Do Gravitational Waves Travel at the Speed of Light?

[Pentcho Valev]

Einstein's general relativity said nothing about the speed of gravitational waves - it was Eddington who discovered, in 1922, that some of them travel at the speed of light:

Arthur Eddington 1922: "The problem of the propagation of disturbances of the gravitational field was investigated by Einstein in 1916, and again in 1918. It has usually been inferred from his discussion that a change in the distribution of matter produces gravitational effects which are propagated with the speed of light; but I think that Einstein really left the question of the speed of propagation rather indefinite. His analysis shows how the co-ordinates must be chosen if it is desired to represent the gravitational potentials as propagated with the speed of light; but there is nothing to indicate that the speed of light appears in the problem, except as the result of this arbitrary choice. [...] Weyl has classified plane gravitational waves into three types, viz.: (1) longitudinal-longitudinal; (2)longitudinal-transverse; (3) transverse-transverse. The present investigation leads to the conclusion that transverse-transverse waves are propagated with the speed of light in all systems of co-ordinates. Waves of the first and second types have no fixed velocity - a result which rouses suspicion as to their objective existence." http://rspa.royalsocietypublishing.org/content/royprsa/102/716/268.full.pdf

Was Eddington's discovery correct? Why do Einsteinians teach that "gravitational waves travel at the speed of light" is a prediction of Einstein's general relativity and don't mention the true discoverer - Eddington?

Pentcho Valev

[Pentcho Valev]

Eddington's discovery that gravitational waves travel at the speed of light was actually his third major hoax. Here are the other two:

Sabine Hossenfelder: "As light carries energy and is thus subject of gravitational attraction, a ray of light passing by a massive body should be slightly bent towards it. This is so both in Newton's theory of gravity and in Einstein's, but Einstein's deflection is by a factor two larger than Newton's. [...] As history has it, Eddington's original data actually wasn't good enough to make that claim with certainty. His measurements had huge error bars due to bad weather and he also might have cherry-picked his data because he liked Einstein's theory a little too much. Shame on him." http://backreaction.blogspot.com/2015/04/a-wonderful-100th-anniversary-gift-for.html

Stephen Hawking: "Einsteins prediction of light deflection could not be tested immediately in 1915, because the First World War was in progress, and it was not until 1919 that a British expedition, observing an eclipse from West Africa, showed that light was indeed deflected by the sun, just as predicted by the theory. This proof of a German theory by British scientists was hailed as a great act of reconciliation between the two countries after the war. It is ionic, therefore, that later examination of the photographs taken on that expedition showed the errors were as great as the effect they were trying to measure. Their measurement had been sheer luck, or a case of knowing the result they wanted to get, not an uncommon occurrence in science." http://www.balajisebookworld.com/Ebooks/a.html

Discover Magazine: "The eclipse experiment finally happened in 1919. Eminent British physicist Arthur Eddington declared general relativity a success, catapulting Einstein into fame and onto coffee mugs. In retrospect, it seems that Eddington fudged the results, throwing out photos that showed the wrong outcome. No wonder nobody noticed: At the time of Einstein's death in 1955, scientists still had almost no evidence of general relativity in action." http://discovermagazine.com/2008/mar/20-things-you-didn.t-know-about-relativity

"Consider the case of astronomer Walter Adams. In 1925 he tested Einstein's theory of relativity by measuring the red shift of the binary companion of Sirius, brightest star in the sky. Einstein's theory predicted a red shift of six parts in a hundred thousand; Adams found just such an effect. A triumph for relativity. However, in 1971, with updated estimates of the mass and radius of Sirius, it was found that the predicted red shift should have been much larger – 28 parts in a hundred thousand. Later observations of the red shift did indeed measure this amount, showing that Adams' observations were flawed. He "saw" what he had expected to see." http://puritanreformed.blogspot.bg/2010/08/fallible-nature-of-supposed-objective.html

"In January 1924 Arthur Eddington wrote to Walter S. Adams at the Mt. Wilson Observatory suggesting a measurement of the "Einstein shift" in Sirius B and providing an estimate of its magnitude. Adams' 1925 published results agreed remarkably well with Eddington's estimate. Initially this achievement was hailed as the third empirical test of General Relativity (after Mercury's anomalous perihelion advance and the 1919 measurement of the deflection of starlight). It has been known for some time that both Eddington's estimate and Adams' measurement underestimated the true Sirius B gravitational redshift by a factor of four." http://adsabs.harvard.edu/abs/2010AAS...21530404H

"...Eddington asked Adams to attempt the measurement. [...] ...Adams reported an average differential redshift of nineteen kilometers per second, very nearly the predicted gravitational redshift. Eddington was delighted with the result... [...] In 1928 Joseph Moore at the Lick Observatory measured differences between the redshifts of Sirius and Sirius B... [...] ...the average was nineteen kilometers per second, precisely what Adams had reported. [...] More seriously damaging to the reputation of Adams and Moore is the measurement in the 1960s at Mount Wilson by Jesse Greenstein, J.Oke, and H.Shipman. They found a differential redshift for Sirius B of roughly eighty kilometers per second." http://adsabs.harvard.edu/full/1980QJRAS..21..246H

Pentcho Valev

[Pentcho Valev]

Even if gravitational waves existed (they don't), LIGO's neutron-star result (GW170817) would still be a fake unless:

(A) in 1922 Arthur Eddington correctly predicted that gravitational waves travel at the speed of light (Einstein's general relativity had said nothing about the speed of gravitational waves).

(B) the gravitational waves experience a delay equal to the Shapiro delay experienced by light.

Given the fact that Eddington was a serial fraudster, I don't think there is any hope for LIGO conspirators.

Pentcho Valev

[Koobee Wublee]

On Friday, December 8, 2017 at 12:40:37 AM UTC-8, Pentcho Valev wrote:

> Einstein's general relativity said nothing about the speed of
> gravitational waves - it was Eddington who discovered, in 1922, that
> some of them travel at the speed of light:

From the null Ricci scalar, one can find one Laplacian and three d’Alembertian equations associated with the weak curvature in spacetime. Of course, the Laplacian equation does not offer any propagating wave, but each of the d’Alembertian equations describe propagating waves. <shrug>

However, unlike the constant that define the speed of light from the permeability and the permittivity of free space deduced from the Maxwell equations, offers no insight on what is should be. It is just assumed to be the speed of light by Einstein dingleberries with no sane reasons. <shrug>

[Pentcho Valev]

The idiocies produced by Einsteinians are immeasurably more monumental than those produced by Flat Earthers:

"In relativity, there is a 'speed of information' - the maximum speed that you can send information from one point to another," says University of Wisconsin-Milwaukee physicist Jolien Creighton, an expert on general relativity and member of the LIGO team that first spotted gravitational waves. Creighton explains that in electromagnetism, when you shake an electron, it creates a change in the electric field that spreads out at the speed of light. Gravity works the same way. Shake a mass and the change in the gravitational field - the gravitational wave - propagates at that same speed. "So the fact that the speed of gravitational waves is equal to the speed of electromagnetic waves is simply because they both travel at the speed of information," Creighton says." http://blogs.discovermagazine.com/d-brief/2017/12/08/speed-of-gravity-light/#.WirbWFWWbIU

Pentcho Valev

[JanPB]

On Friday, December 8, 2017 at 10:52:53 AM UTC-8, Pentcho Valev wrote:
> The idiocies produced by Einsteinians are immeasurably more monumental than those produced by Flat Earthers:

Pentcho, just drop the subject. It's not your thing, you don't get it at all.
Why are you wasting your time on something you have no understanding of?

--
Jan

[Lara Ashline]

Actually you, provable, having a zero understanding in Math Modelling in
Physics. A left-behind. The world is discrete, my friend I love you so
much. I wonder whether you are capable going from a continuous domain over
to a discrete.

[JanPB]

Something's buzzing in my ear.

--
Jan

[Python]

Lara Ashline aka 'nym-shifting' troll:

Dear nym-shifting troll, you are a disgusting piece of shit making a
fool of yourself.

[Omar Shabsigh]

The thing some are talking about all the time: GRAVITATIONAL WAVES. Who said the gravity is a wave?? And why should gravity propagate. Gravity is a force. Do forces propagate or do they just exist? Can the knowledgeable Einsteinian give a good answer?

[Tom Roberts]

On 12/8/17 12/8/17 7:07 PM, Omar Shabsigh wrote:
> The thing some are talking about all the time: GRAVITATIONAL WAVES. Who said
> the gravity is a wave??

Nobody. But in GR, gravity is not a force, it is an aspect of the geometry, the
curvature of spaceTIME. All discussions of gravitational waves are in the
context of GR.

Note that gravity waves are completely different, appear in
Newtonian mechanics, and include such phenomena as ocean waves.
They are waves in continuous bodies drive by gravity.

> And why should gravity propagate.

It doesn't. At least not in GR. The geometry just is. But CHANGES in geometry
can and do propagate. Moreover, there is a class of such changes that are
self-propagating waves, which are known as gravitational waves (a minor
misnomer, as they are really waves in geometry).

> Gravity is a force.

What GOD whispered in your ear and told you this?

In our current best model of gravity, General Relativity, gravity is NOT a
force. See above.

> Do forces propagate or do they just exist?

That depends on the force (which does not include gravity -- see above).

Electromagnetic fields can and do support self-propagating waves, and these
waves can and do interact with charged particles as a force. But it takes both
field and charge to manifest as a force, so it is the FIELD which is waving, and
the WAVES which are propagating, not the force.

The weak and strong nuclear forces also propagate, in the sense that we model
them as "propagators" consisting of gauge bosons. They are very short range, ~
10^-15 meters, and we cannot tell whether there is any sort of wave associated
with them.

Tom Roberts

[Odd Bodkin]

Omar Shabsigh <omar.s...@gmail.com> wrote:
> On Friday, December 8, 2017 at 12:40:37 AM UTC-8, Pentcho Valev wrote:
>> Einstein's general relativity said nothing about the speed of
>> gravitational waves - it was Eddington who discovered, in 1922, that
>> some of them travel at the speed of light:
>>
>> Arthur Eddington 1922: "The problem of the propagation of disturbances
>> of the gravitational field was investigated by Einstein in 1916, and
>> again in 1918. It has usually been inferred from his discussion that a
>> change in the distribution of matter produces gravitational effects
>> which are propagated with the speed of light; but I think that Einstein
>> really left the question of the speed of propagation rather indefinite.
>> His analysis shows how the co-ordinates must be chosen if it is desired
>> to represent the gravitational potentials as propagated with the speed
>> of light; but there is nothing to indicate that the speed of light
>> appears in the problem, except as the result of this arbitrary choice.
>> [...] Weyl has classified plane gravitational waves into three types,
>> viz.: (1) longitudinal-longitudinal; (2)longitudinal-transverse; (3)
>> transverse-transverse. The present investigation leads to the conclusion
>> that transverse-transverse waves are propagated with the speed of light
>> in all systems of co-ordinates. Waves of the first and second types have
>> no fixed velocity - a result which rouses suspicion as to their
>> objective existence

>> " http://rspa.royalsocietypublishing.org/content/royprsa/102/716/268.full.pdf
>>
>> Was Eddington's discovery correct? Why do Einsteinians teach that
>> "gravitational waves travel at the speed of light" is a prediction of
>> Einstein's general relativity and don't mention the true discoverer - Eddington?
>>
>> Pentcho Valev
>
> The thing some are talking about all the time: GRAVITATIONAL WAVES. Who
> said the gravity is a wave?? And why should gravity propagate. Gravity is
> a force. Do forces propagate or do they just exist? Can the knowledgeable
> Einsteinian give a good answer?
>

Force is a an antiquated concept, actually. Gravity is an interaction, like
electromagnetism. Interactions have propagating carriers. Those propagators
often exhibit wave propagation.

--
Odd Bodkin -- maker of fine toys, tools, tables

[Koobee Wublee]

On Friday, December 8, 2017 at 3:56:56 PM UTC-8, JanPB wrote:

> On December 8, 2017 at 12:30:11 PM UTC-8, Lara Ashline wrote:

> > Actually you, provable, having a zero understanding in Math
> > Modelling in Physics. A left-behind. The world is discrete,
> > my friend I love you so much. I wonder whether you are capable
> > going from a continuous domain over to a discrete.
>
> Something's buzzing in my ear.

Do you not read the post you are responding to? Or do you have a mental breakdown? <shrug>

[Koobee Wublee]

On Friday, December 8, 2017 at 6:02:54 PM UTC-8, tjrob137 wrote:

> On 12/8/17, 7:07 PM, Omar Shabsigh wrote:

> > The thing some are talking about all the time: GRAVITATIONAL
> > WAVES. Who said the gravity is a wave??
>

> But in GR, gravity is not a force, it is an aspect of the
> geometry, the curvature of spaceTIME.

Let’s see. If there is no gravitational time dilation (g_00 = 1 instead of 1 – 2 G M / c² / r), there is no gravity regardless how much space is curved. In another words, only the gravitational time dilation manifests gravity and nothing else. With that said, the geodesic equations clearly spells out gravity is a real force acting on anything with non-zero mass. <shrug>

> > And why should gravity propagate.
>
> It doesn't. At least not in GR. The geometry just is. But CHANGES
> in geometry can and do propagate. Moreover, there is a class of
> such changes that are self-propagating waves, which are known as
> gravitational waves (a minor misnomer, as they are really waves
> in geometry).

Notice what Tom is saying is all an opinion with nothing backed up by the mathematics of GR. It is like how one sect interprets a religion versus another. That sometimes led to civil wars even if they all believe in the same god. <shrug>

> > Gravity is a force.
>
> What GOD whispered in your ear and told you this?

Certainly not your god Einstein the nitwit, the plagiarist, and the liar. <shrug>

> In our current best model of gravity, General Relativity, gravity
> is NOT a force.

See above. <shrug>

[Koobee Wublee]

On Friday, December 8, 2017 at 6:32:44 PM UTC-8, Odd Bodkin wrote:

> Force is a an antiquated concept, actually.

This is absolutely nonsense. See the previous post of Koobee Wublee in this thread addressing Tom. <shrug>

> Gravity is an interaction, like electromagnetism.

Interaction associates with a force. <shrug>

> Interactions have propagating carriers.

Not necessarily. <shrug>

> Those propagators often exhibit wave propagation.

PD needs to understand the basics better in order to write textbooks for the next generation of students. <shrug>

[mlwo...@wp.pl]

On Saturday, 9 December 2017 03:32:44 UTC+1, Odd Bodkin wrote:

> Force is a an antiquated concept, actually. Gravity is an interaction, like
> electromagnetism. Interactions have propagating carriers. Those propagators
> often exhibit wave propagation.

And it's true!!!
Because it's "confirmed differently".

[Pentcho Valev]

LIGO's gravitational waves are ripples in spacetime which doesn't exist:

Nima Arkani-Hamed (06:09): "Almost all of us believe that space-time doesn't really exist, space-time is doomed and has to be replaced by some more primitive building blocks." https://www.youtube.com/watch?v=U47kyV4TMnE

Nobel Laureate David Gross observed, "Everyone in string theory is convinced...that spacetime is doomed. But we don't know what it's replaced by." https://www.edge.org/response-detail/26563

What scientific idea is ready for retirement? Steve Giddings: "Spacetime. Physics has always been regarded as playing out on an underlying stage of space and time. Special relativity joined these into spacetime... [...] The apparent need to retire classical spacetime as a fundamental concept is profound..." https://www.edge.org/response-detail/25477

http://negrjp.fotoblog.uol.com.br/images/photo20150819051851.jpg

Pentcho Valev

[Lara Ashline]

Exactly, my old friend, he is trying hard hiding from the public his
incompetencies in Math Modelling etc, by "private terminology", "buzzing"
and so on, to stuff completely Greek to him. I bet he never participated
to conferences in Math Modelling in Physics, he doesn't even know what
those things are, lol. Not even a simple z-transform, I bet he rarely is
using it.

What a shame, for a mathematician, or for a one pretending to be. And for
his friends, as Python, Gary and probebly some other guys. lol, lol, lol
and lol.

Excellent post, Dr. Koobee.

[Odd Bodkin]

Koobee Wublee <koobee...@gmail.com> wrote:
> On Friday, December 8, 2017 at 6:32:44 PM UTC-8, Odd Bodkin wrote:
>
>> Force is a an antiquated concept, actually.
>
> This is absolutely nonsense. See the previous post of Koobee Wublee in
> this thread addressing Tom. <shrug>

Not much of anything you say is worth referring to, KW.

>
>> Gravity is an interaction, like electromagnetism.
>
> Interaction associates with a force. <shrug>

That may have been the way you learned it. It’s antiquated.

>
>> Interactions have propagating carriers.
>
> Not necessarily. <shrug>
>
>> Those propagators often exhibit wave propagation.
>
> PD needs to understand the basics better in order to write textbooks for
> the next generation of students. <shrug>

I have no interest in writing textbooks. I’m busy enough. Maybe PD will, I
don’t know.

[Lara Ashline]

Keep quite, stupid python. Greek for him, greek for you even more.

[mlwo...@wp.pl]

On Saturday, 9 December 2017 19:00:36 UTC+1, Odd Bodkin wrote:
> Koobee Wublee <koobee...@gmail.com> wrote:
> > On Friday, December 8, 2017 at 6:32:44 PM UTC-8, Odd Bodkin wrote:
> >
> >> Force is a an antiquated concept, actually.
> >
> > This is absolutely nonsense. See the previous post of Koobee Wublee in
> > this thread addressing Tom. <shrug>
>
> Not much of anything you say is worth referring to, KW.
>
> >
> >> Gravity is an interaction, like electromagnetism.
> >
> > Interaction associates with a force. <shrug>
>
> That may have been the way you learned it. It’s antiquated.

The EXPERIMENTS!!! The MEASUREMENTS!!! The WISE GURUS!!!!
They all say it's antiquated!!!

[Pentcho Valev]

"But a matter of far greater concern is the conduct of the mainstream physics establishment worldwide. A gigantic fraud was committed right in their midst and in their name, and not a single voice of dissent or criticism or even concern arose. They are in full solidarity with the fraudsters." https://dreamheron.wordpress.com/2017/11/12/ligo-a-man-who-tried-to-save-you-world/

That is normal - we all live in the post-truth world. There WERE mainstream critics but they are now unpersons:

"Withers, however, was already an unperson. He did not exist : he had never existed." https://ebooks.adelaide.edu.au/o/orwell/george/o79n/chapter1.4.html

"Einstein believed in neither gravitational waves nor black holes. [...] Dr Natalia Kiriushcheva, a theoretical and computational physicist at the University of Western Ontario (UWO), Canada, says that while it was Einstein who initiated the gravitational waves theory in a paper in June 1916, it was an addendum to his theory of general relativity and by 1936, he had concluded that such things did not exist. Furthermore - as a paper published by Einstein in the Annals of Mathematics in October, 1939 made clear, he also rejected the possibility of black holes. [...] On September 16, 2010, a false signal - a so-called "blind injection" - was fed into both the Ligo and Virgo systems as part of an exercise to "test ... detection capabilities". At the time, the vast majority of the hundreds of scientists working on the equipment had no idea that they were being fed a dummy signal. The truth was not revealed until March the following year, by which time several papers about the supposed sensational discovery of gravitational waves were poised for publication. "While the scientists were disappointed that the discovery was not real, the success of the analysis was a compelling demonstration of the collaboration's readiness to detect gravitational waves," Ligo reported at the time. But take a look at the visualisation of the faked signal, says Dr Kiriushcheva, and compare it to the image apparently showing the collision of the twin black holes, seen on the second page of the recently-published discovery paper. "They look very, very similar," she says. "It means that they knew exactly what they wanted to get and this is suspicious for us: when you know what you want to get from science, usually you can get it." The apparent similarity is more curious because the faked event purported to show not a collision between two black holes, but the gravitational waves created by a neutron star spiralling into a black hole. The signals appear so similar, in fact, that Dr Kiriushcheva questions whether the "true" signal might actually have been an echo of the fake, "stored in the computer system from when they turned off the equipment five years before"." http://www.thenational.ae/arts-life/the-review/why-albert-einstein-continues-to-make-waves-as-black-holes-collide#full

Natalia Kiriushcheva: "What is shown on this picture? The equation on this page is what will be left from Einstein's equations of General Relativity (GR) after linearization. i.e. after a certain assumption is imposed: the gravitational field is considered weak (is it a correct assumption for two black holes?). Moreover, this equation is similar to the wave equation of the Maxwell theory that (after some additional manipulations) describes propagation of electromagnetic waves in the absence of sources (absence of any source, including a system of two black holes!). Einstein pointed out in this paper that its result is not general, it is valid only under assumption that the gravitational field is weak and only linear coordinate transformations (a linearized version of the general coordinate transformations of GR) can be applied to these (linearized) equations. Einstein also did not predict in this paper "that two celestial bodies in orbit will generate invisible ripples in spacetime that experts call gravitational waves", as BI claims. He was talking about "the system" that radiates energy, without specifying what kind of system it is." https://gravityattraction.wordpress.com/2016/02/18/hard-evidence-of-einsteins-involvement/

James Creswell, Sebastian von Hausegger, Andrew D. Jackson, Hao Liu, Pavel Naselsky, June 27, 2017: "As a member of the LIGO collaboration, Ian Harry states that he "tried to reproduce the results quoted in 'On the time lags of the LIGO signals'", but that he "[could] not reproduce the correlations claimed in section 3". Subsequent discussions with Ian Harry have revealed that this failure was due to several errors in his code. After necessary corrections were made, his script reproduces our results. His published version was subsequently updated. [...] It would appear that the 7 ms time delay associated with the GW150914 signal is also an intrinsic property of the noise. The purpose in having two independent detectors is precisely to ensure that, after sufficient cleaning, the only genuine correlations between them will be due to gravitational wave effects. The results presented here suggest this level of cleaning has not yet been obtained and that the identification of the GW events needs to be re-evaluated with a more careful consideration of noise properties." http://www.nbi.ku.dk/gravitational-waves/gravitational-waves.html

James Creswell, Sebastian von Hausegger, Andrew D. Jackson, Hao Liu, Pavel Naselsky, August 21, 2017: "In view of unsubstantiated claims of errors in our calculations, we appreciated the opportunity to go through our respective codes together - line by line when necessary - until agreement was reached. This check did not lead to revisions in the results of calculations reported in versions 1 and 2 of arXiv:1706.04191 or in the version of our paper published in JCAP. It did result in changes to the codes used by our visitors [LIGO conspirators]. [...] In light of the above, our view should be clear: We believe that LIGO has not yet attained acceptable standards of data cleaning. Since we regard proof of suitable cleaning as a mandatory prerequisite for any meaningful comparison with specific astrophysical models of GW events, we continue to regard LIGO's claims of GW discovery as interesting but premature." http://www.nbi.ku.dk/gravitational-waves/gravitational-waves-comment2.html

Pentcho Valev

[Tom Roberts]

On 12/9/17 12:21 AM, Koobee Wublee wrote:
> On Friday, December 8, 2017 at 6:02:54 PM UTC-8, tjrob137 wrote:
>> But in GR, gravity is not a force, it is an aspect of the geometry, the
>> curvature of spaceTIME.
>

> If there is no gravitational time dilation (g_00 = 1 instead of 1 – 2 G M /
> c² / r), there is no gravity regardless how much space is curved.

Non sequitur. That is completely unrelated to what I said.

Moreover, it is just plain wrong, for any sensible meaning of "gravity".

You mention a coordinate-dependent component of the metric.
If "gravity" means that timelike geodesics are deflected by
a mass, then the value of g_00 is COMPLETELY IRRELEVANT,
because geodesics are independent of coordinates.
IOW: in Schwarzschild spacetime outside the horizon, an
appropriate coordinate transform from the usual Schw.
coordinates can give g_00 = 1; this does not affect the
geodesics, they are still deflected, and gravity is still
present.

> [... further nonsense]

Tom Roberts

[Koobee Wublee]

On Monday, December 11, 2017 at 8:12:06 AM UTC-8, tjrob137 wrote:
> On 12/9/17 12:21 AM, Koobee Wublee wrote:

> > If there is no gravitational time dilation (g_00 = 1 instead of
> > 1 – 2 G M / c² / r), there is no gravity regardless how much
> > space is curved.
>

> You mention a coordinate-dependent component of the metric.

Koobee Wublee always mean the metric is coordinate dependent --- even at elements level. <shrug>

** geometry = metric * coordinate

Where

** geometry = reality, invariant
** coordinate = coordinate system to describe reality
** metric = connection between reality and observation

This is grade school geometry. How come there are so many idiots not able to understand this in the academics? <shrug>

> > In another words, only the gravitational time dilation manifests
> > gravity and nothing else.
>

> If "gravity" means that timelike geodesics are deflected by a mass,
> then the value of g_00 is COMPLETELY IRRELEVANT, because geodesics
> are independent of coordinates.

The term (1 – 2 GM/c² / r) must be invariant since gravitational time dilation must be invariant. <shrug>

> > With that said, the geodesic equations clearly spells out gravity
> > is a real force acting on anything with non-zero mass. <shrug>
>

> in Schwarzschild spacetime outside the horizon, an appropriate
> coordinate transform from the usual Schw. coordinates can give
> g_00 = 1; this does not affect the geodesics, they are still
> deflected, and gravity is still present.

First of all, there is no standard definition as the Schwarzschild coordinate system. Almost of all description is done in the well-defined polar coordinate system. <shrug>

It is hopeless to find a time measurement that will violate the invariance in the quantity (1 – 2 GM/c² / r) in polar coordinate system or (1 – 2 GM/c² / √(x²+y²+z²)) in Cartesian coordinate system. <shrug>

Finally, a photon always deflect the same amount regardless of which coordinate system one chooses to describe the mathematical model of photon deflection. This is all grade school math, Tom. <shrug>

[JanPB]

On Monday, December 11, 2017 at 10:02:29 AM UTC-8, Koobee Wublee wrote:
> On Monday, December 11, 2017 at 8:12:06 AM UTC-8, tjrob137 wrote:
> > On 12/9/17 12:21 AM, Koobee Wublee wrote:
>
> > > If there is no gravitational time dilation (g_00 = 1 instead of
> > > 1 – 2 G M / c² / r), there is no gravity regardless how much
> > > space is curved.
> >
> > You mention a coordinate-dependent component of the metric.
>
> Koobee Wublee always mean the metric is coordinate dependent --- even at elements level. <shrug>

This is twice removed from the correct GR:

1. metric is a bilinear map (of tangent vectors), i.e. it's independent
of coordinates by definition,

2. the _geometry_ of spacetime (i.e. the physics) is NOT the same as
a metric tensor anyway. For example, the following two metric tensors written
with respect to one and the same coordinate system (denoted by (x,y), say)
define the same geometry, and therefore the same physics:

ds^2 = dx^2 + dy^2

and:

ds^2 = dx^2 + 42 dy^2

Two different tensors, same geometry (flat in this case).

> ** geometry = metric * coordinate

Gobbledygook. (Does it mean that metric = geometry/coordinate?)

> Where
>
> ** geometry = reality, invariant
> ** coordinate = coordinate system to describe reality
> ** metric = connection between reality and observation

The third item is gobbledygook.

> This is grade school geometry.

Virtually EVERYTHING Koobee says is WRONG.

> How come there are so many idiots not able to understand this in the academics? <shrug>

"Real knowledge is to know the extent of one's ignorance" -Confucius

> > > With that said, the geodesic equations clearly spells out gravity
> > > is a real force acting on anything with non-zero mass. <shrug>
> >
> > in Schwarzschild spacetime outside the horizon, an appropriate
> > coordinate transform from the usual Schw. coordinates can give
> > g_00 = 1; this does not affect the geodesics, they are still
> > deflected, and gravity is still present.
>
> First of all, there is no standard definition as the Schwarzschild coordinate system. Almost of all description is done in the well-defined polar coordinate system. <shrug>

It's not polar as in "polar coordinates in R^3". The mistake here
is ignoring the fact that solving the EFE assuming the coordinates
(t,r,theta,phi) is ONLY assuming that: i.e. using 3 smooth functions
satisfying certain spherical symmetry constraint, and nothing else
(the mental reflex upon writing things like "(t,r,theta,phi)" is to
assume those coordinates actually mimic ALL aspects of the standard
polar coordinates, like r=0 being a point, etc. etc.)

The rest, which is an absolutely PATHETIC display of Koobee attempting
to "teach" Tom, is hereby mercifully snipped.

Again, I'm not going to discuss this with you, this post is not addressed
to you.

--
Jan

[Lara Ashline]

JanPB wrote:

>> ** geometry = metric * coordinate
>
> Gobbledygook. (Does it mean that metric = geometry/coordinate?)

That's units, not coordinate. But Dr. Koobee is still consistent,
since the [units] supposes coordinates.

For instance, you provable are incompetent in physics one more time

How do I refine a subdomain mesh?
https://se.mathworks.com/matlabcentral/answers/354266-how-do-i-refine-a-subdomain-mesh?requestedDomain=www.mathworks.com

[David (Kronos Prime) Fuller]

JanPB

Without a Coordinate System is just Bulk Modulus = v^2/c^2

Truncated Octahedral Tessellation

http://www.wolframalpha.com/input/?i=cos(x)%2Bcos(y)%2Bcos(z)+%3D+0

cos(x)+cos(y)+cos(z) = 0

e^(-i x)/2 + e^(i x)/2 + e^(-i y)/2 + e^(i y)/2 + e^(-i z)/2 + e^(i z)/2 = 0


http://hyperphysics.phy-astr.gsu.edu/hbase/permot3.html

http://hyperphysics.phy-astr.gsu.edu/hbase/Sound/souspe2.html#c1

http://hyperphysics.phy-astr.gsu.edu/hbase/Sound/imgsou/soundwater.gif

(((6.6740800122e-11)/2 (newtons / (meters^2))) / (3.71295775e-28 (kg / (meter^3))))^0.5 = 299792458 m / s

[Koobee Wublee]

On December 11, 2017, Jan Bielawski the queen of England wrote:

> On December 11, 2017 at 10:02:29 AM UTC-8, Koobee Wublee wrote:

> > Koobee Wublee always mean the metric is coordinate dependent ---
> > even at elements level. <shrug>
>

> 1. metric is a bilinear map (of tangent vectors), i.e. it's
> independent of coordinates by definition,

In the following post, Koobee Wublee has shown you how to derive the metric based on a known coordinate system. <shrug>

https://groups.google.com/d/msg/sci.physics.relativity/o7CSECN9_oI/280vMLFGBwAJ

Clearly, the metric is coordinate dependent to satisfy the invariance in the geometry. Jan is lying once again. <shrug>

> 2. the _geometry_ of spacetime (i.e. the physics) is NOT the same

> as a metric tensor anyway...

Gobbledygook. <shrug>

> > ** geometry = metric * coordinate
>

> > Where
> >
> > ** geometry = reality, invariant
> > ** coordinate = coordinate system to describe reality
> > ** metric = connection between reality and observation
>

> Does it mean that metric = geometry/coordinate?

No, the operator is a dot product of two matrices defined and explained in the post above. <shrug>

> > First of all, there is no standard definition as the
> > Schwarzschild coordinate system. Almost of all description is
> > done in the well-defined polar coordinate system. <shrug>
>
> It's not polar as in "polar coordinates in R^3". The mistake here
> is ignoring the fact that solving the EFE assuming the coordinates

> (t,r,theta,phi) is ONLY assuming that...

Absolutely nonsense. <shrug>

> [rest of gobbledygook snipped]

[David (Kronos Prime) Fuller]

Honorable Judge Wublee is Correct

[JanPB]

On Monday, December 11, 2017 at 11:46:19 AM UTC-8, Lara Ashline wrote:
> JanPB wrote:
>
> >> ** geometry = metric * coordinate
> >
> > Gobbledygook. (Does it mean that metric = geometry/coordinate?)
>
> That's units, not coordinate.

No, it's not "units" either. (I know what Koobee is trying to say, it's
that he doesn't know enough to express himself properly to he fakes it
by hacking his way through).

> But Dr. Koobee is still consistent,
> since the [units] supposes coordinates.
>
> For instance, you provable are incompetent in physics one more time

"I am the queen of England." See how easy it is just to SAY anything? It's
a bit harder to back it up, isn't it.

> How do I refine a subdomain mesh?
> https://se.mathworks.com/matlabcentral/answers/354266-how-do-i-refine-a-subdomain-mesh?requestedDomain=www.mathworks.com

Whatever. You are changing the subject again.

--
Jan

[JanPB]

On Monday, December 11, 2017 at 12:52:50 PM UTC-8, Koobee Wublee wrote:
> On December 11, 2017, Jan Bielawski the queen of England wrote:
> > On December 11, 2017 at 10:02:29 AM UTC-8, Koobee Wublee wrote:
>
> > > Koobee Wublee always mean the metric is coordinate dependent ---
> > > even at elements level. <shrug>
> >
> > 1. metric is a bilinear map (of tangent vectors), i.e. it's
> > independent of coordinates by definition,
>
> In the following post, Koobee Wublee has shown you how to derive the metric based on a known coordinate system. <shrug>
>
> https://groups.google.com/d/msg/sci.physics.relativity/o7CSECN9_oI/280vMLFGBwAJ
>
> Clearly, the metric is coordinate dependent to satisfy the invariance in the geometry. Jan is lying once again. <shrug>

Metric is a tensor. This means coordinate-independent, by definition.

> > 2. the _geometry_ of spacetime (i.e. the physics) is NOT the same
> > as a metric tensor anyway...
>
> Gobbledygook. <shrug>
>
> > > ** geometry = metric * coordinate
> >
> > > Where
> > >
> > > ** geometry = reality, invariant
> > > ** coordinate = coordinate system to describe reality
> > > ** metric = connection between reality and observation
> >
> > Does it mean that metric = geometry/coordinate?
>
> No, the operator is a dot product of two matrices defined and explained in the post above. <shrug>

It's nonsense.

> > > First of all, there is no standard definition as the
> > > Schwarzschild coordinate system. Almost of all description is
> > > done in the well-defined polar coordinate system. <shrug>
> >
> > It's not polar as in "polar coordinates in R^3". The mistake here
> > is ignoring the fact that solving the EFE assuming the coordinates
> > (t,r,theta,phi) is ONLY assuming that...
>
> Absolutely nonsense. <shrug>

Nervertheless, this is how it is. You cannot change it (neither can I).

--
Jan

[Koobee Wublee]

On December 11, 2017, Jan Bielawski the queen of England wrote:
> Koobee Wublee wrote:

> > Koobee Wublee always mean the metric is coordinate dependent ---
> > even at elements level. <shrug>
>

> > https://groups.google.com/d/msg/sci.physics.relativity/o7CSECN9_oI/280vMLFGBwAJ
>
> > Clearly, the metric is coordinate dependent to satisfy the

> > invariance in the geometry. <shrug>

>
> Metric is a tensor. This means coordinate-independent, by definition.

Clearly, Jan does not understand basic geometry. Oh, there is no way to write down the differential equations of the field equations without deciding on a set of coordinate system. Does Jan wish to present otherwise? Please don’t use your dick again. <shrug>

> > ** geometry = metric * coordinate
>
> > Where
>
> > ** geometry = reality, invariant
> > ** coordinate = coordinate system to describe reality
> > ** metric = connection between reality and observation
>

> > the operator is a dot product of two matrices defined and
> > explained in the post above. <shrug>
>
> It's nonsense.

What is nonsense? <shrug>

> > there is no standard definition as the Schwarzschild coordinate
> > system. Almost of all description is done in the well-defined
> > polar coordinate system. <shrug>
>

> Nervertheless, this is how it is. You cannot change it (neither
> can I).

You got that right. We can never change Jan’s ignorance and stupidity. <shrug>

[Koobee Wublee]

On December 11, 2017, Jan Bielawski the queen of England wrote:

> You are changing the subject again.

Koobee Wublee is still waiting for Jan to explain how the Lagrangian is derived to derive the Christoffel symbols. It is clearly Jan does not understand the basics of GR. <shrug>

[Odd Bodkin]

Koobee Wublee <koobee...@gmail.com> wrote:
> On December 11, 2017, Jan Bielawski the queen of England wrote:
>> On December 11, 2017 at 10:02:29 AM UTC-8, Koobee Wublee wrote:
>
>>> Koobee Wublee always mean the metric is coordinate dependent ---
>>> even at elements level. <shrug>
>>
>> 1. metric is a bilinear map (of tangent vectors), i.e. it's
>> independent of coordinates by definition,
>
> In the following post, Koobee Wublee has shown you how to derive the
> metric based on a known coordinate system. <shrug>

You don’t have to derive a metric. Idiot.

>
> https://groups.google.com/d/msg/sci.physics.relativity/o7CSECN9_oI/280vMLFGBwAJ
>
> Clearly, the metric is coordinate dependent to satisfy the invariance in
> the geometry. Jan is lying once again. <shrug>
>
>> 2. the _geometry_ of spacetime (i.e. the physics) is NOT the same
>> as a metric tensor anyway...
>
> Gobbledygook. <shrug>
>
>>> ** geometry = metric * coordinate
>>
>>> Where
>>>
>>> ** geometry = reality, invariant
>>> ** coordinate = coordinate system to describe reality
>>> ** metric = connection between reality and observation
>>
>> Does it mean that metric = geometry/coordinate?
>
> No, the operator is a dot product of two matrices defined and explained
> in the post above. <shrug>
>
>>> First of all, there is no standard definition as the
>>> Schwarzschild coordinate system. Almost of all description is
>>> done in the well-defined polar coordinate system. <shrug>
>>
>> It's not polar as in "polar coordinates in R^3". The mistake here
>> is ignoring the fact that solving the EFE assuming the coordinates
>> (t,r,theta,phi) is ONLY assuming that...
>
> Absolutely nonsense. <shrug>
>
>> [rest of gobbledygook snipped]
>

[JanPB]

On Monday, December 11, 2017 at 5:31:02 PM UTC-8, Koobee Wublee wrote:
> On December 11, 2017, Jan Bielawski the queen of England wrote:
> > Koobee Wublee wrote:
>
> > > Koobee Wublee always mean the metric is coordinate dependent ---
> > > even at elements level. <shrug>
> >
> > > https://groups.google.com/d/msg/sci.physics.relativity/o7CSECN9_oI/280vMLFGBwAJ
> >
> > > Clearly, the metric is coordinate dependent to satisfy the
> > > invariance in the geometry. <shrug>
> >
> > Metric is a tensor. This means coordinate-independent, by definition.
>
> Clearly, Jan does not understand basic geometry. Oh, there is no way to write down the differential equations of the field equations without deciding on a set of coordinate system.

False.

> Does Jan wish to present otherwise? Please don’t use your dick again. <shrug>

I'm rather expensive as a tutor.

--
Jan

[mlwo...@wp.pl]

On Monday, 11 December 2017 20:36:59 UTC+1, JanPB wrote:


> 1. metric is a bilinear map (of tangent vectors), i.e. it's independent
> of coordinates by definition,

And how does your bunch of idiots measure it?

[Koobee Wublee]

On December 11, 2017, Jan Bielawski the queen of England wrote:
> Koobee Wublee wrote:

> > Koobee Wublee always mean the metric is coordinate dependent ---
> > even at elements level. <shrug>
>
> > https://groups.google.com/d/msg/sci.physics.relativity/o7CSECN9_oI/280vMLFGBwAJ
>
> > Clearly, the metric is coordinate dependent to satisfy the
> > invariance in the geometry. <shrug>
>

> > there is no way to write down the differential equations of the
> > field equations without deciding on a set of coordinate system.
>
> False.

What is false? <shrug>

> > Does Jan wish to present otherwise? Please don’t use your
> > dick again. <shrug>
>
> I'm rather expensive as a tutor.

Don’t make Koobee Wublee sick. Nobody is asking you to play a tutor using your dick. <shrug>

> > ** geometry = metric * coordinate
>
> > Where
>
> > ** geometry = reality, invariant
> > ** coordinate = coordinate system to describe reality
> > ** metric = connection between reality and observation
>
> > the operator is a dot product of two matrices defined and
> > explained in the post above. <shrug>

The analogy here is:

** 56 = 7 x 8

Where

** 56 plays the geometry
** 7 plays the metric
** 8 plays the coordinate system

56 can also be written as follows. <shrug>

** 56 = 14 x 4

Where

** 14 plays the metric
** 4 plays the coordinate system

Clearly, 7 is not 14, but Jan the queen of England is claiming it does. Why are Einstein dingleberries such fvcking stupid? <shrug>

[Koobee Wublee]

On Monday, December 11, 2017 at 6:23:39 PM UTC-8, Odd Bodkin wrote:

> Koobee Wublee wrote:

> > In the following post, Koobee Wublee has shown you how to derive
> > the metric based on a known coordinate system. <shrug>
>
> You don’t have to derive a metric.

Idiot! <shrug>

> > https://groups.google.com/d/msg/sci.physics.relativity/o7CSECN9_oI/280vMLFGBwAJ

The above post derives one such metric, imbecile! <shrug>

> > ** geometry = metric * coordinate
>
> > Where
>
> > ** geometry = reality, invariant
> > ** coordinate = coordinate system to describe reality
> > ** metric = connection between reality and observation
>

> > the operator is a dot product of two matrices defined and
> > explained in the post above. <shrug>

Chew on that, moron! <shrug>

[JanPB]

All nonsense. Sorry.

--
Jan

[Odd Bodkin]

I have. It’s wrong. Repeating that you’ve made an incorrect claim does not
make it right. The fact that it’s still on the internet (no one cares) does
not make it less wrong.

[numbernu...@gmail.com]

The Caltech-MIT LIGO detected stellar black hole gravitational waves using relativity (SPACE-TIME) to alter the armature length of Michelson interferometer but the LIGO experiment is based on a constant magnitude of the translational velocity but Einstein's translationtal velocity formed by the earth's daily and yearly motions is not constant. At the surface of the earth, for the time of 6:00 pm, the magnitude of the earth's tangential velocity vector that forms Einstein's translation velocity is 462 m/s (fig 7) and increases to 5,077 m/s at 7:00 pm. At midnight, the translational velocity is 30,462 m/s. The magnitude of Einstein's translational velocity increases from 462 m/s to 30,462 m/s (6:00 pm - 12:00 am) yet the LIGO experiment is based on the constant magnitude of Einstein's translational velocity to form the alteration of the armature length to justify the existence of LIGO gravitational waves. Pentcho Valev mY RUssian friend stop putting in the links which makes it difficult to read.

WEEP

[numbernu...@gmail.com]

Do you need me to explain how the translational velocity affects TIME-SPACE?


or do you need your mommy to hold your collective hands while you take a peeeeeeeeeeeeeeeeeeeeeee.

[numbernu...@gmail.com]

What he is trying to say is what I have posted. Also, all of you must try to write longer sentences that is related to the previous sentence and tell a story so that you and the reader can understand it. Trying to be smart using muddled mathematics in not fooling anyone unless you are a bed wetter or was previously.

[numbernu...@gmail.com]

They're doing the same thing using Michelson's interferometer to justify the ether then since they got away with that using the interferometer to justify gravitational waves. Don't you get it?

[Carl Susumu]

RE: CERN


In the CERN (2017) particle physics experiment, two hadron beams are collided to form subatomic particles but a proton has a mass 1,888 time larger than an electron and have a charge. The positive charges of the protons of the hadron beam would limit the concentration of protons in the hadron beam since like charge protons would repeal. It is unlikely that a high enough concentration of protons can produce a proton beam that could result in the collision of accelerated proton of the hadron beam. A collision implies a particle beam interacting with a metal structure. There is no physical example of an electron, nuclear or proton beam that interaction produces a collision of the entities within the beams when the beams are combined. When two particle beams are combine at perpendicular angles there is never an interaction that depicts a collision of the particles within the beams is occurring. In addition, it does not appear realistic that high energy protons of the hadron beam can be both accelerated and made to propagate on a exact circular path of the 27 km CERN vacuum tube since as the protons propagate through the vacuum tube the velocity of the protons increase which would require magnets that increasing magnetic fields are pointing towards the center of the circular 27 km vacuum tube. The size of the magnets would increase alone the path of the 27 km CERN vacuum tube to produce the increasing magnetic fields required to kept the protons contain in the 27 km circumference path of the CERN vacuum tube but the magnets along the vacuum tube appear to have the same size. Also, when the protons interact during the collision it is more likely that the protons would repel rather then collide and form subatomic particles. The high energy protons of the proton beam are propagating at the velocity of .999c that energy can cut steel which would make it very difficult to contain the CERN hadron beam within the 27 km circular vacuum tube without the protons contacting the outer surface of the vacuum tube enclosure producing a hole in the vacuum tube and eliminating the vacuum. The high energy protons that are propagating through the circular accelerator vacuum tube would have the energy to cut steel which can be tested by placing a .1 thick steel plate in that path of the accelerated proton beam. The accelerated protons at the end of the path through the vacuum tube would produces a hole in the steel test plate is placed in the path of the proton beam when the hadron beam accelerator is activated for approximately 20 seconds a hole would be formed in the steel test plate. Also, when the two hadron beams are combine a 300 $ (used) Geiger counter could be used to determine if the CERN experiment is accurately producing subatomic particles. The ATLAS experiment and Compact Muon Solenoid are the detectors that are used to detect the subatomic particles of the CERN accelerator collision. Also, the Compact Muon Solenoid is a metal structure that weighs 14,000 lb that resemble the Muon detector used the Soudan mine Muon experiment.



Pentcho Valev

[Koobee Wublee]

On December 12, 2017, Jan Bielawski the queen of England wrote:

> On December 11, 2017 at 10:25:20 PM UTC-8, Koobee Wublee wrote:

> > Koobee Wublee always mean the metric is coordinate dependent ---
> > even at elements level. <shrug>
>
> > https://groups.google.com/d/msg/sci.physics.relativity/o7CSECN9_oI/280vMLFGBwAJ
>
> > Clearly, the metric is coordinate dependent to satisfy the
> > invariance in the geometry. <shrug>
>
> > there is no way to write down the differential equations of the
> > field equations without deciding on a set of coordinate system.
>

> > ** geometry = metric * coordinate
>
> > Where
>
> > ** geometry = reality, invariant
> > ** coordinate = coordinate system to describe reality
> > ** metric = connection between reality and observation
>
> > the operator is a dot product of two matrices defined and
> > explained in the post above. <shrug>
>

> All nonsense. Sorry.

Let’s remind the readers how Jan the self-proclaimed queen of England does basic mathematics. Consider a flat geometry described using the Cartesian and the polar coordinate systems. <shrug>

** ds² = [A] * d²[q1] = [B] * d²[q2]

Where

** ds² = flat geometry
** d²[q1] = matrix[(dx dx,dx dy,dx dz),(dy dx,dy dy,dy dz),(dz dx,dz dy,dz dz)]
** d²[q2] = matrix[(dr dr,dr dθ,dr dφ),(dy dr,dθ dθ,dθ dφ),(dφ dr,dφ dθ,dφ dφ)]
** [A] = matrix[(1,0,0),(0,1,0),(0,0,1)]
** [B] = matrix[(1,0,0),(0,r² cos²(θ),0),(0,0,r²)]
** [a] * [b] = a_ij b_ij, dot product of matrices [a] and [b]

Clearly, the matrices d²[q1] and d²[q2] are different due to two distinct and very different coordinate systems. Then, mathematically without any doubt, the other two matrices [A] and [B] are also different despite serving as the metrics in describing the same type of geometry ds². <shrug>

Insisting on [A] equaling to [B], who can Jan Bielawski the self-proclaimed queen of England fool? The fools can only be proper idiots, spacetime imbeciles, relativistic morons, and Einstein dingleberries. These are all con-artists who cannot do basic math. <shrug>

[Edwin Huckabee]

Koobee Wublee wrote:

> Let’s remind the readers how Jan the self-proclaimed queen of England
> does basic mathematics. Consider a flat geometry described using the
> Cartesian and the polar coordinate systems. <shrug>
>
> ** ds² = [A] * d²[q1] = [B] * d²[q2]
>
> Where
>
> ** ds² = flat geometry ** d²[q1] = matrix[(dx dx,dx dy,dx dz),(dy
> dx,dy dy,dy dz),(dz dx,dz dy,dz dz)]
> ** d²[q2] = matrix[(dr dr,dr dθ,dr dφ),(dy dr,dθ dθ,dθ dφ),(dφ dr,dφ
> dθ,dφ dφ)]
> ** [A] = matrix[(1,0,0),(0,1,0),(0,0,1)]
> ** [B] = matrix[(1,0,0),(0,r² cos²(θ),0),(0,0,r²)]
> ** [a] * [b] = a_ij b_ij, dot product of matrices [a] and [b]

Fully consistent, to me.

[Paparios]

As you are a divergent moron troll, of course morons do agree...

[Koobee Wublee]

On Wednesday, December 13, 2017 at 5:32:38 PM UTC-8, Paparios wrote:

Only a proper idiot, spacetime imbecile, covariant nincompoop, relativistic moron, or Einstein dingleberry has problem understanding the basic math. <shrug>

[Edwin Huckabee]

I couldn't say it better by myself. Thanks. It appears to me, these self-
established wannabe relativists are severely illiterates in physics, math
modeling and distributed computation. Domains which constitutes a MUST
before starting pretending understanding Relativity, as a branch of
science.

Funny, all they have are "observations", but no theory capable to explain
it satisfactory.

[Paparios]

Au contraire, moron, liar and ignorant janitor. For years you have demonstrated your absolute lack of logic and mathematic abilities. You are, for years now, a waste of time, the same way as Kenseto and your sucker friend the nymshifting troll..<shrug><shrug>

[Paparios]

You are just a waste of time. Go on and continue to diverge troll and sucking the janitor troll.

[mlwo...@wp.pl]

Said a poor idiot rejecting pythagorean theorem.

[Paparios]

I forgot this other succker troll

[David (Kronos Prime) Fuller]

(Au contraire, moron, liar and ignorant janitor.) LOL ...

[Koobee Wublee]

On Thursday, December 14, 2017 at 5:37:02 AM UTC-8, Paparios wrote:
> Koobee Wublee wrote:

> > Koobee Wublee always mean the metric is coordinate dependent ---
> > even at elements level. <shrug>
>
> > https://groups.google.com/d/msg/sci.physics.relativity/o7CSECN9_oI/280vMLFGBwAJ
>
> > Clearly, the metric is coordinate dependent to satisfy the
> > invariance in the geometry. <shrug>
>
> > there is no way to write down the differential equations of the
> > field equations without deciding on a set of coordinate system.
>
> > ** geometry = metric * coordinate
>
> > Where
>
> > ** geometry = reality, invariant
> > ** coordinate = coordinate system to describe reality
> > ** metric = connection between reality and observation
>
> > the operator is a dot product of two matrices defined and
> > explained in the post above. <shrug>
>

> > Consider a flat geometry described using the Cartesian and the
> > polar coordinate systems. <shrug>
>
> > ** ds² = [A] * d²[q1] = [B] * d²[q2]
>
> > Where
>
> > ** ds² = flat geometry
> > ** d²[q1] = matrix[(dx dx,dx dy,dx dz),(dy dx,dy dy,dy dz),(dz dx,dz dy,dz dz)]
> > ** d²[q2] = matrix[(dr dr,dr dθ,dr dφ),(dy dr,dθ dθ,dθ dφ),(dφ dr,dφ dθ,dφ dφ)]
> > ** [A] = matrix[(1,0,0),(0,1,0),(0,0,1)]
> > ** [B] = matrix[(1,0,0),(0,r² cos²(θ),0),(0,0,r²)]
> > ** [a] * [b] = a_ij b_ij, dot product of matrices [a] and [b]
>

> > Clearly, the matrices d²[q1] and d²[q2] are different due to two
> > distinct and very different coordinate systems. Then,
> > mathematically without any doubt, the other two matrices [A] and
> > [B] are also different despite serving as the metrics in

> > describing the same type of geometry ds². <shrug>

>
> Au contraire, moron, liar and ignorant janitor.

That is not proper to address the Einstein dingleberries including yourself. Please use terms like proper idiots, spacetime imbeciles, covariant nincompoops, or relativistic morons. <shrug>

> For years you have demonstrated your absolute lack of logic and
> mathematic abilities.

Where is your justification for that? <shrug>

> You are, for years now, a waste of time,

For years, you are able to learn and be demystified. <shrug>

> the same way as Kenseto and your sucker friend the nymshifting
> troll..<shrug><shrug>

The nymshifting troll PD aka Odd Bodkin is not Koobee Wublee’s friend. Your babbling here does not contain a single sentence that is anything truthful. <shrug>

[Paparios]

<shrug><shrug>

[Edwin Huckabee]

Paparios wrote:

>> > Only a proper idiot, spacetime imbecile, covariant nincompoop,
>> > relativistic moron, or Einstein dingleberry has problem understanding
>> > the basic math. <shrug>
>>
>> I couldn't say it better by myself. Thanks. It appears to me, these
>> self-
>> established wannabe relativists are severely illiterates in physics,
>> math modeling and distributed computation. Domains which constitutes a
>> MUST before starting pretending understanding Relativity, as a branch
>> of science. Funny, all they have are "observations", but no theory
>> capable to explain it satisfactory.
>
> You are just a waste of time. Go on and continue to diverge troll and
> sucking the janitor troll.

This must be an attempt to elude your frustration, NOT living in an
industrial civilized country. Whatever diploma you might have, it doesn't
reflects education and everything.

[JanPB]

On Wednesday, December 13, 2017 at 3:55:43 PM UTC-8, Koobee Wublee wrote:
> On December 12, 2017, Jan Bielawski the queen of England wrote:
> > On December 11, 2017 at 10:25:20 PM UTC-8, Koobee Wublee wrote:
>
> > > Koobee Wublee always mean the metric is coordinate dependent ---
> > > even at elements level. <shrug>
> >
> > > https://groups.google.com/d/msg/sci.physics.relativity/o7CSECN9_oI/280vMLFGBwAJ
> >
> > > Clearly, the metric is coordinate dependent to satisfy the
> > > invariance in the geometry. <shrug>
> >
> > > there is no way to write down the differential equations of the
> > > field equations without deciding on a set of coordinate system.
> >
> > > ** geometry = metric * coordinate
> >
> > > Where
> >
> > > ** geometry = reality, invariant
> > > ** coordinate = coordinate system to describe reality
> > > ** metric = connection between reality and observation
> >
> > > the operator is a dot product of two matrices defined and
> > > explained in the post above. <shrug>
> >
> > All nonsense. Sorry.
>
> Let’s remind the readers how Jan the self-proclaimed queen of England does basic mathematics. Consider a flat geometry described using the Cartesian and the polar coordinate systems. <shrug>
>
> ** ds² = [A] * d²[q1] = [B] * d²[q2]

Part of Koobee's problem is that his grasp of this is very feeble and he
invents his private contorted notation for everything. This is done in order to
compensate for his lack of understanding the subject. One of such contortions
is exemplified above by the bizarre square-brackets notation invented by him
in order to shoehorn his (false) mantra "tensors are matrices" into the
picture.

> Where
>
> ** ds² = flat geometry
> ** d²[q1] = matrix[(dx dx,dx dy,dx dz),(dy dx,dy dy,dy dz),(dz dx,dz dy,dz dz)]
> ** d²[q2] = matrix[(dr dr,dr dθ,dr dφ),(dy dr,dθ dθ,dθ dφ),(dφ dr,dφ dθ,dφ dφ)]

This is also a cuckoo notation. Incompetence.

> ** [A] = matrix[(1,0,0),(0,1,0),(0,0,1)]
> ** [B] = matrix[(1,0,0),(0,r² cos²(θ),0),(0,0,r²)]
> ** [a] * [b] = a_ij b_ij, dot product of matrices [a] and [b]

"Dot product of matrices" is another sign of incompetence.

> Clearly, the matrices d²[q1] and d²[q2] are different due to two distinct and very different coordinate systems. Then, mathematically without any doubt, the other two matrices [A] and [B] are also different despite serving as the metrics in describing the same type of geometry ds². <shrug>
>
> Insisting on [A] equaling to [B], who can Jan Bielawski the self-proclaimed queen of England fool?

You keep misquoting what I say. Whether you are intentionally LYING or are
simply possessed of an extraordinarily dissipated thinking I cannot tell.

So, here is what I in fact said:

1. Two _different_ metric _tensors_ can be different yet represent the same
geometry.

2. The _same_ metric _tensor_ written in terms of two different coordinate
systems will have different coefficients in each. These coefficients for
rank-2 tensors can be conveniently displayed as a table of numbers (a matrix).

A while ago I posted a graded sequence of examples illustrating this, and
much more: https://groups.google.com/forum/#!msg/sci.physics.relativity/h49c7pfteQQ/jrJqiL7XBAAJ

3. Whenever some fool says things that are easy to SAY in one short phrase
but difficult to refute in a brief post (e.g. "relativity is wrong"), I
sometimes like to say "I am the queen of England" as a demonstration that
it's easy to SAY outrageous things of no consequence in the real world. This
in Koobee's mind transmogrified into "Jan says he is the queen of England".
There was a similar case with one of the resident idiots here who keeps
repeating a lie that I claimed I was like Bach (in terms of talent or some
such nonsense).

Honestly I don't know if this is a mental disorder, just plain lack of
intelligence, inability to focus well enough to read someone else's post
with reasonable understanding, inability to understand a metaphor or simile -
it's odd and somewhat disturbing to see this sort of thing in an adult.

> The fools can only be proper idiots, spacetime imbeciles, relativistic morons, and Einstein dingleberries. These are all con-artists who cannot do basic math. <shrug>

No, you simply don't understand this stuff. I don't know why you chose
as a hobby something you have either no talent for or just no inclination
to actually study it. It's obviously something very strong, as it makes you
assume ludicrous things (like every physicist is an idiot, etc.). But
unless you change your approach RADICALLY, this is going to remain a source of
endless frustration for you. There is NO other way out of the corner
you've painted yourself into.

--
Jan

[Koobee Wublee]

On December 14, 2017, JanPB the self-proclaimed mathematician wrote:

> On December 13, 2017 at 3:55:43 PM UTC-8, Koobee Wublee wrote:

> > Koobee Wublee always mean the metric is coordinate dependent ---
> > even at elements level. <shrug>
>
> > https://groups.google.com/d/msg/sci.physics.relativity/o7CSECN9_oI/280vMLFGBwAJ
>
> > Clearly, the metric is coordinate dependent to satisfy the
> > invariance in the geometry. <shrug>
>
> > there is no way to write down the differential equations of the
> > field equations without deciding on a set of coordinate system.
>
> > ** geometry = metric * coordinate
>
> > Where
>
> > ** geometry = reality, invariant
> > ** coordinate = coordinate system to describe reality
> > ** metric = connection between reality and observation
>
> > the operator is a dot product of two matrices defined and
> > explained in the post above. <shrug>
>

> > Consider a flat geometry described using the Cartesian and the
> > polar coordinate systems. <shrug>
>
> > ** ds² = [A] * d²[q1] = [B] * d²[q2]
>

> > Where
>
> > ** ds² = flat geometry
> > ** d²[q1] = matrix[(dx dx,dx dy,dx dz),(dy dx,dy dy,dy dz),(dz dx,dz dy,dz dz)]
> > ** d²[q2] = matrix[(dr dr,dr dθ,dr dφ),(dy dr,dθ dθ,dθ dφ),(dφ dr,dφ dθ,dφ dφ)]

> > ** [A] = matrix[(1,0,0),(0,1,0),(0,0,1)]
> > ** [B] = matrix[(1,0,0),(0,r² cos²(θ),0),(0,0,r²)]
> > ** [a] * [b] = a_ij b_ij, dot product of matrices [a] and [b]
>

> > Clearly, the matrices d²[q1] and d²[q2] are different due to two
> > distinct and very different coordinate systems. Then,
> > mathematically without any doubt, the other two matrices [A] and
> > [B] are also different despite serving as the metrics in
> > describing the same type of geometry ds². <shrug>

> Part of Koobee's problem is that his grasp of this is very feeble
> and he invents his private contorted notation for everything. This
> is done in order to compensate for his lack of understanding the
> subject.

Let’s see what this proper idiot is bitching about to cover its own incompetence. <shrug>

> One of such contortions is exemplified above by the bizarre
> square-brackets notation invented by him in order to shoehorn
> his (false) mantra "tensors are matrices" into the picture.

Koobee Wublee puts brackets around a letter to indicate it being a matrix, and Koobee Wublee has clearly explained that as so in the post. If you have problems with that, it is solely your problem, spacetime imbecile! <shrug>

> This is also a cuckoo notation. Incompetence.

Inability to understand the simple mathematics demonstrates that Jan the self-proclaimed mathematician is a complete covariant nincompoop. <shrug>

> "Dot product of matrices" is another sign of incompetence.

The operator identified by Koobee Wublee is exactly the one that produces the Ricci scalar, relativistic moron! <shrug>

> > Insisting on [A] equaling to [B], who can Jan Bielawski the
> > self-proclaimed queen of England fool?
>
> You keep misquoting what I say. Whether you are intentionally
> LYING or are simply possessed of an extraordinarily dissipated
> thinking I cannot tell.

Koobee Wublee has expressed the simple mathematics at very clearly. Don’t blame Koobee Wublee for Jan’s own stupidity. Jan the self-proclaimed queen of England is not man enough to own its own incompetence. <shrug>

> So, here is what I in fact said:
>
> 1. Two _different_ metric _tensors_ can be different yet represent
> the same geometry.

Now, Jan is lying. Jan always has claimed all the matrices that describe the same type of geometry are all identical. Koobee Wublee has demonstrated that a metric is nothing but a matrix. <shrug>

> 2. The _same_ metric _tensor_ written in terms of two different
> coordinate systems will have different coefficients in each.
> These coefficients for rank-2 tensors can be conveniently displayed
> as a table of numbers (a matrix).

Not even wrong. Try not to change the topic again. <shrug>

[... rest of round-and-round nonsense snipped]

That is enough for Koobee Wublee’s patience. Koobee Wublee has had enough watching Jan the Einstein dingleberry chasing its own tail. <shrug>

[JanPB]

You missed the point. Which is: the matrices DO NOT BELONG THERE. Symbols like
"dx", "ds^2", etc. denote tensors, not matrices.

[..]

> > > Insisting on [A] equaling to [B], who can Jan Bielawski the
> > > self-proclaimed queen of England fool?
> >
> > You keep misquoting what I say. Whether you are intentionally
> > LYING or are simply possessed of an extraordinarily dissipated
> > thinking I cannot tell.
>
> Koobee Wublee has expressed the simple mathematics at very clearly.

Again, you are not reading the post you are responding to. What I was responding to was
NOT "simple mathematics" but your fabricating a story about me.

> Don’t blame Koobee Wublee for Jan’s own stupidity. Jan the self-proclaimed queen of England

Which, needless to say, I never proclaimed.

> is not man enough to own its own incompetence. <shrug>

Non sequitur. Gobbledygook.

> > So, here is what I in fact said:
> >
> > 1. Two _different_ metric _tensors_ can be different yet represent
> > the same geometry.
>
> Now, Jan is lying. Jan always has claimed all the matrices that describe the same type of geometry are all identical. Koobee Wublee has demonstrated that a metric is nothing but a matrix. <shrug>

Not even wrong. A complete misrepresentation of what I said. Again, what I said was this:
https://groups.google.com/forum/#!msg/sci.physics.relativity/h49c7pfteQQ/jrJqiL7XBAAJ

> > 2. The _same_ metric _tensor_ written in terms of two different
> > coordinate systems will have different coefficients in each.
> > These coefficients for rank-2 tensors can be conveniently displayed
> > as a table of numbers (a matrix).
>
> Not even wrong. Try not to change the topic again. <shrug>

Oh stop this idiocy. Total cuckoo.

--
Jan

[Koobee Wublee]

On December 14, 2017, JanPB the self-proclaimed mathematician wrote:

> Again, you are not reading the post you are responding to.

Cheap shot, Jan. <shrug>

> NOT "simple mathematics" but your fabricating a story about me.

Koobee Wublee is in the process of proving how clueless Jan the self-proclaimed mathematician is. <shrug>

> > Don’t blame Koobee Wublee for Jan’s own stupidity. Jan the

> > self-proclaimed queen of England is not man enough to own its own
> > incompetence. <shrug>
>
> Which, needless to say, I never proclaimed. Non sequitur. Gobbledygook.

Denial, Jan. You have bragged about being a mathematician, the queen of England, and others. <shrug>

> > Jan always has claimed all the matrices that describe the same
> > type of geometry are all identical. Koobee Wublee has
> > demonstrated that a metric is nothing but a matrix. <shrug>
>
> Not even wrong. A complete misrepresentation of what I said. Again,
> what I said was this:
>
> https://groups.google.com/forum/#!msg/sci.physics.relativity/h49c7pfteQQ/jrJqiL7XBAAJ

And the following post said you have been wrong. <shrug>

https://groups.google.com/d/msg/sci.physics.relativity/h49c7pfteQQ/5PdRRGfpBAAJ

> Oh stop this idiocy. Total cuckoo.

Demystification is always painful on your part. <shrug>

[JanPB]

On Thursday, December 14, 2017 at 11:16:01 PM UTC-8, Koobee Wublee wrote:
> On December 14, 2017, JanPB the self-proclaimed mathematician wrote:
>
> > NOT "simple mathematics" but your fabricating a story about me.
>
> Koobee Wublee is in the process of proving how clueless Jan the self-proclaimed mathematician is. <shrug>

You have zero chance of succeeding.

> > > Don’t blame Koobee Wublee for Jan’s own stupidity. Jan the
> > > self-proclaimed queen of England is not man enough to own its own
> > > incompetence. <shrug>

That "its" again.

> > Which, needless to say, I never proclaimed. Non sequitur. Gobbledygook.
>
> Denial, Jan. You have bragged about being a mathematician, the queen of England, and others. <shrug>

I have not "bragged" about being a mathematician, I am one. And about that queen
of England I have already explained it to you. You'll just have to live with your lies looks like.

> Demystification is always painful on your part. <shrug>

There is no "demystification" that you would understand.

--
Jan

[mlwo...@wp.pl]

On Friday, 15 December 2017 09:38:52 UTC+1, JanPB wrote:
> On Thursday, December 14, 2017 at 11:16:01 PM UTC-8, Koobee Wublee wrote:
> > On December 14, 2017, JanPB the self-proclaimed mathematician wrote:
> >
> > > NOT "simple mathematics" but your fabricating a story about me.
> >
> > Koobee Wublee is in the process of proving how clueless Jan the self-proclaimed mathematician is. <shrug>
>
> You have zero chance of succeeding.
>
> > > > Don’t blame Koobee Wublee for Jan’s own stupidity. Jan the
> > > > self-proclaimed queen of England is not man enough to own its own
> > > > incompetence. <shrug>
>
> That "its" again.
>
> > > Which, needless to say, I never proclaimed. Non sequitur. Gobbledygook.
> >
> > Denial, Jan. You have bragged about being a mathematician, the queen of England, and others. <shrug>
>
> I have not "bragged" about being a mathematician, I am one.

You're a mathematician believing "immutable mathematical truth".
150 years after Lobachevsky.
You're a poor mathematician and a poor idiot.

[Tom Roberts]

On 12/15/17 12:14 AM, JanPB wrote:
> Symbols like "dx", "ds^2", etc. denote tensors, not matrices.

Yes. But some people around here, who learned from different textbooks, could
well be confused by this. So I'll explain the three COMPLETELY DIFFERENT
meanings of these symbols. In a textbook they would not be confused, because
tensors would be typeset as bold; here we do not have that ability.

Consider this equation, in the context of GR and a specific 4-D manifold:

ds^2 = g_ij dx^i dx^j (1)

Interpretation A:
The {x^k} are coordinates on a region of the manifold, which means that each
point in the region corresponds to a 4-tuple of coordinate values (x^0, x^1,
x^2, x^3); i, j, and k are indexes which range from 0 to 3. ds is an
infinitesimal real number, ds^2 is its square, and the {dx^k} are real numbers
representing infinitesimal displacements along the corresponding coordinate
axes. The {g_ij} are a set of real functions of the {x^k}, called the COMPONENTS
of the metric tensor. Equation (1) is called "the line element", because it
expresses the notion that for an infinitesimal line with coordinate
displacements {dx^k}, its length will be ds. The metric tensor g is:
g = g_ij e^i e^j
where the {e^k} are the basis co-vector fields of the coordinate system {x^k};
they are labeled with an upper (contravariant) index, but each is a covariant
vector.

Interpretation B:
The {x^k} are coordinates on a region of a 4-D manifold, which means they are
real functions on the manifold (i.e. each of them maps each point of the region
to a real number, in a systematic and continuous manner). Each of the {dx^k} is
a ONE-FORM, because here d is the exterior derivative (a one-form is a rank-1
covariant tensor function on the manifold). The {g_ij} are a set of real
functions on the manifold, called the components of the metric tensor. Equation
(1) is called "the metric tensor" because it defines ds^2 as a symmetric rank-2
tensor; it is obviously a function on the manifold (aka a tensor field on the
manifold). The metric tensor g is:
g = ds^2

Interpretation C:
The {x^k} are coordinates on a region of a 4-D manifold, which means they are
real functions on the manifold (i.e. each of them maps each point of the region
to a real number, in a systematic and continuous manner). Each of the {dx^k} is
a ONE-FORM, because here d is the exterior derivative (a one-form is a rank-1
covariant tensor function on the manifold). g_ij is the metric tensor, in
abstract-index notation (note the absence of {}, because it is a single object,
a rank-2 tensor, not a set of components). ds is an infinitesimal real number,
ds^2 is its square. The right-hand side of (1) is the contraction of three
tensors. The metric tensor is g_ij (not g, because this is abstract index notation).

Many textbooks use interpretation A; Jan is using interpretation B. Wald uses C.

Koobee Wublee uses NONE OF THESE, which is why he is so confused
and makes so many incorrect statements. He thinks the {g_ij} of A
are "the metric tensor, which is a matrix". While it is true that
the {g_ij} form a matrix, they are not at all a tensor -- see above
for what the metric tensor actually is. Some archaic textbooks also
get this wrong and sowed much confusion. In particular:
Weinberg, _Gravitation_and_Cosmology_ **AVOID**
Eisenhardt, _Riemannian_Geometry_ **AVOID**
Here are some books that get it right:
Misner, Thorne, and Wheeler, _Gravitation_
Hawking and Ellis, _The_Large_Scale_Structure_of_Space-Time_
Frankel, _The_Geometry_of_Physics_
Baez and Muniain, _Gauge_Fields,_Knots,_and_Gravity_
Wald, _General_Relativity_

Tom Roberts

[Koobee Wublee]

> > describing the same type of geometry ds². <shrug>
>
> You have zero chance of succeeding. There is no "demystification"
> that...

Not able to understand basic math but insisting on voodoo math is solely of the problems of Jan the self-proclaimed queen of England. We have the records here. Maybe the self-proclaimed mathematician will wake up one day. Nah, too stupid. <shrug>

[Koobee Wublee]

On Friday, December 15, 2017 at 10:01:04 AM UTC-8, tjrob137 wrote:
> On 12/15/17 12:14 AM, JanPB wrote:

> > Symbols like "dx", "ds^2", etc. denote tensors, not matrices.
>
> Yes.

No, nobody is claiming dx and ds² as matrices. For the n’th time, when Koobee Wublee points out a matrix, a bracket is enclosed. <shrug>

> Consider this equation, in the context of GR and a specific 4-D
> manifold:
>
> ds^2 = g_ij dx^i dx^j (1)

This a mathematical model of any geometry in any numbers of dimensions. It is proto-GR first introduced by Riemann. <shrug>

> Interpretation A:
> ds is an infinitesimal real number, and the {dx^k} are real numbers
> corresponding coordinate axes. The {g_ij} are the COMPONENTS of the
> metric tensor. Equation (1) is called "the line element"... The

> metric tensor g is:
> g = g_ij e^i e^j
> where the {e^k} are the basis co-vector fields

This is wrong. e^i e^j information is in g_ij. See a derivation of a metric below. <shrug>

https://groups.google.com/d/msg/sci.physics.relativity/o7CSECN9_oI/280vMLFGBwAJ

> Interpretation B:
> The {g_ij} are the components of the metric tensor. Equation (1) is
> called "the metric tensor". The metric tensor g is:
> g = ds^2

This is also wrong since g cannot be ds². Equation (1) says so. <shrug>

> Interpretation C:

> The right-hand side of (1) is the contraction of three tensors. The

> metric tensor is g_ij (not g...

This is just stupid, absolutely nonsense. <shrug>

Interpretation D (by Koobee Wublee):

Equation (1) describes the geometry, ds, being observed using coordinate system dx^i and metric elements g_ij where it is simpler to group g_ij into the matrix, the metric [g], and to group dx^i dx^j into another matrix, the coordinate correlation matrix d[q]². In doing so, it allows us to introduce a mathematical operator, *, connecting the matrices [g] and d[q]² such that equation (1) is satisfied with the new definitions of matrices [g] and d[q]². See below. <shrug>

** ds² = [g] * d[q]² = [g]_ij d[q]^i d[q]^j

Where

** [g] = matrix representing the metric with elements [g]_ij
** d[q]² = coordinate correlation matrix with elements d[q]^i d[q]^j
** [a] * [b] = [a]_ij [b]^ij, dot product of matrices [a] and [b]

> Many textbooks use interpretation A; Jan is using interpretation
> B. Wald uses C.

All wrong. They don’t reflect the math in equation (1). <shrug>

> Koobee Wublee uses NONE OF THESE,

Thank god that Koobee Wublee is not that fvcking stupid. <shrug>

> which is why he is so confused and makes so many incorrect
> statements.

Interpretation D is the only one accurately describe the mathematics of equation (1). <shrug>

> He thinks the {g_ij} of A are "the metric tensor, which is a matrix".

Not exactly. <shrug>

[JanPB]

On Friday, December 15, 2017 at 10:01:04 AM UTC-8, tjrob137 wrote:

A small correction: the equation (1) is not a pure tensor equation in the abstract index
notation (notice e.g. the indices are upstairs instead of downstairs), it's an expression
in coordinates with g_ij being a bunch of numbers (not a tensor). In the abstract index
notation (1) would be written as:

ds^2 = g_ab = g_ij (dx^i)_a (dx^j)_b

(here "ds^2" denotes a tensor, and the space between ) (dx^i)_a and (dx^j)_b denotes
the tensor product (frequently omitted in print to avoid excessive notational pedantry)).

This complication comes from using coordinates here. Without coordinates ds^2
is simply g_ab and evaluating the dot product of vectors X and Y (say) looks like:

ds^2 = g_ab X^a Y^b

(here "ds^2" is a number and BOTH spaces on the RHS denote tensor products)

The RHS literally means:

g_ab (x) X^a (x) Y^b contracted on both a and b

...where "(x)" is the tensor product. In other words:

(rank-2 covariant) (x) (rank-1 contravariant) (X) (rank-1 contravariant)

...which is a rank-(2,2) tensor, contracted on both indices, resulting in a rank-0 tensor (a
number).

--
Jan

[Koobee Wublee]

On December 16, 2017, Jan Bielawski the queen of England wrote:

> On Friday, December 15, 2017, tjrob137 wrote:

> > Consider this equation, in the context of GR and a specific
> > 4-D manifold:
>
> > ds^2 = g_ij dx^i dx^j (1)
>

> the equation (1) is not a pure tensor equation in the abstract

> index notation. In the abstract index notation (1) would be

> written as:
>
> ds^2 = g_ab = g_ij (dx^i)_a (dx^j)_b
>

> (here "ds^2" denotes a tensor) (dx^i)_a and (dx^j)_b denotes
> the tensor product.

Absolutely nonsense. In the math of GR, g_ab never came up. It is the matrix [g] with elements [g]_ij that are solutions to the field equations. <shrug>

Since ultimately it is the metric represented by the matrix [g] with elements [g]_ij that is the end game, Koobee Wublee has identified [g] that satisfies equation (1):

** ds² = [g] * d[q]² = [g]_ij d[q]^i d[q]^j

Where

** [g] = matrix representing the metric with elements [g]_ij
** d[q]² = coordinate correlation matrix with elements d[q]^i d[q]^j

** [a] * [b] = [a]_ij [b]^ij, dot product of matrices [a] and [b]

Consequently, the 16 differential equations found in the null Ricci tensor ([R] = 0) can be laid out based on the already established coordinate system [q]^i. The solution of ([R] = 0) becomes [g]. The field equations can be written in its matrix form:

** [R] – R [g] / 2 = k [T]

Where

** [R] = matrix representing the Ricci tensor
** R = [g^-1] * [R] = [g^-1]^ij [R]_ij = Ricci scalar
** [g^-1] = inverse of [g]
** [T] = matrix representing the energy momentum tensor
** k = constant (never properly derived)

Despite GR being such garbage with so many self-contradictions, the field equations in the matrix form are beautiful, no? <shrug>

> [rest of garbage not related to GR snipped]

Obviously, Jan the self-proclaimed mathematician is getting very desperate. The queen of England is now resorting to throwing shit in all directions hoping for one or two to stick. <shrug>

[mitchr...@gmail.com]

On Friday, December 8, 2017 at 9:17:36 AM UTC-8, Pentcho Valev wrote:
> Even if gravitational waves existed (they don't), LIGO's neutron-star result (GW170817) would still be a fake unless:
>
> (A) in 1922 Arthur Eddington correctly predicted that gravitational waves travel at the speed of light (Einstein's general relativity had said nothing about the speed of gravitational waves).
>
> (B) the gravitational waves experience a delay equal to the Shapiro delay experienced by light.
>
> Given the fact that Eddington was a serial fraudster, I don't think there is any hope for LIGO conspirators.
>
> Pentcho Valev

How fast does the Earth field orbit the Sun?
Gravity appears to move through itself
below light speed.

Mitchell Raemsch

[JanPB]

On Saturday, December 16, 2017 at 1:29:15 PM UTC-8, Koobee Wublee wrote:
> On December 16, 2017, Jan Bielawski the queen of England wrote:
> > On Friday, December 15, 2017, tjrob137 wrote:
>
> > > Consider this equation, in the context of GR and a specific
> > > 4-D manifold:
> >
> > > ds^2 = g_ij dx^i dx^j (1)
> >
> > the equation (1) is not a pure tensor equation in the abstract
> > index notation. In the abstract index notation (1) would be
> > written as:
> >
> > ds^2 = g_ab = g_ij (dx^i)_a (dx^j)_b
> >
> > (here "ds^2" denotes a tensor) (dx^i)_a and (dx^j)_b denotes
> > the tensor product.
>
> Absolutely nonsense. In the math of GR, g_ab never came up.

Kindly complain to Roger Penrose(*) and Robert Wald(**). Just leave me out of it.

(*) Roger Penrose, Wolfgang Rindler,
Spinors and Space-Time, volume 1
Cambridge University Press, 1987
ISBN 0521337070

(**) Robert Wald,
General Relativity
University of Chicago Press, 1984
ISBN 0226870332

> It is the matrix [g] with elements [g]_ij that are solutions to the field equations. <shrug>

Irrelevant. Off-topic. Read the posts you are responding to.

> Since ultimately it is the metric represented by the matrix [g] with elements [g]_ij that is the end game, Koobee Wublee has identified [g] that satisfies equation (1):

OK, I'm game, here is an irrelevant link: https://www.youtube.com/watch?v=5q6oRi1PIOo

[skipping the rest of Koobee's posting which is likewise totally off-topic]

--
Jan

[Koobee Wublee]

On December 16, 2017, Jan Bielawski the queen of England wrote:

> On December 16, 2017 at 1:29:15 PM UTC-8, Koobee Wublee wrote:
> > On December 16, 2017, Jan Bielawski the queen of England wrote:
> > > On Friday, December 15, 2017, tjrob137 wrote:

> > > > Consider this equation, in the context of GR and a specific
> > > > 4-D manifold:
>
> > > > ds^2 = g_ij dx^i dx^j (1)
>
> > > the equation (1) is not a pure tensor equation in the abstract
> > > index notation. In the abstract index notation (1) would be
> > > written as:
>
> > > ds^2 = g_ab = g_ij (dx^i)_a (dx^j)_b
>
> > > (here "ds^2" denotes a tensor) (dx^i)_a and (dx^j)_b denotes
> > > the tensor product.
>

> > Absolutely nonsense. In the math of GR, g_ab never came up. It

> > is the matrix [g] with elements [g]_ij that are solutions to the
> > field equations. <shrug>
>

> Kindly complain to Roger Penrose(*) and Robert Wald(**). Just leave
> me out of it.

You are the ones throwing their shit around. How can you wash your hands after attempting to mystify others? You have not been behaving like a messenger but a preacher. <shrug>

> > Since ultimately it is the metric represented by the matrix [g]
> > with elements [g]_ij that is the end game, Koobee Wublee has
> > identified [g] that satisfies equation (1):
>

> > ** ds² = [g] * d[q]² = [g]_ij d[q]^i d[q]^j
>
> > Where
>
> > ** [g] = matrix representing the metric with elements [g]_ij
> > ** d[q]² = coordinate correlation matrix with elements d[q]^i d[q]^j
> > ** [a] * [b] = [a]_ij [b]^ij, dot product of matrices [a] and [b]

Oh, before Koobee Wublee forgets, Koobee Wublee wish to point out that when [g] is put through the vigorous grinder of tensor calculus, the coordinate system is already established. That means THOU SHALL NOT FVCK WITH THE COORDINATE SYSTEM AT THE END. Each solution (metric), that there are infinite numbers of them, represents a completely independent type of geometry to the others. GR is utter nonsense. <shrug>

> > Consequently, the 16 differential equations found in the null
> > Ricci tensor ([R] = 0) can be laid out based on the already
> > established coordinate system [q]^i. The solution of ([R] = 0)
> > becomes [g]. The field equations can be written in its matrix
> > form:
>
> > ** [R] – R [g] / 2 = k [T]
>
> Where
>
> > ** [R] = matrix representing the Ricci tensor
> > ** R = [g^-1] * [R] = [g^-1]^ij [R]_ij = Ricci scalar
> > ** [g^-1] = inverse of [g]
> > ** [T] = matrix representing the energy momentum tensor
> > ** k = constant (never properly derived)
>
> > Despite GR being such garbage with so many self-contradictions,

> > the field equations in the matrix form are beautiful, no? <shrug>

>
> Irrelevant. Off-topic. Read the posts you are responding to.

So, Jan the self-proclaimed mathematician has run out of arguments. <shrug>

[JanPB]

On Saturday, December 16, 2017 at 9:48:16 PM UTC-8, Koobee Wublee wrote:
> On December 16, 2017, Jan Bielawski the queen of England wrote:
> > On December 16, 2017 at 1:29:15 PM UTC-8, Koobee Wublee wrote:
> > > On December 16, 2017, Jan Bielawski the queen of England wrote:
> > > > On Friday, December 15, 2017, tjrob137 wrote:
>
> > > > > Consider this equation, in the context of GR and a specific
> > > > > 4-D manifold:
> >
> > > > > ds^2 = g_ij dx^i dx^j (1)
> >
> > > > the equation (1) is not a pure tensor equation in the abstract
> > > > index notation. In the abstract index notation (1) would be
> > > > written as:
> >
> > > > ds^2 = g_ab = g_ij (dx^i)_a (dx^j)_b
> >
> > > > (here "ds^2" denotes a tensor) (dx^i)_a and (dx^j)_b denotes
> > > > the tensor product.
> >
> > > Absolutely nonsense. In the math of GR, g_ab never came up. It
> > > is the matrix [g] with elements [g]_ij that are solutions to the
> > > field equations. <shrug>
> >
> > Kindly complain to Roger Penrose(*) and Robert Wald(**). Just leave
> > me out of it.
>
> You are the ones throwing their shit around.

It's up to you to learn this stuff if you're interested. I gave you a couple of good references.
Your calling it "shit" doesn't change anything in the real world.

> How can you wash your hands after attempting to mystify others? You have not been behaving like a messenger but a preacher. <shrug>

I'm not trying to mystify although it's not easy to write this in a short ASCII post. That's
why I (and Tom) added the references.

> Oh, before Koobee Wublee forgets, Koobee Wublee wish to point out that when [g] is put through the vigorous grinder of tensor calculus, the coordinate system is already established. That means THOU SHALL NOT FVCK WITH THE COORDINATE SYSTEM AT THE END.

No, that's false. You are eternally confused by coordinates and coordinate changes and have
somehow managed to erect an entire fake, nonexistent, universe around this confusion.

> Each solution (metric), that there are infinite numbers of them, represents a completely independent type of geometry to the others. GR is utter nonsense. <shrug>

Incorrect. Sorry, but you won't get anywhere with this by daydreaming.

And I'm leaving aside here the fact that your posts were off-topic. They had nothing to do
with Tom's post.

--
Jan

[Koobee Wublee]

On December 16, 2017, Jan Bielawski the queen of England wrote:
> On December 16, 2017 at 1:29:15 PM UTC-8, Koobee Wublee wrote:
> > On December 16, 2017, Jan Bielawski the queen of England wrote:
> > > On Friday, December 15, 2017, tjrob137 wrote:

> > > > Consider this equation, in the context of GR and a specific
> > > > 4-D manifold:
>
> > > > ds^2 = g_ij dx^i dx^j (1)
>
> > > the equation (1) is not a pure tensor equation in the abstract
> > > index notation. In the abstract index notation (1) would be
> > > written as:
>
> > > ds^2 = g_ab = g_ij (dx^i)_a (dx^j)_b
>
> > > (here "ds^2" denotes a tensor) (dx^i)_a and (dx^j)_b denotes
> > > the tensor product.
>
> > Absolutely nonsense. In the math of GR, g_ab never came up. It
> > is the matrix [g] with elements [g]_ij that are solutions to the
> > field equations. <shrug>
>

> > Since ultimately it is the metric represented by the matrix [g]
> > with elements [g]_ij that is the end game, Koobee Wublee has
> > identified [g] that satisfies equation (1):
>
> > ** ds² = [g] * d[q]² = [g]_ij d[q]^i d[q]^j
>
> > Where
>
> > ** [g] = matrix representing the metric with elements [g]_ij
> > ** d[q]² = coordinate correlation matrix with elements d[q]^i d[q]^j
> > ** [a] * [b] = [a]_ij [b]^ij, dot product of matrices [a] and [b]
>

> > when [g] is put through the vigorous grinder of tensor calculus,
> > the coordinate system is already established. That means THOU

> > SHALL NOT FVCK WITH THE COORDINATE SYSTEM AT THE END. Each

> > solution (metric), that there are infinite numbers of them,
> > represents a completely independent type of geometry to the
> > others. GR is utter nonsense. <shrug>
>

> No, that's false.

What you are referring to and your bibles contain nothing but mysticism that has nothing to do with GR. Thus, your arguments are totally off topic. We have identified your problem --- mysticism. <shrug>

[Edwin Huckabee]

Koobee Wublee wrote:

>> No, that's false.
>
> What you are referring to and your bibles contain nothing but mysticism
> that has nothing to do with GR. Thus, your arguments are totally off
> topic. We have identified your problem --- mysticism. <shrug>

I have to agree. When the 100% empty space and time starts to contract and
dilate, that's where are my limits are located.

[David (Kronos Prime) Fuller]

Edwin Huckabee

You do not seem to have Any idea what space Time is at all.