Higgs boson could also explain the earliest expansion of the Universe

Aug 15, 2014
The influence of the Higgs boson and its field (inset) on cosmological inflation could manifest in the observation of gravitational waves by the BICEP2 telescope (background). Credit: the BICEP2 Collaboration (background); © 2014 Fedor Bezrukov, RIKEN–BNL Research Center (inset)

Fedor Bezrukov from the RIKEN–BNL Research Center and Mikhail Shaposhnikov from the Swiss Federal Institute of Technology in Lausanne propose that the Higgs boson, which was recently confirmed to be the origin of mass, may also be responsible for the mode of inflation and shape of the Universe shortly after the Big Bang. "There is an intriguing connection between the world explored in particle accelerators today and the earliest moments of the existence of the Universe," explains Bezrukov.

The Universe started with a giant explosion known as the Big Bang, and has been expanding ever since. The expansion is balanced such that its shape is flat and not bent, which can only be the case for a very specific distribution of matter density.

The coupling between the Higgs and other fundamental particles provides . In the first moments of the Universe, however, coupling between the Higgs field and gravity accelerated the Universe's expansion. An important parameter for this coupling is the mass of the Higgs boson. Experiments at the Large Hadron Collider at CERN (European Organization for Nuclear Research) have shown that the mass of the Higgs boson is very close to a critical value that separates two possible types of Universe—the stable one we know or a potentially unstable alternate.

Bezrukov and Shaposhnikov have now studied the implications arising from the Higgs mass being near this critical boundary and the impact this has on cosmological inflation. Through theoretical arguments, they found that as the mass of the Higgs approaches the critical value, gravitational waves from the Big Bang become strongly enhanced. The Big Bang is thought to have created many gravitational waves, which act like ripples in space and time, and it is these waves that are amplified for a Higgs of near-critical mass.

Experimentally, the influence of the Higgs boson could have significant implications for the observation of gravitational waves, which had eluded physicists until recently, when analysis of data acquired by the BICEP2 telescope near the South Pole suggested the first signs of gravitational waves in the that fills the Universe (Fig. 1).

The BICEP2 result, however, is far from unequivocal, with continued debate as to whether the incredibly faint signal of could really be detected in this way. The effects of a near-critical Higgs mass could put such debate to rest. "The Higgs mass at the critical boundary could explain the BICEP2 result," Bezrukov explains.

Explore further: Should the Higgs boson have caused our Universe to collapse?

More information: Bezrukov, F. & Shaposhnikov, M. Higgs inflation at the critical point. Physics Letters B 734, 249–254 (2014). DOI: 10.1016/j.physletb.2014.05.074

add to favorites email to friend print save as pdf

Related Stories

Higgs quest deepens into realm of 'New Physics'

Jul 02, 2014

Two years after making history by unearthing the Higgs boson, the particle that confers mass, physicists are broadening their probe into its identity, hoping this will also solve other great cosmic mysteries.

Interview: CERN chief firmer on Higgs boson

Jan 27, 2013

The world should know with certainty by the middle of this year whether a subatomic particle discovered by scientists is a long-sought Higgs boson, the head of the world's largest atom smasher said Saturday.

Recommended for you

Uncovering the forbidden side of molecules

23 hours ago

Researchers at the University of Basel in Switzerland have succeeded in observing the "forbidden" infrared spectrum of a charged molecule for the first time. These extremely weak spectra offer perspectives ...

How Paramecium protozoa claw their way to the top

Sep 19, 2014

The ability to swim upwards – towards the sun and food supplies – is vital for many aquatic microorganisms. Exactly how they are able to differentiate between above and below in often murky waters is ...

User comments : 1246

Adjust slider to filter visible comments by rank

Display comments: newest first

Grallen
2.4 / 5 (9) Aug 15, 2014
"Higgs boson, which was recently confirmed to be the origin of mass" Did I miss this article? I thought that we weren't even sure the particle found was the higgs yet.
Tektrix
5 / 5 (6) Aug 15, 2014
"Higgs boson, which was recently confirmed to be the origin of mass" Did I miss this article? I thought that we weren't even sure the particle found was the higgs yet.


http://home.web.c...gs-boson
Toiea
3.1 / 5 (8) Aug 15, 2014
Universe Shouldn't Be Here, According to Higgs Physics One peer-reviewed study against another one...
EyeNStein
3 / 5 (2) Aug 15, 2014
Or more likely the reverse is true:-
The state and laws of the brand new universe, before the first symmetry (or possibly second symmetry?) broke, determined the mass of the Higgs boson, now that we have mass.

(I suspect that a second broken symmetry may have existed where it was equally likely Anti-matter could have dominated matter at its breaking rather than as in our matter dominated universe?)
Tektrix
5 / 5 (4) Aug 15, 2014
And this- more to the point of your question, Grallen: http://home.web.c...gs-boson
arom
1 / 5 (10) Aug 15, 2014
Fedor Bezrukov …. and Mikhail Shaposhnikov …. propose that the Higgs boson, which was recently confirmed to be the origin of mass, may also be responsible for the mode of inflation and shape of the Universe shortly after the Big Bang …..

The Universe started with a giant explosion known as the Big Bang, and has been expanding ever since. The expansion is balanced such that its shape is flat and not bent, which can only be the case for a very specific distribution of matter density.


This seems to be a valuable idea which could extend the standard model of particle physics theory from the seeming dead end, rather than going to the dreaming theory such as supergravity, string theory, etc. Anyway the conventional the standard model is very complicate and difficult to understand, maybe this simple idea could help ….
http://www.vacuum...=9〈=en
richardwenzel987
5 / 5 (3) Aug 15, 2014
I suspect that the Higgs mass will be PRECISELY at the critical boundary. And so, it will be necessary to explain how that came about, give all the other possible values.
George_Rajna
Aug 16, 2014
This comment has been removed by a moderator.
Nashingun
1 / 5 (8) Aug 16, 2014
What is this... Another bogus interpretation of Higgs boson on BB theory? The last I heard according to CERN Higgs physics explained that after the BANG the universe must have collapsed and not able to sustain.. This is so opposite of reality where the universe is presently sustaining. Yet here we are again with their bogus experiments.
mooster75
4.3 / 5 (6) Aug 16, 2014
What is this... Another bogus interpretation of Higgs boson on BB theory? The last I heard according to CERN Higgs physics explained that after the BANG the universe must have collapsed and not able to sustain.. This is so opposite of reality where the universe is presently sustaining. Yet here we are again with their bogus experiments.

I'm pretty ignorant. The farthest I went with science studies were freshman intro courses. And yet, you make me feel like a genius. Thank you.
Reg Mundy
1.5 / 5 (15) Aug 16, 2014
Does anybody actually believe this stuff? There's no proof, no evidence, it seems mainly conjecture to me! In my experience, when things have to get really complicated to fit known facts, they usually turn out to be wrong, and a simple explanation usually turns out to be right. The statement that the Higgs confers mass on elementary particles has no basis in proven evidence.
Da Schneib
3.8 / 5 (10) Aug 16, 2014
The Universe started with a giant explosion known as the Big Bang, and has been expanding ever since. The expansion is balanced such that its shape is flat and not bent, which can only be the case for a very specific distribution of matter density.
Actually it's not quite flat- it has to be very slightly positively curved. The two most obvious arguments are the existence of Λ (because the universe is expanding) and its nearness to zero (since atoms are not ripped apart by it, nor is anything smaller than a galaxy cluster), and the existence (proven in the lab) of the Casimir force, which again shows a low value close to zero (because otherwise the plates would fly apart at greater distances, rather than being pushed a little bit when they're very close together); Λ is the only good explanation for the Casimir force, and explains dark energy/exponential expansion quite readily.
Da Schneib
4.3 / 5 (16) Aug 16, 2014
Does anybody actually believe this stuff? There's no proof,
Proof is for mathematical theorems. Scientific theories are supported by evidence, which consists of the original thing the hypothesis/es was/were were developed to explain, and the observations it predicted before anyone looked, as well as the lack of any prediction that turned out wrong.

no evidence, it seems mainly conjecture to me!
This is incorrect. The Higgs is now a theory, not a conjecture or hypothesis; it's what the LHC was built to check for, and they found it to the necessary confidence level.

In my experience, when things have to get really complicated to fit known facts, they usually turn out to be wrong, and a simple explanation usually turns out to be right.
And that's why we use Newtonian physics for the GPS satellite orbital adjustments. Oh, wait...
Da Schneib
4.1 / 5 (13) Aug 16, 2014
"Higgs boson, which was recently confirmed to be the origin of mass" Did I miss this article? I thought that we weren't even sure the particle found was the higgs yet.
François Englert and Peter Higgs just won a Nobel Prize in Physics for it in 2013, according to CERN's web site, "for the theoretical discovery of a mechanism that contributes to our understanding of the origin of mass of subatomic particles, and which recently was confirmed through the discovery of the predicted fundamental particle, by the ATLAS and CMS experiments at CERN's Large Hadron Collider."

What we don't know is if it's the Standard Model Higgs or a different one that implies reality that we have not yet definitively seen which is outside the standard model. But that's a detail; the Higgs is real, and it's the origin of mass (which they can tell from the decay modes, which are as predicted by Peter Higgs' theory) and this is according to not one but two experiments.
Da Schneib
3.5 / 5 (8) Aug 16, 2014
I suspect that the Higgs mass will be PRECISELY at the critical boundary. And so, it will be necessary to explain how that came about, give all the other possible values.
This is an amusing and interesting speculation similar to another speculation I saw today from another user on another thread. In support of it, nature always seems to "hide" underlying realities from higher level phenomena; quantum mechanics (QM), classical thermodynamics, and the Fluctuation Theorem (FT) form an excellent example. Classically, the 2LOT predicts that entropy will never be reduced in any closed system, whereas QM doesn't predict entropy at all; the FT shows how they gradually fade from one to the other, resolving the conflict, and predicts the size/speed point at which the changeover occurs, and how the 2LOT fades incrementally away below that point. Also there is the famous Lewis Carrol quote after FitzGerald found that by applying his contraction it fixed the Michelson/Moreley experiment.
Toiea
1 / 5 (12) Aug 16, 2014
Actually, the Universe never exploded. We just observe, how the light gets scattered with density fluctuations of vacuum at distance in similar way, like the ripples at the water surface. Their change of wavelength is interpreted as a metric expansion of space-time with scientific trolls, who cannot imagine any other mechanism, than this one prescribed with their religion. The Higgs field are the density fluctuations of vacuum at smallest scales, so we can say, it's responsible for light scattering at highest energy density observable, i.e. this one which is attributed to universe expansion at the very first moments after its alleged formation.

The religion is not characterized with absence of logical reasoning (after all, even the theologists developed many logical proofs of God existence) - but with unwillingness to think about any other option.
Da Schneib
3.8 / 5 (10) Aug 16, 2014
Scattering doesn't cause a wavelength change, Zephir/Tolea. Certainly not from gamma rays to microwave.
Toiea
1 / 5 (12) Aug 16, 2014
So which effect is responsible for wavelength change at the water surface? A Holy Spirit? Try to surprise me...
Da Schneib
3.8 / 5 (10) Aug 16, 2014
That's not a picture of water waves changing wavelength *during propagation* but of the frequency of oscillation changing *at the source* due to the decreasing energy of the oscillator (water that has been vertically disturbed, generating transverse waves in the water surface driven by momentum conservation after the initial energy input, and damped by friction and dissipation). You'll need to provide a reputable source that claims water waves change wavelength *during propagation* because that's what you're claiming light does.

Hint: you won't find one. To change the wavelength of water waves you need to change their energy, which requires energy to be created or annihilated, which are things that never happen in our universe due to conservation of energy. Water waves can change energy due to input from the wind, which can add or subtract energy according to its direction relative to their direction of propagation. But that's not happening in your picture.
Toiea
1 / 5 (10) Aug 16, 2014
That's not a picture of water waves changing wavelength *during propagation* but of the frequency of oscillation changing *at the source* due to the decreasing energy of the oscillator
This is the remnant of rain droplets. Which change of "source frequency" do you have on mind? I just like how you're trying to twist an apparent physical reality...;-)
You'll need to provide a reputable source that claims water waves change wavelength *during propagation* because that's what you're claiming light does
This is typical attitude for mainstream physicists: what isn't published in Nature, it doesn't exist, despite of many photographs (1, 2). As nonsensically as it sounds, this attitude helps the scientists to continue in their "Duh" research and taking money for trivial research too.
Toiea
1 / 5 (10) Aug 16, 2014
BTW This effect is called a Stokes scattering and the change of wavelength during it is routinely known to everyone, who deals with physics just a bit seriously. The water surface exhibits both anti-Stokes scattering, both Stokes and Rayleigh scattering, depending on the original wavelength of wave scattered. And the vacuum is not different in this matter.
Da Schneib
3.9 / 5 (11) Aug 16, 2014
Do you understand that a raindrop forms the only energy input, but that conservation of momentum causes the elastic water surface to act as a damped oscillator under the influence of gravity and surface tension? Apparently not. A bug walking on water knows more about it than you.

If you don't understand the difference between damped and undamped oscillators, or how a damped oscillator works, what the heck are you doing talking about light?
Toiea
1 / 5 (11) Aug 16, 2014
Damped oscillator doesn't change its wavelength/frequency during its damping - the water surface does. Apparently some other strange effect unknown to contemporary physics gets involved here - which one? A Holy Spirit? Just say it..
Da Schneib
4.1 / 5 (13) Aug 16, 2014
It's actually called Rahman Stokes scattering, and it only happens at specific wavelengths; those wavelengths correspond to energy differences between various vibration modes of molecules, not atoms (for example hydrogen in free space), or between those various modes and the ground state. Free hydrogen atoms cannot get rid of an absorbed photon at any frequency but the one they started with, or one of a small number of frequencies that are well known and can easily be blocked out of the data.

You, however, are alleging that this happens to *all wavelengths*, which is inconsistent with Rahman scattering of any kind. The only scattering the light does is Rayleigh scattering, except at those well-known, easily blocked out frequencies.

Perhaps you should study Rahman Stokes scattering a bit more carefully, Zephir.
Toiea
1 / 5 (11) Aug 16, 2014
It's actually called Rahman
Do you mean Raman or Brahman? Nope, we aren't talking about hydrogen atoms here - just about water surface. Not all words and strange names which you possibly remember from textbooks in connection to scattering will apply here. You're supposed to understand the subject, not to parrote it randomly.
The only scattering the light does is Rayleigh scattering, except at those well-known, easily blocked out frequencies
So why are talking here about Rahman-Brahman scattering of photons here at all? Such a scattering should never ever happen, until light is formed with photons.
Da Schneib
3.7 / 5 (9) Aug 16, 2014
Damped oscillator doesn't change its wavelength/frequency during its damping - the http://image.shut...497.jpg. Apparently some other strange effect unknown to contemporary physics gets involved here - which one? A Holy Spirit? Just say it..
This is incorrect, as can clearly be seen in your picture. You're violating energy conservation again, Zephir.
Da Schneib
3.7 / 5 (9) Aug 16, 2014
Rahman Stokes scattering loses energy from the photon into the molecule it's scattering from; Rahman anti-Stokes scattering gains energy.

You are apparently unaware that Stokes and anti-Stokes scattering are THE two types of Rahman scattering.

It looks like you're talking about stuff you don't actually understand and hoping you'll find something I don't happen to know about that you can pounce on. And you act like everyone does the same thing.

You're doing this because you want to change the subject so you can stop looking like you don't know what you're talking about, but you've jumped from the frying pan into the fire.
Da Schneib
3.7 / 5 (9) Aug 16, 2014
Zephir has confused a spring oscillator, which is designed to minimize damping by being used in air and whose momentum decays only very slowly, with water surface oscillations, which are readily damped by the mass of water underneath, which has much higher friction than air, and which readily transmits energy away from the surface as longitudinal waves within the body of the water.

Furthermore, he also doesn't understand that even a spring oscillator eventually slows down; only by adding energy back in to replace it is the harmonic oscillator kept in time. Zephir has confused a damped oscillator with an undamped oscillator, and real dust and molecular cloud Rahman Stokes scattering with Rayleigh scattering that occurs in intergalactic neutral hydrogen.

He's also forgotten that Rahman scattering only affects frequency lines, which since they are known are readily eliminated by filtering them out.

Finally he's implied that there is a time when light is not photons.
Da Schneib
3.8 / 5 (9) Aug 16, 2014
Another way to look at Rahman Stokes scattering is that it occurs in ground state molecules, whereas Rahman anti-Stokes scattering can occur only in excited molecules. Note that Rahman Stokes scattering can also occur in non-ground state atoms, but anti-Stokes scattering can only occur in them.

And finally, from here: http://en.wikiped...cillator
First sentence: "In real oscillators, friction, or damping, slows the motion of the system. "

We done here?
Da Schneib
3.7 / 5 (9) Aug 16, 2014
And just to complete the picture, note that there is a third type of oscillator, called the driven oscillator. This is where the oscillation receives a "kick" in phase with its oscillation and of the correct energy to offset what it has lost since the last "kick." This is the standard "tank" circuit used in electronics to generate a stable frequency for radio transmission, and also the escapement in a spring-driven watch.
TechnoCreed
4.4 / 5 (7) Aug 17, 2014
Water waves can change energy due to input from the wind, which can add or subtract energy according to its direction relative to their direction of propagation.
Surface waves on water loose energy just because water is not a perfect fluid, it has viscosity... I guess you already know that, but I want to reinforce your point.
Rahman scattering
Take away the 'h' in raman please, it hurt my eyes.

Maybe I should not do that. You are so damn good and I do not want to discredit your work, but it was itching to much.
Da Schneib
4.1 / 5 (9) Aug 17, 2014
Bah. My bad. A man who can't take criticism never learns anything. Thank you. ;) Comes of posting lots of politics and foreign policy and not enough physics and cosmology. I'm fixing that, though. :D

You're right about viscosity; friction between the molecules. You're looking at it classically; I'm looking at it from a molecular standpoint and on down to QM. My father once said to me that a really good teacher knows lots of ways to say the same thing in order to reach as many students as possible.
Toiea
1.4 / 5 (11) Aug 17, 2014
You are so damn good and I do not want to discredit your work
The real expert would never misspell such a name, because it has the "Raman" name before eyes all time.
note that there is a third type of oscillator, called the driven oscillator
This is another syndrome of contemporary physicists, they cannot focus on problem due to quanta of information which they learned at school. So no, this water surface picture DOESN'T illustrate the Ra(h)man scattering, it doesn't illustrate hydrogen molecules, it doesn't illustrate string oscillator, it doesn't illustrate driven oscillator. The only person who is trying to confuse subject with it here is just you, sorry.
You're looking at it classically; I'm looking at it from a molecular standpoint and on down to QM
The water surface is a typical classical system, only really very bad teacher could look at it from "a molecular standpoint and on down to QM".
Toiea
1.4 / 5 (10) Aug 17, 2014
There is another problem of mainstream physicists, they're trying to push their beloved/adored quantum mechanics theory everywhere, despite the system is already very distant from quantum mechanical description already - actually the more, the less they understand it. Does something in real life look like the quantum wave packet? If not, why to drag the quantum mechanics into it? Such an approach will not be insightful, but a misleading instead.

What the formally thinking physicists (and their wannabes like you) are forgetting all the time is, the quantum mechanics is not the only theory describing this world, the general relativity is here too and the general relativity differs from quantum mechanics in 108 orders of magnitude by its predictions. You cannot really know, how many orders of magnitude the water surface is from quantum mechanics behavior already.
Toiea
1.4 / 5 (10) Aug 17, 2014
even a spring oscillator eventually slows down
This is not relevant to problem - the spring oscillator doesn't change its frequency with compare to water surface, which is why it's used in clock. For frequency change of ripples at the water surface the string oscillator is apparently irrelevant and its rising here serves only as an evidence of your apparent confusion.

So, what is responsible for change of frequency of water surface ripples? A viscosity only does change the amplitude of undulations, not a frequency. Is it still a Holy Spirit at the very end?
peabody3000
5 / 5 (2) Aug 17, 2014
"Higgs boson, which was recently confirmed to be the origin of mass" Did I miss this article? I thought that we weren't even sure the particle found was the higgs yet.


yes they confirmed it a few months ago. personally i have to philosophically consider that a grand mistake may possibly have been made but the announcement is at least fully official
Reg Mundy
1.4 / 5 (9) Aug 17, 2014
"Higgs boson, which was recently confirmed to be the origin of mass" Did I miss this article? I thought that we weren't even sure the particle found was the higgs yet.


yes they confirmed it a few months ago. personally i have to philosophically consider that a grand mistake may possibly have been made but the announcement is at least fully official

Yes, it is claptrap, the Higgs is no more the "origin" of mass than any other particle. All particles have mass, even photons (though almost infinitely small)(too many links to include).
Mass of Higgs 125.03+0.26 − 0.27 (stat) +0.13 − 0.15 (sys) GeV/c2
Mass of photon <1×10−18 eV/c2
(courtesy of Wikipedia). So, how many Higgs in a photon?
Egleton
1 / 5 (6) Aug 17, 2014
The Higggs causes mass. What does "causes" mean?
The best explination that I have heard seems to imply some sort of viscous field, which is way off the observed phenomena of inertia.
And then there is inertia. No-one is discussing what causes a hadron to resist a change of momentum.
How about this? There are no "fields". They are mental constructs. No magnetic field, gravitational field and no Higgs field. There are just rule-sets.
If you jump off a tall building you will suffer damage, not because of some putitive "field" but because those are the rules of the game.
http://www.my-big-toe.com/
Toiea
1 / 5 (7) Aug 17, 2014
the Higgs is no more the "origin" of mass than any other particle
In mainstream physics it's a Higgs field and it's responsible only for some 2% (?) of matter in form of W/Z bosons mediating weak force inside of hadrons.
The best explination that I have heard seems to imply some sort of viscous field, which is way off the observed phenomena of inertia.
The mass doesn't manifest itself with braking of particles but with their inertia. The best explanation comes from dense aether model, which considers the Higgs field as a density fluctuations of another massive field, aka aether. Every undulation of such a field exposes more density fluctuations at the place where it is spreading and these density fluctuations are giving it mass. This mechanism applies to all bosons and particle solitons, though.
Reg Mundy
1 / 5 (9) Aug 17, 2014
The Higggs causes mass. What does "causes" mean?
The best explination that I have heard seems to imply some sort of viscous field, which is way off the observed phenomena of inertia.
And then there is inertia. No-one is discussing what causes a hadron to resist a change of momentum.
How about this? There are no "fields". They are mental constructs. No magnetic field, gravitational field and no Higgs field. There are just rule-sets.
If you jump off a tall building you will suffer damage, not because of some putitive "field" but because those are the rules of the game.
http://www.my-big-toe.com/

Very perceptive!
Reg Mundy
1 / 5 (10) Aug 17, 2014
the Higgs is no more the "origin" of mass than any other particle
In mainstream physics it's a Higgs field and it's responsible only for some 2% (?) of matter in form of W/Z bosons mediating weak force inside of hadrons.
The best explination that I have heard seems to imply some sort of viscous field, which is way off the observed phenomena of inertia.
The mass doesn't manifest itself with braking of particles but with their inertia. The best explanation comes from dense aether model, which considers the Higgs field as a density fluctuations of another massive field, aka aether. Every undulation of such a field exposes more density fluctuations at the place where it is spreading and these density fluctuations are giving it mass. This mechanism applies to all bosons and particle solitons, though.

The best explanation for inertia is conservation of angular momentum and gyroscopic effects, as explained in the part of my book dealing with motion - how matter moves.
Toiea
1 / 5 (8) Aug 17, 2014
The best explanation for inertia is conservation of angular momentum and gyroscopic effects
This is a description of few of effects, in which inertia manifests itself. The people knew about all of it before two hundred years - do you really believe, they already understood, what the inertia is just because of it?

The contemporary people are so naive, they're writing a textbooks of classical physics and they present them as a their personal TOE, because they don't know, it's a subject of high school physics. The George_Rajna's posts here are typical example of such attitude.
Da Schneib
4.2 / 5 (5) Aug 17, 2014
even a spring oscillator eventually slows down
the spring oscillator doesn't change its frequency with compare to water surface, which is why it's used in clock. For frequency change of ripples at the water surface the string oscillator is apparently irrelevant and its rising here serves only as an evidence of your apparent confusion.
This is incorrect according to the Wiki article http://en.wikiped...cillator
That article claims and sources reliable evidence to show that in fact all *real* classical oscillators decay in both frequency and amplitude over time unless they are driven.

So, what is responsible for change of frequency of water surface ripples? A viscosity only does change the amplitude of undulations, not a frequency.
This is incorrect, as can be seen from the picture in *your own link*.
Toiea
1 / 5 (9) Aug 17, 2014
This is incorrect, as can be seen from the picture in *your own link*
It just illustrates, another effect (other than viscosity) gets involved. Which effect it is? A Holy Spirit? As you can see, the knowledge of math means nothing, if you cannot understand the phenomena at its intuitive level, which would allow the subsequent construction of formal model.

A hint: in [url=http://i.imgur.com/NqMtkaz.jpg]this picture[/ur] the wavelength is changing in both direction depending on original wavelength. The viscosity, damping of oscillator, etc.. blah blah remains the very same - but the result still changes. What's the reason of this, after then?
Toiea
1 / 5 (8) Aug 17, 2014
Could you imagine some mechanical (spring?) oscillator, which would change its wavelength/frequency with time depending on the original frequency used? If not, you shouldn't drag such a model into your thoughts, as such model will remain misleading: it cannot illustrate the physical situation, which you're trying to describe.
Da Schneib
4.5 / 5 (8) Aug 17, 2014
...they confirmed it a few months ago.

Yes, it is claptrap, the Higgs is no more the "origin" of mass than any other particle.
Better notify the Nobel Prize committee. Good luck with that.

All particles have mass, even photons (though almost infinitely small)(too many links to include).
Photons do not have mass, they have energy, and the energy makes them have momentum. This is basic Special Relativity. The correct equation to calculate the energy for *anything*, including photons as well as tardyons (particles that move slower than the speed of light) is

E² = P²c² + m²c⁴
Where,
E is energy
P is momentum
c is the speed of light, and
m is the mass.

Note that because there are two terms, even if the mass term is zero the energy will be nonzero if momentum term is not. Personally I'm going with Einstein.
Mass of Higgs 125.03+0.26 − 0.27 (stat) +0.13 − 0.15 (sys) GeV/c2
Mass of photon <1×10−18 eV/c2
(courtesy of Wikipedia). So, how many Higgs in a photon?
Da Schneib
4.4 / 5 (7) Aug 17, 2014
Ran out of room; an additional thought, if tachyons exist then physicists expect that they too will follow Einstein's equation above. Since they are expected to have mass that is an imaginary number, the energy is expected to be a complex number.
Da Schneib
4.3 / 5 (6) Aug 17, 2014
The last three lines of the post 2 before this one are not mine; I mistakenly included them in my post and meant to delete them before posting. They are from the same post as the quote in my post. Sorry about that.
Da Schneib
4.3 / 5 (6) Aug 17, 2014
The Higggs causes mass. What does "causes" mean?
Just as electromagnetism and gravity (and for that matter the color and weak nuclear forces) create a field that permeates all of space, which is zero or as close as you can get to it anywhere there is no matter or energy and you are far from any matter or energy, so does the Higgs.

When two electrons, or an electron and a proton, or any other charged particles for that matter, interact by the electromagnetic force, they influence the electric field that fills all of space close to them, which is what "electric charge" means. It falls off in field strength in the familiar inverse square manner familiar to all physics and electronics or electrical students. This field acts upon the field of the other particle, and vice versa, and also on all other electric fields either one encounters.

This is explained by physicists as the creation of virtual particles by the field. Dirac developed this approach and Feynman extended it.
Da Schneib
4.3 / 5 (6) Aug 18, 2014
contd

So just as the electric field creates a region of space where the probability of encountering a virtual photon is enhanced, the Higgs field creates a region of space where the probability of encountering a virtual Higgs particle is enhanced. However, this field's value far from any matter or energy is not zero. As a result, there is a virtual field filling all of space, which any particle that has mass interacts with; this interaction is what causes it to have mass.

The mass of the Higgson (if I may coin a New Word™) is immaterial. It's the interaction with the virtual Higgsons that creates mass. Whether this is gravitational mass, inertial mass, or both is unknown at this time; but now that we can detect it we will find out.
gsvasktg
1 / 5 (4) Aug 18, 2014
A serious conceptual error underlies the higgs boson finding. Space matter density is 3.6 E minus 25 kgs/cum and this is confirmed by an axiomatic theory. The mass of the higgs is due to the impact velocity of the nuclear particles. The mass of the fundamental particle Ne is 9.6 E minus 35 kgs or 53 EV . One cannot measure the water molecule by a throwing a rock into it.
7 Nes create aphoton or quanta in space. see website kapillavastu dot com and pho pdf for a rigourous & detailed workout. The axiomatic theory is vaid for all times as it has its own internal proof
Da Schneib
4.3 / 5 (6) Aug 18, 2014
the Higgs is no more the "origin" of mass than any other particle
In mainstream physics it's a Higgs field and it's responsible only for some 2% (?) of matter in form of W/Z bosons mediating weak force inside of hadrons This is incorrect. It's the interaction with the Higgs field that causes the effects of mass.

The mass doesn't manifest itself with braking of particles but with their inertia.
This is correct but may only deal with inertial mass. How gravitational mass arises awaits a quantum field theory of gravity that is consistent with the Standard Model of Particle Physics.
Da Schneib
4 / 5 (4) Aug 18, 2014
the Higgs is no more the "origin" of mass than any other particle
In mainstream physics it's a Higgs field and it's responsible only for some 2% (?) of matter in form of W/Z bosons mediating weak force inside of hadrons.
The best explination that I have heard seems to imply some sort of viscous field, which is way off the observed phenomena of inertia.
The mass doesn't manifest itself with braking of particles but with their inertia. The best explanation comes from dense aether model, which considers the Higgs field as a density fluctuations of another massive field, aka aether. Every undulation of such a field exposes more density fluctuations at the place where it is spreading and these density fluctuations are giving it mass. This mechanism applies to all bosons and particle solitons, though.

Da Schneib
4.4 / 5 (7) Aug 18, 2014
I waited too long and the edit time was up. Please ignore the previous comment and read this one.
The best explanation for inertia is conservation of angular momentum and gyroscopic effects, as explained in the part of my book dealing with motion - how matter moves.
The Higgs field explains inertial mass, and the gravity field will explain gravitational mass once we prove it exists. The gravity field is difficult to detect and is currently less than our instruments are sensitive enough to find. When we make sensitive enough instruments we will be able to reconcile inertial mass with gravitational mass, as the Equivalence Principle of General Relativity Theory indicates. Then the gravitational field's relation to the Higgs field will be clear, and GRT and QM will be reconciled with one another.
Da Schneib
4.4 / 5 (7) Aug 18, 2014
This is a description of few of effects, in which inertia manifests itself. The people knew about all of it before two hundred years - do you really believe, they already understood, what the inertia is just because of it?
Now we've found the reason for it-- the Higgs field that permeates all of space.

The contemporary people are so naive, they're writing a textbooks of classical physics
The Higgs is not a classical phenomenon. It's a quantum phenomenon. Nobody knew about quanta two hundred years ago. You're making stuff up agaiin.

they present them as a their personal TOE, because they don't know, it's a subject of high school physics.
I've never seen QM and the SM presented as high school physics. I doubt they are in Czechoslovakia either.

Just sayin'.
Da Schneib
4.4 / 5 (7) Aug 18, 2014
This is incorrect, as can be seen from the picture in *your own link*
It just illustrates, another effect (other than viscosity) gets involved.
What effect is that? Reliable links and quotes please.
As you can see, the knowledge of math means nothing, if you cannot understand the phenomena at its intuitive level, which would allow the subsequent construction of formal model.
Math is the language of physics. If you don't understand the math you don't understand the physics. You need at least introductory calculus to understand even classical physics. It's clear you haven't had it and don't.
the wavelength is changing in both direction
This is obviously incorrect looking at the picture. You're making stuff up again.
The viscosity, damping of oscillator... remains the very same - but the result still changes. What's the reason of this, after then?
It only changes downward. You're making stuff up again.
Da Schneib
4.4 / 5 (7) Aug 18, 2014
Could you imagine some mechanical (spring?) oscillator, which would change its wavelength/frequency with time depending on the original frequency used?
Yes. It's in Wikipedia. Already linked above.
Da Schneib
4.5 / 5 (8) Aug 18, 2014
A serious conceptual error underlies the higgs boson finding.

The mass of the higgs is due to the impact velocity of the nuclear particles.
Reliable links and quotes from scholarly peer-reviewed literature please.

What's an "axiomatic theory?" An axiom is not a theory. Please don't make up eclectic terminology and pretend it has real meaning. Thanks.
johanfprins
1 / 5 (7) Aug 18, 2014
"the Higgs boson, which was recently confirmed to be the origin of mass,"

Please refer me to the experimental results that "confirmed" this. I do not know of ANY! All I know is that they picked up a noisy signal at CERN, which could be ANYTHING!!
Watebba
1.3 / 5 (7) Aug 18, 2014
The Higgs field explains inertial mass, and the gravity field will explain gravitational mass once we prove it exists.
We already know about curvature of space-time around massive bodies, which is called so and it's already proven well. On the other hand, it doesn't explain the gravity of massive bodies, it describes it. The derivation of Einstein's field equations depends on Newton's gravitational law in the same way, like for example the ancient Keplers laws - it doesn't explain it.
It only changes downward. You're making stuff up again.
It only changes downward for elastic spring oscillator with viscosity. The water and vacuum can do the opposite for certain range of wavelengths. Did you see this picture? This behavior is for example the reason of observations like these ones 1, 2.
Reg Mundy
1 / 5 (7) Aug 18, 2014
"the Higgs boson, which was recently confirmed to be the origin of mass,"

Please refer me to the experimental results that "confirmed" this. I do not know of ANY! All I know is that they picked up a noisy signal at CERN, which could be ANYTHING!!

Nice to see you again, Johan.
Watebba
1.8 / 5 (4) Aug 18, 2014
Math is the language of physics. If you don't understand the math you don't understand the physics
I can understand the both, but for construction of formal model you should understand the physical one first (the opposite way it's random guessing and it just doesn't work - compare the SUSY, stringy theories). So that the understanding of physics has always a priority before understanding of math. The math is only descriptive language of physical models - no less, no more. And frankly, the physics is full of phenomena, which still have no formal descriptions yet.
refer me to the experimental results that "confirmed" this
For example this one. The scalar boson character of Higgs resonance is already verified well.
Reg Mundy
1 / 5 (8) Aug 18, 2014
The best explanation for inertia is conservation of angular momentum and gyroscopic effects, as explained in the part of my book dealing with motion - how matter moves.
The Higgs field explains inertial mass, and the gravity field will explain gravitational mass once we prove it exists. The gravity field is difficult to detect and is currently less than our instruments are sensitive enough to find. When we make sensitive enough instruments we will be able to reconcile inertial mass with gravitational mass, as the Equivalence Principle of General Relativity Theory indicates. Then the gravitational field's relation to the Higgs field will be clear, and GRT and QM will be reconciled with one another.

I like your optimism. Could you let us know when all this is likely to happen? Meanwhile, I stand by my theory that there is NO GRAVITY, the effect is caused by expansion, and we will never detect a gravitational field, gravitinos, gravitons, gravity waves, Dark Matter, etc.
Watebba
2.3 / 5 (6) Aug 18, 2014
we will never detect a gravitational field
IMO you're just confusing the subject - the existence of gravity field is already proven well for centuries with forces which this field exerts to massive bodies. What for example the observation of gravitational lensing is? It's a detection of gravitational field - one of ways, in which this field manifest itself. Or you can detect the time dilatation with sensitive clock and another aspects of this field.
antialias_physorg
5 / 5 (6) Aug 18, 2014
if you cannot understand the phenomena at its intuitive level, which would allow the subsequent construction of formal model.

You are aware that our 'intuition' is an evolved feature?
Evolved features relate to things that markedly influence immediate survival. Cosmological, gravitational or quantum mechanical effects have not figured as individual selection mechanisms in the past.

So why would you even expect out intuition to be a good tool for getting at these phenomena? That seems nonsensical.
Watebba
1.4 / 5 (5) Aug 18, 2014
So why would you even expect out intuition to be a good tool for getting at these phenomena? That seems nonsensical.
Of course it's not reliable tool, but at the moment when you have no better way how to derive mathematical model, you simply have to use it. For example W. Pauli predicted the existence of neutrinos from missing energy of particle collisions. How he deduced it? Well, just with intuition. No formal model predicted the existence of such a particle before years. For further reading see the recent article here Harry J. Lipkin: Who ordered theorists?:
The best physics I have known was done by experimenters who ignored theorists completely and used their own intuitions to explore new domains where no one had looked before. No theorists had told them where and how to look.
The question rather is, why do you consider nonsensical things, which some others are writing articles about into Physics Today? I'd say, your intuition in this matter isn't very good.
johanfprins
1 / 5 (7) Aug 18, 2014
refer me to the experimental results that "confirmed" this
For example http://blogs.disc...R7HJC6E. The scalar boson character of Higgs resonance is already verified well.


This does not prove that the noise they measured at CERN gives "other particles" mass. It only proves wishfull surreal conjectures. Nothing more!
Watebba
1 / 5 (5) Aug 18, 2014
But this scalar noise would just give the particle mass - even in dense aether model. It's one of straightforward implications of it. The AWT even goes further in it than the contemporary physics and it attributes this noise to mass of all bosons, not just W/Z ones. Even at the water surface the existence of Brownian noise leads to massive behavior of surface solitons. It's evidence of massive nature of vacuum.

All bosons are waves, i.e. deforms of vacuum. Every deform of such vacuum leads into exposing of addition tiny density fluctuations of it, therefore the vacuum tends to be more massive, the higher the specific area, i.e. the frequency of this deform is. If the vacuum would be smooth and empty, then all waves would penetrate mutually in it like the massless ghosts. The existence of Higgs field was designed just to explain the difference.
Watebba
1 / 5 (5) Aug 18, 2014
The contemporary physics is somewhat confused in the matter of Higgs field, as it considers, that the lightweight bosons (gluons and photons) have zero mass, so no Higgs field is required at their energy density scale. IMO this is incorrect, as the gluons are massless only in context of already very dense atom nuclei, from human perspective they're very massive. In AWTthe Higgs field is scale invariant and it applies to all distance scales, including cosmological ones. The visible manifestation of Higgs field at the human observer scale is the CMBR noise, the manifestation of Higgs field at larger scale is the foamy structure of dark matter. The bosons are massless only when they have the same frequency like the underlying field. The bosons of lower frequency than the Higgs field are getting scattered fast with it like tachyons, so we tend to ignore their formation and propagation at distance. This is the reason, why we can observe only massive bosons (CPT symmetry gets broken here).
antialias_physorg
5 / 5 (7) Aug 18, 2014
For example W. Pauli predicted the existence of neutrinos from missing energy of particle collisions. How he deduced it?

No Zeph.
He took the current theory of his time and extrapolated from that. Energy was missing, but it wasn't coming out in the form of photons. So it had to be something massive.
That's the normal way to go about things unless you have evidence that something is seriously weird (as, e.g., in quantum mechanics). Then you need to start thinking in other directions.

The notion of "shut up and calculate" did not arise because scientists were lazy or lacked intuition. It's just the realization that at some point you have no choice but to trust tools other than intuition (in this case the math/physics that matches observation)

Only when that fails, too, can should go with random guessing (which is what intuition then boils down to).
swordsman
3.4 / 5 (5) Aug 18, 2014
Gravity is an electromagnetic effect that has been analyzed in detail. The question is possibly the definition of what mass consists. Are there two types of mass, one of which has no gravitational effects?
vlaaing peerd
5 / 5 (1) Aug 18, 2014
Is there anything else beside a higgsfield to provide mass to ...stuff 'n things?

I didn't the point about the mass of the higgs boson, if the higgsfield causes mass for other particles, what gives the higgs particle its' mass, if it is part of the mass-providing-field itself?
johanfprins
1 / 5 (6) Aug 18, 2014
EM-energy =m*c^2. This has been known since 1905. So why does one need a Higgs boson?Unless you are so crazy that you believe that an EM wave moves within an aether. If the latter is the case, the Doppler formulas would not be the ones that we measure.


Egleton
1.8 / 5 (5) Aug 18, 2014
Intuition is not "just" produced by evolution. Einstein likened it to a beautiful gift. The Rational mind is the faithful servant. We acknowledge the servant but ignore the gift.

https://www.youtu...9WO2B8uI

This whole fields and Higgs stuff fails the Grandmother test. Can you explain it to your grandmother?
Watebba
1 / 5 (5) Aug 18, 2014
This has been known since 1905. So why does one need a Higgs boson?
I already explained it here. If this formula would be valid, then all bosons would be massive. But the scientists so far recognized only W/Z bosons massive - so that they were forced to invent special particle and mechanism, which is responsible for it - just for W/Z bosons.
If the latter is the case, the Doppler formulas would not be the ones that we measure.
Why not? For example, if the ether would behave like the superfluid, then its action to light waves wouldn't be distinguishable from pure vacuum.
Watebba
1 / 5 (4) Aug 18, 2014
We acknowledge the servant but ignore the gift
The intuition can be learned in the same way, like the math formalism. It just requires different methods and attitude.
This whole fields and Higgs stuff fails the Grandmother test
It can be illustrated with water surface analogies. But these analogies lead to conclusions, which aren't consistent with contemporary physics (like the massive gluon). Fortunately it's not a problem of these analogies, but contemporary physics, because the gluons are really massive - if the wouldn't we could detect them at arbitrary distance (between others). Unfortunately the contemporary physics is deeply confused just with its formal models, i.e. the fact, they appear working despite they're based on logically fringe assumptions in similar way, like the epicycles models of medieval era.

This is just the caveat of formal models: they can work for many logical configurations at the same moment.
antialias_physorg
4.2 / 5 (5) Aug 18, 2014
EM-energy =m*c^2. This has been known since 1905. So why does one need a Higgs boson?

Because that doesn't tell you what 'm' actually is. And scientists have wondered about the physical meaning of the c^2 ever since (and still do). There's no question that the formula works and has been very successful. But there is still some explaining to do why it works (and I guess the one who figures that one out will get a Nobel Prize)
Watebba
1 / 5 (4) Aug 18, 2014
For example, whole the Standard model is based on the assumption, the interior of atom nuclei behaves like void empty vacuum. In this extremely dense environment all particles do behave as a much lightweight, than they really are: for example mesons do behave like bosons at the surface of atom nuclei. And the gluons do behave like massless bosons in its volume. They can swim from one side of it into another like photons across Universe. The fact, they cannot leave the atom nuclei doesn't bother anyone here: the whole formal model works and it gives the predictions consistent with observations - so what?

But the consequence (between others) is, you'll need a special explanation for existence heavy W/Z bosons, like the Higgs model. These bosons are so heavy, that even inside of very dense atom nuclei their own mass cannot be ignored so that they represent a deviation from all other gauge bosons. So we need some special particle which would explain this deviation - i.e. the Higgs boson.
Watebba
1 / 5 (4) Aug 18, 2014
So why does one need a Higgs boson? Because that doesn't tell you what 'm' actually is.
Nope, the only reason for Higgs boson introduction into physicis was the nonzero mass of W/Z bosons. If these bosons would appear massless like all other gauge bosons, nobody would care about some Higgs field at all. The physicists do care not least a bit, what the 'm' and another quantities actually are, until their equations are working.
Watebba
1 / 5 (4) Aug 18, 2014
One of common misinterpretations (which even Alphanumeric is spreading here carelessly) is, that the Higgs boson is responsible for the mass of all particles. Actually Higgs boson is only manifestation of Higgs field and the Higgs field is used for explanation of only about one percent of the observable matter. This is another argument against the claim, that the Higgs boson is here for explanation of E=mc^2. It actually introduces a deep dichotomy into explanation of matter with mass energy equivalence instead. You can believe me, I do understand perfectly, where and how the contemporary physics sucks.
Benni
3 / 5 (4) Aug 18, 2014
As you can see, the knowledge of math means nothing, if you cannot understand the phenomena at its intuitive level, which would allow the subsequent construction of formal model. Math is the language of physics. If you don't understand the math you don't understand the physics. You need at least introductory calculus to understand even classical physics. It's clear you haven't had it and don't.


Uh,oh, you just made a colossal blunder making a statement like this, the down voting of anything you post will soon begin as the mathematically challenged find this the most offensive language you can post on this site.

Toiea
1 / 5 (5) Aug 18, 2014
you just made a colossal blunder making a statement like this
It's not evident, which portion of quote you're trying to reply to. Does it actually matter?

BTW Best of all, the Higgs boson is not a prediction of Standard Model, rather violation of it. The Standard Model only requires the presence of Higgs field, but in no equation the mass of Higgs expects a fixed value. Well established "hiearchy problem" implies, that the quantum corrections to SM Lagrangian can make the mass of the Higgs particle arbitrarily large, since virtual particles with arbitrarily large energies are allowed in quantum mechanics. The Higgs boson is important for Higgs field detection, but the arguments for its existence are weak and generally come outside of Standard Model - the Standard Model doesn't require any measurable parameter of Higgs boson for nothing specific. The Higgs field can contribute to mass of W/Z bosons even without some Higgs resonance quite comfortably.
johanfprins
2 / 5 (4) Aug 18, 2014
So why does one need a Higgs boson? Because that doesn't tell you what 'm' actually is.
Nope, the only reason for Higgs boson introduction into physicis was the nonzero mass of W/Z bosons. If these bosons would appear massless like all other gauge bosons, nobody would care about some Higgs field at all. The physicists http://www.youtub...e-DwULM, what the 'm' and another quantities actually are, until their equations are working.


Zephyr: No matter how many pseudonyms you invent, you have been born a moron and will die a moron!
Toiea
1 / 5 (5) Aug 18, 2014
A pseudonyms? You mean arguments? I'm aware, my answers cannot impress both the supporters of Standard Model, both deniers of it. As usually, the truth is somewhere inbetween.
Da Schneib
4 / 5 (4) Aug 18, 2014
"the Higgs boson, which was recently confirmed to be the origin of mass,"

Please refer me to the experimental results that "confirmed" this. I do not know of ANY! All I know is that they picked up a noisy signal at CERN, which could be ANYTHING!!
How about the Nobel Prize Committee? From Higgs' "facts" page on their site: http://www.nobelp...cts.html

Francois Englert shared it with Higgs: http://www.nobelp...cts.html
Da Schneib
4 / 5 (4) Aug 18, 2014
The Higgs field explains inertial mass, and the gravity field will explain gravitational mass once we prove it exists.
We already know about curvature of space-time around massive bodies, which is called so and it's already proven well.
My bad. I thought "graviton field" but typed "gravity field."

It only changes downward. You're making stuff up again.
It only changes downward for elastic spring oscillator with viscosity. The water and vacuum can do the opposite for certain range of wavelengths.
Where does the energy come from? Because there's no free energy. The vacuum cannot create energy from nothing; averaged over macroscopic times and distances, the average energy of the vacuum must be the "zero point" defined by the cosmological constant.

Did you see http://i.imgur.com/NqMtkaz.jpg.
Yes. See the decay in the distance between the ripples? It's quite obvious if you look for it.
Da Schneib
4.2 / 5 (5) Aug 18, 2014
Math is the language of physics. If you don't understand the math you don't understand the physics
I can understand the both, but for construction of formal model you should understand the physical one first (the opposite way it's random guessing and it just doesn't work - compare the SUSY, stringy theories). So that the understanding of physics has always a priority before understanding of math. The math is only descriptive language of physical models - no less, no more. And frankly, the physics is full of phenomena, which still have no formal descriptions yet.
But we're not talking about developing new theories but about understanding what the existing ones say. You need to know the math to understand that thoroughly. On discovering new theories I am not sure I agree either, but that's another subject.

And BTW nice link to the Higgs info.
Da Schneib
4.2 / 5 (5) Aug 18, 2014
Could you let us know when all this is likely to happen?
"Hey Mike, ol' buddy, when you gonna be done with that painting on the ceiling? The Pope is coming for Mass next week and you *know* he hates the smell of paint."

Meanwhile, I stand by my theory that there is NO GRAVITY
"Tomorrow there will be NO WEATHER."
Toiea
1 / 5 (5) Aug 18, 2014
You need to know the math to understand that thoroughly
This is a stance of the lobby of theorists and high-school teachers. Whole the contemporary world is driven with various lobbies (GMO, BigPharma, fossil fuel lobby) - and the formally thinking people doing research are one of them. Whereas the real physicists are saying, you should be able to explain your model to your grandmother for to understand it at all. Apparently the cake of Nature understanding can be eaten from both sides (formal and non-formal one) and I just did choose this simpler easier to understand one (but more difficult to accept). The problem with acceptation of simple and easy logics of AWT and its water surface analogies is, it steals the jobs for many people, who just planned their safe life in "proper" (but pretty slow) progress in explanation of reality. The fact, this reality can be explained at few pages is very annoying situation for these people.
Da Schneib
4.3 / 5 (6) Aug 18, 2014
The visible manifestation of Higgs field at the human observer scale is the CMBR noise
This is incorrect. The CMB is the redshifted signal from the surface of last scattering when the temperature of the universe goes below the mass/energy needed to keep electrons in atoms.
Toiea
1 / 5 (5) Aug 18, 2014
The CMB is the redshifted signal from the surface of last scattering
Which surface are you talking about? The surface of what? This is not even mainstream physics.
Toiea
1 / 5 (5) Aug 18, 2014
BTW I don't want to undermine the significance of math and the relative success of formal physics in the progress of reality understanding of the last century, but from AWT follows, this success was mostly given with accident - the physicists mostly revealed just the portion of Universe, which follows the low-dimensional models well. With increasing distance from human observer scale the character of Universe returns to its intrinsic high-dimensional complex geometry, which poses a real problem for strictly deterministic low-dimensional models of contemporary physics. The problem isn't that the description of this geometry is difficult to describe with formal math, but simply because it's plain noneffective. Nobody will try to derive the exact analytical solution of the fluid inside of turbulent system - but this is exactly, what the quantum gravity theorists are attempting to do with space-time right now - until their money are going, indeed.
Toiea
1 / 5 (5) Aug 18, 2014
If I simplify the situation, during last century the physicists focused to description of pretty regular symmetric systems, like the electron orbital or massive stars (which are formed mostly with electron orbitals). These objects follows low-dimensional physical models so well, it even annoys the physicists, who are looking for signs of "New Physics" there. Unfortunately, once we move from dimensional scale of atom orbitals or massive stars, the situation becomes increasingly complex again and all these well working formal models developed during last century get gradually violated and poorly conditioned again. And the AWT predicts, that with increasing distance from human observer scale the situation will get only worse - we will be forced to change our gnoseologic paradigms for not to lose the contact with observable reality at all.
Da Schneib
4.2 / 5 (5) Aug 18, 2014
You need to know the math to understand that thoroughly
This is a stance of the lobby of theorists and high-school teachers.
Yeah, like Richard Feynman.

Oh, wait...
Da Schneib
4.3 / 5 (6) Aug 18, 2014
The CMB is the redshifted signal from the surface of last scattering
Which surface are you talking about? The surface of what? This is not even mainstream physics.
Of course not, it's cosmology.

http://en.wikiped...Features
Search on "surface of last scattering."
Da Schneib
4.2 / 5 (5) Aug 18, 2014
Stars are not made of electron orbitals, Zeph.
Toiea
1 / 5 (5) Aug 18, 2014
Well, it's visibility scope of the universe. BTW Note that the Universe geometry (FLRW metric) is inversed geometry (Schwarzchild metric) of black holes - they have their "surface of last scattering" too - and they remain stationary. Why the same geometry is considered developing in time with cosmologists? It should be stationary as well.
Toiea
1 / 5 (5) Aug 18, 2014
Stars are not made of electron orbitals
Didn't I say "mostly formed"? It's an example of holographic duality: the objects which are mostly composed of some smaller objects retain the character of these smaller objects. The (behavior of) large groups of people reflect the nature of individual people, the large stars composed mostly of spherical orbitals retain their sphericity (or shape in general).
Da Schneib
4.4 / 5 (7) Aug 18, 2014
Note that the Universe geometry (FLRW metric) is inversed geometry (Schwarzchild metric) of black holes
This is incorrect. The universe would be contracting if it were true, not expanding and not steady state.

... they have their "surface of last scattering" too
Black holes do not have a surface of last scattering.

and they remain stationary.
This is incorrect. They spin. It's one of their only attributes.

Why the same geometry is considered developing in time with cosmologists?
It's not. See above.
Da Schneib
4.2 / 5 (5) Aug 18, 2014
Stars are not made of electron orbitals
Didn't I say "mostly formed"?
That would also be incorrect. Stars are mostly made of protons.

Stars are spherical because of gravity, not because they're "formed (mostly) of orbitals."
johanfprins
1 / 5 (5) Aug 19, 2014
"the Higgs boson, which was recently confirmed to be the origin of mass,"

Please refer me to the experimental results that "confirmed" this. I do not know of ANY! All I know is that they picked up a noisy signal at CERN, which could be ANYTHING!!
How about the Nobel Prize Committee? From Higgs' "facts" page on their site: http://www.nobelp...cts.html


There is no proof whatsoever that mass would not have existed if this excited matter-state (noise) were not observed at CERN. Epicycles fitted the motion of the planets as seen from earth, but this did not prove that epicycles existed: In fact they do not! And I am sure that the "Higgs-field" is NOT required to explain mass. There is no experimental proof for this other than that it fits surreal mathematical symmetries.
Reg Mundy
1 / 5 (5) Aug 19, 2014
EM-energy =m*c^2. This has been known since 1905. So why does one need a Higgs boson?

Because that doesn't tell you what 'm' actually is. And scientists have wondered about the physical meaning of the c^2 ever since (and still do). There's no question that the formula works and has been very successful. But there is still some explaining to do why it works (and I guess the one who figures that one out will get a Nobel Prize)

Hey, my theory explains what mass actually is! Do you think I should notify the Nobel committee?
Reg Mundy
1 / 5 (6) Aug 19, 2014
@Da Schneib
Could you let us know when all this is likely to happen?
"Hey Mike, ol' buddy, when you gonna be done with that painting on the ceiling? The Pope is coming for Mass next week and you *know* he hates the smell of paint."

Meanwhile, I stand by my theory that there is NO GRAVITY
"Tomorrow there will be NO WEATHER."

Now, Daz, you should know better than to provoke me.....
Anyway, you know I am right, so stop wriggling. You were warned not to read the book.
Watebba
1.5 / 5 (4) Aug 19, 2014
I am sure that the "Higgs-field" is NOT required to explain mass
Which alternative explanation do you propose?
my theory explains what mass actually is!
Which alternative explanation do you propose?
Reg Mundy
1 / 5 (6) Aug 19, 2014
Stars are not made of electron orbitals
Didn't I say "mostly formed"?
That would also be incorrect. Stars are mostly made of protons.

Stars are spherical because of gravity, not because they're "formed (mostly) of orbitals."

Stars are spherical because the matter within them is expanding and peer pressure forces a spherical shape. There ain't no gravity!
Watebba
1 / 5 (4) Aug 19, 2014
Stars are mostly made of protons
When we talk about shape of stars, the volume ratio is important here.
Stars are spherical because the matter within them is expanding and peer pressure forces a spherical shape
Of course, but I'm disputing the sphericity of objects as a general measure of distance scale. At the human scale nearly nothing is really spherical (with exception of human made artifacts). With increasing distance from human observer scale the sphericity of objects increases first, then decreases again. The distance scale at which the symmetry of objects gets highest is also the scope of the best validity of 4D space-time based physics (quantum mechanics and general relativity).
antialias_physorg
4.2 / 5 (5) Aug 19, 2014
Hey, my theory explains what mass actually is! Do you think I should notify the Nobel committee?

Currently you have only claims what your alleged theory can do. In this you are like religious nuts: they also claim they have a god - but fail to show it. Unless you show what you have to peer review: who cares?

At the human scale nearly nothing is really spherical

Erm...maybe because spherical isn't a sensible shape in a non-uniform environment with a predominant force direction?
johanfprins
1 / 5 (5) Aug 19, 2014
I am sure that the "Higgs-field" is NOT required to explain mass

Which alternative explanation do you propose?


I do not have to propose an alternative since Einstein already solved this in 1905 with E=m*c^2.

When you Lorentz-transform the dimensions of a moving electron from the IRF in which it is stationary into the one you observe it as being moving with a speed v, the electron becomes longer and has a de Broglie wavelength. This proves that the mass of the electron is electromagnetic energy. Thus, for a stationary electron with rest-mass m(0) one has EM energy h*(nu)=m(0)*c^2. When it moves, it has energy m*c^2 where m>m(0); so that m*c^2=T+m(0)*c^2 where T=kinetic energy.

When it is trapped around around a nucleus it has a STATIONARY energy m*c^2 where m is smaller than m(0).;I.e. m*c^2=m(0)*c^2minusU, where U is the ionization-energy (purely potentialo energy). Sucha n electron-wave has NO kinetic-energy whatsoever. It can have quantum fluctuations though.
Watebba
1.3 / 5 (4) Aug 19, 2014
I do not have to propose an alternative since Einstein already solved this in 1905 with E=m*c^2.
So why the gluon and photons are massless and the W/Z bosons are massive? BTW formal formula i not an explanation - just a regression. The gravitational law is not an explanation of gravity - only description.
Erm...maybe because spherical isn't a sensible shape in a non-uniform environment with a predominant force direction?
This is just the point: at the human observer scale the Universe is full of hyperdimensional forces, which are violating the inverse square law. The irregularity of objects is therefore a manifestation of extradimensions of space-time.
johanfprins
1 / 5 (5) Aug 19, 2014
I do not have to propose an alternative since Einstein already solved this in 1905 with E=m*c^2.
So why the gluon and photons are massless and the W/Z bosons are massive?


If you slam electrons and protons together you will get matter waves with higher energies, which will decay back into lower mass-energy protons, electrons, neutrino's and photons (light-waves); since the latter are all EM-energy. A photon-WAVE is NOT massless: It has a centre-of mass. This centre-of-mass moves with c within all IRF's and can thus not be stationary: Therefore a photon-wave does not have REST-mass. Gluons, and W/Z bosons are interpretations of excited EM-waves, all of which have mass-energy: Some with rest-mass others not with rest-mass . The interpretation that they are "particles" that cause force, is probably a figment of imagination that lives within halucinary minds.
Toiea
1 / 5 (3) Aug 19, 2014
Gluons, and W/Z bosons are interpretations of excited EM-waves, all of which have mass-energy: Some with rest-mass others not with rest-mass
That was deep, really. May the Force be with you.
Da Schneib
4.2 / 5 (5) Aug 19, 2014
There is no proof
Physics doesn't have "proof." Proof is mathematics, not physics. Physics has theories, and the theory of the Higgs and its responsibility for the quantum parameter we call "mass" have been accepted as correct by the overwhelming majority of professional physicists.

There is no experimental proof for this other than that it fits surreal mathematical symmetries.
That would be "experimental evidence" in physics. There is no proof in physics.

In fact, the experimental evidence of the detection of the Higgs at the CERN LHC is comprehensive, impressive, and sufficient to convince the physics community that the Higgs has been seen. Its decay modes, which two different detector experiments have confirmed, are the last piece of what you call "proof" required.
johanfprins
1 / 5 (6) Aug 19, 2014
There is no proof
Physics doesn't have "proof." Proof is mathematics, not physics.
Utter BULLSHIT that can only come from a seriously demented mind!

Physics has theories, and the theory of the Higgs and its responsibility for the quantum parameter we call "mass" have been accepted as correct by the overwhelming majority of professional physicists.
Without ANY definite experimental proof!!

There is no experimental proof for this other than that it fits surreal mathematical symmetries.
That would be "experimental evidence" in physics. There is no proof in physics.
If there is NO proof in physics why did they build an acceleotor in CERN for billions of dollars which they STOLE from taxpayers pockets?

johanfprins
1 / 5 (5) Aug 19, 2014
There is no proof
Physics doesn't have "proof." Proof is mathematics, not physics. Physics has theories, and the theory of the Higgs and its responsibility for the quantum parameter we call "mass" have been accepted as correct by the overwhelming majority of professional physicists.
So you are arguing that physics is determined by majority-vote? LOL!!

There is no experimental proof for this other than that it fits surreal mathematical symmetries.
That would be "experimental evidence" in physics. There is no proof in physics.

In fact, the experimental evidence of the detection of the Higgs at the CERN LHC is comprehensive, impressive, and sufficient to convince the physics community that the Higgs has been seen. Its decay modes, which two different detector experiments have confirmed, are the last piece of what you call "proof" required.


You claim thre is "no proof in physics" and in the very next breath you claim "proof" at CERN! LOL! Are you sane?
Da Schneib
5 / 5 (3) Aug 19, 2014
If you don't understand the difference between physics and math I can't help. Good luck.

So you are arguing that physics is determined by majority-vote?
You are claiming I'm making the Appeal to Authority Fallacy. However, it's not a fallacy if the authority cited really is expert in the subject at hand. Are you arguing the majority of practicing professional physicists are not experts on physics?

Really?
Reg Mundy
1 / 5 (6) Aug 19, 2014
@johan

The interpretation that they are "particles" that cause force, is probably a figment of imagination that lives within halucinary minds.
While generally I agree with you, I'm glad you used the word "probably" in there, as EXPERIMENTAL EVIDENCE splits equally between particle and wave. I suggest that light is actually a wave but when we measure it using time, because time itself is quantum, everything we measure with it at a very small scale APPEARS quantum to us. And a quantum, or packet, of light is as much a particle as any other particle and is composed like them of gyrating sub-particles each of which is composed of lesser particles and so ad infinitum. The Higgs is just another particle and this "conferring mass" is conjecture at best or rubbish.
Reg Mundy
1 / 5 (6) Aug 19, 2014
@DAz
If you don't understand the difference between physics and math I can't help. Good luck.

So you are arguing that physics is determined by majority-vote?
You are claiming I'm making the Appeal to Authority Fallacy. However, it's not a fallacy if the authority cited really is expert in the subject at hand. Are you arguing the majority of practicing professional physicists are not experts on physics?

Really?

Maths deals with models, physics deals with reality. When"practicing professional physicists " pursue mathematical models with no basis in reality, the result is billions of dollars spent looking for gravity waves, devising equations for string theory in eleven dimensions, etc.
Da Schneib
5 / 5 (2) Aug 19, 2014
Particles can actually be seen with the naked eye shining by the light of photons they emit and absorb. You trap them with a Penning Trap (http://en.wikiped...ng_trap) and excite them with a laser. This was done in Seattle at the University of Washington in 1984 by Hans Dehmelt. You can read about it in The Infamous Boundary by David Wick. Dehmelt shared a Nobel Prize for his work with Penning Traps.

Here is an excerpt from the announcement made at the time:
Here, right now, in a little cylindrical domain... in the center of our Penning Trap resides positron Priscilla, who has been giving spontaneous and command performances of her quantum jump ballets for the last three months.


For that matter, if your eyes are well-accustomed to the dark and you are into the scotopic vision domain, your eye can see a single photon.

So denying particles is pretty much contrary to reality.
Uncle Ira
5 / 5 (5) Aug 19, 2014
Maths deals with models, physics deals with reality. When"practicing professional physicists " pursue mathematical models with no basis in reality, the result is billions of dollars spent looking for gravity waves, devising equations for string theory in eleven dimensions, etc.


@ Reg-Skippy I don't think that is right. You got it mixed up Cher. I'm not a professional physicist like you aren't either so maybe we should ask one of the professional physics-Skippys if that is what think think about what they are doing.
Da Schneib
5 / 5 (3) Aug 19, 2014
Maths deals with models, physics deals with reality.
This is incorrect. Physics deals with models of reality. These models are called conjectures, hypotheses, theories, and Laws of Nature. All of them are mathematical, but math is not physics; it's the *language* of physics.

When"practicing professional physicists " pursue mathematical models with no basis in reality
All conjectures have a basis in reality. There is no reason to make a conjecture unless you see something and go, "Gee, that's funny." Isaac Asimov said that's how most great discoveries work in science.

When they figure out a way to test the conjecture, then it's promoted to a hypothesis. When they test it, if it passes the test, then it becomes a theory. This is called the Scientific Method. Later, because of the math in a successful theory, a conjecture outside the scope of that theory might arise based on the theory's math. The source of the conjecture is immaterial. It's the same process either way.
Reg Mundy
1 / 5 (5) Aug 19, 2014
@Daz
This is incorrect. Physics deals with models of reality.

Crap! Physics deals with reality. Demonstrable experiments, real observations, actual measurements.

When they figure out a way to test the conjecture, then it's promoted to a hypothesis.

Which is of course where modern mainstream physics falls down. Too much conjecture being promoted before it passes any meaningful test.
RealityCheck
1 / 5 (6) Aug 19, 2014
WARNING: Real Science based on Objectively Observable Reality. All BBangers read further at own risk!
The CMB is the redshifted signal from the surface of last scattering...
That is the single most misleading continuing BB-Inflation/Expansion etc hypothesis/conjecture-based Interpretation made by professional physicists today. Their 'surface of last scattering' is entirely derived from BB scenarios, not actually confirmed in reality. In fact, every processing/evolving E-M/Gravitational 'feature/body' (from micro to macro scales) has its own type of 'surface of last scattering' from which a certain range of radiations may be detected as the CMB we 'see' now/here given sufficient travel opportunity from 'sources'. Eg, there are innumerable Black Hole features where radiation leaving the close vicinity of event horizons is gravity-red-shifted to a 'CMB-like spectrum' of Microwave lengths/frequencies. Same for massive accretion disk & plasma jet gravity-red-shift.

Rethink it all.:)
Toiea
1 / 5 (3) Aug 19, 2014
This is incorrect. Physics deals with models of reality.
Crap! Physics deals with reality. Demonstrable experiments, real observations, actual measurements.
Well said. The physics is experimental science and it always was. Just the decadent physicists (of recent decades) attempted to transform it into mess separated from reality. "It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong." - Richard P. Feynman
Captain Stumpy
5 / 5 (7) Aug 20, 2014
"It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong." - Richard P. Feynman
@toiea/wataba/zephir
I am SO glad you posted that quote
I want you to SEE SOME EXPERIMENTS THAT PROVE SOMETHING:
http://arxiv.org/...1284.pdf
http://exphy.uni-...2009.pdf

Notice that both of these EXPERIMENTS prove your precious awt is DEAD
this is where you accept reality and accept, per your own quote that I re-posted, that the experiments do NOT match your philosophy, and thus your philosophy is WRONG

just wanted to POINT OUT THE FLAW to you so that you will remember it more clearly, because I am using your own argument for proof. Perhaps now we can sink it for good?

THANKS
johanfprins
1 / 5 (5) Aug 20, 2014
So you are arguing that physics is determined by majority-vote?
You are claiming I'm making the Appeal to Authority Fallacy. However, it's not a fallacy if the authority cited really is expert in the subject at hand. Are you arguing the majority of practicing professional physicists are not experts on physics?


In Galileo's time the majority explained planetary motion in terms of epicycles. So according to your argument the majority was correct and Galileo was wrong? Are you really as stupid as all this or are you just a troll? The majority of "experts'' can be wrong, as they have been time and again: Remember tectonic plates?

Really?
johanfprins
1 / 5 (5) Aug 20, 2014
Particles can actually be seen with the naked eye shining by the light of photons they emit and absorb. You trap them with a Penning Trap (http://en.wikiped...ng_trap) and excite them with a laser. This was done in Seattle at the University of Washington in 1984 by Hans Dehmelt. You can read about it in The Infamous Boundary by David Wick. Dehmelt shared a Nobel Prize for his work with Penning Traps.
How do you know that they are "particles" before they ABSORB light-energy? When a wave is trapped within a cavity, it MUST form a stationary wave that fills the WHOLE cavity. When you shine light on these waves you change the BOUNDARY CONDITIONS, and the waves collapse onto localised waves. Quantum mechanics is quite adamant about the fact that when you measure, what you measure changes. So this experiment does NOT prove the existence of particles AT ALL!!

So denying particles is pretty much contrary to reality.
BULLSHIT. You are not posting logic.
johanfprins
1 / 5 (5) Aug 20, 2014
While generally I agree with you, I'm glad you used the word "probably" in there, as EXPERIMENTAL EVIDENCE splits equally between particle and wave. I suggest that light is actually a wave but when we measure it using time, because time itself is quantum, everything we measure with it at a very small scale APPEARS quantum to us.
Nope! An electron is always a wave which acts as an antenna that can absorb light. The fraction of light that an antenna can absorb is determined by the physical properties of the antenna: As any radio engineer can attest to. An electron has energy m*c^2=h*(nu)" ALWAYS. This is where Plancks's constant comes into the picture. A trapped electron has an energy m*c^2=h*(nu) which is smaller than its rest-mass energy m(0)*c^2=h*(nu0). Threfore it can be ionized when it gains energy h*(nu0)-h*(nu)=h*(nuEM), where (nuEM) is the frequency of the light wave required. Thus h comes from the properties of the electrons: NOT from a light-wave being quantized!
johanfprins
1 / 5 (5) Aug 20, 2014
"It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong." - Richard P. Feynman


That is NOT the full story: Even when it fits known experimental data it could still be wrong! If you are a competent physicist you expect that it might be possible to discover new data that the existing theory cannot model. If you do not keep this option open, you should get the hell out of physics research. Particle physicists fudge away what they do not like, using renormalization, to "doctor" up their numbers. This is not physics but FRAUD!
Reg Mundy
1.8 / 5 (5) Aug 20, 2014
@johan
Nope! An electron is always a wave which acts as an antenna that can absorb light.

Why are you talking about electrons in particular when I was talking about photons? If you are choosing to focus on electrons, why electrons rather than any of the other "particles"? But I'll play along. For electrons, how do you explain the results of Young's experiment en.wikipedia.org/wiki/Double-slit_experiment where electrons behave as PARTICLES? If it looks like a duck, walks like a duck, and quacks........
johanfprins
1 / 5 (5) Aug 20, 2014
@johan
Nope! An electron is always a wave which acts as an antenna that can absorb light.

Why are you talking about electrons in particular when I was talking about photons? If you are choosing to focus on electrons, why electrons rather than any of the other "particles"? But I'll play along. For electrons, how do you explain the results of Young's experiment en.wikipedia.org/wiki/Double-slit_experiment where electrons behave as PARTICLES? If it looks like a duck, walks like a duck, and quacks........


They do NOT behave like particles, since by definition particles cannot diffract EVER! The electron is an EM-wave moving with a speed v that is less than c. ALL waves change their shapes and sizes when the boundary conditions change. When the electron-WAVE reaches the double slits, NEW boundary conditions are encountered, so that the wave morphs and MOVE THROUGH BOTH SLITS. On the other side of the slits the two lobes that moved through th slits interfere and form ...
johanfprins
1 / 5 (6) Aug 20, 2014
.... a diffracted wave-front that approaches the screen in which there is a distribution of atomic-sized absorbers or entities with which each wave can interact. Only ONE of the absorbers can absorb the incoming single electron wave. Thus, as soon as the wave resonates with one of the absorbers, it HAS TO COLLAPSE IN SIZE TO BE ABSORBED AND THUS LEAVES A SPOT.

Since resonance is higher where the identical impinging waves have their highest intensities, the spots accumulate after many electron-waves to give the intensity of the identical waves. There is NO VOODOO probabilities involved except for the NORMAL probability one has for a roulette-wheel, caused in this case by resonance, which allows the incoming waves to collapse into different absorbers.

If you are a moron and try and check through which slit a single electron-wave has moved, your detector collapses the diffracted wave BEFORE it reaches the screen, AND THEREFORE YOUR DIFFRACTION PATTERN DISAPPEARS.

NO Particles!!!

johanfprins
1 / 5 (6) Aug 20, 2014
Why in God's name do theoretical physicists want to believe in Voodoo? Why can they not realise that Niels Bohr had a mental problem when he argued that "if you think you understand quantum mechanics you do not understand it"! This is bullshit!

All mainstream physicists from Bohr to Feynman have confessed that "they do not understand quantum mechanics". But when Schroedinger and de Broglie, and other competent physicists, argue that it can be explained totally in terms of wave and wave interactions, they are villified and mocked, since "one can only understand QM if you do not understand it". What a load of claptrap. To understand physics you must first confess that you are unable to understand it: If you do not do so, you do not understand it!

Theoretical physics is thus based on the absurd!
Watebba
1 / 5 (3) Aug 20, 2014
The contemporary science shares many similarities with Holy Church of medieval era. The scientists maintain their own chorals, initiation periods for their priests (no one else can publish in science), the explanations, why we shouldn't research something, because it doesn't belong into subject of science and even the claims, some things cannot be understood at all in similar way, like St. Thomas Aquinas excluded the option of understanding of God in advance.
Reg Mundy
1 / 5 (6) Aug 20, 2014
@johan
Sorry. J, I should have been more specific, rather than quoting Young in general, this is the part to which I refer:-
versions of the experiment that include detectors at the slits find that each detected photon passes through one slit (as would a classical particle), and not through both slits (as would a wave).
Particles!
(Actually, I don't believe either explanation. I think an electron (and a photon) are that part of a wave we perceive in a quantum of time, which to us is the same as a particle.)
Toiea
1.5 / 5 (4) Aug 20, 2014
IMO the wave-particle duality has its tangible physical representation: why not to handle the electron as a pin-point particle, which is surrounded with its wake wave like the boat or the fish at the water surface? Such a mutually connected system would fit both the attitude of particle physicists, both the people, who just want to see the waves everywhere.
Toiea
1 / 5 (4) Aug 20, 2014
I think an electron (and a photon) are that part of a wave we perceive in a quantum of time, which to us is the same as a particle
IMO it applies to photon only (and still for quite long-wavelenght ones). Only microwave photons do behave like / are identical with their own deBroglie wave. The photons of increasing frequency are gradually separated more and more from their deBroglie wave like energetically more dense solitons.

Such an interpretation fits well the results of double slit experiment, in which the long-wavelength photons interfere with slit rather like the homogeneous classical wave, whereas the short-wavelength photons manifest itself like the better or worse distinguished spots or sparks at the target. There must be a seamless transition in behavior of both types of photons with increasing frequency.

Toiea
1 / 5 (4) Aug 20, 2014
The mainstream physics ignores all thougts about actual shape of photons consequentially, but we already know, that the long wavelenght photons behave and interfere like the classical wave, the X-ray photons behave like the diffuze clouds inside of Wilson cloud chamber and the gamma ray photons spread like the well localized sparks across space. Therefore the appearance of photons at the target during double slit experiment corresponds well with their bulk behavior inside of detectors. Maybe we should less speculate and better observe the artifacts, which we already know well (or at least believe so).
johanfprins
1 / 5 (6) Aug 21, 2014
versions of the experiment that include detectors at the slits find that each detected photon passes through one slit (as would a classical particle), and not through both slits (as would a wave). Particles!
The detector CANNOT determine through which slit the photon has moved since it can ONLY detect the photon-wave by COLLAPSING it.
Thus when you have two detectors behind the slits and think that you measure through which slit the wave has moved, you are arguing like an idiot. The wave moves through BOTH slits but can only be detected by ONE of the detectors. It thus collapses into the detector with which it resonates FIRST.
Thus by sending many photon-waves THROUGH BOTH SLITS, half of them will collapse within one detector and half within the other detector. THIS DOES NOT TELL YOU THAT THE WAVE MOVED THROUGH A SINGLE SLIT.
It is unbelievable that the mainstream theoretical physicists could have been so stupid for nearly 100 years. No wonder they are hunting surreal Higgs!
johanfprins
1 / 5 (6) Aug 21, 2014
@ Reg Mundy

Actually, I don't believe either explanation. I think an electron (and a photon) are that part of a wave we perceive in a quantum of time, which to us is the same as a particle.


There is no quantum in time: There is only a quantum in EM wave-energy. Time is continuous and absolute (not relative as Einstein has incorrectly argued). See the draft of my recent book entitled: EINSTEIN=GENIUS: But a genius sometimes blunders:
https://www.resea...=prf_act

Watebba
1 / 5 (3) Aug 21, 2014
There is no quantum in time
It's not quite so simple. You're apparently supporter of abstract observational perspective, in which space always represents a continuum and all particles are waves of it. But for observer at the water surface the wave behaves so only up to certain minimal level, bellow which all waves will get scattered with Brownian noise, i.e. density fluctuations of underwater. The water still behave like the continuum, unfortunately for observer of water surface, who has its surface ripples only available for observation such an information is accessible and quite abstract. All the available physical experiments would indicate the fragmentation of time arrow and surface ripples into particles instead. In AWT the 3D observable reality observable through density fluctuations of vacuum foam behaves in the same way, like the above simplified 2D water surface analogy and the uncertainty of quantum mechanics is already an evidence of the fragmentation of time arrow.
Reg Mundy
1.8 / 5 (5) Aug 21, 2014
@johan
The detector CANNOT determine through which slit the photon has moved since it can ONLY detect the photon-wave by COLLAPSING it.
What about electrons?
There is no quantum in time: There is only a quantum in EM wave-energy. Time is continuous and absolute

Beg to differ. Time is quantum. Unless you can prove it isn't.....
Reg Mundy
1 / 5 (4) Aug 21, 2014
@johan
Liked your abstract. Agree with your assessment of modern physicists, who have been subverted by mathematical models which fail the traditional PHYSICS tests of reality and experimental evidence. However, disagree with your clock having to keep a million different times for a million observers, clock only keeps one time and each observer perceives a different time. Great fun!
Watebba
1 / 5 (4) Aug 21, 2014
@johan Liked your abstract
J.F.Prins is definitely an interesting and controversial personality of contemporary physics - in some areas he acts as a true progressive, on the other hand when it gets into his math and its naive extrapolations, then he becomes a more conservative, than the mainstream physics (which already has its math developed into much more complex & deeper level so that its extrapolations are more close to AWT reality). In particular regarding particle-wave duality the mainstream science ocupies a much more insightful stance. For example by Aspect's experiments the particle-wave duality not only can be observed, but it manifest itself at the same moment in context of single experiment. Which is nothing strange if you take a look at the double slit experiment: its result appears both like being composed of dots, bot like the wave interference pattern - at the same moment.
johanfprins
1 / 5 (6) Aug 21, 2014
@johan
The detector CANNOT determine through which slit the photon has moved since it can ONLY detect the photon-wave by COLLAPSING it.
What about electrons?
An electron is also an electromagnetic wave. This follows from relativity (read my book). An electron-wave also adapts its shape and size when the boundary conditions change. The Copenhagen baboons called this behaviour "quantum jumps" of "particles".
There is no quantum in time: There is only a quantum in EM wave-energy. Time is continuous and absolute

Beg to differ. Time is quantum. Unless you can prove it isn't.....
It is not necessary for ME to disprove your statement. The onus is on YOU to prove that it is so. It is your statement for which there is NO experimental evidence whatsoever.
johanfprins
1 / 5 (6) Aug 21, 2014
@johan
Liked your abstract. Agree with your assessment of modern physicists, who have been subverted by mathematical models which fail the traditional PHYSICS tests of reality and experimental evidence. However, disagree with your clock having to keep a million different times for a million observers, clock only keeps one time and each observer perceives a different time. Great fun!

Read the correct derivation of the Lorentz equations in chapter 6: You will then see that the million times deduction is absolutely correct. And, Oh yes, first read chapter 4, so that you can understand what "the principle of relativity" actually is. Einstein, Lorentz, Poincare, and all mainstream modern theoretical physicists up to the present have NEVER understood Galileo's genius!

A relativistic transformation of physics CAN NEVER be covariant.
Watebba
2 / 5 (4) Aug 21, 2014
An electron is also an electromagnetic wave. This follows from relativity (read my book).
The electron exhibits a weak charge, which is NOT an electromagnetic wave. It's particularly because the transverse waves of vacuum undulate inside the deep density gradient, which is formed with its spreading, so that inside of this gradient they do change into longitudinal wave. This longitudinal wave (actually inner portion of Mobius loop of photon) manifest itself as a weak charge of electron at distance.

The ability of theorists, who are able to follow their pet concepts deeply outside of their validity scope is nothing exceptional in physics - for example the relativity theorists are insisting on fixed speed of light trough vacuum even at the moment, when this light is already revolving the black hole at place. From their intrinsic perspective they're indeed still correct, but from perspective of any other outer observer the light is already sitting at place and it doesn't move at all
Watebba
2 / 5 (4) Aug 21, 2014
The similar conceptual problem for contemporary physicists represents the dark matter, which we can imagine like the composite emergent effect of many tiny but temporal gravitational lenses. Locally the light indeed follows these lenses exactly according the rules of relativistic aberration - but when this lens disappears faster than the photon can encircle it in symmetric way, then the result indeed breaks the Gauss theorem of Maxwell electrodynamics and the path of light will get the character of magnetic monopole. Macroscopically such an environment of many temporal gravitational lenses violates the relativity after then. It's interpretation therefore will depend on what we could actually observe and measure in it - not the abstract ideal behavior of light.
johanfprins
3.3 / 5 (7) Aug 21, 2014
@ Watebba, Zephyr or whatever.

Anybody who argues that an EM wave moves within an aether, is so lost that it is a waste of time to argue with him/her. The Doppler-shifts for a wave moving within an aether is TOTALLY different from the Doppler shifts for a wave that moves with the SAME speed c within ALL inertial refrence frames. Experimental data proves that the Doppler-shifts for electromagntic waves are NOT those of waves moving within an aether!

Regards,
Johan
Captain Stumpy
4.8 / 5 (6) Aug 21, 2014
@ Watebba, Zephyr or whatever.

Anybody who argues that an EM wave moves within an aether, is so lost that it is a waste of time to argue with him/her. The Doppler-shifts for a wave moving within an aether is TOTALLY different from the Doppler shifts for a wave that moves with the SAME speed c within ALL inertial refrence frames. Experimental data proves that the Doppler-shifts for electromagntic waves are NOT those of waves moving within an aether!

Regards,
Johan
@johanfprins
yes, it is zephir (Watebba is zephir)
5 stars for this comment! great one

Next time add something like THIS for zephir to chew on: http://arxiv.org/...1284.pdf

THANKS
Toiea
1 / 5 (3) Aug 21, 2014
The Doppler-shifts for a wave moving within an aether is TOTALLY different from the Doppler shifts for a wave that moves with the SAME speed c within ALL inertial reference frames
You're right, but the trick is in the way of observation of this motion. At the water surface we can walk around river and we can observe its ripples with using of light waves, which are indeed much faster and which they mediate another reference frames for us. But in vacuum we have no such option available and we are forced to observe the spreading of its waves with the very same kind of waves.

For to get the AWT analogies, you should live at the water surface like the blind waterstrider and you should observe the waves, measure their wavelength and timing frequency with just, well with surface waves again - pretty consequentially. Which may be indeed difficult to imagine for someone - but IMO it's still better to have some illustrative & faithful physical analogy for relativity, than none at all.
Toiea
1 / 5 (3) Aug 21, 2014
For example, this is the picture of AWT analogy of the relativistic length contraction at the water surface. If we would observe the boat with light waves, then we will observe nothing unusual and the boat could achieve the arbitrary speed. But if we could use only the surface ripples for its observation, then the whole notion of boat will become reduced to the pair of wake waves at the bow and stern of boat. And we can measure the length of the boat only as a distance between them. Under such a situation the perceived length (and motion direction vector) of boat will depend on relative speed of boat pretty much. As we can imagine easily, under such a situation we can never observe the boat moving faster than the wake wave itself.
johanfprins
2.3 / 5 (3) Aug 21, 2014
@ Captain Stumpy,

Thanks for the reference: I will read it with interest.

I see he/she has just now posted the same old crap under the name Toiea. He/she just cannot understand that a water-wave moves within an aether called "water" and that the Doppler formulas FOR ALL WAVES MOVING WITHIN A MEDIUM CANNOT EVER BE WHAT IS MEASURED FOR EM-WAVES!!

I wish this person would go back to primary school and learn a bit of physics!
Toiea
1 / 5 (4) Aug 21, 2014
the Doppler formulas FOR ALL WAVES MOVING WITHIN A MEDIUM CANNOT EVER BE WHAT IS MEASURED FOR EM-WAVES
Well, the situation is exactly the opposite. The EM-Wave motion is equivalent to motion of all transverse waves in any (material) environment. Actually, until these waves are perfectly transverse, the exact nature of environment doesn't matter here at all - their reference frame remains undefined anyway.

The tiny capillary waves at the water surface are rather close to true transverse waves and for such a waves the water behaves like the thin elastic membrane driven with surface tension only. No motion or reference frame of underwater gets actually detectable here, because it has no meaning for transverse wave. The weak residual reference frame drag coming from Brownian noise can be eliminated with using of waves of the same effective wavelength - and this is the main trick of the zero result of this study.
johanfprins
1 / 5 (5) Aug 21, 2014
BTW: Thre is NOT ANY RELATIVISTIC LENGTH CONTRACTION: Einstein transformed from the WRONG IRF into the WRONG IRF. When you do the transformation correctly, YOU GET LENGTH DILATION: THE ROD BECOMES LONGER!!
Toiea
1 / 5 (4) Aug 21, 2014
When you do the tranmsformation correctly, YOU GET LENGTH DILATION: THE ROD BECOMES LONGER!
That's nice, but this is not what we can actually measure from extrinsic perspective. The length contraction is actually required for to maintain the light speed invariance - it's the straightforward consequence of the Lorentz invariance postulate. Without lenght contraction the light emanated with object in motion would propagate with superluminal speed.

animation

You can believe me, my logics is straightforward, consistent and undeniable. It's based only and just on geometric assumptions, as expressed with various pictures and animations. So it cannot be a subject of accidental inversion of observational perspectives and similar stuffs, which are typical for formal derivations of mainstream physics (including yours).
Da Schneib
3.7 / 5 (3) Aug 21, 2014
This is incorrect. Physics deals with models of reality.

Crap! Physics deals with reality. Demonstrable experiments, real observations, actual measurements.
What about the theorists? You're only paying attention to the experimenters. The experimenters don't know what to look for, how to look for it, or anything else without the theorists telling them, and if the experimenters see something strange it's the theorists who look for an explanation. You've left out half of physics and called it "not physics." This is silliness.

When they figure out a way to test the conjecture, then it's promoted to a hypothesis.

Which is of course where modern mainstream physics falls down. Too much conjecture being promoted before it passes any meaningful test.
Are you suggesting that adding a test that experimentalists can check to a conjecture is somehow invalid? How then do you suggest hypotheses be formed? You will cripple physics. Again, more silliness.
Da Schneib
3.7 / 5 (3) Aug 21, 2014
So you are arguing that physics is determined by majority-vote?
...it's not a fallacy if the authority cited really is expert in the subject at hand. Are you arguing the majority of practicing professional physicists are not experts on physics?
In Galileo's time the majority explained planetary motion in terms of epicycles.
Irrelevant. In Galileo's time the scientific method of the time did not involve testing hypotheses, or even devising tests for conjectures.

You haven't answered my question.
Toiea
1 / 5 (3) Aug 21, 2014
The experimenters don't know what to look for, how to look for it, or anything else without the theorists telling them
Apparently at least one guy would disagree with you in this matter. You can count me a second one, BTW.
"I have no patience with social scientists, historians, and philosophers who insist that the "scientific method" is doing experiments to check somebody's theory. The best physics I have known was done by experimenters who ignored theorists completely and used their own intuitions to explore new domains where no one had looked before. No theorists had told them where and how to look".
Da Schneib
3.7 / 5 (3) Aug 21, 2014
Particles can actually be seen with the naked eye shining by the light of photons they emit and absorb.
How do you know that they are "particles" before they ABSORB light-energy?
Because charge, mass, energy, lepton count, electron family count, and angular momentum are all conserved quantities, and light doesn't get absorbed by empty space.

When a wave is trapped within a cavity, it MUST form a stationary wave that fills the WHOLE cavity.
The wave isn't trapped in a cavity. The particle is trapped in a magnetic field.

When you shine light on these waves you change the BOUNDARY CONDITIONS, and the waves collapse onto localised waves.
Yes. The electron absorbs the photon, then re-emits it in a random direction.

contd
Da Schneib
3 / 5 (2) Aug 21, 2014
Quantum mechanics is quite adamant about the fact that when you measure, what you measure changes. So this experiment does NOT prove the existence of particles AT ALL!!
1. Then what's absorbing and emitting the light?
2. Your argument proves that experimenting is useless. Are you serious?
RealityCheck
1 / 5 (6) Aug 21, 2014
Hi Schneib. :)
...light doesn't get absorbed by empty space.
Careful, mate. Conventional mainstream theory has it that photons emitted away from just above the EH of black holes 'fade to nothing', and their energy/perturbations/oscillations of 'field/vacuum' are effectively subsumed back into 'diffuse' vacuum-energy state/content. :)
The experimenters don't know what to look for, how to look for it, or anything else without the theorists telling them, and if the experimenters see something strange it's the theorists who look for an explanation
Again, careful. Most often, it has been 'pure research' experimentalists who have discovered 'phenomena' which had not been previously treated/explained in the current professional theory-set. Once these experimentalists have got enough to go on, the theorists add conjectures/hypotheses for FURTHER experiments designed to 'tease out' SPECIFIC aspects. At no stage did theory precede original experimenters 'pure research' efforts. :)
Da Schneib
3.7 / 5 (3) Aug 21, 2014
IMO the wave-particle duality has its tangible physical representation: why not to handle the electron as a pin-point particle, which is surrounded with its wake wave like the boat or the http://i.imgur.com/LVKGkjo.gif? Such a mutually connected system would fit both the attitude of particle physicists, both the people, who just want to see the waves everywhere.
You have re-discovered the "wave packet" analogy, but be aware that this is only a teaching tool, not the true state of affairs.
Da Schneib
3.7 / 5 (3) Aug 21, 2014
There is no quantum in time: There is only a quantum in EM wave-energy.
It is not yet known if time or space are quantized. If they are it would contradict relativity, but possibly only on the smallest scales of time and size.

Time is continuous and absolute (not relative as Einstein has incorrectly argued).
There is no absolute time. You are arguing against Special Relativity Theory, which is a foundational theory that has been proven literally millions of times. College physics students prove it every day.
Da Schneib
3.7 / 5 (3) Aug 21, 2014
A relativistic transformation of physics CAN NEVER be covariant.
What does this even mean? A transform changes one set of coordinates into another, not one physical fact into another. How can the original coordinates not be covariant with the new ones, should the underlying physical quantity of them both change?
Da Schneib
3 / 5 (2) Aug 21, 2014
Experimental data proves that the Doppler-shifts for electromagntic waves are NOT those of waves moving within an aether!
How can we experiment on aether if it doesn't exist?
Da Schneib
3.7 / 5 (3) Aug 21, 2014
No theorists had told them where and how to look".
So they had no education in physics?

Really?
Da Schneib
3.7 / 5 (3) Aug 21, 2014
...light doesn't get absorbed by empty space.
Conventional mainstream theory has it that photons emitted away from just above the EH of black holes 'fade to nothing'
This is incorrect. In fact, current theory has it that energy from inside the black hole can quantum tunnel past the event horizon if it's sufficiently close, and if this happens then two particles must be created to preserve conservation of momentum and other more esoteric laws of nature. One of the quanta then escapes, and the other falls back into the EH.

Please detail where exactly you believe this means a photon is being absorbed into empty space.

At no stage did theory precede original experimenters 'pure research' efforts.
So experimenters don't test hypotheses?

Really?

Toiea
1 / 5 (2) Aug 21, 2014
You have re-discovered the "wave packet" analogy, but be aware that this is only a teaching tool, not the true state of affairs
How can you know, what the "true state of affairs" is?
RealityCheck
1 / 5 (5) Aug 21, 2014
Hi Schneib. :)
...current theory has it that energy from inside the black hole can quantum tunnel past the event horizon if it's sufficiently close, and if this happens then two particles must be created to preserve conservation of momentum and other more esoteric laws of nature. One of the quanta then escapes, and the other falls back into the EH.
Careful. That scenario wasn't what I was referring to. I said any radiation away from the vicinity of EH. Ie, as in 'gravitational redshifting' per current theory. Ok?

Please detail where exactly you believe this means a photon is being absorbed into empty space.
My 'detail' on that aspect is in my ToE, so not at liberty to expand. However, current quantum theory of vacuum energy/processes etc is close enough for the caution I made earlier re possibility for vacuum to absorb photonic energy.

So experimenters don't test hypotheses?
Who said that? Please read again my comments re 'pure research' and 'specific aspect' etc :)
Da Schneib
4.3 / 5 (6) Aug 21, 2014
You have re-discovered the "wave packet" analogy, but be aware that this is only a teaching tool, not the true state of affairs
How can you know, what the "true state of affairs" is?
Because a million physicists say so and have confirmed it by experiment.
Da Schneib
4.2 / 5 (5) Aug 21, 2014
I said any radiation away from the vicinity of EH. Ie, as in 'gravitational redshifting' per current theory.
Radiation away from the EH of a BH propagates as photons per current theory.

My 'detail' on that aspect is in my ToE...
Not interested in your ToE. Sticking with the mainstream.

So experimenters don't test hypotheses?
Who said that?
It's an inescapable implication of what you said. Sorry, not responsible for your words. If you choose not to back them up I will simply ignore anything further on the subject as more politics.
johanfprins
1 / 5 (3) Aug 22, 2014
When you do the transformation correctly, YOU GET LENGTH DILATION: THE ROD BECOMES LONGER!
That's nice, but this is not what we can actually measure from extrinsic perspective. The length contraction is actually required for to maintain the light speed invariance - it's the straightforward consequence of the Lorentz invariance postulate.
No it is NOT. You are misinterpretating Galileo's principle of relativity which demands that a relativistic coordinate transformation of physics equations is not invariant.

Without lenght contraction the light emanated with object in motion would propagate with superluminal speed.
BULLSHIT!!! You just do not understand mathematics. You should consider going back to kindergarten.

You can believe me, my logics is straightforward, consistent and undeniable. It's based only and just on geometric assumptions, as expressed with various pictures and animations
You have not got a clue what logic is all about! Ducks farting?
johanfprins
1 / 5 (3) Aug 22, 2014
So you are arguing that physics is determined by majority-vote?
...it's not a fallacy if the authority cited really is expert in the subject at hand. Are you arguing the majority of practicing professional physicists are not experts on physics?
In Galileo's time the majority explained planetary motion in terms of epicycles.
Irrelevant. In Galileo's time the scientific method of the time did not involve testing hypotheses, or even devising tests for conjectures.
Are you sure you are not Zephyr? Testing was largely ignored, until Galileo brought it back in full force. It is exactly for this why Galileo was nearly burnt at the stake. You are now arguing that we should go back to the time when hypotheses were not tested: Since the hypothesis that Higgs causes mass-energy is untestable, you want to argue that tests are not required.

You haven't answered my question.
Which question?
johanfprins
1 / 5 (3) Aug 22, 2014
When a wave is trapped within a cavity, it MUST form a stationary wave that fills the WHOLE cavity.
The wave isn't trapped in a cavity. The particle is trapped in a magnetic field. A cavity can be generated by using either electric- or magnetic-fields. The wave morphs to adapt to these boundary conditions: There are no particles involved anywhere or at any time.

When you shine light on these waves you change the BOUNDARY CONDITIONS, and the waves collapse onto localised waves.
Yes. The electron absorbs the photon, then re-emits it in a random direction.
What is the electron before it absorbs a photon? Within an ideally perfect metal it is a stationary wave that fills the whole metal (a delocalised wave). When you shine light, the boundary conditions change and the wave collpases into a localised wave, which can be ejected if the frequency of the light-wave is high enough.

johanfprins
1 / 5 (2) Aug 22, 2014
Quantum mechanics is quite adamant about the fact that when you measure, what you measure changes. So this experiment does NOT prove the existence of particles AT ALL!!
1. Then what's absorbing and emitting the light?
The electron-wave absorbs the light and since its energy increases it has to change size and shape: It can thus either collapse or inflate. If you are a baboon you call this a "quantum jump" by a "particle".
2. Your argument proves that experimenting is useless. Are you serious?

Where in God's name does it prove that? You must be the person who Mark Twain met and commented: "Today I have met a man who knows more things that are not so than any other man I have ever met!"
johanfprins
1 / 5 (2) Aug 22, 2014
There is no quantum in time: There is only a quantum in EM wave-energy.
It is not yet known if time or space are quantized. If they are it would contradict relativity, but possibly only on the smallest scales of time and size.
Thus thre is NO PROOF that time can be qunatized: It is at present just a conjecture coming from a halucinating mind!

me is continuous and absolute (not relative as Einstein has incorrectly argued).
There is no absolute time. You are arguing against Special Relativity Theory, which is a foundational theory
Wrong again!! I am not arguing against the Special Theory of Relativity, but against the wrong interpretation of this theory. It does NOT require time to be anything else but absolute. In fact, by assuming that a single event can occur at different times is absurd! Or that two simultaneous events occur at different times is also absurd, since simultaneous means THE SAME TIME!! It cannot ever mean anything else!
johanfprins
1 / 5 (2) Aug 22, 2014
A transform changes one set of coordinates into another, not one physical fact into another.
Wrong again! When you do a relativistic coordinate transformation it DOES change one physical fact into another. This is what Galileo argued already 400 years ago. It is a pity that Lorentz, Poincare, Einstein, Minkowski and all the modern theoretical physicists have been too stupid to understand Galileo.

How can the original coordinates not be covariant with the new ones, should the underlying physical quantity of them both change?
The coordinates on their own might be, but covariance does not refer to coordinates but to the equations of physics. And the equations of physics can NEVER be covariant under a relativistic coordinate transformation. They are only covariant if the origins of the two coordinate systems do NOT move relative to one another.

See: https://www.resea...=prf_act



johanfprins
1 / 5 (2) Aug 22, 2014
Experimental data proves that the Doppler-shifts for electromagntic waves are NOT those of waves moving within an aether!
How can we experiment on aether if it doesn't exist?


Are you really so stupid? Air is the aether of sound waves. and the formulas that can be verified to be valid for sound waves, ARE NOT THE SAME as the formulas for light waves. You can only obtain the correct formulas for light-waves when you derive the Doppler-formulas by assuming that there is NO AETHER! You cannot obtain these formulas IN ANY OTHER MANNER!!
johanfprins
1 / 5 (1) Aug 22, 2014
You have re-discovered the "wave packet" analogy, but be aware that this is only a teaching tool, not the true state of affairs
How can you know, what the "true state of affairs" is?
Because a million physicists say so and have confirmed it by experiment.


Any person using this argument is not even a physicist's arsehole!!

The guiding principle in physics MUST BE that no matter how well and for how long a theory has withstood the test of time, a new fact could emerge at any time to prove that the theory has been flawed all along. If a person is not willing to accept this, he/she should stay the hell out of physics (or any science for that matter).

You should rather become a priest in some fundamentalist dogmatic religion and stop posting rubbish on a physics forum!
Reg Mundy
1 / 5 (3) Aug 22, 2014
This is incorrect. Physics deals with models of reality.

Crap! Physics deals with reality. Demonstrable experiments, real observations, actual measurements.
What about the theorists? You're only paying attention to the experimenters. The experimenters don't know what to look for, how to look for it, or anything else without the theorists telling them, and if the experimenters see something strange it's the theorists who look for an explanation. You've left out half of physics and called it "not physics." This is silliness.

Would you mind reading what I actually said, rather than what you imagined I said.
Reg Mundy
1 / 5 (3) Aug 22, 2014
Which is of course where modern mainstream physics falls down. Too much conjecture being promoted before it passes any meaningful test.
Are you suggesting that adding a test that experimentalists can check to a conjecture is somehow invalid? How then do you suggest hypotheses be formed? You will cripple physics. Again, more silliness.

Ditto. The point being that current mainstream physics accepts as PROVEN hypotheses which have no experimental proof, e.g the constant assertion that Dark Matter exists as if it was fact, the constant references to gravity waves as if they have been proven to exist, and so on. I don't want to "cripple physics", I love physics! I just want it to return to a reasonable level of sanity!
Uncle Ira
4 / 5 (4) Aug 22, 2014
I love physics! I just want it to return to a reasonable level of sanity!


@ Reg-Skippy, then why you don't go to a science school like the real physicist-Skippys so you can make a theory that peoples can understand? Until you go to science school nobody is going to pay any attention to ideas except to call you silly.
Watebba
2.3 / 5 (3) Aug 22, 2014
Because a million physicists say so and have confirmed it by experiment
Which experiment disproves the insight, that the deBroglie wave is the wake wave of particle in vacuum? Just this insight was confirmed with Couder & Fort experiments. And which physicist denies it? IMO you're completely confused.
johanfprins
1 / 5 (2) Aug 22, 2014
Because a million physicists say so and have confirmed it by experiment
Which experiment disproves the insight, that the deBroglie wave is the wake wave of particle in vacuum? Just this insight was confirmed with Couder & Fort experiments. And which physicist denies it? IMO you're completely confused.


I deny it. The Couder & Fort experiments are water-waves. Water acts as an aether; while for electromagnetic waves there is NO AETHER!!!!!!! The Doppler effect measured for EM waves proves that these waves do not move in an aether! Can you not get it through your bony head?
Toiea
1 / 5 (1) Aug 22, 2014
The Couder & Fort experiments are water-waves. Water acts as an aether
Well and the waves in Couder & Fort experiments act like the quantum waves in vacuum. Take it or leave it.
johanfprins
1 / 5 (1) Aug 22, 2014
The Couder & Fort experiments are water-waves. Water acts as an aether
Well and the waves in Couder & Fort experiments act like the quantum waves in vacuum. Take it or leave it.

I will gladly leave it since it is unproven BULLSHIT that comes, without any experimental proof, from a demented mind.
Da Schneib
5 / 5 (4) Aug 22, 2014
Irrelevant. In Galileo's time the scientific method of the time did not involve testing hypotheses, or even devising tests for conjectures.
Testing was largely ignored, until Galileo brought it back in full force.
What's "brought it back" mean?

It is exactly for this why Galileo was nearly burnt at the stake.
This is incorrect. Galileo was permitted to publish a book consisting of a dialog regarding heliocentrism vs. geocentrism; unfortunately, he named the querent "Simplicio," which is a translation from Latin of "Simplicius," but also has a bad connotation in Italian of "simpleton," and he inadvertently put Pope Urban's words in Simplicio's mouth. Urban had formerly been a supporter of Galileo, but this angered him, and he ordered Galileo to be tried for heresy, with results we're all familar with.

However, this had nothing to do with testing or not testing hypotheses, nor with devising tests for a conjecture to upgrade it to a hypothesis.
Da Schneib
5 / 5 (4) Aug 22, 2014
You are now arguing that we should go back to the time when hypotheses were not tested:
This is incorrect and is a massive distortion of my position. If you're just going to make stuff up about me to try to get me riled up, I'll stop paying attention to you.

Since the hypothesis that Higgs causes mass-energy is untestable, you want to argue that tests are not required.
I am arguing nothing of the kind. In fact, the Higgs was predicted not merely to be the field that causes mass attributes of quanta, but also to have many other characteristics, such as a lower mass limit, a particular spin, particular decay modes, particular half-lives for these different modes, etc., etc. All of these other qualities have been confirmed. The Higgs is now the Higgs Theory, and Higgs won a Nobel Prize in Physics for it.

You're making stuff up again.

Which question?
Are you arguing the majority of practicing professional physicists are not experts on physics?
Whydening Gyre
5 / 5 (2) Aug 22, 2014
Galileo was permitted to publish a book consisting of a dialog regarding heliocentrism vs. geocentrism; unfortunately, he named the querent "Simplicio," which is a translation from Latin of "Simplicius," but also has a bad connotation in Italian of "simpleton," and he inadvertently put Pope Urban's words in Simplicio's mouth. Urban had formerly been a supporter of Galileo, but this angered him, and he ordered Galileo to be tried for heresy, with results we're all familar with.

LOL, the Da!
I was not familiar with that little bit of historical nuance, but I love it!
Da Schneib
5 / 5 (1) Aug 22, 2014
Quantum mechanics is quite adamant about the fact that when you measure, what you measure changes. So this experiment does NOT prove the existence of particles AT ALL!!
1. Then what's absorbing and emitting the light?
The electron-wave absorbs the light
What electron wave? We're talking about at the edge of a black hole's event horizon.

2. Your argument proves that experimenting is useless. Are you serious?

Where in God's name does it prove that?
Nowhere in Bog's name; Bog isn't involved. You said:
So this experiment does NOT prove the existence of particles AT ALL!!
implying experimenting is useless.
johanfprins
1 / 5 (3) Aug 22, 2014
What's "brought it back" mean?
Archimedes started it but was afterwards ignored. Galileo brought it back! It seems you do not know anything about the history of physics. Tsk! Tsk!

This is incorrect. Galileo was permitted to publish a book consisting of a dialog regarding heliocentrism vs. geocentrism; unfortunately, he named the querent "Simplicio," etc. etc. etc.
I know this history better than a thick-skulled bonehead like you can EVER do. Please read before opening that useless orifice in your upper extremity!
https://www.resea...=prf_act

However, this had nothing to do with testing or not testing hypotheses, nor with devising tests for a conjecture to upgrade it to a hypothesis.
Oh yes it has! You are just too STUPID to understand it!!
johanfprins
1 / 5 (4) Aug 22, 2014
You are now arguing that we should go back to the time when hypotheses were not tested:
This is incorrect and is a massive distortion of my position. If you're just going to make stuff up about me to try to get me riled up, I'll stop paying attention to you.
Oooh! How worried I am! An IDIOT, hiding like a criminal behind a pseudonym, is going to ignore me!! I am mortally insulted: Boo-hoo-hoo!

the Higgs was predicted not merely to be the field that causes mass attributes of quanta, but also to have many other characteristics, such as a lower mass limit, a particular spin, particular decay modes, particular half-lives for these different modes, etc., etc. All of these other qualities have been confirmed. The Higgs is now the Higgs Theory, and Higgs won a Nobel Prize in Physics for it.


There is NO EXPERIMENT THAT PROVES that if there is NOT the noise seen at CERN there will not be rest-mass. Only a moron will claim that the latter has been proved!!

johanfprins
2 / 5 (4) Aug 22, 2014
Which question?
Are you arguing the majority of practicing professional physicists are not experts on physics?


I am arguing that "experts" have been wrong many times in history, and that a person who thinks that a coterie of "experts" cannot ever be wrong is a certifiable moron: Just like you are!!! And will always be until you die!
johanfprins
1 / 5 (4) Aug 22, 2014
So this experiment does NOT prove the existence of particles AT ALL!!
implying experimenting is useless.


No, the incorrect interpretation of experimental results, as in this case, is useless. If you do not have the brains to correctly interpret what you see, as in your case, you should rather keep your trap shut!!
RealityCheck
1 / 5 (4) Aug 22, 2014
Hi Schneib. :)
Careful.
Radiation away from the EH of a BH propagates as photons per current theory.
Look up current theory 'interpretations/explanations' re 'infinitely redshifted' light trying to escape from too close vicinity of EH. Where does the energy go when such photons are redshifted so extremally that their photonic energy effectively goes to nil? When the e-m field 'perturbation feature' of such a 'photon' is effectively 'damped to practically zero', then the field has effectively absorbed the 'oscillatory energy' back into its quantum vacuum ground state 'background e-m energy content'.

Sticking with the mainstream.
Who suggested otherwise? I refrained from detailing from my ToE, didn't I; and kept to current conventional quantum vacuum theory for your benefit.

It's an inescapable implication of what you said
You jumping to misconstrued inferences was the problem, not what I said.

So careful, mate, not to sound so 'certain' all the time, ok? Cheers. :)
mikep608
1 / 5 (3) Aug 23, 2014
What is this... Another bogus interpretation of Higgs boson on BB theory? The last I heard according to CERN Higgs physics explained that after the BANG the universe must have collapsed and not able to sustain.. This is so opposite of reality where the universe is presently sustaining. Yet here we are again with their bogus experiments.

If the universe collapsed, what fueled its expantion after the collapse? That's a rhetorical question. It doesn't have an answer because the Big Bong never happened. someone thinks there is a RELIC micro wave abckground, so then assumed an entire story of the Big Bong right down to what happened to a trillionth of a second after it started. They can lie about this because they know there is a lot of "smart" stoopid people around.
Da Schneib
3 / 5 (4) Aug 23, 2014
time is continuous and absolute (not relative as Einstein has incorrectly argued).
There is no absolute time. You are arguing against Special Relativity Theory, which is a foundational theory
Wrong again!!
What "again?" What do you claim was the first time?

I am not arguing against the Special Theory of Relativity
You are arguing against one of its postulates by arguing that "Einstein [was] incorrect" about absolute spacetime, and about relative time and space.

but against the wrong interpretation of this theory.
The postulates of SRT are not "interpretations."
Da Schneib
3 / 5 (4) Aug 23, 2014
It does NOT require time to be anything else but absolute.
This is incorrect. Space and time are relative; only spacetime is absolute. The first is a postulate of relativity; the second is a result of it.

On Earth.

In fact, by assuming that a single event can occur at different times is absurd!
Actually, it's irrelevant if it's "absurd;" it's correct. The order of events can be different between observers in inertial frames, moving differently. Which order is correct? Answer: both.

Or that two simultaneous events occur at different times is also absurd, since simultaneous means THE SAME TIME!! It cannot ever mean anything else!
You don't understand relativity.
johanfprins
1 / 5 (3) Aug 23, 2014
I am not arguing against the Special Theory of Relativity
You are arguing against one of its postulates by arguing that "Einstein [was] incorrect" about absolute spacetime, and about relative time and space. Please quote the postulate by Einstein that time is not absolute but relative. Einstein formulated two postulates, and not one of them contains this as a postulate. Clearly you do not know your physics. Have you had some instruction in logic and physics? If you have you should sue your professors, since they are then just as incompetent as Minkowski had been.

but against the wrong interpretation of this theory.
The postulates of SRT are not "interpretations."
I repeat again: Where in Einstein's postulates is it claimed that time is relative and not absolute? You are promoting an urban myth.
johanfprins
1 / 5 (4) Aug 23, 2014
It does NOT require time to be anything else but absolute.
This is incorrect. Space and time are relative; only spacetime is absolute. The first is a postulate of relativity; the second is a result of it.
The FIRST is NOT a part of Einstein's postulates: And space-time cannot exist since the Lorentz-equations are not isometric within a 4D manifold.

In fact, by assuming that a single event can occur at different times is absurd!
Actually, it's irrelevant if it's "absurd;" it's correct.
It is obviously absurd.
The order of events can be different between observers in inertial frames, moving differently. Which order is correct? Answer: both.
WRONG! It only means that an observer cannot "see" a moving event at the actual coincident time that it occurs within his AND the moving reference-frame. This is a result of the Doppler effect: Not of the relativity of time: The latter is absurdly impossible!

It is YOU who wants to believe Voodoo!
Da Schneib
3 / 5 (4) Aug 23, 2014
A transform changes one set of coordinates into another, not one physical fact into another.
Wrong again!
You're making stuff up again. I quoted the definition of "transform."

When you do a relativistic coordinate transformation it DOES change one physical fact into another. This is what Galileo argued already 400 years ago.
Where?

It is a pity that Lorentz, Poincare, Einstein, Minkowski and all the modern theoretical physicists have been too stupid to understand Galileo.
It's a pity you don't understand Lorentz, Poincare, Einstein, and Minkowsky.

How can the original coordinates not be covariant with the new ones, should the underlying physical quantity of them both change?
We were not discussing covariance. We were discussing transforms. And you claimed transforms change the underlying physical reality, an obvious gaffe. See the quote above.
Da Schneib
3 / 5 (4) Aug 23, 2014
The coordinates on their own might be, but covariance does not refer to coordinates but to the equations of physics.
The equations of phyics give coordinates as some of their results, and those coordinates are covariant. That's the point of writing the physical equations. Your statement is meaningless.

And the equations of physics can NEVER be covariant under a relativistic coordinate transformation.
This is incorrect. The Lorentz and Poincare symmetries show the invariance of relativistic transforms from one inertial coordinate system to another.

They are only covariant if the origins of the two coordinate systems do NOT move relative to one another.
This is incorrect according to the relativity postulate of SRT. You're making stuff up again.

See: https://www.resea...=prf_act
Why would I bother? You deny relativity.
Da Schneib
3 / 5 (4) Aug 23, 2014
Experimental data proves that the Doppler-shifts for electromagntic waves are NOT those of waves moving within an aether!
How can we experiment on aether if it doesn't exist?
Air is the aether of sound waves
Please quote reliable scientific literature that confirms this claim. Otherwise I will dismiss it as more of your made-up fantasies.
johanfprins
1 / 5 (3) Aug 23, 2014
A transform changes one set of coordinates into another, not one physical fact into another.
Wrong again!
You're making stuff up again. I quoted the definition of "transform."
And I quoted the definition of a relativistic transform as it follows logically from Salviati's arguments. You obviously are still a Simplicious!

It's a pity you don't understand Lorentz, Poincare, Einstein, and Minkowsky.
They did not understand what they were doing wrong: Why must I agree with what they did wrong?

We were not discussing covariance. We were discussing transforms. And you claimed transforms change the underlying physical reality, an obvious gaffe.
This is exactly what a relativistic transform does. When you "look out" into another IRF in which physics is occurring, you do not see a covariant transformation of the latter physics equations into your IRF. This is what relativity is all about my dear Simplicious! Have you ever heard of the Cariolus-force?
johanfprins
1.3 / 5 (3) Aug 23, 2014
Experimental data proves that the Doppler-shifts for electromagntic waves are NOT those of waves moving within an aether!
How can we experiment on aether if it doesn't exist?
Air is the aether of sound waves
Please quote reliable scientific literature that confirms this claim. Otherwise I will dismiss it as more of your made-up fantasies.
Are you able to derive the Doppler formulas for say sound waves in air? Are you able to derive the Doppler formulas for light-waves from the Lorentz transformation? It seems you are too incompetent to do this simple physics.

Now, if you have enough brains, which you obviously do not have, you can look these formulas up and you will see that they are different for sound-waves and for light-waves, since the Lorentz tranformation is based on the fact the there is NOT an aether. There are many experiments which measured the Doppler shift of light-waves that proved that the formulas for NO AETHER apply.
Da Schneib
3 / 5 (4) Aug 23, 2014
What's "brought it back" mean?
Archimedes started it...
This is incorrect. Archimedes claimed that objects of different mass accelerate at different rates. Galileo proved this wrong.

You're making stuff up again.

This is incorrect. Galileo was permitted to publish a book consisting of a dialog regarding heliocentrism vs. geocentrism
I know this history better
This is obviously incorrect since you claim that Galileo was prosecuted for heliocentrism.

You're making stuff up again.

However, this had nothing to do with testing or not testing hypotheses, nor with devising tests for a conjecture to upgrade it to a hypothesis.
Your answer contained only insults. I reiterate my original claim which you have not successfully refuted.
Da Schneib
3 / 5 (4) Aug 23, 2014
Johan, if you can't properly quote you're obviously incompetent to have this discussion. Stop making stuff about what I said up or I will not respond to you again other than to mark everything you say with a 1.
johanfprins
1 / 5 (2) Aug 23, 2014
The coordinates on their own might be, but covariance does not refer to coordinates but to the equations of physics.
When using the Lorentz transformation, not even the coordinates are isometric and therefore not even the coordinates are covariant. Furthermore, even when the coordinates are isometric, as in the case of a Galilean tranformation, the physical equations are NOT covariant, since apparent forces have to be taken into account. It is YOUR statements that are meaningless even though it has been the mainstream interpretation by "your experts" (morons) for more than 100 years.

They are only covariant if the origins of the two coordinate systems do NOT move relative to one another.
This is incorrect according to the relativity postulate of SRT. There is no such postulate. Furthermore, Simplicious, a postulate cannot overrule the fundamentals of mathematics.

You deny relativity.
. I do NOT. I just reject the wrong interpretation of STR.

johanfprins
1 / 5 (2) Aug 23, 2014
Johan, if you can't properly quote you're obviously incompetent to have this discussion. Stop making stuff about what I said up or I will not respond to you again other than to mark everything you say with a 1.
You are already consistently doing the latter, Simplicious!

I have not made up any stuff. Why must I quote to you what any person with a knowledge of physics can easily find within elementary textrbooks on physics? Can you not look up the Doppler formulas for sound waves and for light-waves? Can you not see that they are different? Can you not find experiments that proved that the Doppler formulas for light-waves are those that must be derived by assuming that there is not an aether. Can you not look up these elementary facts which any undergarduate student is capable of doing?
johanfprins
1 / 5 (2) Aug 23, 2014
What's "brought it back" mean?
Archimedes started it...
This is incorrect. Archimedes claimed that objects of different mass accelerate at different rates. Galileo proved this wrong.
I have not calimed that Archemedes was correct in all his beliefs, but that he was able to do experiments and make conclusions from his experiments. You are deliberately obfuscating.

This is incorrect. Galileo was permitted to publish a book consisting of a dialog regarding heliocentrism vs. geocentrism
I know this history better
This is obviously incorrect since you claim that Galileo was prosecuted for heliocentrism.
Galileo was forced to reject his heresy of promoting heliocentrism. You obviously do not know ANYTHING! My God, where do you come from?

I reiterate my original claim which you have not successfully refuted.
Which original claim? You have claimed many things which are blatantly absurd and thus wrong!
Toiea
1 / 5 (3) Aug 23, 2014
You guys didn't apparently move in your "discussion" (?) not at least a bit. IMO it's mostly because johanfprins represents a typical anarchist of former era, who bases his argumentation to dismissal of everything, what the contemporary physics achieved and Da Schneib is a typical crackpot basher of former era. The only connecting point of both of you, i.e. the dismissal of aether model illustrates this too. It forces Johan to dismiss even the few progressive findings of mainstream physics of recent years. It illustrates that the scientists who are progressive in the area of their expertise (i.e. the superconductivity) may still remain conservative in another areas.
johanfprins
3 / 5 (2) Aug 23, 2014
Toiea, Zephyr what ave ou!

No physics training! No physics knopwledge: The MASTER BLLSHIITER!!
Captain Stumpy
5 / 5 (4) Aug 23, 2014
and Da Schneib is a typical crackpot basher of former era
@zephir
even though you are the ultimate crackpot "basher" from a former era, i think you have this wrong.
Johan is anti QM in many ways, and does not like relativity's certain aspects, whereas Da Schneib is pushing modern science and it's modern interpretations based upon the evidence that he believes is true, to which I also agree with.
Johan has not made me a convert based upon arguments here because there is no links to evidence and empirical studies, whereas Da Schneib is posting about modern science and there is plenty of proof of that to find. I can almost follow along in certain science books I have.

neither support aether because it is a DEAD THEORY
and THAT is based upon ACTUAL EVIDENCE http://arxiv.org/...1284.pdf

http://exphy.uni-...2009.pdf
johanfprins
1 / 5 (3) Aug 23, 2014
Dear Captain Stumpy,

Please make sure of your facts before posting lies!

I am NOT anti QM or anti Relativity: I am only anti interpretations which are absurd nonsense. I do not think it is possible to convert any person who can believe absurditites are real, as you and the mainstream theoretical physicists are doing.

Da Schneib is posting absurd interpretation. It is impossible for two events to occur simultaneously at two different times. Just as it is impossible for an elephant to use a piece of confetti as toilet paper. It is also impossible for "particles" to diffract; which means that a photon and an electron must each be a wave.

Why are people preventing me from publishing when I can prove that de Broglie waves follow logically from Maxweell's equations; and when I prove that when a moving event occurs simultaneously with a stationary event, a stationary observer cannot see this event at the actual simultaneous instant that it is occurring. Why believe in Voodoo?
Captain Stumpy
5 / 5 (5) Aug 23, 2014
I am NOT anti QM or anti Relativity: I am only anti interpretations which are absurd nonsense.
@Johan
I dont consider it a lie or derogatory. and I didn't say you were ANTI-QM or ANTI relativity... I said
anti QM in many ways, and does not like relativity's certain aspects
and that is NOT A LIE. you do NOT like the modern interpretations
so jumping to a conclusion and getting defensive about it is not helping. it is FACTUALLY true, and so is the REST of my comment.
It was not a SLAM either, in fact... that is ONE reason I don't rate (most of) your posts (until I can know more)

I don't agree with you on a lot of things, and I agree with you on a lot of things, but until I can see some empirical evidence which sways my opinion, I think the modern interpretations are more factually valid than yours. and they DEFINITELY have produced more EVIDENCE than you have supporting their interpretations
(QM is HUGELY successful too.. which hurts your argument)
Captain Stumpy
5 / 5 (4) Aug 23, 2014
cont'd
Please make sure of your facts before posting lies
So don't get angry and lash out. convince me with empirical evidence of your interpretation from the same peer reviewed sources that I would require of anyone else. which is only argument from a strong and empirically proven position anyway
if I don't understand something I will ask for clarity
Da Schneib is posting absurd interpretation
No, Da Schneib is posting a MODERN interpretation, and until you can produce evidence that is better than the evidence used to validate the modern interpretations, then I will continue to AGREE with him.

THAT is nothing but LOGICAL IMHO... (especially considering the HUGE success of QM)
and like science, I can change my mind given evidence to the contrary. so get some and quit being so pissy
and leave out BOOKS unless they are online references that can be read and studied for a factual refute point-by-point
otherwise you are pulling a reg-mundy

THANKS
Captain Stumpy
5 / 5 (4) Aug 23, 2014
Why are people preventing me from publishing when I can prove that de Broglie waves follow logically from Maxweell's equations
@johan
this is not something that we can advise on NOR can we help with. UNTIL you can show proof and convince a peer reviewed panel, it is also not considered evidence, just like zephir's conjectures
Why believe in Voodoo?
Well, for one thing, they happens to be the most successful theories in existence to date, with the most proof and the most evidence, and the most working functional technology based upon it, etc

why NOT believe that the voodoo is real when supported with THAT kind of background?

to reiterate: until you give the same level of proof behind empirically proven "Voodoo" then NO ONE will be able to accept your comments as proof, much like your argument with Furlong elsewhere with him supporting modern physics

just bring that evidence to the table so EVERYONE can see it
it helps

PEACE
johanfprins
1 / 5 (4) Aug 23, 2014
. I said
anti QM in many ways, and does not like relativity's certain aspects
and that is NOT A LIE.
IT IS A CRIMINAL DISTORTION OF THE TRUTH OF WHICH YOU SHOULD BE ASHAMED OFF IF YOU HAVE ANY INTERGRITY: But I must admit that it is doubtfull that you have any!

I don't agree with you on a lot of things, and I agree with you on a lot of things, but until I can see some empirical evidence which sways my opinion, I think the modern interpretations are more factually valid than yours.
] I do not think you are capable of interpreting factual evidence even when it jumps up and bites your nose off.
and they DEFINITELY have produced more EVIDENCE than you have supporting their interpretations
another lie by an incompetent!
(QM is HUGELY successful too.. which hurts your argument)
It has ONLY been successful in Chemistry and Solid State Physics, where the absurd interpretations are irrelevant! You are too stupid to know this of course!
johanfprins
1 / 5 (4) Aug 23, 2014
So don't get angry and lash out. convince me with empirical evidence of your interpretation from the same peer reviewed sources
If you do not realise that "peer review" is at present the same as in the time of Galileo: Namely Censorship of any new ideas, then you are an even bigger FOOL than I have thought that you are.
that I would require of anyone else. which is only argument from a strong and empirically proven position anyway
Where is the empirical eveidence that the noise seen at CERN gives matter mass? It does NOT exist but you accept that it does since the reviewers are dishonestly claiming this

No, Da Schneib is posting a MODERN interpretation, and until you can produce evidence that is better than the evidence used to validate the modern interpretations, then I will continue to AGREE with him.
There exists NO EVIDENCE whatsoever that his "modern" interpretation is NOT absurd. He even admitted that it is absurd: ONLY a FOOL like YOU will ignore this!
Captain Stumpy
5 / 5 (3) Aug 23, 2014
You are too stupid to know this of course!
@johan
of course
that is why you are so insanely successful and rich and widely known with your publications and conclusions about the non-reality of modern QM interpretations
IT IS A CRIMINAL DISTORTION OF THE TRUTH
Nope. and it is NOT a lie. You do NOT like modern QM or some interpretations of QM/relativity. how is that criminal?
if it is, please feel free to litigate. my address is freely available as I've been known by Stumpy for decades
I do not think you are capable
personal conjecture not supported by evidence
another lie
all you gotta do is prove it. feel free to link the scads of material here... we will wait
IF YOU HAVE ANY INTERGRITY: But I must admit that it is doubtfull that you have any!
at least I can admit when I am wrong. you seem to think you are infallible.

tell you what.
PROVE ME WRONG
LINK YOUR EMPIRICAL EVIDENCE
WE'LL COMPARE IT TO MODERN PHYSICS EVIDENCE
simple solution, johan
Captain Stumpy
5 / 5 (3) Aug 23, 2014
If you do not realise that "peer review"...Censorship of any new ideas
@johan
IOW - you are the victim of a global conspiracy against you, right? okee dokee then. lets all let johan, rc, reg and zeph publish. they seem to be equals in this regard
Where is the empirical eveidence that the noise seen at CERN gives matter mass?
in all honesty, I don't think there is any, but I don't know there isn't either. There was a PREDICTION about the Higgs, and the results from CERN only prove that prediction to a high degree of accuracy. I can't state anything further than that
There exists NO EVIDENCE whatsoever that his "modern" interpretation is NOT absurd
lack of evidence is not proof of anything unless you are trying to establish there is a LACK of evidence

PRESENT EVIDENCE SUPPORTING YOUR CONJECTURE
until then, it is CONJECTURE
not PROOF
johanfprins
1 / 5 (4) Aug 23, 2014
THAT is nothing but LOGICAL IMHO... (especially considering the HUGE success of QM)
A further proof that a collossal FOOL: Where have I EVER claimed that QM did not lead to spectacular successes? You are building up fairytales in your mind since you are too STUPID to think and argue logically.
I can change my mind given evidence to the contrary. so get some and quit being so pissy
You will NEVER be able to change your mind since you are too stupid and dishonest to do this!

and leave out BOOKS unless they are online references that can be read and studied for a factual refute point-by-point
You see what a DISHONEST LIAR you are: I have given you references to my books which you can download and read for a factual refute point-by-point. But of course you KNOW that you are TOO STUPID to understand physics and wait for a referee to think for you: Only then you will blindly follow like the zombie-moron you are.
Captain Stumpy
5 / 5 (3) Aug 23, 2014
ONLY a FOOL like YOU will ignore this
Where have I EVER claimed that QM did not lead to spectacular successes?
@johan
now, where did I ever say that you DID make that claim? better put your glasses back on and learn to read, old guy
you like to denigrate others without ANY evidence just because they don't agree with you...
well, johan, that is one of the definitions of a TROLL

shall we just start downvoting you and reporting your comments for being off topic and TROLLING whenever you decide to eschew polite conversation and be derogatory?

I am giving you the chance to produce EVIDENCE of the same caliber as the EVIDENCE supporting modern physics

use it

otherwise... you are just trolling
Captain Stumpy
5 / 5 (3) Aug 23, 2014
You will NEVER be able to change your mind since you are too stupid and dishonest to do this!
@johan
you don't know me at all, therefore this is PERSONAL CONJECTURE WITHOUT EVIDENCE and TROLLING
You see what a DISHONEST LIAR you are:
PLEASE PROVIDE PROOF
I have not make ONE factually incorrect statement here... so PROVE I am a liar..

you know what, johan, maybe I should just IGNORE your proof like you are doing above
then start reporting you as a troll

maybe this is just an exercise in futility?
I ASKED FOR PROOF and you want to RANT because you are old and set in your ways
try using your own brain and re-reading the above
learn to comprehend instead of attack because I have a different opinion

PROVE YOUR COMMENTS OR BE TREATED LIKE A TROLL
johanfprins
1 / 5 (4) Aug 23, 2014
this is not something that we can advise on NOR can we help with. UNTIL you can show proof and convince a peer reviewed panel, it is also not considered evidence,
Bose could not find such a panel: Fortunately for him Einstein was still alive. The person who argued Tectonic Plates could also not find such a peer-review panel;T hey wereboth ignored since thje maiority of people on earth are collossal idiots like you, who are NOT ABLE TO THINK FOR THEMSELVES.
Well, for one thing, they happens to be the most successful theories in existence to date, with the most proof and the most evidence, and the most working functional technology based can convince anybody with common sense:
Obvioulsy an IDIOT like YOU are excluded.
why NOT believe that the voodoo is real when supported with THAT kind of background?
Since there is NO BACKGROUND that supports it. ONLY an IDIOT like YOU will think that such background exists
johanfprins
1 / 5 (4) Aug 23, 2014
just bring that evidence to the table so EVERYONE can see itit helps
I have done so but you are not willing to download it and read it unless another IDIOT like you recommends to you that you should do so. Well I suppose I cannot blame you nsince you have proved OVER and OVER again that you are far TOO STUPID to9 understand physics and logic!
Captain Stumpy
5 / 5 (4) Aug 23, 2014
collossal idiots like you, who are NOT ABLE TO THINK FOR THEMSELVES.

Obvioulsy an IDIOT like YOU are excluded.

ONLY an IDIOT like YOU will think that such background exists
@johan
and we are to assume, by your posts here on a pop-sci site and NOT in a peer reviewed journal with impact in physics, that you are correct based only upon your opinion and posted conjecture?

sorry johanny....
you are TROLLING and your suppositions are not supported by anything but your own flawed conspiracy minded conjecture
johanfprins
1 / 5 (5) Aug 23, 2014
You are too stupid to know this of course!
@johan
of course
that is why you are so insanely successful and rich and widely known with your publications and conclusions about the non-reality of modern QM interpretations
Thank you for realisng that I am successfull. At leasr you got one thing right; EVEN though you are still an idiot.
Nope. and it is NOT a lie. You do NOT like modern QM or some interpretations of QM/relativity. how is that criminal?
It is a blatant lie since I consider QM and STR ase the most important breakthroughs in physics. I am just appalled by the fcat that people lose all common sense when interpreting the consequences of these breakthroughs. In your case it is not really surprizing since you have proved consistently on this forum that you are an idiot!
tell you what.
PROVE ME WRONG
LINK YOUR EMPIRICAL EVIDENCE
WE'LL COMPARE IT TO MODERN PHYSICS EVIDENCE


I have done this many times on this forum! Don't blame me, Cap IDIOT!
johanfprins
1 / 5 (5) Aug 23, 2014
IOW - you are the victim of a global conspiracy against you, right?
Here comes the trademark of the UTTER scoundrel! Bringing up conspiracies. WHERE in FUCK's NAME have I ever claimed a conspiracy? You are the lowest of the lowest pieces if excrement who ever paraded as a human being! You need psychiatric help!
in all honesty, I don't think there is any, but I don't know there isn't either. There was a PREDICTION about the Higgs, and the results from CERN only prove that prediction to a high degree of accuracy.
No it does NOT. They could not give an accurate value for the energy, and they do not even know whether there are one or a million of these noises. To predict that at higher energies you will get more resonances does not prove that these noises are responsible for mass-energy. They claim the latter without doing any experiment that can falsify this claim: Since they know that the world is populated by IDIOTS like you that will swallow absurd bullshit!
johanfprins
1 / 5 (5) Aug 23, 2014
PRESENT EVIDENCE SUPPORTING YOUR CONJECTURE
until then, it is CONJECTURE
not PROOF
I have done this OVER and OVER and OVER on this forum by giving refrences for further reading whichYOU refuse to do. YOU are the one who wants to beleive mainstream CONJECTURES, like wave-particle duality, time dilation, Higgs boson giving mass etc. Then you ask ME for proof while this is exactly what I have been doing all along!

You are demented! have you got ANY tertiary qualifications? I doubt it!

Signing off to cool off!
What an utter moron you are!!
Captain Stumpy
5 / 5 (4) Aug 23, 2014
WHERE in FUCK's NAME have I ever claimed a conspiracy?
@johan
I would say here applies
If you do not realise that "peer review" is at present the same as in the time of Galileo: Namely Censorship of any new ideas
you suggest you can't get published because "it's everyone else censoring you"
poor baby
You are the lowest of the lowest pieces if excrement who ever paraded as a human being!
I must have hit the nail on the head to get you to respond thus! Thanks! means i am headed in the right direction
They claim the latter without doing any experiment that can falsify this claim
WOW! I didn't know you had evidence falsifying the CERN findings! PLEASE LINK IT!
I have done this OVER and OVER and OVER on this forum
when you make a comment, prove it, even if it is over and over. I am a newbie, remember?
have you got ANY tertiary qualifications?
only a couple bach degree's. nothing big. I'm an investigator, which requires that.

Captain Stumpy
5 / 5 (4) Aug 23, 2014
YOU are the one who wants to beleive mainstream CONJECTURES
@johan
because they've given proof of concept, and proof of experiment which you've not provided
Then you ask ME for proof while this is exactly what I have been doing all along!
just because you are old and think everyone looks alike doesn't mean you've EVER proved ANYTHING to ME
that is YOUR FALLACY... when you can recognize that, we can move on
Signing off to cool off!
take your time.

and as for cooling off... the question I would ask is: WHY did you get so angry?
was it because I was pointing out the truth and you didn't like it?
because that is what it is going to look like to anyone reading this from the top.

sorry johanny... your actions above are considered an "epic failure" in today's lingo.

tell you what. I will leave... I don't want to feel responsible for your infarction. and given how you reacted to being PROVEN wrong, you are headed for one.

Captain Stumpy
5 / 5 (3) Aug 23, 2014
OOPS
almost forgot... @johanfprins
you posted 54 total times
you have THREE links above... ALL THREE ARE TO YOUR OWN BOOK, which is paywalled, and so it cannot be downloaded for free here
OR TO YOUR PROFILE on researchgate
Which means that you totally pulled a reg mundy here and referenced YOURSELF in the comments above, providing NO empirical data other than self-reference!

sorry johan, self gratification is NOT proof, more like pseudoscience
http://sci-ence.o...-flags2/

as you can see... it is one HUGE RED FLAG of Quackery
you seem to show a FEW of those red flags
and you get ESPECIALLY MAD when proved wrong in your own words and in black and white...

So I am not replying to you anymore as you are a QUACK
an UBER TROLL
and likely will have an infarction due to blood pressure rise or stress caused by being proven a loser troll PSEUDOSCIENCE QUACK on this thread...

BYE BYE JOHAN
nice chatting with you
Whydening Gyre
5 / 5 (2) Aug 23, 2014
Thus spake the Lord -
Unh-hunh, don't pull me into this one - again.....
johanfprins
1 / 5 (4) Aug 24, 2014
Only people who have something to hide post under pseudonyms. TROLLS thus always do. I do not. So I am not a troll
Only people who are criminals spout about physics without revealing their competency in this subject. I do not do so since my credentials and CV is on internet for all to see. Thus I am not a criminal.
Only people who are incompetent depend on peer review in order to decide what is valid. I am competent enough to decidefor myself what is relevant and what not. I do not require faceless peers to form my opinions.
Only scoundrels are willing toact as peer reviewers while remaining anonymous. I always asked the journals for which I did reviewing, for example Phys. Rev. :etter, Phys. Rev B, Phil Mag, etc. NOT to keep me anonymous. I am not a faceless scoundrel.

I wish that people who post on this forum, would be willing to display the same integrity: But I suppose that this type of honesty in physics have been totally lost during the 20th century after 1927.
Reg Mundy
1 / 5 (4) Aug 24, 2014
@johan
You are wasting your time with Cap'n Grumpy, he is purely a shit stirrer, never contributes anything worthwhile to this forum, specialises in insulting anyone who tries to introduce anything which does not conform to mainstream science, and accuses everybody who disagrees with him of being a troll to disguise the fact that he is a troll.
Just ignore him, as he and Ira are sock puppets whose only purpose in life is to cause trouble.
johanfprins
1 / 5 (4) Aug 24, 2014
@ Reg Mundy,

Thanks for the advice. I have been disgusted all along with how he also treats you. He does not have to agree with you, but should not just attack you when you are thinking for yourself. He should study your ideas and then pointedly post well-argued reasons on what he disagrees. But what do you expect when, at present, editors of "peer-reviewed" journals act in the same brainless manner! Physics is in grave trouble since 1927, and it is getting worse!
Captain Stumpy
4 / 5 (4) Aug 25, 2014
Only people who have something to hide post under pseudonyms
@johanny
sorry, tardie... Truck Captain is my RANK and Stumpy IS MY official call sign used for more than 24 years... and a TROLL is someone who posts inflammatory, extraneous, or off-topic messages in an online community (such as a newsgroup, forum, chat room, or blog) with the deliberate intent of provoking readers... I would say that your posts above are proof of this!
https://en.wikipe...ernet%29
you post to fight about your BELIEFS

you don't like mainstream science. we get it
DEAL WITH IT

neither YOU nor REG "think for yourself"
you are promoting PSEUDOSCIENCE
WHICH is WHY I argue against it!

if there is no links to proof- empirical evidence from a reputable peer reviewed source with an impact in the area of the topic, then IT IS PSEUDOSCIENCE

therefore JOHAN and REG are nothing more than WORDY ACERBIC PSEUDOSCIENCE TROLLS

conversation over, unless you got PROOF
sorry boys
you fail
johanfprins
1 / 5 (4) Aug 25, 2014
@ Captain Stumpy,

I have posted proof all over where I could: See my publcations on ResearhGate. I cannot help it that you do not have any grey matter between your ugly ears, and therefore refuses to see proof, even when you fall into it face down! .It is no use to argue with a brainless idiot like you! So F-OFF!!

On ReseachGate I get agreements from qualified scientists who do not hide like criminals behind anonymity. They do not have to, since their qualifications are there for all to see.
Watebba
1 / 5 (3) Aug 25, 2014
Guys, such a way of "discussion" is boring and it leads nowhere. Stumpy doesn't want to argue and his only argument is "IT'S A PSEUDOSCIENCE". And Johan doesn't want to argue and his only argument is "IT'S A RESEARCH GATE". Actually both of you are doing Argument from authority. If you want to enjoy the enjoyable and educational discussion, you should only argue here. If you have no arguments, then the Socratic discussion is over. How it comes, the modern seemingly educated people cannot dispute at all?
Captain Stumpy
5 / 5 (3) Aug 25, 2014
modern seemingly educated people cannot dispute at all
there is NO dispute
johanny is using self reference to his own book which we have to pay for to understand his "argument" which is PSEUDOSCIENCE RED FLAG 1
there is no "argument from authority" because johanny is arguing from a failed fallacious position with NO peer reviewed empirical data from a reputable source with an impact in physics... he is arguing from the standpoint of "buy my book"
IOW - he is pulling a Reg and TROLLING/SPAMMING
there is NO proof with that
When there is a peer reviewed study posted, then it will be a "discussion" between qualified educated people, but johanny just likes to argue to get his dander up and play the "righteous indignation" card- which might work if he was arguing from a solid point and not from fallacy/self promotion
so, johanny! to borrow a phrase from YOU
It is no use to argue with a brainless idiot like you! So F-OFF

until then he is a PSEUDOSCIENCE TROLL
Reg Mundy
1 / 5 (4) Aug 25, 2014
@ Reg Mundy,

Thanks for the advice. I have been disgusted all along with how he also treats you. He does not have to agree with you, but should not just attack you when you are thinking for yourself. He should study your ideas and then pointedly post well-argued reasons on what he disagrees. But what do you expect when, at present, editors of "peer-reviewed" journals act in the same brainless manner! Physics is in grave trouble since 1927, and it is getting worse!

See what I mean about Grumpy? A complete dickhead. People like him would prevent any new idea not already accepted by mainstream science from ever being considered.
I have played a game with him on another thread - I post one comment, he posts half a dozen all the same, insults and rubbish. I then post another comment. Back he comes again, same old tripe. I am trying to wear him out, he must spend hours typing in the same old tosh, or maybe he just copies and pastes it - it rarely changes. A real live MORON!
thefurlong
5 / 5 (5) Aug 25, 2014
@Reg Mundy
"I post one comment, he posts half a dozen all the same, insults and rubbish. I then post another comment. Back he comes again, same old tripe"
Number of postings from Reg Mundy so far: 20
Number of postings from Captain Stumpy so far: 16
Number of times Reg Mundy has actually made an observation verified by objective verification: 0.

Also, allow me to point out that Captain has been contending with several crackpots, whereas, your posts pretty much consist of either glad-handing Johan Prins, or, as usual spouting unfounded nonsense of which you seem to be the sole purveyor.

Also, I would be remiss if I did not also note that you have yet to respond to my last point here: http://phys.org/n...ght.html
thefurlong
5 / 5 (5) Aug 25, 2014
@Reg Mundy
Let me be more specific.
You did not respond to the point I made to your claims about the Heisenberg uncertainty principle not providing large enough energy fluctuations to for everything to return under gravity. The inevitable return of everything under the influence of a gravitational field is one of the predictions of your "theory," and it is overwhelmingly incorrect.
Aligo
1 / 5 (5) Aug 25, 2014
People like him would prevent any new idea not already accepted by mainstream science from ever being considered. I have played a game with him on another thread
Most of opponents of new ideas are recruited from quite insignificant zealots, as Einstein recognized and noted before years. Against his relativity the book "Hundert Autoren gegen Einstein" (A Hundred Authors Against Einstein) has been published in 1931. Nobody of you would probably remember these authors today: a solely insignificant teachers and publicists were signed under that book. The really significant scientists only rarely engage in these fights at publics, being smart enough for it.
johanfprins
1 / 5 (4) Aug 26, 2014
NO disputejohanny is using self reference to his own book which we have to pay
My latest book "EINSTEIN=GENIUS: But a genius sometimes blunders" is available on ResearchGate; and when in final format will remain free of charge on the internet, and at cost from a publisher.

Not that you will download and read it since you are, just like Zephyr (and all his aliases) NOT capable of thinking for yourself. At least Zephyr tries to do so but is just too brainless to realise what logic is all about. You also do not realise what logic is all about, and therefore needs a TROLL who acts as an editor or a reviewer to think for you! So to repeat: F-OFF!!
Reg Mundy
1 / 5 (5) Aug 26, 2014
@Reg Mundy
Let me be more specific.
You did not respond to the point I made to your claims about the Heisenberg uncertainty principle not providing large enough energy fluctuations to for everything to return under gravity. The inevitable return of everything under the influence of a gravitational field is one of the predictions of your "theory," and it is overwhelmingly incorrect.

Last I remember is your failure to supply the correct answer to your own challenge to calculate a simple compound interest problem. If I make a serious reply to this question, no doubt you will waffle off into your own silly world like you did last time I tried to engage you in some serious debate. And if you reply to this, please stick to English not some obscure language you can take refuge in! Twerp!
thefurlong
5 / 5 (4) Aug 26, 2014
Last I remember is your failure to supply the correct answer to your own challenge to calculate a simple compound interest problem.

Hahah. All you had to do was click on the link, silly. That's why I provided it. It was my next to last post.
But since you are so characteristically lazy and your memory sucks, let me quote what I last said to you in relation to your so-called expansion theory:

When the final speed is much larger than 0, you need to impart enough energy to get it not only to stop, but to turn around. You cannot apply the uncertainty principle to this because the energy required is far more than the energy fluctuations that this principle can provide. Gravity is an inverse square law. The energy it can provide at increasing large radii gets EXTREMELY SMALL very quickly. Now, I have not included any equations, so that should be enough for you to understand.

Does this jog your conveniently short memory?
Reg Mundy
1 / 5 (5) Aug 26, 2014
@furbrain
Just out of interest, your previous posting to your last provides numbers of comments submitted by myself and Cap'n Grumpy. My comment about this referred to ANOTHER THREAD, not this one, which you would have understood if you had either read my comment properly or interpreted it correctly, both of which would seem to be beyond your capabilities.
No matter, I will address your latest sally.
From the point of view of the travelling object, it is stationary and the Earth is moving away from it. However, according to your espoused theory of "gravity" it experiences a diminishing force which, no matter how small, causes it to accelerate in the direction of Earth (in the absence of other forces). Get your head round that, Einstein!
thefurlong
5 / 5 (5) Aug 26, 2014
Just out of interest, your previous posting to your last provides numbers of comments submitted by myself and Cap'n Grumpy. My comment about this referred to ANOTHER THREAD

Man, that thread really scarred you, didn't it, Mr. Ed. I'd be scarred too, if someone showed me just how diminutive my intellect was in that manner.
Get your head round that, Einstein!

Hahahaha. What? What is there to get my head around?
We've been over this. The acceleration causes the relative speed to approach a limit. In a whole family of cases, that limit is WELL ABOVE 0. It's like starting with 2, removing 1/2, then iteratively removing half of what you did before. You will never get below 1.
Reg Mundy
1 / 5 (5) Aug 26, 2014
@furbrain Ah, you mean like the man running away from the bullet. When the bullet reaches where he was, he has moved on, so the bullet then crosses the intervening distance, but again he has moved on, not very much but finite. Every time the bullet gets where he was, he has moved on, ad infinitum, so the bullet never reaches him. To prove this theory, get someone to fire a bullet at you as you run away. According to your theory, the bullet will never reach you.....Try it, and if you are right I will concede defeat. If I don't hear from you again, either I am right or you are too much of a fink to rely on your own theory and have wimped out of this simple test. Better still, try it on Cap'n Grumpy first, nobody will miss him either!
Reg Mundy
1 / 5 (5) Aug 26, 2014
@furbrain
It's like starting with 2, removing 1/2, then iteratively removing half of what you did before. You will never get below 1.

OK, start with 2, remove half, half of 2 is 1, so 2 -1 leaves 1. Remove half, that leaves 0.5, remo... hang on a minute, you say I will never get below 1! Oh, Furbrain, you really are a lamentable mathematician. I suggest sticking to the simple things in life, and avoid the headaches....
thefurlong
5 / 5 (5) Aug 26, 2014
Ah, you mean like the man running away from the bullet. When the bullet reaches where he was, he has moved on, so the bullet then crosses the intervening distance, but again he has moved on, not very much but finite.

THE BULLET IS MOVING AT CONSTANT SPEED, YOU NITWIT! Of course you can't outrun it if you are moving slower than it!

Of course, even in that scenario, if you happen to be moving faster than the bullet, you can easily out-run it.

Ahhh, my brain is melting from the profusion of stupidity in your comment.

Look, you can easily check to see that your logic is severely flawed. Start in the middle of a room. Move one third the distance to the wall in 1 second. Then, move one ninth the distance, and so on, so that you are iteratively moving 1/3 of how much you moved last time. You will find that you cannot travel more than 1/2 the distance to the wall. You nincompoop.
thefurlong
5 / 5 (5) Aug 26, 2014
OK, start with 2, remove half, half of 2 is 1, so 2 -1 leaves 1. Remove half, that leaves 0.5, remo...

LOL! Reading is fundamental.
I said,
It's like starting with 2, removing 1/2

Not "start with 2, and remove half of two".

So, the process goes:
2 - 1/2 = 3/2
3/2 - 1/4 = 5/4
5/4 - 1/8 = 9/8

So, your fraction is n+1/n, which approaches 1 as n approaches infinity. Silly Reg!
Captain Stumpy
4.2 / 5 (5) Aug 26, 2014
Ask a TROLL to provide empirical data, and this is what you get:
Not that you will download and read it
NOT capable of thinking for yourself
You also do not realise what logic is all about, and therefore needs a TROLL who acts as an editor or a reviewer to think for you
So to repeat: F-OFF!!
or they attack like this
no doubt you will waffle off into your own silly world
My latest book "EINSTEIN=GENIUS: But a genius sometimes blunders" is available on ResearchGate
I put johanny's "read my book" comments with reg because that is reg's typical reply

so... I ask for proof and I get self reference...
and NO empirical data from a peer reviewed reputable journal with an impact in the subject.

but because I want PROOF of comment, I am the one who is wrong because I "can't think for myself" - HA! what a CROCK!
http://sci-ence.o...-flags2/

gotta love the TROLLS logic, right?

GOOD WORK, Furlong!
Reg Mundy
1 / 5 (5) Aug 26, 2014
@furbrain
OK, start with 2, remove half, half of 2 is 1, so 2 -1 leaves 1. Remove half, that leaves 0.5, remo...

LOL! Reading is fundamental.
I said,
It's like starting with 2, removing 1/2

Not "start with 2, and remove half of two".

So, the process goes:
2 - 1/2 = 3/2
3/2 - 1/4 = 5/4
5/4 - 1/8 = 9/8

So, your fraction is n+1/n, which approaches 1 as n approaches infinity. Silly Reg!

Depends on the way I read your question, furbrain, Typical of your sloppy phraseology. 50% of people reading it will take my interpretation, although I knew what you meant to say. You should try to be more explicit with your English.
Reg Mundy
1 / 5 (5) Aug 26, 2014
@furbrain

THE BULLET IS MOVING AT CONSTANT SPEED, YOU NITWIT! Of course you can't outrun it if you are moving slower than it!.


Ah, the application of common sense! Yet you refuse to apply the same logic to an object moving away from Earth that is ALWAYS subject to your "gravity" no matter how far away it travels and must therefore eventually start moving BACK TOWARDS EARTH! And you call me a nincompoop!
Reg Mundy
1 / 5 (5) Aug 26, 2014
@furbrain
Oh, by the way, a word of advice.
I see Cap'n Grumpy is cosying up to you. I advise caution, as he will be suggesting you hang about outside fire stations with him shortly, and similar activities. Of course, you may welcome his attentions, especially if you have a tendency to bowl from the pavilion end as I suspect from some of your posts, but if you don't, keep a tight grip on your trousers.
Whydening Gyre
5 / 5 (5) Aug 26, 2014
@furbrain
Oh, by the way, a word of advice.
I see Cap'n Grumpy is cosying up to you. I advise caution, as he will be suggesting you hang about outside fire stations with him shortly, and similar activities. Of course, you may welcome his attentions, especially if you have a tendency to bowl from the pavilion end as I suspect from some of your posts, but if you don't, keep a tight grip on your trousers.

Holy crap! Has it come down to gay bashing?!?!
thefurlong
5 / 5 (5) Aug 26, 2014
Yet you refuse to apply the same logic to an object moving away from Earth that is ALWAYS subject to your "gravity" no matter how far away it travels and must therefore eventually start moving BACK TOWARDS EARTH!

It is remarkable how stubbornly you avoid actually responding to the arguments I make. It really gets tiresome. You accuse Captain of not saying anything of substance, but in the two comments you left in response to my own, you have offered no rebuttals. All I see are commentary on my use of English, which has nothing to do with science or math, and a reiteration of "everything must always return under gravity," which was your stance to begin with (as such, it is a form of begging the question).
There is no response to the asymptotic sequence I provided for you, and no response to my overall point that the acceleration from gravity is like this asymptotic sequence. It accelerates some objects so that their velocities approach a finite speed that can be >> than 0.
Captain Stumpy
5 / 5 (5) Aug 26, 2014
Holy crap! Has it come down to gay bashing?!?!
@Whyde
more likely wishful thinking and an offer...

I don't swing that way. sorry reg.
Some techniques are comically simple. Emotionally charged, yet evidence-free, accusations of scams, fraud and cover-ups are common. While they mostly lack credibility, such accusations may be effective at polarising debate and reducing understanding.
from here: http://phys.org/n...firstCmt

just more evidence supporting the TROLL moniker earned by reg et al
It is remarkable how stubbornly you avoid actually responding to the arguments I make.


KEEP UP THE GREAT WORK @Furlong!

thefurlong
5 / 5 (6) Aug 26, 2014
especially if you have a tendency to bowl from the pavilion end as I suspect from some of your posts

And...as usual, your hypothesis is wayyyyy off base. I am not going to even bother asking you for evidence for that one.

So let's see....things you have done in your comments so far...you have
1) gratuitously broached the subject of my sexuality, as if that has anything to do with the topic
2) implied that Captain complementing me is somehow a sexual advance
3) suggested that homosexual practices are something one should be ashamed of
4) accused me of not communicating properly in English, when the whole point of having a sufficient vocabulary is to communicate ideas as precisely as possible

Things you have not done
1) Refute any of the points I made
2) Offered any argument but a rehashing of your original stance.

I suspect that you will continue along this trajectory, or just give up as you have done before. After all, you are "The Laziest Crackpot in Existence" (TM)
Whydening Gyre
5 / 5 (6) Aug 26, 2014
As a practicing lesbian (unfortunately trapped in a man's body) I greatly resent the implication that one's sexual orientation in any way affects their ability to reason....
thefurlong
5 / 5 (5) Aug 26, 2014
Depends on the way I read your question, furbrain, Typical of your sloppy phraseology. 50% of people reading it will take my interpretation, although I knew what you meant to say. You should try to be more explicit with your English.

Also, my question to you is, if you knew what I meant to say, why even bother misrepresenting my argument? You know what that's called? The straw man fallacy.

Look, either you were being dense, and couldn't make the connection that because one of the possibilities was correct, I probably meant that one (I can't even accuse you of being pedantic there, because a pedant would have taken what I wrote literally and deduced the correct meaning), or you were being purposely deceptive. Is there another possibility that I am missing? If not, I know you won't admit to being dense, so, if you are logical, you should admit to being deceptive. And if you are deceptive in such an obvious way here, why should I bother reading your book?
thefurlong
5 / 5 (6) Aug 26, 2014
As a practicing lesbian (unfortunately trapped in a man's body) I greatly resent the implication that one's sexual orientation in any way affects their ability to reason....

As a human being, I greatly resent the implication that one's sexual orientation in any way affects their ability to reason.

Identifying one's sexuality/gender identity is no more valid than identifying one's eye color. It is a neutral characteristic--or at least that's how it should be.
Reg Mundy
1 / 5 (5) Aug 26, 2014
@furbrain
Oh, by the way, a word of advice.
I see Cap'n Grumpy is cosying up to you. I advise caution, as he will be suggesting you hang about outside fire stations with him shortly, and similar activities. Of course, you may welcome his attentions, especially if you have a tendency to bowl from the pavilion end as I suspect from some of your posts, but if you don't, keep a tight grip on your trousers.

Holy crap! Has it come down to gay bashing?!?!

Of course not! To each his own, I say. Its a free world (or should be). I was merely pointing out the creepy nature of the Cap'n sidling up to furbrain, who is demonstrably naïve in many ways.
Reg Mundy
1 / 5 (5) Aug 26, 2014
As a practicing lesbian (unfortunately trapped in a man's body) I greatly resent the implication that one's sexual orientation in any way affects their ability to reason....

As a human being, I greatly resent the implication that one's sexual orientation in any way affects their ability to reason.

Identifying one's sexuality/gender identity is no more valid than identifying one's eye color. It is a neutral characteristic--or at least that's how it should be.

C'mon, furbrain, I've never accused you of being a human being!
thefurlong
5 / 5 (6) Aug 26, 2014
Of course not! To each his own, I say. Its a free world (or should be). I was merely pointing out the creepy nature of the Cap'n sidling up to furbrain, who is demonstrably naïve in many ways.


Two words, which you are using there, demonstrably and naive, do not mean what you suggest they mean.

Also, the fact that you bring up sexuality at all belies your supposed accepting stance of non cis-gendered sexuality.
thefurlong
5 / 5 (6) Aug 26, 2014
As a practicing lesbian (unfortunately trapped in a man's body) I greatly resent the implication that one's sexual orientation in any way affects their ability to reason....

As a human being, I greatly resent the implication that one's sexual orientation in any way affects their ability to reason.

Identifying one's sexuality/gender identity is no more valid than identifying one's eye color. It is a neutral characteristic--or at least that's how it should be.

C'mon, furbrain, I've never accused you of being a human being!

And you have still not actually offered a rebuttal to my points, Reg. Funny how it works out that way. Normally, I do not see eye to eye with Johan F. Prins, but I believe that his "duck farting" metaphor aptly describes what you are currently doing. Maybe you could, I don't know, actually respond to the points I made? Or you could keep doing the functional equivalent of calling me a poopyhead--because that's science.
Captain Stumpy
5 / 5 (5) Aug 26, 2014
Of course not! To each his own, I say. Its a free world (or should be). I was merely pointing out the creepy nature of the Cap'n sidling up to furbrain, who is demonstrably naïve in many ways.
I call BS on this one
first off, I am a lesbian
second off, you wrote the comment to elicit a response (in this case, to denigrate another and to establish that said person was involved in questionable activities according to your personal faith) which would be defined as TROLLING

it is also defined as GAY BASHING
it can also be defined as hostility and prejudice

therefore you continually substantiate my own claims above with your continued refusal to present evidence for your position and that you are a TROLL

P.S. I happen to AGREE with Furlong because he uses his brains and doesn't keep saying "read my book"

LOGIC and INTELLIGENCE from Furlong
denigration, redirection, hostility and ignorance (as well as stupidity) from reg
Hmmm... not a hard choice
Reg Mundy
1 / 5 (5) Aug 26, 2014
@furbrain

There is no response to the asymptotic sequence I provided for you, and no response to my overall point that the acceleration from gravity is like this asymptotic sequence. It accelerates some objects so that their velocities approach a finite speed that can be >> than 0.

The asymptotic sequence you provide is of no more relevance to the gravity problem than the bullet/man sequence. Yet you reiterate the example yet again! What is the significance of the half? Why not use 2/3 instead? Then your equation becomes:-
2 -2/3 = 1 1/3
1 1/3 - 4/9 = 8/9
So, as a model for the situation of the object flying away from Earth, it is meaningless. It defies common sense, like the bullet/man mathematical model which also tends asymptotically to infinity.
Captain Stumpy
5 / 5 (5) Aug 26, 2014
As a practicing lesbian (unfortunately trapped in a man's body) I greatly resent the implication that one's sexual orientation in any way affects their ability to reason....
@Whyde
(I am a practicing lesbian trapped in a mans body too)
I COMPLETELY AGREE, which is why I reported those posts
And you have still not actually offered a rebuttal to my points, Reg. Funny how it works out that way. Normally, I do not see eye to eye with Johan F. Prins, but I believe that his "duck farting" metaphor aptly describes what you are currently doing.
@Furlong
I prefer either:
MONKEY POO FLING argument

OR
the PIGEON CHESS argument

thefurlong
5 / 5 (6) Aug 26, 2014
The asymptotic sequence you provide is of no more relevance to the gravity problem than the bullet/man sequence. Yet you reiterate the example yet again!


And the crackpot deluge continues. Keep throwing crap. Maybe some of it will stick.

So, I take this as meaning you accept that such asymptotic behavior can arise in everyday situations. And yes, you're technically correct, using an exponential series is not related to the inverse squared law, but if you do accept that asymptotic behavior is valid in everyday reality, then you should accept the formal derivation of escape velocity that I provided here http://phys.org/n...ng.html.

But of course, you aren't actually interested in arguing. If you were, you would provide formal arguments for your claims, which you have consistently been unable or unwilling to do--and this wouldn't be such a problem with me if you did act like you were some authority on a subject on which you know very little.
Whydening Gyre
5 / 5 (2) Aug 26, 2014
Okay. That all out o the way, I have a prob with the first statement of the article -

"... the Higgs boson, which was recently confirmed to be the origin of mass..."

I thought what was confirmed was it's existence - with the conferring of mass still in the hypothesys stage...
thefurlong
5 / 5 (5) Aug 26, 2014
@Reg Mundy
And this is why, I would like to remind you that here, http://phys.org/n...ng.html, I stopped trying to argue with you, and, diverted my efforts to objectively demonstrating your incompetence, which you deftly complied with by, 1) mistaking a continual growth, non-homogenous differential word problem, with a compound interest problem, 2) confusing numerical error with conceptual error, 3) not understanding just what all those numbers at the end were (they served for some very cheap laughs, I must say). and 4) Responding to those numbers by converting your text to ascii.

Now, I have been calling you Mr. Ed, for reasons you may not be able to fathom, but maybe, instead, I will call you Clever Hans. That more aptly describes you mastery of mathematics and science.
Reg Mundy
1 / 5 (5) Aug 26, 2014
@furbrain
So, you agree that I'm "technically correct", do you. Does that mean you are admitting you are wrong? Or are you going to blame semantics yet again?
Meanwhile, I have never argued against the mathematical model of an asymptotic equation, nor that it NEVER applies in reality. I am arguing against applying it to "gravity", which continues to act at any distance, so that insisting on an escape velocity is rather like insisting that the bullet never reaches the man in my example. Common sense says it does, mathematical model says it doesn't. You can't have it both ways (except in your universe, which seems to confuse models with reality).
thefurlong
5 / 5 (4) Aug 26, 2014
So, you agree that I'm "technically correct", do you

About one thing, which would be obvious to anyone with a brain since I am clearly using an analogy in place of a formal argument.
Does that mean you are admitting you are wrong?

Only if you don't understand how logic works.
Or are you going to blame semantics yet again?

What was that, pot?
thefurlong
5 / 5 (4) Aug 26, 2014
Meanwhile, I have never argued against the mathematical model of an asymptotic equation, nor that it NEVER applies in reality

Oh, good, so then you can drop that silly straw man argument about a bullet, since it doesn't apply. I am glad to see we are making progress in this conv--
I am arguing against applying it to "gravity", which continues to act at any distance, so that insisting on an escape velocity is rather like insisting that the bullet never reaches the man in my example.

Dammit, Reg!

Ok, fine, what do you have to explain why solving the differential equation for two gravitating bodies somehow does not apply at infinity?
Common sense says it does, mathematical model says it doesn't.

Common sense is a really poor substitute for empirical reasoning. That the sun revolved around the Earth was common sense. That flies spontaneously generated from meat was common sense. Got anything else?
thefurlong
5 / 5 (4) Aug 26, 2014
You can't have it both ways (except in your universe, which seems to confuse models with reality).

Can't have what both ways? As I suggested, common sense is a really bad compass to use for understanding the universe. Common sense did not give us Ampere's law. Common sense did not give us the Rydberg formula. Common sense did not help us solve the ultraviolet catastrophe. Sorry Reg, but that isn't an argument.

The math is right there--and not just the math. The evidence is right there, too.

http://en.wikiped.../1980_E1
http://en.wikiped.../2000_U5
http://en.wikiped...4_(NEAT)
http://en.wikiped.../2009_R1
http://en.wikiped.../1956_R1
http://en.wikiped...(LONEOS)

These comets all conform to our expectations of gravitational behavior. They all appear to be following hyperbolic trajectories.

So, you are just wrong. But who am I kidding, you'll just true believe it all away anyway.
Whydening Gyre
5 / 5 (2) Aug 27, 2014
Not quite sure I'm with ya on the "common sense" statement, 1/8...
I always thought empirical reasoning WAS common sense....

Anyway, I may not have any common sense, but I know it when I see it...
thefurlong
5 / 5 (4) Aug 27, 2014
Not quite sure I'm with ya on the "common sense" statement, 1/8...
I always thought empirical reasoning WAS common sense....

Far from it. Common sense is a collection of, often subjective, ad-hoc rules of thumb that humans have developed as a mechanism for living in the present age without doing the dirty work of testing hypotheses. These are rules that are held to be widely accepted and understood, even if they are incorrect. Such rules, by their nature, are not often held to scrutiny because they are taken for granted.

In contrast, empirical reasoning is reasoning based on actual examination.

Common sense is saying, "I know that the earth is round because it's what I was taught in grade school." Empiricism is saying, "I know the earth is round because I have either examined the evidence from peer-reviewed scientific literature or have conducted experiments or measurements to test the hypothesis out myself."

(to be continued)
thefurlong
5 / 5 (4) Aug 27, 2014
(continued)
I suspect that Reg Mundy is invoking common sense because he was taught things like "what goes up must come down.", and rules of thumb like approximating gravity at the surface of the earth as a constant acceleration of 9.8m/s^2. I don't think he has bothered asking himself the conditions in which one could have a force field that, despite pulling on an object, does not necessitate its return. We do not need the feeble force of gravity for this. We can set up initial conditions in which two oppositely charged objects that are placed in a vacuum chamber will continuously fly away from each other despite being attracted. This is because, again, the coloumb force is an inverse square law. To put it succinctly, it gets too small too quickly with distance, to force an object to completely slow down.
Whydening Gyre
5 / 5 (1) Aug 27, 2014
Well, I've always found it good rule of thumb to empirically examine (where possible) everything I've been "told"..:-)
My "empirical examination" is then also my rationalization method...
BTW, I'm not taking a "reg" attitude on this...
Whydening Gyre
5 / 5 (1) Aug 27, 2014
one could have a force field that, despite pulling on an object, does not necessitate its return.

It's kinetic energy might be greater than the "attractive" energy
We can set up initial conditions in which two oppositely charged objects that are placed in a vacuum chamber will continuously fly away from each other despite being attracted.

Only because they had a prior kinetic/momentum "direction" to start with.
Given that perfect vacuum setup, if you had each of those particles sitting perfectly still to start, given enuff time they will eventually "sense" each other (personally, I think it's "magnetism", but that's for another discussion) and ever so slowly "gravitate" (begin moving) towards each other. Might take a long time, but acceleration will build.
Reg Mundy
1 / 5 (5) Aug 28, 2014
one could have a force field that, despite pulling on an object, does not necessitate its return.

It's kinetic energy might be greater than the "attractive" energy
We can set up initial conditions in which two oppositely charged objects that are placed in a vacuum chamber will continuously fly away from each other despite being attracted.

Only because they had a prior kinetic/momentum "direction" to start with.
Given that perfect vacuum setup, if you had each of those particles sitting perfectly still to start, given enuff time they will eventually "sense" each other (personally, I think it's "magnetism", but that's for another discussion) and ever so slowly "gravitate" (begin moving) towards each other. Might take a long time, but acceleration will build.

Wow! A convert!
Reg Mundy
1 / 5 (5) Aug 28, 2014
You can't have it both ways (except in your universe, which seems to confuse models with reality).

Can't have what both ways? As I suggested, common sense is a really bad compass to use for understanding the universe. Common sense did not give us Ampere's law. Common sense did not give us the Rydberg formula. Common sense did not help us solve the ultraviolet catastrophe. Sorry Reg, but that isn't an argument.

The math is right there--and not just the math. The evidence is right there, too.

http://en.wikiped.../1980_E1

These comets all conform to our expectations of gravitational behavior. They all appear to be following hyperbolic trajectories.

So, you are just wrong. But who am I kidding, you'll just true believe it all away anyway.

Just how are all these links relevant? Comets move from being influenced by one body to the influence of another, in our Solar system mainly the Sun. Are you saying they leave our solar system and never return?
Reg Mundy
1 / 5 (5) Aug 28, 2014
@Reg Mundy
And this is why, I would like to remind you that here, http://phys.org/n...ng.html, I stopped trying to argue with you, and, diverted my efforts to objectively demonstrating your incompetence, which you deftly complied with by, 1) mistaking a continual growth, non-homogenous differential word problem, with a compound interest problem, 2) confusing numerical error with conceptual error, 3) not understanding just what all those numbers at the end were (they served for some very cheap laughs, I must say). and 4) Responding to those numbers by converting your text to ascii..

1. Anyone with sense will calculate population growth using my method. You're too far up your own ass to see that.
2. Your attention to detail is sadly lacking, hence your multiple arithmetical errors and stupid answers.
3. There was little difference between those numbers and your answers!
4. Just to show I knew what you were doing - changing the goalposts 'cos you'd lost...
Reg Mundy
1 / 5 (5) Aug 28, 2014
@furbrain
OK, lets get down to brass tacks. Write out the equation for a bullet leaving the Earth at an initial 25,000 mph (more than the commonly-accepted escape velocity of 22,000 mph), assuming that the Earth is the only body in the universe. Solve the equation to tell me at what distance the bullet completely escapes the influence of Earth's "gravity". That is the point at which it will have "escaped".
thefurlong
5 / 5 (5) Aug 28, 2014
Anyone with sense will calculate population growth using my method. You're too far up your own ass to see that.

Do you know what the difference is between a continuous growth problem and a compound interest problem?
I am butthurt from Furlong embarrassing me like that

I know, Reg, I know.
There was little difference between those numbers and your answers!

Yep, they were very similar. What's a few orders of magnitude when you don't even know what an equation is?
Just to show I knew what you were doing - changing the goalposts 'cos you'd lost...

Ahh, well, if you knew what I was doing, then I am sure you will be able to tell me what those numbers meant. You still haven't told me the password.
thefurlong
5 / 5 (5) Aug 28, 2014
OK, lets get down to brass tacks. Write out the equation for a bullet leaving the Earth at an initial 25,000 mph (more than the commonly-accepted escape velocity of 22,000 mph), assuming that the Earth is the only body in the universe. Solve the equation to tell me at what distance the bullet completely escapes the influence of Earth's "gravity". That is the point at which it will have "escaped".

Escape velocity is not the velocity at which an object escapes a gravitational field, you vacuous crackpot. Gravitational fields are infinite (as far as we can tell). It is the velocity at which, despite the pull of the gravitational field, the object will keep on going (though continuously slowing down asymptotically).
Now, we've been over this several times, and you have yet to explain why a bullet leaving Earth at a speed above escape velocity can't approach a final speed that is larger than 0.
Whydening Gyre
5 / 5 (4) Aug 28, 2014
Wow! A convert!

Don't think so...
Da Schneib
5 / 5 (3) Aug 29, 2014
...I quoted the definition of a relativistic transform as it follows logically from Salviati's arguments. You obviously are still a Simplicious!
And I quoted directly from the Minkowsky spacetime interval formula: d² = x² + y² + z² - c²t²

We were not discussing covariance. We were discussing transforms. And you claimed transforms change the underlying physical reality, an obvious gaffe.
This is exactly what a relativistic transform does.
No, it's not. Period. Transforms change between coordinate systems; they do not change reality. This is basic physics.

When you "look out" into another IRF in which physics is occurring, you do not see a covariant transformation of the latter physics equations into your IRF.
Then all IRFs are not equivalent, the Second Postulate of Special Relativity Theory. You are now directly denying SRT (and consequently GRT as well).
Da Schneib
5 / 5 (2) Aug 29, 2014
Johan, it's Jack Sarfatti. http://en.wikiped...Sarfatti
Da Schneib
4 / 5 (4) Aug 29, 2014
When using the Lorentz transformation, not even the coordinates are isometric and therefore not even the coordinates are covariant.
See the Minkowski formula.
Da Schneib
5 / 5 (2) Aug 29, 2014
Anybody ever notice all of Johan's posts are 1000 characters long so as to cause the most annoyance when trying to reply to them?

Same as RealityCheck's.

Just sayin'.
Da Schneib
5 / 5 (3) Aug 29, 2014
Okay. That all out o the way, I have a prob with the first statement of the article -

"... the Higgs boson, which was recently confirmed to be the origin of mass..."

I thought what was confirmed was it's existence - with the conferring of mass still in the hypothesys stage...
Good question.

The answer is, they wrote out the theory (that is, did the mathematics) behind Peter Higgs' idea, and figured out a whole bunch of characteristics that such a boson should have. As a result of knowing that, they can sort through large numbers (Billions? Millions for certain) of collisions looking for what they should see if such a particle forms in the quark-gluon plasma created at direct impact (not a very likely event, but if you watch long enough anything possible is inevitable- and even how often it happens is more grist for the mill, confirming another prediction about how the Higgs should behave.

contd
Da Schneib
5 / 5 (3) Aug 29, 2014
Half-life (usually called "lifetime" short for "average lifetime") of a particle is a major clue to its makeup, as are not only the number of different decay modes, but the half-life of each mode. Other major clues are the total spins of the particles it breaks down into, and the total charges of each of the four forces it carries).

They looked. They saw it. Not just one thing. A multitude of things. They didn't see one Higgs. They saw a number large enough that they can declare they saw the Higgs with 5 sigma results at an appropriate confidence level to cause their reviewers to allow their announcement to be published. It's the sum total of all those Higgs bosons they saw, each interaction analyzed down to the fly's eyelash (and actually a lot smaller than that). And for each of more than a handful of parameters (I don't think more than a dozen, and I think half that or less, but that's just my opinion). There's not a lot of wriggle room there.
Da Schneib
5 / 5 (3) Aug 29, 2014
I'm going to point out yet again that the fallacy of the Appeal to Authority is only a fallacy if the authority cited is not, in fact, an expert. Otherwise it's perfectly valid. I would accept, for example, the opinion of a computer programmer that s/he has found a bug. I would not accept, for example, that same computer programmer's view that the Earth is flat.

Misrepresenting an appeal to a valid expert is often confused with the Appeal to Authority fallacy by... well, let's not sling pejoratives around, shall we? Oh, wait, too late...
RealityCheck
1 / 5 (6) Aug 29, 2014
Hi Schneib. :)
Anybody ever notice all of Johan's posts are 1000 characters long so as to cause the most annoyance when trying to reply to them?

Same as RealityCheck's.

Just sayin'.

Why keep making such personality cult innuendoes/ assumptions/conclusions like that, mate?

Every time you make such fallacious 'connections/interpretations' like that it makes your scientific credibility plummet since no real scientist would make such obviously silly speculations. Oh, I forgot, that is what the 'modern scientist' is 'trained' to do nowadays, make silly claims they can't support with actual facts, just 'interpretative license' (shades of BICEP2 exercise 'method/claims' using 'interpretative license').

Mate, don't make such silly UNscientific posts/innuendoes if you want to be taken seriously, ok?

I am RealityCheck, not Johan or anyone else. If I conduct an Internet Experiment using another TEMPORARY username, I'll let all know immediately afterwards, as usual.

Good luck!:)
thefurlong
5 / 5 (3) Aug 29, 2014
Johan, it's Jack Sarfatti. http://en.wikiped...Sarfatti

Who's Jack Sarfatti?
Da Schneib
5 / 5 (3) Aug 29, 2014
A non-mainstream physicist who's not a crank. He's a big proponent of the Alcubierre drive. http://en.wikiped...re_drive He got fooled by Uri Gellar, but after he'd seen Randi do the same trick he decided he'd been tricked and said so.
Da Schneib
5 / 5 (2) Aug 29, 2014
So you do it to be irritating.

Just checking, "RC."
RealityCheck
1 / 5 (4) Aug 29, 2014
Hi Schneib. :)
So you do it to be irritating.
I don't know what others do, but I just compose a post and if it (usually) is too long for the text limit here, I then 'edit down' by removing some non-critical words like 'a' or 'the' etc until it 'fits'....and I submit. I don't complain when others do it too and I have to work-around it. :)

Just checking, "RC."
Next time, just ask me instead of making silly innuendoes like that which make you look trollish personality-gossiping/baiting 'social media twit' rather than serious scientist (which I feel you have great potential for becoming IF you drop 'mutual 5s' gang-voting practice and motivations/distractions and concentrate purely on the science matters not the source/personality).

By the way, I may have read somewhere, that Johan F Prins is South African (and 75---is that correct, Johan? :) ).

Whereas I am Aussie and 64 (---soon 65 :) ). BTW, how old are you, mate?

Left plenty room this time, just for you! :)
johanfprins
1 / 5 (5) Aug 29, 2014
When using the Lorentz transformation, not even the coordinates are isometric and therefore not even the coordinates are covariant.
See the Minkowski formula.

Exactly!! Minkowski recklessly assumed that the coordinates x,y,z,t are isometric while it is not the case. Minkowski's space time cannot exist in reality since it violates the most basic rules of linear mathematics. For example. they are not linearly independent which they MUST be in order to be able to define an isometry.
johanfprins
1 / 5 (5) Aug 29, 2014
...I quoted the definition of a relativistic transform as it follows logically from Salviati's arguments. You obviously are still a Simplicious!
And I quoted directly from the Minkowsky spacetime interval formula: d² = x² + y² + z² - c²t²
Which proves what a SOT you are. For such a "space-time" interval to be able to define an isometry, one MUST have that wnen d=0 THAT x=y=z=t=0. This is not the case in Minkowski space and therefore it violates the most basic rule of linear algebra.

No, it's not. Period. Transforms change between coordinate systems; they do not change reality. This is basic physics.
More evidence of what a SOT you are. When you transform the equations of physics, you get effects that seem to be caused by forces which do not exist. Thus by looking from your IRF into another IRF in which the physics is actually occurring, the physics you see becomes distorted. This is what relativity is all about.

continued

johanfprins
1 / 5 (5) Aug 29, 2014
When you "look out" into another IRF in which physics is occurring, you do not see a covariant transformation of the latter physics equations into your IRF.
Then all IRFs are not equivalent, the Second Postulate of Special Relativity Theory.
Equivalence of IRF's means, since the time of Galileo's Salviati, that the SAME experiment can be done in any IRF, and the same results will be obtained. This does NOT mean that when looking from outside into the IRF in which the experiment is being done you will see the same results as an observer with the experiment is seeing. Thus, the transformation of the physics equations is NOT covariant for a relativistic coordinate transformation.

You are now directly denying SRT (and consequently GRT as well).
No I am not "denying" SRT but only pointing out that a relativistic coordinate transformation does not leave the physics equations to be covariant. Yes GRT and Dirac's equation for an electron are suspect.
johanfprins
1 / 5 (5) Aug 29, 2014
By the way, I may have read somewhere, that Johan F Prins is South African (and 75---is that correct, Johan? :) ).
I am actually 72: But it does not matter since I still have a sane and sound mind. In fact my mind is better than it was when I was studying at University. The only problem is that I have passed my three score years plus 10, and am thus living on injury time.

johanfprins
1 / 5 (5) Aug 29, 2014
The answer is, they wrote out the theory (that is, did the mathematics) behind Peter Higgs' idea, and figured out a whole bunch of characteristics that such a boson should have.
This is circumstantial evidence: Not proof that noise at CERN gives other "particles" rest-mass. This is not the way physics must be done, since a conclusion is reached for which NOBODY can design an experiment that can falsify it, if it is false: Which it indeed is, since the origin of rest-mass follows simply from Einstein's Special Theory of Relativity.

Any EM-wave which cannot move with the speed of light, automatically has rest-mass since such an enity MUST be stationary within an inertial reference-frame. A single ligh-wave cannot be stationary within any IRF, unless it is trapped within a reflecting cavity, and therefore a free light-wave has no rest-mass. An electron-wave cannot move with speed c since it MUST be stationary within an IRF: Therefore it has rest-mass energy!
Reg Mundy
1 / 5 (5) Aug 29, 2014

Escape velocity is not the velocity at which an object escapes a gravitational field, you vacuous crackpot. Gravitational fields are infinite (as far as we can tell). It is the velocity at which, despite the pull of the gravitational field, the object will keep on going (though continuously slowing down asymptotically).
Now, we've been over this several times, and you have yet to explain why a bullet leaving Earth at a speed above escape velocity can't approach a final speed that is larger than 0.

You really are stupid. As long as gravity provides ANY effect, the object MUST accelerate towards Earth if there are no other influences. Eventually, it must start to move towards Earth. Your asymptotic model is not applicable, it is only a model, and does not reflect REALITY. Unless, of course, the laws of gravity are wrong......
Reg Mundy
1 / 5 (5) Aug 29, 2014
@Daz
A non-mainstream physicist who's not a crank. He's a big proponent of the Alcubierre drive. http://en.wikiped...re_drive He got fooled by Uri Gellar, but after he'd seen Randi do the same trick he decided he'd been tricked and said so.

To quote from your link "Although the metric proposed by Alcubierre is mathematically valid in that it is consistent with the Einstein field equations, it may not be physically meaningful or indicate that such a drive could be constructed." This is a good example of where mathematical equations fail to pass the reality test. It's very similar to the equation deriving escape velocity under the "laws of gravity" as imagined by Newton, Einstein, etc., but try explaining that to furbrain.
thefurlong
5 / 5 (5) Aug 29, 2014
As long as gravity provides ANY effect, the object MUST accelerate towards Earth if there are no other influences.

Oh, hey, you said something correct. Now, if you understand what acceleration means then maybe you can see how ridiculous your position i--
Eventually, it must start to move towards Earth.

Ok, Clever Hans. Time for calc refresher (or more likely, primer in your case) 101. When the second derivative (acceleration) points in some direction, the 0th derivative (position) can continue to move in the opposite direction. All it means is that the velocity continues to move in the direction of acceleration. If the velocity is pointing in the opposite direction, then *gasp* the position moves in the direction of the velocity. And, if the 3rd derivative, knowns as "jerk", (OMG there are 3rd derivatives?) happens to, say, be moving in the direction opposite to acceleration, then the magnitude of acceleration will get increasingly small (To be continued)
thefurlong
5 / 5 (5) Aug 29, 2014
(continued)
Likewise, if the 4th derivative governs how the 3rd changes, and so on.
In short, you inconceivably uneducated wonder, VELOCITY AND ACCELERATION ARE TWO DIFFERENT THINGS. Therefore, you arrogant, prokaryote of a thinker, just because you are accelerated in the direction of something, it does not mean you will move in that direction.

Here are the only cases in which you must eventually move in the direction of acceleration: if your 3rd, 4th, 5th,..., and so on, derivatives are all 0 (in other words, you have a parabola). This is not even physics. It's just basic, 300+ year old math. And guess what determines all these derivatives? Differential equations (which, in physics ARE DECIDED BY PHYSICAL LAWS)! And the equation of gravity is such that all those derivatives make the magnitude of acceleration grow so small so quickly that even velocity will not necessarily start pointing in the direction of acceleration. You moronic gnat.
thefurlong
5 / 5 (5) Aug 29, 2014
@Johan F. Prins
Which proves what a SOT you are. For such a "space-time" interval to be able to define an isometry, one MUST have that wnen d=0 THAT x=y=z=t=0. This is not the case in Minkowski space and therefore it violates the most basic rule of linear algebra.

If (Ox, Oy, Oz) is the origin, then (x-Ox)^2 + (y-Oy)^2 + (z-Oz)^2 which we know as euclidean distance, is an invariant of Galilean transformation. Let's suppose that we wish to find a similar invariant of the Lorentz transformation. I can not only show, but derive from the Lorentz transformation, that (x-Ox)^2 + (y-Oy)^2 + (z-Oz)^2 - c^2(t-Ot)^2 is the ONLY such possible invariant relating a point to the origin. As such it is the only possible way for us to assign a distance between a space-time point and the origin. If I start from the Lorentz transformation, and without invoking length contraction and time dilation, arrive at this invariant, will that help?
johanfprins
1 / 5 (5) Aug 29, 2014
@Johan F. Prins
If (Ox, Oy, Oz) is the origin, then (x-Ox)^2 + (y-Oy)^2 + (z-Oz)^2 which we know as euclidean distance, is an invariant of Galilean transformation.
Correct for the Galilean COORDINATE TRANSFORMATION. But even for the Galilean coordinate transformation the equations for the laws of physics are NOT covariant. They are ONLY covariant when the origins of the two systems DO NOT move relative to one another!
Let's suppose that we wish to find a similar invariant of the Lorentz transformation. I can not only show, but derive from the Lorentz transformation, that (x-Ox)^2 + (y-Oy)^2 + (z-Oz)^2 - c^2(t-Ot)^2 is the ONLY such possible invariant relating a point to the origin.
You can only "derive" this by stupidly dividing by ZERO: As both Einstein and Minkowski have done. If you know ANYTHING about mathematics you will know that dividing by ZERO is a NO! NO! This is so since you then cannot have one-to-one relationships!
thefurlong
5 / 5 (5) Aug 29, 2014
Correct for the Galilean COORDINATE TRANSFORMATION. But even for the Galilean coordinate transformation the equations for the laws of physics are NOT covariant. They are ONLY covariant when the origins of the two systems DO NOT move relative to one another!

I don't understand your point. If you accept the Lorentz Transformation, then you accept that any invariant of it holds. I don't know why you are talking about covariance.
You can only "derive" this by stupidly dividing by ZERO

That's not correct. I derived it without dividing by 0, using calculus-based arguments. And I can show you how, too. If Einstein and Minkowski did it by dividing by zero, they did things the hard way.
Da Schneib
5 / 5 (3) Aug 29, 2014
No science, RC.
Da Schneib
5 / 5 (4) Aug 29, 2014
When using the Lorentz transformation, not even the coordinates are isometric and therefore not even the coordinates are covariant.
See the Minkowski formula.

Exactly!! Minkowski recklessly assumed that the coordinates x,y,z,t are isometric
So did Einstein. So did Lorentz. And I'm not sure you're using that word (isometric) correctly. How about if you tell us what YOU mean when you say it.
Da Schneib
5 / 5 (5) Aug 29, 2014
When you "look out" into another IRF in which physics is occurring, you do not see a covariant transformation of the latter physics equations into your IRF.
Then all IRFs are not equivalent, the Second Postulate of Special Relativity Theory.
Equivalence of IRF's means, since the time of Galileo's Salviati...
We're talking about Einstein, Lorentz, Minkowsky, and Poincare. This is relativity physics, not Galilean/Newtonian mechanics. Try to keep up.

The answer is, they wrote out the theory (that is, did the mathematics) behind Peter Higgs' idea, and figured out a whole bunch of characteristics that such a boson should have.
This is circumstantial evidence
The results are now 7 sigma. http://www.extrem...gs-boson

There is still an outside chance it could be something other than a Higgs, but if it is it's something we've never seen before and will immediately lead to new physics.
Da Schneib
5 / 5 (3) Aug 29, 2014
{Jack's} a big proponent of the Alcubierre drive.
This is a good example of where mathematical equations fail to pass the reality test.
No, it's a case just like frame dragging, where physical observations have not YET confirmed the solution. Gravity Probe B confirmed frame dragging. Data will confirm or deny the Alcubierre solution. Absence of evidence is not evidence of absence. <- basic logic
thefurlong
5 / 5 (5) Aug 29, 2014
The results are now 7 sigma. http://www.extrem...gs-boson

Wow, I didn't realize they had made such progress! All right, humanity! Huzzah for some of us actually doing something about understanding the universe instead of armchair griping over mathematical descriptions we don't understand, about physical laws we don't understand!
johanfprins
1 / 5 (4) Aug 29, 2014
I don't understand your point. If you accept the Lorentz Transformation, then you accept that any invariant of it holds.
OK! Let us first forget about the invariance of physics equations. What do you understand about "invariance" of a coordinate transformation in an n-dimensional manifold? It must mean that the n coordinates of an n-dimensional point and tne transformed n-coordinates MUST reference the SAME n-dimensional point. The Lorentz transformation does NOT do this. If, the distances between points in an n-dimensional manifold are not changed by the coordinate transformation, then the transformation is isometric. This is also not the case for the Lorentz transformation!

I derived it without dividing by 0, using calculus-based arguments. And I can show you how, too.
I know all the so-called derivations: They are all wrong since they assume that a relativistic transformation of the physics has an inverse transformation OF THE same physics! WRONG!
Da Schneib
5 / 5 (4) Aug 29, 2014
@furlong, great exposition of basic mechanics and the physical meaning of derivatives as applied to those mechanics. And a good job explaining precisely where Reg is wrong again. Applause. Also a nice job on Johan.
johanfprins
1 / 5 (5) Aug 29, 2014
So did Einstein. So did Lorentz. And I'm not sure you're using that word (isometric) correctly. How about if you tell us what YOU mean when you say it.
It is in all textbooks DICKHEAD! It means that in an n-dimensional manifold the the position-vector from the origin to a point P remains invariant even though its coordinates change when they are transformed by an isometric transfrmation. Wow! You are really stupid!!
johanfprins
1 / 5 (5) Aug 29, 2014
Equivalence of IRF's means, since the time of Galileo's Salviati... ]We're talking about Einstein, Lorentz, Minkowsky, and Poincare. This is relativity physics, not Galilean/Newtonian mechanics. Try to keep up.
You are so stupid that it is amazing that you can write and read. The principle of relativity has remained the same as Galileo has defined it: This is the reason why c MUST be the same within ALL IRF's. Einstein, Poincare etc. were too stupid to fully understand Galileo.

Signing off for now!

The answer is, they wrote out the theory (that is, did the mathematics) behind Peter Higgs' idea, and figured out a whole bunch of characteristics that such a boson should have. This is circumstantial evidence
The results are now 7 sigma.
7 sigma that this noise is responsible for mass? STOP posting BULLSHIT!! There is no proof and NEVER will be any experimental proof that mass is caused by this noise!
thefurlong
4.2 / 5 (5) Aug 29, 2014
What do you understand about "invariance" of a coordinate transformation in an n-dimensional manifold?

But this has nothing to do with what I am saying. I am making a very simple observation. If you want to find a scalar quantity that depends on position and time that does not change when we apply the Lorentz transform, then your options are limited. If you set out to find this quantity in 1+1 dimensions, then you will inevitably arrive at the Minkowski Space-Time interval.
I know all the so-called derivations

How do you know? I found my derivation without consulting any textbooks or any other sources of information. My derivation came purely from my own brain.

They are all wrong since they assume that a relativistic transformation of the physics has an inverse transformation OF THE same physics!

My derivation does not invoke the inverse Lorentz transform. All it does is rely on the definition of the Lorentz transform.
RealityCheck
1 / 5 (5) Aug 29, 2014
Hi Schneib. :)
No science, RC.
How long has it been since you were last reminded that I had withdrawn from detailed science discussion to avoid risk of my ToE insights/details being plagiarized, mate? :)

And how long has it been since you were last reminded that I'll come in as necessary to respond to troll posts like yours which ignore the fact that I would not have to post at all if YOU stopped making such inane and unscientific comments about the person like that, ie about me?

If you concentrate on science discussions you are having, and remember THIS latest reminder about the above facts, then you would be better respected as a scientist, and not come across as a 'twitter twit' troll who keeps making inane references to people who have stopped discussing detailed science here and elsewhere?

So if you'll let sleeping dogs lay, you won't be coming across as absent-minded twitter twit again, will you?

Bye, good luck in your science discussions, Schneib, everyone! :)
johanfprins
1 / 5 (5) Aug 29, 2014
But this has nothing to do with what I am saying. I am making a very simple observation. If you want to find a scalar quantity that depends on position and time that does not change when we apply the Lorentz transform, then your options are limited. If you set out to find this quantity in 1+1 dimensions, then you will inevitably arrive at the Minkowski Space-Time interval.
Only when you violate the rules of mathematics!
How do you know? I found my derivation without consulting any textbooks or any other sources of information. My derivation came purely from my own brain.
After having folowed your posts on this forum, it is the best proof that it MUST be wrong!

My derivation does not invoke the inverse Lorentz transform. All it does is rely on the definition of the Lorentz transform.
I will humor your insanity: Give me this "definition" and I will return to it tomorrow. It is late here in South Africa.

Good night for now!
thefurlong
4.3 / 5 (6) Aug 29, 2014
@furlong, great exposition of basic mechanics and the physical meaning of derivatives as applied to those mechanics. And a good job explaining precisely where Reg is wrong again. Applause. Also a nice job on Johan.


Thanks. I must say that I admire your taking Zephyr to task on Raman scattering and his silly ideas. It's so frustrating, though. It's bad enough that these people have fundamentally broken their own minds to the point where no amount of mathematical/physical/empirical evidence will persuade them out of their silly notions, since they have taken, as one of their postulates, that peer reviewed science is inevitably wrong, but I don't understand why they feel it necessary to paint themselves as experts on subjects that they can't be bothered to understand. I mean, do they do the same thing to their plumber or their electrician? Why are experts in other fields not subject to the same degree of faux skepticism?
Da Schneib
5 / 5 (5) Aug 29, 2014
If (Ox, Oy, Oz) is the origin, then (x-Ox)^2 + (y-Oy)^2 + (z-Oz)^2 which we know as euclidean distance, is an invariant of Galilean transformation.
Correct for the Galilean COORDINATE TRANSFORMATION. But even for the Galilean coordinate transformation the equations for the laws of physics are NOT covariant. They are ONLY covariant when the origins of the two systems DO NOT move relative to one another!
Only if you leave out time.

Let's suppose that we wish to find a similar invariant of the Lorentz transformation. I can not only show, but derive from the Lorentz transformation, that (x-Ox)^2 + (y-Oy)^2 + (z-Oz)^2 - c^2(t-Ot)^2 is the ONLY such possible invariant relating a point to the origin.
You can only "derive" this by stupidly dividing by ZERO
Or if you claim that time is not a dimension.

Oops.
thefurlong
5 / 5 (6) Aug 29, 2014
After having folowed your posts on this forum, it is the best proof that it MUST be wrong!

Johan, I have gotten you twice to admit that you were fundamentally incorrect about some relativistic concept. I am not saying that to be spiteful, but only to point out that of the two of us, only one of us has turned out to be unequivocally wrong on at least not just one, but two occasions, so what you just wrote above is unfounded.
I will humor your insanity: Give me this "definition" and I will return to it tomorrow. It is late here in South Africa.

You meant "derivation", right?

Da Schneib
5 / 5 (5) Aug 29, 2014
It means that in an n-dimensional manifold the the position-vector from the origin to a point P remains invariant even though its coordinates change when they are transformed by an isometric transfrmation.
Precisely. And space and time form a 4-dimensional continuum. That's a Postulate of Relativity called the Continuum Postulate.
Uncle Ira
5 / 5 (7) Aug 29, 2014
@ Everybody. Well I tell you one thing that is true. If that Skippy-couyon spend half the time on those toes of his as he does running around the interweb defending his dishonor they would have been fixed up years ago.

Why he not just put on the shoes and let his toes go like that and be done with them. Ol Ira probably don't spend 10 or 9 minutes a week worrying about my toes, and I can't remember the last time I felt the urge to mention them on an interweb comment place.

@ Really-Skippy you been talking about those toes of yours for 9 or 8 months now. What's up with your toes? Toes don't have anything to do with Higgs-Skippy's bosons that this article is about.

Do better Cher, people starting to think you are one weird couyon.
RealityCheck
1 / 5 (5) Aug 29, 2014
Hi thefurlong. :)

Johan is referencing your own words/statement re 'definition'; as follows...

All it does is rely on the definition of the Lorentz transform.


Perhaps you meant to say 'derivation' there?

Carry on, guys; these particular discussions are very interesting to observe! Thanks. :)

Cheers, and bye for now....and enjoy your discussions, all! :)

Da Schneib
5 / 5 (3) Aug 29, 2014
Equivalence of IRF's means, since the time of Galileo's Salviati...
]We're talking about Einstein, Lorentz, Minkowsky, and Poincare. This is relativity physics, not Galilean/Newtonian mechanics.
The principle of relativity has remained the same as Galileo has defined it This is incorrect. The principle of relativity is now in four dimensions, not three as Galileo and Newton defined it.

...they wrote out the theory (that is, did the mathematics) behind Peter Higgs' idea, and figured out a whole bunch of characteristics that such a boson should have.
This is circumstantial evidence
The results are now 7 sigma. 7 sigma that this noise is responsible for mass? ... There is no proof and NEVER will be any experimental proof that mass is caused by this noise! It fits the Standard Model of Particle Physics, which just happens to be the most carefully confirmed theory in all of physics and has been for decades. As I said, try to keep up.
Da Schneib
5 / 5 (4) Aug 29, 2014
Johan is referencing your own words/statement re 'definition'; as follows...
All it does is rely on the definition of the Lorentz transform.
Perhaps you meant to say 'derivation' there?
No, he meant "definition." His statements are derived from that definition.

Thought you were leaving.
Da Schneib
5 / 5 (3) Aug 29, 2014
Now everybody go look up "couyon." Snicker.

Nice to see you, Ira.

Gotta go, got places to do and things to be. Catch you later.
RealityCheck
1 / 5 (5) Aug 29, 2014
Hi Schneib. :)
No, he meant "definition." His statements are derived from that definition.
Then that is what Johan said he was going to follow up with furlong in the morning, isn't it? Ie, the 'definition' itself, as mentioned by furlong? As distinct from 'derivations'?

Thought you were leaving.
I am reading through interesting threads. I noted an inconsistency and asked for clarification by furlong. You have a problem with that now? :)

And stop using pedantic issues like 'leaving' as basis for inane commentary, else you will continue to come across as a twitter twit lame-o troll. Bye for now. :)
thefurlong
5 / 5 (3) Aug 29, 2014
Hi thefurlong. :)

Johan is referencing your own words/statement re 'definition'; as follows...

All it does is rely on the definition of the Lorentz transform.


Perhaps you meant to say 'derivation' there?

No, I meant to say definition. The Lorentz transformation is defined a certain way, and I use that definition to derive the minkowski space-time interval.
thefurlong
5 / 5 (5) Aug 29, 2014
Hi thefurlong. :)

Hi, RealityCheck
Then that is what Johan said he was going to follow up with furlong in the morning, isn't it? Ie, the 'definition' itself, as mentioned by furlong? As distinct from 'derivations'?

Johan, occasionally, like me, has finger slips. He already knows the definition of the Lorentz transformation, so it is more probable that he meant "derivation."
RealityCheck
1 / 5 (5) Aug 29, 2014
Hi Schneib. :)
Now everybody go look up "couyon." Snicker.

Nice to see you, Ira.

Gotta go, got places to do and things to be. Catch you later.
This is why you still come across as a biased twitter twit rather than any sort of fairminded true scientist. You accept NO SCIENCE posts from your friend BOT-operator trolls, and you encourage UNscientific downvoting tactics irrespective of science content posted by others, and you STILL carry on personality cult crap and encourage it in your mindless troll 'friends'. No wonder 'modern science' and 'modern scientists' are having such a hard time living down the 'reputation' which you idiotic 'defenders of mainstream' are giving them without their permission. With 'firnds' like you, modern scienc/scientists need no 'enemies' at all!

Get over yourself, mate, if you want to become a real objective/dispassionate objective observer/scientist/commenter when you 'grow up'. Good luck with that, mate; seems like you'll need plenty! :)
RealityCheck
1 / 5 (5) Aug 29, 2014
Hi thefurlong. :)

Johan is referencing your own words/statement re 'definition'; as follows...

All it does is rely on the definition of the Lorentz transform.


Perhaps you meant to say 'derivation' there?

No, I meant to say definition. The Lorentz transformation is defined a certain way, and I use that definition to derive the minkowski space-time interval.


Ok, mate. In that case, I will await Johan's intended follow-up on your 'definition' as meant by you. This is a very interesting discussion!

Thanks for the reply on that point. Cheers. :)
Da Schneib
5 / 5 (4) Aug 30, 2014
RC, I'm biased against people who claim mainstream physicists are idiots.

Get over it.
johanfprins
1 / 5 (4) Aug 30, 2014
@furlong, great exposition of basic mechanics and the physical meaning of derivatives as applied to those mechanics. And a good job explaining precisely where Reg is wrong again. Applause. Also a nice job on Johan
What nice job? I have not seen this great exposition of furlong anywhere on this thread. Are your minnows now praising you even though you have not posted anything?

johanfprins
1 / 5 (5) Aug 30, 2014
Correct for the Galilean COORDINATE TRANSFORMATION. But even for the Galilean coordinate transformation the equations for the laws of physics are NOT covariant
The Galilean transformation gives you forces that do not exist. Please show me how you remove these forces by assuminfg that time is a fourth dimension?

Let's suppose that we wish to find a similar invariant of the Lorentz transformation. I can not only show, but derive from the Lorentz transformation, that (x-Ox)^2 + (y-Oy)^2 + (z-Oz)^2 - c^2(t-Ot)^2 is the ONLY such possible invariant relating a point to the origin.
You can only "derive" this by stupidly dividing by ZERO
Or if you claim that time is not a dimension. Time cannot be a fourth coordinate that can be isometrically transformed, since this would require that the same single event must occur SIMULTANEOUSLY at two or more different times, which is ABSURD. Simultaneous means THE SAME INSTANT IN TIME.
johanfprins
1 / 5 (5) Aug 30, 2014
Johan, I have gotten you twice to admit that you were fundamentally incorrect about some relativistic concept.
YOU DID NOT! I admitted to making two mathematical errors; I never admitted that your logic of time-dilation is correctr. Nothing that is absurd can be correct physics. Have you heard of reductio ad absurdum?
I am not saying that to be spiteful, but only to point out that of the two of us, only one of us has turned out to be unequivocally wrong
You have been wrong overal since you are defending the absurd!
so what you just wrote above is unfounded.
In your naive childish manner you argue that once a person has bravely admitted to a mistake, he must be wrong on every aspect ad infinitum? This must be the reason why mainstream physicists like you will rather die than admit that they made a mistake!
You meant "derivation", right?
No I did not: You said that you have a "definition" which proves that what you claim is correct. I want your definition!
Da Schneib
5 / 5 (3) Aug 30, 2014
The Lorentz Transform says some of your time is my space (and vice versa) if we're moving relative to one another. It says:

x' = (t - vx/c²)/√(1-v²/c²)
Where,
x' is the x observed in the frame you're converting to
x is the observer's frame's x
t is the observer's frame's time
v is the observer's frame's velocity
c is the speed of light

See there? It says space (x) turns into time + space (t - vx/c²). Right there in black and white. Time is a dimension.

If you believe Lorentz. If you don't you're on your own.

And BTW there's your definition, johan.
johanfprins
1 / 5 (5) Aug 30, 2014
It means that in an n-dimensional manifold the the position-vector from the origin to a point P remains invariant even though its coordinates change when they are transformed by an isometric transfrmation.
Precisely. And space and time form a 4-dimensional continuum. That's a Postulate of Relativity called the Continuum Postulate.

I can postulate that horses have wings: But it does not give them wings. If you postulate something that is absurd, you are not jus wrong but also insane.

If you have four coordinates x,y,z,w so that d^2=x^2+y^2+z^2+w^2, and you find that d CAN ONLY be ZERO when x=y=z=w=0: ONLY THEN you have a four-dimensional continuum which can be isometricaly transformed. If you set w=ict, you do not have that d can only be zero when x=y=z=ict=0. It can be zero for non-zero values of x,y,z,ict.

The most inportant rule in elementary linear algebra demands that in the latter case these coordinates CANNOT define a 4D continuum. space-time=absurdity.
Da Schneib
5 / 5 (3) Aug 30, 2014
Yes but you can't derive relativity from postulating horses have wings, johan.

Also, there is no simultaneity. It simply can't exist, if the speed of light is constant for every observer.
johanfprins
1 / 5 (4) Aug 30, 2014
The principle of relativity has remained the same as Galileo has defined it
This is incorrect. The principle of relativity is now in four dimensions, not three as Galileo and Newton defined it.
You are making up physics asd it suits your insanity. Nowhere has Einstein claimed this in his two postulates. is firts postulate is in fact a re-affirmiation of Galileo's principle of relativity, and the second is not really another postulate since it flows logcally from the first. Now Da Schneib is redefining Einstein's own postulates: LOL!

It fits the Standard Model of Particle Physics, which just happens to be the most carefully confirmed theory in all of physics and has been for decades.
The motion of the planets fitted the standard model of epicycles which just happens to be the most carefully theory for 2000 years. So according to you we must reject the sun-centered universe? There is no impeccable experimental proof that the Higgs causes mass.
Da Schneib
5 / 5 (3) Aug 30, 2014
Also it looks like you forgot Minkowski again. Let me remind you:

d² = x² + y² + z² - t²c²

See that minus there? That makes time a hyperbolic dimension. You're *adding* "w" not subtracting it, which means it's a space-like dimension, not a time-like dimension.
johanfprins
1 / 5 (4) Aug 30, 2014
Hi thefurlong. :)

Johan is referencing your own words/statement re 'definition'; as follows...

All it does is rely on the definition of the Lorentz transform.


Perhaps you meant to say 'derivation' there?

No, I meant to say definition. The Lorentz transformation is defined a certain way, and I use that definition to derive the minkowski space-time interval.
Why are you then not doing this?

I will be back later!
Da Schneib
5 / 5 (3) Aug 30, 2014
The principle of relativity is now in four dimensions, not three as Galileo and Newton defined it.
You are making up physics asd it suits your insanity.
So you've never actually seen the Lorentz Transform equations before.

Thanks for admitting it. You can find them here: http://en.wikiped...irection
Da Schneib
5 / 5 (3) Aug 30, 2014
He did do it, johan. His definition is the Lorentz Transform and his outcome you have seen, and the derivation you have also seen. Why are you asking dumb questions instead of learning something?
Da Schneib
5 / 5 (3) Aug 30, 2014
Here are the postulates of relativity: http://casa.color...ate.html

That's by Andrew Hamilton, a Professor in the Astrophysics department at the University of Colorado at Boulder. He's teaching Cosmology this Autumn; in the past couple years he's taught General Relativity, Atomic and Molecular Physics, and Black Holes. What classes have you taught recently, johan?
Reg Mundy
1 / 5 (5) Aug 30, 2014
(continued)
In short, you inconceivably uneducated wonder, VELOCITY AND ACCELERATION ARE TWO DIFFERENT THINGS. Therefore, you arrogant, prokaryote of a thinker, ......and so on...
.. You moronic gnat.

Steady on, furbrain! I'm beginning to think you do not respect my intellect! I am beginning to understand that you disagree with me slightly on a couple of the more trivial aspects of our discussion. However, as you are so slow to comprehend what I am saying, I will forgive you, I know its hard for someone as dim as you to follow logic.
Lets get to the nub of this. You say that, according to the laws of gravity, there is an escape velocity. I say the laws of gravity are WRONG! You obviously have not understood where I am coming from, and have been vehemently striving to prove that I AM RIGHT! You see, every object that has mass also has a Schwartzchild radius (see en.wikipedia.org/wiki/Schwarzschild_radius). Now, who is correct, you or Schwarzschild? Guess you lose, furbrain!
johanfprins
1 / 5 (5) Aug 30, 2014
RC, I'm biased against people who claim mainstream physicists are idiots.

Get over it.


And I detest people who argue that mainstream physicits can never be wrong. Such people should not be doing physics since the most important rule in physics is that a researcher must accept that whatever he/she believes in today, no matter how long the theory or model have apparently withstood the test of time, can at any time be proved to have been wrong all along. If you do not approach physics with this attitude you are a traitor agsinst everything that physics is supposed to be.
Reg Mundy
1 / 5 (5) Aug 30, 2014
(continued)
In short, you inconceivably uneducated wonder, VELOCITY AND ACCELERATION ARE TWO DIFFERENT THINGS. Therefore, you arrogant, prokaryote of a thinker, ......and so on...
.. You moronic gnat.

Steady on, furbrain! I'm beginning to think you do not respect my intellect! I am beginning to understand that you disagree with me slightly on a couple of the more trivial aspects of our discussion. However, as you are so slow to comprehend what I am saying, I will forgive you, I know its hard for someone as dim as you to follow logic.
Lets get to the nub of this. You say that, according to the laws of gravity, there is an escape velocity. I say the laws of gravity are WRONG! You obviously have not understood where I am coming from, and have been vehemently striving to prove that I AM RIGHT! You see, every object that has mass also has a Schwartzchild radius (see en.wikipedia.org/wiki/Schwarzschild_radius). Now, who is correct, you or Schwarzschild? Guess you lose, furbrain!
Reg Mundy
1 / 5 (5) Aug 30, 2014
@Daz
{Jack's} a big proponent of the Alcubierre drive.
This is a good example of where mathematical equations fail to pass the reality test.
No, it's a case just like frame dragging, where physical observations have not YET confirmed the solution. Gravity Probe B confirmed frame dragging. Data will confirm or deny the Alcubierre solution. Absence of evidence is not evidence of absence. <- basic logic

No, Daz, its not just like frame dragging.
Incidentally, why not apply your logic to escape velocity? No evidence there either, as nothing has ever escaped Earth's influence WITHOUT FALLING UNDER THE INFLUENCE OF OTHER OBJECTS - so there is no proof for furbrain's model or mine.
johanfprins
1 / 5 (4) Aug 30, 2014
The Lorentz Transform says some of your time is my space (and vice versa) if we're moving relative to one another.
That is your interpretation which is WRONG
It says:

x' = (t - vx/c²)/√(1-v²/c²)
=G*(t-vx/c^2)

As far as know the Lorentz-transformation from the IRF x,y,z into the IRF x',y',z'
is given by:

x'=G*(x-vt) and
t'=G*(t-vx/c^2)

Why must I argue relativity with NITWIT like you who does not even know what the Lorentz equations are? When you differentiatethe first equation by t, you obtain, since x does not change with time, that;

dx'/dt=v'=G*(-v). In other words you observe the entity that is stationary at x, to be moving with an apparent speed v'. If this enrity has a rest-mass m(0) it is moving with an apparent momentum p'=-G*(m(0)v), which has an appparent relativistic mass m=G*m(0)This is the correct interpretation, NOT time dilation on a moving clock.

johanfprins
1 / 5 (4) Aug 30, 2014
Yes but you can't derive relativity from postulating horses have wings, johan.
Excatly! So why are you making equally absurd postulates and calimng that you can derive relativity from them?

Also, there is no simultaneity. It simply can't exist, if the speed of light is constant for every observer.
So Einstein was bullshitting when he claimed in his train experiment that two lightning strikes can hit the embankment simultaneously? Why do you not stick with Einstein's postulates but make your own absurd ones.
johanfprins
1 / 5 (4) Aug 30, 2014
Also it looks like you forgot Minkowski again. Let me remind you:

d² = x² + y² + z² - t²c²

See that minus there? That makes time a hyperbolic dimension. You're *adding* "w" not subtracting it, which means it's a space-like dimension, not a time-like dimension.


This is eaxctly the point: If you can subtract w^2, w cannot be an extra coordinates. Calling it a hyperbolic dimension cannot change this mathematical fact! You make me think about the little girl who was buying ice cream for her and her little brother who wanted vanilla. When the vedor told her he is out of vanilla, she knew she had a crisis on hand. She ordered another flavour and when she handed if to her brother she said: Here is your pink vanilla.
Reg Mundy
1 / 5 (4) Aug 30, 2014
@Daz
@furlong, great exposition of basic mechanics and the physical meaning of derivatives as applied to those mechanics. And a good job explaining precisely where Reg is wrong again. Applause. Also a nice job on Johan.

Are you reading the same website as I am? Where is this proof that "Reg is wrong"? In all the jousts twixt me and furbrain, he has admitted being incorrect three times, and that does not include three wrong answers to a problem he himself posed! Just because I like to torment him with spurious arguments to drive him wild, doesn't mean I am wrong. Meanwhile, I can't wait to see his response to my last post! I expect the expletives to be choice!
johanfprins
1 / 5 (4) Aug 30, 2014
The principle of relativity is now in four dimensions, not three as Galileo and Newton defined it.
You are making up physics asd it suits your insanity.
So you've never actually seen the Lorentz Transform equations before.
them here: http://en.wikiped...irection


This derivation is wrong since it assumes that the time within the two IRF's are simultaneously different for the SAME SINGLE EVENT in space. There is your horses with wings. A SINGLE event can only occur at a SINGLE instant in time t' on ALL clocks within both IRF's. However, an observer at 0 moving relative to this event occurring at the time t', cannot see it at the instant that it acttually occurs. He sees it ocuurring at a different time t, which is also the same on all clocks within both IRF's. The reason for this is the Doppler shift in the frequency of the light emitted by the event which is moving relative to him with speed c.
johanfprins
1 / 5 (4) Aug 30, 2014
He did do it, johan. His definition is the Lorentz Transform and his outcome you have seen, and the derivation you have also seen. Why are you asking dumb questions instead of learning something?


How can his definition be the Lorentz transform from which he claims that he is deriving these equations? Are you really so insane?

That's by Andrew Hamilton, a Professor in the Astrophysics department at the University of Colorado at Boulder.
Thank God he is not my professor. He should be sued for lying and misleading his students.

The correct derivation and interpretation of the Lorentz equations can be found in my latest book (see chapter 6): Einstein=Genius: But a genius sometimes blunders. I have updated it today and it can be found here:
https://www.resea...=prf_act

If you want to prove me wrong, read this and use mathematics and logic to do so, instead of just spouting mainstream dogma like a moron!

thefurlong
5 / 5 (3) Aug 30, 2014
Hi thefurlong. :)

Johan is referencing your own words/statement re 'definition'; as follows...

All it does is rely on the definition of the Lorentz transform.


Perhaps you meant to say 'derivation' there?

No, I meant to say definition. The Lorentz transformation is defined a certain way, and I use that definition to derive the minkowski space-time interval.
Why are you then not doing this?

I will be back later!

Gotta have a day trip. Going to scatter my dad's ashes. I'll try to provide you with a derivation soon.
johanfprins
3 / 5 (4) Aug 30, 2014
Gotta have a day trip. Going to scatter my dad's ashes. I'll try to provide you with a derivation soon.
My deepest sympathy. It is a sad day when you loose your father. Even 30 years later, as in my case, it is still overwhelming.

Regards,
Johan
Da Schneib
5 / 5 (4) Aug 30, 2014
RC, I'm biased against people who claim mainstream physicists are idiots.


And I detest people who argue that mainstream physicits can never be wrong.
Show where I ever argued any such thing.
Da Schneib
5 / 5 (3) Aug 30, 2014
@Daz
{Jack's} a big proponent of the Alcubierre drive.
This is a good example of where mathematical equations fail to pass the reality test.
No, it's a case just like frame dragging, where physical observations have not YET confirmed the solution. Gravity Probe B confirmed frame dragging. Data will confirm or deny the Alcubierre solution. Absence of evidence is not evidence of absence. <- basic logic

No, Daz, its not just like frame dragging.
Sure it is, at least in the fashion I stated: frame dragging was derived from the Einstein field equations for gravity early in the 20th century, but not proven by experiment until Gravity Probe B a decade or so back. The Alcubierre solution is likewise derived from the Einstein field equations but has not yet been tested.

Simple as that.
Da Schneib
5 / 5 (3) Aug 30, 2014
The Lorentz Transform says some of your time is my space (and vice versa) if we're moving relative to one another.
That is your interpretation which is WRONG
It's not an "interpretation." It's the physical meaning of the equation. The t for time is right there, and so's the x for space, and you subtract, as you must always do instead of adding when you deal with time, because time is not a circularly symmetric dimension like x, y, or z. It's hyperbolic. There is another form of the Lorentz transform that uses hyperbolic trig. You can look it up on the internet. Search on "relativity rapidity."

An "interpretation" is for complicated math; this is direct dead simple algebra, which is apparently outside your ken.
Da Schneib
5 / 5 (3) Aug 30, 2014
Yes but you can't derive relativity from postulating horses have wings, johan.
Excatly! So why are you making equally absurd postulates and calimng that you can derive relativity from them?
This is incorrect. Einstein made the postulates. Are you claiming Einstein was stupid?

Einstein was bullshitting
Oh, so now Einstein was lying.

Gotcha.
Da Schneib
5 / 5 (3) Aug 30, 2014
Also it looks like you forgot Minkowski again. Let me remind you:

d² = x² + y² + z² - t²c²

See that minus there? That makes time a hyperbolic dimension. You're *adding* "w" not subtracting it, which means it's a space-like dimension, not a time-like dimension.


This is eaxctly the point: If you can subtract w^2, w cannot be an extra coordinates.
Why?

You're making another claim you can't back up. You say this, and then do a bunch of handwaving, saying it's "mathematically impossible," yet there it is, and it's been confirmed by experiment better than any theory but the Standard Model. For a hundred and nine years, it's been confirmed by every experiment to test it. Tens of thousands of them, and if you count the times it's been tested in the lab by students, millions.

And you claim all those experiments are "wrong."

Pardon my skepticism.
Da Schneib
5 / 5 (3) Aug 30, 2014
So you've never actually seen the Lorentz Transform equations before.
...them here: http://en.wikiped...irection
This derivation is wrong since it assumes that the time within the two IRF's are simultaneously different for the SAME SINGLE EVENT in space.
No, it doesn't assume it at all. The meaninglessness of simultaneity is a direct result derived from the observed constancy of the speed of light for all observers regardless of their state of motion. A circularly symmetric dimension cannot do this; only a hyperbolically symmetric one can.

Furthermore, you're calling Lorentz and Einstein and Poincare and Minkowski wrong again.
Da Schneib
5 / 5 (2) Aug 30, 2014
He did do it, johan. His definition is the Lorentz Transform and his outcome you have seen, and the derivation you have also seen.
How can his definition be the Lorentz transform from which he claims that he is deriving these equations?
It's self-evident, if you stop believing weird stuff like Einstein lying and Lorentz, Minkowski, and Poincare being wrong when SRT has been tested for almost a hundred and ten years.
Da Schneib
5 / 5 (3) Aug 30, 2014
Going to scatter my dad's ashes.
Condolences. I hope he lived well.
Reg Mundy
1 / 5 (2) Aug 30, 2014
@thefurlong
Peace, brother, I withdraw my provocations. You will need much time to recover.
Reg Mundy
1 / 5 (5) Aug 30, 2014
@Daz
No, Daz, its not just like frame dragging.
Sure it is, at least in the fashion I stated: frame dragging was derived from the Einstein field equations for gravity early in the 20th century, but not proven by experiment until Gravity Probe B a decade or so back. The Alcubierre solution is likewise derived from the Einstein field equations but has not yet been tested.
Simple as that.

Let's not get into semantics and perspectives otherwise you will be claiming a duck and a walrus are the same 'cos they are both animals.
Regrettably (from your point of view) the effect ascribed to frame-dragging as per GRT derivation and demonstrated by GP-B are exactly the same results derived from expansion theory treating the Earth and the probe as a composite object (which they are unless the distance between them renders effect miniscule as per "gravity").If theory of gravity and GRT is correct, you are correct. If gravity does not exist, I am correct. Can you prove gravity exists?
Aligo
3 / 5 (2) Aug 30, 2014
If gravity does not exist, I am correct. Can you prove gravity exists?
Try to jump from some bridge and levitate there - the higher, the better evidence of your ideas.
Da Schneib
5 / 5 (2) Aug 30, 2014
No, Daz, its not just like frame dragging.
Sure it is, at least in the fashion I stated

Let's not get into semantics
There's no semantics about it.

You're claiming one solution to the Einstein field equations is just fine, but the other isn't.

Good luck with that.
Da Schneib
5 / 5 (2) Aug 30, 2014
If gravity does not exist, I am correct. Can you prove gravity exists?
Try to jump from some bridge and levitate there - the higher, the better evidence of your ideas.
An excellent suggestion, Aligo. Unfortunately people who don't "believe in" gravity generally learn better a little too late...

And actually it's not the fall that kills you, it's the sudden acceleration at the end. Oh, wait, he doesn't believe in acceleration either. My bad.
johanfprins
1 / 5 (5) Aug 31, 2014
RC, I'm biased against people who claim mainstream physicists are idiots.


And I detest people who argue that mainstream physicits can never be wrong.
Show where I ever argued any such thing.
Go back to all your posts and you will see thatn this is your approach all along. You are really the most stupid person I have EVER come across in my life Even Zephyr looks like a genius compared to you!
johanfprins
1 / 5 (5) Aug 31, 2014
because time is not a circularly symmetric dimension like x, y, or z. It's hyperbolic.
EXACTLY!! And therefore it CANNOT be an extra dimension. There is no such a thing as a hyperbolic dimension: It is just as impossible as "postulating flying horses".

There is another form of the Lorentz transform that uses hyperbolic trig. You can look it up on the internet. Search on "relativity rapidity."
I do not have to look it up since I know the theory of relativity far better than a nincompoop like you ever will. The fact that you use hyperbolic trig. is PROOF that time is NOT an extra dimension and CAN NEVER be!

An "interpretation" is for complicated math; this is direct dead simple algebra, which is apparently outside your ...
The correct interpretation follows from simple math, not complicated "hyperbolic trig". Simple linear algebra DEMANDS that coordinates must be LINEARLY INDEPENDENT in order to define dimensions, This is grade 10 algebra, YOU bloody FOOL
johanfprins
1 / 5 (5) Aug 31, 2014
Yes but you can't derive relativity from postulating horses have wings, johan.
Excatly! So why are you making equally absurd postulates and calimng that you can derive relativity from them?
This is incorrect. Einstein made the postulates. Are you claiming Einstein was stupid?
Where did Einstein make this postulate in 1905? Einstein was a genius but he also blundered as he himsefl admitted. One of his biggest blunders was to accept Minkowski's space-time, after he at first wisely opposed it!

Einstein was bullshitting
Oh, so now Einstein was lying.
Another illustration of how brainless you are. YOU stated that simultaneity is impossible. I then pointed out that Einstein postulated that simultaneity is possible and then asked YOU the question whether YOU are claiming that Einstein was bullshitting. With your low IQ you now claim that I was claiming that Einstein was lying! It is YOUR claim that Einstein must have been lying, you bloody
Da Schneib
5 / 5 (2) Aug 31, 2014
RC, I'm biased against people who claim mainstream physicists are idiots.
And I detest people who argue that mainstream physicits can never be wrong.
Show where I ever argued any such thing.
Go back to all your posts and you will see thatn this is your approach all along.
Quote it. Or admit you're making stuff up again.
Da Schneib
5 / 5 (2) Aug 31, 2014
because time is not a circularly symmetric dimension like x, y, or z. It's hyperbolic.
EXACTLY!! And therefore it CANNOT be an extra dimension.
That's like saying a dog can't be a dog because it has a head. This is called a "non sequitur," literally, "it does not follow." Or in spicier terminology, "that dog don't hunt."
Da Schneib
5 / 5 (1) Aug 31, 2014
...you can't derive relativity from postulating horses have wings, johan.
Excatly!
Then why did you say it? Because you *can* derive relativity from *Einstein's* postulates.

Claiming all postulates are the same is incorrect. Logically speaking.
Da Schneib
5 / 5 (2) Aug 31, 2014
Where did Einstein make this postulate in 1905?
Linked and quoted. You're making stuff up again, johan.

It is YOUR claim that Einstein must have been lying,
Quote it or admit you're making stuff up yet again, johan.

That's a right nasty habit you got there, son.
johanfprins
1 / 5 (5) Aug 31, 2014
This is eaxctly the point: If you can subtract w^2, w cannot be an extra coordinates.
Why?
If you know grade 10 linear algebra you would not have asked such a stupid question. The answer is straightforward: Since there is a minus in front of w^2, you can have that d=0, even though not all the coordinates x,y,z, and w have to be zero. In order for these coordinates to define FOUR separtae dimensions ONE MUST HAVE that x=y=z=w=0 when d=0.This is called "linear independence", you fool!

For a hundred and nine years, it's been confirmed by every experiment to test it. Tens of thousands of them, and if you count the times it's been tested in the lab by students, millions.
What has been tested to prove that d is only zero when x=y=z=t=0? I know of NO such tests: Thus time cannot be a fourth dimension EVER!!!

Pardon my skepticism.

You shoudl not apologise for your scepticism but for your blatant stupidity!!
Da Schneib
5 / 5 (3) Aug 31, 2014
If you know grade 10 linear algebra you would not have asked such a stupid question.
So you don't have an answer. Because if you don't know any plain ol' ordinary algebra, not to mention hyperbolic trigonometry, you sure the heck don't know any fancy linear algebra.

I didn't figure you did, I just wanted to see you admit it.

Thanks.
johanfprins
1 / 5 (5) Aug 31, 2014
No, it doesn't assume it at all.
Have EVER read Einstein's original manuscript in which he does this? Obviously not since tyou are too stupid
The meaninglessness of simultaneity is a direct result derived from the observed constancy of the speed of light for all observers regardless of their state of motion.
The fact that an observer cannot record a moving event at the actual instant that it is occurring does NOT make simultaneity meaningless, you bloody idiot
A circularly symmetric dimension cannot do this;
I have not stated this anywhere since time IS NOT AN EXTRA dimension, you idiot!
only a hyperbolically symmetric one can.
There is no such thing as a hyperboic extra dimension since such a coordinate is NOT linearly independent from x,y, and z.

Furthermore, you're calling Lorentz and Einstein and Poincare and Minkowski wrong again.

They did not understand the implications of the principle of relativity. They should have studied Galileo bette
johanfprins
1 / 5 (5) Aug 31, 2014
He did do it, johan. His definition is the Lorentz Transform and his outcome you have seen, and the derivation you have also seen.
How can his definition be the Lorentz transform from which he claims that he is deriving these equations?
It's self-evident, if you stop believing weird stuff like Einstein lying and Lorentz, Minkowski, and Poincare being wrong when SRT has been tested for almost a hundred and ten years.

You cannot use the equations that you have to derive as a definition of what you have to derive: MORON!
Da Schneib
5 / 5 (3) Aug 31, 2014
No, it doesn't assume it at all.
Have EVER read Einstein's original manuscript in which he does this?
I've read where he explains it in terms obvious enough even for people like you who are innumerate. The name of the book is "Relativity," and the author is Albert Einstein.

Sonny, I have the math to understand SRT fairly completely, and to keep up when a real relativist starts talking. You get lost about the time they start squaring things. So you shouldn't pretend because everyone who cares, knows.
Da Schneib
5 / 5 (2) Aug 31, 2014
You cannot use the equations that you have to derive as a definition of what you have to derive
Well, that's a Good Thing because he didn't. Perhaps you'd care to actually show us where he did instead of groundlessly claiming it.

You're making stuff up again, johan.
johanfprins
1 / 5 (4) Aug 31, 2014
His definition is the Lorentz Transform and his outcome you have seen, and the derivation you have also seen.
How can his definition be the Lorentz transform from which he claims that he is deriving these equations?
It's self-evident, if you stop believing weird stuff like Einstein lying
It is YOU who claimed that Einstein was lying to assume simultaneous lightning strikes: NOT I!! ASSHOLE!!
and Lorentz, Minkowski, and Poincare being wrong when SRT has been tested for almost a hundred and ten years.
Nowhere have I claimed that SRT is wrong, you moron. I am claiming that its interpretation is wrong. It is accepted that Lorentz's interpretation based on length contraction and local time is wrong (even though there is no experiment that can prove him wrong) so there is no experimental-reason why Einstein and Minkopski's interpretations could not also be wrong: And in fact they are wrong since a time-coordinate CANNOT be an extra dimension, EVER!
Da Schneib
5 / 5 (1) Aug 31, 2014
It is YOU who claimed that Einstein was lying to assume simultaneous lightning strikes:
Where?

Quote it or admit tacitly that you are making stuff up again. And trying to cover it up with insults.
johanfprins
1 / 5 (4) Aug 31, 2014
RC, I'm biased against people who claim mainstream physicists are idiots.
And I detest people who argue that mainstream physicits can never be wrong.
Show where I ever argued any such thing.
Go back to all your posts and you will see thatn this is your approach all along.
Quote it. Or admit you're making stuff up again.


RC, I'm biased against people who claim mainstream physicists are idiots.
And I detest people who argue that mainstream physicits can never be wrong.
Show where I ever argued any such thing.
Go back to all your posts and you will see thatn this is your approach all along.
Quote it. Or admit you're making stuff up again.

See your argumenst about Lorentz, Einstein, Poincare just above. You are arguing that they can never be wrong. And then work your way up through all your idiotic posts; in which you repeated this mantra over and over and over again!.
Da Schneib
5 / 5 (1) Aug 31, 2014
jonny, quote it or it never happened.

You're making stuff up again, jonny boy.
johanfprins
1 / 5 (5) Aug 31, 2014
because time is not a circularly symmetric dimension like x, y, or z. It's hyperbolic.
EXACTLY!! And therefore it CANNOT be an extra dimension.
That's like saying a dog can't be a dog because it has a head
Nope, this is saying that an animal CAN be a dog when it has hoofs and it brays. YOU are saying that all animals who have heads are dogs!

This is called a "non sequitur," literally, "it does not follow." Or in spicier terminology, "that dog don't hunt."
And you are the expert in posting non sequitur's!! But obviousy too stupid to realise that you are accusing other people of doing what you are actually doing: IDIOT!!

See corecvtion to CAN above!
Da Schneib
5 / 5 (2) Aug 31, 2014
Nope, this is saying that an animal CAN be a dog when it has hoofs and it brays. YOU are saying that all animals who have heads are dogs!
Quote it or it never happened, jonny boy.

I'm not the one claiming someone said "scientists are never wrong" when what they really said is they detest people who always claim they are.

You quoted it yourself, then claimed I said something that's just not in that quote.

You're making stuff up again, jonny child.
Reg Mundy
1 / 5 (5) Aug 31, 2014
If gravity does not exist, I am correct. Can you prove gravity exists?
Try to jump from some bridge and levitate there - the higher, the better evidence of your ideas.

That is equally evidence for expansion theory, the effect is identical. Even Einstein believed"there is no difference between gravity and acceleration", he was more right than he suspected, there is NO GRAVITY, it is all acceleration.
Reg Mundy
1 / 5 (5) Aug 31, 2014
If gravity does not exist, I am correct. Can you prove gravity exists?
Try to jump from some bridge and levitate there - the higher, the better evidence of your ideas.
An excellent suggestion, Aligo. Unfortunately people who don't "believe in" gravity generally learn better a little too late...

And actually it's not the fall that kills you, it's the sudden acceleration at the end. Oh, wait, he doesn't believe in acceleration either. My bad.

Of course I believe in acceleration, you dolt! Its all about acceleration! But acceleration (i.e. force) is always the result of an interaction between two or more masses, i.e. contact, electromagnetic, electrostatic, but NO GRAVITY!
Da Schneib
5 / 5 (2) Aug 31, 2014
Well, now, Reg, actually there *is* a difference between gravity and acceleration. The difference is, gravity always is a spherical field (or one that conforms to the shape of the matter that originates it- for example an oblong asteroid generates an oblong gravity field) whereas true acceleration does not cause apparently curved paths.

If two objects fall in a real gravity field, they will move closer to one another, eventually enough that it can be measured in a real experiment; however, if two objects fall under the influence of a linear acceleration of their environment they will not.

However, this test fails as a *local* experiment, and thus is not included under the rubric of local gravity experiments which cannot determine the difference between acceleration and gravity.

And I'm sorry but I have a great deal of trouble believing in electrostatic forces that happen without electrostatic charges, Reg.

Just sayin'.
johanfprins
1 / 5 (5) Aug 31, 2014
Excatly!
Then why did you say it? Because you *can* derive relativity from *Einstein's* postulates.
Because Einsteinn DID NOT POSTULTE THAT HORSES HAVE WINGS!! YOU IDIOT.

Einstein gave two postulates: The first was just a re-affirmation of Galileo's principle of relativity (although he framed it in a very obscure manner; probably because he did not want to give Galileo any credit) and the second "postulate" was to claim that the speed of light is the same within all inertial refrence frames.

This should not have been a postulate since it flows logically from Galileo's principle of relativity: If hiswere not so. one would be able to measure from within an IRF whether this IRF is moving or not. And if this is possible one will have to reject Galileo's principle of relativity as being correct.

I derived the Lorentz equations from Galileo's principle of relativity; and its corrolary of light speed.. The derivation proves that there cannot be a space-time manifold.
Da Schneib
5 / 5 (2) Aug 31, 2014
Excatly!
Then why did you say it? Because you *can* derive relativity from *Einstein's* postulates.
Because Einsteinn DID NOT POSTULTE THAT HORSES HAVE WINGS!!
Well of course he didn't. He postulated spacetime as a continuum, he postulated that IRFs exist, he postulated that IRFs are all equally valid, and he postulated that the speed of light is constant for all observers.

And out came SRT.

That's the story.

I have linked a University of Colorado at Boulder Professor who teaches relativity who says so. Who do you have?
johanfprins
1 / 5 (5) Aug 31, 2014
Where did Einstein make this postulate in 1905?
Linked and quoted. You're making stuff up again, johan.
Where has it been "linked and quoted" .YOU are the one making up stuff and lying through your nose!

It is YOUR claim that Einstein must have been lying,
Quote, it or admit you're making stuff up yet again, johan. I alreday explained it in detai that you claimed that simultaneity cannot exist, while Einstein claimed that it can. So YOU are claiming that Einstein must have been lying! MORON!

That's a right nasty habit you got there, son.
This habit is YOURS MORON!!
Da Schneib
5 / 5 (1) Aug 31, 2014
johanfprins
1 / 5 (5) Aug 31, 2014
If you know grade 10 linear algebra you would not have asked such a stupid question.
So you don't have an answer. Because if you don't know any plain ol' ordinary algebra, not to mention hyperbolic trigonometry, you sure the heck don't know any fancy linear algebra.
You see what a low down son-of-a-bitch you are. You only post the first sentence and not my effiort to explain it to you! So I will post it again:

"The answer is straightforward: Since there is a minus in front of w^2, you can have that d=0, even though not all the coordinates x,y,z, and w have to be zero. In order for these coordinates to define FOUR separtae dimensions ONE MUST HAVE that x=y=z=w=0 when d=0.This is called "linear independence", you fool!"

johanfprins
1 / 5 (4) Aug 31, 2014
You cannot use the equations that you have to derive as a definition of what you have to derive
Well, that's a Good Thing because he didn't. Perhaps you'd care to actually show us where he did instead of groundlessly claiming it
Who is "he"? I am quting your insaqne post in which you wrote down a WRONG equation for the Lorentz transformation, and then, like thyen idiot you are, claimed that it defines then Lorentz equations.

You're making stuff up again, johan.

It is YOU who are lying and obfuscating again since you see it as your holy task in life to defend m,ainstream dogma to the hilt. You just do not habe the open mind required, or the brains to analyse physics objectively!
Da Schneib
5 / 5 (2) Aug 31, 2014
Yep, always quotin' and linkin' sources and proving what I say is correct.

I must irritate the heck out of you.

:D
Da Schneib
5 / 5 (2) Aug 31, 2014
Who is "he"?
furlong.
johanfprins
1 / 5 (4) Aug 31, 2014
It is YOU who claimed that Einstein was lying to assume simultaneous lightning strikes:
Where?
Did you or did you not state that simultaneity is impossible? You did state this. I pointed out that Einstein assumed simultaneous lightning strikes. So if your claim that simultaneity is impossible is correct you must be stating that Einstein was WRONG! And thus a bullshitter!

Quote it or admit tacitly that you are making stuff up again. And trying to cover it up with insults. [/qHow many time must I quote and explain this before it will penetrate your brainless bony skull?]
Da Schneib
5 / 5 (2) Aug 31, 2014
Here's the quote, directly from the master himself:

Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative.


Oh, look there, you made stuff up again. Einstein didn't say they were simultaneous. He said they were simultaneous *relative to the embankment* not the train.

Link to the chapter: http://www.bartle...3/9.html

There you have it.

Now stop making stuff up, jonny boy.
johanfprins
1 / 5 (5) Aug 31, 2014
Nope, this is saying that an animal CAN be a dog when it has hoofs and it brays. YOU are saying that all animals who have heads are dogs!
Quote it or it never happened, jonny boy.
This is an euphimism for what you are stating that time can be fourth dimension. Hell you ARE STUPID!!

I'm not the one claiming someone said "scientists are never wrong" when what they really said is they detest people who always claim they are.
Give me examples of people "who ALWAYS claim they are wrong": YOU are making stuff up. I have never met a person in my life who ALWAYS claim that mainstream physicistsare wrong. Only an infantile moronic person will makwe such a stupid statement.

You have consistently argued that mainstream scientists can never bbe wrong, and therefore absurd conclusions like time being a fourth dimension, wave particle duality, Higgs boson etc. should NEVER be questioned. You calim that they have been proved without quaoting falsifiable experiments.

Da Schneib
5 / 5 (2) Aug 31, 2014
You have consistently argued that mainstream scientists can never bbe wrong
Never happened. You can't quote it; you have tried and failed.

You're making stuff up again, jonny.
johanfprins
1 / 5 (5) Aug 31, 2014
Because Einsteinn DID NOT POSTULTE THAT HORSES HAVE WINGS!!
Well of course he didn't. He postulated spacetime as a continuum,
HE DID NOT!!!!!! He resisted this stupidity of Minkowski for 10 years before desperately grabbing at straws to complete his general theory of realtivity.
he postulated that IRFs exist,
No he di not: Galileo and Newton already did this many years before Einstein did this.
he postulated that IRFs are all equally valid,
Galileo postulated this NOT Einstein you fool!
and he postulated that the speed of light is constant for all observers.
He need not have postulated this since it follows logically from Galileo's relativity!

And out came SRT.
Obviously SRT follows from Galileo's relativity: But NOT space-time as if time is a fourth dimension.: And in addition Einstein's derivation of the Lorentz equations violates the rules of mathematics! Einstein struggled with mathematics.

Da Schneib
5 / 5 (3) Aug 31, 2014
Give me examples of people "who ALWAYS claim they are wrong":
You said,
Einstein was bullshitting


We done here?
Da Schneib
5 / 5 (3) Aug 31, 2014
{Einstein} postulated spacetime as a continuum,
HE DID NOT!!!!!!
Over here, crazy jonny. Over there, Nobel Laureate Albert Einstein.

Hmmmm, who to believe...

Snicker.
johanfprins
1 / 5 (5) Aug 31, 2014
http://casa.color...ate.html

If you want to give references please do not choose a professor who is obviously a moron. He teaches his students that the first postulate of the special theory of relativity is that there IS a space-time continuum in which time defines a fourth dimension. Firsly Einstein never used this postulate, and secondly, it is impossible since time is not linearly independent from x,y,and z. Any grade 10 pupil can verify that in such a case a fourh coordinate cannot define a fourth dimension! This guy is just as big a moron as you are. Maybe it is YOU who are self-refrencing and then hiding like then criminal you are behind anonymity?
johanfprins
1 / 5 (5) Aug 31, 2014
Who is "he"?
furlong.

As far as I remember the furlong posted that he will be back to do so. YOU ARE mentally disturbed you know!
Aligo
1 / 5 (3) Aug 31, 2014
{Einstein} postulated spacetime as a continuum
Actually Einstein was unpleasantly surprised with 4D space-time model, published with his teacher H. Minkowski in 1911. Einstein didn't like him, so he could hardly use Minkowski model as an postulate in special relativity published in 1905. It's true, he adopted this model later in development of general relativity. It's also true, that the space-time is considered a smooth continuum in relativity, which introduces a bias at large scales where the tiny quantum fluctuations of space-time cannot be neglected.
johanfprins
1 / 5 (5) Aug 31, 2014
Here's the quote, directly from the master himself:


Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative.
Correct! This proves that you are a liar. You made the sweeping statement that simultaneity is not possible. You did not state that two events that occur simultaneous (so simultaneity IS possible), cannot be OBSERVED to be simultaneous by an observer moving relative to the events. By not OBSERVING the events to be simultaneous obviously does not mean that the events cannot be simultaneous as you so stupidly argued.

For example oif two lights at different distances from me simultaneously send out two light-pulses to me I will NOT be able to see the light-pulses simultaneously. This does NOT mean that the light pulses were NOT emitted simultaneously, you moron!.

johanfprins
1 / 5 (4) Aug 31, 2014
You have consistently argued that mainstream scientists can never be wrong
Never happened. You can't quote it; you have tried and failed.

You're making stuff up again, jonny.
Show me a post in which you admitted that an aspect of mainstream science MIGHT be wrong. Stop being such a blatant liar!
johanfprins
1 / 5 (4) Aug 31, 2014
Give me examples of people "who ALWAYS claim they are wrong":
You said, Einstein was bullshitting
You are taking this out of context since I asked YOU whther your sweeping statemnt that simultaneity is impossible means that Einstein is bullshitting. Furthermore I dio claim that Einstein made mistakes and I specify these. I do not ALWAYS claim that mainstream physicists are wrong. You are like an infant that cannot argue logically since you make sweeping statements which muddy the water of logic.
We done here?

We could be as soon as you stop posting claptrap!
johanfprins
1 / 5 (5) Aug 31, 2014
{Einstein} postulated spacetime as a continuum,
HE DID NOT!!!!!!
Over here, crazy jonny. Over there, Nobel Laureate Albert Einstein.
Over where? Reference please! Einstein did not postulate this, Minkpwski did and Einstein resisted this postulate for 10 years, and then fatally blundered by accepting it against his better instincts.

Hmmmm, who to believe...Snicker.
Definitely not you and your moronic professor in Boulder. My grandson is considering to go to this University. I will advise him to avoid this University like the pest.
Aligo
1 / 5 (3) Aug 31, 2014
Try to think for example about Shapiro delay. The light passing through gravitational lens (or gravitational wave) gets delayed, so that the light cannot propagate through vacuum with fixed speed. If you place many temporal gravitational lenses or waves into path of light, then the light will get delayed even when the space-time will remain macroscopically flat. At the case of many frequent gravitational waves/lenses at the quantum scale this effect even leads into quantum uncertainty. Therefore the postulate of fixed speed of light doesn't say nothing special about actual speed of light propagation through vacuum, once you have not the lensing of space-time well defined and/or once you cannot describe it from inside of each lens. The net effects of the dynamic field of many tiny gravitational lenses/wave would curve and delay the propagation of light rather arbitrarily, despite the special relativity will remain strictly maintained in context of every temporal gravitational lens/wave.
Aligo
1 / 5 (2) Aug 31, 2014
J.F.Prins is fundamental here more than Schneib and mainstream physics, as he's trying to apply the fixed light speed postulate even to quantum scale, where just this postulate gets violated with quantum uncertainty in most apparent way. The uncertainty in spreading of quantum mechanics signals just means, that the information mediated with light arriwes to observer in random moment due to complex dynamic character of space-time / vaccum at the quantum scales. It's like to observe the pollen grains floating at the water surface with water surface ripples - because of density fluctuations of Brownian noise in underwater, these waves will spread randomly and the path of grains will fluctuate not only because these grains are already wildly moving, but also because of myriads of tiny Shapiro delays, which would affect their spreading there. Got it?
johanfprins
1 / 5 (4) Aug 31, 2014
J.F.Prins is fundamental here more than Schneib and mainstream physics, as he's trying to apply the fixed light speed postulate even to quantum scale, where just this postulate gets violated with quantum uncertainty in most apparent way. The uncertainty in spreading of quantum mechanics signals just means,


Aligo, Zephyrr or whoever you are: There is no uncertainty "in the spreading of EM-waves" in free space. Matter-waves and light-waves are both EM waves. Each of these waves moves through free-space following a classical path, since each single wave has a centre-of-mass owing to E=m*c^2. Obviously when a light-wave moves through a gravitational field that surrounds a matter-wave, it's speed slows down. But the latter has nothing to do with space-time, or curved space-time. It cannot have anything to do with any model that assumes that time is a fourth-dimension, since the latter is impossible.
Uncle Ira
5 / 5 (4) Aug 31, 2014
Definitely not you and your moronic professor in Boulder. My grandson is considering to go to this University. I will advise him to avoid this University like the pest.


@ johnpringle-Skippy, that is a cheap shot Cher because the Professor-Skippy isn't here to explain how it could be you that is the moron. What that Professor-Skippy ever done to you to make you so mad with him? It would have been the civilized thing for you do to just say you got the different ideas than him and you think he might be mistaking the way things are.

Oh yeah, I almost forget me. Are you always in the really bad mood and grumpy? Or just when you come here? If is just when you come here you should take some of that advisement you passing along to your grandson and try to stay away because I don't think being so grumpy all the time is good for the health.
Aligo
3 / 5 (2) Aug 31, 2014
Obviously when a light-wave moves through a gravitational field..., its speed slows down. But the latter has nothing to do with... curved space-time.
Of course it has, until the gravitational field is just curved space-time. This curving can be observed during solar eclipses, as Eddington did in 1921 and it's apparently related to gravity field of Sun. Look, you're denying the whole contemporary physics with your deductions and you're still calling me a crackpot, when I'm trying to explain it here instead?
johanfprins
1 / 5 (5) Aug 31, 2014
@ johnpringle-Skippy, that is a cheap shot Cher because the Professor-Skippy isn't here to explain how it could be you that is the moron. What that Professor-Skippy ever done to you to make you so mad with him? It would have been the civilized thing for you do to just say you got the different ideas than him and you think he might be mistaking the way things are.
The veryt first time that you posted some truth in your usual stupid ass way.

However I would not like such a professor teaching my grandson, since he does not understand the difference between a postulate and an interpretation: And to start off a course in special relativity by claiming that space-time is the postulate on which Einstein based his theory, proves that he is not competent to teach physics at a university.

johanfprins
1 / 5 (4) Aug 31, 2014
Obviously when a light-wave moves through a gravitational field..., its speed slows down. But the latter has nothing to do with... curved space-time

Of course it has, until the gravitational field is just curved space-time. This curving can be observed
It is not "curving of space-time" it is refraction of light within the gravitational field which always exist around a matter-wave.
re denying the whole contemporary physics with your deductions
I am not denying "the whole contemporary physics", I am just claiming that the modelling and explanation for certain aspects of it it are based on absurd Voodoo; which is not possible.
and you're still calling me a crackpot,
Yes you are since you want to belive that the absurd can be real
when I'm trying to explain it here instead? You do not have the brains to explain anything without claiming the absurd, like ducks farting in a pond!!
Reg Mundy
1 / 5 (5) Aug 31, 2014
@Daz
Well, now, Reg, actually there *is* a difference between gravity and acceleration. The difference is, gravity always is a spherical field (or one that conforms to the shape of the matter that originates it- for example an oblong asteroid generates an oblong gravity field) whereas true acceleration does not cause apparently curved paths.

If two objects fall in a real gravity field, they will move closer to one another, eventually enough that it can be measured in a real experiment; however, if two objects fall under the influence of a linear acceleration of their environment they will not.......
And I'm sorry but I have a great deal of trouble believing in electrostatic forces that happen without electrostatic charges, Reg.
.

This is a load of chunter and nothing to do with what I have said. The bit about electrostatics leaves me breathless at your perversity. You cannot really be so stupid, so must be essaying the obtuse.
Reg Mundy
1 / 5 (5) Aug 31, 2014
@Aligo
Try to think for example about Shapiro delay. The light passing through gravitational lens (or gravitational wave) gets delayed, so that the light cannot propagate through vacuum with fixed speed.

Whoa there, laddie! Gravitational lens - no proof of existence. Gravitational wave - no proof of existence. Light cannot propagate through vacuum with fixed speed - are you serious? This is no way to start a comment, with baseless assertions you cannot justify.
Da Schneib
5 / 5 (4) Aug 31, 2014
http://casa.color...ate.html

If you want to give references please do not choose a professor who is obviously a moron.
LOL

So, how many classes in relativity did you teach this year?

Just askin'.
Da Schneib
5 / 5 (4) Aug 31, 2014
{Einstein} postulated spacetime as a continuum
Actually Einstein was unpleasantly surprised with 4D space-time model, published with his teacher H. Minkowski in 1911.
This is impossible. He couldn't have realized the relativity of simultaneity without it.
Da Schneib
5 / 5 (4) Aug 31, 2014
Here's the quote, directly from the master himself:
Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train? We shall show directly that the answer must be in the negative.
Correct! This proves that you are a liar. You made the sweeping statement that simultaneity is not possible.
No, I didn't. I said simultaneity is meaningless. What is simultaneous for one observer might not be for another. It depends on their mutual states of motion, and the states of motion of the two events.

This is relativity 101. Basic, about as basic as it gets. I knew about this when I was twelve, johan.
Da Schneib
5 / 5 (4) Aug 31, 2014
You have consistently argued that mainstream scientists can never be wrong
Never happened. You can't quote it; you have tried and failed.

You're making stuff up again, jonny.
Show me a post in which you admitted that an aspect of mainstream science MIGHT be wrong. Stop being such a blatant liar!
The Alcubierre drive. You'll also find my approach to the Lorentz transform rather non-traditional; I use hyperbolic trig. Except when I have to deal with cranks.
Da Schneib
5 / 5 (4) Aug 31, 2014
{Einstein} postulated spacetime as a continuum,
HE DID NOT!!!!!!
Sure he did:
the world of physical phenomena which was briefly called "world" by Minkowski is naturally four-dimensional in the space-time sense.
Chapter 17 of "Relativity," Albert Einstein, 1920. http://www.bartle.../17.html

We done here?
Da Schneib
5 / 5 (4) Aug 31, 2014
Give me examples of people "who ALWAYS claim they are wrong":
You said, Einstein was bullshitting
You are taking this out of context
Sorry, I don't play that game. You said it. Either own it or be a man and admit you made a mistake.
Da Schneib
5 / 5 (4) Aug 31, 2014
I am not denying "the whole contemporary physics", I am just claiming that the modelling and explanation for certain aspects of it it are based on absurd Voodoo
Yeah, like for instance the entire General and Special Relativity Theories. That's not "a different opinion." That's denial of the foundation of modern physics. For example, the Standard Model of Particle Physics, often referred to as "the SM," explicitly incorporates SR. So does astrophysics. So does cosmology. And the relativity of simultaneity is a direct result of SR.
Da Schneib
5 / 5 (4) Aug 31, 2014
This is a load of chunter
Another claim without any evidence to back it up.

And by the way, acceleration is not a force. It's the result of a force.
Da Schneib
5 / 5 (4) Aug 31, 2014
Gravitational lens - no proof of existence.
Let's start with Eddington's journey to South Africa to view a star close to the Sun during an eclipse. Next, go look up the "Einstein Cross." That will do for evidence, I'd say.

From the Wikipedia article on Einstein's Cross:
QSO 2237+0305 is a gravitationally lensed quasar that sits directly behind ZW 2237+030, Huchra's Lens.
johanfprins
1 / 5 (4) Aug 31, 2014
http://casa.color...ate.html

If you want to give references please do not choose a professor who is obviously a moron.
LOL

So, how many classes in relativity did you teach this year?
Many, many many over 50 years. I am now obviously retired and not teaching anymore. How many classes has this obviously stupid upstart taught during his lifetime as a professor? What is his age by the way? He is obviously not well trained in physics. Did he also study in Boulder, and is therefore still there vomiting the same crap they have fed him on?

Just askin'.
Just answering!
johanfprins
1 / 5 (5) Aug 31, 2014
{Einstein} postulated spacetime as a continuum
Actually Einstein was unpleasantly surprised with 4D space-time model, published with his teacher H. Minkowski in 1911.
This is impossible. He couldn't have realized the relativity of simultaneity without it.

One of them few times that Zephyr posted sense. It is refreshing to know that although he, just like you, embrace absurdity as reality, he has some sane moments: You NEVER have!
Da Schneib
5 / 5 (4) Aug 31, 2014
Here's Hamilton's first paper, from 1982: http://adsabs.har...98...59H

It's titled, "Gravitational spin precession in binary systems," a nice little exercise in relativistic astrophysics. Since then he has published over a hundred and forty papers in the peer-reviewed scholarly literature.

This was published at the earliest while he was in his mid-20s, so that would make him nearly fifty. He has two kids, and one of them is a graduate student in neuroscience. The other is a filmmaker and clothing designer.

Maybe you should have read his home page a bit more.
Uncle Ira
5 / 5 (4) Aug 31, 2014
The veryt first time that you posted some truth in your usual stupid ass way.


And you tell the GREAT BIG LIE there Skippy. Let me tell you the one thing, I always tell the truth as I see it. And I tell you the one another thing, I am not stupid. I am the engineer on the towboat. I bet I could beat you playing the chess, I am the USCF master level player in the tournaments. I may be might be ignorant on some things, but I am not stupid.

So you still not answer the question I give you. Are you always so grumpy? Or only when you come to the physorg? Some thing got you so mad that you always using the bad words to call names to people here and if you do that everywhere, where peoples can get at you I wonder if you do not get slapped a lot. But since they can't get at you here at the physorg I suspect you do not do that everywhere unless you one of those weird peoples who like to get slapped.
Da Schneib
5 / 5 (4) Aug 31, 2014
No science, johan.
johanfprins
1 / 5 (5) Aug 31, 2014
Give me examples of people "who ALWAYS claim they are wrong":
You said, Einstein was bullshitting
Even if this is the case, it does not mean that I have claimed that the mainstream is always wrong
You are taking this out of context
Sorry, I don't play that game. You said it. Either own it or be a man and admit you made a mistake.

I just stated what YOU implied! It is YOU who have stated that "simultaneity is impossible", and I just pointed out that this implies that Einstein was bullshitting. It is YOU, with your sweeping statement, who really stated that Einstein must have been a bullshitter: I just pointed out that you have stated this. Obviously by using one of your trademark, typically insane, claims! You need psychiatric help urgently. BTW what is your qualifications: I suspect that you failed grade 5?
Da Schneib
5 / 5 (4) Aug 31, 2014
Give me examples of people "who ALWAYS claim they are wrong":
You said, Einstein was bullshitting
Even if this is the case, it does not mean that I have claimed that the mainstream is always wrong
The mainstream is SRT. You say SRT is wrong. That's the end of that.

You are taking this out of context
Sorry, I don't play that game. You said it. Either own it or be a man and admit you made a mistake.
I just stated what YOU implied!
Nope. Sorry, that dog don't hunt.

You're making stuff up again, jonny boy.
johanfprins
1 / 5 (5) Aug 31, 2014
Yeah, like for instance the entire General and Special Relativity Theories. That's not "a different opinion."
Please prove that this is so in special relativity! I do not find special relativity wrong, except for the absurd, insane claim that it defines time as a fourth dimension, which any grade 10 pupil can prove CANNOT EVER be the case!.

That's denial of the foundation of modern physics.
Any physics based on time as a fourth dimension is insane physics. If you call this the "foundation opf physics" then you are a bigger MORON than I have EVER imagined.

For example, the Standard Model of Particle Physics, often referred to as "the SM," explicitly incorporates SR.
No it DOES NOT! It incorpoartes the insane absurdity of time as a fourth dimension: The latter is not relativity but crap!
. And the relativity of simultaneity is a direct result of SR.
But not when it is explained in terms ot time being a fourth dimension: Then it is crap!
Aligo
3 / 5 (4) Aug 31, 2014
Actually Einstein was unpleasantly surprised with 4D space-time model, published with his teacher H. Minkowski in 1911. ...This is impossible. He couldn't have realized the relativity of simultaneity without it.
Minkowski called Einstein "Lazy dog" and Einstein didn't like him just because of it. When Minkowski published his space-time formulation of special relativity, Einstein just said: "Since the mathematicians have invaded the theory of relativity, I do not understand it myself anymore.". Einstein's derivation doesn't use Minkowski space, but Minkowski space and Einstein's method produce the same results. Since the Minkowski space method is easier to use and more generalizable, people prefer it. To be sure, Minkowski was just elaborating on an idea of Poincare here. The Poincare group combines rotations in space-time (the Lorentz group) with translations in space-time.
johanfprins
1.5 / 5 (4) Aug 31, 2014
Gravitational lens - no proof of existence.
Let's start with Eddington's journey to South Africa to view a star close to the Sun during an eclipse.
I did not know he came to South Africa to do that: Are youn all there?
Da Schneib
5 / 5 (2) Aug 31, 2014
Yeah, like for instance the entire General and Special Relativity Theories. That's not "a different opinion."
Please prove that this is so in special relativity!
That what is so?
I do not find special relativity wrong, except for the absurd, insane claim that it defines time as a fourth dimension, which any grade 10 pupil can prove CANNOT EVER be the case!.
Then you're not as smart as a 10th grader, since you haven't proved it yet.
That's denial of the foundation of modern physics.
Any physics based on time as a fourth dimension is insane physics.
So now SRT is "insane physics."

Deny all you like, jonny boy, but the fact of the matter is if you don't accept the Continuum Postulate you don't accept SRT or GRT. It's just that simple.
Da Schneib
5 / 5 (2) Aug 31, 2014
Gravitational lens - no proof of existence.
Let's start with Eddington's journey to South Africa to view a star close to the Sun during an eclipse.
I did not know he came to South Africa to do that
My bad, I had forgotten it was Principe.
Da Schneib
5 / 5 (3) Aug 31, 2014
contd
For example, the Standard Model of Particle Physics, often referred to as "the SM," explicitly incorporates SR.
No it DOES NOT! It incorpoartes the insane absurdity of time as a fourth dimension
Among other things from SR, yes, yes it does. Thanks for admitting it.

And the relativity of simultaneity is a direct result of SR.
But not when it is explained in terms ot time being a fourth dimension
The Lorentz Transform describes space and time as interchangeable in black and white, converting time in one IRF to space+time in another, explicitly, in the equations of the transform. You're not even fighting just relativity; you're denying physics that's been established since the nineteenth century.
johanfprins
1 / 5 (5) Aug 31, 2014
Here's Hamilton's first paper, from 1982: http://adsabs.har...98...59H

It's titled, "Gravitational spin precession in binary systems," a nice little exercise in relativistic astrophysics. Since then he has published over a hundred and forty papers in the peer-reviewed scholarly literature.

This was published at the earliest while he was in his mid-20s, so that would make him nearly fifty. He has two kids, and one of them is a graduate student in neuroscience. The other is a filmmaker and clothing designer.

Maybe you should have read his home page a bit more.
A person who claims that Einstein's first postulate is continuous space-time is a moron! Anybody wo knows physics will confirm that thisIS NOT WHAT EINSTEIN CLAIMED when he published his manuscript in 1905..
johanfprins
1 / 5 (5) Aug 31, 2014
No science, johan.

You are too stupid to even recognise science even if you fall with your ugly fat face into it. What are your qualifications? The fact that you hide behind anonymity and not post your CV is admitting that you are a nincompoop, who knows bloody nothing; and ho are just TROLLING. How about coming out of the closst as a operson with INTEGRITY will do. YOU STINK TO HIGH HEAVEN!!!
Da Schneib
5 / 5 (3) Aug 31, 2014
Actually Einstein was unpleasantly surprised with 4D space-time model, published with his teacher H. Minkowski in 1911.
This is impossible. He couldn't have realized the relativity of simultaneity without it.
Minkowski called Einstein "Lazy dog" and Einstein didn't like him just because of it.
He sure was nice to him in "Relativity." He credits him with putting him on the path to GRT. He may not have liked it at first but when all was said and done he accepted it and even gave proper credit for it. And used it in explaining relativity for everyone, which is what his book "Relativity" was, after all, all about.

To be sure, Minkowski was just elaborating on an idea of Poincare here. The Poincare group combines rotations in space-time (the Lorentz group) with translations in space-time.
Correct.
Da Schneib
5 / 5 (3) Aug 31, 2014
Here's Hamilton's first paper, from 1982...
A person who claims that Einstein's first postulate is continuous space-time is a moron! Anybody wo knows physics will confirm that thisIS NOT WHAT EINSTEIN CLAIMED when he published his manuscript in 1905.
So what? It's what he said in "Relativity." And it's how it's derived in modern physics. Welcome to the 20th century.

Oh, wait...
johanfprins
1 / 5 (5) Aug 31, 2014
The mainstream is SRT. You say SRT is wrong. That's the end of that.
I did not! I stated that the mainstream interpretation of SRT in terms of a four-dimensional spave-time is wrong; as any 10th grader who has done linear algebra can attest to. Not the postulates of SRT! So stop lying you backward bastard.

You're making stuff up again, jonny boy.
I am NOT making up stuff: I am just applying the impeccable rules of linear algebra. ou cannot fudge mathematics to suit your hallucinations!
Da Schneib
5 / 5 (1) Aug 31, 2014
Better, Zeph. You get a "5" for that one.
Da Schneib
5 / 5 (3) Aug 31, 2014
The mainstream is SRT. You say SRT is wrong. That's the end of that.
I did not! I stated that the mainstream interpretation of SRT in terms of a four-dimensional spave-time is wrong
First, it's not an "interpretation." It's a direct result of the math of SRT. Second, you're denying Einstein's own statements about it. Third, you're denying the entire modern approach to SRT and GRT, and in doing so denying almost all of modern physics, which is based on SRT and GRT.

You're making stuff up again, jonny boy.
I am NOT making up stuff: I am just applying the impeccable rules of linear algebra
First, you don't need linear algebra for relativity. You can use it, and it leads some interesting places, but it's not required.

Second, what you made up is that you didn't claim Einstein was "bulls----ing." Your own words.
johanfprins
1 / 5 (5) Aug 31, 2014
That what is so?
That I stated that special reltivity is wrong! It is correct: It is only the insane interpretation that it means that a four-dimensional space-time exists that is WRONG! Even Einstein did not accept this absurd interpretation for at least 10 years and he DID NOT USE IT AS A POSTULATE LIKE YOUR STUPID PROFESSOR AT BOULDER IS CLAIMING!!!.
Then you're not as smart as a 10th grader, since you haven't proved it yet
Anybody who knows lineaa algebra wil realise that I have proved this over and over and over again on this thread". For a 5th grader like you it must be difficult [q Any physics based on time as a fourth dimension is insane physics.
So now SRT is "insane physics."
NOT SRT but the insane interpretation of SRT; namely that it defines a space-time!

Da Schneib
5 / 5 (3) Aug 31, 2014
That I stated that special reltivity is wrong!
If you say the relativity of simultaneity is wrong, you are saying relativity is wrong. Simple as that.

Anybody who knows lineaa algebra wil realise that I have proved this over and over and over again on this thread
Where? Quote please. Or else you're making stuff up again.

NOT SRT but the insane interpretation of SRT that it defines a space-time!
It's not an "interpretation." It's a direct mathematical result of SRT. For the fourth or fifth time.
Aligo
1 / 5 (2) Aug 31, 2014
It's a direct mathematical result of SRT. For the fourth or fifth time.
Four dimensional space-time is a postulate of SR. You just linked it yourself as a Continuum postulate.
johanfprins
1 / 5 (5) Aug 31, 2014
Among other things from SR, yes, yes it does.
It does NOT. Only a moron who does not understand 10th grade linera algebra will claim tha it does.

he Lorentz Transform describes space and time as interchangeable in black and white,
No it does NOT. It only shows that time must also be transformed: NOT that time and space is interchangeable you moron. If it were to be so Lorentz would hvae picked it up when he and Poincare discovered these equations. Theyu DID not and NOR did Einstein!!
converting time in one IRF to space+time in another
BULLSHIT!!
You're not even fighting just relativity; you're denying physics that's been established since the nineteenth century.
That is what YOU are doing you fool!
Da Schneib
4 / 5 (4) Aug 31, 2014
Four dimensional space-time is a postulate of SR. You just linked it yourself as a Continuum postulate.
After showing SRT is correct in millions of experiments, we have extremely strong evidence that its postulates must be correct. If they were wrong the experiments would have failed.
Da Schneib
5 / 5 (3) Aug 31, 2014
The Lorentz Transform describes space and time as interchangeable in black and white,
No it does NOT.
x' = (t - vx/c²)/√(1-v²/c²), johan.

You're making stuff up again.
johanfprins
1 / 5 (5) Aug 31, 2014
A person who claims that Einstein's first postulate is continuous space-time is a moron! Anybody wo knows physics will confirm that thisIS NOT WHAT EINSTEIN CLAIMED when he published his manuscript in 1905.
So what? It's what he said in "Relativity." And it's how it's derived in modern physics.
Einstein grabbed at straws against his better knowledge and betrayed himself. The fact is that this is NOT the postulate on which relativity is based. All relativity is based on a single concept which was formulated by Galileo. Anybody who brings in any other postulates is a moron!

Da Schneib
5 / 5 (3) Aug 31, 2014
Einstein said himself that Minkowski's work led him to GRT.

You're making stuff up again, johan.
johanfprins
1 / 5 (5) Aug 31, 2014
First, it's not an "interpretation." It's a direct result of the math of SRT.
Incorrect mathematics, yes. This means that it is WRONG since the four coordinates ARE NOT LINEARLY INDEPENDENT!!! Can you Not get this through your THICK skull?

Second, you're denying Einstein's own statements about it.
Einstein rejected space-time for more than 10 years. He made a mistake to change his mind.
Third, you're denying the entire modern approach to SRT and GRT,
I admit that I reject all physics which are based on impossible mathematics like space-time and renormalisation. This does NOT mean that I reject the wholw "modern approach" en toto. Only an idiot like YOU will make such a sweeping statement. Are you totally incapable of arguing anything without making sweeping statements? Are you really so infantile in your thinking?
Da Schneib
5 / 5 (4) Aug 31, 2014
This means that it is WRONG since the four coordinates ARE NOT LINEARLY INDEPENDENT!
That has nothing to do with it. You're trying to hide behind linear algebra. It's pretty transparent, you know. Besides, if they're not independent, then they're dependent; and if they're dependent then time is a dimension.

Oops.

Second, you're denying Einstein's own statements about it.
Einstein rejected space-time for more than 10 years.
So what?

Third, you're denying the entire modern approach to SRT and GRT,
I admit that I reject all physics which are based on impossible mathematics like space-time and renormalisation.
So you reject Feynman too? Your denial of physics is even worse than I thought. Classic cargo cult science.

This does NOT mean that I reject the wholw "modern approach" en toto.
Yes, it unfortunately does.

And you're making stuff up again, johan.
RealityCheck
1 / 5 (5) Aug 31, 2014
Hi Schneib. :)
...it's not an "interpretation." It's a direct result of the math of SRT
Careful mate! As all professors should remind their students from time to time, "time" is an IMAGINARY graphing/maths 'device' for representing relativities between different stages of analytical aspects being treated mathematically/graphically. Space-time "Co-ordinate Frames' are IMAGINARY constructs also, as any real mathematician-physicist will tell you. As such, the ONLY real EFFECTIVE PHYSICAL DIMENSIONS are the SPATIAL dimensions...and everything else is ABSTRACTLY/METHEMATICALLY DERIVED from MOTION in those THREE ONLY spatial dimensions.

Einstein made quite clear when he finally opted for a 'space-time' ANALYTICAL CONSTRUCT----which, in Einstein's Leyden Address, he made quite clear that it removed the last vestiges of PHYSICAL PROPERTIES/MECHANISMS from space-----in preference for an ABSTRACTION based in imaginary maths/graphing REPRESENTATIONAL algorithms/modeling for CALCULATIONS/PREDICTIONS and NOT "EXPLANATIONS" of the underlying physical properties/mechanisms as such.

So in your further discussion with Johan, be careful in depending too much on your 'certainty' if your 'physical understandings' are based on such imaginary graphing/maths 'abstraction devices', because they were EXPRESSLY designed to only calculate, not actually explain, the physical properties/mechanisms in reality from which such a 'construct' was ABSTRACTED for maths/graphs 'represenations/interpretations/calculations sequencing/relationships etc.

Ok? Good luck. :)
johanfprins
1 / 5 (5) Aug 31, 2014
If you say the relativity of simultaneity is wrong, you are saying relativity is wrong.
I have not said that. I have said that your sweeping claim (how esle from an infantile mind) that simultaneity is impossible is wrong. Simple as that!

Anybody who knows lineaa algebra wil realise that I have proved this over and over and over again on this thread
Where? Quote please. Or else you're making stuff up again. Go back to my precvious posts. In fact you at fisrt ignored the proof, and I then reposted it: And you are still ignoring it" You bloody MORON!

NOT SRT but the insane interpretation of SRT that it defines a space-time!
It's not an "interpretation." It's a direct mathematical result of SRT. For the fourth or fifth time. It is NOT a direct mathematical result of SRT arsehole. Minkowski was an incompetent mathematician who violated the rules of mathematics. I can understand why Einstein bunked his lectures!
johanfprins
1 / 5 (5) Aug 31, 2014
Four dimensional space-time is a postulate of SR. You just linked it yourself as a Continuum postulate.
After showing SRT is correct in millions of experiments, we have extremely strong evidence that its postulates must be correct. If they were wrong the experiments would have failed.

I have not questioned the postulat,e as formulated by Galileo, on which ALL relativity is based. I am just questioning your stupid insistence that Einstein derived SRT from the "postulate" of a "space-time continuum. This is a LIE which can only come from a personn like you who has NO INTEGRITY! And who apprently teaches this nonsense at Boulder!
johanfprins
1 / 5 (5) Aug 31, 2014
Einstein said himself that Minkowski's work led him to GRT.
Yes it did in the end when Einstein grabbed at straws. That is why Einstein's GTR is wrong! As we all know that it must be since it cannot be EVER unified with quantum mechanics.

You're making stuff up again, johan.

YOU are the one who are embracing absurdity as reality!
Da Schneib
5 / 5 (4) Aug 31, 2014
If you say the relativity of simultaneity is wrong, you are saying relativity is wrong.
I have not said that.
This does NOT mean that the light pulses were NOT emitted simultaneously
Oops. You just claimed absolute time.
I have said that your sweeping claim ... that simultaneity is impossible is wrong.
But I didn't say it's impossible. I said it's meaningless. You're making things up again, johan.
Anybody who knows lineaa algebra wil realise that I have proved this over and over and over again on this thread
Where? Quote please.
Go back to my precvious posts.
Quote or it never happened. You're making things up again, johan.
NOT SRT but the insane interpretation of SRT that it defines a space-time!
It's not an "interpretation." It's a direct mathematical result of SRT.
It is NOT a direct mathematical result of SRT.
Linked and quoted. You're making stuff up again, johan.

contd
johanfprins
1 / 5 (5) Aug 31, 2014
The Lorentz Transform describes space and time as interchangeable in black and white,
No it does NOT. x' = (t - vx/c²)/√(1-v²/c²), johan.
This is NOT a Lorentz equation, as I have already pointed out above. So please ask somebody to show you where to look up the Lorentz equations before postin claptrap, as you consistently are doing!

Da Schneib
5 / 5 (4) Aug 31, 2014
I have not questioned the postulat,e as formulated by Galileo, on which ALL relativity is based.
Einstein himself says Galilean relativity is wrong, and deduces it from the observed constancy of the speed of light. Chapter 4.
The laws of the mechanics of Galilei-Newton can be regarded as valid only for a Galileian system of co-ordinates.


I am just questioning your ... insistence that Einstein derived SRT from the "postulate" of a "space-time continuum.
So now you claim relativity contains neither space nor time.

Builda de runway an de CARGO will come from de sky. Car-GO. Car-Go. Car-GO.

And who apprently teaches this nonsense at Boulder!
LOL
Da Schneib
5 / 5 (4) Aug 31, 2014
Einstein said himself that Minkowski's work led him to GRT.
Yes it did in the end when Einstein grabbed at straws. That is why Einstein's GTR is wrong! As we all know that it must be since it cannot be EVER unified with quantum mechanics.
Sure it can. We already know of at least three promising approaches, and Garret Lisi may have found a fourth.

Also, you're denying relativity again, johan. Quote, "Einstein's GTR(sic) is wrong!" end quote.

The Lorentz Transform describes space and time as interchangeable in black and white,

No it does NOT.
x' = (t - vx/c²)/√(1-v²/c²), johan.
This is NOT a Lorentz equation
Sigh. http://en.wikiped...ormation

You're making stuff up again, johan.
johanfprins
1 / 5 (5) Aug 31, 2014
This means that it is WRONG since the four coordinates ARE NOT LINEARLY INDEPENDENT!
That has nothing to do with it. You're trying to hide behind linear algebra
I am not hiding! I am just insisting on using correct mathematics.
It's pretty transparent, you know. Besides, if they're not independent, then they're dependent; and if they're dependent then time is a dimension.
Oh my God. PLEAASE go and get some instruction in mathematics and coordinate transformations. You obviously know NOTHING.

I think I should rather say goodbye for now since I feel sick. Stupidity is bad but consistent stupidity makes me want to vomit!!

Da Schneib
5 / 5 (4) Aug 31, 2014
That has nothing to do with it. You're trying to hide behind linear algebra
I am not hiding! I am just insisting on using correct mathematics.
The Lorentz Transform is correct mathematics, whether it's in linear form, algebraic form, or trigonometric form. You're making stuff up again, johan.

if they're not independent, then they're dependent; and if they're dependent then time is a dimension.
Oh my God. PLEAASE go and get some instruction in mathematics and coordinate transformations. You obviously know NOTHING.
Another claim with no evidence. You're making stuff up again, johan.
Da Schneib
5 / 5 (4) Aug 31, 2014
I think I should rather say goodbye for now since I feel sick. Stupidity is bad but consistent stupidity makes me want to vomit!!
Translation, you're getting your a-- kicked and you're running away.

Bye, jonny boi.
RealityCheck
1 / 5 (5) Aug 31, 2014
Hi Schneib. :)
Better, Zeph. You get a "5" for that one.
No science there, mate. :)

Practice what you preach, Schneib.

Anyhow, what makes you think that your/others' ratings of 5s or 1s carries any weight in a discussion such as this? Drop that irrelevance if you want to come across as a real scientist, hey?

BTW, you never did answer this question: How old are you, Schneib? And what is your own status in the field of physics?

Like you like to say:
Just askin'.


Oh, and did you catch my caution to you above, posted about 40 minutes ago. For your own good, take a good slow read of it before proceeding in your current vein with Johan, mate. Good luck. :)
Da Schneib
5 / 5 (3) Aug 31, 2014
No science, RC, and a direct untruth about Zeph's post, too.

Minkowski was just elaborating on an idea of Poincare
RealityCheck
1 / 5 (5) Aug 31, 2014
Hi Schneib. :)
No science, RC, and a direct untruth about Zeph's post, too.

Minkowski was just elaborating on an idea of Poincare


But the point was: What has your rating got to do with it either way, mate?

And I pointed out that you may have missed my caution to you earlier, regarding the actual nature of the "time" and "space-time construct/abstraction as per Einstein's clear observations about same in his Leyden Address.

That was to do with actual Einstein's science, not just your/others' opinions re same, hey? :)

And you still did not answer my other two questions. How old are you? What is your own standing in the field of physics?

Thanks. :)
Da Schneib
5 / 5 (4) Aug 31, 2014
Sorry, I don't share personal information with people I don't like.
RealityCheck
1 / 5 (6) Aug 31, 2014
Hi Schneib. :)
Sorry, I don't share personal information with people I don't like.


What's "like" got to do with it? This is not some 'social media' forum, it's a science discussion forum. and just saying your age and professional status without any 'personal details' like name/address/institution etc is NOT going to endanger your privacy or your life, mate!

Go on, don't be so 'precious' and coy about it. How old are you, and what status (without any further personal details) have you in this field? Then at least your interlocutors may get a feel for where your attitude/certainty etc may be coming from, hey?

Go on, be a sport, sport! :)
Uncle Ira
4.3 / 5 (6) Aug 31, 2014
Oh, and did you catch my caution to you above, posted about 40 minutes ago.


He was probably ignoring the way everybody else does. That's all you do around here when you aren't defending your dishonor from international troll bot moderator mafia gang.

And I pointed out that you may have missed my caution to you earlier,


Boy is one slow couyon Really-Skippy. I guessed you missed everybody always ignores your cautions. And when you tell them not to talk to each other too.

And you still did not answer my other two questions.


Maybe like everybody else he noticed that you never answer any questions from peoples and doesn't think he is obliged too answer yours.

How old are you?]/q]

Why you ask him that? You some kind of pervert or something?

What is your own standing in the field of physics?


That's the one peoples have asked you about 1000 or 900 times and never you answered. He should make you wait that long too him.
RealityCheck
1 / 5 (6) Aug 31, 2014
Poor poor Ira-BOT-dummy.

Certain 'bigmouth' trolls/egomaniacs ignore my cautions at their own peril to their own understanding/credibility. Just look how they came a cropper when they ignored my caution re BICEP2 flawed 'work/paper' publish-or-perish 'exercise'. You poor dummies never learn because your own ego and bigmouthed dumbass arrogance gets in the way. Like above again, hey? And I DID say exactly what I have been doing 'from scratch' as an independent objective researcher into the Universal Physics; and did so more than once now, you poor poor Ira-BOT-dummy. Sad.
Reg Mundy
1 / 5 (5) Aug 31, 2014
Gravitational lens - no proof of existence.
Let's start with Eddington's journey to South Africa to view a star close to the Sun during an eclipse.
I did not know he came to South Africa to do that
My bad, I had forgotten it was Principe.

Eddington's photos are equally explained by expansion theory, read up on it. So, again, where is your proof that "gravity" exists? Newton dreamt it up, no doubt whilst under the influence, to explain why objects seem to be attracted to each other. No doubt if he had said it was because fairies push them, he would have been laughed at, so he invented this magical invisible unstoppable undetectable unshieldable new force called gravity. Am I the only person who would have laughed at him?
Now, stop being a prat and detail proof of "gravity"'s existence. I await with bated breath....
Reg Mundy
1 / 5 (5) Aug 31, 2014
@RC
Why are you wasting your time (and everybody elses) by responding to Irate? You'll be jousting with Cap'n Grumpy next!
RealityCheck
1 / 5 (5) Aug 31, 2014
Hi Reg. :) To be strictly objective, I have to state that Einstein was correct to characterize 'gravity' EFFECT OBSERVED as a consequence of the mass 'conditioning' its surroundings. Unfortunately, Einstein did not actually identify the physical MECHANISM which 'couples' space to mass and vice-versa. Fortunately my ToE 'from scratch' DOES identify the energy-space nature/process/mechanism so as to produce mutual acceleration between energy-space 'features'.

Just so you know that; and that it comes from an impartial someone who has no 'sides' of his own in all of this....only that of the objective physical reality itself. I am taking the risk of telling you/everyone this now because it may save you and others a lot of angst, time and trouble if you knew that now rather than wait for my ToE to come out. Good luck. :)

PS: re the Ira-BOT-Idiot, it's the occasional correction for his own good and that of his 'friends'. Just being humane is all! Cheers. :)
Uncle Ira
5 / 5 (6) Aug 31, 2014
ignore my cautions at their own peril .


See Really-Skippy that's why everybody thinks you have the really bad mental condition.

look how they came a cropper when they ignored my caution


From here it looks the world still spinning, and all those really Scientist-Skippys still have their jobs. What you got Cher, for all your warnings and cautions? From here it looks like all you got is peoples making the BIG fun with you.

from scratch


Well that must have been the wrong place to start from Skippy, because so far you have gotten to point of postums on the physorg. Not counting all the other places where you got the boot for issuing too many cautions and warning and such like.

Universal Physics


Bet you really do good there too. That sounds like the physics school I go to. Down here we call him it interweb forum commenting place on the physorg.Going to the same physics school as ol Ira ain't nothing to brag about (And edit don't work with the Ira.)
Da Schneib
5 / 5 (4) Aug 31, 2014
Just look how they came a cropper when they ignored my caution re BICEP2 flawed 'work/paper' publish-or-perish 'exercise'.
As I recall you got pwnt on BICEP2. It turned out you were quoting the publicist, not the scientists. Remember?
RealityCheck
1 / 5 (5) Aug 31, 2014
Hi Schneib. :)
Just look how they came a cropper when they ignored my caution re BICEP2 flawed 'work/paper' publish-or-perish 'exercise'.
As I recall you got pwnt on BICEP2. It turned out you were quoting the publicist, not the scientists. Remember?

Not so. The 'scientists' issued a statement afterwards still 'standing by' their 'claims/work'. Remember?

How one-eyed do they make 'scientists' out your way, mate? You obviously are not an objective mind, but an ego-driven arrogant mindless 'parroter' of what others tell you but you will not think out for yourself.

Anyhow, did you read my earlier caution and Einstein's clear statement of what 'time' and 'space-time' etc are/are not? :)

If you won't even get that much past your biased blinkers, then you will continue to fail at being in any way 'scientific' or 'objective' in either research or understanding of the physical (not imaginary) things that underlay the universal phenomena set observed. Good luck with that. :)
Da Schneib
5 / 5 (4) Aug 31, 2014
Just look how they came a cropper when they ignored my caution re BICEP2 flawed 'work/paper' publish-or-perish 'exercise'.
As I recall you got pwnt on BICEP2. It turned out you were quoting the publicist, not the scientists.
Not so.
So.

Anyhow, did you read my earlier caution and Einstein's clear statement of what 'time' and 'space-time' etc are/are not?
No, I read Einstein's book.
RealityCheck
1 / 5 (4) Aug 31, 2014
Hi Schneib. :)
So.
I just pointed out you/others were misunderstanding what 'time' and 'space-time' etc are/are not, again, and referenced Einstein; and you come back with that lame-o juvenile non-counter straight from the 'school playground'? Just how old (or should that be 'young') are you? Face it, you have just been pwned on this, mate; and you know it. :) :)

No, I read Einstein's book.
How about his Leyden Address where he states it clearly and unambiguously, without any chance of you misunderstanding it; like you/others obviously have been, despite purportedly "reading his book". Face it, you've been pwned on this one as well. :)

Good luck with your bias and denial, 'scientist'. :)
Da Schneib
5 / 5 (4) Aug 31, 2014
So.
I just pointed out you/others were misunderstanding what 'time' and 'space-time' etc are/are not, again, and referenced Einstein
out of context. And from prior work that was later superceded. Oops.

No, I read Einstein's book.
How about his Leyden Address where he states it clearly and unambiguously, without any chance of you misunderstanding it; like you/others obviously have been, despite purportedly "reading his book".
Link and quote or it never happened.
Da Schneib
5 / 5 (3) Aug 31, 2014
Einstein's Leyden address, concluding paragraph:
Recapitulating, we may say that according to the general theory of relativity space is endowed with physical qualities; in this sense, therefore, there exists an ether. According to the general theory of relativity space without ether is unthinkable; for in such space there not only would be no propagation of light, but also no possibility of existence for standards of space and time (measuring-rods and clocks), nor therefore any space-time intervals in the physical sense. But this ether may not be thought of as endowed with the quality characteristic of ponderable media, as consisting of parts which may be tracked through time. The idea of motion may not be applied to it.
Einstein's "aether" is not your "aether."
RealityCheck
1 / 5 (4) Aug 31, 2014
Hi Schneib. :)
out of context.
I was calling attention to the lameness of your dismissive attitude in general, not only in the BICEP2 case (which left all you 'experts' redfaced because you didn't take my caution/advice seriously), because here we had a situation where you were given the LeydenAddress as a source for you to check re 'time' and 'space-time' etc abstractions, but you have that "So" attitude even as you are being set straight about THAT to.

And from prior work that was later superceded. Oops.
Rewrite the history of what went down re BICEP2 too, to suit your 'one-eye' versions, why don't ya!

How about his Leyden Address where he states it clearly and unambiguously, without any chance of you misunderstanding it; like you/others obviously have been, despite purportedly "reading his book".
Link and quote or it never happened.
You denying Einstein ever gave his Leyden Address, now!?

Wiki/Google it and read/understand it yourself.
Da Schneib
5 / 5 (4) Aug 31, 2014
out of context.
I was calling attention to the lameness
Personally I find quoting out of context pretty lame.

You denying Einstein ever gave his Leyden Address, now!?
No, it just doesn't say what you claim. I just quoted it and proved that.

You're making stuff up again, RC.
RealityCheck
1 / 5 (4) Aug 31, 2014
Hi Schneib. :) I'm not talking of any particular 'aether'. It has to do with the underlying physical entity/nature of the physically real universal phenomena. Period.

You quoted part of it.

Did you read the part where he explains that by modeling it all via the space-time 'abstraction', it effectively takes away the last vestiges of physical properties/mechanisms aspects from whatever it is that underlies the universal physical reality?
Da Schneib
5 / 5 (4) Aug 31, 2014
Quote and link or it never happened, RC. Welcome to the Internets, n00b.

Also, I just quoted Einstein saying that there isn't any "physical entity/...physically real... phenomena(sic)" involved with his "aether" concept.

You're making stuff up again, RC.
Uncle Ira
5 / 5 (4) Aug 31, 2014
You denying Einstein ever gave his Leyden Address, now!?


Oh great wise and wondrous Really-Skippy. He just give the quote from him. Now do realize why everybody thinks you have the mental conditions? Do you diligence better Skippy, you making it too easy for ol Ira to make the BIG fun on you.

Wiki/Google it and read/understand it yourself.


You are one slow p'tit boug, why you tell him to google him, when is the one postum the parts of it. Maybe you should check with the care givers there at the mental condition place and ask if you missed your medicines on time.
Da Schneib
5 / 5 (4) Aug 31, 2014
Note most particularly, BTW, the use of "space-time intervals" in the quote from the Leyden address. It's clearly apparent from this that Einstein has at this point accepted Minkowski's work and even incorporated it into his thinking.
RealityCheck
1 / 5 (5) Aug 31, 2014
Hi Schneib. :)
I just quoted Einstein saying that there isn't any "physical entity/...physically real... phenomena(sic)" involved with his "aether" concept.
Dear me, just how lazy and confirmation-biasedreading/researching do they teach 'modern scientists' to BE nowadays where you were 'taught', mate? You went straight for the 'concluding paragraph' that suited your biased/misunderstanding, and missed entirely where Einstein gives the REASON WHY the 'space-time' modeling ABSTRACTION leaves behind all physical property/mechanism 'considerations', and just treats 'space-time' as a math/analysis CALCULATION/CONSTRUCT modeling tool/algorithm etc. with the 'time' being the 'imaginary' part of the 'space-time' construct. Your own referenced Andrew Hamilton says:
the postulate implies that the time dimension behaves in many ways as if it were an IMAGINARY spatial dimension.


See? It's merely an abstract/derived analytical/graphing convenience/term, not a real physical dimension.
RealityCheck
1 / 5 (5) Aug 31, 2014
Ok, for all those who are too lazy to find the most relevant bit (for this issue) for themselves, here is the relevant excerpt from Einstein's Leyden Address...
More careful reflection teaches us, however, that the special theory
of relativity does not compel us to deny ether. We may assume the
existence of an ether; only we must give up ascribing a definite
state of motion to it, i.e. we must by abstraction take from it the
last mechanical characteristic which Lorentz had still left it.


...where Einstein in the last part of that last sentence quoted above says how abstraction will thereafter ignore the question of 'ether' and any properties/mechanisms it may possess....and so models the observed phenomena/measurements in a 'space-time' mode so that no actual identification/explanation of actual physically real properties/mechanisms are even attempted in such abstract modeling.

Get it now, Schneib? :)

PS: Schneib, can you stop your 'friend' drooling all over the forum floor!
Whydening Gyre
5 / 5 (5) Aug 31, 2014
Wow, Schneib! You are like Errol Flynn, fencing with 2 or 3 guys all at the same time!
Da Schneib
5 / 5 (4) Aug 31, 2014
You went straight for the 'concluding paragraph' that suited your biased/misunderstanding
Einstein summed up and said you're full of it.

You're making stuff up again, RC.

Your own referenced Andrew Hamilton says:
the postulate implies that the time dimension behaves in many ways as if it were an IMAGINARY spatial dimension.
See? It's merely an abstract/derived analytical/graphing convenience/term, not a real physical dimension.
"Imaginary" has a completely different meaning than you're giving it here. Apparently you don't have enough math to know that.

Quadrature is calculated using imaginary numbers. Does that mean it is "imaginary?" I guarantee you'll find out how "imaginary" it is if you overload your transformer with an inductive load without correcting the quadrature with a big capacitor! You will definitely let the smoke out of your transformer, and probably your equipment, too.

You're making stuff up again, RC.
Da Schneib
5 / 5 (4) Aug 31, 2014
...where Einstein in the last part of that last sentence quoted above says how abstraction will thereafter ignore the question of 'ether' and any properties/mechanisms it may possess....and so models the observed phenomena/measurements in a 'space-time' mode so that no actual identification/explanation of actual physically real properties/mechanisms are even attempted in such abstract modeling.
Nice word salad. Does it come with dressing?

The Lorentz Transform turns space into time and vice versa. What this means is that if we are in relative motion in two IRFs, some of what you see as "space" looks to me like "time" and vice versa. There's no abstract about it; fixed velocity is a fixed rotation in 4D spacetime, and you can actually see it due to foreshortening. Furthermore, you can see it slowing the passage of time from the POV of an observer. Acceleration, that is changing velocity, denotes a changing rotation in 4D spacetime. These facts are clear from the Transform.
Uncle Ira
5 / 5 (4) Aug 31, 2014
PS: Schneib, can you stop your 'friend' drooling all over the forum floor!


Let explain the one thing for you, I am not standing on the floor. It's a deck. And let explain one another thing to you like I already have 3 or 2 dozens times before. I am an independant operator and I come in here alone, and I go out of here alone. And one more another thing. It ain't my fault you say the silly stuffs for me to notice and make BIG fun with. If you wouldn't say them I wouldn't have any material to work with.

Now I am going to make you happy Cher. I got to go do the start-um-up thing because we going to get moving in about an hour. So I'll be really busy and won't be able to make the fun with you for a few hours from now.

Oh yeah I almost forget, my super-dooper stupid people detector and bad karma vote machine is not working, so it may take some time before I can get around to giving you the bad karma points. I know you work hard for them but you got to be patient child, I'll get to it as soon as I can.

So do your diligence better while I'm working the engines. Ol Ira is signing off from St Louis and getting ready to go south with the tow. Laissez les bons temps rouler Really-Skippy

P.S. You can pretend like the silly looking pointy cap is really a "Smart-Scientist-Mr-Wizzard" hat while I'm gone if you want to, but don't get to upset if everybody can still tell you got him for being the class couyon. Let him at the door on your way out so the next couyon won't have to go without one to wear..
Da Schneib
5 / 5 (4) Aug 31, 2014
Laissez les bons temps rouler yourself, there, Ira!

Have a quiet trip. See you again soon.
Da Schneib
5 / 5 (4) Aug 31, 2014
Wow, Schneib! You are like Errol Flynn, fencing with 2 or 3 guys all at the same time!
Just having a good time and surfing some while I load up CDs into iTunes for import into my iPod! :D

Hope you're picking up on some of this stuff, Whyde. Ask a few questions and make me feel useful, will ya? I'm getting tired of edumacating all these, ummm, let's say, folks with offbeat ideas.
RealityCheck
1 / 5 (5) Aug 31, 2014
Hi Schneib. :)
Einstein summed up and...
...and explained WHY in the earlier bit I quoted for you, ie....
...we must by abstraction take from it the
last mechanical characteristic
Why do you not acknowledge that explanation behind his 'summation' view?

"Imaginary" has a completely different meaning than you're giving it here. Apparently you don't have enough math to know that.
That's the point, mate! It IS maths/analysis construct 'convenience' concept/entity, not real physical effective dimension like spatial. Imaginary. ie, abstract, as in 'space-time' construct is abstract according to Einstein himself above.

Come on, mate; don't be stubborn. I posted the quote you asked for. Einstein says abstraction modeling via 'space-time' construct avoids necessity to treat/identify/explain any actual physical effective properties/mechanisms.

It's right there, in black and white, from Einstein's mouth. How long do you think you can keep denying it to yourself? Not healthy.
RealityCheck
1 / 5 (5) Aug 31, 2014
Hi Whyde. :) It's not a game. It's serious that 'self professed experts' such as Schneib won't acknowledge what I posted from Einstein's own Leyden Address; and that Scneib keeps making ego-driven arrogant and unscientific smartass comebacks (ably assisted by his 'special friend' trailing along behind him and drooling all over the floor) which make a mockery of his 'scientific' pretenses/parrotings in this instance. If you think he is doing 'science discourse' in this instance (as he seems to be thinking he is doing) then it's too sad for 'modern science and scientists' reputation.

Keep objective, and stay unimpressed by such smartass comebacks like those he is doing now which ignore the very science facts he ASKED for! Cheers! :)
Da Schneib
5 / 5 (4) Aug 31, 2014
Einstein summed up...
...and explained WHY in the earlier bit I quoted for you, ie....
...we must by abstraction take from it the last mechanical characteristic
Why do you not acknowledge that explanation behind his 'summation' view?
Because he's talking about making the aether abstract and you're pretending he's talking about making all of physics abstract.

"Imaginary" has a completely different meaning than you're giving it here.
It IS maths/analysis construct 'convenience' concept/entity, not real physical effective dimension like spatial. Imaginary. ie, abstract, as in 'space-time' construct is abstract according to Einstein himself above.
No, imaginary as in imaginary numbers.

You're making stuff up again, RC.

I posted the quote you asked for.
And it doesn't say what you claimed.
RealityCheck
1 / 5 (5) Aug 31, 2014
Hi Schneib. :)
Because he's talking about making the aether abstract and you're pretending he's talking about making all of physics abstract.
Just what do you think Einstein is saying when he says clearly...
...we must by abstraction take from it the last mechanical characteristic
He is naturally talking of an abstract math/analytical modeling construct (such as 'space-time'), which would make only that physics modeled using such a 'space-time' abstraction also an abstraction. Any 'standalone' analyses NOT invoking abstraction of 'space-time' will be what they are in their own applicable specific context.

No, imaginary as in imaginary numbers.
Like I said, a math entity/concept 'time' abstraction in analysis construct, not actually attempting to depict an actual real physically effective dimension in its own right. A convenient imaginary analytical abstraction.

And it doesn't say what you claimed
What 'claimed'? It's there in black and white. You're in denial.
Da Schneib
4.2 / 5 (5) Aug 31, 2014
Because he's talking about making the aether abstract and you're pretending he's talking about making all of physics abstract.
Just what do you think Einstein is saying when he says clearly...
...we must by abstraction take from it the last mechanical characteristic
"It" is the aether.

You're making stuff up again, RC.

No, imaginary as in imaginary numbers.
Like I said, a math entity/concept 'time' abstraction in analysis construct, not actually attempting to depict an actual real physically effective dimension in its own right. A convenient imaginary analytical abstraction.
Nope. Just imaginary numbers.

You're making stuff up again, RC.
RealityCheck
1 / 5 (5) Aug 31, 2014
"It" is the aether.
Yes. And its abstractly replaced by 'space-time' analytical convenience which no longer capable of actually treating the underlying physical properties/mechanisms of the underlying universal reality....irrespective of what one calls that.

Nope. Just imaginary numbers.
So you agree, 'time dimension' is just a 'derived' imaginary number term/concept used in abstract maths/geometrical analysis/modeling using the abstraction of 'space-time' replacement for underlying universal physical dimensions/motions.

Good. Glad you have those two things straight at last.

PS: If you keep up with your sloganeering phrases like "You're making things up again...", then your claim to being a 'scientist' will be even further eroded with every iteration of such inane juvenile 'devices' for evasion/disparagement while evading/denying the facts. For your own reputation's sake, quit it quick, mate. :)
Da Schneib
5 / 5 (4) Aug 31, 2014
"It" is the aether.
Yes. And its abstractly replaced
No. It's rendered unnecessary.

You're making stuff up again, RC.

Nope. Just imaginary numbers.
So you agree, 'time dimension' is just a 'derived' imaginary number term/concept
No. Imaginary doesn't mean what you're trying to pretend it does, in this context.

You're making stuff up again, RC.

And if you don't like me saying that, then stop making stuff up.
Da Schneib
5 / 5 (4) Aug 31, 2014
BTW, what you're doing with Einstein's Leyden address is exactly the same thing the religious nutjobs are doing who claim Einstein "believed in Bog." The Gob or Dog or whatever that Einstein believed in isn't an imaginary dude in the sky. Similarly, Einstein's aether has nothing to do with yours.
thefurlong
5 / 5 (5) Aug 31, 2014
@Johan, Da Schneib, and Reg and any others
Thank you for your sympathies. It is remarkable how quickly people discard their vitriol in respect for personal loss. I am touched that people I regularly insult and accuse of being crazy can still say such kind things to me (not including Da Schneib). It hits home that you are not just text, but human beings.
Out of respect for that, permit me to extend an olive branch.
One thing I like about Whydening Gyre is that she usually attempts to not engage in hostile behavior. OTOH, it is too easy for me to do this.
I propose a truce. Reg and Johan, I will read your responses and ignore the slights retroactively in my responses. If you are respectful to me, I will remain respectful to you.
Da Schneib
5 / 5 (4) Sep 01, 2014
Unfortunatley, furlong, I think they all think that telling them they're wrong is an "insult." But good luck.
RealityCheck
1 / 5 (5) Sep 01, 2014
Hi Schneib. :)
No. It's rendered unnecessary.
Sure, if one wants to 'settle for' more abstraction instead of eventual physical explanation.

GR 'space-time' model cannot, by dint of being abstraction, ever identify/explain mechanisms/properties of underlying universal energy-space. Obviously, it can abstractly DESCRIBE/PREDICT mathematically/analytically OBSERVED/MEASURED MOTIONS/EFFECTS....but it cannot physically identify/explain underlying mechanisms/properties/entities involved in physical reality of space and motion therein.

Imaginary doesn't mean what you're trying to pretend it does, in this context.
Your referenced lecturer...
http://casa.color...ate.html
...says:
Amongst other strange consequences, the postulate implies that the time dimension behaves in many ways as if it were an imaginary spatial dimension.
'Time dimension' doesn't exist in its own right AS a dimension (like spatial), it can 'behave' abstractly/imaginary in analysis. :)
RealityCheck
1 / 5 (5) Sep 01, 2014
Hi Schneib. :)
BTW, what you're doing with Einstein's Leyden address is exactly the same thing the religious nutjobs are doing who claim Einstein "believed in Bog." The Gob or Dog or whatever that Einstein believed in isn't an imaginary dude in the sky. Similarly, Einstein's aether has nothing to do with yours.
That's not very nice, mate. I am atheist since age nine, and that is the last thing I would do.

But you seem to have no boundaries as to what distractions and personal crap you will employ to avoid the obvious. So it seems you are the one guilty of 'reading what you want to read' in things (and avoid what you want to deny).

Again, not talking of any aether(s), only of whatever is the underlying fundamental universal context from which everything arises, whatever one calls it.

Abstraction of 'space-time' effectively gives up on ever being able to actually identify physical properties/entities/mechanisms relating to that, just as Einstein's Leyden Excerpt admits clearly!
thefurlong
5 / 5 (6) Sep 01, 2014
On to the derivation. I will use 1+1 dimensions for simplicity
Start with
x' = (x - vt)*g
t' = (t - xv/c^2)*g
where g is the Lorentz factor
We want an invariant, f(x,t) such that
1) f(x',t') = f(x,t)

I will make one further assumption, which is that f is differentiable.

Now, f(x',t') = f((x-vt)*g,(t-xv/c^2)*g).

I will take advantage of this by assuming that v is really small.
If v is small, g can be approximated by 1. I assume you accept this for now. If not, let me know.
Now, since v is small, using the chain rule, the expression on the right hand side can be approximated by
f(x,t) - df/dx(x,t)vt - df/dt(x,t)xv/c^2
But f(x,t) = f(x',t'), so,
f(x,t) = f(x,t) - df/dx(x,t)vt - df/dt(x,t)xv/c^2
Thus,
2) t*df/dx = -(x/c^2)*df/dt
(to be continued)
thefurlong
5 / 5 (6) Sep 01, 2014
(continued)
So, I have just arrived at a partial differential equation that f must satisfy.
Now, to solve it, it helps to find where f must be constant.
In other words, I want curves u, v, such that the total differential, df(u,v) = du*df/dx(u,v) + dv*df/dt(u,v) = 0.
Well, df/dx(u,v) =-(u/(vc^2))*df/dt from equation 2,
So, -du*(u/(vc^2))*df/dt(u,v)+ dv*df/dt(u,v) = 0
We can divide both sides by df/dt(u,v), which is not 0, and we are left with
du*(u/(vc^2)) = dv
separating the variables,
u*du = v*c^2*dv
Integrating both sides, we find that, in general
u^2/2 = c^2v^2/2 + k^2/2, where k^2/2 is some constant.
Rewriting this,

3) u^2 - c^2v^2 = k^2.
Now, u and v are dummy variables. We know the domain of f actually x and t, so we can rewrite 3) as
4) x^2 - c^2t^2 = k^2

Now, this enables us to express what form f must take.
First, observe that because of equation 4,
5) f(x,t) = f(sqrt(x^2-c^2t^2),0s)
(to be continued)
thefurlong
5 / 5 (6) Sep 01, 2014
(continued)
Hence in order for f to satisfy 1), f must satisfy equation 5.
But this is the same as saying that f(x,t) = H(sqrt(x^2 - c^2t^2)), where H is some arbitrary function.
Therefore the only possible invariants for the Lorentz transformation are those that are a function of x^2-c^2t^2. THEY MUST DEPEND ON THE MINKOWSKI DISTANCE TO THE ORIGIN. (You don't have to call it distance, if you don't want to).
This also means that if we wish to define an analogous notion of distance under the Lorentz transformation without using time, we are out of luck. Of course, we can have this notion, but nobody will any longer be able to agree on how far away two things are from each other, which is definitely not what happens with Galilean relativity. This is the only nontrivial metrical quantity everyone in every inertial rest frame can agree on.
Thus ends my derivation. Please let me know if you have questions.
Da Schneib
5 / 5 (4) Sep 01, 2014
No. It's rendered unnecessary.
Sure, if one wants to 'settle for' more abstraction instead of eventual physical explanation.
The physical explanation of EM radiation is the exchange of photons. The physical explanation of gravity is the curvature of space. Sounds pretty concrete to me.

underlying universal energy-space.
There isn't any such thing.

Imaginary doesn't mean what you're trying to pretend it does, in this context.
'Time dimension' doesn't exist in its own right AS a dimension (like spatial), it can 'behave' abstractly/imaginary in analysis.
You can quote that as many times as you like and it still won't mean "imaginary" as in not part of reality, it will always mean "imaginary" as in numbers.

As far as time being a dimension, the Lorentz Transform mixes time with space, which could not happen if time weren't a dimension. You can't mix dimensions with other things; only with other dimensions.
Da Schneib
5 / 5 (4) Sep 01, 2014
That's not very nice, mate. I am atheist since age nine, and that is the last thing I would do.
Then how come you're doing it?

Again, not talking of any aether(s), only of whatever is the underlying fundamental universal context from which everything arises, whatever one calls it.
Spacetime, and the other seven dimensions if you like string physics.

Abstraction of 'space-time' effectively gives up on ever being able to actually identify physical properties/entities/mechanisms relating to that, just as Einstein's Leyden Excerpt admits clearly!
It "admits" nothing, and says nothing like what you claim. Einstein is talking about the aether concept in that excerpt, not spacetime or anything else. You're just taking what he said out of context, like the religious nutjobs do. Apparently you miss religion, so you make a religion out of physics instead.
Whydening Gyre
4.8 / 5 (4) Sep 01, 2014
One thing I like about Whydening Gyre is that she usually attempts to not engage in hostile behavior. OTOH, it is too easy for me to do this.

Thanks for the compliment 1/8 mile. However, for the record, I am a practicing male lesbian. Really...:-) Married twice with kids and everything...:-)
I avoid confrontational crap because it's contrary to my "everything works if you let it" philosophy of life.
I am sorry to hear about your father. The scattering of ashes has deep connotations that I respect immensely and commend you for.
That said, I understand I do not hold a candle to most of you guys in the math training department, so what I am able to wade through, I translate to my own medium - 3d visual. (You wouldn't believe the "test lab" I have built in my head - it's immense!...:-)
Anyway, anything after that, I ask that you all help me a little in further translation...:-)

Thanks in advance,
Chuck Geier
Reg Mundy
1 / 5 (4) Sep 01, 2014
@Daz

The physical explanation of EM radiation is the exchange of photons. The physical explanation of gravity is the curvature of space. Sounds pretty concrete to me.

underlying universal energy-space.

There isn't any such thing.

So "curvature of space" is fine, but there is no such thing as "universal energy-space"? Your logic as usual is difficult to underestimate. Statements such as this, where you pick and choose what is real and what is imaginary/impossible, all without a shred of evidence either way, are a pretty good indication of what is wrong with the mainstream science you espouse.
Reg Mundy
1 / 5 (3) Sep 01, 2014
This is a load of chunter
Another claim without any evidence to back it up.

And by the way, acceleration is not a force. It's the result of a force.

Who said acceleration is a force? You are trying to put words in my mouth. Put them in your own instead. You can't have acceleration without a force, and all forces result in acceleration unless opposed by other forces. Ain't no GRAVITY! All forces are the result of interactions between masses either by contact or magnetism or electrostatic if charged. Get round that, dickhead!
RealityCheck
1 / 5 (5) Sep 01, 2014
Hi Schneib. :)
No. It's rendered unnecessary
Yes, unnecessary for the purposes of abstraction and maths convenience. :)

But when real physical properties and mechanisms are required to be identified and explained in reality, then there is inescapable necessity for some underlying real physical universal entity/substrate which can effect the features and phenomena properties/mechanics observed around you. By 'space-time' abstraction (as Einstein called it) one may 'settle for' maths fantasies without real physical explanation of properties and mechanisms (like still absent from 'space-time' maths abstraction models), but there comes a point when abstraction is 'not enough' for those scientists who want the explanation for gravity and the rest in reality effective physical property/mechanism terms, not just fairy-tale-maths terms. :)

Get real, mate. Eventually most of us, even 'scientists' must 'grow up and leave fairy-tales behind', yes? :)
RealityCheck
1 / 5 (5) Sep 01, 2014
Hi Schneib. :)
The physical explanation of EM radiation is the exchange of photons
It's all ABSTRACT maths 'space-time model', not actual explanation of physical mechanical properties (as Einstein's Leyden Address clearly stated was the purpose of such abstraction, rather than pursuing mechanical properties explanations of actual physical underlying universal substrate whence all energy-space features/processes arise and evolve/devolve as observed all around us).

The physical explanation of gravity is the curvature of space.[q/]Since when has the standard 'space-time' model actually explained the underlying mechanisms for coupling 'photon' to 'space' and 'matter'? Never. Just describe the effects/values as observed, not yet explained as to underlying mechanism responsible for 'abstract math modeling of space-time 'curving' etc.

Believing that 'abstraction based maths models are reality' seems to colour and blinker your own critical faculties. Fairy-tales are not reality. ok?
RealityCheck
1 / 5 (5) Sep 01, 2014
Hi Schneib. :)

underlying universal energy-space.
There isn't any such thing.
Now you are contradicting the basic entities of Quantum Mechanics and General Relativity, Schneib?:)

Space and Energy is all that can be objectively physically really observed/treated in QM and GR; and GR model is abstract (as space-time constructs); while QM models are statistical teatments of observed energy-space quanta/interactions.

I coined "energy-space" as a physically observable retro-replacement, for reality-based ToE theorizing from scratch. Because final, complete and consistent reality-based ToE must necessarily be couched/explained in reality-based mechanical properties/behaviours of real physical entities (not maths-abstraction 'models').

Others are increasingly using 'energy-space' too!

Only reality-based 'energy-space' context ToE can ever be complete consistent in all respects, and identify/explain GRAVITY nature/mechanism (which 'space-time' can't deliver, ever). :)
RealityCheck
1 / 5 (5) Sep 01, 2014
Hi Schneib. :)
As far as time being a dimension, the Lorentz Transform mixes time with space, which could not happen if time weren't a dimension. You can't mix dimensions with other things; only with other dimensions.
It's MATHS, Schneib, not reality. Just because one can mix terms in a certain way according to certain maths definitions/rules, doesn't give 'time' real/hierarchical 'dimensional' status/independence as spatial dimensions. :)

You seem to believe all sorts of made-up maths-fairy-tale 'rules and devices' are 'real' physical things/processes! That way lay madness/frustration from UNreal fairy-worlds of your own making according to made-up rules and entities, not the actual reality.

I'm not 'mixing' abstract analytical 'time' convenience with real spatial dimnsions, mate. You are. And you have only your maths abstractions to 'justify' doing so, even though real physicists (including Einstein) realize 'time' is merely a convenient abstraction, as in 'space-time' model.
RealityCheck
1 / 5 (5) Sep 01, 2014
Hi Schneib. :)
Spacetime, and the other seven dimensions if you like string physics.
Again: What has personal subjective "like" or "dislike" got to do with anything in science?

And just because mathematical-physicists extend the 'space-time' ABSTRACTION modeling fantasy technique to include a further number of abstraction 'dimensions', does not make it all 'real' in underlying property/mechanics 'energy-space' dimensional terms. Stop tacitly 'equating' fantasies with reality, mate. It won't end well for your mind's sanity if you keep doing that while pretending to an 'objective intellect'. :)

...and says nothing like what you claim. Einstein is talking about the aether concept in that excerpt, not spacetime or anything else.
Rational reading/comprehension would extrapolate as effectively includin ALL underlying substrates, not just 'luminiferous aether' back then.

And removal-by-ABSTRACTION of ALL mechanical properties is what 'space-time' did. Don't deny the obvious. :)
Da Schneib
5 / 5 (4) Sep 02, 2014
And by the way, acceleration is not a force. It's the result of a force.
Who said acceleration is a force?
You.
Da Schneib
5 / 5 (4) Sep 02, 2014
underlying universal energy-space.
There isn't any such thing.
So "curvature of space" is fine, but there is no such thing as "universal energy-space"?
Yep. The difference is there's evidence for curvature of space.

See Eddington at Principe.
Da Schneib
5 / 5 (4) Sep 02, 2014
RC, aether is so abstract it doesn't exist.

Get it?
RealityCheck
1 / 5 (5) Sep 02, 2014
Hi Schneib. :)
RC, aether is so abstract it doesn't exist. Get it?
Why so obsessed with 'old' aether? We've all moved on, to 'new' aether concepts like Dark Energy, Dark Matter, Higgs Field, Quantum Vacuum 'fluctuations' and, of course, my REAL PHYSICAL mechanically effective UNIFYING "energy-space' in/from which all such differentiated/emergent 'field/aether' phenomena/entities evolve.

If you think abstract 'space-time' concept is 'real', then maybe you should take note of antialias_physorg, who has used my REAL 'energy-space' to enhance the reality-basis of his own explanations. :)

Other mainstream physicists/scientists increasingly eschewing abstract mathematical 'space-time', and using my "energy-space" construct in order to enhance reality-basis for their explanations; eg, Robert Hogan:
http://www.livesc...sed.html
...and if we shake it too much, we could go into this new energy space,...

Reality is always best, mate! :)
Da Schneib
4.2 / 5 (5) Sep 02, 2014
Because there isn't any aether but old aether.
RealityCheck
1 / 5 (5) Sep 02, 2014
Hi Schneib. :)
Because there isn't any aether but old aether.
Can't you move on from that old 'version' of underlying energy-space aspect?

Learn to evolve in your thinking/arguments, and encompass the hints I have been providing about "energy-space" consistent reality physics, which is slowly but surely displacing "space-time" abstractions models; else you will be left behind when the new cosmological physical paradigm resulting from the reality-based ToE arrives. Coming soon. :)

Anyhow, until then, enjoy your polite, tolerant and objective open discussions; and good luck and good thinking to you/everyone.

Cheers all real science/scientists for now! :)
Da Schneib
5 / 5 (3) Sep 02, 2014
So is space-energy like space-aliens?
RealityCheck
1 / 5 (5) Sep 02, 2014
Hi Schneib. :)
So is space-energy like space-aliens?
No, you're thinking of 'space-time' based sci-fi 'fantasy entities' again.

And you just ignored my above references to other mainstreamers increasingly using real 'energy-space' terms/concepts for real explanations.

Why do you do that?

Is 'denial' some new 'rule' for 'modern scientists' like you? Is this some 'new scientific method' you 'fantasy-mathematical-physists' are trying to institute in place of the current respected one?

How old are you, and what is your actual qualification which makes you think you have the right to make such inane one-liner evasions while pretending to be 'scientific'?

Your reputation is in jeopardy more and more, from your own penchant for lame evasions in lieu of proper understanding. Do better, mate, for your own reputation's sake. :)
Da Schneib
5 / 5 (3) Sep 02, 2014
No, I'm thinking what you're saying sounds like space-aliens.
RealityCheck
1 / 5 (5) Sep 02, 2014
Hi Schneib. :)
No, I'm thinking what you're saying sounds like space-aliens.
But you're not 'thinking', you're 'kneejerking' and 'denying' and 'evading' the points made.

Do you understand that other mainstream physicists/science commentators are ALSO increasingly using my "energy-space" terms/concepts to make their explanations more reality-referential and mechanism-based, rather than the usual 'space-time' abstractions that can NEVER, by dint of being abstract-only, actually identify/explain any entity/mechanism...ie, as currently is the problem for gravity 'explanations' dependent on the abstract math 'space-time' analysis/modeling construct?

Do you think these other mainstreamer physicists/commentators using my 'energy-space' terms/concepts also sound like they are spouting science fiction when they do so in preference to using 'space-time' abstractions?
Reg Mundy
1 / 5 (5) Sep 02, 2014
And by the way, acceleration is not a force. It's the result of a force.
Who said acceleration is a force?
You.

Out of context. The implication was that you can't have one without the other not that one is the other. Why do you wilfully misunderstand?
Reg Mundy
1 / 5 (5) Sep 02, 2014
@Daz
No, I'm thinking what you're saying sounds like space-aliens.

Yeah, right on, that's about as likley as "curvature in the space-time continuum" and other such rot. You tell him!
RealityCheck
1.8 / 5 (5) Sep 02, 2014
Hi Reg, Schneib, everyone. Unless one heads their reply with the username one is responding to, readers find it difficult to keep cross-conversations straight as to who is talking to whom; and thus may be confusing/mistaking who is actually being 'addressed' in your post/reply. May I suggest making it your habit (as I have made it mine) to mention your interlocutor(s) by name at the beginning of every respective response to respective interlocutor? This would help greatly all casual readers as well as involved parties to a complex cross-exchange! It would be greatly appreciated! Thanks, everyone. :)

PS: Hi Reg Mundy. I see you just did 'address' your post that time. Thanks! :)
thefurlong
5 / 5 (7) Sep 02, 2014
@RealityCheck
It's MATHS, Schneib, not reality. Just because one can mix terms in a certain way according to certain maths definitions/rules, doesn't give 'time' real/hierarchical 'dimensional' status/independence as spatial dimensions. :)


This is not actually an argument, though. Understanding of Physics cannot exist without math of some kind. The underlying math is supposed to describe reality to our best understanding. In fact, the best of these mathematics come from postulates about what we observe about the world, with the postulates of relativity counted among these. And these postulates lead directly to time behaving like an extra dimension. You don't have to call it that if you don't want to, but that doesn't change its behavior.

Dismissing a physical argument because "it's just math" is like dismissing a historical argument because "it's just English." It's a non-argument.
Whydening Gyre
5 / 5 (4) Sep 02, 2014
Out of context. The implication was that you can't have one without the other not that one is the other. Why do you wilfully misunderstand?

@reg
Wrong. you can have force without acceleration. However you wouldn't SEE evidence of the force, so it would appear that is non-existent, as well.
my REAL PHYSICAL mechanically effective UNIFYING "energy-space' in/from...

@rc
YOUR mechanism?
With 7 billion brains constantly running on this planet, I doubt you are the only one...
Perhaps you recall the drunken musings of an American you talked to back in 1975...
Whydening Gyre
5 / 5 (4) Sep 02, 2014
This is not actually an argument, though. Understanding of Physics cannot exist without math of some kind. The underlying math is supposed to describe reality to our best understanding.

IT describes reality as we know it quite well. But, like a weatherman predicting the weather for a given day - it doesn't always predict so well... Not because the math is wrong - it's the human interpretation...
In fact, the best of these mathematics come from postulates about what we observe about the world, with the postulates of relativity counted among these.

Agreed. I love the fact that ALL things are relative to one another. INCLUDING Relativity.
(cont)
Whydening Gyre
5 / 5 (4) Sep 02, 2014
And these postulates lead directly to time behaving like an extra dimension. You don't have to call it that if you don't want to, but that doesn't change its behavior.

this is a truth that begs no argument.

Dismissing a physical argument because "it's just math" is like dismissing a historical argument because "it's just English." It's a non-argument.

I prefer, "Well, it's simple English, Sire..." :-)
thefurlong
5 / 5 (5) Sep 02, 2014
@Reg Mundy
Lets get to the nub of this. You say that, according to the laws of gravity, there is an escape velocity. I say the laws of gravity are WRONG!

What do you mean by that? Are you saying that gravity does not behave like an inverse square law?
You see, every object that has mass also has a Schwartzchild radius (see en.wikipedia.org/wiki/Schwarzschild_radius). Now, who is correct, you or Schwarzschild? Guess you lose, furbrain!

What does Schwarzchild radius have to do with this conversation?
Whydening Gyre
5 / 5 (4) Sep 02, 2014
You see, every object that has mass also has a Schwartzchild radius (see en.wikipedia.org/wiki/Schwarzschild_radius). Now, who is correct, you or Schwarzschild? Guess you lose, furbrain!

What does Schwarzchild radius have to do with this conversation?

@1/8th
I believe he is of the impression that gravity does not exist without the presence of mass...
thefurlong
5 / 5 (5) Sep 02, 2014
IT describes reality as we know it quite well. But, like a weatherman predicting the weather for a given day - it doesn't always predict so well... Not because the math is wrong - it's the human interpretation...

Sure. Science isn't perfect. It probably never will be. However, there is no evidence that viewing time as an extra dimension is a flawed interpretation--at least there is no objective evidence. In other words, there is no evidence that unambiguously implies that this interpretation is incorrect.
Whydening Gyre
5 / 5 (4) Sep 02, 2014
IT describes reality as we know it quite well. But, like a weatherman predicting the weather for a given day - it doesn't always predict so well... Not because the math is wrong - it's the human interpretation...

Sure. Science isn't perfect. It probably never will be. However, there is no evidence that viewing time as an extra dimension is a flawed interpretation--at least there is no objective evidence. In other words, there is no evidence that unambiguously implies that this interpretation is incorrect.

@1/8th
Apologies if you thought I inferred a flaw. In this particular case, I think it is correct. It's just further PREDICTIONS of it become relative(:-) to interpretation. Math, like English, is subject to contextual interpretation...
thefurlong
5 / 5 (7) Sep 02, 2014
@Reg Mundy
All forces are the result of interactions between masses either by contact or magnetism or electrostatic if charged. Get round that, dickhead!

Actually that isn't true. The majority of forces, at least in everyday interactions, are remote. Even the force between the ground and your shoes is actually a remote interaction. If you were to zoom into the junction of your shoes and the ground, you would see space between the (most of the) atoms of your shoes and the atoms of the ground. In this case, the force comes from the pauli-exclusion principle and the electromagnetic interaction. Your energy state is too low to permit the electrons in your shoes, and those in the ground to actually come anywhere near each other (relatively speaking).
Essentially, there isn't any such thing as a "contact" force.
Da Schneib
4.3 / 5 (6) Sep 02, 2014
No, I'm thinking what you're saying sounds like space-aliens.
But you're not 'thinking', you're 'kneejerking' and 'denying' and 'evading' the points made.
You haven't made any points. You just keep saying stuff like "space-energy" and "aether." The first is made-up BS, and the second is BS people stopped believing after the 19th Century.
Da Schneib
4.3 / 5 (6) Sep 02, 2014
Out of context.
Nope. Now you're making stuff up too.
Da Schneib
4.2 / 5 (5) Sep 02, 2014
No, I'm thinking what you're saying sounds like space-aliens.
Yeah, right on, that's about as likley as "curvature in the space-time continuum" and other such rot. You tell him!
We can measure the curvature of spacetime. And we have. And it's not just "curved;" It's curved by *precisely the amount* predicted by Einstein.
RealityCheck
1 / 5 (4) Sep 02, 2014
Hi Schneib. :) Before any further ado, can you please oblige by making it a habit to address your posts by indicating the name of the person you are quoting/responding to? It helps everyone reading/following the cross-conversations to avoid mistaking who said what to whom etc. Thanks! :)
Da Schneib
4.2 / 5 (5) Sep 02, 2014
You'll figure it out eventually.
RealityCheck
1 / 5 (5) Sep 02, 2014
Hi thefurlong. :)
Understanding of Physics cannot exist without math of some kind. The underlying math is supposed to describe reality to our best understanding.
Understanding is ultimately 'intuitive'. It is the Communication of such understanding via mathematics/abstractions.

And when math goes beyond reality (witness String Theory's myriad Un-real interpretations/alternatives), it helps to stop for a reality-check before proceeding further. Remember: "Reality first. Maths/Abstractions second. Always." :)
...the postulates of relativity....lead directly to time behaving like an extra dimension.
Yes, 'time' "behaves like" an extra 'dimension"...but ONLY within the 'artificial' context, rules and interpretations of 'space-time' math/abstract 'modeling' construct, not in objectively observable reality itself having ONLY SPATIAL dimensions for MOTIONs (ie, evolving energy-space processes/features) within that spatial extent.

Analytical 'dimension', not real dimension. :)
RealityCheck
2 / 5 (4) Sep 02, 2014
PS: @thefurlong...Please accept my sincere, if belated, condolences regarding your father. :(
Reg Mundy
1 / 5 (4) Sep 02, 2014
@WG
Out of context. The implication was that you can't have one without the other not that one is the other. Why do you wilfully misunderstand?


@reg
Wrong. you can have force without acceleration. However you wouldn't SEE evidence of the force, so it would appear that is non-existent, as well.
Gimme an instance. why don't cha?
Reg Mundy
1 / 5 (5) Sep 02, 2014
@Reg Mundy
All forces are the result of interactions between masses either by contact or magnetism or electrostatic if charged. Get round that, dickhead!

Actually that isn't true. The majority of forces, at least in everyday interactions, are remote. Even the force between the ground and your shoes is actually a remote interaction. If you were to zoom into the junction of your shoes and the ground, you would see space between the (most of the) atoms of your shoes and the atoms of the ground. In this case, the force comes from the pauli-exclusion principle and the electromagnetic interaction. Your energy state is too low to permit the electrons in your shoes, and those in the ground to actually come anywhere near each other (relatively speaking).
Essentially, there isn't any such thing as a "contact" force.

You have just succintly described the contact force........
Da Schneib
4.2 / 5 (5) Sep 02, 2014
@WG
Out of context. The implication was that you can't have one without the other not that one is the other. Why do you wilfully misunderstand?


@reg
Wrong. you can have force without acceleration. However you wouldn't SEE evidence of the force, so it would appear that is non-existent, as well.
Gimme an instance. why don't cha?
Squeeze a steel bar in your hand. Nothing moves. Did you exert a force?
Reg Mundy
1 / 5 (5) Sep 02, 2014
No, I'm thinking what you're saying sounds like space-aliens.
Yeah, right on, that's about as likley as "curvature in the space-time continuum" and other such rot. You tell him!
We can measure the curvature of spacetime. And we have. And it's not just "curved;" It's curved by *precisely the amount* predicted by Einstein.

Rubbish! You think curvature has been measured in an imaginary thing! You then imagine it must be correct! Space is the place we inhabit. Time is our mechanism of moving from one state of the cosmos (and all the particles in it) to another as dictated by the laws of physics.
Reg Mundy
1 / 5 (5) Sep 02, 2014
@WG
Out of context. The implication was that you can't have one without the other not that one is the other. Why do you wilfully misunderstand?


@reg
Wrong. you can have force without acceleration. However you wouldn't SEE evidence of the force, so it would appear that is non-existent, as well.
Gimme an instance. why don't cha?
Squeeze a steel bar in your hand. Nothing moves. Did you exert a force?

Several forces, actually, net result zero except for compression of the steel bar. (My, what a strong boy you are....)
Da Schneib
4.3 / 5 (6) Sep 02, 2014
No, I'm thinking what you're saying sounds like space-aliens.
Yeah, right on, that's about as likley as "curvature in the space-time continuum" and other such rot. You tell him!
We can measure the curvature of spacetime. And we have. And it's not just "curved;" It's curved by *precisely the amount* predicted by Einstein.

Rubbish!
Eddington measured it in Principe on 29 May 1919. This is nearly a century ago. Try to keep up.
Da Schneib
4.2 / 5 (5) Sep 02, 2014
@WG
Out of context. The implication was that you can't have one without the other not that one is the other. Why do you wilfully misunderstand?


@reg
Wrong. you can have force without acceleration. However you wouldn't SEE evidence of the force, so it would appear that is non-existent, as well.
Gimme an instance. why don't cha?
Squeeze a steel bar in your hand. Nothing moves. Did you exert a force?

Several forces, actually, net result zero except for compression of the steel bar. (My, what a strong boy you are....)
So there's your instance of a force without acceleration. What now, sport?
RealityCheck
1 / 5 (5) Sep 02, 2014
Hi Whyde. :)
...you can have force without acceleration.
Rock sitting on ground is under 'accelerative effect' from resistance of em forces involved. Even though rock not 'moving downwards anymore' due to gravitational acceleration force, it's still 'experiencing' forces. The MACRO scale 'acceleration/energies' forces are balanced/dissipated at the quantum scale of never-ceasing em-perturbations/fluctuations.
With 7 billion brains constantly running on this planet, I doubt you are the only one...
Professional theorists number in millions, but haven't delivered 'coupling/mechanism' explanation how mass-energy 'curves surroundings' to produce objectively observed, only mathematically/abstractly modeled/described, 'gravity effect'.

It only takes ONE 'original thinker' to have a-haa! moment/insight.

History of science replete with diversions/misunderstandings/herd mentality etc....but it's EASY once the a-haa! is had by said ONE mind. Remember that other ONE mind, Einstein? :)
RealityCheck
1 / 5 (5) Sep 02, 2014
Hi Schneib.
You'll figure it out eventually.

Maybe if you stopped obsessing about 'aether' and quit defaulting to 'one-liner' evasions in lieu of comprehending points made, you too might "figure it out eventually". But judging by your continuing arrogant egotistical attitude and personality cult way of 'doing/understanding science', it is more likely that you will need my upcoming ToE to finally get you off your shaky-legged and unheeding high-horse you have imagined for yourself! Good luck with that. Anyhow, no more time to spend on your tactics/antics etc. Must get back to my work.

Bye for now, Schneib, thefurlong, everyone...take care until we 'speak' again! :)
Uncle Ira
3.9 / 5 (7) Sep 02, 2014
History of science replete with diversions/misunderstandings/herd mentality etc....


And crankpots like you with the mental condition.

but it's EASY once the a-haa! is had by said ONE mind.


So far it's looks like all you got is ahhaahahaahaahoooi. Cher you should do better.

Remember that other ONE mind, Einstein? :)


But the Einstein-Skippy went to science school and learned his science stuffs there. All you done is go to the interweb you call the Universal Physics and got that mostly wrong. If you went to the real science school you would be able to do better diligence and not just argue about words that you keep changing and banging around like the pin-ball balls.

Really-Skippy, you need to do your diligence better matey-Skippy.
RealityCheck
1 / 5 (5) Sep 02, 2014
Poor poor Ira-BOT, wrong again. I went to university to study Science (Nuclear Physics especially) and Computer-Programming/Economics/Psychology; and then continued my own further researches as scrupulously independent and objective observer/theoretician and problem-solver/inventor. You on the other hand have nothing to offer anyone except your BOT-operated ratings-sabotaging/skewing on internet sites. Your 'friends' must be proud of your achievements, or they wouldn't be your 'friends', hey? Poor poor Ira-BOT. Bye.
Uncle Ira
4.3 / 5 (6) Sep 02, 2014
I went to university to study Science (Nuclear Physics especially) and Computer-Programming/Economics/Psychology;


No offense Cher, but I think you are telling another one of those GREAT BIG LIES. If it is not the lie, then you wasted your money.

But my money is on it being the GREAT BIG LIE because I can't see you sitting in a university school classroom very long before the professor asked you to leave (after wearing the silly looking pointy cap didn't do you no good) so he can do his diligence and teach the student-Skippys who came there to learn.

You on the other hand have nothing to offer anyone


According to all the fan mails I get from the physorg I think that might be wrong. From here it looks like your fan mail is in the Zephir-Skippy category. But the Reg-Skippy seems to like you.

I also got the nice family and the good honest job to do. I don't have to play scientist-Skippy at the home for people with the mental condition.

Bye.


Yeah, sure.
Whydening Gyre
5 / 5 (4) Sep 02, 2014
@WG
Out of context. The implication was that you can't have one without the other not that one is the other. Why do you wilfully misunderstand?


@reg
Wrong. you can have force without acceleration. However you wouldn't SEE evidence of the force, so it would appear that is non-existent, as well.
Gimme an instance. why don't cha?
Squeeze a steel bar in your hand. Nothing moves. Did you exert a force?

Thanks, Schneib...
Whydening Gyre
5 / 5 (6) Sep 02, 2014
Space is the place we inhabit. Time is our mechanism of moving from one state of the cosmos (and all the particles in it) to another as dictated by the laws of physics.

You call it "state", I call that rubbish double-speak.
"Time" is our word/label for the monitoring of sequential movement of 3d object through 3d space.
Stop arguing what is actually observed. And stop adding "expansion" crap as a viable description of that motion. You're making more of it than is actually there.
RealityCheck
1 / 5 (5) Sep 02, 2014
Hi Whyde. :) I just logging out when I refreshed and noted the new posts.

Careful, mate. It's more subtle, and goes on deeper than the macro-scale obvious. The 'acceleration' from the applied macro forces are then redistributed to all the INTERIOR 'accelerations' going on at the inter/intra atomic levels throughout the steel and the hands for as long as the 'squeezing' forces are applied. Think 'heat'; which is a function of 'accelerations' of constituent parts at whatever scale involved. And then think heat radiation (accelerating photons away from the seemingly 'static' macro-masses involved). Remember, it's not as straightforward as it seems. Just because the macro-scale BODY 'doesn't move' as a direct result of the 'squeezing force' accelerations from hand-surface/body-to-bar-surface-body', it does not mean that something ELSE at the smaller scales within those two bodies is not 'accelerating' due to deformation of molecular/atomic 'electron-orbitals-accelerations'. Good luck.: )
Whydening Gyre
5 / 5 (6) Sep 02, 2014
It only takes ONE 'original thinker' to have a-haa! moment/insight.

History of science replete with diversions/misunderstandings/herd mentality etc....but it's EASY once the a-haa! is had by said ONE mind. Remember that other ONE mind, Einstein.
Like I said. ONE in 7 billion? Statistically not probable.
Again, remember that drunk American in 1975...
RealityCheck
1 / 5 (5) Sep 02, 2014
Poor poor Ira-BOT. Talk about Psychology!

The BOT-operating site/ratings-saboteur says "No offense"? LOL

And then goes on to accuse others of "BIG LIE"? LOL

Hooeee! We have a real doozie of a 'case study' there, folks!

Poor poor Ira-BOT. It didn't know that the lecturer asked me to help out some 'slow' learners in his class by taking them aside after classes to give them some pointers; because the lecturer saw that I was always ahead of what the lecturer was covering, and that the 'slow' learners needed some timely insights from me which they apparently were missing because the lecturer had no time to review just for them.

Poor poor self-deluding BIG LIE and BOT-CON Uncle Ira. Perfect example of BOT-operating moron psychology 'personified'. What a 'case study' it would make 'in person', hey folks! :)
Whydening Gyre
5 / 5 (6) Sep 02, 2014
Poor poor Ira-BOT. Talk about Psychology!

The BOT-operating site/ratings-saboteur says "No offense"? LOL

And then goes on to accuse others of "BIG LIE"? LOL

Hooeee! We have a real doozie of a 'case study' there, folks!

Poor poor Ira-BOT. It didn't know that the lecturer asked me to help out some 'slow' learners in his class by taking them aside after classes to give them some pointers; because the lecturer saw that I was always ahead of what the lecturer was covering, and that the 'slow' learners needed some timely insights from me which they apparently were missing because the lecturer had no time to review just for them.

Poor poor self-deluding BIG LIE and BOT-CON Uncle Ira. Perfect example of BOT-operating moron psychology 'personified'. What a 'case study' it would make 'in person', hey folks! :)

You need to lighten up, Francis...
RealityCheck
1 / 5 (5) Sep 02, 2014
Hi Whyde. :)
Like I said. ONE in 7 billion? Statistically not probable.
Mate, deduct from that 7 bn, all those in 'subsistence/survival pre-occupation'; all those those in 'social media pre-occupied'; all those in 'drunken/stoned pre-occupied'; all those in 'trolling/stalking psychopathy pre-occupied'; and the just plain dumbass criminal morons in 'Ira-BOT-type pre-occupation'.

That leaves only millions of minds with opportunity/capability to pursue professional physics theorizing!

And as indicated by their current collective inability to identify/explain actual coupling/mechanism for observed/described gravity effect, these millions are all 'self-deluding-maths/abstractions pre-occupied'.

Hence the ONE 'original mind' isn't among that millions. :)
RealityCheck
1 / 5 (5) Sep 02, 2014
Hi Whyde. :)
You need to lighten up, Francis...
Already there, mate. But science discourse isn't a 'social media sport' for morons and trolls to 'conduct/skew' on the internet as they please, now is it? Or is that what some people think it is? :)

Anyhow, who's "Francis"?

And why not try telling Ira to stop his BOT-TROLLING idiocy instead? That would be the logical/scientific solution.

Stay loose! Cheers. :)
johanfprins
1 / 5 (5) Sep 03, 2014
@ The Furlong,

I have not been back to thism thread since a brainless troll like Da Schneib makes me sick. Add to that Ira's attempts to be cute by wording his posts as if he is an idiotic baby, then it is just too much for a sane mind to endure. I thus just scanned this thread to see whether you are back and posted what you promised. I saw you started off by assuming what you have to prove: That is circular reasoning. Instead of getting into a long argument with you I suggest that we go back to basics.

In most textbooks it is assumed that when 0 and 0' coincides the time on both clocks is set to zero, AND a spherical wavefront is emitted which moves away from 0 with a speed c and also moves away from 0' with a speed c. Do you agree that after a time t on the clock at 0 the radius of the sphere around 0 must be r=ct: And do you agree that after a time t' on the clock at 0' he radius around 0' must be r'=ct'?

continued
johanfprins
1 / 5 (4) Sep 03, 2014
@ The Furlong

Do you agree that the radius r around 0 must have the same length along all directions, And do you agreen that the radius r' around 0' must have the same length along all directions? Thus, this must also be the case along the Y-direction, so that y=ct AND y'=ct'. This demands that if t' is not the same as t, one MUST have that y and y' must be different: However according to the Lorntz equations one MUST have tha y=y': Thus one must have that t=t'. The times must be the SAME on both clocks for ever after.

The Lorentz-transformed time is thus a different time on both clocks than the time at which the event ACTUALLY occurs. One must thus derive the Lorentz equations by starting off from t=t'. When you do this you do not need to divide by zero as Minkowski has done.

The correct derivation is given in section 6 of my book which is freely available on the internet for everybody to read. https://www.resea...=prf_act

continued
johanfprins
1 / 5 (4) Sep 03, 2014
@ The Furlong

Will you please do me the courtesy of downloading the updated version of my book: "EINSTEIN=GENIUS: But a genius sometimes blunders". If you have any questions please contact me directly at my e-mail johanprins@cathodixx.com so that we can converse like adults are supposed to do; without having to also read the comments by brainless trolls..
thefurlong
5 / 5 (5) Sep 03, 2014
I saw you started off by assuming what you have to prove: That is circular reasoning.

Please explain. My proposal to you was that the minkowski distance to the origin could be derived from the Lorentz Transformation without dividing by 0. I started from the Lorentz transformation, and ended up at x^2-c^t^2 just as promised.

I'll respond to your other stuff later (got a big day of presenting a useful algorithm (and positive results thereof) to people with money ahead of me), but I don't see what it has to do with my derivation of minkowski distance to the origin from the Lorentz transformation.
johanfprins
1 / 5 (4) Sep 03, 2014
My proposal to you was that the minkowski distance to the origin could be derived from the Lorentz Transformation without dividing by 0. I started from the Lorentz transformation, and ended up at x^2-c^t^2 just as promised.


First define what you mean by origin! What is the coordinates of the origin and are they unique? You will find they are not unique since during your derivation you are dividing by zero. A 4D manifold can ONLY have a unique origin, if the origin is the only 4D point which has a zero 4D position vector s(vector)=0 so that x=y=z=w=0. If you have points for which you have zero position vectors while x,y,z,and w are not zero, then you do not have a unique origin and then you are dividing by zero to derive what you have been deriving.
thefurlong
5 / 5 (5) Sep 03, 2014
You will find they are not unique since during your derivation you are dividing by zero.

Let's take this one claim at a time. Where do you say I am dividing by 0 in my derivation?
Captain Stumpy
5 / 5 (5) Sep 03, 2014
You will find they are not unique since during your derivation you are dividing by zero.

Let's take this one claim at a time. Where do you say I am dividing by 0 in my derivation?

@thefurlong
Just got back... just heard about your father.
My sincere condolences. If you need anything... you can PM me at http://saposjoint.net/ or at http://www.sciforums.com/ (profile same as here: TruckCaptainStumpy)

PEACE
thefurlong
5 / 5 (6) Sep 03, 2014
You will find they are not unique since during your derivation you are dividing by zero.

Let's take this one claim at a time. Where do you say I am dividing by 0 in my derivation?

@thefurlong
Just got back... just heard about your father.
My sincere condolences. If you need anything... you can PM me at http://saposjoint.net/ (profile same as here: TruckCaptainStumpy)

PEACE

Thanks, Captain. I appreciate your sympathy.
johanfprins
1 / 5 (4) Sep 03, 2014
Let's take this one claim at a time. Where do you say I am dividing by 0 in my derivation?


Firstly you do not define what you mean by an invariant: Then you assume that there MUST be an invariant and then derive an invariant that is not there by fudging the mathematics.

This is similar to what Einstein did! He assumed that the time transformation of the Lorentz equations must be time-dilation (although he did not state that this is his major assumption), and then he "derived" the Lorentz equations by equating two equations with one another, which look similar, while each is zero. When you do this you are dividing by zero. He then "derived" time-dilation from his assumption that the time transformation must be time-dilation: Circular reasoning!

I explain this very well in section 6 of ny book. Why must I reproduce it here where I am limited by the number of words per post, and have difficulty in typing the mathematics, if you can simply download it; and read it?
thefurlong
5 / 5 (6) Sep 03, 2014
Firstly you do not define what you mean by an invariant

Sorry, I did not explicitly tell you the definition. If T is a transform of a point, P, then an invariant of T, F is a function, such that F(T(P)) = F(P). In other words, F takes the same value regardless of whether we have transformed P with T.
This is similar to what Einstein did! He assumed that the time transformation of the Lorentz equations must be time-dilation (although he did not state that this is his major assumption),

I didn't derive the Lorentz transformation at all. I took it as given. Then, using that, I derived x^2 - c^2t^2 = 0.
and then he "derived" the Lorentz equations by equating two equations with one another, which look similar, while each is zero.

This doesn't follow. Please address the actual steps that I took in my derivation. Tell me the step where I actually divided by 0, please.
johanfprins
1 / 5 (3) Sep 03, 2014
Sorry, I did not explicitly tell you the definition. If T is a transform of a point, P, then an invariant of T, F is a function, such that F(T(P)) = F(P). In other words, F takes the same value regardless of whether we have transformed P with T.
This is ONLY valid it your transform T is a single-point to a single point transformation WITH AN INVERSE: WHICH THE LORENTZ TRANSFORMATION IS NOT! No relativistic transformation of ANY physics equation has an inverse.If you believe this, you do not understand the principle of relativity.

I didn't derive the Lorentz transformation at all. I took it as given. Then, using that, I derived x^2 - c^2t^2 = 0.
Why not? The whole argument has been that in the derivations used by Einstein and Minkowski, they divided by zero. Thus what are you trying to prove if you do not show that their derivations sid not involve this?

continue
johanfprins
1 / 5 (3) Sep 03, 2014
Please address the actual steps that I took in my derivation. Tell me the step where I actually divided by zero
This will require from me to each you the mathematics which you do not WANT to understand.
Why do you not rather for once answer my simple questions above; Let me repeat:
1. When 0 amd 0' coincide, and a spherical wavefront is emitted from this coincident point at time t=t'=0, does this wavefront simultaneously spread out with speed c around BOTH 0 and 0'? YES or NO?
2. Does this mean that one can write around 0 that r=ct and around 0' that r'=ct'? YES or NO?
3. Is r around 0 the same in ALL directions, and r' around 0' the same in ALL directions? YES or NO?
3. Does this mean that along the Y direction (perpendicular to the motion) one must have that y=ct and y'=ct'? YES or NO?
4. Does the Lorentz transform demand that y must be equal to y'? YES or NO?
5. Does this mean that y=ct=y'=ct'? YES or NO?
6 Does this mean that t=t'? YES or NO?
Please answer!
thefurlong
5 / 5 (6) Sep 03, 2014
@johanfprins
Then you assume that there MUST be an invariant and then derive an invariant that is not there by fudging the mathematics.

I didn't catch this. Actually, I don't assume that there MUST be an invariant. I assume there MIGHT be an invariant and derive its properties should it exist. Remember, I am only setting about the task of DERIVING x^2-c^t^2.

Technically, I didn't prove that this function is actually an invariant of the Lorentz transformation. You could even argue that I didn't even prove that it exists (existence theorems can be quite hairy)! In actuality, I proved something related--that if a function F is an invariant of the Lorentz transformation in 1+1 dimensions, then it MUST completely depend on the value x^2-c^2t^2.

In order to really prove x^2-c^2t^2 is an invariant, I would have to show that x'^2 -c^2t'^2 = x^2-c^2t^2, where the primed coordinates are the transformed one.
thefurlong
4.3 / 5 (6) Sep 03, 2014
This is ONLY valid it your transform T is a single-point to a single point transformation WITH AN INVERSE: WHICH THE LORENTZ TRANSFORMATION IS NOT! No relativistic transformation of ANY physics equation has an inverse.If you believe this, you do not understand the principle of relativity.

This has nothing to do with my argument. I do not take the inverse transformation anywhere in my derivation. Please show me where you think I do that.
Why not? The whole argument has been that in the derivations used by Einstein and Minkowski, they divided by zero. Thus what are you trying to prove if you do not show that their derivations sid not involve this?

Because I am not Einstein or Minkowski. I am not using their derivations. I am using my own. Please tell me which step of my derivation involves dividing by 0.
thefurlong
5 / 5 (6) Sep 03, 2014
Please address the actual steps that I took in my derivation. Tell me the step where I actually divided by zero

This will require from me to each you the mathematics which you do not WANT to understand.

How do you know that, if you won't try? Just point out where I divided by 0, please. I would like to point out that we have at least twice disagreed on some mathematical point, and then arrived at a mutual agreement about it. I am confident we can do that again.
thefurlong
5 / 5 (6) Sep 03, 2014
Why do you not rather for once answer my simple questions above;

Because I told you that I wanted to take this one point at a time. Otherwise, we will be overwhelmed by all the places in which our worldviews differ.

It isn't fair of you to accuse me of making a mistake, but not tell me where I did it. A professor I had did this same thing to me because he assumed that since I was the student, I was automatically wrong, even when I protested that I had checked my work several times over for the error. I turned out to be right, once he bothered to actually check his own work.

If I am satisfied with my own derivation and have checked it several times (which I have), then I cannot be expected to find the flaw in my own argument. If you wish to correct me, then you will have to show me where the flaw is. You can't just tell me I've made a flaw and expect me to automatically believe you are correct. That's not how convincing works.
johanfprins
1 / 5 (6) Sep 03, 2014
It isn't fair of you to accuse me of making a mistake, but not tell me where I did it.
On the contrary, it is devious of you to tell me that you will derive the Lorentz equations without dividing by zero and then accept the equiations as given, and then obfuscating everything by using wrong mathematics; which requires from me to waste my time in order to explain to you step by step what the rules of elementary linear algebra and coordinate transformations are.

Your teachers in high school should already have done this. It is incredible that you are obviously doing physics without even understanding what linearly independent coordinates must be. Your professors in mathematics must have been jus as incompetent as the one that Einstein has had in Zurich; namely Minkowski.

I have asked you time and again to read my derivation of the Lorentz equations in which I do not divide by zero; but you just refuse. If this is your game, please stop wasting my time.
thefurlong
5 / 5 (6) Sep 03, 2014
On the contrary, it is devious of you to tell me that you will derive the Lorentz equations without dividing by zero and then accept the equiations as given, and then obfuscating everything by using wrong mathematics;

I didn't do this at all. Here is what I wrote to you in the beginning,
If I start from the Lorentz transformation, and without invoking length contraction and time dilation, arrive at this invariant, will that help?

I didn't tell you I was going to derive the Lorentz transformation. I said explicitly, as you can see, that I was going to start from it and derive x^2-c^2t^2. And that's what I did!
johanfprins
1 / 5 (4) Sep 03, 2014
If I start from the Lorentz transformation, and without invoking length contraction and time dilation, arrive at this invariant, will that help?
You did not arrive at an invariant since an invariant is ONLY POSSIBLE when the coordinates of a coordianate transformations are lineraly independent!! The coordinates x, y,.z and ct ARE NOT liner4aly independent. Thus you must FIRST prove that they are loneraly independent before you can claim that anything is invariant under such a coordinate tranformation.
[q I said explicitly, as you can see, that I was going to start from it and derive x^2-c^2t^2. And that's what I did! Well then I misunderstood you! But to conclude that two expressions in different coordinates are invariant, you must first prove that the coordinates are isometric, and you have not proved this. Please answer my 6 SIMPLE questions above and you will see that x^2-c^2t^2 and x'^2-c^2t'^2 are NOT invariant AND CAN NEVER BE!!!
johanfprins
1 / 5 (4) Sep 03, 2014
Since (x,t) and(x',t') are NOT the coordinates of the SAME point in a manifold They are two different non-coincident points!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!11
thefurlong
5 / 5 (6) Sep 03, 2014
You did not arrive at an invariant since an invariant is ONLY POSSIBLE when the coordinates of a coordianate transformations are lineraly independent!! The coordinates x, y,.z and ct ARE NOT liner4aly independent....

None of this comment has anything to do with my derivation of x^2-c^2t^2 from the Lorentz transformation Please tell me the step where I divided by 0. You haven't done that yet. I would greatly appreciate it if you would.
thefurlong
5 / 5 (6) Sep 03, 2014
@johanfprins
Also, do you acknowledge, then, that I have't been trying to derive the Lorentz contraction this thread?
thefurlong
5 / 5 (6) Sep 03, 2014
@johanfprins
I really do want to respond to the points you made regarding linear dependence, and your other questions you asked me, but I don't want to, until I am certain that you and I agree on what we are debating about, and in order for that to happen, you need to tell me the step where I divided by 0.

With all due respect, I am going to keep asking you to tell me where I divided by 0 until you actually give me the answer, and not respond to anything else. I am not asking you to explain anything, or to reteach me mathematics. I am merely asking you for the step where I divided by 0. That's all I want. It should be easy for you to tell me where I did.
johanfprins
1 / 5 (5) Sep 03, 2014
You did not arrive at an invariant since an invariant is ONLY POSSIBLE when the coordinates of a coordianate transformations are lineraly independent!! The coordinates x, y,.z and ct ARE NOT liner4aly independent....

None of this comment has anything to do with my derivation of x^2-c^2t^2 from the Lorentz transformation Please tell me the step where I divided by 0. You haven't done that yet. I would greatly appreciate it if you would.


After you have answered my six questions above. Once you have done this and see what the result is you should br able to find your mistake by yourself, unless you are really so incredibly stupid that you are beyond redemption.
johanfprins
1 / 5 (4) Sep 03, 2014
@johanfprins
Also, do you acknowledge, then, that I have't been trying to derive the Lorentz contraction this thread?

Then I misunderstood you since this is what the argument has been about: NOT substituting the Lorentz transformation into x^2+y^2+z^2-c^2t^2 and finding that it leadas to x'^2+y'^2+z'^2-c^2t'^2. We ALL know that this is rthe case: And if you know mathematics you will know that this DOES NOT PROVE ANY INVARIANCE WHATSOEVER! I have repeatedly told you this on this thread, and as you can verify yourself by answering my 6 question, x,y,z,t and x',y',z',t' are NOT the coordinates of the SAME point within a 4D manifold: As any grade 10 pupil in school can tell you.
johanfprins
1 / 5 (4) Sep 03, 2014
@johanfprins
I really do want to respond to the points you made regarding linear dependence, and your other questions you asked me, but I don't want to, until I am certain that you and I agree on what we are debating about, and in order for that to happen, you need to tell me the step where I divided by 0.
As I have stated, I do not have the time to go step by step through your incorrect mathematics just because you are too incompetent to understand linear algebra.

With all due respect, I am going to keep asking you to tell me where I divided by 0 until you actually give me the answer, and not respond to anything else.
Like Einstein and Minkowski you have assumed that x^2-ct^2 can be linearly related to x'^2-c^2t'^2. THIS IS WHERE YOU ALL ARE DIVIDING BY ZERO.
I am not asking you to explain anything, or to reteach me mathematics.
I cannot do this unless you are able to understand linear independence: First answer my questions.
thefurlong
5 / 5 (5) Sep 03, 2014
As I have stated, I do not have the time to go step by step through your incorrect mathematics just because you are too incompetent to understand linear algebra.

You don't have to go through step by step. Just give me the quote at which I go from not having divided by 0 to having divided by 0.
johanfprins
1 / 5 (4) Sep 03, 2014
On to the derivation. I will use 1+1 dimensions for simplicity
Start with
x' = (x - vt)*g
t' = (t - xv/c^2)*g
where g is the Lorentz factor
We want an invariant, f(x,t) such that
1) f(x',t') = f(x,t)

Already here you are going wrong by NOT stating that x,t and x',t' MUST be the coordinates of the SAME point. If they are not the coordinates of the SAME point the function is the same BUT it is NOT an invariant of the coordinate transformation: Jeesh, are you REALLY so stupid!! You ARE wasting my time my boy! You must first prove that x,t and x',t' are the coordinates of the SAME point in the manifold (in this case 2D)! So for F-sake, first prove this. I am getting tired of your nonsensical arguments.

thefurlong
5 / 5 (5) Sep 03, 2014
Like Einstein and Minkowski you have assumed that x^2-ct^2 can be linearly related to x'^2-c^2t'^2. THIS IS WHERE YOU ALL ARE DIVIDING BY ZERO.

What? What do you mean by "linearly related"? Show me the step where I assume that. Just give me the quote, please.
johanfprins
1 / 5 (3) Sep 03, 2014
Like Einstein and Minkowski you have assumed that x^2-ct^2 can be linearly related to x'^2-c^2t'^2. THIS IS WHERE YOU ALL ARE DIVIDING BY ZERO.

What? What do you mean by "linearly related"? Show me the step where I assume that. Just give me the quote, please.


Go through your derivation and note that your initial assumption is that the Lorentz coordinate transformation gives that x,t and x',t' are the coordinates of the same point in 2D "space-time". This is where you are already dividing by zero, since in order to derive this, one has to set two expressions which are separately equal to zero equal to one another. I already knew in primary school that this is the same as dividing by zero.

Will you now answer my questions please?

I am signing off for now. It is late in SA!
thefurlong
5 / 5 (6) Sep 03, 2014
Already here you are going wrong by NOT stating that x,t and x',t' MUST be the coordinates of the SAME point.

Well, it isn't what I asked for, but it's a start. Thank you!
Yes, I didn't state that they are coordinates of the same point, but this is implicit.

Let me be clear about something. This conversation, is not about how to derive the Lorentz transform, or that it is even correct. It is about finding an invariant of a transform that gives a single point TWO DIFFERENT COORDINATES without dividing by 0.
If they are not the coordinates of the SAME point the function is the same BUT it is NOT an invariant of the coordinate transformation

I honestly don't know what you are saying there. Please rephrase.
Jeesh, are you REALLY so stupid!! You ARE wasting my time my boy!

That's strike one. If you are going to be disrespectful, when I have not been disrespectful to you, this conversation is over.
thefurlong
5 / 5 (5) Sep 03, 2014
Go through your derivation and note that your initial assumption is that the Lorentz coordinate transformation gives that x,t and x',t' are the coordinates of the same point in 2D "space-time".

Yes, they do.
This is where you are already dividing by zero, since in order to derive this, one has to set two expressions which are separately equal to zero equal to one another. I already knew in primary school that this is the same as dividing by zero.

What two expressions? So, what you are saying is that I am relying on an assumption that you claim can only be arrived at by dividing by 0. In other words, in my derivation, I didn't derive by 0, but I am using an assumption, which you think requires division by 0. This means I didn't actually divide by 0.

No, I don't have to prove that division by 0 is not required for that initial assumption any more that one is required to prove that primes exist before proving Fermat's little theorem. (to be continued)
thefurlong
5 / 5 (5) Sep 03, 2014
(continued)
They are two different problems. I started with an assumption you think is flawed. Whether you accept that assumption is beside the point. If you can admit that I did not divide by 0 in my derivation, which I did not, then we can move on. If you can tell me you agree with me on this assessment, we can move on, and I can start to answer your questions. If you don't agree, then you still have not answered my question. I await your (respectful) response.
Reg Mundy
1 / 5 (4) Sep 03, 2014
@Daz
@WG
Out of context. The implication was that you can't have one without the other not that one is the other. Why do you wilfully misunderstand?


@reg
Wrong. you can have force without acceleration. However you wouldn't SEE evidence of the force, so it would appear that is non-existent, as well.
Gimme an instance. why don't cha?
Squeeze a steel bar in your hand. Nothing moves. Did you exert a force?

Several forces, actually, net result zero except for compression of the steel bar. (My, what a strong boy you are....)
So there's your instance of a force without acceleration. What now, sport?

I say that all (net) force produces acceleration, and that all acceleration is caused by force. You come up with crap about squeezing lumps of steel. Either you are exceedingly stupid, or are wilfully obtuse. In either case, I have lost interest in trying to have a logical discussion with you.
Whydening Gyre
5 / 5 (2) Sep 03, 2014
Go through your derivation and note that your initial assumption is that the Lorentz coordinate transformation gives that x,t and x',t' are the coordinates of the same point in 2D "space-time".

Yes, they do.[

Still not quite getting how they can be the same coordinate points... Aren't they describing oppositely?
Whydening Gyre
5 / 5 (2) Sep 03, 2014
I say that all (net) force produces acceleration, and that all acceleration is caused by force. You come up with crap about squeezing lumps of steel. Either you are exceedingly stupid, or are wilfully obtuse. In either case, I have lost interest in trying to have a logical discussion with you.

All acceleration caused by force? Well, duh...
That acceleration requires force? Well, duh...
Those were stupidly obvious statements in a single sentence...
All you're trying to do is re-describe and thusly reinvent, the wheel...
thefurlong
5 / 5 (3) Sep 03, 2014
Go through your derivation and note that your initial assumption is that the Lorentz coordinate transformation gives that x,t and x',t' are the coordinates of the same point in 2D "space-time".

Yes, they do.[

Still not quite getting how they can be the same coordinate points... Aren't they describing oppositely?

I took this to mean that they are coordinates that describe the same space-time point, but in different inertial reference frames.
Whydening Gyre
5 / 5 (3) Sep 04, 2014
Still not quite getting how they can be the same coordinate points... Aren't they describing oppositely?

I took this to mean that they are coordinates that describe the same space-time point, but in different inertial reference frames.

Okay, I got that.
But (maybe) a dumb question then... How could they then still be the same coordinates?
thefurlong
5 / 5 (6) Sep 04, 2014
Still not quite getting how they can be the same coordinate points... Aren't they describing oppositely?

I took this to mean that they are coordinates that describe the same space-time point, but in different inertial reference frames.

Okay, I got that.
But (maybe) a dumb question then... How could they then still be the same coordinates?

They aren't, unless they describe the origin, x=0m,y=0m,z=0m,t=0s.
johanfprins
1 / 5 (3) Sep 04, 2014
Yes, they do.
The conclusion that they do, follows by dividing by zero. Thus to accept "that they do" you start off of by incorporating this division in your derivation: Thus you are dividing by zero!!

What two expressions?
x-ct=0 and x'-ct'=0.
So, what you are saying is that I am relying on an assumption that you claim can only be arrived at by dividing by 0.
Exactly
In other words, in my derivation, I didn't derive by 0...
By using an assumption that is based on dividing by zero YOU ARE dividing by zero to get the result that you derive.

Einstein's book: RELATIVITY: The Special and General Theory: p. 226:. He wrote the two equations above (x'-ct')=lambda*(x-ct). This means that lambda=0/0: Thus he is dividing by zero in order to claim that x,t and x',t' are coordinates of the SAME point in a 2D manifold. Deriving the LT correctly, x,t and x',t' are NOT the coordinates of the same point.
Da Schneib
5 / 5 (4) Sep 04, 2014
x,y,z,t and x',y',z',t' are NOT the coordinates of the SAME point within a 4D manifold
Actually, they *are* the coordinates of the same point in two different IRFs. That's what you're transforming with a transform like, in this example, the Lorentz Transform.
Da Schneib
5 / 5 (4) Sep 04, 2014
On to the derivation. I will use 1+1 dimensions for simplicity
Start with
x' = (x - vt)*g
t' = (t - xv/c^2)*g
where g is the Lorentz factor
We want an invariant, f(x,t) such that
1) f(x',t') = f(x,t)
Already here you are going wrong by NOT stating that x,t and x',t' MUST be the coordinates of the SAME point.
Errr, then what's f(x',t') = f(x,t) mean? That equals sign means they're the same point to me. I'm not sure what you're looking at that causes you to claim this. Please explain when and how = does not mean "equals." Thanks in advance.
johanfprins
1 / 5 (4) Sep 04, 2014
x,y,z,t and x',y',z',t' are NOT the coordinates of the SAME point within a 4D manifold
Actually, they *are* the coordinates of the same point in two different IRFs. That's what you're transforming with a transform like, in this example, the Lorentz Transform.
That is where physics has gone wrong in 1905, and nobody ever picked it up.
They are NOT the coordinates within two different IRF's (coordinate systems) of the same point in a 4D manifold. This is so since Einstein's derivation is based on the assumption that these coordinates are isometric within a 4D manifold, which they are NOT!!! And to obtain this wrong interpretation both Einstein and Minkowski divided by zero. I have just now quoted directly from Einstein's book which PROVES that Einstein divided bt zero. This was a bad habit of Einstein, since he did this on more than one occasion.
johanfprins
1 / 5 (4) Sep 04, 2014
On to the derivation. I will use 1+1 dimensions for simplicity
Start with
x' = (x - vt)*g
t' = (t - xv/c^2)*g
where g is the Lorentz factor
We want an invariant, f(x,t) such that
1) f(x',t') = f(x,t)
Already here you are going wrong by NOT stating that x,t and x',t' MUST be the coordinates of the SAME point.
Errr, then what's f(x',t') = f(x,t) mean?
It means that you have a function that gives you the same value at two different positions at two different times.
That equals sign means they're the same point to me.
Do you want to claim that when a function at two different points have the same value at these points these points MUST be the same point?
I'm not sure what you're looking at that causes you to claim this. Please explain when and how = does not mean "equals."
Here "equals" mean that you have the same value at TWO different points in the 2D manifold! See section 6 of my book EINSTEIN=GENIUS: But sometimes a genius blunders.
Da Schneib
5 / 5 (6) Sep 04, 2014
@Daz
@WG
Out of context. The implication was that you can't have one without the other not that one is the other. Why do you wilfully misunderstand?


@reg
Wrong. you can have force without acceleration. However you wouldn't SEE evidence of the force, so it would appear that is non-existent, as well.
Gimme an instance. why don't cha?
Squeeze a steel bar in your hand. Nothing moves. Did you exert a force?

Several forces, actually, net result zero except for compression of the steel bar. (My, what a strong boy you are....)
So there's your instance of a force without acceleration. What now, sport?

I say that all (net) force produces acceleration, and that all acceleration is caused by force.
Where do you claim you said, "net" before bringing it into the discussion to cover your error in not saying it in the first place?

You said, "force causes acceleration." You never mentioned *net* force.
Da Schneib
5 / 5 (6) Sep 04, 2014
Go through your derivation and note that your initial assumption is that the Lorentz coordinate transformation gives that x,t and x',t' are the coordinates of the same point in 2D "space-time".

Yes, they do.[

Still not quite getting how they can be the same coordinate points... Aren't they describing oppositely?
No. That's the definition of a transform: describing the same point in two different coordinate systems. When we talk about the Lorentz Transform and Special Relativity Theory, we're talking not only about the same point in space, but the same point in time, and transforming it from one observer's POV to another. Those two observers are, by definition in SRT, both in IRFs. To describe the transform between an IRF and an accelerated frame, or between two accelerated frames, requires not SRT but GRT.
Da Schneib
5 / 5 (5) Sep 04, 2014
By using an assumption that is based on dividing by zero YOU ARE dividing by zero to get the result that you derive.
This is incorrect. It's like saying because you breathe air just like an elephant, you're an elephant. Non sequitur, to use the Latin.

Einstein's book: RELATIVITY: The Special and General Theory: p. 226:. He wrote the two equations above (x'-ct')=lambda*(x-ct). This means that lambda=0/0
Sorry, I don't see the derivation for (x'-ct')=lambda*(x-ct) ∴ lambda=0/0. Lambda appears to be the Lorentz transform to me.

Thus he is dividing by zero in order to claim that x,t and x',t' are coordinates of the SAME point
But it's a transform. Of course they're the same point; that's implicit. If you're trying to transform between different points, you're making an error by using a transform in the first place. Transforms do not translate between different points; they translate the same point into different coordinate systems (such as IRFs)
Da Schneib
5 / 5 (4) Sep 04, 2014
On to the derivation. I will use 1+1 dimensions for simplicity
Start with
x' = (x - vt)*g
t' = (t - xv/c^2)*g
where g is the Lorentz factor
We want an invariant, f(x,t) such that
1) f(x',t') = f(x,t)
Already here you are going wrong by NOT stating that x,t and x',t' MUST be the coordinates of the SAME point.
Errr, then what's f(x',t') = f(x,t) mean?
It means that you have a function that gives you the same value at two different positions at two different times.
That equals sign means they're the same point to me.
Do you want to claim that when a function at two different points have the same value at these points these points MUST be the same point?
I'm not sure what you're looking at that causes you to claim this. Please explain when and how = does not mean "equals."
Here "equals" mean that you have the same value at TWO different
Stop. "Equals" doesn't mean "different."
johanfprins
1 / 5 (5) Sep 04, 2014
Go through your derivation and note that your initial assumption is that the Lorentz coordinate transformation gives that x,t and x',t' are the coordinates of the same point in 2D "space-time".

Yes, they do.[

Still not quite getting how they can be the same coordinate points... Aren't they describing oppositely?
No. That's the definition of a transform:
This is the definition of an isometric transform. The Lorentz transformation is NOT isometric.
When we talk about the Lorentz Transform and Special Relativity Theory, we're talking not only about the same point in space, but the same point in time, and transforming it from one observer's POV to another.
WRONG!! If you derive the LT correctly you will see that you are talking about TWO different points in space at two different times. What you are claiming is not physically possible and also not mathematically possible since it requires division by zero!
johanfprins
1 / 5 (5) Sep 04, 2014
By using an assumption that is based on dividing by zero YOU ARE dividing by zero to get the result that you derive
This is incorrect.
It is perfectly correct. If you start from an assumption that is derived by dividing by zero, then your whole derivation is obtained by dividing with zero!
Non sequitur
I do not think that you understand this word.

Sorry, I don't see the derivation for (x'-ct')=lambda*(x-ct) ∴ lambda=0/0. Lambda appears to be the Lorentz transform to me.
x'-ct'=0 and x-ct=0; thus lamda=0/0. Dividing zero with zero is ALWAYS meaningless. So how can this be the Lorentz transform?

But it's a transform. Of course they're the same point; that's implicit.
No it is not implicit. In mathematics there are many different transforms of coordinates and only a subset of them are isometric so that the coordinates before the transform and the coordinates after the tranform are coincident. This is not the case for the Lorentz transform.
johanfprins
1 / 5 (5) Sep 04, 2014
If you're trying to transform between different points, you're making an error by using a transform in the first place. Transforms do not translate between different points; they translate the same point into different coordinate systems (such as IRFs)
Only when the transform is isometric which the Lorentz transform IS NOT. I think that you should go for a course in mathematics, since you are obviously out of your league here!

Stop. "Equals" doesn't mean "different."
"Equals " means that the value of the function is the same for x,t and for x',t': It does not prove that x,t and x',t' are coordinates for the same point in so called "space-time". And in fact space-time is not physically possible as you will see when you read section 6 of my book: EINSTEIN=GENIUS: But a genius sometimes blunders. https://www.resea...=prf_act
Da Schneib
5 / 5 (5) Sep 04, 2014
Go through your derivation and note that your initial assumption is that the Lorentz coordinate transformation gives that x,t and x',t' are the coordinates of the same point in 2D "space-time".

Yes, they do.[

Still not quite getting how they can be the same coordinate points... Aren't they describing oppositely?
No. That's the definition of a transform:
This is the definition of an isometric transform. The Lorentz transformation is NOT isometric.
No, it's not. It's a definition of ALL transforms. That's what transforms do, period. If you're claiming something else is a "transform," you're using the noun incorrectly (at least in physics).
Da Schneib
5 / 5 (5) Sep 04, 2014
If you start from an assumption that is derived by dividing by zero, then your whole derivation is obtained by dividing with zero!
Non sequitur
I do not think that you understand this word.
I think my reductio ad absurdum made it clear I do. I think you failed to quote it because it did.

Sorry, I don't see the derivation for (x'-ct')=lambda*(x-ct) ∴ lambda=0/0. Lambda appears to be the Lorentz transform to me.
x'-ct'=0 and x-ct=0; thus lamda=0/0.
But they're not being divided. That's an equals sign, not a division sign. Division signs look like this / or this ÷. Not like this =.

But it's a transform. Of course they're the same point; that's implicit.
No it is not implicit.
Yes, it is. That's the meaning of "transform." If you want to it for something else, you're not talking about physics any more.
Reg Mundy
1 / 5 (4) Sep 04, 2014
@Daz

I say that all (net) force produces acceleration, and that all acceleration is caused by force.
Where do you claim you said, "net" before bringing it into the discussion to cover your error in not saying it in the first place?

You said, "force causes acceleration." You never mentioned "net".

You know what I meant. You are just being perverse, i.e. wasting both our time.
Da Schneib
5 / 5 (3) Sep 04, 2014
If you're trying to transform between different points, you're making an error by using a transform in the first place. Transforms do not translate between different points; they translate the same point into different coordinate systems (such as IRFs)
Only when the transform is isometric which the Lorentz transform IS NOT.
No. It's the definition of ALL transforms, isometric or not. You do not transform points into one another unless you're talking about the same point in different coordinate systems. This is basic physics.

Stop. "Equals" doesn't mean "different."
"Equals" means that the value of the function is the same for x,t and for x',t'
They can't be unless they're the same coordinate system in which case the transform is

x'=x
y'=y
z'=z
t'=t
Reg Mundy
1 / 5 (4) Sep 04, 2014
I say that all (net) force produces acceleration, and that all acceleration is caused by force. You come up with crap about squeezing lumps of steel. Either you are exceedingly stupid, or are wilfully obtuse. In either case, I have lost interest in trying to have a logical discussion with you.

All acceleration caused by force? Well, duh...
That acceleration requires force? Well, duh...
Those were stupidly obvious statements in a single sentence...
All you're trying to do is re-describe and thusly reinvent, the wheel...

I make a statement as a basis for a logical conclusion, and you argue with it and then later say I am re-inventing the wheel, which I assume means you now agree with it. Well, duh...Wanker!
Da Schneib
5 / 5 (4) Sep 04, 2014
@Daz

I say that all (net) force produces acceleration, and that all acceleration is caused by force.
Where do you claim you said, "net" before bringing it into the discussion to cover your error in not saying it in the first place?

You said, "force causes acceleration." You never mentioned "net".

You know what I meant. You are just being perverse, i.e. wasting both our time.
Not in any way shape form or fashion. If you mean net forces cause accelerations, I'll think about that. A photon being absorbed by an electron doesn't necessarily accelerate the electron; if it's in an orbital, it just kicks the electron to a higher energy suborbital or orbital. And only if it's the right frequency of photon to be the exact energy necessary to put it in an exact orbital of that atom.

Now if you make the common error of thinking orbitals are like orbits and electrons like planets, then you might think so, but actually they're more like resonances, like harmonics.
Reg Mundy
1 / 5 (4) Sep 04, 2014
@Daz
Sorry, I don't see the derivation for (x'-ct')=lambda*(x-ct) ∴ lambda=0/0. Lambda appears to be the Lorentz transform to me.

x'-ct'=0 and x-ct=0; thus lamda=0/0.

But they're not being divided. That's an equals sign, not a division sign. Division signs look like this / or this ÷. Not like this =.
By every rule of mathematics I know, x=yz means (x/z)=y
Thus lamda=0/0
You are just being perverse, i.e. wasting both your time, my time, and Johans time.
Da Schneib
5 / 5 (4) Sep 04, 2014
Sorry, I don't see the derivation for (x'-ct')=lambda*(x-ct) ∴ lambda=0/0. Lambda appears to be the Lorentz transform to me.
x'-ct'=0 and x-ct=0; thus lamda=0/0.
But they're not being divided. That's an equals sign, not a division sign. Division signs look like this / or this ÷. Not like this =.
By every rule of mathematics I know, x=yz means (x/z)=y
What's that got to do with anything? I don't see any division at all. There is no division in
(x'-ct')=lambda*(x-ct)
at all. Can you point to where you claim there is division in
(x'-ct')=lambda*(x-ct)
please?
johanfprins
1 / 5 (4) Sep 04, 2014
This is the definition of an isometric transform. The Lorentz transformation is NOT isometric.
No, it's not. It's a definition of ALL transforms. That's what transforms do, period. A tranform is a matrix-operator that connects coordinates. Matrices do not all have inverses and do not necessarily generated coordinates that are coincident with the coordinates that they transform. This is well-known mathematics which any dude who have half a brain understands!
If you're claiming something else is a "transform," you're using the noun incorrectly (at least in physics).
No I am not: YOU are assuming that ALL transforms MUST be isometric so that the coordinates before the transform coincides on a one to one basis with the coordinates after the transformation. This just proves that you you do not know any mathematics or physics. You are just a troll.
Da Schneib
5 / 5 (3) Sep 04, 2014
This is the definition of an isometric transform. The Lorentz transformation is NOT isometric.
No, it's not. It's a definition of ALL transforms. That's what transforms do, period.
A tranform is a matrix-operator that connects coordinates. Matrices do not all have inverses
But transforms between coordinate systems in real space and time do.
johanfprins
1 / 5 (4) Sep 04, 2014
I think my reductio ad absurdum made it clear I do. I think you failed to quote it because it did.
You are far too stupid to understand what reductio ad absurdum means: Why should I quote bullshit?

Sorry, I don't see the derivation for (x'-ct')=lambda*(x-ct) ∴ lambda=0/0. Lambda appears to be the Lorentz transform to me.
x'-ct'=0 and x-ct=0; thus lamda=0/0.
But they're not being divided. That's an equals sign, not a division sign. Division signs look like this / or this ÷. Not like this =.
Another proof that you do not even understand KinderGarten arithmatic. Let us just use = since you are too stupid. Then the equation states that 0=lambda*0 SEE THE = SIGN?). This equation is valid FOR ANY value of lambda, and therefore lambda cannot be a specific transformation EVER!!

No it is not implicit.
Yes, it is. That's the meaning of "transform." No it is not!!! Please stop displaying your ignorance: TROLL!
Da Schneib
5 / 5 (4) Sep 04, 2014
I think my reductio ad absurdum made it clear I do. I think you failed to quote it because it did.
You [don't] understand what reductio ad absurdum means
Sure I do; it means I take your claims and prove that they lead to an absurd conclusion.

Sorry, I don't see the derivation for (x'-ct')=lambda*(x-ct) ∴ lambda=0/0. Lambda appears to be the Lorentz transform to me.
x'-ct'=0 and x-ct=0; thus lamda=0/0.
But they're not being divided. That's an equals sign, not a division sign. Division signs look like this / or this ÷. Not like this =.
...you do not even understand KinderGarten arithmatic.
Sure I do. I know what = means. I can also spell arithmetic properly.
johanfprins
1 / 5 (4) Sep 04, 2014
This is the definition of an isometric transform. The Lorentz transformation is NOT isometric.
No, it's not. It's a definition of ALL transforms. That's what transforms do, period.
A transform is a matrix-operator that connects coordinates. Matrices do not all have inverses
But transforms between coordinate systems in real space and time do.


Only when it is not a relativistic coordinate transformation (RCT): A RCT transforms the PHYSICS that is occurring within a specific inertial reference frame into another reference frame: It gives you the altered physics you will see when looking INTO an IRF in which the physics is ACTUALLY occurring. Only a FOOL like you will conclude that this RCT physics can be transformed back into the coordinate system in which it is actually occurring.
Da Schneib
5 / 5 (4) Sep 04, 2014
This is the definition of an isometric transform.
The Lorentz transformation is NOT isometric.
No, it's not.
You are misquoting me. I will not respond to a post that contains a lie about what I said.

Sorry about that. It's my policy.
johanfprins
1 / 5 (5) Sep 04, 2014
This is the definition of an isometric transform.
The Lorentz transformation is NOT isometric.
No, it's not.
You are misquoting me. I will not respond to a post that contains a lie about what I said.

Sorry about that. It's my policy.

I hope that you will keep to your promise not to respond by posting the usual bullshit that only YOU are capable of posting! Good riddance! TROLL!
Da Schneib
5 / 5 (3) Sep 04, 2014
You still haven't shown where the dividing by zero happens, johan buddy.

I don't think it happens at all. I think you're making it up, just like you're making up quotes and putting them in my mouth.
johanfprins
1 / 5 (4) Sep 04, 2014
You still haven't shown where the dividing by zero happens, johan buddy.

I don't think it happens at all. I think you're making it up, just like you're making up quotes and putting them in my mouth.


YOU are the biggest IDIOT on this earth. If you have an equation that x=lambda*y, then as any kid in grade 2 knows, you obtain that lambda=x/y: Thus you obtain lambda by DIVIDING you KLUTZ!!! Thus if x=0 and y=0, as Einstein admitted in his book, then you MUST have that lambda=0/0: Thus you are dividing by ZERO. Lambda cannot be the Lorentz transform EVER! So please F-OFF!
thefurlong
5 / 5 (6) Sep 04, 2014
@johanfprins
[The conclusion that they do, follows by dividing by zero. Thus to accept "that they do" you start off of by incorporating this division in your derivation: Thus you are dividing by zero!!

Actually, no. That's like saying "You used money that was obtained from selling drugs to buy a television. Therefore, you sold drugs." One does not follow from the other, because you could have obtained the money from someone else who sold drugs.

If I use an assumption, it means that a-priori, it is taken to be true. In fact, proof by contradiction ALWAYS starts out by assuming the thing to be proven false as true. If the reasoning is correct (for example, no divisions by 0 occur), then it will produce the desired contradiction. We can get to whether the assumption itself is true or not, but accusing me of dividing by 0 is just not correct. Now, is my reasoning correct or not, or did I actually divide by 0?
thefurlong
5 / 5 (6) Sep 04, 2014
What two expressions?
x-ct=0 and x'-ct'=0.

Thank you.
Einstein's book: RELATIVITY: The Special and General Theory: p. 226:. He wrote the two equations above (x'-ct')=lambda*(x-ct). This means that lambda=0/0

No. That doesn't mean that lambda = 0/0. I have a very trivial counter example for you.

2*0 = 0
3*0 = 0

But by your logic, 2 = 0/0 = 3. But, obviously, 2 != 3, and yet, it is correct for me to write 2*0 = 0 and 3*0 = 0.
Did Einstein, afterwards, use lambda*(x - ct)/(x' - ct') ?
If he did, then there might be a problem.

However, I would like to point out that, mathematically, 0/0 is very different from division by 0. It depends on the context. Division by 0 is where you take a non-zero number and divide it by 0. This is never valid.

However, 0/0 depends on the context. For example, in the case of derivatives, dy and dx can always be considered to have a 0 value, unless you take dy/dx, which is just differentiation.
thefurlong
5 / 5 (6) Sep 04, 2014
@Reg Mundy
By every rule of mathematics I know, x=yz means (x/z)=y

That's not true at all. As I said to Johan,
0 = 0*2.
0 = 0*3.

But that doesn't mean that 2=0/0 and 3 = 0/0. That would mean that 2 = 3, which is incorrect. In fact, 0 has no multiplicative inverse for this very reason.
Thus lamda=0/0

If this is the only reason you and Johan have for thinking that Einstein divided by 0, then you can stop griping about it. It is perfectly reasonable to write x+y=z+w when x+y=0 and z+w=0. I'll show you:
2 + -2 = 0
3 + -3 = 0

So, 2+-2 = 3+-3
Hence, -1 = -1
AND
1 = 1
See! There is absolutely no contradiction, and we arrived at not just one, but two correct conclusions.
Reg Mundy
1 / 5 (4) Sep 04, 2014
@Daz
Sorry, I don't see the derivation for (x'-ct')=lambda*(x-ct) ∴ lambda=0/0. Lambda appears to be the Lorentz transform to me.
x'-ct'=0 and x-ct=0; thus lamda=0/0.
But they're not being divided. That's an equals sign, not a division sign. Division signs look like this / or this ÷. Not like this =.
By every rule of mathematics I know, x=yz means (x/z)=y
What's that got to do with anything? I don't see any division at all. There is no division in
(x'-ct')=lambda*(x-ct)
at all. Can you point to where you claim there is division in
(x'-ct')=lambda*(x-ct)
please?

Lets write (x'-ct')=lambda*(x-ct) as (x'-ct')/(x-ct)=lambda shall we? I don't think that is too big a strain on your arithmetical ability, though I might be wrong.....
Then when both (x'-ct')=0 and (x-ct)=0, 0/0=lambda
Can you see a division there? Oh, you can't.......
See what I mean about you being deliberately obtuse? Now get back to your more sensible questions, or shut up.
Whydening Gyre
5 / 5 (4) Sep 04, 2014
I say that all (net) force produces acceleration, and that all acceleration is caused by force. You come up with crap about squeezing lumps of steel. Either you are exceedingly stupid, or are wilfully obtuse. In either case, I have lost interest in trying to have a logical discussion with you.

All acceleration caused by force? Well, duh...
That acceleration requires force? Well, duh...
Those were stupidly obvious statements in a single sentence...
All you're trying to do is re-describe and thusly reinvent, the wheel...

I make a statement as a basis for a logical conclusion, and you argue with it and then later say I am re-inventing the wheel, which I assume means you now agree with it. Well, duh...Wanker!

Didn't say you WERE re-inventing the wheel, just TRYING...
johanfprins
1 / 5 (4) Sep 04, 2014
That's like saying "You used money that was obtained from selling drugs to buy a television. Therefore, you sold drugs."
Well somebody must have sold drugs. So it is true that you used drug-money to buy a TV.
One does not follow from the other, because you could have obtained the money from someone else who sold drugs.
So what? It is still drug money!

If I use an assumption, it means that a-priori, it is taken to be true.
Well this is like making the a priori assumption that the money you used has not been obtained from selling drugs. It does not change the fact that the money has been obtained by selling drugs.

In fact, proof by contradiction ALWAYS starts out by assuming the thing to be proven false as true.
And in this case you seem to get a nice TV as if you have used clean money; but you did not; since x,t and x',t' are NOT the coordinates for the SAME point in a 2D manifold. Einstein reached this conclusion by deriving by ZERO.
continued
johanfprins
1 / 5 (4) Sep 04, 2014
Thus by starting from an impossible result which has been obtained by dividing by zero, you incorporate dividing by zero in your derivation and thus ends up with the wrong conclusion: You did not prove invariance, since such a proof requires that x,t and x',t' must be coincident coordinates; which they are NOT.

If the reasoning is correct (for example, no divisions by 0 occur), then it will produce the desired contradiction.
Not in this case since you start off by assuming that a contradiction obtained by dividing by zero is correct, and then you get a result which is still an impossible absurd contradition. You have NOT proved any invariance as you are claiming since for this to be true x,t and x',t' must be the coordinates for the same point which they are NOT. If you derive the LT by NOT dividing by zero, the result is that x,t and x',t' are two different points in space and two different times on ALL the synchronized clocks in the Universe.

johanfprins
1 / 5 (2) Sep 04, 2014
No. That doesn't mean that lambda = 0/0. I have a very trivial counter example for you.

2*0 = 0
3*0 = 0

But by your logic, 2 = 0/0 = 3. But, obviously, 2 != 3, and yet, it is correct for me to write 2*0 = 0 and 3*0 = 0.
Exactly: This is why only an idiot will multiply zero with a number and then believe that you can obtain this number by dividing zero into zero. This is what Einstein has done. Nobody in his right mind will equate two expressions which are both zero UNLESS you can prove that you have two numbers, like the ratio of two differentials which approach zero in the LIMIT. In this case this is NOT SO!!

Did Einstein, afterwards, use lambda*(x - ct)/(x' - ct') ?
If he did, then there might be a problem.
YES!!!! This is what he did and what has been done for over 100 years.

It depends on the context. Division by 0 is where you take a non-zero number and divide it by 0. This is never valid.
Neither is taking 0 and dividing by zero unless...... contin
johanfprins
1 / 5 (4) Sep 04, 2014
you have two non-zero expressions which you divide and can prove that as these numbers simultaneously approachees zero, you obtain a definite result in the limit. This has not been proved by Einstein or anybody else EVER when deriving the Lorentz equation incorrectlty, as Einstein has done.

However, 0/0 depends on the context. For example, in the case of derivatives, dy and dx can always be considered to have a 0 value, unless you take dy/dx, which is just differentiation.
Please, I obviously know mathematics far better than you will ever know it. So do not try and reach me mathematics; since you are incapable to even understand that coordinates must be linearly independent in order to span a manifold!
thefurlong
5 / 5 (4) Sep 04, 2014
That's like saying "You used money that was obtained from selling drugs to buy a television. Therefore, you sold drugs."
Well somebody must have sold drugs. So it is true that you used drug-money to buy a TV.
One does not follow from the other, because you could have obtained the money from someone else who sold drugs.
So what? It is still drug money!

No, you misunderstand what I am saying. Let's leave emotions and morality out of this. I am using logic.
If you buy a tv with money that was used to buy drugs, it is incorrect to say that the person who bought the tv sold drugs to do so. It does not logically follow.

The transaction to buy a tv was legal. The transaction to sell drugs wasn't. Hence the person who used money from selling drugs to buy a tv was not doing anything illegal.

Indeed, even if I use an initially incorrect assumption, it does not mean my conclusion or proof is incorrect. (to be continued)
johanfprins
1 / 5 (3) Sep 04, 2014
@Reg Mundy
By every rule of mathematics I know, x=yz means (x/z)=y

That's not true at all. As I said to Johan,
0 = 0*2.
0 = 0*3.

ETC. ETC. ETC.


You are grabbing at straws. Einstein and NOBODY after him has proved that (x-ct)=0 and x'-ct'=0, are limiting expressions for two NON-ZERO expressions. Thus, by writing that (x-ct)=lambda*(x'-ct') you are dividing by zero. The fact that Einstein fudged his way through to get the Lorentz equations have been fortuitous. It only seems to work since Einsteuin believed the absurdity that two synchronized clocks can keep different time rates when they move relative to one another. In fact he a priori assumed time dilation to occur. The latter impossibility which violates the principle of relativity, made the division by zero seem palatable. It is NOT palatable: It is absurd and WRONG.

I have treated ALL these issues in detail in my book.
thefurlong
5 / 5 (4) Sep 04, 2014
(continued)
For example, suppose I started with faulty assumption that 8 is a prime.

Then, I said therefore, (8 mod 3) != 0, and therefore
(8 mod 3)*10 != 0, then my proof would be correct. This is because it doesn't rely on the part that was a contradiction. The incorrect part is assuming 8 is a prime to begin with, but there was no mistake made afterwards.
johanfprins
1 / 5 (3) Sep 04, 2014
I am using logic.
No you are not.
If you buy a tv with money that was used to buy drugs, it is incorrect to say that the person who bought the tv sold drugs to do so.
I did not state that that person himself sold drugs: I am saying that he used drug money to buy a TV.

The transaction to buy a tv was legal. The transaction to sell drugs wasn't. Hence the person who used money from selling drugs to buy a tv was not doing anything illegal.
WRONG. If you buy an item with illegal money you are guitly of a crime, since the onus is on YOU to check that the money you spend has been obtained legally.

Indeed, even if I use an initially incorrect assumption, it does not mean my conclusion or proof is incorrect. (to be continued)
You start of with the wrong assumption and threfore end with a conclusion which seems correct but is wrong. It is similar to epicycles, which are impossible, but did give the motion of the planets as observed from earth.
thefurlong
5 / 5 (5) Sep 04, 2014
@johanfprins
Did Einstein, afterwards, use lambda*(x - ct)/(x' - ct') ?
If he did, then there might be a problem.
YES!!!! This is what he did and what has been done for over 100 years.

Where?

That's not true at all. As I said to Johan,
0 = 0*2.
0 = 0*3.

ETC. ETC. ETC.


You are grabbing at straws. Einstein and NOBODY after him has proved that (x-ct)=0 and x'-ct'=0, are limiting expressions for two NON-ZERO expressions. Thus, by writing that (x-ct)=lambda*(x'-ct') you are dividing by zero

You don't seem to have understood my point.
Do you acknowledge that
2*0 = 0, and 3*0 = 0?
And further, that
2*0 = 3*0?
Have you divided by 0 in either case?
If not, then why would you continue to say that setting x - ct = lambda*(x'-ct') is equivalent to dividing by 0? That makes no sense.

The only way he would have divided by 0 is if he then actually divided through by either the LHS or the RHS. Please show me where he did this.
Whydening Gyre
5 / 5 (3) Sep 04, 2014
You guys are just doing this to mess with my head first thing in the morning, aren't ya...?
johanfprins
1 / 5 (3) Sep 04, 2014
(continued)
For example, suppose I started with faulty assumption that 8 is a prime.

Then, I said therefore, (8 mod 3) != 0, and therefore
(8 mod 3)*10 != 0, then my proof would be correct. This is because it doesn't rely on the part that was a contradiction. The incorrect part is assuming 8 is a prime to begin with, but there was no mistake made afterwards.


There was a mistake made afterwards since you came to the conclusion that you have proved an invariance; which you have not done since such an invariance demands that x,y and x',t'
must be coincident coordinates: WHICH THEY ARE NOT!

If you start from an absurdity which you belive is not an absurdity and carry it logically through youust end up with an absurdity. Which is exactly where you ended up: But you want to believe that the latter absurdity is not an absurdity. So how can I help you if you embrace absurdity as reality?
thefurlong
5 / 5 (3) Sep 04, 2014
The transaction to buy a tv was legal. The transaction to sell drugs wasn't. Hence the person who used money from selling drugs to buy a tv was not doing anything illegal.

WRONG. If you buy an item with illegal money you are guitly of a crime, since the onus is on YOU to check that the money you spend has been obtained legally.

I can't find anything on this after searching for 10 minutes. Can you provide me with an actual source that confirms that it is illegal to buy a tv with money used from drugs that you obtained from someone else?
thefurlong
5 / 5 (4) Sep 04, 2014
There was a mistake made afterwards since you came to the conclusion that you have proved an invariance; which you have not done since such an invariance demands that x,y and x',t'

Not at all. Here. Let me define an arbitrary transformation. Take the position x and map it to -x. So, T(x) = -x

Now, I want f(T(x)) = f(x)
So
f(-x) = f(x)
Therefore, if f is symmetric, then f is an invariant of T.
Therefore,
f(x) = x^2 is an invariant of T(x) = -x.

This says nothing about where T came from. I just assume it exists.
Think of me using the Lorentz transform in that way. It doesn't matter where it came from--at least not in the context of my derivation.