Quarks, the elementary particles 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 *Physical Review Letters* (Vol. 104:13).

The findings reduce the uncertainty of the quark 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 strong force. 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 proton -- 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 theoretical physics right now," Lepage said. "Indeed it is unclear why quarks have mass at all." He added that the new Large Hadron Collider 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.

**Explore further:**
Physicists plan quark conference

## Alizee

May 03, 2010## hard2grep

If there are particles not affected by gravity floating around, would we not be looking at a large discrepancy??? hmm...

## shavera

## Alizee

May 03, 2010## Alizee

May 03, 2010## Aliensarethere

## Alizee

May 03, 2010## Aliensarethere

## TheWalrus

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## Alizee

May 03, 2010## Alizee

May 03, 2010## Alizee

May 03, 2010## brentrobot

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

## daywalk3r

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 :)

## daywalk3r

## daywalk3r

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

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

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.

A harmonic wave is equivalent to a delocalized photon.

they both obey maxwell's equations and hence have the same speed

## Parsec

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.

## Titto

## johanfprins

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

## ZeroX

## srikkanth_kn

## 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.

## ZeroX

## ZeroX

## Neodim

## ZeroX

http://www.scienc...1725.htm

## johanfprins

ZeroX:

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: 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

Correction: I meant a COHERENT harmonic wave

## baudrunner

That works for the higher order elements in the periodic table as well. The most massive elements have the shortest shelf-life.

## johanfprins

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## Skeptic_Heretic

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## Alizee

May 04, 2010## johanfprins

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

## johanfprins

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

## Alizee

May 05, 2010## johanfprins

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

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?

A clue: Note that my e-mail is at cathodixx (note the double x, which means two surfaces)

## Alizee

May 05, 2010## smd

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## Alizee

May 05, 2010## johanfprins

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

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

Anybody who can understand this gobblydook needs help.

## Alizee

May 05, 2010## Alizee

May 05, 2010## Alizee

May 05, 2010## hard2grep

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## Alizee

May 05, 2010## johanfprins

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

You do not really think one should waste time trying to engage you in logical discourse, do you?

## Alizee

May 05, 2010## Alizee

May 05, 2010## Alizee

May 05, 2010## Alizee

May 05, 2010## Alizee

May 05, 2010## daywalk3r

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

## Alizee

May 05, 2010## Alizee

May 05, 2010## johanfprins

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

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

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

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

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!

## ZeroX

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

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

Typical is it not!

## johanfprins

Exactly! This is why it is based on virtual reality!

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 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

Of course it does. If you interpret a wave-intensity as a "probability" instead of a mass-energy you will obviously have a contradiction!

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)

Are you going to nominate me? Thanks!

## Alizee

May 06, 2010## Alizee

May 06, 2010## Alizee

May 06, 2010## daywalk3r

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## johanfprins

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!

Please stop punting a ridiculous theory under different names: Alizee, Seneca? Xerox? and what have you!!!

## Alizee

May 06, 2010## Alizee

May 06, 2010## johanfprins

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?

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

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## johanfprins

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

Yes you have: It it is known as "common sense"!

## Alizee

May 06, 2010## johanfprins

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## Alizee

May 06, 2010## Alizee

May 06, 2010## johanfprins

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## daywalk3r

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## Alizee

May 06, 2010## Alizee

May 06, 2010## Alizee

May 06, 2010## johanfprins

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

Thank you.

## Alizee

May 07, 2010## Alizee

May 07, 2010## johanfprins

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## Alizee

May 08, 2010## Alizee

May 08, 2010## Alizee

May 08, 2010## Shootist

Yeah, flat earth, with four corners, 6000 years old.

## johanfprins

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## Alizee

May 08, 2010## Alizee

May 08, 2010## Alizee

May 08, 2010## zslewis91

## Alizee

May 08, 2010## johanfprins

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.

Correct! Because they do not understand what is really going on. Therefore they paddle along with impossible concepts like "dense ether theory"!

## johanfprins

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## Alizee

May 09, 2010## Alizee

May 09, 2010## Alizee

May 09, 2010## Alizee

May 09, 2010## johanfprins

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## Alizee

May 09, 2010## johanfprins

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## johanfprins

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

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

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

## johanfprins

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

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

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

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

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

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".

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

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

## johanfprins

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

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

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

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

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

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

## Skeptic_Heretic

I'm sorry, can you say pseudoscience? Quantum motion is not directly observable, and definitely not by use of base human senses.

## ZeroX

http://www.youtub...aWCXkSC0

## johanfprins

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

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

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

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## Alizee

May 10, 2010## Skeptic_Heretic

No, I evaluate and prefer to not interject when it is unwarranted. It has become warranted.

## Alizee

May 10, 2010## johanfprins

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## johanfprins

Give me such an incontrovertible proof. You cannot.

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## johanfprins

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## johanfprins

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## johanfprins

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## johanfprins

Thus you have TWO uncertainty intervals per electron? WOW!!

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## johanfprins

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## johanfprins

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## johanfprins

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## johanfprins

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

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## Alizee

May 10, 2010## johanfprins

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.

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

This remark explains a lot about your incapability to argue physics in a logical manner above: It is beneath contempt.

## ZeroX

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

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

http://vergil.che...e20.html

## johanfprins

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.

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

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

## Skeptic_Heretic

That's a really big misunderstanding to have. One that effectively invalidates the majority of what you've said above.

## johanfprins

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## johanfprins

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## johanfprins

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

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

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

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## Alizee

May 12, 2010## johanfprins

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

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

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

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## JayK

## Alizee

May 13, 2010## Alizee

May 13, 2010## Alizee

May 13, 2010## JayK

## Alizee

May 13, 2010## Alizee

May 13, 2010## Alizee

May 13, 2010## Skeptic_Heretic

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## Alizee

May 13, 2010## daywalk3r

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

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## johanfprins

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

## johanfprins

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## ZeroX

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

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

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

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

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

I agree that string and quantum loop theory fall in the same category as your aether theory: It is all just claptrap!

## ZeroX

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

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

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

If you know elementary physics you will know that liquid helium and Brownian motion have NOTHING in common!

## ZeroX

May 14, 2010## johanfprins

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

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## johanfprins

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.

Thank God for small mercies!

## Alizee

May 15, 2010## Alizee

May 15, 2010## Alizee

May 15, 2010