A minor real-world exercise for those who fashion themselves to be physicists
Eric Gisse
started posting here - that they have a better collective grasp of
physics than the lowly people who actually study it, I figure its'
about time to bring in a piece of the real thing for observation.
I will not give the conventional explanation nor enough information
for an uneducated observer to determine the "orthodox" answer to this
problem.
A 1uCi source [was much stronger - 6 e-folds tones anything down] is
placed on a small electromagnetic drive unit. A signal is sent out by
the controlling [and ancient! I haven't seen Windows 3.1 in 15 fucking
years] computer to a drive unit that makes the oscillator move in a
sawtooth wave pattern of constant acceleration.
My sketched out version of the apparatus is here:
http://yfrog.com/eisdc10237j - It is actually rather hard to take a
text-capturing photo even with full control over the camera.
As can be seen from the notes, the oscillator is moving back and forth
at 8.8mm/s. This is somewhat important - it should be considered. What
is also important is that there is a net magnetic moment for Iron, and
our old friend the Zeeman effect comes into play. But which Zeeman
effect? What magnetic fields?
The sensor is a gas based scintillating detector (geiger counter) fed
by a 2kV source. Whatever the counter sees gets fed into a pre-
amplifier, then into an amplifier, then finally into the multi-channel
analyzer inside the thousand year old system. The spectrum of
radiation was initially poked through to determine the appropriate
settings for the single channel analyzer, in order to constrain its'
attention to the part that I give a shit about.
After determining the appropriate window settings and giving myself a
moderate gamma dose by moving lead blocks around to make sure the
source and detector were aligned, everything got re-buried and ignored
until periodically baby-sat by me or the lab professor. Not like
anything needs to be done, anyway.
After 3 days of operation, the data was pulled from the system, parsed
using a few Linux commands, fed into gnuplot, and shown as such:
http://yfrog.com/06mossbauerwp - For the interested, I'll explain
exactly how I produced the plot if asked. gnuplot is awesome. Another
week of data collection ended yesterday, so I'll wander up and grab
the data and post it later this afternoon as well. Not like the
resonant points are /that/ hard to miss on what is currently obtained
at any rate...
What I would like to see the USENET braintrust accomplish is a
quantitative description of the events that are being seen. Or even a
convincing qualitative explanation.
Modern physics has a unique and well understood answer for this
problem. Does anyone have a differing opinion?
koobee...@gmail.com
Nobody gives a fvck about your playing house pretending to go to
college doing all these useless experiments at tax payers’ expense ---
not even your fellow Einstein Dingleberries. <shrug>
Eric Gisse
Its' ok - physics is hard. I don't expect you to understand.
Dorn.Strich
From the contrived nature of the problem, it was probably the thesis
problem he could not complete and failed and still puzzles him to this
day.
PD
Ah, so it isn't really about the science at all, is it?
It's about your tax dollars being spent on things you don't approve of?
Eric Gisse
In what way is the problem contrived, David? Is there a part you would
like help understanding?
Dorn.Strich
Look up the meaning of contrived. You'll have your answer.
Benj
> Listen, Gisse the college drop-out, the troll, and the lair.
Lair? Hmm. Wonder what that means?
But yeah, Gisse, nobody does give a fvck about your stupid experiment,
especially when you haven't the gumption to even produce a decent
drawing (you mean *I* have to go photoshop your pencil scribblings to
see what you are trying to say?) or some photos of the apparatus, but
still expect us to read your mind.
Well read this Gisse, A "Geiger counter" doesn't produce any spectrum
at all! If you got one then probably all that the windoze 3.1 (a truly
fine OS by Vista standards, by the way) system was collecting was
hum! I trust you really meant "proportional counter" but hey, why
should I waste time trying to teach you how to talk? I've got more
Burger King floors to mop!
George Hammond
[Hammond]
10 seconds on Google turns up a complete explanation of
your output graph with the 6 peaks and an explanation of it
as being due to Zeeman hyperfine splitting caused by the
ferromagnetc internal magnetic field of the iron sample...
the 6 peak hyperfine split is typical of iron apparently.
see:
http://en.wikipedia.org/wiki/M%C3%B6ssbauer_spectroscopy
PS I spent several years working with microwave (atomic
rotational) spectroscopy and there we used the Stark Effect
(hyperfine) line splitting phenomenon to modulate the atomic
rotational lines so we could use a "phase locked amplifier"
locked at the Stark modulation frequency to vastly reduce
S/N and increase detection sensitivity.... this gave me the
clue as to what you're playing with.
===================================
HAMMOND'S PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
Casey Bennetto mp3 God=G_uv folk song:
http://interrobang.jwgh.org/songs/hammond.mp3
==================================
doug
George Hammond wrote:
Most experimenters want to increase S/N.
Eric Gisse
> On Apr 3, 1:09 pm, koobee.wub...@gmail.com wrote:
>
> > Listen, Gisse the college drop-out, the troll, and the lair.
>
> Lair? Hmm. Wonder what that means?
Wolves live inside me.
>
> But yeah, Gisse, nobody does give a fvck about your stupid experiment,
> especially when you haven't the gumption to even produce a decent
> drawing (you mean *I* have to go photoshop your pencil scribblings to
> see what you are trying to say?) or some photos of the apparatus, but
> still expect us to read your mind.
No, that's why I wrote all those words to go with the somewhat blurry
photograph.
Is there a part you need help understanding?
>
> Well read this Gisse, A "Geiger counter" doesn't produce any spectrum
> at all! If you got one then probably all that the windoze 3.1 (a truly
> fine OS by Vista standards, by the way) system was collecting was
> hum! I trust you really meant "proportional counter" but hey, why
> should I waste time trying to teach you how to talk? I've got more
> Burger King floors to mop!
Since you only seem content to whine, I guess you can go back to
mopping.
Eric Gisse
[...]
Correct. I'm more interested to see if people like Porat or Strich can
come up with that information.
George Hammond
[Hammond]
I'm sure strik9 can, he's a war veteran and they're even
more desperate and resourceful than physicists. The litmus
test is <doug> the stalker who constantly brags about being
a phd in physics and is so generally unaware and ignorant I
find it incredible that he even got out of high school.
koobee...@gmail.com
Well, is it not hard to fathom that I do not appreciate anything
wasteful, non-productive? In case if you have not noticed the dire
economic situation we are facing because of the idiot *bankers*.
Speaking of these idiot bankers, there were at least 3 ancient Chinese
philosophers living 2,000-2,500 years ago. They were:
** Lao Zhu who founded Taoism – Basically occult in nature
** Confucius who founded Confucianism – Dictating all to obey
authority and the authority to be fair
** Mo Zhi who preached liberty/peace, dignity, and responsibility
It was Confucianism that won out. The philosophy actually works very
well in an ideal world, but in real life, the authority always becomes
corrupted. The check and balance is thus missing. That breeds a
whole culture of sheeples. Einstein Dingleberries are fine examples.
In my opinion, Mo Zhi’s teachings did offer a better cultural
advancement for the entire people and united peoples. In modern day
concept, liberty means free of enslavement from these idiot bankers
who insist on burden the working people with debts. Dignity means to
be proud of your works whatever it is --- perfection at the finest
level. Responsibility means self-advancement whatever is deemed
worthy. I do see some traces of these traits in countries like the US
and Japan, but a full embracement adopted independently by a people
rediscovering Mo Zhi’s teachings ended in total annihilation. The
corrupted capitalists had to unite with the also very corrupted
totalitarians to destroy this culture that these evil allies knew the
eventual ends of their reigns if the vanquished were allowed to
proliferate. The aftermath was full of lies and deceits about this
conflict. Hey, I am certain you know who I am talking about, but you
can never be sure.
Just imagine if Mo Zhi’s teachings did win out against Confucianism
more than 2000 years ago, there is only one united peoples of China/
world.
Sue...
<< Abstract. Attention is drawn to an error in
the paper of Champeney and Moon. The simplest
interpretation of this experiment and that of
Moon et al. is that time dilatation is a
consequence of acceleration and not of
relative velocity. >>
http://www.iop.org/EJ/abstract/0370-1328/86/3/126
Sue...
doug
So since I can point out all your silly mistakes, what does
that say about you.
doug
George Hammond wrote:
> X-No-Archive: Yes
> [Hammond]
> Your atuteness in posting about typographical errors has
> become legendary on Usenet.
You are ignoring the fact that I have been pointing out
the silly mistakes in your delusional "scientific" "theory"
of "god". It is humorous to see you flail about in your
wild imagination claiming things like that the Greeks
and Romans were too stupid to know the difference between
12 and 13 and so on.
> Since you brag constantly about having a phd in physics
> and have such phenomenal secretarial skills, I suggest you
> pursue a career as a male secretary.... you could become the
> world's first male secretary with a phd in physics... and as
> a side benefit you could even keep your nite job as a Usenet
> stalker!
For a good laugh, read the complete nonsense in these links:
George Hammond
>
>
>George Hammond wrote:
>
snip
>>>>
>>>>[Hammond]
>>>> I'm sure strik9 can, he's a war veteran and they're even
>>>>more desperate and resourceful than physicists. The litmus
>>>>test is <doug> the stalker who constantly brags about being
>>>>a phd in physics and is so generally unaware and ignorant I
>>>>find it incredible that he even got out of high school.
>>>
>>>So since I can point out all your silly mistakes, what does
>>>that say about you.
>>>
>>>
>>
>> [Hammond]
>> Your atuteness in posting about typographical errors has
>> become legendary on Usenet.
>
>You are ignoring the fact that I have been pointing out
>the silly mistakes in your delusional "scientific" "theory"
>of "god". It is humorous to see you flail about in your
>wild imagination claiming things like that the Greeks
>and Romans were too stupid to know the difference between
>12 and 13 and so on.
>
>
[Hammond]
I got your number slimebag... we both know it.... and you
can't stay away from me because of it.
It isn't that 12 is NOT equal to 13 that bothers you....
it's the fact that 12 is NEARLY equal to 13 that is
aggravating the shit out of you... in fact it's literally
HAUNTING you.
See slimebag.... the fact is you're a physicist... even
if a second rate and sleazy one.... you have that little
inextinguishable spark of sensitivity to scientific truth
that you can't escape.... it HAUNTS you and drives you nuts.
What you know is that some other physicist has discovered
that because the human brain is anatomically CUBIC and that
a cube as any physicist knows has 13 symmetry axes....that
this would cause every psychometric correlation matrix, no
matter what the size, to have 13 eigenvectors corresponding
to the 13 symmetry axes of the brain. Fact is, 50 years of
computerized linear algebra (Factor Analysis) in Psychometry
has ALREADY experimentally discovered this although of
course they have no idea WHY.
However, Hammond has discovered, proved and published WHY
and the paper was peer reviewed by 5 professorial level
experts in psychometrics and unanimously recommended for
publication in a premier academic journal (Hammond 1994).
Hammond has shown that this explains the "12 OLYMPIAN
GODS" of ancient history (the Egypto-Greco-Roman
pantheon)... and what YOU KNOW is that this is a HAUNTING
and INESCAPABLE SCIENTIFIC CLUE of historic proportions.
You KNOW THIS because it is instinctively structurally
obvious to you as a PhD Physicist... and the fact not only
HAUNTS you, but it also arouses your sense of GUILT AND
RESPONSIBILITY as a physicist.
What you know is that this simple scientific explanation
is a momentous scientific fact of historical proportions...
and knowing that is what is haunting you every time you hear
the name Hammond.
Nice to watch you sweat sucker... nice to know you're
aware that the end is nigh.... that yes, someone has
discovered the world's first scientific proof of God... and
that a slimebag like YOU is squirming because YOU
unfortunately happen to know enough physics, and psychology
apparently, to ACTUALLY REALIZE that Hammond has struck the
mother lode.
Keep squirming sucker... nice to see you sweat! Think
of all the billions of suffering humans you are failing to
relieve by not speaking out... squirm and sweat some more,
I'm enjoying every minute of it.... you so richly deserve
it!
PD
> On Apr 3, 11:47 am, PD <TheDraperFam...@gmail.com> wrote:
>
> > On Apr 3, 1:09 pm, koobee.wub...@gmail.com wrote:
> > > Nobody gives a fvck about your playing house pretending to go to
> > > college doing all these useless experiments at tax payers’ expense ---
> > > not even your fellow Einstein Dingleberries. <shrug>
>
> > Ah, so it isn't really about the science at all, is it?
> > It's about your tax dollars being spent on things you don't
> > approve of?
>
> Well, is it not hard to fathom that I do not appreciate anything
> wasteful, non-productive?
Yes, I know -- it's common for engineers to see absolutely no value in
the R side of R&D. That's why they're engineers -- they see value in D
and none in R. Fortunately, there are checks and balances.
> In case if you have not noticed the dire
> economic situation we are facing because of the idiot *bankers*.
And what does the activity of making money from money have to do with
the worth of fundamental research.
Wow, you and Potter should be having this conversation. He lives in
China. He's retired. He prefers D to R. He's a nutjob. You have a lot
in common.
PD
koobee...@gmail.com
> On Apr 4, 2:25 am, koobee.wub...@gmail.com wrote:
> > Well, is it not hard to fathom that I do not appreciate anything
> > wasteful, non-productive?
>
> Yes, I know -- it's common for engineers to see absolutely no value in
> the R side of R&D. That's why they're engineers -- they see value in D
> and none in R. Fortunately, there are checks and balances.
You are wrong once again. There is plenty of R and D in the
engineering world. Engineers see the values in both R and D. <shrug>
> > In case if you have not noticed the dire
> > economic situation we are facing because of the idiot *bankers*.
>
> And what does the activity of making money from money have to do with
> the worth of fundamental research.
None. <shrug>
> > Speaking of these idiot bankers, there were at least 3 ancient Chinese
> > philosophers living 2,000-2,500 years ago. They were:
>
> > ** Lao Zhu who founded Taoism – Basically occult in nature
SR and GR are both occult in actuality. <shrug>
Actually, I have been to China a couple times. Although both trips
were business, they were still culturally rewarding for me. <shrug>
Physics is a study to model the world/universe based on mathematics.
The physicists seem to have done that horribly after embracing the
occult in SR and GR. SR is falsified by the twins paradox right from
the bat. It makes no sense to address experimental results. GR is
falsified by an infinite numbers of unique solutions to the field
equations that are static, spherically symmetric, and asymptotically
flat. Believing in the nonsense of SR and GR is like practicing
occult where
** FAITH IS THEORY
** MYSTICISM IS WISDOM
** IGNORANCE IS KNOWLEDGE
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** BELIEVING IS LEARNING
** LYING IS TEACHING
<shrug>
Eric Gisse
[...]
> GR is
> falsified by an infinite numbers of unique solutions to the field
> equations that are static, spherically symmetric, and asymptotically
> flat.
Five years running and you still can't explain how they are unique if
they are all related by a coordinate transformation.
[snip spew]
koobee...@gmail.com
1) Geometry is invariant regardless what choice of coordinate system
you choose.
2) Nincompoop.
3) Coordinate system alone cannot possibly describe the geometry.
4) Nitwit.
5) After settling on a coordinate system to describe the invariant
geometry, it has to incorporate the metric to describe the geometry.
6) Idiot.
7) To describe the flat spacetime geometry, there are two coordinate
systems commonly utilized. The metric under the Euclidean coordinate
system is drastically different from the metric under the spherically
symmetric polar coordinate system.
8) Moron.
9) There is no room for Confucian thinking in science haunted the
development of physics in the past 100 years.
10) Vegetable.
11) Gisse is a college drop-out, a troll, and a liar who worships
Einstein the nitwit, the plagiarist, and the liar.
12) <shrug>
13) After waking up about an hour ago, go back to sleep.
14) Dream about
Eric Gisse
> This post is adopting Uncle Al’s style.
Uncle Al's style requires one to be educated. You are not educated.
>
> 1) Geometry is invariant regardless what choice of coordinate system
> you choose.
Codified by the object called a "tensor".
You do not understand this.
>
> 2) Nincompoop.
>
> 3) Coordinate system alone cannot possibly describe the geometry.
Codified by the object called a "tensor".
You do not understand this.
>
> 4) Nitwit.
>
> 5) After settling on a coordinate system to describe the invariant
> geometry, it has to incorporate the metric to describe the geometry.
The metric is the geometry.
You do not understand this.
>
> 6) Idiot.
>
> 7) To describe the flat spacetime geometry, there are two coordinate
> systems commonly utilized. The metric under the Euclidean coordinate
> system is drastically different from the metric under the spherically
> symmetric polar coordinate system.
Its' the same metric in two different coordinate systems. They are the
same.
You do not understand this.
[snip rest]
koobee...@gmail.com
> On Apr 6, 2:39 pm, koobee.wub...@gmail.com wrote:
> > This post is adopting Uncle Al’s style.
>
> Uncle Al's style requires one to be educated. You are not educated.
That is amusing indeed.
> > 1) Geometry is invariant regardless what choice of coordinate system
> > you choose.
>
> Codified by the object called a "tensor".
Nonsense! Mysticism!
> > 2) Nincompoop.
>
> > 3) Coordinate system alone cannot possibly describe the geometry.
>
> Codified by the object called a "tensor".
Nonsense! Mysticism!
In differential geometry, a tensor is nothing more than a multi-
dimensional matrix. Most are 3-by-3 or 4-by-4 depending on discussing
space or spacetime. Riemann tensor is a 3-by-3-by-3-by-3 or 4-by-4-
by-4-by-4 matrix. <shrug>
> > 4) Nitwit.
>
> > 5) After settling on a coordinate system to describe the invariant
> > geometry, it has to incorporate the metric to describe the geometry.
>
> The metric is the geometry.
Nonsense! Mysticism!
The metric is merely a connection allowing the choice of coordinate
system to describe the geometry. The metric is not the geometry.
<shrug>
> > 6) Idiot.
>
> > 7) To describe the flat spacetime geometry, there are two coordinate
> > systems commonly utilized. The metric under the Euclidean coordinate
> > system is drastically different from the metric under the spherically
> > symmetric polar coordinate system.
>
> Its' the same metric in two different coordinate systems. They are the
> same.
For example to describe flat space, in Euclidean coordinate system (x,
y, z), the metric is [(1, 0, 0), (0, 1, 0), (0, 0, 1)], and in common
spherically symmetric polar coordinate system (r, Longitude,
Latitude), the metric is [(1, 0, 0), (0, r^2 cos^2(Latitude), 0), (0,
0, r^2)]. The former has a determinant of 1 while the latter (r^4
cos^2(Latitude)). The metrics are different because of different
choices in coordinate systems where the geometry is always invariant.
<shrug>
That explains why Gisse is a college drop-out, a troll, and a liar who
worships Einstein the nitwit, the plagiarist, and the liar. <shrug>
Eric Gisse
> On Apr 6, 4:29 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Apr 6, 2:39 pm, koobee.wub...@gmail.com wrote:
> > > This post is adopting Uncle Al’s style.
>
> > Uncle Al's style requires one to be educated. You are not educated.
>
> That is amusing indeed.
>
> > > 1) Geometry is invariant regardless what choice of coordinate system
> > > you choose.
>
> > Codified by the object called a "tensor".
>
> Nonsense! Mysticism!
Only to the uneducated. Read a book on tensor analysis.
>
> > > 2) Nincompoop.
>
> > > 3) Coordinate system alone cannot possibly describe the geometry.
>
> > Codified by the object called a "tensor".
>
> Nonsense! Mysticism!
>
> In differential geometry, a tensor is nothing more than a multi-
> dimensional matrix.
Incorrect. A tensor is a multilinear object that exists on a manifold,
while a matrix is an ordered collection of numbers.
> Most are 3-by-3 or 4-by-4 depending on discussing
> space or spacetime. Riemann tensor is a 3-by-3-by-3-by-3 or 4-by-4-
> by-4-by-4 matrix. <shrug>
>
> > > 4) Nitwit.
>
> > > 5) After settling on a coordinate system to describe the invariant
> > > geometry, it has to incorporate the metric to describe the geometry.
>
> > The metric is the geometry.
>
> Nonsense! Mysticism!
>
> The metric is merely a connection allowing the choice of coordinate
> system to describe the geometry. The metric is not the geometry.
> <shrug>
Except the metric is the geometry. Open a textbook on differential
geometry.
Plus the metric is not the same as a connection. Open a textbook on
differential geometry.
If you don't want to talk about differential geometry that's fine. But
don't come at me whining about your personal beliefs because there's
no point in arguing about them.
>
> > > 6) Idiot.
>
> > > 7) To describe the flat spacetime geometry, there are two coordinate
> > > systems commonly utilized. The metric under the Euclidean coordinate
> > > system is drastically different from the metric under the spherically
> > > symmetric polar coordinate system.
>
> > Its' the same metric in two different coordinate systems. They are the
> > same.
>
> For example to describe flat space, in Euclidean coordinate system (x,
> y, z), the metric is [(1, 0, 0), (0, 1, 0), (0, 0, 1)], and in common
> spherically symmetric polar coordinate system (r, Longitude,
> Latitude), the metric is [(1, 0, 0), (0, r^2 cos^2(Latitude), 0), (0,
> 0, r^2)]. The former has a determinant of 1 while the latter (r^4
> cos^2(Latitude)). The metrics are different because of different
> choices in coordinate systems where the geometry is always invariant.
> <shrug>
Why are you talking about the determinant? The determinant is not a
scalar quantity.
Try this: g_uv g^uv
It will always be equal to the number of dimensions. In every
coordinate system. If what you were saying were true, this would be
false. Is it false, or do you need to invent a reason for why I'm
wrong even though the counterexample is staring you in the face?
A honest person would simply work through a few calculations and give
me the name of the textbook he is basing his claims off of. But since
you refuse to do any calculations and even more strongly refuse to
name even one textbook that agrees with you, it seems quite obvious
that you are dishonest.
>
> That explains why Gisse is a college drop-out, a troll, and a liar who
> worships Einstein the nitwit, the plagiarist, and the liar. <shrug>
If you believed what you were saying you wouldn't fall for the "troll"
every time. So either you are lying or you are stupid - which is it?
George Hammond
koobee...@gmail.com wrote:
>
>For example to describe flat space, in Euclidean coordinate system (x,
>y, z), the metric is [(1, 0, 0), (0, 1, 0), (0, 0, 1)], and in common
>spherically symmetric polar coordinate system (r, Longitude,
>Latitude), the metric is [(1, 0, 0), (0, r^2 cos^2(Latitude), 0), (0,
>0, r^2)]. The former has a determinant of 1 while the latter (r^4
>cos^2(Latitude)). The metrics are different because of different
>choices in coordinate systems where the geometry is always invariant.
><shrug>
>
[Hammond]
LOL
The "metric tensor" is a tensor, and as you have already
admitted the two metrics are related by the simple Cartesian
to Spherical coordinate transformation. Therefore by the
MATHEMATCAL DEFINITION of a tensor the two metrics are
ABSOLUTELY PROVEN to be identical. QED.
Go jump in a lake.
George Hammond
<jow...@gmail.com> wrote:
>
>
> So either you are lying or you are stupid - which is it?
>
[Hammond]
He's neither... he's what is called "ignorant" which
means he's too stupid to realize that he's smart.
koobee...@gmail.com
> On Apr 6, 9:22 pm, koobee.wub...@gmail.com wrote:
> > Nonsense! Mysticism!
>
> Only to the uneducated. Read a book on tensor analysis.
Speaking from a college drop-out. <shrug>
> > Nonsense! Mysticism!
>
> > In differential geometry, a tensor is nothing more than a multi-
> > dimensional matrix.
>
> Incorrect. A tensor is a multilinear object that exists on a manifold,
> while a matrix is an ordered collection of numbers.
Wrong, in differential geometry, a tensor is nothing more than a multi-
dimensional matrix. <shrug>
> > Most are 3-by-3 or 4-by-4 depending on discussing
> > space or spacetime. Riemann tensor is a 3-by-3-by-3-by-3 or 4-by-4-
> > by-4-by-4 matrix. <shrug>
The college drop-out just would not understand that the tensor is
nothing more a multi-dimensional matrix for all practical mathematical
operations. <shrug>
> > Nonsense! Mysticism!
>
> > The metric is merely a connection allowing the choice of coordinate
> > system to describe the geometry. The metric is not the geometry.
> > <shrug>
>
> Except the metric is the geometry.
The metric is not and cannot be the geometry for the reasons I have
told many time before. <shrug>
> Open a textbook on differential geometry.
I have understood the subject well. There is no need to open any
books written by Einstein Dingleberries to promote matheMagics.
<shrug>
> Plus the metric is not the same as a connection. Open a textbook on
> differential geometry.
The metric connects the choice of coordinate system to describe the
invariant geometry. The metric must be a connection. <shrug>
> If you don't want to talk about differential geometry that's fine.
I do. I just do not want to talk about illogics and matheMagics.
<shrug>
> But
> don't come at me whining about your personal beliefs because there's
> no point in arguing about them.
You are the one who come to me with your illogics and matheMagics.
<shrug>
> > For example to describe flat space, in Euclidean coordinate system (x,
> > y, z), the metric is [(1, 0, 0), (0, 1, 0), (0, 0, 1)], and in common
> > spherically symmetric polar coordinate system (r, Longitude,
> > Latitude), the metric is [(1, 0, 0), (0, r^2 cos^2(Latitude), 0), (0,
> > 0, r^2)]. The former has a determinant of 1 while the latter (r^4
> > cos^2(Latitude)). The metrics are different because of different
> > choices in coordinate systems where the geometry is always invariant.
> > <shrug>
>
> Why are you talking about the determinant?
Can you show [(1, 0, 0), (0, 1, 0), (0, 0, 1)] is the same as [(1, 0,
0), (0, r^2 cos^2(Latitude), 0), (0, 0, r^2)]?
> The determinant is not a scalar quantity.
So, the college drop-out has no clue to what the determinant is.
<shrug>
> [snipped irrelevant nonsense]
>
> > That explains why Gisse is a college drop-out, a troll, and a liar who
> > worships Einstein the nitwit, the plagiarist, and the liar. <shrug>
>
> If you believed what you were saying you wouldn't fall for the "troll"
> every time. So either you are lying or you are stupid - which is it?
After all that whining, Gisse remains a college drop-out, a troll, and
koobee...@gmail.com
> On Mon, 6 Apr 2009, koobee.wub...@gmail.com wrote:
> >For example to describe flat space, in Euclidean coordinate system (x,
> >y, z), the metric is [(1, 0, 0), (0, 1, 0), (0, 0, 1)], and in common
> >spherically symmetric polar coordinate system (r, Longitude,
> >Latitude), the metric is [(1, 0, 0), (0, r^2 cos^2(Latitude), 0), (0,
> >0, r^2)]. The former has a determinant of 1 while the latter (r^4
> >cos^2(Latitude)). The metrics are different because of different
> >choices in coordinate systems where the geometry is always invariant.
> ><shrug>
> LOL
The reverend is getting maniacal.
> The "metric tensor" is a tensor,
Of course, the metric is a tensor which is a matrix. <shrug>
> and as you have already
> admitted the two metrics are related by the simple Cartesian
> to Spherical coordinate transformation.
What transformation? The reverend is lying again. <shrug>
> Therefore by the
> MATHEMATCAL DEFINITION of a tensor the two metrics are
> ABSOLUTELY PROVEN to be identical. QED.
The reverend is as fvcked in his logic as usual. <shrug>
> Go jump in a lake.
After you.
George Hammond
koobee...@gmail.com wrote:
>
>> >For example to describe flat space, in Euclidean coordinate system (x,
>> >y, z), the metric is [(1, 0, 0), (0, 1, 0), (0, 0, 1)], and in common
>> >spherically symmetric polar coordinate system (r, Longitude,
>> >Latitude), the metric is [(1, 0, 0), (0, r^2 cos^2(Latitude), 0), (0,
>> >0, r^2)]. The former has a determinant of 1 while the latter (r^4
>> >cos^2(Latitude)). The metrics are different because of different
>> >choices in coordinate systems where the geometry is always invariant.
>> ><shrug>
>
>> LOL
>
>
>> and as you have already
>> admitted the two metrics are related by the simple Cartesian
>> to Spherical coordinate transformation.
>
>What transformation? The reverend is lying again. <shrug>
>
[Hammond]
"What transformation?"... you're kiddin... every freshman
knows it:
x = rcos(phi)cos(theta)
y = rcos(phi)sin(theta)
z = rsin(phi)
where phi is the latitude and theta is the longitude
.
.
Anybody who would bother to argue with you is a moron....
sorry, I'm not a moron. Get lost.
koobee...@gmail.com
> koobee.wub...@gmail.com wrote:
> >What transformation? The reverend is lying again. <shrug>
>
> "What transformation?"... you're kiddin... every freshman
> knows it:
>
> x = rcos(phi)cos(theta)
> y = rcos(phi)sin(theta)
> z = rsin(phi)
>
> where phi is the latitude and theta is the longitude
> .
> .
> Anybody who would bother to argue with you is a moron....
> sorry, I'm not a moron. Get lost.- Hide quoted text -
You still have not shown any transformation.
You have to prove the metric represented by the matrix [(1, 0, 0), (0,
1, 0), (0, 0, 1)] is the same as the metric represented by the matrix
[(1, 0, 0), (0, r^2 cos^2Latitude, 0), (0, 0, r^2)]. Any junior high
school children will tell you that is indeed fruitless.
Don’t come back until you have figured out.
PD
ROFLMAO.
Eric Gisse
> On Apr 7, 2:48 pm, George Hammond <Nowhe...@notspam.com> wrote:
>
> > koobee.wub...@gmail.com wrote:
> > >What transformation? The reverend is lying again. <shrug>
>
> > "What transformation?"... you're kiddin... every freshman
> > knows it:
>
> > x = rcos(phi)cos(theta)
> > y = rcos(phi)sin(theta)
> > z = rsin(phi)
>
> > where phi is the latitude and theta is the longitude
> > .
> > .
> > Anybody who would bother to argue with you is a moron....
> > sorry, I'm not a moron. Get lost.- Hide quoted text -
>
> You still have not shown any transformation.
Probably because it is a trivial exercise that anyone who dares argue
about GR should be able to understand.
>
> You have to prove the metric represented by the matrix [(1, 0, 0), (0,
> 1, 0), (0, 0, 1)] is the same as the metric represented by the matrix
> [(1, 0, 0), (0, r^2 cos^2Latitude, 0), (0, 0, r^2)]. Any junior high
> school children will tell you that is indeed fruitless.
Is the stupidest junior high school student the standard of education
in your mind?
>
> Don’t come back until you have figured out.
So does that mean we can enjoy blissful silence on your part?
George Hammond
koobee...@gmail.com wrote:
>On Apr 7, 2:48�pm, George Hammond <Nowhe...@notspam.com> wrote:
>> koobee.wub...@gmail.com wrote:
>
>> >What transformation? �The reverend is lying again. �<shrug>
>>
>> � "What transformation?"... you're kiddin... every freshman
>> knows it:
>>
>> x = rcos(phi)cos(theta)
>> y = rcos(phi)sin(theta)
>> z = rsin(phi)
>>
>> where phi is the latitude and theta is the longitude
>> .
>> .
>> Anybody who would bother to argue with you is a moron....
>> sorry, I'm not a moron. �Get lost.- Hide quoted text -
>
>You still have not shown any transformation.
>
>
[Hammond]
Whaddaru blind? The transformation is posted above and
is practically the world's most well known elementary
coordinate transformation.
>
>
>You have to prove the metric represented by the matrix [(1, 0, 0), (0,
>1, 0), (0, 0, 1)] is the same as the metric represented by the matrix
>[(1, 0, 0), (0, r^2 cos^2Latitude, 0), (0, 0, r^2)]. Any junior high
>school children will tell you that is indeed fruitless.
>
>Don�t come back until you have figured out.
>
>
[Hammond]
Who do you think you're kiddin Oooby Dooby? In your
previous post above you ADMIT that both metrics are the
"flat space metric":
===========Koobic Kwoatta==============
For example to describe flat space, in Euclidean coordinate
system (x,y, z), the metric is [(1, 0, 0), (0, 1, 0), (0, 0,
1)], and in common spherically symmetric polar coordinate
system (r, Longitude, Latitude), the metric is [(1, 0, 0),
(0, r^2 cos^2(Latitude), 0), (0,0, r^2)].
=======end Koobic Kwote================
[Hammond]
I don't have to PROVE they're IDENTICAL METRICS, you've
already admitted that by stating they are BOTH identically
the "flat space" metric.
What are you doing; playing word games or something?
Go jump in a lake.
koobee...@gmail.com
> koobee.wub...@gmail.com wrote:
> >You still have not shown any transformation.
>
> Whaddaru blind? The transformation is posted above and
> is practically the world's most well known elementary
> coordinate transformation.
You still have not shown any transformation. <shrug>
> >You have to prove the metric represented by the matrix [(1, 0, 0), (0,
> >1, 0), (0, 0, 1)] is the same as the metric represented by the matrix
> >[(1, 0, 0), (0, r^2 cos^2Latitude, 0), (0, 0, r^2)]. Any junior high
> >school children will tell you that is indeed fruitless.
>
> >Don t come back until you have figured out.
>
> Who do you think you're kiddin Oooby Dooby? In your
> previous post above you ADMIT that both metrics are the
> "flat space metric":
No. <shrug>
> ===========Koobic Kwoatta==============
> For example to describe flat space, in Euclidean coordinate
> system (x,y, z), the metric is [(1, 0, 0), (0, 1, 0), (0, 0,
> 1)], and in common spherically symmetric polar coordinate
> system (r, Longitude, Latitude), the metric is [(1, 0, 0),
> (0, r^2 cos^2(Latitude), 0), (0,0, r^2)].
> =======end Koobic Kwote================
That is correct. <shrug>
> I don't have to PROVE they're IDENTICAL METRICS, you've
> already admitted that by stating they are BOTH identically
> the "flat space" metric.
I also said the choice of coordinate for each case is different. That
is the linchpin. <shrug>
> What are you doing; playing word games or something?
No, I don’t bluff or play games. <shrug>
> Go jump in a lake.
After you.
koobee...@gmail.com
> On Apr 7, 2:04 pm, koobee.wub...@gmail.com wrote:
> > You still have not shown any transformation.
>
> Probably because it is a trivial exercise that anyone who dares argue
> about GR should be able to understand.
Wrong. <shrug>
> > You have to prove the metric represented by the matrix [(1, 0, 0), (0,
> > 1, 0), (0, 0, 1)] is the same as the metric represented by the matrix
> > [(1, 0, 0), (0, r^2 cos^2Latitude, 0), (0, 0, r^2)]. Any junior high
> > school children will tell you that is indeed fruitless.
>
> Is the stupidest junior high school student the standard of education
> in your mind?
No, try look at the mirror for a change. <shrug>
> > Don’t come back until you have figured out.
>
> So does that mean we can enjoy blissful silence on your part?
So, what is the answer from a college drop-out, a troll, and a liar?
Eric Gisse
> On Apr 7, 4:28 pm, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Apr 7, 2:04 pm, koobee.wub...@gmail.com wrote:
> > > You still have not shown any transformation.
>
> > Probably because it is a trivial exercise that anyone who dares argue
> > about GR should be able to understand.
>
> Wrong. <shrug>
Are you seriously asking this newsgroup to teach you how to transform
the metric from one coordinate system to another?
[...]
koobee...@gmail.com
> Are you seriously asking this newsgroup to teach you how to transform
> the metric from one coordinate system to another?
No. Try again.
Hint: Gisse is a college drop-out, a troll, and a liar.
<shrug>
George Hammond
[Hammond]
Dude;.... converting from inches to centimeters is a
coordinate transformation:
x' =2.54 x
But OBVIOUSLY the METRICAL PROPERTY (e.g. Pythagorean
theorem) of (flat) space space remains UNCHANGED whether you
use inches or centimeters.
YOU DON'T SEEM TO UNDERSTAND THIS....
PERHAPS YOU HAVE A THINKING DISORDER...?
koobee...@gmail.com
> koobee.wub...@gmail.com wrote:
> >I also said the choice of coordinate for each case is different. That
> >is the linchpin. <shrug>
>
> Dude;.... converting from inches to centimeters is a
> coordinate transformation:
That is correct. <shrug>
> x' =2.54 x
Your simple coordinate transformation does not apply. <shrug>
> But OBVIOUSLY the METRICAL PROPERTY (e.g. Pythagorean
> theorem) of (flat) space space remains UNCHANGED whether you
> use inches or centimeters.
BUT ALSO OBVIOUSLY, THE METRIC WHEN USING THE CARTESIAN COORDIANTE
SYSTEM HAS TO BE DIFFERENT FROM THE METRIC WHEN USING THE POLAR
COORDIANTE SYTEM IN DESCRIBING FLAT SPACE. <SHRUG>
> YOU DON'T SEEM TO UNDERSTAND THIS....
I AM AFRAID YOU ARE THE ONE WHO DOES NOT UNDERSTAND THIS SIMPLE
LOGIC. <SHRUG>
> PERHAPS YOU HAVE A THINKING DISORDER...?
YOU ARE WRONG AGAIN. THE REVEREND IS BETTER TO SPEND TIME CONVERTING
EINSTEIN DINGLEBERRIES TO WHATEVER. AFTER ALL, EINSTEIN DINGLEBERRIES
ARE ALREADY MYSTIFIED BY NONSENSE. <SHRUG>
George Hammond
koobee...@gmail.com wrote:
>On Apr 7, 11:35 pm, George Hammond wrote:
>> koobee.wub...@gmail.com wrote:
>
>> >I also said the choice of coordinate for each case is different. That
>> >is the linchpin. <shrug>
>>
>> Dude;.... converting from inches to centimeters is a
>> coordinate transformation:
>
>That is correct. <shrug>
>
>> x' =2.54 x
>
>Your simple coordinate transformation does not apply. <shrug>
>
>> But OBVIOUSLY the METRICAL PROPERTY (e.g. Pythagorean
>> theorem) of (flat) space space remains UNCHANGED whether you
>> use inches or centimeters.
>
>BUT ALSO OBVIOUSLY, THE METRIC WHEN USING THE CARTESIAN COORDIANTE
>SYSTEM HAS TO BE DIFFERENT FROM THE METRIC WHEN USING THE POLAR
>COORDIANTE SYTEM IN DESCRIBING FLAT SPACE. <SHRUG>
>
>
[Hammond]
You're full of it. You're trying to tell us the metric
is "different" if we convert it from centimeters to
inches.... that's the stupidest thing I've ever heard.
The next thing you'll be telling us is that the metric is
"different" if we write it in blue ink instead of
black. <TIC>
>
>
>
>YOU ARE WRONG AGAIN. THE REVEREND IS BETTER TO SPEND TIME CONVERTING
>EINSTEIN DINGLEBERRIES TO WHATEVER. AFTER ALL, EINSTEIN DINGLEBERRIES
>ARE ALREADY MYSTIFIED BY NONSENSE. <SHRUG>
>
[Hammond]
Sorry to disappoint you Fuckward I'm not a reverend, I'm
a physicist. I have no credentials in divinity and I am an
M.S. in physics. <TIC>
Don Stockbauer
You guys are so behind the times. Einstein's upgraded from a
Dingleberry to a Blackberry. But he got hungry and ate it.
Eric Gisse
> On Apr 7, 11:35 pm, George Hammond wrote:
>
> > koobee.wub...@gmail.com wrote:
> > >I also said the choice of coordinate for each case is different. That
> > >is the linchpin. <shrug>
>
> > Dude;.... converting from inches to centimeters is a
> > coordinate transformation:
>
> That is correct. <shrug>
>
> > x' =2.54 x
>
> Your simple coordinate transformation does not apply. <shrug>
>
> > But OBVIOUSLY the METRICAL PROPERTY (e.g. Pythagorean
> > theorem) of (flat) space space remains UNCHANGED whether you
> > use inches or centimeters.
>
> BUT ALSO OBVIOUSLY, THE METRIC WHEN USING THE CARTESIAN COORDIANTE
> SYSTEM HAS TO BE DIFFERENT FROM THE METRIC WHEN USING THE POLAR
> COORDIANTE SYTEM IN DESCRIBING FLAT SPACE. <SHRUG>
Prove it.
Calculate g_uv g^uv in the metric in Euclidean coordinates, and in
Spherical coordinates.
sal
> My sketched out version of the apparatus is here:
>
> http://yfrog.com/eisdc10237j - It is actually rather hard to take a
> text-capturing photo even with full control over the camera.
Set the camera to overexpose by about a stop. Otherwise it'll think the
paper is medium gray and expose to make it come out that way. (This may
be a separate setting if you're using flash -- some cameras have two
exposure settings, one for flash, one for everything else.)
Set it to mild telephoto, to cut vignetting and pincushion.
Try it with and without flash (take an extra shot). Using flash makes it
easier to get a sharp result, but you may need to angle the paper slightly
to avoid a hot spot in the middle due to specular reflection off the paper.
Finally, run it through Gimp to correct the perspective, punch up the
contrast, and do an unsharp mask to bring out the print a bit. If you
shot close up and it's pincushioned you can also fix that in Gimp.
The result will still look bad but will be somewhat more legible.
Alternatively, use a scanner :-)
--
Nospam becomes physicsinsights to fix the email
koobee...@gmail.com
> Koobee Wublee wrote:
> >BUT ALSO OBVIOUSLY, THE METRIC WHEN USING THE CARTESIAN COORDIANTE
> >SYSTEM HAS TO BE DIFFERENT FROM THE METRIC WHEN USING THE POLAR
> >COORDIANTE SYTEM IN DESCRIBING FLAT SPACE. <SHRUG>
>
> You're full of it. You're trying to tell us the metric
> is "different" if we convert it from centimeters to
> inches.... that's the stupidest thing I've ever heard.
The problem with you is that you do not understand that:
** The geometry is invariant regardless of which coordinate system
you choose.
** The metric is not the geometry. The metric is an connection to
describe the geometry using the already selected choice of coordinate
system.
<shrug>
> The next thing you'll be telling us is that the metric is
> "different" if we write it in blue ink instead of
> black. <TIC>
Therefore, the metric must be different for each different coordinate
system to describe the same geometry. Again, here are two examples to
describe the geometry of flat spacetime:
** Linearly rectangular coordinate system
Coordinate system = (x, y, z)
Metric = [(1, 0, 0), (0, 1, 0), (0, 0, 1)]
** Spherically symmetric polar coordinate system
Coordinate system = (r, Longitude, Latitude)
Metric = [(1, 0, 0), (0, r^2 cos^2(Latitude), 0), (0, 0, r^2)]
The coordinates are different, and the metric must also be different.
<shrug>
> >YOU ARE WRONG AGAIN. THE REVEREND IS BETTER TO SPEND TIME CONVERTING
> >EINSTEIN DINGLEBERRIES TO WHATEVER. AFTER ALL, EINSTEIN DINGLEBERRIES
> >ARE ALREADY MYSTIFIED BY NONSENSE. <SHRUG>
>
> Sorry to disappoint you Fuckward I'm not a reverend, I'm
> a physicist. I have no credentials in divinity and I am an
> M.S. in physics. <TIC>
You behave like a reverend. <shrug> You are the reverend of these
newsgroups trying to convert the zealots of SR and GR into whatever.
You are at the right place to do so because Einstein Dingleberries
think:
** FAITH IS THEORY
** MYSTICISM IS WISDOM
** IGNORANCE IS KNOWLEDGE
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** BELIEVING IS LEARNING
** LYING IS TEACHING
<shrug>
koobee...@gmail.com
> On Apr 7, 10:58 pm, koobee.wub...@gmail.com wrote:
> > BUT ALSO OBVIOUSLY, THE METRIC WHEN USING THE CARTESIAN COORDIANTE
> > SYSTEM HAS TO BE DIFFERENT FROM THE METRIC WHEN USING THE POLAR
> > COORDIANTE SYTEM IN DESCRIBING FLAT SPACE. <SHRUG>
>
> Prove it.
I just did in many posts in this thread. <shrug>
> Calculate g_uv g^uv in the metric in Euclidean coordinates, and in
> Spherical coordinates.
There are both 3 in 3-d. g_uv g^uv has no significance in
differential geometry. This delusion explains why Gisse is a college
drop-out, a troll, and a liar. <shrug>
Eric Gisse
> On Apr 8, 1:59 am, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Apr 7, 10:58 pm, koobee.wub...@gmail.com wrote:
> > > BUT ALSO OBVIOUSLY, THE METRIC WHEN USING THE CARTESIAN COORDIANTE
> > > SYSTEM HAS TO BE DIFFERENT FROM THE METRIC WHEN USING THE POLAR
> > > COORDIANTE SYTEM IN DESCRIBING FLAT SPACE. <SHRUG>
>
> > Prove it.
>
> I just did in many posts in this thread. <shrug>
You asserted - you did not prove.
Just because it looks different does not mean it is different.
>
> > Calculate g_uv g^uv in the metric in Euclidean coordinates, and in
> > Spherical coordinates.
>
> There are both 3 in 3-d.
You claim the metric is "different" in a new coordinate system.
How can it always give the same answer if the metric is "different"?
> g_uv g^uv has no significance in
> differential geometry.
No significance? It is the trace of the delta function.
[snip insults]
koobee...@gmail.com
> On Apr 8, 8:28 am, koobee.wub...@gmail.com wrote:
> > I just did in many posts in this thread. <shrug>
>
> You asserted - you did not prove.
Don’t blame your lack of aptitude on me. <shrug>
> Just because it looks different does not mean it is different.
I don’t think we need to go on. You are indeed retarded. <shrug>
That is why Gisse is a college drop-out, a troll, and a liar. The
only thing the college drop-out, the troll, and the liar to do is to
worship Einstein the nitwit, the plagiarist, and the liar. <shrug>
Eric Gisse
I find it interesting that the vast majority of your posts these days
are just copied and pasted from a text file of insults.
You claim that the Euclidean metric in Cartesian and Spherically
symmetric coordinates are "different" yet you claim that g_uv g^uv = 3
in either case. By what standard are they "different" ?
koobee...@gmail.com
> On Apr 8, 9:24 am, koobee.wub...@gmail.com wrote:
> > That is why Gisse is a college drop-out, a troll, and a liar. The
> > only thing the college drop-out, the troll, and the liar to do is to
> > worship Einstein the nitwit, the plagiarist, and the liar. <shrug>
>
> [whining crap snipped]
> You claim that the Euclidean metric in Cartesian and Spherically
> symmetric coordinates are "different" yet you claim that g_uv g^uv = 3
> in either case. By what standard are they "different" ?
Let’s say you have two numbers ‘6’, and ‘7’.
** g_uv g^uv for ‘6’ is 1 (u, v = 0, g_uv = 6).
** g_uv g^uv for ‘7’ is 1 (u, v = 0, g_uv = 7).
g_uv in this example only has 1 element.
Therefore, according to your illogic, 6 = 7. You are indeed a retard
fit to be a college drop-out, a troll, and a liar. <shrug>
PD
I would imagine KW would say that the vectors described by
(x,y) = (1,1)
and
(r,theta) = (sqrt(2),pi/4)
are different because
sqrt(2) =/= 1
and
pi/4 =/= 1.
Interesting standard, but that seems to be his.
PD
George Hammond
[Hammond]
You're fulla crap. The diagonal of a football field is
111.8 yards whether you compute it using Cartesian
Coordinates or using Spherical Coordinates.
You don't know what you're talking about.
koobee...@gmail.com
> I would imagine KW would say that the vectors described by
> (x,y) = (1,1)
> and
> (r,theta) = (sqrt(2),pi/4)
> are different because
> sqrt(2) =/= 1
> and
> pi/4 =/= 1.
Your imagination is indeed delusional. The above two vectors are
indeed identical. <shrug>
> Interesting standard, but that seems to be his.
As a professor, you should know that the metric is not the same as the
vector. <shrug>
koobee...@gmail.com
> koobee.wub...@gmail.com wrote:
> >Therefore, the metric must be different for each different coordinate
> >system to describe the same geometry. Again, here are two examples to
> >describe the geometry of flat spacetime:
>
> >** Linearly rectangular coordinate system
>
> > Coordinate system = (x, y, z)
> > Metric = [(1, 0, 0), (0, 1, 0), (0, 0, 1)]
>
> >** Spherically symmetric polar coordinate system
>
> > Coordinate system = (r, Longitude, Latitude)
> > Metric = [(1, 0, 0), (0, r^2 cos^2(Latitude), 0), (0, 0, r^2)]
>
> You're fulla crap.
It looks like the reverend and the college drop-out are meant for each
other. <shrug>
> The diagonal of a football field is
> 111.8 yards whether you compute it using Cartesian
> Coordinates or using Spherical Coordinates.
That is correct. This is geometry, and it is invariant under any
choice of coordinate ssytem. <shrug>
> You don't know what you're talking about.
The metric is not geometry. Time to time again, you have failed to
understand this very simple concept. <shrug>
George Hammond
[Hammond]
Yes, that's exactly the stupidity he's arguing.
He can't figure out why the diagonal of a Football field is
111.8 yards regardless of whether the opposite corners are
located using Cartesian or Polar coordinates.
In short, he doesn't know what a metric is, or what it does,
or why it "specifies the geometry" of the space
independantly of the coordinate system used.
Why for Pete's sake, he thinks the Euclidean metric is
"different" if it is expressed in INCHES rather than in
CENTIMETERS.
Baahh, ha, ha, ha....aaah, ha, ha, has haaa....
George Hammond
[Hammond]
You're fulla crap. The "geometry" you loosly refer to is
RIGOROUSLY defined as the DISTANCE between any two points
and this DISTANCE is INDEPENDENT of the coordinate system
being used. For instance INCHES and CENTIMETERS are two
different coordinant systems (x'=2.54x). If you're going to
sit there and endlessly argue that the METRIC expressed in
INCHES is "different" than the same metric expressed in
CENTIMETERS then you're obviously a CERTIFIABLE INCOMPETENT.
koobee...@gmail.com
> >The metric is not geometry. Time to time again, you have failed to
> >understand this very simple concept. <shrug>
>
> You're fulla crap. The "geometry" you loosly refer to is
> RIGOROUSLY defined as the DISTANCE between any two points
> and this DISTANCE is INDEPENDENT of the coordinate system
> being used. For instance INCHES and CENTIMETERS are two
> different coordinant systems (x'=2.54x). If you're going to
> sit there and endlessly argue that the METRIC expressed in
> INCHES is "different" than the same metric expressed in
> CENTIMETERS then you're obviously a CERTIFIABLE INCOMPETENT.
The reverend has confused himself by choosing the metric and the
English systems. Using the established choice of coordinate system,
the metric describes how the geometry at a coordinate location changes
with respect to that particular point’s surrounding points. Since the
application of the metric and the English systems are implied by the
reverend to be flat, the metric for each is the same. The reverend
needs to compare the linearly rectangular versus the spherically
symmetric polar coordinate systems to understand this point. In doing
so, the different metrics are stunningly apparent. Gee! Is the
reverend also retarded? <shrug>
George Hammond
[Hammond]
Firstly I have no credentials or aspirations in Theology,
am not an ordained minister and am not therefore a
"reverend".
Secondly I am an M.S. in Physics and am therefore
properly referred to as a "physicist" and my CV is posted on
my website.
Finally, your pointless and incoherent statement above
makes no scientific sense whatsoever since:
The METRIC of a space uniquely and completely
defines the "geometry" of the space irregardless
what coordinate system used.
It has been amply demonstrated that you are unable to
comprehend this. I therefore suggest you return to the
skilled trades where you belong.
PD
Of course not. A vector is tensor of rank {1 0}.
The metric is a tensor of different rank.
However, the stretch should not be too hard for you to follow.
PD
PD
http://en.wikibooks.org/wiki/Real_analysis/Metric_Spaces
Now, if you would be so kind -- just as you have sworn that you are
the ONLY person to understand curvature since Riemann -- PLEASE tell
me that you are the only person since Frechet to understand metric
spaces, and that the world has it all wrong.
PD
Eric Gisse
> On Apr 8, 10:33 am, Eric Gisse <jowr...@gmail.com> wrote:
>
> > On Apr 8, 9:24 am, koobee.wub...@gmail.com wrote:
> > > That is why Gisse is a college drop-out, a troll, and a liar. The
> > > only thing the college drop-out, the troll, and the liar to do is to
> > > worship Einstein the nitwit, the plagiarist, and the liar. <shrug>
>
> > [whining crap snipped]
> > You claim that the Euclidean metric in Cartesian and Spherically
> > symmetric coordinates are "different" yet you claim that g_uv g^uv = 3
> > in either case. By what standard are they "different" ?
>
> Let’s say you have two numbers ‘6’, and ‘7’.
No. Answer the question before moving onto an example designed to be
asinine.
You claim that the Euclidean metric in Cartesian and Spherically
symmetric coordinates are "different" yet you claim that g_uv g^uv = 3
in either case. By what standard are they "different" ?
[snip rest]
Eric Gisse
[...]
> Interesting standard, but that seems to be his.
>
> PD
Fuck knows.
The only thing I can tell for sure is that he doesn't want to get into
any more technical arguments with me. Back in Oct/Nov he thought he
was "calling my bluff" when I offered to work through the derivation
of the field equations, and within two weeks I had built up mocking
material for years on end.
No more arguments. Just cut and paste insults that morph over time to
reflect his current feelings about me.
koobee...@gmail.com
> On Apr 8, 2:01 pm, koobee.wub...@gmail.com wrote:
> > As a professor, you should know that the metric is not the same as the
> > vector. <shrug>
>
> Of course not. A vector is tensor of rank {1 0}.
Wrong again. A vector is not a matrix or a tensor of rank-1.
However, a vector can be written as a matrix under some special
circumstances, but care must still be exercised when manipulating this
matrix because all unit vectors are left out. <shrug>
For example, using the spherically symmetric polar coordinate system,
we have:
** (r, Longitude, Latitude) = The coordinate.
** s/ = r r^ = The vector
Where
** r^ = Radial unit vector
To derive the metric from the vector above, we start the task by
taking the derivative as follows.
ds\ = dr r^ + r dr^ = dr r^ + r cos(Latitude) dLongitude Longitude^ +
r dLatitude Latitude^
Where
** dr^ = cos(Latitude) dLongitude Longitude^ + dLatitude Latitude^
** Longitude^ = Unit vector along the longitude
** Latitude^ = Unit vector along the latitude
Then, the dot product of the vector with itself becomes the following.
ds^2 = dr^2 + r^2 cos^2(Latitude) dLongitude^2 + r^2 dLatitude^2
Thus, the metric is [(1, 0, 0), (0, r^2 cos^2(Latitude), 0), (0, 0,
r^)].
You can do the same for using linearly rectangular coordinate system.
In doing so, you will get a different metric.
<shrug>
> The metric is a tensor of different rank.
The metric is merely a square matrix. It is rank 2. <shrug>
> However, the stretch should not be too hard for you to follow.
Yes, I know that. Remember that I am the only one to have understood
Riemann’s description of curved space. <shrug>
koobee...@gmail.com
> koobee.wub...@gmail.com wrote:
> >The reverend has confused himself by choosing the metric and the
> >English systems. Using the established choice of coordinate system,
> >the metric describes how the geometry at a coordinate location changes
> >with respect to that particular point’s surrounding points. Since the
> >application of the metric and the English systems are implied by the
> >reverend to be flat, the metric for each is the same. The reverend
> >needs to compare the linearly rectangular versus the spherically
> >symmetric polar coordinate systems to understand this point. In doing
> >so, the different metrics are stunningly apparent. Gee! Is the
> >reverend also retarded? <shrug>
>
> Firstly I have no credentials or aspirations in Theology,
> am not an ordained minister and am not therefore a
> "reverend".
The reverend is too modest. <shrug>
> Secondly I am an M.S. in Physics and am therefore
> properly referred to as a "physicist" and my CV is posted on
> my website.
The reverend is too delusional. <shrug>
> Finally, your pointless and incoherent statement above
> makes no scientific sense whatsoever since:
>
> The METRIC of a space uniquely and completely
> defines the "geometry" of the space irregardless
> what coordinate system used.
This is just not true. There is no possible way to define the
geometry without using a coordinate system. You must choose a
coordinate system to do so. In doing so, the metric connecting your
choice of coordinate system to describe the invariant geometry must be
only unique to your choice of coordinate system. <shrug>
> It has been amply demonstrated that you are unable to
> comprehend this. I therefore suggest you return to the
> skilled trades where you belong.
The reverend still cannot understand that geometry is not the metric,
and vice versa.
It looks like the reverend is also a victim of Orwellian education in
his pursue of theological connection in physics where:
hanson
>
"Gergi/Grodi/Georgi aka Hemorrhoid Hammond aka
"George Hammond" <Nowh...@notspam.com> wrote:
> koobee...@gmail.com wrote:
>> Grodi Georgi Hammond wrote:
>>> Koobee Wublee wrote:
>>
You're full of it. that's the stupidest thing I've ever heard.
>>
Koobee wrote:
The problem with you is that you do not understand that:
>>
Hemorrhoid Hammond wrote:
You're fulla crap.
==================================
http://georgi.com/scientific_Poof_of_god
Hemorrhoid's mirror site:
http://Poof-of-god.on-the-freehustling.com
GOD=G_uv (a G_oofy Poof by God George)
http://intense-Bang.jwgh.org/hammorroid-songs.
==================================
>
hanson wrote:
ahahah... WOW!... Grodi, did they let you out again, or
does the up-coming full moon make your hemmies itch?
... and either case makes you repeat your g_uv crap? ..
>
But listen, Grodi, here's the reason why:
#=# Nobody is born religious. #=#
#=# Religion is an acquired disease #=#
#=# Religion is a tool used by the few to fuck the many #=#
>
Grodi, it does appear more & more that you try to be one of
those bastards of "the few who wish to fuck the many"...
But very slim pickens for you, so far, huh, Grodi?.... ahahaha...
Yet, thanks for the laughs.... ahahaha... ahahahanson
George Hammond
[Hammond]
Wow.... a slam dunk!
The metric g_uv is defined via differential geometry by
"parallel displacement" such that the metric and only the
metric gives the true distance between two arbitrary points
in an arbitray space:
distance^2 = g_uv dx^u dx^v
As a direct result of this construction it turns out that:
g_uv g^uv = THE NUMBER OF DIMENSIONS
and this result which is obviously coordinate independent is
a property of the METRIC TENSOR ONLY*.
This then, is an actual mathematical PROOF that the
METRIC of an arbitrary space expressed in Cartesian
coordinates is "IDENTICAL" to the metric of said space
expressed in Spherical coordinates.
*Dirac says:
"The ONLY place where the number of dimensions
appears in tensor formalism is in the equation:
g^m_m = number of dimensions "
(Dirac: General Theory of Relativity, Ch. 15)
koobee...@gmail.com
> The metric g_uv is defined via differential geometry by
> "parallel displacement" such that the metric and only the
> metric gives the true distance between two arbitrary points
> in an arbitray space:
>
> distance^2 = g_uv dx^u dx^v
The equation above is very correct. However, your interpretation is
illogical. The geometry is described by ds^2, and there is no way in
hell that g_uv can represent the geometry independent of the
coordinate system (dx^u dx^v). Your mathematics even indicate so.
<shrug>
> As a direct result of this construction it turns out that:
>
> g_uv g^uv = THE NUMBER OF DIMENSIONS
Yes, that is all it means and nothing else. <shrug>
> and this result which is obviously coordinate independent is
> a property of the METRIC TENSOR ONLY*.
No, just re-examine the first equation you wrote down. You are indeed
retarded, aren’t you?
> This then, is an actual mathematical PROOF that the
> METRIC of an arbitrary space expressed in Cartesian
> coordinates is "IDENTICAL" to the metric of said space
> expressed in Spherical coordinates.
This is absolutely nonsense. Your mathematics does not agree with
your inconsistent statement above. <shrug>
> *Dirac says:
>
> "The ONLY place where the number of dimensions
> appears in tensor formalism is in the equation:
>
> g^m_m = number of dimensions "
What does that really mean again?
> (Dirac: General Theory of Relativity, Ch. 15)
Who gives a damn? The discussion and logic involved only requires
junior high
George Hammond
[Hammond]
Nonsense.... the metrical equation:
--------------------------------------------
g_uvg^uv = number of dimensions
----------------------------------------------
is MATHEMATICALLY INDEPENDENT of coordinate systems
and ONLY the metric tensor has this property BECAUSE it is
by construction the quantity which specifies completly the
"geometry" of the space.
You're all through bozo... try growing a beard if you
want to fool someone.
PD
> On Apr 8, 3:52 pm, PD <TheDraperFam...@gmail.com> wrote:
>
> > On Apr 8, 2:01 pm, koobee.wub...@gmail.com wrote:
> > > As a professor, you should know that the metric is not the same as the
> > > vector. <shrug>
>
> > Of course not. A vector is tensor of rank {1 0}.
>
> Wrong again. A vector is not a matrix or a tensor of rank-1.
Neither is a tensor a matrix. That's precisely the point. A tensor is
defined by its transformation properties, and that is definable
without a coordinate system at all, and therefore without any
reference to a matrix at all.
In conventional analysis (though the analog in differential geometry
is there as well), a tensor has a rank {a b}, where a indicates the
number of contravariant indeces and b indicates the number of
covariant indeces. A contravariant vector is one of these objects
where a=1 and b=0. A 1-form, or a covariant vector, is one of these
objects where a=0 and b=1.
If you sniff and tell me that this is impossible and that you
understand better than anyone what a tensor is, I'll be more than
amused.
PD
koobee...@gmail.com
> On Apr 8, 11:06 pm, koobee.wub...@gmail.com wrote:
> > > Of course not. A vector is tensor of rank {1 0}.
>
> > Wrong again. A vector is not a matrix or a tensor of rank-1.
>
> Neither is a tensor a matrix. That's precisely the point.
You said a vector is a tensor. You sound like you are contradicting
yourself. You are just trolling now. There is no need to continue.
> [snipped the rest of contradictory remarks]
koobee...@gmail.com
> koobee.wub...@gmail.com wrote:
> >> distance^2 = g_uv dx^u dx^v
>
> >The equation above is very correct. However, your interpretation is
> >illogical. The geometry is described by ds^2, and there is no way in
> >hell that g_uv can represent the geometry independent of the
> >coordinate system (dx^u dx^v). Your mathematics even indicate so.
> ><shrug>
> --------------------------------------------
> g_uvg^uv = number of dimensions
> ----------------------------------------------
>
> is MATHEMATICALLY INDEPENDENT of coordinate systems
> and ONLY the metric tensor has this property BECAUSE it is
> by construction the quantity which specifies completly the
> "geometry" of the space.
So is 6 * 1/6 = 1, and 7 * 1/7 = 1. <shrug>
> You're all through bozo... try growing a beard if you
> want to fool someone.
The reverend and Gisse the college drop-out, the troll, and the liar
are both extremely retarded. <shrug>
PD
> On Apr 9, 9:04 am, PD <TheDraperFam...@gmail.com> wrote:
>
> > On Apr 8, 11:06 pm, koobee.wub...@gmail.com wrote:
> > > > Of course not. A vector is tensor of rank {1 0}.
>
> > > Wrong again. A vector is not a matrix or a tensor of rank-1.
>
> > Neither is a tensor a matrix. That's precisely the point.
>
> You said a vector is a tensor.
A vector is not a matrix. Neither is a tensor a matrix.
A vector is indeed a tensor of rank {1 0}. Do you need one or several
math references regarding this, or are you content to tell yourself
you know better than all of them anyway, being an engineer and all?
> You sound like you are contradicting
> yourself.
I'm not. I explained further in the part that you declined to continue
reading.
Eric Gisse
[...]
> The reverend and Gisse the college drop-out, the troll, and the liar
> are both extremely retarded. <shrug>
It must then be extremely difficult for you to explain why the both of
us manage to show you as an idiot on a regular basis.
http://groups.google.com/group/sci.physics.relativity/browse_frm/thread/bda48e445c49364e
George Hammond
[Hammond]
Na, na, na..... 6*1/6 does NOT EQUAL 4 which is the
number of dimensions of spacetime.
You're missing Gisse's superb observation that the
equation:
g_uvg^uv=4=number of dimensions
is always true regardless of what willy nilly coordinates
are used. This fact PROVES that the metric written in
Cartesian coordinates must be IDENTICAL to the metric
written in Spherical coordinates..... thus proving that your
assertion to the contrary is nothing but a semantic snow job
with no physical meaning!
Jeff▲Relf
and I don't much care .. but, maybe, George knows his shit.
Mensanator
> Rarely do I understand what George Hammond is talking about,
> and I don't much care .. but, maybe, George knows his shit.
And then again, perhaps not. He never got his PhD.
George Hammond
<Jeff...@Seattle.inValid> wrote:
>Rarely do I understand what George Hammond is talking about,
>and I don't much care .. but, maybe, George knows his shit.
>
>
[Hammond]
Yuh... you're a fast kid who's not afraid of Physics and
you're a survivor.... that's why you're one of the first to
know it; same reason I discovered it!
Eventually the entire Physics community is going to find
out about it.... then we got a real problem... and I'm
talking about not giving the psychologists a nervous
breakdown, the Theologians hysteria or precipitating a
global Religious conflict.
So like said, keep the discovery of a scientific proof of
God under your hat until the Physics department can figure
out how to midwife the event. But... if Barack Obama can
step up to world peace, Physics ought to be able to handle
the scientific discovery of God.