Masses of common quarks are revealed

May 03, 2010 By Anne Ju
Quarks exist in a soup of other quarks, antiquarks and gluons within a proton or neutron. Determining their mass has been difficult due to the strong force that binds them together. (Christine Davies/University of Glasgow)

(PhysOrg.com) -- A research group co-founded by Cornell physics professor G. Peter Lepage has calculated the mass of the three lightest and, therefore, most elusive quarks: up, down and strange.

Quarks, the that make up protons and neutrons, have been notoriously difficult to nail down -- much less weigh -- until now. A research group co-founded by Cornell physics professor G. Peter Lepage has calculated, with a razor-thin margin of error, the mass of the three lightest and, therefore, most elusive quarks: up, down and strange.

The work of Lepage, the Harold Tanner Dean of the College of Arts and Sciences, and collaborators from several international institutions, is published online (March 31) and in print in (Vol. 104:13).

The findings reduce the uncertainty of the masses by 10 to 20 times down to a few percent. Scientists have known the mass of a proton for almost a century, but getting the mass of the individual quarks inside has been an ongoing challenge. The quarks are held together by the so-called strong force -- so powerful that it's impossible to separate and study them. They exist in a soup of other quarks, antiquarks and gluons, which are another type of particle.

To determine the quark masses, Lepage explained, it was necessary to fully understand the . They tackled the problem with large supercomputers that allowed them to simulate the behavior of quarks and gluons inside such particles as protons.

Quarks have an astonishingly wide range of masses. The lightest is the up quark, which is 470 times lighter than a proton. The heaviest, the t quark, is 180 times heavier than a -- or almost as heavy as an entire atom of lead.

"So why these huge ratios between masses? This is one of the big mysteries in right now," Lepage said. "Indeed it is unclear why quarks have mass at all." He added that the new in Geneva was built to address this question.

According to their results, the up quark weighs approximately 2 mega electron volts (MeV), which is a unit of energy, the down quark weighs approximately 4.8 MeV, and the strange quark weighs in at about 92 MeV.

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User comments : 161

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Alizee
May 03, 2010
This comment has been removed by a moderator.
hard2grep
1.8 / 5 (15) May 03, 2010
Of course they have mass. pretty much everything has mass; if you can name it, it should have mass. If you can't name it, then it does not exist. How would a black hole be possible unless this is true.
If there are particles not affected by gravity floating around, would we not be looking at a large discrepancy??? hmm...
shavera
4.3 / 5 (8) May 03, 2010
hard2grep: photons and gluons are massless.
Alizee
May 03, 2010
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Alizee
May 03, 2010
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Aliensarethere
not rated yet May 03, 2010
What are you guys (Alizee and hard2grep) ? Theoretical physics professors ? :-)
Alizee
May 03, 2010
This comment has been removed by a moderator.
Aliensarethere
3.3 / 5 (6) May 03, 2010
Special relativity does say that a particle with non-zero rest mass can never reach the speed of light. So light can't have mass. Einstein described the duality of light in 1905, which he got the Nobel prize for in 1919
TheWalrus
2.3 / 5 (11) May 03, 2010
Blah blah blah if photons have no mass how can they exist yadda yadda yadda...

I love it when high school drop-outs with no background in physics try to explain what the real scientists should have said.
Alizee
May 03, 2010
This comment has been removed by a moderator.
Question
1 / 5 (2) May 03, 2010
TheWalrus: does energy have mass?
Why couldn't EMR have a rest mass but no mass when traveling at the speed of light?
Alizee
May 03, 2010
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Alizee
May 03, 2010
This comment has been removed by a moderator.
Alizee
May 03, 2010
This comment has been removed by a moderator.
Bloodoflamb
5 / 5 (7) May 03, 2010
The most commonly quoted upper limit on the photon mass is ~10^-14eV (~10^-50kg), though other considerations place an upper limit of 3*10^-27eV (10^-63kg).

Please also note that there is a difference between rest mass and effective mass. Rest mass is the mass (energy) that an object takes on when it is not moving. Effective mass is E/c^2=[m^2+(p/c)^2]^(1/2). Photons, as far as we can tell have energy E=pc, and if you have a non-moving photon, you don't actually have a photon.
brentrobot
1 / 5 (3) May 03, 2010
The question you should be asking about photons and mass is whether or not photons can have a gravitational effect on matter?

For example if you have two one kilo mirrors separated by one meter, and a bunch of photons bouncing between them with an energy equivalent to one kilo of matter, how much does the whole thing weigh when put on a scale?

2 kilos or 3 kilos?

Alternatively if you have a super laser emit 1 kilogram worth of photons, your laser just decreased in mass by one kilo, resulting in a change in the curvature of spacetime.

Did the laser just emit a buttload of gravitons along with the photons? Or did the photons emit gravitons as they were escaping the laser? Did a strong gravity wave just leave the laser along with the light? Or is the "gravity wave" trapped inside the photons?

Maybe one of you guys can give me an adequate explanation.
Dr_McFoo
4.2 / 5 (5) May 04, 2010
Einstein didn't get the Nobel prize for relativity. He got it for the photoelectric effect
daywalk3r
3 / 5 (18) May 04, 2010
All this talk about "mass", though obviously none of the talkers has a clue, what it really represents..

Mass can be looked upon as "energy density". And therefor - anything that has energy potential, has a "mass potential" aswell.

A photon is an energy carrier (eg. has energy potential), so in this regard, it has to have "mass", sort of..

A gluon is a "virtual particle", virtualized (postulated) to fill a gap. So wether it has mass or not, is - at least for now - just subject of mere speculation.

This should answer a couple of the questions posted above aswell :)
Bloodoflamb
4.2 / 5 (5) May 04, 2010
The question you should be asking about photons and mass is whether or not photons can have a gravitational effect on matter?
Yes. Photon energy is part of the gravitational energy-stress tensor, and induces gravitational effects on matter as well as on other photons.
All this talk about "mass", though obviously none of the talkers has a clue, what it really represents..
Err... I clearly went to lengths to differentiate between rest mass (which photons do not have) and effective mass (m=E/c^2).
daywalk3r
2.8 / 5 (18) May 04, 2010
If you excite atom nuclei by X-ray photon, then the excited atom would become heavier by E=mc^2 equation, which can be detected for example by atomic mass spectrometry. This could serve as an experimental evidence, photon really transfers mass (i.e. matter) - not just momentum.
Only the part of the photon(s) which gets "captured", eg. not reflected/radiated, etc. :)
Bloodoflamb
3.7 / 5 (6) May 04, 2010
Only the part of the photon(s) which gets "captured", eg. not reflected/radiated, etc. :)
This is an obtuse way of putting it. Photons are, by their very definition, quantized with respect to their interaction with charged particles. A photon isn't made up of 'parts,' so to speak, and can't get 'partially absorbed.'
daywalk3r
2.7 / 5 (19) May 04, 2010
Err... I clearly went to lengths to differentiate between rest mass (which photons do not have) and effective mass (m=E/c^2).
Yea yea.. You are clearly one of the "brighter ones" in here :) Though, you are partially wrong with that statement aswell :-P

Even though photons do not directly exhibit rest mass, they still have what I would call a "rest-mass potential". As soon as they get captured (by space-curvature, eg. black hole), this potential will automaticaly and directly contribute to the rest mass of the captor.

And apart from that, I don't really like definitions based on the "planck stuff", as they tend to (quite literally) "break apart" at small enough scales :-P
daywalk3r
2.8 / 5 (20) May 04, 2010
A photon isn't made up of 'parts,' so to speak, and can't get 'partially absorbed.'
That's the planck stuff I was talking about above ^^ Funny :)

Photons are, by their very definition, quantized with respect to their interaction with charged particles.
Guess we both would be quite surprised, how far that "very definition" is, from what the very definition will be in a couple of decades/centuries :)

But even this definition has allready some very truth to it - it means, that if you have an infinitesimally small "charged particle", then you would have infinitesimally small photons/quanta as well :)
plokolp
5 / 5 (1) May 04, 2010
Zero rest mass of photon is very speculative. Special relativity has nothing to say about it, because it doesn't recognize/predict photons at all - photon is concept of quantum mechanics.


Of course special relativity cannot predict it's postulates : http://en.wikiped...ativity, that the speed of light is absolute.
Also, since there is no rest frame of the photon, you cannot define a rest mass in the same sense.

The photon is not just a concept of quantum mechanics. The photon concept is a result of the combination of the classical equations of electrodynamics with quantization. Also, with just classical electrodynamics, you CAN show that the statistics of light trapped in a box is the same as that of trapped particles of some average energy.

Speed of harmonic light waves shouldn't be confused with speed of photons


A harmonic wave is equivalent to a delocalized photon.
they both obey maxwell's equations and hence have the same speed
Parsec
4 / 5 (4) May 04, 2010
The experimental evidence for non-zero rest mass of photons is quite extensive today. For example, if photons would be completely massless, they couldn't interact mutually like particles. But from high-power laser experiments follows, the crossection of mutual photon-photon interactions is nonzero.

http://adsabs.har...96h3602L

The standard model demands that carriers of forces with infinite range have a zero rest mass. This includes photons( carriers of the the electromagnetic force), and gravitons. While its possible that the standard model is incorrect, it has been experimentally verified to so many decimal places, its pretty unlikely in such a fundamental assumption.
frajo
5 / 5 (1) May 04, 2010
Simmillia simillibus observatur
Which textbooks have you been reading? Not only is your understanding of physics a Greek catastrophe but also your Latin. "Similia similibus curentur" (yes, it's a Latin plural) is the credo/slogan of homeopathy, i.e. of pseudoscience, quackery, and "cruel deception".
http://en.wikiped...meopathy
frajo
5 / 5 (1) May 04, 2010
What are you guys (Alizee and hard2grep) ? Theoretical physics professors ? :-)
Obviously one guy with several accounts - formerly seneca/broglia, now shockr/hard2grep. Have a look at those who permanently rate him high.
Titto
1 / 5 (10) May 04, 2010
The only scientists I do respect is the ones that believe in Christ!!!
johanfprins
1 / 5 (3) May 04, 2010
A light wave can only have mass when it has inertia. Inertia means that there must be an inertial reference frame within which a light wave is stationary. Thus, the stationary light waves within a cavity (black-body radiation) all add mass to the cavity: This mass cannot come from "photon-particles" flying around with speed c within the cavity, since such entities do not have inertia.

When an atomic electron absorbs a "photon-wave", this wave collapses and its speed becomes zero relative to the inerial refrence frame of the atom. The "photon-wave" thus entangles and add mass to the original electron wave, which then has to morph into a higher energy wave. This morphing has erroneously been called a quantum jump.

This mechanism also happens when a light pulse is stopped within Bose-Einstein Condensate; and can only happen when the matter wave is a single holistic wave that is in instantaneous contact with itself over its whole intensity distribution in three-dimensional space.
ZeroX
2 / 5 (4) May 04, 2010
is the credo/slogan of homeopathy
The fact, something follows from Google in the same query doesn't mean, it's the same thing. It's different slogan, which has nothing to do with homeopathy. For example alchemists used the slogan "Similia similibus solvuntur", which is used in organic chemistry up to this day.
ZeroX
1 / 5 (2) May 04, 2010
frajo
5 / 5 (1) May 04, 2010
The Standard Model has nothing to do with such difference, as it deals with much higher energies inside of atom nuclei, for which some subtle difference in photon rest mass is completelly neglibible (QCD QED).
"ZeroX" stands for zero-day account, I assume? Obviously, you shared your physics (and English) textbooks with the seneca-broglia complex.
On topic: No, the standard model of particle physics does not deal exclusively with higher energies.

P.S.: Well, he just deleted his comment. :)
srikkanth_kn
1 / 5 (3) May 04, 2010
So many comments posted, I have a hard time even to scroll down the page. Guys, We are yet to perform and analyse the experiments proposed in LHC. Wait for the results. Please. Higgs Boson, the mass giving god particle is yet to reveal itself.
ZeroX
1 / 5 (4) May 04, 2010
..the photon is not just a concept of quantum mechanics. The photon concept is a result of the combination of the classical equations of electrodynamics with quantization...
Well, and the quantization is just a concept of quantum mechanics... again. The photon concept doesn't require classical electrodynamics at all, it can be derived solely from quantum mechanics as a particle mediating quanta of energy.
A harmonic wave is equivalent to a delocalized photon
Delocalized photon is an oxymoron, as such photon would fill whole observable universe. And vice-versa: when wave is confined into limited space, it cannot be harmonic anymore. You're just confusing well understood concepts.
ZeroX
1 / 5 (2) May 04, 2010
..the standard model of particle physics does not deal exclusively with higher energies..
I didn't say, it deals EXCLUSIVELLY with high energies, I just said, the rest mass of photon at range 10-62 kg cannot affect Standard Model predictions in measurable way, so such assumption cannot be interpreted as a fundamental flaw of SM. Such rest mass could affect the photon behavior at cosmic scales, for example.
ZeroX
1 / 5 (2) May 04, 2010
In adition, the estimation of the rest mass of photon would depend on the fact, whether you're assuming vacuum as a flat 4D space-time or space-time filled by CMB noise (which is effectivelly slightly higher-dimensional). Such curved space-time has non-zero mass/energy density, so that even the photon of rest mass in 10E-17 kg range would spread through it as effectivelly massless particle.
Neodim
1 / 5 (1) May 04, 2010
The free quark exists?
ZeroX
1 / 5 (2) May 04, 2010
Only the heaviest ones and they're unstable - the lifetime of top quark is only in 5E-25 second range.

http://www.scienc...1725.htm
johanfprins
1 / 5 (4) May 04, 2010
srikkanth kn:
Higgs Boson, the mass giving god particle is yet to reveal itself.
It does not exist but they will probably find it after $20 billion, just as they found the vector bosons.
ZeroX:
Delocalized photon is an oxymoron, as such photon would fill whole observable universe. And vice-versa: when wave is confined into limited space, it cannot be harmonic anymore. You're just confusing well understood concepts.

A harmonic wave's momentum has no uncertainty so that p=k. This is true of all "photons" no matter what their length seems to be to an observer. It is possible because a "photon" has no inertia. Therefore it is in effect infintely long even though it can collapse to seem smaller within the inertial refrence frame of the observer.
Neodim:
The free quark exists?
Not even a confined quark exists. Just like the Higg's boson does not exist. All these "particles" are derived by using nonsense like order parameters and spontaneous symmetry breaking!
johanfprins
1 / 5 (3) May 04, 2010
A harmonic wave's momentum has no uncertainty so that p=k.

Correction: I meant a COHERENT harmonic wave
hush1
1 / 5 (3) May 04, 2010
No infinite set is compact. If closed and bounded sets (intervals)in a space work for the non-communicative Standard Theory of Particle Physics community, more power to them. :)
baudrunner
1 / 5 (3) May 04, 2010
Particles that have a lot of mass have a lot of motion at very high velocities. Interesting manifestation of the paradox principle. Very little motion means very stable configuration. Very much high-velocity motion means a great struggle for survival. Less presence - more stable. More presence - less stable.

That works for the higher order elements in the periodic table as well. The most massive elements have the shortest shelf-life.
johanfprins
1.6 / 5 (7) May 04, 2010
Particles that have a lot of mass have a lot of motion at very high velocities. Interesting manifestation of the paradox principle. Very little motion means very stable configuration. Very much high-velocity motion means a great struggle for survival. Less presence - more stable. More presence - less stable.

Please stop confusing mass with REST MASS. When talking about the mass of a "particle" it is its rest mass. Any mass above that is a relativistic effect relative to an observer who is not moving with the "particle". It is a kind of REAL illusion; just like the slowing of a clock which does not REALLY happen within the refrence frame within which the clock is actually stationary.
Mr_Man
May 04, 2010
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Skeptic_Heretic
4 / 5 (5) May 04, 2010
What are you guys (Alizee and hard2grep) ? Theoretical physics professors ? :-)

No, it's one guy/girl with access to a lot of physics papers and a large clipboard by which to copy and paste excerpts that it thinks agree with it's philosophical view of the universe.

Wanna see a trick, ask it to do math.
Alizee
May 04, 2010
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Alizee
May 04, 2010
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johanfprins
1.5 / 5 (8) May 05, 2010
Frankly, what these mainstream physicists could really get from introduction of some Theory Of Everything? Most of them would only risk their jobs. This is why they're adhering to their math models so much - it keeps them separated them from the feedback of publicity in similar way, like the usage of Latin words between doctors and lawyers.

Amen. The theoretical physics church-sect has since 1927 become worse than any fundamentalist religion can EVER hope to be. The time has come to clean the rot!
ZeroX
1.8 / 5 (5) May 05, 2010
Well, mainstream opponents should reconcile their theories first - or there will be any objective reason for change of situation.
johanfprins
1 / 5 (3) May 05, 2010
Well, mainstream opponents should reconcile their theories first - or there will be any objective reason for change of situation.

I have done so for 10 years and reconciled my theories to the point of being self-consistent (see my website) but the "practioners" block everything that is not mainstream. They have not learned what Einstein meant when he reponded to the book "100 authorities against Einstein" by saying "why 100 authorities when a single small fact should have been sufficient!"

Try and get these knuckleheads to discuss the single small fact that according to Born's interpretation of the intensity of a wave function the most "probable position" to find an electron is usually where the wave function has zero intensity! No ways will they even consider the implications of such a single small fact!

So stop spouting claptrap XeroX
Bloodoflamb
1 / 5 (1) May 05, 2010
I have done so for 10 years and reconciled my theories to the point of being self-consistent (see my website) but the "practioners" block everything that is not mainstream. ...

Try and get these knuckleheads to discuss the single small fact that according to Born's interpretation of the intensity of a wave function the most "probable position" to find an electron is usually where the wave function has zero intensity!
? One can construct wavefunctions without ever bringing them into a spatial representation. And this is no less legitimate a way of doing physics so long as one can still make predictions that are testable. What a thing `is, in and of itself,' is not the fundamental question in physics. The fundamental question is how to create the most complete and objective (in the sense of being quantitative) description of the universe. How one interprets those descriptions is a matter of preference.

Though I would like to see your website, if you want to post it or PM a link.
Alizee
May 05, 2010
This comment has been removed by a moderator.
johanfprins
1 / 5 (3) May 05, 2010
Really? Can you demonstrate it?

It is simple to do it by yourself. The simplest calculation is to derive standing Schroedinger waves in a box. The lowest energy one (n=1) has a maximum which corresponds to the most probable position (mpp) of the electron. But when calculating the mpp for n=2 you will find it is at a node where the intensity distribution is zero. This is so for all the solutions for which n is even. Another interesting calculation is to calculate the mpp for n=3. It corresponds to a maximum: BUT there are now THREE maxima with the same intensity: So why should only one of them be the mpp? It is of course a mockery to derive Heisenberg's "uncertainty" for position around a mpp of zero or one for which there are equal probabilities at other positions. Why this simple fact has not been picked up over the past 80 years astonishes me.

What mortifies me is that I have accepted it for many years myself!
johanfprins
1 / 5 (3) May 05, 2010
The fundamental question is how to create the most complete and objective (in the sense of being quantitative) description of the universe. How one interprets those descriptions is a matter of preference.

Unfortunately this is the inheritance the Copenhagenists left us, with which I disagree fully. Why would you follow this route if you can interpret your description of the universe to be comaptible with cause and effect and classical mechanics?

Though I would like to see your website, if you want to post it or PM a link.

A clue: Note that my e-mail is at cathodixx (note the double x, which means two surfaces)
Alizee
May 05, 2010
This comment has been removed by a moderator.
smd
1 / 5 (1) May 05, 2010
I'm not a physicist so please don't ridicule me for asking the following question.

The article states that the quarks measured were those comprising protons. At the same time, it states (third paragraph from the end) that "Quarks have an astonishingly wide range of masses. The lightest is the up quark, which is 470 times lighter than a proton. The heaviest, the t quark, is 180 times heavier than a proton -- or almost as heavy as an entire atom of lead."

My question is therefore how a quark can have a greater mass than a proton, much less an entire atom of a heavy element.
Alizee
May 05, 2010
This comment has been removed by a moderator.
Alizee
May 05, 2010
This comment has been removed by a moderator.
johanfprins
1 / 5 (4) May 05, 2010
It doesn't - in quantum mechanics all three maxima together have the same probability of particle - from this the various shapes of orbital lobes follow. The same result follows from time independent Schroedinger equation, which you told us, you're respecting - or not?

Yes I respect the concept of a time-independent matter-wave but not three or more "uncertainties" per wave. It violates Heisenberg's original postulate. Neither do I respect an "uncertainty" in position around a position at which the wave has ZERO intensity. All harmonic waves have their intensities equal to their energies. So why should a matter-wave be different? The energy of a matter-wave is its mass and its "most probable position" is OBVIOUSLY ITS CENTRE OF MASS.

Please can you for once use your own brain and not post irrelevant references to websites! Or is this just too much for you?

johanfprins
1 / 5 (3) May 05, 2010
My question is therefore how a quark can have a greater mass than a proton, much less an entire atom of a heavy element.

Congratulations! This is a brilliant question and further proof that we need people like you with common sense to prod the physics community. Except that "quarks" are nonsense the fact is that all so-called "fundamental particles" are excited states of fundamental waves: They are light waves, electron-waves. proton-waves and neutrino-waves. Excited states are NOt fundamental "building" blocks of matter and never will be!
johanfprins
1.6 / 5 (7) May 05, 2010
So I presume, the particle generations of quarks could be modeled by nested vortex fluctuations of sufficiently dense elastic fluid, supercritical vapor in particular.

Anybody who can understand this gobblydook needs help.
Alizee
May 05, 2010
This comment has been removed by a moderator.
Alizee
May 05, 2010
This comment has been removed by a moderator.
Alizee
May 05, 2010
This comment has been removed by a moderator.
hard2grep
1 / 5 (3) May 05, 2010
In my non-educated view, I find that there are particles, and waves. Particles have mass. Waves are the transmission of information. If you can detect a particle, then it is a mass.

Judging by the resposes to this article, I would say that this is important research that needs to be done to resolve all the conflicts above. forgive my simplicity, but I am new.
Alizee
May 05, 2010
This comment has been removed by a moderator.
Alizee
May 05, 2010
This comment has been removed by a moderator.
johanfprins
1.6 / 5 (7) May 05, 2010
In my non-educated view, I find that there are particles, and waves. Particles have mass. Waves are the transmission of information. If you can detect a particle, then it is a mass.

A matter-wave has an intensity which is equal to its mass; since mass is energy: Therefore it has a centre-of mass. A body with a centre of mass moves as if it is a particle: Thus a wave with mass-energy can also move as if it is a "particle" which it is NOT. When such a wave encounters boundary conditions which forces it change its shape and size, wave properties, as we know it, appear. The fact is that it has all along been a wave. There is NO "wave-particle duality" or complementarity since "particles" just do not exist.
It is really so simple: CAN NOBODY UNDERSTAND THAT JUPITER MOVES AS IF IT IS A POINT PARTICLE WITH ALL ITS MASS AT ITS CENTRE OF MASS? EVEN ARCHIMEDES KNEW THAT THIS WOULD BE THE CASE! WHAT HAS BEEN WRONG WITH BOHR?
johanfprins
2 / 5 (8) May 05, 2010
Dear Alizee,
You do not really think one should waste time trying to engage you in logical discourse, do you?
Alizee
May 05, 2010
This comment has been removed by a moderator.
Alizee
May 05, 2010
This comment has been removed by a moderator.
Alizee
May 05, 2010
This comment has been removed by a moderator.
frajo
5 / 5 (1) May 05, 2010
My question is therefore how a quark can have a greater mass than a proton, much less an entire atom of a heavy element.
According to the standard model of particle physics, there are six different quarks. Protons (and neutrons) are formed of three quarks each of which is either an "up" or a "down" quark. The masses of these "up" and "down" quarks are considerably smaller than the masses of the remaining four quarks ("strange", "charme", "top", "bottom"). Thus, the four heavy quarks are not constituents of the proton (or the neutron).
Alizee
May 05, 2010
This comment has been removed by a moderator.
Alizee
May 05, 2010
This comment has been removed by a moderator.
daywalk3r
3 / 5 (18) May 05, 2010
..it could explain the mysterious gamma ray flashes, which can travel across whole Universe without dispersion. The explanation of such stability is much easier,if we imagine such flash as a dense swarm of photons,which travel in collective way like boids or like individual vortices in vortex ring.
Thats all nice and dandy, but in the case of photons, we are talking about EM radiation. We can "imagine" all we want, but thinking that photons are like individual particles, which at the moment of their release (radiation) are shot in a specific direction, is just plainly wrong. Photons are RADIATED.

In this regard, it's much propper to imagine the photonic "wave-front" as the surface of an expanding baloon, where the wall thickness coresponds to radiation intensity. No matter how big the balloon is, there will allways be a trace of the radiation, just with dropping intensity as the distance from centre increases. So perfectly non-dispersive propagation of G-rays is simply not real
daywalk3r
2.9 / 5 (17) May 05, 2010
There is NO "wave-particle duality" or complementarity since "particles" just do not exist.
..the particle-wave duality is simply experimentally proven stuff.
Yes, it indeed is.. And it describes how stuff BEHAVES, and not what it exactly IS, nor what it is made of. Though I think the problem might lie rather in the definition of what a "particle" actually is..
Alizee
May 05, 2010
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Alizee
May 05, 2010
This comment has been removed by a moderator.
johanfprins
1 / 5 (3) May 06, 2010
@Alizee,
All I can say after reading your ideas which are all based on unproven speculation is to quote Mark Twain: "Today I have met a man who knows more things that are not so than any other man I have ever met".

Light in vacuum does not have rest mass while an electron-wave has. Therefore a "free electron" MUST be able to be stationary within its own inertial refrence frame. This in turn demands that it MUST have a centre of mass which is stationary within this same inertial refrence frame; or else all the laws of physics have to be re-invented: Also the Schroedinger equation. Since the centre-of-mass MUST be stationary there cannot be any uncertainty relationship between position and momentum. Heisenberg's uncertainty relationship for position and momentum is just the normal relationship for ANY harmonic wave since such a wave exists within both position and k-space. Therefore Planck's constant appears on both sides of Heisenberg's relationship: And therefore it cancels out.
ZeroX
1 / 5 (4) May 06, 2010
... your ideas which are all based on unproven speculation...
These ideas are just based on removal of all aditional speculations by using of Occam's razor principle. All observable objects appear like pin-point particles from sufficiently distant perspective and universe is a random gas of these objects.

..electron-wave has rest mass ...therefore a "free electron" MUST be able to be stationary within its own inertial refrence frame..
This still doesn't say, this inertial reference frame must remain stationary too. Because vacuum is particle gas, every particle in it is attacked by collisions of surrounding particles like pollen grain in water excerting Brownian motion. We can observe it at the case of lightweight atoms, like those of liquid hellium - such liquid never freeze at room pressure, even at absolute zero. It can serve as a tangible evidence, reference frame of particles isn't complete stationary.
johanfprins
2.1 / 5 (7) May 06, 2010
This still doesn't say, this inertial reference frame must remain stationary too.


This remark is incontrovertible proof that you do not understand the most important foundation stones which had been laid by Galileo, Newton and Einstein, and on which ALL physics MUST be based.

An inertial reference frame is defined as a reference frame within which one CANNOT perform ANY physics-experiment to determine whether the reference frame is moving or not. If your physics requires that it must be moving, as you have just indicated that it must, it is wrong: Unless you have new laws that can replace those of Galileo, Newton and Einstein.
ZeroX
2 / 5 (4) May 06, 2010
I see - you're just mixing insintric perspective of relativity and exsintric perspective of quantum mechanic. Now you're trying to refute uncertainty postulate by theorems of SR - and some halfeducated forum troll even upvotes you for this. This is simply funny...:-)

It's just the accidental story of yours (extrapolation of your figth against BCS theory, I presume), you're fighting against QM by using of SR postulates. I could refute special relativity by strict adherence to quantum mechanics postulates instead without problem.

I indeed know, special relativity is deterministic theory, so it violates indeterminism of special relativity and vice-versa - but such insight is just a trivial finding. Sorry, but you simply CANNOT extrapolate SR theorems to QM phenomena and vice-versa. These two theories aren't consistent and one violates the another one.
johanfprins
1 / 5 (3) May 06, 2010
Sorry, but you simply CANNOT extrapolate SR theorems to QM phenomena and vice-versa. These two theories aren't consistent and one violates the another one.

The only reason why you find them not consistent is because the Copenhagen interpretation of QM violates all the equations ever derived in physics. Even Schroedingere's equation accepts that an electron has a rest mass: Rest mass means that such an entity must have an inertial reference frame within which it is stationary. And such a body CANNOT have uncertainties in the position and momentum of its centre-of-mass. Schroedinger's equation and SR are totally consistent as soon as one drops the "probability-interpretation" for QM. In fact the wave-equation for mater-waves is then also consistent with Einstein's gravity.

The "troll" who upvoted me obviously understands far more than you are capable of understanding!
frajo
4 / 5 (4) May 06, 2010
some halfeducated forum troll even upvotes you
You don't use your real name. You change your nick every fortnight. You don't understand physics or mathematics. But you're calling someone names because he downrates you. What a miserable sight.
ZeroX
1 / 5 (2) May 06, 2010
you don't use your real name
Can you prove, "Frajo" or "JayK" is your real name?
.because he downrates you..
Nope, because he upovotes crackpot ideas without any experimental evidence.

Copenhagen interpretation of QM doesn't care about center of mass at all - it's probabilistic interpretation of QM and as such it fits all the observations, which fall into realm of QM. It was confirmed by experimental tests of Bell's inequality. And it fits dense aether model of vacuum well.
ZeroX
1 / 5 (2) May 06, 2010
Schroedinger's equation and SR are totally consistent as soon as one drops the "probability-interpretation" for QM
This is nonsense, as the probability interpretation itself is just an intepretation of Schroedinger's equation. The consistency of some equations with another ones doesn't depend on their intepretations.
the wave-equation for mater-waves is then also consistent with Einstein's gravity
I doubt it, because from Schroedinger's equation of free particle wave follows, this particle wave would expand into infinity - whereas from Einstein's relativity follows, it should collapse into singularity like geon.

In fact, QM doesn't recognize "Einstein's gravity" at all and the existence of gravity cannot be derived from it - until you demonstrate the opposite (and deserve Nobel price in such way).
johanfprins
1 / 5 (3) May 06, 2010
You don't use your real name. You change your nick every fortnight. You don't understand physics or mathematics. But you're calling someone names because he downrates you. What a miserable sight.

Typical is it not!
johanfprins
1 / 5 (3) May 06, 2010
openhagen interpretation of QM doesn't care about center of mass at all

Exactly! This is why it is based on virtual reality!
it's probabilistic interpretation of QM and as such it fits all the observations, which fall into realm of QM.

It most certainly does not. As I have already pointed out on this forum, this interpretation leads to the conclusion that the "most probable position" for an electron is at a position where the probability to find an electron is excatly ZERO!
It was confirmed by experimental tests of Bell's inequality. And it fits dense aether model of vacuum well.

It was NOT confirmed by tests of Bell's inequality at all. How? Of course it fits your aether model since this model is also virtual reality.
johanfprins
1 / 5 (3) May 06, 2010
The consistency of some equations with another ones doesn't depend on their intepretations

Of course it does. If you interpret a wave-intensity as a "probability" instead of a mass-energy you will obviously have a contradiction!
In fact, QM doesn't recognize "Einstein's gravity" at all and the existence of gravity cannot be derived from it

Only because of the stupidity of the probability interpretation. When you accept that the intensity of a harmonic matter wave is its energy, as is the case for ALL harmonic waves ever discovered; you will realise that this intensity represents the mass of the wave-entity AND the curvature of space around the mass. (the latter curvature is wrongly interpreted in terms of probability as "tunnelling tails)
- until you demonstrate the opposite (and deserve Nobel price in such way

Are you going to nominate me? Thanks!
Alizee
May 06, 2010
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Alizee
May 06, 2010
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Alizee
May 06, 2010
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daywalk3r
2.8 / 5 (16) May 06, 2010
Copenhagen interpretation of QM doesn't care about center of mass at all
Aswell as it doesn't care about physical reality.
- it's a probabilistic interpretation of QM and as such it fits all the observations, which fall into realm of QM.
It fits observations about as much as the 50:50 probability "rule" fits the "coin-throw" experiments.. Statisticaly speaking, it indeed fits (surprise surprise), but sadly has little to nothing to say about what's really going on..

The probability interpretation was allways just a crude placeholder - sufficient for basic calculaions and statistics - but we are nearing a breaking point in physics, where it either will need to go and make room for something more "sophisticated", or we stay in caves for some longer.

Though we will never be able to predict the outcome of infinite-complexity systems with a 100% certainity, we at least need to push the probability approach into "deeper levels", somewhat..
Alizee
May 06, 2010
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johanfprins
1 / 5 (3) May 06, 2010
This is quite common situation in other theories, too. For example the center of mass of Jupiter-Sun (or whole solar system) lies outside of surface of Sun. Which basically means, with respect to Newtonian physics all planets are revolving just an empty place of space.

So you agree with me that the "most probable position" of a matter wave is really its centre of mass! Thanks this is exactly what I am trying to say all along. It has NOTHING to do with probability!

Aether theory proposes one of the ways: I believe, it could be possible to demonstrate by solution

Please stop punting a ridiculous theory under different names: Alizee, Seneca? Xerox? and what have you!!!
Alizee
May 06, 2010
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Alizee
May 06, 2010
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johanfprins
1 / 5 (3) May 06, 2010
Probabilistic interpretation of QM is quite OK, simply because our Universe IS of probabilistic nature. Even simple water droplet demonstrates it

Please STOP deliberately to confuse "normal probability" used when modelling many bodies with QM probability involving a SINGLE entity. You are not REAALY as stupid as this, ARE YOU?
It indeed does. The particle-wave duality was proposed just on the background of famous double-slit experiments. It's correspondence principle is even very insightful

Nonsense: The fact that the double slit diffraction pattern disappears when a moren tries to measure through which slit the "particle" came proves that there was a wave which moved through both slits which is then modified when being measured behind the slits. This is the way ALL waves react: When you change the boundary conditions the wave morphs into a new cofiguration.
johanfprins
1.7 / 5 (6) May 06, 2010
This is just a dilemma: how the location of probability distribution, which is the center of mass can not have nothing to do with probability? I'm afraid, you're trapped in semantic tautology

I wish you would take a course in simple logic and stop throwing words around which have nothing to do with the experimental facts!
Alizee
May 06, 2010
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johanfprins
1 / 5 (3) May 06, 2010
As we know, formal math cannot describe even the motion of three gravitating bodies reliably - but it can describe the probabilistic distribution of their center of mass.

If you are not able to realise that this probability has NOTHING in common with QM probability, then the time has come for you to find a job as a street sweeper!
johanfprins
1 / 5 (3) May 06, 2010
I never proposed anything, which couldn't be modeled at least conceptually by using of few trivial tools: water surface, soap foam, supercritical fluid, mercury droplets... did I forget something?

Yes you have: It it is known as "common sense"!
Alizee
May 06, 2010
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johanfprins
1 / 5 (3) May 06, 2010
I presume, in near future these experiments will become as iconic, as the M-M experiment, for example.

It is an obtuse demonstration that individual waves can form a macro-wave by entangling to, in the process, lose their individual existences. This is exactly what i have achieved with the electron-waves I have extracted from a diamond, which then form a superconducting holistic wave through which electron-charges are teleported. It has NOTHING to do with your concepts on so called aether theory. Sorry to disappoint you!
Alizee
May 06, 2010
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Alizee
May 06, 2010
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Alizee
May 06, 2010
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johanfprins
1 / 5 (3) May 06, 2010
I presume, we could still model your superconductivity by single particle, which basically disproves the "holistic wave teleportation" mechanism of yours.

I am not even going to waste my time to try and argue with you over "galaxies" and what have you. Nonetheless, "single particles" do not exist: ONLY waves where matter waves have rest masses and centres of mass and light waves in vacuum have no rest masses. Only when a light wave moves through a material does it acquire inertia and thus part of its energy becomes rest mass. That is why these waves are also modelled in terms of complex wave amplitudes; just like matter waves!

Bye until tomorrow!
Alizee
May 06, 2010
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daywalk3r
3 / 5 (16) May 06, 2010
Heisenberg's uncertainty principle can be extended to infinite set of non-commuting,still perfectly well measurable quantities. It basically says, "as far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality".
That's a nice Einstein quote, which I can only agree with.

I have no issues with "uncertainity" aswell, because as I have wrote before allready, you can not be 100% certain in a system with infinite-complexity. Any attempt to precisely pin-point a physical property of an entity by the means of mathematics is bound to fail - as there will allways be a margin of error.

I think that is exactly what Einstein was trying to express with the above quoted sentence, and it fits nicely with the basic Heisenberg interpretation.

Though problems arise when this kind of concept gets interpreted in a (certainly) wrong way, and one tries to build physical reality based on statistical "principles".
Alizee
May 06, 2010
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Alizee
May 06, 2010
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Alizee
May 06, 2010
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Alizee
May 06, 2010
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johanfprins
1 / 5 (3) May 07, 2010
@ Alizee,
Stop overwhelming this forum with irrelevant nonsense. When talking about Copenhagen uncertainty, it is the so-called "inbuilt" uncertainty for a single electron according to which a measurement of its position, even if it could be done with 100% certainty, causes an infinite uncertainty in its momentum; and vice versa. It is the latter interpretation that violates the most fundamental law on which ALL physics has been formulated: namely Galileo's inertia; This concept has been quantified by Newton as rest mass. Thus any entity with rest mass cannot have such inbuilt uncertainty in position and momentum. It is this misinterpretation which makes it impossible to reconcile QM with Einstein's gravity. The statistical parameters and uncertainties, which you are obfuscating the argument with, are what one normally expects when one cannot measure with 100% accuracy and when working with many entities. The latter HAS NOTHING to do with Born's and Heisenberg's wrong interpretation
smd
not rated yet May 07, 2010
My question is therefore how a quark can have a greater mass than a proton, much less an entire atom of a heavy element.
According to the standard model of particle physics, there are six different quarks. Protons (and neutrons) are formed of three quarks each of which is either an "up" or a "down" quark. The masses of these "up" and "down" quarks are considerably smaller than the masses of the remaining four quarks ("strange", "charme", "top", "bottom"). Thus, the four heavy quarks are not constituents of the proton (or the neutron).


Thank you.
Alizee
May 07, 2010
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Alizee
May 07, 2010
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johanfprins
1.7 / 5 (6) May 08, 2010
This uncertainty is not "inbuilt", but experimentally observed


What is observed is proof that an electron is NEVER a particle but ALWAYS a wave with a centre-of mass. An extended body with a centre-of-mass can move like a particle even though it is not a particle. A wave with a centre-of-mass does NOT violate Galileo's inertia nor Newtons definition of rest mass. Thus it is fully consistent with ALL physics from Galileo to Schroedinger! It is not necessary to invoke inane concepts like "wave-particle duality" or "complementarity".
Alizee
May 08, 2010
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Alizee
May 08, 2010
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Alizee
May 08, 2010
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Alizee
May 08, 2010
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Shootist
1 / 5 (3) May 08, 2010
The only scientists I do respect is the ones that believe in Christ!!!


Yeah, flat earth, with four corners, 6000 years old.
johanfprins
1 / 5 (3) May 08, 2010
This is merely a semantical problem, Johan: we cannot imagine particle without wave packet concept and we cannot consider wave without some particle environment on the background, which is mediating these waves

I do not have the time to read through the rest of your meanderings: So I will only read what I have quoted and try to get you to reason logically for a change.

How and from what do you construct your "wave packet"? Schroedinger's equation for a free electron (V=0) does NOT give a wave packet but a time-independent wave with the same intensity which fills an infinite space. Such an electron is obviously nonsense.
Alizee
May 08, 2010
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Alizee
May 08, 2010
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Alizee
May 08, 2010
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Alizee
May 08, 2010
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zslewis91
1 / 5 (4) May 08, 2010
hey "johanfprins" sounds like you wanna blow Alizee...hey "johanfprins" why dont you just get over it, you strike me as the type of "hay-seed", dull red Crayola Crayon.. that wishes he was just a litte brighter, and hates bright red at the top of the box.....hey "johanfprins" get over it.... your not the brightest in the box...BOO HOO BOO HOO..big cry baby..hey Alizee, thanks for keeping it real...hey johanfprins..GET A LIFE...Then a clue..D-BAG!
Alizee
May 08, 2010
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johanfprins
1.5 / 5 (6) May 09, 2010
Quantum mechanics cannot explain, why all particles doesn't evaporate, general relativity cannot explain, why all particles will collapse into black holes.

Alizee, you see why one cannot have an intelligent conversation with you. You start of with general statements that you cannot prove; and then proceed to confuse the issues. "Quntum mechanics " cannot explain is a nonsense statement. It can mean that it will never be explainable or that the present crop of theoretical physicists are just incapable of getting their act together. I believe it is the last and that you are part of this confusion.
This is why scientists are combining postulates of relativity and quantum mechanics into various hybrid theories (Dirac theory, string theory, LQG theory) for to get more realistic predictions...

Correct! Because they do not understand what is really going on. Therefore they paddle along with impossible concepts like "dense ether theory"!
johanfprins
1 / 5 (3) May 09, 2010
Wave packet (a soliton) is formed in spreading of wave through inhomogeneous environment, which becomes more dense under influence of energy wave like soap foam during shaking. There you can see simulation in Java..

So electrons consist of "soap foam". Please become real and talk sense. I have NEVER seen soap foam used to model bound electrons. The Schroedinger equation is the best approximation we have at present. So I am asking you how this equation can model a "wave packet" in free space and then you avoid it by blowing soap bubbles.

The fact is that the shape and size of ANY wave are determined by its boundary conditions. What are the boundary conditions in space which causes a wave packet? Wave packets only form from many electron waves superposing within conductors when one applies an electric field. The latter causes the boundary conditions which form them. How is this done for a solitary electron-wave obtained from Schr. Eq. (a single wave) in space?
Alizee
May 09, 2010
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Alizee
May 09, 2010
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Alizee
May 09, 2010
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Alizee
May 09, 2010
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Alizee
May 09, 2010
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johanfprins
1 / 5 (3) May 09, 2010
From Schroedinger or Dirac equation simply follows, free particle will expand into infinite volume gradually - this is just a bare fact.


The reason why this supposedly happens is because these equations are solved without stipulating boundary conditions. Any such solution is NOT physics. Furthermore by then superposing the possible "boundaryless" solutions as if they are all electrons is obviously nonsensiscal. Both Schr. and Dirac's equations are SINGLE ELECTRON equations so that just ONE of the possible solutions can be the electron! That means that a Dirac electron MUST have an energy of minus infinity!

When stipulating the correct boundary conditions there is no "spreading" whatsoever; just as it must be!

All derivations obtained by ignoring boundary conditions are voodoo. This is why ALL quantum field models are just plain nonsense.
Alizee
May 09, 2010
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Alizee
May 09, 2010
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johanfprins
1 / 5 (3) May 09, 2010
@Alizee,

Why you have to bring in concepts that cannot be tested experimentally to create a virtual reality you WANT to believe in, is beyond my comprehension.

It is just a simple fact that if you have to integrate once you have one constant which mathematics cannot supply, but has to be supplied by common sense. If you have to integrate twice, you have two constants etc. Stipulating these boundary conditions IS physics, avoiding them is "beautiful mathematics' which Dirac propagated, but has NOTHING to do with physics whatsoever. If you derive a solution which requires an infinite space within our universe, which is NOT infinite, as you guys have been doing consistently for 80 years now, you are not doing physics. I do not know what to call it other than pure nonsense!

Schroedinger's wave equation requires THREE integration constants: One for time and TWO for space. So please, do some physics, and give me the two space constants for a solitary erlectron in space.
Alizee
May 09, 2010
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johanfprins
1 / 5 (3) May 10, 2010
This is quite normal, in physics all solutions are approximate. This doesn't mean, all these equations and solutions are "pure nonsense".


They are pure physics nonsense up to the point that one specifies the boundary conditions and check whether it makes physical sense.

A simple example which illustrates this fact is London's first equation for superconductors. The London brothers derived from it that the current must be a constant when the electric-field is zero. This constant is a boundary condition after the equation has been integrated. The conclusion that the current must be a constant is mathematically correct but not yet physics. To derive the physics involved one must specify the boundary condition: i.e. the constant. When doing this for a superconductor between two contacts, one finds that the only physically possible value for this constant is ZERO. But in text books it is argued that the London bros. have given a reason why a non-zero current can flow.
johanfprins
1 / 5 (3) May 10, 2010
This applet is all based on Schroedinger equation and its solution is quite rigorous.


As usual you are not answering what has been asked, but try and sideline the argument by referring to websites which have NOTHING to do with the original argument.

Let us try again. We have Schroedinger's equation and want to use it to model a solitary electron in space. In all text book it is assumed that one can put the potential energy term equal to zero. It is then derived that the electron can be represented by a single, infinite wave with only one space boundary condition which determines its intensity. This is NOT possible because the space coordinates have to be integrated twice. So I am asking again: Where is the other boundary condition?
ZeroX
1 / 5 (2) May 10, 2010
..In all text book it is assumed that one can put the potential energy term equal to zero. ...
Not at all - if you would do it, then you would solve particle in potential box actually. But for free particle such condition doesn't exist - there is no boundary or area, where you should put zero potential arbitrarily.

This is why another conditions are used: a combination of Dirichlet condition (which is asumming, potential differential is always zero at the sufficient distance from the center of wave packet) and an integral condition (which is assumming, the probability of the occurence of the wave packet over whole integration interval is unitary - we don't know exactly, where the particle is, but we can be perfectly sure, it will all be there).

You should learn a bit about solving of differential equations - putting constants on the boundary isn't far the only way, how to fit the solution.
ZeroX
1 / 5 (2) May 10, 2010
The Dirichlet condition is just the biggest trouble in physically relevant solution, because the potential differential can be never zero - even at the infinite distance from particle due the omnipresent fluctuations of vacuum. But this isn't problem of Schrodinger equation, because the potential fluctuations in vaccum cannot be derived from it in any way. And these fluctuations could be extremelly large at short distances, albeit their average value appears zero all the time.
johanfprins
1 / 5 (3) May 10, 2010
You should learn a bit about solving of differential equations - putting constants on the boundary isn't far the only way, how to fit the solution.


I know about the Dirichlet condition etc. It is a fudge, just like the phase angle used to model superconduction is a fudge. It is an arbitrary boundary condition which cannot be used when you put the potential energy equal to zero. When doing the latter there is NO PHYSICAL reason why the intensity must drop off to zero at large distances. You can just as well assume that you do have a particle in a box. The Dirichlet criterion is just another way of introducing this same inconsistency.

What you should do is to learn basic mathematics and logic.
johanfprins
1 / 5 (3) May 10, 2010
even at the infinite distance from particle due the omnipresent fluctuations of vacuum.


This is another fudge: There is no exclusive experimental proof whatsoever that it is the vacuum fluctuating. All the arguments and experimental results used to postulate this stupidity can be explained in a better fashion without ending up with an infinite vacuum energy and renormalisation. When you get infinities in your calculations it is mathematics telling you that you are wrong!

Why did you decide to now start posting as ZeroX, Alizee? Have you got a split personality. However, one does need one to believe in your arguments!
ZeroX
1 / 5 (2) May 10, 2010
..I know about the Dirichlet condition etc. It is a fudge..
LOL - it's the same math, like the Schrodinger equation derivation itself. If you don't believe in logic under which such condition was introduced into math, then you shouldn't believe even in Schrodinger equation itself, because its derivation was based on the same logic.
..there is no exclusive experimental proof whatsoever that it is the vacuum fluctuating. ..

For example liquid hellium never freeze at room pressure, even at zero temperature, because its atoms are moving wildly.

It could be explained in similar way, like the Brownian motion of pollen grain in water, whose fluctuations keep these grains in eternal motion - even at the completelly calm watter surface.

I'm posting like ZeroX, because I'm using a different computer by now and I forget the password. I don't require people to believe in my arguments, only to understand them. This discussion shouldn't be based on belief and religion.
ZeroX
1 / 5 (2) May 10, 2010
.. there is NO PHYSICAL reason why the intensity must drop off to zero at large distances...
This is just why the Dirichlet condition is used at the case of free particle. We can assume, the probability density of free particle doesn't change in sufficient distance from particle - which is exactly, what the whole Dirichlet condition is about.

If you don't believe in vacuum fluctuations, then there is no objective reason, why just this condition should be violated in some way. Until you demonstrate some counter-example, indeed.

As you can see, just the fact, the free particle doesn't expand in vacuum into infinity could serve as an (indirect) proof of vacuum fluctuations from perspective of steady-state solution of Schrodinger equation. Everything fits perfectly, here.
johanfprins
1 / 5 (3) May 10, 2010
For example liquid hellium never freeze at room pressure, even at zero temperature, because its atoms are moving wildly.


Where was liquid helium cooled to absolute zero temperature so that you can make such a statement?

The reason why the He-atoms, which by the way are not "atoms" anymore when they form a superfluid (each is a single holistic wave) is that Heisenberg's relationship for energy and time allows such holistic waves to change their energies for short time intervals: i.e. by quantum fluctuations. This is the same way in which superconduction occurs. Thus in a sense you are correct that the energy is supplied by the "vacuum" but not by the "infinitely" large vacuum energy (quantum foam) as modelled by you.
johanfprins
1 / 5 (3) May 10, 2010
the probability density of free particle doesn't change in sufficient distance from particle - which is exactly....


This is where we will never agree since the intensity distribution of a matter wave is IN MOST CASES not commensurate with a "probability distribution".

As you can see, just the fact, the free particle doesn't expand in vacuum into infinity could serve as an (indirect) proof of vacuum fluctuations from perspective of steady-state solution of Schrodinger equation. Everything fits perfectly, here.


It does not because you are not specifying the reference frame in which you are modelling the electron: Thus violating Einstein's Special Relativity.

The reason why an electron wave does not spread in vacuum is that it MUST be a stationary, time-independent wave which does not change its shape and size with time within its inertial refrence frame.

There exists a non-zero potential energy which ties the wave down within its inertial reference frame.
ZeroX
1 / 5 (2) May 10, 2010
The reason why the He-atoms form a superfluid.. is that Heisenberg's relationship for energy and time allows such holistic waves to change their energies for short time intervals..
versus
..since the centre-of-mass must be stationary there cannot be any uncertainty relationship between position and momentum. ...

You see - suddenly Heissenberg's uncertainty principle appears good enough for you...;-) Of course, we should ask further, why some uncertainty principle should be valid at all? Which/where is the physical origin of this uncertainty?
ZeroX
1 / 5 (2) May 10, 2010
..the reason why an electron wave does not spread in vacuum is that it MUST be a stationary, time-independent wave which does not change its shape
This is just an circular reasoning - circulus in probando. Why this wave MUST be a stationary? Because Mr. Prins believes so?
johanfprins
1 / 5 (3) May 10, 2010
You see - suddenly Heissenberg's uncertainty principle appears good enough for you...;-) Of course, we should ask further, why some uncertainty principle should be valid at all? Which/where is the physical origin


Nice try: But you will note that I have NOT called it an "uncertainty" relationship because that is NOT what it is. It is a resonance- relationship which allows a holistic wave to vary its energy by an interval (del)E as long as it is not for longer than (del)t. You can easily derive this fact from the width's of atomic emission lines. In fact you can apply this to ALL waves when they resonate: Also to a light wave being received by a radio antenna. This is where it comes from: Purely from wave behaviour which has been well-known LONG before quantum mechanics came along.

johanfprins
1 / 5 (2) May 10, 2010
To proceed:
Similarly Heisenberg's relationship for position and momentum is NOT an uncertainty relationship for the position and momentum of a "particle". It is purely the sizes of the wave in position and k-space when it is a standing wave. This has also been known LONG before quantum mechanics. So why should this be the case FOR ALL harmonic waves EVER KNOWN but not for an electron wave? Furthermore why should the total intensity of ALL harmonic waves EVER KNOWN be equal to their energies, but when it comes to a harmonic electron wave it suddenly is not so? Obviously IT MUST STILL be so so that the intensity is the mass of the wave. It is simple and it shows that their is NO DICHOTOMY between quantum mechanics and classical physics at all!
johanfprins
1 / 5 (3) May 10, 2010
This is just an circular reasoning - circulus in probando. Why this wave MUST be a stationary? Because Mr. Prins believes so?


Because Galileo said so and Newton quantified the fact that an entity with mass MUST be stationary within its own inertial reference frame. This is why it is so: And this is why the use of mass in any equation (and thus also Schroedinger's wave equation) mmandates that it must be so. This is why interpreting the sizes of a wave in position and k-spaces as "uncertainties" in position and momentum invalidates ALL physics which came before Heisenberg, Born and Bohr touted the absurd probability interpretation. It is also so that when jettisoning the probability interpretation, classical mechanics, quantum mechanics and Einstein's gravity dovetails without any major problems.
ZeroX
1 / 5 (2) May 10, 2010
..I have NOT called it an "uncertainty" relationship because that is NOT what it is...
But you're using it for explanation of the ethernal motion of atoms in liquid helium. You're saying, quantum wave must always remains stable, but your's saying too, such wave should be able to change their energies for tiny time intervals. And we can find much more semantic controversies in the intepretation of yours.
Because Galileo said so and Newton quantified the fact that an entity with mass MUST be stationary within its own inertial reference frame
As I told you already, uncertainty principle enables to violate the stable position of these reference frames by quantum fluctuations of vacuum. When density of environment around massive object is fluctuating, for light waves, which are affected by such density fluctations such object is moving anyway. You cannot provide stable inertial frame for small object in dynamic vacuum.
Skeptic_Heretic
not rated yet May 10, 2010
This entire discussion can be wrapped up in "there's a force interaction that we're not aware of".

For example, on the macro scale we see black holes have a "small" size and a "large" size. The difference between the two is due to the interactions between the attractive forces of gravity upon matter and the repulsive forces of radiation escape pressure. We very well could be looking at a rehash on the micro scale with a different set of forces, or a corresponding analogy to the large forces on a smaller scale.
ZeroX
1 / 5 (2) May 10, 2010
The Galileo / Newtonian mechanics is relevant only for time averaged observations of reference frame in simmilar way, like the laws of thermodynamics are valid for bulk environment. Inside of gas the temperature of molecules fluctuates wildly in the same way, like the energy density in vacuum for wave packets. Because you cannot ensure, energy density changes for particle wave packet as a whole, you can observe, the particle effectivelly dances at place - even inside of potential box.

http://tinyurl.com/36fql36

Such dance is nothing special, the same motion exhibit the density fluctuation inside of dense gas. And the above animation is the result of solution of Schrodinger equation inside of potential hole, i.e. with providing all integration constants, which you're requiring. As you can see, the resulting wave not only changes its energy, but the location too. It's actually moving like pollen grain under Brownian noise.
ZeroX
1 / 5 (2) May 10, 2010
This entire discussion can be wrapped up in "there's a force interaction that we're not aware of".
The quantum motion is observable even by naked eye at the case of Brownian motion and special mixtures of fluid and solids (carbon disulphide and colloid sulphur, for example). We could observe it in fluid hellium even at zero temperature. This is not so minute effect, if you know, where to look for it. If we could use microwave or maybe deep infrared glowes, we could see, even empty vacuum is full of tiny density fluctuations, like hot air above camp fire. In fact, the sensitivity of our sensoric organs is just a one step from observation of quantum phenomena at common life in simmilar way, like from observation of graviational lensing and polarization of CMB at cosmic scales. We aren't separated very much from direct observation of curvature of our Universe on both Planck, both cosmologic scales. It's possible, some animals could observe these effects directly.
Skeptic_Heretic
5 / 5 (1) May 10, 2010
The quantum motion is observable even by naked eye

I'm sorry, can you say pseudoscience? Quantum motion is not directly observable, and definitely not by use of base human senses.
ZeroX
1 / 5 (3) May 10, 2010
I'm sorry, can you say pseudoscience?
Brownian motion is an example of quantum motion. mediated by particle of environment. You can observe it for example by motion of tiny bubbles in superfluid hellium. You can even track & record the wave motion of single electron in such way. And the quantum motion of individual atoms in boson condensate is observable directly, because these atoms are glowing like tiny dots under laser light.

http://www.youtub...aWCXkSC0
johanfprins
1.7 / 5 (6) May 10, 2010
But you're using it for explanation of the ethernal motion of atoms in liquid helium. You're saying, quantum wave must always remains stable,

The ability of a wave to resonate when required to do so when the boundray conditions require it to do so has nothing to do with the uncertainty of a point particle which requires it to be described by a wave whose intensity is a probability distribution. That holistic waves can jump around at low temperatures owing to quantum fluctuations does not mean there is any uncertainty involved at all.
There is not a most probable position of the wave around which a particle is fluctuating. The wave has a centre of mass and quantum fluctuations causes the wave to move like any other object with a centre-of-mass when it is vibrated. So there is no uncertainty involved.
johanfprins
2.1 / 5 (7) May 10, 2010
As I told you already, uncertainty principle enables to violate the stable position of these reference frames by quantum fluctuations of vacuum.;


It is the centre of mass of the wave that moves when the wave moves NOT the inertial reference frame. So you are again talking nonsense. The Copenhagen interpretation of uncertainties does not come into the picture anywhere!

You are confusing ordinary statistical behaviour like Brownian movement with the statistical behaviour that has been postulated by Bohr, Born and Heisenberg. If you do not understand the difference you should again read a book on elementary quantum mechanics.
Skeptic_Heretic
not rated yet May 10, 2010
If you do not understand the difference you should again read a book on elementary quantum mechanics.

As two scientists who have engaged in these sorts of conversations before, from me to you, you now know that the being known as zerox/alizee/slotin/alexa/etc... is merely a wanna be sophist and woefully ignorant of the reality of quantum mechanics and standard physical concepts. Why have you spent so much time trying to exemplify your addendum to the body of prior work when this slug can't understand the prior work, let alone, the addendum? Is it your love of teaching and knowledge or your want for a good fight that leads you to commit this repeated self insanity mechanism?
johanfprins
2.1 / 5 (7) May 10, 2010
Why have you spent so much time trying to exemplify your addendum to the body of prior work when this slug can't understand the prior work, let alone, the addendum? Is it your love of teaching and knowledge or your want for a good fight that leads you to commit this repeated self insanity mechanism?


No not a good fight. What worries me at present is that the physics community is more willing to classify a person as a crank than trying to understand: Yes there are many cranks around and yes since the advent of desktop computers they are flooding the system with wrong insights: However, I believe that it is incumbent on a scientist to argue and explain the science to everybody in as simple a fashion as possible. If we are not willing to do this we might be throwing out the baby with the bathwater in some cases. I have seen this happen too often lately to shirk reponsibilty in the same way. Even a fool can make a point or ask a relevant question from which one can learn.
Alizee
May 10, 2010
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Alizee
May 10, 2010
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Skeptic_Heretic
not rated yet May 10, 2010
which Prof. Prins is fighting against here all the time. You probably missed the whole point of discussion.

No, I evaluate and prefer to not interject when it is unwarranted. It has become warranted.
Alizee
May 10, 2010
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johanfprins
1 / 5 (3) May 10, 2010
Currently both probabilistic interpretation of QM, both uncertainty principle, both particle wave duality are concepts well supported by number of experiments. Can you provide some experiment, which violates the above concepts?

This is of course a lie! I have given incontrovertible proof that the probability interpretation is wrong. As already mentioned many times in this forum, a probability distribution for the position of an electron for which the "most probable position" to find the electron is at a position where the intensity of the distribution is ZERO cannot be a probability distribution of an electron's position. As Einstein said: Only a single small fact is required to prove that a theory is wrong.
The proof is incontrovertible!
Alizee
May 10, 2010
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johanfprins
1 / 5 (3) May 10, 2010
Every other proof is controvertible, you know...

Give me such an incontrovertible proof. You cannot.

It isn't. It's a probability distribution of the occurrence of particle


I assume that what you are saying is that it does not require a most probable position. If not, around which position do you calculate Heisenberg's "uncertainty" interval for position? Or does this uncertainty not apply for a wave for which the most probable position is zero? Thus Heisenberg's relationship is null and void for most electrons?
Alizee
May 10, 2010
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johanfprins
1 / 5 (3) May 10, 2010
Frankly, did you really spent whole rest of your life in fight for the nonexistence of probability of electron at the place, where such probability is zero?


This is a stupid remark which do not really deserve a response. It seems you are at the end of your tether and is now becoming abusive instead of explaining how an uncertainty interval of position can be present around a position at which one can NEVER find an electron.
Alizee
May 10, 2010
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johanfprins
1 / 5 (3) May 10, 2010
Have look at this picture .. It illustrates the electron, which revolves atom nuclei along many paths. As the result, a two-lobe orbital is formed, analogous to common p-orbitals


What you have done is to prove my case: i.e. that the wave-intensity cannot be a probability distribution which defines a Heisenberg "uncertainty" interval; or else the highest intensity would have had to be at the position where you can NEVER find an electron. Thanks for proving so eloquently exactly what I am saying.
Alizee
May 10, 2010
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johanfprins
1 / 5 (3) May 10, 2010
LOL, why not? Such P-orbitals with two lobes are defining the shape of most molecules.


Exactly! This is why the wave-intensities of "electrons" are real waves and the "most probable position" as interpreted by the Copenhagen interpretation is the centre-of mass of the wave. The latter can be at a position where the mass-intensity is zero. The most probable position cannot be at a position where a "probability distribution" has zero-intensity.

You keep on proving me right; Namely that the probability interpretation, wave-particle duality and complementarity are all unphysical: In fact these concepts are just plain nonsense! Thanks again! It is unbelievable that physicists could have believed in such claptrap for nearly 80 years.
Alizee
May 10, 2010
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johanfprins
1 / 5 (3) May 10, 2010
It isn't. For p-orbital the most probable positions of electrons are at the end of both lobes -

Thus you have TWO uncertainty intervals per electron? WOW!!
this is why elongated molecules are formed. If it would be center of mass, then the resulting molecule would be spherically symmetric - which violates experiments.

Not true. To form bonds from p-orbitals you have hybridisation. Seeing that it is impossible to instruct you in the simple physics of atomic orbitals I will not even try to tell you about chemistry. It is obviously far beyond you capabiliy!
Alizee
May 10, 2010
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johanfprins
1 / 5 (3) May 10, 2010
They're all supported by experiments and observations,


I ask again: Which experiments and observations "support" this nonsense.

Again, you are just proving that you know NOTHING about chemical bonding: So let us not meander into that field which you are obviously unable to EVER understand!
Alizee
May 10, 2010
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johanfprins
1 / 5 (3) May 10, 2010
Of course they're. There exist many orbitals with even higher symmetry - with three lobes, four lobes, etc.


You are just wasting everybody's time since two or more uncertainty intervals per electron TOTALLY invalidates the probability interpretation of the wave intensity. If you cannot see this, you should rather stop flooding forums like this one with claptrap.
Alizee
May 10, 2010
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johanfprins
1 / 5 (3) May 10, 2010
You just proved, you have no better arguments, again. How do you want to explain existence of angled molecules, if the probability of all electron waves would sit just at the center of their mass, like at the case of spherical wave?


I told you that I am not going to discuss chemical bonding with you since you clearly are incapable of understanding it. Obviously during cnemical bonding the centre-of-mass of the bond does NOT have to coincide with the nuclei. It is simple to derive that for a covalent bond the centre-of-mass of the boson-wave now fallsmidway between the nuclei. Possibly too difficult for you to compreend! But this does not change the fact that for all electron waves around a nucleus the centre-of-mass must coincide with the centre-of mass of the nucleus.
Alizee
May 10, 2010
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johanfprins
1 / 5 (3) May 10, 2010
do you want to explain the existence of angled molecules, if the probability of all electron waves would ...

Can you NOT get it through your skull that the intensity of an electron wave is NOT a probability distribution but the distibution of the wave's mass? And that any mass-distribution has a centre of mass! When an angled molecule is held together by covalent bonds the centre-of mass of each bond is situated at midpoint between the two nuclei it is bonding! Good God man you really cannot be so stupid can you?
johanfprins
1 / 5 (3) May 10, 2010
Of course not. But the probability of these electron waves will not be at the center of mass of the nucleus, thus allowing a places with zero probability around it.


There is NO PROBABILITY involved whatsoever!!!!!! There is no harmonic wave in the universe which has an intensity distribution which is a probability distribution. The intensities of ALL harmonic waves have ALWAYS been their energies and will ALWAYS be their energies!

Good night! I pray that you try and do some thinking by tomorrow; if you are capabl;e of it.
Alizee
May 10, 2010
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frajo
2.3 / 5 (3) May 10, 2010
I'm posting like ZeroX, because I'm using a different computer by now and I forget the password.
And in the discussion of the article "New evidence for quantum Darwinism found in quantum dots" you have written seven comments all of which have been voted on through the Alizee account. Using more than one account in order to get better ratings for oneself is misuse.
Alizee
May 10, 2010
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johanfprins
1 / 5 (3) May 11, 2010
Not just mass, but charge and spin too.

Obviously the charge is also distributed within the wave and the charge-centre must coincide with the centre-of-mass. Charge is, however NOT energy but mass is: Therefore the intensity of the wave is its mass distribution. The wave does not have "spin": It has a magnetic moment just like a light-wave has a magnetic component.

Intensity is intensity, energy is energy - these two quantities cannot be mixed.


For ALL fields and wave-fields EVER KNOWN the intensity is the energy of the field. Now you come along and state that they cannot be the same!!! I think you need to read an elementary physics-book on fields.
johanfprins
1 / 5 (3) May 11, 2010
The usage of physorg threads for off-topic personal comments is misuse too and in addition it's prohibited in guideline section EXPLICITLY


This remark explains a lot about your incapability to argue physics in a logical manner above: It is beneath contempt.
ZeroX
1.7 / 5 (6) May 11, 2010
..For ALL fields and wave-fields EVER KNOWN the intensity is the energy of the field.
Quantum mechanics recognizes only total energy Hamiltonian, frequency & amplitude of wave function, and it's square (probability distribution) as a measurable quantities.

Intensity is nothing, what I could define/compare with experiment in conection to Schrodinger equation and some interpretations of it. For example, intensity of radiation flux is flow of energy per unit of time. If intensity is both energy, both mass distribution, then I'm really confused by your terms.
johanfprins
1 / 5 (3) May 11, 2010
Quantum mechanics recognizes only total energy Hamiltonian, frequency & amplitude of wave function, and it's square (probability distribution) as a measurable quantities.


Not "quantum mechanics" but the Copenhagen interpretation which is only valid in Alice's (or is it Alexa's) wonderland. You are NOT contributing when you just quote dogma which is obviously wrong since it is not based on reality and therefore cannot dovetail with classical mechanics and Einstein's gravity.
ZeroX
2.1 / 5 (7) May 11, 2010
These terms are insintric part of Schrodinger equation. You aparently believe, only Copenhagen interpretation is responsible for probabilistic interpetation of Schrodinger equation. But this intepretation is an imanent part of postulates of quantum mechanics. And it doesn't recognize something like "intensity".

http://vergil.che...e20.html
johanfprins
1.7 / 5 (6) May 11, 2010
But this intepretation is an imanent part of postulates of quantum mechanics.

Do you want to tell me that the word "imanent" makes these postulates holy dogma which was brought down from Mount Sinai by Moses?
You are wasting my time and everybody's time on this forum. Let us agree to disagree. I maintain that the present interpretation of quantum mechanics is wrong. You maintain that it is correct. You kept on promising that you will quote experimental results to prove your point but you never do. I have better things to do than to argue with a person who never answrs points raised but repeats wrong dogma in a mantra-type fashion.
And it doesn't recognize something like "intensity"

So when you calculate (psi)(psi)* you are not calculating an intensity field. Really if talking nonsense was a spot you would have been a leopard!

ZeroX
2.3 / 5 (6) May 11, 2010
Do you want to tell me that the word "imanent" makes these postulates holy dogma
Well, postulates are defining theory. You cannot change postulates of relativity theory, which Einstein invented. You can only propose alternative theory by using of your own postulate set and demonstrate, it's equivalent to the original theory. Do you have done it? I doubt it.
So when you calculate (psi)(psi)* you are not calculating an intensity field?
Intensity of what? Nope, the product (psi)(psi) is probability density.

http://tinyurl.com/3ank6b7

Sorry proffesor - with full respect, it's just you, who didn't understood the rules. You needn't to convince me about postulates of QM, because I didn't invented quantum mechanics, so I cannot change its postulates. If you believe, these postulates are saying something different, then the publicly awailable sources - then it's your problem, not mine and it's you, who is wasting the time of other readers, not me.

I hope, I'm clear by now.
ZeroX
2.6 / 5 (5) May 11, 2010
Frankly, I don't see a huge difference between "intensity field" (intensity of what?) and the "probability distribution". With the only exception, the later one is is well defined term, which has an experimental support. You're just confusing well established terms.
Skeptic_Heretic
not rated yet May 11, 2010
Frankly, I don't see a huge difference between "intensity field" (intensity of what?) and the "probability distribution".

That's a really big misunderstanding to have. One that effectively invalidates the majority of what you've said above.
johanfprins
1.8 / 5 (5) May 11, 2010
Frankly, I don't see a huge difference between "intensity field" (intensity of what?) and the "probability distribution". With the only exception, the later one is is well defined term, which has an experimental support.


I did not want to respond to your assinine comments anymore. But I am getting tired that you claim experimental support without being able to supply it! PLEASE NOTE that a "postulate" is not an experimental proof!
Alizee
May 11, 2010
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johanfprins
1.8 / 5 (5) May 11, 2010
Without it it's still just a semantic twaddling. Maybe my comments may sound assinine for someone - but for me they're quite legitimate.

Semantics? You yourself has said clearly that the probability interpretation cannot be reconciled with classical physics. The interpretation that the intensities of all waves are their energies immediately reconciles quantum mechanics with classiocal mechanics and Einstein's gravity, since one then has that both light and matter only exist of waves AND that one can model all interactions, also the photo-electric effect purely in terms of the known behaviour and interactions of waves. Particle interactions are not at all required. Is this twaddle? The only twaddle in physics has been wave-particle duality, probability waves, and complementarity. And of course your so-called ether models!
Alizee
May 11, 2010
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johanfprins
1.8 / 5 (5) May 12, 2010
I presented a few classical experiments, which are demonstrating it.


You fooled me! I have dilligently looked in your comments for experimental evidence to support your ideas and found none to date!

The experiment that you just quoted above proves that "single particles" do not exist: Only waves. To obtain diffraction you must have a holistic wave which can split up into fragments which stay in immediate contact with each other. A particle can never diffract since, by definition of the properties of a particle, it cannot move through two slits simultaneously.

ZeroX
2.1 / 5 (7) May 12, 2010
..A particle can never diffract since, by definition of the properties of a particle, it cannot move through two slits simultaneously.
I see - I think I know, where is your problem in understanding by now.

Well, the particles doesn't move through two slits simultaneously. Believe it or not, this doesn't change the result of double slit experiment. What is spreading through both slits is just a deBroglie wave of vacuum, which is created around particles like wave around duck swimming at the river surface.

http://www.tinyurl.cz/oy7

Particle itself indeed remains quite tiny and well localized.
Skeptic_Heretic
not rated yet May 12, 2010
What is spreading through both slits is just a deBroglie wave of vacuum

Pseudoscience again and a telling statement of ignorance on your part.

How exactly have you proven that a "wave of vaccuum" can have the characteristics of photons?

How exactly have you established that by passing through the barrier there is non-localized decay (of light...)?

Lastly, do you know what you're talking about?

johanfprins
2.3 / 5 (6) May 12, 2010
What is spreading through both slits is just a deBroglie wave of vacuum, which is created around particles like wave around duck swimming at the river surface.

De Broglie wave of vacuum? Funny wave of vacuum which gives a double-slit diffraction pattern even though Alizee KNOWS that the "duck" only swims through one slit; but then do not interfere to give diffraction when Alizee confirms that the duck only swims through one slit.
Alizee
May 12, 2010
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Alizee
May 12, 2010
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johanfprins
1 / 5 (3) May 13, 2010
If electron would be the wave, like the photon is, it wouldn't create a dots at the target - it would form only diffusive patterns.


The photon-waves also create dots. The fact is that a wave changes its shape and size when its boundary conditions change. Within a screen the entities which can interact and absorb an impinging wave are of atomic size: This mandates that the impinging wave, no matter what its size before reaching the screen, must "collapse" into a smaller wave which then causes the spot. This happens for for both a photon-wave and an electron-wave. Thus "particles" are not needed to explain "spots" on the screen: They are caused by typical and classically well-known behaviour of waves.
ZeroX
2.3 / 5 (6) May 13, 2010
...the photon-waves also create dots..
Yes, but there is a seamless transition from diffusive interference pattern created by low frequency waves to dot pattern of short wavelength photons, which are composed of well distingushed spots. It's evident, we are dealing with two levels of waves - those inside of particles, which are very shortwavelength ones, and diffusive waves of vacuum environment, which surrounds them and which have nothing to do with waves inside (so called deBroglie wave). While wavelength of photon can change in many orders of magnitude, it's quite imposibble, the difraction of photons itself will be responsible for all these difraction patterns at double slit of milimeter size.

There is an interesting point too, gamma ray photons are spreading like massive particles (for example electrons), surrounded by its own deBroglie wave and gravity field., because their interference patterns are of the same character.
Skeptic_Heretic
5 / 5 (1) May 13, 2010
I didn't use the "decay" word - so I don't understand, why are you asking for it.

If you don't understand why I'm asking for it you have no idea what a DeBroglie wave is. Especially if you think they can be composed of vaccuum.

johanfprins
1.8 / 5 (5) May 13, 2010
It's evident, we are dealing with two levels of waves - those inside of particles, which are very shortwavelength ones, and diffusive waves of vacuum environment, which surrounds them and which have nothing to do with waves inside (so called deBroglie wave).


Nonsense!! A wave is a wave, and its wavelength
obviously determines what the smallest volume can be into which it can collapse. The fact that light waves with long wavelengths make larger spots is caused by the fact that only absorbers of larger size can collapse them than light-waves with shorter wavelengths.

The fact that gamma-waves spread like electron-waves is further proof that both entities are waves with short wavelengths and not particles at all.
Alizee
May 13, 2010
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JayK
1.8 / 5 (5) May 13, 2010
Just to make it clear, Alizee and ZeroX are the same person and it is trying to push Aether Theory over all other current science. Arguing with it is pointless, it just uses that as an excuse to spew Aether crap all over the comment thread.
Alizee
May 13, 2010
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Alizee
May 13, 2010
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Alizee
May 13, 2010
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JayK
2 / 5 (4) May 13, 2010
Yep, it is all a big massive conspiracy of the Holy Church of Jesus-Einstein that nobody considers Aether theory to be anything other than the product of damaged pseudo-intellectuals clamoring for attention.
Alizee
May 13, 2010
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Alizee
May 13, 2010
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Alizee
May 13, 2010
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Skeptic_Heretic
5 / 5 (2) May 13, 2010
I'm not saying, dense aether theory is something extraordinary intellectual stuff - on the contrary, its main power is in its conceptual simplicity.


We established that the Universe wasn't composed of heavenly water a long time ago. Since then we've actually gone to space and proven that is wasn't a heavenly water.

There is no "dense aether".

Conceptual simplicity is akin to creationist cosmology.
Alizee
May 13, 2010
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Alizee
May 13, 2010
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daywalk3r
3.1 / 5 (19) May 13, 2010
Physicists already know, vacuum is not a homogeneous stuff, it's full of subtle density fluctuations, i.e. tiny space-time curvatures in general relativity sense, or quantum foam in quantum mechanics sense.
Perfect vacuum does not exist, nor does a perfectly isolated system. And those "fluctuations" might aswell be caused by the gazillions of stuff that is flying through it every second, from every direction, permanently.

Every "(energy) density fluctuation" (as you like to call it) causes a space-time curvature fluctuation. So every, even the tinniest single energy wave/particle/whatnot, that flies through it, causes a fluctuation of some sort & amplitude. Interactions between them the more.

There you have your quantum (soap?) foam.. Nothing more than just an interferrence pattern, and no hocus-pocus stuff popping out of nowhere from nothing in "empty" vacuum involved.
daywalk3r
3.1 / 5 (19) May 13, 2010
geometrodynamic theory based on emergence concept
Interesting.. Then why does it sometimes seem to be emerging from a goulash of various pseudo/pop-science theories (like real goulash emerges from water, meat, popatoes and paprika), rather than from geometrodynamics? :-P

Serriously.. I've seen you so many times trying to "adapt" various popular (pseudo)scientific gimmicks and stuff, even from theories/hypotheses that are mutually totally incompatible, that by now, it's really hard for me to believe a word from that quote above..
ZeroX
May 14, 2010
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johanfprins
1 / 5 (2) May 14, 2010
I would understand the stance of people, if they would have a better models for main concepts of quantum mechanics or relativity already - but they have none. They simply don't want and refute to understand it - they're adhere on formal regressions based on few formal postulates. Their future development is conditioned by deeper understanding of underlying reality.

The simplest and consistent model is that both matter and light consist solely of waves. All interactions between matter and light can then be modelled by solving differential wave equations. So the problem lies with YOU Alizee, Alexa, Seneca, XeroX or whatever personality you really have!
SO:
For half-educated trolls like Alizee, Xerox, Seneca, Alexa etc. the meaning of life is to prohibit the spreading of analogies and connections by all means possible. They're useless parasites of human society. What these people afraid of - the understanding of reality doesn't harm anyone?
ZeroX
2.6 / 5 (5) May 14, 2010
...interesting.. Then why does it sometimes seem to be emerging from a goulash of various pseudo/pop-science theories...rather than from geometrodynamics?
Because geometrodynamics is matematical description of situation, whereas the intuitive models are understandable. I've nothing against math and I'm using it in my work often. But when environment becomes fuzzy and turbulent, then the mathematical description of it brings no significant advantage over intuitive models. Of course, it still enables army of mathematicians to keep their jobs, which is why these people are fighting about relevancy of these models so much.
johanfprins
1 / 5 (2) May 14, 2010
In fact, one of my motivations is to make the understanding of science, mainstream physics in particular as unintellectual, as possible.


Your model has succeeded in this repect since only a person with no intellect at all will not see that you are more of the track than mainstream physics can EVER HOPE TO BE.
ZeroX
May 14, 2010
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ZeroX
2.6 / 5 (5) May 14, 2010
..only a person with no intellect at all will not see that you are more of the track than mainstream physics can EVER HOPE TO BE.
In its consequence, it could help the mainstream scientists, too. They're developing gazillions of theories and occasionally losing orientations in it. I'm eager collector of various theories, so I can imagine, how the people, who are specializing and have no time to follow ideas of other people must be disoriented.

In this way, many mainstream physicists have literally no chance to recognize emergent logical patterns in fast growing pile of new ideas because of limited speed of information spreading inside of human society. Therefore the establishing of more general and fuzzy meta-theories could be understood as a sort of undeniable physical process.

The people would require meta-theories for to navigate in growing ocean of information faster in the same way, like they're undulating along foamy density fluctuations (branes) through vacuum.
johanfprins
1 / 5 (2) May 14, 2010
Dense aether theory isn't another religion, it's about finding of logical connections.


You succeeded in fooling me!! I have not seen a SINGLE logical connection in any of your propaganda for your aether theory. Or do you believe like Joseph Goebbels did that by just repeating the illogical enough people will believe that it is true?
ZeroX
2.8 / 5 (5) May 14, 2010
For example:

1) gas contains density fluctuations => in dense gas these fluctuations appear like membranes of foam or water surface

2) vacuum is dense gas => vacuum contains membranes of foam, which appear like watter surface

3) swimming duck forms a ripples around it at water surface => massive particle forms a ripples around it at the surfaces of foam.

Are you able to follow this logic - or do you see some hole in it?
johanfprins
1 / 5 (2) May 14, 2010
2) vacuum is dense gas => vacuum contains membranes of foam, which appear like watter surface


There is no experimental proof for this assertion. Starting off with postulates which cannot be proved experimentally usually creates nonsense; as your model is doing;

OK, OK, I have closed my eyes and saw little ducks swimming past creating ripples in a foam of infinite density. I think I need a psychiatrist now!
ZeroX
2.6 / 5 (5) May 14, 2010
BTW If string or loop quantum gravity theorists would be a little more clever, they could made their theories a much more familiar for laymans and successful in predictions, which they desperately lacking by now.

Of course they didn't do it, because it would enable the people to understand many trivialities, too. From the same reasons, druids and shamans didn't explained their tricks to the rest of society, even at the case, when they had logical explanations for it. They would lost their mystery, power and... jobs.

In this way modern physics catched itself into trap of its own positivism.
johanfprins
1 / 5 (2) May 14, 2010
BTW If string or loop quantum gravity theorists would be a little more clever, they could made their theories a much more familiar for laymans and successful in predictions, which they desperately lacking by now.


I agree that string and quantum loop theory fall in the same category as your aether theory: It is all just claptrap!
ZeroX
2.6 / 5 (5) May 14, 2010
string and quantum loop theory fall in the same category as your aether theory: It is all just claptrap!
Why? Because you have said so?
..vacuum is dense gas ... there is no experimental proof for this assertion...
Does light travel in transversal waves like energy through dense gas? Did we talk about eternal motion of fluid helium like about Brownian motion of pollen grains in fluid? etc...
johanfprins
2 / 5 (4) May 14, 2010
Why? Because you have said so?


Because there are no experimental data which support these theories in any manner whatsoever.
There is not even related experimental data from which one can argue that they might have any merit. That is why!
johanfprins
2.6 / 5 (5) May 14, 2010
Does light travel in transversal waves like energy through dense gas?


It does not since if it did Einstein's Special Relativity would be wrong.

Did we talk about eternal motion of fluid helium like about Brownian motion of pollen grains in fluid? etc...


If you know elementary physics you will know that liquid helium and Brownian motion have NOTHING in common!
ZeroX
May 14, 2010
This comment has been removed by a moderator.
johanfprins
2 / 5 (4) May 14, 2010
Why not? The zero point energy of eternal movement of atoms can be attributed to Brownian motion of particles in dense fluid in the role of vacuum quite easily. Without it you would have no explanation for such motion.


If you want to believe such nonsense you are welcome to do so. The helium atoms move by quantum fluctuations which are allowed by Heisenberg's relationship for energy and time. They are not buffeted by non-existent "particles" constituting a so-calle aether.

I am not even going to try and reason with you about Einstein's special teory of relativity. It is clear that you do not understand it and do not want to understand it.

So good luck to you!! Maybe you should contact Mark MacCutcheon: He still believes that the planets are being "pushed" around their orbits.
ZeroX
2 / 5 (4) May 14, 2010
..the helium atoms move by quantum fluctuations which are allowed by Heisenberg's relationship for energy and time..
Heisenberg's relationship is theorem of QM derived from experiments. BTW It's a product of the same Copenhagen's interpretation of QM, which you're fighting against..;-)

The same is valid about invariance of light speed in relativity. In this theory the invariance of light speed is one of postulates as a part of Lorentz symmetry. In dense aether theory it's a theorem, i.e. a consequence of transversal character of energy spreading in common dense massive environments.

In dense aether theory everything has its own very good logical reason, including the zero-point energy, uncertainty principle and/or invariance of light speed to the motion of environment. This theory proposes testable connection with observable reality for all phenomena and concepts of formal theories, so I'm not required to believe in any postulate, which hasn't support in everyday reality.
Alizee
May 14, 2010
This comment has been removed by a moderator.
johanfprins
1.8 / 5 (4) May 14, 2010
Heisenberg's relationship is theorem of QM derived from experiments. BTW It's a product of the same Copenhagen's interpretation of QM, which you're fighting against..;-)


It is NOT!!! It is a relationship that is NOT an "uncertainty" relationship. This is where the Copenhagenists went off the rails. It is purely a resonance relationship.

I'm finishing with discussion here - you can continue in it therein:


Thank God for small mercies!
Alizee
May 15, 2010
This comment has been removed by a moderator.
Alizee
May 15, 2010
This comment has been removed by a moderator.
frajo
5 / 5 (1) May 15, 2010
Because this thread is overly long and it has nothing to do with quarks anyway, I'm finishing with discussion here
19 hours later: two further comments by the same user.
Just to demonstrate how serious he is.
Alizee
May 15, 2010
This comment has been removed by a moderator.
hush1
1 / 5 (1) Jun 05, 2010
Children (of all ages) use words that only have meaning to them and to no one else.
The words can be common to all.

Neologism

"for exact understanding exact language is necessary."

(Gurdjieff to Ouspensky)

You need no judgment or defense.
You need to be understood.

You need to understand others do not understand you.

I need to know you know a language - a language that has meaning - beyond the language you understand.

My understanding is irrelevant.
Everyone understanding is relevant except mine.

Nothing is so wonderful it can not be repeated.
Including mutual understanding.