply has sci.physics.research removed. It is a priviledged
newsgroup for the priests to comfort themselves into believing
‘PRIESTHOOD IS SCIENCE’.
Eric Gisse
> On Jul 18, 4:07 am, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > On repeating Einstein's rotating disk experiment,
>
> Einstein the nitwit, the plagiarist, and the liar was totally wrong
> about this.
You discredit yourself every time you repeat these little idiocies.
[snip]
Koobee Wublee
> You discredit yourself every time you repeat these little idiocies.
And what have you found so contradicting to you being a multi-year
super-senior?
In the meantime, my post has not been addressed in any intelligent
responses.
On Jul 18, 4:07 am, Chalky <chalkys...@bleachboys.co.uk> wrote:
> On repeating Einstein's rotating disk experiment,
Einstein the nitwit, the plagiarist, and the liar was totally wrong
about this.
The emitter is rotating around the detector in a perfect circle. It
Where
Where
Where
Ps. This reply has sci.physics.research removed. It is a newsgroup
for the priests to comfort themselves into believing ‘PRIESTHOOD IS
SCIENCE’.
The Orwellian educational system demands:
** MYSTICISM IS WISDOM
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** FAITH IS THEORY
** LYING IS TEACHING
** BELIEVING IS LEARNING
** IGNORANCE IS TRUTH
Androcles
"Koobee Wublee" <koobee...@gmail.com> wrote in message
news:38b877e0-45dc-49c6...@u6g2000prc.googlegroups.com...
** MYSTICISM IS WISDOM
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** FAITH IS THEORY
** LYING IS TEACHING
** BELIEVING IS LEARNING
** IGNORANCE IS TRUTH
Very applicable to the fuckheaded aetherialist Koobee Wankley.
Igor
Frankly, I prefer to put my faith into people who actually understand
how to tranform domains. Not in people so full of themselves that
they think they can ignore basic algebra. But then education has
never been a top priority with you.
Chalky
> On 18 Jul, 12:07, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
>
>
> > On repeating Einstein's rotating disk experiment, I find redshift of
> > rim
> > = sqrt(1 - [v/c]^2) = sqrt(1 - [wr/c]^2)
> > = sqrt(1 - integral[ w^2 r]dr/[c]^2)
> > = sqrt(1 - 2delphi/c]^2)
> > where w = angular velocity, and del phi is difference in gravitational
> > potential energy from observer located at middle of disk.
>
> > Clearly this is only equal to the traditional gravitational Doppler
> > shift factor (1 - delphi/c^2), to a first approximation.
>
> > Is there any more subtle relativistic effect here that I have missed?
>
> > If not, does the general principle thus establish that the following
> > statement is, consequently, a general law of nature?
>
> > "The gravitational Doppler shift factor of 1 - del phi/c^2, is only
> > generally valid to a first approximation."
>
> This seems to me to be "Special" and not "General" Relativity. The
> disc is not gravitational, it is constrained to spin by quantum
> electrodynamics NOT by gravity/curvature of space. What we must do is
> take the instantaneous Lorenz Transformation at a point on the rim.
>
>
> This is quite a spectacular example. Blue rather than red shift
> produces Vacuum UV/Soft X rays, but the principle is the same.
Your derivation indicates a misunderstanding of SR too. The Lorentz
transforms do NOT tell the high velocity observers that their local
physics is different. That would violate the Special Principle.
Furthermore it appears to violate the basic tenets of Quantum
Mechanics as well. The electron is bound by quantum uncertainty so you
can't really talk about what the electron 'sees' classically, let
alone apply the Lorentz transforms to this to decide that it will now
see local physics differently because it is moving.
Chalky
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.
SCNR. ; )
Chalky
> On 18 Jul, 12:07, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > On repeating Einstein's rotating disk experiment, I find redshift of
> > rim
> > = sqrt(1 - [v/c]^2) = sqrt(1 - [wr/c]^2)
> > = sqrt(1 - integral[ w^2 r]dr/[c]^2)
> > = sqrt(1 - 2delphi/c]^2)
> > where w = angular velocity, and del phi is difference in gravitational
> > potential energy from observer located at middle of disk.
>
> > Clearly this is only equal to the traditional gravitational Doppler
> > shift factor (1 - delphi/c^2), to a first approximation.
>
> > Is there any more subtle relativistic effect here that I have missed?
>
> > If not, does the general principle thus establish that the following
> > statement is, consequently, a general law of nature?
>
> > "The gravitational Doppler shift factor of 1 - del phi/c^2, is only
> > generally valid to a first approximation."
>
> 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.
> it is constrained to spin by quantum
> electrodynamics NOT by gravity/curvature of space. What we must do is
> take the instantaneous Lorenz Transformation at a point on the rim.
>
> http://groups.google.co.uk/group/sci.physics/browse_frm/thread/55ba5d...
Then your disagreement is with Einstein.
See Einstein A. (1920) Relativity, the special and the general theory.
Ch 18 general principle
Ch 20 general postulate
Ch 23 rotating disk experiment
Koobee Wublee
Frankly, ignorant Igor, I don’t give a damn what your faith is.
<shrug>
As usual, Gisse’s, Androcle’s, and your responses have nothing to do
with my posts. Oh, well. How do you feel to be in bed with Androcles
in denying the Aether?
I have done my transformations in accordance to the rules of
mathematics. You do not understand these. Instead, you put your
faith in the priests. That is OK. That is your life under the
Orwellian control where:
** MYSTICISM IS WISDOM
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** FAITH IS THEORY
** LYING IS TEACHING
** BELIEVING IS LEARNING
** IGNORANCE IS TRUTH
<shrug>
Koobee Wublee
> "Koobee Wublee" <koobee.wub...@gmail.com> wrote in message
> Very applicable to the fuckheaded aetherialist Koobee Wankley.
In your dreams, there is no Aether.
In reality, the Aether is all around you. <shrug>
In your life, you are under the Orwellian control for so long where:
Androcles
Very applicable to the fuckheaded aetherialist Wanker Wublee.<shrug>
Koobee Wublee
> Very applicable to the fuckheaded aetherialist Wanker Wublee.<shrug>
The Aether-denier nitwit, Androcles, is in need of his medication
again. Listen:
“Dad, this is Wendy. Please stop denying the Aether. Please stop
being an Einstein-Dingleberry for a change. Only doing that, I can
contact you.”
In the meantime, the serious discussions of gravitational red shift
and transverse Doppler blue shift so impolitely interrupted by Gisse,
Igor, and Androcles shall continue.
The original post my mine which still meets no serious challenges is
presented below.
On Jul 21, 11:35 pm, "Androcles" wrote:
> Very applicable to the fuckheaded aetherialist Wanker Wublee.<shrug>
The Aether-denier nitwit, Androcles, is in need of his medication
again. Listen:
“Dad, this is Wendy. Please stop denying the Aether. Please stop
being an Einstein-Dingleberry for a change. Only doing that, I can
contact you.”
In the meantime, the serious discussions of gravitational red shift
and transverse Doppler blue shift so impolitely interrupted by Gisse,
Igor, and Androcles shall continue.
The original post my mine which still meets no serious challenges is
presented below.
Androcles
On Jul 21, 11:35 pm, "Androcles" wrote:
> Very applicable to the fuckheaded aetherialist Wanker Wublee.<shrug>
The Aether-denier nitwit, Androcles, is in need of his medication
again.
----------------------------------------------------------------------------------
The aether assertive fuckwit, Koobee Wublee, has no evidence for his faith
and in need of sense.
He even plagiarises the kook Roberts' shrug as if it meant anything.
Igor
Your religion is apparently contrarianism. Stick a fork in you.
You're done.
Eric Gisse
[...]
> Your religion is apparently contrarianism. Stick a fork in you.
> You're done.
More correct than you think.
Koobee Wublee
> The aether assertive fuckwit, Koobee Wublee, has no evidence for
> his faith and in need of sense.
What faith?
> He even plagiarises the kook Roberts' shrug as if it meant anything.
So, the nitwit Androcles has run out of argument. He now calls
shrugging a unique invention of Professor Roberts. The best place for
Androcles is to hide under a rock. <shrug>
Koobee Wublee
This is how a serious scientific discussion degenerates into when
confronted with the point-blank argument. So, the priests of SR and
GR with their own tails between their legs just sent these two clowns
Igor and Gisse to disrupt the discussion. This has been done so many
times in the past. In the mean time, the real issue still has to be
addressed.
On Jul 18, 4:07 am, Chalky wrote:
> On repeating Einstein's rotating disk experiment,
Einstein the nitwit, the plagiarist, and the liar was totally wrong
Androcles
| On Jul 22, 1:05 am, "Androcles" wrote:
|
| > The aether assertive fuckwit, Koobee Wublee, has no evidence for
| > his faith and in need of sense.
|
| What faith?
Your blind faith in a stupid aether for which you have no evidence, of
course.
How does a ring laser gyroscope work, fuckwit?
Koobee Wublee
> 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 in which it is described in general as follows.
ds^2 = c^2 T dt^2 (1 – K / r) – dr^2 / (1 – K / r) – r^2 dO^2
Where
** K, T = Integration constants
** dO^2 = r^2 cos^2(Latitude) dLongitude^2 + r^2 dLatitude^2
To impose the limiting case to Newtonian mechanics, the integration
constants must be fudged into the following to do so.
** T = 1
** K = 2 G M / c^2
Of course, the (2 G M /c^2 / r) term smells like twice the Newtonian
potential. In particular, if you work out the mathematics of the
Lagrangian method confined to the Schwarzschild metric, you will find
the following Euler Lagrange equation associated with time.
(1 – 2 U) / (1 – B^2) = Constant
Where
** B^2 c^2 = (dr/dt)^2 / (1 – 2 U)^2 + r^2 (dO/dt)^2 / (1 – 2 U)
** U = G M / c^2 / r
The constant can be interpreted as the following.
Constant = E / (m c^2)
Where
** E = Observed energy
** m = intrinsic mass of the particle
Under weak gravitation and low speed, you have the Newtonian
conservation of energy as described below.
B2 – 2 U = constant
Or
B^2 / 2 – U = constant
Where
** B^2 c^2 ~ (dr/dt)^2 + r^2 (dO/dt)^2
Why is this simple mathematical result so surprising to you?
Koobee Wublee
> Your blind faith in a stupid aether for which you have no evidence, of
> course.
The Aether is predicted by electromagnetism. <shrug>
> How does a ring laser gyroscope work, fuckwit?
It works without violating the existence of the Aether. <shrug>
Remember that you are merely a programmer to code what the algorithms
dictated to you by your control system engineers (and thus your
superiors). It is no wonder that you reject the Aether without
knowing your own proper hierarchy. <shrug>
By the way, do you hear what I hear? Wendy is calling you again from
the other side. “Dad, please do not deny the Aether. Understand
electromagnetism and accept the existence of the Aether.”
Eric Gisse
> On Jul 22, 6:04 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Jul 22, 3:11 pm, Igor <thoov...@excite.com> wrote:
> > > Your religion is apparently contrarianism. Stick a fork in you.
> > > You're done.
>
> > More correct than you think.
>
It is never a serious scientific discussion with you. You are
delusional if you think otherwise - just read all slanders and lies
you write about Einstein, your broad insults against physics, etc,
etc. That doesn't even touch on the major faults in your arguments
which you can never even address.
Androcles
On Jul 22, 11:30 pm, "Androcles" wrote:
> Your blind faith in a stupid aether for which you have no evidence, of
> course.
The Aether is predicted by electromagnetism. <shrug>
---------------------------------------------------------------------
Bullshit. <shrug>
> How does a ring laser gyroscope work, fuckwit?
| It works without violating the existence of the Aether. <shrug>
--------------------------------------------------------------------
Koobee Wublee
> It is never a serious scientific discussion with you.
Ah, finally the troll speaks the truth. You are just trolling. That
is very despicable on your part.
> You are
> delusional if you think otherwise - just read all slanders and lies
> you write about Einstein,
I have written no lies about Einstein the nitwit, the plagiarist, and
the liar. Since Einstein is your messiah on the religions of SR and
GR, you are just deeply hurt by the truth. <shrug>
Oh, I want to add more about Einstein’s study on the Newtonian law of
gravity. As I have already pointed out, by 1908 or so Einstein
finally understood Newton’s law of gravity given him the benefit of
doubt. It only took Einstein more than 10 years to do so. Einstein-
Dingleberries would call that feat a sign of great intellectual might
in him. Some like myself would call that a pathetic achievement of a
nitwit. Some would call Einstein’s reverse-engineering and re-
discovery of the principle of equivalence as a true genius in the
making. Some like myself would just call that plagiarism to add up to
piles of plagiarized works he had accumulated. <shrug>
> your broad insults against physics, etc,
What insults. The truth is not insult even if the truth hurts.
<shrug>
Since you are hurt by reading what I have written, deep down you must
know that I am right. The truth shall triumph over the Orwellian lies
where:
** MYSTICISM IS WISDOM
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** FAITH IS THEORY
** LYING IS TEACHING
** BELIEVING IS LEARNING
** IGNORANCE IS TRUTH
** STUPIDITY IS INTELLECT
> etc. That doesn't even touch on the major faults in your arguments
> which you can never even address.
Mind you that we are not discussing my personal bad habits. <shrug>
Androcles
Typical Aetherialist preaching, at least Gisse knows you are a fucking idiot
even if he doesn't know Einstein was.
How does a ring laser gyroscope work, clueless shit-for-brains?
Eric Gisse
> On Jul 23, 12:23 am, Eric Gisse wrote:
>
> > It is never a serious scientific discussion with you.
>
> Ah, finally the troll speaks the truth. You are just trolling. That
> is very despicable on your part.
>
> > You are
> > delusional if you think otherwise - just read all slanders and lies
> > you write about Einstein,
>
> I have written no lies about Einstein the nitwit, the plagiarist, and
> the liar. Since Einstein is your messiah on the religions of SR and
> GR, you are just deeply hurt by the truth. <shrug>
....and this is why it is never a serious scientific discussion with
you.
[snip]
Ian Parker
> On Jul 21, 6:41 pm, Ian Parker <ianpark...@gmail.com> wrote:
>
>
>
>
>
> > On 18 Jul, 12:07, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > > On repeating Einstein's rotating disk experiment, I find redshift of
> > > rim
> > > = sqrt(1 - [v/c]^2) = sqrt(1 - [wr/c]^2)
> > > = sqrt(1 - integral[ w^2 r]dr/[c]^2)
> > > = sqrt(1 - 2delphi/c]^2)
nal
> > > potential energy from observer located at middle of disk.
>
> > > Clearly this is only equal to the traditional gravitational Doppler
> > > shift factor (1 - delphi/c^2), to a first approximation.
>
> > > Is there any more subtle relativistic effect here that I have missed?
>
> > > If not, does the general principle thus establish that the following
> > > statement is, consequently, a general law of nature?
>
> > > "The gravitational Doppler shift factor of 1 - del phi/c^2, is only
> > > generally valid to a first approximation."
>
> > 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.
>
by Tensors. We do not observe the Universe revolving round us. If you
are on a disc you observe your surroundings rotating. You and your
surroundings have an instantanous Lorenz transformation.
I think that the confusion is due to a faulty understanding of Mach's
principle which relativity does not affect.
> > it is constrained to spin by quantum
> > electrodynamics NOT by gravity/curvature of space. What we must do is
> > take the instantaneous Lorenz Transformation at a point on the rim.
>
> >http://groups.google.co.uk/group/sci.physics/browse_frm/thread/55ba5d...
>
> Then your disagreement is with Einstein.
>
> See Einstein A. (1920) Relativity, the special and the general theory.
>
> Ch 18 general principle
> Ch 20 general postulate
> Ch 23 rotating disk experiment- Hide quoted text -
>
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 =96(1-
v*v/c*c) If you integrate your centrifugal force, you will get =96(1-v*v/
c*c).
Something is wrong though. Of we move a mass to the rim energy will
increase. In fact if the rotation speed remains constant, energy has
to be supplied to the axle as we move towards the rim. Thinking of it
like this it is clear we don't have the "sealed" system of gravitating
bodies. We have a system which we supply energy to via the axle and
can take energy out, also via the axle when a mass moves from the rim
to the center.
- Ian Parker
Koobee Wublee
Well, I cannot believe this post would pass through the moderator of
sci.physics.research. However, somehow, ‘-‘ is replaced by ‘=96’.
<shrug>
http://groups.google.com/group/sci.physics.research/msg/db71e0c3c966f2c6
Chalky
encoding. Please check that your posts are in plain ASCII to avoid
similar transmission problems. -ik ]
On Jul 23, 10:08 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> ds^2 = c^2 T dt^2 (1 =96 K / r) =96 dr^2 / (1 =96 K / r) =96 r^2 dO^2
> (1 =96 2 U) / (1 =96 B^2) = Constant
> ** B^2 c^2 = (dr/dt)^2 / (1 =96 2 U)^2 + r^2 (dO/dt)^2 / (1 =96 2 U)
> B2 =96 2 U = constant
> B^2 / 2 =96 U = constant
>
> Where
>
> ** B^2 c^2 ~ (dr/dt)^2 + r^2 (dO/dt)^2
>
> Why is this simple mathematical result so surprising to you?
Sorry, I don't understand what you mean by 1=96 or by 2=96 so I don't
understand what your question is.
Koobee Wublee
> On Jul 23, 10:08 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> [ Mod. note: Things like =96, etc. come from the quoted-printable MIME
> encoding. Please check that your posts are in plain ASCII to avoid
> similar transmission problems. -ik ]
> > ds^2 = c^2 T dt^2 (1 =96 K / r) =96 dr^2 / (1 =96 K / r) =96 r^2 dO^2
> > (1 =96 2 U) / (1 =96 B^2) = Constant
> > ** B^2 c^2 = (dr/dt)^2 / (1 =96 2 U)^2 + r^2 (dO/dt)^2 / (1 =96 2 U)
> > B2 =96 2 U = constant
> > B^2 / 2 =96 U = constant
>
> > Where
>
> > ** B^2 c^2 ~ (dr/dt)^2 + r^2 (dO/dt)^2
>
> > Why is this simple mathematical result so surprising to you?
>
> Sorry, I don't understand what you mean by 1=96 or by 2=96 so I don't
> understand what your question is.
That is understandable that you cannot understand it when my post was
distorted with ‘-‘ replaced by ‘=96’. This I have already logged the
concern. Please see the link below.
http://groups.google.com/group/sci.physics.relativity/msg/61ca758433261af2?hl=en
I don’t think this is deliberate by the moderator of
sci.physics.research, but I am still very surprised that this post
stating that the Einstein field equations actually have an infinite
solutions would even get passed moderation. I thank the moderator
whoever he or she is. Anyhow, for a non-distorted publication of my
post, please visit sci.physics.relativity.
http://groups.google.com/group/sci.physics.relativity/msg/f42dddd7a3dc9470?hl=en
Androcles
|
http://groups.google.com/group/sci.physics.relativity/msg/f42dddd7a3dc9470?hl=en
That won't establish your crank aether or how a ring laser gyroscope works,
for which you have no equations.
Koobee Wublee
> That won't establish your crank aether or how a ring laser gyroscope works,
> for which you have no equations.
Off topic. You need to wait for your turn. <shrug>
Unless you want to pay me to solve your problem for you, parasite
Androcles. Better yet, just shut up and code whatever your control
system engineer tells you to code.
Androcles
| On Jul 23, 11:37 pm, "Androcles" wrote:
|
| > That won't establish your crank aether or how a ring laser gyroscope
works,
| > for which you have no equations.
|
| Off topic.
Rotating disk, fuckhead. Redshift, fuckhead. You won't establish your crank
Chalky
>
> > 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.
>
> 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?
Koobee Wublee
> 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 = c^2 (1 – K / u) dt^2 – dr^2 (du/dr)^2 / (1 – K / u) – u^2 dO^2
Where
** K = Integration constant
** u^3 = r^3 + K^3
** dO^2 = 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 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – 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 = c^2 dt^2 / (1 + K / r) – dr^2 (1 + K / r) – (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.physics.relativity/msg/7dd4e0a10f3d816f?hl=en
> 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.physics.relativity/msg/f42dddd7a3dc9470?hl=en
> Why should I be surprised by that?
Since you raised the issue, I thought you were surprised. That is
all. <shrug>
Eric Gisse
[...]
> Thus, they are all valid solutions.
That's because they are all the SAME solution.
[...]
Koobee Wublee
You don't understand mathematics. So, go back enjoy yourself being a
multi-year super senior. <shrug>
Pmb
>You don't understand mathematics. So, go back enjoy yourself being a
>multi-year super senior. <shrug>
Have you ever studied math in collage for 4 years or more?
Pete
Chalky
[quoted abuse etc snipped]
>Chalky wrote
> > I find redshift of rim
> > = sqrt(1 - [v/c]^2) = sqrt(1 - [wr/c]^2)
> > = sqrt(1 - integral[ w^2 r]dr/[c]^2)
> > = sqrt(1 - 2delphi/c]^2)
> > where w = angular velocity, and del phi is difference in gravitational
> > potential energy from observer located at middle of disk.
>
> The frequency shift of light is actually blue shifted in according to
> the following.
>
> 1 / sqrt(1 – B^2) ~ 1 + B^2 / 2
>
> Where
>
> ** B c = v
> ** ~ if (1 >> B)
No. It is red shifted. Clocks on the rapidly moving rim are seen to go
slower (by others), which means atomic clocks are seen to go slower.
Thus it takes longer for inertial observers to see a single cycle of
the emitted wave. This means RED shift, since the observed wavelength
is longer.
> In the meantime, the gravitational red shift is fudged into the
> following by the priests preferred to be called physicists.
>
> Sqrt(1 – 2 U) ~ 1 - U
>
> Where
>
> ** U = G M / c^2 / r
> ** ~ if (1 >> U)
>
> > Clearly this is only equal to the traditional gravitational Doppler [...]
>
> Now, please go back and re-do your analysis. Show the priests how
> you
> can justify (+ B^2 / 2, blue shift) with (- U, red shift).
If you want to submit your argument at the moderated group, I will
respond as above there too.
Chalky
It would have been better for Koobee Wublee to quote my post typo
corrected version of this, so I re-insert the most important
correction below.
On Jul 21, 11:31 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> On Jul 18, 4:07 am, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > On repeating Einstein's rotating disk experiment,
> > I find redshift of rim
is given by 1/(1 + z) =
> > sqrt(1 - [v/c]^2) = sqrt(1 - [wr/c]^2)
> > = sqrt(1 - integral[ w^2 r]dr/[c]^2)
> > = sqrt(1 - 2delphi/c]^2)
> > where w = angular velocity, and del phi is difference in gravitational
> > potential energy from observer located at middle of disk.
Please read this in conjunction with my prior response over red/blue
shift confusion
Daryl McCullough
>
>On Jul 24, 3:19=A0pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
>
>[...]
>
>> Thus, they are all valid solutions.
>
>That's because they are all the SAME solution.
That's a concept that Koobee has never been able to understand:
a change of coordinates does *not* make a different solution.
Two solutions are only different if there is no coordinate
change that can convert one into the other.
--
Daryl McCullough
Ithaca, NY
Chalky
Sorry, that should have read (1 - 2GM/R)^-1/2
> http://groups.google.com/group/sci.physics.relativity/msg/7dd4e0a10f3...
>
> > 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.physics.relativity/msg/f42dddd7a3d...
>
> > Why should I be surprised by that?
>
> Since you raised the issue, I thought you were surprised. That is
> all. <shrug>
Relieved, more like.
Eric Gisse
You can transform any solution you have written into another by the
appropriate composition of coordinate transformations. It isn't my
fault _YOU_ do not understand the mathematics.
Its' like I've been pointing out - you can't explain the difference
between Euclidean space in Cartesian coordinate and Euclidean space in
Spherical coordinates. They are the same, and you can't explain why
Schwarzschild is special other than it fits into your delusions.
Chalky
wrote:
Thanks for that. This puts everything in context, especially when
combined with his now diagnosed conceptual colour blindness.
Medically, color blindness is (typically) inability to distinguish red
and green. Apparently, Oobee 1 Konublee has the same problem with red
and blue
:-)
Chalky
> 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.
Koobee Wublee
> On Jul 22, 8:39 am, Koobee Wublee wrote
> > The frequency shift of light is actually blue shifted in according to
> > the following.
>
> > 1 / sqrt(1 – B^2) ~ 1 + B^2 / 2
>
> > Where
>
> > ** B c = v
> > ** ~ if (1 >> B)
>
> No. It is red shifted. Clocks on the rapidly moving rim are seen to go
> slower (by others), which means atomic clocks are seen to go slower.
> Thus it takes longer for inertial observers to see a single cycle of
> the emitted wave. This means RED shift, since the observed wavelength
> is longer.
You do not understand the Lorentz transform. <shrug>
> > In the meantime, the gravitational red shift is fudged into the
> > following by the priests preferred to be called physicists.
>
> > Sqrt(1 – 2 U) ~ 1 - U
>
> > Where
>
> > ** U = G M / c^2 / r
> > ** ~ if (1 >> U)
>
> If you want to submit your argument at the moderated group, I will
> respond as above there too.
Since no off-party-line thoughts will be tolerated with that moderated
group, it merely becomes a heaven for the cowards who cannot face
Koobee Wublee in a gentlemanly discussion. <shrug>
> > > On repeating Einstein's rotating disk experiment,
> > > I find redshift of rim
>
> is given by 1/(1 + z) =
This is just plain wrong. You need to apply the Lorentz transform to
this problem. The transverse Doppler shift is blue not red. Time
dilation in this case translates to a blue shift not red. <shrug>
The source of your confusion is the understanding of the Lorentz
transform. <shrug>
Eric Gisse
> On Jul 25, 3:44 am, Chalky wrote:
>
>
>
> > On Jul 22, 8:39 am, Koobee Wublee wrote
> > > The frequency shift of light is actually blue shifted in according to
> > > the following.
>
> > > 1 / sqrt(1 – B^2) ~ 1 + B^2 / 2
>
> > > Where
>
> > > ** B c = v
> > > ** ~ if (1 >> B)
>
> > No. It is red shifted. Clocks on the rapidly moving rim are seen to go
> > slower (by others), which means atomic clocks are seen to go slower.
> > Thus it takes longer for inertial observers to see a single cycle of
> > the emitted wave. This means RED shift, since the observed wavelength
> > is longer.
>
> You do not understand the Lorentz transform. <shrug>
Then how come you are the one who keeps trying to apply it to
situations where it does not apply?
>
> > > In the meantime, the gravitational red shift is fudged into the
> > > following by the priests preferred to be called physicists.
>
> > > Sqrt(1 – 2 U) ~ 1 - U
>
> > > Where
>
> > > ** U = G M / c^2 / r
> > > ** ~ if (1 >> U)
>
> > If you want to submit your argument at the moderated group, I will
> > respond as above there too.
>
> Since no off-party-line thoughts will be tolerated with that moderated
> group, it merely becomes a heaven for the cowards who cannot face
> Koobee Wublee in a gentlemanly discussion. <shrug>
How is chanting "Einstein the nitwit, the plagiarist, and the liar" in
any way, shape, or form a "gentlemanly discussion"?
>
> > > > On repeating Einstein's rotating disk experiment,
> > > > I find redshift of rim
>
> > is given by 1/(1 + z) =
>
> This is just plain wrong. You need to apply the Lorentz transform to
> this problem. The transverse Doppler shift is blue not red. Time
> dilation in this case translates to a blue shift not red. <shrug>
Except the Lorentz transform isn't valid here.
>
> The source of your confusion is the understanding of the Lorentz
> transform. <shrug>
The gods of irony are generous this day.
Koobee Wublee
> Sorry, that should have read (1 - 2GM/R)^-1/2
There is no need for you to correct and apologize for a simple error
like that. It is very simple for any scholar familiar with the
subject to correct as he reads along. <shrug>
> >http://groups.google.com/group/sci.physics.relativity/msg/7dd4e0a10f3...
>
> >http://groups.google.com/group/sci.physics.relativity/msg/f42dddd7a3d...
> Relieved, more like.
As I said, the source of your confusion lies in your lack of
understanding to the Lorentz transform. You need to work on that. I
cannot help you further. <shrug>
Koobee Wublee
> You can transform any solution you have written into another by the
> appropriate composition of coordinate transformations.
I have done no such transformation. Your confusion lies in your faith
in the metric as a tensor in which you have failed to understand the
following series of logical concept.
** The geometry must be invariant.
** The coordinate along cannot determine what the geometry is.
** The metric along cannot determine what the geometry is.
** To describe the geometry, you must specify your choice of
coordinate system and the metric.
This should be a very simple concept where a 10-year-old without first
having his mind polluted by the alchemists of modern differential
geometry should have no problem understanding. <shrug>
Another confusion among the priests of GR is the way Schwarzschild
derived the original solution to the field equations. See
Schwarzschild’s own paper provided in the link above.
If the choice of coordinate system results in a determinant of -1 on
the metric, the Ricci tensor can be drastically simplified. Thus,
Schwarzschild set up to transform the spherically symmetric polar
coordinate system into one such coordinate system that would result in
a determinant of -1 on the metric. Doing that, he found the solution
to this new coordinate system. After that, being a good mathematician
as he was, he properly translated the new coordinate system back to
the spherically symmetric polar coordinate system.
Despite the complexity of the metric associated with the spherically
symmetric polar coordinate system, the metric associated with the new
coordinate system somewhat resembles the Schwarzschild metric to be
derived by Hilbert a year or two later. In doing so, Hilbert realized
there an infinite number of solutions to the field equations. The
field equations are just utterly nonsense. That is why he just walked
away and allowed Einstein the nitwit, the plagiarist, and the liar who
does not know anything better to inundate himself in all these credits
that he does not deserve.
Well, Schwarzschild did make a mistake in translating rectangular
coordinate system to spherically symmetric polar coordinate system in
the very beginning of the paper, but that does not have any impact on
the rest of the paper.
> It isn't my
> fault _YOU_ do not understand the mathematics.
I know that it is not your fault. However, it is no my fault either
that you cannot understand why the metric is merely an ordinary 3-by-3
matrix in 3-dimensional space or 4-by-4 in 4-dimensional spacetime.
> Its' like I've been pointing out - you can't explain the difference
> between Euclidean space in Cartesian coordinate and Euclidean space in
> Spherical coordinates.
Well, all these solutions I have presented are all ONLY VALID in the
symmetrically spherical polar coordinate system and nothing else.
This shows you cannot distinguish among coordinate systems. <shrug>
> They are the same, and you can't explain why
> Schwarzschild is special other than it fits into your delusions.
You tell me why you think the Schwarzschild metric is special among
all the valid solutions to the field equations. You still have no
answer other than you were fed with this crap even before you know how
to wipe your own *ss. <shrug>
Koobee Wublee
> "Koobee Wublee" wrote:
> news:c3d4beb0-84db-4751...@s21g2000prm.googlegroups.com...
>
> >You don't understand mathematics. So, go back enjoy yourself being a
> >multi-year super senior. <shrug>
>
> Have you ever studied math in collage for 4 years or more?
What have I done to you? Why are you so insulting?
I thought the whiner in you has promised not to read any of my posts.
This is the fourth time that you have broken your own oath. Your
words mean nothing.
Koobee Wublee
> Thanks for that. This puts everything in context, especially when
> combined with his now diagnosed conceptual colour blindness.
> Medically, color blindness is (typically) inability to distinguish red
> and green. Apparently, Oobee 1 Konublee has the same problem with red
> and blue
Do you not know of the following equation?
E = m c^2 / sqrt(1 – B^2)
When we observe a moving particle, the observed energy or the observed
mass increases according to (1 / sqrt(1 – B^2)).
Do you not know blue light is observed to have more energy than red
light?
That is why transverse Doppler effect under the Lorentz transform is
blue shifted.
Are you still color blind?
Thus, SR time dilation does not translate to red shift as in the case
of gravitational time dilation. <shrug>
Koobee Wublee
> On Jul 25, 9:19 am, Koobee Wublee wrote:
> > You do not understand the Lorentz transform. <shrug>
>
> Then how come you are the one who keeps trying to apply it to
> situations where it does not apply?
You do not understand the Lorentz transform.
> > Since no off-party-line thoughts will be tolerated with that moderated
> > group, it merely becomes a heaven for the cowards who cannot face
> > Koobee Wublee in a gentlemanly discussion. <shrug>
>
> How is chanting "Einstein the nitwit, the plagiarist, and the liar" in
> any way, shape, or form a "gentlemanly discussion"?
It is the truth. Truth hurts. <shrug>
> > This is just plain wrong. You need to apply the Lorentz transform to
> > this problem. The transverse Doppler shift is blue not red. Time
> > dilation in this case translates to a blue shift not red. <shrug>
>
> Except the Lorentz transform isn't valid here.
It helps if you understand the Lorentz transform first. <shrug>
> > The source of your confusion is the understanding of the Lorentz
> > transform. <shrug>
>
> The gods of irony are generous this day.
Right, go back to sleep. Isn’t 10:30 too early for you --- the
pathetic life style of a multi-year super-senior?
Eric Gisse
Irrelevant. The fact that such transformations exist [you have been
given them in the past] is enough to prove that the solutions are
equivalent.
[snip]
Eric Gisse
> On Jul 25, 7:19 am, Chalky wrote:
>
> > Thanks for that. This puts everything in context, especially when
> > combined with his now diagnosed conceptual colour blindness.
> > Medically, color blindness is (typically) inability to distinguish red
> > and green. Apparently, Oobee 1 Konublee has the same problem with red
> > and blue
>
> Do you not know of the following equation?
>
> E = m c^2 / sqrt(1 – B^2)
>
> When we observe a moving particle, the observed energy or the observed
> mass increases according to (1 / sqrt(1 – B^2)).
That is neither "observed energy" and especially not "observed mass".
It is simply the conserved energy associated with geodesic motion in
Schwarzschild.
Chalky
> On Jul 25, 7:19 am, Chalky wrote:
>
> > Thanks for that. This puts everything in context, especially when
> > combined with his now diagnosed conceptual colour blindness.
> > Medically, color blindness is (typically) inability to distinguish red
> > and green. Apparently, Oobee 1 Konublee has the same problem with red
> > and blue
>
> Do you not know of the following equation?
>
> E = m c^2 / sqrt(1 – B^2)
>
> When we observe a moving particle, the observed energy or the observed
> mass increases according to (1 / sqrt(1 – B^2)).
>
> Do you not know blue light is observed to have more energy than red
> light?
>
> That is why transverse Doppler effect under the Lorentz transform is
> blue shifted.
Oh dear. Light from a transversely moving body does NOT travel faster
and thus get heavier.
No wonder you don't want to respond again in a moderated newsgroup and
no wonder I don't want to respond again in this one.
Eric Gisse
> On Jul 25, 10:23 am, Eric Gisse wrote:
>
> > On Jul 25, 9:19 am, Koobee Wublee wrote:
> > > You do not understand the Lorentz transform. <shrug>
>
> > Then how come you are the one who keeps trying to apply it to
> > situations where it does not apply?
>
> You do not understand the Lorentz transform.
Then explain how the Lorentz transform applies on non-Minkowski
manifolds and how it applies when there is acceleration.
>
> > > Since no off-party-line thoughts will be tolerated with that moderated
> > > group, it merely becomes a heaven for the cowards who cannot face
> > > Koobee Wublee in a gentlemanly discussion. <shrug>
>
> > How is chanting "Einstein the nitwit, the plagiarist, and the liar" in
> > any way, shape, or form a "gentlemanly discussion"?
>
> It is the truth. Truth hurts. <shrug>
Just because you say so doesn't mean it is the truth.
Need I remind you that you are incapable of proving the area of a
sphere is equal to 4pi r^2?
>
> > > This is just plain wrong. You need to apply the Lorentz transform to
> > > this problem. The transverse Doppler shift is blue not red. Time
> > > dilation in this case translates to a blue shift not red. <shrug>
>
> > Except the Lorentz transform isn't valid here.
>
> It helps if you understand the Lorentz transform first. <shrug>
Then explain how the Lorentz transform applies on non-Minkowski
manifolds and how it applies when there is acceleration.
>
> > > The source of your confusion is the understanding of the Lorentz
> > > transform. <shrug>
>
> > The gods of irony are generous this day.
>
> Right, go back to sleep. Isn’t 10:30 too early for you --- the
> pathetic life style of a multi-year super-senior?
The only pathetic one here is you. All you do is insult those who are
better than you.
Spaceman
> On Jul 23, 10:08 pm, Ian Parker <ianpark...@gmail.com> 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
Koobee Wublee
> On Jul 25, 10:03 am, Koobee Wublee wrote:
> > You do not understand the Lorentz transform.
>
> Then explain how the Lorentz transform applies on non-Minkowski
> manifolds and how it applies when there is acceleration.
Who said this scenario apply to non-Minkowski spacetime?
> > It is the truth. Truth hurts. <shrug>
>
> Just because you say so doesn't mean it is the truth.
Just because you are ignorant, it does not mean it is false.
> Need I remind you that you are incapable of proving the area of a
> sphere is equal to 4pi r^2?
This proves your ignorance. <shrug>
> > It helps if you understand the Lorentz transform first. <shrug>
>
> Then explain how the Lorentz transform applies on non-Minkowski
> manifolds and how it applies when there is acceleration.
Who said this scenario apply to non-Minkowski spacetime?
> > Right, go back to sleep. Isn’t 10:30 too early for you --- the
> > pathetic life style of a multi-year super-senior?
>
> The only pathetic one here is you. All you do is insult those who are
> better than you.
** MYSTICISM IS WISDOM
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** FAITH IS THEORY
** LYING IS TEACHING
** BELIEVING IS LEARNING
** IGNORANCE IS TRUTH
** STUPIDITY IS INTELLECT
You cannot even get a degree singing along the party line. That
explains your lack of amplitude. <shrug>
Koobee Wublee
> Koobee Wublee wrote:
> > Do you not know of the following equation?
>
> > E = m c^2 / sqrt(1 – B^2)
>
> > When we observe a moving particle, the observed energy or
> > the observed mass increases according to (1 / sqrt(1 – B^2)).
>
> > Do you not know blue light is observed to have more energy
> > than red light?
>
> > That is why transverse Doppler effect under the Lorentz
> > transform is blue shifted.
>
> > Are you still color blind?
>
> > Thus, SR time dilation does not translate to red shift as in
> > the case of gravitational time dilation. <shrug>
>
> Oh dear. Light from a transversely moving body does NOT travel faster
> and thus get heavier.
Oh dear. You are totally ignorant. This is a fine example of
‘IGNORANCE IS INTELLECT’. Have you not hear of the following?
E = h f
> No wonder you don't want to respond again in a moderated newsgroup and
> no wonder I don't want to respond again in this one.
Most of my posts are rejected by the moderator due to me not singing
along the party line. I maintain what I actually see. That is the
emperor has no clothes on. <shrug>
Koobee Wublee
> On Jul 25, 9:59 am, Koobee Wublee wrote:
> > Do you not know of the following equation?
>
> > E = m c^2 / sqrt(1 – B^2)
>
> > When we observe a moving particle, the observed energy or the observed
> > mass increases according to (1 / sqrt(1 – B^2)).
>
> That is neither "observed energy" and especially not "observed mass".
> It is simply the conserved energy associated with geodesic motion in
> Schwarzschild.
The Schwarzschild metric does not apply in this case. The spacetime
is still flat using the spherically symmetric polar coordinate
system. You need to learn something about what coordinate systems
are. <shrug>
Eric Gisse
> On Jul 25, 12:14 pm, Eric Gisse wrote:
>
> > On Jul 25, 9:59 am, Koobee Wublee wrote:
> > > Do you not know of the following equation?
>
> > > E = m c^2 / sqrt(1 – B^2)
>
> > > When we observe a moving particle, the observed energy or the observed
> > > mass increases according to (1 / sqrt(1 – B^2)).
>
> > That is neither "observed energy" and especially not "observed mass".
> > It is simply the conserved energy associated with geodesic motion in
> > Schwarzschild.
>
> The Schwarzschild metric does not apply in this case. The spacetime
> is still flat using the spherically symmetric polar coordinate
> system. You need to learn something about what coordinate systems
> are. <shrug>
It is not flat. There are nonzero components of the Riemann tensor and
the Ricci scalar is nonzero.
Koobee Wublee
<snipped fermented diarrhea>
Eric Gisse
[...]
> You cannot even get a degree singing along the party line. That
> explains your lack of amplitude. <shrug>
"amplitude" ?
Try again.
Koobee Wublee
> On Jul 25, 1:12 pm, Koobee Wublee wrote:
> > The Schwarzschild metric does not apply in this case. The spacetime
> > is still flat using the spherically symmetric polar coordinate
> > system. You need to learn something about what coordinate systems
> > are. <shrug>
>
> It is not flat. There are nonzero components of the Riemann tensor and
> the Ricci scalar is nonzero.
The spacetime can be described as follows in the spherically symmetric
polar coordinate system.
ds^2 = c^2 dt^2 – dr^2 – r^2 dO^2
It looks like Minkowski spacetime to me. <shrug>
Koobee Wublee
Aptitude.
Eric Gisse
> On Jul 25, 12:10 pm, Eric Gisse wrote:
>
> <snipped fermented diarrhea>
Remember this the next time you whine nobody will treat you to a nice
scientific discussion.
>
> I have done no such transformation. Your confusion lies in your faith
> in the metric as a tensor in which you have failed to understand the
> following series of logical concept.
I never said you did - just that the transformation existed. I have
shown them to you before.
>
> ** The geometry must be invariant.
Meaningless. Why can't you use the proper terminology?
>
> ** The coordinate along cannot determine what the geometry is.
The language is once again huddled in the corner, crying. Stop hitting
it.
>
> ** The metric along cannot determine what the geometry is.
Yes, it can. Riemannian geometry is a manifold with a metric and a
torsionless connection, which in turn is a function of the metric.
Every quantity in Riemannian geometry is defined in terms of the
metric.
You know so little for someone who thinks so highly of himself.
>
> ** To describe the geometry, you must specify your choice of
> coordinate system and the metric.
For what value of "describe" ?
Saying R^a_bcd = 0 is coordinate independent, and that defines special
relativity. Of course if you want to have an actual coordinate
representation you kinda need coordinates. However, the coordinates
are irrelevant.
Remember my example of Euclidean geometry in Cartesian then Spherical
coordinates? You still can't explain the difference but you sure can
copy and paste real nice!
>
> This should be a very simple concept where a 10-year-old without first
> having his mind polluted by the alchemists of modern differential
> geometry should have no problem understanding. <shrug>
Nobody is taught differential geometry until their early twenties if
at all. Regardless, where did you learn your version of differential
geometry?
>
> Another confusion among the priests of GR is the way Schwarzschild
> derived the original solution to the field equations. See
> Schwarzschild’s own paper provided in the link above.
What link? Oh that's right, you are copying and pasting this from your
little book o' rants.
[snip irrelevancies]
> > It isn't my
> > fault _YOU_ do not understand the mathematics.
>
> I know that it is not your fault. However, it is no my fault either
> that you cannot understand why the metric is merely an ordinary 3-by-3
> matrix in 3-dimensional space or 4-by-4 in 4-dimensional spacetime.
Since you can't tell me the difference between a matrix and a tensor,
you are unfit to judge.
>
> > Its' like I've been pointing out - you can't explain the difference
> > between Euclidean space in Cartesian coordinate and Euclidean space in
> > Spherical coordinates.
>
> Well, all these solutions I have presented are all ONLY VALID in the
> symmetrically spherical polar coordinate system and nothing else.
> This shows you cannot distinguish among coordinate systems. <shrug>
That's because all they are is spherical polar coordinates. The only
difference is one is written in an unknown function of r, one is
written in regular isotropic coordinates, and another is written in
terms of an offset radial coordinate.
How many years have you spent whining about a subject you know nothing
about? Looks to be at least 4 so far.
>
> > They are the same, and you can't explain why
> > Schwarzschild is special other than it fits into your delusions.
>
> You tell me why you think the Schwarzschild metric is special among
> all the valid solutions to the field equations. You still have no
> answer other than you were fed with this crap even before you know how
> to wipe your own *ss. <shrug>
Again? Maybe you'll remember this time.
The isotropic Schwarzschild coordinates are special because the radial
coordinate is defined such that surfaces of constant r and t have area
4 pi r^2. Nothing more.
Or do you mean why the Schwarzschild /solution/ is special among other
solutions? Well, it isn't. I have no way of telling because you are
horribly imprecise with your language.
Eric Gisse
> On Jul 25, 2:19 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Jul 25, 1:12 pm, Koobee Wublee wrote:
> > > The Schwarzschild metric does not apply in this case. The spacetime
> > > is still flat using the spherically symmetric polar coordinate
> > > system. You need to learn something about what coordinate systems
> > > are. <shrug>
>
> > It is not flat. There are nonzero components of the Riemann tensor and
> > the Ricci scalar is nonzero.
>
> The spacetime can be described as follows in the spherically symmetric
> polar coordinate system.
>
> ds^2 = c^2 dt^2 – dr^2 – r^2 dO^2
>
> It looks like Minkowski spacetime to me. <shrug>
It is. Except you did not define your terms and you use 1-B^2 to be
your personal little substitute for elements of Schwarzschild on a
regular basis. Don't blame me because I got used to your language.
Eric Gisse
Now try an original insult. Move beyond your delusions about my
personal & professional life, lies and slander about Einstein, and
general fantasy that you know what you are talking about. I want /
effort/ !
Since you can't justify your presence here with a good set of
knowledge, I want to see a real good putdown! I want fresh material!
C'mon!
Koobee Wublee
> Now try an original insult. Move beyond your delusions about my
> personal & professional life, lies and slander about Einstein, and
> general fantasy that you know what you are talking about. I want /
> effort/ !
All my comments are not meant as insults but as point blank assessment
of the dire situation you are facing.
> Since you can't justify your presence here with a good set of
> knowledge, I want to see a real good putdown! I want fresh material!
> C'mon!
Let’s see. You have no college degrees; you are incapable of
understanding simple mathematics such as what coordinate systems are;
you are drowning in the diarrhea that you were fed with where:
** MYSTICISM IS WISDOM
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** FAITH IS THEORY
** LYING IS TEACHING
** BELIEVING IS LEARNING
** IGNORANCE IS KNOWLEDGE
<shrug>
Koobee Wublee
> On Jul 25, 1:23 pm, Koobee Wublee wrote:
> > The spacetime can be described as follows in the spherically symmetric
> > polar coordinate system.
>
> > ds^2 = c^2 dt^2 – dr^2 – r^2 dO^2
>
> > It looks like Minkowski spacetime to me. <shrug>
>
> It is. Except you did not define your terms and you use 1-B^2 to be
> your personal little substitute for elements of Schwarzschild on a
> regular basis. Don't blame me because I got used to your language.
You represent an absolute failure in our educational system. <shrug>
You do not understand the basics, and it shows in your absurd remarks
as well as your lack of a college degree after so many years being a
super-senior. <shrug>
Eric Gisse
Nope. Just the typical sad pathetic bullshit you have been spouting
repeatedly, nothing new or interesting. You aren't even an interesting
crank.
Koobee Wublee
> On Jul 25, 1:20 pm, Koobee Wublee wrote:
> Remember this the next time you whine nobody will treat you to a nice
> scientific discussion.
You have been whining for years snipping the stuff that you cannot
understand. Is there any more original excuses such as “Oh, my dog
licked my face, and I do not remember what I have read in the last few
days.”? <shrug>
> > I have done no such transformation. Your confusion lies in your faith
> > in the metric as a tensor in which you have failed to understand the
> > following series of logical concept.
>
> I never said you did - just that the transformation existed. I have
> shown them to you before.
The absurd basis of your entire argument to support the myth of the
metric being a tensor lies in these accusations of creative
transformations I have done. So, you are conveniently playing dumb.
That is fine. You are really so. <shrug>
> > ** The geometry must be invariant.
>
> Meaningless. Why can't you use the proper terminology?
You have failed to understand the meaning of that. <shrug>
> > ** The coordinate along cannot determine what the geometry is.
>
> The language is once again huddled in the corner, crying. Stop hitting
> it.
The language is as point blank as it can be. If you still do not
understand, just say so.
> > ** The metric along cannot determine what the geometry is.
>
> Yes, it can. Riemannian geometry is a manifold with a metric and a
> torsionless connection, which in turn is a function of the metric.
> Every quantity in Riemannian geometry is defined in terms of the
> metric.
Again, you are degrading Riemann. All Riemann pointed is the
following.
ds^2 = g_ij dq^i dq^j
The geometry ds^2 must be invariant where there is a unique [g] (the
metric matrix) with elements g_ij for each choice of coordinate
system. <shrug>
> You know so little for someone who thinks so highly of himself.
It seems to be mike and myself are the only ones who actually
understand what Riemann was saying. <shrug>
> > ** To describe the geometry, you must specify your choice of
> > coordinate system and the metric.
>
> For what value of "describe" ?
It makes no sense as usual. <shrug>
> Saying R^a_bcd = 0 is coordinate independent, and that defines special
> relativity. Of course if you want to have an actual coordinate
> representation you kinda need coordinates. However, the coordinates
> are irrelevant.
That is the problem. It is impossible to describe anything you see
without specifying what coordinate systems you are using. This should
be a very basic axiom in the study of geometry. This means you are so
inept of a student, or the educational system is totally screwed up.
For what I have seen, it is both. <shrug>
> Remember my example of Euclidean geometry in Cartesian then Spherical
> coordinates? You still can't explain the difference but you sure can
> copy and paste real nice!
Keep babbling about your fantasy world. <shrug>
> > This should be a very simple concept where a 10-year-old without first
> > having his mind polluted by the alchemists of modern differential
> > geometry should have no problem understanding. <shrug>
>
> Nobody is taught differential geometry until their early twenties if
> at all. Regardless, where did you learn your version of differential
> geometry?
Oh, basic geometry such as shapes etc are already exposed to kids as
young as two years old. You are just making excuses to be so
ignorant.
> > Another confusion among the priests of GR is the way Schwarzschild
> > derived the original solution to the field equations. See
> > Schwarzschild’s own paper provided in the link above.
>
> What link? Oh that's right, you are copying and pasting this from your
> little book o' rants.
>
> [snip irrelevancies]
You are a troll. <shrug>
> > I know that it is not your fault. However, it is no my fault either
> > that you cannot understand why the metric is merely an ordinary 3-by-3
> > matrix in 3-dimensional space or 4-by-4 in 4-dimensional spacetime.
>
> Since you can't tell me the difference between a matrix and a tensor,
> you are unfit to judge.
I have told you so in the past. You did not understand. You never
will. It shows that you are still a multi-year super-senior. <shrug>
> > Well, all these solutions I have presented are all ONLY VALID in the
> > symmetrically spherical polar coordinate system and nothing else.
> > This shows you cannot distinguish among coordinate systems. <shrug>
>
> That's because all they are is spherical polar coordinates.
I have already said that. <shrug>
> The only
> difference is one is written in an unknown function of r, one is
> written in regular isotropic coordinates, and another is written in
> terms of an offset radial coordinate.
Now, why is this r unknown in the spherically symmetric polar
coordinate system? See how you do not even understand what coordinate
systems are. <shrug>
> How many years have you spent whining about a subject you know nothing
> about? Looks to be at least 4 so far.
Yeah, speaking from a multi-year super-senior. You are also very
delusional. <shrug>
> > You tell me why you think the Schwarzschild metric is special among
> > all the valid solutions to the field equations. You still have no
> > answer other than you were fed with this crap even before you know how
> > to wipe your own *ss. <shrug>
>
> Again? Maybe you'll remember this time.
>
> The isotropic Schwarzschild coordinates are special because the radial
> coordinate is defined such that surfaces of constant r and t have area
> 4 pi r^2. Nothing more.
Oh, you call that special in the world of curved space or spacetime?
You are so ignorant.
> Or do you mean why the Schwarzschild /solution/ is special among other
> solutions? Well, it isn't. I have no way of telling because you are
> horribly imprecise with your language.
What is so imprecise about my language, multi-year super-senior? You
are just looking for excuses to cover your ignorance. <shrug>
Koobee Wublee
> On Jul 25, 11:03 pm, Koobee Wublee wrote:
> > All my comments are not meant as insults but as point blank assessment
> > of the dire situation you are facing.
>
> > > Since you can't justify your presence here with a good set of
> > > knowledge, I want to see a real good putdown! I want fresh material!
> > > C'mon!
>
> > Let’s see. You have no college degrees; you are incapable of
> > understanding simple mathematics such as what coordinate systems are;
> > you are drowning in the diarrhea that you were fed with where:
>
> > ** MYSTICISM IS WISDOM
> > ** PLAGIARISM IS CREATIVITY
> > ** CONJECTURE IS REALITY
> > ** FAITH IS THEORY
> > ** LYING IS TEACHING
> > ** BELIEVING IS LEARNING
> > ** IGNORANCE IS KNOWLEDGE
>
> > <shrug>
>
> Nope. Just the typical sad pathetic bullshit you have been spouting
> repeatedly, nothing new or interesting. You aren't even an interesting
> crank.
Oh, the multi-year super-senior has his feeling hurt. What a pathetic
life!
Koobee Wublee
> Typical Aetherialist preaching, at least Gisse knows you are a fucking idiot
> even if he doesn't know Einstein was.
It sounds like Androcles and Eric Gisse are getting married. Please
don’t invite my to your wedding. I think Massachusetts and California
will allow same-sex or same-scumbag marriage. Better take that
advantage.
> How does a ring laser gyroscope work, clueless shit-for-brains?
If you don’t know how a ring laser gyroscope works, you are truly what
you have described yourself as a clueless sh*t-for-brain. <shrug>
Koobee Wublee
> On Jul 23, 8:39 am, Koobee Wublee wrote:
> > I have written no lies about Einstein the nitwit, the plagiarist, and
> > the liar. Since Einstein is your messiah on the religions of SR and
> > GR, you are just deeply hurt by the truth. <shrug>
>
> ....and this is why it is never a serious scientific discussion with
> you.
All my posts are very serious even the part describing Einstein as a
nitwit, a plagiarist, and a liar. Nobody so far has seriously tried
to challenge that. If you do, speak your case in a convincing matter,
don’t whine about how much fermented diarrhea of Einstein you have
swallowed.
Koobee Wublee
> Rotating disk, fuckhead. Redshift, fuckhead. You won't establish your crank
> aether or how a ring laser gyroscope works, for which you have no equations.
So, Androcles the nitwit is stuck in his rotating disk. <shrug>
Just because you do not understand a subject, you do not have to deny
that subject. <shrug>
Eric Gisse
> On Jul 25, 2:34 pm, Eric Gisse wrote:
>
> > On Jul 25, 1:20 pm, Koobee Wublee wrote:
> > Remember this the next time you whine nobody will treat you to a nice
> > scientific discussion.
>
> You have been whining for years snipping the stuff that you cannot
> understand. Is there any more original excuses such as “Oh, my dog
> licked my face, and I do not remember what I have read in the last few
> days.”? <shrug>
>
> > > I have done no such transformation. Your confusion lies in your faith
> > > in the metric as a tensor in which you have failed to understand the
> > > following series of logical concept.
>
> > I never said you did - just that the transformation existed. I have
> > shown them to you before.
>
> The absurd basis of your entire argument to support the myth of the
> metric being a tensor lies in these accusations of creative
> transformations I have done. So, you are conveniently playing dumb.
> That is fine. You are really so. <shrug>
...accusations?
You raise the denial of relativity uncomplicated mathematics to an
ART. Unfortunately the denial of coordinate transformations is an
inferior class of stupidity to denying two negatives multiplied
together is a positive, however you got spaceshit beat for bile and
jealousy.
>
> > > ** The geometry must be invariant.
>
> > Meaningless. Why can't you use the proper terminology?
>
> You have failed to understand the meaning of that. <shrug>
Inventing your own language to describe a well defined field is silly.
Fortunately for you, you are not above being silly.
>
> > > ** The coordinate along cannot determine what the geometry is.
>
> > The language is once again huddled in the corner, crying. Stop hitting
> > it.
>
> The language is as point blank as it can be. If you still do not
> understand, just say so.
Oh I understand what you are /hinting/ at, but since you can't explain
what you are trying to say in the standard language of the field I
have to play these little guessing games. Makes it a lot easier to
play your little denial and goalpost shifting games.
>
> > > ** The metric along cannot determine what the geometry is.
>
> > Yes, it can. Riemannian geometry is a manifold with a metric and a
> > torsionless connection, which in turn is a function of the metric.
> > Every quantity in Riemannian geometry is defined in terms of the
> > metric.
>
> Again, you are degrading Riemann. All Riemann pointed is the
> following.
>
> ds^2 = g_ij dq^i dq^j
>
> The geometry ds^2 must be invariant where there is a unique [g] (the
> metric matrix) with elements g_ij for each choice of coordinate
> system. <shrug>
For future reference, invariant does not mean "looks the same".
>
> > You know so little for someone who thinks so highly of himself.
>
> It seems to be mike and myself are the only ones who actually
> understand what Riemann was saying. <shrug>
Which works of Riemann did you study, then? How about some literature
references?
>
> > > ** To describe the geometry, you must specify your choice of
> > > coordinate system and the metric.
>
> > For what value of "describe" ?
>
> It makes no sense as usual. <shrug>
>
> > Saying R^a_bcd = 0 is coordinate independent, and that defines special
> > relativity. Of course if you want to have an actual coordinate
> > representation you kinda need coordinates. However, the coordinates
> > are irrelevant.
>
> That is the problem. It is impossible to describe anything you see
> without specifying what coordinate systems you are using.
Sometimes but not always - look at the example I just gave you. That's
a coordinate independent way of stating something.
> This should
> be a very basic axiom in the study of geometry. This means you are so
> inept of a student, or the educational system is totally screwed up.
> For what I have seen, it is both. <shrug>
Third option: You have failed to understand a concept that is just as
important as it is basic, in that the coordinates _do not matter_.
Sure you need /some/ coordinates to actually write down what the
metric is, but the coordinates don't matter. Hand me a different set
of coordinates and I can put the metric in that form and it won't
matter because the quantities will be the same in every coordinate
system.
>
> > Remember my example of Euclidean geometry in Cartesian then Spherical
> > coordinates? You still can't explain the difference but you sure can
> > copy and paste real nice!
>
> Keep babbling about your fantasy world. <shrug>
Inability to address the point noted. You cannot explain the
difference between the two coordinate representations of the same
thing, and completely fail to grasp that when applied to general
relativity. Must be the autism.
>
> > > This should be a very simple concept where a 10-year-old without first
> > > having his mind polluted by the alchemists of modern differential
> > > geometry should have no problem understanding. <shrug>
>
> > Nobody is taught differential geometry until their early twenties if
> > at all. Regardless, where did you learn your version of differential
> > geometry?
>
> Oh, basic geometry such as shapes etc are already exposed to kids as
> young as two years old. You are just making excuses to be so
> ignorant.
"Basic geometry", as you call it, is nothing more than Euclidean
geometry. The foundations of which are not seriously taught until -
again - the student is in the early 20's. Regardless, "basic geometry"
is /not/ differential geometry. Try to understand the subjects you
criticize.
>
> > > Another confusion among the priests of GR is the way Schwarzschild
> > > derived the original solution to the field equations. See
> > > Schwarzschild’s own paper provided in the link above.
>
> > What link? Oh that's right, you are copying and pasting this from your
> > little book o' rants.
>
> > [snip irrelevancies]
>
> You are a troll. <shrug>
Then stop responding. You've called me a troll dozens of times now,
but your compulsion prevents you from letting me have the last word.
>
> > > I know that it is not your fault. However, it is no my fault either
> > > that you cannot understand why the metric is merely an ordinary 3-by-3
> > > matrix in 3-dimensional space or 4-by-4 in 4-dimensional spacetime.
>
> > Since you can't tell me the difference between a matrix and a tensor,
> > you are unfit to judge.
>
> I have told you so in the past. You did not understand. You never
> will. It shows that you are still a multi-year super-senior. <shrug>
I'll let you in on a little secret - this stuff isn't taught at the
undergraduate level. Don't let that stop you from whining about this
irrelevant "multi-year super-senior" bullshit though!
>
> > > Well, all these solutions I have presented are all ONLY VALID in the
> > > symmetrically spherical polar coordinate system and nothing else.
> > > This shows you cannot distinguish among coordinate systems. <shrug>
>
> > That's because all they are is spherical polar coordinates.
>
> I have already said that. <shrug>
So why are you even raising the point?
>
> > The only
> > difference is one is written in an unknown function of r, one is
> > written in regular isotropic coordinates, and another is written in
> > terms of an offset radial coordinate.
>
> Now, why is this r unknown in the spherically symmetric polar
> coordinate system? See how you do not even understand what coordinate
> systems are. <shrug>
Explain how "r" is defined by spherical symmetry. Then explain what
"r" is - I think I know what you are talking about, but why assume?
>
> > How many years have you spent whining about a subject you know nothing
> > about? Looks to be at least 4 so far.
>
> Yeah, speaking from a multi-year super-senior. You are also very
> delusional. <shrug>
>
> > > You tell me why you think the Schwarzschild metric is special among
> > > all the valid solutions to the field equations. You still have no
> > > answer other than you were fed with this crap even before you know how
> > > to wipe your own *ss. <shrug>
>
> > Again? Maybe you'll remember this time.
>
> > The isotropic Schwarzschild coordinates are special because the radial
> > coordinate is defined such that surfaces of constant r and t have area
> > 4 pi r^2. Nothing more.
>
> Oh, you call that special in the world of curved space or spacetime?
> You are so ignorant.
It isn't special, it's just preferred.
>
> > Or do you mean why the Schwarzschild /solution/ is special among other
> > solutions? Well, it isn't. I have no way of telling because you are
> > horribly imprecise with your language.
>
> What is so imprecise about my language, multi-year super-senior? You
> are just looking for excuses to cover your ignorance. <shrug>
Everything. I can't tell if you do it on purpose because it leaves you
such latitude for changing your argument or because you simply do not
know better. It matters not which it is because you have had the
proper terminology explained to you many times over the years and you
simply choose not to learn.
Eric Gisse
How could someone who cannot even do an integral hurt my feelings?
The only insult you think you have doesn't bother me in the least,
even though you repeat it all the time. Keep repeating it for years on
end though - it just shows how sad and immature you are.
Eric Gisse
Scat play already? Jeeze - at least take me out to dinner first.
Androcles
"Koobee Wublee" <koobee...@gmail.com> wrote in message
news:529321be-e1b0-470a...@i24g2000prf.googlegroups.com...
Chalky
> Ps. This reply has sci.physics.research removed.
Untrue. I have just noticed it has sci.physics substituted for
sci.physics.research to ensure that nothing valid at SPR is included
in the SP discussion.
Well done.
By this ploy you have managed to prove by example the old adage that
empty vessels make most sound.
> It is a priviledged
> newsgroup for the priests to comfort themselves into believing
> ‘PRIESTHOOD IS SCIENCE’.
No, it is a mechanism anyone can use to prevent you from swamping
every discussion with nonsense and sufficient abuse to start a flame
war with everyone who rises to the bait.
You have already discovered you can get published there if you
(accidentally) fail to say something stupid.
Chalky
I should have said here "You have already discovered you can get
published there if you (accidentally) fail to say something *too*
stupid."
The point of discussion is to resolve differences in understanding,
not to suppress them.
Chalky
> On 22 Jul, 01:05, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > On Jul 21, 6:41 pm, Ian Parker <ianpark...@gmail.com> 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.physics.foundations/browse_thread/thread/0d10c7aacfc6038e#
I suggest you read it very carefully. As I have already pointed out,
your argument contradicts Einstein.
Chalky
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 23, 10:08pm, Ian Parker <ianpark...@gmail.com> wrote:
> On 22 Jul, 01:05, Chalky <chalkys...@bleachboys.co.uk> 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.
Chalky
wrote:
> >> increase.
>
> > 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).
Koobee Wublee
> Accoding to John Norton, Divine Albert did not lie about the
> Michelson-Morley experiment; only "later writers" did:
>
> http://philsci-archive.pitt.edu/archive/00001743/02/Norton.pdf
> John Norton: "Einstein regarded the Michelson-Morley experiment as
> evidence for the principle of relativity,
Einstein was a nitwit. Voigt has already shown the null results of
the MMX prove the principle of relativity wrong. See the Voigt
transform. In doing so, the absolute frame of reference must exist
and thus the Aether. <shrug>
> whereas later writers almost
> universally use it as support for the light postulate of special
> relativity......
Later writers had a sip of Einstein’s fermented diarrhea, and the rest
is history.
> THE MICHELSON-MORLEY EXPERIMENT IS FULLY COMPATIBLE
> WITH AN EMISSION THEORY OF LIGHT THAT CONTRADICTS THE LIGHT
> POSTULATE."
Yes, that is true, but the emission theory does not agree with
electromagnetism. <shrug>
> Yet Divine Albert seems to be the original liar:
Yes, Einstein was a nitwit, a plagiarist, and a liar. <shrug>
> http://query.nytimes.com/gst/abstract.html?res=9806EFDD113FEE3ABC4152...
> The New York Times, April 19, 1921
>
> "Michelson showed that relative to the moving co-ordinate system K1,
> the light traveled with the same velocity as relative to K, which is
> contrary to the above observation. How could this be reconciled?
> Professor Einstein asked."
Voigt answered it in 1887. That is 18 years before Einstein’s
plagiarized work of 1905 papers.
Koobee Wublee
> 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.
Whatever you say, but the bottom line is that the transverse Doppler
is blue shifted under the Lorentz transform. <shrug>
> Thanks also for responding via the moderated newsgroup instead of via
> the rogue sub-thread created by Kubee Wooblie removing SPResearch from
> the loop.
See what I mean. The priests at sci.physics.research allow such a
personal attack on me. My name is black-listed. Only occasionally, a
new moderator would allow me to go through.
> 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).
Eric Gisse the multi-year super-senior does not even have a bachelor’s
degree. He is merely a malicious troll. However, I am not surprised
that his trollish posts would be published by the priests in that good-
old-boy newsgroup. <shrug>
By the way, changing sci.physics.research to sci.physics is my doing
not the priests at sci.physics.research.
Koobee Wublee
> > No he doesn't.
>
> Who doesn't what? You are not making sense.
This observer on the rotating disk. Gee, that was your own scenario.
I guess it is not required to have the publisher of
sci.physics.research to be coherent. <shrug>
> > If we are in a gravitational field we can describe it
> > by Tensors.
>
> We can describe it any way we like, that works.
Oh, you are side-stepping the central party-line theme. Be careful or
you will find yourself to be disallowed as a member of harem in
sci.physics.research. <shrug>
> > We do not observe the Universe revolving round us.
>
> Of course we DO! (what planet are you on?)
This is a philosophic point. There is no right or wrong. <shrug>
> 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.
Oh, really. What if your observation is distorted?
> > 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
Thus, according to the mathematics of the Lorentz transform, you end
up with a blue shift in transverse Doppler. <shrug>
> > 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,
What are you talking about slowing down? In your own scenario, you
have already decided the disk rotating at a certain speed. So, the
flip-flop in you is a role model of the priests at
sci.physics.research. So be it. <shrug>
> But, yes, the situation is not identical to the gravitational fields
> _induced by mass_
No, it is not. <shrug>
Your Messiah, Einstein the nitwit, the plagiarist, and the liar never
said so even after he re-discovered the principle of equivalence after
finally understood the Newtonian law of gravity. <shrug>
> 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.
Although I can not speak for Mr. Parker, your statement is full of
insult.
> You seem to be going one step beyond though by now denying that the
> ladder ever existed.
Again, insult like this should never be published by a properly
moderated group. <shrug>
Koobee Wublee
> On Jul 26, 12:25 pm, Chalky < wrote:
> > Untrue. I have just noticed it has sci.physics substituted for
> > sci.physics.research to ensure that nothing valid at SPR is included
> > in the SP discussion.
Hmmm... Let me know when your cranial elevator ever goes to the top
floor.
> > Well done.
> > By this ploy you have managed to prove by example the old adage that
> > empty vessels make most sound.
<shrug> Whatever that supposes to mean.
> > No, it is a mechanism anyone can use to prevent you from swamping
> > every discussion with nonsense and sufficient abuse to start a flame
> > war with everyone who rises to the bait.
> > You have already discovered you can get published there if you
> > (accidentally) fail to say something stupid.
>
> I should have said here "You have already discovered you can get
> published there if you (accidentally) fail to say something *too*
> stupid."
Well, I thank new moderators without chancing my works. <shrug>
> The point of discussion is to resolve differences in understanding,
> not to suppress them.
The sole job of the priests at sci.physics.research is to proliferate
the following Orwellian teachings.
Chalky
> On Jul 27, 11:51 am, Chalky wrote:
> > Thanks also for responding via the moderated newsgroup instead of via
> > the rogue sub-thread created by Kubee Wooblie removing SPResearch from
> > the loop.
>
> See what I mean. The priests at sci.physics.research allow such a
> personal attack on me.
It is not a personal attack it is an objective statement of fact
> My name is black-listed.
No it isn't they have already given you two bites of the cherry there.
> Only occasionally, a
> new moderator would allow me to go through.
You only occasionally forget to lace your postings with things like
"Einstein the liar, plagiarist & twit". If I said Koobee Wublee the
liar, plagiarist & twit, that posting would be rejected too (despite
being closer to the mark).
>
> > 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).
>
> Eric Gisse the multi-year super-senior does not even have a bachelor’s
> degree. He is merely a malicious troll. However, I am not surprised
> that his trollish posts would be published by the priests in that good-
> old-boy newsgroup. <shrug>
See what I mean?
> By the way, changing sci.physics.research to sci.physics is my doing
> not the priests at sci.physics.research.
I know. That is what I said above, and you called it a personal attack
on you. You need a reality check.
Chalky
> On Jul 27, 11:51 am, Chalky wrote:
>
> > On Jul 23, 10:08pm, Ian Parker wrote:
> > > No he doesn't.
>
> > Who doesn't what? You are not making sense.
>
> This observer on the rotating disk. Gee, that was your own scenario.
I said that observer could... The response is semantically incorrect
in that context.
In context, Ian could have been saying No Einstein doesn't say...
[snip]
> > 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.
>
> Although I can not speak for Mr. Parker, your statement is full of
> insult.
FYI Kierkegaard is a widely respected, influential, and well regarded
philosopher.
Koobee Wublee
> On Jul 28, 6:51 am, Koobee Wublee wrote:
> > See what I mean. The priests at sci.physics.research allow such a
> > personal attack on me.
>
> It is not a personal attack it is an objective statement of fact
I am just demonstrating that personal attack is all subjective. It is
merely an excuse to silent some certain scholars not conformed to the
party-line of thinking. <shrug>
> > My name is black-listed.
>
> No it isn't they have already given you two bites of the cherry there.
Yes, when the new moderators let their guards down, it could happen.
There is no way to pass through the old-timers such as Igor Kronkite
or somebody. <shrug>
> > Only occasionally, a
> > new moderator would allow me to go through.
>
> You only occasionally forget to lace your postings with things like
> "Einstein the liar, plagiarist & twit".
Einstein was long dead. All description of him cannot be construed as
a personal attack. There are merely remarks about the historical
character named Einstein who was a nitwit, a plagiarist, and a liar.
<shrug>
Furthermore, Einstein now fills the role of the messiah-ship of the
religions of SR and GR. The ones feel hurt by the truth are the
zealous believer in the religions of SR and GR. So, whatever I point
out correctly that Einstein was a nitwit, a plagiarist, and a liar,
that should also not be construed as personal insults. <shrug>
> If I said Koobee Wublee the
> liar, plagiarist & twit, that posting would be rejected too (despite
> being closer to the mark).
That would consider personal insult.
> > Eric Gisse the multi-year super-senior does not even have a bachelor’s
> > degree. He is merely a malicious troll. However, I am not surprised
> > that his trollish posts would be published by the priests in that good-
> > old-boy newsgroup. <shrug>
>
> See what I mean?
What did you mean?
> > By the way, changing sci.physics.research to sci.physics is my doing
> > not the priests at sci.physics.research.
>
> I know. That is what I said above, and you called it a personal attack
> on you. You need a reality check.
When I said ‘Einstein was a nitwit, a plagiarist, and a liar’, and you
call that a personal attack on you. You need a reality check.
<shrug>
There is no personal attack on my part. Only your ego is deeply
bruised. Ouch! The truth hurts. <shrug>
Eric Gisse
[snip]
> > > Eric Gisse the multi-year super-senior does not even have a bachelor’s
> > > degree. He is merely a malicious troll. However, I am not surprised
> > > that his trollish posts would be published by the priests in that good-
> > > old-boy newsgroup. <shrug>
>
> > See what I mean?
>
> What did you mean?
Your inability to recognize that you are being a disgusting person is
why your posts bounce out of moderated newsgroups.
[snip]
Koobee Wublee
> Your inability to recognize that you are being a disgusting person is
> why your posts bounce out of moderated newsgroups.
There is nothing out of you except personal attacks. Go back to
sleep, it is only one o’clock. In the meantime,
> > Eric Gisse the multi-year super-senior does not even have a bachelor’s
> > degree. He is merely a malicious troll. However, I am not surprised
> > that his trollish posts would be published by the priests in that good-
> > old-boy newsgroup. <shrug>
>
> See what I mean?
What did you mean?
> > By the way, changing sci.physics.research to sci.physics is my doing